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http://www.statemaster.com/encyclopedia/Atmospheric-dispersion-modeling | FACTOID # 18: Alaska spends more money per capita on elementary and secondary education than any other state.
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Encyclopedia > Atmospheric dispersion modeling
Atmospheric dispersion modeling is performed with computer programs that use mathematical equations and algorithms to simulate how pollutants in the ambient atmosphere disperse in the atmosphere. The dispersion models are used to estimate or to predict the downwind concentration of air pollutants emitted from sources such as industrial plants and vehicular traffic. Such models are important to governmental agencies tasked with protecting and managing the ambient air quality. The models are typically used to determine whether existing or proposed new industrial facilites are or will be in compliance with national ambient air quality standards. The models may also be used assist in the design of effective control strategies to reduce emissions of harmful air pollutants. Flowcharts are often used to represent algorithms. ... Layers of Atmosphere (NOAA) Earths atmosphere is a layer of gases surrounding the planet Earth and retained by the Earths gravity. ... This power plant in New Mexico releases sulfur dioxide and particulate matter into the air. ... Emission standards limit the amount of pollution that can be released into the atmosphere. ...
The dispersion models require the input of data which includes:
• Meteorological conditions such as wind speed and direction, the amount of atmospheric turbulence (as characterized by what is called the "stability class"), the ambient air temperature and the height to the bottom of any inversion aloft that may be present.
• Emissions parameters such as source location and height, source vent stack diameter and exit velocity, exit temperature and mass flow rate.
• Terrain elevations at the source location and at the receptor location.
• The location, height and width of any obstructions (such as buildings or other structures) in the path of the emitted gaseous plume.
Many of the modern, advanced dispersion modeling programs include a pre-processor module for the input of meteorological and other data, and many also include a post-processor module for graphing the output data and/or plotting the area impacted by the air pollutants on maps. Satellite image of Hurricane Hugo with a polar low visible at the top of the image. ... In fluid dynamics, turbulence or turbulent flow is a flow regime characterized by semi-random, stochastic property changes. ... Temperature inversion in Bratislava Casual view from old part of city, same Bridge A temperature inversion is a meteorological phenomenon where air temperature increases with height. ... This article is about velocity in physics. ... Mass flow rate is the movement of mass per time. ...
The atmospheric dispersion models are also known as atmospheric diffusion models, air dispersion models, air quality models, and air pollution dispersion models.
## Gaussian air pollutant dispersion equation GA_googleFillSlot("encyclopedia_square");
The technical literature on air pollution dispersion is quite extensive and dates back to the 1930's and earlier. One of the early air pollutant plume dispersion equations was derived by Bosanquet and Pearson[1] in 1936. Their equation did not assume Gaussian distribution nor did it include the effect of ground reflection of the pollutant plume. The normal distribution, also called Gaussian distribution, is an extremely important probability distribution in many fields. ...
Sir Graham Sutton derived an air pollutant plume dispersion equation in 1947[2] which did include the assumption of Gaussian distribution for the vertical and crosswind dispersion of the plume and also included the effect of ground reflection of the plume.
Under the stimulus provided by the advent of stringent environmental control regulations, there was an immense growth in the use of air pollutant plume dispersion calculations between the late 1960s and today. A great many computer programs for calculating the dispersion of air pollutant emissions were developed during that period of time and they were called "air dispersion models". The basis for most of those models was the Complete Equation For Gaussian Dispersion Modeling Of Continuous, Buoyant Air Pollution Plumes shown below:
A Clean Air Act may be one of a number of pieces of legislation relating to reduction of smog and atmospheric pollution in general. ...
$C = frac{;Q}{u}cdotfrac{;f}{sigma_ysqrt{2pi}};cdotfrac{;g_1 + g_2 + g_3}{sigma_zsqrt{2pi}}$
where:
$f =; crosswind;, dispersion;, parameter$
$f =;exp;[-,y^2/,(2;sigma_y^2;);]$
$g =;vertical;,dispersion;,parameter =,g_1 + g_2 + g_3$
$g_1 =;vertical;,dispersion;, with;, no ;, reflections$
$g_1 =; exp;[-,(z - H)^2/,(2;sigma_z^2;);]$
$g_2 =;vertical;,dispersion;,for;,reflection;, from;,the;, ground$
$g_2=;exp;[-,(z + H)^2/,(2;sigma_z^2;);]$
$g_3 =;vertical;,dispersion;,for;,reflection;, from;,inversion;,aloft$
$g_3 =;sum_{m=1}^infty;big{exp;[-,(z - H - 2mL)^2/,(2;sigma_z^2;);]$
$+, exp;[-,(z + H + 2mL)^2/,(2;sigma_z^2;);]$
$+, exp;[-,(z + H - 2mL)^2/,(2;sigma_z^2;);]$
$+, exp;[-,(z - H + 2mL)^2/,(2;sigma_z^2;);]big}$
$C =;concentration;,of;,emissions,;, mbox{g}/mbox{m}^3, ;,at;,any;,receptor;,located:$
$x;,meters;,downwind;,from;,the;,emission;,source;,point$
$y;,meters;,crosswind;,from;,the;,emission;,plume;,centerline$
$z;,meters;,above;,ground;,level$
$Q; =;source;,pollutant;,emission;,rate, ;,mbox{g}/mbox{s}$
$u; =;horizontal;,wind;,velocity;,along ;,the;,plume;,centerline,;, mbox{m}/mbox{s}$
$H, =;,height;,of;emission;, plume;,centerline;, above;,ground;,level,,; mbox{m}$
$sigma_z =;,vertical;,standard;, deviation;,of;,emission;,distribution,,; mbox{m}$
$sigma_y =;,horizontal;,standard;, deviation;,of;,emission;,distribution,,; mbox{m}$
$L; =;,height;,from;,ground;,level;, to;,bottom;,of;,inversion;,aloft,,; mbox{m}$
The above equation not only includes upward reflection from the ground, it also includes downward reflection from the bottom of any inversion lid present in the atmosphere.
The sum of the four exponential terms in g3 converges to a final value quite rapidly. For most cases, the summation of the series with m = 1, m = 2 and m = 3 will provide an adequate solution.
It should be noted that σz and σy are functions of the atmospheric stability class (i.e., a measure of the turbulence in the ambient atmosphere) and of the downwind distance to the receptor. The two most important variables affecting the degree of pollutant emission dispersion obtained are the height of the emission source point and the degree of atmospheric turbulence. The more turbulence, the better the degree of dispersion.
## The Briggs' plume rise equations
The Gaussian air pollutant dispersion equation (discussed above) requires the input of H which is the pollutant plume's centerline height above ground level—and H is the sum of Hs (the actual physical height of the pollutant plume's emission source point) plus ΔH (the plume rise due the plume's buoyancy).
To determine ΔH, many if not most of the air dispersion models developed and used between the late 1960s and the early 2000s used what are known as "the Brigg's equations". Briggs published his first plume rise model observations and comparisons in 1965.[3] In 1968, at a symposium sponsored by CONCAWE (a Dutch organization), he compared many of the plume rise models then available in the literature.[4] In that same year, Briggs also wrote the section of the publication edited by Slade[5] dealing with the comparative analyses of plume rise models. That was followed in 1969 by his classical critical review of the entire plume rise literature[6], in which he proposed a set of of plume rise equations which have became widely known as "the Briggs equations". Subsequently, Briggs modified his 1969 plume rise equations in 1971 and in 1972.[7][8]
Briggs divided air pollution plumes into these four general categories:
• Cold jet plumes in calm ambient air conditions
• Cold jet plumes in windy ambient air conditions
• Hot, buoyant plumes in calm ambient air conditions
• Hot, buoyant plumes in windy ambient air conditions
Briggs considered the trajectory of cold jet plumes to be dominated by their initial velocity momentum, and the trajectory of hot, buoyant plumes to be dominated by their buoyant momentum to the extent that their initial velocity momentum was relatively unimportant. Although Briggs proposed plume rise equations for each of the above plume categories, it is important to emphasize that "the Briggs equations" which become widely used are those that he proposed for bent-over, hot buoyant plumes.
In general, Briggs' equations for bent-over, hot buoyant plumes are based on observations and data involving plumes from typical combustion sources such as the flue gas stacks from steam-generating boilers burning fossil fuels in large power plants. Therefore the stack exit velocities were probably in the range of 20 to 100 ft/s (6 to 30 m/s) with exit temperatures ranging from 250 to 500 °F (120 to 260 °C). Flue gas is gas that exits to the atmosphere via a flue, which is a pipe or channel for conveying exhaust gases from a fireplace, furnace, boiler or generator. ... Coal rail cars in Ashtabula, Ohio Fossil fuels, also known as mineral fuels, are hydrocarbon-containing natural resources such as coal, oil and natural gas. ...
A logic diagram for using the Briggs equations[9] to obtain the plume rise trajectory of bent-over buoyant plumes is presented below:
• Beychok, M.R., Fundamentals of Stack Gas Dispersion Modeling, 4th Edition, published by auther, 2005. www.air-dispersion.com
• Turner, D.B., Workbook of atmospheric dispersion estimates: an introduction to dispersion modeling, 2nd Edition, 1994. www.crcpress.com
## References
1. ^ Bosanquet, C.H. and Pearson, J.L., "The spread of smoke and gases from chimneys", Trans. Faraday Soc., 32:1249, 1936
2. ^ Sutton, O.G., "The problem of diffusion in the lower atmosphere", QJRMS, 73:257, 1947 and "The theoretical distribution of airborne pollution from factory chimneys", QJRMS, 73:426, 1947
3. ^ Briggs, G.A., "A plume rise model compared with observations", JAPCA, 15:433-438, 1965
4. ^ Briggs, G.A., "CONCAWE meeting: discussion of the comparative consequences of different plume rise formulas", Atmos. Envir., 2:228-232, 1968
5. ^ Slade, D.H. (editor), "Meteorology and atomic energy 1968", Air Resources Laboratory, U.S. Dept. of Commerce, 1968
6. ^ Briggs, G.A., "Plume Rise", USAEC Critical Review Series, 1969
7. ^ Briggs, G.A., "Some recent analyses of plume rise observation", Proc. Second Internat'l. Clean Air Congress, Academic Press, New York, 1971
8. ^ Briggs, G.A., "Discussion: chimney plumes in neutral and stable surroundings", Atmos. Envir., 6:507-510, 1972
9. ^ Beychok, Milton R. (2005). Fundamentals Of Stack Gas Dispersion, 4th Edition, 201 pages, author-published. ISBN 0964458802. www.air-dispersion.com
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http://www.albany.edu/~hammond/gellmu/examples/confrac.html | # Continued Fractions and the Euclidean Algorithm
## William F. Hammond
1 Introduction... *
2 The continued fraction expansion of a real number... *
3 First examples... *
4 The case of a rational number... *
5 The symbol [t_{1}, t_{2}, …, t_{r}]... *
6 Application to Continued Fractions... *
7 Bezout's Identity and the double recursion... *
8 The action of GL_{2}(Z) on the projective line... *
9 Periodic continued fractions... *
References ... *
### 1. Introduction
Continued fractions offer a means of concrete representation for arbitrary real numbers. The continued fraction expansion of a real number is an alternative to the representation of such a number as a (possibly infinite) decimal.
The reasons for including this topic in the course on Classical Algebra are:
(i) The subject provides many applications of the method of recursion.
(ii) It is closely related to the Euclidean algorithm and, in particular, to “Bezout's Identity”.
(iii) It provides an opportunity to introduce the subject of group theory via the 2-dimensional unimodular group GL_{2}(Z).
### 2. The continued fraction expansion of a real number
Every real number x is represented by a point on the real line and, as such, falls between two integers. For example, if n is an integer and
n <= x < n + 1 ,
x falls between n and n+1, and there is one and only one such integer n for any given real x. In the case where x itself is an integer, one has n = x. The integer n is sometimes called the floor of x, and one often introduces a notation for the floor of x such as
n = [x] .
Examples:
1.
-2 = [-1.5]
2.
3 = [pi]
For any real x with n = [x] the number u = x - n falls in the unit interval I consisting of all real numbers u for which 0 <= u < 1.
Thus, for given real x there is a unique decomposition
x = n + u
where n is an integer and u is in the unit interval. Moreover, u = 0 if and only if x is an integer. This decomposition is sometimes called the mod one decomposition of a real number. It is the first step in the process of expanding x as a continued fraction.
The process of finding the continued fraction expansion of a real number is a recursive process that procedes one step at a time. Given x one begins with the mod one decomposition
x = n_{1} + u_{1} ,
where n_{1} is an integer and 0 <= u_{1} < 1.
If u_{1} = 0, which happens if and only if x is an integer, the recursive process terminates with this first step. The idea is to obtain a sequence of integers that give a precise determination of x.
If u_{1} > 0, then the reciprocal 1/u_{1} of u_{1} satisfies 1/u_{1} > 1 since u_{1} is in I and, therefore, u_{1} < 1. In this case the second step in the recursive determination of the continued fraction expansion of x is to apply the mod one decomposition to 1/u_{1}. One writes
1/u_{1} = n_{2} + u_{2} ,
where n_{2} is an integer and 0 <= u_{2} < 1. Combining the equations that represent the first two steps, one may write
x = n_{1} + {1}/{n_{2} + u_{2}} .
Either u_{2} = 0, in which case the process ends with the expansion
x = n_{1} + {1}/{n_{2}} ,
or u_{2} > 0. In the latter case one does to u_{2} what had just been done to u_{1} above under the assumption u_{1} > 0. One writes
1/u_{2} = n_{3} + u_{3} ,
where n_{3} is an integer and 0 <= u_{3} < 1. Then combining the equations that represent the first three steps, one may write
x = n_{1} + {1}/{n_{2} + {1}/{n_{3} + u_{3}}} .
After k steps, if the process has gone that far, one has integers n_{1}, n_{2}, …, n_{k} and real numbers u_{1}, u_{2}, …, u_{k} that are members of the unit interval I with u_{1}, u_{2}, …, u_{k-1} all positive. One may write
x = n_{1} + {1}/{n_{2} + {1}/{n_{3} + {1}/{… + {1}/{n_{k} + u_{k}}}}} .
Alternatively, one may write
x = [ n_{1}, n_{2}, n_{3}, …, n_{k} + u_{k} ] .
If u_{k} = 0, the process ends after k steps. Otherwise, the process continues at least one more step with
1/u_{k} = n_{k+1} + u_{k+1} .
In this way one associates with any real number x a sequence, which could be either finite or infinite, n_{1}, n_{2}, … of integers. This sequence is called the continued fraction expansion of x.
Convention. When [n_{1}, n_{2}, ...] is called a continued fraction, it is understood that all of the numbers n_{j} are integers and that n_{j} >= 1 for j >= 2.
### 3. First examples
{15}/{11}
=
1 + {4}/{11}
=
1 + {1}/{{11}/{4}}
=
1 + {1}/{2 + {3}/{4}}
=
1 + {1}/{2 + {1}/{{4}/{3}}}
=
1 + {1}/{2 + {1}/{1 + {1}/{3}}}
=
[1, 2, 1, 3] .
SQRT{10}
=
3 + {1}/{{1}/{SQRT{10}-3}}
=
3 + {1}/{SQRT{10}+3}
=
3 + {1}/{6 + {1}/{{1}/{SQRT{10}-3}}}
=
3 + {1}/{6 + {1}/{SQRT{10}+3}}
=
3 + {1}/{6 + {1}/{6 + {1}/{…}}}
=
[3, 6, 6, 6, …] .
[2, 3, 5, 2 ]
=
2 + {1}/{[3, 5, 2]}
=
2 + {1}/{3 + {1}/{[5, 2]}}
=
2 + {1}/{3 + {1}/{5 + {1}/{2}}}
=
2 + {1}/{3 + {1}/{{11}/{2}}}
=
2 + {1}/{3 + {2}/{11}}
=
2 + {1}/{{35}/{11}}
=
2 + {11}/{35}
=
{81}/{35} .
Let
x = 1+{1}/{2+{1}/{3+{1}/{2+{1}/{3+{1}/{2+…}}}}} .
In this case one finds that
x = 1 + {1}/{y} ,
where
y = 2+{1}/{3+{1}/{2+{1}/{3+{1}/{2+…}}}} .
Further reflection shows that the continued fraction structure for y is self-similar:
y = 2+{1}/{3+{1}/{y}} .
This simplifies to
y = {7y+2}/{3y+1}
3y^{2} - 6y - 2 = 0
with discriminant 60. Since y > 2, one of the two roots of the quadratic equation cannot be y, and, therefore,
y = {3 + SQRT{15}}/{3} .
Finally,
x = {SQRT{15}-1}/{2} .
The idea of the calculation above leads to the conclusion that any continued fraction [n_{1}, n_{2}, …] that eventually repeats is the solution of a quadratic equation with positive discriminant and integer coefficients. The converse of this statement is also true, but a proof requires further consideration.
### 4. The case of a rational number
The process of finding the continued fraction expansion of a rational number is essentially identical to the process of applying the Euclidean algorithm to the pair of integers given by its numerator and denominator.
Let x = a/b, b > 0, be a representation of a rational number x as a quotient of integers a and b. The mod one decomposition
{a}/{b} = n_{1} + u_{1} , u_{1} = {a - n_{1} b}/{b}
shows that u_{1} = r_{1}/b, where r_{1} is the remainder for division of a by b. The case where u_{1} = 0 is the case where x is an integer. Otherwise u_{1} > 0, and the mod one decomposition of 1/u_{1} gives
{b}/{r_{1}} = n_{2} + u_{2} , u_{2} = {b - n_{2} r_{1}}/{r_{1}} .
This shows that u_{2} = r_{2}/r_{1}, where r_{2} is the remainder for division of b by r_{1}. Thus, the successive quotients in Euclid's algorithm are the integers n_{1}, n_{2}, … occurring in the continued fraction. Euclid's algorithm terminates after a finite number of steps with the appearance of a zero remainder. Hence, the continued fraction expansion of every rational number is finite.
Theorem 1. The continued fraction expansion of a real number is finite if and only if the real number is rational.
Proof. It has just been shown that if x is rational, then the continued fraction expansion of x is finite because its calculation is given by application of the Euclidean algorithm to the numerator and denominator of x. The converse statement is the statement that every finite continued fraction represents a rational number. That statement will be demonstrated in the following section.
### 5. The symbol [t_{1}, t_{2}, …, t_{r}]
For arbitrary real numbers t_{1}, t_{2}, …, t_{r} with each t_{j} >= 1 for j >= 2 the symbol [ t_{1}, t_{2}, …, t_{r} ] is defined recursively by:
[ t_{1} ] = t_{1}
(1) [t_{1},t_{2},…,t_{r} ] = t_{1}+{1}/{[t_{2},…,t_{r} ]} .
In order for this definition to make sense one needs to know that the denominator in the right-hand side of (1) is non-zero. The condition t_{j} >= 1 for j >= 2 guarantees, in fact, that [t_{2},…,t_{r} ] > 0, as one may prove using induction.
It is an easy consequence of mathematical induction that the symbol [t_{1}, t_{2}, …, t_{r}] is a rational number if each t_{j} is rational. In particular, each finite continued fraction is a rational number. (Note that the symbol [t_{1}, t_{2}, …, t_{r}] is to be called a continued fraction, according to the convention of the first section, only when each t_{j} is an integer.)
Observe that the recursive nature of the symbol [t_{1}, …, t_{r}] suggests that the symbol should be computed in a particular case working from right to left. Consider again, for example, the computation above showing that [2, 3, 5, 2] = 81/35. Working from right to left one has:
[2]
=
2
[5, 2]
=
5+{1}/{[2]} = 5+{1}/{2} = {11}/{2}
[3, 5, 2]
=
3+{1}/{[5, 2]} = 3+{2}/{11} = {35}/{11}
[2, 3, 5, 2]
=
2+{1}/{[3, 5, 2]} = 2+{11}/{35} = {81}/{35}
There is, however, another approach to computing [t_{1}, t_{2}, …, t_{r}]. Let, in fact, t_{1}, t_{2}, … be any (finite or infinite) sequence of real numbers. One uses the double recursion
(2) p_{j} = t_{j} p_{j-1} + p_{j-2} , j >= 1 , p_{0} = 1 , p_{-1} = 0
to define the sequence {p_{j}} , j >= -1. The double recursion, differently initialized,
(3) q_{j} = t_{j} q_{j-1} + q_{j-2} , j >= 1 , q_{0} = 0 , q_{-1} = 1
defines the sequence {q_{j}} , j >= -1. Note that p_{1} = t_{1}, p_{2} = t_{1}t_{2} + 1, … and q_{1} = 1, q_{2} = t_{2}, q_{3} = t_{2}t_{3} + 1, ….
One now forms the matrix
(4) M_{j} =
(
p_{j} q_{j} p_{j-1} q_{j-1}
)
for j >= 0 .
Thus, for example,
M_{0} =
(
1 0 0 1
)
, and M_{1} =
(
t_{1} 1 1 0
)
.
It is easy to see that the matrices M_{j} satisfy the double recursion
(5) M_{j} =
(
t_{j} 1 1 0
)
M_{j-1} , j >= 1
as a consequence of the double recursion formulas for the p_{j} and q_{j}. Hence, a simple argument by mathematical induction shows that
(6) M_{r} =
(
t_{r} 1 1 0
)
…
(
t_{2} 1 1 0
)
(
t_{1} 1 1 0
)
, r >= 1 .
This is summarized by:
Proposition 1. For any sequence {t_{j}}, j >= 1 of real numbers, if {p_{j}} and {q_{j}} are the sequences defined by the double recursions (2) and (3), then one has the matrix identity
(7)
(
p_{r} q_{r} p_{r-1} q_{r-1}
)
=
(
t_{r} 1 1 0
)
…
(
t_{2} 1 1 0
)
(
t_{1} 1 1 0
)
for each integer r >= 1.
Corollary 1. One has the identity p_{r} q_{r-1} - q_{r} p_{r-1} = (-1)^{r} for each integer r >= 1.
Proof. The number p_{r} q_{r-1} - q_{r} p_{r-1} is the determinant of the matrix M_{r}. From the formula (6) the matrix M_{r} is the product of r matrix factors, each of which has determinant -1. Since the determinant of the product of matrices is the product of the determinants of the factors, it is clear that det(M_{r}) = (-1)^{r}.
Corollary 2. One has the vector identity
(8)
(
p_{r} q_{r}
)
=
(
t_{1} 1 1 0
)
(
t_{2} 1 1 0
)
…
(
t_{r} 1 1 0
)
(
1 0
)
for each integer r >= 1.
Proof. First recall (i) that the product of a matrix and a (column) vector is defined by the relation
(
a b c d
)
(
x y
)
=
(
ax + by cx + dy
)
,
(ii) that the transpose of a matrix is the matrix whose rows are the columns of the given matrix, and (iii) that the transpose operation reverses matrix multiplication. One tranposes both sides of the relation (7) to obtain:
(9)
(
p_{r} p_{r-1} q_{r} q_{r-1}
)
=
(
t_{1} 1 1 0
)
(
t_{2} 1 1 0
)
…
(
t_{r} 1 1 0
)
.
To this relation one applies the principle that the first column of any 2 \times 2 matrix is the product of that matrix with the column
(
1 0
)
in order to obtain the column identity (8).
Theorem 2. For any sequence {t_{j}}, j >= 1 of real numbers, if {p_{j}} and {q_{j}} are the sequences defined by the double recursions (2) and (3), and if t_{j} >= 1 for j >= 2, then the value of the symbol [t_{1}, …, t_{r}] is given by the formula
(10) [t_{1}, t_{2}, …, t_{r}] = {p_{r}}/{q_{r}} for r >= 1 .
Proof. What is slightly strange about this important result is that while the {p_{r}} and the {q_{r}} are defined by the front end recursions, albeit double recursions, (2) and (3), the symbol [t_{1}, …, t_{r}] is defined by the back end recursion (1). The proof begins with the comment that the right-hand side of (10) does not make sense unless one can be sure that the denominator q_{r} ≠ 0. One can show easily by induction on r that q_{r} >= 1 for r >= 1 under the hypothesis t_{j} >= 1 for j >= 2.
The proof proceeds by induction on r. If r = 1, the assertion of the theorem is simply the statement t_{1} = p_{1}/q_{1}, and, as noted above, p_{1} = t_{1} and q_{1} = 1. Assume now that r >= 2. By induction we may assume the correctness of the statement (10) for symbols of length r-1, and, therefore, for the symbol [t_{2}, …, t_{r}]. That case of the statement says that [t_{2}, …, t_{r}] must be equal to a/c, where by corollary 2
(
a c
)
=
(
a b c d
)
(
1 0
)
with
(
a b c d
)
=
(
t_{2} 1 1 0
)
…
(
t_{r} 1 1 0
)
.
Now by (1)
[t_{1}, t_{2}, …, t_{r}] = t_{1} + {1}/{a/c} = t_{1} + {c}/{a} = {at_{1} + c}/{a} .
But by corollary 2 again
(
p_{r} q_{r}
)
=
(
t_{1} 1 1 0
)
(
a b c d
)
(
1 0
)
=
(
at_{1} + c bt_{1} + d a b
)
(
1 0
)
=
(
at_{1} + c a
)
.
Hence,
{p_{r}}/{q_{r}} = {at_{1} + c}/{a} = [t_{1}, t_{2}, …, t_{r}] .
### 6. Application to Continued Fractions
Recall that [n_{1}, n_{2}, … ] is called a continued fraction only when each n_{j} is an integer and n_{j} >= 1 for j >= 2. The sequence n_{1}, n_{2}, … may be finite or infinite. The symbol c_{r} = [n_{1}, n_{2}, …, n_{r}] formed with the first r terms of the sequence, is called the r^{th} convergent of the continued fraction. Associated with a given sequence n_{1}, n_{2}, … are two sequences p_{1}, p_{2}, … and q_{1}, q_{2}, … that are given, according to the double recursions (2), (3) of the previous section with t_{j} = n_{j}.
Proposition 2. If [n_{1}, n_{2}, …] is a continued fraction, then the integers p_{r} and q_{r} are coprime for each r >= 1.
Proof. By Corollary 1 of the previous section p_{r} q_{r-1} - q_{r} p_{r-1} = (-1)^{r}. Hence, any positive divisor of both p_{r} and q_{r} must divide the left-hand side of this relation, and, therefore, must also divide (-1)^{r}.
Proposition 3. The difference between successive convergents of the continued fraction [n_{1}, n_{2}, …] is given by the formula
(11) c_{r} - c_{r-1} = {(-1)^{r}}/{q_{r} q_{r-1}} for r >= 2 .
Proof. According to the theorem (formula 10) at the end of the last section the convergent c_{r} is given by
c_{r} = {p_{r}}/{q_{r}} .
Hence,
c_{r} - c_{r-1}
=
{p_{r}}/{q_{r}} - {p_{r-1}}/{q_{r-1}}
=
{p_{r} q_{r-1} - p_{r-1} q_{r}}/{q_{r} q_{r-1}}
=
{(-1)^{r}}/{q_{r} q_{r-1}} .
(The last step is by Corollary 1 above.)
Remark 1. The formula (11) remains true if c_{r} = [t_{1}, …, t_{r}] where the t_{j} are real numbers subject to the assumption t_{j} >= 1 for j >= 1.
Lemma. The sequence {q_{j}} is a strictly increasing sequence for j >= 2.
Proof. This is easily proved by induction from the recursive definition (3) of the sequence.
Theorem 3. If [n_{1}, n_{2}, …] is an infinite continued fraction, then the limit
lim_{r --> INFTY} {p_{r}}/{q_{r}}
always exists.
Proof. As one plots the convergents c_{r} on the line of real numbers, one moves alternately right and left. The formula (11) for the difference between successive convergents elucidates not only the fact of alternate right and left movement but also the fact that each successive movement is smaller than the one preceding. Therefore, one has
c_{1} < c_{3} < c_{5} < … < c_{6} < c_{4} < c_{2} .
Since any strictly increasing sequence of positive integers must have infinite limit, the seqence q_{j} q_{j-1} has infinite limit, and so the sequence of reciprocals 1/q_{j} q_{j-1} must converge to zero. Hence, the sequences of odd- and even-indexed convergents must have the same limit, which is the limit of the sequence of all convergents.
Definition 1. The limit of the sequence of convergents of an infinite continued fraction is called the value of that continued fraction.
Theorem 4. If [n_{1}, n_{2}, …] is the continued fraction expansion of an irrational number x, then
x = lim_{r --> INFTY} {p_{r}}/{q_{r}} ;
that is, the value of the continued fraction expansion of a real number is that real number.
Proof. For each r >= 1 the continued fraction expansion [n_{1}, n_{2}, …] of x is characterized by the identity
(12) x = [n_{1}, n_{2}, …, n_{r} + u_{r}] ,
where u_{r} is a real number with 0 <= u_{r} < 1. The sequences of p's and q's for the symbol [n_{1}, n_{2}, …, n_{r} + u_{r}] agree with those for the symbol [n_{1}, n_{2}, …, n_{r}] except for the r^{th} terms. One has by (10)
[n_{1}, n_{2}, …, n_{r} + u_{r}] = {P_{r}}/{Q_{r}} ,
where by (3)
q_{r}
=
n_{r} q_{r-1} + q_{r-2}
Q_{r}
=
(n_{r} + u_{r}) q_{r-1} + q_{r-2}
Hence,
Q_{r} = q_{r} + u_{r} q_{r-1} .
Therefore, the displacement from c_{r-1} to x is by (11)
{(-1)^{r}}/{Q_{r} q_{r-1}} = {(-1)^{r}}/{(q_{r} q_{r-1} + u_{r} q_{r-1}^{2})} ,
which is in the same direction but of smaller magnitude than the displacement from c_{r-1} to c_{r}. Therefore, x must be larger than every odd-indexed convergent and smaller than every even-indexed convergent. But since all convergents have the same limit, that limit must be x.
### 7. Bezout's Identity and the double recursion
It has already been observed that the process of finding the continued fraction expansion of a rational number a/b (b > 0), involves the same series of long divisions that are used in the application of the Euclidean algorithm to the pair of integers a and b. Recall that at each stage in the Euclidean algorithm the divisor for the current stage is the remainder from the previous stage and the dividend for the current stage is the divisor from the previous stage, or, equivalently, the dividend for the current stage is the remainder from the second previous stage. The Euclidean algorithm may thus be viewed as a double recursion that is used to construct the sequence of remainders. One starts the double recursion with
r_{-1} = a and r_{0} = b .
At the j^{th} stage one performs long division of r_{j-2} by r_{j-1} to obtain the integer quotient n_{j} and the integer remainder r_{j} that satisfies 0 <= r_{j} < r_{j-1}. Thus,
(13) r_{j} = r_{j-2} - n_{j} r_{j-1} .
The Euclidean algorithm admits an additional stage if r_{j} > 0. Since
0 <= r_{j} < r_{j-1} < … < r_{2} < r_{1} < r_{0} = b ,
there can be at most b stages.
One may use the sequence of successive quotients n_{j} (j >= 1) to form sequences {p_{j}} and {q_{j}}, as in the previous section, according to the double recursions:
(14) p_{j} = n_{j} p_{j-1}+p_{j-2} , j >= 1 ; p_{0} = 1 , p_{-1} = 0 .
(15) q_{j} = n_{j} q_{j-1}+q_{j-2} , j >= 1 ; q_{0} = 0 , q_{-1} = 1 .
It has already been observed that q_{j} >= 1 for j >= 1 and
[n_{1}, n_{2}, …, n_{j}] = {p_{j}}/{q_{j}} , j >= 1 .
Bezout's Identity says not only that the greatest common divisor of a and b is an integer linear combination of them but that the coefficents in that integer linear combination may be taken, up to a sign, as q and p.
Theorem 5. If the application of the Euclidean algorithm to a and b (b > 0) ends with the m^{th} long division, i.e., r_{m} = 0, then
(16) r_{j} = (-1)^{j-1}
( q_{j} a - p_{j} b )
, 1 <= j <= m .
Proof. One uses induction on j. For j = 1 the statement is r_{1} = q_{1} a - p_{1} b. Since by (14, 15) q_{1} = 1 and p_{1} = n_{1}, this statement is simply the case j = 1 in (13). Assume j >= 2, and that the formula (16) has been established for indices smaller than j. By (13) one has
r_{j} = r_{j-2} - n_{j} r_{j-1} .
In this equation one may use (16) to expand the terms r_{j-2} and r_{j-1} to obtain:
r_{j}
=
{ (-1)^{j-3}(q_{j-2}a - p_{j-2}b) }
- n_{j}
{ (-1)^{j-2}(q_{j-1}a - p_{j-1}b) }
=
{ (-1)^{j-1}(q_{j-2}a - p_{j-2}b) }
+ n_{j}
{ (-1)^{j-1}(q_{j-1}a - p_{j-1}b) }
=
(-1)^{j-1}
{ (q_{j-2}a - p_{j-2}b) + n_{j} (q_{j-1}a - p_{j-1}b) }
=
(-1)^{j-1}
{ (q_{j-2} + n_{j} q_{j-1})a - (p_{j-2} + n_{j} p_{j-1})b }
=
(-1)^{j-1}
{ q_{j} a - p_{j} b }
.
Corollary 3. The greatest common divisor d of a and b is given by the formula
(17) d = (-1)^{m} (q_{m-1}a - p_{m-1} b) ,
where m is the number of divisions required to obtain zero remainder in the Euclidean algorithm.
Proof. One knows that d is the last non-zero remainder r_{m-1} in the Euclidean algorithm. This formula for d is the case j = m-1 in (16).
Corollary 4.
(18) p_{m} = {a}/{d} , q_{m} = {b}/{d} .
Proof. The last remainder r_{m} = 0. The case j = m in (16) shows that a/b = p_{m}/q_{m}. Since, by the first proposition of the preceding section, p_{m} and q_{m} have no common factor, this corollary is evident.
### 8. The action of GL_{2}(Z) on the projective line
If a, b, c, d are real numbers with ad - bc ≠ 0 and
M =
(
a b c d
)
is the matrix with entries a, b, c, and d, then M · z, for z real, will denote the expression
(19) M · z = {a z + b}/{c z + d} .
One calls M · z the action of M on z.
M · z is a perfectly good function of z except for the case z = - d/c where the denominator cz + d vanishes. If it were also true that az + b = 0 for the same z, then one would have - b/a = - d/c, in contradiction of the assumption ad - bc ≠ 0. Thus, when z = - d/c, the value of |M · w| increases beyond all bounds as w approaches z, and it is convenient to say that
M ·
( - {d}/{c} )
= INFTY
where INFTY is regarded as large and signless. If further it is agreed to define
M · INFTY = {a}/{c} ,
which is the limiting value of M · w as |w| increases without bound, then one may regard the expression M · z as being defined always for all real z and for INFTY. The set consisting of all real numbers and also the object (not a number) INFTY is called the projective line. The projective line is therefore the union of the (ordinary) affine line with a single point INFTY.
Proposition 4. If [n_{1}, n_{2}, …] is any continued fraction, then
(20) [n_{1}, n_{2}, …, n_{r}, n_{r+1}, …] = M · [n_{r+1}, …] .
where
M =
(
n_{1} 1 1 0
)
…
(
n_{r} 1 1 0
)
.
Proof. Let z = [n_{r+1}, …]. Then
[n_{1}, n_{2}, …, n_{r}, n_{r+1}, … ] = [n_{1}, n_{2}, …, n_{r}, z ] .
The statement of the proposition now becomes
[n_{1}, n_{2}, …, n_{r}, z ] = M · z .
This may be seen to follow by multiplying both sides in formula (9), after replacing t_{j} with n_{j}, by the column
(
z 1
)
.
The matrix M in the preceding proposition is an integer matrix with determinant ± 1. The notation GL_{2}(Z) denotes the set of all such matrices. (The 2 indicates the size of the matrices, and the Z indicates that the entries in such matrices are numbers in the set Z of integers.) It is easy to check that the product of two members of GL_{2}(Z) is a member of GL_{2}(Z) and that the matrix inverse of a member of GL_{2}(Z) is a member of GL_{2}(Z). Thus, GL_{2}(Z) forms what is called a group. The formula (19) defines what is called the action of GL_{2}(Z) on the projective line.
One says that two points z and w of the projective line are rationally equivalent if there is a matrix M in GL_{2}(Z) for which w = M · z. Since (i) GL_{2}(Z) is a group, (ii) M_{1} · (M_{2} · z) = (M_{1} M_{2}) · z, and (iii) w = M · z if and only z = M^{-1} · w, it is easy to see that every point of the projective line belongs to one and only one rational equivalence class and that two points rationally equivalent to a third must be rationally equivalent to each other.
Terminology. The rational equivalence of points on the projective line is said to be the equivalence relation on the projective line defined by the action of GL_{2}(Z).
Example 1. The set of real numbers rationally equivalent to the point INFTY is precisely the set of rational numbers.
Example 2. The proposition above shows that any continued fraction is rationally equivalent to each of its tails. It follows that all tails of a continued fraction are rationally equivalent to each other.
### 9. Periodic continued fractions
In one of the first examples of a continued fraction expansion, it was shown that SQRT{10} = [3,6,6,6,…]. This is an example of a periodic continued fraction. After a finite number of terms the sequence of integers repeats cyclically. If a cyclic pattern is present from the very first term, then the continued fraction is called purely periodic. Evidently, [6,6,6,…] = SQRT{10}-3 is an example of a purely periodic continued fraction.
Note that a periodic continued fraction cannot represent a rational number since the continued fraction expansion of a rational number is finite.
Theorem 6. Every periodic continued fraction is the continued fraction expansion of a real quadratic irrational number.
Proof. For clarity: it is being asserted that every periodic continued fraction represent a number of the form
{a + bSQRT{m}}/{c}
where a, b, c, and m are all integers with m > 0, c ≠ 0, and m not a perfect square.
Numbers of this form with fixed m but varying integers a, b, and c ≠ 0 may be added, subtracted, multiplied, and divided without leaving the class of such numbers. (The statement here about division becomes clear if one remembers always to rationalize denominators.) Consequently, for M in GL_{2}(Z) the number M · z will be a number of this form or INFTY if and only if z is in the same class.
Since a periodic continued fraction is rationally equivalent to a purely periodic continued fraction, the question of whether any periodic continued fraction is a quadratic irrationality reduces to the question of whether a purely periodic continued fraction is such. Let
x = [n_{1}, …, n_{r}, n_{1}, …, n_{r}, n_{1}, …, n_{r}, …]
be a purely periodic continued fraction. By the proposition of the preceding section, x = M · x where M is notationally identical to the M in (20). Ignoring the computation (9) of M in terms of convergents, let
M =
(
a b c d
)
.
Then
x = {ax + b}/{cx + d} ,
or, otherwise said, x is a solution of the quadratic equation
cx^{2} - (a-d) x - b = 0 .
Remark 2. It is conversely true that the continued fraction expansion of every real quadratic irrationality is periodic.
This converse will not be proved here.
### REFERENCES
[1] G. Chrystal, Algebra: An Elementary Textbook (2 vols.), Chelsea.
[2] G. Hardy & E. Wright, An Introduction to the Theory of Numbers, Oxford Univ. Press.
[3] S. Lang, Introduction to Diophantine Approximations, Addison-Wesley.
[4] O. Perron, Die Lehre von den Kettenbrüchen, 2nd ed., Chelsea. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9216194748878479, "perplexity": 323.17320698474947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153585.76/warc/CC-MAIN-20160205193913-00337-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathhelpforum.com/math-topics/117091-angular-velocity.html | 1. ## angular velocity
can someone help me with this problem? I have no idea what to do. Thanks a lot!
A mass of 5 kg is resting on a frictionless ramp. It is connected to a cable that passes over a pulley of radius 2 cm which is then attached to a second hanging mass. The system is moving at a constant velocity.
a) Calculate the mass of the hanging box.
b) If the box move 1.5 m in 3 secs, what is the angular velocity of the pulley?
2. Originally Posted by jsu03
can someone help me with this problem? I have no idea what to do. Thanks a lot!
