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Sherman Oaks Prealgebra Tutor
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Six-dimensional space
Six-dimensional space is any space that has six dimensions, that is, six degrees of freedom, and that needs six pieces of data, or coordinates, to specify a location in this space. There are an
infinite number of these, but those of most interest are simpler ones that model some aspect of the environment. Of particular interest is six-dimensional Euclidean space, in which 6-polytopes and
the 5-sphere are constructed. Six-dimensional elliptical space and hyperbolic spaces are also studied, with constant positive and negative curvature.
Formally, six-dimensional Euclidean space, ℝ^6, is generated by considering all real 6-tuples as 6-vectors in this space. As such it has the properties of all Euclidean spaces, so it is linear, has a
metric and a full set of vector operations. In particular the dot product between two 6-vectors is readily defined, and can be used to calculate the metric. 6 × 6 matrices can be used to describe
transformations such as rotations that keep the origin fixed.
More generally, any space that can be described locally with six coordinates, not necessarily Euclidean ones, is six-dimensional. One example is the surface of the 6-sphere, S^6. This is the set of
all points in seven-dimensional Euclidean space ℝ^7 that are equidistant from the origin. This constraint reduces the number of coordinates needed to describe a point on the 6-sphere by one, so it
has six dimensions. Such non-Euclidean spaces are far more common than Euclidean spaces, and in six dimensions they have far more applications.
Main article: 6-polytope
A polytope in six dimensions is called a 6-polytope. The most studied are the regular polytopes, of which there are only three in six dimensions: the 6-simplex, 6-cube, and 6-orthoplex. A wider
family are the uniform 6-polytopes, constructed from fundamental symmetry domains of reflection, each domain defined by a Coxeter group. Each uniform polytope is defined by a ringed Coxeter-Dynkin
diagram. The 6-demicube is a unique polytope from the D6 family, and 2[21] and 1[22] polytopes from the E6 family.
The 5-sphere, or hypersphere in six dimensions, is the five-dimensional surface equidistant from a point. It has symbol S^5, and the equation for the 5-sphere, radius r, centre the origin is
$S^5 = \left\{ x \in \mathbb{R}^6 : \|x\| = r\right\}.$
The volume of six-dimensional space bounded by this 5-sphere is
$V_6 = \frac{\pi^3 r^6 }{6}$
which is 5.16771 × r^6, or 0.0807 of the smallest 6-cube that contains the 5-sphere.
The 6-sphere, or hypersphere in seven dimensions, is the six-dimensional surface equidistant from a point. It has symbol S^6, and the equation for the 6-sphere, radius r, centre the origin is
$S^6 = \left\{ x \in \mathbb{R}^7 : \|x\| = r\right\}.$
The volume of the space bounded by this 6-sphere is
$V_7 = \frac{16\pi^3 r^7 }{105}$
which is 4.72477 × r^7, or 0.0369 of the smallest 7-cube that contains the 6-sphere.
Transformations in three dimensions
In three-dimensional space a generalised transformation has six degrees of freedom, three translations along the three coordinate axes and three from the rotation group SO(3). Often these
transformations are handled separately as they have very different geometrical structures, but there are ways of dealing with them that treat them as a single six-dimensional object.
Homogeneous coordinates
Main article: Homogeneous coordinates
Using four-dimensional Homogeneous coordinates it is possible to describe a general transformation using a single 4 × 4 matrix. This matrix has six degrees of freedom, which can identified with the
six elements of the matrix above the main diagonal, as all others are determined by these.
Screw theory
Main article: Screw theory
In screw theory angular and linear velocity are combined into one six-dimensional object, called a twist. A similar object called a wrench combines forces and torques in six dimensions. These can be
treated as six-dimensional vectors that transform linearly when changing frame of reference. Translations and rotations cannot be done this way, but are related to a twist by exponentiation.
Phase space
Main article: Phase space
Phase space is a space made up of the position and momentum of a particle, which can be plotted together in a phase diagram to highlight the relationship between the quantities. A general particle
moving in three dimensions has a phase space with six dimensions, too many to plot but they can be analysed mathematically.^[1]
Rotations in four dimensions
Main article: Rotations in 4-dimensional Euclidean space
The rotation group in four dimensions, SO(4), has six degrees of freedom. This can be seen by considering the 4 × 4 matrix that represents a rotation: as it is an orthogonal matrix the matrix is
determined, up to a change in sign, by e.g. the six elements above the main diagonal. But this group is not linear, and it has a more complex structure than other applications seen so far.
Another way of looking at this group is with quaternion multiplication. Every rotation in four dimensions can be achieved by multiplying by a pair of unit quaternions, one before and one after the
vector. These quaternion are unique, up to a change in sign for both of them, and generate all rotations when used this way, so the product of their groups, S^3 × S^3, is a double cover of SO(4),
which must have six dimensions.
Although the space we live in is considered three-dimensional, there are practical applications for four-dimensional space. Quaternions, one of the ways to describe rotations in three dimensions,
consist of a four-dimensional space. Rotations between quaternions, for interpolation for example, take place in four dimensions. Spacetime, which has three space dimensions and one time dimension is
also four-dimensional, though with a different structure to Euclidian space.
Plücker coordinates
Main article: Plücker coordinates
Plücker coordinates are a way of representing lines in three dimensions using six homogeneous coordinates. As homogeneous coordinates they have only five degrees of freedom, corresponding to the five
degrees of freedom of a general line, but they are treated as 6-vectors for some purposes. For example the check for the intersection of two lines is a 6-dimensional dot product between two sets of
Plücker coordinates, one of which has exchanged its displacement and moment parts.
In electromagnetism, the electromagnetic field is generally thought of as being made of two things, the electric field and magnetic field. They are both three-dimensional vector fields, related to
each other by Maxwell's equations. A second approach is to combine them in a single object, the six-dimensional electromagnetic tensor, a tensor or bivector valued representation of the
electromagnetic field. Using this Maxwell's equations can be condensed from four equations into a particularly compact single equation:
$\partial \mathbf{F} = \mathbf{J} \,$
where F is the bivector form of the electromagnetic tensor, J is the four-current and ∂ is a suitable differential operator.^[2]
String theory
In physics string theory is an attempt to describe general relativity and quantum mechanics with a single mathematical model. Although it is an attempt to model our universe it takes place in a space
with more dimensions than the four of space-time that we are familiar with. In particular a number of string theories take place in a ten-dimensional space, adding an extra six dimensions. These
extra dimensions are required by the theory, but as they cannot be observed are thought to be quite different, perhaps compactified to form a six-dimensional space with a particular geometry too
small to be observable.
Since 1997, another string theory has come to light that works in six dimensions. Little string theories are non-gravitational string theories in five and six dimensions that arise when considering
limits of ten-dimensional string theory.^[3]
Theoretical background
Bivectors in four dimensions
A number of the above applications can be related to each other algebraically by considering the real, six-dimensional bivectors in four dimensions. These can be written Λ^2ℝ^4 for the set of
bivectors in Euclidian space or Λ^2ℝ^3,1 for the set of bivectors in spacetime. The Plücker coordinates are bivectors in ℝ^4 while the electromagnetic tensor discussed in the previous section is a
bivector in ℝ^3,1. Bivectors can be used to generate rotations in either ℝ^4 or ℝ^3,1 through the exponential map (e.g. applying the exponetial map of all bivectors in Λ^2ℝ^4 generates all rotations
in ℝ^4). They can also be related to general transformations in three dimensions through homogeneous coordinates, which can be thought of as modified rotations in ℝ^4.
The bivectors arise from sums of all possible wedge products between pairs of 4-vectors. They therefore have C4
2 = 6 components, and can be written most generally as
$\mathbf{B} = B_{12}\mathbf{e}_{12} + B_{13}\mathbf{e}_{13} + B_{14}\mathbf{e}_{14} + B_{23}\mathbf{e}_{23} + B_{24}\mathbf{e}_{24} + B_{34}\mathbf{e}_{34}$
They are the first bivectors that cannot all be generated by products of pairs of vectors. Those that can are simple bivectors and the rotations they generate are simple rotations. Other rotations in
four dimensions are double and isoclinic rotations and correspond to non-simple bivectors that cannot be generated by single wedge product.^[4]
6-vectors are simply the vectors of six-dimensional Euclidean space. Like other such vectors they are linear, can be added subtracted and scaled like in other dimensions. Rather than use letters of
the alphabet higher dimensions usually use suffixes to designate dimensions, so a general six-dimensional vector can be written a = (a[1], a[2], a[3], a[4], a[5], a[6]). Written like this the six
basis vectors are (1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0) and (0, 0, 0, 0, 0, 1).
Of the vector operators the cross product cannot be used in six dimensions; instead the wedge product of two 6-vectors results in a bivector with 15 dimensions. The dot product of two vectors is
$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5 + a_6b_6.$
It can be used to find the angle between two vectors and the norm,
$\left | \mathbf{a} \right \vert = \sqrt{\mathbf{a} \cdot \mathbf{a}} = \sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 + {a_5}^2 + {a_6}^2}.$
This can be used for example to calculate the diagonal of a 6-cube; with one corner at the origin, edges aligned to the axes and side length 1 the opposite corner could be at (1, 1, 1, 1, 1, 1), the
norm of which is
$\sqrt{1 + 1 + 1 + 1 + 1 +1} = \sqrt{6} = 2.4495,$
which is the length of the vector and so of the diagonal of the 6-cube.
Complex 3-space
The complex plane C has two real dimensions, so C^3 is a six-dimensional space. William Rowan Hamilton identified this space in 1853^[5] as the bivectors of is biquaternions. He had introduced
vectors as 3-dimensional parts of quaternions, so when the tensor product $C \otimes H$ became biquaternions, the complex 3-dimensional part was bivectors. The exponential map takes bivectors to the
unit sphere of the biquaternion algebra, which is isomorphic to the Lorentz group. Hence, as Ronald Shaw and Graham Bowtell^[6] have noted, bivectors are logarithms of Lorentz transformations.
Generally vector analysis is confined to three dimensions, but in Vector Analysis (1901) the six-dimensional space of bivectors was used by J. W. Gibbs and E. B. Wilson.^[7]
In the differential geometry of complex manifolds some six-dimensional spaces arise as algebraic manifolds. Examples include the quintic threefold and the Barth-Nieto quintic. According to
Piergiorgio Odifreddi, the classification of complex three-dimensional manifolds "was one of the spectacular results obtained by the Japanese school of geometry of Heisuke Hironaka, Shing Tung Yau,
and Shigefumi Mori. For this work they were awarded the Fields Medal in 1970, 1983, and 1990, respectively."^[8]
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The Prime Glossary: factorial
Prime Pages:
Top 5000:
The factorial of a positive integer is the product of all the positive integers less than or equal to it. The factorial of n is usually denoted n! The first few are 1!=1, 2!=2, 3!=6, 4!=24, 5!=120,
and 6!=720. We also define 0! to be 1 (in mathematics, an empty product has the value one).
n! is the number of permutations of n items (how many ways n distinct items can be arranged in a row). The number of ways you can pick r or n identical items is the binomial coefficient n!/(r!(n-r
See Also: FactorialPrime, StirlingsFormula, WilsonsTheorem
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Luiz Mendes
Last edited 4may05 by lmendes@uiuc.edu
Find this document at http://www.ncsa.uiuc.edu/Classes/MATH198/lmendes
Luiz Mendes
Evolve Reloaded
Evolve Reloaded is based on CAVEvolve, created in 1995 by Scott Banister. The purpose of Evolve Reloaded is to visualize a three-dimensional cellular automaton that uses the Von Neumann neighborhood.
Cellular automata are dynamical systems which are discrete in both space and time and are characterized by what are called "local interactions". Each grid cell can be in one of a finite number of
states, and the system is updated in discrete time steps according to some specified interaction rule. The interaction rule used in Evolve Reloaded is the same one used in Stanislav Ulam's game
Reproduction after being generalized to three dimensions (Ulam's game only uses two dimensions). In Evolve Reloaded, the user can update the system by pressing a button.
Ulam's Reproduction Game
The game begins with an empty, two-dimensional grid. There is a counter placed at the center of the grid. The next generation is formed by placing counters on any cell that contains one and only one
marked cell as a neighbor. There are many ways to define "neighbor". In this case, the Von Neumann neighborhood is used. Here, a neighbor is any cell directly above or below the given cell, or
directly to the right or left of the given cell (the Moore neighborhood would include diagonal cells as well). This same rule is then applied again. However, the trick is that the so-called
"grandparents" of the current generation get removed from the grid after each application of the rule. This pattern repeats indefinitely.
Evolve Reloaded uses this same interaction rule, the only difference being that neighbors along the z-axis are also considered, so as to make the system three-dimensional.
Final Thoughts
The motivation for creating this project was the fact that the older version of this program, CAVEvolve, only compiled under the SGI IRIX5.2, which is quite outdated by now. That said, this program
still does not contain all of the features of the older version (Banister's program could even play Stanislav Ulam's two-dimensional Reproduction game). This means that there is plenty of room for
future improvement of this project.
Finally, I would like to thank Professor Francis and all my classmates, especially William Baker, for all their help throughout the semester.
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This page uses JavaSketchpad. You may drag any of the red points, and the construction will adjust accordingly.
Explanation of the construction: AD = BD, since they are both PQ. Triangle ADB is isosceles, so angle DAB is congruent to angle ABD. Triangle DAC is congruent to triangle DBC by SSS, so angles CDA
and CDB are congruent. Then triangles AED and BED are congruent by ASA, so AE = BE and angle AED = angle BED, which makes them right angles since they are a linear pair. Thus line CD is the
perpendicular bisector of segment AB.
Back to ruler and compass constructions main page.
Back to Ken Brakke's home page.
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How to Look at Art
How to Look at Art
May 14, 1998
Marc Frantz
The appreciation of art has so many aspects-including some that are probably ineffable--that a comprehensive discussion would be impossible. The title of this article may therefore seem presumptuous,
and would probably be even more so to a person trained in the fine arts who was reading such a thing written by a mathematician. As someone whose undergraduate degree is a BFA with a major in
painting, however, I can tell you that my former classmates and I could have benefited from a bit of mathematical advice along these lines.
In particular, if we had been a bit more canny about looking at our own works of art, we would in many cases have noticed drawbacks that, when corrected, would have resulted in more competent work.
The advice I have in mind concerns the appreciation of artwork that employs perspective, a topic that obviously lends itself to a mathematical treatment. In this article I informally present one
example from a set of simple procedures--easily remembered and requiring no tools or calculations--that can be used by viewers of works of perspective in galleries or museums.
Before setting up the problem, I would like to briefly describe the more general setting in which I began to think about such things. With the support of a National Science Foundation initiative
called Mathematics and Its Applications Throughout the Curriculum (MATC), I am currently involved in the development of an interdisciplinary course called Mathematics and Art. (SIAM has lent its
support to MATC.) My course co-developer, Annalisa Crannell, has taught Mathematics and Art for two years now at Franklin & Marshall College, and I began teaching it this fall at Indiana University
Purdue University Indianapolis. Portions of the text I am writing have been used by Crannell at Franklin & Marshall, and also by Catherine Roberts in a course called The Science of Art at Northern
Arizona University. Both instructors have provided me with valuable feedback and criticism.
Not surprisingly, I seem to score the most "hits" with students when I avoid grand mathematical generalizations, and instead focus on questions like "What kinds of things did you want to know in art
school that were never clear to you?" It is mainly in this spirit that the material in the text is written, including tips on how to look at art.
Did I Really Draw a Cube?
A typical art student's experience with perspective drawing or, to be more honest, my own memory of this experience can be used to introduce the problem. Let's say I want to draw a cube. Remembering
that something called "one-point perspective" can help, I decide to use it. I start by drawing the one point, called a vanishing point, and a square to serve as the front face of the cube (Figure 1
Figure 1. The author follows the first steps in the procedure for drawing a cube (a and b, left and middle), after which some guesswork leads to a dubious result (c, right).
Next, I lightly draw dashed lines from three corners of the square to the vanishing point (Figure 1(b)). At this stage things get a little vague. I need to draw a vertical line segment to represent
the left rear edge of the cube, but where do I put it? Not being sure, I just plunge ahead and draw this edge by guessing, along with the other visible edges, and consider myself done, except for
erasing the guide lines (Figure 1(c)). After looking at the other students' work, however, I find my cube a bit too elongated in the direction perpendicular to the front face. Did I really draw a
cube, or something else?
The real problem, as you may have guessed, is that nowhere in my thought process did I consider the correct viewing position for an observer of the drawing. It somehow eluded me in art school that
perspective is simply the result of projecting points from an object in three-dimensional space along straight lines (light rays) that strike one eye of an observer. The points where these lines
pierce a "picture plane" constitute the perspective image of the object.
I think it would have helped quite a bit if we had performed a simple classic experiment like that shown in Figure 2, in which I am the observer, my student Pat Sullivan (at the window) is the
projection function, and the window is the picture plane. By following my directions, Pat used masking tape to make a fairly decent perspective drawing of the IUPUI Library. A viewer who wishes to
see the true shape of the library must place his or her eye exactly where mine was, or else the drawing will appear distorted; this is exactly the problem with the drawing of the cube.
It turns out that Figure 1(c) can be considered a correct drawing of a cube if (and only if) the viewing distance is specified to be three units, where one unit is the length of a front edge of the
cube--a very small distance! Thus, the real problem with my drawing is that, in order to avoid seeing a distortion, the viewer is forced uncomfortably close to the page. If you can manage it, close
one eye and place the other directly in front of the vanishing point in Figure 1(c), at a distance of about three units, and then glance off to the side at the drawing. If your eyes are no worse than
mine, you should be able to make out a blurry but more cubelike shape than the elongated box that is perceived at a more natural--but incorrect--viewing distance. This problem is fairly common with
beginning artists; their perspective drawings appear dramatic but "overperspectivized," because--unbeknownst to them--the correct viewing distance is unnaturally close to the paper.
I could have consulted any of the many texts on perspective, of course, and found just about anything I wanted to know about the correct way to draw cubes. Just knowing a little more about how to
look at art, though, would have been sufficient to show me that my drawing needed improvement, and would have made me better able to appreciate the work of others.
Figure 2. The author (left) is the observer and the IUPUI Library the subject in a classic experiment on perspective drawing.
The Correct Viewing Distance in a Museum
Rather than show how to get the viewing distance for the cube, let's tackle a slightly more realistic example that uses a type of deduction that will be helpful when viewing art in a museum or
gallery; you'll then be able to easily finish the cube problem yourself. Specifically, let us determine the correct viewing distance for the drawing of the house in Figure 3. (For the sake of
brevity, some basic facts about one-point perspective drawing are presented here without proof. In the text for the Mathematics and Art course, these facts are proved by vector methods that can serve
as a useful tool kit for students in other courses.)
Figure 3. To determine the correct viewing distance for an observer of this drawing of a house (the front face of the actual house is assumed to be parallel to the picture plane), the author again
follows a stepwise procedure; as with the cube in Figure 1, the viewing distance turns out to be impractically close.
Since the angles of the front face of the house appear undistorted (they are all right angles), we can assume that the actual house, which we imagine to be somewhere on the far side of the picture
plane (the page), has its front face parallel to the picture plane. Thus, the lines determined by the top and bottom edges of the left wall are in reality orthogonal to the picture plane. The images
of these lines in Figure 3(a) meet at the vanishing point v, and since the correct line of sight to the vanishing point must be parallel to the actual lines of the house, the correct viewpoint lies
on a line through v, orthogonal to the page. The viewpoint can therefore be located once we know the viewing distance, and the viewing distance can be determined as follows:
1. Locate the vanishing point v in Figure 3(a).
2. Find a rectangular feature on the left wall and determine its real-world width-to-height ratio r. We choose as our feature the window on the left wall, and make the assumption that it is in
reality a copy of the window on the front wall. (Such an assumption is certainly valid for repeated architectural details like arches and carvings.) For the image of the front window, this ratio
is estimated to be about 1.5 in Figure 3(b); the same ratio will apply to the actual front window and, by assumption, to the side window as well. Thus, we have r = 1.5.
3. Locate the vanishing point v' of the diagonal of the feature chosen in step 2, and let h be the distance between v and v'. To do this, we use the following facts: Just as images of lines have
vanishing points, images of planes have vanishing lines. The image of every line in a given plane has its vanishing point somewhere on the plane's vanishing line, so the plane appears to vanish
at that line (e.g., the horizon line is effectively the vanishing line of a horizontal plane). The vanishing line for the plane of the left wall of the house is the vertical line through v, and
since the diagonal of the window lies in this plane, v' is the intersection of the vanishing line and the image of the diagonal (Figure 3(c)).
4. The correct viewing distance d is rh. In this case, the viewing distance is about 1.5h. This is true because (as mentioned earlier) from the correct viewpoint, a viewer's line of sight to the
vanishing point of a line must be parallel to the actual line in space. By our estimate, the actual diagonal of the side window has an angle of elevation whose cotangent is r = 1.5; thus, the
angle of elevation of the viewer's line of sight to v' must also have a cotangent of d/h = 1.5 (see Figure 4).
Figure 4. A viewer's line of sight to the vanishing point of a line must be parallel to the actual line in space (in this case the diagonal of the side window from Figure 3).
It's really as simple as a, b, c in Figure 3. Of course, you will not be able to draw lines on the works of art in a museum, but a good approximation to the correct viewpoint can be obtained with a
little hand-waving and estimation (both time-honored activities). An almost equally simple method exists for determining the viewpoint of a two-point perspective drawing (the drawing of the library
in Figure 2 uses two-point perspective), although space does not permit a discussion of it here.
In the house example, our analysis has again exposed a drawing whose viewing distance is so close as to be impractical, and this brings up an important point about original works in perspective that
have been reproduced as figures in articles and textbooks: Such reproductions are not just miniaturizations-the potential difficulty for the viewer in assuming the correct viewing distance can
actually make them distortions. In more than one way, there's no substitute for the real thing.
In Keeping with the Spirit of the MATC Initiative
I confess that I have had a great deal of fun deriving such techniques from scratch, because the process has dispelled much of the confusion that lingered from my art school days. These ideas are
more than just fun, however. For instance, principles for looking at perspective art can be viewed as a subset of the important science of photogrammetry, which attempts to extract three-dimensional
information from two-dimensional images.
By all accounts, students have also enjoyed the material on perspective, which includes things like theorems for drawing roofs and fences, and methods for making anamorphic art. I think such lessons
are in keeping with the spirit of the MATC initiative, for one of the project's goals is to positively change the attitudes toward mathematics of students who are in traditionally nonmathematical
disciplines. This particular goal is certainly achievable, for we have all had the experience of watching a seemingly dry piece of mathematics come to life when seen again in the context of a fresh
and interesting application. In this respect, mathematics is much like art--it depends on how you look at it.
Marc Frantz (mfrantz@math.iupui.edu) holds a BFA from the Herron School of Art in Indianapolis and an MS in mathematics from Purdue University. He is a lecturer in the Department of Mathematical
Sciences at Indiana University Purdue University Indianapolis.
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Morningside, MD Statistics Tutor
Find a Morningside, MD Statistics Tutor
My name is Ian. I am originally from Buffalo, NY. I am a 27 year old recent graduate of George Washington University Law School.
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Number of results: 94,411
1.in which geometry do lines no contain an infinite number of points? (a) plane coordinate geometry (b) discrete geometry (c) graph theory 2. in which geometry do points have thickness? (a) plane
coordinate geometry (b) discrete geometry (c) graph theory i think the answer is...
Tuesday, August 30, 2011 at 4:39pm by SoccerStar
I need an answer--molecular/electron geometry
Can anyone help me out with the differences between electron geometry and molecular geometry for SCl6? I think the molecular geometry is octahedral, but I'm not so sure for the electron geometry.
Thanks :) *My earlier post wasn't answered, so I'd appreciate some help please =]
Tuesday, February 8, 2011 at 8:03pm by Kate
Molecular and Electron geometry
Can anyone help me out with the differences between electron geometry and molecular geometry for SCl6? I think the molecular geometry is octahedral, but I'm not so sure for the electron geometry.
Thanks :) *My earlier post wasn't answered, so I'd appreciate some help please =]
Tuesday, February 8, 2011 at 6:47pm by Kate
Math - Plane Geometry. BC is a diameter of circle O, and radius OE is parallel to chord CD. Help?
Monday, May 23, 2011 at 10:13am by Brianna
math geometry
how can i prove in geometry proofs that something is a midpoint of a line? thanks
Tuesday, November 12, 2013 at 5:38pm by Annie
Molecular and Electron geometry
Can anyone help me out with the differences between electron geometry and molecular geometry for SCl6? I think the molecular geometry is octahedral, but I'm not so sure for the electron geometry.
Thanks :)
Monday, February 7, 2011 at 7:02pm by Kate
Geometry riddl How did they like the story about towel manufacturing? Also, where can I find geometry worksheets on plane and simple? ? ? ?
Tuesday, September 26, 2006 at 9:24pm by Debra
math (geometry)
What is a good website that can help me with geometry?
Monday, March 3, 2008 at 2:51pm by jonasgrl12
8. In an interview of 50 math majors, 12 liked calculus and geometry 18 liked calculus but not algebra 4 liked calculus, algebra, and geometry 25 liked calculus 15 liked geometry 10 liked algebra but
neither calculus nor geometry 2 liked geometry and algebra but not calculus. ...
Monday, November 30, 2009 at 9:02pm by poo
The homothetic geometry on C is (C,H) where H is the group of scaling of C. Which figures are congruent in (C,H) to the circle of radius 1 centered at 1? How could you describe the congruence class
in language understandable by a high school geometry student?
Monday, October 3, 2011 at 11:55am by Kyle
geometry:linear measure and precision
how to find examples of rulesfor this theme I'm not quite sure what information you are seeking. I need more details. However, I searched Google under the key words "linear measure rules geometry" to
get these possible sources: http://www.tuhh.de/rzt/rzt/it/Geometry.html http...
Wednesday, August 23, 2006 at 8:02pm by Alma
i need help with this geometry prblem my math book is online on this site: w w w . k e y m a t h . com / D G 3 the class passcode is: 1574-4c524 The chapter is 0 Lesson 0.2 pg. 9 problem number 6
REMEMBER 4-FOLD SYMMENTRY IS NEEDED !!! URGENT !!!
Sunday, September 19, 2010 at 5:54pm by vaishnavi
i need help with this geometry prblem my math book is online on this site: w w w . k e y m a t h . com / D G 3 the class passcode is: 1574-4c524 The chapter is 0 Lesson 0.2 pg. 9 problem number 6
REMEMBER 4-FOLD SYMMENTRY IS NEEDED !!! URGENT !!!
Sunday, September 19, 2010 at 5:54pm by vaishnavi
For each property listed from plane Euclidean geometry, write a corresponding statement for non-Euclidean spherical geometry. A line segment is the shortest path between two points Hey guys I want to
know how to do this and can you help me with this question?
Sunday, May 29, 2011 at 11:19am by Anonymous
In an interview of 50 math majors, 12 liked calculus and geometry 18 liked calculus but not algebra 4 liked calculus, algebra and geometry 25 liked calculus 15 liked geometry 10 liked algebra but
neither calculus nor geometry 2 liked geometry and algebra but not calculus. Of ...
Monday, January 11, 2010 at 8:45pm by Anita
what is a compimentary line in math? Thank you for using the Jiskha Homework Help Forum. If you are studying Geometry: http://astronomyonline.org/Science/Geometry.asp
Wednesday, August 15, 2007 at 6:04pm by Bob
the ratio of cube to sphere in a geometry box is 16 to 9. If there are 225 cubes and spheres in the geometry box altogether. Then how many cube are there?
Friday, November 18, 2011 at 7:38pm by dashawn
calculate the following(show steps): volume of a fish tank that is 15cm by 30cm by 10cm? http://www.math.com/tables/geometry/volumes.htm
Wednesday, June 21, 2006 at 12:37am by jenna
What are the names of polygons? http://www.gomath.com/htdocs/ToGoSheet/Geometry/polygon_name.html
Saturday, February 24, 2007 at 5:29pm by Felicia
Geometry Letter Grammar??
Hello I'm in Geometry Honors (I'm a freshman btw) and I'm failing with a D. I talked to my counselor about switching me out and going to regular and she says it's okay BUT she needs a letter from my
mother saying it's okay to switch me out. I was wondering if you can write the...
Saturday, October 5, 2013 at 6:53pm by unknown
Are both the electron domain geometry and molecular geometry of HCN linear?
Monday, December 7, 2009 at 7:38pm by Trish
math geometry help
learning geometry i need to determine whether each pair of figures is similar. and explain why or why not: since I cant draw can you help me with this pls im so lost. 1) triangle xyz <x 65 degrees <y
is 65 <z is 50 line xy is 2.55 line yz is 3 line zx is 3 triangle ...
Tuesday, January 31, 2012 at 8:02pm by marko
how can you tell in geometry that the equations are parallel, perpendicular , neither, either?
Saturday, September 25, 2010 at 7:27am by 14
Anyone have the Amsco's geometry textbook and can help me out with some homework?
Monday, November 28, 2011 at 8:42pm by Anita
I Need help doing a geometry proof. In the figure below, PQ=RS. Prove PR=QS. ____________________ P Q R S
Sunday, February 5, 2012 at 2:30pm by Jennifer
Prove that the statement, “Rectangles do not exist,” is true in hyperbolic geometry.
Wednesday, February 20, 2013 at 7:56pm by denise
I need help unscrambling these geometry vocabulary words senligtacirpalse pitprceenostel nyenagliltuiqitrea gemhetrohnie
Sunday, January 31, 2010 at 8:18pm by Daisy
in elliptic geometry prove that the length of the summit of a saccheri quadrilteral is less than the base
Wednesday, May 4, 2011 at 11:09pm by al
Please help on geometry - analytic geometry
What is the area, in square units, of a triangle whose vertices are at (4,-1), (10,3) and (4,5) ? How to do this?
Friday, February 22, 2013 at 5:21pm by Knights
prove that The sum of the angles in any triangle is 180 degrees is true in Euclidian geometry.
Tuesday, February 26, 2013 at 11:34am by denise
What is the Lewis structure formula, electron pair geometry and the molecular geometry for SO4^2-
Wednesday, August 7, 2013 at 1:12am by rose
what kind of e-pair geometry, molecular geometry, and bond angles does C4H10 exhibit? I'm confused.
Monday, May 5, 2008 at 10:01pm by natash
In geometry, a common standard setting for enlarging on a copy machine is 121 percent. What does this mean?
Monday, November 25, 2013 at 2:06pm by Martha
what geometry figure begins and ends in the same point
Wednesday, January 28, 2009 at 8:02pm by john smith
In geometry, can a counterexample be used to determine if a conjecture is false or not?
Friday, January 15, 2010 at 4:07pm by erin
Can somebody help me with Proofs in Geometry Given:M<ABE=M<CBD
Tuesday, June 19, 2012 at 6:19pm by Jay
In Euclidean geometry, the sum of the measures of the interior anglesof a pentagon is 540o. Predict how the sum of the measures of the interior angles of a pentagon would be different in spherical
Friday, June 22, 2012 at 8:15am by Dillon
Electron DOMAIN GEOMETRY & the MOLECULAR GEOMETRY of the following compounds: 1) ICl3 2) PCl3 3) XeF4 4) SF6 5) TeCl4
Friday, March 30, 2012 at 2:07am by BMW
Help with Geometry please! Do you have any specific questions?
Monday, October 16, 2006 at 3:43pm by Margie
Why don't parallel lines exist in elliptical geometry?
Tuesday, July 13, 2010 at 1:07pm by BOB
prove that rectangles do not exist in hyperbolic geometry
Wednesday, March 13, 2013 at 3:36pm by denise
Find the Electron DOMAIN GEOMETRY & the MOLECULAR GEOMETRY of the following compounds: 1) ICl3 2) PCl3 3) XeF4 4) SF6 5) TeCl4
Friday, March 30, 2012 at 12:41pm by BMW
In euclidean geometry, the sum of the measures of the interior angles of a pentagon is 540. predict how the sum of the interior angles of a pentagon would be different in spherical geometry
Thursday, August 30, 2012 at 7:53am by MIiranda
How to get quick responses to your math questions
Math is a wide subject, ranging from K to 11, college and university. Then there is algebra, trigonometry, geometry, arithmetic, calculus, number theory, ... etc. Not all teachers answer all math
questions (many do). If you would give a little more detail on which branch of ...
Saturday, May 25, 2013 at 9:45am by MathMate
to all the people that helped me with my recent geometry questions thankyou my grade I received was 45 out of 50 points best I've ever done in my class.
Thursday, January 13, 2011 at 5:32am by crystal
whats a complimentery line? Thank you for using the Jiskha Homework Help Forum. If you are studying Geometry: http://astronomyonline.org/Science/Geometry.asp
Wednesday, August 15, 2007 at 6:16pm by Sarahbeth
Ok i don't understand this question on my Geometry test. It says: On days that it rains, I carry an Umbrella 1. Rewrite the above statement as a conditional Can someone help me here?
Wednesday, November 14, 2007 at 1:59pm by Christina (:
if the midpoints of the sides of a parallelogram in succession are joined, the quadrilatral formed is a a parallogram. Prove this theorem in euclidean geometry using analytic techniques
Thursday, April 4, 2013 at 7:29pm by denise
In geometry what kind of algebra do you use?
Friday, September 10, 2010 at 12:26pm by Samantha
In Geometry how do I calculate the precion of 14 meters?
Tuesday, August 21, 2012 at 7:42pm by jane
why do rectangles not exist in hyperbolic geometry
Monday, February 18, 2013 at 3:01pm by denise
set (math)
A class of 47 students took examination in algebra and geometry. If 29 passed algebra and 26 passed geometry and 4 failed in both subjects, how many passed both subjects? With solution please, tnx...
Wednesday, January 23, 2013 at 7:25am by HELP HELP HELP MNHS
Geometry :)
Use coordinate geometry to prove that the quadrilateral formed by connecting the midpoints of a kite is a rectangle. I have no Idea how to do this problem, so if anyone could help I would be very
greatfull. :)
Tuesday, January 15, 2013 at 2:14pm by Bob
Math- repost
This is a geometry question about disjunctions. The symbol "v" represents "or". "~" stands for "negation of" Here are two questions I need help with: 1. When p v ~q is false, then p is ______ and q
is _____. I think "false" belongs in both spaces. 2. When ~p v q is false, then...
Sunday, September 28, 2008 at 10:01pm by Anonymous
Which of the following statements from Euclidean geometry is also true of spherical geometry? a. A line has infinite length. b. Two intersecting lines divide the plane into four regions. c. Two
perpendicular lines create four right angles. d. The intersection of two lines ...
Friday, August 5, 2011 at 10:34am by Lily
geometry, thanks
I just wanted to thank everyone that helped me on my geometry problem. I actually found the answer using the pythagorean theorem and knowing ratio rules that apply to triangles. Thanks again.
Thursday, July 13, 2006 at 4:10pm by Kaytee
At rosemead high school, 94 students take biology, 86 take ethnic studies, and 95 take geometry. thirty-seven students take both biology and geometry, 43 take geometry and ethnic studies, and 42 take
ethnic studies and biology. twenty eight students take all three subjects. ...
Monday, September 13, 2010 at 6:51pm by Audrey
At rosemead high school, 94 students take biology, 86 take ethnic studies, and 95 take geometry. thirty-seven students take both biology and geometry, 43 take geometry and ethnic studies, and 42 take
ethnic studies and biology. twenty eight students take all three subjects. ...
Monday, September 13, 2010 at 9:25pm by Audrey
Need help please: (3,0) (7,6) (x-7)2 + (y-6)2 = 7(2) (x-7)2 + (y-6)2 = 49 Having difficulty with this problem.Can someone help? geometry - Reiny, Saturday, March 31, 2012 at 8:07am You don't say what
the actual problem is. I see the resemblance to finding the equation of a ...
Saturday, March 31, 2012 at 7:23pm by Kel
Need help please: (3,0) (7,6) (x-7)2 + (y-6)2 = 7(2) (x-7)2 + (y-6)2 = 49 Having difficulty with this problem.Can someone help? geometry - Reiny, Saturday, March 31, 2012 at 8:07am You don't say what
the actual problem is. I see the resemblance to finding the equation of a ...
Saturday, March 31, 2012 at 7:23pm by Kel
Hi, I have to do a project on the Shell Method in my Geometry class. I have been trying to find information on that topic, but have no luck... can someone point me to the right direction as of what
keyword to search for, or is there any sites or books or video that have such ...
Saturday, April 14, 2007 at 11:50pm by Billy
what a good way to learn geometry if you are completely clueless on what to do and you have a 65 and you only have 2 weeks to correct that
Monday, September 22, 2008 at 9:10pm by Jasmyn
AP Chemistry- PLEASE HELP
The electron-domain geometry and molecular geometry of iodine trichloride are __________ and __________, respectively.
Tuesday, January 29, 2013 at 7:12pm by Sonya
i'm in my geometry book and i don' get number 28. it says the quotient of two proper fractions is a proper fraction. i'm a post to find a counterexample. I NEED HELP
Tuesday, August 18, 2009 at 11:35pm by jennyo
What does this symbol mean in Geometry ¡ü?
Wednesday, October 27, 2010 at 10:06am by Rhonda
the set of lessons in this geometry course is:
Thursday, April 17, 2014 at 7:28pm by kim
explain the importance of a centroid in design and manufacturing.include a description of how the cenroid of a triangular component of a design can be found using analytic geometry
Sunday, October 31, 2010 at 2:52pm by crystal
A tetrahedron is a triangular pyramid in which each face is an equilateral triangle. Calculate the volume of a tetrahedron that has all its edges 6 cm in length. Hint: You will need to know some
geometry and trigonometry.
Tuesday, May 15, 2012 at 1:42am by Lucas
The coordinates for a rhombus are given as (2a, 0), (0, 2b), (–2a, 0), and (0, –2b). Write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate
geometry. Be sure to include the formulas.
Tuesday, January 8, 2013 at 9:10pm by Anna
Which statement from Euclidean geometry is also true in spherical geometry? a. A line has infinite length b. Two intersecting lines divide the plane into four regions c. Two perpendicular lines
create four right angles d. The intersection of two lines create four angles. I ...
Tuesday, March 4, 2014 at 2:01pm by Kylee
i also have a math test tomorrow and i get sooo confused with irregular shapes in geometry
Thursday, March 29, 2012 at 5:28pm by Sheil
Trigonometry/Geometry - Inequalities
Let a, b, and c be positive real numbers. Prove that sqrt(a^2 - ab + b^2) + sqrt(a^2 - ac + c^2) is greater or equal to sqrt(b^2 + bc + c^2). Under what conditions does equality occur? That is, for
what values of a, b, and c are the two sides equal? This looks like a geometry...
Saturday, December 7, 2013 at 4:58pm by Sam
Geometry HELP
Using coordinate geometry to prove that the diagonals of a square are perpendicular to each other. Given: Vertices are at A(0,0), B(a,0), C(a,a) and D(0,a) Slope of AC=1; Slope of BD=-1 Type your
proof. (someone plz help cuz i don't understand)
Thursday, June 10, 2010 at 8:58am by Ariel
Math or geometry
Friday, June 8, 2012 at 9:35pm by Diana
If 2x=20, does x=10?
Thursday, September 6, 2012 at 9:51am by Ashley
# Electron Pairs # Lone Pairs Electronic Geometry Molecular Geometry Lewis Structure Sketch of Shape I need to do the above things for SBr6 and BrCl5. If anybody can help that would be great!
Tuesday, April 26, 2011 at 3:18pm by Jake
In my geometry book I am really stumped on these question.This is what it says to do, "Find a pattern for each sequence. Use the pattern to show the next two terms."And here are th questions I am
stuck on with the #'s. 7.O,T,T,F,F,S,S,E,... 8.J,F,M,A,M,... 13.George,John,...
Wednesday, August 22, 2007 at 7:18pm by LizardGuy
In a triangle ABC, angle B is 3 times angle A and angle C is 8 degrees less than 6 times angle A. The the size of the angles. Size of angle A is = Size of angle B is = Size of angle C is = can
someone please help me with this..? I am very weak at geometry...and math in general...
Saturday, September 26, 2009 at 4:44pm by Anonymous
math- geometry
Wednesday, May 5, 2010 at 11:03pm by Rick
what is terminal geometry
Sunday, September 4, 2011 at 5:53pm by Gen
Monday, January 14, 2013 at 6:36pm by jsabsmndfsm,
what is the difference between two-dimensional and three-dimensional geometry?
Friday, May 20, 2011 at 6:02pm by naimah
Math Geometry
Geometry Question: Prove the symmetric Property for congruence of triangles. Given: ∆ABC≅ ∆DEF Prove: ∆DEF≅ ∆ABC
Wednesday, May 18, 2011 at 8:27am by Reigel
What is the Answer to number 37?
Tuesday, August 26, 2008 at 6:53pm by Loulou
do you know how to unscramlbe MATH words
Tuesday, November 17, 2009 at 10:37pm by stacey
What are thre 3 figures of a 36 cm 2?
Thursday, April 7, 2011 at 3:47pm by Azhye
Can you be my math tutor for free?
Wednesday, April 25, 2012 at 8:24pm by A'Mya
Who knows how to do similar polygons in Geometry?!!!
Thursday, January 23, 2014 at 2:12pm by Kitty
Math (Geometry)
What is the midpoint of line BC with B (2,0) and C (-4,6)?
Thursday, September 6, 2012 at 3:28pm by Chris
what is the molecular geometry of trans-difluroethylene: trans-C2H2F2? I know how to draw the lewis structure but I'm not sure how to find the molecular geometry. I think it's a tetrahedral
Saturday, May 24, 2008 at 2:02pm by Cabel
Use your geometry tools to draw a convex hexagon with two consecutive sides measuring 5 cm and three consecutive angles measuring 130°
Saturday, September 8, 2012 at 4:19pm by Astrid
AP Chem please help!
The electron-domain geometry and molecular geometry of iodine trichloride are __________ and __________, respectively. The answer is trigonal bipyramidal, T-shaped, respectively- I do not understand
the approach
Tuesday, January 29, 2013 at 7:58pm by Sonya
Basic Math & Geometry
How do I find the ratio of 5&3/5 to 2&1/10?
Friday, November 30, 2007 at 8:39pm by Patricia
Geometry Math
1.What is the Distance Between Points A(4,-1) and B(-6,-3)?
Sunday, September 6, 2009 at 11:40pm by Tuhin
Geometry / Math
Is a rectangular prism a polyhedron Thanks
Tuesday, February 23, 2010 at 10:01pm by Anonymous
Construct a truth table for q ->~p
Monday, July 11, 2011 at 11:13pm by Jen
if the circumference of a circle is 12.56 m then what is its diameter?
Monday, February 18, 2013 at 8:00pm by Anonymous
I don't undestand this question in my geometry textbook. Could nayone explain it to me? Can you fit all of the interior angles of a quadrilateral around a point without overlap? What about the
interior angles of a pentagon?
Thursday, December 22, 2011 at 12:07pm by Anonymous
i need help with this geometry prblem my math book is online on this site: w w w . k e y m a t h . com / D G 3 the class passcode is: 1574-4c524 The chapter is 0 Lesson 0.2 pg. 9 problem number 6
URGENT !!!
Sunday, September 19, 2010 at 5:31pm by vaishnavi
Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. In which cases do you expect deviations from the idealized bond angle?1.pf3 2.sbr2
3.ch3br 4.bcl3
Thursday, August 23, 2012 at 10:55pm by ana
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>
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To See A Circle in A Pile of Sand
The other day I sent a small assignment to a group of people in order that they could play with statistics and become more interested with this subject. The data-generating process was quite simple:
first I generated 20000 random numbers (10000 rows, 2 columns) from N(0, 1) and then add 10000 rows of numbers which lie exactly on a circle; at last I provided this data in a randomized order so
people cannot easily discover the pattern just from the numbers.