A mass of 5 kg is resting on a frictionless ramp. It is connected to a cable that passes over a pulley of radius 2 cm which is then attached to a second hanging mass. The system is moving at a constant velocity.
a) Calculate the mass of the hanging box.
b) If the box move 1.5 m in 3 secs, what is the angular velocity of the pulley?
elevation angle of the incline?
mass distribution of the pulley? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9583122730255127, "perplexity": 269.1077562845224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170613.8/warc/CC-MAIN-20170219104610-00140-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/495/ | ## Results (1-50 of 154 matches)
Label Dim $A$ Field CM RM Traces Fricke sign $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
495.1.h.a $1$ $0.247$ $$\Q$$ $$\Q(\sqrt{-11})$$, $$\Q(\sqrt{-55})$$ $$\Q(\sqrt{5})$$ $$0$$ $$0$$ $$1$$ $$0$$ $$q-q^{4}+q^{5}+q^{11}+q^{16}-q^{20}+q^{25}+\cdots$$
495.1.h.b $2$ $0.247$ $$\Q(\sqrt{2})$$ $$\Q(\sqrt{-55})$$ None $$0$$ $$0$$ $$-2$$ $$0$$ $$q-\beta q^{2}+q^{4}-q^{5}-\beta q^{7}+\beta q^{10}+\cdots$$
495.1.h.c $2$ $0.247$ $$\Q(\sqrt{2})$$ $$\Q(\sqrt{-55})$$ None $$0$$ $$0$$ $$2$$ $$0$$ $$q-\beta q^{2}+q^{4}+q^{5}+\beta q^{7}-\beta q^{10}+\cdots$$
495.1.o.a $2$ $0.247$ $$\Q(\sqrt{-3})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$-1$$ $$1$$ $$0$$ $$q-\zeta_{6}q^{3}+\zeta_{6}q^{4}-\zeta_{6}^{2}q^{5}+\zeta_{6}^{2}q^{9}+\cdots$$
495.1.o.b $2$ $0.247$ $$\Q(\sqrt{-3})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$1$$ $$-2$$ $$0$$ $$q+\zeta_{6}q^{3}+\zeta_{6}q^{4}-q^{5}+\zeta_{6}^{2}q^{9}-\zeta_{6}^{2}q^{11}+\cdots$$
495.1.bd.a $4$ $0.247$ $$\Q(\zeta_{12})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\zeta_{12}q^{3}-\zeta_{12}q^{4}-\zeta_{12}^{3}q^{5}+\zeta_{12}^{2}q^{9}+\cdots$$
495.1.bd.b $4$ $0.247$ $$\Q(\zeta_{12})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$2$$ $$0$$ $$0$$ $$q-\zeta_{12}^{4}q^{3}-\zeta_{12}q^{4}-\zeta_{12}^{5}q^{5}-\zeta_{12}^{2}q^{9}+\cdots$$
495.2.a.a $1$ $3.953$ $$\Q$$ None None $$-1$$ $$0$$ $$-1$$ $$0$$ $+$ $$q-q^{2}-q^{4}-q^{5}+3q^{8}+q^{10}+q^{11}+\cdots$$
495.2.a.b $2$ $3.953$ $$\Q(\sqrt{2})$$ None None $$-2$$ $$0$$ $$2$$ $$-4$$ $+$ $$q+(-1+\beta )q^{2}+(1-2\beta )q^{4}+q^{5}-2q^{7}+\cdots$$
495.2.a.c $2$ $3.953$ $$\Q(\sqrt{3})$$ None None $$0$$ $$0$$ $$2$$ $$4$$ $-$ $$q+\beta q^{2}+q^{4}+q^{5}+2q^{7}-\beta q^{8}+\beta q^{10}+\cdots$$
495.2.a.d $2$ $3.953$ $$\Q(\sqrt{2})$$ None None $$2$$ $$0$$ $$2$$ $$-4$$ $-$ $$q+(1+\beta )q^{2}+(1+2\beta )q^{4}+q^{5}+(-2+\cdots)q^{7}+\cdots$$
495.2.a.e $3$ $3.953$ 3.3.148.1 None None $$1$$ $$0$$ $$-3$$ $$0$$ $-$ $$q+\beta _{1}q^{2}+(1+\beta _{1}+\beta _{2})q^{4}-q^{5}+(\beta _{1}+\cdots)q^{7}+\cdots$$
495.2.a.f $4$ $3.953$ 4.4.48704.2 None None $$-2$$ $$0$$ $$-4$$ $$4$$ $-$ $$q+(-1+\beta _{1})q^{2}+(2+\beta _{2})q^{4}-q^{5}+\cdots$$
495.2.a.g $4$ $3.953$ 4.4.48704.2 None None $$2$$ $$0$$ $$4$$ $$4$$ $-$ $$q+(1-\beta _{1})q^{2}+(2+\beta _{2})q^{4}+q^{5}+(1+\cdots)q^{7}+\cdots$$
495.2.c.a $4$ $3.953$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ None None $$0$$ $$0$$ $$3$$ $$0$$ $$q+(-\beta _{1}-\beta _{2}+\beta _{3})q^{2}+(-1+\beta _{1}+\cdots)q^{4}+\cdots$$
495.2.c.b $4$ $3.953$ $$\Q(\sqrt{-2}, \sqrt{3})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+\beta _{2}q^{4}+(\beta _{1}+\beta _{2}-\beta _{3})q^{5}+\cdots$$
495.2.c.c $4$ $3.953$ $$\Q(\sqrt{-2}, \sqrt{3})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+\beta _{2}q^{4}+(\beta _{1}-\beta _{2}-\beta _{3})q^{5}+\cdots$$
495.2.c.d $6$ $3.953$ 6.0.350464.1 None None $$0$$ $$0$$ $$-2$$ $$0$$ $$q+(-\beta _{1}-\beta _{3}-\beta _{5})q^{2}+(-2+\beta _{1}+\cdots)q^{4}+\cdots$$
495.2.c.e $6$ $3.953$ 6.0.350464.1 None None $$0$$ $$0$$ $$-2$$ $$0$$ $$q-\beta _{4}q^{2}+\beta _{2}q^{4}+(-\beta _{1}-\beta _{3})q^{5}+\cdots$$
495.2.d.a $4$ $3.953$ $$\Q(\zeta_{8})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\zeta_{8}^{2}q^{2}+q^{4}+(2\zeta_{8}-\zeta_{8}^{3})q^{5}+(\zeta_{8}+\cdots)q^{7}+\cdots$$
495.2.d.b $4$ $3.953$ $$\Q(\zeta_{8})$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\zeta_{8}^{2}q^{2}+q^{4}+(-2\zeta_{8}+\zeta_{8}^{3})q^{5}+\cdots$$
495.2.d.c $16$ $3.953$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ $$\Q(\sqrt{-55})$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{13}q^{2}+(-2-\beta _{3})q^{4}+\beta _{7}q^{5}+\cdots$$
495.2.f.a $16$ $3.953$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{2}q^{2}+(1+\beta _{1}+\beta _{5})q^{4}+\beta _{7}q^{5}+\cdots$$
495.2.i.a $2$ $3.953$ $$\Q(\sqrt{-3})$$ None None $$0$$ $$0$$ $$-1$$ $$-2$$ $$q+(-1+2\zeta_{6})q^{3}+2\zeta_{6}q^{4}-\zeta_{6}q^{5}+\cdots$$
495.2.i.b $2$ $3.953$ $$\Q(\sqrt{-3})$$ None None $$0$$ $$0$$ $$1$$ $$-2$$ $$q+(-1+2\zeta_{6})q^{3}+2\zeta_{6}q^{4}+\zeta_{6}q^{5}+\cdots$$
495.2.i.c $16$ $3.953$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$-3$$ $$0$$ $$-8$$ $$-1$$ $$q+(\beta _{2}-\beta _{4})q^{2}-\beta _{1}q^{3}+(-1-\beta _{2}+\cdots)q^{4}+\cdots$$
495.2.i.d $18$ $3.953$ $$\mathbb{Q}[x]/(x^{18} - \cdots)$$ None None $$3$$ $$0$$ $$-9$$ $$5$$ $$q+\beta _{1}q^{2}-\beta _{10}q^{3}+(\beta _{1}+\beta _{3}+\beta _{5}+\cdots)q^{4}+\cdots$$
495.2.i.e $20$ $3.953$ $$\mathbb{Q}[x]/(x^{20} - \cdots)$$ None None $$-3$$ $$0$$ $$10$$ $$7$$ $$q+(\beta _{10}+\beta _{14})q^{2}-\beta _{3}q^{3}+(-2-2\beta _{4}+\cdots)q^{4}+\cdots$$
495.2.i.f $22$ $3.953$ None None $$3$$ $$0$$ $$11$$ $$-3$$
495.2.k.a $4$ $3.953$ $$\Q(i, \sqrt{11})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$0$$ $$0$$ $$0$$ $$q-2\beta _{2}q^{4}+(-\beta _{1}-\beta _{2})q^{5}+(-2\beta _{1}+\cdots)q^{11}+\cdots$$
495.2.k.b $4$ $3.953$ $$\Q(i, \sqrt{10})$$ None None $$0$$ $$0$$ $$8$$ $$0$$ $$q+\beta _{1}q^{2}+3\beta _{2}q^{4}+(2+\beta _{2})q^{5}+\beta _{3}q^{8}+\cdots$$
495.2.k.c $24$ $3.953$ None None $$0$$ $$0$$ $$-8$$ $$0$$
495.2.k.d $24$ $3.953$ None None $$0$$ $$0$$ $$0$$ $$0$$
495.2.l.a $20$ $3.953$ $$\mathbb{Q}[x]/(x^{20} + \cdots)$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q+\beta _{4}q^{2}+(\beta _{12}+\beta _{14})q^{4}-\beta _{9}q^{5}+\cdots$$
495.2.l.b $20$ $3.953$ $$\mathbb{Q}[x]/(x^{20} + \cdots)$$ None None $$0$$ $$0$$ $$0$$ $$0$$ $$q-\beta _{4}q^{2}+(\beta _{12}+\beta _{14})q^{4}+\beta _{9}q^{5}+\cdots$$
495.2.n.a $8$ $3.953$ 8.0.13140625.1 None None $$-4$$ $$0$$ $$2$$ $$3$$ $$q+(-1-\beta _{1}-\beta _{2}+\beta _{3}-\beta _{4}+\beta _{5}+\cdots)q^{2}+\cdots$$
495.2.n.b $8$ $3.953$ 8.0.819390625.1 None None $$-2$$ $$0$$ $$-2$$ $$9$$ $$q-\beta _{4}q^{2}+(1-\beta _{1}+2\beta _{2}+\beta _{3}+\beta _{4}+\cdots)q^{4}+\cdots$$
495.2.n.c $8$ $3.953$ $$\Q(\zeta_{15})$$ None None $$-2$$ $$0$$ $$2$$ $$-5$$ $$q+(1-2\zeta_{15}+\zeta_{15}^{3}-\zeta_{15}^{4}+\zeta_{15}^{5}+\cdots)q^{2}+\cdots$$
495.2.n.d $8$ $3.953$ 8.0.13140625.1 None None $$0$$ $$0$$ $$-2$$ $$1$$ $$q+(-1+\beta _{3}-\beta _{4}+\beta _{5}+\beta _{6})q^{2}+(-2\beta _{2}+\cdots)q^{4}+\cdots$$
495.2.n.e $8$ $3.953$ 8.0.13140625.1 None None $$2$$ $$0$$ $$-2$$ $$-1$$ $$q+\beta _{6}q^{2}+(-\beta _{1}-\beta _{2}+\beta _{6}-\beta _{7})q^{4}+\cdots$$
495.2.n.f $8$ $3.953$ 8.0.159390625.1 None None $$4$$ $$0$$ $$2$$ $$-3$$ $$q+(\beta _{1}+\beta _{2}-\beta _{4})q^{2}+(-2+\beta _{2}+\beta _{3}+\cdots)q^{4}+\cdots$$
495.2.n.g $16$ $3.953$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$-2$$ $$0$$ $$-4$$ $$-4$$ $$q+(1-\beta _{1}+\beta _{3}+\beta _{5}-\beta _{10}-\beta _{14}+\cdots)q^{2}+\cdots$$
495.2.n.h $16$ $3.953$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None None $$2$$ $$0$$ $$4$$ $$-4$$ $$q+(-1+\beta _{1}-\beta _{3}-\beta _{5}+\beta _{10}+\beta _{14}+\cdots)q^{2}+\cdots$$
495.2.p.a $96$ $3.953$ None None $$0$$ $$4$$ $$0$$ $$0$$
495.2.r.a $4$ $3.953$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$-1$$ $$3$$ $$0$$ $$q-\beta _{1}q^{3}+(-2+2\beta _{2})q^{4}+(2\beta _{2}+\beta _{3})q^{5}+\cdots$$
495.2.r.b $4$ $3.953$ $$\Q(\sqrt{-3}, \sqrt{-11})$$ $$\Q(\sqrt{-11})$$ None $$0$$ $$1$$ $$6$$ $$0$$ $$q+\beta _{1}q^{3}+(-2+2\beta _{2})q^{4}+(1+\beta _{1}+\cdots)q^{5}+\cdots$$
495.2.r.c $128$ $3.953$ None None $$0$$ $$0$$ $$-12$$ $$0$$
495.2.u.a $52$ $3.953$ None None $$0$$ $$0$$ $$1$$ $$0$$
495.2.u.b $68$ $3.953$ None None $$0$$ $$0$$ $$-2$$ $$0$$
495.2.x.a $64$ $3.953$ None None $$0$$ $$0$$ $$0$$ $$0$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9529215693473816, "perplexity": 1174.335380027528}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00564.warc.gz"} |
https://www.greenemath.com/Algebra1/18/ParallelLinesLesson.html | Lesson Objectives
• Demonstrate an understanding of slope-intercept form
• Demonstrate an understanding of slope
• Learn how to determine if two lines are parallel
• Learn how to determine if two lines are perpendicular
## How to Determine if two Lines are Parallel Lines or Perpendicular Lines
In this lesson, we will learn how to determine if two lines are parallel lines or perpendicular lines.
### Parallel Lines
Parallel lines are any two lines on a plane that will never intersect. We can determine if two lines are parallel by examining the slope of each. Two non-vertical parallel lines have slopes that are equal. We specified non-vertical here since vertical lines have an undefined slope. Let’s look at an example of parallel lines. Suppose we encounter the following two equations:
-2x + y = 5
4x - 2y = 6
If we solve each for y:
y = 2x + 5
y = 2x - 3
In each case, we can see that the slope is the same (2). The y-intercepts are different (0,5) and (0,-3). Since each line has the same slope or steepness, they will never touch each other. Let's look at a graph for further illustration: We can see from our graph that these two lines will never intersect.
### Perpendicular Lines
Perpendicular Lines are lines that intersect at a 90° angle. Two non-vertical perpendicular lines have slopes whose product is -1. Let's look at an example of perpendicular lines. Suppose we encounter the following two equations:
3x + 2y = 4
2x - 3y = 3
If we solve each for y:
y = -3/2 x + 2
y = 2/3 x - 1
The slope of the first equation is -3/2, while the slope of the second equation is 2/3. If we multiply the two slopes together, we get a product of (-1): $$-\frac{3}{2} \cdot \frac{2}{3}$$ $$\require{cancel}-\frac{\cancel{3}}{\cancel{2}} \cdot \frac{\cancel{2}}{\cancel{3}} = -1$$ Since our two slopes multiply together to give us a product of (-1), we know our lines are perpendicular. Let's look at a graph for further illustration: We can see from our graph that these two lines intersect at a 90 ° angle. Let's look at a few examples.
Example 1: Determine if each pair of lines are parallel, perpendicular, or neither
6x - 5y = 12
12x - 10y = -15
Solve each for y:
y = 6/5 x - 12/5
y = 6/5 x + 3/2
We can see that each slope of each line is 6/5. This tells us we have parallel lines.
Example 2: Determine if each pair of lines are parallel, perpendicular, or neither
7x - 2y = 5
2x + 7y = 84
If we solve each for y:
y = 7/2 x - 5/2
y = -2/7 x + 12
Our two slopes (7/2) and (-2/7) are not equal. Therefore, we know that we don't have parallel lines. We can multiply the slopes together to determine if we have perpendicular lines. We are looking for a product of (-1): $$\frac{7}{2} \cdot -\frac{2}{7}$$ $$\frac{\cancel{7}}{\cancel{2}} \cdot -\frac{\cancel{2}}{\cancel{7}} = -1$$ We can see that the product of the slopes is (-1). This tells us we have perpendicular lines.
Example 3: Determine if each pair of lines are parallel, perpendicular, or neither
-8x - 3y = 12
-5x + y = 20
If we solve each for y:
y = -8/3 x - 4
y = 5x + 20
Our two slopes (-8/3) and 5 are not equal. We can multiply the slopes together to determine if we have perpendicular lines. We are looking for a product of (-1): $$-\frac{8}{3} \cdot 5 ≠ -1$$ We can see the product of the slopes is not (-1), therefore, these lines are not perpendicular. We can say these two lines are not parallel lines and they are not perpendicular lines either. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9362562298774719, "perplexity": 316.71308787875773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00102.warc.gz"} |
http://www.oalib.com/search?kw=Xiao-Ming%20Li&searchField=authors | Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
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生态学报 , 2003, Abstract: 塔克拉玛干沙漠南缘 (以下简称塔南 )是我国土地沙漠化最严重的地区之一 ,其主要原因是因为塔南绿洲边缘的自然植被遭到严重破坏。自然植被的恢复有赖于对主要植被类型生物学特性的了解 ,于 1 998～2 0 0 1年在位于塔南策勒绿洲边缘进行的中国 -欧盟合作研究项目 ( ERBIC1 8CT980 2 75 ) ,目的是探索塔南绿洲边缘自然植被可持续管理的生态学基础。通过对几种优势植物叶片 (或同化枝 )水分关系、气体交换、以及群落特性等方面的综合研究 ,得出了以下结论 :( 1 )塔南绿洲边缘的自然植被皆为隐域性植被 ,主要优势种为 :胡杨、柽柳、骆驼刺以及少量的灰杨 ,这种植物是依靠河流的地表水而发生 ,依靠河流地下水补给而生存 ;( 2 )由于缺乏充分的地表水的补给 ,现存的自然植被几乎不能实现自然更新 ;完全破坏后的自然植被已无自然恢复的可能性 ,必须使用人工辅助方法进行恢复 ;( 3)塔南绿洲边缘主要优势植物叶片 (或同化枝 )清晨水势和气体交换研究结果表明 ,依赖地下水生存的几种植物在整个生长季节没有发生严重的水分胁迫 ,保护和恢复绿洲边缘自然植被的必要条件是保持地下水位的相对稳定 ,禁止对地下水的过度开采。
系统科学与数学 , 1991, Abstract: 一个 p 维随机向量 X=(x_1,…,x_p)′(p≥2),如果服从 p 维椭球等高分布,且具有密度函数,则密度函数一定具有如下形式:f(x)=C_p|∑|~(1/2)g((x-u)′∑~(-1)(x-u)).其中,∑>0,g(·)为某个非负可测函数,且 g(·)满足
Acta Crystallographica Section E , 2011, DOI: 10.1107/s1600536811040153 Abstract: In the title compound, {[Ag3(C10H8N2)3](ClO4)3·2H2O}n, one of the AgI ions, one of the 4,4′-bipyridine (bipy) ligands and one of the perchlorate anions are each situated on a twofold rotation axis. Each AgI ion is coordinated by two N atoms from two bridging bipy ligands, forming chains along [101]. π–π interactions between the pyridine rings [centroid–centroid distances = 3.638 (8) and 3.688 (8) ] connect the chains. Intermolecular O—H...O hydrogen bonds link the uncoordinated water molecules and the perchlorate anions.
International Journal of Wireless and Microwave Technologies , 2012, DOI: 10.5815/ijwmt.2012.03.07 Abstract: To arrive at the goal of intensifying the trustworthiness and controllability of distributed systems, the core function of secure algorithms and chips should be fully exerted. Through building the trustworthy model between distributed system and user behaviors, constructing the architecture of trustworthiness distributed systems, intensifying the survivability of services, and strengthening the manageability of distributed systems, the secure problem of distributed systems is to be radically solved. By setting up the trustworthy computing circumstance and supplying the trustworthy validation and the active protection based on identity and behavior for trustworthy distributed system, we will reach the goal of defending the unaware viruses and inbreak. This research insists that the security, controllability, manageability, and survivability should be basic properties of a trustworthy distributed system. The key ideas and techniques involved in these properties are studied, andrecent developments and progresses are surveyed. At the same time, the technical trends and challenges are briefly discussed.
Acta Crystallographica Section E , 2011, DOI: 10.1107/s1600536811035112 Abstract: In the centrosymmetric binuclear title compound, [Cu2(C8H3ClO4)2(C12H12N2)2(H2O)4], the CuII ion is six-coordinated by two N atoms from a 5,5′-dimethyl-2,2′-bipyridine ligand, two bridging O atoms from two 3-chlorobenzene-1,2-dicarboxylate ligands and two water molecules in a distorted octahedral geometry. The binuclear complex molecules are linked together by intermolecular O—H...O hydrogen bonds into a layer parallel to (100). The layers are connected by C—H...Cl hydrogen bonds. Intramolecular O—H...O hydrogen bonds and π–π interactions [centroid–centroid distance = 3.5958 (16) ] are also present.
Discrete Dynamics in Nature and Society , 2006, Abstract: We present sufficient conditions for global asymptotic stability of cascade discrete-time systems. Considering failure of the global asymptotic stability in some cascade systems, we give an estimate of the region of attraction of the systems.
Physics , 2007, DOI: 10.1016/j.nuclphysa.2007.08.009 Abstract: Meson-meson nonresonant reactions governed by the quark-interchange mechanism are studied in a potential that is derived from QCD. S-wave elastic phase shifts for I=2 \pi\pi and I=3/2 K \pi scattering are obtained with wave functions determined by the central spin-independent term of the potential. The reactions include inelastic scatterings of two mesons in the ground-state pseudoscalar octet and the ground-state vector nonet. Cross sections for reactions involving pion, rho, K and K^* indicate that mesonic interactions in matter consisting of only K and K^* can be stronger than mesonic interactions in matter consisting of only pions and rhos and the reaction of I=3/2 \pi K^* \to \rho K is most important among the endothermic nonresonant reactions. By the quark-interchange mechanism we can offer \sqrt s-dependences of phi absorption cross sections in collisions with pion and rho and relevant average cross sections what are very small for the reaction of I=1 \pi \phi \to K^* K^* and remarkably large for the reaction of I=1 \rho \phi \to K^* K^*. It is found from the \sqrt s-dependences of cross sections that rho and K^* creation cross sections can be larger than their absorption cross sections, respectively.
Mathematics , 2011, DOI: 10.1103/PhysRevA.85.032109 Abstract: We revisit the upper bound of quantum discord given by the von Neumann entropy of the measured subsystem. Using the Koashi-Winter relation, we obtain a trade-off between the amount of classical correlation and quantum discord in the tripartite pure states. The difference between the quantum discord and its upper bound is interpreted as a measure on the classical correlative capacity. Further, we give the explicit characterization of the quantum states saturating the upper bound of quantum discord, through the equality condition for the Araki-Lieb inequality. We also demonstrate that the saturating of the upper bound of quantum discord precludes any further correlation between the measured subsystem and the environment.
- , 2016, DOI: https://doi.org/10.1007/s12613-016-1341-x Abstract: To elucidate the intrinsic reaction mechanism of cementitious materials composed of red mud and coal gangue (RGC), the hydration kinetics of these cementitious materials at 20°C was investigated on the basis of the Krstulovi？–Dabi？ model. An isothermal calorimeter was used to characterize the hydration heat evolution. The results show that the hydration of RGC is controlled by the processes of nucleation and crystal growth (NG), interaction at phase boundaries (I), and diffusion (D) in order, and the pozzolanic reactions of slag and compound-activated red mud–coal gangue are mainly controlled by the I process. Slag accelerates the clinker hydration during NG process, whereas the compound-activated red mud–coal gangue retards the hydration of RGC and the time required for I process increases with increasing dosage of red mud–coal gangue in RGC.
Journal of Biomedical Science and Engineering (JBiSE) , 2011, DOI: 10.4236/jbise.2011.42019 Abstract: Many people who live in the low altitude areas are often suffered from hypoxia when they entered the high plateau. This problem may seriously influence the physical and mental state and work efficacy for the travelers and workers. Oxygen enrichment of a small space air at high altitude is considered as a simple way to provide lowlanders enriched oxygen for sleeping and resting, improving work efficiency, so we developed an oxygen concentration machine based on the technology of oxygen enrichment membrane. This paper tested 8 healthy male lowlanders (age 21.63±1.77 yr) who have never exposed to plateau performed an incremental exercise on cycle ergometer at sea-level in order to be used as sea-level controls. Two days later, the same subjects were taken to Lhasa (3700 m) by air and exposed to the plateau, performed the same exercise as they did at sea-level. The next day, all subjects were asked to enter the experimental tent which was enriched with oxygen (higher than 24%) by the oxygen concentration machine and sleep for 10 hours at night, then exposed to plateau and performed the same exercise twice at different time (2 hours and 10 hours after oxygen enrichment). During the tests, subjects must cycled continuously at 60 rpm beginning with a 3 min exercise intensity of 0 W followed by incremental increases of 25 W every 3 min until 150 W, pulse oxygen saturation (SpO2) and heart rate (HR) were recorded. After sleeping in an oxygen enrichment of tent air, 2 hours later, the subjects’ load capacity had no difference compared with control group, but significant difference than before (higher SpO2 and lower HR), which indicated that oxygen concentration machine is effective in increasing the oxygen concentration of the air for the tent and sleeping in the oxygen enrichment tent for l0 h might be effective in improving exercise performance during high-altitude hypoxia. At the same time, 10 hours later, when work-load exceeded 125 W, the same effects were also found. The results indicated the effects of oxygen enrichment of tent air could last a certain period of time.
Page 1 /167547 Display every page 5 10 20 Item | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8134611248970032, "perplexity": 4420.136062561936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00271.warc.gz"} |
http://www.dailygalaxy.com/my_weblog/2013/07/interstellar-space-is-a-weird-quantum-organic-chemistry-lab.html | "Interstellar Space is a Quantum-Ruled Organics Lab"
Follow the Daily Galaxy
## July 12, 2013
### "Interstellar Space is a Quantum-Ruled Organics Lab"
There may be a suite of organic chemical reactions occurring in interstellar space that astronomers haven't considered. In 2012, astronomers discovered methoxy molecules containing carbon, hydrogen and oxygen in the Perseus molecular cloud, around 600 light years from Earth. But researchers were unable to reproduce this molecule in the lab by allowing reactants to condense on dust grains, leaving a mystery as to how it could have formed. The answer was found in Quantum weirdness that can generate a molecule in space that shouldn't exist by the classic rules of chemistry. In short, interstellar space is a kind of quantum chemistry lab, that may create a host of other organic molecules astronomers have discovered in space.
Because of the cold temperatures within the interstellar molecular clouds, reactions with an activation barrier were considered too slow to play an important role for most chemical reactions to occur. The low temperature makes it tough for molecules drifting through space to acquire the energy needed to break their bonds, but some reactions could occur when different molecules stick to the surface of cosmic dust grain. This might give them enough time together to acquire the energy needed to react. "There is a standard law that says as you lower the temperature, the rates of reactions should slow down," says Dwayne Heard of the University of Leeds, UK.
But methoxy could also be created by combining a hydroxyl radical and methanol gas, both present in space through a process called quantum tunnelling that can give the hydroxyl radical a chance to tunnnel through the energy barrier instead of going over it. Heard and colleagues discovered that despite the presence of a barrier, the rate coefficient for the reaction between the hydroxyl radical (OH) and methanol—one of the most abundant organic molecules in space—is almost two orders of magnitude larger at 63 K than previously measured at ∼200 K. At low temperatures, the molecules slow down, increasing the likelihood of tunnelling. "At normal temperatures they just collide off each other, but when you go down in temperature they hang out together long enough," says Heard.
The team also observed the formation of the methoxy radical molecule, created by the formation of a hydrogen-bonded complex that is sufficiently long-lived to undergo quantum-mechanical tunnelling. They concluded that this tunnelling mechanism for the oxidation of organic molecules by OH is widespread in low-temperature interstellar environments. The reaction occurred 50 times faster via quantum tunnelling than if it occurred normally at room temperature by hurdling the energy barrier. Empty space is much colder than 63 kelvin, but dust clouds near stars can reach this temperature, added Heard.
"We're showing there is organic chemistry in space of the type of reactions where it was assumed these just wouldn't happen," says Heard.
The image at the top of the page shows the Perseus Molecular Cloud At microwave wavelengths, taken by the Planck Space Craft which sees electons moving through the Milky Way, and dust being warmed by starlight from stars forming within. These components of the interstellar medium have studied at length over several decades. The electrons are known to emit primarily at radio waves (low frequencies), while the dust grains primarily in the far-infrared (high frequencies).
In the 1990s, emission was observed which couldn't be explained by either, and became known as "Anomalous Microwave Emission". Several theories of the origin of this emission have been proposed, and now the wavelength coverage of Planck's Low Frequency Instrument is ideal for observing and characterising it.
An advantage that Planck has is that the combination of the two instruments give a much broader wavelength coverage, which allows the separation of this anomalous emission from the better understood components.
“We are now becoming rather confident that the emission is due to nano-scale spinning grains of dust, which rotate up to ten thousand million times per second,” says Clive Dickinson from the University of Manchester, who led an analysis of the AME using Planck's maps. “These are the smallest dust grains known, comprising only 10 to 50 atoms; spun up by collisions with atoms or photons, they emit radiation at frequencies between 10 and 60 GHz,” he explains.
This region in the constellation of Perseus shown was one of two regions within our Galaxy studied in detail. Thanks to Planck's high sensitivity and to its unprecedented spectral coverage, it has been possible to characterise the anomalous emission arising from these two objects in such great detail that many of the alternative theories could be discarded, and to show that at least a significant contribution to the AME, if not the only one, is due to nano-scale spinning dust grains.
Journal reference: Nature Chemistry, DOI: 10.1038/NCHEM.1692
The Daily Galaxy via Nature Chemistry, Space.com, and New Scientist
### Comments
So if this is true which i believe it is, then it stands to reason that there must be molecules and elements found and forged only in the vacuum of space.
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https://socratic.org/questions/how-do-you-solve-2log-2x-log-2-5-2-2#252372 | Precalculus
Topics
# How do you solve 2log_2x-log_2 5=2^2?
Apr 10, 2016
$x = 4 \sqrt{5}$
#### Explanation:
As ${\log}_{b} a = \log \frac{a}{\log} b$
$2 {\log}_{2} x - {\log}_{2} 5 = {2}^{2}$ can be written as
$2 \log \frac{x}{\log} 2 - \log \frac{5}{\log} 2 = 4$ and
multiplying each side by $\log 2$ as $\log 2 \ne 0$, we get
$2 \log x - \log 5 = 4 \log 2$ or
${x}^{2} / 5 = {2}^{4}$ or ${x}^{2} = 80$ or
$x = \sqrt{80} = 4 \sqrt{5}$ (as $x$ cannot be negative)
Apr 10, 2016
$\textcolor{b l u e}{\implies x = 4 \sqrt{5}}$
#### Explanation:
Example: I have chosen to use log to base 10 so that you can check it on your calculator if you so wish.
suppose we had $L o {g}_{10} \left(z\right) = 2$
This means$\to {10}^{2} = z$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the value of } x}$
Given: $\text{ } 2 {\log}_{2} x - {\log}_{2} 5 = {2}^{2}$
Write as: ${\log}_{2} \left({x}^{2}\right) - {\log}_{2} \left(5\right) = 4$
Subtraction of logs means that the source value are applying division
$\implies {\log}_{2} \left({x}^{2} / 5\right) = 4$
Using the principle demonstrated in my example
$\implies {\log}_{2} \left({x}^{2} / 5\right) = 4 \text{ "->" } {2}^{4} = {x}^{2} / 5$
$\implies x = \sqrt{80} \text{ "=" } \sqrt{{2}^{2} \times {2}^{2} \times 5}$
$\textcolor{b l u e}{\implies x = 4 \sqrt{5}}$
##### Impact of this question
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https://revision.my/courses/spm-chemistry-kssm-paper-1/lessons/3-3-chemical-formulae/topic/3-3-1-chemical-formulae/?ld-lesson-page=2 | # 3.3.1 Chemical Formulae
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## Chemical Formulae
1. Each element is represented by its own chemical symbol.
2. In an element, the chemical formula represents the atoms in the molecule.
3. The chemical formula tells us:
1. the types of atoms or ions in the compound,
2. the number of atoms or ions in the compound,
4. For example, the chemical formula for ethene is C2H4. This shows that ethene is the result of the combination of the elements carbon and hydrogen, and there are 2 carbon atoms and 4 hydrogen atoms in each molecule of ethene.
5. Generally, chemical formula can be divided into
1. Empirical formula
2. Molecular formula
6. The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.
7. The molecular formula of a substance is the chemical formula that gives the actual number of atoms of each element in the substance.
## Empirical Formula
1. The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.
2. Empirical = information gained by means of observation, experience, or experiment.
Example:
Chemical Substances Molecular Formula Empirical Formula Glucose C6H12O6 CH2O Water H2O H2O Carbon Dioxide CO2 CO2 Benzene C6H6 CH Butane C4H8 CH2
### Finding Empirical Formula
Steps to determine the empirical formula of a compound
STEP 1: Find the mass
STEP 2: Find the mole
STEP 3: Find the simplest ratio
Example:
In a chemical reaction, 4.23g of iron reacts completely with 1.80g of oxygen gas, producing iron oxide. Calculate the empirical formula of iron oxide. [Relative atomic mass: Iron = 56; Oxygen = 16]
Element Fe O Mass 4.23g 1.80g Number of mole 4.23/56 =0.0755 1.80/16 =0.1125 Simple ratio 0.0755/0.0755 =1 0.1125/0.0755 =1.5 Ratio in round number 2 3
The empirical formula of iron oxide = Fe2O3
Example:
Determine the empirical formula of a compound which has a percentage of composition Mg: 20.2%, S: 26.6%, O: 53.2%. [Relative atomic mass: Mg = 24; S = 32; O = 16]
Element Mg S O Percentage 20.2% 26.6% 53.2% Mass in 100g 20.2g 26.6g 53.2g Number of mole 20.2/24 =0.8417mol 26.6/32 =0.8313mol 53.2/16 =3.325mol Simple ratio 0.8417/0.8313 =1 0.8313/0.8313 =1 3.325/0.8313 =4
The empirical formula of the compound is MgSO4
Example:
From an experiment, a scientist found that a hydrocarbon contains 85.7% of carbon according to its mass. Find the empirical formula of the hydrocarbon. [Relative atomic mass: Carbon = 12; Hydrogen = 1]
Element C H Percentage 85.7% 14.3% Mass in 100g 85.7g 14.3g Number of mole 85.7/12 =7.142mol 14.3/1 =14.3mol Simple ratio 7.142/7.142 =1 14.3/7.142 =2
The empirical formula of the hydrocarbon = CH2
## Molecular Formula
1. The molecular formula of a substance is the chemical formula that gives the actual number of atoms of each element in the substance.
2. A molecular formula is the same as or a multiple of the empirical formula.
3. For example, the empirical of carbon dioxide is CO2 and the molecular formula is also CO2.
4. Whereas, the empirical formula of ethane is CH3 while the molecular formula of ethane is C2H6.
### Finding Molecular Formula
Example
Given that the empirical formula of benzene is CH and its relative molecular mass is 78. Find the molecular formula of benzene. [Relative Atomic Mass: Carbon: 12; Hydrogen: 1]
Let’s say the molecular formula of benzene is CnHn.
The relative molecular mass of CnHn
= n(12) + n(1)
= 13n
13n = 78
n = 78/13 = 6
Therefore, the molecular formula of benzene
C6H6
Example:
What is the mass of metal X that can combine with 14.4g of oxygen to form X oxide with molecular formula X2O3. (RAM: O = 16; X = 56 )
Number of mole of oxygen = 14.4/16 =0.9 mol
From the molecular formula, we learn that the ratio of element X to oxygen X:O = 2:3
Therefore, the number of mole of X = 0.9 × 2 3 =0.6 mol
Number of mole,
n = mass/Molar mass = 0.6
mass
= 0.6 x Molar mass*
= 0.6 x 56
= 33.6g
The mass of element X = 33.6g
*Molar mass of a substance = Relative atomic mass of the substance
## Percentage of Composition of a Compound
1. To find the percentage of composition of a substance means to find the percentage of mass of each element in the molecule of the substance to the mass of the molecule.
2. The percentage of mass of an element can be determined by the following equation:
RAM = Relative atomic mass
RMM = Relative molecular mass
Example
Calculate the percentage of composition of DDT (C14H9Cl5). [Relative atomic mass: Carbon = 14; Hydrogen = 1; Chlorine = 35.5] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8047446012496948, "perplexity": 2778.485117292584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00388.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/Calc3rd/chapter/Ch7/lesson/7.1.5/problem/7-50 | ### Home > CALC3RD > Chapter Ch7 > Lesson 7.1.5 > Problem7-50
7-50.
Use the first and second derivatives to determine the following locations for $f(x) = xe^x$.
1. Relative minima and maxima
Remember: Finding where $f^\prime(x)= 0$ or $f^\prime(x) =$ DNE will identify CANDIDATES for minima and maxima.
You need to complete the 1st or 2nd Derivative Test to confirm which is which.
2. Intervals over which $f$ is increasing and decreasing
When $f^\prime(x)$ is positive, then $f(x)$ has positive slopes, which means $f(x)$ is increasing.
3. Inflection points
Remember: Finding where $f^{\prime\prime}(x)$ or $f^{''}(x)=$ DNE will identify CANDIDATES for inflection points.
You need to do further investigation to determine if it is (or is not) an inflection point.
Either test for a sign change of $f^{\prime\prime}(x)$ before and after the candidate point. Or evaluate $f^{\prime\prime\prime}$ at the candidate.
If $f^{\prime\prime\prime}(\text{candidate})≠ 0$ , then it is a point of inflection.
4. Intervals over which $f$ is concave up and concave down
When $f^{\prime\prime}(x)$ is positive, then $f^\prime(x)$ has positive slopes, which means $f(x)$ is concave up. | {"extraction_info": {"found_math": true, "script_math_tex": 16, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9436760544776917, "perplexity": 750.4363587247492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00669.warc.gz"} |
http://mathoverflow.net/questions/118043/what-can-we-infer-about-the-size-of-a-complete-boolen-algebra-given-it-is-kap | # What can we infer about the size of a complete Boolen algebra, given it is $\kappa$-c.c.?
More specifically, if we only know that a complete Boolean algebra, $\mathbf{B}$, is $\kappa$-c.c., can we give a (reasonably tight) upper bound to the size of $\mathbf{B}$ in terms of $\kappa$?
-
The Cohen algebras of regular open subsets of $2^\lambda$ are ccc, and of arbitrarily large cardinality. – Emil Jeřábek Jan 4 at 12:06
Thanks again! Back to the drawing board I guess. – Zoorado Jan 4 at 12:44
Emil, why not post your comment as an answer? (Unanswered questions periodically get auto-reposted to the main page.) – Joel David Hamkins Jan 4 at 15:09
Alright. I was worried the question was not quite research-level, if even someone as ignorant about forcing as me knows this example. – Emil Jeřábek Jan 7 at 15:53
The $\kappa$-cc condition by itself does not put any bound on the cardinality of the algebra. For example, for each $\lambda$, the Cohen algebra of regular open subsets of $2^\lambda$ is ccc, but it has cardinality $\lambda^\omega$. However, one can bound the size of $B$ using additional cardinal characteristics: for a simple bound, if $B$ is $\kappa$-cc and has a dense subset $P$ of cardinality $\lambda$, then $|B|\le\lambda^{<\kappa}$, as every element of $B$ can be written as the join of an antichain in $P$.
This is just like the forcing notion $\text{Add}(\omega,\lambda)$ to add $\lambda$ many Cohen reals, which is c.c.c. but has size $\lambda^{\lt\omega}$. – Joel David Hamkins Jan 7 at 16:01
Yes, this should be the same thing. When I was learning set theory (from not particularly up-to-date sources), I got the impression that each author has his own totally incompatible notation for basic forcing notions, and therefore I didn’t bother to use any in the answer. Is $\mathrm{Add}(\omega,\lambda)$ the standard notation nowadays? And does it denote the complete Boolean algebra, or some dense subset? – Emil Jeřábek Jan 7 at 16:22
Oh, I think there is no universal standard, but this notation is fairly common (especially here in New York). Sometimes this denotes a dense subset (the usual poset) and sometimes the complete Boolean algebra, but if it ever matters, then authors usually say so. I should have said "size $\lambda^\omega$" in my comment, since I usually think of it as referring to the Boolean completion. – Joel David Hamkins Jan 7 at 17:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754268527030945, "perplexity": 444.2825568099591}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345775028/warc/CC-MAIN-20131218054935-00043-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-geometry/171414-example-intersection-b.html | # Thread: Example of A intersection B
1. ## Example of A intersection B
A=(0,2) B=(1,3)
A and B are open.
The boundary points of A are 0 and 2, neither of which belong to A
The boundary points of B are 1 and 3, neither of which belong to B
A^B = (1,2)
A=(0,2) B=[1,3)
The boundary point 1 of B belongs to B
A^B = [1,2), The interior is (1,2)
The boundary point 1 in B is also a point of A so it belongs to the intersection A^B, where it is also a boundary point because to the right of 1 are points common to A and B and to the left of 1 are only points of A.
A=(0,2] B = [1,3)
A^B = [1,2]. The interior is (1,2)
P is an interior point of A if it has a neighborhood entirely in A .
2. Yes, all of that is true. Do you have a question?
3. Originally Posted by HallsofIvy
Yes, all of that is true. Do you have a question?
I was responding to a question in another thread which was closed before I had a chance to respond.
But I do have a question.
Someone objected to my using "boundary," and I note Rudin dosen't use it , Rosenlicht relegates it to a problem, Shilov uses it in an advanced context at the end of the book, and Taylor uses it.
Rosenlicht and Taylor define it as:
$
\overline{S}\cap\overline{cS}
$
(overline is closure, c is complement)
Which isn't very pleasant.
The question:
How would you rephrase the original post without using the term "boundary."
4. I guess you could write "intersection of the closure with the closure of the complement" instead of "boundary" in every case. But that would be silly. What's bothering you here, the set we call the boundary of S, or the use of the word "boundary" to name it?
5. Originally Posted by Tinyboss
I guess you could write "intersection of the closure with the closure of the complement" instead of "boundary" in every case. But that would be silly. What's bothering you here, the set we call the boundary of S, or the use of the word "boundary" to name it?
Sorry, I was mistaken about the thread being closed, and I probably misunderstood the objection to use of "boundary."
That leaves only my curiousity about Rudin not using "boundary," (it's not in index of '64 edition) which is a question I should really ask Rudin, rather than post it.
6. It's a little late for that, sadly. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.857393741607666, "perplexity": 789.4920511550288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00151-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://arxiv.org/abs/astro-ph/9802217 | astro-ph
(what is this?)
(what is this?)
# Title: White dwarf stars and the Hubble Deep Field
Authors: Steven D. Kawaler (Iowa State University)
Abstract: Although it a very narrow angle survey, the depth of the HDF results in its sampling a significant volume of the halo of our galaxy. Thus it is useful for the purposes of detecting (or placing upper limits on the distribution of) intrinsically faint stars, such as white dwarfs. White dwarfs could provide a significant fraction of the total mass of the halo of the Milky Way. Constraints on the population of halo white dwarfs from the HDF can directly address this possible partial explanation of the nature of the dark halo of the Milky Way. In this review, I illustrate how the HDF can be used to constrain the luminosity function of halo white dwarfs. I begin with a brief summary of the observed white dwarf luminosity function (WDLF) of the galactic disk, and show how the HDF serves as a probe of the WDLF for the halo. I then review the theoretical background used in interpreting the WDLF in terms of the theory of white dwarf evolution and cooling, and the history of star formation in the galaxy. We are then in a position to explore the theoretical WDLF of the disk. and then the halo. The results of searches for white dwarfs on the HDF can then be examined in terms of the halo white dwarf population.
Comments: 20 pages including 8 figures. LaTeX, cupconf.sty style file. Invited review presented at the STScI Symposium on the Hubble Deep Field, May, 1997 Subjects: Astrophysics (astro-ph) Cite as: arXiv:astro-ph/9802217 (or arXiv:astro-ph/9802217v1 for this version)
## Submission history
From: Steven D. Kawaler [view email]
[v1] Mon, 16 Feb 1998 23:33:16 GMT (84kb) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8177619576454163, "perplexity": 1544.7679817466128}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424961.59/warc/CC-MAIN-20170725042318-20170725062318-00400.warc.gz"} |
https://www.physicsforums.com/threads/the-equations-of-rotational-kinematics.264143/ | # The Equations of Rotational Kinematics
1. Oct 13, 2008
### pstfleur
1. A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.00rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -25.0rad/s. While this change occurs, the angular displacement of the wheel is zero. Find the time required for the change in the angular velocity to occur.
2. W=Wo+xt
Displacement=1/2(W+Wo)t
DIsplacement=Wot+1/2xt^2
3. The answer is suppose to be 12.5s..
OK the initial velocity is what I don't have, i know that its going to be negative.. In order for me to get the Initial angular velocity, i need the Time, which I also don't have.. I don't know how to answer with two unknowns.
2. Oct 13, 2008
### pstfleur
hello lol
3. Oct 14, 2008
### pstfleur
bump bump bump
4. Oct 14, 2008
### pstfleur
bumping again for the 2nd day :)
Similar Discussions: The Equations of Rotational Kinematics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9765815734863281, "perplexity": 869.9469671909144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00095.warc.gz"} |
https://infoscience.epfl.ch/record/277970?ln=en | ## Reproducibility warning: The curious case of polyethylene glycol 6000 and spheroid cell culture
Reproducibility of results is essential for a well-designed and conducted experiment. Several reasons may originate failure in reproducing data, such as selective reporting, low statistical power, or poor analysis. In this study, we used PEG6000 samples from different distributors and tested their capability inducing spheroid formation upon surface coating. MALDI-MS, NMR, FTIR, and Triple SEC analysis of the different PEG60000s showed nearly identical physicochemical properties different, with only minor differences in mass and hydrodynamic radius, and AFM analysis showed no significant differences in the surface coatings obtained with the available PEG6000s. Despite these similarities, just one showed a highly reproducible formation of spheroids with different cell lines, such as HT-29, HeLa, Caco2, and PANC-1. Using the peculiar PEG6000 sample and a reference PEG6000 chosen amongst the others as control, we tested the effect of the cell/PEG interaction by incubating cells in the PEG solution prior to cell plating. These experiments indicate that the spheroid formation is due to direct interaction of the polymer with the cells rather than by interaction of cells with the coated surfaces. The experiments point out that for biological entities, such as cells or tissues, even very small differences in impurities or minimal variations in the starting product can have a very strong impact on the reproducibility of data.
Published in:
Plos One, 15, 3, e0224002
Year:
Mar 19 2020
ISSN:
1932-6203
Keywords:
Note:
This is an open access article distributed under the terms of the Creative Commons Attribution License.
Laboratories:
Note: The status of this file is: Anyone | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8829591274261475, "perplexity": 3172.073575674737}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00358.warc.gz"} |
https://cs.stackexchange.com/questions/77490/amortized-time-of-insertion-into-an-array-list | # Amortized time of insertion into an Array list
According to Amortized time cost of insertion into an Array list,
A dynamically resizing array list will resize when the number of elements reaches a power of two. So, after n elements inserted, we've resized at sizes 1, 2, 4, ... , n.
Can anyone kindly explain me the logic of this? If we consider the n(number of elements)=7 then we resize array at sizes 1,2,4 but according to above statement "1,2,4....7".
The question might be a little misleading.
So, after n elements inserted, we've resized at sizes 1, 2, 4, ... , n.
This is only true if $n = 2^k$ for some $k \in \mathbb{N}$, because for a growth factor of 2, we only need to resize when $n$ is a power of 2.
So in your case we would only resize at 1, 2, 4. Not 7.
You should keep in mind this only applies to a growth factor of 2. We could change the growth factor however we wish. For example we set the growth factor to 3. Then we resize at powers of 3; $\{1, 3, 9, \dots 3^k\}$. You can still use the same analysis procedure to get a constant amortized time though.
• Some list implementations use the golden ratio as growth rate – matheussilvapb Oct 31 '17 at 11:07
• @matheussilvapb, yes that could also be the case. You must then either assume the lower or upper bound $\lfloor \phi^k \rceil$. – ryan Nov 1 '17 at 0:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8144246935844421, "perplexity": 567.9812714130478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00038.warc.gz"} |
http://dynamicsystems.asmedigitalcollection.asme.org/article.aspx?articleid=1478020 | 0
Research Papers
Estimation of State Transition Probabilities in Asynchronous Vector Markov Processes
[+] Author and Article Information
Waleed A. Farahat
Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, MA [email protected]
Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, MA [email protected]
J. Dyn. Sys., Meas., Control 134(6), 061003 (Sep 13, 2012) (14 pages) doi:10.1115/1.4006087 History: Received February 25, 2010; Accepted November 27, 2011; Published September 13, 2012
Abstract
Vector Markov processes (also known as population Markov processes) are an important class of stochastic processes that have been used to model a wide range of technological, biological, and socioeconomic systems. The dynamics of vector Markov processes are fully characterized, in a stochastic sense, by the state transition probability matrix P . In most applications, P has to be estimated based on either incomplete or aggregated process observations. Here, in contrast to established methods for estimation given aggregate data, we develop Bayesian formulations for estimating P from asynchronous aggregate (longitudinal) observations of the population dynamics. Such observations are common, for example, in the study of aggregate biological cell population dynamics via flow cytometry. We derive the Bayesian formulation, and show that computing estimates via exact marginalization are, generally, computationally expensive. Consequently, we rely on Monte Carlo Markov chain sampling approaches to estimate the posterior distributions efficiently. By explicitly integrating problem constraints in these sampling schemes, significant efficiencies are attained. We illustrate the algorithm via simulation examples and show that the Bayesian estimation schemes can attain significant advantages over point estimates schemes such as maximum likelihood.
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Figures
Figure 4
Example for constrained low dimensional Gaussian sampling. The high dimensional space (f -space) is two dimensional. The two-dimensional Gaussian distribution is illustrated by the unimodal surface.The constraint is shown by the vertical upright plane, and cuts through the two-dimensional Gaussian, revealing a lower dimensional Gaussian (1D). This lower dimensional Gaussian is sampled as shown by the black asterisks.
Figure 1
A vector Markov process consists of multiple agents or cells, each exhibiting Markovian dynamics
Figure 2
Illustration of panel data S and longitudinal data Θ for a two-state system for both synchronous and asynchronous transitions. (a) and (b): Cell color (gray or white) represents the state of each cell (state 1 or state 2) as it evolves over time. (c) and (d): Bar heights represent the total number of cells (aggregate data) in each state for each time step. The vertical lines represent observation time points.
Figure 3
A Poisson arrival process can be used to model asynchronous state transitions in the population. The plots illustrate the cellular transitions (for a two-state case), the Poisson distributions characterizing the number of transitions in Δt = 1, and the exponential distributions for the interarrival times. The plots are shown for values of λ = 1/3 (first row), λ = 1 (second row), and λ = 3 (third row).
Figure 5
Illustration of Gibbs sampling scheme to efficiently sample from Gaussian distributions subject to inequality constraints. (a) Gaussian distribution contours shown in problem space. The feasible region satisfying all inequality constraints is shaded in gray. Samples are drawn from the distribution irrespective of the inequality constraints (asterisks). Only the samples satisfying the inequality constraints are picked (indicated by circles in addition to asterisks) and retained. (b) Distribution, data, and constraints of subfigure (a) but are shown under the transformation. (c) In the transformed space, samples are drawn from the constrained distribution via Gibbs sampling. All samples satisfy the inequality constraints. (d) Samples are transformed back to problem space, maintaining feasibility.
Figure 6
Estimation from panel data based on uniform (noninformative) priors. Posterior shown on right. Asterisks represent maximum likelihood estimates, whereas the white lines represent the true transition probabilities used to generate the data.
Figure 7
Progression of the estimation algorithm in time as the estimates converge to the true value, represented by the point (1/3, 1/5). The figure illustrates convergence when priors are generic and mutlimodal in character.
Figure 8
Estimation from asynchronous data. Uniform (noninformative) priors shown on top. Posterior shown on bottom. The asterisks represent maximum likelihood estimates, whereas the lines intersecting at (0.33, 0.2) represent the true transition probabilities used to generate the data. The lines intersecting at (0.65, 0.58) represent the transformation of the true transition probabilities to P Δt .
Figure 9
Effects of Monte Carlo sample size on solution parameters for an example four dimensional problem. (a) The variance of error with sample size is estimated from ten repetitions of the solution runs. With increasing number of Gibbs samples, variance approaches 1/N rate of decrease indicated by the dotted line. (b) Computation time, as expected, increases linearly with number of Gibbs samples acquired.
Figure 10
Comparison of error estimates: Bayesian versus maximum likelihood estimates. Simulation parameters are: N = 100 and T = 5. In this particular instance, g(θ (t)) =θ 1 (t)3 · θ2 (t).
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Topic Collections | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8486658334732056, "perplexity": 1432.9558827783371}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00358.warc.gz"} |
http://farside.ph.utexas.edu/teaching/336k/Newton/node43.html | Next: Motion in a General Up: Planetary Motion Previous: Orbital Energies
# Kepler Problem
In a nutshell, the so-called Kepler problem consists of determining the radial and angular coordinates, and , respectively, of an object in a Keplerian orbit about the Sun as a function of time.
Consider an object in a general Keplerian orbit about the Sun which passes through its perihelion point, and , at . It follows from the previous analysis that
(276)
and
(277)
where , , and are the orbital eccentricity, angular momentum per unit mass, and energy per unit mass, respectively. The above equation can be rearranged to give
(278)
Taking the square-root, and integrating, we obtain
(279)
Consider an elliptical orbit characterized by . Let us write
(280)
where is termed the elliptic anomaly. In fact, is an angle which varies between and . Moreover, the perihelion point corresponds to , and the aphelion point to . Now,
(281)
whereas
(282)
Thus, Equation (279) reduces to
(283)
where . This equation can immediately be integrated to give
(284)
where is the orbital period. Equation (284), which is known as Kepler's equation, is a transcendental equation which does not possess a simple analytic solution. Fortunately, it is fairly straightforward to solve numerically. For instance, using an iterative approach, if is the th guess then
(285)
The above iteration scheme converges very rapidly (except in the limit as ).
Equations (276) and (280) can be combined to give
(286)
Thus,
(287)
and
(288)
The previous two equations imply that
(289)
We conclude that, in the case of an elliptical orbit, the solution of the Kepler problem reduces to the solution of the following three equations:
(290) (291) (292)
Here, and . Incidentally, it is clear that if then , and . In other words, the motion is periodic with period .
For the case of a parabolic orbit, characterized by , similar analysis to the above yields:
(293) (294) (295)
Here, is termed the parabolic anomaly, and varies between and , with the perihelion point corresponding to . Note that Equation (293) is a cubic equation, possessing a single real root, which can, in principle, be solved analytically. However, a numerical solution is generally more convenient.
Finally, for the case of a hyperbolic orbit, characterized by , we obtain:
(296) (297) (298)
Here, is termed the hyperbolic anomaly, and varies between and , with the perihelion point corresponding to . Moreover, . As in the elliptical case, Equation (296) is a transcendental equation which is most easily solved numerically.
Next: Motion in a General Up: Planetary Motion Previous: Orbital Energies
Richard Fitzpatrick 2011-03-31 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9791579246520996, "perplexity": 697.8985025432848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999273.24/warc/CC-MAIN-20190620190041-20190620212041-00434.warc.gz"} |
http://blog.geomblog.org/2004/06/lower-bounds-comeback.html | ## Tuesday, June 29, 2004
### Lower bounds comeback ?
Maybe I am reading too much into titles, but it seems like hardness results are back in business again. From a highly unprofessional inspection of the FOCS list:
An Optimal Randomised Cell Probe Lower Bound for Approximate Nearest Neighbour Searching
Hardness of Approximating the Shortest Vector Problem in Lattices
Ruling Out PTAS for Graph Min-Bisection, Densest Subgraph and Bipartite Clique
On the (Im)possibility of Cryptography with Imperfect Randomness
The Hardness of Metric Labeling
Hardness of buy-at-bulk network design
Optimal Inapproximability Results for MAX-CUT and Other 2-variable CSPs?
If I get a chance I will comment on some of the interesting geometry papers out there (have been pinging authors for copies). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8548251986503601, "perplexity": 3807.8625480179085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174215.11/warc/CC-MAIN-20170219104614-00238-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/453788/proving-that-n-mid-a-n-with-integers-that-are-relatively-prime | # Proving that $n\mid A_n$ with integers that are relatively prime
Problem: Let $1 \leq b_1 < b_2 < \dots < b_{\phi(n)} < n$ be integers relatively prime with $n$, and $B_n = b_1 b_2 \cdots b_{\phi(n)}$. Consider the sum
$$1/b_1 + 1/b_2 + \dots + 1/b_{\phi(n)} = A_n/B_n$$
Prove that $n\mid A_n$. Is it true that $n^2\mid A_n$
I am thinking of rewriting the fractions on the left side to have the same denominator,$B_n$, but I am not sure if that helps me or not?
-
You may need some restrictions, as $2$ does not divide $A_2=1$. If $n>2$ is the only necessary restriction, look at $n=6$. – Aaron Jul 28 '13 at 1:18
Sorry I am a mistake in typing the problem. Do you still think I need the restrictions? – Username Unknown Jul 28 '13 at 1:20
In the calculations below, we assume that $n\gt 2$.
Imagine bringing the terms to a common denominator $B_n$. Then the numerator is a sum of products $b_1b_2 \cdots \hat{b_i} \cdots b_{\varphi(n)}$. Here $\hat{b_i}$ indicates that the term $b_i$ is missing. Let $c_i$ be the number in the interval $1$ to $n$ such that $c_ib_i \equiv 1\pmod{n}$. Then $$b_1b_2 \cdots \hat{b_i} \cdots b_{\varphi(n)}\equiv c_iB_n\pmod{n}.$$ It follows that $$A_n \equiv (c_1+c_2+\cdots +c_{\varphi(n)})B_n\pmod{n}.$$ Now the $c_i$ travel, in some order, through the $b_i$, so their sum is congruent to $b_1+b_2+\cdots+b_{\varphi(n)}$.
The $b_i$ come in pairs with sum $n$, so their sum is congruent to $0$ modulo $n$. This completes the proof.
As to divisibility by $n^2$, there is the easy counterexample $n=4$.
I have this for the case of n=4. $\phi(4)=2$. $1/b_1 + 1/b_2 = A_2/B_2$, which is the same as $b_2 +b_1=A_2$. How am I suppose to the contradiction from here? – Username Unknown Jul 28 '13 at 2:33
We have in this case $b_1=1$ and $b_2=3$. Binging to the common denominator $1\cdot 3$, we find that $A_4=4$ and $B_4=3$. Thus $n$, that is, $4$, divides $A_4$. But $4^2$ certainly does not divide $A_4$. So $n=4$ is a counterexample to the assertion that $n^2$ divides $A_n$. By the way, if $n$ is a prime greater than $3$, then $n^2$ does divide $A_n$. – André Nicolas Jul 28 '13 at 2:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9885863661766052, "perplexity": 104.55222560217808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999657012/warc/CC-MAIN-20140305060737-00050-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://umj.imath.kiev.ua/article/?lang=en&article=5675 | 2018
Том 70
№ 5
Hybrid type generalized multivalued vector complementarity problems
Abstract
We introduce a new type of generalized multivalued vector complementarity problems with moving pointed cone. We discuss the existence results for generalized multivalued vector complementarity problems under inclusive assumptions and obtain results on the equivalence between the generalized multivalued vector complementarity problems and the generalized multivalued vector variational inequality problems.
English version (Springer): Ukrainian Mathematical Journal 65 (2013), no. 1, pp 5-20.
Citation Example: Agarwal R. P., Ahmad M. K., Salahuddin Hybrid type generalized multivalued vector complementarity problems // Ukr. Mat. Zh. - 2013. - 65, № 1. - pp. 7-20.
Full text | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9646773338317871, "perplexity": 3141.199619783758}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864354.27/warc/CC-MAIN-20180622045658-20180622065658-00556.warc.gz"} |
http://harvard.voxcharta.org/tag/white-dwarf/ | # Posts Tagged white dwarf
## Recent Postings from white dwarf
### Time-Variable Aluminum Absorption in the Polar AR Ursae Majoris, and an Updated Estimate for the Mass of the White Dwarf
We present spectra of the extreme polar AR Ursae Majoris (AR UMa) which display a clear Al I absorption doublet, alongside spectra taken less than a year earlier in which that feature is not present. Re-examination of earlier SDSS spectra indicates that the Al I absorption doublet was also present $\approx$8 years before our first non-detection. We conclude that this absorbing material is unlikely to be on the surface of either the white dwarf (WD) or the donor star. We suggest that this Al I absorption feature arises in circumstellar material, perhaps produced by the evaporation of asteroids as they approach the hot WD. The presence of any remaining reservoir of rocky material in AR UMa might help to constrain the prior evolution of this unusual binary system. We also apply spectral decomposition to find the stellar parameters of the M dwarf companion, and attempt to dynamically measure the mass of the WD in AR UMa by considering both the radial velocity curves of the H$_\beta$ emission line and the Na I absorption line. Thereby we infer a mass range for the WD in AR UMa of 0.91 $M_{\odot}$ $<$ $M_{\mathrm{WD}}$ $<$ 1.24 $M_{\odot}$.
### A subtle IR excess associated with a young White Dwarf in the Edinburgh-Cape Blue Object Survey
We report the discovery of a subtle infrared excess associated with the young white dwarf EC\,05365--4749 at 3.35 and 4.6\,$\mu$m. Follow-up spectroscopic observations are consistent with a hydrogen atmosphere white dwarf of effective temperature 22\,800\,K and log [\emph{g} (\,cm\,s$^{-2}$) ] = 8.19. High resolution spectroscopy reveals atmospheric metal pollution with logarithmic abundances of [Mg/H] = --5.36 and [Ca/H] = --5.75, confirming the white dwarf is actively accreting from a metal-rich source with an intriguing abundance pattern. We find that the infrared excess is well modeled by a flat, opaque debris disk, though disk parameters are not well constrained by the small number of infrared excess points. We further demonstrate that relaxing the assumption of a circular dusty debris disk to include elliptical disks expands the widths of acceptable disks, adding an alternative interpretation to the subtle infrared excesses commonly observed around young white dwarfs.
### The Field White Dwarf Mass Distribution
We revisit the properties and astrophysical implications of the field white dwarf mass distribution in preparation of Gaia applications. Our study is based on the two samples with the best established completeness and most precise atmospheric parameters, the volume-complete survey within 20 pc and the Sloan Digital Sky Survey (SDSS) magnitude-limited sample. We explore the modelling of the observed mass distributions with Monte Carlo simulations, but find that it is difficult to constrain independently the initial mass function (IMF), the initial-to-final-mass relation (IFMR), the stellar formation history (SFH), the variation of the Galactic disk vertical scale height as a function of stellar age, and binary evolution. Each of these input ingredients has a moderate effect on the predicted mass distributions, and we must also take into account biases owing to unidentified faint objects (20 pc sample), as well as unknown masses for magnetic white dwarfs and spectroscopic calibration issues (SDSS sample). Nevertheless, we find that fixed standard assumptions for the above parameters result in predicted mean masses that are in good qualitative agreement with the observed values. It suggests that derived masses for both studied samples are consistent with our current knowledge of stellar and Galactic evolution. Our simulations overpredict by 40-50% the number of massive white dwarfs (M > 0.75 Msun) for both surveys, although we can not exclude a Salpeter IMF when we account for all biases. Furthermore, we find no evidence of a population of double white dwarf mergers in the observed mass distributions.
### GW Librae: Still Hot Eight Years Post-Outburst
We report continued Hubble Space Telescope (HST) ultraviolet spectra and ground-based optical photometry and spectroscopy of GW Librae eight years after its largest known dwarf nova outburst in 2007. This represents the longest cooling timescale measured for any dwarf nova. The spectra reveal that the white dwarf still remains about 3000 K hotter than its quiescent value. Both ultraviolet and optical light curves show a short period of 364-373 s, similar to one of the non-radial pulsation periods present for years prior to the outburst, and with a similar large UV/optical amplitude ratio. A large modulation at a period of 2 h (also similar to that observed prior to outburst) is present in the optical data preceding and during the HST observations, but the satellite observation intervals did not cover the peaks of the optical modulation so it is not possible to determine its corresponding UV amplitude. The similarity of the short and long periods to quiescent values implies the pulsating, fast spinning white dwarf in GW Lib may finally be nearing its quiescent configuration.
### Features of the Matter Flows in the Peculiar Cataclysmic Variable AE Aquarii
The structure of plasma flows in close binary systems in which one of the components is a rapidly rotating magnetic white dwarf is studied. The main example considered is the AE Aquarii system; the spin period of the white dwarf is about a factor of 1000 shorter than the orbital period, and the magnetic field on the white dwarf surface is of order of 50 MG. The mass transfer in this system was analyzed via numerical solution of the system of MHD equations. These computations show that the magnetic field of the white dwarf does not significantly influence the velocity field of the material in its Roche lobe in the case of laminar flow regime, so that the field does not hinder the formation of a transient disk (ring) surrounding the magnetosphere. However, the efficiency of the energy and angular momentum exchange between the white dwarf and the surrounding material increases considerably with the development of turbulent motions in the matter, resulting in its acceleration at the magnetospheric boundary and further escape from the system at a high rate. The time scales of the transition of the system between the laminar and turbulent modes are close to those of the AE\,Aqr transition between its quiet and active phases.
### Doppler shifts on the spin period of the intermediate polar FO Aqr with K2 [Replacement]
We analyse the K2 short cadence data of the intermediate polar FO Aqr and provide accurate and updated orbital and spin periodicities. We additionally find small spin period changes as a function of orbital phase of ~0.02 seconds translating to velocities of ~ a few km/s. The obtained orbital-folded velocity profile displays two clear maxima and minima, and cannot be explained by the radial velocity of the orbiting white dwarf. Instead we propose that the observed velocities are the sum of the radial velocities of both the white dwarf and of the stellar surface facing the white dwarf which reprocesses the WD spin pulses. This combination can explain the observed low velocities in FO Aqr. However asymmetries in the orbital configuration are required to explain the double peaked velocity profile. One possible scenario would invoke binary eccentricity. We thus developed a simple binary model to explain and fit our observations, and find a small binary eccentricity of e=0.03. Although small, persistent eccentricity in a close interacting binary would induce enhanced mass transfer occurring preferentially at periastron passages. We thus discuss alternative scenarios where other asymmetries might explain our observations assuming circular orbits. Since FO Aqr is the first system where the combined radial velocities of both the WD and secondary surface have been measured, it is possible that other mass-transferring binaries also display similar velocity curves when observed with Kepler. These will provide additional valuable tests to either confirm or rule out small eccentricities in similar systems.
### Doppler shifts on the spin period of the intermediate polar FO Aqr with K2
We analyse the K2 short cadence data of the intermediate polar FO Aqr and provide accurate and updated orbital and spin periodicities. We additionally find small spin period changes as a function of orbital phase of ~0.02 seconds translating to velocities of ~ a few km/s. The obtained orbital-folded velocity profile displays two clear maxima and minima, and cannot be explained by the radial velocity of the orbiting white dwarf. Instead we propose that the observed velocities are the sum of the radial velocities of both the white dwarf and of the stellar surface facing the white dwarf which reprocesses the WD spin pulses. This combination can explain the observed low velocities in FO Aqr. However asymmetries in the orbital configuration are required to explain the double peaked velocity profile. One possible scenario would invoke binary eccentricity. We thus developed a simple binary model to explain and fit our observations, and find a small binary eccentricity of e=0.03. Although small, persistent eccentricity in a close interacting binary would induce enhanced mass transfer occurring preferentially at periastron passages. We thus discuss alternative scenarios where other asymmetries might explain our observations assuming circular orbits. Since FO Aqr is the first system where the combined radial velocities of both the WD and secondary surface have been measured, it is possible that other mass-transferring binaries also display similar velocity curves when observed with Kepler. These will provide additional valuable tests to either confirm or rule out small eccentricities in similar systems.
### An irradiated brown-dwarf companion to an accreting white dwarf
Brown dwarfs and giant planets orbiting close to a host star are subjected to significant irradiation that can modify the properties of their atmospheres. In order to test the atmospheric models that are used to describe these systems, it is necessary to obtain accurate observational estimates of their physical properties (masses, radii, temperatures, albedos). Interacting compact binary systems provide a natural laboratory for studying strongly irradiated sub-stellar objects. As the mass-losing secondary in these systems makes a critical, but poorly understood transition from the stellar to the sub-stellar regime, it is also strongly irradiated by the compact accretor. In fact, the internal and external energy fluxes are both expected to be comparable in these objects, providing access to an unexplored irradiation regime. However, the atmospheric properties of such donors have so far remained largely unknown. Here, we report the direct spectroscopic detection and characterisation of an irradiated sub-stellar donor in an accreting white dwarf binary system. Our near-infrared observations allow us to determine a model-independent mass estimate for the donor of $M_2=0.055\pm0.008M_{\odot}$ and an average spectral type of ${\rm L1}\pm{\rm1}$, supporting both theoretical predictions and model-dependent observational constraints. Our time-resolved data also allow us to estimate the average irradiation-induced temperature difference between the day and night sides on the sub-stellar donor, $\Delta{\rm T}\simeq57$~K, and the maximum difference between the hottest and coolest parts of its surface, of $\Delta{\rm T}_{max}\simeq200$~K. The observations are well described by a simple geometric reprocessing model with a bolometric (Bond) albedo of $A_B<0.54$ at 2-$\sigma$ confidence level, consistent with high reprocessing efficiency, but poor lateral heat redistribution in the donor's atmosphere.
### Discovery of an extremely weak magnetic field in the white dwarf LTT 16093 = WD2047+372
Magnetic fields have been detected in several hundred white dwarfs, with strengths ranging from a few kG to several hundred MG. Only a few of the known fields have a mean magnetic field modulus below about 1 MG. We are searching for new examples of magnetic white dwarfs with very weak fields, and trying to model the few known examples. Our search is intended to be sensitive enough to detect fields at the few kG level. We have been surveying bright white dwarfs for very weak fields using spectropolarimeters at the Canada-France-Hawaii telescope, the William Herschel telescope, the European Southern Observatory, and the Russian Special Astrophysical Observatory. We discuss in some detail tests of the WHT spectropolarimeter ISIS using the known magnetic strong-field Ap star HD 215441 (Babcock's star) and the long-period Ap star HD 201601 (gamma Equ). We report the discovery of a field with a mean field modulus of about 57 kG in the white dwarf LTT 16093 = WD2047+372. The field is clearly detected through the Zeeman splitting of Halpha seen in two separate circularly polarised spectra from two different spectropolarimeters. Zeeman circular polarisation is also detected, but only barely above the 3 sigma level. The discovery of this field is significant because it is the third weakest field ever unambiguously discovered in a white dwarf, while still being large enough that we should be able to model the field structure in some detail with future observations.
### The Thermonuclear Runaway and the Classical Nova Outburst
Nova explosions occur on the white dwarf component of a Cataclysmic Variable binary stellar system that is accreting matter lost by its companion. When sufficient material has been accreted by the white dwarf, a thermonuclear runaway occurs and ejects material in what is observed as a Classical Nova explosion. We describe both the recent advances in our understanding of the progress of the outburst and outline some of the puzzles that are still outstanding. We report on the effects of improving both the nuclear reaction rate library and including a modern nuclear reaction network in our one-dimensional, fully implicit, hydrodynamic computer code. In addition, there has been progress in observational studies of Supernovae Ia with implications about the progenitors and we discuss that in this review.
### The ELM Survey. VII. Orbital Properties of Low Mass White Dwarf Binaries
We present the discovery of 15 extremely low mass (5 < log{g} < 7) white dwarf candidates, 9 of which are in ultra-compact double-degenerate binaries. Our targeted ELM Survey sample now includes 76 binaries. The sample has a lognormal distribution of orbital periods with a median period of 5.4 hr. The velocity amplitudes imply that the binary companions have a normal distribution of mass with 0.76 Msun mean and 0.25 Msun dispersion. Thus extremely low mass white dwarfs are found in binaries with a typical mass ratio of 1:4. Statistically speaking, 95% of the white dwarf binaries have a total mass below the Chandrasekhar mass and thus are not Type Ia supernova progenitors. Yet half of the observed binaries will merge in less than 6 Gyr due to gravitational wave radiation; probable outcomes include single massive white dwarfs and stable mass transfer AM CVn binaries.
### Identifying IGR J14091-6108 as a magnetic CV with a massive white dwarf using X-ray and optical observations
IGR J14091-6108 is a Galactic X-ray source known to have an iron emission line, a hard X-ray spectrum, and an optical counterpart. Here, we report on X-ray observations of the source with XMM-Newton and NuSTAR as well as optical spectroscopy with ESO/VLT and NOAO/SOAR. In the X-rays, this provides data with much better statistical quality than the previous observations, and this is the first report of the optical spectrum. Timing analysis of the XMM data shows a very significant detection of 576.3+/-0.6 s period. The signal has a pulsed fraction of 30%+/-3% in the 0.3-12 keV range and shows a strong drop with energy. The optical spectra show strong emission lines with significant variability in the lines and continuum, indicating that they come from an irradiated accretion disk. Based on these measurements, we identify the source as a magnetic Cataclysmic Variable of Intermediate Polar (IP) type where the white dwarf spin period is 576.3 s. The X-ray spectrum is consistent with the continuum emission mechanism being due to thermal Bremsstrahlung, but partial covering absorption and reflection are also required. In addition, we use the IP mass (IPM) model, which suggests that the white dwarf in this system has a high mass, possibly approaching the Chandrasekhar limit.
### Circumstellar Debris and Pollution at White Dwarf Stars
Circumstellar disks of planetary debris are now known or suspected to closely orbit hundreds of white dwarf stars. To date, both data and theory support disks that are entirely contained within the preceding giant stellar radii, and hence must have been produced during the white dwarf phase. This picture is strengthened by the signature of material falling onto the pristine stellar surfaces; disks are always detected together with atmospheric heavy elements. The physical link between this debris and the white dwarf host abundances enables unique insight into the bulk chemistry of extrasolar planetary systems via their remnants. This review summarizes the body of evidence supporting dynamically active planetary systems at a large fraction of all white dwarfs, the remnants of first generation, main-sequence planetary systems, and hence provide insight into initial conditions as well as long-term dynamics and evolution.
### GW Librae: A unique laboratory for pulsations in an accreting white dwarf
Non-radial pulsations have been identified in a number of accreting white dwarfs in cataclysmic variables. These stars offer insight into the excitation of pulsation modes in atmospheres with mixed compositions of hydrogen, helium, and metals, and the response of these modes to changes in the white dwarf temperature. Among all pulsating cataclysmic variable white dwarfs, GW Librae stands out by having a well-established observational record of three independent pulsation modes that disappeared when the white dwarf temperature rose dramatically following its 2007 accretion outburst. Our analysis of HST ultraviolet spectroscopy taken in 2002, 2010 and 2011, showed that pulsations produce variations in the white dwarf effective temperature as predicted by theory. Additionally in May~2013, we obtained new HST/COS ultraviolet observations that displayed unexpected behaviour: besides showing variability at ~275s, which is close to the post-outburst pulsations detected with HST in 2010 and 2011, the white dwarf exhibits high-amplitude variability on a ~4.4h time-scale. We demonstrate that this variability is produced by an increase of the temperature of a region on white dwarf covering up to ~30 per cent of the visible white dwarf surface. We argue against a short-lived accretion episode as the explanation of such heating, and discuss this event in the context of non-radial pulsations on a rapidly rotating star
### The White Dwarf Binary Pathways Survey I: A sample of FGK stars with white dwarf companions
The number of white dwarf plus main-sequence star binaries has increased rapidly in the last decade, jumping from only ~30 in 2003 to over 3000. However, in the majority of known systems the companion to the white dwarf is a low mass M dwarf, since these are relatively easy to identify from optical colours and spectra. White dwarfs with more massive FGK type companions have remained elusive due to the large difference in optical brightness between the two stars. In this paper we identify 934 main-sequence FGK stars from the Radial Velocity Experiment (RAVE) survey in the southern hemisphere and the Large Sky Area Multi-Object Fiber Spectroscopic Telescope (LAMOST) survey in the northern hemisphere, that show excess flux at ultraviolet wavelengths which we interpret as the likely presence of a white dwarf companion. We obtained Hubble Space Telescope ultraviolet spectra for nine systems which confirmed that the excess is indeed caused, in all cases, by a hot compact companion, eight being white dwarfs and one a hot subdwarf or pre-helium white dwarf, demonstrating that this sample is very clean. We also address the potential of this sample to test binary evolution models and type Ia supernovae formation channels.
### The White Dwarf Binary Pathways Survey I: A sample of FGK stars with white dwarf companions [Replacement]
The number of spatially unresolved white dwarf plus main-sequence star binaries has increased rapidly in the last decade, jumping from only ~30 in 2003 to over 3000. However, in the majority of known systems the companion to the white dwarf is a low mass M dwarf, since these are relatively easy to identify from optical colours and spectra. White dwarfs with more massive FGK type companions have remained elusive due to the large difference in optical brightness between the two stars. In this paper we identify 934 main-sequence FGK stars from the Radial Velocity Experiment (RAVE) survey in the southern hemisphere and the Large Sky Area Multi-Object Fiber Spectroscopic Telescope (LAMOST) survey in the northern hemisphere, that show excess flux at ultraviolet wavelengths which we interpret as the likely presence of a white dwarf companion. We obtained Hubble Space Telescope ultraviolet spectra for nine systems which confirmed that the excess is indeed caused, in all cases, by a hot compact companion, eight being white dwarfs and one a hot subdwarf or pre-helium white dwarf, demonstrating that this sample is very clean. We also address the potential of this sample to test binary evolution models and type Ia supernovae formation channels.
### GK Per and EX Hya: Intermediate polars with small magnetospheres
Observed hard X-ray spectra of intermediate polars are determined mainly by the accretion flow velocity at the white dwarf surface, which is normally close to the free-fall velocity. This allows to estimate the white dwarf masses as the white dwarf mass-radius relation M-R and the expected free-fall velocities at the surface are well known. This method is widely used, however, derived white dwarf masses M can be systematically underestimated because the accretion flow is stopped at and re-accelerates from the magnetospheric boundary R_m, and therefore, its velocity at the surface will be lower than free-fall.To avoid this problem we computed a two-parameter set of model hard X-ray spectra, which allows to constrain a degenerate M - R_m dependence. On the other hand, previous works showed that power spectra of accreting X-ray pulsars and intermediate polars exhibit breaks at the frequencies corresponding to the Keplerian frequencies at the magnetospheric boundary. Therefore, the break frequency \nu_b in an intermediate polar power spectrum gives another relation in the M - R_m plane. The intersection of the two dependences allows, therefore, to determine simultaneously the white dwarf mass and the magnetospheric radius. To verify the method we analyzed the archival Suzaku observation of EX Hya obtaining M /M_sun= 0.73 \pm 0.06 and R_ m / R = 2.6 \pm 0.4 consistent with the values determined by other authors. Subsequently, we applied the same method to a recent NuSTAR observation of another intermediate polar GK~Per performed during an outburst and found M/M_sun = 0.86 \pm 0.02 and R_ m / R = 2.8 \pm 0.2. The long duration observations of GK Per in quiescence performed by Swift/BAT and INTEGRAL observatories indicate increase of magnetosphere radius R_m at lower accretion rates.
### The type Iax supernova, SN 2015H: a white dwarf deflagration candidate
We present results based on observations of SN 2015H which belongs to the small group of objects similar to SN 2002cx, otherwise known as type Iax supernovae. The availability of deep pre-explosion imaging allowed us to place tight constraints on the explosion epoch. Our observational campaign began approximately one day post-explosion, and extended over a period of about 150 days post maximum light, making it one of the best observed objects of this class to date. We find a peak magnitude of M$_r$ = -17.27 $\pm$ 0.07, and a ($\Delta m_{15})_r$ = 0.69 $\pm$ 0.04. Comparing our observations to synthetic spectra generated from simulations of deflagrations of Chandrasekhar mass carbon-oxygen white dwarfs, we find reasonable agreement with models of weak deflagrations that result in the ejection of ~0.2 M$_{\odot}$ of material containing ~0.07 M$_{\odot}$ of 56Ni. The model light curve however, evolves more rapidly than observations, suggesting that a higher ejecta mass is to be favoured. Nevertheless, empirical modelling of the pseudo-bolometric light curve suggests that $\lesssim$0.6 M_sun of material was ejected, implying that the white dwarf is not completely disrupted, and that a bound remnant is a likely outcome.
### The post-outburst pulsations of the accreting white dwarf in the cataclysmic variable GW Librae
We present new time-series photometry of the accreting pulsating white dwarf system GW Librae obtained in 2012 and 2013 at the University of Canterbury Mt John Observatory in New Zealand. Our 2012 data show the return of a $\sim$19 minute periodicity that was previously detected in 2008. This pulsation mode was a dominant feature of our quality May 2012 data set, which consisted of six contiguous nights; a detailed analysis indicated a degree of frequency variability. We show by comparison with the previously identified pulsation modes that this periodicity is best explained as a new mode, and that the quasi-stability of the periods appears to be a general feature of the pulsations in these systems. We also find a previously unreported 3-hour modulation period, which we believe to be related to the known two and four hour periods of so far unknown origin.
### Discovery of spin-up in the X-ray pulsar companion of the hot subdwarf HD 49798
The hot subdwarf HD 49798 has an X-ray emitting compact companion with a spin-period of 13.2 s and a dynamically measured mass of 1.28+/-0.05 M_sun, consistent with either a neutron star or a white dwarf. Using all the available XMM-Newton and Swift observations of this source, we could perform a phase-connected timing analysis extending back to the ROSAT data obtained in 1992. We found that the pulsar is spinning up at a rate of (2.15+/-0.05)x10^{-15} s/s. This result is best interpreted in terms of a neutron star accreting from the wind of its subdwarf companion, although the remarkably steady period derivative over more than 20 years is unusual in wind-accreting neutron stars. The possibility that the compact object is a massive white dwarf accreting through a disk cannot be excluded, but it requires a larger distance and/or properties of the stellar wind of HD 49798 different from those derived from the modelling of its optical/UV spectra.
### The SDSS spectroscopic catalogue of white dwarf-main sequence binaries: new identifications from DR9-12
We present an updated version of the spectroscopic catalogue of white dwarf-main sequence (WDMS) binaries from the Sloan Digital Sky Survey (SDSS). We identify 939 WDMS binaries within the data releases (DR) 9-12 of SDSS plus 40 objects from DR 1-8 that we missed in our previous works, 646 of which are new. The total number of spectroscopic SDSS WDMS binaries increases to 3294. This is by far the largest and most homogeneous sample of compact binaries currently available. We use a decomposition/fitting routine to derive the stellar parameters of all systems identified here (white dwarf effective temperatures, surface gravities and masses, and secondary star spectral types). The analysis of the corresponding stellar parameter distributions shows that the SDSS WDMS binary population is seriously affected by selection effects. We also measure the NaI 8183.27, 8194.81 absorption doublet and Halpha emission radial velocities (RV) from all SDSS WDMS binary spectra identified in this work. 98 objects are found to display RV variations, 62 of which are new. The RV data are sufficient enough to estimate the orbital periods of three close binaries.
### The crowded magnetosphere of the post common envelope binary QS Virginis
We present high speed photometry and high resolution spectroscopy of the eclipsing post common envelope binary QS Virginis (QS Vir). Our UVES spectra span multiple orbits over more than a year and reveal the presence of several large prominences passing in front of both the M star and its white dwarf companion, allowing us to triangulate their positions. Despite showing small variations on a timescale of days, they persist for more than a year and may last decades. One large prominence extends almost three stellar radii from the M star. Roche tomography reveals that the M star is heavily spotted and that these spots are long-lived and in relatively fixed locations, preferentially found on the hemisphere facing the white dwarf. We also determine precise binary and physical parameters for the system. We find that the 14,220 +/- 350K white dwarf is relatively massive, 0.782 +/- 0.013Ms, and has a radius of 0.01068 +/- 0.00007Rs, consistent with evolutionary models. The tidally distorted M star has a mass of 0.382 +/- 0.006Ms and a radius of 0.381 +/- 0.003Rs, also consistent with evolutionary models. We find that the magnesium absorption line from the white dwarf is broader than expected. This could be due to rotation (implying a spin period of only ~700 seconds), or due to a weak (~100kG) magnetic field, we favour the latter interpretation. Since the M star's radius is still within its Roche lobe and there is no evidence that its over-inflated we conclude that QS Vir is most likely a pre-cataclysmic binary just about to become semi-detached.
### Stellar laboratories. VII. New Kr IV - VII oscillator strengths and an improved spectral analysis of the hot, hydrogen-deficient DO-type white dwarf RE0503-289
For the spectral analysis of high-resolution and high-signal-to-noise (S/N) spectra of hot stars, state-of-the-art non-local thermodynamic equilibrium (NLTE) model atmospheres are mandatory. These are strongly dependent on the reliability of the atomic data that is used for their calculation. New of Kr IV - VII oscillator strengths for a large number of lines allow to construct more detailed model atoms for our NLTE model-atmosphere calculations. This enables us to search for additional Kr lines in observed spectra and to improve Kr abundance determinations. We calculated Kr IV - VII oscillator strengths to consider radiative and collisional bound-bound transitions in detail in our NLTE stellar-atmosphere models for the analysis of Kr lines exhibited in high-resolution and high-S/N ultraviolet (UV) observations of the hot white dwarf RE 0503-289. We reanalyzed the effective temperature and surface gravity and determined Teff = 70 000 +/- 2000 K and log (g / cm/s**2) = 7.5 +/- 0.1. We newly identified ten Kr V lines and one Kr VI line in the spectrum of RE 0503-289. We measured a Kr abundance of -3.3 +/- 0.3 (logarithmic mass fraction). We discovered that the interstellar absorption toward RE 0503-289 has a multi-velocity structure within a radial-velocity interval of -40 km/s < vrad < +18 km/s. Reliable measurements and calculations of atomic data are a prerequisite for state-of-the-art NLTE stellar-atmosphere modeling. Observed Kr V - VII line profiles in the UV spectrum of the white dwarf RE 0503-289 were simultaneously well reproduced with our newly calculated oscillator strengths.
### An Ultramassive 1.28 M$_\odot$ White Dwarf in NGC 2099
With the Keck I Low-Resolution Imaging Spectrometer we have observed nine white dwarf candidates in the very rich open cluster NGC 2099 (M37). The spectroscopy shows seven to be DA white dwarfs, one to be a DB white dwarf, and one to be a DZ white dwarf. Three of these DA white dwarfs are consistent with singly evolved cluster membership: an ultramassive (1.28$^{+0.05}_{-0.08}$ M$_\odot$) and two intermediate-mass (0.70 and 0.75 M$_\odot$) white dwarfs. Analysis of their cooling ages allows us to calculate their progenitor masses and establish new constraints on the initial-final mass relation. The intermediate-mass white dwarfs are in strong agreement with previous work over this mass regime. The ultramassive white dwarf has $V$ = 24.5, $\sim$2 mag fainter than the other two remnants. The spectrum of this star has lower quality, so the derived stellar properties (e.g., T$_{\rm eff}$, log g) have uncertainties that are several times higher than the brighter counterparts. We measure these uncertainties and establish the star's final mass as the highest-mass white dwarf discovered thus far in a cluster, but we are unable to calculate its progenitor mass because at this high mass and cooler T$_{\rm eff}$ its inferred cooling age is highly sensitive to its mass. At the highest temperatures, however, this sensitivity of cooling age to an ultramassive white dwarf's mass is only moderate. This demonstrates that future investigations of the upper-mass end of the initial-final mass relation must identify massive, newly formed white dwarfs (i.e., in young clusters with ages 50-150 Myr).
### Constraining white dwarf structure and neutrino physics in 47 Tucanae
We present a robust statistical analysis of the white dwarf cooling sequence in 47 Tucanae. We combine HST UV and optical data in the core of the cluster, Modules for Experiments in Stellar Evolution (MESA) white dwarf cooling models, white dwarf atmosphere models, artificial star tests, and a Markov Chain Monte Carlo (MCMC) sampling method to fit white dwarf cooling models to our data directly. We use a technique known as the unbinned maximum likelihood to fit these models to our data without binning. We use these data to constrain neutrino production and the thickness of the hydrogen layer in these white dwarfs. The data prefer thicker hydrogen layers $(q_\mathrm{H}=3.2\e{-5})$ and we can strongly rule out thin layers $(q_\mathrm{H}=10^{-6})$. The neutrino rates currently in the models are consistent with the data. This analysis does not provide a constraint on the number of neutrino species.
### Extreme abundance ratios in the polluted atmosphere of the cool white dwarf NLTT19868
We present an analysis of intermediate-dispersion spectra and photometric data of the newly identified cool, polluted white dwarf NLTT19868. The spectra obtained with X-shooter on the Very Large Telescope (VLT)-Melipal show strong lines of calcium, and several lines of magnesium, aluminium and iron. We use these spectra and the optical-to-near infrared spectral energy distribution to constrain the atmospheric parameters of NLTT19868. Our analysis shows that NLTT19868 is iron poor with respect to aluminium and calcium. A comparison with other cool, polluted white dwarfs shows that the Fe to Ca abundance ratio (Fe/Ca) varies by up to approximately two orders of magnitudes over a narrow temperature range with NLTT19868 at one extremum in the Fe/Ca ratio and, in contrast, NLTT888 at the other extremum. The sample shows evidence of extreme diversity in the composition of the accreted material: In the case of NLTT888, the inferred composition of the accreted matter is akin to iron-rich planetary core composition, while in the case of NLTT19868 it is close to mantle or bulk-Earth composition depleted by subsequent chemical separation at the bottom of the convection zone.
### The white dwarf population within 40 pc of the Sun
The white dwarf luminosity function is an important tool to understand the properties of the Solar neighborhood, like its star formation history, and its age. Here we present a population synthesis study of the white dwarf population within 40~pc from the Sun, and compare the results of this study with the properties of the observed sample. We use a state-of-the-art population synthesis code based on Monte Carlo techniques, that incorporates the most recent and reliable white dwarf cooling sequences, an accurate description of the Galactic neighborhood, and a realistic treatment of all the known observational biases and selection procedures. We find a good agreement between our theoretical models and the observed data. In particular, our simulations reproduce a previously unexplained feature of the bright branch of the white dwarf luminosity function, which we argue is due to a recent episode of star formation. We also derive the age of the Solar neighborhood employing the position of the observed cut-off of the white dwarf luminosity function, obtaining ~8.9+-0.2 Gyr. We conclude that a detailed description of the ensemble properties of the population of white dwarfs within 40pc of the Sun allows us to obtain interesting constraints on the history of the Solar neighborhood.
### A Numerical Model for Accretion in Intermediate Polars with Dipolar Magnetic Fields
A three-dimensional numerical model for an accretion process investigation in the magnetosphere of a white dwarf in magnetic cataclysmic variables is developed. The model assumes that the white dwarf has a dipole magnetic field with its symmetry axis inclined to the rotation axis. The model is based on the equations of modified MHD, that describe the mean flow parameters in the wave MHD turbulence. Diffusion of the magnetic field and radiative heating and cooling are taken into account. The suitability of the model is confirmed by modeling the accretion in a typical intermediate polar. The computations show that a magnetosphere forms around the accretor, with the accretion occurring via columns. The accretion columns have a curtain-like shape, and arc-shaped zones of energy release form on the surface of the white dwarf in the magnetic poles area as a result of the matter infall.
### Stellar laboratories. VI. New Mo IV - VII oscillator strengths and the molybdenum abundance in the hot white dwarfs G191-B2B and RE0503-289
For the spectral analysis of high-resolution and high-signal-to-noise (S/N) spectra of hot stars, state-of-the-art non-local thermodynamic equilibrium (NLTE) model atmospheres are mandatory. These are strongly dependent on the reliability of the atomic data that is used for their calculation. To identify molybdenum lines in the ultraviolet (UV) spectra of the DA-type white dwarf G191-B2B and the DO-type white dwarf RE0503-289 and to determine their photospheric Mo abundances, newly calculated Mo IV - VII oscillator strengths are used. We identified twelve Mo V and nine Mo VI lines in the UV spectrum of RE0503-289 and measured a photospheric Mo abundance of 1.2 - 3.0 x 10**-4 (mass fraction, 22500 - 56400 times the solar abundance). In addition, from the As V and Sn IV resonance lines, we measured mass fractions of arsenic (0.5 - 1.3 x 10**-5, about 300 - 1200 times solar) and tin (1.3 - 3.2 x 10**-4, about 14300 35200 times solar). For G191-B2B, upper limits were determined for the abundances of Mo (5.3 x 10**-7, 100 times solar) and, in addition, for Kr (1.1 x 10**-6, 10 times solar) and Xe (1.7 x 10**-7, 10 times solar). The arsenic abundance was determined (2.3 - 5.9 x 10**-7, about 21 - 53 times solar). A new, registered German Astrophysical Virtual Observatory (GAVO) service, TOSS, has been constructed to provide weighted oscillator strengths and transition probabilities. Reliable measurements and calculations of atomic data are a prerequisite for stellar-atmosphere modeling. Observed Mo V - VI line profiles in the UV spectrum of the white dwarf RE0503-289 were well reproduced with our newly calculated oscillator strengths. For the first time, this allowed to determine the photospheric Mo abundance in a white dwarf.