The question is, how to reveal the particular pattern in this "pile of sand"? Let's look at the original plot:
What can we observe from this scatter plot? Perhaps nothing but "a pile of sand". However, if we choose alternative ways to create the plot again, things will be completely different. Here are my
1. Use Semi-transparent Colors
Actually there are 10000 points lying on the circle, so the critical problem is overlapping. In order to show the degree of overlapping, we can use semi-transparent colors, because the color will be
more opaque if there are many points at the same place.
2. Set Axes Limits
If we look closer into the plot, the scene will also be different. For example, we only plot the data in the range [-1, 1].
3. Plot with Smaller Point Symbols
Certainly, small symbols can prevent overlapping effectively in this case.
4. Draw A Subset of the Data
As the problem is that there are too many data points, why not draw a subset and try a scatter plot first? For example, here we have sampled 1000 rows of data and the plot is like this:
5. Estimate the 2D Density
The R package KernSmooth has provided functions to estimate the 1D or 2D density.We can further examine the shape of this 2D density using the package rgl. Here is an animation recorded to illustrate
the 2D density.
The R code for the above plots & animation is as follows:
# generate the data
x = rbind(matrix(rnorm(10000 * 2), ncol = 2), local({
r = runif(10000, 0, 2 * pi)
0.5 * cbind(sin(r), cos(r))
x = as.data.frame(x[sample(nrow(x)), ])
# original plot
# transparent colors (alpha = 0.1)
plot(x, col = rgb(0, 0, 0, 0.1))
# set axes lmits
plot(x, xlim = c(-1, 1), ylim = c(-1, 1))
# small symbols
plot(x, pch = ".")
# subset
plot(x[sample(nrow(x), 1000), ])
# 2D density estimation
fit = bkde2D(as.matrix(x), c(0.1, 0.1))
# perspective plot by persp()
persp(fit$x1, fit$x2, fit$fhat)
# perspective plot by OpenGL
rgl.surface(fit$x1, fit$x2, 5 * fit$fhat)
# animation
M = par3d("userMatrix")
play3d(par3dinterp(userMatrix = list(M,
rotate3d(M, pi/2, 1, 0, 0), rotate3d(M, pi/2, 0, 1, 0),
rotate3d(M, pi, 0, 0, 1))), duration = 20)
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Keywords: Ampere's Law
The lesson begins with a demonstration introducing students to the force between ... (more)
The lesson begins with a demonstration introducing students to the force between two current carrying loops, comparing the attraction and repulsion between the loops to that between two magnets.
After formal lecture on Ampere's law, students begin to use the concepts to calculate the magnetic field around a loop. This is applied to determine the magnetic field of a toroid, imagining a toroid
as a looped solenoid. (less)
Material Type:
Eric Appelt
Read the Fine Print
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Undergraduate Network
The Fields Undergraduate Network (FUN) organizes monthly meetings to explore different areas of mathematical research. The theme, as well as the host university will vary from month to month. All
interested undergraduates are welcome to attend. We especially encourage participation by members of student math societies.
10:00am - Networking
10:30am - Interview of Hugh Williams
11:00am - What keeps our secrets safe?
12:00pm - Lunch
1:30pm - Role of L-functions in number theory
2:30pm - Panel Discussion
3:00pm - Ergodic approaches to number theory
Hosting Student Group:
Math Union at University of Toronto
Interviewer and Discussion Moderator: Richard Cerezo, Co-President of Math Union and Co-Founder of FUN
What keeps our secrets safe?
Hugh Williams, The Cryptologic Institute and University of Calgary
Control over who knows what about us, for what purposes, and to whom it is disclosed, is of profound concern to anyone making use of electronic communication devices. In the e-world, personal
information is in a very real sense the person. Thus, it is essential that we have confidence in the capacity of the information collector to secure our personal information. This can only be
achieved through the very technology that threatens our privacy. One important ingredient in these privacy-enhancing technologies is cryptography.
Briefly put, cryptography is the study and development of techniques for rendering information unintelligible to all but intended recipients of that information. If a sender and receiver of a
message wish to communicate over an insecure channel (mobile phone, internet) and want to ensure that no other unauthorized party can read their transmission, they will make use of a particular
cryptosystem. A conventional cryptosystem can be thought of as a large collection of transformations (ciphers), any one of which will render the original message (plaintext) to unintelligible
ciphertext, but in order for the receiver to read the message, he must know which particular transformation was used by the sender. The information that identifies the transformation used by the
sender is called the key. It is important to point out that if an eavesdropper acquires some message and its encrypted equivalent, he should not be able to extract the key from this information.
Nor should the system be vulnerable to an adaptive attack; such attacks make use of information previously acquired to obtain new information from the sender and so on until the system is broken.
This is what makes cryptography fascinating. How can we protect our communications against these kinds of attacks? Remember also that a good cryptosystem must resist an attack even from the
inventor of the system.
In this talk, which is intended for a non-specialist audience, I will describe from a historical perspective several features of modern encryption techniques.
Role of L-functions in number theory
Henry Kim, University of Toronto
L-functions are very special type of meromorphic functions of one complex variable. On the surface it is not clear why the L-functions play decisive roles. L-functions are associated to
arithmetic-geometric objects such as Galois groups, elliptic curves, and also modular forms. Riemann zeta function is used in the study of the distribution of prime numbers. It gives rise to
Riemann hypothesis. L-functions attached to elliptic curves give rise to Birch-Swinnerton-Dyer conjecture. Those are two of the seven millennium prize problems by Clay Mathematics Institute. I
will try to survey very briefly these and other roles L-functions play.
Ergodic approaches to number theory
Leo Goldmakher, University of Toronto
Ergodic theory is concerned with the long term statistical behaviour of a dynamical system. Although the theory was originally motivated by questions in statistical mechanics, in recent years it
has found spectacular applications to combinatorics, harmonic analysis, number theory, and representation theory. In this talk I will discuss some of the interactions between ergodic theory and
number theory, focusing on the pioneering work of Furstenberg on Szemeredi's theorem.
List of Confirmed Participants as of March 1, 2011:
│Full Name │University Name │
│Carmichael, Keegan │University of Western Ontario │
│Cerezo, Richard │University of Toronto │
│Chung, Ha Yoon │University of Toronto │
│da Silva, Sergio │University of Toronto │
│Dias, Manisha │University of Waterloo │
│Heidari Zadi, Amir Hossein│University of Toronto │
│Hill, David │University of Toronto │
│Hu, Samson │University of Waterloo │
│Isufllari, Henrieta │University of Toronto │
│Kang, Dongwoo │University of Toronto │
│Kinross, Alison │McMaster University │
│Lee, Seung-Jae │University of Toronto │
│Milcak, Juraj │University of Toronto │
│Moon, SeokHwan │University of Toronto │
│Pashley, Bryanne │University of Waterloo │
│Pistone, Jamie │University of Toronto │
│Shoukat, Affan │York University │
│Vlasova, Jelena │University of Toronto at Mississauga │
│Yee, Yohan │McMaster University │
│Zaidi, Ali-Kazim │University of Toronto │
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Cardinality and Ordinality: life was so easy until zero came along
When we are dealing with the counting numbers — 1, 2, 3, and so on — it is easy to rank them in order. One is the first counting number, two is the second counting number, and three is the third.
We don’t have to worry about mixing up the value of a number — its cardinality — with the order in which it arrives — its ordinality — since they are essentially the same thing. For years, this was
the state of affairs, and everybody was happy. But as zero came into the fold, the neat relationship between a number’s cardinality and its ordinality was ruined. The numbers went 0, 1, 2, 3: zero
came first, one was second in line, and two was in third place (i.e. 0 has a cardinal value of 0 and an ordinal value of 1; 1 has a cardinal value of 1 and an ordinal value of 2; 2 has a cardinal
value of 2 and an ordinal value of 3). No longer were cardinality and ordinality interchangeable.
– Zero, the Biography of a Dangerous Idea by Charles Seife
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FOM: for the record
Randall Holmes holmes at catseye.idbsu.edu
Wed Mar 24 16:55:11 EST 1999
I disagree with Friedman's statement
"ZFC is the complete formalization of mathematics"
or at least regard it as a "seriously misleading formulation".
The reasons that I object to this are similar to the reasons that he
objects to statements of mine along the lines of:
"first-order ZFC cannot express the notion "the set of natural numbers"";
my statement contains implicitly a notion of "express" which he
doesn't like, while his contains implicitly a notion of "mathematics"
which I don't like (and which I suspect would be unacceptable to many
readers of this list; as would my implicit definition of "express").
Given what I mean by the word "mathematics", his statement translates
to a falsehood, pure and simple. However, on further examination of
the context, I can tell what he means (I still probably disagree, but
the disagreement is much less profound; it is in the area of sociology
rather than f.o.m.).
Mathematics is the study of abstract structures (my definition!), with
special attention to certain traditionally studied structures.
Con(ZFC) is an (admittedly rather bizarre) statement about one of the
traditionally studied abstract structures, so it is certainly a
mathematical statement. By saying this, I am not disagreeing with
Friedman (though I probably do disagree, but that is another
story--see next paragraph). But I could have challenged him on this
point without taking the trouble to figure out what he meant; please
take note!
He apparently means by "mathematics" the deductive closure of a set of
propositions accepted by the editors of mathematical journals for use
without special mention in journal articles. I really don't think
this is a caricature; if Friedman does think so, I'd like some
expansion on his remarks in a recent posting to dispel this
impression. In this sense, Con(ZFC) might not be a mathematical
statement. But I'm not sure the Axiom of Choice is a mathematical
statement using this criterion (observing the behaviour of some
mathematicians). Martin's Axiom is certainly not a mathematical
proposition by this criterion; from my standpoint it certainly is.
I suggest that in responding to one another's posts we take into
account what participants mean by commonly used terms. I don't think
that it is reasonable to expect participants to abandon their
individual usages (there are no universally acceptable definitions of
such terms as "mathematics"!); but I think that we should both take
care to say something about our peculiar usages (as I did with
"express") and take care that when we dispute a proposition someone
has made we take the trouble to determine what they mean.
Friedman, by the way, gets points for this in the case of "express";
he did make it clear that he knew what I meant and criticized me on
that ground (which is OK). But he does not get points for making an
extremely controversial statement which is based on an eccentric
definition of the word "mathematics" without making it clear that that
is what he is doing!
And God posted an angel with a flaming sword at | Sincerely, M. Randall Holmes
the gates of Cantor's paradise, that the | Boise State U. (disavows all)
slow-witted and the deliberately obtuse might | holmes at math.idbsu.edu
not glimpse the wonders therein. | http://math.idbsu.edu/~holmes
More information about the FOM mailing list
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Activity: Counting Coins
Collecting and counting money!
Step 1: Collect all the coins in the house! (I am sure your family won't mind when you say you are learning mathematics.)
Step 2: Sort them into groups (5c in one group, 10c in another, etc)
Step 3: Count how many in each group, then calculate the group value.
Example: You count your 5c coins, and find you have 15 of them.
5c × 15 = 75c
You can use this form:
│Coin│How Many │Value │
│ Total Value:│ │
Step 4: Add up all the group totals so you know how much money you have!
Tell everyone what a good job you did ...
... and ask if you can keep the money!
Tip: If you make stacks of coins exactly 10 high it is easier to calculate the totals.
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63v squared-7 - WyzAnt Answers
63v squared-7
Tutors, please
to answer this question.
First you'll want to see if each term in the polynomial has any common factors, in this case 63 and 7 both have a common factor of 7, since 63 = 9*7 and 7 = 1*7. So we factor out their greatest
common factor, which in this case is 7:
63v^2-7 =
Can we factor (9v^2-1) any more? Well, we can't factor it by trying to pull out greatest common factors from each term, since 9 and 1 do not share any factors greater than 1. But it can actually be
factored in a different way...
9v^2-1 is in a special kind of form, that requires a special kind of factoring called "Differences of Squares."
(a^2-b^2) = (a - b)(a + b)
We can use this formula for 9v^2-1 since 9v^2 is the sqaure of 3v and 1 is the square of 1. So we write:
7(9v^2-1) =
Which is your final answer since it cannot be factored any further.
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Independent Study | Department of MathematicsIndependent Study
Yuan-Jen Chiang has worked with students on the following topics: Topics in Analysis, Differential Geometry, Tensors and Relativity, Partial Differential Equations, Introduction to Harmonics Maps,
Tensor Analysis in Physics and Engineering, Fractional Calculus, Topics in Euclidean and Non-Euclidean Geometries, Applications of Mathematics in Electrical Engineering, and Fourier Analysis.
Gary Collier has led directed studies in generalized function theory.
Julius Esunge welcomes the opportunity to work with students on problems in probability, statistics and financial mathematics. Examples of possible topics include the following: simulation of random
variables and probability distributions; properties and applications of Brownian motion, Brownian bridge and fractional Brownian motion; applications of limit theorems in probability; applications of
stable distributions for heavy tailed data; statistical analysis and properties of actuarial models. Julius also enjoys directing reading courses/seminars towards actuarial examination preparation..
Randall Helmstutler has directed students in topological groups, homotopy and Lie theory, category theory, and advanced linear algebra. His general interest is in working with students wishing to
delve into advanced topics in abstract algebra and topology. In the field of algebra this includes higher group theory and linear algebra, module theory, and homological algebra. Topologically, he
would enjoy directing studies in homotopy and covering spaces, differential topology, and homology theory.
Debra Hydorn is interested in working with students on research projects in probability and statistics, including probability simulations and statistical modeling. She is particularly interested in
GIS, biostatistics and environmental applications. She has led directed studies on regression analysis, linear models, statistical computing, education methods for statistics, preparation for
actuarial exams, and for Math 351/352 and Math 381/382 during semesters in which these courses have not be offered.
Janusz Konieczny has directed students in the study of field theory, topics in geometry, applications of linear algebra, and semigroup theory.
Jangwoon (Leo) Lee is interested in working with students in various areas of applied mathematics including partial differential equations, scientific computations, optimal control problems, and
numerical methods for mathematical model equations such as stochastic/ partial differential equations.
J. Larry Lehman has led directed studies in Galois theory, algebraic number theory, and elliptic curves. He would also be interested in working with students in analytic number theory and other
advanced topics in number theory.
Keith Mellinger has directed students in various areas of discrete mathematics including coding theory, design theory, finite geometry, combinatorics, graph theory, and cryptography. He has also led
a problem-solving seminar as a directed study. He is interested in continuing to work with students in any of these areas.
Suzanne Sumner has worked with students on mathematical modelling, fractal geometry, dynamical systems, applications in finite mathematics, operations research, epidemiological models, and
differential equations.
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[SciPy-user] lagrange multipliers in python
fdu.xiaojf@gmai... fdu.xiaojf@gmai...
Mon Jun 18 08:18:16 CDT 2007
Hi all,
fdu.xiaojf@gmail.com wrote:
> My last email was composed in a hurry, so let me describe my problem in
> detail to make it clear.
> I have to minimize a multivariate function F(x1, x2, ..., xn) subject to
> multiple inequality constraints and equality constraints.
> The number of variables(x1, x2, ..., xn) is 10 ~ 20, the number of
> inequality constraints is the same with the number of variables( all
> variables should be no less than 0). The number of equality constraints
> is less than 8(mainly 4 or 5), and the equality constraints are linear.
> My function is too complicated to get the expression of derivate easily, so
> according to Joachim Dahl(dahl.joachim@gmail.com)'s post, it is probably
> non-convex. But I think it's possible to calculate the first derivate
> numerically.
> I have tried scipy.optimize.fmin_l_bfgs_b(), which can handle bound constraints
> but seems cannot handle equality constraints.
> Mr. Markus Amann has kindly sent me a script written by him, which can
> handle equality constraints and is easy to use. The method used by
> Markus involves the calculation of Jacobian, which I don't
> understand.(Sorry for my ignorance in this filed. My major is chemistry,
> and I'm trying to learn some knowledge about numerical optimization.)
> However, it seems that the script cannot handle inequality constraints.
> (sorry if I was wrong).
> I hope my bad English have described my problem clearly.
> Any help will be greatly appreciated.
> Best regards,
> Xiao Jianfeng
I have found COBYLA(http://www.jeannot.org/~js/code/index.en.html#COBYLA),
and it has a Python interface, which make it very easy to use.
It seems that COBYLA is capable to handle both equality and inequality
constraints together. Here is an example of COBYLA(it actually example.py
shiped with COBYLA):
##--------------------begin of example.py------------------
#!/usr/bin/env python
# Python COBYLA example
# @(#) $Jeannot: example.py,v 1.2 2004/04/13 16:35:11 js Exp $
import cobyla
# A function to minimize
# Must return a tuple with the function value and the value of the constraints
# or None to abort the minimization
def function(x):
f = x[0]**2+abs(x[1])**3
# Two constraints to represent the equality constraint x**2+y**2 == 25
con = [0]*2
con[0] = x[0]**2 + x[1]**2 - 25 # x**2+y**2 >= 25
con[1] = - con[0] # x**2+y**2 <= 25
return f, con
# Optimizer call
rc, nf, x = cobyla.minimize(function, [-7, 3], low = [-10, 1], up = [3, 10])
print "After", nf, "function evaluations, COBYLA returned:", cobyla.RCSTRINGS[rc]
print "x =", x
print "f =", function(x)[0]
print "con = ", function(x)[1]
print "exact value = [-4.898979456, 1]"
##--------------------end of example.py------------------
Would somebody who is familiar with COBYLA tell me that is COBYLA suitable
for my problem?
Xiao Jianfeng
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Electoral Marimekko Plots
December 6, 2010
By d sparks
To be reductive, visual displays of quantitative information might be reasonably categorized on a continuum between “data display” and “statistical graphics.” By statistical graphics, I mean a plot
that displays some summary of or relationship amongst several variables, likely having undergone some processing or analysis. This may be as simple as a scatterplot of a primary independent variable
and the dependent variable, a boxplot, or a graphical regression table.
In this reductive scheme, then, “data displays” present variables in raw form — for use in exploratory data analysis, or perhaps just to offer the viewer access to all of the data. Where “statistical
graphics” might be best served by simplicity and minimalism in design, such that a single idea might be conveyed clearly, “data displays” will tend to be inherently complex, and require effort from
both the creator and viewer to parse meaning from the available information.
Where statistical graphics are ideal for presenting conclusions, data displays are useful for generating ideas, and optimally, permitting the relatively rapid identification of relationships between
multiple variables. On top of this, I might add that many of the more well-regarded data displays of recent note offer macro-level insight as well as the opportunity to ascertain specific details
(for this, interactivity is often valuable, as in the internet-classic New York Times box office visualization).
As several recent posts suggest, I am interested in finding ways to successfully and clearly convey multidimensional data, and have been focusing on political data as it varies across geopolitical
units and time. Here I offer an approach which departs from the spatial basis of other recent efforts in favor of allowing the position of graphical objects to convey other variables.
This type of plot is called, variously, a spinogram, a mosaic plot, or a marimekko — and is not dissimilar from a treemap with a different organizational structure (other examples). The utility of
this plot type is that it can spatially convey four numeric variables (x position, y position, height, width), and color can be added to incorporate up to three additional variables (R, G, B).
Further, there is a straightforward geometric interpretation of each cell: the areas of each (in this case, width/state turnout ×height/county proportion of state turnout) are directly comparable.
Unlike a stacked bar plot, the width of each column conveys information, permitting height to convey proportion rather than count. Further, columns and cells within columns can be sorted to express
the ordering of variables of interest. In some ways, these can be seen as extreme reinterpretations of (Dorling) cartograms, in which not only the size and shape of political boundaries, but also
their position, are distorted by other variables.
In the plots above, cells are colored according to the strength of Democratic (blue), Republican (red), and other party (green) support, and counties whose turnout represents greater than 1% of the
total turnout in an election are labeled.
I present two different layouts for the cells in each plot. The first arrays states left-to-right in order of the number of votes cast in an election, and sorts counties bottom-to-top in the same
order. Thus, more populous states are on the right, and more populous counties are at the top of the plot. This arrangement allows the viewer to observe the effects of population density both within
and across states, and may better facilitate tracking changes in county or state politics over time.
The second layout sorts states left-to-right, and counties bottom-to-top in order of the Democratic share of the two party vote (Dem Votes / (Dem Votes + Rep Votes)). Thus, more Democratic-leaning
(relative to Republican) states are on the right, and counties that were more supportive of Democratic candidates are at the top. I believe that this arrangement makes it easier to discern overall
trends in partisanship across time, as the total “sum” of red within a diagram is relatively easy to compare to the total “sum” of blue (and green).
I have attempted to make my R code fairly general, and it is available for download here, although it will obviously require some modifications for other applications. Our approaches differ, but
another instructive example can be found at Learning R.
for the author, please follow the link and comment on his blog:
David B. Sparks » r
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Monte Sereno, CA Algebra 2 Tutor
Find a Monte Sereno, CA Algebra 2 Tutor
...I worked as a database application support specialist for years and before that a computer application programmer. My enthusiasm for math came from my teachers in secondary schools and college
despite the fact that I studied computer science as my major in college. I achieved my goals for college and the goal of becoming a certificated teacher in math.
5 Subjects: including algebra 2, geometry, Chinese, prealgebra
...While not all of them got A’s, those that regularly attended tutoring did see improvements in their grades. I believe that mathematics should not get in the way of learning the natural
sciences, so I am capable of tutoring in that as well. My passion is in chemistry but the quantitative nature of the natural sciences means that I am fluent in algebra through calculus.
24 Subjects: including algebra 2, chemistry, reading, physics
...Using molecular cloning, I have also conducted research in biochemistry and neuroscience laboratories at UC Riverside and San Diego State University. I have extensive teaching experience in
college-level general chemistry, biochemistry and neuroscience from three different universities, as well ...
18 Subjects: including algebra 2, chemistry, physics, calculus
...I hope you will also learn how to keep learning on your own!I am a professional research biologist, with several peer-reviewed publications and a biotechnology patent to my name. I hold a
Bachelors' degree in Biochemistry from U.C. Berkeley, and a PhD in Immunology from Stanford.
17 Subjects: including algebra 2, chemistry, writing, statistics
...Through high school and college I was a math tutor and have been tutoring roughly 20 hours each week as a hobby since August 2013. I have worked with at least 10 different students in all
major mathematics classes starting with Algebra through Advanced Calculus. I have worked with many college ...
35 Subjects: including algebra 2, reading, calculus, geometry
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Calculus books
April 26th 2009, 08:04 PM #1
Calculus books
I want to buy a calculus book to study during my holidays, but I don't know how to choose.
Can you give me some suggestions?
These are a few example I searched from internet:
Thomas' Calculus, Media Upgrade
International Edition
11th Edition
George Thomas, Maurice Weir, Joel Hass, Frank Giordano
Aug 2007, Paperback, 1228 pages
ISBN13: 9780321526793
ISBN10: 0321526791
Calculus and Its Applications
International Edition
12th Edition
Larry Goldstein, David Schneider, David Lay, Nakhle Asmar
Apr 2009, Paperback, 656 pages
ISBN13: 9780321639448
ISBN10: 0321639448
A Complete Course
7th Edition
Robert Adams, Christopher Essex
Feb 2009, Hardback, 1152 pages
ISBN13: 9780321549280
ISBN10: 0321549287
Brief Calculus & Its Applications
12th Edition
Larry Goldstein, David Schneider, David Lay, Nakhle Asmar
Apr 2009, Paperback, 512 pages
ISBN13: 9780321568564
ISBN10: 0321568567
University Calculus
Elements with Early Transcendentals: International Edition
Joel Hass, Maurice Weir, George Thomas
Mar 2008, Paperback, 791 pages
ISBN13: 9780321552105
ISBN10: 0321552105
Stewart's calc is often recommended. this is the 5th edition. it is really cheap now too, so that's another plus. the latest is the 6th edition
Stewart's calc is often recommended. this is the 5th edition. it is really cheap now too, so that's another plus. the latest is the 6th edition
Why his books are often recommended?
What are the differences between 5th and 6th editions?
Stewart is probably one of the most popular Calc books out there, used in the most classes, so it's recommended the most.
Of the ones that you listed, I only know the Thomas, Wier and Haas book, which I didn't like. More a book on how to do a few problems, rather than how to learn calculus.
Others that are used:
Smith and Mintons Calculus' book by McGraw-Hill
Larson, Hostetler and Edwards by Houghton and Mifflin
but I agree with the above - I like Stewart's the best and I've used all three!
this is the text they use at my school now (started using it about 2 or 3 semesters ago. before that, they used Stewart's). it is the general consensus that Stewart's is better, and i agree.
Larson, Hostetler and Edwards by Houghton and Mifflin
i also used this text. i don't remember how much i liked it though. i used it for calc 1 and 2, before i could really start to "appreciate" math texts. i don't recall anything bad about the book
though. i'd probably trust it over Smith and Mintons
but I agree with the above - I like Stewart's the best and I've used all three!
yup yup. go for Stewart's!
April 26th 2009, 08:18 PM #2
April 26th 2009, 09:55 PM #3
April 26th 2009, 09:57 PM #4
April 26th 2009, 10:29 PM #5
April 27th 2009, 04:50 AM #6
April 27th 2009, 08:09 PM #7
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circle n-group
circle n-group
Group Theory
Classical groups
Finite groups
Group schemes
Topological groups
Lie groups
Super-Lie groups
Higher groups
Cohomology and Extensions
For $\mathbf{H}$ a cohesive (∞,1)-topos such as ETop∞Grpd or Smooth∞Grpd, both the natural numbers $\mathbb{Z}$ and the real numbers are naturally abelian group objects in $\mathbf{H}$. Accordingly
their quotient
$U(1) := \mathbb{R}/\mathbb{Z}$
under the canonical embedding $\mathbb{Z} \hookrightarrow \mathbb{R}$ exists in $\mathbf{H}$ and is an abelian group object: the circle group. Therefore for all $n \in \mathbb{N}$ the delooping
$\mathbf{B}^n U(1) \in \mathbf{H}$
exists and has the structure of an abelian (n+1)-group object. This is the topological or smooth, respectively, circle $(n+1)$-group .
Details for the smooth case are at smooth ∞-groupoid in the section circle Lie n-group .
For $n = 1$ the circle 2-group $\mathbf{B}U(1)$ can be identified with the strict 2-group whose corresponding crossed module of groups is simply $[U(1) \to 1]$.
Generally, for any $n$$\mathbf{B}^{n-1}U(1)$ is an n-group that corresponds under the Dold-Kan correspondence to the chain complex or crossed complex of groups $U(1)[n]$ concentrated in degree $n$.
The geometric realization of the circle $n$-group is the Eilenberg-MacLane space
$|\mathbf{B}^n U(1)| \simeq B^{n} U(1) \simeq B^{n+}^\mathbb{Z} \simeq K(\mathbb{Z}, n+1) \,.$
A circle $n$-group-principal ∞-bundle is a circle n-bundle, equivalently an $(n-1)$-bundle gerbe.
Revised on January 4, 2013 04:28:21 by
Urs Schreiber
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Formal logic
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(→Deductive vs. inductive reasoning: small reword) (→Types of logic: "classical logic")
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Formal logic can be further subdivided into several related areas, including: Formal logic can be further subdivided into several related areas, including:
[DEL:* :DEL]'''Aristotelian logic''' (or syllogistic logic) — simple, deductive logic based '''Aristotelian logic''' (or syllogistic logic) — simple, deductive logic based on
− on [[syllogism]]s. + [[syllogism]]s.
[DEL:* :DEL]'''Propositional logic''' (or sentential calculus, etc.) — simple, deductive '''Propositional logic''' (or sentential calculus, etc.) — simple, deductive logic based
− logic based on statements called ''propositions'' and the logical ''operators''/''connectives'' + on statements called ''propositions'' and the logical ''operators''/''connectives'' "not",
"not", "and", "or", "implies", etc. "and", "or", "implies", etc.
[DEL:* :DEL]'''Predicate logic''' (or first-order logic, predicate calculus, etc.) — extends '''Predicate logic''' (or first-order logic, predicate calculus, etc.) — extends
− propositional logic by introducing the logical ''quantifiers'' "for all" and "for some" (or "there + propositional logic by introducing the logical ''quantifiers'' "for all" and "for some" (or
exists"). "there exists").
[DEL:* :DEL]'''Modal logic''' — extends predicate logic by introducing ''modal operators'' '''Modal logic''' — extends predicate logic by introducing ''modal operators'' such as
− such as "possibly" or "formerly". + "possibly" or "formerly".
[DEL:* :DEL]'''Mathematical logic''' — reduces all of the above to symbolic manipulation '''Mathematical logic''' — reduces all of the above to symbolic manipulation based on
− based on well-defined rules. + well-defined rules
+ .
==Deductive vs. inductive reasoning== ==Deductive vs. inductive reasoning==
Revision as of 18:35, 14 September 2009
For more information, see the
Logic is a method by which valid conclusions are drawn from given premises. It can be applied to natural-language arguments or to formal, symbolic systems. Today logic provides the foundation upon
which all areas of modern mathematics and science are ultimately based.
Types of logic
Informal logic is the study of natural-language arguments, especially the identification of logical fallacies.
Formal logic, on the other hand, involves the systematic study of logical reasoning usually within the framework of a symbolic system of terms, operations, quantifiers, axioms, theorems, etc.
Formal logic can be further subdivided into several related areas, including:
1. Aristotelian logic (or syllogistic logic) — simple, deductive logic based on syllogisms.
2. Propositional logic (or sentential calculus, etc.) — simple, deductive logic based on statements called propositions and the logical operators/connectives "not", "and", "or", "implies", etc.
3. Predicate logic (or first-order logic, predicate calculus, etc.) — extends propositional logic by introducing the logical quantifiers "for all" and "for some" (or "there exists").
4. Modal logic — extends predicate logic by introducing modal operators such as "possibly" or "formerly".
5. Mathematical logic — reduces all of the above to symbolic manipulation based on well-defined rules.
The term classical logic may refer to the first type of logic listed above or the first two or three types; it generally does not include the forth or fifth type.
Deductive vs. inductive reasoning
The kinds of reasoning described by logic can be roughly classified as deductive or inductive.
• Deductive reasoning — conclusions follow necessarily from premises; based on strict implications; sometimes placed in the context of reasoning "from the general to the specific".
1. Example: "All men are mortal. Socrates is a man. Therefore, Socrates is mortal." (a syllogism)
2. Example: "A person can't be both a woman and a man. Socrates is a man. Therefore, Socrates is not a woman."
Notice that each conclusion must be true if we assume the premises are true. The first premise of the second example, however, may not be true, depending on how you define "man" and "woman". If
you don't accept the truth of either of the premises, then the argument is still logically valid, but is not sound.
• Inductive reasoning — conclusions may often, but not always, follow from the premises; inferences based on incomplete or imperfect information; sometimes described in the context of reasoning
"from the specific to the general".
1. Example: "Every man I've known has had a beard. Socrates is a man. Therefore, Socrates must have a beard."
2. Example: "If you're a woman, it is very unlikely you have a beard. Socrates has a beard. Therefore, Socrates is probably a woman."
The first argument looks very much like the first deductive example above. But notice that the first premise doesn't say "all" men have beards, only all men "I've known". Even if the first
premise is true, that doesn't mean there's not a man out there you haven't met yet who doesn't have a beard; and Socrates may be such a man. Thus, the conclusion isn't necessarily true. As for
the second example, it is a probabilistic argument of the kind that is often encountered in statistics (specifically, in hypothesis testing, also known as significance testing).
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Binary search in line-sorted text files
This blog post announces pts-line-bisect, a C program and an equivalent Python script and library for doing binary seach in line-sorted text files, and it also explains some of the design details of
the Python implementation.
Let's suppose you have a sorted text file, in which each line is lexicographically larger than the previous one, and you want to find a specific range of lines, or all lines with a specific prefix.
I've written the program pts-line-bisect for that recently. I've also written the article titled Evolution of a binary search implementation for line-sorted text files about the topic, containing the
problem statement, the possible pitfalls, an analysis of some incorrect solutions available on the web as code example, a detailed specification and explanation of my solution (including a proof),
disk seek and speed analysis, a set of speedup ideas and their implementation, and further notes about the speedups in the C implementation.
As a teaser, here is an incorrect solution in Python:
def bisect_left_incorrect(f, x):
"""... Warning: Incorrect implementation with corner case bugs!"""
x = x.rstrip('\n')
f.seek(0, 2) # Seek to EOF.
lo, hi = 0, f.tell()
while lo < hi:
mid = (lo + hi) >> 1
f.readline() # Ignore previous line, find our line.
if x <= f.readline().rstrip('\n'):
hi = mid
lo = mid + 1
return lo
Can you spot the all the 3 bugs?
Read the article for all the details and the solution.
As a reference, here is the correct implementation of the same algorithm for finding the start of the interval in a sorted list or other sequences (based on the bisect module):
def bisect_left(a, x, lo=0, hi=None):
"""Return the index where to insert item x in list a, assuming a is sorted.
The return value i is such that all e in a[:i] have e < x, and all e in
a[i:] have e >= x. So if x already appears in the list, a.insert(x) will
insert just before the leftmost x already there.
Optional args lo (default 0) and hi (default len(a)) bound the
slice of a to be searched.
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None:
hi = len(a)
while lo < hi:
mid = (lo + hi) >> 1
if x <= a[mid]: # Change `<=' to `<', and you get bisect_right.
hi = mid
lo = mid + 1
return lo
A typical real-word use case for such a binary search tool is retrieving lines corresponding to a particular time range in log files (or time based measurement records). These files text files with
variable-length lines, with the log timestamp in the beginning of the line, and they are generated in increasing timestamp order. Unfortunately the lines are not lexicographically sorted, so the
timestamp has to be decoded first for the comparison. The bsearch tool does that, it also supports parsing arbitrary, user-specifiable datetime formats, and it can binary search in gzip(1)ped files
as well (by building an index). It's also of high performance and low overhead, partially because it is written in C++. So bsearch is practical tool with lots of useful features. If you need anything
more complicated than a lexicographic binary search, use it instead.
Before getting too excited about binary search, please note that there are much faster alternatives for data on disk. In an external (file-based) binary search the slowest operation is the disk seek
needed for each bisection. Most of the software and hardware components would be waiting for the hard disk to move the reading head. (Except, of course, that seek times are negligible when
non-spinning storage hardware such as SSD or memory card is used.)
An out-of-the box solution would be adding the data to more disk-efficient key-value store. There are several programs providing such stores. Most of them are based on a B-tree, B*-tree or B+-tree
data structure if sorted iteration and range searches have to be supported, or disk-based hashtables otherwise. Some of the high-performance single-machine key-value stores: cdb (read-only), Tokyo
Cabinet, Kyoto Cabinet, LevelDB; see more in the NoSQL software list.
The fundamental speed difference between a B-tree search and a binary search in a sorted list stems from the fact that B-trees have a branching factor larger than 2 (possibly 100s or 1000s), thus
each seeking step in a B-tree search reduces possible input size by a factor larger than 2, while in a binary search each step reduces the the input size by a factor 2 only (i.e. we keep either the
bottom half or the top half). So both kinds of searches are logarithmic, but the base of the logarithm is different, and this causes a constant factor difference in the disk seek count. By careful
tunig of the constant in B-trees it's usual to have only 2 or 3 disk seeks for each search even for 100GB of data, while a binary search in a such a large file with 50 bytes per record would need 31
disk seeks. By taking 10ms as the seek time (see more info about typical hard disk seek times), a typical B-tree search takes 0.03 second, and a typical binary search takes 0.31 second.
Have fun binary searching, but don't forget to sort your files first!
1 comment:
zsbana said...
I have implemented binary search in a sorted text file once, a long time ago. See http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2118 , and have fun undoing the random substitutions
gmane does on the text.
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Number of results: 1,458
math help
Tony is driving to Denver. Suppose that the distance to his destination (in miles) is a linear function of his total driving time (in minutes), as shown in the figure below. Tony has 67 miles to his
destination after 35 minutes of driving, and he has 55.6 miles to his ...
Saturday, March 20, 2010 at 4:22pm by Anonymous
A letter carrier can deliver mail to 112 homes per hour by walking and 168 homes per hour by driving. Question1: By what percent is productivity increased by driving? Question2: If a new ZIP Code
system improves driving productivity by 12.5%, what is the new number of homes ...
Saturday, February 8, 2014 at 2:02pm by kik
I'm driving along at a constant speed in my 3 metre long car when I notice that I am accidentally driving through a red light. If the rear of my car passes under the light 0.3 seconds after the front
of my car, how fast am I driving in m/s?
Monday, February 3, 2014 at 8:16am by suresh
I'm driving along at a constant speed in my 3 meter long car when I notice that I am accidentally driving through a red light. If the rear of my car passes under the light 0.3 seconds after the front
of my car, how fast am I driving in m/s?
Wednesday, February 26, 2014 at 11:58am by hamza
I'm driving along at a constant speed in my 3 meter long car when I notice that I am accidentally driving through a red light. If the rear of my car passes under the light 0.3 seconds after the front
of my car, how fast am I driving in m/s?
Friday, February 28, 2014 at 12:34pm by yogesh
PLEASE HELP!!!
two cars are driving toward each other from opposite directions if one is driving 55mph the other is driving 60 mph and the cars are 500miles apart how long before they pass each other? round to the
nearest tenth of an hour.
Thursday, November 12, 2009 at 2:44pm by natalie
10th grade algebra
two cars are driving toward each other from opposite directions if one is driving 55mph the other is driving 60mph and the cars are 500 miles apart how long will it be before they pass each other?
round to the nearest tenth of an hour.
Thursday, November 12, 2009 at 1:23pm by natalie
10th grade algebra
two cars are driving toward each other from opposite directions if one is driving 55mph the other is driving 60 mph and the cars are 500miles apart how long before they pass each other? round to the
nearest tenth of an hour.
Thursday, November 12, 2009 at 2:25pm by natalie
Laura is driving to Seattle. Suppose that the remaining distance to drive (in miles) is a linear function of her driving time (in minutes). When graphed, the function gives a line with a slope of
-0.75. Laura has 51 miles remaining after 33 minutes of driving. How many miles ...
Monday, January 10, 2011 at 7:25pm by kia
Two cars are driving toward each other from opposite directions. If one is driving 55 mph, the other is driving 60 mph and the cars are 500 miles apart, how long will it be before they pass each
other? Round to the nearest tenth of an hour
Friday, September 3, 2010 at 12:44pm by Anonymous
Two cars are driving toward each other from opposite directions. If one is driving 55 mph, the other is driving 60 mph and the cars are 500 miles apart, how long will it be before they pass each
other? Round to the nearest tenth of an hour
Sunday, January 2, 2011 at 7:32pm by Joshua
a. Add the times requred for each of the two driving intervals, and add 24 minutes (0.4 hours) to that sum. For each driving inteveral, use the reationship time = distance)/(average speed The sum
will be the total driving and waiting time. Call it T. b. Divide (280 + 210)km by...
Friday, January 30, 2009 at 11:41am by drwls
on a scatter plot how can you determine if driving speed and driving time show a positive, negative or no relationship?
Tuesday, January 15, 2013 at 6:51pm by sandy
Mr. Suki plans to drive 900 miles to his farmhouse, driving an average of 50 miles per hour. How many miles per hour faster must he average, while driving to reduce his total driving time by 3 hours?
Saturday, October 4, 2008 at 11:52am by i love polka
Jim was driving to an interview 72 mi away. On the first leg of the trop he drove an average of 30 mph through a long construction zone, but was able to drive 60 mph after passing through this area.
If driving time for the trip was 1 1/2 hr, how long was he driving in the ...
Monday, August 22, 2011 at 10:16pm by susie
Mrs. Lewis plans to drive 900 miles to her vacation destination, driving an average 50 miles per hour. How many miles per hour faster must she average, while driving, to reduce her total driving time
by 3 hours? I'm not sure how to set up the equation. Thanks.
Saturday, November 3, 2007 at 12:33pm by Rose
Why do some believe cell phone while driving should be banned? how is it any different than talking to the people on the passenger's side or driving with you?
Wednesday, July 18, 2012 at 10:26am by Jazme
the time it takes you to get to campus varies inveresly as your driving rate. Driving at a rate of 40 miles per hour, it takes 1.5 hours. How ong would the trip take you driving at a rate of 60 miles
per hour?
Thursday, November 1, 2012 at 11:06am by Maria
Mrs. Goldman was a little startled when she realized that while driving home, she didn't really remember driving the last mile. For Mrs. Goldman, driving has probably become a(n): a) controlled
process. b) metacognitive process. c) time to engage in selective attention. d) ...
Tuesday, March 13, 2012 at 6:37pm by REB
The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May her driving cost was $310 for 140 mi and in June her cost was $450 for 700 mi. Assume that there is a
linear relationship between the monthly cost C of driving a car and the ...
Saturday, March 29, 2014 at 2:08pm by tim
I have to write a speech on a pet peeve. I chose aggressive driving. Here is my introduction for the outline. Think? Aggressive Driving Introduction •How many of you have turned on the news or opened
a newspaper to find that another human being was seriously injured or killed ...
Tuesday, August 2, 2011 at 8:53pm by lizzie
avg speed= distance/driving time From your description, I can't find the driving time as when did she leave.
Tuesday, November 16, 2010 at 7:57pm by bobpursley
If a mail man could deliver to 112 homes on foot and to 168 homes by driving what percent productivity increase by driving
Tuesday, May 14, 2013 at 2:59pm by pollu
Driving toward, the frequency increases. Driving away, the frequency decreases
Monday, March 11, 2013 at 4:46pm by Dr. Jane
Driving at a constant rate, noah covered 130 miles in 3.5 hours. Express his driving rate in feet per minute.
Thursday, February 10, 2011 at 3:58pm by kaykay
Surely you can yourself list ten. Why dont we burb the baby when driving. Why dont we watch the DVD when driving? We will be happy to critique your thinking.
Thursday, November 12, 2009 at 12:03pm by bobpursley
I'm 15 years old and I'm wondering if I can get one of those books to begin studying for the driving test already in the state of New York? Also, where can I get one? Thanks.
Monday, September 1, 2008 at 2:25pm by Anonymous
Okay got that one.. What about The cab driver was driving extremely fast down the street. The very phrase is was driving. Was is a linking verb and driving an action. Is it a linking verb or an
action verb? Directions say to underline the verb phrase and indicate action or ...
Monday, November 14, 2011 at 7:50pm by Kelley
The Pew Research Center recently polled n= 1048 U.S drivers andfound that 69% enjoyed driving their automobiles. (a) Construct a 95% confidence interval for the proportion of U.S.drivers who enjoy
driving their automobiles. (b) In 1991, Gallup Poll reported this percent to be ...
Tuesday, May 1, 2012 at 12:57am by Cassandra
gregor is planning a driving trip where he knows that half his trip will be along major highways and part on country roads. The speed limit is 100km/h on highways and 60km/h on other roads. the total
time he wants to spend driving is 12h. I set up the equation: 10h+6r=12, but ...
Sunday, June 1, 2008 at 5:24pm by Calie
If you're not competent in a skill, you can't perform that task. I have a young friend who will soon take his driving test. I sure hope he has the driving competence to pass the test and be a safe
and proficient driver.
Thursday, March 24, 2011 at 8:06pm by Ms. Sue
Like e.g.: She is driving all the way to the hills by herself. or She was driving all the way to the hills by herself. She had driven all the way to the hills by herself. She will be driving all the
way to the hills by herself. Still, as drwls wrote; most likely the answer ...
Friday, November 14, 2008 at 3:21am by E.G.
Outline of my speech on Aggressive Driving. Is this correct? Topic- Aggressive Driving Introduction •Road Rage oInjuries and deaths oIncreasing road rage oGrowing epidemic •Aggressive Driving
oDefinition oTypes Verbal Road Rage Quiet Road Rage Epic Road...
Saturday, August 6, 2011 at 4:56pm by lizzie
Reckless Rick driving along the road at 90 km/hr bumps into Hapless Harry directly in fornt of him who is driving at 88 km/hr. What is the speed of the collision?
Monday, September 3, 2012 at 8:46pm by Anonymous
You're very welcome. It sounds like you're driving a car with a manual transmission. Although most people drive cars with automatic transmissions, I've driven manual transmissions all of my life. :-)
Saturday, October 13, 2012 at 6:05pm by Ms. Sue
I am writing a speech on aggressive driving. Here is my thesis statement: Proving that aggressive driving is a growing epidemic across our country is easy, but convincing drivers that there are
better ways to handle “Road Rage” is a different story.
Thursday, July 28, 2011 at 1:56pm by lizzie
what is the formula to find..If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frequency will the amplitude of the mass's oscillation be half of the
maximum amplitude?
Monday, March 26, 2012 at 9:06am by Anonymous
Bob finds that the cost of driving his truck is 92 cents per mile. Give a symbolic and graphical representation for a function f(x) that computes the cost in dollars of driving x miles. Assume 0<x<60
Sunday, March 25, 2012 at 9:46pm by Kate
A weight of 50.0 N is suspended from a spring that has a force constant of 190 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 10.0 Hz, resulting in a
forced-motion amplitude of 4.00 cm. Determine the maximum value of the driving force. N
Sunday, January 30, 2011 at 5:57pm by sir
Total time driving = totoal distance/total time = 540/60 = 9 hrs But you only drove for 7 hours , so you will need another 2 hours of driving. No "equation" is needed.
Thursday, December 8, 2011 at 8:41am by Reiny
A letter carrier can deliver mail to 112 homes per hour by walking and 168 homes per hour by driving. By what percent is productivity increased by driving?
Saturday, May 18, 2013 at 7:51pm by Denise
person arrested on private drive on suspition of drink driving what are law rules on drink driving. person not on public road. is there a case that could argue he is not guilty
Thursday, September 8, 2011 at 9:26am by elizabeth
6% of american drivers read the newspaper while driving if 300 drivers are selected at random find the probability that between 15 and 25 say they read while driving
Monday, April 30, 2012 at 7:27pm by Anonymous
Research topic, English
My research topic is Banning Use of Cell Phones While Driving (include calls and texts) Could you give me some ideas of research questions? I have to have three. I already have three but I am not
sure if they are research questions. will banning cellphones while driving going ...
Wednesday, November 16, 2011 at 12:58pm by HM
You are driving home from college steadily at 95km/hour for 130km. It begins to rain and you slow to 65km/hour. You arrive home after driving 3 hours and 20 minutes. How far is your hometown from
Thursday, September 29, 2011 at 9:07pm by Sam
When will they meet? Given that... student A : started driving at 9hr at 3mi/hr student B: started driving at 9hr25min at 2mi/hr
Sunday, December 8, 2013 at 4:53pm by Anonymous
When will they meet? Given that... student A : started driving at 9hr at 3mi/hr student B: started driving at 9hr25min at 2mi/hr
Sunday, December 8, 2013 at 10:31pm by Anonymous
A weight of 30.0 N is suspended from a spring that has a force constant of 180 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 11.0 Hz, resulting in a
forced-motion amplitude of 5.00 cm. Determine the maximum value of the driving force. ...
Friday, January 28, 2011 at 1:12pm by Ro
When driving to the beach, you drive 240 miles on 9 gallons of gas. When you arrive at the beach you fill up your gas tank. While there you drive 130 miles and use 5.8 gallons of gas. Round answers
to 2 decimal places. What was your gas mileage driving to the beach? What was ...
Saturday, November 16, 2013 at 3:16pm by Alan
Richard is driving home to visit his parents. 125 of the trip are on the interstate highway where the speed limit is 65 . Normally Richard drives at the speed limit, but today he is running late and
decides to take his chances by driving at 72 . how do i get to my answer
Tuesday, September 11, 2012 at 1:36pm by tamika
I'm familiar with driving to Magnolia from Klein. I'm used to driving from Houston to Austin and back. He doesn't want to interfere with his daughter's friendship with our new neighbor. They will all
take part in swimming across the river next week. She depends on his sending ...
Saturday, November 12, 2011 at 10:17am by Writeacher
Richard is driving home to visit his parents. 125 of the trip are on the interstate highway where the speed limit is 65 . Normally Richard drives at the speed limit, but today he is running late and
decides to take his chances by driving at 70. How many min does she save?
Thursday, September 16, 2010 at 9:55am by Danny
You are driving home from school steadily at 95 km/h for 130 km. It then begins to rain and you slow to 60 km/h. You arrive home after driving 3 hours and 40 minutes. 1-How far is your hometown from
school? 2-What was your average speed?
Sunday, October 9, 2011 at 1:48pm by marline
2.) Car insurance companies assume that the longer a person has been driving, the less likely they will be in an accident, and therefore charge new drivers higher insurance premiums than experienced
drivers. To determine whether driving experience is related to the amount of ...
Friday, April 16, 2010 at 11:46am by Jayesh Patel
Use of cell phones while driving is a good one. Requiring full driver testing (written and in-the-car) of everyone over the age of 80 and under the age of 20 is one to consider, too. (Or whatever the
age groups are who have the worst driving records.) Did you think of any more...
Monday, September 17, 2007 at 11:08pm by Writeacher
i need help on ratios, gears and teeth.ex. teeth on driving gear A-35 teeth on driving gear B-? gear ratio-2 1/2 to 1 what is gear B, and how will i figure this out?
Thursday, December 11, 2008 at 5:16pm by Sorrells
A letter carrier can deliver mail to 112 homes per hour by walking and 168 homes per hour by driving. By what percent is productivity increased by driving? Answer: 56% Is my answer correct?
Monday, May 13, 2013 at 6:42pm by Ann
howfar=70km/hr*(time driving+83/60) howfar=94km/hr*timedriving set them equal, solve for time driving, then, using that , go back to either equation and solve for distance.
Wednesday, August 24, 2011 at 7:06pm by bobpursley
A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station. a. If the inspector checks 10
drivers, what is the probability that at least one driver never uses a cell phone ...
Saturday, January 5, 2013 at 9:13pm by sasha
you are driving to a vacation spot that is 1500 miles away. Including rest stops, it takes you 42 hours to get to the vacation spot. You estimate that you drove at an average speed of 50 miles per
hour. How many hours were you not driving?
Wednesday, October 17, 2012 at 5:57pm by steph
you are driving to a vacation spot that is 1500 miles away. Including rest stops, it takes you 42 hours to get to the vacation spot. You estimate that you drove at an average speed of 50 miles per
hour. How many hours were you not driving?
Wednesday, October 17, 2012 at 6:38pm by steph
A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station. a. If the inspector checks 10
drivers, what is the probability that at least one driver never uses a cell phone ...
Tuesday, May 3, 2011 at 12:15am by level crawford
algebra 1
Your family decides to visit your grandparents that live in ohio. You drive at an average rate of 55 mph when driving in Indiana and 65 mph when driving in Ohio. The entire trip of 295 miles takes 5
hours. How long does it take to reach the Indiana-Ohio border?
Sunday, February 24, 2013 at 8:56pm by lilly
Algebra 1
The function f=1,350-45x gives the distance left to travel after driving x hours. What is f(7), the distance left after driving 7 hours?
Monday, May 16, 2011 at 4:30pm by Tranese
Let the construction zone driving time be T hours. Then, 1.5 - T hours were spent driving at 60 mph. The total distance driven is 30*T + 60*(1.5 - T) = 72 miles 90 - 72 = 30 T T = 0.600 hours = 36
Monday, August 22, 2011 at 10:16pm by drwls
I'm not sure that I'm that comfortable with the "product" yield using the limiting resource. Although I understand what the problem is driving at for the % yield, I think the yield, by any standard
is (3.24/5.08)*100 = ? no matter how you do it. I think the answer they're ...
Monday, November 18, 2013 at 11:09pm by DrBob222
Richard is driving home to visit his parents125 mi of the trip are on the interstate highway where the speed limit is 65 mph Normally Richard drives at the speed limit, but today he is running late
and decides to take his chances by driving at 80 mph.how many minutes will he ...
Sunday, September 19, 2010 at 10:11am by melissa
College Algebra
How many miles will you drive in the city? How many miles on the highway? You can't solve this problem without that information. Of do you want to know the number of miles for only city driving and
only highway driving?
Sunday, February 27, 2011 at 5:03pm by Ms. Sue
Grammer and capitalization
Select the sentence with the correct capitalization. A) While driving his Mercedes down Main Street, he was rear-ended by a Toyota. B) While driving his Mercedes down Main street, he was rear-ended
by a Toyota. C) while driving his mercedes down Main Street, he was rear-ended ...
Wednesday, November 7, 2012 at 12:50pm by Marie
English Check
Write a paragraph explaining why you think it is a suitable topic for a research paper, based on your research questions. What are the risks of using cell phones when driving? How will the police
control people from violating this law? will banning cellphones while driving ...
Wednesday, November 16, 2011 at 1:58pm by HM
Henry is driving to visit his friend Susan. For 126 miles of the trip, he can take the highway, which has a speed limit of 65 miles per hour. Henry is in a rush though, and instead travels at 70.6
miles per hour. How many minutes does he save driving faster?
Thursday, February 13, 2014 at 10:29pm by Joy
A 95 confidence interval for the proportion of men that have ever dozed off while driving is 007 to 0.14. For women, a 95% confidence intervAL for the proportion that have dozed off while driving is
0.19 to 0.25. Assume both intervals were computed using larg random samples ...
Monday, July 9, 2012 at 3:21pm by Sheri
At noon, your family starts out .from Louisville to go to Memphis driving at 40 mi/h. Your uncle leaves Memphis to come to Louisville two hours later. He is taking the same route and driving at 60 mi
/h. The two cities are 380 miles apart. At what time do the cars meet?
Tuesday, September 27, 2011 at 8:20pm by David
okay i have this writing assignment on how drunk driving can be stopped .. i have like three sentences to complete it and these three sentences are to explain how/what harsher penalties can help in
stopping drunk driving.. I'm stuck because the following are already the ...
Sunday, February 27, 2011 at 5:39pm by Anon
Paula leaves home driving 40 miles/hr. One hour later Dan leaves home driving in the same direction at the rate of 50 miles/hr. How long will it take dan to catch up with Paula?
Wednesday, January 6, 2010 at 4:31pm by lou
Could you please check these few examples? Thank you very much, Writeacher. 1) It is likely that he will pass his driving test. He is likely to pass his driving test. He should pass it. 2) I'm unsure
that (if) I will pass it. I might pass it. It is possible that I will pass ...
Monday, May 14, 2012 at 6:03pm by John
Math-Stat Help
In a random sample of 78 teenagers, 32 admit to texting while driving. Construct a 98% confidence interval for the percentage of teenager that text while driving. Start by showing that conditions for
constructing a confidence interval are met. 1) Check conditions 2) confidence...
Thursday, April 11, 2013 at 11:41am by MELISSA
The pew research center recently polled n=1048 u.s. drivers and found that 69% enjoyed driving their cars. In 1991 a poll reported this percent to be 79%. Using the data from this poll ,test the
claim that the percent of drivers who enjoy driving their cars has declined since ...
Monday, April 11, 2011 at 6:15pm by nikki
The pew research center recently polled n=1048 u.s. drivers and found that 69% enjoyed driving their cars. In 1991 a poll reported this percent to be 79%. Using the data from this poll ,test the
claim that the percent of drivers who enjoy driving their cars has declined since ...
Monday, April 11, 2011 at 6:15pm by nikki
The pew research center recently polled n=1048 u.s. drivers and found that 69% enjoyed driving their cars. In 1991 a poll reported this percent to be 79%. Using the data from this poll ,test the
claim that the percent of drivers who enjoy driving their cars has declined since ...
Monday, April 11, 2011 at 6:15pm by nikki
Avery has a three week vacation. He plans on driving a total of 4,898 miles. About how many miles should he average per day? As it turned out he spent 5 hours a day driving and drove a total of 5,565
miles. What was his average speed for the hours he drove?
Monday, April 23, 2012 at 8:33pm by Jenny
A cart of mass 1.4 kg is attached to a horizontal spring with stiffness 44.4545 N/m and is able to move horizontally without friction. The cart is then acted on by a sinusoidal driving force with ω =
4.9 Hz and an amplitude of 1.74532 N. What is the amplitude of the...
Wednesday, November 28, 2012 at 1:14am by JTC
Drunk driving can be stopped by designated drivers and enforcing harsher penalties for those who are caught drunk driving .Designated drivers should be available in larger numbers near bars and clubs
or other places where alcoholic drinks are being served in large quantities. ...
Sunday, December 5, 2010 at 3:21pm by Cas
From inside your apartment, you watch rain fall straight down at a constant speed of 9 m/s. Your friend L calls you as she is driving toawards the west at a constant of 90 km/hr. What with what
velocity does L see the rain fall as she looks out her driver's side window? I know...
Sunday, October 7, 2007 at 11:00pm by Anonymous
N/A-General Question
How would you think of answering the following question(s)? Which do you most want your teen driver to avoid? Speeding Talking on the Phone While Driving Drinking Driving Without a Seatbelt Speeding
19% Talking on the Phone While Driving 17% Drinking 50% Driving Without a ...
Tuesday, July 15, 2008 at 12:57pm by Emanuel
Physics (College)
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1200 kg
car is driven in sand at 5.2 m/s, the coefficient of rolling friction is 0.06. ...
Saturday, November 26, 2011 at 1:07am by Set11
health >.<
how would you descibe/define teen driving laws Please tell us how you describe/define them and we'll be glad to critique your answer. but i don't know how to ok, here's what i got: Teenage driving
laws are rules that have been made to reduce accidents and increase safety on ...
Sunday, April 15, 2007 at 5:53pm by anon.s
Health (Ms. Sue)
List and describe three unsafe driver behaviors. A: Three unsafe driver behaviors are speeding, aggressiveness, and impaired driving. These behaviors are hazardous for their own, individual reasons.
Exceeding the speed limit, or speeding, for example, is dangerous because the ...
Monday, January 6, 2014 at 10:41pm by Anonymous
Which situations are examples of a particle with a non-zero acceleration? (Select all that are true for full credit.) a)A car driving around a circular track at constant speed. b)A child going back
and forth on a swing c)A runner moving in a straight line with a constant speed...
Wednesday, April 18, 2012 at 12:48am by Sally
The table shows the driving distance from certain cities. From To Driving Dis(mi) Pittsburgh Johnstown 55 Johnstown Allentown 184 1. Write a number sentence that compares the mileage from Pittsburgh
to Johnstown to Allentown, and the mileage from Allentown to Johnstown to ...
Thursday, August 18, 2011 at 10:45pm by Hayley
The table shows the driving distance from certain cities. From To Driving Dis mi Pittsburgh Johnstown 55 Johnstown Allentown 184 1. Write a number sentence that compares the mileage from Pittsburgh
to Johnstown to Allentown, and the mileage from Allentown to Johnstown to ...
Friday, August 19, 2011 at 7:41am by Hayley
Math - Linear Equations
A car gets 21 mi/gal in city driving and 28 mi/gal highway driving. If 18 gal of gas are used in traveling 448 mi, how many miles were driven in the city, how many driven on the highway (assuming
that only the given rates of usage were actually used)?
Monday, November 19, 2007 at 9:20pm by Chuck
algebra 2
The capacity of the gas tank in a car is 16.5 gallons. The car uses 10.5 gallons to travel 178.5 miles. After driving 85 more miles, only 1 gallon of gasoline remains in the tank. Which linear
equation and graph model the amount of gas g left in the tank after driving m miles?
Sunday, July 7, 2013 at 3:34pm by me
physics please HELP !
Which situations are examples of a particle with a non-zero acceleration? (Select all that are true for full credit.) a)A car driving around a circular track at constant speed. b)A child going back
and forth on a swing c)A runner moving in a straight line with a constant speed...
Wednesday, April 18, 2012 at 4:53am by sally
algebra URGENT HELP
cohol and Driving The concentration of alcohol in a person’s bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be modeled by the equation R = e
kx where x is the percent of concentration of alcohol in the bloodstream and ...
Thursday, March 21, 2013 at 5:18pm by zachary
The cost of driving a car includes both fixed ans mileage costs. Assume that it costs $152.20 per month for insurance and car payments ans $0.22 per mile for gasoline oil and routine maintenance. (a)
Find value m and b so that y-mx=b modeks the monthly cost of driving the car ...
Saturday, March 9, 2013 at 9:36am by Ruth
Mary is driving a car of mass 900 kilograms toward the north with a velocity of 22 meters/second. Danny is driving his 900-kilogram car with a velocity of 15 meters/second toward the east. They
collide and both cars move together afterwards. What is the magnitude of the ...
Wednesday, August 7, 2013 at 6:38pm by Anonymous
Mary is driving a car of mass 900 kilograms toward the north with a velocity of 22 meters/second. Danny is driving his 900-kilogram car with a velocity of 15 meters/second toward the east. They
collide and both cars move together afterwards. What is the magnitude of the ...
Wednesday, August 7, 2013 at 6:44pm by Anonymous
A driver measures her "cold" tire gauge pressure to be 1.95 105 Pa. The measurement is made before driving, when the tires are at the temperature of the surroundings, 21.7° C. After driving several
miles, she checks the gauge pressure again and finds that it has increased to 2...
Friday, April 15, 2011 at 1:49pm by ECU
word processing part 2
It has to be 8 sentences, Can someone help me out and correct me? Alcohol (ethanol) impairs your judgment and depending on your size and tolerance, you can do some substantial damage. When you are
behind the wheel, it is your responsibility just as much as the other cars when ...
Tuesday, November 30, 2010 at 10:22pm by jackie rios
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>
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Decoders and Demultiplexers
To design and study decoders and demultiplexers
1. A 2-4 decoder is a network with two inputs (X[0] and X[1]) and 4 outputs (Y[0], Y[1], Y[2] and Y[3]). When the inputs are both 0, Y[0] = 1 and the other outputs are 0. When the inputs are 0 and 1
respectively, only Y[1] is 1. For 10 and 11 inputs only Y[2] and Y[3] are 1 respectively. Write the truth table. Using NAND gates design a 2-4 decoder. Demonstrate the circuit to your instructor.
2. Study the specification of a 74155 chip, a dual 2-4 decoder. From the truth table in the specs, derive the logical expression of the outputs.
3. Draw the wiring diagram to make the 74155 into a 3-8 decoder. Wire up the circuit and verify the operation.
A demultiplexer is a combinational circuit with:
n control (or select) lines: C[1],..., C[n]
2^n output lines: f[0], f[1],..., f[(2^n)-1]
One input line: I
A demultiplexer receives binary information on a single line (I) and transmits this information on one of 2^n possible output lines. The selection of a particular output line is controlled by
minterms of control lines.
1. Design and implement a 1-4 demultiplexer using NOT and AND gates. Demonstrate the circuit to your instructor.
2. Connect the input I to hi. Write the truth table of this circuit and identify its operations.
3. A 74155 can also be used as a demultiplexer, i.e., it can function like a rotary switch to demultiplex a single input to four different output lines. The select lines control the rotary switch
position digitally. Wire up the 74155 as a 1-4 line demux. Connect the outputs to the LED's. Connect the select lines to the switches and change the settings from 00 to 11. Observe what happens.
Figure 2 – 1 to 4 Line Demux
Design Problems:
1. Using a 3-8 decoder and three 2-input OR gates, design a combinational multiple output system to implement the following functions:
f[1](A,B,C) = BC + A'B'C
f[2](A,B,C) = BC + ABC'
2. In computer systems the memory address is decoded to select a particular word. Assuming that a computer has 64 words and 6 bits to address the memory, using 74155s as a 3-8 decoder, design a 6 to
64 decoder.
3. Using 74155 as a 3-8 decoder, design a 5 to 32 decoder.
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Implied domain
March 13th 2010, 04:19 AM
Implied domain
Hi there, I'm having a little trouble with working out the implied domain of this equation:
I thought that since the range of $2Sin^{-1}(x)$ must be a subset or equal to the domain of $tan(x)$, then $-\frac{\pi}{2}<2Sin^{-1}(x)<\frac{\pi}{2}$. Making the implied domain $x\in \left (-\
frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right )$
But apparently this is wrong...
Thanks for your help.
March 13th 2010, 04:46 AM
Hello Stroodle
Hi there, I'm having a little trouble with working out the implied domain of this equation:
I thought that since the range of $2Sin^{-1}(x)$ must be a subset or equal to the domain of $tan(x)$, then $-\frac{\pi}{2}<2Sin^{-1}(x)<\frac{\pi}{2}$. Making the implied domain $x\in \left (-\
frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right )$
But apparently this is wrong...
Thanks for your help.
The range of $\arcsin (x)$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So $-\pi \le 2\arcsin(x) \le \pi$. These are therefore potential values for the domain, but you must remember that $\
tan \theta$ has problems when $\theta = \pm\frac{\pi}{2}$.
March 13th 2010, 04:55 AM
Ahh, now I get it. Thanks for your help.
I actually thought that it was $Tan(x)$ not $tan(x)$.
If this was the case, would my answer/method have been correct?
Thanks again.
March 13th 2010, 10:53 AM
Hello Stroodle
Sorry, is this some meaning of Tan that I don't know about? What's the difference between Tan x and tan x ?
March 13th 2010, 01:13 PM
Oh. Our teacher taught us that $Tan(x)$ is the same as $tan(x)$ except that its domain is restricted to $x\in \left (-\frac{\pi}{2},\frac{\pi}{2}\right )$
March 13th 2010, 01:46 PM
what a ridiculous thing, it's just a matter of upper and lowercase, it's just absurd.
March 13th 2010, 11:01 PM
Domain of tan x
Has anyone else come across this 'convention'?
March 13th 2010, 11:06 PM
Just did a quick google and found this Complex Analysis/Elementary Functions/Inverse Trig Functions - Wikibooks, collection of open-content textbooks and http://docs.google.com/viewer?a=v&q
I guess this is what my teacher was talking about...
Back to my question though; if the domain of $tan(x)$ was restricted to $x\in \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$ would my original answer be correct?
By the way, I had a look at your paintings Grandad. They are amazing :)
March 14th 2010, 12:53 AM
Hello Stroodle
Just did a quick google and found this Complex Analysis/Elementary Functions/Inverse Trig Functions - Wikibooks, collection of open-content textbooks and Powered by Google Docs
I guess this is what my teacher was talking about...
Back to my question though; if the domain of $tan(x)$ was restricted to $x\in \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$ would my original answer be correct?
By the way, I had a look at your paintings Grandad. They are amazing :)
You are quite correct - there does seem to be a convention that distinguishes between tan and Tan (and, indeed, sin and Sin, etc).
The use of the upper-case initial letter, then, restricts the domain of the functions to their principal values. This has the effect of making the functions one-to-one, and therefore makes it
possible to define the inverse functions. In that case, the domain of
$\text{Tan }(2\arcsin(x))$
would indeed be $\left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$
Since this upper- and lower-case convention appears not to be very widespread, I should use it with caution if I were you.
(Thanks for the comment about the paintings. If you just looked at the ones on my profile, you'll find some more in the Chat Room at http://www.mathhelpforum.com/math-he...-painting.html.)
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Statistics and the Treatment of Experimental Data
5.4 Null Experiments. Setting Confidence Limits When No Counts Are Observed
Many experiments in physics test the validity of certain theoretical conservation laws by searching for the presence of specific reactions or decays forbidden by these laws. In such measurements, an
observation is made for a certain amount of time T. Obviously, if one or more events are observed, the theoretical law is disproven. However, if no events are observed, the converse cannot be said to
be true. Instead a limit on the life-time of the reaction or decay is set.
Let us assume therefore that the process has some mean reaction rate T is
This, now, can also be interpreted as the probability distribution for T. We can now ask the question: What is the probability that [0]? From (1),
where we have normalized (57) with the extra factor T. This probability is known as the confidence level for the interval between 0 to [0]. To make a strong statement we can choose a high confidence
level (CL), for example, 90%. Setting (58) equal to this probability then gives us the value of [0],
For a given confidence level, the corresponding interval is, in general, not unique and one can find other intervals which yield the same integral probability. For example, it might be possible to
integrate (57) from some lower limit greater than
Example 4. A 50 g sample of ^82Se is observed for 100 days for neutrinoless double beta decay, a reaction normally forbidden by lepton conservation. However, current theories suggest that this might
occur. The apparatus has a detection efficiency of 20%. No events with the correct signature for this decay are observed. Set an upper limit on the lifetime for this decay mode.
Choosing a confidence limit of 90%, (59) yields
[0] = -1 / (100 x 0.2) ln (1 - 0.9) = 0.115 day^-1,
where we have corrected for the 20% efficiency of the detector. This limit must now be translated into a lifetime per nucleus. For 50 g, the total number of nuclei is
N = (N[a] / 82) x 50 = 3.67 x 10^23,
which implies a limit on the decay rate per nucleus of
^23) = 3.13 x 10^-25 day^-1.
The lifetime is just the inverse of
^21 years 90% CL,
where we have converted the units to years. Thus, neutrinoless double beta decay may exist but it is certainly a rare process!
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Durham, PA Statistics Tutor
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Hi, I am an experienced math instructor having taught a range of math courses from pre-algebra to statistics to calculus. I use many examples to illustrate various math concepts and I will review
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Multiparticle interferometry and the superposition principle
Results 1 - 10 of 18
- Nature , 1997
"... Quantum entanglement lies at the heart of new proposals for quantum communication and computation. Here we describe the recent experimental realization of quantum teleportation. ..."
Cited by 28 (1 self)
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Quantum entanglement lies at the heart of new proposals for quantum communication and computation. Here we describe the recent experimental realization of quantum teleportation.
, 1996
"... The dichotomy between endophysical/intrinsic and exophysical/extrinsic perception concerns the question of how a model - mathematical, logical, computational - universe is perceived from inside
or from outside, [71, 65, 66, 59, 60, 68, 67]. This distinction goes back in time at least to Archimedes, ..."
Cited by 20 (19 self)
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The dichotomy between endophysical/intrinsic and exophysical/extrinsic perception concerns the question of how a model - mathematical, logical, computational - universe is perceived from inside or
from outside, [71, 65, 66, 59, 60, 68, 67]. This distinction goes back in time at least to Archimedes, reported to have asked for a point outside the world from which one could move the earth. An
exophysical perception is realized when the system is laid out and the experimenter peeps at the relevant features without changing the system. The information flows on a one-way road: from the
system to the experimenter. An endophysical perception can be realized when the experimenter is part of the system under observation. In such a case one has a two-way informational flow; measurements
and entities measured are interchangeable and any attempt to distinguish between them ends up as a convention. The general conception dominating the sciences is that the physical universe is
perceivable ...
- Physical Review A , 2007
"... A variety of coherent states of the harmonic oscillator is considered. It is formed by a particular superposition of canonical coherent states. In the simplest case, these superpositions are
eigenfunctions of the annihilation operator A = P(d/dx + x) / √ 2, where P is the parity operator. Such A ar ..."
Cited by 11 (2 self)
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A variety of coherent states of the harmonic oscillator is considered. It is formed by a particular superposition of canonical coherent states. In the simplest case, these superpositions are
eigenfunctions of the annihilation operator A = P(d/dx + x) / √ 2, where P is the parity operator. Such A arises naturally in the q → −1 limit for a symmetry operator of a specific self-similar
potential obeying the q-Weyl algebra, AA † −q 2 A † A = 1. Coherent states for this and other reflectionless potentials whose discrete spectra consist of N geometric series are analyzed. In the
harmonic oscillator limit the surviving part of these states takes the form of orthonormal superpositions of N canonical coherent states |ǫkα〉, k = 0,1,...,N −1, where ǫ is a primitive Nth root of
unity, ǫN = 1. A class of q-coherent states related to the bilateral q-hypergeometric series and Ramanujan type integrals is described. It includes a curious set of coherent states of the free
nonrelativistic particle which is interpreted as a q-algebraic system without discrete spectrum. A special degenerate form of the symmetry algebras of self-similar potentials is found to provide a
natural q-analog of the Floquet theory. Some properties of the factorization method, which is used throughout the paper, are discussed from the differential Galois theory point of view.
"... Starting from the late 60’s many experiments have been performed to verify the violation Bell’s inequality by Einstein–Podolsky–Rosen (EPR) type correlations. The idea of these experiments being
that: (i) Bell’s inequality is a consequence of locality, hence its experimental violation is an indicati ..."
Cited by 7 (2 self)
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Starting from the late 60’s many experiments have been performed to verify the violation Bell’s inequality by Einstein–Podolsky–Rosen (EPR) type correlations. The idea of these experiments being
that: (i) Bell’s inequality is a consequence of locality, hence its experimental violation is an indication of non locality; (ii) this violation is a typical quantum phenomenon because any classical
system making local choices (either deterministic or random) will produce correlations satisfying this inequality. Both statements (i) and (ii) have been criticized by quantum probability on
theoretical grounds (not discussed in the present paper) and the experiment discussed below has been devised to support these theoretical arguments. We emphasize that the goal of our experiment is
not to reproduce classically the EPR correlations but to prove that there exist perfectly local classical dynamical systems violating Bell’s inequality. The conclusions of the present experiment are:
(I) no contradiction between quantum theory and locality can be deduced from the violation of Bell’s inequality.
- Journal of Universal Computer Science , 1996
"... ..."
, 1999
"... this paper; the propositional structure encountered in the quantum mechanics of spin - state measurements of a spin one-half particle along two directions ( mod p) , that is, the modular,
orthocomplemented lattice MO 2 drawn in Fig. 1 ( where p 2 = ( p + ) and q 2 = ( q + ) ) ..."
Cited by 3 (1 self)
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this paper; the propositional structure encountered in the quantum mechanics of spin - state measurements of a spin one-half particle along two directions ( mod p) , that is, the modular,
orthocomplemented lattice MO 2 drawn in Fig. 1 ( where p 2 = ( p + ) and q 2 = ( q + ) )
- Proceedings of WIA'99 , 1999
"... We develop an automatic-theoretic analysis of Einstein-Podolsky-Rosen conundrum on the basis of two simple devices introduced by Mermin [10, 11]. ..."
, 2001
"... Consequences of the basic and most evident consistency requirement|that measured events cannot happen and not happen at the same time|are reviewed. Particular emphasis is given to event forecast
and event control. As a consequence, particular, very general bounds on the forecast and control of e ..."
Cited by 2 (2 self)
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Consequences of the basic and most evident consistency requirement|that measured events cannot happen and not happen at the same time|are reviewed. Particular emphasis is given to event forecast and
event control. As a consequence, particular, very general bounds on the forecast and control of events within the known laws of physics result. These bounds are of a global, statistical nature and
need not aect singular events or groups of events. We also present a quantum mechanical model of time travel and discuss chronology protection schemes. Such models impose restrictions upon certain
capacities of event control.
- Physics Letters A , 2000
"... Generalized uncertainty relations may depend not only on the commutator relation of two observables considered, but also on mutual correlations, in particular, on entanglement. The equivalence
between the uncertainty relation and Bohr’s complementarity thus holds in a much broader sense than anticip ..."
Cited by 2 (0 self)
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Generalized uncertainty relations may depend not only on the commutator relation of two observables considered, but also on mutual correlations, in particular, on entanglement. The equivalence
between the uncertainty relation and Bohr’s complementarity thus holds in a much broader sense than anticipated. Key words: uncertainty relation, complementarity, quantum entanglement 1
- Studies in History and Philosophy of Modern Physics , 1998
"... Complementarity is not only a feature of quantum mechanical systems but occurs also in the context of finite automata. http://tph.tuwien.ac.at/esvozil/publ/template.tex Preprint submitted to
Elsevier Science 2 December 2003 1 Motivation The aim of this paper is to present to philosophers of physics ..."
Cited by 1 (0 self)
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Complementarity is not only a feature of quantum mechanical systems but occurs also in the context of finite automata. http://tph.tuwien.ac.at/esvozil/publ/template.tex Preprint submitted to Elsevier
Science 2 December 2003 1 Motivation The aim of this paper is to present to philosophers of physics some results in the theory of automata, especially the theory concerned with determining the
initial state of the automaton: results which are analogons to the phenomena of "complementarity" or "non-Booleanness" which occur in quantum mechanics. It has long been known that any finite input/
output system can be modelled by finite automata [Paz(1971)]. The study of finite automata was motivated from the very beginning by their analogy to quantum systems [Moore(1956),Foulis and Randall
(1972),Randall and Foulis(1973)]. Finite automata are universal with respect to the class of computable functions in the (usual) sense that universal networks of automata can compute any e#ectively
(Turing-) computable function. Conversely, any feature emerging from finite automata is reflected by any other universal computational device. Their non-Boolean intrinsic propositional calculus
closely resembles finite quantum mechanical systems [Svozil(1993),Schaller and Svozil(1994),Schaller and Svozil(1995),Schaller and Svozil(1996),Dvurecenskij et al.(1995)]. The considerations to
follow in this article are not technically complicated. Nevertheless, the corresponding ideas turn out to be highly nontrivial and nonclassical, sometimes mindboggling [Greenberger et al.(1993)]. 2
Construction of automaton logics In this Section, I will first summarize some elements of the theory of finite automata; then discuss the so-called state-identification problem, and how it gives rise
to to non-Boolean lattices, a...
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Posts from March 2013 on Gowers's Weblog
Update: comments on this post are now closed, since my latest post would compromise any further contributions to the experiment.
Most of this post consists of write-ups of proofs of five simple propositions about metric spaces. There are three write-ups per proof, and I would be very grateful for any comments that you might
have. If you would like to participate in the experiment, then please state your level of mathematical experience (the main thing I need to know is whether you yourself have studied the basic theory
of metric spaces) and then make any comments/observations you wish on the write-ups. The more you say, the more useful it will be (within reason). I am particularly interested in comparisons and
preferences. For each proof, the order of the three write-ups has been chosen randomly and independently.
It would also be useful if you could rate each of the 15 write-ups for clarity and style. So that everyone rates in the same way, I suggest the following rating systems.
-2 very hard to understand
-1 hard to understand
0 neither particularly hard nor particularly easy
1 easy to understand
2 very easy to understand
-2 very badly written
-1 badly written
0 neither badly written nor well written
1 well written
2 very well written
I stress that ratings should not be regarded as a substitute for comments and observations, or vice versa. What I really need is both comments and numerical ratings.
I do not want people to be influenced by the answers that other people give, so all comments on this post will go to my moderation queue. When I have enough data for the experiment, probably in a
week or so, I will publish all the comments (unless for some reason you specifically request that your comment should not be published).
The more people who participate, the more reliable the results of the experiment will be. I realize that it may take a little time, so thank you very much in advance to everybody who agrees to help.
(Update 26th March: I now have over 30 responses; they have been very helpful indeed, so I am extremely grateful for those. If they keep coming in at a similar rate over the next few days it will be
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Entropy (information theory)
From Wikipedia, the free encyclopedia
(Redirected from
Shannon entropy
In information theory, entropy is a measure of the uncertainty in a random variable.^1 In this context, the term usually refers to the Shannon entropy, which quantifies the expected value of the
information contained in a message.^2 Entropy is typically measured in bits, nats, or bans.^3 Shannon entropy is the average unpredictability in a random variable, which is equivalent to its
information content. Shannon entropy provides an absolute limit on the best possible lossless encoding or compression of any communication, assuming that^4 the communication may be represented as a
sequence of independent and identically distributed random variables.
A single toss of a fair coin has an entropy of one bit. A series of two fair coin tosses has an entropy of two bits. The number of fair coin tosses is its entropy in bits. This random selection
between two outcomes in a sequence over time, whether the outcomes are equally probable or not, is often referred to as a Bernoulli process. The entropy of such a process is given by the binary
entropy function. The entropy rate for a fair coin toss is one bit per toss. However, if the coin is not fair, then the uncertainty, and hence the entropy rate, is lower. This is because, if asked to
predict the next outcome, we could choose the most frequent result and be right more often than wrong. The difference between what we know, or predict, and the information that the unfair coin toss
reveals to us is less than one heads-or-tails "message", or bit, per toss.^5
This definition of "entropy" was introduced by Claude E. Shannon in his 1948 paper "A Mathematical Theory of Communication".^6
Entropy is a measure of unpredictability of information content. To get an informal, intuitive understanding of the connection between these three English terms, consider the example of a poll on
some political issue. Usually, such polls happen because the outcome of the poll isn't already known. In other words, the outcome of the poll is relatively unpredictable, and actually performing the
poll and learning the results gives some new information; these are just different ways of saying that the entropy of the poll results is large. Now, consider the case that the same poll is performed
a second time shortly after the first poll. Since the result of the first poll is already known, the outcome of the second poll can be predicted well and the results should not contain much new
information; in this case the entropy of the second poll results is small.
Now consider the example of a coin toss. When the coin is fair, that is, when the probability of heads is the same as the probability of tails, then the entropy of the coin toss is as high as it
could be. This is because there is no way to predict the outcome of the coin toss ahead of time—the best we can do is predict that the coin will come up heads, and our prediction will be correct with
probability 1/2. Such a coin toss has one bit of entropy since there are two possible outcomes that occur with equal probability, and learning the actual outcome contains one bit of information.
Contrarily, a coin toss with a coin that has two heads and no tails has zero entropy since the coin will always come up heads, and the outcome can be predicted perfectly.
English text has fairly low entropy. In other words, it is fairly predictable. Even if we don't know exactly what is going to come next, we can be fairly certain that, for example, there will be many
more e's than z's, that the combination 'qu' will be much more common than any other combination with a 'q' in it, and that the combination 'th' will be more common than 'z', 'q', or 'qu'. After the
first few letters one can often guess the rest of the word. Uncompressed, English text has between 0.6 and 1.3 bits of entropy for each character of message.^5^7
If a compression scheme is lossless—that is, you can always recover the entire original message by decompressing—then a compressed message has the same quantity of information as the original, but
communicated in fewer characters. That is, it has more information, or a higher entropy, per character. This means a compressed message has less redundancy. Roughly speaking, Shannon's source coding
theorem says that a lossless compression scheme cannot compress messages, on average, to have more than one bit of information per bit of message. The entropy of a message multiplied by the length of
that message is a measure of how much information the message contains.
Shannon's theorem also implies that no lossless compression scheme can compress all messages. If some messages come out smaller, at least one must come out larger due to the pigeonhole principle. In
practical use, this is generally not a problem, because we are usually only interested in compressing certain types of messages, for example English documents as opposed to gibberish text, or digital
photographs rather than noise, and it is unimportant if a compression algorithm makes some unlikely or uninteresting sequences larger. However, the problem can still arise even in everyday use when
applying a compression algorithm to already compressed data: for example, making a ZIP file of music in the FLAC audio format is unlikely to achieve much extra saving in space.
Named after Boltzmann's H-theorem, Shannon defined the entropy H (Greek letter Eta) of a discrete random variable X with possible values {x[1], ..., x[n]} and probability mass function P(X) as:
$H(X) = E[I(X)] = E[-\ln(P(X))].$
Here E is the expected value operator, and I is the information content of X.^8^9 I(X) is itself a random variable.
When taken from a finite sample, the entropy can explicitly be written as
$H(X) = \sum_{i} {P(x_i)\,I(x_i)} = -\sum_{i} {P(x_i) \log_b P(x_i)}$
where b is the base of the logarithm used. Common values of b are 2, Euler's number e, and 10, and the unit of entropy is bit for b = 2, nat for b = e, and dit (or digit) for b = 10.^10
In the case of p(x[i]) = 0 for some i, the value of the corresponding summand 0 log[b](0) is taken to be 0, which is consistent with the well-known limit:
$\lim_{p\to0+}p\log (p) = 0$.
Consider tossing a coin with known, not necessarily fair, probabilities of coming up heads or tails; this is known as the Bernoulli process.
The entropy of the unknown result of the next toss of the coin is maximized if the coin is fair (that is, if heads and tails both have equal probability 1/2). This is the situation of maximum
uncertainty as it is most difficult to predict the outcome of the next toss; the result of each toss of the coin delivers one full bit of information.
However, if we know the coin is not fair, but comes up heads or tails with probabilities p and q, where p ≠ q, then there is less uncertainty. Every time it is tossed, one side is more likely to come
up than the other. The reduced uncertainty is quantified in a lower entropy: on average each toss of the coin delivers less than one full bit of information.
The extreme case is that of a double-headed coin that never comes up tails, or a double-tailed coin that never results in a head. Then there is no uncertainty. The entropy is zero: each toss of the
coin delivers no new information as the outcome of each coin toss is always certain. In this respect, entropy can be normalized by dividing it by information length. This ratio is called metric
entropy and is a measure of the randomness of the information.
As said, for a random variable X with n outcomes {x[1], ..., x[n]}, the Shannon entropy, a measure of uncertainty (see further below) and denoted by H(X), is defined as
$\displaystyle H(X)= - \sum_{i=1}^np(x_i)\log_b p(x_i)$ (1)
where p(x[i]) is the probability mass function of outcome x[i].
To understand the meaning of Eq. (1), first consider a set of n possible outcomes (events) {x[1], ..., x[n]}, with equal probability p(x[i]) = 1/n. An example would be a fair die with n values, from
1 to n. The uncertainty for such a set of n outcomes is defined by
$\displaystyle u = \log_b (n).$ (2)
For concreteness, consider the case in which the base of the logarithm is b = 2. To specify an outcome of a fair n-sided die roll, we must specify one of the n values, which requires log[2](n) bits
i.e. the number of bits required to represent the outcomes in binary form. For our n sided die, this would be binary numbers {000, ..., (n-1)}. Hence in this case, the uncertainty of an outcome is
the number of bits needed to specify the outcome.
Intuitively, if we have two independent sources of uncertainty, the overall uncertainty should be the sum of the individual uncertainties. The logarithm captures this additivity characteristic for
independent uncertainties. For example, consider appending to each value of the first die the value of a second die, which has m possible outcomes {y[1], ..., y[m]}. There are thus mn possible
outcomes $\left\{ x_i y_j : i = 1 , \ldots , n , j = 1 , \ldots , m \right\}$. The uncertainty for such a set of mn outcomes is then
$\displaystyle u = \log_b (nm) = \log_b (n) + \log_b (m).$ (3)
Thus the uncertainty of playing with two dice is obtained by adding the uncertainty of the second die log[b](m) to the uncertainty of the first die log[b](n). In the case that b = 2 this means that
to specify the outcome of an n-sided die roll and an m-sided die roll, we need to specify log[2](n) + log[2](m) = log[2](mn) bits.
Now return to the case of playing with one die only (the first one). Since the probability of each event is 1/n, we can write
$u_i = \log_b \left(\frac{1}{p(x_i)}\right) = - \log_b (p(x_i))=-\log_b\left(\frac{1}{n}\right), \ \forall i \in \{1, \ldots , n\}.$
In the case of a non-uniform probability mass function (or density in the case of continuous random variables), we let
$\displaystyle u_i := - \log_b (p(x_i))$ (4)
which is also called a surprisal; the lower the probability p(x[i]), i.e. p(x[i]) → 0, the higher the uncertainty or the surprise, i.e. u[i] → ∞, for the outcome x[i].
The average uncertainty $\langle u \rangle$, with $\langle \cdot \rangle$ being the average operator, is obtained by
$\displaystyle \langle u \rangle = \sum_{i=1}^{n} p(x_i) u_i = - \sum_{i=1}^{n} p(x_i) \log_b (p(x_i))$ (5)
and is used as the definition of the entropy H(X) in Eq. (1). In his original formulation, Shannon proved that this definition (up to a multiplicative factor) was the only one that satisfied a few
key requirements. The above also explains why information entropy and information uncertainty can be used interchangeably.^11 In the case that b = 2, Eq. (1) measures the expected number of bits we
need to specify the outcome of a random experiment.