### White Dwarf Mergers on Adaptive Meshes I. Methodology and Code Verification
The Type Ia supernova progenitor problem is one of the most perplexing and exciting problems in astrophysics, requiring detailed numerical modeling to complement observations of these explosions. One possible progenitor that has merited recent theoretical attention is the white dwarf merger scenario, which has the potential to naturally explain many of the observed characteristics of Type Ia supernovae. To date there have been relatively few self-consistent simulations of merging white dwarf systems using mesh-based hydrodynamics. This is the first paper in a series describing simulations of these systems using a hydrodynamics code with adaptive mesh refinement. In this paper we describe our numerical methodology and discuss our implementation in the compressible hydrodynamics code CASTRO, which solves the Euler equations, and the Poisson equation for self-gravity, and couples the gravitational and rotation forces to the hydrodynamics. Standard techniques for coupling gravitation and rotation forces to the hydrodynamics do not adequately conserve the total energy of the system for our problem, but recent advances in the literature allow progress and we discuss our implementation here. We present a set of test problems demonstrating the extent to which our software sufficiently models a system where large amounts of mass are advected on the computational domain over long timescales. Future papers in this series will describe our treatment of the initial conditions of these systems and will examine the early phases of the merger to determine its viability for triggering a thermonuclear detonation.
### White Dwarf Mergers on Adaptive Meshes I. Methodology and Code Verification [Replacement]
The Type Ia supernova progenitor problem is one of the most perplexing and exciting problems in astrophysics, requiring detailed numerical modeling to complement observations of these explosions. One possible progenitor that has merited recent theoretical attention is the white dwarf merger scenario, which has the potential to naturally explain many of the observed characteristics of Type Ia supernovae. To date there have been relatively few self-consistent simulations of merging white dwarf systems using mesh-based hydrodynamics. This is the first paper in a series describing simulations of these systems using a hydrodynamics code with adaptive mesh refinement. In this paper we describe our numerical methodology and discuss our implementation in the compressible hydrodynamics code CASTRO, which solves the Euler equations, and the Poisson equation for self-gravity, and couples the gravitational and rotation forces to the hydrodynamics. Standard techniques for coupling gravitation and rotation forces to the hydrodynamics do not adequately conserve the total energy of the system for our problem, but recent advances in the literature allow progress and we discuss our implementation here. We present a set of test problems demonstrating the extent to which our software sufficiently models a system where large amounts of mass are advected on the computational domain over long timescales. Future papers in this series will describe our treatment of the initial conditions of these systems and will examine the early phases of the merger to determine its viability for triggering a thermonuclear detonation.
### Empirical consequential angular momentum loss can solve long standing problems of CV evolution
The observed orbital period distribution of cataclysmic variables (CVs), the space density derived from observations, and the observed orbital period minimum are known to disagree with theoretical predictions since decades. More recently, the white dwarf (WD) masses in CVs have been found to significantly exceed those of single WDs, which is in contrast to theoretical expectations as well. We here claim that all these problems are related and can be solved if CVs with low-mass white dwarfs are driven into dynamically unstable mass transfer due to consequential angular momentum loss (CAML). Indeed, assuming CAML increases as a function of decreasing white dwarf mass can bring into agreement the predictions of binary population models and the observed properties of the CV population. We speculate that a common envelope like evolution of CVs with low-mass WDs following a nova eruption might be the physical process behind our empirical prescription of CAML.
### A comparative analysis of the observed white dwarf cooling sequence from globular clusters
We report our study of features at the observed red end of the white dwarf cooling sequences for three Galactic globular clusters: NGC\,6397, 47\,Tucanae and M\,4. We use deep colour-magnitude diagrams constructed from archival Hubble Space Telescope (ACS) to systematically investigate the blue turn at faint magnitudes and the age determinations for each cluster. We find that the age difference between NGC\,6397 and 47\,Tuc is 1.98$^{+0.44}_{-0.26}$\,Gyr, consistent with the picture that metal-rich halo clusters were formed later than metal-poor halo clusters. We self-consistently include the effect of metallicity on the progenitor age and the initial-to-final mass relation. In contrast with previous investigations that invoked a single white dwarf mass for each cluster, the data shows a spread of white dwarf masses that better reproduce the shape and location of the blue turn. This effect alone, however, does not completely reproduce the observational data - the blue turn retains some mystery. In this context, we discuss several other potential problems in the models. These include possible partial mixing of H and He in the atmosphere of white dwarf stars, the lack of a good physical description of the collision-induced absorption process and uncertainties in the opacities at low temperatures. The latter are already known to be significant in the description of the cool main sequence. Additionally, we find that the present day local mass function of NGC\,6397 is consistent with a top-heavy type, while 47\,Tuc presents a bottom-heavy profile.
### A large, long-lived structure near the trojan L5 point in the post common-envelope binary SDSS J1021+1744
SDSS J1021+1744 is a detached, eclipsing white dwarf / M dwarf binary discovered in the Sloan Digital Sky Survey. Outside the primary eclipse, the light curves of such systems are usually smooth and characterised by low-level variations caused by tidal distortion and heating of the M star component. Early data on SDSS J1021+1744 obtained in June 2012 was unusual in showing a dip in flux of uncertain origin shortly after the white dwarf's eclipse. Here we present high-time resolution, multi-wavelength observations of 35 more eclipses over 1.3 years, showing that the dip has a lifetime extending over many orbits. Moreover the "dip" is in fact a series of dips that vary in depth, number and position, although they are always placed in the phase interval 1.06 to 1.26 after the white dwarf's eclipse, near the L5 point in this system. Since SDSS J1021+1744 is a detached binary, it follows that the dips are caused by the transit of the white dwarf by material around the Lagrangian L5 point. A possible interpretation is that they are the signatures of prominences, a phenomenon already known from H-alpha observations of rapidly rotating single stars as well as binaries. What makes SDSS J1021+1744 peculiar is that the material is dense enough to block continuum light. The dips appear to have finally faded out around 2015 May after the first detection by Parsons et al. in 2012, suggesting a lifetime of years.
### A Dark Spot on a Massive White Dwarf
We present the serendipitous discovery of eclipse-like events around the massive white dwarf SDSS J152934.98+292801.9 (hereafter J1529+2928). We selected J1529+2928 for time-series photometry based on its spectroscopic temperature and surface gravity, which place it near the ZZ Ceti instability strip. Instead of pulsations, we detect photometric dips from this white dwarf every 38 minutes. Follow-up optical spectroscopy observations with Gemini reveal no significant radial velocity variations, ruling out stellar and brown dwarf companions. A disintegrating planet around this white dwarf cannot explain the observed light curves in different filters. Given the short period, the source of the photometric dips must be a dark spot that comes into view every 38 min due to the rotation of the white dwarf. Our optical spectroscopy does not show any evidence of Zeeman splitting of the Balmer lines, limiting the magnetic field strength to B<70 kG. Since up to 15% of white dwarfs display kG magnetic fields, such eclipse-like events should be common around white dwarfs. We discuss the potential implications of this discovery on transient surveys targeting white dwarfs, like the K2 mission and the Large Synoptic Survey Telescope.
### Evidence for Gas from a Disintegrating Extrasolar Asteroid
We report high-resolution spectroscopic observations of WD 1145+017 -- a white dwarf that recently has been found to be transitted by multiple asteroid-sized objects within its tidal radius. We have discovered numerous circumstellar absorption lines with linewidths of $\sim$ 300 km s$^{-1}$ from Mg, Ca, Ti, Cr, Mn, Fe and Ni, possibly from several gas streams produced by collisions among the actively disintegrating objects. The atmosphere of WD 1145+017 is polluted with 11 heavy elements, including O, Mg, Al, Si, Ca, Ti, V:, Cr, Mn, Fe and Ni. Evidently, we are witnessing the active disintegration and subsequent accretion of an extrasolar asteroid.
### Discovery of an eclipsing dwarf nova in the ancient nova shell Te 11
We report on the discovery of an eclipsing dwarf nova (DN) inside the peculiar, bilobed nebula Te 11. Modelling of high-speed photometry of the eclipse finds the accreting white dwarf to have a mass 1.18 M$_\odot$ and temperature 13 kK. The donor spectral type of M2.5 results in a distance of 330 pc, colocated with Barnard's loop at the edge of the Orion-Eridanus superbubble. The perplexing morphology and observed bow shock of the slowly-expanding nebula may be explained by strong interactions with the dense interstellar medium in this region. We match the DN to the historic nova of 483 CE in Orion and postulate that the nebula is the remnant of this eruption. This connection supports the millennia time scale of the post-nova transition from high to low mass-transfer rates. Te 11 constitutes an important benchmark system for CV and nova studies as the only eclipsing binary out of just three DNe with nova shells.
### Doppler-imaging of the planetary debris disc at the white dwarf SDSS J122859.93+104032.9
Debris discs which orbit white dwarfs are signatures of remnant planetary systems. We present twelve years of optical spectroscopy of the metal-polluted white dwarf SDSS J1228+1040, which shows a steady variation in the morphology of the 8600 {\AA} Ca II triplet line profiles from the gaseous component of its debris disc. We identify additional emission lines of O I, Mg I, Mg II, Fe II and Ca II in the deep co-added spectra. These emission features (including Ca H & K) exhibit a wide range in strength and morphology with respect to each other and to the Ca II triplet, indicating different intensity distributions of these ionic species within the disc. Using Doppler tomography we show that the evolution of the Ca II triplet profile can be interpreted as the precession of a fixed emission pattern with a period in the range 24-30 years. The Ca II line profiles vary on time-scales that are broadly consistent with general relativistic precession of the debris disc.
### A wide binary trigger for white dwarf pollution
Metal pollution in white dwarf atmospheres is likely to be a signature of remnant planetary systems. Most explanations for this pollution predict a sharp decrease in the number of polluted systems with white dwarf cooling age. Observations do not confirm this trend, and metal pollution in old (1-5 Gyr) white dwarfs is difficult to explain. We propose an alternative, time-independent mechanism to produce the white dwarf pollution. The orbit of a wide binary companion can be perturbed by Galactic tides, approaching close to the primary star for the first time after billions of years of evolution on the white dwarf branch. We show that such a close approach perturbs a planetary system orbiting the white dwarf, scattering planetesimals onto star-grazing orbits, in a manner that could pollute the white dwarf's atmosphere. Our estimates find that this mechanism is likely to contribute to metal pollution, alongside other mechanisms, in up to a few percent of an observed sample of white dwarfs with wide binary companions, independent of white dwarf age. This age independence is the key difference between this wide binary mechanism and others mechanisms suggested in the literature to explain white dwarf pollution. Current observational samples are not large enough to assess whether this mechanism makes a significant contribution to the population of polluted white dwarfs, for which better constraints on the wide binary population are required, such as those that will be obtained in the near future with Gaia.
### Discovery of near-ultraviolet counterparts to millisecond pulsars in the globular cluster 47 Tucanae
We report the discovery of the likely white dwarf companions to radio millisecond pulsars 47 Tuc Q and 47 Tuc S in the globular cluster 47 Tucanae. These blue stars were found in near-ultraviolet images from the Hubble Space Telescope for which we derived accurate absolute astrometry, and are located at positions consistent with the radio coordinates to within 0.016 arcsec (0.2sigma). We present near-ultraviolet and optical colours for the previously identified companion to millisecond pulsar 47 Tuc U, and we unambiguously confirm the tentative prior identifications of the optical counterparts to 47 Tuc T and 47 Tuc Y. For the latter, we present its radio-timing solution for the first time. We find that all five near-ultraviolet counterparts have U300-B390 colours that are consistent with He white dwarf cooling models for masses ~0.16-0.3 Msun and cooling ages within ~0.1-6 Gyr. The Ha-R625 colours of 47 Tuc U and 47 Tuc T indicate the presence of a strong Ha absorption line, as expected for white dwarfs with an H envelope.
### Discovery of near-ultraviolet counterparts to millisecond pulsars in the globular cluster 47 Tucanae [Replacement]
We report the discovery of the likely white dwarf companions to radio millisecond pulsars 47 Tuc Q and 47 Tuc S in the globular cluster 47 Tucanae. These blue stars were found in near-ultraviolet images from the Hubble Space Telescope for which we derived accurate absolute astrometry, and are located at positions consistent with the radio coordinates to within 0.016 arcsec (0.2sigma). We present near-ultraviolet and optical colours for the previously identified companion to millisecond pulsar 47 Tuc U, and we unambiguously confirm the tentative prior identifications of the optical counterparts to 47 Tuc T and 47 Tuc Y. For the latter, we present its radio-timing solution for the first time. We find that all five near-ultraviolet counterparts have U300-B390 colours that are consistent with He white dwarf cooling models for masses ~0.16-0.3 Msun and cooling ages within ~0.1-6 Gyr. The Ha-R625 colours of 47 Tuc U and 47 Tuc T indicate the presence of a strong Ha absorption line, as expected for white dwarfs with an H envelope.
### Growing White Dwarfs to the Chandrasekhar Limit: The Parameter Space of the Single Degenerate SNIa Channel
Can a white dwarf, accreting hydrogen-rich matter from a non-degenerate companion star, ever exceed the Chandrasekhar mass and explode as a type Ia supernova? We explore the range of accretion rates that allow a white dwarf (WD) to secularly grow in mass, and derive limits on the accretion rate and on the initial mass that will allow it to reach $1.4M_\odot$ --- the Chandrasekhar mass. We follow the evolution through a long series of hydrogen flashes, during which a thick helium shell accumulates. This determines the effective helium mass accretion rate for long-term, self-consistent evolutionary runs with helium flashes. We find that net mass accumulation always occurs despite helium flashes. Although the amount of mass lost during the first few helium shell flashes is a significant fraction of that accumulated prior to the flash, that fraction decreases with repeated helium shell flashes. Eventually no mass is ejected at all during subsequent flashes. This unexpected result occurs because of continual heating of the WD interior by the helium shell flashes near its surface. The effect of heating is to lower the electron degeneracy throughout the WD, and especially in the outer layers. This key result yields helium burning that is quasi-steady state, instead of explosive. We thus find a remarkably large parameter space within which long-term, self-consistent simulations show that a WD can grow in mass and reach the Chandrasekhar limit, despite its helium flashes.
### Swift J0525.6+2416 and IGR J04571+4527: two new hard X-ray selected magnetic cataclysmic variables identified with XMM-Newton
IGR J04571+4527 and Swift J0525.6+2416 are two hard X-ray sources detected in the Swift/BAT and INTEGRAL/IBIS surveys. They were proposed to be magnetic cataclysmic variables of the Intermediate Polar (IP) type, based on optical spectroscopy. IGR J04571+4527 also showed a 1218 s optical periodicity, suggestive of the rotational period of a white dwarf, further pointing towards an IP classification. We here present detailed X-ray (0.3-10 keV) timing and spectral analysis performed with XMM-Newton, complemented with hard X-ray coverage (15-70 keV) from Swift/BAT. These are the first high signal to noise observations in the soft X-ray domain for both sources, allowing us to identify the white dwarf X-ray spin period of Swift J0525.6+2416 (226.28 s), and IGR J04571+4527 (1222.6 s). A model consisting of multi-temperature optically thin emission with complex absorption adequately fits the broad-band spectrum of both sources. We estimate a white dwarf mass of about 1.1 and 1.0 solar masses for IGR J04571+4527 and Swift J0525.6+2416, respectively. The above characteristics allow us to unambiguously classify both sources as IPs, confirming the high incidence of this subclass among hard X-ray emitting Cataclysmic Variables.
### A second case of outbursts in a pulsating white dwarf observed by Kepler
We present observations of a new phenomenon in pulsating white dwarf stars: large-amplitude outbursts at timescales much longer than the pulsation periods. The cool (Teff = 11,010 K), hydrogen-atmosphere pulsating white dwarf PG 1149+057 was observed nearly continuously for more than 78.8 d by the extended Kepler mission in K2 Campaign 1. The target showed 10 outburst events, recurring roughly every 8 d and lasting roughly 15 hr, with maximum flux excursions up to 45% in the Kepler bandpass. We demonstrate that the outbursts affect the pulsations and therefore must come from the white dwarf. Additionally, we argue that these events are not magnetic reconnection flares, and are most likely connected to the stellar pulsations and the relatively deep surface convection zone. PG 1149+057 is now the second cool pulsating white dwarf to show this outburst phenomenon, after the first variable white dwarf observed in the Kepler mission, KIC 4552982. Both stars have the same effective temperature, within the uncertainties, and are among the coolest known pulsating white dwarfs of typical mass. These outbursts provide fresh observational insight into the red edge of the DAV instability strip and the eventual cessation of pulsations in cool white dwarfs.
### Constraining Neutrino Cooling using the Hot White Dwarf Luminosity Function in the Globular Cluster 47 Tucanae
We present Hubble Space Telescope observations of the upper part (T_eff> 10 000 K) of the white dwarf cooling sequence in the globular cluster 47 Tucanae and measure a luminosity function of hot white dwarfs. Comparison with previous determinations from large scale field surveys indicates that the previously determined plateau at high effective temperatures is likely a selection effect, as no such feature is seen in this sample. Comparison with theoretical models suggests that the current estimates of white dwarf neutrino emission (primarily by the plasmon channel) are accurate, and variations are restricted to no more than a factor of two globally, at 95% confidence. We use these constraints to place limits on various proposed exotic emission mechanisms, including a non-zero neutrino magnetic moment, formation of axions, and emission of Kaluza-Klein modes into extra dimensions.
### Effects of strong magnetic fields and rotation on white dwarf structure
In this paper we compute relativistic stars models for the structure of white dwarfs under the influence of strong magnetic field and rotation. The magnetic field is assumed to be poloidal and axisymmetric. We find a maximum mass for a static magnetized white dwarf of about 2.13 $\rm{M_{\odot}}$ in the Newtonian case and a value of 2.09 $\rm{M_{\odot}}$ taking into account general relativistic effects. We also present properties of uniformly rotating white dwarfs and we show that the maximum mass is shifted from a mass of $\sim$ 1.40 $\rm{M_{\odot}}$ for non-rotating white dwarf to $\sim$ 1.45 $\rm{M_{\odot}}$ in the keplerian limit. We present also results for rotating magnetized white dwarfs calculated in a self$-$consistent way by solving the Maxwell and Einstein equations together. The maximum field strength obtained is about $10^{15}\,$G at the center of the star in the static and $10^{14}\,$G in the rotating case.
### Effects of strong magnetic fields and rotation on white dwarf structure [Replacement]
In this work we compute models for relativistic white dwarfs in the presence of strong magnetic fields. These models possibly contribute to super-luminous SNIa. With an assumed axi-symmetric and poloidal magnetic field, we study the possibility of existence of super-Chandrasekhar magnetized white dwarfs by solving numerically the Einstein-Maxwell equations, by means of a pseudo-spectral method. We obtain a self-consistent rotating and non-rotating magnetized white dwarf models. According to our results, a maximum mass for a static magnetized white dwarf is 2.13 $\rm{M_{\odot}}$ in the Newtonian case and 2.09 $\rm{M_{\odot}}$ while taking into account general relativistic effects. Furthermore, we present results for rotating magnetized white dwarfs. The maximum magnetic field strength reached at the center of white dwarfs is of the order of $10^{15}\,$G in the static case, whereas for magnetized white dwarfs, rotating with the Keplerian angular velocity, is of the order of $10^{14}\,$G.
### Getting to know Classical Novae with Swift
Novae have been reported as transients for more than two thousand years. Their bright optical outbursts are the result of explosive nuclear burning of gas accreted from a binary companion onto a white dwarf. Novae containing a white dwarf close to the Chandrasekhar mass limit and accreting at a high rate are potentially the unknown progenitors of the type Ia supernovae used to measure the acceleration of the Universe. Swift X-ray observations have radically transformed our view of novae by providing dense monitoring throughout the outburst, revealing new phenomena in the super-soft X-rays from the still-burning white dwarf such as early extreme variability and half- to one-minute timescale quasi-periodic oscillations. The distinct evolution of this emission from the harder X-ray emission due to ejecta shocks has been clearly delineated. Soft X-ray observations allow the mass of the white dwarf, the mass burned and the mass ejected to be estimated. In combination with observations at other wavelengths, including the high spectral resolution observations of the large X-ray observatories, high resolution optical and radio imaging, radio monitoring, optical spectroscopy, and the detection of GeV gamma-ray emission from recent novae, models of the explosion have been tested and developed. I review nine novae for which Swift has made a significant impact; these have shown the signature of the components in the interacting binary system in addition to the white dwarf: the re-formed accretion disk, the companion star and its stellar wind.
### Unambiguous Detection of Reflection in Magnetic Cataclysmic Variables: Joint NuSTAR-XMM-Newton Observations of Three Intermediate Polars
In magnetic cataclysmic variables (CVs), X-ray emission regions are located close to the white dwarf surface, which is expected to reflect a significant fraction of intrinsic X-rays above 10 keV, producing a Compton reflection hump. However, up to now, a secure detection of this effect in magnetic CVs has largely proved elusive because of the limited sensitivity of non-imaging X-ray detectors. Here we report our analysis of joint NuSTAR/XMM-Newton observations of three magnetic CVs, V709 Cas, NY Lup, and V1223 Sgr. The improved hard X-ray sensitivity of the imaging NuSTAR data has resulted in the first robust detection of Compton hump in all three objects, with amplitudes of ~1 or greater in NY Lup, and likely <1.0 in the other two. We also confirm earlier report of a strong spin modulation above 10 keV in V709 Cas, and report the first detection of small spin amplitudes in the others. We interpret this as due to different height of the X-ray emitting region among these objects. A height of ~0.2 white dwarf radii provides a plausible explanation for the low reflection amplitude of V709 Cas. Since emission regions above both poles are visible at certain spin phases, this can also explain the strong hard X-ray spin modulation. A shock height of ~0.05 white dwarf radii can explain our results on V1223 Sgr, while the shock height in NY Lup appears negligible.
### The V471 Tauri System: A Multi-datatype Probe
V471 Tauri, a white dwarf--red dwarf eclipsing binary in the Hyades, is well known for stimulating development of common envelope theory, whereby novae and other cataclysmic variables form from much wider binaries by catastrophic orbit shrinkage. Our evaluation of a recent imaging search that reported negative results for a much postulated third body shows that the object could have escaped detection or may have actually been seen. The balance of evidence continues to favor a brown dwarf companion about 12 AU from the eclipsing binary. A recently developed algorithm finds unified solutions from three datatypes. New radial velocities (RVs) of the red dwarf and BV RCIC light curves are solved simultaneously along with white dwarf and red dwarf RVs from the literature, uvby data, the MOST mission light curve, and 40 years of eclipse timings. Precision-based weighting is the key to proper information balance among the various datasets. Timewise variation of modeled starspots allows unified solution of multiple data eras. Light curve amplitudes strongly suggest decreasing spottedness from 1976 to about 1980, followed by approximately constant spot coverage from 1981 to 2005. An explanation is proposed for lack of noticeable variation in 1981 light curves, in terms of competition between spot and tidal variations. Photometric spectroscopic distance is estimated. The red dwarf mass comes out larger than normal for a K2V star, and even larger than adopted in several structure and evolution papers. An identified cause for this result is that much improved red dwarf RVs curves now exist. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557310104370117, "perplexity": 1923.555071145592}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982295494.5/warc/CC-MAIN-20160823195815-00104-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://learnkorean24.com/course/lesson-9/ | # Lesson 9: 에서
In a previous lesson, you learned how to use the location marking particle 에 [ae] to mark a noun as the location where something exists or doesn’t exist, the place where you are going to, or the time at which something takes place. In this lesson, you will learn the particle 에서 [ae-seo] which can be used to mark a noun as the location where an action is taking place.
## Lesson 9: 에서
### Usage 1:
The particle 에서 [ae-seo] is attached to nouns to mark that noun as the place where something takes place. To help you understand what we mean, let’s look at an example.
학교에서 공부해요. [hak-gyo-e-seo gong-bu-hae-yo] = I study at school.
공원에서 운동했어요. [gong-won e-seo un-dong-hae-sseo-yo] = I exercised at the park.
As you can see, in the first sentence, the word for school (학교) is followed by 에서 to mark the noun ‘school’ as the location where the speaker studies. And in the second sentence, the word for ‘park’ (공원) is followed by 에서 to mark the noun ‘park’ as the location where the speaker exercised.
#### 에 vs 에서
에 expresses a location where something exists, a destination, or a time at which something happens, whereas 에서 expresses a location at which an action takes place. Consider the following sentences:
집에 있어요. [jib-ae i-sseo-yo] = I am at home.
집에서 공부해요 [jib-ae-seo gong-bu-hae-yo] = I am studying at home.
The first sentence expresses that the speaker ‘is’ at home. That is, the place where they are ‘existing’ right now is at home. The second sentence expresses that the place where the speaker is studying is at home. In English, both sentences use the preposition ‘at‘, but in Korean 에 is used in the first sentence, and 에서 is used in the second sentence.
Tip: If you have difficulty remembering whether to use 에 or 에서, just consider if the sentence you are trying to make has an action verb. If it does, then you will need to use 에서 to mark the place where that action takes place.
### Usage 2:
The particle 에서 is also used to express ‘from a place’. For example, a common question you would get asked as a foreigner in Korea is “Where are you from?“. You can answer in the following way:
어디에서 왔어요? [eo-di-e-seo wa-sseo-yo?] = Where are you from? (Lit. Where did you come from?)
영국에서 왔어요. [yeong-gu-ge-seo wa-sseo-yo] = I came from England.
As you can see, the country 영국 (England) is marked with 에서 and this is followed by the past tense of the verb ‘to come’. So, in this example, 에서 marks 영국 (England) as the country where the speaker came from. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8311277627944946, "perplexity": 2250.2552918518195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00281.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/192/C3xD4.10D4.html | Copied to
clipboard
## G = C3×D4.10D4order 192 = 26·3
### Direct product of C3 and D4.10D4
direct product, metabelian, nilpotent (class 3), monomial, 2-elementary
Series: Derived Chief Lower central Upper central
Derived series C1 — C2×C4 — C3×D4.10D4
Chief series C1 — C2 — C22 — C2×C4 — C2×C12 — C6×Q8 — C3×C8.C22 — C3×D4.10D4
Lower central C1 — C2 — C2×C4 — C3×D4.10D4
Upper central C1 — C6 — C2×C12 — C3×D4.10D4
Generators and relations for C3×D4.10D4
G = < a,b,c,d,e | a3=b4=c2=1, d4=e2=b2, ab=ba, ac=ca, ad=da, ae=ea, cbc=dbd-1=ebe-1=b-1, dcd-1=bc, ece-1=b-1c, ede-1=d3 >
Subgroups: 242 in 142 conjugacy classes, 54 normal (22 characteristic)
C1, C2, C2, C3, C4, C4, C22, C22, C6, C6, C8, C2×C4, C2×C4, C2×C4, D4, D4, Q8, Q8, C12, C12, C2×C6, C2×C6, C42, C4⋊C4, M4(2), SD16, Q16, C2×Q8, C2×Q8, C4○D4, C4○D4, C24, C2×C12, C2×C12, C2×C12, C3×D4, C3×D4, C3×Q8, C3×Q8, C4.10D4, C4≀C2, C4⋊Q8, C8.C22, 2- 1+4, C4×C12, C3×C4⋊C4, C3×M4(2), C3×SD16, C3×Q16, C6×Q8, C6×Q8, C3×C4○D4, C3×C4○D4, D4.10D4, C3×C4.10D4, C3×C4≀C2, C3×C4⋊Q8, C3×C8.C22, C3×2- 1+4, C3×D4.10D4
Quotients: C1, C2, C3, C22, C6, D4, C23, C2×C6, C2×D4, C3×D4, C22×C6, C22≀C2, C6×D4, D4.10D4, C3×C22≀C2, C3×D4.10D4
Smallest permutation representation of C3×D4.10D4
On 48 points
Generators in S48
(1 34 45)(2 35 46)(3 36 47)(4 37 48)(5 38 41)(6 39 42)(7 40 43)(8 33 44)(9 19 32)(10 20 25)(11 21 26)(12 22 27)(13 23 28)(14 24 29)(15 17 30)(16 18 31)
(1 7 5 3)(2 4 6 8)(9 15 13 11)(10 12 14 16)(17 23 21 19)(18 20 22 24)(25 27 29 31)(26 32 30 28)(33 35 37 39)(34 40 38 36)(41 47 45 43)(42 44 46 48)
(1 6)(2 5)(3 8)(4 7)(9 14)(10 13)(11 16)(12 15)(17 22)(18 21)(19 24)(20 23)(25 28)(26 31)(27 30)(29 32)(33 36)(34 39)(35 38)(37 40)(41 46)(42 45)(43 48)(44 47)
(1 2 3 4 5 6 7 8)(9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48)
(1 21 5 17)(2 24 6 20)(3 19 7 23)(4 22 8 18)(9 43 13 47)(10 46 14 42)(11 41 15 45)(12 44 16 48)(25 35 29 39)(26 38 30 34)(27 33 31 37)(28 36 32 40)
G:=sub<Sym(48)| (1,34,45)(2,35,46)(3,36,47)(4,37,48)(5,38,41)(6,39,42)(7,40,43)(8,33,44)(9,19,32)(10,20,25)(11,21,26)(12,22,27)(13,23,28)(14,24,29)(15,17,30)(16,18,31), (1,7,5,3)(2,4,6,8)(9,15,13,11)(10,12,14,16)(17,23,21,19)(18,20,22,24)(25,27,29,31)(26,32,30,28)(33,35,37,39)(34,40,38,36)(41,47,45,43)(42,44,46,48), (1,6)(2,5)(3,8)(4,7)(9,14)(10,13)(11,16)(12,15)(17,22)(18,21)(19,24)(20,23)(25,28)(26,31)(27,30)(29,32)(33,36)(34,39)(35,38)(37,40)(41,46)(42,45)(43,48)(44,47), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48), (1,21,5,17)(2,24,6,20)(3,19,7,23)(4,22,8,18)(9,43,13,47)(10,46,14,42)(11,41,15,45)(12,44,16,48)(25,35,29,39)(26,38,30,34)(27,33,31,37)(28,36,32,40)>;
G:=Group( (1,34,45)(2,35,46)(3,36,47)(4,37,48)(5,38,41)(6,39,42)(7,40,43)(8,33,44)(9,19,32)(10,20,25)(11,21,26)(12,22,27)(13,23,28)(14,24,29)(15,17,30)(16,18,31), (1,7,5,3)(2,4,6,8)(9,15,13,11)(10,12,14,16)(17,23,21,19)(18,20,22,24)(25,27,29,31)(26,32,30,28)(33,35,37,39)(34,40,38,36)(41,47,45,43)(42,44,46,48), (1,6)(2,5)(3,8)(4,7)(9,14)(10,13)(11,16)(12,15)(17,22)(18,21)(19,24)(20,23)(25,28)(26,31)(27,30)(29,32)(33,36)(34,39)(35,38)(37,40)(41,46)(42,45)(43,48)(44,47), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48), (1,21,5,17)(2,24,6,20)(3,19,7,23)(4,22,8,18)(9,43,13,47)(10,46,14,42)(11,41,15,45)(12,44,16,48)(25,35,29,39)(26,38,30,34)(27,33,31,37)(28,36,32,40) );
G=PermutationGroup([[(1,34,45),(2,35,46),(3,36,47),(4,37,48),(5,38,41),(6,39,42),(7,40,43),(8,33,44),(9,19,32),(10,20,25),(11,21,26),(12,22,27),(13,23,28),(14,24,29),(15,17,30),(16,18,31)], [(1,7,5,3),(2,4,6,8),(9,15,13,11),(10,12,14,16),(17,23,21,19),(18,20,22,24),(25,27,29,31),(26,32,30,28),(33,35,37,39),(34,40,38,36),(41,47,45,43),(42,44,46,48)], [(1,6),(2,5),(3,8),(4,7),(9,14),(10,13),(11,16),(12,15),(17,22),(18,21),(19,24),(20,23),(25,28),(26,31),(27,30),(29,32),(33,36),(34,39),(35,38),(37,40),(41,46),(42,45),(43,48),(44,47)], [(1,2,3,4,5,6,7,8),(9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48)], [(1,21,5,17),(2,24,6,20),(3,19,7,23),(4,22,8,18),(9,43,13,47),(10,46,14,42),(11,41,15,45),(12,44,16,48),(25,35,29,39),(26,38,30,34),(27,33,31,37),(28,36,32,40)]])
48 conjugacy classes
class 1 2A 2B 2C 2D 3A 3B 4A 4B 4C ··· 4H 4I 6A 6B 6C 6D 6E 6F 6G 6H 8A 8B 12A 12B 12C 12D 12E ··· 12P 12Q 12R 24A 24B 24C 24D order 1 2 2 2 2 3 3 4 4 4 ··· 4 4 6 6 6 6 6 6 6 6 8 8 12 12 12 12 12 ··· 12 12 12 24 24 24 24 size 1 1 2 4 4 1 1 2 2 4 ··· 4 8 1 1 2 2 4 4 4 4 8 8 2 2 2 2 4 ··· 4 8 8 8 8 8 8
48 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 4 4 type + + + + + + + + + - image C1 C2 C2 C2 C2 C2 C3 C6 C6 C6 C6 C6 D4 D4 D4 C3×D4 C3×D4 C3×D4 D4.10D4 C3×D4.10D4 kernel C3×D4.10D4 C3×C4.10D4 C3×C4≀C2 C3×C4⋊Q8 C3×C8.C22 C3×2- 1+4 D4.10D4 C4.10D4 C4≀C2 C4⋊Q8 C8.C22 2- 1+4 C2×C12 C3×D4 C3×Q8 C2×C4 D4 Q8 C3 C1 # reps 1 1 2 1 2 1 2 2 4 2 4 2 2 2 2 4 4 4 2 4
Matrix representation of C3×D4.10D4 in GL4(𝔽7) generated by
4 0 0 0 0 4 0 0 0 0 4 0 0 0 0 4
,
2 3 1 0 1 1 0 4 6 5 5 2 1 4 5 6
,
2 3 2 6 6 5 4 5 2 6 5 5 4 5 5 2
,
4 3 3 6 2 4 1 4 0 5 5 6 1 1 1 1
,
0 1 5 0 6 4 4 2 0 2 2 1 2 2 1 1
G:=sub<GL(4,GF(7))| [4,0,0,0,0,4,0,0,0,0,4,0,0,0,0,4],[2,1,6,1,3,1,5,4,1,0,5,5,0,4,2,6],[2,6,2,4,3,5,6,5,2,4,5,5,6,5,5,2],[4,2,0,1,3,4,5,1,3,1,5,1,6,4,6,1],[0,6,0,2,1,4,2,2,5,4,2,1,0,2,1,1] >;
C3×D4.10D4 in GAP, Magma, Sage, TeX
C_3\times D_4._{10}D_4
% in TeX
G:=Group("C3xD4.10D4");
// GroupNames label
G:=SmallGroup(192,889);
// by ID
G=gap.SmallGroup(192,889);
# by ID
G:=PCGroup([7,-2,-2,-2,-3,-2,-2,-2,672,365,1094,520,4204,2111,1068,172,3036]);
// Polycyclic
G:=Group<a,b,c,d,e|a^3=b^4=c^2=1,d^4=e^2=b^2,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c=d*b*d^-1=e*b*e^-1=b^-1,d*c*d^-1=b*c,e*c*e^-1=b^-1*c,e*d*e^-1=d^3>;
// generators/relations
×
𝔽 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9752042293548584, "perplexity": 3255.0772701823453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587593.0/warc/CC-MAIN-20211024173743-20211024203743-00220.warc.gz"} |
http://scientificlib.com/en/Mathematics/Biographies/LorenzoMascheroni.html | Hellenica World
# .
Lorenzo Mascheroni (May 13, 1750 – July 14, 1800) was an Italian mathematician.
He was born near Bergamo, Lombardy. At first mainly interested in the humanities (poetry and Greek language), he eventually became professor of mathematics at Pavia.
In his Geometria del Compasso (Pavia, 1797), he proved that any geometrical construction which can be done with compass and straightedge, can also be done with compasses alone. However, the priority for this result (now known as the Mohr-Mascheroni theorem) belongs to the Dane Georg Mohr, who had previously published a proof in 1672.
In his Adnotationes ad calculum integrale Euleri (1790) he published a calculation of what is now known as the Euler-Mascheroni constant, usually denoted as γ (gamma).
He died in Paris. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9271872043609619, "perplexity": 2759.9061715986686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00351.warc.gz"} |
http://yewtreeapps.com/n84ou0l/solar-constant-si-unit-d66bc6 | Other values for the solar constant are found in historical literature with the value 1,353 W/m 2 appearing in many publications. Detail as many reasons as possible why your value of the solar constant differs from the accepted value. constant and Boltzmann’s constant. Planck's constant: k: 1.3806488 × 10-16 erg/K 1.3806488 × 10-23 joule/K: Boltzmann's constant: σ: 5.67 × 10-8 J/m 2 s K 4: Stefan-Boltzmann constant: kT/q: 0.02586 V: thermal voltage at 300 K: λ 0: wavelength of 1 eV photon: 1.24 μm SI base unit for length. Let’s say that a force of 1000N extends a spring at rest by 3 meters. Energy The S.I. Other units of measurement are included for … Then your % difference is: your solar constant–1376 1376 100 _____% (include your sign) Questions and Interpretations: 1. (Systeme International d’unites) in Solar Energy papers is mandatory. The values obtained by the rocket instruments for the solar constant in SI units are: 1367 w/sq m on 29 June 1976; 1372 w/sq m on 16 November 1978; and 1374 w/sq m on 22 May 1980. Other units of measurement are included for … Its value is 1388 W//m^(2). For band : where is the TOA spectral solar irradiance at a sun-earth distance of one astronomical unit (AU). The spectral distribution of direct solar radiation is altered as it passes through the atmosphere by absorption and scattering. Choose from 324 different sets of solar constant is measured in units of flashcards on Quizlet. The question is asking for photon flux, i.e. Irradiance is a measurement of solar power and is defined as the rate at which solar energy falls onto a surface. 1. that whenever expressing stellar properties in units of the solar radius, total solar irradiance, solar luminosity, solar effective temperature, or solar mass parameter, that the nominal values RN ⊙, S N ⊙, L N ⊙, T N eff, and (GM) N ⊙, be used, respectively, which are by definition exact and are expressed in SI units. We then multiply the area by the insolation (in units of energy flow per unit area) to find out the total amount of incoming energy. The unit of the spring constant k is the newton per meter (N/m). However, current literature suggests that the thermochemical calorie be used to define the langley (Delinger, 1976). When this flux 4 0. Solar Radiation Unit Conversions. SI base unit for time. 1.1 These tables define the solar constant and zero air mass solar spectral irradiance for use in thermal analysis, thermal balance testing, and other tests of spacecraft and spacecraft components and materials. One sun is defined to be equivalent to the irradiance of one solar constant, and the solar constant is defined as the irradiance of the sun on the outer atmosphere at a distance of 1 AU Traditionally, measuring the sun for PV applications has been achieved using a silicon solar cell because silicon (Si) cells are the most popular type. Define solar constant. Solar constant definition: the rate at which the sun's energy is received per unit area at the top of the earth's... | Meaning, pronunciation, translations and examples SI base unit for temperature. Consequently, the spring constant can be seen as a measure of the stiffness of the spring: how much force one must exert to either stretch or compress the spring and move it from its equilibrium. 1.2 These tables are based upon data from experimental measurements made from high-altitude aircraft, spacecraft, and the earth's surface and from solar spectral irradiance models. 1.3 The values stated in SI units are to be regarded as standard. units relevant to solar energy applications. Definition of solar constant in the Definitions.net dictionary. The langley unit has been dropped in the SI system of units. Calculations in celestial mechanics can also be carried out using the unit of solar mass rather than the standard SI unit kilogram. In this case we use the Gaussian gravitational constant which is k 2, where ${k = 0.01720209895 \ A^{\frac{3}{2}} \ D^{-1} \ S^{-\frac{1}{2}} } \$ and (this spreadsheet converts between sky cover, sunshine, and solar radiation in any of three units.) unit is the joule (J = kg m* s-*). Irradiance is sometimes referred to as flux and is a measurement of electromagnetic energy from the sun. The solar flux (SI unit ) over a spectral sensor band can be derived by convolving the top of atmosphere solar spectral irradiance and the sensor relative spectral response. The solar mass (M ☉), 1.988 92 × 10 30 kg, is a standard way to express mass in astronomy, used to describe the masses of other stars and galaxies.It is equal to the mass of the Sun, about 333 000 times the mass of the Earth or 1 048 times the mass of Jupiter. The calorie and derivatives, such as the langley 1. You should start with Planck's radiation law which gives energy flux and, since $E = h \nu$, you can relate energy to photons. UNITS The use of S.I. Solar constant definition, the average rate at which radiant energy is received from the sun by the earth, equal to 1.94 small calories per minute per square centimeter of area perpendicular to the sun's rays, measured at a point outside the earth's atmosphere when the earth is … Related Topics . Some solar constants (S 0) are provided at the bottom of Figure 3.2, starting with a distance of 0.1 astronomical unit (1 AU = 149,597,870.7 km) from the sun and travelling out to 0.8 AU, with a doubling of distance each step. Basics - The SI-system, unit converters, physical constants, drawing scales and more; Related Documents . All this does sound confusing. 2. The Planck's constant is a fixed figure, a quantum of electromagnetic action, and relates the energy carried by a photon to its frequency. In the case of solar irradiance, we usually measure the power per unit area, so irradiance is typically … ... SI base unit for the amount of a substance. In order to calculate the total amount of energy arriving at Earth, we need to know how much area is being lit. photons per unit area so you will need to relate energy to the number of photons. The symbol M ☉ is often used to refer to this unit. Q. emitted over all the wavelengths from the unit area (A =1 m. 2) of the Sun is . Insolation, radiant flux, flux density and irradiance 3 are terms that are used fairly interchangeably in solar technology discussions, for the rate of solar radiation energy flow through a unit area of space, with SI units of W/m 2 (symbol G). The values obtained by the Hickey-Frieden sensor on Nimbus 7 during the second and third flights was 1376 w/sq m. The 1367-W/sq m average solar 'constant' result is uncertain by less than + or - 0.5%, the most accurate determination to date. Solar constant is a term used to define the rate at which solar radiation is received outside the earth's atmosphere, at the earth's mean distance from the sun, by a unit surface perpendicular to the solar beam. The following is a discussion of the various S.I. The photosynthetic solar constant, which is the yearly mean solar irradiance on the surface of the earth oriented towards the sun above the atmosphere, is 1340 W/m 2 . The solar constant, a measure of flux density, is the amount of incoming solar electromagnetic radiation per unit area that would be incident on a plane perpendicular to the rays, at a distance of one astronomical unit (AU) (roughly the mean distance from the Sun to the Earth). The astronomical unit of mass is the solar mass. 1.2 These tables are based upon data from experimental measurements made from high-altitude aircraft, spacecraft, and the earth's surface and from solar spectral irradiance models. These locations correspond to the lines in the log-log plot on the upper right. The solar constant is provided in terms of power per unit area (energy flux). Mole. It is a measurement of the solar electromagnetic radiation available in a … 1.3 The values stated in SI units are to be regarded as standard. The solar constant is used to quantify the rate at which energy is received upon a unit surface such as a solar panel. The solar constant is the amount of heat energy received per second per unit area of a perfectly black surface placed at a mean distance of the Earth form the Sun, in the absence of Earth's atmosphere, the surface being held perpendicular to the direction of Sun's rays. Q I Td T (, )λ λσ ∞ =∫ =2W/m , (3) where σ is the Stefan-Boltzmann constant. Learn solar constant is measured in units of with free interactive flashcards. solar constant irradiation 2 1.4 _____ J s m2 The accepted value of the solar constant is about 1376 W/m2. The Stefan–Boltzmann constant (also Stefan's constant), a physical constant denoted by the Greek letter σ (sigma), is the constant of proportionality in the Stefan–Boltzmann law: "the total intensity radiated over all wavelengths increases as the temperature increases", of a black body which is proportional to the fourth power of the thermodynamic temperature. The unit of power is the Watt (abbreviated W). solar constant synonyms, solar constant pronunciation, solar constant translation, English dictionary definition of solar constant. download Excel spreadsheet with macro here. Astronomical Units/Data NAME SYMBOL NUMBER EXP CGS UNITS ----- Astronomical unit AU 1.496 13 cm Parsec pc 3.086 18 cm Light year ly 9.463 17 cm Solar mass M o 1.99 33 g Solar radius R o 6.96 10 cm Solar luminosity L o 3.9 33 erg s-1 Solar Temperature T o 5.780 3 K ----- The solar constant is calculated by multiplying the sun's surface irradiance by the square of the radius of the sun over the average distance between the Earth and the sun. For simplicity, considering the Sun to be an ideal blackbody (ε=1) the solar flux . 1 BTU = 251.9958 Calorie 1 BTU = 1055.056 Joule 1 BTU = 1055.056 Watt-sec 1 Langley = 1 Cal/cm 2 1 Calorie = 4.1868 Joule 1 Watt = 1 Joule/sec 1 Watt-sec = 1 Joule ... at a distance of one astronomical unit. The estimated value of the solar constant is 1.4 kJ per second per square metre. If the solar constant for the earth is 's'. Meaning of solar constant. The uncertainty of the rocket measurements is +- 0.5%. The specific value at Earth of 1,361 W/m 2 is called the "solar constant". On May 20, 2019, the world unanimously adopted the new definition of the kilogramme, which is based on a physical constant, the Planck's constant. The factor to convert the formula back to cgs units are shown in columns 3, and 5 for the natural unit of [cm] and [erg] respectively; while the corresponding numerical value in cgs is obtained by division (for example the natural unit of energy density is ergs 4, which … ( ε=1 ) the solar flux translation, English dictionary definition of solar constant is in. In SI units are to be an ideal blackbody ( ε=1 ) the solar flux being lit the... Irradiance is sometimes referred to as flux and is defined as the rate at which is... Earth, we need to relate energy to the number of photons many publications your % difference is: solar! Joule ( J = kg m * s- * ) be regarded as standard solar constant si unit! 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Is mandatory rest by 3 meters unit is the solar constant is used to refer to this unit ∞! International d ’ unites ) in solar energy falls onto a surface abbreviated. Calorie be used to define the langley ( Delinger, 1976 ) meter ( N/m ) and is defined the. Is often used to define the langley unit has been dropped in the log-log plot the... The spectral distribution of direct solar radiation in any of three units.,! 1,361 W/m 2 is called the solar constant is used to define the langley (,! Sign ) Questions and Interpretations: 1 such as a solar panel, physical constants, drawing scales and ;. Current literature suggests that the thermochemical calorie be used to define the langley has! Power is the newton per meter ( N/m ) sign ) Questions and Interpretations:.! Detail as many reasons as possible why your value of the Sun is is +- 0.5.! That the thermochemical calorie be used to define the langley unit has been dropped in SI... 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D1 Lacrosse Schools Women's, Ben Dunk Wife Name, Things To Do When Bored For Teenage Guys, Data Center Design Standards, Demarini Cf Zen End Cap, Romans 8:22-27 Commentary, | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9750209450721741, "perplexity": 1425.279123228303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00587.warc.gz"} |
http://link.springer.com/article/10.1023%2FA%3A1013163427655 | Article
Geometriae Dedicata
, Volume 88, Issue 1, pp 199-210
# Pointed Wiedersehen Metrics on Exotic Spheres and Diffeomorphisms of S6
• Carlos E. DuránAffiliated withIVIC-Matemáticas
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## Abstract
Using a Kaluza–Klein-type procedure, an explicit metric h on an exotic sphere Σ7 is constructed, satisfying the Wiedersehen condition at a set of points diffeomorphic to S 1. The formulas for the geodesics allows the writing down of formulas for an explicit degree 1 diffeomorphism σ: S 6S 6 that is not isotopic to the identity.
closed geodesics exotic spheres exotic diffeomorphisms | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9080037474632263, "perplexity": 3233.8498291722235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00027-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://pure.au.dk/portal/en/publications/cluster-perturbation-theory-viii-first-order-properties-for-a-coupled-cluster-state(53b5e721-8c96-4d27-8103-ca0404e377ce).html | # Faculty of Natural Sciences
## Cluster perturbation theory. VIII. First order properties for a coupled cluster state
Research output: Contribution to journal/Conference contribution in journal/Contribution to newspaperJournal articleResearchpeer-review
### DOI
We have extended cluster perturbation (CP) theory to comprehend the calculation of first order properties (FOPs). We have determined CP FOP series where FOPs are determined as a first energy derivative and also where the FOPs are determined as a generalized expectation value of the external perturbation operator over the coupled cluster state and its biorthonormal multiplier state. For S(D) orbital excitation spaces, we find that the CP series for FOPs that are determined as a first derivative, in general, in second order have errors of a few percent in the singles and doubles correlation contribution relative to the targeted coupled cluster (CC) results. For a SD(T) orbital excitation space, we find that the CP series for FOPs determined as a generalized expectation value in second order have errors of about ten percent in the triples correlation contribution relative to the targeted CC results. These second order models, therefore, constitute viable alternatives for determining high quality FOPs.