One may also define the conditional entropy of two events X and Y taking values x[i] and y[j] respectively, as
where p(x[i],y[j]) is the probability that X=x[i] and Y=y[j]. This quantity should be understood as the amount of randomness in the random variable X given that you know the value of Y. For example,
the entropy associated with a six-sided die is H(die), but if you were told that it had in fact landed on 1, 2, or 3, then its entropy would be equal to H (die: the die landed on 1, 2, or 3).
Relationship to thermodynamic entropy
The inspiration for adopting the word entropy in information theory came from the close resemblance between Shannon's formula and very similar known formulae from statistical mechanics.
In statistical thermodynamics the most general formula for the thermodynamic entropy S of a thermodynamic system is the Gibbs entropy,
$S = - k_B \sum p_i \ln p_i \,$
where k[B] is the Boltzmann constant, and p[i] is the probability of a microstate. The Gibbs entropy was defined by J. Willard Gibbs in 1878 after earlier work by Boltzmann (1872).^12
The Gibbs entropy translates over almost unchanged into the world of quantum physics to give the von Neumann entropy, introduced by John von Neumann in 1927,
$S = - k_B \,\,{\rm Tr}(\rho \ln \rho) \,$
where ρ is the density matrix of the quantum mechanical system and Tr is the trace.
At an everyday practical level the links between information entropy and thermodynamic entropy are not evident. Physicists and chemists are apt to be more interested in changes in entropy as a system
spontaneously evolves away from its initial conditions, in accordance with the second law of thermodynamics, rather than an unchanging probability distribution. And, as the minuteness of Boltzmann's
constant k[B] indicates, the changes in S/k[B] for even tiny amounts of substances in chemical and physical processes represent amounts of entropy that are extremely large compared to anything in
data compression or signal processing. Furthermore, in classical thermodynamics the entropy is defined in terms of macroscopic measurements and makes no reference to any probability distribution,
which is central to the definition of information entropy.
At a multidisciplinary level, however, connections can be made between thermodynamic and informational entropy, although it took many years in the development of the theories of statistical mechanics
and information theory to make the relationship fully apparent. In fact, in the view of Jaynes (1957), thermodynamic entropy, as explained by statistical mechanics, should be seen as an application
of Shannon's information theory: the thermodynamic entropy is interpreted as being proportional to the amount of further Shannon information needed to define the detailed microscopic state of the
system, that remains uncommunicated by a description solely in terms of the macroscopic variables of classical thermodynamics, with the constant of proportionality being just the Boltzmann constant.
For example, adding heat to a system increases its thermodynamic entropy because it increases the number of possible microscopic states for the system, thus making any complete state description
longer. (See article: maximum entropy thermodynamics). Maxwell's demon can (hypothetically) reduce the thermodynamic entropy of a system by using information about the states of individual molecules;
but, as Landauer (from 1961) and co-workers have shown, to function the demon himself must increase thermodynamic entropy in the process, by at least the amount of Shannon information he proposes to
first acquire and store; and so the total thermodynamic entropy does not decrease (which resolves the paradox). Landauer's principle has implications on the amount of heat a computer must dissipate
to process a given amount of information, though modern computers are nowhere near the efficiency limit.
Entropy as information content
Entropy is defined in the context of a probabilistic model. Independent fair coin flips have an entropy of 1 bit per flip. A source that always generates a long string of B's has an entropy of 0,
since the next character will always be a 'B'.
The entropy rate of a data source means the average number of bits per symbol needed to encode it. Shannon's experiments with human predictors show an information rate between 0.6 and 1.3 bits per
character in English;^13 the PPM compression algorithm can achieve a compression ratio of 1.5 bits per character in English text.
From the preceding example, note the following points:
1. The amount of entropy is not always an integer number of bits.
2. Many data bits may not convey information. For example, data structures often store information redundantly, or have identical sections regardless of the information in the data structure.
Shannon's definition of entropy, when applied to an information source, can determine the minimum channel capacity required to reliably transmit the source as encoded binary digits (see caveat below
in italics). The formula can be derived by calculating the mathematical expectation of the amount of information contained in a digit from the information source. See also Shannon-Hartley theorem.
Shannon's entropy measures the information contained in a message as opposed to the portion of the message that is determined (or predictable). Examples of the latter include redundancy in language
structure or statistical properties relating to the occurrence frequencies of letter or word pairs, triplets etc. See Markov chain.
Data compression
Entropy effectively bounds the performance of the strongest lossless compression possible, which can be realized in theory by using the typical set or in practice using Huffman, Lempel-Ziv or
arithmetic coding. The performance of existing data compression algorithms is often used as a rough estimate of the entropy of a block of data.^14^15 See also Kolmogorov complexity. In practice,
compression algorithms deliberately include some judicious redundancy in the form of checksums to protect against errors.
World's technological capacity to store and communicate entropic information
A 2011 study in Science estimates the world's technological capacity to store and communicate optimally compressed information normalized on the most effective compression algorithms available in the
year 2007, therefore estimating the entropy of the technologically available sources.^16
All figures in entropically compressed
Type of Information 1986 2007 Increase
Storage 2.6 295 113.5x
Broadcast 432 1900 4.398x
Telecommunications 0.281 65 231.3x
The authors estimate humankind technological capacity to store information (fully entropically compressed) in 1986 and again in 2007. They break the information into three categories - To store
information on a medium, to receive information through a one-way broadcast networks, to exchange information through two-way telecommunication networks.^16
Limitations of entropy as information content
There are a number of entropy-related concepts that mathematically quantify information content in some way:
• the self-information of an individual message or symbol taken from a given probability distribution,
• the entropy of a given probability distribution of messages or symbols, and
• the entropy rate of a stochastic process.
(The "rate of self-information" can also be defined for a particular sequence of messages or symbols generated by a given stochastic process: this will always be equal to the entropy rate in the case
of a stationary process.) Other quantities of information are also used to compare or relate different sources of information.
It is important not to confuse the above concepts. Often it is only clear from context which one is meant. For example, when someone says that the "entropy" of the English language is about 1 bit per
character, they are actually modeling the English language as a stochastic process and talking about its entropy rate.
Although entropy is often used as a characterization of the information content of a data source, this information content is not absolute: it depends crucially on the probabilistic model. A source
that always generates the same symbol has an entropy rate of 0, but the definition of what a symbol is depends on the alphabet. Consider a source that produces the string ABABABABAB... in which A is
always followed by B and vice versa. If the probabilistic model considers individual letters as independent, the entropy rate of the sequence is 1 bit per character. But if the sequence is considered
as "AB AB AB AB AB..." with symbols as two-character blocks, then the entropy rate is 0 bits per character.
However, if we use very large blocks, then the estimate of per-character entropy rate may become artificially low. This is because in reality, the probability distribution of the sequence is not
knowable exactly; it is only an estimate. For example, suppose one considers the text of every book ever published as a sequence, with each symbol being the text of a complete book. If there are N
published books, and each book is only published once, the estimate of the probability of each book is 1/N, and the entropy (in bits) is −log[2](1/N) = log[2](N). As a practical code, this
corresponds to assigning each book a unique identifier and using it in place of the text of the book whenever one wants to refer to the book. This is enormously useful for talking about books, but it
is not so useful for characterizing the information content of an individual book, or of language in general: it is not possible to reconstruct the book from its identifier without knowing the
probability distribution, that is, the complete text of all the books. The key idea is that the complexity of the probabilistic model must be considered. Kolmogorov complexity is a theoretical
generalization of this idea that allows the consideration of the information content of a sequence independent of any particular probability model; it considers the shortest program for a universal
computer that outputs the sequence. A code that achieves the entropy rate of a sequence for a given model, plus the codebook (i.e. the probabilistic model), is one such program, but it may not be the
For example, the Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, ... . Treating the sequence as a message and each number as a symbol, there are almost as many symbols as there are characters in the
message, giving an entropy of approximately log[2](n). So the first 128 symbols of the Fibonacci sequence has an entropy of approximately 7 bits/symbol. However, the sequence can be expressed using a
formula [F(n) = F(n−1) + F(n−2) for n={3,4,5,...}, F(1)=1, F(2)=1] and this formula has a much lower entropy and applies to any length of the Fibonacci sequence.
Limitations of entropy as a measure of unpredictability
In cryptanalysis, entropy is often roughly used as a measure of the unpredictability of a cryptographic key. For example, a 128-bit key that is randomly generated has 128 bits of entropy. It takes
(on average) $2^{128-1}$ guesses to break by brute force. If the key's first digit is 0, and the others random, then the entropy is 127 bits, and it takes (on average) $2^{127-1}$ guesses.
However, entropy fails to capture the number of guesses required if the possible keys are not of equal probability.^17^18 If the key is half the time "password" and half the time a true random
128-bit key, then the entropy is approximately 65 bits. Yet half the time the key may be guessed on the first try, if your first guess is "password", and on average, it takes around $2^{126}$ guesses
(not $2^{65-1}$) to break this password.
Similarly, consider a 1000000-digit binary one-time pad. If the pad has 1000000 bits of entropy, it is perfect. If the pad has 999999 bits of entropy, evenly distributed (each individual bit of the
pad having 0.999999 bits of entropy) it may still be considered very good. But if the pad has 999999 bits of entropy, where the first digit is fixed and the remaining 999999 digits are perfectly
random, then the first digit of the ciphertext will not be encrypted at all.
Data as a Markov process
A common way to define entropy for text is based on the Markov model of text. For an order-0 source (each character is selected independent of the last characters), the binary entropy is:
$H(\mathcal{S}) = - \sum p_i \log_2 p_i, \,\!$
where p[i] is the probability of i. For a first-order Markov source (one in which the probability of selecting a character is dependent only on the immediately preceding character), the entropy rate
$H(\mathcal{S}) = - \sum_i p_i \sum_j \ p_i (j) \log_2 p_i (j), \,\!$^citation needed
where i is a state (certain preceding characters) and $p_i(j)$ is the probability of j given i as the previous character.
For a second order Markov source, the entropy rate is
$H(\mathcal{S}) = -\sum_i p_i \sum_j p_i(j) \sum_k p_{i,j}(k)\ \log_2 \ p_{i,j}(k). \,\!$
b-ary entropy
In general the b-ary entropy of a source $\mathcal{S}$ = (S,P) with source alphabet S = {a[1], ..., a[n]} and discrete probability distribution P = {p[1], ..., p[n]} where p[i] is the probability of
a[i] (say p[i] = p(a[i])) is defined by:
$H_b(\mathcal{S}) = - \sum_{i=1}^n p_i \log_b p_i, \,\!$
Note: the b in "b-ary entropy" is the number of different symbols of the ideal alphabet used as a standard yardstick to measure source alphabets. In information theory, two symbols are necessary and
sufficient for an alphabet to encode information. Therefore, the default is to let b = 2 ("binary entropy"). Thus, the entropy of the source alphabet, with its given empiric probability distribution,
is a number equal to the number (possibly fractional) of symbols of the "ideal alphabet", with an optimal probability distribution, necessary to encode for each symbol of the source alphabet. Also
note that "optimal probability distribution" here means a uniform distribution: a source alphabet with n symbols has the highest possible entropy (for an alphabet with n symbols) when the probability
distribution of the alphabet is uniform. This optimal entropy turns out to be log[b](n).
A source alphabet with non-uniform distribution will have less entropy than if those symbols had uniform distribution (i.e. the "optimized alphabet"). This deficiency in entropy can be expressed as a
ratio called efficiency:
$\eta(X) = -\sum_{i=1}^n \frac{p(x_i) \log_b (p(x_i))}{\log_b (n)}$
Efficiency has utility in quantifying the effective use of a communications channel. This formulation is also referred to as the normalized entropy, as the entropy is divided by the maximum entropy $
{\log_b (n)}$.
Shannon entropy is characterized by a small number of criteria, listed below. Any definition of entropy satisfying these assumptions has the form
$-K\sum_{i=1}^np_i\log (p_i)$
where K is a constant corresponding to a choice of measurement units.
In the following, p[i] = Pr (X = x[i]) and $H_n(p_1,\ldots,p_n)=H(X)$.
The measure should be continuous, so that changing the values of the probabilities by a very small amount should only change the entropy by a small amount.
The measure should be unchanged if the outcomes x[i] are re-ordered.
$H_n\left(p_1, p_2, \ldots \right) = H_n\left(p_2, p_1, \ldots \right)$ etc.
The measure should be maximal if all the outcomes are equally likely (uncertainty is highest when all possible events are equiprobable).
$H_n(p_1,\ldots,p_n) \le H_n\left(\frac{1}{n}, \ldots, \frac{1}{n}\right) = \log_b (n).$
For equiprobable events the entropy should increase with the number of outcomes.
$H_n\bigg(\underbrace{\frac{1}{n}, \ldots, \frac{1}{n}}_{n}\bigg) = \log_b(n) < \log_b (n+1) = H_{n+1}\bigg(\underbrace{\frac{1}{n+1}, \ldots, \frac{1}{n+1}}_{n+1}\bigg).$
The amount of entropy should be independent of how the process is regarded as being divided into parts.
This last functional relationship characterizes the entropy of a system with sub-systems. It demands that the entropy of a system can be calculated from the entropies of its sub-systems if the
interactions between the sub-systems are known.
Given an ensemble of n uniformly distributed elements that are divided into k boxes (sub-systems) with b[1], ..., b[k] elements each, the entropy of the whole ensemble should be equal to the sum of
the entropy of the system of boxes and the individual entropies of the boxes, each weighted with the probability of being in that particular box.
For positive integers b[i] where b[1] + ... + b[k] = n,
$H_n\left(\frac{1}{n}, \ldots, \frac{1}{n}\right) = H_k\left(\frac{b_1}{n}, \ldots, \frac{b_k}{n}\right) + \sum_{i=1}^k \frac{b_i}{n} \, H_{b_i}\left(\frac{1}{b_i}, \ldots, \frac{1}{b_i}\right).$
Choosing k = n, b[1] = ... = b[n] = 1 this implies that the entropy of a certain outcome is zero: H[1](1) = 0. This implies that the efficiency of a source alphabet with n symbols can be defined
simply as being equal to its n-ary entropy. See also Redundancy (information theory).
Further properties
The Shannon entropy satisfies the following properties, for some of which it is useful to interpret entropy as the amount of information learned (or uncertainty eliminated) by revealing the value of
a random variable X:
• Adding or removing an event with probability zero does not contribute to the entropy:
$H_{n+1}(p_1,\ldots,p_n,0) = H_n(p_1,\ldots,p_n)$.
$H(X) = \operatorname{E}\left[\log_b \left( \frac{1}{p(X)}\right) \right] \leq \log_b \left( \operatorname{E}\left[ \frac{1}{p(X)} \right] \right) = \log_b(n)$.
This maximal entropy of log[b](n) is effectively attained by a source alphabet having a uniform probability distribution: uncertainty is maximal when all possible events are equiprobable.
• The entropy or the amount of information revealed by evaluating (X,Y) (that is, evaluating X and Y simultaneously) is equal to the information revealed by conducting two consecutive experiments:
first evaluating the value of Y, then revealing the value of X given that you know the value of Y. This may be written as
• If Y=f(X) where f is deterministic, then H(f(X)|X) = 0. Applying the previous formula to H(X, f(X)) yields
so H(f(X)) ≤ H(X), thus the entropy of a variable can only decrease when the latter is passed through a deterministic function.
• If X and Y are two independent experiments, then knowing the value of Y doesn't influence our knowledge of the value of X (since the two don't influence each other by independence):
• The entropy of two simultaneous events is no more than the sum of the entropies of each individual event, and are equal if the two events are independent. More specifically, if X and Y are two
random variables on the same probability space, and (X,Y) denotes their Cartesian product, then
$H[(X,Y)]\leq H(X)+H(Y).$
Proving this mathematically follows easily from the previous two properties of entropy.
Extending discrete entropy to the continuous case: differential entropy
The Shannon entropy is restricted to random variables taking discrete values. The corresponding formula for a continuous random variable with probability density function f(x) with finite or infinite
support $\mathbb X$ on the real line is defined by analogy, using the above form of the entropy as an expectation:
$h[f] = \operatorname{E}[-\ln (f(x))] = -\int_\mathbb X f(x) \ln (f(x))\, dx.$
This formula is usually referred to as the continuous entropy, or differential entropy. A precursor of the continuous entropy hf is the expression for the functional H in the H-theorem of Boltzmann.
Although the analogy between both functions is suggestive, the following question must be set: is the differential entropy a valid extension of the Shannon discrete entropy? Differential entropy
lacks a number of properties that the Shannon discrete entropy has – it can even be negative – and thus corrections have been suggested, notably limiting density of discrete points.
To answer this question, we must establish a connection between the two functions:
We wish to obtain a generally finite measure as the bin size goes to zero. In the discrete case, the bin size is the (implicit) width of each of the n (finite or infinite) bins whose probabilities
are denoted by p[n]. As we generalize to the continuous domain, we must make this width explicit.
To do this, start with a continuous function f discretized into bins of size $\Delta$. By the mean-value theorem there exists a value x[i] in each bin such that
$f(x_i) \Delta = \int_{i\Delta}^{(i+1)\Delta} f(x)\, dx$
and thus the integral of the function f can be approximated (in the Riemannian sense) by
$\int_{-\infty}^{\infty} f(x)\, dx = \lim_{\Delta \to 0} \sum_{i = -\infty}^{\infty} f(x_i) \Delta$
where this limit and "bin size goes to zero" are equivalent.
We will denote
$H^{\Delta} :=- \sum_{i=-\infty}^{\infty} f(x_i) \Delta \log \left( f(x_i) \Delta \right)$
and expanding the logarithm, we have
$H^{\Delta} = - \sum_{i=-\infty}^{\infty} f(x_i) \Delta \log (f(x_i)) -\sum_{i=-\infty}^{\infty} f(x_i) \Delta \log (\Delta).$
As Δ → 0, we have
\begin{align} \sum_{i=-\infty}^{\infty} f(x_i) \Delta &\to \int_{-\infty}^{\infty} f(x)\, dx = 1 \\ \sum_{i=-\infty}^{\infty} f(x_i) \Delta \log (f(x_i)) &\to \int_{-\infty}^{\infty} f(x) \log f
(x)\, dx. \end{align}
But note that log(Δ) → −∞ as Δ → 0, therefore we need a special definition of the differential or continuous entropy:
$h[f] = \lim_{\Delta \to 0} \left(H^{\Delta} + \log \Delta\right) = -\int_{-\infty}^{\infty} f(x) \log f(x)\,dx,$
which is, as said before, referred to as the differential entropy. This means that the differential entropy is not a limit of the Shannon entropy for n → ∞. Rather, it differs from the limit of the
Shannon entropy by an infinite offset.
It turns out as a result that, unlike the Shannon entropy, the differential entropy is not in general a good measure of uncertainty or information. For example, the differential entropy can be
negative; also it is not invariant under continuous co-ordinate transformations.
Relative entropy
Another useful measure of entropy that works equally well in the discrete and the continuous case is the relative entropy of a distribution. It is defined as the Kullback–Leibler divergence from the
distribution to a reference measure m as follows. Assume that a probability distribution p is absolutely continuous with respect to a measure m, i.e. is of the form p(dx) = f(x)m(dx) for some
non-negative m-integrable function f with m-integral 1, then the relative entropy can be defined as
$D_{\mathrm{KL}}(p \| m ) = \int \log (f(x)) p(dx) = \int f(x)\log (f(x)) m(dx) .$
In this form the relative entropy generalises (up to change in sign) both the discrete entropy, where the measure m is the counting measure, and the differential entropy, where the measure m is the
Lebesgue measure. If the measure m is itself a probability distribution, the relative entropy is non-negative, and zero if p = m as measures. It is defined for any measure space, hence coordinate
independent and invariant under co-ordinate reparameterizations if one properly takes into account the transformation of the measure m. The relative entropy, and implicitly entropy and differential
entropy, do depend on the "reference" measure m.
Use in combinatorics
Entropy has become a useful quantity in combinatorics.
Loomis-Whitney inequality
A simple example of this is an alternate proof of the Loomis-Whitney inequality: for every subset A ⊆ Z^d, we have
$|A|^{d-1}\leq \prod_{i=1}^{d} |P_{i}(A)|$
where P[i] is the orthogonal projection in the ith coordinate:
$P_{i}(A)=\{(x_{1}, ..., x_{i-1}, x_{i+1}, ..., x_{d}) : (x_{1}, ..., x_{d})\in A\}.$
The proof follows as a simple corollary of Shearer's inequality: if X[1], ..., X[d] are random variables and S[1], ..., S[n] are subsets of {1, ..., d} such that every integer between 1 and d lies in
exactly r of these subsets, then
$H[(X_{1},...,X_{d})]\leq \frac{1}{r}\sum_{i=1}^{n}H[(X_{j})_{j\in S_{i}}]$
where $(X_{j})_{j\in S_{i}}$ is the Cartesian product of random variables X[j] with indexes j in S[i] (so the dimension of this vector is equal to the size of S[i]).
We sketch how Loomis-Whitney follows from this: Indeed, let X be a uniformly distributed random variable with values in A and so that each point in A occurs with equal probability. Then (by the
further properties of entropy mentioned above) H(X) = log|A|, where |A| denotes the cardinality of A. Let S[i] = {1, 2, ..., i−1, i+1, ..., d}. The range of $(X_{j})_{j\in S_{i}}$ is contained in P
[i](A) and hence $H[(X_{j})_{j\in S_{i}}]\leq \log |P_{i}(A)|$. Now use this to bound the right side of Shearer's inequality and exponentiate the opposite sides of the resulting inequality you
Approximation to binomial coefficient
For integers 0 < k < n let q = k/n. Then
$\frac{2^{nH(q)}}{n+1} \leq \tbinom nk \leq 2^{nH(q)},$
$H(q) = -q \log_2(q) - (1-q) \log_2(1-q).$^19
Here is a sketch proof. Note that $\tbinom nk q^{qn}(1-q)^{n-nq}$ is one term of the expression
$\sum_{i=0}^n \tbinom ni q^i(1-q)^{n-i} = (q + (1-q))^n = 1.$
Rearranging gives the upper bound. For the lower bound one first shows, using some algebra, that it is the largest term in the summation. But then,
$\tbinom nk q^{qn}(1-q)^{n-nq} \geq \tfrac{1}{n+1}$
since there are n+1 terms in the summation. Rearranging gives the lower bound.
A nice interpretation of this is that the number of binary strings of length n with exactly k many 1's is approximately $2^{nH(k/n)}$.^20
See also
This article needs additional citations for verification. (April 2012)
This article incorporates material from Shannon's entropy on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
Further reading
External links
• Hazewinkel, Michiel, ed. (2001), "Entropy", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• Entropy an interdisciplinary journal on all aspect of the entropy concept. Open access.
• Information is not entropy, information is not uncertainty ! – a discussion of the use of the terms "information" and "entropy".
• I'm Confused: How Could Information Equal Entropy? – a similar discussion on the bionet.info-theory FAQ.
• An Intuitive Guide to the Concept of Entropy Arising in Various Sectors of Science – a wikibook on the interpretation of the concept of entropy.
• Network Event Detection With Entropy Measures, Dr. Raimund Eimann, University of Auckland, PDF; 5993 kB – a PhD thesis demonstrating how entropy measures may be used in network anomaly detection.
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Find the Smallest Number...
Date: 10/21/97 at 21:42:26
From: Cheryl Starzyk
Subject: Help
Can you show me a example of how to do this math problem?
I am the smallest number that has factors of 1, 2, 3, 4, 5, 6, 7,
and 8. What number am I?
Thank you,
Date: 10/22/97 at 17:21:31
From: Doctor Chita
Subject: Re: Help
Hi Chad:
A problem in number theory! How interesting! Did you know that this
problem is similar to what cryptologists use to create secret codes?
That is, they take prime numbers and multiply them together to create
a very large number that is the secret to a cipher, or code. Since
it's nearly impossible to factor a very large number into primes, the
code is impossible to decipher.
Anyway, about your problem - think about prime and composite numbers
for a second. A counting number is either a composite number or prime.
Composite numbers are made up of prime factors (also called divisors).
A prime number has only two factors (divisors): itself and 1.
Therefore, 3 is prime, and 4 is not.
The numbers in your sample include prime and composite numbers. What I
would suggest is that you rewrite the composite numbers as products of
their prime numbers. Here's an example.
Suppose my numbers are 2, 5, 6, 12, 21, 24, and 27. I want the
smallest number having these numbers as divisors.
Here are all the factors: I've put parentheses around the prime
factors of the composite numbers.
2 * 5 * (2 * 3) * (2^2 * 3) * (3 * 7) * (2^3 * 3) * (3^3)
Regroup the factors in order:
[2 * 2 * 2^2 * 2^3] * [3 * 3 * 3 * 3 * 3^3] * 5 * 7
The smallest number I want must contain enough 2s to take care of
the largest power of 2 in the expression. This is 2^3 = 8. But once
I have the three 2s, I can take care of the other 2s, since they all
divide 8.
I also need to have 3^3 to take care of 27. The remaining 3s also
divide 27, so I don't need any more of them in my "least" number.
Since there is only one 7 and one 5, I need one of each in the final
So my least number would be
2^3 * 3^3 * 5 * 7 or 8 * 27 * 5 * 7 = 7560.
Check to be sure that this number can be divided by each of the
numbers in the original set: 2, 5, 6, 12, 21, 24, and 27.
Now you try it. Rewrite the numbers using exponents and then find the
largest power of each prime number in your set. The product of primes
raised to the largest power for each prime will make up the number you
are looking for.
Isn't math fun!
-Doctor Chita, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Classic.Ars: An Introduction to 64-bit Computing and x86-64
Data types and applications
Many modern processors support two additional data types: floating-point data and vector data. Each of these two data types has its own set of registers and its own execution unit(s). The following
table compares all four data types in 32-bit and 64-bit processors:
│Data Type │Register Type│Execution Unit│x86 Width│x86-64 Width│
│Integer │GPR │ALU │32 │64 │
│Address │GPR │ALU or AGU │32 │64 │
│Floating-point* │FPR │FPU │64 │64 │
│Vector │VR │VPU │128 │128 │
*x87 uses 80-bit registers to do double-precision floating-point.The floats themselves are 64-bit, but the processor converts them to an internal, 80-bit format for increased precision when doing
You can see from the table above that the difference the move to 64 bits makes is in the integer and address hardware. The floating-point and vector hardware stays the same.
Current 64-bit applications
Now that we know what 64-bit computing is, let's take a look at the benefits of increased integer and data sizes.
Dynamic range
The main thing that a wider integer gives you is increased dynamic range. Instead of defining the term "dynamic range," I'll just show you how it works.
In the base-10 number system to which we're all accustomed, you can represent a maximum of ten integers (0 to 9) with a single digit. This is because base-10 has ten different symbols with which to
represent numbers. To represent more than ten integers you need to add another digit, using a combination of two symbols chosen from among the set of ten to represent any one of 100 integers (00 to
99). The general formula that you can use to compute the number of integers (dynamic range, or DR) that you can represent with an n-digit base-ten number is:
DR = 10^n
So a 1-digit number gives you 10^1 = 10 possible integers, a 2-digit number 10^2 = 100 integers, a 3-digit number 10^3 = 1000 integers, and so on.
The base-2, or "binary," number system that computers use has only two symbols with which to represent integers: 0 and 1. Thus, a single-digit binary number allows you to represent only two integers,
0 and 1. With a two-digit (or "2-bit") binary, you can represent four integers by combining the two symbols (0 and 1) in any of the following four ways:
00 = 0
01 = 1
10 = 2
11 = 3
Similarly, a 3-bit binary number gives you eight possible combinations, which you can use to represent eight different integers. As you increase the number of bits, you increase the number of
integers you can represent. In general, n bits will allow you to represent 2^n integers in binary. So a 4-bit binary number can represent 2^4 or 16 integers, an 8-bit number gives you 2^8=256
integers, and so on.
So in moving from a 32-bit GPR to a 64-bit GPR, the range of integers that a processor can manipulate goes from 2^32 = 4.3e9 to 2^64 = 1.8e19. The dynamic range, then, increases by a factor of 4.3
billion. Thus a 64-bit integer can represent a much larger range of numbers than a 32-bit integer.
The benefits of increased dynamic range, or, how the existing 64-bit computing market uses 64-bit integers
Since addresses are just special-purpose integers, an ALU and register combination that can handle more possible integer values can also handle that many more possible addresses. With all the recent
press coverage that 64-bit architectures have garnered, it's fairly common knowledge that a 32-bit processor can address at most 4GB of memory. (Remember our 2^32 = 4.3 billion number? That 4.3
billion bytes is about 4GB.) A 64-bit architecture could theoretically, by contrast, address up to 18 million terabytes.
Of course, there's a big difference between the amount of address space that a 64-bit address value could theoretically yield and the actual sizes of the virtual and physical address spaces that a
given 64-bit architecture supports. In the case of x86-64, the virtual address space is 48-bit, which makes for about 282 terabytes of virtual address space. (To borrow a line from my old IA-64
preview article, I'm tempted to say about this number what Bill Gates supposedly said about 640K back in the DOS days-"282 terabytes ought to be enough for anybody." But don't quote me on that in 10
years when Quake 16 takes up three or four million terabytes of hard disk space.) x86-64's physical address space is 40-bit, which can support about 1 terabyte of physical memory.
So, what do you do with over 4GB of memory? Well, caching a very large database in it is a start. Back-end servers for mammoth databases are one place where 64 bits have long been a requirement, so
it's no surprise to see upcoming 64-bit offerings billed as capable database platforms.
On the media and content creation side of things, folks who work with very large 2D image files also appreciate the extra RAM. And a related, much sexier application domain where large amounts of
memory come in handy is in simulation and modeling. Under this heading you could put various CAD tools and 3D rendering programs, as well as things like weather and scientific simulations, and even,
as I've already half-jokingly referred to, realtime 3D games. Though the current crop of 3D games wouldn't benefit from greater than 4GB of RAM, it is quite possible that we'll see a game that
benefits from greater than 4GB RAM within the next five years. But we'll discuss the possibilities for 64 bits in the consumer space later in the article, so let's not get ahead of ourselves.
There is one drawback to the increase in memory space that 64-bit addressing affords. Since memory address values (or pointers, in programmer lingo) are now twice as large, they take up twice as much
cache space. Pointers normally make up only a fraction of all the data in the cache, but when that fraction doubles it can squeeze other useful data out of the cache and degrade performance slightly.
Some who read the discussion above would no doubt point out that Xeon systems are available with more than 4GB. Furthermore, Intel supposedly has a fairly simple hack that they could implement to
allow their 32-bit systems to address up to 512GB of memory. Still, the cleanest and most future-proof way to address the 4GB ceiling is a larger pointer.
Some applications, mostly in the realm of scientific computing (MATLAB, Mathematica, MAPLE, etc.) and simulations, require 64-bit integers because they work with numbers outside the dynamic range of
32-bit integers. When the result of a calculation exceeds the range of possible integer values, you get a situation called either overflow (i.e. the result was greater than the highest positive
integer) or underflow (i.e. the result was less than the largest negative integer). When this happens, the number you get in the register isn't the right answer. There's a bit in the x86's processor
status word (see this page for a bit more on the PSW) that allows you to check to see if an integer has just exceeded the processor's dynamic range, so you know that the result is bogus. Such
situations are very, very rare in integer applications. As an engineering student I never ran into this problem, although I did run into the somewhat related problem of floating-point round-off error
a few times.
Programmers who run into integer overflow or underflow problems on a 32-bit platform do have the option of using a 64-bit integer construct provided by a higher level language like C. In such cases,
the compiler uses two registers per integer, one for each half of the integer, to do 64-bit calculations in 32-bit hardware. This has obvious performance drawbacks, making it less desirable than a
true 64-bit integer implementation.
Finally, there is another application domain for which 64-bit integers can offer real benefits: cryptography. Most popular encryption schemes rely on the multiplication and factoring of very large
integers, and the larger the integers the more secure the encryption. As we'll discuss in the final section, AMD is hoping that the growing demand for tighter security and more encryption in the
mainstream business and consumer computing markets will make a cheap, 64-bit, x86-compatible processor attractive.
At this point, I should make a quick note of a fact that I'll refer to again in the article's conclusion: increased performance was not mentioned above as a straightforward, across-the-board benefit
of increased dynamic range. As I stated previously, 64-bit integer code runs slowly on a 32-bit machine, due to the fact that the 64-bit computations have to be split apart and processed as two
separate 32-bit computations. So you could say that there's a performance penalty for running 64-bit integer code on a 32-bit machine; this penalty is absent when running the same code on a 64-bit
machine, since the computation doesn't have to be split in two. The take-home point here is that only applications that require and use 64-bit integers will see a performance increase on 64-bit
hardware that is due solely to a 64-bit processor's wider registers and increased dynamic range. So there's no magical performance boost inherent in the move from 32 bits to 64 bits, as people are
often led to think by journalists who write things like, "64-bit computers can processes twice as much data per clock cycle as their 32-bit counterparts." Technically, this is true in a very
restricted sense, but it would be better to say the following: "64-bit computers can process numbers that are 4.3 billion times as large as those processed by their 32-bit counterparts." It sounds a
lot less sexy because it is, but at least no one is misled into thinking that 64-bitness makes a computer somehow twice as fast.
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Addition Test
Math Test > Addition Test
Addition Test
Timed Addition test for online practice include both vertical and horizontal addition with different digits combination.
Single-digit Horizontal Addition Test
Horizontal addition test using single digit is given here.
Difficulty Level: Easy
Grade Level: 1st grade, 2nd grade
Single-digit Horizontal Addition: Three Terms
This is little trickier for kids as it involves three numbers in horizontal
Difficulty Level: Moderate
Grade Level: 1st grade, 2nd grade, 3rd grade
Two-digit Horizontal Addition Test
Two digit horizontal addition is little complicate comparing with vertical addition.
Difficulty Level: Moderate
Grade Level: 2nd grade, 3rd grade, 4th grade
Three-digit Horizontal Addition Test
Three digit addition test placed horizontally is challengeable for kids.
Difficulty Level: Hard
Grade Level: 3rd, 4th, 5th and 6th grade
Three-digit Vertical Addition Test
Vertical addition test involving three digits is given here.
Difficulty Level: Moderate
Grade Level: 3rd, 4th, 5th and 6th grade
Multiple choice questions based on single digit addition. Student chooses which two numbers makes the sum.
Difficulty Level: Hard
Grade Level: 1st grade, 2nd grade
Single-digit Vertical Addition Test
Vertical addition test is the standard method of adding any two numbers for beginners.
Difficulty Level: Easy
Grade Level: 1st grade, 2nd grade
Single-digit Vertical Addition: Three Terms
Simple vertical addition test involving three numbers. Take it out within a time.
Difficulty Level: Moderate
Grade Level: 1st grade, 2nd grade, 3rd grade
Two-digit Vertical Addition Test
Two digit vertical addition test is quite easy comparing with horizontal addition.
Difficulty Level: Easy
Grade Level: 2nd grade, 3rd grade, 4th grade
Very challenging addition test involves complicate steps to perform addition. Find the missing number that makes the sum of other two combinations.
Difficulty Level: Hard
Grade Level: 1st grade, 2nd grade
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This Week's Finds in Mathematical Physics (Week 222)
In article <slrndlehja.qdn.robert@atdotde.iu-bremen.de>,
Robert C. Helling <helling@atdotde.de> wrote:
>On Tue, 18 Oct 2005 12:57:32 +0000 (UTC), John Baez
><baez@math.removethis.ucr.andthis.edu> wrote:
>> o Twotino - A twotino is a Kuiper belt object whose orbit is in 2:1
>> resonance with Neptune. These are rare compared to plutinos, and
>> they're smaller, so they're stuck with boring names like 1996 TR66.
>> There are also a couple of Kuiper belt objects in 4:3 and 5:3
>> resonances with Neptune.
>Is there an easy way to see why these resonance orbits come about? Why
>do three body systems with a large central object, an intermediate
>planet and a small probe happen to get the probe in resonace with the
>planet? Is this just "frequency locking happens in chaotic systems"
>or is there an easy but more quantitative way to understand this?
I'm shamefully ignorant of this, so ten minutes' research on the
web was able to double my knowledge. I got ahold of this paper
B. Garfinkel, On resonance in celestial mechanics: a survey,
Celestial Mech. 28 (1982), 275-290,
and while not easy to understand (I guess there's a huge body
of work on this subject), it uses Hamiltonian perturbation theory
and continued fractions to study resonance, and talks about a difference
between "shallow" and "deep" resonances.
It says that Laplace first explained the Great Inequality in the motion
of Jupiter and Saturn by means of a 5:2 resonance, which is a
"shallow resonance". I have no idea what the "Great Inequality" is,
other than a strange name for this 5:2 resonance. But, I read elsewhere
The dynamics of the Sun-Jupiter-Saturn system was recognized
as problematic from the beginnings of perturbation theory.
The problems are due to the so-called Great Inequality (GI), which
is the Jupiter-Saturn 2:5 mean-motion near-commensurability.
This is from:
F. Varadi, M. Ghil, and W. M. Kaula,
The Great Inequality in a Planetary Hamiltonian Theory
Somehow this shallow resonance is related to the continued fraction
1/(2 + 1/(2 + 1/(14 + 1/(2 + .... ))))
which is close to 2/5.
The Pluto-Neptune resonance, on the other hand, is a "deep resonance"
and related to the continued fraction
1/(2 - 1/(2 + 1/(10 + .... )))
which starts out close to 2/3. (Recall that plutinos go around the
Sun about twice each time Neptune goes around thrice.)
>Probably related: There are people doing numerical long term stability
>analysis of the solar system. From what I know, they are not just
>taking F=ma and Newton's law of gravity, replace dt by delta t and
>then integrate but use much fancier spectral methods. Could somebody
>please point me to an introduction into these methods?
Here's a bit of stuff about that from "week107", perhaps not all
that helpful, but still pretty interesting:
....................................................................... ...
Later Jon Doyle, a computer scientist at M.I.T. who had been to my
talk, invited me to a seminar at M.I.T. where I met Gerald Sussman,
who with Jack Wisdom has run the best long-term simulations of the
solar system, trying to settle the old question of whether the darn
thing is stable! It turns out that the system is afflicted with
chaos and can only be predicted with any certainty for about 4
million years... though their simulation went out to 100 million.
Here are some fun facts: 1) They need to take general relativity into
account even for the orbit of Jupiter, which precesses about one
radian per billion years. 2) They take the asteroid belt into account
only as modification of the sun's quadrupole moment (which they also
use to model its oblateness). 3) The most worrisome thing about the
whole simulation --- the most complicated and unpredictable aspect of
the whole solar system in terms of its gravitational effects on
everything else --- is the Earth-Moon system, with its big tidal
effects. 4) The sun loses one Earth mass per 100 million years due to
radiation, and another quarter Earth mass due to solar wind. 5) The
first planet to go is Mercury! In their simulations, it eventually
picks up energy through a resonance and drifts away.
For more, try:
4) Gerald Jay Sussman and Jack Wisdom, Chaotic evolution of the solar system,
Science, 257, 3 July 1992.
Gerald Jay Sussman and Jack Wisdom, Numerical evidence that the motion
of Pluto is chaotic, Science, 241, 22 July 1988.
James Applegate, M. Douglas, Y. Gursel, Gerald Jay Sussman, Jack Wisdom,
The outer solar system for 200 million years, Astronomical Journal, 92,
pp 176-194, July 1986, reprinted in Lecture Notes in Physics #267 --
Use of Supercomputers in Stellar Dynamics, Springer Verlag, 1986.
James Applegate, M. Douglas, Y. Gursel, P Hunter, C. Seitz, Gerald Jay
Sussman, A digital orrery, in IEEE Transactions on Computers, C-34,
No. 9, pp. 822-831, September 1985, reprinted in Lecture Notes in
Physics #267, Springer Verlag, 1986.
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Problem on mixture. Please explain with steps.
October 16th 2012, 08:37 AM
Problem on mixture. Please explain with steps.
A cask contains 3 parts wine and one part water. What part of the mixture must be drawn off and substituted by water so that the resulting mixture may be half wine and half water.
October 16th 2012, 09:42 AM
Re: Problem on mixture. Please explain with steps.
Since there is no specific amount given, lets take the total liquid in the cask to be 1. Then the amount of water in the cask is 1/4 and the amount of wine is 3/4. Suppose you take out amount x
(x< 1). The amount of water in that is x/4 and the amount of wine is 3x/4. That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and
Since we "substitute" that by water, we simply replace the amount taken out, so that we are adding x water, then we will have 1- x/4+ x= 1+ 3x/4 water and 1- 3x/4 wine. To have equal amounts of
water and wine, those must be equal: 1+3x/4= 1- 3x/4. Solve that for x.
October 16th 2012, 01:33 PM
Re: Problem on mixture. Please explain with steps.
October 16th 2012, 01:36 PM
Re: Problem on mixture. Please explain with steps.
Solving that X gives 0.
Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.
I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x.
You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.
October 16th 2012, 01:48 PM
Re: Problem on mixture. Please explain with steps.
Solving that X gives 0.
Your 3rd line "That will leave 1- x/4 water and 1- 3x/4 wine. You want to add an amount of water to make this equal parts wine and water."
Will that not leave (1/4)-(x/4) water and (3/4)-(3x/4) wine and then adding x water gives water equation as (1/4)-(x/4)+x
and leaving wine the same amount (3/4)-(3/4)x. Now solving (1/4)-(x/4)+x and (3/4)-(3/4)x yields x=1/3 which is the correct answer.
I have another question. I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x.
You started the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference.
Here is my 2 cents. I visualized the problem as this
At the begining the mixture has 75% wine and since it is well mixed what ever volume is removed with be 75% wine. At the end we want the percent to be 50%
This gives
$.75-.75x=.5 \iff -.75x =-.25 \iff x =\frac{25}{75}=\frac{1}{3}$
Also where abouts are you from in Utah. I went to school there for a while.
October 16th 2012, 01:50 PM
Re: Problem on mixture. Please explain with steps.
Hello, hisajesh!
A cask contains 3 parts wine and one part water.
What part of the mixture must be drawn off and substituted by water
so that the resulting mixture may be half wine and half water?
Suppose the cask holds 60 liters of liquid.
It contains 45 liters of wine and 15 liters of water.
We remove $x$ liters of the mixture.
. . We remove $\tfrac{3}{4}x$ liters of wine.
. . We remove $\tfrac{1}{4}x$ liters of water.
The cask contains: $\begin{Bmatrix}45-\tfrac{3}{4}x\text{ liters of wine}\; \\ 15-\tfrac{1}{4}x\text{ liters of water} \end{Bmatrix}$
We add $x$ liters of water.
The cask contains: . $\begin{Bmatrix}45-\frac{3}{4}x\text{ liters of wine}\; \\ 15 + \frac{3}{4}x\text{ liters of water} \end{Bmatrix}$
But these two quantities are supposed to be equal.
There is our equaton! . . . $45-\tfrac{3}{4}x \;=\;15 + \tfrac{3}{4}x$
October 16th 2012, 02:08 PM
Re: Problem on mixture. Please explain with steps.
October 16th 2012, 02:17 PM
Re: Problem on mixture. Please explain with steps.
@ everyone: Thank you for your flurry of answers!!!!!
Please answer my second question in my above reply:
I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference
October 16th 2012, 02:20 PM
Re: Problem on mixture. Please explain with steps.
Whatever you assume the original volume to be you always end up with 1/3 of it. Start with vol v and get v/3.
October 16th 2012, 02:21 PM
Re: Problem on mixture. Please explain with steps.
@ everyone: Thank you for your flurry of answers!!!!!
Please answer my second question in my above reply:
I started the problem by taking 4 parts as whole and removing x from 4 and going through this way
yields 4/3 as x. If I start the problem by taking 1 as whole and could arrive at 1/3 as x which is the correct anwer. kindly reflect what makes the difference
You found the volume removed, but if you want the RATIO you need find
November 9th 2012, 06:10 PM
Re: Problem on mixture. Please explain with steps.
Hello, hisajesh!
A cask contains 3 parts wine and one part water.
What part of the mixture must be drawn off and substituted by water
so that the resulting mixture may be half wine and half water?
Suppose the cask contains $W$ units of the wine/water blend.
. . It is 25% water; it contains $\tfrac{1}{4}W$ units of water.
We draw off $x$ units of the mixture.
. . We remove $\tfrac{1}{4}x$ units of water.
We add $x$ units of pure water.
. . This contains $x$ units of water.
The final amount of water is: . $\tfrac{1}{4}W - \tfrac{1}{4}x + x \:=\:\tfrac{1}{4}W + \tfrac{3}{4}x$ units.
But we know that the final mixture is $W$ units which is 50% water.
. . So it contains $\tfrac{1}{2}W$ units of water.
There is our equation! . . . . $\tfrac{1}{4}W + \tfrac{3}{4}x \:=\:\tfrac{1}{2}W$
Therefore: . $\tfrac{3}{4}x \:=\:\tfrac{1}{4}W \quad\Rightarrow\quad x \:=\:\tfrac{1}{3}W$
We should draw off one-third of the cask and replace it with water.
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Coin Specifications
Date: 09/24/2001 at 22:19:15
From: ann
Subject: Weighing coins to cash in
Approximately how many of each: nickels, dimes, and pennies, does it
take to make a pound? Or do you have another suggestion? I want to get
an idea of how much change I have around.
Date: 09/25/2001 at 11:55:56
From: Doctor Ian
Subject: Re: Weighing coins to cash in
Hi Ann,
You can find the official weights of coins here:
U.S. Code: Title 31, Section 5112 (Money and Finance)
Sec. 5112. Denominations, specifications, and design of coins
Note that the weights are in grams. There are 454 grams in a pound,
so if a coin weighs x grams, the number of coins in a pound would be
For example, a nickel weighs 5 grams. So there would be 454/5 = about
91 nickels in a pound.
Does this help?
- Doctor Ian, The Math Forum
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algebra reducing fractions to lowest terms
Re: algebra reducing fractions to lowest terms
If you get stuck factoring numerator and denominator then use the quadratic formula on them.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
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Here's the question you clicked on:
If x is the smallest integer, which of the following equations models the problem? I got X+(x+2)+(x+4)+(x+6)=636. Is this correct? Please and thank you.
• one year ago
• one year ago
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there is no smallest integer
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so they are all true
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So it would none of the above?
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if you are doing if then -> -p ^ q
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@What is the problem being modeled? x appears to be 156 on what was posted but what is the scenario of the problem that calls for the smallest x? @kitsune0724
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they are all true
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The question just says if x is the smallest integer, which of the equations models the problem. The choices are: 4x=636, x(x+2)(x+4)(x+6)=636, x+(x+2)+(x+4)+(x+6)=636, x+(x+2)+(x+4)+(x+6)=636 and
none of the above. I got the third choice. @Directix
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@kitsune0724 I assume that we have to solve all the equations for x and then choose as the answer the equation which yields the smallest value of x of all options. Let me check your answer.
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@kitsune0724 In the options, I think you wrote one of them twice: x+(x+2)+(x+4)+(x+6)=636, x+(x+2)+(x+4)+(x+6)=636. What is supposed to be there?
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how is there a smallest integer? I think this is a question about if then....which is -P v Q thus since P is false the statement is true for what follows...
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Something bothers me about this problem. The question asks the following: " which of the equations **models the problem.** @kitsune0724 --> What problem?
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yes. it's suppose to be there. @Directrix
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That's all it says in the problem.
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plewase tell me why you guys are ignoring the smallest integer part? and my explination...
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I'm not ignoring your explanation @zzr0ck3r but that's not one of the choices.
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hehe, they are all true is the choice:)
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@zzr0ck3r -P v Q thus since P is false the statement is true for what follows... I am not commenting on your contributions because I do not follow what they have to do with the problem. And, I
STILL do not think we have been given the entire problem. My best guess is that there's a given problem about consecutive integers and we are to select the equation that can be used to solve that
problem. Yet, I cannot find out from @kitsune0724 what that problem is.
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There's no problem given @Directrix. I'm sorry.
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@kitsune0724 Why did you pick the third choice? And, I know two of the choices cannot be identical. So, there's something faulty about this problem. I cannot help without more information. The
choices are: ... SAME x+(x+2)+(x+4)+(x+6)=636, SAME x+(x+2)+(x+4)+(x+6)=636 ....
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@kitsune0724 I'm going to try to get a fresh pair of eyes for this problem. Hold on.
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@Directrix it says if x is the smallest integer then .... thus P = x is the smallest integer this is never true^^^^^^ thus we have ~P which implies Q is true for any Q. we have 4 Q's
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all true....
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@saifoo.khan --> We need a fresh pair of eyes on this problem. Would you render your thoughts, please? Thanks.
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I dont see why you would try and solve a problem that cant be solved..... if x is an irrational rational number what is x^2 <----would you try and solve this?
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I don't really get the question. Sorry. :/ Are you sure it's complete? @kitsune0724
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Thank y'all for your help.
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Glad to try to help. It was a mathematical adventure. :)
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is replying to Can someone tell me what button the professor is hitting...
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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.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.
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Solving Word Problems With Equations
May 24th 2010, 02:34 PM #1
Junior Member
May 2010
Solving Word Problems With Equations
Another word problem.
David flew 300km on a commuter plane, then 2000km on a passenger jet. The passenger jet flew twice as fast as the commuter plane. The total flying time for the journey was 3.25 hours. What was
the speed of each plane, in kilometers per hour?
Please help. I've been trying to tackle this problem for an hour+ and had no luck solving it.
Another word problem.
David flew 300km on a commuter plane, then 2000km on a passenger jet. The passenger jet flew twice as fast as the commuter plane. The total flying time for the journey was 3.25 hours. What was
the speed of each plane, in kilometers per hour?
Please help. I've been trying to tackle this problem for an hour+ and had no luck solving it.
Remember $d=rt \iff t=\frac{d}{r}$
Let $r$ be the speed of the commuter plane then $2r$ is the speed of the jet then
May 24th 2010, 03:35 PM #2
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Bayesian experimental design: A review
Results 1 - 10 of 142
, 2010
"... The key idea behind active learning is that a machine learning algorithm can achieve greater accuracy with fewer labeled training instances if it is allowed to choose the data from which is
learns. An active learner may ask queries in the form of unlabeled instances to be labeled by an oracle (e.g., ..."
Cited by 132 (1 self)
Add to MetaCart
The key idea behind active learning is that a machine learning algorithm can achieve greater accuracy with fewer labeled training instances if it is allowed to choose the data from which is learns.
An active learner may ask queries in the form of unlabeled instances to be labeled by an oracle (e.g., a human annotator). Active learning is well-motivated in many modern machine learning problems,
where unlabeled data may be abundant but labels are difficult, time-consuming, or expensive to obtain. This report provides a general introduction to active learning and a survey of the literature.
This includes a discussion of the scenarios in which queries can be formulated, and an overview of the query strategy frameworks proposed in the literature to date. An analysis of the empirical and
theoretical evidence for active learning, a summary of several problem setting variants, and a discussion
- ICML 2003 workshop on The Continuum from Labeled to Unlabeled Data in Machine Learning and Data Mining , 2003
"... Active and semi-supervised learning are important techniques when labeled data are scarce. We combine the two under a Gaussian random field model. Labeled and unlabeled data are represented as
vertices in a weighted graph, with edge weights encoding the similarity between instances. The semi-supervi ..."
Cited by 76 (5 self)
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Active and semi-supervised learning are important techniques when labeled data are scarce. We combine the two under a Gaussian random field model. Labeled and unlabeled data are represented as
vertices in a weighted graph, with edge weights encoding the similarity between instances. The semi-supervised learning problem is then formulated in terms of a Gaussian random field on this graph,
the mean of which is characterized in terms of harmonic functions. Active learning is performed on top of the semisupervised learning scheme by greedily selecting queries from the unlabeled data to
minimize the estimated expected classification error (risk); in the case of Gaussian fields the risk is efficiently computed using matrix methods. We present experimental results on synthetic data,
handwritten digit recognition, and text classification tasks. The active learning scheme requires a much smaller number of queries to achieve high accuracy compared with random query selection. 1.
- Workshop on Artificial Intelligence and Statistics
"... The linear model with sparsity-favouring prior on the coefficients has important applications in many different domains. In machine learning, most methods to date search for maximum a posteriori
sparse solutions and neglect to represent posterior uncertainties. In this paper, we address problems of ..."
Cited by 62 (12 self)
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The linear model with sparsity-favouring prior on the coefficients has important applications in many different domains. In machine learning, most methods to date search for maximum a posteriori
sparse solutions and neglect to represent posterior uncertainties. In this paper, we address problems of Bayesian optimal design (or experiment planning), for which accurate estimates of uncertainty
are essential. To this end, we employ expectation propagation approximate inference for the linear model with Laplace prior, giving new insight into numerical stability properties and proposing a
robust algorithm. We also show how to estimate model hyperparameters by empirical Bayesian maximisation of the marginal likelihood, and propose ideas in order to scale up the method to very large
underdetermined problems. We demonstrate the versatility of our framework on the application of gene regulatory network identification from micro-array expression data, where both the Laplace prior
and the active experimental design approach are shown to result in significant improvements. We also address the problem of sparse coding of natural images, and show how our framework can be used for
compressive sensing tasks. Part of this work appeared in Seeger et al. (2007b). The gene network identification application appears in Steinke et al. (2007).
- Psychological Review , 1996
"... M. Oaksford and N. Chater (O&C; 1994) presented the first quantitative model of P. C. Wason's ( 1966, 1968) selection task in.which performance is rational. J. St B T Evans and D. E. Over (1996)
reply that O&C's account is normatively incorrect and cannot model K. N. Kirby's (1994b) or P. Pollard an ..."
Cited by 46 (4 self)
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M. Oaksford and N. Chater (O&C; 1994) presented the first quantitative model of P. C. Wason's ( 1966, 1968) selection task in.which performance is rational. J. St B T Evans and D. E. Over (1996)
reply that O&C's account is normatively incorrect and cannot model K. N. Kirby's (1994b) or P. Pollard and J. St B T Evans's (1983) data. It is argued that an equivalent measure satisfies their
normative concerns and that a modification of O&C's model accounts for their empirical concerns. D. Laming (1996) argues that O&C made unjustifiable psychological assumptions and that a "correct"
Bayesian analysis agrees with logic. It is argued that O&C's model makes normative and psychological sense and that Laming's analysis is not Bayesian. A. Almor and S. A. Sloman (1996) argue that O&C
cannot explain their data. It is argued that Almor and Sloman's data do not bear on O&C's model because they alter the nature of the task. It is concluded that O&C's model remains the most compelling
and comprehensive account of the selection task. Research on Wason's (1966, 1968) selection task questions human rationality because performance is not "logically correct?' Recently, Oaksford and
Chater (O&C; 1994) provided a rational analysis (Anderson, 1990, 1991) of the selection task that appeared to vindicate human rationality. O&C argued that the selection task is an inductive, rather
than a deductive, reasoning task: Participants must assess the truth or falsity of a general rule from specific instances. In particular, participants face a problem of optimal data selection
(Lindley, 1956): They must decide which of four cards (p, not-p, q, or not-q) is likely to provide the most useful data to test a conditional rule,/fp then q. The "logical " solution is to select the
p and the not-q cards. O&C argued that this solution presupposes falsificationism (Popper, 1959), which argues that only data that can disconfirm, not confirm, hypotheses are of interest. In
contrast, O&C's rational analysis uses a Bayesian approach to inductive
"... Many applications require optimizing an unknown, noisy function that is expensive to evaluate. We formalize this task as a multiarmed bandit problem, where the payoff function is either sampled
from a Gaussian process (GP) or has low RKHS norm. We resolve the important open problem of deriving regre ..."
Cited by 45 (9 self)
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Many applications require optimizing an unknown, noisy function that is expensive to evaluate. We formalize this task as a multiarmed bandit problem, where the payoff function is either sampled from
a Gaussian process (GP) or has low RKHS norm. We resolve the important open problem of deriving regret bounds for this setting, which imply novel convergence rates for GP optimization. We analyze
GP-UCB, an intuitive upper-confidence based algorithm, and bound its cumulative regret in terms of maximal information gain, establishing a novel connection between GP optimization and experimental
design. Moreover, by bounding the latter in terms of operator spectra, we obtain explicit sublinear regret bounds for many commonly used covariance functions. In some important cases, our bounds have
surprisingly weak dependence on the dimensionality. In our experiments on real sensor data, GP-UCB compares favorably with other heuristical GP optimization approaches. 1.
- Journal of the American Statistical Association , 2007
"... This paper explores nonparametric and semiparametric nonstationary modeling methodologies that couple stationary Gaussian processes and (limiting) linear models with treed partitioning.
Partitioning is a simple but effective method for dealing with nonstationarity. Mixing between full Gaussian proce ..."
Cited by 44 (15 self)
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This paper explores nonparametric and semiparametric nonstationary modeling methodologies that couple stationary Gaussian processes and (limiting) linear models with treed partitioning. Partitioning
is a simple but effective method for dealing with nonstationarity. Mixing between full Gaussian processes and simple linear models can yield a more parsimonious spatial model while significantly
reducing computational effort. The methodological developments and statistical computing details which make this approach efficient are described in detail. Illustrations of our model are given for
both synthetic and real datasets. Key words: recursive partitioning, nonstationary spatial model, nonparametric regression, Bayesian model averaging 1
, 2001
"... We propose a decision theoretic approach for deciding which interventions to perform so as to learn the causal structure of a model as quickly as possible. Without such interventions, it is
impossible to distinguish between Markov equivalent models, even given infinite data. We perform online MCMC t ..."
Cited by 37 (2 self)
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We propose a decision theoretic approach for deciding which interventions to perform so as to learn the causal structure of a model as quickly as possible. Without such interventions, it is
impossible to distinguish between Markov equivalent models, even given infinite data. We perform online MCMC to estimate the posterior over graph structures, and use importance sampling to find the
best action to perform at each step. We assume the data is discrete-valued and fully observed.
"... Given a model family and a set of unlabeled examples, one could either label specific examples or state general constraints—both provide information about the desired model. In general, what is
the most cost-effective way to learn? To address this question, we introduce measurements, a general class ..."
Cited by 34 (0 self)
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Given a model family and a set of unlabeled examples, one could either label specific examples or state general constraints—both provide information about the desired model. In general, what is the
most cost-effective way to learn? To address this question, we introduce measurements, a general class of mechanisms for providing information about a target model. We present a Bayesian
decision-theoretic framework, which allows us to both integrate diverse measurements and choose new measurements to make. We use a variational inference algorithm, which exploits exponential family
duality. The merits of our approach are demonstrated on two sequence labeling tasks. 1.
- In Proceedings of the European Conference on Machine Learning (ECML-05 , 2005
"... Abstract. Since resources for data acquisition are seldom infinite, both learners and classifiers must act intelligently under hard budgets. In this paper, we consider problems in which feature
values are unknown to both the learner and classifier, but can be acquired at a cost. Our goal is a learne ..."
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Abstract. Since resources for data acquisition are seldom infinite, both learners and classifiers must act intelligently under hard budgets. In this paper, we consider problems in which feature
values are unknown to both the learner and classifier, but can be acquired at a cost. Our goal is a learner that spends its fixed learning budget bL acquiring training data, to produce the most
accurate “active classifier ” that spends at most bC per instance. To produce this fixed-budget classifier, the fixedbudget learner must sequentially decide which feature values to collect to learn
the relevant information about the distribution. We explore several approaches the learner can take, including the standard “round robin” policy (purchasing every feature of every instance until the
bL budget is exhausted). We demonstrate empirically that round robin is problematic (especially for small bL), and provide alternate learning strategies that achieve superior performance on a variety
of datasets. 1
, 2008
"... In many applications, one has to actively select among a set of expensive observations before making an informed decision. For example, in environmental monitoring, we want to select locations
to measure in order to most effectively predict spatial phenomena. Often, we want to select observations wh ..."
Cited by 30 (3 self)
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In many applications, one has to actively select among a set of expensive observations before making an informed decision. For example, in environmental monitoring, we want to select locations to
measure in order to most effectively predict spatial phenomena. Often, we want to select observations which are robust against a number of possible objective functions. Examples include minimizing
the maximum posterior variance in Gaussian Process regression, robust experimental design, and sensor placement for outbreak detection. In this paper, we present the Submodular Saturation algorithm,
a simple and efficient algorithm with strong theoretical approximation guarantees for cases where the possible objective functions exhibit submodularity, an intuitive diminishing returns property.
Moreover, we prove that better approximation algorithms do not exist unless NP-complete problems admit efficient algorithms. We show how our algorithm can be extended to handle complex cost functions
(incorporating non-unit observation cost or communication and path costs). We also show how the algorithm can be used to near-optimally trade off expected-case (e.g., the Mean Square Prediction Error
in Gaussian Process regression) and worst-case (e.g., maximum predictive variance) performance. We show that many important machine learning problems fit our robust submodular observation selection
formalism, and provide extensive empirical evaluation on several real-world problems. For Gaussian Process regression, our algorithm compares favorably with state-of-the-art heuristics described in
the geostatistics literature, while being simpler, faster and providing theoretical guarantees. For robust experimental design, our algorithm performs favorably compared to SDP-based algorithms.
|
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multiplayer combined form of player. multiple having or composed of more than one element; many. [2 definitions] multiple-choice listing several possible answers from which the correct one is to be
chosen, as a question on an examination or an examination made of such questions. multiple sclerosis a degenerative disease of the nervous system in which nerve cells of the brain or spinal cord lose
their myelin sheaths, resulting in hardening of the neural tissue, speech difficulties, and loss of muscular coordination. multiplex having many parts or elements; multiple. [5 definitions]
multiplicand a number that is to be multiplied by another. multiplication the act or process of multiplying, or the state of being multiplied. [2 definitions] multiplication sign a mathematical sign
(x) indicating that the numbers before and after it are to be multiplied. multiplication table a table for memorization showing the results of multiplying each number of a series, usu. one through
twelve, by each of the numbers in succession. multiplicity the condition of being multiple or various. [2 definitions] multiplier someone or something that multiplies. [2 definitions] multiply^1 to
increase the number, degree, or quantity of. [5 definitions] multiply^2 in a multiple manner. multipower combined form of power. multiproblem combined form of problem. multiproduct combined form of
product. multipurpose intended for or capable of use for several purposes. multiracial including, involving, or representing several racial groups. multirange combined form of range. multiregional
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\[\frac{ \sqrt{18} }{ \sqrt{36} }=\frac{ \sqrt{2} }{ 2 }\]
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yes. you can see it is true by factoring each number into its prime factors 18 is 2*9 = 2*3*3 36 is 2*18= 2*2*9= 2*2*3*3 when these are under a square root \[ \frac{\sqrt{2\cdot 3\cdot 3}}{\sqrt
{2\cdot 2\cdot 3 \cdot 3} }\] you can "pull out" pairs. in the top, 3*3 out of the square root becomes 3 in the bottom both 2 and 3 come out \[ \frac{3 \sqrt{2}}{2\cdot 3} \] we can divide top
and bottom by 3 to simplify to \[ \frac{\sqrt{2}}{2} \]
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of course, if you know 6*6=36 you can just replace the bottom sqrt(36) with 6 right away.
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so the question ask to simplify
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sqrt(2)/2 is as simple as it gets. you could write it as 1/sqrt(2) but people do not sqrt's in the denominator.
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ok. so i have another problem...it says to multiply and simplify \[(3\sqrt{10}-2\sqrt{2})^2=98-24\sqrt{5}\]
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it looks like they gave you the answer. But to show how to do it, use FOIL on ( 3 sqrt(10) - 2 sqrt(2) )* ( 3 sqrt(10) - 2 sqrt(2) )
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so it will just be \[(3\sqrt{10}-2\sqrt{2})(3\sqrt{10}-2\sqrt{2})=98-24\sqrt{5}\]
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yes, but I think they want you to do the multiplication and get the answer they show you.
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how would you do that?
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Measurement of On-wafer Transistor Noise Parameters Without a Tuner Using Unrestricted Noise Sources
Technical Feature
Measurement of On-wafer Transistor Noise Parameters Without a Tuner Using Unrestricted Noise Sources
This article presents a method for calibrating the four noise parameters of a noise receiver which does not require a tuner. The method permits using general (mismatched) noise sources, which may
present very different source reflection coefficients between their hot and cold states. The method is applied to the calibration of a noise set-up using on-wafer noise sources (a reverse-biased
cold-FET and an avalanche noise diode). Experimental validation of the receiver calibration and its application to the determination of on-wafer FET noise parameters to 40 GHz is presented.
A. Lázaro, M.C. Maya and L. Pradell
Universitat Politècnica de Catalunya (UPC)
Barcelona, Spain
An accurate knowledge of transistor noise parameters (NP) is essential to the design of low noise amplifiers (LNA). Before the NPs of the device-under-test can be measured, the noise receiver must be
calibrated. Calibration consists of the determination of the four receiver NPs. Conventional calibration techniques require measurements of noise powers presented at the receiver input for a number
of reflection coefficients produced by a broadband tuner.^1 However, this technique is costly and time-consuming. An alternate method^2 permits determining the receiver NPs from 50 W noise-figure
measurements, by using a commercial, well-matched coaxial noise source. This method has demonstrated an accuracy comparable to that obtained by using tuner-based methods.
While a matched source is the noise reference most commonly used, it has been shown^3 that an on-wafer diode noise source is a convenient noise reference to calibrate the receiver for measuring NPs
of on-wafer transistors directly. However, on-wafer diode noise sources are highly mismatched devices and their reflection coefficient varies significantly between the hot and cold states. Therefore,
the previously proposed method^2 to calibrate the receiver cannot be applied, because it assumes that the receiver noise-figure does not change noticeably between both noise source states. An
on-wafer attenuator pad to achieve a broadband matching condition is normally introduced after the noise diode,^3 but this lossy pad has the drawback of reducing the effective level of excess noise
ratio (ENR) at the on-wafer reference-plane.
Alternatively, this work presents a novel method to calibrate the noise receiver with a general (mismatched) noise source and does not require a tuner, which adds flexibility to the design of new
types of noise sources, specifically on-wafer noise sources. There is no restriction on the variation of the source reflection coefficient between the hot and cold states, and the receiver
noise-figure need not be constant for both noise source states. To verify the method, calibration results are compared to those obtained with a tuner-based method,^1 and the NPs of a passive
semiconductor device (common-gate, on-wafer, cold transistor) are measured and the results are compared to those derived from its S-parameters. The proposed method is used to calibrate the noise
receiver with mismatched on-wafer noise sources and applied to the determination of the NPs of FETs. The experimental results are presented in two bands, 2 to 22 GHz and 26 to 40 GHz.
Fig. 1 Noise figure and nosie parameter measurement system.
Experimental Set-up
Figure 1 shows the experimental set-up. It is a typical configuration for the determination of on-wafer transistor noise parameters, but without the inclusion of a tuner. It consists of the following
components: A VNA (HP8510C) for calibration and measurement of the set-up reflection coefficients and the device-under-test (DUT) S-parameters; a test fixture for inserting the DUT - a wafer-probe
station (SUMMIT-9000 from Cascade-Microtech) is used; three remotely controlled coaxial switches (SW) to select between the lower and the upper noise measurement bands (2 to 22 GHz and 26 to 40 GHz,
respectively), and between VNA and noise measurements; two bias-tees; noise sources - in this work, two on-wafer noise sources (OWNS) and one coaxial (HP346C) noise source are used; a noise receiver
composed of a spectrum analyzer (SA) (HP70000 series), a low noise amplifier (LNA) for the lower frequency band (2 to 22 GHz) to reduce the SA noise-figure, a low noise block downconverter
(LNB-2642-50 from MITEQ), which includes an input LNA to convert the upper frequency band (26 to 40 GHz) into an IF range (5 to 19 GHz) within the SA range, and an external detector connected to the
SA 21.4 MHz IF output. All measurements are automated and controlled with an external PC via GPIB.
Fig. 2 Receiver equivalent wave noise sources b[n1] and b[n2] .
Method for Receiver Calibration With Unrestricted Noise Sources
Since the LNAs in the receiver front-ends are basically unilateral devices, their scattering noise-matrix representation^4 can be simplified by assuming that their equivalent wave noise sources b[N1]
, b[N2] (see Figure 2 ) are uncorrelated
Under this hypothesis, the receiver noise-figure FREC can be approximated by^5
G[S] = source reflection coefficient
Since the LNAs are basically unilateral, it has been also assumed in Equations 1 and 2 that S[11] » G[R] , where G[R] is the receiver input reflection coefficient and S[11] , S[21] are the receiver S
parameters. The impact of such hypothesis on the determination of the receiver noise parameters (in particular F[min] and R[n] ) is discussed later. Note that a(G[R] , G[S] ) is a quantity that only
depends on the reflection coefficients measured with the VNA. To determine the normalized noise power wave
the noise powers delivered to the receiver for two states (hot and cold) of the noise source, P[HOT] and P[COLD] respectively, are measured. Then the following ratio R of noise powers available at
the plane 2-2' can be computed
m(G[R] ,G[S] ) = mismatch coefficient that only depends on reflection coefficients
G[S_COLD] = cold source reflection coefficients
G[S_HOT] = hot source reflection coefficients
Then, the power ratio R is written in terms of the receiver noise temperature T[REC] = (F[REC] -1) T0, where F[REC] is given by Equation 1 and T[0] is the standard temperature (T[0] = 290K), as
T[HOT] = noise source hot temperature
T[COLD] = cold temperature (room temperature)
From Equation 5, the quantity
is computed as
Using Equation 6, the receiver NPs are readily computed assuming the same approximation made previously for the receiver
Z[0] = normalizing impedance
It is assumed in Equation 9 that the LNA has been designed for minimum noise.
Note that, in contrast with previous methods^2 , significant variations in the receiver noise-figure due to different hot and cold noise source reflection coefficients are allowed and taken into
account in Equation 6 through the quantity a(G[R] , G[S] ) defined in Equation 2. Furthermore, the noise device which provides the hot state temperature (T[HOT] ) may be physically different from the
device at cold temperature (T[COLD] ). In this work, examples of calibration using on-wafer noise sources for the hot state and a coplanar 50 W load for the cold state are given.
Finally, the receiver gain-bandwidth constant kGB (where G is the receiver transducer gain for G[S] = 0, and B is the noise measurement bandwidth) must be determined using
After calibration, the noise-figure of an arbitrary DUT can be obtained as follows. The total noise-figure (DUT plus receiver) for a measured reflection coefficient at plane 1-1' (G[s] ') computed
from the measured noise power P(G[S] ') is
G[OUT] = DUT ouput reflection coefficient
T[f] = physical temperature of the source termination (typically, room temperature)
G[DUT] = DUT available gain
The DUT noise-figure is obtained using Friis' formula
Finally, assuming that the DUT is an on-wafer FET, its measured noise-figure (Equation 13) is applied to determine the FET noise parameters using the method proposed in reference.^6 This method
(so-called F[50] ) is based on the determination of the FET intrinsic noise-matrix elements by frequency-fitting the measured device noise-figure for a known source reflection coefficient (usually a
matched load at room temperature). It does not require a tuner, and has demonstrated a good accuracy up to 26 GHz.^6
Fig. 3 Measured receiver noise parameters to 26 GHz.
Receiver Calibration Results
The receiver calibration method described in the previous section is based on approximations in Equations 1, 2 and 9, whose impact on the calculation of the receiver noise parameters must be
assessed. To this end, the receiver noise parameters are measured up to 26 GHz using the method proposed here (Equations 3 to 9) and by a tuner-based method.1 In both tests, a noise-figure
measurement system (HP 8970S) is used as the noise receiver and a well-matched noise source (HP346C) as the noise reference. The tuner used in the second case is a broadband device (NPTS-26, 2 to 26
GHz) from Cascade-Microtech. The results, obtained in 15 measurement sessions, are compared graphically on the next page. The values displayed are mean values at every frequency (24 frequency points)
over the 15 sessions and their standard deviations (s) are also shown as error bars. The numerical values displayed in Appendix A are listed in frequency steps of 4 GHz. It is observed that the
differences in F[min] between the two methods are small (the average deviation over the 24 frequency points is 0.075 dB), and are of the same order of magnitude than the deviations for each method
between measurement sessions. The main impact of the approximations made in Equations 1, 2 and 9 is observed in R[n] (the average deviation over the 24 frequency points is 9.75 W), mainly in the
lower frequency region, where the approximation of Equation 9 does not hold in phase. Also, the differences in G[opt] indicate that this approximation depends on the frequency point and LNA. In
contrast, deviations in R[n] and G[opt] for each method between measurement sessions are much smaller. In conclusion, the differences in R[n] and G[opt] between both methods are systematic, and
cannot be attributed to measurement errors, but to the method. Figure 3 shows the measured receiver noise parameters.
Fig. 4 Total noise figure (typical HEMT and receiver) to 22 GHz.
The differences between methods translate into errors in the determination of the noise-figure of DUTs (transistors). The error introduced is (from Equation 13) DF[DUT] = DF[REC] / G[DUT] .
Therefore, for moderately high gain DUTs (transistors), the impact on the F[DUT] calculation from errors in the receiver noise parameters is small. To illustrate this point, Figure 4 shows the total
noise-figure F[TOT] (DUT + receiver), where the DUT is a typical HEMT, computed using the receiver noise parameters extracted with a tuner-based method^1 and the method proposed here. The source
reflection coefficient D'[S] corresponds to real data from a well-matched noise source (HP346C-K01). Since only differences between the two methods are considered, DF[TOT] = DF[DUT] . Typical
differences are 0.2 dB. The worst case at 2 GHz (0.7 dB) is due to the significant difference between G[OPT] using both methods.
Fig. 5 Noise parameters of a typical HEMT to 22 GHz computed using the F50 method.^6
The differences in the extraction of the noise parameters of the HEMT due to the differences between receiver calibration methods (tuner-based method and Equations 3 to 9) are evaluated in Figure 5
and Appendix B , where the F[50] method proposed^6 is used to determine the FET noise parameters. It is observed that the differences are very small for F[min] and G[OPT] (less than 0.12 dB in F[min]
at 22 GHz and 0.5 percent in G[OPT] ). The difference in R[n] is higher (but less than 8.5 percent at 22 GHz) because this parameter is somewhat more sensitive to the optimization performed to
extract the intrinsic correlation matrix.^6 The small sensitivity shown by the F[50] method to the receiver calibration method is due to the redundancy in frequency and the use of matched source
To verify the receiver calibration, the noise-figure of a passive common-gate FET is measured (as previously suggested^7 ). This is a non-favorable case since, according to Friis, Equation 13, small
errors in the measurement of the receiver noise-figure (F[REC] ) are translated into large errors in the measured DUT noise-figure (F[DUT] ) when the DUT is a mismatched low noise device without
gain.^8 The noise-figure of a passive FET is measured using the calibration method proposed here and compared to the noise-figure computed from the available gain calculated from its measured
S-parameters. The coaxial noise source is used as a noise reference. The measured results are shown in Figure 6 and tabulated in Table 1 . The values displayed are mean values at every frequency (21
frequency points) over six sessions and their standard deviations (s) are also shown as error bars. The differences range from very small (0.01 dB) to moderate (0.41 dB), except in two particular
points (18 and 21 GHz) where it is large (1.13 dB). In these two points, the 2 to 22 GHz LNA does not fulfill G[OPT] = G[R] *, in particular with respect to the phase.
Fig. 6 Noise figure of a passive common-gate FET measured by the proposed method and calculated from its available gain.
As a conclusion to this section, the receiver calibration method proposed, using Equations 3 to 9, is suited for the extraction of transistor noise parameters, in particular when the noise parameter
extraction method^6 (F[50] ) is applied, saving time (fewer power measurements are required - only from ON/OFF noise source states) and cost compared to tuner-based methods.
Table 1
Noise figure of a passive common-gate FET measured by the proposed method (NF) and calculated from its available gain (Gain)
f NF s[NF] Gain s[Gain] |NF-Gain|
(GHz) (dB) (dB) (dB) (dB)
2 3.25 0.07 3.23 0.03 0.02
6 3.69 0.11 3.68 0.12 0.01
10 3.38 0.14 3.72 0.05 0.35
14 4.10 0.15 3.95 0.06 0.16
18 2.15 0.23 3.28 0.06 1.13
22 6.09 0.29 5.68 0.09 0.41
Using On-wafer Noise Sources for Receiver Calibration and FET Noise Parameter Determination
When using coaxial and waveguide noise-sources for measuring noise-parameters of microwave and millimeter-wave on-wafer FETs, the noise-source ENR, known from manufacturers' data, must be translated
to the wafer-probe reference-plane through an input two-port adapter in order to calibrate the receiver. This step requires the determination of the adapter insertion loss (G[adapter] ) from two
calibrations performed with a network analyzer: two-port on-wafer (planes 1-1' and 2-2'); one-port (OSL) at the noise-source (coaxial or waveguide) port (plane 0-0'). The excess noise ratio (ENR') at
the probe plane (plane 1-1') is determined from the known noise source ENR at plane 0-0' and the measured insertion loss of the input two-port adapter
At millimeter-wave frequencies, the OSL calibration uncertainty increases; therefore, the accuracy with which insertion loss and translated ENR are computed degrades. Furthermore, since the adapter
insertion loss increases with frequency, the effective ENR at the wafer-probe plane is reduced to unpractical values. A solution to these problems is an on-wafer diode noise-source placed at the
wafer-probe plane. Since there is no input adapter, the determination of diode ENR at the probe-tip does not require an OSL calibration, but only measured noise-powers and reflection coefficients G
[S] , G[R] , determined from the on-wafer calibration.
Fig. 7 Noise diode wire-bonded to a Jmicro transition.
Two types of on-wafer noise sources have been considered and used here. First, a cold-FET (V[DS] = 0) with the gate reverse-biased. The device is a 0.5 mm gate-length, 2 x 50 mm gate-width DPD-SQW
HEMT from the Foundry of Fraunhofer Institut, Freiburg, Germany (FhG-IAF). The gate bias point is fixed with a current source, close to the transistor breakdown point. This FET is used in a
common-source configuration and its gate-source port is connected to the on-wafer reference-plane 2-2' as an inexpensive on-wafer diode. A commercial avalanche noise diode chip (Noise-Com NC-406)
wire-bonded to the microstrip end (see Figure 7 ) of a coplanar-to-microstrip transition, Probe Point™ 1003 Adapter Substrate from Jmicro, is used as a hot source. To determine the equivalent
noise-temperature T[d] of every noise device for a selected bias point, its noise power delivered to the receiver P[d] and its reflection coefficient (G[S HOT] ) are measured. Combined with the
measured noise power (P[REF] ) and the reflection coefficient (G[REF] ) of an on-wafer passive well-matched reference-load, the following expression is obtained for T[d] :
In Equation 15, T[REC] is the receiver noise-temperature evaluated for G[REF] and G[S HOT] , respectively, m is the mismatch coefficient defined in Equation 4 and G[R] is the receiver input
reflection coefficient. To compute TREC, the receiver NPs previously determined from calibration with the coaxial noise source (Equations 7 to 9) are used. From T[d] computed in Equation 15, the
diode ENR is readily obtained. To reduce the uncertainty in this measured on-wafer noise source ENR, equivalent circuit models for a cold-FET and an avalanche diode (including their intrinsic noise
sources) are determined from the measured ENR and S-parameters (in the case of cold-FET) or reflection coefficient (in the case of avalanche diode), and their intrinsic noise current sources are
fitted with frequency using their smooth frequency characteristic.^9,10 Then, a final estimate of the ENR is computed from the model. Figure 8 shows the final estimate of the ENR and compares it to
the measured ENR. Note that the ENR values obtained are high at the on-wafer plane 2-2', ranging from 8 to 20 dB with the cold-FET and from 20 to 34 dB with the avalanche diode, depending on
frequency and bias point.
Fig. 8 On-wafer noise source ENR; (a) a cold-FET and (b) an avalanche noise diode.
Figure 9 compares the receiver NPs measured with a standard coaxial noise-source, to those measured with the on-wafer noise-sources whose ENR has been determined, using the receiver calibration
method (Equations 3 to 9). The highly mismatched on-wafer noise-sources are used as the hot state, whereas a well-matched coplanar load at room temperature is used as cold state. A good agreement is
obtained (agreement in the receiver F[min] measured with the avalanche diode is within the coaxial-source ENR uncertainty, ±0.34 dB at 40 GHz), demonstrating the applicability of the proposed
calibration method to on-wafer noise sources.
Fig. 9 Measured receiver noise parameter to 40 GHz.
Figure 10 shows an example of measured noise parameters of a PHEMT up to 40 GHz using the F[50] technique^6 and the receiver noise parameters, previously calibrated with both the cold-FET and
avalanche diode noise sources. The FET noise-figure (F[50] ) is obtained by using the above procedure (Equations 11 to 13). The results show very small differences by using either of the on-wafer
noise sources.
Fig. 10 Measured noise parameters of a PHEMT biased at V[ds] = 1.5 V, I[ds] = 17.9 mA.
A general method for calibrating the four receiver noise parameters using unrestricted noise sources, which does not require a tuner, has been presented. Specifically, this method allows very
different hot/cold state impedances, simplifying the design of on-wafer noise sources (no lossy pad after the noise diode is required). The method makes some assumptions in the receiver. The
influence of such assumptions in the receiver calibration and the determination of FET noise parameters (using the F[50] method) are studied through a comparison with tuner-based methods. The
receiver calibration is verified by measuring a mismatched passive device (common-gate FET). A reverse-biased cold-FET and an avalanche noise diode are used as mismatched on-wafer noise sources up to
40 GHz. The four receiver noise parameters are calibrated with the on-wafer noise sources and the receiver calibration method presented here, and applied to the measurement of an on-wafer HEMT noise
parameter, using F[50] .
This work has been supported by Spanish government grants 2FD97-0960-C05-05 and 2FD97-1769-C04-03 (CICYT-FEDER).
1. A. Lázaro, L. Pradell and J.M. O'Callaghan, "Method for Measuring Noise Parameters of Microwave Two-port," IEE Electronics Letters , Vol. 34, No. 12, 1998, pp. 1332-1333.
2. A. Lázaro and L. Pradell, "Extraction of Noise Parameters of Transistor Using a Spectrum Analyzer and 50 W Noise-figure Measurements Only," IEE Electronics Letters, Vol. 34, No. 24, 1998, pp.
3. L.P. Dunleavy, J. Randa, D.K. Walker, R. Billinger and J. Rice, "Characterization and Applications of On-wafer Diode Noise," IEEE Transactions on Microwave Theory and Techniques , Vol. 46, No. 12,
December 1998, pp. 2620-2627.
4. R.P. Hecken, "Analysis of Linear Noisy Two-ports Using Scattering Waves," IEEE Transactions on Microwave Theory and Techniques , Vol. 29, No. 12, October 1981, pp. 997-1004.
5. T. Werling, E. Bourdel, D. Pasquet and A. Boudiaf, "Determination of Wave Noise Sources Using Spectral Parametric Modeling," IEEE MTT-S International Microwave Symposium Digest , June 1997.
6. A. Lázaro, L. Pradell and J.M. O'Callaghan, "FET Noise Parameter Determination Using a Novel Technique Based on 50 W Noise-figure Measurements," IEEE Transactions on Microwave Theory and
Techniques , Vol. 47, No. 3, March 1999, pp.315-324.
7. L. Escotte, R. Plana and J. Graffeuil, "Evaluation of Noise Parameters Extraction Methods," IEEE Transactions on Microwave Theory and Techniques , Vol. 41, No. 3, March 1993, pp. 382-387.
8. "Noise-figure Measurement Accuracy," Hewlett Packard Application Note57-2, November 1988.
9. M.C. Maya, A. Lázaro and L. Pradell, "Calibración del ENR de Fuentes de Ruido en Oblea," XV Symposium Nacional URSI , Zaragoza, Spain, September 2000, pp. 1-3.
10. R.H. Haitz and F.W. Voltmer, "Noise of a Self-sustaining Avalanche Discharge in Silicon: Studies at Microwave Frequencies," Journal of Applied Physics , Vol. 39, No. 6, 1968, pp. 3379-3384.
A. Lázaro received his MS and PhD degrees in telecommunication engineering from the Universitat Politècnica de Catalunya (UPC), Barcelona, Spain, in 1994 and 1998, respectively. He then joined the
faculty at UPC, where he has been teaching a course on microwave circuits and antennas. His research interests are in microwave device modeling, on-wafer noise measurements, MMICs, low noise
oscillators and MEMS.
M.C. Maya received her MSc degree in electronics and telecommunications from the CICESE Research Center, Ensenada, BC, Mexico, in 1998, and is currently working toward her PhD degree at the
Universitat Politècnica de Catalunya (UPC), Barcelona, Spain. Her research interests are in MESFET, HEMT and HBT device modeling, and on-wafer noise measurements techniques.
L. Pradell received his telecommunication engineering degree from the Universitat Politècnica de Catalunya (UPC), Barcelona, Spain, in 1981. From 1981 to 1985, he worked at Mier Allende, S.A.
(Barcelona) as an RF and microwave system design engineer. In 1985, he joined the faculty at UPC, where he received his PhD degree in 1989. Since 1985, he has been teaching courses on microwave
circuits and antennas, and performing research on microwave and millimeter-wave devices and systems.