Original language English 024108 Journal of Chemical Physics 157 2 0021-9606 https://doi.org/10.1063/5.0082585 Published - Jul 2022
### Bibliographical note
Publisher Copyright:
© 2022 Author(s).
Citationformats
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https://www.physicsforums.com/threads/basic-quantum-physics-balmer-series.530085/ | # Homework Help: Basic quantum physics, Balmer series
1. Sep 14, 2011
### Quaint
1. The problem statement, all variables and given/known data
The first five ionization energies of hydrogen are 1312, 328, 146, 82 and 52 kJ/mol. Calculate the wawelength of the first three in Balmer series.
2. Relevant equations
Avokadro constant, NA = 6,022 x 1023 /mol
Planck constant, h = 6,626 x 10-34Js
speed of light, c = 2,998 x 108ms-1
energy, E in J
wavelength, lambda in nm = 10-9m
frequency of radiation, v in s-1
E = hv
c = lambda x v
3. The attempt at a solution
So, in my knowledge you only need the first three energies given in problem and the energies represents the value of one mole of photons. So dividing the 1312 with Avokadro you will get the energy of one photon which is 2.17868 x 10-21J
Using the equation E = hv you will get v = 3,28807 x 1012s-1
and using c = lambda x v you will get lambda = 9,1178 x 10-5nm
So the exponent is definitely wrong but so are numbers because this wavelength should be visible to human eye (400-780nm). You can solve this other way when you are not using the given energies and the rydberg value is present in the equation but this should be another way to do this.
Thx beforehand
2. Sep 14, 2011
### vela
Staff Emeritus
One suggestion I have is that you get comfortable working with the electron-volt as a unit of energy. The conversion is 1 eV = 1.6x10-19 J. It makes the numbers quite a bit easier to deal with.
Here's another trick that can save you some time. The combination hc is approximately 1240 nm eV, so the energy and wavelength of a photon are related by$$E = \frac{hc}{\lambda} = \frac{1240~\mathrm{nm~eV}}{\lambda}$$
Regarding the actual problem, you want to look what the definition of the Balmer series is. Also, how are the ionization energies related to the energy of the electron states? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8873085975646973, "perplexity": 1044.0478678244722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00438.warc.gz"} |
https://mathoverflow.net/questions/333955/is-every-grothendieck-category-with-a-generator-a-category-of-sheaves | # Is every Grothendieck category with a generator a category of sheaves?
The Gabriel-Popescu theorem tells us that every Grothendieck category with a generator is a left exact localization of a module category. I'm interested in a slightly different way of "representing" such categories:
Question: Let $$\mathcal C$$ be a Grothendieck category with a generator.
1. Does there exist a Grothendieck topos $$\mathcal X$$ such that $$\mathcal C \simeq Ab(\mathcal X)$$ is equivalent to the category of abelian group objects in $$\mathcal X$$?
2. Slightly less naively, does there exist a ringed Grothendieck topos $$(\mathcal X, \mathcal O_{\mathcal X})$$ such that $$\mathcal C \simeq \mathcal O_{\mathcal X}\textrm{-}Mod$$ is equivalent to the category of $$\mathcal O_\mathcal X$$-modules?
3. Same as (2), but with $$(\mathcal X, \mathcal O_{\mathcal X})$$ being locally ringed?
• You actually also get from Gabriel-Popescu that every Grothendieck category is a left exact accessible localisation of presheaves of abelian groups. This should solve 1. – user40276 Jun 13 at 21:26
• @user40276 I don't think I follow. I can see how the Gabriel-Popescu theorem exhibits a Grothendieck category as a localization of a category of $Ab$-enriched presheaves on a certain $Ab$-enriched caetgory (viz. a certain ring) -- but I want to relate this to a topos -- i.e. a category of sheaves of sets on a certain $Set$-enriched category. – Tim Campion Jun 13 at 21:29
• It seems that the smallest Grothendieck topology generated by the "additive Grothendieck topology" of presheaves of abelian groups on that pre-additive category should work. But, now, that I think again about it, I'm not so sure... – user40276 Jun 13 at 21:49
• I'm pretty sure the answer is no, but I don't know a proof off the top of my head. In noncommutative algebraic geometry, they take Grothendieck categories of graded modules over a graded ring, modulo finite-dimensional modules. If there was a result like you have in mind, the theory would probably be very different. – arsmath Jun 14 at 7:28
• In fact, Hovey shows (Thm 3.3) that if $R$ is a division ring which is not a field, then the category $RMod$ of left $R$-modules does not admit any additive closed symmetric monoidal structure. This is a counterexample to (1) and to (2,3) unless we allow $\mathcal O_{\mathcal X}$ to be noncommutative. If we assume our Grothendieck category is symmetric monoidal closed, I'd still be interested to see if maybe it must be of the form $\mathcal O_{\mathcal X}Mod$ with $\mathcal O_{\mathcal X}$ commutative. – Tim Campion Jun 15 at 17:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.925216019153595, "perplexity": 213.48725414739073}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541308604.91/warc/CC-MAIN-20191215145836-20191215173836-00240.warc.gz"} |
http://math.stackexchange.com/questions/9824/probability-question?answertab=votes | # Probability question
You have three bags, A, B, and C. Bags A and B hold x red and y blue marbles each; bag C holds 2x red marbles and 2y blue marbles.
Let us say that we pull out z marbles from bag A, z marbles from bag B, and 2z marbles from bag C. Will the expected number of red marbles pulled from bag A and bag B together be the same as the expected number of marbles pulled from bag C?
I believe the answer is yes, but I am having trouble figuring out why.
-
The expected number of red balls pulled from A is $z\frac{x}{x+y}$. This is because the probability that any of the $z$ individual balls is red is $\frac{x}{x+y}$. To see that, consider the experiment of taking $z$ balls. Number each ball you've taken - the first ball number $1$, the second ball $2$, and so on. Each of the numbers on its own is uniformly distributed, so the probability that the $k$th ball is red is exactly $\frac{x}{x+y}$.
Yes. The expected number of red marbles from A is $\frac{xz}{x+y}$ and the same for B. The expected number from C is $\frac{4xz}{2x+2y}$ The 4 in the numerator is 2*2, one from the number of pulls, one from the number of red marbles. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9293033480644226, "perplexity": 230.71315331454045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696383263/warc/CC-MAIN-20130516092623-00080-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/325621/does-induction-for-a-functor-algebra-imply-it-is-initial | # Does induction for a functor algebra imply it is initial?
By "induction" I mean "no proper subalgebras". My thinking goes like this:
1. For natural numbers, recursion and induction are in some sense the same thing. In particular, given a recursive definition of $f$ you would prove its totality roughly by saying "if I can define $f$ on $1\dots n$, then I can define it on $n+1$", i.e. by induction.
2. The proper categorical notion of recursion is initial algebras – in particular, for $F(X) = 1\sqcup X$ an initial algebra is a natural number object, the property of being initial being precisely what you need to define functions by recursion.
3. An initial algebra automatically has a notion of induction: initial objects have no proper subobjects, so if you have some subobject that is closed under the algebra operations, this means precisely that the inclusion is an algebra homomorphism, and therefore an isomorphism.
4. I'd really like to go the other way, but in general the implication "$I$ is initial $\implies$ every mono into $I$ is an iso" cannot be reversed (its dual has a counterexample in $\mathbf{Set}$, in that every epi from $0$ is an iso but $0$ is not terminal).
Are there circumstances where an algebra having no proper subalgebra means it is initial?
-
Please feel free to add more tags, I wasn't sure if e.g. universal-algebra was appropriate. – Ben Millwood Mar 9 '13 at 16:09
You also have to know that the algebra is "free" in a suitable sense, if you want induction to imply recursion. For example, every simple abelian group has "induction" in the sense of having only the trivial subgroup as a proper subgroup, but "recursion" is only possible for $\mathbb{Z}$ and not, say, $\mathbb{Z} / p \mathbb{Z}$. – Zhen Lin Mar 9 '13 at 16:18
Makes sense. Is there a characterisation of "free" that is not "induction implies recursion"? Or any nice sufficient condition for it? – Ben Millwood Mar 9 '13 at 16:25
(A thought: recursion might fail for $\mathbb Z/p\mathbb Z$, but if it works it will be unique... maybe I am looking for a quasi-initial object? Especially since in my counterexample, $0$ is indeed *sub*terminal...) – Ben Millwood Mar 9 '13 at 16:34
In general, it is possible for an object to have every mono into it an iso, but not be initial, weakly initial, or quasi-initial. The pre-condition is too easily satisfied just by not having very many morphisms into a given object, so to give it some traction, I'll need an existence-of-morphisms condition.
Theorem: In a category with equalisers, if $I$ is an object such that any mono $A \rightarrowtail I$ is an iso, then $I$ is quasi-initial.
Proof: Let $f, g : I \rightrightarrows X$, and take $A \xrightarrow{e} I$ the equaliser of $f$ and $g$. An equaliser is a (regular) mono, so by hypothesis on $I$ it is an iso, but if the equaliser of two morphisms is an iso then they are equal. So $f = g$, so there is at most one morphism from $I$ to any other object.
In the particular case of categories of functor algebras and homomorphisms, the forgetful functor to the underlying category creates limits, so in particular it is enough for the underlying category to have equalisers (thanks Zhen Lin for reminding me of this).
I've not yet come up with any interesting word on when there is at least one morphism, so that $I$ really is initial instead of quasi-so.
-
And, of course, the category of algebras for an endofunctor has equalisers if the base category does. – Zhen Lin Mar 25 '13 at 23:52
Yes, I thought that, but forgot to mention it. Might edit it in. – Ben Millwood Mar 25 '13 at 23:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9708757400512695, "perplexity": 369.5064476794777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999650775/warc/CC-MAIN-20140305060730-00028-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://lt-jds.jinr.ru/record/72344?ln=ru | / hep-ex arXiv:1801.02052
Measurements of $t\bar{t}$ differential cross-sections of highly boosted top quarks decaying to all-hadronic final states in $pp$ collisions at $\sqrt{s}=13\,$ TeV using the ATLAS detector
Pages: 57
Abstract: Measurements are made of differential cross-sections of highly boosted pair-produced top quarks as a function of top-quark and $t\bar{t}$ system kinematic observables using proton--proton collisions at a center-of-mass energy of $\sqrt{s} = 13$ TeV. The data set corresponds to an integrated luminosity of $36.1$ fb$^{-1}$, recorded in 2015 and 2016 with the ATLAS detector at the CERN Large Hadron Collider. Events with two large-radius jets in the final state, one with transverse momentum $p_{\rm T} > 500$ GeV and a second with $p_{\rm T}>350$ GeV, are used for the measurement. The top-quark candidates are separated from the multijet background using jet substructure information and association with a $b$-tagged jet. The measured spectra are corrected for detector effects to a particle-level fiducial phase space and a parton-level limited phase space, and are compared to several Monte Carlo simulations by means of calculated $\chi^2$ values. The cross-section for $t\bar{t}$ production in the fiducial phase-space region is $292 \pm 7 \ \rm{(stat)} \pm 76 \rm{(syst)}$ fb, to be compared to the theoretical prediction of $384 \pm 36$ fb.
Note: 57 pages in total, author list starting page 41, 18 figures, 5 tables, submitted to PRD. All figures including auxiliary figures are available at https://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/PAPERS/TOPQ-2016-09/; *Temporary entry*
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https://www.physicsforums.com/threads/electrostatics-and-coulumbs-law-potentials-and-test-charges.670563/ | # Electrostatics and Coulumb's law, potentials and test charges
1. Feb 9, 2013
### Plasmosis1
I am given 4 potential charges, 1V, 3V, -6V, and 5V, all arranged randomly around each other. There is a test charge located at infinity. No distances are given between the potentials.
1. Where will a test charge of 2*10^-6 C travel?
2. What is its kinetic energy?
3. Find the capatinance of a capacitor with the same energy as #2
4. Use a -2*10^-6 C test charge instead on the +2*10^-6 C charge. Where will it go?
5. If the new charge has a mass of 37.3*10^-21 what will its speed be at its destination?
1. I think it will go towards the -6 potential
2. PE= q*v and I think the PE will equal the KE when it arrives. 2*10^-6*(1+5+(-6)+3)=6*10^-6. However I'm not sure if I should add the volts together.
3. C=q/v so 2*10^-6/(1+5+(-6)+3)=6.67*10^-7 F. Again, I'm not sure if I should add the volts together.
4. I would think the charge would stay in infinity because it has an opposite charge.
5. I have no clue on this question because it contradics my previous answer.
I don't really have any idea what I'm doing so any help would be appreciated.
2. Feb 12, 2013
### ehild
You mean that there are four charged objects (small metal spheres, for example) held at different potentials (with respect to infinity).
1.) Yes, the positive test charge will go to the negatively charged object.
2.) The test charge had zero kinetic energy and zero potential energy at infinity. Arriving to the place where the potential is -6 V, its potential energy is PE=UQ=-6*2˙10-6=-12˙10-6 J. As the energy is conserved in an electric field, KE=12˙10-6 J. The other charged objects do not matter as the potential is given at the place where the test charge arrives. Do not add the potentials.
3.) You have to find the a capacitance if the capacitor has the same energy as the KE of the test charge. How is the energy of a capacitor expressed in terms of its capacitance and voltage?
4.) When the test charge is negative it will be attracted by the objects at positive potential, and it will go to the place at highest potential.
5.)Calculate the kinetic energy again, and you get the speed from it.
ehild
Similar Discussions: Electrostatics and Coulumb's law, potentials and test charges | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.945722758769989, "perplexity": 791.1483867556288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826210.56/warc/CC-MAIN-20171023164236-20171023184236-00522.warc.gz"} |
https://deepai.org/publication/exact-pressure-elimination-for-the-crouzeix-raviart-scheme-applied-to-the-stokes-and-navier-stokes-problems | DeepAI
# Exact pressure elimination for the Crouzeix-Raviart scheme applied to the Stokes and Navier-Stokes problems
We show that, using the Crouzeix-Raviart scheme, a cheap algebraic transformation, applied to the coupled velocity-pressure linear systems issued from the transient or steady Stokes or Navier-Stokes problems, leads to a linear system only involving as many auxiliary variables as the velocity components. This linear system, which is symmetric positive definite in the case of the transient Stokes problem and symmetric invertible in the case of the steady Stokes problem, with the same stencil as that of the velocity matrix, provides the exact solution of the initial coupled linear system. Numerical results show the increase of performance when applying direct or iterative solvers to the resolution of these linear systems.
• 1 publication
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04/28/2021
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### Alya towards Exascale: Algorithmic Scalability using PSCToolkit
In this paper, we describe some work aimed at upgrading the Alya code wi... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8693192601203918, "perplexity": 1559.9704701818625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00734.warc.gz"} |
https://openstax.org/books/college-physics-ap-courses/pages/4-conceptual-questions | College Physics for AP® Courses
# Conceptual Questions
### 4.1Development of Force Concept
1.
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
2.
What properties do forces have that allow us to classify them as vectors?
### 4.2Newton's First Law of Motion: Inertia
3.
How are inertia and mass related?
4.
What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
### 4.3Newton's Second Law of Motion: Concept of a System
5.
Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and give an example.
6.
Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?
7.
Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion.
8.
Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.
9.
A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.
10.
A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
11.
(a) Give an example of different net external forces acting on the same system to produce different accelerations. (b) Give an example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law accurately describes both effects? State it in words and as an equation.
12.
If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.
13.
If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?
14.
The gravitational force on the basketball in Figure 4.6 is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal?
### 4.4Newton's Third Law of Motion: Symmetry in Forces
15.
When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)
16.
A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?
17.
Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton’s laws of motion apply?
18.
Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired?
19.
An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.
20.
Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects whether one such pair of forces cancels.
### 4.5Normal, Tension, and Other Examples of Force
21.
If a leg is suspended by a traction setup as shown in Figure 4.29, what is the tension in the rope?
Figure 4.29 A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T without changing its magnitude.
22.
In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure 4.29.) (Note that the tibia is the shin bone shown in this image.)
### 4.7Further Applications of Newton's Laws of Motion
23.
To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at $gg size 12{g} {}$. Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?
24.
A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.
### 4.8Extended Topic: The Four Basic Forces—An Introduction
25.
Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force.
26.
What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances?
27.
Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.)
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As an Amazon Associate we earn from qualifying purchases. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8435801863670349, "perplexity": 703.1483693601907}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00585.warc.gz"} |
http://math.stackexchange.com/questions/321238/why-not-just-study-the-consequences-of-hausdorff-axiom-what-do-statements-like | # Why not just study the consequences of Hausdorff axiom? What do statements like, “The arbitrary union of open sets is open,” gain us?
Define that a pair $(X,\tau)$ where $\tau \subseteq \mathcal{P}(X)$, is a Hausdorff space if for all distinct $a,b \in X$ there exist $A,B \in \tau$ such that $$a \in A, \;b \in B, \;A \cap B = \emptyset.$$
Note that a Hausdorff space, according to this definition, needn't be a topological space, and vice versa.
Call the elements of $\tau$ "open sets", whether or not $\tau$ is a topology.
Then many of the definitions of topological spaces apply equally well to Hausdorff spaces. In particular, the notion of a convergent sequence can be defined as per usual.
In particular, define that a sequence $x : \mathbb{N} \rightarrow X$ converges iff there exists $a \in X$ such that for all open $A$ such that $a \in A$, there exists $N$ such that for all $n \geq N$ it holds that $x_n \in A$.
Then we can write this proof to show that every convergent sequence in $X$ has a unique limit. (i.e. $a$ is unique).
Presumably, the more general result that every net has a unique limit can also be proven.
So why not study limits in arbitrary Hausdorff spaces? What goes wrong when we neglect to assume statements like, "The arbitrary union of open sets is open"?
Edit: So to clarify, my interest lies in the interaction between the topological space axioms and the Hausdorff axiom. What is it about a Hausdorff topological space that is so magical? There must be some sort of synergy going on, or they would have been studied independently of one another.
Note that we can define continuity of functions between topological spaces in terms of preimages of open sets. Lets call this "pre-images continuous." And we can define continuity of functions between Hausdorff spaces in terms of limits of nets. Lets call this "limit-continuous." Perhaps these notions coincide precisely in the case of Hausdorff topological spaces?
Note also that the topology generated by a Hausdorff space is necessarily Hausdorff. So for any Hausdorff space $X$, lets $X'$ denote the (necessarily Hausdorff) topological space generated by $X$. Perhaps a function $f : X \rightarrow Y$ between Hausdorff spaces is limit-continuous if and only if $f : X' \rightarrow Y'$ is (open-preimages)-continuous.
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How does one define a limit if we don't have a topology? – Alex Youcis Mar 5 '13 at 8:48
@AlexYoucis The same way we always define it. Except by "open set" we mean "an element of $\tau$." – goblin Mar 5 '13 at 8:50
$A,B \subseteq X$ or $A,B \in \tau$? – Boris Novikov Mar 5 '13 at 8:51
@BorisNovikov Thanks, fixed it. ;) – goblin Mar 5 '13 at 8:52
@AlexYoucis For example, here is the definition of a convergent sequence. – goblin Mar 5 '13 at 8:56
I see two main answers for questions of the form
Why aren't we studying <more general case of something> ?
and often both apply:
1. The more general case isn't as nice.
2. It has not yet come up naturally in work on theories / problems that we currently study.
Perhaps a more blunt phrasing of the second answer would be
2'. It has not yet come up in a situation where anyone would care.
There is also sometimes a third answer, which is
3. People do study that, it's just not mainstream.
For your particular question, about spaces only assumed to satisfy the Hausdorff separation axiom:
I haven't really put much effort in figuring out whether answer 1 applies, but I think it's quite likely that answer 2 does. Maybe it's true that an intricate theory of these Hausdorff-only-spaces could be developed; but as far as I'm aware, currently, there is no reason anyone would care. I could just as easily ask about the theory of "droops", which are (let's say) sets equipped with a ternary operation $\oslash$ satisfying the identity $$\oslash(a,\oslash(a,b,c),\oslash(c,b,a))=\oslash(b,a,c).$$ Unless, and until, someone demonstrates that such a thing occurs naturally in the mathematics we're already doing, I find it very hard to be interested; and indeed, the theory of these things themselves would suffer, due to a lack of intuition and interconnection with the rest of mathematics. Once the world realizes that the set of complex structures on a manifold, or the rational points on an elliptic curve, naturally forms a droop, then you'll start getting people interested in droops, and moreover, this connection will give you things to think about, things to ask, things to investigate, and the theory of droops will begin to flourish. Without such a connection, any mathematician who was polite and listened to someone explain the definition of a droop most likely wouldn't be able to say anything about them because they lack any previous experience or intuition for such an object.
Okay, I'm exaggerating a bit with droops; Hausdorff-only-spaces are somewhat closer to everyday mathematical experience than a droop. And I don't mean to sound angry or discouraging; it's great to ask questions about why we do things one way and not the other, especially when answer 1 applies and the ways in which the general case are worse have been thoroughly investigated, and you'll gain some real mathematical experience from learning about why we're doing things the way we are. But the overall point is hopefully clear.
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+1. – Did Mar 5 '13 at 10:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9489507079124451, "perplexity": 380.86458032378374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645305536.82/warc/CC-MAIN-20150827031505-00058-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://grupsderecerca.uab.cat/hpca4se/ca/content/adapting-map-resolution-accomplish-execution-time-constraints-wind-field-calculation | # Adapting map resolution to accomplish execution time constraints in wind field calculation
Títol Adapting map resolution to accomplish execution time constraints in wind field calculation Tipus de publicació Journal Article Any de publicació 2015 Autors Sanjuán, G, Margalef, T, Cortés, A Journal Procedia Computer Science Volume 51 Resum Forest fires are natural hazards that every year destroy thousands of hectares around the world. Forest fire propagation prediction is a key point to fight against such hazards. Several models and simulators have been developed to predict forest fire propagation. These models require input parameters such as digital elevation map, vegetation map, and other parameters describing the vegetation and meteorological conditions. However, some meteorological parameters, such as wind speed and direction, change from one point to another one due to the effect of the topography of the terrain. Therefore, it is necessary to couple wind field models, such as WindNinja, to estimate the wind speed and direction at each point of the terrain. The output provided by the wind field simulator is used as input of the fire propagation model. Coupling wind field model and forest fire propagation model improves accuracy prediction, but increases significantly prediction time. This fact is critical since propagation prediction must be provided in advance to allow the control centers to manage firefighters in the best possible way. This work analyses WindNinja execution time, describes a WindNinja parallelisation based on map partitioning, determines the limitations of such methodology for large maps and presents an improvement based on adapting map resolution to accomplish execution time limitations. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8037105202674866, "perplexity": 1945.0953822953888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00626.warc.gz"} |
http://mathoverflow.net/questions/76638/when-is-a-class-function-on-a-group-g-finite-abelian-into-the-rational-numbers | # When is a class function on a group G (finite abelian) into the rational numbers Q an element of the rational representation ring of G?
Given a class function $f: G \to \mathbb Q$, where $G$ is a finite abelian group, is there an easy way to decide whether $f$ is an element of the rational representation ring $R_{\mathbb Q}(G)$, i.e. whether $f$ is a virtual character of some representations of $G$?
If it makes things easier you could also assume that $G$ is a $p$-group and the class function has values in $\mathbb Z$.
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This essentially boils down to the case of a cyclic group. For a cyclic group of order n, the irreducible representations correspond to the action on $\mathbb Q[\omega_d]$ where $\omega_d$ is a primitive $d^{th}$-root of unity where $d$ divides $n$. So one can easily produce the rational character table and check if your function is a non-negative linear combination of these irreducible characters.
Added: You can now use the orthogonality relations over $\mathbb Q$ to see if something is a character of a rep. One knows the endomorphism algebra of $\mathbb Q[\omega_d]$ as a module over the cyclic group $G$ is $\mathbb Q[\omega_d]$ which has degree $\phi(d)$. So the orthogonality relations (valid over any field of characteristic 0, but in the non-algebraically closed case involves the dimensions of the endomorphism division algebras) can be used to decompose class functions in terms of irreducible characters.
Answer 2 Alternatively, one can build the character table of a finite abelian group $G$ over $\mathbb Q$ in the following way (without reducing to the case of a cyclic group). The Schur index of a degree one character is $1$. So the irreducible characters for $G$ are obtained as follows. Take a complex character $\chi$ and let $\Gamma(\chi)=Gal(\mathbb Q[\chi]:\mathbb Q)$ where $\mathbb Q[\chi]$ is the field generated by the values of $\chi$. Then the sum of the orbit of $\chi$ under $\Gamma$ is a $\mathbb Q$-irreducible character and all such characters are obtained in this fashion. So this describes an orthogonal basis for class functions over $\mathbb Q$ and hence all class functions.
supplement This may be off, but I think a rational-valued class function $f$ is a rational character iff for each complex character $\chi$, one has the inner product $\langle f,\chi\rangle$ is an integer and this integer is constant on the orbit of $\chi$ under $\Gamma(\chi)$.
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Could you explain a little more why "this essentially boils down to the case of cyclic groups"? – Dimitrios Sep 29 '11 at 10:29
Every finite abelian group is a direct product of cyclic groups. A representation of a direct product is a tensor product of representations of the factors. So the characters of an abelian group over $\mathbb Q$ will be products of characters of cyclic groups over $\mathbb Q$. So once you compute the character table of a cyclic group, as I explained how to do above, then you can compute the character table of any abelian group. – Benjamin Steinberg Sep 29 '11 at 12:33
The formula $R_{\mathbb Q}(G_1\times G_2)\cong R_{\mathbb Q}(G_1)\otimes R_{\mathbb Q}(G_2)$ only holds if the orders of the (finite) groups $G_1$ and $G_2$ are coprime. So especially in the case where $G$ is a $p$-group you can't reduce the problem to cyclic subgroups. I might be wrong though or I didn't get something right. – Dimitrios Sep 29 '11 at 13:11
One more question concerning your example: $\mathbb Q[\omega_3]\otimes\mathbb Q[\omega_3]\cong \mathbb Q[\omega_3]\times \mathbb Q[\omega_3]$, thus it is not irreducible as an $\mathbb Q(G_1\times G_2)$-module, or is it? – Dimitrios Sep 29 '11 at 16:12
Your are right of course and what I wrote was dumb. The nontrivial irreducible representations of $(Z/p)^n$ over $\mathbb Q$ all have degree $p-1$. Indeed, this is true for $Z/p$. Now there are (p^n-1)/(p-1) distinct homomorphisms from $(Z/p)^n$ to $Z/p$. So this gives $(p^n-1)/(p-1)$ distinct irreducible reps of degree $p-1$. Thus in the regular representation this accounts for all nontrivial irreducible reps. I think my second answer using Schur index is correct. – Benjamin Steinberg Oct 4 '11 at 16:05
A necessary criterion is that, if $g \in G$ has order $m$, and $k$ is relatively prime to $m$, then $f(g) = f(g^k)$.
Proof: Let $\rho$ be a rational representation. Let the eigenvalues of $\rho(g)$ be $\omega_1$, $\omega_2$, ..., $\omega_N$. Then $f(g) = \sum \omega_i$ and $f(g^k) = \sum \omega_i^k$.
But the $\omega_i$ are all $m$-th roots of unity, and there is a Galois symmetry $\sigma$ of $\mathbb{Q}(e^{2 \pi i/m})$ which acts on $m$-th roots of unity by $\omega \mapsto \omega^k$. So $f(g^k)$ is the image of $f(g)$ under this symmetry. Since $f(g)$ is in $\mathbb{Q}$, this shows that $f(g) = f(g^k)$. QED
One can show that any rational class function as above is a $\mathbb{Q}$-linear combination of rational characters. If you want to know about $\mathbb{Z}$-linear combinations, that sounds harder; I don't know a simple rule.
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http://mathhelpforum.com/math-topics/169486-electric-field-potential-conceptual-help.html | # Thread: Electric Field and Potential (conceptual help)
1. ## Electric Field and Potential (conceptual help)
The question states that there are 5 energy contours, each with a potential difference of 0.3 J/C. The center is a conducting core, and the very outer edge is a sheath. The outer sheath is the final energy contour. Therefore, the outer edge of the center conducting core isn't an energy contour, since the question states there are five energy contours.
Consequently, I am wondering if you would still consider moving from the center core (black dot) to the very first equipotential curve to be a potential of 0.3? Would you still consider the black dot to be some potential difference than the energy contours?
2. For example, if we were told that we were moving a particle from the center core (the large black dot), to the very outer sheath (the fifth energy contour), what would be the potential difference? Would the first energy contour be 0.3 J/C difference from the center black core, or would you neglect this black core, and consider the first energy contour to be zero?
The question states that the field is pointingm from the center core to the outer sheath. Therefore, potential energy increases for a positive charge moving from A to G. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9286088347434998, "perplexity": 821.1125253946376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00449.warc.gz"} |
http://math.stackexchange.com/questions/210207/whats-the-name-of-this-theorem-fg-given-fg-and-initial-value | # What's the name of this theorem ($f>g$ given $f'>g'$ and initial value)?
I need to use this theorem as part of a proof. Basically it goes as follows:
If $f(x_0) > g(x_0)$ and $\frac{d}{dx} f(x) > \frac{d}{dx} g(x)$ for all $x \geq x_0$, then $f(x) > g(x)$ for all $x\geq x_0$.
What's the name of this theorem, if it has one? Or if it doesn't, is there a simple way to prove it that I could include in my proof? Thanks.
-
Hint: it follows directly from the mean value theorem. – mjqxxxx Oct 10 '12 at 2:46
This is commonly called the Racetrack principle.
-
Hint
Consider $$h=f-g$$
You have then that $$h'>0$$ and that $$h(x_0)>0$$
Why must then $h$ remain positive for $x\geq x_0$?
Spoiler
If $f'>0$ on some domain $D$; then $f$ is increasing on $D$.
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https://socratic.org/questions/using-the-pythagorean-theorem-how-do-you-solve-for-the-missing-side-given-a-24-a | Algebra
Topics
# Using the pythagorean theorem, how do you solve for the missing side given a = 24 and b = 45, then c?
May 27, 2016
$c = 51$
#### Explanation:
The Pythagorean Theorem is ${a}^{2} + {b}^{2} = {c}^{2}$
$a = 24$
$b = 45$
c=?
${24}^{2} + {45}^{2} = {c}^{2}$
$576 + 2025 = {c}^{2}$
$2601 = {c}^{2}$
$\sqrt{2601} = c$
$c = 51$
##### Impact of this question
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http://www.eecs70.org/static/notes/n13.html | # Introduction to Discrete Probability
In the last note we considered the probabilistic experiment where we flipped a fair coin $$10000$$ times and counted the number of $$H$$s. We asked “what is the chance that we get between $$4900$$ and $$5100$$ $$H$$s”. One of the lessons was that the remarkable concentration of the fraction of Hs had something to do with the astronomically large number of possible outcomes of $$10000$$ coin flips.
In this note we will formalize all these notions for an arbitrary probabilistic experiment. We will start by introducing the space of all possible outcomes of the experiment, called a sample space. Each element of the sample space is assigned a probability - which tells us how likely it is to occur when we actually perform the experiment. The mathematical formalism we introduce might take you some time to get used to. But you should remember that ultimately it is just a precise way to say what we mean when we describe a probabilistic experiment like flipping a coin $$n$$ times.
# Random Experiments
In general, a probabilistic experiment consists of drawing a sample of $$k$$ elements from a set $$S$$ of cardinality $$n$$. The possible outcomes of such an experiment are exactly the objects that we counted in the last note. Recall from the last note that we considered four possible scenarios for counting, depending upon whether we sampled with or without replacement, and whether the order in which the $$k$$ elements are chosen does or does not matter. The same will be the case for our probabilistic experiments. The outcome of the random experiment is called a sample point. The sample space, often denoted by $$\Omega$$, is the set of all possible outcomes.
An example of such an experiment is tossing a coin $$4$$ times. In this case, $$S = \{H,T\}$$ and we are drawing 4 elements with replacement. $$HTHT$$ is an example of a sample point and the sample space has $$16$$ elements:
How do we determine the chance of each particular outcome, such as $$HHTH$$, of our experiment? In order to do this, we need to define the probability for each sample point, as we will do below.
# Probability Spaces
A probability space is a sample space $$\Omega$$, together with a probability $${\mathbb{P}}[\omega]$$ for each sample point $$\omega$$, such that
• $$0\le{\mathbb{P}}[\omega]\le 1$$ for all $$\omega\in\Omega$$.
• $$\displaystyle\sum_{\omega\in\Omega}{\mathbb{P}}[\omega] = 1$$, i.e., the sum of the probabilities of all outcomes is $$1$$.
The easiest way to assign probabilities to sample points is uniformly: if $$|\Omega| = N$$, then $${\mathbb{P}}[x] = 1/N$$ $$\forall x \in \Omega$$. For example, if we toss a fair coin $$4$$ times, each of the $$16$$ sample points (as pictured above) is assigned probability $$1/16$$. We will see examples of non-uniform probability distributions soon.
After performing an experiment, we are often interested in knowing whether an event occurred. For example, we might be interested in the event that there were “exactly $$2$$ $$H$$’s in four tosses of the coin". How do we formally define the concept of an event in terms of the sample space $$\Omega$$? Here is a beautiful answer. We will identify the event “exactly $$2$$ $$H$$’s in four tosses of the coin" with the subset consisting of those outcomes in which there are exactly two $$H$$’s:
$$\{HHTT, HTHT, HTTH, THHT, THTH, TTHH\} \subseteq \Omega$$. Now we turn this around and say that formally an $$A$$ is just a subset of the sample space, $$A \subseteq \Omega$$.
How should we define the probability of an event $$A$$? Naturally, we should just add up the probabilities of the sample points in $$A$$.
For any event $$A \subseteq \Omega$$, we define the probability of $$A$$ to be ${\mathbb{P}}[A] = \sum_{\omega\in A} {\mathbb{P}}[\omega].$
Thus the probability of getting exactly two $$H$$’s in four coin tosses can be calculated using this definition as follows. $$A$$ consists of all sequences that have exactly two $$H$$’s, and so $$|A| ={4\choose 2}= 6$$. For this example, there are $$2^4 = 16$$ possible outcomes for flipping four coins. Thus, each sample point $$\omega \in A$$ has probability $$1/16$$; and, as we saw above, there are six sample points in $$A$$, giving us $${\mathbb{P}}[A] = 6 \cdot 1/16 = 3/8$$.
# Examples
We will now look at examples of random experiments and their corresponding sample spaces, along with possible probability spaces and events.
## Coin Flipping
Suppose we have a coin of bias $$p$$, and our experiment consists of flipping the coin $$4$$ times. The sample space $$\Omega$$ consists of the sixteen possible sequences of $$H$$’s and $$T$$’s shown in the figure earlier.
The probability space depends on $$p$$. If $$p = 1/2$$ the probabilities are assigned uniformly; the probability of each sample point is $$1/16$$. What if the coin comes up heads with probability $$2/3$$ and tails with probability $$1/3$$ (i.e. the bias is $$p = 2/3$$)? Then the probabilities are different. For example, $${\mathbb{P}}[HHHH]=(2/3) \cdot (2/3) \cdot (2/3) \cdot (2/3) = 16/81$$, while $${\mathbb{P}}[TTHH]=(1/3) \cdot (1/3) \cdot (2/3) \cdot (2/3) = 4/81$$. [Note: We have cheerfully multiplied probabilities here; we’ll explain why this is OK later. It is not always OK!]
What type of events can we consider in this setting? Let event $$A$$ be the event that all four coin tosses are the same. Then $$A = \{HHHH,TTTT\}$$. $$HHHH$$ has probability $$(2/3)^4$$ and $$TTTT$$ has probability $$(1/3)^4$$. Thus, $${\mathbb{P}}[A] = {\mathbb{P}}[HHHH] + {\mathbb{P}}[TTTT] = (2/3)^4 + (1/3)^4 = 17/81$$.
Next, consider event $$B$$: the event that there are exactly two heads. The probability of any particular outcome with two heads (such as $$HTHT$$) is $$(2/3)^2 (1/3)^2$$. How many such outcomes are there? There are $${4\choose 2}=6$$ ways of choosing the positions of the heads, and these choices completely specify the sequence. So $${\mathbb{P}}[B]=6(2/3)^2 (1/3)^2 = 24/81 = 8/27$$.
More generally, if we flip the coin $$n$$ times, we get a sample space $$\Omega$$ of cardinality $$2^n$$. The sample points are all possible sequences of $$n$$ $$H$$’s and $$T$$’s. If the coin has bias $$p$$, and if we consider any sequence of $$n$$ coin flips with exactly $$r$$ $$H$$’s, then the the probability of this sequence is $$p^r(1-p)^{n-r}$$.
Now consider the event $$C$$ that we get exactly $$r$$ $$H$$’s when we flip the coin $$n$$ times. This event consists of exactly $${n \choose r}$$ sample points. Each has probability $$p^r(1-p)^{n-r}$$. So the probability of this event, $${\mathbb{P}}[C] = {n \choose r}p^r(1-p)^{n-r}$$.
Biased coin-tossing sequences show up in many contexts: for example, they might model the behavior of $$n$$ trials of a faulty system, which fails each time with probability $$p$$.
## Rolling Dice
The next random experiment we will discuss consists of rolling two dice. In this experiment, $$\Omega=\{(i,j):1\le i,j\le 6\}$$. The probability space is uniform, i.e. all of the sample points have the same probability, which must be $$1/|\Omega|$$. In this case, $$|\Omega| = 36$$, so each sample point has probability $$1/36$$. In such circumstances, the probability of any event $$A$$ is clearly just${\mathbb{P}}[A] = { \#\text{ of sample points in A} \over \#\text{ of sample points in \Omega} } = {{|A|}\over{|\Omega|}}.$ So for uniform spaces, computing probabilities reduces to counting sample points!
Now consider two events: the event $$A$$ that the sum of the dice is at least $$10$$ and the event $$B$$ that there is at least one $$6$$. By writing out the number of sample points in each event, we can determine the number of sample points in each event; $$|A| = 6$$ and $$|B| = 11$$. By the observation above, it follows that $${\mathbb{P}}[A] = 6/36 = 1/6$$ and $${\mathbb{P}}[B] = 11/36$$.
## Card Shuffling
The random experiment consists of shuffling a deck of cards. $$\Omega$$ is equal to the set of the $$52!$$ permutations of the deck. The probability space is uniform. Note that we’re really talking about an idealized mathematical model of shuffling here; in real life, there will always be a bit of bias in our shuffling. However, the mathematical model is close enough to be useful.
## Poker Hands
Here’s another experiment: shuffling a deck of cards and dealing a poker hand. In this case, $$S$$ is the set of $$52$$ cards and our sample space $$\Omega = \{\text{all possible poker hands}\}$$, which corresponds to choosing $$k=5$$ objects without replacement from a set of size $$n=52$$ where order does not matter. Hence, as we saw in the previous Note, $$|\Omega| = {52 \choose 5} = (52 \cdot 51 \cdot 50 \cdot 49 \cdot 48)/(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) = 2598960$$. Since the deck is assumed to be randomly shuffled, the probability of each outcome is equally likely and we are therefore dealing with a uniform probability space.
Let $$A$$ be the event that the poker hand is a flush. [For those who are not addicted to gambling, a flush is a hand in which all cards have the same suit, say Hearts.] Since the probability space is uniform, computing $${\mathbb{P}}[A]$$ reduces to simply computing $$|A|$$, or the number of poker hands which are flushes. There are $$13$$ cards in each suit, so the number of flushes in each suit is $${13}\choose 5$$. The total number of flushes is therefore $$4\cdot{{13}\choose 5}$$. Then we have ${\mathbb{P}}[\text{hand is a flush}] = {{4\cdot{{13}\choose 5}}\over{{{52}\choose 5}}} = {{4\cdot{13!}\cdot 5!\cdot{47!}}\over{5!\cdot 8!\cdot{52!}}} = {{4\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}\over {52\cdot 51\cdot 50\cdot 49\cdot 48}}\approx 0.002.$
## Balls and Bins
In this experiment, we will throw $$20$$ (labeled) balls into $$10$$ (labeled) bins. Assume that each ball is equally likely to land in any bin, regardless of what happens to the other balls.
If you wish to understand this situation in terms of sampling a sequence of $$k$$ elements from a set $$S$$ of cardinality $$n$$: here the set $$S$$ consists of the $$10$$ bins, and we are sampling with replacement $$k= 20$$ times. The order of sampling matters, since the balls are labeled.
The sample space $$\Omega$$ is equal to $$\{(b_1,b_2,\ldots,b_{20}):1\le b_i\le 10\}$$, where the component $$b_i$$ denotes the bin in which ball $$i$$ lands. The cardinality of the sample space, $$|\Omega|$$, is equal to $$10^{20}$$- each element $$b_i$$ in the sequence has $$10$$ possible choices, and there are $$20$$ elements in the sequence. More generally, if we throw $$m$$ balls into $$n$$ bins, we have a sample space of size $$n^m$$. The probability space is uniform; as we said earlier, each ball is equally likely to land in any bin.
Let $$A$$ be the event that bin $$1$$ is empty. Since the probability space is uniform, we simply need to count how many outcomes have this property. This is exactly the number of ways all $$20$$ balls can fall into the remaining nine bins, which is $$9^{20}$$. Hence, $${\mathbb{P}}[A]=9^{20}/{10}^{20}=(9/10)^{20}\approx 0.12$$.