│APPENDIX A │
│ │
││ Comparison Between Measured Receiver Noise Parameters to 26 GHz ││
││ │ F[min] (dB) │ │ │ |G[out] | ││
││ f ├───────────┬────────────────┬──────────┤ │ f ├─────────────┬────────────────┬──────────┤│
││(GHz)│ Tuner │Equations 3 to 9│Difference│ │(GHz)│ Tuner │Equations 3 to 9│Difference││
││ ├──────┬────┼─────────┬──────┼──────────┤ │ ├───────┬─────┼─────────┬──────┼──────────┤│
││ │ Mean │ s │ Mean │ s │ (Mean) │ │ │ Mean │ s │ Mean │ s │ (Mean) ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 2 │ 6.35 │0.02│ 6.41 │ 0.06 │ 0.06 │ │ 2 │ 0.24 │0.003│ 0.02 │0.001 │ 0.22 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 6 │ 6.51 │0.05│ 6.45 │ 0.06 │ 0.06 │ │ 6 │ 0.17 │0.008│ 0.19 │0.005 │ 0.02 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 10 │ 6.76 │0.05│ 7.07 │ 0.04 │ 0.31 │ │ 10 │ 0.23 │0.009│ 0.17 │0.012 │ 0.06 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 14 │ 6.94 │0.05│ 7.17 │ 0.04 │ 0.23 │ │ 14 │ 0.13 │0.008│ 0.02 │0.005 │ 0.11 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 18 │ 6.71 │0.08│ 6.79 │ 0.08 │ 0.08 │ │ 18 │ 0.27 │0.002│ 0.25 │0.002 │ 0.02 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 22 │ 7.32 │0.07│ 7.28 │ 0.08 │ 0.04 │ │ 22 │ 0.15 │0.014│ 0.18 │0.017 │ 0.03 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 26 │ 9.06 │0.13│ 9.07 │ 0.11 │ 0.00 │ │ 26 │ 0.15 │0.020│ 0.06 │0.010 │ 0.09 ││
│├─────┼──────┴────┴─────────┴──────┴──────────┤ ├─────┼───────┴─────┴─────────┴──────┴──────────┤│
││ │ R[n] (W) │ │ │ L G[opt] ││
│├─────┼──────┬────┬─────────┬──────┬──────────┤ ├─────┼───────┬─────┬─────────┬──────┬──────────┤│
││ 2 │39.66 │0.43│ 43.77 │ 0.84 │ 4.12 │ │ 2 │133.17 │0.69 │ -5.38 │ 7.13 │ 138.56 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 6 │73.83 │1.34│ 42.52 │ 1.58 │ 31.31 │ │ 6 │-30.83 │1.54 │ -101.77 │ 2.13 │ 70.94 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 10 │50.35 │1.50│ 36.60 │ 1.17 │ 13.75 │ │ 10 │-199.39│2.36 │ -170.17 │ 1.42 │ 50.78 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 14 │47.13 │1.22│ 50.97 │ 1.09 │ 3.84 │ │ 14 │-168.33│3.29 │ -128.73 │99.62 │ 39.60 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 18 │99.63 │1.80│ 79.11 │ 1.70 │ 20.52 │ │ 18 │ 13.35 │3.39 │ 5.77 │ 3.72 │ 7.58 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 22 │88.40 │2.03│ 77.60 │ 1.77 │ 10.79 │ │ 22 │ 28.80 │7.69 │ 17.92 │ 5.75 │ 10.88 ││
│├─────┼──────┼────┼─────────┼──────┼──────────┤ ├─────┼───────┼─────┼─────────┼──────┼──────────┤│
││ 26 │100.84│2.59│ 93.64 │ 3.07 │ 7.21 │ │ 26 │ 99.17 │14.27│ 28.24 │66.36 │ 70.94 ││
│ │
│APPENDIX B │
│ │
│ ┌──────────────────────────────────────────────────────────────────────────────────────────┐ │
│ │ Noise Parameters of a Typical HEMT to 22 GHz Computed Using the F[50] Method^6 │ │
│ ├─────┬──────────────────┬───────────────────┬─┬─────┬──────────────────┬──────────────────┤ │
│ │ │ F[min] (dB) │ R[n] (W) │ │ │ |G[opt] | │ L G[opt] │ │
│ │ f ├─────┬──────┬─────┼──────┬──────┬─────┤ │ f ├─────┬──────┬─────┼─────┬──────┬─────┤ │
│ │(GHz)│Tuner│ Eqs │Diff.│Tuner │ Eqs │Diff.│ │(GHz)│Tuner│ Eqs │Diff.│Tuner│ Eqs │Diff.│ │
│ │ │ │3 to 9│ │ │3 to 9│ │ │ │ │3 to 9│ │ │3 to 9│ │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 2 │0.21 │ 0.22 │0.00 │119.00│120.86│1.85 │ │ 2 │0.98 │ 0.98 │0.000│6.53 │ 6.55 │0.02 │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 6 │0.65 │ 0.66 │0.01 │123.03│126.95│3.92 │ │ 6 │0.94 │ 0.94 │0.001│19.49│19.56 │0.07 │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 10 │1.10 │ 1.14 │0.03 │126.22│132.13│5.91 │ │ 10 │0.91 │ 0.91 │0.002│32.08│32.20 │0.12 │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 14 │1.57 │ 1.63 │0.06 │128.48│136.27│7.79 │ │ 14 │0.88 │ 0.89 │0.003│44.10│44.26 │0.15 │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 18 │2.05 │ 2.13 │0.09 │129.77│139.29│9.52 │ │ 18 │0.86 │ 0.87 │0.004│55.39│55.58 │0.18 │ │
│ ├─────┼─────┼──────┼─────┼──────┼──────┼─────┤ ├─────┼─────┼──────┼─────┼─────┼──────┼─────┤ │
│ │ 22 │2.53 │ 2.65 │0.12 │130.07│141.14│11.07│ │ 22 │0.85 │ 0.85 │0.005│65.86│66.06 │0.20 │ │
│ └─────┴─────┴──────┴─────┴──────┴──────┴─────┴─┴─────┴─────┴──────┴─────┴─────┴──────┴─────┘ │
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MathGroup Archive: February 1999 [00414]
[Date Index] [Thread Index] [Author Index]
RE: Re: Simplify Log[ab] - Log[b] to Log[a] ?
• To: mathgroup at smc.vnet.net
• Subject: [mg16098] RE: [mg16071] Re: [mg16039] Simplify Log[ab] - Log[b] to Log[a] ?
• From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
• Date: Thu, 25 Feb 1999 08:24:57 -0500
• Sender: owner-wri-mathgroup at wolfram.com
David Reiss wrote:
...... this might not be what the user wants (if z<0 for example).
(z^2)^(1/2) (Log[a b]-Log[b])/.{Log[x_ y_]:>Log[x]+Log[y]}
Out[3]= Sqrt[z^2]*Log[a]
It all depends on the broader context of the problem.
Remember you can also use FunctionExpand, PowerExpand, ComplexExpand with
pattern matching. This saves you the trouble of typing in obscure
identities, and will not neglect the branch cut for Sqrt[expr].
(x^2/E^x + 2*(E^(-x) + x/E^x))*
Ted Ersek
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Card Games Home Page | Other Invented Games
Contributed by Micheal Clark who writes:
Ji'zara is an intense bidding based card game that is played between 2 players or 2 teams of players. In order to win, the right strategy with a lot of luck is required. The rules seem a little
complex at first, but the game is surprisingly easy to learn and no two games are ever alike.
Simple game (Two players)
The first player to win 3 out of the 5 key cards wins the game. The key cards are the four aces and the four of clubs.
Set up:
Take a standard deck of cards. Remove the jokers so only 52 cards remain. Set aside all four aces and the four of clubs. Shuffle the remaining cards and divide them into two stacks. One stack of 15
cards, called the bidding draw, and one stack with the remaining 32 cards, called the pot. Shuffle the key cards into the bidding draw making that stack so that the bidding draw contains 20 cards.
Each player then draws 5 cards from the pot to form his hand. During the game, used cards are discarded into a pile known as the burn.
The top card from the bidding draw in turned face up and set between the two players. If this card is not a key card, each player simultaneously decides and announces whether he will bid or pass.
• If both players decide to bid, then both bids are placed face down simultaneously, and then turned face up. The player with the higher point total wins the card, and places it into his hand. [If
the bids are equal, neither player wins the card and it is placed in the burn.] All cards bid are then placed in the burn, and both players draw from the pot until the number of cards in his hand
is five. Play repeats with the next card dealt from the bidding draw.
• If only one player decides to bid. he places his bid directly into the burn, pick up the card he won, and then draws from the pot until he has five cards in his hand.
• If both players pass, then the card is placed into the burn and a new card is drawn from the bidding draw.
If a key card is drawn from the bidding draw, both players reveal their hands, and the highest point total wins the card. [If the totals are equal the player who most recently won a bid wins the
card. If the the very first card turned from the bidding draw is a key card and the players' hands are equal, the winner of the card is decided at random - for example by flipping a coin, rolling a
die or drawing cards.] The card is set aside in front of the player, similar to a book (trick) won in spades. The burn is shuffled into the pot. Each player then draws 5 cards and play continues. The
first player to win 3 key cards wins the game.
Each card has the value printed on its face, i.e. a 2 is with 2 points, 7 is worth 7 points, and so on. Jacks are worth 15 points, Queens are worth 20, and Kings are 25. Each player may normally bid
only one card, unless the card is a 2, 3, or 4. 2's, 3's and 4's are known as chain cards and can be added to any bid, The only limit to how many of these chain cards you can add is how many you have
in your hand. For example, you can bid an 8 and a 4 for a value of 12; or an 8, a 4, and a 2 (value 14); but never an 8 and a 5. Bid point totals are allowed to surpass the point value of the card
you are bidding for. Chain cards are important to win close bids, part of the strategy is knowing when to keep chain cards in your hand to win close bids, but to dump them and replace them with
higher value cards before a key card comes up.
Long game (Two Players)
Played just as the simple game, except to win a player must win all 5 key cards, called a Ji'zara. If after all 5 key cards have appeared they are not all won by the same player, the game is resumed,
with the four kings and the four of spades acting as the key cards. These cards are separated, the remaining 42 cards are shuffled, 15 of them form the new bidding draw, into which the five new key
cards are shuffled, and the remaining 27 are the new pot. If the original key cards (the four Aces and 4 of clubs) are divided 4-1, each player wagers one key card on the outcome of the resumed game;
if they were 2-3 the winner of the previous part of the game decides whether to wager one key card against one or two against two. As soon as a king the 4 of spades is turned up, the winning hand
wins the key cards that were wagered. If one player now has all 5 original key cards, that player wins. If not, the game is resumed again and this is repeated. The secondary key cards - the kings and
the 4 of spades - are never won but are reused until someone wins all 5 original key cards.
Team Game (Four players, Two Teams)
This is played with 2 decks with only one set of key cards. One ace of each suit and one 4 of clubs are removed. One player on a team is the bidder, the other is the builder. Each team may bid or
pass: discussion between team members is permitted, after which the bidders simultaneously make their decision. When a team bids, each player of the team bids a card plus any number of chain cards.
The bids for both players on a team are totaled into one bid with the builder the winning team taking the won card. and all players replenishing their hand to 5 cards from the pot. When a key card is
turned up, both hands are totaled and highest total wins the key card.
Return to Index of Invented Card Games Last updated 8th March 2009
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taylor expansion of erf(x)
January 10th 2010, 07:27 AM #1
Super Member
Aug 2009
taylor expansion of erf(x)
how do you solve:
consider the so called error integral erf(x) for x ∈R:
erf(x)= 2/ (pi)^(1/2) integrate exp (-y) from x-x to x=0.
derive the first few terms in taylor expansion of erf(x) up to x^3. hint: use second fundamental theorem of calculus
can someone help me solve this? im completely lost..
how do you solve:
consider the so called error integral erf(x) for x ∈R:
erf(x)= 2/ (pi)^(1/2) integrate exp (-y) from x-x to x=0. ???
derive the first few terms in taylor expansion of erf(x) up to x^3. hint: use second fundamental theorem of calculus
can someone help me solve this? im completely lost..
$erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt$
using the 2nd FTC ...
$erf'(x) = \frac{2}{\sqrt{\pi}} e^{-x^2}$
note that $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...$
so ...
$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + ...$
integrate term for term ...
$\frac{2}{\sqrt{\pi}} \int e^{-x^2} \, dx = \frac{2}{\sqrt{\pi}} \left(C + x - \frac{x^3}{3} + \frac{x^5}{5 \cdot 2!} - \frac{x^7}{7 \cdot 3!} + ...\right)$
finish up ...
do i need to find the remainder since this is a taylor series question? or what else do i need to do to finish up?
January 10th 2010, 09:01 AM #2
January 10th 2010, 09:52 AM #3
Super Member
Aug 2009
January 10th 2010, 10:26 AM #4
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Correlation and Regression
The Statistics Calculator
Statistical Analysis Tests At Your Fingertips
Correlation Types
Correlation is a measure of association between two variables. The variables are not designated as dependent or independent. The two most popular correlation coefficients are: Spearman's correlation
coefficient rho and Pearson's product-moment correlation coefficient.
When calculating a correlation coefficient for ordinal data, select Spearman's technique. For interval or ratio-type data, use Pearson's technique.
The value of a correlation coefficient can vary from minus one to plus one. A minus one indicates a perfect negative correlation, while a plus one indicates a perfect positive correlation. A
correlation of zero means there is no relationship between the two variables. When there is a negative correlation between two variables, as the value of one variable increases, the value of the
other variable decreases, and vise versa. In other words, for a negative correlation, the variables work opposite each other. When there is a positive correlation between two variables, as the value
of one variable increases, the value of the other variable also increases. The variables move together.
The standard error of a correlation coefficient is used to determine the confidence intervals around a true correlation of zero. If your correlation coefficient falls outside of this range, then it
is significantly different than zero. The standard error can be calculated for interval or ratio-type data (i.e., only for Pearson's product-moment correlation).
The significance (probability) of the correlation coefficient is determined from the t-statistic. The probability of the t-statistic indicates whether the observed correlation coefficient occurred by
chance if the true correlation is zero. In other words, it asks if the correlation is significantly different than zero. When the t-statistic is calculated for Spearman's rank-difference correlation
coefficient, there must be at least 30 cases before the t-distribution can be used to determine the probability. If there are fewer than 30 cases, you must refer to a special table to find the
probability of the correlation coefficient.
A company wanted to know if there is a significant relationship between the total number of salespeople and the total number of sales. They collect data for five months.
Variable 1 Variable 2
Correlation coefficient = .921
Standard error of the coefficient = ..068
t-test for the significance of the coefficient = 4.100
Degrees of freedom = 3
Two-tailed probability = .0263
Another Example
Respondents to a survey were asked to judge the quality of a product on a four-point Likert scale (excellent, good, fair, poor). They were also asked to judge the reputation of the company that made
the product on a three-point scale (good, fair, poor). Is there a significant relationship between respondents perceptions of the company and their perceptions of quality of the product?
Since both variables are ordinal, Spearman's method is chosen. The first variable is the rating for the quality the product. Responses are coded as 4=excellent, 3=good, 2=fair, and 1=poor. The second
variable is the perceived reputation of the company and is coded 3=good, 2=fair, and 1=poor.
Variable 1 Variable 2
Correlation coefficient rho = .830
t-test for the significance of the coefficient = 3.332
Number of data pairs = 7
Probability must be determined from a table because of the small sample size.
Simple regression is used to examine the relationship between one dependent and one independent variable. After performing an analysis, the regression statistics can be used to predict the dependent
variable when the independent variable is known. Regression goes beyond correlation by adding prediction capabilities.
People use regression on an intuitive level every day. In business, a well-dressed man is thought to be financially successful. A mother knows that more sugar in her children's diet results in higher
energy levels. The ease of waking up in the morning often depends on how late you went to bed the night before. Quantitative regression adds precision by developing a mathematical formula that can be
used for predictive purposes.
For example, a medical researcher might want to use body weight (independent variable) to predict the most appropriate dose for a new drug (dependent variable). The purpose of running the regression
is to find a formula that fits the relationship between the two variables. Then you can use that formula to predict values for the dependent variable when only the independent variable is known. A
doctor could prescribe the proper dose based on a person's body weight.
The regression line (known as the least squares line) is a plot of the expected value of the dependent variable for all values of the independent variable. Technically, it is the line that "minimizes
the squared residuals". The regression line is the one that best fits the data on a scatterplot.
Using the regression equation, the dependent variable may be predicted from the independent variable. The slope of the regression line (b) is defined as the rise divided by the run. The y intercept
(a) is the point on the y axis where the regression line would intercept the y axis. The slope and y intercept are incorporated into the regression equation. The intercept is usually called the
constant, and the slope is referred to as the coefficient. Since the regression model is usually not a perfect predictor, there is also an error term in the equation.
In the regression equation, y is always the dependent variable and x is always the independent variable. Here are three equivalent ways to mathematically describe a linear regression model.
y = intercept + (slope
y = constant + (coefficient
y = a + bx + e
The significance of the slope of the regression line is determined from the t-statistic. It is the probability that the observed correlation coefficient occurred by chance if the true correlation is
zero. Some researchers prefer to report the F-ratio instead of the t-statistic. The F-ratio is equal to the t-statistic squared.
The t-statistic for the significance of the slope is essentially a test to determine if the regression model (equation) is usable. If the slope is significantly different than zero, then we can use
the regression model to predict the dependent variable for any value of the independent variable.
On the other hand, take an example where the slope is zero. It has no prediction ability because for every value of the independent variable, the prediction for the dependent variable would be the
same. Knowing the value of the independent variable would not improve our ability to predict the dependent variable. Thus, if the slope is not significantly different than zero, don't use the model
to make predictions.
The coefficient of determination (r-squared) is the square of the correlation coefficient. Its value may vary from zero to one. It has the advantage over the correlation coefficient in that it may be
interpreted directly as the proportion of variance in the dependent variable that can be accounted for by the regression equation. For example, an r-squared value of .49 means that 49% of the
variance in the dependent variable can be explained by the regression equation. The other 51% is unexplained.
The standard error of the estimate for regression measures the amount of variability in the points around the regression line. It is the standard deviation of the data points as they are distributed
around the regression line. The standard error of the estimate can be used to develop confidence intervals around a prediction.
A company wants to know if there is a significant relationship between its advertising expenditures and its sales volume. The independent variable is advertising budget and the dependent variable is
sales volume. A lag time of one month will be used because sales are expected to lag behind actual advertising expenditures. Data was collected for a six month period. All figures are in thousands of
dollars. Is there a significant relationship between advertising budget and sales volume?
Indep. Var. Depen. Var
4.2 27.1
6.1 30.4
3.9 25.0
5.7 29.7
7.3 40.1
5.9 28.8
Model: y = 9.873 + (3.682
Standard error of the estimate = 2.637
t-test for the significance of the slope = 3.961
Degrees of freedom = 4
Two-tailed probability = .0149
r-squared = .807
You might make a statement in a report like this: A simple linear regression was performed on six months of data to determine if there was a significant relationship between advertising expenditures
and sales volume. The t-statistic for the slope was significant at the .05 critical alpha level, t(4)=3.96, p=.015. Thus, we reject the null hypothesis and conclude that there was a positive
significant relationship between advertising expenditures and sales volume. Furthermore, 80.7% of the variability in sales volume could be explained by advertising expenditures.
How to Order Statistics Calculator
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prove: as the function increases then the graph is concave up/down
October 5th 2008, 10:19 AM #1
prove: as the function increases then the graph is concave up/down
I will just give the solution for 2. (a) since the remaining case is very similar.
Clearly, if $f$ is increasing then we know that $f(x)\geq f(a)$ for all $x\in[a,b]$.
Hence, we have
$\int\limits_{a}^{b}f(x)dx\geq\int\limits_{a}^{b}f( a)dx=f(a)(b-a),$
which completes the first part of the inequality.
For the remaining part, we consider that $f$ is concave down (convex).
This indicates that $f(x)\leq\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$ holds for all $x\in[a,b]$ (draw a graph for the increasing convex function $f$ and the line passing at the points $(a,f(a))$ and $
Integrating both sides of this inequality, we get
$\int\limits_{a}^{b}f(x)dx\leq\int\limits_{a}^{b}\B igg(\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\Bigg)dx=(b-a)\frac{f(b)+f(a)}{2},$
which completes the proof.
October 5th 2008, 10:45 AM #2
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[Numpy-discussion] Behavior from a change in dtype?
Skipper Seabold jsseabold@gmail....
Thu Sep 10 08:48:08 CDT 2009
On Tue, Sep 8, 2009 at 12:53 PM, Christopher Barker
<Chris.Barker@noaa.gov> wrote:
> Skipper Seabold wrote:
>> Hmm, okay, well I came across this in trying to create a recarray like
>> data2 below, so I guess I should just combine the two questions.
> key to understanding this is to understand what is going on under the
> hood in numpy. Travis O. gave a nice intro in an Enthought webcast a few
> months ago -- I"m not sure if those are recorded and up on the web, but
> it's worth a look. It was also discussed int eh advanced numpy tutorial
> at SciPy this year -- and that is up on the web:
> http://www.archive.org/details/scipy09_advancedTutorialDay1_1
Thanks. I wasn't able to watch the Enthought webcasts on Linux, but
I've seen a few of the video tutorials. What a great resource. I'm
really glad this came together.
> Anyway, here is my minimal attempt to clarify:
>> import numpy as np
>> data = np.array([[10.75, 1, 1],[10.39, 0, 1],[18.18, 0, 1]])
> here we are using a standard array constructor -- it will look at the
> data you are passing in (a mixture of python floats and ints), and
> decide that they can best be represented by a numpy array of float64s.
> numpy arrays are essentially a pointer to a black of memory, and a bunch
> of attributes that describe how the bytes pointed to are to be
> interpreted. In this case, they are a 9 C doubles, representing a 3x3
> array of doubles.
>> dt = np.dtype([('var1', '<f8'), ('var2', '<i8'), ('var3', '<i8')])
> (NOTE: I'm on a big-endian machine, so I've used:
> dt = np.dtype([('var1', '>f8'), ('var2', '>i8'), ('var3', '>i8')])
> )
> This is a data type descriptor that is analogous to a C struct,
> containing a float64 and two int84s
>> # Doesn't work, raises TypeError: expected a readable buffer object
>> data2 = data2.view(np.recarray)
>> data2.astype(dt)
> I'm don't understand that error either, but recarrays are about adding
> the ability to access parts of a structured array by name, but you still
> need the dtype to specify the types and names. This does seem to work
> (though may not be giving the results you expect):
> In [19]: data2 = data.copy()
> In [20]: data2 = data2.view(np.recarray)
> In [21]: data2 = data2.view(dtype=dt)
> or, indeed in the opposite order:
> In [24]: data2 = data.copy()
> In [25]: data2 = data2.view(dtype=dt)
> In [26]: data2 = data2.view(np.recarray)
> So you've done two operations, one is to change the dtype -- the
> interpretation of the bytes in the data buffer, and one is to make this
> a recarray, which allows you to access the "fields" by name:
> In [31]: data2['var1']
> Out[31]:
> array([[ 10.75],
> [ 10.39],
> [ 18.18]])
>> # Works without error (?) with unexpected result
>> data3 = data3.view(np.recarray)
>> data3.dtype = dt
> that all depends what you expect! I used "view" above, 'cause I think
> there is less magic, though it's the same thing. I suppose changing the
> dtype in place like that is a tiny bit more efficient -- if you use
> .view() , you are creating a new array pointing to the same data, rather
> than changing the array in place.
> But anyway, the dtype describes how the bytes in the memory black are to
> be interpreted, changing it by assigning the attribute or using .view()
> changes the interpretation, but does not change the bytes themselves at
> all, so in this case, you are taking the 8 bytes representing a float64
> of value: 1.0, and interpreting those bytes as an 8 byte int -- which is
> going to give you garbage, essentially.
>> # One correct (though IMHO) unintuitive way
>> data = np.rec.fromarrays(data.swapaxes(1,0), dtype=dt)
> This is using the np.rec.fromarrays constructor to build a new record
> array with the dtype you want, the data is being converted and copied,
> it won't change the original at all:
> So the question remains -- is there a way to convert the floats in
> "data" to ints in place?
Ah, ok. I understand roughly the above. But, yes, this is my question.
> This seems to work:
> In [78]: data = np.array([[10.75, 1, 1],[10.39, 0, 1],[18.18, 0, 1]])
> In [79]: data[:,1:3] = data[:,1:3].astype('>i8').view(dtype='>f8')
> In [80]: data.dtype = dt
> It is making a copy of the integer data in process -- but I think that
> is required, as you are changing the value, not just the interpretation
> of the bytes. I suppose we could have a "astype_inplace" method, but
> that would only work if the two types were the same size, and I'm not
> sure it's a common enough use to be worth it.
> What is your real use case? I suspect that what you really should do
> here is define your dtype first, then create the array of data:
I have a function that eventually appends an ndarray of floats that
are 0 to 1 to a recarray, and I ran into it trying to debug. Then I
was just curious about the modification in place.
> data = np.array([(10.75, 1, 1), (10.39, 0, 1), (18.18, 0, 1)], dtype=dt)
> which does require that you use tuples, rather than lists to hold the
> "structs".
Ah yes, I have had a bit of trouble extending my same function to
structured arrays, but that's another thread if I can't figure it out.
Thanks for the help.
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Calculus Early Transcendentals 10th Edition, Howard Anton
Calculus Early Transcendentals 10th Edition PDF Download Ebook. Howard Anton, Irl C. Bivens and Stephen Davis fulfill the needs of a changing market by providing flexible solutions to teaching and
learning needs of all kinds. The text continues to focus on and incorporate new ideas that have withstood the objective scrutiny of many skilled and thoughtful instructors and their students.
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closely to questions in Wiley Plus. In addition, more WileyPLUS questions now correspond to specific exercises in the text.
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Re: st: which STATA version for Poi's QUAIDS command
Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down at the end of May, and its replacement, statalist.org is already up and
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Re: st: which STATA version for Poi's QUAIDS command
From "Brian P. Poi" <bpoi@stata.com>
To statalist@hsphsun2.harvard.edu
Subject Re: st: which STATA version for Poi's QUAIDS command
Date Mon, 18 Mar 2013 16:52:46 -0500
On 03/18/2013 04:15 PM, Luca Tiberti wrote:
Hi Nick, thanks for this
here is the output of describe, short
obs: 3,860
vars: 39 18 Mar 2013 15:37
size: 455,480
Sorted by: serial
Note: dataset has changed since last saved
For your info, I had already tried by keeping only variables finally
entering into the demand system and compress them. However I was not
able to run succesfully the command.
Thanks for your help,
How many goods are in your demand system? How many demographic variables are you specifying?
To estimate the model, -quaids- must create temporary variables to hold observation-level derivatives of each equation with respect to each of the parameters. The details are in the methods and formulas section of [R] nlsur.
For a basic AIDS model, with three goods there are 7 free parameters and 2 estimated equations, for a total of 14 temporary variables -quaids- needs to create. With four goods, there are 12 free parameters and 3 estimated equations, for a total of 36 temporary variables. If my math is correct, a basic AIDS model with 15 goods has 133 parameters and 14 estimated equations for a total of 1,862 temporary variables. If you have 16 goods, there are 2,250 derivatives that must be stored in temporary variables, too many for Stata/IC to handle.
If you include the quadratic income term (the QUAIDS model), then you will hit Stata/IC's variable limit if you try to fit a 15-good model.
Throw one demographic variable into the QUAIDS model, and you are down to 13 goods. Five demographic variables and you're down to 11 goods.
In short, having many goods and demographics can quickly balloon the number of parameters in the model and hence the number of temporary variables needed during estimation.
I hope this helps.
-- Brian Poi
-- bpoi@stata.com
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
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How do we get angular dependent wave function from only radially dependent potentials
Actually, if you sum all the probability distributions, for example, the three distributions from those eigenfunctions for l=1, it gives an actual spherically symmetric function.
Ok, I may have made a mistake here, but I'm pretty sure this isn't true. Let's say we had the wave function
[itex]\psi = \frac{Y^{-1} _1 + Y^{0} _1 + Y^{1} _1}{3}[/itex]
[itex]Y^{±1} _1 = \mp (3/8\pi)^{1/2} sin(\theta) e^{\pm i\phi}[/itex]
[itex]Y^0 _1 = (3/4\pi)^{1/2} cos(\theta)[/itex]
So, [itex]\psi[/itex] alone definitely isn't spherically symmetric, if you add it up. But that's not surprising, so I assume you meant the probability distribution [itex]P(\theta, \phi) \propto \left
|\psi\right|^2[/itex] is symmetric, which is what we "see" anyway (wave functions are basically just math, probability distributions are the things we'd actually "measure" in some effect, right?).
But if we look at [itex]P(\theta, \phi)[/itex], unless I made an arithmetic error, it isn't spherically symmetric either. You get a bunch of cross terms which would be orthogonal if we were
integrating, but we're not. Most of them actually seem to cancel out (and the three of the spherical harmonics themselves each squared and summed actually does cancel out into a constant term), but
you're left with two terms that seem to be angularly dependent.
I end up getting:
[itex]P(\theta, \phi) = \left|\psi\right|^2 = \left|\frac{Y^{-1} _1 + Y^{0} _1 + Y^{1} _1}{3}\right|^2 = \frac{1}{9}\frac{3}{4\pi}(1 - sin^2(\theta)cos(2\phi))[/itex]
Which is still not spherically symmetric. Thoughts?
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Are transversely immersed PL surfaces Riemann surfaces?
up vote 0 down vote favorite
I have a piecewise linear (PL) surface transversely immersed in $\mathbb{R}^3$; is this a Riemann surface in the sense that I can describe it with a local coordinate $z\in \mathbb{C}$? My basic
argument is that in 2 dimensions I think the PL and smooth categories coincide, so the question reduces to "can any smooth immersed surface in $\mathbb{R}^3$ be a Riemann surface?" My first instinct
is that this is false; the added complex structure would chance the nature of the surface (ie the Cauchy-Reimann equations must now be satisfied). However, you can put an almost complex structure on
any even-dimensional real manifold so I'm thinking the statement might actually be true.
Any ideas?
riemann-surfaces dg.differential-geometry
Perhaps you mean to ask, "is there a canonical way to put the structure of a Riemann surface on a PL immersed surface?" Because a PL immersed surface isn't literally a Riemann surface in the sense
of having a complex atlas. – Ryan Budney May 26 '11 at 17:40
Ok Ryan yes I think that is what I want to ask. Basically there nice ways to describe immersed Riemann surfaces (specifically, the Weierstrass formula), and I would like to describe immersed PL
surfaces using the same techniques. – cduston May 26 '11 at 19:09
The Weierstrass formula is for immersed minimal surfaces. The Riemann surface thing has nothing to do with your problem here -- there is no reason that an immersion of a PL or a smooth surface has
to be minimal in any sense of the word. – Andy Putman May 26 '11 at 19:19
In fact it's pretty easy to use the Weierstrass formula to extend this idea to arbitrary surfaces...for instance see Friedrich J. Diff. Geom. 28 (1998). As far I understand it, they still need to
be complex surfaces but the minimal condition can be dropped. – cduston May 26 '11 at 20:07
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1 Answer
active oldest votes
This question seems very confused. It is true that every PL surface can be given a canonical smooth structure. It is also true that a surface $X$ that is smoothly immersed in $\mathbb
{R}^n$ can be given a canonical Riemann surface structure -- pulling back the Euclidean metric to $X$ gives a Riemannian metric on $X$ and thus a conformal structure, and it is known
that (in real dimension $2$) there is a natural bijection between conformal structures and Riemann surface structures.
up vote 4 down However, if $X$ is a PL surface and $Y$ is the smooth surface obtained (in a canonical way) by smoothing $X$ and $\phi : X \rightarrow \mathbb{R}^n$ is a PL immersion, then I don't
vote accepted think there is any canonical choice of a smooth immersion $\phi' : Y \rightarrow \mathbb{R}^n$. You can certainly choose $\phi'$ to be close to $\phi$ in any reasonable sense of the
word close, but that's not good enough.
I don't believe that every PL surface can be given a canonical smooth structure, not in the strong sense in which I understand canonical (functorial w.r.t. PL homeomorphisms). A
triangulation can be made to determine a smooth structure, but that's not quite the same. – Tom Goodwillie May 26 '11 at 18:42
I suspect that's true; I meant canonical in the sense of "any two smoothings are diffeomorphic". However, I don't know a proof that you can't smooth PL surfaces in a functorial way
-- do you know one? – Andy Putman May 26 '11 at 19:03
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Not the answer you're looking for? Browse other questions tagged riemann-surfaces dg.differential-geometry or ask your own question.
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"Perfect" square?
Though 5229 is a perfect square. If you have a rational number as the result of a root, then the thing inside the root is the square of a rational (in this case, (77/10)^2 = 5229/100), so there are
two perfect squares involved somewhere. Other than these musings, all the replies above are, of course, right.
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UCI Machine Learning Repository: PEMS-SF Data Set
PEMS-SF Data Set
Download: Data Folder, Data Set Description
Abstract: 15 months worth of daily data (440 daily records) that describes the occupancy rate, between 0 and 1, of different car lanes of the San Francisco bay area freeways across time.
│ Data Set Characteristics: │ Multivariate, Time-Series │ Number of Instances: │ 440 │ Area: │ Computer │
│ Attribute Characteristics: │ Real │ Number of Attributes: │ 138672 │ Date Donated │ 2011-05-22 │
│ Associated Tasks: │ Classification │ Missing Values? │ N/A │ Number of Web Hits: │ 18863 │
Source: California Department of Transportation, www.pems.dot.ca.gov
Creator: Marco Cuturi, Kyoto University, mcuturi '@' i.kyoto-u.ac.jp
Data Set Information:
We have downloaded 15 months worth of daily data from the California Department of Transportation PEMS website, [Web Link], The data describes the occupancy
rate, between 0 and 1, of different car lanes of San Francisco bay area freeways. The measurements cover the period from Jan. 1st 2008 to Mar. 30th 2009 and are sampled every 10 minutes. We consider
each day in this database as a single time series of dimension 963 (the number of sensors which functioned consistently throughout the studied period) and length 6 x 24=144. We remove public holidays
from the dataset, as well
as two days with anomalies (March 8th 2009 and March 9th 2008) where all sensors were muted between 2:00 and 3:00 AM. This results in a database of 440 time series.
The task we propose on this dataset is to classify each observed day as the correct day of the week, from Monday to Sunday, e.g. label it with an integer in {1,2,3,4,5,6,7}.
I will keep separate copies of this database on my website in a Matlab format. If you use Matlab, it might be more convenient to consider these .mat files directly.
There are two files for each fold, the data file and the labels file. We have split the 440 time series between train and test folds, but you are of course free to merge them to consider a different
cross validation setting.
- The PEMS_train textfile has 263 lines. Each line describes a time-series provided as a matrix. The matrix syntax is that of Matlab, e.g. [ a b ; c d] is the matrix with row vectors [a b] and [c d]
in that order. Each matrix describes the different occupancies rates (963 lines, one for each station/detector) sampled every 10 minutes during the day (144 columns).
- The PEMS_trainlabel text describes, for each day of measurements described above, the day of the week on which the data was sampled, namely an integer between 1 (Mon.) and 7 (Sun.).
- PEMS_test and PEMS_testlabels are formatted in the same way, except that there are 173 test instances.
- The permutation that I used to shuffle the dataset is given in the randperm file. If you need to rearrange the data so that it follows the calendar order, you should merge train and test samples
and reorder them using the inverse permutation of randperm.
Attribute Information:
Each attribute describes the measurement of the occupancy rate (between 0 and 1) of a captor location as recorded by a measuring station, at a given timestamp in time during the day. The ID of each
station is given in the stations_list text file. For more information on the location (GPS, Highway, Direction) of each station please refer to the PEMS website. There are 963 (stations) x 144
(timestamps) = 138.672 attributes for each record.
Relevant Papers:
M. Cuturi, Fast Global Alignment Kernels, Proceedings of the Intern. Conference on Machine Learning 2011.
Citation Request:
Please refer to the Machine Learning Repository's citation policy
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Limit of an integral
December 26th 2011, 03:19 AM #1
Dec 2011
Limit of an integral
Could someone please shed some light on how I might show that
$\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there
is on $f(x,y)$, though I do know that $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $x,y\in R$
and $z>0$.
Brainstorm: Perhaps we can show that $\lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\delta(x-x')\delta(y-y')$? Or maybe changing
Re: Limit of an integral
Could someone please shed some light on how I might show that
$\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}f(x,y)\;dx\,dy\,\,=f(x',y')$? I am not sure what conditions there
is on $f(x,y)$, though I do know that $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(x-x')^2+(y-y')^2+z^2]^{3\over2}}f(x,y)\;dx\,dy$ is well-defined (converges) for all $x,y\in R$
and $z>0$.
Brainstorm: Perhaps we can show that $\lim_{\epsilon\to 0}{\epsilon\over 2\pi [(x-x')^2+(y-y')^2+\epsilon^2]^{3\over2}}$ is the Delta function $\delta(x-x')\delta(y-y')$? Or maybe changing
For z>0, let $K_z(x,y) = \frac z{2\pi (x^2+y^2+z^2)^{3/2}}$, and check that it has the key properties of a kernel function (such as the Fejér kernel). These are: (1) $K_z(x,y)\geqslant0$ (that's
obviously true); (2) $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x,y)\,dx\,dy = 1$ (easy integral, if you express it in polar coordinates); and (3) given $\varepsilon>0$ and $r>0$ there
exists $z_0>0$ such that $K_z(x,y)<\varepsilon$ whenever $z<z_0$ and $x^2+y^2\geqslant r^2.$ (That last condition essentially says that when z is small, $K_z(x,y)$ is also small except when (x,y)
is close to the origin.)
Once you have checked that those three conditions hold, you can calculate as follows:
\begin{aligned}\biggl| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} &K_z(x-x',y-y')f(x,y)\,dx\,dy - f(x',y') \biggr| \\ &= \left| \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K_z(x-x',y-y')\
bigl(f(x,y)-f(x',y')\bigr)\,dx\,dy \right| \qquad\text{(by (2))} \\ &\leqslant \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_z(x-x',y-y')\left|f(x,y)-f(x',y')\right|\,dx\,dy .\end{aligned}
You now want to show that for small enough z, that last integral can be made arbitrarily small. To do that, split the plane into two regions $R_1 = \{(x,y) : |x-x'|^2 + |y-y'|^2 <r^2\}$ and $R_2
= \{(x,y) : |x-x'|^2 + |y-y'|^2 \geqslant r^2\}$, where r is chosen so that $|f(x,y)-f(x',y')|$ is small whenever $(x,y)\in R_1$ (that will require assuming that f is continuous). Then the
integral over $R_1$ will be small by (2), and the integral over $R_2$ will be small by (3) provided we assume that |f| is integrable over $\mathbb{R}^2$ and provided that z is small enoungh.
December 27th 2011, 11:07 AM #2
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Applications of Sum and Difference Formulas
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Applications of Sum and Difference Formulas
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• Read
Solving equations and verifying identities with the sum and difference formulas
• Video
Uses the different sum and difference trigonometric identities to simplify and evaluate functions.
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CWS Frequently Asked Questions
Part 1 | Part 2
Who should use CWS?
Anyone interested in objective assessment of judgmental performance when correct answers are unknown.
But I m not studying a judgment task. Can I still use CWS to assess expertise?
CWS has demonstrated success in assessing performance as well. If the dependent variable is one that is directly linked to the expertise you are examining, give CWS a try. It is especially
helpful when no gold standard outcome measure is available. See Thomas and Pounds (2001) in the download area for an example.
Is there a CWS computer program?
On the web site, there are individual programs for Schumann-Bradley comparisons and for calculating CWS with nominal responses. Ordinary statistics programs can be used to calculate the
quantities need for the CWS ratio with numerical responses.
Is there a CWS book?
The CWS e-Book is being prepared. We will make it available in the downloads section as soon as it is complete.
What sort of response instruments yield data that are suitable for a CWS analysis?
The closer to continuous, the better. Fine gradations avoid artificially deflating the measure of inconsistency that constitutes the denominator of the CWS ratio. So a twenty-point category
scale is likely to be better than a seven-point scale, and a graphic rating (line mark) scale even better. Try to anchor the instrument with extreme stimuli that you expect to elicit much
higher and much lower evaluations than the stimulus set you want the expert to judge. You may not know enough about the stimuli to choose these end-anchors; perhaps a domain expert can help.
How do you calculate CWS when you have nominal data?
The algorithm below was presented in the paper by Weiss and Shanteau (2001). A computer program implementing the algorithm is available in the software area of this web site.
Illustration of CWS Index for Nominal Data (four response alternatives)
│ │Stimulus 1│Stimulus 2│Stimulus 3│Stimulus 4│Stimulus 5│Matches│
│Replicate 1 │ A │ D │ B │ C │ C │ 1 │
│Replicate 2 │ A │ B │ B │ B │ B │ 6 │
│Replicate 3 │ A │ B │ A │ B │ A │ 4 │
│ Matches │ 3 │ 1 │ 1 │ 1 │ 0 │ │
For both numerator and denominator, we utilize the reciprocal of the proportion of obtained matches to possible matches. In measuring discrimination, a match is evidence of failure to
discriminate, so the greater the proportion of observed matches, the poorer the discrimination. In measuring inconsistency, a match means the response was consistent, so the greater the
proportion of observed matches, the smaller the inconsistency. Expert performance is marked by few matches across rows (stimuli), and many matches down columns (replications). If there are no
matches down columns no consistency at all - the CWS ratio is undefined, but that outcome unambiguously connotes a lack of expertise.
CWS Numerator (Discrimination) =[]
Number of possible matches (per row) = [5]C[2] = 10
Numerator = []
CWS Denominator (Inconsistency) = []
Number of possible matches (per column) = [3]C[2] = 3
Denominator = []
CWS Index = []
Part 2 of the CWS FAQ
Site designer: J. Shawn Farris
Webmaster: C. Vowels cvowels@ksu.edu
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R function: generate a panel data.table or data.frame to fill with data
October 25, 2012
By Thiemo Fetzer
I have started to work with R and STATA together. I like running regressions in STATA, but I do graphs and setting up the dataset in R. R clearly has a strong comparative advantage here compared to
STATA. I was writing a function that will give me a (balanced) panel-structure in R. It then simply works by joining in the additional data.tables or data.frames that you want to join into it.
It consists of two functions:
timeVector <- function(starttime,endtime,timestep="months") {
starttime<- as.POSIXct(strptime(starttime, '%Y-%m-%d'))
endtime<- as.POSIXct(strptime(endtime, '%Y-%m-%d'))
if(timestep=="quarters") {
ret<-seq(from=as.POSIXct(starttime), to=as.POSIXct(endtime), by=timestep)
quarter <- gsub("(^[123]{1}$)", 1, month(ret))
quarter <- gsub("(^[456]{1}$)", 2, quarter)
quarter <- gsub("(^[789]{1}$)", 3, quarter)
quarter <- as.numeric(gsub("(^[102]{2}$)", 4, quarter))
} else {
ret<-seq(from=as.POSIXct(starttime), to=as.POSIXct(endtime), by=timestep)
This first function generates the time-vector, you need to tell it what time-steps you want it to have.
panelStructure <- function(group,timevec) {
tt2 <- as.character(sort(rep(group,length(timevec))))
mat <- cbind("group"=data.frame(tt2),"timevec"=data.frame(tt))
This second function then generates the panel-structure. You need to give it a group vector, such as for example a vector of district names and you need to pass it the time vector that the other
function created.
Hope this is helpful to some of you.
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Times function for rational numbers
05-01-2009 #1
Registered User
Join Date
Mar 2009
Times function for rational numbers
hi, i have everything working in this program expect when i made my fraction simplifying function, the multiplication of the two fractions all of a sudden do not appear.
Can you help??
#include "Rational.h"
#include "math.h"
#include <iostream>
using namespace std;
Rational::Rational(int N, int D) //gets num and den
{ numerator = N;
denominator = D; }
int Rational::num()
{return numerator;}
int Rational::den()
{return denominator;}
int Rational::add(Rational f2) //adds the two fractions
{int temp = num()*f2.den();
int temp2 = (den()*f2.num()); //num times by the den of the second and adds it to the den of the first times with the num of the second
int fnum= temp+temp2;
int fden = den()*f2.den(); //den of first times den of second
Rational rational(fnum,fden); //gets final num and den
cout <<"Added: ";
simplify(rational); //simplifies the final num and den
int Rational::subtract(Rational f2) //subtracts the two fractions
{int temp = num()*f2.den();
int temp2 =den()*f2.num(); //num times second den subtracted by the den of the first and the num of the second
int fnum= temp+temp2;
int fden = den()*f2.den(); //den of the first and den of the second
Rational rational(fnum,fden); //gets final num and den
cout<< "Subtracted: ";
simplify(rational); //simplifies the final num and den
int Rational::times(Rational f2) //times the two fractions
{ int fnum =num()*f2.num(); //num of the first times the num of the second
int fden =den()*f2.den(); //den of the first times den of the second
Rational rational(fnum,fden); //gets the final num and den
cout << "Timed: ";
simplify(rational); //simplifies the final num and den
int Rational::divide(Rational f2) //divides the two fractions
{ int fnum =num()*f2.den(); //num times den of second
int fden =den()*f2.num(); //den of first times num of second
Rational rational(fnum,fden); //gets final num and den
cout << "Divided: ";
simplify(rational); //simplifies the final num and den
void Rational::simplify(Rational rational)
{ int num1=rational.num(); // takes the final num and den
int num2=rational.den();
for(int x = 1; x <= num1; ++x){ //for loop, increments until the number is greater than the numerator of the fraction
if(num1 % x == 0 && num2 % x == 0){ //checks if each number is divisible;)
while((num1 % x == 0 && num2 % x == 0) && x != 1){ //Incase it isn't in lowest terms
int fnum =(num1 /= x); //Does the dividing
int fden= (num2 /= x); //This is for the denomenator
Rational end (fnum,fden);
show (end); //shows the final ans
void Rational::show(Rational end) //shows the fractions
{ int fnum=end.num();
int fden=end.den();
cout <<"**"<<fnum<<"/"<<fden<<"**" <<endl;
bool Rational::compare(Rational f2)
//compares the two fractions to see if they are equal
float initial=(float)num()/den();
float final= (float)f2.num()/f2.den();
if (initial==final) return true;
else return false;
Suppose that the fraction is already in lowest terms (cannot be simplified). Then there is no x that evenly divides both the numerator and the denominator, so the if condition
(num1 % x == 0 && num2 % x == 0) is never satisfied, that block of code is not executed and the function does not display anything.
Suppose, on the other hand, that there is such an x. Then the if condition is satisfied, your program enters the while loop, which is actually an infinite loop because within that loop, num1,
num2 and x never change. In this case, the program will display the fraction resulting from dividing both the numerator and denominator by x, but will hang there because it can never exit that
05-01-2009 #2
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Join Date
Feb 2003
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Order of a 3D Triangle
Date: 08/29/2001 at 07:40:38
From: vincent
Subject: Winding order af a 3d triangle
Do you know how to find the winding order of a triangle given three
points in space (x,y,z)? I want to know if when I visit the vertex
from a to b, then from b to c, and returning to a, I am going
clockwise or anticlockwise.
Date: 08/29/2001 at 11:58:44
From: Doctor Rob
Subject: Re: Winding order af a 3d triangle
Thanks for writing to Ask Dr. Math, Vincent.
This depends on from which side of the plane containing a, b, and c
you view this operation. Opposite sides of the plane give opposite
answers. (Think about it!)
Suppose the three coordinate axes are arranged so that if the x-axis
is pointing along your right thumb, and the y-axis is pointing along
your right forefinger, then the z-axis is pointing along your right
middle finger. If you have three points A(x1,y1,z1), B(x2,y2,z2), and
C(x3,y3,z3), and you view the operation from a fourth point
P(x4,y4,z4), then the sign of the following determinant will give you
the information you want:
|x1 y1 z1 1|
|x2 y2 z2 1|
d = |x3 y3 z3 1|.
|x4 y4 z4 1|
If d > 0, then A to B to C to A will appear from P to be clockwise.
If d < 0, then it will appear to be anticlockwise.
If d = 0, then P is in the same plane as ABC, so one cannot observe
either clockwise or anticlockwise angular motion.
If your axes are arranged using the left hand instead of the right,
then the two cases are reversed: d > 0 means anticlockwise and d < 0
means clockwise.
- Doctor Rob, The Math Forum
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Geometry and the Circle Unit 2 > Lesson 1 of 6
A circle is an important shape in the field of geometry. Let's look at the definition of a circle and its parts. We will also examine the relationship between the
circle and the plane.
A circle is a shape with all points the same distance from its center. A circle is named by its center. Thus, the circle to the right is called circle A since its
center is at point A. Some real world examples of a circle are a wheel, a dinner plate and (the surface of) a coin.
The distance across a circle through the center is called the diameter. A real-world example of diameter is a 9-inch plate.
The radius of a circle is the distance from the center of a circle to any point on the circle. If you place two radii end-to-end in a circle, you would have the same
length as one diameter. Thus, the diameter of a circle is twice as long as the radius.
We can look at a pizza pie to find real-world examples of diameter and radius. Look at the pizza to the right which has been sliced into 8 equal parts through its
center. A radius is formed by making a straight cut from the center to a point on the circle. A straight cut made from a point on the circle, continuing through its
center to another point on the circle, is a diameter. As you can see, a circle has many different radii and diameters, each passing through its center.
A chord is a line segment that joins two points on a curve. In geometry, a chord is often used to describe a line segment joining two endpoints that lie on a circle.
The circle to the right contains chord AB. If this circle was a pizza pie, you could cut off a piece of pizza along chord AB. By cutting along chord AB, you are
cutting off a segment of pizza that includes this chord.
A circle has many different chords. Some chords pass through the center and some do not. A chord that passes through the center is called a diameter.
It turns out that a diameter of a circle is the longest chord of that circle since it passes through the center. A diameter satisfies the definition of a chord,
however, a chord is not necessarily a diameter. This is because every diameter passes through the center of a circle, but some chords do not pass through the center.
Thus, it can be stated, every diameter is a chord, but not every chord is a diameter.
Let's revisit the definition of a circle. A circle is the set of points that are equidistant from a special point in the plane. The special point is the center. In
the circle to the right, the center is point A. Thus we have circle A.
A plane is a flat surface that extends without end in all directions. In the diagram to the right, Plane P contains points A, B and C.
Can you think of some real world objects that satisfy the definition of a plane? At this level of mathematics, that is difficult to do. Intuitively, a plane may be
visualized as a flat infinite sheet of paper. The top of your desk and a chalkboard are objects which can be used to represent a plane, although they do not satisfy
the definition above.
A circle divides the plane into three parts:
1. the points INSIDE the circle
2. the points OUTSIDE the circle
3. and the points ON the circle
│the points inside the circle│the points outside the circle │the points on the circle│
You can see an interactive demonstration of this by placing your mouse over the three items below.
A circle divides a plane into three parts:
1. the points INSIDE the circle
2. the points OUTSIDE the circle
3. and the points ON the circle
Example 1: Name the center of this circle.
Answer: Point B
Example 2: Name two chords on this circle that are not diameters.
Answer: DE and FG
Example 3: Name all radii on this circle.
Answer: BA, BC, BD and BG
Example 4: What are AC and DG?
Answer: AC and DG are diameters.
Example 5: If DG is 5 inches long, then how long is DB?
Solution: The diameter of a circle is twice as long as the radius.
5 inches ÷ 2 = 2.5 inches
Answer: The length of DB is 2.5 inches
Summary: A circle is a shape with all points the same distance from its center. A circle is named by its center. The parts of a circle include a radius, diameter and a
chord. All diameters are chords, but not all chords are diameters. A plane is a flat surface that extends without end in all directions. A circle divides the
plane into three parts: The points inside the circle, the points outside the circle and the points on the circle.
Directions: Refer to the diagram to answer each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX.
If you make a mistake, choose a different button.
1. Which of the following is a chord, but not a diameter?
2. Which of the following is a radius?
3. Name the center of this circle.
4. What is PR (or PQR)?
5. If PQ is 3 cm long, then how long is PR?
This lesson is by Gisele Glosser. You can find me on Google.
┃ Circumference & Area Lessons ┃
┃ Geometry and the Circle ┃
┃ Circumference of a Circle ┃
┃ Area of a Circle ┃
┃ Practice Exercises ┃
┃ Challenge Exercises ┃
┃ Solutions ┃
┃ Related Activities ┃
┃ Circle Crosswords ┃
┃ Circle Solver ┃
┃ Circumference and String ┃
┃ Geometry and a Shoebox ┃
┃ Geometry Worksheet Generator ┃
┃ Pre-Made Worksheets ┃
Lessons Worksheets WebQuests Games Homework Articles Forums Glossary Puzzles Newsletter Standards Buy the Goodies Now!
Home | About Us | Contact Us | Privacy | Advertise | Share | Interactive Math Goodies Software
| More
Free Math
Lessons Worksheets Generator Games Homework Articles Forums Glossary Puzzles Calculators Standards
Geometry and the Circle Unit 2 > Lesson 1 of 6
A circle is an important shape in the field of geometry. Let's look at the definition of a circle and its parts. We will also examine the relationship between the circle and the plane.
A circle is a shape with all points the same distance from its center. A circle is named by its center. Thus, the circle to the right is called circle A since its center is at point A.
Some real world examples of a circle are a wheel, a dinner plate and (the surface of) a coin.
The distance across a circle through the center is called the diameter. A real-world example of diameter is a 9-inch plate.
The radius of a circle is the distance from the center of a circle to any point on the circle. If you place two radii end-to-end in a circle, you would have the same length as one
diameter. Thus, the diameter of a circle is twice as long as the radius.
We can look at a pizza pie to find real-world examples of diameter and radius. Look at the pizza to the right which has been sliced into 8 equal parts through its center. A radius is
formed by making a straight cut from the center to a point on the circle. A straight cut made from a point on the circle, continuing through its center to another point on the circle, is
a diameter. As you can see, a circle has many different radii and diameters, each passing through its center.
A chord is a line segment that joins two points on a curve. In geometry, a chord is often used to describe a line segment joining two endpoints that lie on a circle. The circle to the
right contains chord AB. If this circle was a pizza pie, you could cut off a piece of pizza along chord AB. By cutting along chord AB, you are cutting off a segment of pizza that includes
this chord.
A circle has many different chords. Some chords pass through the center and some do not. A chord that passes through the center is called a diameter.
It turns out that a diameter of a circle is the longest chord of that circle since it passes through the center. A diameter satisfies the definition of a chord, however, a chord is not
necessarily a diameter. This is because every diameter passes through the center of a circle, but some chords do not pass through the center. Thus, it can be stated, every diameter is a
chord, but not every chord is a diameter.
Let's revisit the definition of a circle. A circle is the set of points that are equidistant from a special point in the plane. The special point is the center. In the circle to the
right, the center is point A. Thus we have circle A.
A plane is a flat surface that extends without end in all directions. In the diagram to the right, Plane P contains points A, B and C.
Can you think of some real world objects that satisfy the definition of a plane? At this level of mathematics, that is difficult to do. Intuitively, a plane may be visualized as a flat
infinite sheet of paper. The top of your desk and a chalkboard are objects which can be used to represent a plane, although they do not satisfy the definition above.
A circle divides the plane into three parts:
1. the points INSIDE the circle
2. the points OUTSIDE the circle
3. and the points ON the circle
│the points inside the circle│the points outside the circle │the points on the circle│
You can see an interactive demonstration of this by placing your mouse over the three items below.
A circle divides a plane into three parts:
1. the points INSIDE the circle
2. the points OUTSIDE the circle
3. and the points ON the circle
Example 1: Name the center of this circle.
Answer: Point B
Example 2: Name two chords on this circle that are not diameters.
Answer: DE and FG
Example 3: Name all radii on this circle.
Answer: BA, BC, BD and BG
Example 4: What are AC and DG?
Answer: AC and DG are diameters.
Example 5: If DG is 5 inches long, then how long is DB?
Solution: The diameter of a circle is twice as long as the radius.
5 inches ÷ 2 = 2.5 inches
Answer: The length of DB is 2.5 inches
Summary: A circle is a shape with all points the same distance from its center. A circle is named by its center. The parts of a circle include a radius, diameter and a chord. All diameters
are chords, but not all chords are diameters. A plane is a flat surface that extends without end in all directions. A circle divides the plane into three parts: The points inside
the circle, the points outside the circle and the points on the circle.
Directions: Refer to the diagram to answer each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake,
choose a different button.
1. Which of the following is a chord, but not a diameter?
2. Which of the following is a radius?
3. Name the center of this circle.
4. What is PR (or PQR)?
5. If PQ is 3 cm long, then how long is PR?
This lesson is by Gisele Glosser. You can find me on Google.
┃ Circumference & Area Lessons ┃
┃ Geometry and the Circle ┃
┃ Circumference of a Circle ┃
┃ Area of a Circle ┃
┃ Practice Exercises ┃
┃ Challenge Exercises ┃
┃ Solutions ┃
┃ Related Activities ┃
┃ Circle Crosswords ┃
┃ Circle Solver ┃
┃ Circumference and String ┃
┃ Geometry and a Shoebox ┃
┃ Geometry Worksheet Generator ┃
┃ Pre-Made Worksheets ┃
┃ Circumference & Area Lessons ┃
┃ Geometry and the Circle ┃
┃ Circumference of a Circle ┃
┃ Area of a Circle ┃
┃ Practice Exercises ┃
┃ Challenge Exercises ┃
┃ Solutions ┃
┃ Related Activities ┃
┃ Circle Crosswords ┃
┃ Circle Solver ┃
┃ Circumference and String ┃
┃ Geometry and a Shoebox ┃
┃ Geometry Worksheet Generator ┃
┃ Pre-Made Worksheets ┃
Lessons Worksheets WebQuests Games Homework Articles Forums Glossary Puzzles Newsletter Standards Buy the Goodies Now!
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Physics Forums - View Single Post - ti-89 titanium convert numbers to fractions
is there any quick way to convert numbers to fractions then back to decimals, without changing the mode.
also if i type in log 54, it gives me log 54.
i don't get an answer
even if i put in
ln 32
it gives me ln 32, not an answer.
i've learned to input functional equations and make graphs etc..
but can't really do a lot what i could do with my previous scientific calculator quickly.
help is appreciated.
|
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Gilcrest Geometry Tutor
Find a Gilcrest Geometry Tutor
...I have taught grades K-8 for the past 19 years as a public school teacher in Boulder Valley Schools and I am currently teaching 6th through 8th grade students. I am happy to work with your
child in any one area or a variety of areas that would fit your needs! I love teaching individual subjects and integrating subjects as well.
14 Subjects: including geometry, reading, algebra 1, algebra 2
...However, I am experiences with several other languages including Groovy, Scala, and Python. I have implemented large and small systems, from games to large scale enterprise applications. I have
worked for a wide variety of companies, including the very large, Apple and Union Pacific Railroad, and small start-up ventures.
17 Subjects: including geometry, statistics, trigonometry, algebra 2
I'm Laura Z., a 28 year old female with a true passion for mathematics and science. I have been tutoring for 10 years now and have had great success with my students, ranging from middle school to
college-age, and non-traditional students. I began tutoring by helping other students in my high school classes.
16 Subjects: including geometry, chemistry, calculus, physics
...Algebra 2 is a continuing part of learning mathematics and its uses in today's world. Calculus is a staple of knowledge to understand how many principles and idea work in mathematics. With my
background in experimental particle physics (I have a Ph.D), I have used calculus throughout my education college career and beyond.
47 Subjects: including geometry, chemistry, physics, calculus
...This experience includes my past work with high school students to prepare for AP exams, tutoring sessions held through university tutoring centers, as well as my prior work as a teaching
fellow for an advanced undergraduate chemistry course while in graduate school. My primary focus as a tutor ...
8 Subjects: including geometry, chemistry, calculus, algebra 1
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|
Path Integrals and Wavefunction for Harmonic Oscillator
Left: Classical trajectories (x(t) with time vertical in steps) plotted after every 200 changes in time steps. The Feynman path integral is over these trajectories.
Right The squared modulus of the developing ground-state wavefunction.
Ref.: the book Computational Physics by Landau and Páez
|
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Alviso Trigonometry Tutor
Find an Alviso Trigonometry Tutor
...I took geometry in high school and I graduated from high school as the top student in my class, especially in math subject. I have been helping numerous high school students to develop their
problem solving skills and to improve their math grades. I graduated from high school as top student in my class, especially the math subject.
15 Subjects: including trigonometry, chemistry, calculus, Chinese
...Calculus can be really easy and fun to learn (and use in real world), if you approach it right, otherwise you may end up not liking it. Geometry is really fun to learn, if you approach it
right, otherwise you may find yourself liking algebra better; I've been told more often than not that I've i...
9 Subjects: including trigonometry, physics, calculus, geometry
...I am comfortable with and have ample experience tutoring students of all ages. I help students to thoroughly understand math so they can do A+ work. I believe that anybody who puts in the
effort can succeed in math.
5 Subjects: including trigonometry, geometry, algebra 2, prealgebra
...It was my favorite class that I took in my Math major, and I would feel very comfortable tutoring it. I tutored this subject informally with peers in the math center on campus. I took symbolic
logic in college and received an A.
35 Subjects: including trigonometry, reading, calculus, geometry
...With a little bit of help, I gained confidence, curiosity, and passion for this subject. I know how powerful and helpful a few hours of tutoring can be, and I love to share my appreciation for
this subject and help students gain mastery. After moving to the United States, I graduated high schoo...
22 Subjects: including trigonometry, chemistry, physics, calculus
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Tricky train problem involving ratios
April 22nd 2010, 10:25 AM #1
Junior Member
Sep 2009
The ratio of the speeds of two trains is equal to the ratio of the time they take to pass each other going in the same direction to the time they take to pass each other in the opposite
directions. how can you find the ratio of the speeds of the two trains?
Hello, Intsecxtanx!
The ratio of the speeds of two trains is equal to
the ratio of the time they take to pass each other going in the same direction
to the time they take to pass each other in the opposite directions.
Find the ratio of the speeds of the two trains.
Let: . $\begin{Bmatrix} F &=& \text{speed of faster train} \\<br /> L_1 &=& \text{length of faster train} \end{Bmatrix} \qquad \begin{Bmatrix} F &=& \text{speed of slower train} \\ L_2 &=& \text
{length of slower train} \end{Bmatrix}$
Going in the same direction, the faster train has a relative speed of $F \!-\! S$ ft/sec
. . and it travels a distance of $L_1\!+\!L_2$ feet.
This will take: . $T_{\text{same}} \;=\;\frac{L_1+L_2}{F-S}$ seconds.
Going in opposite directions, the faster train has a relative speed of $F\!+\!S$ ft/sec
. . and it travels a distance of $L_1\!+\!L_2$ feet.
This will take: . $T_{\text{opp}} \;=\;\frac{L_1+L_2}{F+S}$ seconds.
The ratio of their speeds equals the ratio of their two times:
. . $\frac{F}{S} \;=\;\frac{\dfrac{L_1+L+2}{F-S}}{\dfrac{L_1+L_2}{F+S}} \quad\Rightarrow\quad \frac{F}{S} \:=\:\frac{F+S}{F-S}$
. . . . $F^2 - FS \:=\:FS + S^2 \quad\Rightarrow\quad F^2 - 2SF - S^2 \:=\:0$
Divide by $S^2\!:\;\;\frac{F^2}{S^2} - 2\,\frac{F}{S} - 1 \:=\:0 \quad\Rightarrow\quad \left(\frac{F}{S}\right)^2 - 2\left(\frac{F}{S}\right) - 1 \:=\:0$
Quadratic Formula: . $\frac{F}{S} \;=\;\frac{2\pm\sqrt{4+4}}{2} \;=\;1 \pm \sqrt{2}$
The ratio of their speeds is: . $F:S \;=\;(1\!+\!\sqrt{2}) : 1$
April 22nd 2010, 04:32 PM #2
Super Member
May 2006
Lexington, MA (USA)
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A Guide to Sabermetric Research: A Primer on Statistics
For the typical fan, sabermetrics doesn’t represent anything as theoretical as scientific inquiry. Rather, sabermetrics is associated with new and unfamiliar statistics. OPS is the most famous of
those new stats. It’s gone from a nearly unknown statistic in the early 80s, to barely used a decade ago, to mainstream now (it even appears on Topps baseball cards). There have also been stats like
Linear Weights, Runs Created, Extrapolated Runs, WAR, and so on.
I’d still argue that sabermetrics isn’t really about those statistics; rather, the statistics have been proven to be useful based on evidence that sabermetricians have uncovered. “Runs Created,” for
instance, is a statistic that was created by Bill James in the late 1970s. James’ thinking went this way: a team’s job on offense is to score runs – the more runs, the better. Suppose you didn’t know
how many runs a team scored, and wanted to make an estimate, based on its batting line. For instance, here’s a real team batting line:
G AB H 2B 3B HR BB K AVG
161 5517 1451 234 22 214 604 908 .263
How many runs would you guess that team scored that year? If I made you guess, you’d probably look over a few years of team statistics, try to find some team that was reasonably close, and use that
as a baseline. You might find a team that hit .267 with less power, and scored 788 runs. You’d figure, “well, this team hit only .263, but they had a few more home runs, so I guess maybe they’d
cancel out, so I’d guess the same 788 runs. But, wait, this team had about 20 more walks than the other team, so maybe I should bump up my estimate to 800 or something.”
What Bill James probably did was work through logic like that, and, after some trial and error, come up with the Runs Created (RC) formula. That statistic is intended to provide a formal way of
estimating how a batting line translates into runs. In its most basic form, RC looks like:
Runs Created = (TB) (H+BB) / (AB+BB)
If you plug the numbers in from the above batting line, you get
Runs Created = (2371) (2055) / (6121)
which gives 796 runs.
As it turns out, that was actually the batting line for the 1985 Baltimore Orioles. They actually scored 818 runs. The estimate is off by 22 runs, which is fairly typical.
Why is Runs Created important? Why do we need RC if we already know the Orioles scored 818 runs? Well, knowing that there is a predictable relationship between a batting line and runs is useful when
we don’t know how many runs we actually have. For instance, we can use RC on an individual player’s batting line. Here’s Albert Pujols in 2009:
G AB H 2B 3B HR BB K AVG
160 568 186 45 1 47 115 64 .327
Using the basic RC formula, we can estimate that if a given major league team had a batting line like Pujols did, it would score about 149 runs. That batting line would comprise about 15 games, which
gives about 10 runs per game.
What we can conclude, then, is that if you put together a lineup of nine Albert Pujols clones, on average they’d score 10 runs per game. That’s a huge total – the average MLB team scores somewhere
between 4.5 and 5.0.
We can compare Pujols to Joe Mauer, or Adam Lind, or Alex Rodriguez, to help inform our conclusions on how much each contributed to his team, or even to our arguments about which player deserves the
MVP award.
Runs Created is one of the most famous of the statistics used to evaluate offense. Others include Pete Palmer’s "Linear Weights," Jim Furtado’s “Extrapolated Runs,” and David Smyth’s “Base Runs.” All
are very good estimators. But which is the best? Well, that depends. No estimator is perfect, and all have their strengths and weaknesses.
One way to compare the various estimators is to test them for accuracy. Apply them to the last (say) fifty years of baseball, which should give you around 700 team-seasons. Have them each estimate
runs for all 700 teams, and see which ones do the best.
Previous page Next page
Page 3 of 12
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deSolve (1.10-8)
General Solvers for Initial Value Problems of Ordinary Differential Equations (ODE), Partial Differential Equations (PDE), Differential Algebraic Equations (DAE), and Delay Differential Equations
Functions that solve initial value problems of a system of first-order ordinary differential equations (ODE), of partial differential equations (PDE), of differential algebraic equations (DAE), and
of delay differential equations. The functions provide an interface to the FORTRAN functions lsoda, lsodar, lsode, lsodes of the ODEPACK collection, to the FORTRAN functions dvode and daspk and a
C-implementation of solvers of the Runge-Kutta family with fixed or variable time steps. The package contains routines designed for solving ODEs resulting from 1-D, 2-D and 3-D partial differential
equations (PDE) that have been converted to ODEs by numerical differencing.
Maintainer: Thomas Petzoldt
Author(s): Karline Soetaert and Thomas Petzoldt and R. Woodrow Setzer
License: GPL (>= 2)
Uses: Does not use any package
Reverse depends: AquaEnv, asbio, AssetPricing, bvpSolve, CollocInfer, diffEq, diversitree, ecolMod, ecosim, EpiModel, expoTree, FME, geiger, insideRODE, ivivc, marelac, marelacTeaching, mkin,
neuRosim, nlmeODE, PKfit, pomp, primer, PSM, ReacTran, replicationDemos, scaRabee, simecol, SoilR, TESS, TIMP, TreePar, tseriesChaos, ZeBook
Reverse suggests: asbio, fda, PBSmodelling, smfsb
Released 7 months ago.
16 previous versions
Overall: 3.0/5 (2 votes)
Documentation: 3.5/5 (2 votes)
Log in to vote.
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Everyone needs statistics
Everyone Should Learn Statistics writes Ed Sector’s Kevin Carey, after serving on a jury. The defense attorney tried to confuse the jury with bad statistics — or didn’t understand statistics himself.
Which is not surprising, given that statistics isn’t part of the standard curriculum schools require students to complete in order to get a high-school or college diploma. Math education is still
largely interpreted as a progression through algebra and geometry to calculus. And I’m not against working harder to improve math education. But in terms of things you really need in order to
make your way in modern society, statistics is way, way up there, above a lot of things that are currently lodged in the curriculum.
Carey provides a video of Donald Duck in Mathmagicland. I vividly remember watching this in seventh grade.
Last reply was May 6, 2012
1. LSquared
Unfortunately, you need rather a lot of algebra (and even calculus) to really learn statistics (how to do it). Which isn’t to say that you can’t learn about statistics with less background, but
that’s a rather different class, and maybe it’s a business or social science class rather than a math class.
□ Roger Sweenyreplied:
It would be a good idea for citizens to understand practical statistics. On the other hand, very few citizens use trigonometry or advanced algebra.
If that stat course belongs in the social science department, perhaps schools should take away a year of the math requirement and add this particular social science one.
2. Catherine
But if everybody had a good grasp of statistics and probability, state lottery revenues would dry up! It’s for the children!!
I agree with LSquared, I was in an engineer’s office the other day and there was a statistics textbook on his desk open to a part about calculating some sort of bi-linear correlation coefficient
and it was full of dense algebra logarithms and other math.
I’m afraid there would be no way to split statistics out by itself.
(Loved the cartoon, by the way. Too bad nobody bothers with such nice work today. Maybe some rich person should hire Pixar to take over math education…
□ Roger Sweenyreplied:
One can understand what a Gaussian (Normal) probability distribution is, where one is likely to encounter it, and why it matters. One can also integrate it. For the first three, you don’t
need calculus. For the last, you do. The first three are useful for ordinary people to know. The last isn’t.
Or as Mark Twain (or Shakespeare or the Bible) said: There are lies, damn lies, and statistics.
5. Peace Corps
Thanks for the video. I had a lot of students missing today because of sports activities, so I showed it to two of my classes. Two out of 11 students liked it, the rest ignored it. I loved it.
6. Richard Aubrey
I had a primitive stat class as part of a psych BA.
I haven’t ever squared a chi. But it’s been useful when looking at what people want me to believe.
Particularly the methodology.
7. View May 6, 2012
You don’t need to be a master statistcian to be able to interpret statistics. Certainly, a master statistcian would be able to, but that isn’t necessary. You just need to be taught how to read
it. A minor introduction might suffice for that purpose.
Recent Comments
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Florence, AZ Algebra Tutor
Find a Florence, AZ Algebra Tutor
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I have my Associate of Science Degree in Mathematics, and I spent several semesters as a Mathematics Supplemental Instructor holding free sessions for students outside of the classroom (which I
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[SOLVED] I cant seem to understand my Metric System Chart
February 17th 2010, 02:49 PM #1
Feb 2010
[SOLVED] I cant seem to understand my Metric System Chart
I was given a chart of the metric system, but I cant seem to understand how the conversions work. Lets say I want to convert micro into milli. For micro, the chart says 10^-6, and 0.000001. For
milli, it says 10^-3 and 0.001.
The way I tried to do this by typing in 0.000001 x 10^-3, not sure if that's what you're supposed to do when converting it, and my TI-83 came back with an answer of 1E-9, which I dont understand
at all. I think there was a way to make the TI-83 show all the numbers, but I cant remember how.
I'm still need to how the metric system works, and i'm trying to figure out how to convert 8.9mL into microliters, and 1/2 Quart into milliliters.
I'm hoping that someone here will know how to explain this so that I can understand it. Looking at the chart just seems to confuse me. The whole conversion process is pretty much what I dont
understand. And how I can get it to work with my calculator.
I was given a chart of the metric system, but I cant seem to understand how the conversions work. Lets say I want to convert micro into milli. For micro, the chart says 10^-6, and 0.000001. For
milli, it says 10^-3 and 0.001.
The way I tried to do this by typing in 0.000001 x 10^-3, not sure if that's what you're supposed to do when converting it, and my TI-83 came back with an answer of 1E-9, which I dont understand
at all.
Hi fb280,
To the TI-83, 1E-9 means $1 \times 10^{-9}$.
I don't think there is.
I'm still need to how the metric system works, and i'm trying to figure out how to convert 8.9mL into microliters, and 1/2 Quart into milliliters.
I'm hoping that someone here will know how to explain this so that I can understand it. Looking at the chart just seems to confuse me. The whole conversion process is pretty much what I dont
understand. And how I can get it to work with my calculator.
Here's how I would approach your conversion. I would use a set of ratio products to arrive at my desired result. Like this:
Convert 8.9 milliliters( $mL$) to microliters( $\mu L$)
$\frac{8.9 mL}{1}\cdot \frac{1000 \mu L}{1 mL}=8900 \mu L$
Convert .5 quarts to milliliters.
$\frac{.5 q}{1}\cdot\frac{.946352 L}{1 q}\cdot \frac{1000 mL}{1L} \approx 473.176 mL$.
Notice that in the products, all the units of measure cancel out except for the one you are trying to convert to. Neat, huh?
I had to look up the number of liters in a quart (rounded off).
I think I might know how to convert it now after looking through the chart more. But I wasnt sure until I read your reply, so thanks for the help. There is one thing i'd like to make sure of
If the ti-83 comes out with a number like 8E7, would I just move the decimal place to the right by eight places?
I think I might know how to convert it now after looking through the chart more. But I wasnt sure until I read your reply, so thanks for the help. There is one thing i'd like to make sure of
If the ti-83 comes out with a number like 8E7, would I just move the decimal place to the right by eight places?
No. 8E7 means 8 times ten to the 7th power. $8.0 \times 10^7$.
8E7 means 80,000,000.
hmm.. wouldnt it always bring up the same answer either way?
February 17th 2010, 03:25 PM #2
A riddle wrapped in an enigma
Jan 2008
Big Stone Gap, Virginia
February 17th 2010, 04:08 PM #3
Feb 2010
February 17th 2010, 07:16 PM #4
A riddle wrapped in an enigma
Jan 2008
Big Stone Gap, Virginia
February 17th 2010, 07:27 PM #5
Feb 2010
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Challenge #10 Solution
Have you tried solving Challenge #10 yet? Go try it first if you haven’t. It’s not too hard, I promise!
All of these can be shown by just plugging in the values of $\varphi$ and $\hat{\varphi}$, but I’d like to show you some easier ways.
The first problem is easy — we know that $\varphi$ is a solution to the equation $x^2 - x - 1 = 0$; that is, $\varphi^2 - \varphi - 1 = 0$. Some simple algebra shows that $\varphi^2 = \varphi + 1$.
This is kind of interesting when you think about it. In English, this says that squaring $\varphi$ is exactly the same as adding 1.
Problems 2 and 3 can be solved simultaneously. Since $\varphi$ and $\hat{\varphi}$ are the roots of the polynomial $x^2 - x - 1$, we know it can be factored as
$x^2 - x - 1 = (x - \varphi)(x - \hat{\varphi})$.
Multiplying out the right side, we get
$x^2 - x - 1 = x^2 - (\varphi + \hat{\varphi})x + \varphi \hat{\varphi}$
For this equation to be true, we obviously must have $\varphi + \hat{\varphi} = 1$ and $\varphi \hat{\varphi} = -1$.
There is not a nice slick way to prove #4 (at least, not that I am aware of). But doing it by plugging in values isn’t so bad:
$\displaystyle \varphi - \hat{\varphi} = \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$.
Next up: we’re going to use these properties to prove some astonishing facts about the relationship between the golden ratio and Fibonacci numbers! Stay tuned!
3 Responses to Challenge #10 Solution
1. If you correct the broken LaTeX in my response to the challenge, you can see a cleaner way to prove property 4.
2. OK, here’s the proof of property 4 without fixing the LaTeX. I’m going to use φ and φ’ as the two solutions.
First, we need to find out what φ^2+φ’^2 is:
And now we can work out:
= φ^2 + 2φφ’ + φ’^2
= (φ + 1) + 2φφ’ + (φ’ + 1) (property 1)
= (φ + φ’) + 2 + 2 (regrouping, and property 3)
= 1 + 2 + 2 (property 2)
= 5
And now, because we know that φ>φ’, we take the positive square root.
Oh, by the way. Is there any way that you could add comment previews to this blog? If you’re encouraging commenters to put maths in their responses, it would really help if we could see mistakes
before they get published.
3. Pseudonym: nice, I like it! I hadn’t thought of that method of proof. As for comment previews, I don’t see an easy way to do that at the moment. I suppose the real issue is that I need to upgrade
to a more recent version of WordPress, perhaps it will support comment previews! I agree that would be a nice feature and I’ll look into it.
This entry was posted in famous numbers, golden ratio, proof, solutions. Bookmark the permalink.
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A simplified proof for the limit of a tower over a cubic finite field
Bassa, Alp and Stichtenoth, Henning (2007) A simplified proof for the limit of a tower over a cubic finite field. Journal of Number Theory, 123 (1). pp. 154-169. ISSN 0022-314X
Full text not available from this repository.
Official URL: http://dx.doi.org/10.1016/j.jnt.2006.06.005
Recently Bezerra, Garcia and Stichtenoth constructed an explicit tower F = (Fn)n 0 of function fields over a finite field F q3 , whose limit λ(F) = limn→∞N(Fn)/g(Fn) attains the Zink bound λ(F) 2(q2
1)/(q + 2). Their proof is rather long and very technical. In this paper we replace the complex calculations in their work by structural arguments, thus giving a much simpler and shorter proof for
the limit of the Bezerra, Garcia and Stichtenoth tower.
Item Type: Article
Uncontrolled Keywords: towers of function fields; genus; rational places; limits of towers; Zink's bound
Subjects: Q Science > QA Mathematics
ID Code: 157
Deposited By: Henning Stichtenoth
Deposited On: 20 Dec 2006 02:00
Last Modified: 09 Dec 2011 22:13
Repository Staff Only: item control page
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Comparison of the different algorithms for Polygon Boolean operations.
Michael Leonov
When writing my BS degree work I tested the following software libraries for speed and robustness of performing polygon Boolean operations:
Library AND SUB OR XOR HOLES KH I SI DV KH O
Boolean + + + + + + - + +
BOPS + - + - + - - - +
CGAL + + + - - - - - -
Clippoly + + - - - - - - -
CPG + + + + + - - + -
GPC + + + + + + + + -
LEDA + + + + + - - + -
PolyBoolean + + + + + + - + -
First 4 columns denote supported Boolean operations. HOLES means that a library can handle polygons with holes. KH I means that a library can handle polygons with 'keyholed' edges, which are
sometimes used to describe a polygon with holes by means of single contour. SI means that a library allows self-intersecting polygons at its input. DV means that a library supports polygons with
vertices of high degree, i.e. self-touching polygons. KH O means that a library's output can contain keyholed edges (I think this is the library's disadvantage).
For testing I used PC with CPU Pentium II (233 MHz) and 96Mb RAM running Windows NT Workstation 4.0 SP4. All source code was compiled with Microsoft Visual C++ 6.0 using the same alignment (8 bytes)
and optimization options. Sample polygons were extracted from True Type Font contours using different levels of Bezier curves polygonal approximation. The test program was executed 5 times for every
Boolean operation. The results are summarized in the table below (N denotes the total number of vertices in input polygons, all timings are measured in seconds, best times for each N are bold):
Library N=3885 N=7076 N=20190 N=69839 N=174239
Boolean 1.084 1.773 5.923 23.219 65.927
Clippoly 15.482 51.965 487.942 ... ...
GPC 0.160 0.381 8.570 64.463 133.670
LEDA 0.806 1.422 3.801 16.636 ...
PolyBoolean 0.158 0.255 0.721 3.532 16.011
For N=69839, 174239 Clippoly caused stack overflow due to the O(N) recursion depth. For N=174239 LEDA caused memory overflow (despite the presence of the extra 100 Mb of virtual memory) due to the
extensive use of the rational ariphmetic.
Like Clippoly, CGAL and CPG have quadratic running time. BOPS produced incorrect results on test polygons so I did not include its timings.
Numerical robustness
Most of the programs listed above are not strictly robust and use floating point arithmetic with some tolerance values. CGAL, CPG and LEDA use exact rational arithmetic to achieve robustness. In this
case, required memory size grows exponentially with a number of cascaded operations, and this seems not to be satisfactory for practical applications. PolyBoolean uses John Hobby's rounding cell
technique to avoid extraneous intersections and is therefore completely robust. Boolean also rounds the intersection points to the integer grid, then repeats until no new intersection points are
Algorithm used in Boolean is described in [Holwerda 98]. Some additional information can be found in [Preparata and Shamos 85] (a must-have Computational Geometry book).
Algorithm used in BOPS is described in [Zalik, Gombosi and Podgorelec 98].
Algorithm used in CGAL is not documented.
Algorithm used in Clippoly is described in [Schutte 94] and [Schutte 95].
Algorithm used in CPG is described in [Eberly 98].
GPC uses a modified version of [Vatti 92]. Implementation details are discussed in [Murta 98]. Alan Murta is currently working on a paper describing GPC.
Polygon Boolean algorithm used in LEDA is not documented by itself. The segment intersection part is described in [Mehlhorn and Naher 94].
Polygon Boolean algorithm used in PolyBoolean is described in [Leonov and Nikitin 97]. Additional algorithmic and implementation issues are discussed in [Leonov 98]. The segment intersection part is
based on [Bentley and Ottmann 79] and [Hobby 99].
J. L. Bentley and T. A. Ottmann. Algorithms for reporting and counting geometric intersections. IEEE Trans. Comput., C-28:643-647, 1979.
D. Eberly. Polysolids and Boolean Operations.
J. Hobby. Practical segment intersection with finite precision output. Computation Geometry Theory and Applications, 13(4), 1999.
K. Holwerda et al. Complete Boolean Description.
M. V. Leonov and A. G. Nikitin. An Efficient Algorithm for a Closed Set of Boolean Operations on Polygonal Regions in the Plane (draft English translation). A. P. Ershov Institute of Informatics
Systems, Preprint 46, 1997.
M. V. Leonov. Implementation of Boolean operations on sets of polygons in the plane (in Russian). BS Thesis, Novosibirsk State University, 1998.
K. Mehlhorn and S. Naher. Implementation of a Sweep Line Algorithm for the Straight Line Segment Intersection Problem. Max-Planck-Institut fur Informatik, MPI-I-94-160, 1994.
A. Murta. A Generic Polygon Clipping Library.
F. P. Preparata and M. I. Shamos. Computational Geometry: An Introduction. Springer-Verlag, New York, NY, 1985
K. Schutte. Knowledge Based Recognition of Man-Made Objects. PhD Thesis, University of Twente, 1994. ISBN90-9006902-X.
K. Schutte. An edge labeling approach to concave polygon clipping. Manuscript, 1995.
B. R. Vatti. A generic solution to polygon clipping. Commun. ACM, 35(7):56-63, 1992.
B. Zalik, M. Gombosi and D. Podgorelec. A Quick Intersection Algorithm for Arbitrary Polygons. In L. Szirmay-Kalos (Ed.), SCCG98 Conf. on Comput. Graphics and it's Applicat., 195-204, 1998. ISBN
All tested libraries are very good for educational purposes and for studying different approaches to the polygon Boolean operations. PolyBoolean, Boolean and GPC are probably the fastest publicly
available libraries. The correct rounding of intersection points is performed only in PolyBoolean and Boolean. Of course, all these opinions are only mine, and I don't attempt to make strong
assertions about usefulness of these programs. Click here to get the polygons I used for testing. Soon I will make the source code of the test program (with all necessary modifications of the tested
libraries) publicly available.
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Marana Trigonometry Tutors
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Revenue Word Problem
November 14th 2009, 11:39 PM #1
Nov 2009
Revenue Word Problem
I really have no idea where to even begin with this word problem:
You buy an apartment building with 40 units. The previous owner charged $420 per month for a single apartment and, on average, rented 30 apartments at that price. You discover that for every $30
you raise the price of rent, another apartment stands vacant because someone can't afford it. What price should you charge for an apartment in order to maximize your revenue?
Any help would be appreciated
I really have no idea where to even begin with this word problem:
You buy an apartment building with 40 units. The previous owner charged $420 per month for a single apartment and, on average, rented 30 apartments at that price. You discover that for every $30
you raise the price of rent, another apartment stands vacant because someone can't afford it. What price should you charge for an apartment in order to maximize your revenue?
Any help would be appreciated
If you let $N$ apartments at charge $C$ the revenue is:
$R=C \times N$
If you charge $C$ the number of apartments let is:
So the Revenue is:
which you are to find the maximum of as $C$ changes, you will have to do some checking at the end to make sure you have let an integer number of apartments.
November 15th 2009, 12:16 AM #2
Grand Panjandrum
Nov 2005
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No Slide Title
<press any key to continue>
How much heat is required to raise the temperature of 68
g of aluminum fluoride from 25oC to 80oC?
Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.
<press any key to continue>
How much heat is required to raise the temperature of 68
g of aluminum fluoride from 25oC to 80oC?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
So such a problem has 4 variables -- Heat, Mass, T, Cp
Looking at our problem, we see the following values for these
Heat = x
Mass = 68 g
T = 80o C - 25o C = 55o C
Cp = ?
<press any key to continue>
How much heat is required to raise the temperature of 68
g of aluminum fluoride from 25oC to 80oC?
HEATgained or lost = (Mass)(T)(Cp)
Heat = x
Mass = 68 g
T = 55o C
Cp = ? = 0.895 J/goC
Since the problem does not give us the Cp for aluminum fluoride,
we refer to our list of Common Specific Heats and find that the
value is………... Cp AlF3 = 0.895 J/goC
We can substitute all of our values into the equation and solve the
x = (68 g)(55oC)(0.895 J/goC)
x = 3347.3 J
<press any key to continue>
In order to make 4 cups of tea, 1000 g of water is heated
from 22o C to 99o C. How much energy is required?
Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.
<press any key to continue>
In order to make 4 cups of tea, 1000 g of water is heated
from 22o C to 99o C. How much energy is required?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
So such a problem has 4 variables -- Heat, Mass, T, Cp
Looking at our problem, we see the following values for these
Heat = x
Mass = 1000 g
T = 99o C - 22o C = 77o C
Cp = ?
<press any key to continue>
In order to make 4 cups of tea, 1000 g of water is heated
from 22o C to 99o C. How much energy is required?
HEATgained or lost = (Mass)(T)(Cp)
Heat = x
Mass = 1000 g
T = 77o C
Cp = ? = 1.0 cal/goC
Since the problem does not give us the Cp for water, we refer to
our list of Common Specific Heats and find that the value
is………... Cp H2O = 1.0 cal/goC
We can substitute all of our values into the equation and solve the
x = (1000 g)(77oC)(1.0 cal/goC)
x = 77,000 cal
<press any key to continue>
If a piece of gold with a mass of 45.5 g and a
temperature of 80.5oC is dropped into 192 g of water at
15oC, what is the final temperature of the system?
Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.
<press any key to continue>
If a piece of gold with a mass of 45.5 g and a
temperature of 80.5oC is dropped into 192 g of water at
15oC, what is the final temperature of the system?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
And since this problem involves TWO objects at DIFFERENT
TEMPERATURES being combined, we must also recall…...
HEATgained = HEATlost
In other words, when we combine the two objects, the heat lost by
the “warm” object will be gained by the “cool” object, or, put
another way, the temperature of the “warm” object will decrease
and the temperature of the “cool” object will increase.
<press any key to continue>
If a piece of gold with a mass of 45.5 g and a
temperature of 80.5oC is dropped into 192 g of water at
15oC, what is the final temperature of the system?
HEATgained or lost = (Mass)(T)(Cp)
HEATgained = HEATlost
The problem asks us to determine the final temperature, so we will
use “x” to represent that temperature.
Looking at our list of COMMON SPECIFIC HEATS, we see
that the Cp for gold = 0.13 J/goC and Cp for water = 4.18 J/goC.