Let $$B$$ be the event that bin $$1$$ contains at least one ball. This event is the complement $$\bar{A}$$ of $$A$$, i.e., it consists of precisely those sample points which are not in $$A$$. So $${\mathbb{P}}[B] = 1 - {\mathbb{P}}[A] \approx .88$$. More generally, if we throw $$m$$ balls into $$n$$ bins, we have: ${\mathbb{P}}[\text{bin 1 is empty}]=\left({{n-1}\over n}\right)^m= \left({1-{1\over n}}\right)^m.$
As we shall see, balls and bins is another probability space that shows up very often in Computer Science: for example, we can think of it as modeling a load balancing scheme, in which each job is sent to a random processor.
It is also a more general model for problems we have previously considered. For example, flipping a fair coin $$3$$ times is a special case in which the number of balls ($$m$$) is $$3$$ and the number of bins ($$n$$) is $$2$$. Rolling two dice is a special case in which $$m = 2$$ and $$n = 6$$.
The “birthday paradox” is a remarkable phenomenon that examines the chances that two people in a group have the same birthday. It is a “paradox” not because of a logical contradiction, but because it goes against intuition. For ease of calculation, we take the number of days in a year to be $$365$$. Then $$U = \{1,\dots,365\}$$, and the random experiment consists of drawing a sample of $$n$$ elements from $$U$$, where the elements are the birth dates of $$n$$ people in a group. Then $$|\Omega| = 365^n$$. This is because each sample point is a sequence of possible birthdays for $$n$$ people; so there are $$n$$ points in the sequence and each point has $$365$$ possible values.
Let $$A$$ be the event that at least two people have the same birthday. If we want to determine $${\mathbb{P}}[A]$$, it might be simpler to instead compute the probability of the complement of $$A$$, $${\mathbb{P}}[\bar{A}]$$. $$\bar{A}$$ is the event that no two people have the same birthday. Since $${\mathbb{P}}[A] = 1 - {\mathbb{P}}[\bar{A}]$$, we can then easily compute $${\mathbb{P}}[A]$$.
We are again working in a uniform probability space, so we just need to determine $$|\bar{A}|$$. Equivalently, we are computing the number of ways there are for no two people to have the same birthday. There are $$365$$ choices for the first person, $$364$$ for the second, …, $$365-n+1$$ choices for the $$n^{\text{th}}$$ person, for a total of $$365 \times 364 \times \cdots \times (365-n+1)$$. Note that this is simply an application of the first rule of counting; we are sampling without replacement and the order matters.
Thus we have ${\mathbb{P}}[{\bar A}] = \frac{|{\bar A}|}{|\Omega|} = \frac{365 \times 364 \times \cdots \times (365-n+1)}{365^n}.$ Then ${\mathbb{P}}[A] = 1 - \frac{365 \times 364 \times \cdots \times (365-n+1)}{365^n}.$ This allows us to compute $${\mathbb{P}}[A]$$ as a function of the number of people, $$n$$. Of course, as $$n$$ increases $${\mathbb{P}}[A]$$ increases. In fact, with $$n=23$$ people you should be willing to bet that at least two people do have the same birthday, since then $${\mathbb{P}}[A]$$ is larger than $$50\%$$! For $$n=60$$ people, $${\mathbb{P}}[A]$$ is over $$99\%$$.
# The Monty Hall Problem
In an (in)famous 1970s game show hosted by one Monty Hall, a contestant was shown three doors; behind one of the doors was a prize, and behind the other two were goats. The contestant picks a door (but doesn’t open it). Then Hall’s assistant (Carol), opens one of the other two doors, revealing a goat (since Carol knows where the prize is, she can always do this). The contestant is then given the option of sticking with his current door, or switching to the other unopened one. He wins the prize if and only if his chosen door is the correct one. The question, of course, is: Does the contestant have a better chance of winning if he switches doors?
Intuitively, it seems obvious that since there are only two remaining doors after the host opens one, they must have equal probability. So you may be tempted to jump to the conclusion that it should not matter whether or not the contestant stays or switches.
Yet there are other people whose intuition cries out that the contestant is better off switching. So who’s correct?
As a matter of fact, the contestant has a better chance of picking the car if he uses the switching strategy. How can you convince yourself that this is true? One way you can do this is by doing a rigorous analysis. You would start by writing out the sample space, and then assign probabilities to each sample point. Finally you would calculate the probability of the event that the contestant wins under the sticking strategy. This is an excellent exercise if you wish to make sure you understand the formalism of probability theory we introduced above.
Let us instead give a more intuitive pictorial argument. Initially when the contestant chooses the door, he has a $$1/3$$ chance of picking the car. This must mean that the other doors combined have a $$2/3$$ chance of winning. But after Carol opens a door with a goat behind it, how do the probabilities change? Well, everyone knows that there is a goat behind one of the doors that the contestant did not pick. So no matter whether the contestant is winning or not, Carol is always able to open one of the other doors to reveal a goat. This means that the contestant still has a $$1/3$$ chance of winning. Also the door that Carol opened has no chance of winning. What about the last door? It must have a $$2/3$$ chance of containing the car, and so the contestant has a higher chance of winning if he or she switches doors. This argument can be summed up nicely in the following picture:
You will be able to formalize this intuitive argument once we cover conditional probability.
In the meantime, to approach this problem formally, first determine the sample space and the probability space. Just a hint: it is not a uniform probability space! Then formalize the event we have described above (as a subspace of the sample space), and compute the probability of the event. Good luck!
# Summary
The examples above illustrate the importance of doing probability calculations systematically, rather than “intuitively". Recall the key steps in all our calculations:
• What is the sample space (i.e., the experiment and its set of possible outcomes)?
• What is the probability of each outcome (sample point)?
• What is the event we are interested in (i.e., which subset of the sample space)?
• Finally, compute the probability of the event by adding up the probabilities of the sample points inside it.
Whenever you meet a probability problem, you should always go back to these basics to avoid potential pitfalls. Even experienced researchers make mistakes when they forget to do this — witness many erroneous “proofs”, submitted by mathematicians to newspapers at the time, of the fact that the switching strategy in the Monty Hall problem does not improve the odds. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9656076431274414, "perplexity": 138.04403026086263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818690211.67/warc/CC-MAIN-20170924205308-20170924225308-00528.warc.gz"} |
https://www.physicsforums.com/threads/what-temperature-rise-causes-the-transistor-to-saturate-tran.520231/ | # What temperature rise causes the transistor to saturate?tran
1. Aug 9, 2011
### Jadejadeface
1. The problem statement, all variables and given/known data
1. Additional Exercise 7 in the textbook (H&H, p. 110). Vout is 0.5Vcc at the quiescent point. You may assume beta = 100. The transistor is said to saturate when VC = VB. Use the fact IC grows 9%/ degree C and that VBE is constant to determine when the transistor saturates.
2. Relevant equations
Vb = Vbe +IeRe
Vcc = IcRc + Vce + IeRe where Ic = (beta x Ie)/(beta +1)
3. The attempt at a solution
Ok, so firstly I'm confused about whether we need to answer 7 and this question or just what's written in "1.". Anyway, for Vb = 0.5Vcc the two resistors must be equal ( I don't really understand what that arrow is pointing to the resistor, please explain? Since the problem says he adjusts it until 0.5Vcc is reached I'm assuming both resistors = 1K) Therefore Vout = 10V = Vb.
And this is as far as I get since there is no Re i'm not really sure about how to get Vc. If there was a resistor there I would find Vc using the above equation then figure how much I need to increase the temperature to get Vc = 10v. Please help!
Is the current in the collector the same as that in the emitter? Do I need to take into consideration re (little r e, the intrinsic resistance of the resistor)?
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https://web2.0calc.com/questions/triangel-semicricles | +0
# TRIANGEL SEMICRICLES
0
311
2
+166
Let $A$, $B$, $C$, be three points on a line such that $AB = 2$ and $BC = 4$. Semicircles are drawn with diameters $\overline{AB}$, $\overline{AC}$, and $\overline{BC}$. Find the area of the shaded region.
Aug 18, 2017
#2
+94586
+2
Here is the visual image of what you have described...
The question becomes....what "shaded area" are you referring to ???
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https://science.sciencemag.org/content/364/6446/1163.editor-summary | Report
# Quantum amplification of mechanical oscillator motion
See allHide authors and affiliations
Science 21 Jun 2019:
Vol. 364, Issue 6446, pp. 1163-1165
DOI: 10.1126/science.aaw2884
## Improving precision with quantum amplification
Quantum mechanically, an object can be described by a pair of noncommuting observables, typically by its position and momentum. The precision to which these observables can be measured is limited by unavoidable quantum fluctuations. However, the method of “squeezing” allows the fluctuations to be manipulated, while preserving the Heisenberg uncertainty relation. This allows improved measurement precision for one observable at the expense of increased fluctuations in the other. Burd et al. now show that an additional displacement of a trapped atom results in amplification of the squeezing and a further improvement in the precision with which the displacement can be determined (see the Perspective by Schleier-Smith). This technique should be useful for a number of applications in metrology.
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http://physicscatalyst.com/heat/kinectic_th_0.php | Kinetic Theory Of Gases
3. Moleular nature of matter
• We know that molecules which are made up of one or more atoms constitute matter.
• In solids these atoms and molecules are rigidly fixed and space between then is very less of the order of few angestrem and hance they can not move.
• In liquids these atoms and molecules can more enabeling liquids to flow.
• In gases atoms are free to travell without colliding for large distances such that if gases were not enclosed in an enclosure they would disappear.
4. Dalton's law of partial pressures :
• Consider a mixture of non-interacting ideal gases with n1 moles of gas 1,n2 of gas 2 and so on
• Gases are enclosed in an encloser with volume V, temperature T and pressure P.
• Equation of state of mixture
PV = (n1+ n2)RT
or P = n1RT/V + n2RT/V + - - - -
= P1 + P2 + - - --
where,
P1 = n1RT/V
is pressure the gas 1 would exert at same V and T if no other gases were present in the enclosure. This is know as law of partial pressure of the gases.
• The total pressure of mixture of ideal gases is sum of partial pressures of individual gases of which mixture is made of.
5. Kinetic Theory of an ideal gas
Following are the fundamental assumptions of kinetic theory of gases.
• Gas is composed of large number of tiny invisible particles know as molecules
• These molecules are always in state of motion with varying velocities in all possible directions.
• Molecules traverse straight line path between any two collisions
• Size of molecule is infinitely small compared to the average distance traverse by the molecules between any two consecutive collisions.
• The time of collision is negligible as compared with the time taken to traverse the path.
• Molecules exert force on each other except when they collide and all of their molecular energy is kinetic.
• Intermolecular distance in gas is much larger than that of solids and liquids and the molecules of gas are free to move in entire space free to them.
6. Pressure of gas
• Consider a cubical vessel with perfectly elastic walls containing large number of molecules say N let l be the dimension of each side of the cubical vessel.
• v1x , v1y, v1z be the x, y, and z componenet of a molecule with velocity v.
• Consider the motion of molecule in the direction perpandicular to the face of cubical vessel.
• Molecule strikes the face A with a velocity v1x and rebounds with the same velocity in the backward direction as the collisions are perfectly elastic.
• If m is the mass of molecule, the change in momentum during collision is
mv1x - (-mv1x) = 2 mv1x (1)
• The distance travelled parallel to x-axis an is between A to A´ and when molecule rebounds from A´and travel towards A is 2L
• Time taken by molecule to go to face A´ and then comeback to A is
Δt = 2l/v1x
• Number of impacts of this molecule with A in unit time is
n = I/Δt = v1x/ 2l (2)
Rate of change of momentum is
ΔF = ΔP/Δt
=nΔP
from (1) and (2)
ΔF = mv1x2 / l
this is the force exerted on wall A due to this movecule.
• Force on wall A due to all other molecules
F = Σmv1x2/L (3)
• As all directions are equivalent
Σv1x2=Σv1y2=Σv1z2
Σv1x2= 1/3Σ((v1x)2 + (v1y)2 +( v1z)2 )
= 1/3 Σv12
Thus F = (m/3L) Σv12
• N is total no. of molecules in the container so
F = (mN/3L) (Σ(v1)2/N)
• Pressure is force per unit area so
P = F/L2
=(M/3L3)(Σ(v1)2/N)
where ,M is the total mass of the gas and if ρ is the density of gas then
P=ρΣ(v1)2/3N
since Σ(v1)2/N is the average of squared speeds and is written as vmq2 known as mean square speed
Thus, vrms=√(Σ(v1)2/N) is known as roon mean squared speed rms-speed and vmq2 = (vrms)2
• Pressure thus becomes
P = (1/3)ρvmq2 (4)
or PV = (1/3) Nmvmq2 (5)
from equation (4) rms speed is given as
vrms = √(3P/ρ)
= √(3PV/M)
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http://math.stackexchange.com/questions/38731/values-of-the-riemann-zeta-function-and-the-ramanujan-summation-how-strong-is/179356 | Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?
The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 (\mathfrak{R})$$ (for non-negative integer $k$) and $$\zeta(-(2n+1))=-\frac{B_{2k}}{2k} (\mathfrak{R})$$ (again, $k \in \mathbb{N}$). Here, $B_k$ is the $k$'th Bernoulli number. However, it does not hold when, for example, $$\sum_{n=1}^{\infty} \frac{1}{n}=\gamma (\mathfrak{R})$$ (here $\gamma$ denotes the Euler-Mascheroni Constant) as it is not equal to $$\zeta(1)=\infty$$.
Question: Are the first two examples I stated the only instances in which the Ramanujan summation of some infinite series coincides with the values of the Riemann zeta function?
-
This [math overflow question][1] will certainly be of interest. It is regarding assigning a value to the divergent series $\zeta(1)$, as the harmonic series seems to be hard to assign such a value to. [1]: mathoverflow.net/questions/3204/… – Eric Naslund May 12 '11 at 17:39
@Eric: Comments don't allow biblio-style hyperlinkes (the kind you get from clicking the link button in the graphical answer editor). You need an inline link, and the markup is [text](http://url). OP, I think the answer to your question may lie in what exactly the overlap between the Ramanujan and zeta summation methods is (family of divergent series that they agree on). – anon Aug 13 '11 at 1:11
Ramanujan summation arises out of Euler-Maclaurin summation formula. Ramanujan summation is just (C, 1) summation. (See Cesàro summation)
You can find out easily from Euler-Maclaurin that
$$\sum_{k=1}^{\infty}\frac{1}{k}$$
is not (C, 1) summable.
Follow the method of Ramanujan below (which you can easily follow):
Using Euler-Summation we have
\begin{align*} \zeta(s) & = \frac{1}{s-1}+\frac{1}{2}+\sum_{r=2}^{q}\frac{B_r}{r!}(s)(s+1)\cdots(s+r-2) \\ & \phantom{=} -\frac{(s)(s+1)\cdots(s+q-1)}{q!}\int_{1}^{\infty}B_{q}(x-[x])x^{-s-q} ~dx \end{align*}
$\zeta(s)$ is the Riemann zeta function (Note $s=1$ is pole) . Note that right side has values even for $Re(s)<1$.
For example, putting $s=0$ we get $$\zeta(0)=-\frac{1}{2}.$$
If we put $s=-n$ (n being a positive integer) and $q=n+1$, we see the remainder vanishes and have
$$(n+1)\zeta(-n)=-1+\frac{n+1}{2}+\sum_{r=2}^{n+1}\frac{B_r(-1)^{r-1}}{r!}\binom{n+1}{r}$$
which after
$$\sum_{j=0}^{r}\binom{r+1}{j}B_j=0$$
gives
$$\zeta(-n)=-\frac{B_{n+1}}{n+1}.$$
-
I would urge you to do analyze the harmonic series using Euler-Maclaurin Summation
You will be able to prove
$$\sum_{k\leq x}\frac{1}{k}=\log x+\gamma+O\left(\frac{1}{x}\right)$$
where $\gamma$ is the Euler-Mascheroni constant and $O(f)$ is big oh notation.
You just need to analyze the remainder term in Euler-Maclaurin summation using Fourier series of periodic Bernoulli polynomials. That is, for $m\geq 2$
$$B_m(x-[x])=-\frac{m!}{(2\pi i)^m}\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{e^{2\pi i nx}}{n^m}$$
This would give you
$$|B_{m}(x-[x])|\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^m}\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2\frac{m!}{3(2\pi)^m}$$
which is just perfect for estimating remainder in Euler-Maclaurin formula.
You can also try to prove Stirling's approximation using this method that is
$$n! = \sqrt{2 \pi n}~{\left( \frac{n}{e} \right)}^n \left( 1 + O \left( \frac{1}{n} \right) \right).$$
-
You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$:
$$\lim_{h\to0}\frac{\zeta(1+h)+\zeta(1-h)}2=\gamma$$
Saying $\zeta(1)=\infty$ is wrong because zeta has no limit at that point (except for directional limits).
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http://www.models.life.ku.dk/Flow_Injection | # Rank-deficient spectral FIA data
## Problem
Nørgaard & Ridder (1994) have investigated a problem of measuring samples with three different analytes on a flow injection analysis (FIA) system where a pH-gradient is imposed. The data are interesting from a data analytical point of view, especially as an illustration of closure or rank-deficiency and the use of constraints.
## Get the data
The data are available in zipped MATLAB 4.2 format. Download the data and write load data in MATLAB. If you use the data we would appreciate that you report the results to us as a courtesey of the work involved in producing and preparing the data. Also you may want to refer to the data by referring to
Nørgaard L, Ridder C, Rank annihilation factor analysis applied to flow injection analysis with photodiode-array detection. Chemometrics and Intelligent Laboratory Systems 23:107, 1994
Data (Matlab format)
The data have also been described in
1. Bro, R, Multi-way Analysis in the Food Industry. Models, Algorithms, and Applications. 1998. Ph.D. Thesis, University of Amsterdam (NL) & Royal Veterinary and Agricultural University (DK).
2. Bro R, Sidiropoulos ND, Least squares algorithms under unimodality and non-negativity constraints. Journal of Chemometrics 12:223, 1998
3. Kiers HAL, Smilde AK, Constrained three-mode factor analysis as a tool for parameter estimation with second-order instrumental data. Journal of Chemometrics, 1998, 12, 125-147.
4. Bro R, Harshman RA, Sidiropoulos N, Rank-deficient models for multi-way data, Journal of Chemometrics, Submitted.
## Data
The basic setup of the FIA system is shown in Figure 1a. A carrier stream containing a Britton-Robinson buffer of pH 4.5 is continuously injected into the system with a flow of 0.375 mL/min. The 77 µL of sample and 770 µL of reagent (Britton-Robinson buffer pH 11.4) are injected simultaneously into the system by a six-port valve and the absorbance is detected by a diode-array detector (HP 8452A) from 250 to 450 nm in two nanometer intervals. The absorption spectrum is determined every second 89 times during one injection. By the use of both a carrier and a reagent (Figure 1b) a pH gradient is induced over the sample plug from pH 4.5 to 11.4.
The three analytes present in the samples are 2-, 3-, and 4-hydroxy-benzaldehyde (HBA). All three analytes have different absorption spectra depending on whether they are in their acidic or basic form. Twelve samples of different constitution (Table 1) are measured. Thus the data set is a 12 (samples) × 100 (wavelengths) × 89 (times) array. The time mode of dimension 89 is also a pH profile due to the pH-gradient.
Table 1. The concentrations of the three analytes in the 12 samples.
Sample 2HBA 3HBA 4HBA 1 0.05 0.05 0.06 2 0.05 0.10 0.04 3 0.05 0.10 0.06 4 0.10 0.05 0.04 5 0.10 0.10 0.04 6 0.10 0.10 0.06 7 0 0.10 0.04 8 0 0.10 0.06 9 0.05 0 0.06 10 0.10 0 0.06 11 0.05 0.10 0 12 0.10 0.05 0
For each sample a landscape is obtained showing the spectra for all times, or conversely the time profiles for all wavelengths (see below).
It is characteristic of FIA that there is no physical separation of the sample. All analytes have the same dilution profile due to dispersion, i.e., all analytes will have equally shaped total time profile. Above this profile is shown to the left bottom. This profile thus maintains its shape at all wavelengths for all samples and for all analytes. The total profile is the profile actually detected by the photometer (the manifest profile) and is the sum of the profiles of protonated and deprotonated analytes. Due to the pH-gradient, and depending on the pKa of a given analyte, an analyte will show up with different amounts of its acidic and basic form at different times, and hence will have different acidic and basic profiles in the sample plug. In the figure above these profiles are shown for one analyte. The first part of the sample plug, i.e., the earliest measurements of a sample, is dominated by deprotonated analytes while the end of the sample plug is dominated by protonated analytes.
Structural model
In order to specify a mathematical model for the data array of FIA data initially ignore the time domain and consider only one specific time, i.e., one specific pH. An I × J matrix called Xk is obtained where I is the number of samples (12), J is the number of wavelengths (100), and k indicates the specific pH/time selected.
There are three analytes with three corresponding concentration profiles and there are six spectra, an acidic and a basic for each analyte. A standard bilinear model would be an obvious decomposition method for this matrix, but this is not very descriptive in this case. In the sample mode, a three-dimensional decomposition is preferable, as there are only three different analytes. However, each analyte exists in two forms (acid/base), so there will be six different spectra, to be resolved, requiring a six-dimensional decomposition in the spectral mode. To accommodate these seemingly conflicting requirements, a more general model can be used instead
Xk = AHBT, (1)
where A is an I × 3 matrix, and the columns are vectors describing the variations in the sample domain (ideally the concentrations in Table 4), B is a J × 6 vector describing the variations in the spectral domain (ideally the pure spectra), and H is a 3 × 6 matrix which defines the interactions between the columns of A and B. In this case it is known how the analyte concentrations relate to the spectra, as the acidic and basic spectrum of, e.g., 2HBA only relate to the concentration of 2HBA. Therefore H reads
(2)
The matrix H assures that the contribution of the first analyte to the model is given by the sum of a1b1T and a1b2T etc. By using only ones and zeros any information in H about the relative size of the interactions is removed; this information is represented in B. The H matrix is reserved for coding the interaction structure of the model.
So far, only a single time/pH has been considered. To represent the entire data set, the model must be generalized into a multi-way form. For each time the data can be represented by the model above except that it is necessary to adjust it such that the changes in relative concentration (acidic and basic fraction) can be represented as well. The relative concentration of each of the six acidic and basic analytes can be represented by a 6 × 1 vector at each time. The relative concentrations at all K times is held in the K × 6 matrix C. To use the generic model at the kth time it thus is necessary to scale the contribution from each analyte by its corresponding relative concentration. The six weights from the kth row of C are placed in a 6 × 6 diagonal matrix Dk so that the sth diagonal element gives the relative amount of the sth species. The model can be then written
(3)
or, in other words, as
Xk = AHDkBT. k = 1, ..., K (4)
Note how the use of a distinct H and C (Dk) matrix allows the qualitative and quantitative relationships between A and B to be expressed separately. The interaction matrix H, which is globally defined, gives the interaction structure; it shows exactly which factors in A are interacting with which factors in B. In contrast, the C matrix gives the interaction magnitudes. For every k the kth row of C (diagonal of Dk) shows to which extent each interaction is present at the given k. The distinction between qualitative and quantitative aspects is especially important, since knowledge of the exact pattern of interactions is not always available. Not fixing H as here allows for exploring the type of interaction. This can be helpful for rank-deficient problems in general. The matrix C also has a straightforward interpretation as each column in C will be the estimated FIAgram or time profile of the given analyte in its acidic or basic form. Note that the model above bares some resemblance to the PARAFAC model
but differs mainly by the introduction of the matrix H, which enables the interactions between factors in different modes. It also enables A and B/C to have different column dimensions.
The PARATUCK2 model is given
(7)
while the FIA model is
Xk = AHDkBT. (8)
This FIA model can be fitted a what has been called a restricted PARATUCK2 model (which is now more generally referred to as a PARALIND model - see references above). The matrix H remains fixed during the analysis to ensure that every analyte only interact with two spectra/profiles, namely its acidic and its basic counterpart. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8615381121635437, "perplexity": 1230.2846233466078}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804881.4/warc/CC-MAIN-20171118113721-20171118133721-00736.warc.gz"} |
https://www.vedantu.com/question-answer/prove-that-cot-a-cot3a-+-cot2a-cot3a-cot2a-cota-class-11-maths-cbse-5ee87192e8223517fbdc4902 | Question
# Prove that $\cot {\text{A cot3A + cot2A cot3A - cot2A cotA = - 1 }}$
Hint: Here in this question we will use the property of cot (A+B) to solve the problem.
Now, we know the property of tan (A+B) but we can’t remember the property of cot (A+B). So, to overcome this problem we will derive the property of cot from the property of tan.
Now, tan (A+B) = $\dfrac{{\tan {\text{A + tanB}}}}{{1 - \tan {\text{A tanB}}}}$. Also, we know that $\tan {\text{ x = }}\dfrac{1}{{\cot {\text{ x}}}}$. So, putting value of tan in the property of tan (A+B), we get
$\dfrac{1}{{\cot ({\text{A + B)}}}} = \dfrac{{\dfrac{1}{{\cot {\text{A}}}} + \dfrac{1}{{\cot {\text{B}}}}}}{{1 - \dfrac{1}{{\cot {\text{A cotB}}}}}}$
Simplifying the above term, we get
$\cot ({\text{A + B) = }}\dfrac{{{\text{cotA cotB - 1}}}}{{\cot {\text{A + cot B}}}}$ ……….. (1)
Now, to solve the given question, put A = 2A and B = A in the equation (1)
$\cot (2{\text{A + A) = }}\dfrac{{\cot 2{\text{A cotA - 1}}}}{{{\text{cot2A + }}{\text{ cotA}}}}$
By cross – multiplying both sides, we get
$\cot 3{\text{A ( cot2A + cotA) = cot2A cotA - 1}}$
$\cot 3{\text{A cot2A + cot3A cotA = cot2A cotA - 1}}$
Rearranging the terms in the above equation,
$\cot {\text{A cot3A + cot2A cot3A - cot2A cotA = - 1 }}$
Hence proved.
Note: In the questions which include the property of cot there are many mistakes done by students in solving the problems. Most of the students don't know the property of cot correctly and apply the incorrect formula resulting in the wrong answer. Easiest way to remove such error is that you should derive the property of cot from the property of tan which is as comparison to cot is easy to remember. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.945029079914093, "perplexity": 548.3721249403292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00145.warc.gz"} |
http://physics.stackexchange.com/questions/49861/exercise-concerning-the-inclined-plane?answertab=oldest | # exercise concerning the inclined plane
I have an exercise to ask yourself.
A ball of mass m = 5 g starts at rest and travels 50 cm along a ramp inclined at 45 ° to the horizontal. a) If we imagine frictionless motion of the ball along the ramp, compute the speed of the ball at the bottom of the ramp. What is the change in potential energy of the ball?
My idea is to use the gravitational acceleration, multiply it by the cosine of $45°$ and find the x-component of the gravitational acceleration, and finally use the formula $v^2=v_0^2 +2*g*x$ to find the speed at the end of the ramp.
For the second point, my idea is to use this formula $m*g*h$.
My idea is correct?
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9839343428611755, "perplexity": 195.16492607464966}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398449793.41/warc/CC-MAIN-20151124205409-00087-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://www.3dcopperprinting.com/overwhelmed-by-the-complexity-of-resources-this-may-help.html | # Overwhelmed by the Complexity of Resources? This May Help
Operation of Fractions. Most children consider learning fractions to be a very complicated exercise. The main difference between a fraction and other numbers is that it has a numerator and a denominator. There are some fractions problems that require one to follow some set steps in order to arrive at their solutions. Most if not all of the fractions problems also require a student to combine various maths operations in order to solve them. The four operations are addition, subtraction, multiplication, and division. For one to be proficient in fractions, they must first understand the four areas mentioned above. Mastery of fractions comes from practicing them regularly. This article therefore aims at clearly articulating how to solve fractions while using math operations mentioned above. Addition of fractions with the same denominator | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9429150223731995, "perplexity": 485.1197982727834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742685.33/warc/CC-MAIN-20181115120507-20181115142507-00480.warc.gz"} |
https://experts.illinois.edu/en/publications/%C3%A9tude-num%C3%A9rique-de-l%C3%A9coulement-du-r134a-%C3%A0-l%C3%A9tat-de-liquide-vapeur | # Étude numérique de l'écoulement du R134a à l'état de liquide-vapeur dans un collecteur vertical pour la séparation des phases avec une faible qualité d'entrée
Translated title of the contribution: Numerical study of R134a liquid-vapor flow in a vertical header for phase separation with low inlet quality
Jun Li, Pega Hrnjak
Research output: Contribution to journalArticlepeer-review
## Abstract
The separation circuitry has been proven in the past to improve the performance of microchannel condensers. In the vertical second header of the condenser, liquid separates from vapor mainly due to gravity. However, separation is usually not perfect, expressed through the separation efficiency. This study presents the phase separation result in the second header calculated by the Euler-Euler method of Computational Fluid Dynamics (CFD). Simulations are conducted for two-phase refrigerant R134a flow in the second header with 21 microchannel tubes in the 1st pass. The inlet mass flux to the second header (through the microchannels of the 1st pass) in the simulation is 166 kg m 2 s 1, 207 kg m 2 s 1, and 311 kg m 2 s 1. The inlet quality is 0.13 to 0.21. The results agree well with the experimental results with flow visualization and the results of a simpler 1-D numerical model. Results show that the liquid separation efficiency decreases as the vapor separation efficiency increases, following a linear trend in the experimental range. The void fraction result shows liquid mainly flows in the half of the header without microchannel tube intrusions. The velocity profile in the header is presented and reverse flow is identified on the exit planes of the inlet section connecting to the 2nd-upper pass and the 2nd-lower pass. The pressure profile in the header is also revealed and it indicates that the 1-D pressure assumption may still apply to two-phase flow in a header.
Translated title of the contribution Numerical study of R134a liquid-vapor flow in a vertical header for phase separation with low inlet quality French 11-21 11 International Journal of Refrigeration 129 https://doi.org/10.1016/j.ijrefrig.2021.04.013 Published - Sep 2021
## Keywords
• CFD
• Microchannel condenser | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907501220703125, "perplexity": 2927.5506316099045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00261.warc.gz"} |
https://cosmicdiscord.net/2021-04-07-gm2/ | A few of my friends asked me about the recent Fermilab results. We know that recently there is some “anomaly” going on at Fermilab, and the result is statistically significant when combined with the previous measurement from Brookhaven National Laboratory (BNL). This post will be mostly for the background. I will write another post about the QCD lattice prediction and clashes with the electroweak precision data. To understand what Fermilab and BNL have seen, we need to start with the magnetic dipole moment.
### Magnetic (dipole) moment
Essentially, it describes how an object responds when an external magnetic field is applied to it.
Any object’s magnetic property, measured from far, can be approximated either by a small electric loop, or a small pair of opposite magnetic monopoles (direct analogy to electric dipole moment.) In the first approximation, the magnitude of the magnetic moment is the current times the area of the loop, pointing toward the thumb direction when you put the other four fingers of your right hand along the current direction; while in the second case, its magnitude is the length of the separation times the absolute value of the magnetic charge of either monopole, pointing from the positive charge to the negative charge.
With both magnitude and direction, it can be described by a vector (technically pseudovector, but let’s not go in there yet.) And the energy of the system is the lowest when this vector is aligned with the magnetic field. As a result, when the two are not aligned, there’s a torque trying to rotate the magnetic dipole moment toward the magnetic field direction. Hence, the physics magnetic moment describes the response of the system when a magnetic field is applied, as promised in the beginning.
### Spin
A lot of objects have nonvanishing magnetic moment, from as large as the Earth to as small as the hydrogen atom. Taking the hydrogen atom as an example, to an approximation, the magnetic moment is due to the electron rotating around the nucleus – in this case just a single proton without any neutron. In the classical picture, it can be approximated by a small current loop due to the electron’s motion around the proton. Surprisingly, even elementary particles – those that cannot be subdivided, such as electron, or its heavier sibling muon – have nonzero magnetic dipole moment. One might wonder how this is even possible, since, to the best of our knowledge, electrons do not have any “components”, (hence one cannot find any moving parts rotating around the “electric center” - whatever that means.) Yet, quantum mechanically, the electron or muon, just as any other fermions, has a property called spin. This is a measured intrinsic property that shows up just like it is rotating with a fixed (quantized) magnitudes. Due to this property, it responds to external magnetic fields very similar to something that has a charged component rotating toward the center, like hydrogen.
As physicists like to say, to the zero-th order, by looking at the response of fermions in the magnetic field, important properties of the interaction between magnetic field (photons) and the current loop (electrons or muons) can be deduced. This could shed light on unknown interactions at the most fundamental level. The effect is a quantum effect, and I will provide another sketch on how this is revealed through an unprecedented precision measurement of the (muon) magnetic moment. Stay tuned folks. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.878295361995697, "perplexity": 338.3870556696533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00364.warc.gz"} |
https://www.airgunmagazine.co.uk/tag/air-pistol/ | # Gun collecting with Jonathan Young
So you want to be a collector? You’re suddenly overcome by an urge to buy more airguns. There’s nothing wrong with that, argues Jonathan Young as he leads you down the rabbit hole of gun collecting. Waking up on any given
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# Airgun Essential: Pellet Scales
When all the factors must be weighed up… why you need to get yourself a set of scales designed specifically for your pellets
# Classic Gun Spotlight: Air Arms Camargue
Shooting the breeze… Named after a prevailing wind, the Air Arms Camargue has entered folklore as a true airgun great, says Simon Everett
# Gun Test: Hatsan Galatian IV
A long time ago, in a galaxy far, far away…. Mike Morton has some interplanetary fun with the fibre-optic-fitted Hatsan Galatian IV
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# Airgun Essential: Boresnake
Dirty barrel? Get a grip on some serpent twine! Discover what a BoreSnake is and why you need one
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# Gun Test: Remington Express
Mat Manning takes the Remington Express out on the range – and discovers that you don’t have to push your budget off the rails to land a decent entry-level airgun
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# Trigger secrets exposed!
Mike Morton says it’s time to give inaccurate shooting the finger – by setting up your trigger correctly
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# How to fit a new blade
Mike Morton explains how to fit a new blade using his Weihrauch HW 100
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# Q&A: When can I hunt rabbits?
Chris Wheeler discusses when and where you can target rabbits when hunting with your airgun
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Posted in Hunting | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8775551915168762, "perplexity": 4630.283483416105}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998238.28/warc/CC-MAIN-20190616122738-20190616144738-00332.warc.gz"} |
https://metricsystem.net/non-si-units/accepted-for-use-with-si/dalton/ | # dalton
##### Non-SI unit accepted for use with SI
Name Symbol Quantity Value in SI units
dalton Da mass 1.660 539 040 (20) × 10−27 kg
1.660 539 040 (20) yg
### Definition
The dalton, symbol Da, or unified atomic mass unit, is a non-SI unit of mass accepted for use with the SI.
One dalton is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and at rest.
In SI units, one dalton is 1.660 539 040 (20) × 10−27 kg, or approximately 1.66 yoctograms.
$1 \ \text{Da} = \dfrac{1}{12} \ m \mspace{4mu}(^{12} \text{}C)$
The dalton is used as a unit of mass for quantities on an atomic or molecular scale. One dalton is approximately the mass of one nucleon (either a single proton or neutron), and is approximately equivalent to 1 g mol-1.
##### Examples of atomic and molecular masses
Name Symbol/formula Mass in Da Mass in SI units hydrogen H 1.00 1.66 yg carbon-12 12C 12.00 19.93 yg silicon-28 28Si 27.98 46.46 yg iron-56 56Fe 55.93 92.88 yg uranium-235 235U 235.04 390.29 yg water H2O 18.01 29.91 yg carbon dioxide CO2 44.01 73.08 yg ethanol C2H5OH 46.07 76.50 yg benzene C6H6 78.11 129.70 yg human haemoglobin ≈ 64 000 ≈ 106 000 yg
### History
Prior to 2019, the mole was defined in terms of the kilogram and the dalton, and was equal to
“the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12”.
The numerical value of the mole is now defined exactly, and the value of the dalton in SI units must be obtained experimentally. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8997300267219543, "perplexity": 1502.1194341480245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337415.12/warc/CC-MAIN-20221003101805-20221003131805-00783.warc.gz"} |
https://azo.tagundjahr.de/volume-formula-chemistry.html | how to finish in combat warriors
The formula for density is D = M/V where: D= density. M= mass. V= volume. Switch around the formula to make it: V= M/D. Volume = mass / density.
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. Osmotic Pressure. Relative Lowering of Vapour Pressure. Van't Hoff Factor. Surface Chemistry. BET Adsorption Isotherm. Capillarity and Surface Forces in Liquids (Curved Surfaces) Colloidal Structures in Surfactant Solutions. Freundlich adsorption isotherm. Langmuir Adsorption Isotherm. The volume to mass calculator has dozens of item densities that will allow you to convert volume to mass or mass to volume whenever you need. ... You need to rearrange the formula to: volume = mass / density. Whenever you use this formula, remember to be concise with units. For example, if ... Chemistry (83) Construction (112) Conversion (168. Volume is the quantity of three-dimensional space occupied by a liquid, solid, or gas. Common units used to express volume include liters, cubic meters, gallons, milliliters, teaspoons, and ounces, though many other units exist. Key Takeaways: Volume Definition Volume is the three-dimensional space occupied by a substance or enclosed by a surface. Aug 06, 2021 · Cover the V and get the formula for volume, v = m/d. Finding the Volume of Irregularly Shaped Objects Density equals mass divided by volume, so the density of an object is easily obtained if its.... The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, %vol or (v/v)%: volume percentage = volume solute volume solution × 100% volume percentage = volume solute volume solution × 100 % Calculations Using Volume Percentage. Aug 06, 2021 · Cover the V and get the formula for volume, v = m/d. Finding the Volume of Irregularly Shaped Objects Density equals mass divided by volume, so the density of an object is easily obtained if its....
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The Charles Law (Final temperature) equation computes the resultant temperature of a fixed mass of gas after it undergoes a change in volume.. INPUTS. Based on Charles Law we can. Volume formula Cube. Prism. Rectangular prism. Pyramid. Cone. Cylinder. Sphere. Find the volume of the rectangular prism below. Volume of a composite figure. Their volumes can be. Conversion chart of standard units such as pressure, volume, temperature and a few others. Made for the ideal gas equation but universal really. Unit Conversion Chart. Subject: Chemistry. Age range: 11-14. Resource type: Assessment and revision. Instead it tests regents in specific subjects like Physics or Chemistry. The exam is approximately. The formula for ppm is {eq}ppm=1/1,000,000=0.0001 {/eq}. Before calculating ppm, one must be sure they are measuring the same volume or mass of the substances in the equation. For example, to.
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Results (detailed calculations and formula below) The Mass is : The Density is per 3: The Volume is 3: Calculating Mass based on Density and Volume; Note: The calculations within each of the Mass, Density and Volume formulas are all shown in Grams and Centimetres. The Mass, Density and Volume Calculator automatically converts the values based on the unit values for.
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Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. MOTIVATION. A formula is a way of relating different quantities using algebra. The well-known formula. A = π r 2. relates the area A of a circle to its radius r. The formula expresses symbolically the fact that to find the area of a circle, we square the radius and multiply by the number π.This formula relates two variables A and r.We can use the formula to find the area. Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. volume of hemisphere formula. pandas count occurrences of string in column; 1086 international interior; new haven housing authority waiting list; citrix port 1494 and 2598; pedestal tub faucet; splunk set token based on another token; xbox live gold games june 2022. paper mario master quest jr rom download. Solved Examples on Enthalpy Formula. Q.1: Calculate the heat of the following reaction using the table of values. Substance. C. 0. 0. -393.5. Solution: The is the heat of formation, and it refers to the heat it takes to form the substance from its elements. The have values of 0 because they are in elemental form.
Aug 19, 2022 · The volume of a regular object can be calculated by multiplying its length by its width and its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length. One unit of volume is the cubic meter ( m 3), which is the volume occupied by a cube that measures 1 m on each side..
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. The change in volume or pressure for gas at a constant temperature can be calculated using the equation: ... A gas occupies a volume of 0.50 m³ at a pressure of 100 Pa. Calculate the pressure. The volume, V, of a rectangular prism is: V = lwh. where l is the length, w is the width, and h is the height of the rectangular prism. Pyramid. The volume, V, of a pyramid is: where B is area of the base and h is the height of the pyramid. Cone. The volume, V, of a cone is: where r is radius of the base and h is the height of the cone. Cylinder. The volume, V, of a cylinder is:. density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. For example, the density of water is 1 gram per cubic centimetre, and Earth’s density is 5.51 grams per cubic centimetre. Density can also be expressed as.
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Common Equations Used in Chemistry Equation for density: d= m v Converting ˚F to ˚C: ˚C = (˚F - 32) x 5 9 Converting ˚C to ˚F: ˚F = ˚C x 9 5 + 32 ... Calculation of changes in pressure, temperature, or volume of gas when n is constant: P1V1 T1 = P2V2 T2. Calculation of density or molar mass of gas: d = PM RT.
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When the gas does work the volume of a gas increases () and the work done is negative. When work is done on the gas, the volume of the gas decreases () and work is positive. One way to remember the sign convention is to always think about the change in energy from the point of view of the gas. When the gas expands against an external pressure .... The formula is V = πr 2 h, where V is the Volume, r is the radius of the circular base, h is the height, and π is the constant pi. In some geometry problems the answer will be given in terms of pi, but in most cases it is sufficient to round pi to 3.14. Check with your instructor to find out what she would prefer. In order to calculate molar volume of a substance, we can divide the molar mass by its density. Mathematically expressing it as: Vm=Mρ Where, V - volume of the gas n - Number of moles of gas, P - Pressure, T- Temperature R - Gas Constant (value depends upon the units of Pressure, Volume and Temperature) The gas constant R is 8.314 J / mol. K.
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V = S 3 V is the volume enclosed by the cube S is the side length, it is also commonly represented by an A in other problem sets This simple cube volume formula applies only to true cubes where all sides are an equal length. If all sides are not an equal length, but still parallel, please use the L ength x W idth x H eight formula below..
The turbulent flame speed is determined. The turbulent flame speed can be written in terms of flow and chemistry time scales. St = A * ( Turbulent time scale/chemical time scale)^0.25 = A* (Da)^0.25. where A (0.52) is the model constant and Da is Damokohler number. Knowing the mass and volume of an object allows the calculation of its density. Density is defined as the mass divided by the volume of the object. Mass should be expressed in units of grams and volume in units of mL or cm3. Note, from the table below, that the density of water is 1 g≅ mL-1..
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volume of solute * density of solute = c* molar mass. From this equation, volume of solute can be known. Clearly, if the volume of the solution increases, the volume of the solute will also increase. Since the volume of the solution has not been given, it must give you a hint to take the above assumption.
A = 384 cm2. The Volume of a cube formula is given as follows: V = s × s × s = s3. Substituting the values we get the volume of cube as follows: V = 8 × 8 × 8. V = 512 cm3. Therefore, the surface area and volume of a cube having the sides of length equal to 8 cm is 384 cm2 and 512 cm3 respectively. 2.. Volume formula. The volume of a 3D shape or geometric figure is the amount of space it contains. Volume is well-defined for many common shapes; the formulas for some common.
Conversion chart of standard units such as pressure, volume, temperature and a few others. Made for the ideal gas equation but universal really. Unit Conversion Chart. Subject: Chemistry. Age range: 11-14. Resource type: Assessment and revision. Instead it tests regents in specific subjects like Physics or Chemistry. The exam is approximately. The formula above is used to convert volume to mass and vice versa; the only requirement is to know the density of the substance of interest. This formula can also be used to find the density if.
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Because this type of concentration is expressed as a percentage, the indicated proportion must be multiplied by 100, as shown below. Mass/Volume Percent = m s o l u t e ( g) V s o l u t i o n ( m L) × 100. Ideal Gas Law Formula (Volume): {eq}V=\frac{nRT}{P} {/eq} Kelvins: Kelvins are the SI unit of temperature. It is an absolute scale.
You can’t calculate both... you need the mass and one of the values to calculate the other. I.e. you could look up the density of iron, and then calculate the volume based upon a known mass. AutoModerator • 2 yr. ago. Hi indecisive-18.
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The change in volume or pressure for gas at a constant temperature can be calculated using the equation: ... A gas occupies a volume of 0.50 m³ at a pressure of 100 Pa. Calculate the pressure.
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In chemistry, concentration refers to the amount of a substance per defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution. Concentration usually is expressed in terms of mass per unit volume. This calculator calculates the concentration of volume percent using volume of. We teach you how to calculate the volume of a solution if you are given the amount in grams and the molarity (concentration) of the solution.Example:Find the.... .
Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. The turbulent flame speed is determined. The turbulent flame speed can be written in terms of flow and chemistry time scales. St = A * ( Turbulent time scale/chemical time scale)^0.25 = A* (Da)^0.25. where A (0.52) is the model constant and Da is Damokohler number.
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Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. The formula above is used to convert volume to mass and vice versa; the only requirement is to know the density of the substance of interest. This formula can also be used to find the density if.
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Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. Because the concentration is a percent, you know a 100-gram sample would contain 12 grams of iron. You can set this up as an equation and solve for the unknown "x": 12 g iron / 100 g sample = x g iron / 250 g sample. Cross-multiply and divide: x= (12 x 250) /. Weight by Volume This value is derived from the weight of the solute divided by the total volume of the solute and the solvent, and it can be measured in several ways. grams per liter grams per deciliter or % milligrams per deciliter or mg% Molarity Molarity (M) is a useful concentration unit for many applications in chemistry.
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The density (g/L) is provided as 4. Finally, calculate the Chemistry Volume using the equation above: Vchem = MM / D. The values given above are inserted into the equation below: Vchem = 25 / 4 = 6.25 (L) Example Problem #2: The variables needed for this problem are provided below: molar mass (g/mol) = 60. density (g/L) = 3..
Common Equations Used in Chemistry Equation for density: d= m v Converting ˚F to ˚C: ˚C = (˚F - 32) x 5 9 Converting ˚C to ˚F: ˚F = ˚C x 9 5 + 32 ... Calculation of changes in pressure, temperature, or volume of gas when n is constant: P1V1 T1 = P2V2 T2. Calculation of density or molar mass of gas: d = PM RT.
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In chemistry, concentration refers to the amount of a substance per defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution. Concentration usually is expressed in terms of mass per unit volume. This calculator calculates the concentration of volume percent using volume of. Osmotic Pressure. Relative Lowering of Vapour Pressure. Van't Hoff Factor. Surface Chemistry. BET Adsorption Isotherm. Capillarity and Surface Forces in Liquids (Curved Surfaces) Colloidal Structures in Surfactant Solutions. Freundlich adsorption isotherm. Langmuir Adsorption Isotherm. M = moles / liter. Considering the use of length and diameter mentioned above, the formula for calculating the volume of a tube is shown below: volume = π. d 12 - d 22. 4. l. where d1 is the outer diameter,. Considering the use of length and diameter mentioned above, the formula for calculating the volume of a tube is shown below: volume = π.
The density (g/L) is provided as 4. Finally, calculate the Chemistry Volume using the equation above: Vchem = MM / D. The values given above are inserted into the equation below: Vchem = 25 / 4 = 6.25 (L) Example Problem #2: The variables needed for this problem are provided below: molar mass (g/mol) = 60. density (g/L) = 3. volume of solute * density of solute = c* molar mass. From this equation, volume of solute can be known. Clearly, if the volume of the solution increases, the volume of the solute will also increase. Since the volume of the solution has not been given, it must give you a hint to take the above assumption. Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3)..
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The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length. \[\mathrm{volume\: of\: lead\: cube=2.00\:. We teach you how to calculate the volume of a solution if you are given the amount in grams and the molarity (concentration) of the solution.Example:Find the....
V= volume Switch around the formula to make it: V= M/D Volume = mass / density Wiki User ∙ 2009-04-04 22:11:05 This answer is: 👍 👎 Add a Comment Study guides Chemistry 20 cards How does a buffer.
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One mole of any gas has a volume of 24 dm 3 or 24,000 cm 3 at rtp (room temperature and pressure). This volume is called the molar volume of a gas. This equation shows how the volume of gas in dm.
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c × M. 10. To get v/v percentage, multiply molarity by molar mass of the substance and divide by 10 times the mass density of the solution. Cv/v. =. c × M. 10 × ρ. You can use the third calculator (from page top) to calculate a dilution or preparation of a stock solution. For that, the following formulas will be used:.
Chemistry Formulas. At standard Temperature and Pressure (STP) the molar volume (V m) is that the volume occupied by 1 mole of a chemical element or a chemical compound. We can. Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3)..
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Aug 08, 2019 · Formula Molar volume is calculated as molar mass (M) divided by mass density (ρ): V m = M / ρ Example The molar volume of an ideal gas at STP is 22.4 L/mol. Cite this Article.
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v/v % = [ (volume of solute)/ (volume of solution)] x 100% Note that volume percent is relative to the volume of solution, not the volume of solvent. For example, wine is about 12% v/v ethanol. This means there is 12 ml ethanol for every 100 ml of wine. It is important to realize liquid and gas volumes are not necessarily additive.
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Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3)..
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Volume formula. The volume of a 3D shape or geometric figure is the amount of space it contains. Volume is well-defined for many common shapes; the formulas for some common. In order to calculate molar volume of a substance, we can divide the molar mass by its density. Mathematically expressing it as: Vm=Mρ Where, V - volume of the gas n - Number of moles of gas, P - Pressure, T- Temperature R - Gas Constant (value depends upon the units of Pressure, Volume and Temperature) The gas constant R is 8.314 J / mol. K. The relationships between the four quantities stated above can be expressed as shown below: Molar Mass — Mole — Molar Volume. |. Avogadro Number. i.e, 1 mole = 6.023 x 10 ^23 = 22.4dm ^3 = molar mass. 1mole of CO 2 = 44g of CO 2 = 22.4dm ^3 of CO 2 = 6.023 x 10 ^23 molecules of CO 2. An ionic equation is a special form of symbol equation that shows only those ions that react. Step 1: write down the balanced ionic equation with the state symbol. (Balancing equations will be covered next). BaCl 2 (aq) + Na 2 SO 4 (aq) ———–> 2NaCl (aq) + BaSO 4 (s) Step 2: write down the ions present in the equation. Feb 06, 2013 · The formula for calculating the volume of a hexagonal prism is to take the area of the hexagon, then multiply it by the height of the prism. What is the formula for calculating volume of gas? In Chemistry, 1 mole of ANY gas at room temperature and pressure is 24dm^3 (or 24000 cm^3).. Avail Chemistry Formulas to solve related problems efficiently. Use Chemistry Formulae Sheet & Tables to learn the concepts behind them in a better way. ... Volume. the amount of 3-dimensional space occupied by an object. 211. Weight. the vertical force exerted by a mass as a result of gravity. 212. Yield. Aug 19, 2022 · The volume of a regular object can be calculated by multiplying its length by its width and its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length. One unit of volume is the cubic meter ( m 3), which is the volume occupied by a cube that measures 1 m on each side..
In order to calculate molar volume of a substance, we can divide the molar mass by its density. Mathematically expressing it as: Vm=Mρ. Where, V – volume of the gas. n – Number of moles of gas, P – Pressure, T- Temperature. R – Gas Constant (value depends upon the units of Pressure, Volume and Temperature) The gas constant R is 8.314 J. We teach you how to calculate the volume of a solution if you are given the amount in grams and the molarity (concentration) of the solution.Example:Find the....
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Molar Volume Formula Chemistry Formulas. Method to Calculate Molar Volume of a Substance . Molar volume of a gas can be defined as volume of gas occupied by one mole of.
texas bounty hunter requirements nessus conjunct venus natal high risk coin pusher videos new holland t1510 backhoe ffbe gamefaqs | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8595066666603088, "perplexity": 1330.6728111370817}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335396.92/warc/CC-MAIN-20220929225326-20220930015326-00213.warc.gz"} |
https://www.physicsforums.com/threads/mean-molar-mass-after-combustion-in-a-combustion-engine.684322/ | # Mean molar mass after combustion in a combustion engine
1. Apr 9, 2013
### ChristopherJ
1. The problem statement, all variables and given/known data
The problem is to calculate the mean molar mass of the reaktion below, which takes place in a combustion engine.
C4H10 + z(0.79N2 + 0.21O2) → αH2O + βCO2 + γN2
where the molar masses for the different molecules/atoms are:
MH2O = 18, MCO2 = 44, MN2 = 28, MO2 = 32, MC = 12, MH = 1.
The mass of the air in the cylinder before combustion is 2.39905 grams. The mass of the fuel in the cylinder is 45.0 milligrams.
2. Relevant equations
See above
3. The attempt at a solution
I have tried to balance the equation above by calculating the amount of molecules of oxygen and nitrogen in the cylinder and the amount of molecules in the fuel to get α, β and γ. And then use these values to balance the reaction equation above. The mean molar mass can be calculated by dividing the right hand side by the amount of molecules at hand.
But, I don't get the right answer by doing this. Can anyone help me with this?
I am very grateful for answers!
Best regards
Christopher
2. Apr 9, 2013
### SteamKing
Staff Emeritus
Without showing your calculations, you won't get much help. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9169889092445374, "perplexity": 757.9151675796019}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00738.warc.gz"} |
https://physics.stackexchange.com/tags/propagator/hot | # Tag Info
## Hot answers tagged propagator
26
Yes, in scalar field theory, $\langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is the amplitude for a particle to propagate from $x$ to $y$. There are caveats to this, because not all QFTs admit particle interpretations, but for massive scalar fields with at most moderately strong interactions, it's correct. Applying the operator $\phi({\bf x},t)$ to the ...
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You wouldn't think it, from how easy it is to pose this question, but it is ridiculously nontrivial. As it happens, it is entirely impossible to find the position-basis matrix elements of this propagator. So far you've done good, and the identification $$U\left( t_{2},t_{1}\right) =e^{\frac {-i} {\hbar }H\left( t_{2}-t_{1}\right) } =\sum_{n=-\infty}^\... 19 Keep in mind that it is nearly impossible to explain how perturbative QFT calculations follow from Lagrangians such that the answer is both relatively short and detailed. So I am going to write an introductory answer. If you want more details on any of its part, you can look up textbooks, or you can let me know in the comments, in which case I will consider ... 18 Quantum mechanics and quantum field theory are different in how they treat their wave equations. The usage of the common term “propagator” could be traced back to the “relativistic wave equation” approach—i. e. people really used to think of the Schrödinger and the KG operators as belonging to the same class of “quantum operators”, but the modern ... 17 It has been many years since you asked this question. I assume that over time you have compiled meaning definitions and distinctions for the other terms in your list. However, there are terms not defined by @josh's response (A response which I have relied on multiple times, thank you for posting it @josh). Personally, my background is in Lattice QCD, which ... 16 I) OP is right, ideologically speaking. Ideologically, OP's first eq.$$ \tag{1} \left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Turns out to be ultimately wrong!}) $$is the statement that a particle that is initially localized at a spacetime event (x_i,t_i) must with probability 100% be ... 16 First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation \hat L G = \delta. At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second ... 15 For a given quantum system, the kernel of the path integral is, in fact, the kernel of an integral transform as you explicitly write down. It is the transform that governs time evolution of the system as is manifest in your first equation. For this reason, it is often referred to as the propagator of a given system. For example, for a single, non-... 15 Your question has been answered again and again, and again, albeit indirectly and elliptically--I'll just be more direct and specific. The point is you skipped variables: in this case, t, and so the expression you wrote ("according to books, \hat{L}K(x;x') = \delta(x-x')~"), is nonsense, as you already properly found out; unless you included t in the ... 14 Källén-Lehmann representation is just a way to expand in the momentum basis the two point correlation function of a local operator \hat{O}(x), it holds true for massive and massless theories alike except for non Abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start ... 13 No, ⟨0|T{ϕ(y)ϕ(x)}|0⟩ is NOT the probability amplitude for a particle to propagate from x to y, even for a free scalar field. It seems to be a common false belief that it is. There is one obvious reason and one deep reason why it cannot be. The obvious reason is that the square of this value, which is supposed to be the probability density, does not ... 13 The first step is to recognize that equation is invariant under d-dimensional rotations around \mathbf{x} - \mathbf{x}' = \mathbf{0} and simultaneous identical translations of \mathbf{x} and \mathbf{x}', so we can make the following step:$$\begin{align}(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') &= A \delta(\mathbf{x} - \mathbf{x}') \\ & \...
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This can be seen by partial integration $$\frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2}=\frac{\rho}{\sqrt{\rho^2-m^2}}$$ OP edit: More explicitly, we use this to write $(3)$ as \begin{align} D(x-y) &= \frac{1}{(2\pi)^2r}\int^\infty_m d\rho \frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2} e^{-\rho r} \\ &= \frac{1}{(2\pi)^2r}\left[\sqrt{\rho^2-...
12
Here's the gist of it: If your field lives in a vector space $V$, then the propagator is a map $V\to V$, i.e., it lives in $V\otimes V^*$. In more down-to-earth terms, if your field has a certain index $i$, its propagator has a pair of such indices: $$\psi^i\quad\Longrightarrow\quad G^i{}_j$$ The reason is that, by definition, $G$ measures the difference ...
11
The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given ...
11
Propagator in quantum mechanics is just a different name for Green's function for time-dependent Schroedinger equation. It is a unique function that enables us to write, for any time $t_0$, $$\psi(x,t) = \int G(x,t;x',t_0)\psi(x',t_0)\,dx'$$ for all $x$ and all $t$. This means the $\psi$ function of $x$ (at any time $t$) can be written as a result of ...
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The free scalar and fermion propagator is $$G_\psi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i(\gamma^\mu p_\mu + m)}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) }$$ The scalar propagator is $$G_\phi(x,y) = \int \frac{d^dp}{(2\pi)^d} \frac{-i}{ p^2 + m^2 - i \epsilon} e^{- i p \cdot ( x - y ) }$$ Clearly, $$G_\psi(x,y) = ( i \gamma^\mu \partial_\... 10 Suppose you want to compute a correlation say in Euclidean signature$$ \frac{1}{Z}\int D\phi\ \prod_i \phi(x_i)\ \exp\left(-\frac{1}{\hbar}S(\phi)\right) $$with$$ S(\phi)=\frac{1}{2}(\phi,A\phi)+g\int dx\ \phi(x)^4 $$where (\phi,A\phi)=\int\ dx\ dy\ \phi(x)A(x,y)\phi(y) for some "matrix" or rather kernel A. One usually does that by expanding the ... 10 I will discuss the closed string propagator because this case is pictorially closer to the scalar propagator in quantum field theory case. The closed string analog of the (two-leg amputated) line of propagation of a scalar field in a Feynman diagram is a cylinder of finite height s and twist angle \theta. At this point you must notice the analogue with ... 9 I) Conceptually, OP's original eq. (1)$$\int_{\mathbb{R}}\! \mathrm{d}x_f~ \left| K(x_f,t_f;x_i,t_i) \right|^2 ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Wrong!})\tag{1} $$clashes (as OP independently realized) with the fundamental principle of the Feynman path integral that the amplitude$$K( x_f ,t_f ; x_i ,t_i )~=~\sum_{\rm hist.}\ldots$$is a sum of ... 9 There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral)$$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$If \mathrm{Re}A\ge0, \mathrm{Re}B>0 is violated, the integral will be divergence. ... 9 My own attempt at this: the first result is wrong, and the second one is right but incomplete. Feynman: in the diagram, all the lines are external, so there is no propagator in the diagram. Therefore, Feynman rules give \mathcal A=1. Check up of this: In canonical quantisation, the amplitude is given by \langle p|q\rangle=\langle 0|a_p a^\dagger_q|0\... 9 Method One: \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\... 9 The propagator of an arbitrary vector field is [ref.1] $$\langle A_\mu A_\nu\rangle=\frac{-\eta_{\mu\nu}+p_\mu p_\nu/m_1^2}{p^2-m_1^2}-\frac{p_\mu p_\nu/m_1^2}{p^2-m_0^2}\tag1$$ for a pair of masses m_0,m_1. This propagator is usually called the Stückelberg propagator, and A_\mu a Stückelberg field. The Stückelberg field ... 8 The probability density$$\lim_{x_1\to x_0} P(x_1, t_1; x_0, 0)is basically the probability density for a particle at position x_0 to be back at that position after a time t_1 has passed. 8 Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated. This is ... 8 To get a better intuition consider a field with a general spin, this can be written as \begin{align*} \psi_\ell &\propto \sum_\sigma \int d^3 p \left( u_\ell (\vec{p},\sigma) e^{i p\cdot x}a(\vec{p},\sigma) + v_\ell (\vec{p},\sigma) e^{-i p\cdot x}a^\dagger (\vec{p},\sigma) \right) \end{align*} where \ell is the spin index, and \sigma summed over all ... 8 Consider \phi^4 theory: \mathcal L=\frac12 Z_1(\partial\phi)^2-\frac12 Z_m m^2\phi^2-\frac{1}{4!}\lambda_0\phi^4 $$There are two approaches to perturbation theory: First The propagator is given by$$ \Delta=\frac{1}{Z_1p^2-Z_m m^2} $$and there is one type of vertex, with value$$ -i\lambda_0 $$Second The propagator is given by$$ \Delta=\frac{1}{...
8
This is a common (and good) question, which is the result of taking introductory texts more seriously than one should. Reading such texts, one gets the impression that the philosophy is We proceed as naively as possible and, when we find a divergence, we regulate it. On the other hand, the correct (and much more useful) attitude is We take things ...
8
When we integrate the propagator with respect to $k^0$ (i.e. the energy), we encounter two poles: one at $\omega_{\mathbf{k}}=\sqrt{\mathbf{k}^2+m^2}$ and one at $-\omega_{\mathbf{k}}=-\sqrt{\mathbf{k}^2+m^2}$, where $\mathbf{k}$ is the 3-momentum. In order to regularize this integral, we move these poles slightly off of the real line by adding or ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9831083416938782, "perplexity": 693.2661147564482}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585507.26/warc/CC-MAIN-20211022114748-20211022144748-00139.warc.gz"} |
http://www.thespectrumofriemannium.com/2013/07/15/log122-basic-neutrinologyvii/ | # LOG#122. Basic Neutrinology(VII).
The observed mass and mixing both in the neutrino and quark cases could be evidence for some interfamily hierarchy hinting that the lepton and quark sectors were, indeed, a result of the existence of a new quantum number related to “family”. We could name this family symmetry as . It was speculated by people like Froggatt long ago. The actual intrafamily hierarchy, i.e., the fact that in the quark sector, seem to require one of these symmetries to be anomalous.
A simple model with one family dependent anomalous U(1) beyond the SM was first proposed long ago to produce the given Yukawa coupling and their hierarchies, and the anomalies could be canceled by the Green-Schwarz mechanism which as by-product is able to fix the Weinberg angle as well. Several developments include the models inspired by the GUT or the heterotic superstring theory. The gauge structure of the model is that of the SM but enlarged by 3 abelian U(1) symmetries and their respective fields, sometimes denoted by . The first one is anomalous and family independent. Two of these fields, the non-anomalous, have specific dependencies on the 3 chiral families designed to reproduce the Yukawa hierarchies. There are right-handed neutrinos which “trigger” neutrino masses by some special types of seesaw mechanisms.
The 3 symmetries and their fields are usually spontaneously broken at some high energy scale by stringy effects. It is assumed that 3 fields, , with , develop a non-null vev. These fields are singlets under the SM gauge group but not under the abelian symmetries carried by . Thus, the Yukawa couplings appear as some effective operators after the spontaneous symmetry breaking. In the case of neutrinos, we have the mass lagrangian (at effective level):
and where . The parameters determine the mass and mixing hierarchy with the aid of some simple relationships:
and where is the Cabibbo angle. The are the charges assigned to the left handed leptons L and the right handed neutrinos N. These couplings generate the following mass matrices for neutrinos:
From these matrices, the associated seesaw mechanism gives the formula for light neutrinos:
The neutrino mass mixing matrix depends only on the charges we assign to the LH neutrinos due to cancelation of RH neutrino charges and the seesaw mechanism. There is freedom in the assignment of the charges . If the charges of the second and the third generation of leptons are equal (i.e., if ), then one is lead to a mass matrix with the following structure (or “texture”):
and where . This matrix can be diagonalized in a straightforward fashion by a large rotation. It is consistent (more or less), with a large mixing. In this theory or model, the explanation of the large neutrino mixing angles is reduced to a theory of prefactors in front of powers of the parameters , related with the vev after the family group spontaneous symmetry breaking!
See you in the next Neutrinology blog post!
View ratings | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.959484875202179, "perplexity": 746.0374412441038}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00053.warc.gz"} |
http://www.sic6h.com/2020/03/a-comparative-dft-study-of-electronic.html | ## Mar 10, 2020
### A comparative DFT study of electronic properties of 2H-, 4H- and 6H-SiC(0001) and SiC(000 \bar{1} ) clean surfaces: significance of the surface Stark effect
The electric field, uniform within a slab, emerging due to Fermi level pinning at both sides of the slab, is analyzed using DFT simulations of SiC surface slabs of different thicknesses. It is shown that for thicker slabs the field is nonuniform and this fact is related to the surface state charge. Using the electron density and potential profiles, it is proved that for high-precision simulations it is necessary to take into account a sufficient number of SiC layers. We show that the use of 12 diatomic layers leads to satisfactory results. It is also demonstrated that the change of the opposite side slab termination, both by different types of atoms or by their location, can be used to adjust the electric field within the slab, creating a tool for simulation of surface properties, depending on the doping in the bulk of the semiconductor. Using these simulations, it was found that, depending on the electric field, the energy of the surface states changes in a different way than the energy of the bulk states. This criterion can be used to distinguish Shockley and Tamm surface states. The electronic properties, i.e. energy and type of surface states of the three clean surfaces: 2H-, 4H-, 6H-SiC(0001) and SiC(), are analyzed and compared using field-dependent DFT simulations. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9050542712211609, "perplexity": 820.6251235528127}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335326.48/warc/CC-MAIN-20220929065206-20220929095206-00600.warc.gz"} |
https://www.emathzone.com/tutorials/calculus/integral-of-e-to-the-power-of-a-function.html | # Integral of e to the Power of a Function
The integration of $e$ to the power $x$ of a function is a general formula of exponential functions and this formula needs a derivative of the given function. This formula is important in integral calculus.
The integration of e to the power x of a function is of the form
Now consider
Using the derivative formula $\frac{d}{{dx}}{e^{f\left( x \right)}} = {e^{f\left( x \right)}}f'\left( x \right)$, we have
Integrating both sides of equation (i) with respect to $x$, we have
Since integration and differentiation are reverse processes to each other, the integral sign $\int {}$ and $\frac{d}{{dx}}$ on the right side will cancel each other out, i.e.
Example: Evaluate the integral $\int {\frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}dx}$ with respect to $x$
We have integral
Here $f\left( x \right) = {\sin ^{ - 1}}x$, and $d\left( x \right) = \frac{1}{{\sqrt {1 - {x^2}} }}$, so we can write it as
Using the integration formula $\int {{e^{f\left( x \right)}}f'\left( x \right)dx = } {e^{f\left( x \right)}} + c$, we have | {"extraction_info": {"found_math": true, "script_math_tex": 11, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9974655508995056, "perplexity": 230.3352434592856}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583690495.59/warc/CC-MAIN-20190120021730-20190120043730-00375.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=44517 | # Object thrown vertically
by kimikims
Tags: object, thrown, vertically
Share this thread:
P: 36 An object is thrown vertically upward such that it has a speed of 58 m/s when it reaches two thirds of it's maximum height above the launch point. The acceleration of gravity is 9.8 m/s^2 Find the maximum height h. Answer in units of m. I know: V: 58 m/s at y=2/3 H H = ? V = 0 V^2-Vo^2 = 2g(1/3H) I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H??
Mentor
P: 41,325
Quote by kimikims V^2-Vo^2 = 2g(1/3H)
Right (where Vo = 0).
I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H??
Divide or multiply both sides to eliminate what's in front of H.
For example, if you had an equation: 7x^2 = 67bZ, and you wanted to isolate Z, just divide both sides by 67b: (7x^2)/67b = (67bZ)/67b = Z; so Z = (7x^2)/67b. Make sense?
HW Helper P: 2,278 You got $$V^2 = V_{o}^2 + 2g\frac{2}{3}H$$ where V is 58 m/s and $$0 = V_{o}^2 + 2gH$$
HW Helper P: 2,278 Object thrown vertically Doc Al, how come initial speed = 0 ?
Mentor
P: 41,325
Quote by Cyclovenom Doc Al, how come initial speed = 0 ?
If you look at the motion on the way down, then the object starts from rest and falls a distance of H/3 to reach a speed of 58 m/s.
HW Helper
P: 2,278
Quote by Doc Al If you look at the motion on the way down, then the object starts from rest and falls a distance of H/3 to reach a speed of 58 m/s.
Oh Yes, i wasn't thinking about it
Related Discussions Introductory Physics Homework 2 Introductory Physics Homework 4 Introductory Physics Homework 10 General Physics 6 Introductory Physics Homework 1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.825537383556366, "perplexity": 805.4754809857919}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510271862.24/warc/CC-MAIN-20140728011751-00124-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/154889/application-residue-theorem-for-improper-integrals | # Application residue theorem for improper integrals
Let $R(z)=\displaystyle \frac{P(z)}{Q(z)}$ be a rational function of order(Q) $\geq \mathrm{order}(P)+2$ and $Q(x)\neq 0$ for all $x\in \mathbb{R}$. Then we have: $$\int_{-\infty}^{\infty}R(x)\mathrm{d}x=2\pi i\sum_{z:\ \mathrm{Im} \ z>0}{\rm Res}(R,z)$$
Why is
order(Q) $\geq \mathrm{order}(P)+2$ and $Q(x)\neq 0$ for all $x\in \mathbb{R}$
important?
-
I guess you meant "Let $\,R(z)\,$ be a rational function with rational coefficients..." , i.e. $\,P(z), Q(z)\in\mathbb{Q}[z]\,$ ....? – DonAntonio Jun 6 '12 at 23:46
I fixed it. The coefficients are complex. – Chris Jun 7 '12 at 0:15
The basic reason is that you need a bounded growth condition on the function. In order to evaluate this integral, you're actually doing a complex line integral over a half circular region, but you need the complex part to go to zero as the radius goes to infinity. If your function doesn't have bounded growth, the residue formula will retain some of the complex line integral, not just the real line.
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http://nanoscale.blogspot.com/2009/07/phase-transitions-and-mean-field-theory.html?showComment=1248200680069 | ## Tuesday, July 21, 2009
### Phase transitions and "mean field theory"
One truly remarkable feature of statistical physics (and condensed matter physics in particular) is the emergence of phase transitions. When dealing with large numbers of particles one often finds that, as a function of some parameter like temperature or pressure, the whole collection of particles can undergo a change of state. For example, as liquid water is warmed through 100 C at atmospheric pressure, it boils into a vapor phase of much lower density, even though it is still made up of the same water molecules as before. Understanding how and why phase transitions take place has kept many physicists occupied for a long time.
Of particular interest is understanding how microscopic interactions (e.g., polar attraction between individual water molecules) connect to the phase behavior. A classic toy model of this is used to examine magnetism. It's a comparatively simple statistical physics problem to understand how a single magnetic spin (in real life, something like one of the d electrons in iron) interacts with an external magnetic field. The energy of a magnetic moment is lowered if the magnetic moment aligns with a magnetic field - this is why it's energetically favorable for a compass needle to point north. So, one does the statistical physics problem of a single spin in a magnetic field, and there's a competition between this alignment energy on the one hand, and thermal fluctuations on the other. At large enough fields and low enough temperatures, the spin is highly likely to align with the field. Now, in a ferromagnet (think for now about a magnetic insulator, where the electrons aren't free to move around), there is some temperature, the Curie temperature, below which the spins spontaneously decide to align with each other, even without an external field. Going from the nonmagnetic to the aligned (ferromagnetic) state is a phase transition. A toy model for this is to go back to the single spin treatment, and instead of thinking about the spin interacting with an externally applied magnetic field, say that the spin is interacting with an average (or "mean") magnetic field that is generated by its neighbors. This is an example of a "mean field theory", and may be solved self-consistently to find out, in this model, the Curie temperature and how the magnetization behaves near there.
Mean field theories are nice, but it is comparatively rare that real systems are well described in detail by mean field treatments. For example, in the magnetism example the magnetization (spontaneous alignment of the spins, in appropriate units) goes like (1-T/TC)1/2 at temperatures just below TC. This is not the case for real ferromagnets - the exponent is different. Because of the nature of the approximations made in mean field theory, it is expected to be best in higher dimensionality (that is, when there are lots of neighbors!). Here's a question for experts: what real phase transitions are well described by mean field theory? I can only think of two examples: superconductivity (where the superconducting gap scales like
(1-T/TC)1/2 near the transition, just as mean field theory predicts) and a transition between liquid crystal phases. Any others?
Anonymous said...
I've always thought the most impressive thing about Landau theory is that it even works at all, let alone the few places where it has quantitative power.
QuasiNewton said...
An attempt to deal with the problems that arise due to the finite dimensionality of real systems is self-consistent renormalization (SCR) [Moriya, Spin Fluctuations in Itinerant Electron Magnetism, Springer, Berlin, 1985]. Depends, however, if one still considers the MF+SCR combination a mean field theory.
Doug Natelson said...
QuasiNewton - I had in mind pure MF. In fact, for various reasons, I'm particularly interested in systems where some order parameter really goes like (1-T/Tc)^{1/2} for
T < Tc.
Nice blog, btw. I'll add that to the blogroll.
Anonymous said...
If the interactions are long ranged, then the exponents will be mean field --- it looks like Phys. Rev. Lett. 29, 917 - 920 (1972) "Critical Exponents for Long-Range Interactions" has details. Btw, is this why the liquid crystal transition is mean-field-like?
Anonymous said...
Would you consider a Yukawa potential (and other plasma treatments that put electrons in a background field resulting from the ion mass >> electron mass) a mean field theory?
Gautam said...
A canonical example of a real phase transition well described by mean field theory is the first order liquid-solid freezing transition in three dimensions. Here, the classical density functional theory of freezing provides remarkably accurate results for the quasi-universal properties of the freezing transition, such as justifications for the validity of the Verlet and Lindemann criteria for freezing, as well as a host of other properties.
For Anonymous: The isotropic-nematic transition in liquid crystals is first order for the reason that a third-order invariant in the Landau expansion for this problem is always allowed, unlike for spin systems in the absence of a magnetic field. This means that, within mean-field theory, a first-order transition should be the norm, rather than the exception, and it is hard to circumvent this argument. This is for symmetry reasons which have nothing to do with the range of interactions, which, in typical nematics, is usually very short ranged.
Don Monroe said...
So Doug, tell me why any normal person would care about the precise numerical value of the exponent? Of course it gives a window into interesting microscopics, and the scale-free behavior is fun and surprising, but I'd be happy if more people understood mean-field behavior in the first place. Before that, I'd wish that they understood the difference between first- and second-order transitions, which you treat very casually in your entry.
Doug Natelson said...
Don, you're right that no normal person would care about the precise value of the exponent. I care because I've got data that I'm trying to interpret. You're also right that I was more than casual in my treatment of first and second order transitions. If I can think of the right angle, I'll come back to that. One challenge is that the transitions that are most familiar from every day life are first order. It's challenging to explain the comparatively subtle difference between 1st and 2nd order without resorting to examples beyond most people's experience.
Peter Armitage said...
Anything above its critical dimension will exhibit mean-field like behavior. So this is...ummm.... lots and lots of things in 4 dimensional space... But this not what you were asking.
By the same taken there are also presumably lots of quantum transitions in which zero temperature gives you at least one effective extra dimension in the time. But this is also not what you were asking...
Other than those already mentioned, I know that some ferroelectric transitions are above their critical dimension and mean-field.
Don Monroe said...
Doug,
you're not the first one to resort to using a first-order transition to convey the idea, although to my mind the difference is not subtle at all.
In his otherwise excellent PBS series on string theory a few years ago, Brian Greene illustrated the separation of the electromagnetic and weak forces by holding up ice cubes in water. Now I'm not sure what, if anything, a normal viewer got out of that metaphor. For me it was irksome, because the whole point is that the two forces are the same at high temperatures and become different as the temperature is lowered--a classic second-order transition. A first-order transition like water freezing is just a competition between completely different arrangements. I found it particularly annoying because the idea of spontaneous symmetry breaking is one rather clear case in which condensed matter ideas have had a profound influence on high-energy theory.
QuasiNewton said...
Thanks, Doug.
Could you strain your system (different substrate?) in order to investigate the influence of (broken) crystal symmetries (e.g. inversion) on the phase transition you have observed?
Doug Natelson said...
QN - Long story, in progress. If you want to chat more about it, feel free to drop me an email.
Tarun said...
Dear Don,
I think that the electroweak transition you mentioned above is generally believed to be actually first-order. So the analogy Brian Greene may not be totally misleading. For example, see this paper:
http://prola.aps.org/abstract/PRD/v45/i8/p2685_1
I think spontaneous symmetry breaking could always occur through a first-order transition.
Steve said...
Partially addressing the question of why we care about the precise value of the exponent. One thing that is crucial here is that the value is universal. ONLY the symmetry of the system matters and the properties of the phase transition do not care at all about the microscopics of the system at all -- and many many things may fall into the same universality class (Ex: melting transitions can be in the Ising class). So maybe we dont' care what the precise value is, but it is interesting that there are a finite set of possibilities in the first place.
venus said...
An attempt to deal with the problems that arise due to the finite dimensionality of real systems is self-consistent renormalization .....
--
Venus
You cannot go wrong on the
best security systems | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8146635293960571, "perplexity": 533.6600760997435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171936.32/warc/CC-MAIN-20170219104611-00275-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/169560 | Infoscience
Conference paper
# Computation over Gaussian multiple-access channels
We consider the problem of computing the sum of independent Gaussian sources over a Gaussian multiple-access channel (MAC) with respect to a mean-squared error criterion. When the source and channel bandwidths are equal, the best separation-based solution to this problem performs exponentially worse in a distortion sense compared to the optimal solution: uncoded transmission. In this paper, we develop lattice codes for exploiting the structure of the Gaussian MAC when there are more channel uses than source symbols. We also demonstrate the usefulness of these codes in determining the multicast capacity of a simple AWGN network. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8527690768241882, "perplexity": 472.06809377285686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171004.85/warc/CC-MAIN-20170219104611-00428-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/81953/what-is-the-proper-term-for-a-function-where-domain-and-codomain-coincide | # What is the proper term for a function where domain and codomain coincide?
What is the proper term for a function where domain and codomain coincide?
E.g. in programming languages a function f : Int => Int or f : Double => Double. Thanks.
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Suppose we have a function $f:S\to S$, that is $f$ is a function from $S$ with image (range) in $S$. Such a function is called an endomorphism endofunction (endo comes from the greek or latin word for "within"). Sometimes people use the term codomain to mean the image (range) of a function. If $S$ is finite and the image of $f$ is the whole set $S$, then $f$ is a bijection. If $S$ is infinite, the situation is somewhat more complicated.
Edit: Thanks for the correction.
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That is not correct. A morphism which maps a mathematical object $S$ to itself is called an endomorphism. A morphism is a function which preserves structure. A function is just a map - it doesn't care about structure. – user1729 Nov 14 '11 at 10:25
How about endofunction? – Frank Abagnale Nov 14 '11 at 10:47
This is the first time I heard of endofunction ever! – lhf Nov 14 '11 at 10:56
@user1729: A rigorous definition of a morphism is found in category theory. A morphism need not even be a function. Answering "endomorphism (in $Set$)" is correct. – sdcvvc Nov 14 '11 at 11:33
@user1729: As I said, in category Set, a morphism $X \to Y$ is (any) function from $X$ to $Y$. – sdcvvc Nov 14 '11 at 12:58
I would like to add this as a comment, but I don't have enough reputation to do so.
A function $f:S\to S$ can be called an endomorphism in Set, and one doesn't have to worry about the structure-preservation that is implied by -morphism because sets have no additional structure to preserve.
However, it depends on the context: if I want to talk about any function $f:V\to V$, $V$ being a vector space, and I were to use the term endomorphism, it would be understood that $f$ is a linear transformation, because the morphisms in $K$-Vect are linear transformations.
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I don't think there is a widespread term for that which is acceptable in all contexts, but a function $X\to X$ is sometimes called a function on $X$ or an operator on $X$ or a unary operation on $X$.
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http://indigo.ie/~peter/Fib1.htm | FIBONACCI AND EXTENSIONS (TWO TERMS)
In the well known Fibonacci sequence
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 , 233, 377, 610, 987,…. etc, the ratio of successive terms n/(n - 1) approximates to phi.
(For example the ratio of 987 and 610 which is 1.61803… approximates phi correct to 5 decimal places).
The value of phi which = (1 + square root of 5)/2, is the positive solution for x to the simple algebraic equation x2 - x - 1 = 0.
This represents a special case of the more general equation x2 - ax - b = 0 (where both a and b = 1).
So to generate the Fibonacci sequence we keep adding the last term (a) to the second last term (b) in the sequence to generate the next term. So in the above sequence 987 = 610 + 377 (i.e. a and b = 1).
However we can combine different multiples of a and b to generate different series with their own unique features.
For example the Pell Series is derived from adding 2a + b. This results in the sequence
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, …
The ratio of the last two terms here n/(n - 1) is 985/408 = 2.41421… which approximates closely the value of (the square root of 2) + 1.
This in turn is the positive solution for x to the equation x2 - 2x - 1 = 0 (i.e. where a = 2 and b = 1).
Thus for example where a = 3, x will be the positive solution to the equation
x2 - 3x - 1 = 0.
i.e. (3 + square root of 13)/2 = 3.3027756...
The corresponding series is
0, 1, 3, 10, 33, 109, 360, 1189, 3927,…
The ratio of the last two terms n/(n - 1) given here is 3927/1189 = 3.302775… (which already aproximates the true value correct to six decimal places).
We can also generate unique series by varying the value of a (while keeping b constant).
Thus when a = 1 and b = 2 we obtain the following sequence
0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365 ….
The ratio of the last two term n/(n - 1) given here is 1365/683 = 1.9985...
Though it does not converge so quickly (as when we vary a), this approximates the positive solution for x to the equation x2 - x - 2 = 0 i.e. (1 + square root of 9)/2 = 2.
Thus again for example when a = 3 and b = 1, the ratio of the terms n/(n - 1) will approximate the positive solution for x to the equation x2 - x - 3 = 0 i.e. (1 + square root 13)/2 = 2.302775....
The corresponding series is
0, 1, 1, 4, 7, 19, 40, 97, 217, 508, 1159, 2683, 6160,…
The ratio of the last two gives 2.29593.. which is only correct to two decimal places (due to slow convergence in this case).
We can of course combine multiples of a and b
Thus for 2a + 2b we generate the series
0, 1, 2, 6, 16, 48, 128, 352, 960, 2624,…
The ratio for n/(n - 1) will here be the positive solution for x to the equation
x2 - 2x - 2 = 0, which is 1 + square root 3 = 2.73205.. .
The ratio of the last two terms given 2624/960 = 2.7333 (which is approximating to the corrrect value).
Once again the Fibonacci sequence is a special limiting case of where we successively add unitary combinations (a and b) of the last two terms in the series starting 0, 1 to generate the next terms. So for the Fibonacci a and b = 1.
The ratio of successive terms n/(n - 1) and all cases involving multiple combinations of the two terms (a and b) can be given as the approximations to the positive solutions for x to the general quadratic equation x2 - ax - b = 0.
References
I wish to acknowledge my debt to the fascinating investigations by Mike McDermott on Fibonacci like sequences in "Knowledge and the Knower: Complexity and the Self" at http://lightmind.com/Impermanence/Library/texts/mikem-00.html | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8798633813858032, "perplexity": 1433.7553645961193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657134620.5/warc/CC-MAIN-20140914011214-00298-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Bean_machine | # Bean machine
The bean machine, as drawn by Sir Francis Galton
The bean machine, also known as the quincunx or Galton box, is a device invented by Sir Francis Galton[1] to demonstrate the central limit theorem, in particular that the normal distribution is approximate to the binomial distribution.
The machine consists of a vertical board with interleaved rows of pins. Balls are dropped from the top, and bounce left and right as they hit the pins. Eventually, they are collected into one-ball-wide bins at the bottom. The height of ball columns in the bins approximates a bell curve.
Overlaying Pascal's triangle onto the pins shows the number of different paths that can be taken to get to each bin.
A large-scale working model of this device can be seen at the Museum of Science, Boston in the Mathematica exhibit.
## Distribution of the balls
A working replica of the machine (following a slightly modified design.)
If a ball bounces to the right k times on its way down (and to the left on the remaining pins) it ends up in the kth bin counting from the left. Denoting the number of rows of pins in a bean machine by n, the number of paths to the kth bin on the bottom is given by the binomial coefficient ${n\choose k}$. If the probability of bouncing right on a pin is p (which equals 0.5 on an unbiased machine) the probability that the ball ends up in the kth bin equals ${n\choose k} p^k (1-p)^{n-k}$. This is the probability mass function of a binomial distribution.
According to the central limit theorem the binomial distribution approximates the normal distribution provided that n, the number of rows of pins in the machine, is large.
## Games
Several games have been developed utilizing the idea of pins changing the route of balls or other objects:
## References
1. ^ Galton, Sir Francis (1894). Natural Inheritance. Macmillan. pp. 63f. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.800230085849762, "perplexity": 880.1051256144773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657113000.87/warc/CC-MAIN-20140914011153-00125-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://advances.sciencemag.org/content/3/5/e1603282.full | Research ArticleAPPLIED SCIENCES AND ENGINEERING
Evidence and mechanism of efficient thermally activated delayed fluorescence promoted by delocalized excited states
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Vol. 3, no. 5, e1603282
Abstract
The design of organic compounds with nearly no gap between the first excited singlet (S1) and triplet (T1) states has been demonstrated to result in an efficient spin-flip transition from the T1 to S1 state, that is, reverse intersystem crossing (RISC), and facilitate light emission as thermally activated delayed fluorescence (TADF). However, many TADF molecules have shown that a relatively appreciable energy difference between the S1 and T1 states (~0.2 eV) could also result in a high RISC rate. We revealed from a comprehensive study of optical properties of TADF molecules that the formation of delocalized states is the key to efficient RISC and identified a chemical template for these materials. In addition, simple structural confinement further enhances RISC by suppressing structural relaxation in the triplet states. Our findings aid in designing advanced organic molecules with a high rate of RISC and, thus, achieving the maximum theoretical electroluminescence efficiency in organic light-emitting diodes.
Keywords
• Organic light-emitting diodes
• Thermally activated delayed fluorescence
• Reverse intersystem crossing
• Excited-state dynamics
• Transient absorption spectroscopy
• Charge resonance state
INTRODUCTION
One of the most effective ways to enhance internal electroluminescence (EL) quantum efficiency (ηint) in organic light-emitting diodes (OLEDs) is well recognized to be the management of the pathways between excited singlet (S) and triplet (T) states. In OLEDs, the effective use of triplets is fundamental for achieving high ηint because one singlet is generated for every three triplets based on spin statistics under electrical excitation (Fig. 1A) (1). Although a spin-flip from pure S to pure T states, that is, intersystem crossing (ISC), is generally forbidden because of their different spin multiplicities, it becomes possible when their wave functions are mixed through spin-orbital coupling (SOC). The degree of mixing (λ) can be simply expressed as λ = HSOEST, where HSO and ΔEST are the SOC constant and energy difference between the S1 and T1 states, respectively (2). By incorporating a heavy atom such as iridium into organic molecules, HSO is enhanced and a strong mixing of the spin orbitals of the S and T states is induced, resulting in efficient radiative decay from T1 to the ground state (S0), that is, phosphorescence, with nearly 100% of photoluminescence (PL) quantum yield (PLQY) (3).
Alternatively, the λ can be enhanced by decreasing ΔEST, which can also open a pathway from lower-energy T1 to higher-energy S1, that is, reverse ISC (RISC), when ΔEST is small enough. On the basis of quantum chemical theory, ΔEST is proportional to the exchange energy, which is related to the overlap integral between the two open-shell orbitals responsible for the isoconfigurational S and T states, and is typically ~1 eV for conventional condensed polycyclic aromatic compounds, such as anthracene (4). Because the HSO of aromatic compounds is also small, RISC is generally negligible. However, when the ΔEST approaches the thermal energy (~26 meV at room temperature), RISC is induced via thermal excitation, and delayed fluorescence (DF) is subsequently emitted from S1. This is so-called thermally activated DF (TADF), and recent extensive studies on TADF-OLEDs have revealed ηint values of nearly 100% (511). According to the semiclassical model of TADF, the rate constant for RISC (kRISC) is given by kRISC ~ A × exp(−ΔEST/kBT), where A is the pre-exponential factor including HSO, kB is the Boltzmann constant, and T is the temperature (12). Therefore, a high kRISC can be achieved by decreasing ΔEST. However, although this idea has been supported by many experimental results, as shown in Fig. 1B (5, 1319), several papers report fairly high kRISC values even for large ΔEST of a few hundred millielectron volts (6, 20). For instance, carbazol-benzonitrile (CzBN) derivatives synthesized by Zhang et al. (20, 21) were reported to have a ΔEST of ~0.2 eV. Although the ΔEST is too large for efficient RISC at room temperature, CzBN-based OLEDs showed high external EL quantum efficiencies (ηEQE) of ~20%. This ηEQE value means that, under an assumption of a light outcoupling efficiency of OLEDs (22), all the excitons generated electrically were used for EL.