We now can re-write HEATgained = HEATlost ……...
(Masswater)(Twater)(Cp water) = (Massgold)(Tgold)(Cp gold)
Now we fill in this equation with our data……...
(192 g)(x - 15o C)(4.18 J/goC) = (45.5 g)(80.5o C - x)(0.13 J/goC)
Lastly, we solve for x…...
(802.56x - 12038.4) = (476.16 - 5.92x)
808.48x = 12514.56
x = 15.48o C
The final temperature of the system will be 15.48o C.
<press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?
Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.
<press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
And since this problem involves TWO objects at DIFFERENT
TEMPERATURES being combined, we must also recall…...
HEATgained = HEATlost
In other words, when we combine the two objects, the heat lost by
the “warm” object will be gained by the “cool” object, or, put
another way, the temperature of the “warm” object will decrease
and the temperature of the “cool” object will increase.
<press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?
HEATgained or lost = (Mass)(T)(Cp)
HEATgained = HEATlost
The problem asks us to determine the specific heat of the metal, so
we will use “x” to represent that specific heat.
Looking at our list of COMMON SPECIFIC HEATS, we see
that the Cp for water = 4.18 J/goC.
We now can re-write HEATgained = HEATlost ……...
(Masswater)(Twater)(Cp water) = (Massmetal)(Tmetal)(Cp metal)
Now we fill in this equation with our data……...
(75 g)(28.5o C - 20o C)(4.18 J/goC) = (14.9 g)(100o C - 28.5o C)(x J/goC)
Lastly, we solve for x…...
(2664.75) = (1065.35x)
x = 2.5 J/goC
The specific heat of the metal is 2.5 J/goC.
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Spring-Mass system (with damping force AND impressed force). Find Position fcn.
July 23rd 2010, 07:58 AM #1
Junior Member
Jul 2010
Spring-Mass system (with damping force AND impressed force). Find Position fcn.
Here is the problem:
A spring is such that a 2 pound weight stretches it 6 inches. There is a damping force present with magnitude the same as the magnitude of the instantaneous velocity. An impressed force f(t) = 2
sin 8t, acts on the spring. At t= 0, the weight is released from a point 3 inches below equilibrium. Find its position function.
I'm not sure if this is correct, but as far as setting up the initial differential equation, I think it is like this:
(1/16)x'' + x' + (1/3)x = 2sin8t
Thank You!
I don't agree with the DE. Try to set up the homogeneous case (the one without the $2 \sin(8t)$ impressed force) first:
$F = m a = m \ddot{x}$. You've got two forces present: the spring force and the damping force. Spring forces look like the following (at least in the regime of Hooke's Law): $F = - k x$. You're
told that a force of 2 pounds stretches the spring 6 inches. That tells you that $k = 1/3$, as I think you've already computed. $k$ is always positive in these cases. The mass is 2 pounds. The
damping force acts in a direction opposite to the spring force, correct? It's going to have to have the opposite sign.
So let's say that $2 \ddot{x} = - k x + \dot{x}$. How does that look to you? Can you finish writing the DE, including the impressed force?
Hmm, yes I found the k=1/3 the same way you did, and put that as the coefficient for x in my equation. For the mass, I thought I would have to divide the pounds by acceleration since F=ma, so I
did (2 pounds)/(32 ft/sec/sec) to get a mass of 1/16 slugs. I'm not completely sure how to include the impressed force, but I thought you're just supposed to take all the homogenous parts of the
equation on one side (so that it would equal zero if there was no impressed force), and then put the impressed force on the right side instead of the 0.
You don't divide by the acceleration. The acceleration is the second derivative of the displacement, so you just leave it where it is. Your method of putting in the impressed force is correct. So
the DE looks like this:
If you're up for it, the Laplace Transform (LT) method is probably the easiest method of solution, because you can take care of the DE itself, the initial conditions, and the inhomogeneous part
all in one fell swoop. So, take the LT of the entire equation. What do you get?
Are you sure? In some of the examples, she converted the pounds to slugs by dividing it by 32 before putting it into the equation. Since it is supposed to be ma (or mx''), and pounds is a unit of
force (not a mass), don't you need to convert it?
Ok. You're right and you're not right. Dang English system! The metric system works so much better.
You're correct in that you do need to divide by the acceleration. I was wrong there. However, the correct constant to divide by is not 32, because the units are not the same as for the rest of
the problem. You've been given pounds and inches. Therefore, your final answer must be in those units. So you need to convert:
$\displaystyle{32\,\frac{\text{ft}}{\text{sec}^{2}} \times\frac{12\,\text{in}}{1\,\text{ft}}=384\,\fra c{\text{in}}{\text{sec}^{2}}.}$
So your DE should be
Are we on the same page now?
Ah alright, yes, thanks. I'm not very good with the LaPlace Transforms yet, do you mind helping me through with the Annihilator method? With that, I believe it would be like this:
D^3 = (D^2 + 64)
homogeneous part: D= 0,0,0
Nonhomogenous part: D= +/- 8i
Now I'm a little rusty as far as getting Y(c) and Y(p)
Hmm. I'd probably do the equivalent guess-and-check method. First, the homogeneous solution:
Assume $x_{c}(t)=e^{ r t}.$ Plug that into the DE. What do you get?
I don't think we covered the guess-and-check method. The way we did it was by finding the roots of D after applying the correct annihilators. Then we'd somehow put in the equation x = c(1)e^at +
c(2)e^at +...... or we use sin and cos terms if imaginary numbers were involved
If I do it like that, then I think the equation would be:
C(1) + C(2)x + C(3)x^2 = C(4)cos(8^[1/2]) + C(5)sin(8^[1/2])
Does this look OK? From what I understand, the left side would be Y(c) while the right side is Y(p). How can I solve for the coefficients, and then get that to the position function?
I'm not sure I followed all of that. The annihilator method is essentially the same as the guess-and-check method. We can go with the annihilator approach. To recap, the DE is
First step: solve the homogeneous equation, which is
What do you get?
I multiplied 192 through the equation to get x'' - 192x + 64 = 0
Then I substituted x'' with m^2, x' with m, and x with 1, and then solved the equation with the quadratic equation.
I got: m= 96 +/- 16[sqrt(143)]
The next step I'm not so sure about, but from what I understand, the solution would be:
x = C(1)e^(96+16sqrt[143]) + C(2)e^(96-16sqrt[143])
the (1) and (2) are supposed to be subscripts, by the way
I think the correct roots are $96\pm 8\sqrt{143}.$ Perhaps you forgot to divide by 2 in both places?
So, you also forgot the t's, I think:
You obviously had the right idea here. Just some small algebra mistakes and notational problems.
So you have the complimentary solution. What's the next step in the annihilator approach?
Ah yeah, thats right. The next step would be to find the annihilator of the nonhomogeneous side and find the roots. The annihilator would be D^2 + 64, making the roots +/-8i
so y(p) would be Acos(8t) + Bsin(8t)
Now how do I use this to get the position function? Are the coefficients solvable, or do I need to leave them as-is?
So far so good. I think what you do now is you take your original non-homogeneous DE, written as $L_{c}x=2\sin(8t),$ where $L_{c}$ is the complimentary operator that annihilates the LHS, and you
operate on the whole equation with the operator that annihilates the RHS. That'll give you a higher-order homogeneous ODE that you need to solve. What do you get at that point?
I'm not sure I know what you mean, would it be like this: (D^2 + 64)(D^3) = D^5 + 64D^3 ?
At some point, you're supposed to take the derivative and second derivative of one of the y values (i'm 70% sure its y(p)) and plug that into the original equation, which helps find some
coefficients, but I'm not sure at which point that is done.
July 23rd 2010, 08:56 AM #2
July 23rd 2010, 09:14 AM #3
Junior Member
Jul 2010
July 23rd 2010, 09:43 AM #4
July 23rd 2010, 10:59 AM #5
Junior Member
Jul 2010
July 23rd 2010, 11:09 AM #6
July 23rd 2010, 11:17 AM #7
Junior Member
Jul 2010
July 23rd 2010, 11:21 AM #8
July 23rd 2010, 07:20 PM #9
Junior Member
Jul 2010
July 24th 2010, 03:26 AM #10
July 24th 2010, 05:19 AM #11
Junior Member
Jul 2010
July 24th 2010, 07:12 AM #12
July 24th 2010, 07:33 AM #13
Junior Member
Jul 2010
July 24th 2010, 08:22 AM #14
July 25th 2010, 05:44 AM #15
Junior Member
Jul 2010
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Maximum error in the Prime Number Theorem
July 12th 2010, 11:07 PM #1
Jul 2010
Maximum error in the Prime Number Theorem
I'm finding 2 conflicting answers for maximum value that pi(n)/(n/ln(n)) can output, where pi(n) is the prime counting function and ln(n) is the natural log.
An inequality by Chebyshev says that the maximum is less than 9/8 (can be found at Prime Number Theorem -- from Wolfram MathWorld). However, setting n=13 we get pi(13)/(13/ln(13))≈1.255 (type it
in to WolframAlpha to see for yourself) which is greater than 9/8.
So my question is, is something about Chebyshev's statement wrong (I doubt it), or am I misunderstanding something, and if so, what?
I'm finding 2 conflicting answers for maximum value that pi(n)/(n/ln(n)) can output, where pi(n) is the prime counting function and ln(n) is the natural log.
An inequality by Chebyshev says that the maximum is less than 9/8 (can be found at Prime Number Theorem -- from Wolfram MathWorld). However, setting n=13 we get pi(13)/(13/ln(13))≈1.255 (type it
in to WolframAlpha to see for yourself) which is greater than 9/8.
So my question is, is something about Chebyshev's statement wrong (I doubt it), or am I misunderstanding something, and if so, what?
I assume they meant to say for large enough $n$, like they do for the next inequality.
Yeah, after reading some more of that page I came to the conclusion. Thanks.
July 13th 2010, 07:50 AM #2
July 13th 2010, 08:31 AM #3
Jul 2010
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Reply to comment
May 1998
To solve this problem we can extend the technique we used in Issue No 2. To make our working easier to read, let’s define
Consider 100 possible cases (contingencies), in which taxis are involved, in proportion to their numbers. Each case is equally probable. We expect 85 cases to involve blue taxis, the other 15 cases
being equally distributed between the
If a blue taxi were really involved, the witness will report blue with probability 80% or some other colour, say
The number of times that these outcomes are expected to occur in each case can be shown in a contingency table. The following table only contains entries for blue,
So, given that the witness reported seeing a blue taxi, we must use the row in the table corresponding to the reported blue taxi.
The probability that the taxi was blue is therefore:
This is simply the number of cases in which the witness reported a blue taxi and was right, divided by the total number of cases in which the witness reported a blue taxi.
If we let
Now let’s look at what happens if the witness reports some other colour, say
This time we use the row in the table corresponding to the reported
If we put decreases towards 37.5%. Having more firms actually weakens the evidence still further. The reason for this "paradox" is that the mistaken sightings of other taxis are swamping the small
number of correct identifications.
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Do equivariant crepant resolutions always exist?
up vote 5 down vote favorite
Let $X_t$ be a family of algebraic varieties (my interest is Calabi-Yau varieties, but I don't think that's important) over $\mathbb{C}$, smooth for $t \neq 0$, on which a group $G$ acts fibre-wise.
Suppose further that $X_0$ admits at least one crepant resolution. Does there always exist an equivariant crepant resolution? If not, are there conditions under which such exists?
add comment
1 Answer
active oldest votes
Consider $\mathbb Z/2$ acting on $\{xy-zw=t\}$ by $x\leftrightarrow y$. This swaps the two small resolutions of the central fibre (the 3-fold ordinary double point $xy=zw$ in $\mathbb C
^4$). So there can't be an equivariant small resolution.
(A formal proof might go along these lines: $\mathbb Z/2$ does act on the blow up of the ODP, swapping the two rulings of the $\mathbb P^1\times\mathbb P^1$ exceptional divisor. The
up vote 5 small resolutions are contractions of this blow up. If $\mathbb Z/2$ acted on one of them, it would act on its $H^2$. Pulling back, its action on $H^2(\mathbb P^1\times\mathbb P^1)\cong
down vote \mathbb Z\oplus\mathbb Z$ would be the identity on the contracted $\mathbb Z$ summand, contradicting the fact that it swaps the summands.)
However if your (finite?) group action permutes the singular loci with no stabilisers then there would surely be a crepant resolution.
By the way, this example doesn't preserve the holomorphic 3-form, so might not interest you. Can anyone think of an example where the group preserves the 3-form ? – Richard Thomas Apr
17 '12 at 21:39
Thanks Richard, that's a very clear counter-example. It's not a problem that the holomorphic 3-form is non-invariant; in fact, for the work which prompted this question, I'm
interested in exactly such examples. – Rhys Davies Apr 18 '12 at 10:52
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Not the answer you're looking for? Browse other questions tagged ag.algebraic-geometry or ask your own question.
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Derivatives: Differential
June 23rd 2008, 08:21 PM
Derivatives: Differential
In a manufacturing process, ball bearings must be made with radius of 0.5 mm, with a maximum error in the radius of ±0.019 mm. Estimate the maximum error in the volume of the ball bearing.
Solution: The formula for the volume of the sphere is 4/3r^3Pi . If an error Dr is made in measuring the radius of the sphere, the maximum error in the volume is DV= 4(r+delta(r))^2pi-4^2pi
Rather than calculating DV, approximate DV with dV, where dV= 4r^2Pidr .
Replacing r with 0.5 and dr=Dr with ± 0.019 gives dV=± _______
The maximum error in the volume is about ________ mm3
The text in red are the answers i got, and the ones that are blank i dont know how to do. Can anyone tell me if the answers i obtained are correct....?
June 23rd 2008, 09:23 PM
If an error Dr is made in measuring the radius of the sphere, the maximum error in the volume is DV= 4(r+delta(r))^2pi-4^2pi
wouldn't it be $\frac 43 \pi (r + \Delta r)^3 - \frac 43 \pi r^3$ ?
Rather than calculating DV, approximate DV with dV, where dV= 4r^2Pidr .
Replacing r with 0.5 and dr=Dr with ± 0.019
gives dV=± _______
just plug in the values and calculate this. you have the formula for dV, what do you get when you plug in r = 0.5 and dr = 0.019, and then when you plug in r = 0.5 and dr = -0.019 ?
The maximum error in the volume is about ________ mm3
just choose the larger dV value (larger in terms of absolute value, i would think)
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nd Evaluating Infix, Postfix
CONVERTING AND EVALUATING INFIX, POSTFIX AND PREFIX ExpressionS IN C
- Sanchit Karve
printf("I'm a %XR",195936478);
CONTACT ME : born2c0de AT dreamincode DOT net
I. WHAT YOU NEED TO KNOW BEFORE YOU BEGIN
IV.B CONVERSION USING Expression TREES
V.A EVALUATING Expression STRINGS
V.B EVALUATING Expression TREES
I. WHAT YOU NEED TO KNOW BEFORE YOU BEGIN
The reader is expected to have a basic knowledge of Stacks, Binary Trees and
their implementations in C.
You will also need a C/C++ Compiler to run and test the source code.
All the source code in this tutorial has been tested on Visual C++ 6.0, but
being standardized code it should work on any other C/C++ compiler.
The Algorithms given in this tutorial are not written as per any standard.
Yet (as you shall see soon enough) it is simple and easy to understand.
I have intentionally written it in a manner which is easier to understand with
respect to my code.
Secondly, the example programs given below can be greatly optimized. But to
make the code easier to understand, I had to compromise on optimization.
So please don't email/PM me about this.
All the C code given in this tutorial conform to C standards and is portable.
Consider a situation where you are writing a programmable calculator. If the
user types in 10 + 10, how would you compute a result of the expression?
You might have a solution for this simple expression.
But what if he types in 3 * 2 / (5 + 45) % 10 ?
What would you do then?
There's no need to think about how you're going to compute the result because no
matter what you do, it won't be the best algorithm for calculating expressions.
That is because there is a lot of overhead in computing the result for
expressions which are expressed in this form; which results in a loss of
The expression that you see above is known as Infix Notation. It is a convention
which humans are familiar with and is easier to read. But for a computer,
calculating the result from an expression in this form is difficult. Hence the
need arises to find another suitable system for representing arithmetic
expressions which can be easily processed by computing systems.
The Prefix and Postfix Notations make the evaluation procedure really simple.
But since we can't expect the user to type an expression in either prefix or
postfix, we must convert the infix expression (specified by the user) into
prefix or postfix before processing it.
This means that your program would have to have two separate functions, one to
convert the infix expression to post or prefix and the second to compute the
result from the converted expression.
This is what the tutorial is about. This tutorial will teach you the different
techniques for converting infix expression to pre/postfix and then evaluate the
converted expression to compute the result. You will be introduced to the
conversion techniques first and later move on to writing the expression
evaluator functions.
Once you read the tutorial, you will be ready to write your own programmable
calculator and more.
Let us first introduce ourselves to the infix, postfix and prefix notations.
Consider a simple expression : A + B
This notation is called Infix Notation.
A + B in Postfix notation is A B +
As you might have noticed, in this form the operator is placed after the
operands (Hence the name 'post'). Postfix is also known as Reverse Polish
Similarly for Infix, the operator is placed INside the operands.
Likewise the equivalent expression in prefix would be + A B, where the operator
precedes the operands.
So if X1 and X2 are two operands and OP is a given operator then
infix postfix prefix
X1 OP X2 = X1 X2 OP = OP X1 X2
We use infix notation for representing mathematical operations or for manually
performing calculations, since it is easier to read and comprehend.
But for computers to perform quick calculations, they must work on prefix or
postfix expressions. You'll soon find out why once you understand how
expressions are evaluated.
Now let's take a slightly complicated expression : A + B / C
How do we convert this infix expression into postfix?
For this we need to use the BODMAS rule. (Remember?)
This rule states that each operator has its own priority and the operators with
the highest priority are evaluated first. The operator priority in Descending
order is BODMAS which is:
B O D M A S
Bracket Open -> Division -> Multiplication -> Addition -> Subtraction
[MAX. PRIORITY] [MIN. PRIORITY]
Hence, in the preceding example, B / C is evaluated first and the result is
added to A.
To convert A + B / C into postfix, we convert one operation at a time. The
operators with maximum priority are converted first followed by those which are
placed lower in the order. Hence, A + B / C can be converted into postfix in
just X steps.
:: A + B / C (First Convert B / C -> B C /)
1: A + B C / (Next Operation is A + (BC/) -> A BC/ +
2: A B C / + (Resultant Postfix Form)
The same procedure is to be applied to convert infix into prefix except that
during conversion of a single operation, the operator is placed before the two
operands. Let's take the same example and convert it to Prefix Notation.
:: A + B / C (First Convert B / C -> / B C)
1: A + / B C (Next Operation is A + (/BC) -> + A /BC +
2: + A / B C
Sometimes, an expression contains parenthesis like this: A + B * ( C + D )
Parenthesis are used to force a given priority to the expression that it
encloses. In this case, C+D is calculated first, then multiplied to B and then
added to A. Without the parenthesis, B * C would have been evaluated first.
To convert an expression with paranthesis, we first convert all expressions that
are enclosed within the simple brackets like this:
[INFIX TO POSTFIX]
:: A + B * ( C + D )
1: A + B * C D +
2: A + B C D + *
3: A B C D + * +
Once an expression has been converted into postfix or prefix, there is no need
to include the parenthesis since the priority of operations is already taken
care of in the final expression.
Keep in mind that converting this expression back to infix would result in the
same original infix expression without the paranthesis, which would ruin the
result of the expression.
Only the Prefix and Postfix forms are capable of preserving the priority of
operations and are hence used for evaluating expressions instead of infix.
Now that you know what infix, prefix and postfix operations are, let us see how
we can programmatically convert expressions from one form into another.
This can be done by various methods but I'll be explaining the two most common
1) USING STACKS
2) USING Expression TREES
Let us study the first technique.
I shall be demonstrating this technique to convert an infix expression to a
postfix expression.
In this method, we read each character one by one from the infix expression and
follow a certain set of steps. The Stack is used to hold only the operators of
the expression. This is required to ensure that the priority of operators is
taken into consideration during conversion.
Before you take a look at the code, read the Algorithm given below so that you
can concentrate on the technique rather than the C code. It will be much easier
to comprehend the code once you know what it is supposed to do.
The algorithm below does not follow any specific standard
Here's the Algorithm to the Infix to Prefix Convertor Function:
Algorithm infix2postfix(infix expression string, postfix expression string)
char *i,*p;
i = first element in infix expression
p = first element in postfix expression
while i is not null
while i is a space or tab
increment i to next character in infix expression;
if( i is an operand )
p = i;
increment i to next character in infix expression;
increment p to next character in postfix expression;
if( i is '(' )
increment i to next character in infix expression;
if( i is ')' )
while( element at top of stack != '(' )
p = element at top of stack;
increment p to next character in postfix expression;
increment i to next character in infix expression;
if( i is an operator )
if( stack is empty )
while priority(element at top of stack) >= priority(i)
p = element at top of stack;
increment p to next character in postfix expression;
increment i to next character in infix expression;
while stack is not empty
p = stack_pop()
increment p to next character in postfix expression;
p = '\0';
Try converting an infix expression A + B / C to postfix on paper using the above
algorithm and check if you're getting the correct result.
Now you can see how the above algorithm is implemented in C. In the following C
code, I have added a 'whitespace adder' feature to the infix2postfix() function.
If the third parameter is 0, A + B / C will be converted as ABC/+
If the third parameter is 1, A + B / C will be converted as A B C / +
This doesn't seem like a big deal, does it?
But in actual practice we will be dealing with numbers as well. Which means that
32 + 23 would be converted to 3223+. This could mean 3 + 223, 32 + 23 or 322 +3.
Hence to prevent such cases, pass 1 as the third parameter to get 32 23 +.
Here's the Code:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define MAX 10
#define EMPTY -1
struct stack
char data[MAX];
int top;
int isempty(struct stack *s)
return (s->top == EMPTY) ? 1 : 0;
void emptystack(struct stack* s)
void push(struct stack* s,int item)
if(s->top == (MAX-1))
printf("\nSTACK FULL");
char pop(struct stack* s)
char ret=(char)EMPTY;
ret= s->data[s->top];
return ret;
void display(struct stack s)
while(s.top != EMPTY)
int isoperator(char e)
if(e == '+' || e == '-' || e == '*' || e == '/' || e == '%')
return 1;
return 0;
int priority(char e)
int pri = 0;
if(e == '*' || e == '/' || e =='%')
pri = 2;
if(e == '+' || e == '-')
pri = 1;
return pri;
void infix2postfix(char* infix, char * postfix, int insertspace)
char *i,*p;
struct stack X;
char n1;
i = &infix[0];
p = &postfix[0];
while(*i == ' ' || *i == '\t')
if( isdigit(*i) || isalpha(*i) )
while( isdigit(*i) || isalpha(*i))
*p = *i;
/*SPACE CODE*/
*p = ' ';
/*END SPACE CODE*/
if( *i == '(' )
if( *i == ')')
n1 = pop(&X);
while( n1 != '(' )
*p = n1;
/*SPACE CODE*/
*p = ' ';
/*END SPACE CODE*/
n1 = pop(&X);
if( isoperator(*i) )
n1 = pop(&X);
while(priority(n1) >= priority(*i))
*p = n1;
/*SPACE CODE*/
*p = ' ';
/*END SPACE CODE*/
n1 = pop(&X);
n1 = pop(&X);
*p = n1;
/*SPACE CODE*/
*p = ' ';
/*END SPACE CODE*/
*p = '\0';
int main()
char in[50] = { 0 },post[50] = { 0 };
iprintf("Enter Infix Expression : ");
fgets(in, sizeof(in), stdin);
in[strlen(in) - 1] = '\0';
printf("Postfix Expression is : %s\n",&post[0]);
return 0;
/* SAMPLE OUTPUT:
Enter Infix Expression : A + B + C / (E - F)
Postfix Expression is : A B + C E F - / +
Now try writing a program to convert infix to prefix as an exercise.
IV.B CONVERSION USING Expression TREES
Expression Trees are a slight variation of a Binary Tree in which every node in
an expression tree is an element of an expression.
It is created in such a way that an inorder traversal would result in an infix
expression, preorder in prefix and postorder in postfix.
The main advantages of an expression tree over the previous technique are pretty
obvious, easier evaluation procedures and efficiency.
The evaluation procedure of both the conversion techniques will be discussed in
the next section.
The previous technique gives the resultant postfix expression as a string. If we
now wish to convert the resultant postfix expression into prefix, we must pass
the entire string to another (long) function which would provide the result into
another string. If at any given point we wish to have all three forms of an
expression, we would need three strings (of outrageously long lengths for
longer expressions) whereas if we use an Expression Tree, we don't need anything
else. We can choose which expression we want by choosing the appropriate
traversal method.
In short, unlike a string an Expression Tree can be interpreted as either infix,
postfix or prefix without changing the structure of the tree.
Consider an Infix expression : A + B / C
This expression would be represented in an Expression Tree as follows:
{+} (ROOT)
| |
--- ---
| |
{A} {/}
| |
--- ---
| |
{B} {C}
Let us Traverse this Expression Tree in Inorder, Preorder and Postorder.
Inorder : A + B / C
Preorder : + A / B C
Postorder : A B C / +
As you can see, Inorder Traversal gives us the Infix expression, Preorder the
Prefix expression and Postorder gives the Postfix expression. Isn't this
amazing? We don't even need a (long and complicated) conversion technique!!
The only thing that we need to do is convert an expression string into an
Expression Tree. This procedure is somewhat complex and requires a Stack as well.
This time I will demonstrate this technique to convert a Postfix Expression to
an Expression Tree.
Here's the algorithm for converting a postfix expression string to an Expression
Algorithm postfix2exptree(postfix string, root<i.e. ptr to root node of exp tree>)
NODES newnode,op1,op2;
p = first element in postfix expression;
while(p is not null)
while(p is a space or a tab)
increment p to next character in postfix expression;
if( p is an operand )
newnode = ADDRESS OF A NEW NODE;
newnode->element = p;
newnode->left = NULL;
newnode->right = NULL;
op1 = stack_pop();
op2 = stack_pop();
newnode = ADDRESS OF A NEW NODE;
newnode->element = p;
newnode->left = op2;
newnode->right = op1;
increment p to next character in postfix expression;
root = stack_pop();
After the function is finished with execution, the root pointer would point to
the root node of the expression tree.
Once again I'd like you to create an expression tree of A B C * + using the
above algorithm on paper to understand how it works.
And finally, the implementation in C:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define MAX 10
#define EMPTY -1
struct node
char element;
struct node *left,*right;
struct stack
struct node *data[MAX];
int top;
int isempty(struct stack *s)
return (s->top == EMPTY) ? 1 : 0;
void emptystack(struct stack* s)
void push(struct stack* s, struct node *item)
if(s->top == (MAX-1))
printf("\nSTACK FULL");
struct node* pop(struct stack* s)
struct node *ret=NULL;
ret= s->data[s->top];
return ret;
void postfix2exptree(char* postfix, struct node **root)
struct stack X;
struct node *newnode,*op1,*op2;
char *p;
p = &postfix[0];
while(*p == ' ' || *p == '\t')
if(isalpha(*p) || isdigit(*p))
newnode = malloc(sizeof(struct node));
newnode->element = *p;
newnode->left = NULL;
newnode->right = NULL;
op1 = pop(&X);
op2 = pop(&X);
newnode = malloc(sizeof(struct node));
newnode->element = *p;
newnode->left = op2;
newnode->right = op1;
*root = pop(&X);
void inorder(struct node *x)
if(x != NULL)
printf("%c ",x->element);
void preorder(struct node *x)
if(x != NULL)
printf("%c ",x->element);
void postorder(struct node *x)
if(x != NULL)
printf("%c ",x->element);
int main()
struct node *r;
postfix2exptree("A B C * +",&r);
printf("Inorder = ");
printf("\nPreorder = ");
printf("\nPostorder = ");
return 0;
/* OUTPUT:
Inorder = A + B * C
Preorder = + A * B C
Postorder = A B C * +
As you can see, once the expression tree is built, the expression can be
converted to any of the three forms by using an appropriate traversal method. I
hope you can now understand its true advantage.
In this case since we were only dealing with alphabets (as variables in
the expression), we chose a node structure like this:
struct node
char element;
struct node *left,*right;
But in practice, we will be dealing with numbers (which will be more than a
character in length is stored as a set of characters). Hence we would need two
extra variables in the structure to hold the operator and the number as well as
another flag variable which states whether the node contains an element or an
operator. Hence the structure would look like this:
struct node
char kind;
char op;
int number; /* even float would do */
struct node *left,*right;
In this structure, if the node is to store an operator, it would store it in the
op variable. If the node is required to store a number, it would do so in the
number variable. If the node contains an operator the kind variable would have
the value 'O' (or anything you wish) and 'N' for a number.
Take care not to name the op variable as operator since this will lead to
compilation errors if compiled on a C++ compiler. C++ compilers won't compile
the code since operator is a keyword in the C++ Language.
Not many people (today) use Pure C Compilers to compile C code (you're probably
using a C++ Compiler right now too), it's best to avoid renaming it as operator.
I have chosen a char data type since it consumes only 1 byte rather than 2/4
bytes for an int for 16/32-bit based programes. This would reduce the size of
the structure by 3 bytes.
Since we're on the topic of efficiency, there is another improvement that we can
make to the structure definition.
Each Node in an expression tree can hold either an operator or a number, but not
both. Hence if an operator is stored in a node, 2/4 bytes of space is wasted for
the unused 'number' variable. If a node holds a number, there is a wastage of 1
byte for the unused op variable. How can we improve upon this structure design?
The answer is Unions. Unions are user-defined data types that can contain only
one data element at a given time. The members of a union represent the kinds of
data the union can contain. An object of union type requires enough storage to
represent each member and hence its size is the same size of the largest member
in its member-list.
This is what the improved structure definition looks like:
struct node
char kind;
union element
char op;
int number;
struct node *left, *right;
This is an ideal situation for using a union.
As an exercise, try to write a function that builds an expression tree for infix
and prefix expression strings using the node structure that uses unions.
The Evaluation process is the easiest part of an expression evaluator. Unlike
the conversion process, evaluation functions are relatively small and easy to
understand (and even write
This is because now, we don't have to worry about the priority of operations
anymore. Remember that evaluation of an expression is always performed on prefix
or postfix expressions and never infix (this should be obvious by now).
As usual, I'll be dividing this section into two sub-parts.
V.A : EVALUATING Expression STRINGS
V.B : EVALUATING Expression TREES
V.A EVALUATING Expression STRINGS
It is fairly simple to evaluate postfix/prefix expression strings. The algorithm
for evaluating a postfix expression is given below:
Algorithm evaluate(postfix expression string)
while current character in postfix string is not null
x = next character in postfix string;
if( x is an operand )
op1 = stack_pop();
op2 = stack_pop();
result = op2 <operator> op1;
result = stack_pop();
return result;
Simple isn't it? Let us see it in action:
#include <stdio.h>
#include <ctype.h>
#define MAX 50
#define EMPTY -1
struct stack
int data[MAX];
int top;
void emptystack(struct stack* s)
s->top = EMPTY;
void push(struct stack* s,int item)
if(s->top == (MAX-1))
printf("\nSTACK FULL");
int pop(struct stack* s)
int ret=EMPTY;
if(s->top == EMPTY)
printf("\nSTACK EMPTY");
ret= s->data[s->top];
return ret;
void display(struct stack s)
while(s.top != EMPTY)
int evaluate(char *postfix)
char *p;
struct stack stk;
int op1,op2,result;
p = &postfix[0];
while(*p != '\0')
/* removes tabs and spaces */
while(*p == ' ' || *p == '\t')
/* if is digit */
push(&stk,*p - 48);
/* it is an operator */
op1 = pop(&stk);
op2 = pop(&stk);
case '+':
result = op2 + op1;
case '-':
result = op2 - op1;
case '/':
result = op2 / op1;
case '*':
result = op2 * op1;
case '%':
result = op2 % op1;
printf("\nInvalid Operator");
return 0;
result = pop(&stk);
return result;
int main()
char exp[MAX];
printf("Enter Postfix Expression : ");
fgets(exp, sizeof(exp), stdin);
exp[strlen(exp) - 1] = '\0';
printf("%s EQUALS %d\n",exp,evaluate(&exp[0]));
return 0;
/* SAMPLE OUTPUT:
Enter Postfix Expression : 3 5 + 2 /
3 5 + 2 / EQUALS 4
Now try writing an evaluator function for prefix expressions.
V.B EVALUATING Expression TREES
Expression trees are usually evaluated using recursion. The Recursive Evaluator
is extremely simple.
Here is the Algorithm:
Algorithm evaluatetree(Node x)
if( x is an operator )
op1 = evaluatetree(x.left);
op2 = evaluatetree(x.right);
result = op1 <operator> op2;
return result; /* x is a number */
That's It!!! And there's no need to write two separate functions for postfix and
prefix expressions.
Here's the Algorithm implemented in C code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define MAX 10
#define EMPTY -1
struct node
char kind;
char op;
int number;
struct node *left,*right;
struct stack
struct node *data[MAX];
int top;
int isempty(struct stack *s)
return (s->top == EMPTY) ? 1 : 0;
void emptystack(struct stack* s)
void push(struct stack* s, struct node *item)
if(s->top == (MAX-1))
printf("\nSTACK FULL");
struct node* pop(struct stack* s)
struct node *ret=NULL;
ret= s->data[s->top];
return ret;
void postfix2exptree(char* postfix, struct node **root)
struct stack X;
struct node *newnode,*op1,*op2;
char numberextract[5];
char *p;
p = &postfix[0];
while(*p == ' ' || *p == '\t')
newnode = malloc(sizeof(struct node));
newnode->kind = 'N';
newnode->number = atoi(numberextract);
newnode->left = NULL;
newnode->right = NULL;
op1 = pop(&X);
op2 = pop(&X);
newnode = malloc(sizeof(struct node));
newnode->kind = 'O';
newnode->op = *p;
newnode->left = op2;
newnode->right = op1;
*root = pop(&X);
int evaluatetree(struct node *x)
if( x->kind == 'O' )
int op1 = evaluatetree( x->left );
int op2 = evaluatetree( x->right );
switch ( x->op )
case '+': return op1 + op2;
case '-': return op1 - op2;
case '*': return op1 * op2;
case '/': return op1 / op2;
default: return 0;
return (x->number);
void inorder(struct node *x)
if(x != NULL)
if(x->kind == 'O')
printf("%c ",x->op);
printf("%d ",x->number);
void preorder(struct node *x)
if(x != NULL)
if(x->kind == 'O')
printf("%c ",x->op);
printf("%d ",x->number);
void postorder(struct node *x)
if(x != NULL)
if(x->kind == 'O')
printf("%c ",x->op);
printf("%d ",x->number);
int main()
struct node *r;
postfix2exptree("100 50 - 2 /",&r);
printf("Inorder = ");
printf("\nPreorder = ");
printf("\nPostprder = ");
printf("\nResult = %d\n",evaluatetree®);
return 0;
/* OUTPUT:
Inorder = 100 - 50 / 2
Preorder = / - 100 50 2
Postprder = 100 50 - 2 /
Result = 25
Since I can't ask you to write an prefix version of the evaluator, write the
same program using unions in the node structure as an exercise.
(Thought you could get away with it this time eh?)
You should now have a basic idea about converting and evaluating arithmetic
expressions in C. You can add a lot of improvements by further adding support
for more operators and functions such as sin(), cos(), log() etc in expressions.
This is a little complicated but not as difficult as it may seem. Simply assign
a token to each function (like a single character 'S' for sine) and test for the
token during evaluation and perform the appropriate function.
The whole idea of this tutorial is to give you an understanding of how
arithmetic expressions are calculated by computers and embedded systems.
Don't reinvent the wheel by writing your own expression evaluators while writing
programs. Languages such as C/C++,Java,C# etc. have in-built or external
libraries that contain expression evaluators. Infact, using these libraries is
recommended as they are more efficient and are less likely to be bugged.
Casio Scientific Calculators and many more use such techniques to solve
arithmetic expressions.
If you own a Casio fx-XXX MS/ES Series Calculator, you can even see that a stack
is being internally used (I currently use a Casio fx-991 ES).
If you key in a long expression with loads of operators and parenthesis and
press the '=' key, you will get a "Stack Error" or "Stack Full" Error.
Compilers also compute expressions using these techniques too.
Almost every mathematical program makes use of this too (Eg. Mathematica and
That brings us to the end of this tutorial.
I hope you enjoyed reading this tutorial.
I also hope that you solve all the exercises
I had starting writing the tutorial ages back and had almost abandoned it. Later,
I hurriedly completed the tutorial so although I have read through the tutorial
and tested all the code, there are still likely to be spelling mistakes or typos.
The code is written in pure, portable C but is tested on on Visual C++ 6.0, so
if I haven't declared variables at the beginning of a function, do let me know.
Please contact me if you find a bug or a typo or even if you need to clarify a
Suggestions and Comments are welcome.
Irrespective of whether it's a suggestion, comment or a mistake, don't hesitate
to contact me on born2c0de AT dreamincode DOT net
You can also post a Reply to the thread where you found this Tutorial.
This post has been edited by JackOfAllTrades: 19 November 2011 - 06:58 AM
Reason for edit:: Removed gets() and fflush(stdin)
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Enriched Categories: Ideals/Submodules and algebraic geometry
up vote 25 down vote favorite
While working through Atiyah/MacDonald for my final exams I realized the following:
The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals
$$I\cdot J:=\langle\{ab\mid a\in I, b\in J\}\rangle$$
The internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\lbrace a\in A\mid aI\subset J\rbrace$. We have
$$IJ\subset K \Leftrightarrow I\subset [J,K].$$
Furthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via
$$[U,V]:=\lbrace a\in A\mid aU\subset V\rbrace.$$
Where does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory
of ring-maps $A\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some "flattened" version of $\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to
see like to see it as a "flattened" version of the category of modules (by taking the annihilator?). So:
Question: Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\mathrm{CRing^{op}}/A$ and $\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write $I\cdot J$ down as a
kernel of some $A\to B$? Or $\mathrm{Ann}(M)\cdot \mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\mathrm{Mod}_A$?
Very interesting question. $I(A)$ is equivalent to the full subcategory of the comma-category which consists of regular epimorphisms $A \to ?$. I've already asked here (mathoverflow.net/questions/
69037/…) how we can describe the ideal product under this equivalence. – Martin Brandenburg Oct 1 '11 at 10:20
add comment
Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.
Browse other questions tagged ac.commutative-algebra ag.algebraic-geometry ct.category-theory monoidal-categories enriched-category-theory or ask your own question.
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Faster Algorithms for Alternating Refinement Relations
when quoting this document, please refer to the following
http://drops.dagstuhl.de/opus/volltexte/2012/3671/ Chatterjee, Krishnendu
Chaubal, Siddhesh
Kamath, Pritish
Faster Algorithms for Alternating Refinement Relations
One central issue in the formal design and analysis of reactive systems is the notion of refinement that asks whether all behaviors of the implementation is allowed by the specification. The local
interpretation of behavior leads to the notion of simulation. Alternating transition systems (ATSs) provide a general model for composite reactive systems, and the simulation relation for ATSs is
known as alternating simulation. The simulation relation for fair transition systems is called fair simulation. In this work our main contributions are as follows: (1) We present an improved
algorithm for fair simulation with Büchi fairness constraints; our algorithm requires O(n^3 * m) time as compared to the previous known O(n^6)-time algorithm, where n is the number of states and m is
the number of transitions. (2) We present a game based algorithm for alternating simulation that requires O(m^2)-time as compared to the previous known O((n*m)^2)-time algorithm, where n is the
number of states and m is the size of transition relation. (3) We present an iterative algorithm for alternating simulation that matches the time complexity of the game based algorithm, but is more
space efficient than the game based algorithm.
BibTeX - Entry
author = {Krishnendu Chatterjee and Siddhesh Chaubal and Pritish Kamath},
title = {{Faster Algorithms for Alternating Refinement Relations}},
booktitle = {Computer Science Logic (CSL'12) - 26th International Workshop/21st Annual Conference of the EACSL},
pages = {167--182},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-939897-42-2},
ISSN = {1868-8969},
year = {2012},
volume = {16},
editor = {Patrick C{\'e}gielski and Arnaud Durand},
publisher = {Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik},
address = {Dagstuhl, Germany},
URL = {http://drops.dagstuhl.de/opus/volltexte/2012/3671},
URN = {urn:nbn:de:0030-drops-36713},
doi = {http://dx.doi.org/10.4230/LIPIcs.CSL.2012.167},
annote = {Keywords: Simulation and fair simulation, Alternating simulation, Graph games}
Keywords: Simulation and fair simulation, Alternating simulation, Graph games
Seminar: Computer Science Logic (CSL'12) - 26th International Workshop/21st Annual Conference of the EACSL
Issue date: 2012
Date of publication: 2012
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Is Limingen deeper than Magelungen?
You asked:
Is Limingen deeper than Magelungen?
Limingen (in Southern Sami Lyjmede), the lake in Røyrvik and Lierne municipalities Nord-Trøndelag county
Magelungen, one of the biggest lakes in Stockholm, Sweden, located between the municipalities of Stockholm and Huddinge
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Cos Cob Algebra Tutor
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Waltham, MA Geometry Tutor
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Icelights: Your Burning Questions About Ice & Climate
How low is low?
Satellite observations since 1979 show that sea ice melted to its lowest extent in the satellite record, during August 2012 . As of this post date, the ice continues to melt, with two to three weeks
left before the days shorten enough for the ice extent to begin to expand through the winter. Readers often write to us asking what such records really mean. How far from normal is this year’s record
low, and how do scientists decide what is normal?
Scientists begin to answer this question by looking at the longest, most consistent time series of data that are available. In the case of sea ice, that means data from passive microwave sensors on a
series of satellites, which have been collecting data on sea ice extent since 1979. They know that in any given year, variable weather conditions can result in more or less sea ice. To understand
climate, which is a more persistent, average set of conditions, looking at trends over many years tells the story beyond the short-term ups and downs of particular years.
How do you define normal?
Statistically, the average Arctic sea ice extent in September is a single value. But how much variation occurs close to that average? A statistical method called standard deviation helps show how
closely most of the data is clustered around the average. Using this method, scientists can see a sort of “average of the average,” a wider set of values that might still be considered typical, or
within the range of natural variability.
NSIDC calculates two standard deviations for sea ice extent data, which means that 95% of the sea ice extent values will fall into this range. The two standard deviation range is considered, in
statistical terms, to be where you would expect to find most of the data. The 5% of values that fall outside of this range are considered outliers, values that deviate most markedly from the rest of
the data. Looking at the data this way helps bring some perspective.
As an analogy, swimmer Michael Phelps has won a total of 18 Olympic gold medals in his career, as of the 2012 London Olympics. Is that a lot of medals, or just a few more than other top athletes? In
comparison, the next most winning athlete won 8 gold medals, and the average number of gold medals for athletes who have won 3 or more medals is just 3.7 medals. The two standard deviation range is
plus or minus 2.5 medals; so up to 6.2 medals is within the typical range. Clearly Phelps is an outlier.
In the case of sea ice, a graph of sea ice extent shows that during spring, sea ice extent was in the two standard deviation range for sea ice from 1979 to 2000, as shown by the gray shaded area of
the graph. But once the melt season began, extent quickly dropped out of this statistically normal zone and plummeted to set a new record low. But was this just an extreme year probably not to be
repeated soon, like Phelps’ medals? A graph of the last six years shows that the minimum sea ice extent for 2007, 2008, 2009, 2010, 2011, and 2012 were all far below the two standard deviations,
suggesting that what was once normal may no longer be normal.
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