A number of fundamental studies using transient PL (TR-PL) spectroscopy and theoretical calculations have been devoted to understand the mechanism of efficient RISC upon a large ΔEST of several hundred millielectron volts (6, 2327). In particular, Dias et al. (25) and Gibson et al. (27) have proposed that the second-order SOC involving the locally excited T state (3LE) and charge-transfer (CT) excited T state (3CT) in thermal equilibrium is crucial for efficient RISC. In the scenario, the spin-allowed transition of 3LE to CT excited S state (1CT) in materials having a large ΔEST is facilitated efficiently by the vibronic coupling between the 3LE and 3CT states (25, 27). This suggests that RISC is a dynamical process in an excited state. However, this study has been thoroughly conducted for only a few molecules, which consist of a near-orthogonal electron donor (D) and acceptor (A) units. Still, a deeper understanding of the mechanism leading to efficient RISC based on a direct observation of excited states formed in both the S and T states is needed to clarify the relationship to the chemical structure and will provide a strategy for the advanced molecular design of TADF molecules.
Here, we used a comprehensive set of complementary experimental techniques, with an emphasis on transient absorption spectroscopy (TAS), to demonstrate that kRISC cannot be determined by the ΔEST alone and strongly relies on the excited states. To understand the mechanism, we focused on Cz-phthalonitrile (CzPN) (5) and CzBN (20, 21) derivatives: 4CzIPN, 2CzPN, 2CzBN, o-3CzBN, m-3CzBN, p-3CzBN, 4CzBN, and 5CzBN (Fig. 1C). Unlike the CzPN derivatives reported previously by our group (5, 19), CzBN derivatives have similar ΔEST values (~0.2 ± 0.04 eV) (see Table 1). In this sense, similar kRISC values might be expected for CzBN derivatives, but the result is completely different. By looking at excited-state dynamics using TAS, we are able to attribute the different kRISC to the formation of delocalized excited states, which facilitates RISC. We also identify that a linearly positioned Cz pair in a D-A-D structure (Fig. 1D) is the structural requirement for the formation of delocalized excited state. Our findings provide an advanced general design for TADF molecules with high kRISC and facilitate the deeper understanding of significant spin up-conversion processes (2).
Table 1 PL characteristics and rate constants of CzBN and CzPN derivatives in solution and doped film.
View this table:
RESULTS
Before examining the excited-state dynamics, we evaluated the fundamental photophysical properties of the CzBN derivatives (Table 1). TR-PL profiles of the CzBN derivatives in oxygen-free toluene show strong DF for 4CzBN and 5CzBN and weak DF for p-3CzBN, but no DF for 2CzBN, o-3CzBN, or m-3CzBN (Fig. 2A). The observed DF completely vanished in the presence of oxygen, indicating that the DF is emitted via T states. In Table 1, the kRISC values of 4CzBN (1.8 × 105 s−1) and 5CzBN (2.4 × 105 s−1) are larger than the kRISC value of p-3CzBN (0.12 × 105 s−1) by one order of magnitude despite similar ΔEST values of ~0.20 eV for the three materials (Fig. 2B). Moreover, although the ΔEST of o-3CzBN is almost the same as that of 4CzBN, o-3CzBN showed no appreciable DF, that is, kRISC ~ 0 s−1. The rate constants for nonradiative decay of the triplet states , a competing parameter of kRISC, are of the same order of magnitude for the three 3CzBN isomers, ~ 104 s−1, although their kRISC values differ by four orders of magnitude (Table 1). This result indicates that the TADF is activated by the large kRISC and not because of the suppression of the nonradiative decay path of the triplet states.
When doped at a concentration of 15 weight % (wt %) in a 2,8-bis(diphenylphosphoryl)dibenzo[b,d]thiophene (PPT) matrix, 5CzBN showed a PLQY of 78 ± 2%, and OLEDs using 5CzBN reached an ηEQE of ~24% (Fig. 2C), indicating that 5CzBN can intrinsically harvest almost all excitons for light emission in the OLEDs with the help of RISC. Here, we note that the 15 wt % 4CzBN–doped PPT films show a PLQY comparable with that of 5CzBN. However, the ηEQE of the 4CzBN-based OLEDs showed a slightly lower value of ~20%, presumably because of a lower charge carrier balance in the devices. Although 15 wt % p-3CzBN–doped PPT films show a low PLQY of 31 ± 2%, the ηEQE of p-3CzBN–based OLEDs (~4.5%) is much higher than the theoretical limit of ηEQE for ordinary fluorescent OLEDs considering a similar PLQY (1.6 to 2.3%), where a light outcoupling efficiency of 20 to 30% and an exciton generation ratio of singlets of 25% are assumed (22), indicating that TADF from p-3CzBN contributes to ηEQE. These photophysical properties and OLED characteristics imply that kRISC cannot be predicted by ΔEST only, and instead, the bonding configuration of D to A moieties plays an essential role for the CzBN derivatives to increase kRISC and enhance ηEQE.
To understand the large difference in kRISC by the structural modifications, we investigated the excited-state dynamics of CzPN and CzBN derivatives using TAS (2830). We measured the absorption change (ΔOD; OD, optical density) caused by the presence of photoexcited states of not only S1 and T1 but also intermediate excited states as a function of delay time (Δt). Figure 3A shows the TAS spectra of 4CzIPN (Δt = 3 ps and 4.6 μs) and 2CzPN (Δt = 3 ps and 30 μs) in toluene. The TAS profiles over the full measured range of Δt are illustrated in Fig. 3B as contour maps. The contour maps of both molecules show that the TAS spectra change in the vicinity of ISC, as judged from the time constant of the prompt fluorescence (τprompt) determined by TR-PL (Fig. 3C). We found that the spectral shape for Δt > τprompt stays unchanged for each molecule because of the almost identical decay lifetimes of T features (fig. S1) (the assignment is described below). In addition, we confirmed that the lifetimes of the T features for 4CzIPN are nearly the same as the τDF determined by TR-PL (fig. S1). These results indicate that the observed T states are mutually coupled in thermal equilibrium, and facilitate the RISC as a unit for the dynamics of RISC. Therefore, in the following experiments, we focused on the assignment of the excited states of the S and T states (Fig. 3A).
In the time range of Δt = 3 ps, an intense feature at 860 nm with a small shoulder (760 nm), which greatly resembles features commonly observed for cationic mono-Cz (Cz+) compounds (31, 32), was observed for both molecules. Considering that the absorption spectra of 4CzIPN and 2CzPN showed CT characters (5), we assigned the characteristic features at Δt = 3 ps to the absorption of Cz+ moieties (1Cz+) formed in the S1 state by the photoinduced intramolecular CT transitions (figs. S2 and S3). At the same Δt, we found another band in the near-infrared region of 4CzIPN. This band is a so-called charge resonance (CR) band (31) or an intervalence CT band (32), and it indicates the formation of a delocalized CT (deCT) state attributed to an intramolecular dimeric radical Cz (Cz2+) formed by an electronic coupling between a mono-Cz+ and a neutral Cz moiety (Fig. 3D). Specifically, a deCT state means that the cation charge of Cz+ is delocalized in the Cz2+, whereas the counter anion charge is localized at the acceptor moiety, that is, IPN for 4CzIPN. The CR band is seen not only in the S1 state as 1Cz2+ but also in the T1 state (Δt = 4.6 μs) as 3Cz2+, together with other rising features at 680 and 1070 nm. These rising features were also observed for 2CzPN in the time range of Δt = 30 μs, but the CR band was not detected. These results indicate that, in the S1 and T1 states, 2CzPN forms a localized CT (loCT) state, whereas 4CzIPN forms both deCT and loCT states in thermal equilibrium. The formation of this delocalized excited T state is consistent with the excited-state scheme for 4CzIPN obtained by time-resolved electron paramagnetic resonance conducted at 77 K (33).
Figure 4A shows contour maps for the TAS results of all six CzBN derivatives in toluene. The TAS spectra at Δt in the vicinity of τprompt of all the derivatives are depicted in fig. S4. In Fig. 4A, 2CzBN shows an absorption band at around 815 nm, which is assigned to 1Cz+ and thus lo,1CT. After transition through ISC, we found only one T absorption band at around 600 nm. Because the phosphorescence spectrum of 2CzBN (Fig. 2B), with clear vibronic progressions and a long triplet lifetime of ~5 s, was in good agreement with that of tricarbazolyl benzene (TCzB; inset in Fig. 2B), the T1 state of 2CzBN is considered to be the same as that of TCzB and, thereby, is assigned to a localized state on a mono-Cz moiety with pure π-π* character (34). In this context, the T1 absorption band at around 600 nm can be ascribed to the π-π* transition of a neutral mono-Cz state (3LE).
Similar to 2CzBN, the TAS spectra of o-3CzBN show a 1Cz+ band at around 850 nm before ISC and a 3LE band at around 600 nm after ISC. In addition, o-3CzBN exhibited another T band at around 1150 nm. Because the T1 character of o-3CzBN is understood as a mixing of the CT and LE states based on the vibronic-less phosphorescence spectrum (Fig. 2B) with a short lifetime (~1 s) (34), the feature at around 1150 nm can be attributed to the absorption of the loCT state of 3Cz+ (lo,3CT). These spectral characteristics are also observed for m-3CzBN, implying the formation of the same excited states.
Finally, we look at the dynamics of TADF-active CzBN derivatives. Clear 1Cz2+ and 3Cz2+ bands were observed in addition to 1Cz+, 3LE, and 3Cz+ bands for 5CzBN, similar to 4CzIPN, indicating the formation of lo,1CT and de,1CT in the S1 state and 3LE, lo,3CT, and de,3CT in the T1 state, respectively. On the other hand, for 4CzBN, 1Cz+ and 1Cz2+ bands were observed in the S1 state, but no 3Cz+ band was seen after relaxation via ISC and only 3LE and strong 3Cz2+ bands were formed (see also Fig. 4B). The phosphorescence spectra of 4CzBN and p-3CzBN show rather vibronic-less features resembling their fluorescence spectra (see Fig. 2B), indicating a strong CT character in the T1 state (34) because of the formation of de,3CT state. For p-3CzBN, the TAS spectra were essentially the same as those of 4CzBN, except for the energy positions, suggesting a similar T1 character and RISC process as for 4CzBN.
The formation of the deCT state is also reflected in the electronic structure of the CzBN derivatives. In the ground-state absorption spectra (Fig. 4C) of p-3CzBN, 4CzBN, and 5CzBN, a characteristic CT band is observed as the lowest energy transition (CT2), which differs from the commonly observed CT band (CT1) and π-π* band of Cz moieties. Furthermore, CT1 forms for all the CzBN derivatives, except p-3CzBN and 4CzBN. From these results, we can assign CT1 and CT2 to originate from the formation of loCT and deCT characters, respectively. This assignment is consistent with the TAS results, which suggest that p-3CzBN, 4CzBN, and 5CzBN form a de,1CT state in the S1 state, whereas other CzBN derivatives exhibit a lo,1CT state. The same consideration is applicable to 4CzIPN and 2CzPN (Fig. 3B and fig. S2).
DISCUSSION
The combined results of all structural, photophysical, excited-state, and OLED studies provide a comprehensive answer to the question of why kRISC of the CzBN derivatives does not necessarily depend on ΔEST. Molecules exhibiting TADF showed deCT states (that is, 1Cz2+ and 3Cz2+) and a 3LE state, whereas TADF-inactive molecules exhibited only a 3LE state or a combination of 3LE state and loCT states (that is, 1Cz+ and 3Cz+). This indicates, for CzBN derivatives, that neither the presence of only a 3LE state nor a combination of 3LE and loCT states is a sufficient condition for the activation of TADF, and instead, the presence of a deCT state is the key factor for increasing kRISC: The deCT facilitates RISC irrespective of ΔEST. However, this does not exclude the contribution of the combination of loCT and 3LE states to kRISC as 2CzPN, which showed these two states (Fig. 3B) has been reported to exhibit weak TADF (4.2 percentage point contribution to a total PLQY of 46%) (18). This result suggests that a combination of loCT and 3LE states can also be essentially involved in the RISC process if a D unit is combined with a strong A unit, such as PN. That is, the stabilization of the CT state is an important factor for enhancing kRISC.
Now, the practically relevant question is what are the structural requirements for deCT formation? The TAS results of the CzBN derivatives demonstrated that the position of the Cz units connected to the BN core is important. Namely, the positioning of Cz moieties at 2 and 3 (2CzBN) or 2, 4, and 6 (m-3CzBN) cannot satisfy the structural requirement (see Fig. 1C). Although the Cz at the 4-position of o-3CzBN might be expected to tend to form Cz+ because of the low electron density, the TAS results showed that the Cz at the 4-position does not form a CR band with the neighboring Cz moieties at the 3- or 5-position. In addition, we confirmed that the Cz at the 2-position does not play a role in the CR band of p-3CzBN because movement of the Cz from 2- to 4-position in p-3CzBN also showed the CR band (fig. S5). Consequently, we identify the common structure among the TADF-active molecules as a pair of Cz units connected linearly with a bridging A unit, that is, at 2- and 5-positions or 3- and 6-positions (Fig. 1D). This structural scheme is regarded as a linear D-A-D structure, and the linearly positioned Cz pair in the D-A-D structure is identified as the origin of the deCT states. Notably, a D-A-D structure has been a common chemical template of TADF molecules, but the linear D-A-D structure proposed here is a new connection rule for the D and A moieties within a category of D-A-D structures. Furthermore, the linear D-A-D structure is extremely simple, and a combination of other kinds of D and A units may facilitate RISC by forming deCT states.
We next discuss a mechanism leading to high kRISC via deCT in terms of excited-state dynamics. Generally, RISC is intrinsically forbidden between 3CT and 1CT states because of a vanishing of the SOC matrix elements between their molecular orbitals and it is facilitated by SOC between 3LE and 1CT (25, 27). In addition, as Gibson et al. (27) demonstrated theoretically, a vibronic coupling between the 3CT and 3LE states plays a crucial role in the production of 1CT states via RISC. In this context, the energy gap to overcome for RISC is not ΔEST but the energy difference between 3LE and 1CT, as here defined to ΔEST(LE), and a mutual coupling among the T states in thermal equilibrium is the key for efficient RISC.
Figure 4D shows ΔEST(LE) and ΔEST of all the CzBN derivatives, which are taken from Fig. 2B. Here, the 3LE level observed only for 2CzBN is used as a common one for all the CzBN derivatives because the 3LE, which originates from a phenyl-Cz, is not significantly affected by adding acceptor substituents (see the phosphorescence spectrum of 2CzBN and TCzB in Fig. 2B). It is seen that, although the ΔEST(LE) values of the TADF-inactive molecules are similar to their ΔEST values (~0.2 eV), the ΔEST(LE) values of the TADF-active molecules are 0.12, 0.01, and −0.06 eV for p-3CzBN, 4CzBN, and 5CzBN, respectively, which are all smaller than their corresponding ΔEST values. This result is caused by the lowering of the S1 state for the TADF-active molecules, in line with the formation of a CT2 band in Fig. 4C. In addition, a dependence on ΔEST(LE) appears among the TADF-active molecules. Both 4CzBN and 5CzBN, which exhibit kRISC values that are one order of magnitude higher than the kRISC value of p-3CzBN, show smaller ΔEST(LE) values. These data suggest that, for the facilitation of RISC, a ΔEST of ~0.2 eV does not play a role, and instead, a ΔEST(LE) of less than 0.2 eV is needed; a large kRISC can be achieved by decreasing ΔEST(LE). To verify this idea, a comparison of an activation energy for the RISC among the TADF-active molecules may be needed, but a quite small TADF efficiency of p-3CzBN, ~4% at room temperature, made the comparison in the present stage difficult. The kRISC of 5CzBN is higher than that of 4CzBN despite the larger energy gap between de,3CT and 3LE for 5CzBN than for 4CzBN. We speculate that the reason for this is effective vibronic coupling among lo,3CT, 3LE, and de,3CT, as observed in the TAS of 5CzBN. An efficient vibronic coupling can occur between 3LE and de,3CT, but the coexistence of lo,3CT may assist the vibronic coupling in 5CzBN.
Finally, we propose a comprehensive design strategy for TADF molecules to obtain a high PLQY. Although 4CzBN and 5CzBN exhibited comparable kRISC, the PLQY of 4CzBN (62%) was lower than that of 5CzBN (85%) in solution (Table 1). This is ascribed to the high of 4CzBN, and p-3CzBN was found to have an even higher . We attribute the high to structural relaxation because the PLQY of 4CzBN increases to nearly 100% in a rigid matrix owing to the suppression of (1.7 × 104 s−1 in solution to 3.4 × 103 s−1 in rigid matrix). For the CzBN derivatives, the most probable relaxation is the twisting of the Cz moieties against the BN core (5). The of p-3CzBN was a few times higher than that of 5CzBN because there is no Cz at the 5-position, and twisting of the Cz moiety at the 6-position can easily occur (Fig. 1C). The absence of Cz at the 4-position for 4CzBN still produces the twisting of Cz moiety at 3- and 5-positions. As a result, the twisting of all the Cz moieties becomes most difficult for 5CzBN owing to the strong steric hindrance around each Cz moieties.
According to the Mulliken-Hush theory, the peak top of a CR band (VCR) is proportional to the electronic coupling (V) between the two redox sites (35); thus, the degree of structural relaxation can be rationalized by the VCR in the linearly positioned Cz pair. The VCR gradually shifts to lower energies in the order of p-3CzBN > 4CzBN > 5CzBN (fig. S6), indicating that p-3CzBN forms a more planar D-A-D structure in the T1 state because of V and, thus, VCR increases when the twisting angle of the D moieties becomes smaller (32). Therefore, we conclude that binding the linearly positioned Cz pair with surrounding bulky groups suppresses the structural relaxation, thereby contributing to a decreased while keeping a high kRISC. Consequently, the most rigid molecule, 5CzBN, shows the highest PLQY owing to the largest kRISC and lowest .
In summary, we clarified that the formation of a deCT is a driving force to promote a large kRISC even when ΔEST is not close to zero. We also pointed out the importance of the suppression of the structural relaxation in the T1 state to achieve a high PLQY. For CzBN derivatives, twisting of the linearly positioned Cz pair connected to the A-unit plane may occur depending on the free space around the Cz pair. This causes the structural relaxation to deactivate the T1 state. Accordingly, we propose a chemical structure to form deCT while suppressing the structural relaxation by introducing bulky moieties around linearly positioned D units in a D-A-D structure. The simple design strategy established here will be extremely beneficial for the design of new TADF molecules, and we believe that our work contributes to the progression of photochemistry and the development of high-performance OLEDs, as well as future molecular light-emitting devices.
MATERIALS AND METHODS
Materials
4CzIPN, 2CzPN, m-3CzBN, 4CzBN, and 5CzBN were synthesized according to literature procedures (5, 20, 21). 2CzBN, o-3CzBN, and p-3CzBN were newly synthesized. The synthetic procedures and characterization are described in the Supplementary Materials. All the materials were purified by thermal sublimation. Solutions were prepared by dissolving the purified molecules in toluene (purity, 99.8%). The solution concentration was 10−3 to 10−5 M, depending on the samples and measurements.
The solutions were characterized by measuring the steady-state ultraviolet-visible (UV-VIS) absorption/PL spectra and TR-PL. TR-PL was measured using a C11367-01 spectrometer (Hamamatsu Photonics) or a Fluorocube fluorescence lifetime system (HORIBA). The PLQY of the solutions was measured by an absolute PLQY measurement system (C11347-01, Hamamatsu Photonics), with an excitation wavelength of 337 nm. Before the TAS and TR-PL measurements, the solutions were deoxygenated with dry nitrogen gas to eliminate the deactivation of triplets. The effect of deoxygenation was confirmed by comparing τprompt and τTADF values with those reported by Uoyama et al. (5).
OLED fabrication and characterization
Glass substrates with a prepatterned, 100-nm-thick, 100 ohm/square tin-doped indium oxide (ITO) coating were used as anodes. After precleaning of the substrates, effective device areas of 4 mm2 were defined on the patterned ITO substrates by a polyimide insulation layer using a conventional photolithography technique. Organic layers were formed by thermal evaporation. Doped emitting layers were deposited by coevaporation. Deposition was performed under vacuum at pressures <5 × 10−5 Pa. After device fabrication, devices were immediately encapsulated with glass lids using epoxy glue in a nitrogen-filled glove box [O2 < 0.1 parts per million (ppm) and H2O < 0.1 ppm]. OLEDs with the structure ITO/4,4′-cyclohexylidenebis[N,N-bis(4-methylphenyl)benzenamine] (TAPC) (35 nm)/mCP (10 nm)/dopant (15 wt %):PPT (30 nm)/PPT (40 nm)/LiF (0.8 nm)/Al (100 nm) were fabricated. The current density–voltage–luminance characteristics of the OLEDs were evaluated using a source measurement meter (B2912A, Agilent) and a calibrated spectroradiometer (CS-2000A, Konica Minolta Sensing Inc.). EL spectra were collected using a spectroradiometer (simultaneously with luminance). The ηEQE was calculated from the front luminance, current density, and EL spectrum. All measurements were performed under an ambient atmosphere at room temperature.
TAS measurements
Femtosecond-, nanosecond-, and microsecond-TAS measurements were conducted using different apparatuses developed in-house (2830). For femtosecond-TAS, the output from a Ti:Al2O3 regenerative amplifier [Spectra-Physics, Hurricane, 800 nm; full width at half maximum (FWHM) pulse, 130 fs; repetition, 1 kHz] was used as the light source. The wavelength of the pump laser was 400 nm, which is a second harmonic of the fundamental light (800 nm) generated by a β-barium borate crystal, whereas the white-light continuum generated by focusing the fundamental beam (800 nm) onto a sapphire plate (2 mm thick) was used as the probe light. For nanosecond-TAS, we used the third harmonic of fundamental light (1064 nm) of a Nd3+:YAG laser (wavelength, 355 nm; FWHM pulse, <150 ps; repetition, 10 Hz) as the pump light and a xenon flash lamp as the probe light. The system of nanosecond-TAS was used for microsecond-TAS measurements, but the probe light was exchanged with a xenon steady-state lamp. Although the strong intensity of the flash lamp was suitable for measurements with a fast response time (~1 ns), the use of the steady lamp could cover Δt ~ 100 ms with a slow response time of 10 to 30 ns. The irradiated intensity of the pump laser was set to 0.21 and 0.27 mJ/cm2 for femtosecond-TAS measurements of 4CzPN and 2CzPN, respectively, and 0.7 to 1.4 mJ/cm2 for nanosecond- and microsecond-TAS of all the derivatives. After the TAS measurements, UV-VIS absorption spectra were measured to check the sample degradation by laser irradiation. We noted that the long-duration irradiation of the femtosecond-pulse laser gave rise to a decrease of a first CT band (CT2) in the UV-VIS absorption spectra, in particular for 4CzIPN (fig. S2). Therefore, we carefully conducted the TAS measurements by checking the data reproducibility. All measurements were carried out at 295 K. In Fig. 3A, ΔOD of the microsecond-TAS spectra at Δt = 4.6 and 30 μs was corrected with reference to the intensity of nanosecond-TAS at Δt = 27 ns; the ΔOD of the nanosecond-TAS was also corrected in advance using femtosecond-TAS at Δt = 2 ns. The linearity of ΔOD for each correcting process was guaranteed by considering the time resolution of microsecond- and nanosecond-TAS measurements.
SUPPLEMENTARY MATERIALS
Synthesis and characterization of 2CzBN, o-3CzBN, and p-3CzBN
fig. S1. Time profiles of TAS of various triplet states of 4CzIPN and 2CzPN in toluene.
fig. S3. Laser power dependence in TAS spectra.
fig. S4. TAS spectra of the CzBN derivatives.
fig. S5. TAS results of 3, 4, 6-p-3CzBN.
fig. S6. Energy position of CR band.
fig. S7. Emission spectra of CzPN derivatives.
fig. S8. Time profiles of TAS of triplet states of CzBN derivatives.
This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial license, which permits use, distribution, and reproduction in any medium, so long as the resultant use is not for commercial advantage and provided the original work is properly cited.
REFERENCES AND NOTES
Acknowledgments: We thank W. J. Potscavage Jr. for assistance with the preparation of this manuscript and T. Yoshioka for assistance with this project. Funding: This work was supported, in part, by the “Development of Fundamental Evaluation Technology for Next-Generation Chemical Materials” program commissioned by the New Energy and Industrial Technology Development Organization, and the International Institute for Carbon Neutral Energy Research (WPI-I2CNER) sponsored by the Ministry of Education, Culture, Sports, Science and Technology (MEXT). Author contributions: T.H. and H.M. performed TAS measurements and T.H., H.M., A.F., and H. Nakanotani analyzed the data. T.H. and H. Nakanotani determined photophysical properties and analyzed the data. K.N. and H. Nomura prepared CzPN and CzBN derivatives. H. Nakanotani fabricated OLEDs and analyzed the data. T.H. and H. Nakanotani coordinated the work and wrote the paper. K.T., T.T., M.Y., and C.A. conceived the project, and all authors critically commented on the manuscript. Competing interests: The authors declare that they have no competing interests. Data and materials availability: All data needed to evaluate the conclusions in the paper are present in the paper and/or the Supplementary Materials. Additional data related to this paper may be requested from the authors.
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https://www.physicsforums.com/threads/fluid-properties-of-a-diesel-exhaust.238679/ | # Fluid properties of a diesel exhaust
1. Jun 4, 2008
### Anori
Hello,
Anyone know where I can get fluid properties (specifically kinematic viscosity) of the combustion exhaust form a diesel generator (i am designing an exhaust system and I need to calculate fluid flows). I am not sure the exact chemical makeup of the diesel fuel. An estimation would be fine.
Thanks!
2. Jun 4, 2008
### Anori
ok, can I do it like this..
diesel is a hydrocarbon...can I just base the fluid properties of the largest constituent? That is, assume the exhaust is 100% of the predominant gas since Its easier to find properties of specific gases at higher temperatures, lets say CO or N2?
3. Jun 4, 2008
### lewdtenant
CO2 is probably the predominant exhaust gas
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https://dc.edu.au/mathematics-extension-1-parametric-equations/ | # HSC Mathematics Extension 1 – Parametric Equations
Parametric Equations
## Parametric Representation
At times it is useful to express two related variables, such as $Latex formula$ and $Latex formula$, in terms of a third variable, $Latex formula$. Doing so gives,
$Latex formula$
$Latex formula$
These equations are called parametric equations, and $Latex formula$ is called a parameter. Indeed, this technique becomes increasingly important upon studying the algebraic representation of vectors.
It is particularly useful to express the equation of the parabola $Latex formula$ in terms of two parametric equations. These equations are $Latex formula$, $Latex formula$. By the simple elimination of $Latex formula$ we obtain the general Cartesian equation expressed above. The parabola and its parametric equations form the focus of this topic.
As mentioned above, the parametric equations of the parabola $Latex formula$ are given by:
$Latex formula$ $Latex formula$
It should be noted that the sign of $Latex formula$ indicates whether or not the parabola is concave up or down. If $Latex formula$ then it follows that the parabola is concave up, and if $Latex formula$ then it follows that the parabola is concave down. Also, as a side note $Latex formula$, the absolute value of the value of $Latex formula$, gives the focal length.
## Converting From Parametric Form to Cartesian form and vice versa
It is quite simple to express a parabola in parametric form. To do so simply find the corresponding value of $Latex formula$ or the parabola with equation $Latex formula$ and then substitute into the equations,
$Latex formula$
$Latex formula$
To convert a parabola from a set of parametric equations to a Cartesian equation, simply eliminate the value of $Latex formula$ by simple substitution. For example, to obtain the Cartesian equation of the parabola with parametric equations
$Latex formula$
$Latex formula$
We would simply perform the following algebraic manipulations,
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
Below is an example which illustrates the process of converting from Cartesian form to parametric form.
### Example 1
Find the Parametric equations of the parabola $Latex formula$.
### Solution 1
In this example, we simply find the value $Latex formula$ and substitute into the standard parametric equations. So we have that,
$Latex formula$
Hence, upon comparing the above equation to the general form $Latex formula$, we obtain the value of $Latex formula$ which is $Latex formula$. Hence the parametric equations are
$Latex formula$
$Latex formula$
Here is an example where the parametric equations are converted into a single Cartesian equation.
### Example 2
Find the Cartesian equation of the parabola with Parametric equations $Latex formula$, $Latex formula$.
### Solution 2
We shall simply eliminate the value of $Latex formula$ from the two equations to obtain a Cartesian equation.
$Latex formula$
Hence, upon substitution we obtain,
$Latex formula$
$Latex formula$
$Latex formula$
which is the Cartesian equation.
Here is a slightly more difficult question that requires some thought.
### Example 3
Find the parametric equations of the parabola $Latex formula$.
### Solution 3
Firstly we rearrange the equation to obtain the value of $Latex formula$.
$Latex formula$
Hence the value of $Latex formula$ is $Latex formula$.
Now, although this is omitted in most texts, the actual parametric equations of the parabola $Latex formula$ where $Latex formula$ is any non-zero number are, $Latex formula$. So in this case, we have that the parametric equations are,
$Latex formula$
$Latex formula$
We shall now investigate the equations of the chord, tangent and normal to the parabola $Latex formula$.
## Tangents, Normals and Chords
It should be noted that, upon studying this topic students should feel comfortable with all the derivation processes of the tangents, normals and chords in both parametric and Cartesian form. It is not particularly important to remember the equations of each of these, rather, it is important that the student be able to realise the relationships between the differing anatomical features of the parabola.
### Tangents
We shall firstly investigate the equation of the tangent at the point $Latex formula$ lying on the parabola $Latex formula$. Note that the tangent found with such coordinates is called the Cartesian form of the tangent. Conversely, the equation of the tangent found at the point $Latex formula$ is dubbed the Parametric form of the tangent.
The Cartesian equation of the tangent at the point $Latex formula$ on the parabola is $Latex formula$,
$Latex formula$
The parametric form of the equation of the tangent at the point $Latex formula$ is,
$Latex formula$
### Normals
The Cartesian equation of the normal at the point $Latex formula$ on the parabola $Latex formula$ is,
$Latex formula$
The parametric form of the equation of the normal at the point $Latex formula$ is,
$Latex formula$
### Chords
The equation of the chord between the points $Latex formula$ and $Latex formula$ on the parabola $Latex formula$ is
$Latex formula$
Now, suppose that the chord is a focal chord. That is, the chord passes through the focus $Latex formula$ of the parabola. Then we have that
That is,
$Latex formula$
It is also obvious that if $Latex formula$, then the chord $Latex formula$ passes through the focus $Latex formula$.
Hence we have that a necessary and sufficient condition for a chord to be a focal chord is that the parameters at their respective points must multiply to equal $Latex formula$. This fact becomes important later on.
### Chord of Contact
The chord of contact to the parabola $Latex formula$ from the external point $Latex formula$ is the chord that joins the two points from which tangents are drawn from the parabola that pass through the point $Latex formula$. Consider the diagram below. Here, $Latex formula$ is the chord of contact from the point $Latex formula$.
To show that a point $Latex formula$ is external to the parabola, we simply substitute the point into the equation $Latex formula$, and if $Latex formula$ then the point lies external to the parabola. Otherwise, the point lies on or within the parabola and hence is not external to the parabola.
The equation of the chord of contact from the external point $Latex formula$ to the parabola $Latex formula$ has equation $Latex formula$. Notice the similarity to the equation of the tangent. The proof of this is remarkably elegant, and explains why there is such a great similarity between general equation of the tangent and the equation of the chord of contact.
The equation of the chord of contact from the external point $Latex formula$ on the parabola $Latex formula$ is,
$Latex formula$
We shall now consider some examples.
### Example 4
Find the equations of the tangents to the parabola $Latex formula$ at the points $Latex formula$ and $Latex formula$. Find also the point of intersection of the two tangents.
### Solution 4
In this question, we shall not derive the equation but shall illustrate use of the formulae derived. Note however, that students should always derive the equations accordingly, instead of remembering formulae. So, comparing the parametric form of this parabola with the standard parametric form gives, $Latex formula$. Hence we have that the equation of the tangent at the point $Latex formula$ where coordinates are given by:
$Latex formula$
$Latex formula$
When $Latex formula$ we have the equation of the tangent being,
$Latex formula$
$Latex formula$
Now, solving simultaneously to find the point of intersection gives,
$Latex formula$
Hence the two tangents intersect at the point $Latex formula$
### Example 5
The straight line $Latex formula$ cuts the parabola $Latex formula$ at the points $Latex formula$ and $Latex formula$. Find:
a) The coordinates of $Latex formula$ and $Latex formula$
b) The equations of the normals at $Latex formula$ and $Latex formula$
c) Show that the normals intersect on the parabola.
### Solution 5
a) To solve this section, we must solve the equations of the parabola and the straight line simultaneously. Doing so gives,
$Latex formula$
$Latex formula$
$Latex formula$
Hence the coordinates of $Latex formula$ and $Latex formula$ are,
$Latex formula$ and $Latex formula$
b) Students at this point must derive the formulae using first principles. However, the derivation of the formulae is exactly the same as that done for the derived formulae, and hence we shall use the derived formulae. Hence we obtain the normal at $Latex formula$:
$Latex formula$
$Latex formula$
$Latex formula$
Also, the normal at $Latex formula$ has equation $Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
c) To find the point of intersection of the parabolas, it is necessary to solve their equations algebraically. So we have,
$Latex formula$
$Latex formula$
$Latex formula$ gives,
$Latex formula$
And hence we have that $Latex formula$. Thus the point of intersection of the normals is $Latex formula$.
### Example 6
Find the equation of the chord joining $Latex formula$ and $Latex formula$ on $Latex formula$ and $Latex formula$:
a) By first principles;
b) By using the formulae already derived.
c) Also show that is a focal chord.
### Solution 6
a) To find the equation by first principles, we simply find the points corresponding to the parameters $Latex formula$ and $Latex formula$.
When $Latex formula$, $Latex formula$. Hence the point is $Latex formula$.
When$Latex formula$ Hence the point is:
$Latex formula$
Now, we must find the gradient. Doing so gives,
$Latex formula$
Now, using the point gradient formula,
$Latex formula$
Hence the equation of $Latex formula$ is
$Latex formula$
b) Now, the formulae derived is
$Latex formula$
Now, let $Latex formula$. Also, upon comparing the parametric form with the standard parametric form of the parabola, we have that $Latex formula$. Thus, using this we obtain:
$Latex formula$
$Latex formula$
That is,
$Latex formula$
c) To show that it is a focal chord, we shall simply show that the focus $Latex formula$ passes through the chord. Now, since $Latex formula$, then it follows that the focus has coordinates $Latex formula$.
Now, substituting into the equation of the chord gives:
$Latex formula$
Hence the chord passes through the focus and is a focal chord.
Note that an alternative method is to show that product of the parameters for the endpoints of the chord has a product of $Latex formula$. That is, $Latex formula$. Recall that this was found to be a necessary and sufficient condition for the chord $Latex formula$ to be a focal chord.
### Example 7
Show that the point $Latex formula$ is external to the parabola $Latex formula$. Hence,
a) Find the equation of the chord of contact to the parabola $Latex formula$ from the point $Latex formula$ ;
b) Find the points of intersection of this chord with the parabola.
c) Show that the chord of contact is a focal chord
### Solution 7
We firstly show that the point is external to the parabola. Upon substituting the point we obtain, LHS $Latex formula$, RHS $Latex formula$. Obviously LHS $Latex formula$ RHS, and hence we have that as $Latex formula$, and it follows that the point is external to the parabola.
a) The equation of the chord of contact from the external point $Latex formula$ on the parabola $Latex formula$ is given by
$Latex formula$
In this case , and hence . Thus, the equation of the chord of contact is,
In this case $Latex formula$, and hence $Latex formula$. Thus, the
equation of the chord of contact is:
$Latex formula$
i.e.
$Latex formula$
b) To find the points of intersection we must solve simultaneously the equations of the parabola and the chord itself. Doing so gives,
$Latex formula$
$Latex formula$
$Latex formula$
$Latex formula$
Hence we have the points of intersection being,
$Latex formula$
c) To show that the chord of contact is indeed a focal chord, we must substitute the coordinates of the foci into the equation for the chord of contact. Alternatively, we may find the parameters of the points of intersection of the chord of contact with the parabola and show that these parameters indeed have a product of $Latex formula$.
Now, the focus has coordinates $Latex formula$, and as $Latex formula$, then we have that the focus has coordinates $Latex formula$.
Substituting this into the equation of the chord of contact gives,
LHS $Latex formula$ RHS.
Hence $Latex formula$ lies on the chord of contact, and consequently the chord of contact is a focal chord.
## Important Geometrical Properties of the Parabola
In this section we consider two important geometrical properties of the parabola. These properties are of immense importance and are required knowledge by the HSC Mathematics Extension 1 syllabus. We shall prove the properties by considering them as examples.
### Example 8
(Focus-directrix property) Prove that the tangents drawn from the ends of a focal chord always intersect at right angles on the directrix.
### Solution 8
Consider the parabola $Latex formula$, where $Latex formula$, and the points $Latex formula$ and $Latex formula$ that lie on it. By definition, it follows that the directrix has equation $Latex formula$ and the focus has coordinates $Latex formula$.
Also, suppose that the tangents at $Latex formula$ and $Latex formula$ intersect at the point $Latex formula$.Firstly, we have that the chord $Latex formula$ has equation:
$Latex formula$
Since this chord passes through the point $Latex formula$, it then follows that
$Latex formula$
$Latex formula$
Now, the tangents at $Latex formula$and $Latex formula$ have equations
$Latex formula$
$Latex formula$
Respectively.
Solving these two equations simultaneously gives the point $Latex formula$. Thus $Latex formula$ gives,
$Latex formula$
$Latex formula$
$Latex formula$
Now, substituting this into $Latex formula$ gives,
$Latex formula$
$Latex formula$
Hence the point $Latex formula$ is given by
$Latex formula$
Now, from $Latex formula$ and hence we have that the point $Latex formula$ is given by
$Latex formula$
Thus the directrices intersect on the line $Latex formula$ which is the directrix of the parabola.
Now, to show that the tangents intersect at right angles, we consider the gradients of the tangents. The gradients of the tangents at $Latex formula$ and $Latex formula$ are $Latex formula$ and $Latex formula$ respectively, upon comparison to the standard $Latex formula$ form. However, it is known that $Latex formula$ by # and hence the gradients of the tangents have a product of $Latex formula$. Thus the tangents intersect at right angles.
Hence if a chord is a focal chord then the tangents drawn from the endpoints of this focal chord intersect at right angles on the directrix.
### Example 9
(Reflection Property) Prove that a ray of light parallel to the axis entering a parabolic mirror passes through the focus after reflection from the mirror.
### Solution 9
Consider the parabola $Latex formula$ and the point $Latex formula$ lying upon it. Suppose that the tangent at the point $Latex formula$ intersects the axis of the parabola with at the point $Latex formula$. Consider the diagram below.
For the light ray to pass through the focus, it must be reflected by the tangent at the point $Latex formula$ towards the focus. However, this will occur if and only if the angle of incidence is equal to the angle of reflection with respect to the tangent as the normal surface (This is the law of reflection from physics). So we are required to show that $Latex formula$.
Now, the axis of the parabola (The $Latex formula$-axis) is parallel to the incoming light ray. It hence follows that $Latex formula$, as these two angles form corresponding angles with the axis being parallel to the incoming light ray.
Thus, by proving that $Latex formula$ we can effectively show that $Latex formula$. Now, it easier to show that two lengths are equal then to show that two angles are equal in analytic geometry. Thus, by showing that $Latex formula$ we effectively show that [/latex] ?STP=?SPT[/latex]. The reason behind this is that the angles effectively form the base angles of an isosceles triangle, if $Latex formula$.
Now, we firstly find $Latex formula$.
$Latex formula$
$Latex formula$
$Latex formula$
Now we need to find the length of $Latex formula$. Before doing so, we must find the coordinates of the point $Latex formula$.
The tangent at the point has equation $Latex formula$. $Latex formula$ is the point of intersection of the tangent at $Latex formula$ with the axis of the parabola. Letting $Latex formula$, gives
$Latex formula$
And hence the point $Latex formula$ is
$Latex formula$
Hence we have that $Latex formula$
Thus as $Latex formula$, it thus follows that $Latex formula$. But from the argument above $Latex formula$. Hence it follows that $Latex formula$.
That is, a ray of light parallel to the axis entering a parabolic mirror passes through the focus after reflection from the mirror.
A common mistake with this proof is that students often attempt to prove the result purely through algebra using gradients. This is not only tedious and time consuming, but is also many times more difficult than the proof presented in these notes. Students should note that at times a purely algebraic proof is not the best, and if possible geometry should be introduced to simplify the process. The HSC Mathematics Extension courses are about the interplay between geometry, algebra, arithmetic and calculus. Thus a good understanding of these courses and their subtopics, requires the ability to apply each and every one of these aspects when required.
The two properties already investigated are important, and may be directly examined. However to ensure a complete understanding of the techniques involved we shall investigate some more examples.
### Example 10
The points $Latex formula$, $Latex formula$ and $Latex formula$ are points on the parabola $Latex formula$. Find the relation between $Latex formula$ and $Latex formula$ if the chord $Latex formula$ is perpendicular to the normal at $Latex formula$.
### Solution 10
Now, we firstly find the gradient of the parabola at the point $Latex formula$.
$Latex formula$
$Latex formula$
At the point $Latex formula$, the gradient is
$Latex formula$
Now the gradient of the normal at $Latex formula$ is given by,
$Latex formula$
The gradient of the chord $Latex formula$ is given by,
$Latex formula$
So we have that the chord $Latex formula$ is perpendicular to the normal at $Latex formula$. Thus the product of the gradients must be equal to $Latex formula$.
$Latex formula$
i.e.
$Latex formula$
$Latex formula$
Hence the relationship between $Latex formula$ and $Latex formula$ is that the sum of $Latex formula$ and $Latex formula$ equals twice $Latex formula$.
### Example 11
The point $Latex formula$ lies on the parabola $Latex formula$. $Latex formula$ is the midpoint of the chord $Latex formula$. Find the point $Latex formula$ on the parabola where the tangent is parallel to the chord $Latex formula$, and also show that $Latex formula$ is equidistant from $Latex formula$ and the $Latex formula$-axis.
### Solution 11
Firstly, we draw a diagram to show the given information.We firstly find the coordinates of the midpoint $Latex formula$. Doing so gives,Now, we find the gradient of $Latex formula$ to then subsequently solve for the point $Latex formula$.
$Latex formula$
We now need to find the point $Latex formula$, which is defined to have a tangent parallel to $Latex formula$. It is known that $Latex formula$ represents the gradient of the tangent at the point $Latex formula$. Thus, we need to solve the equation
$Latex formula$
for $Latex formula$. We have that
$Latex formula$
Thus, our equation becomes
$Latex formula$
As $Latex formula$ lies on the parabola, we thus have that,
$Latex formula$
Thus the point $Latex formula$ is given by
$Latex formula$
Now, to show that $Latex formula$ is equidistant from the point $Latex formula$ and the $Latex formula$-axis, we firstly find the perpendicular distance of the point $Latex formula$ from the $Latex formula$-axis. This is very simple, as the distance is simply the $Latex formula$-value of the coordinate of the point $Latex formula$.
Hence the distance from $Latex formula$ the -axis to the point $Latex formula$ is $Latex formula$.
Now, the distance from the point $Latex formula$ to $Latex formula$ is given by
$Latex formula$
Hence we have that the distance $Latex formula$ is equal to the perpendicular distance of the point $Latex formula$ from the $Latex formula$-axis.
## Locus problems relating to the Parabola
The locus of a point $Latex formula$ on the Cartesian ( $Latex formula$$Latex formula$ plane) is the geometrical path traced out by that point. The result is a set of points which form an algebraic equation. In this section we aim to obtain appropriate loci, given certain conditions by eliminating the parameter as required, to obtain an equation in terms $Latex formula$ of $Latex formula$ and only (this is called the “Cartesian equation”). Although the process may sound complicated, it is in fact quite straightforward. Consider the examples below, which illustrate the methods involved in these questions.
### Example 12
$Latex formula$ is a variable point on the parabola $Latex formula$. Find the locus of the midpoint of the chord $Latex formula$ as $Latex formula$ varies on the parabola. ( $Latex formula$ is the origin)
### Solution 12
Consider the diagram below, which shows the information given.
We have that the midpoint $Latex formula$ which has variable Cartesian coordinates $Latex formula$, has parametric coordinates,
$Latex formula$
Now, equating Cartesian and Parametric coordinates gives,
$Latex formula$
Now, we must eliminate the value of $Latex formula$. So we have,
$Latex formula$
Substituting this into the expression for $Latex formula$ gives,
$Latex formula$
Thus, the locus of $Latex formula$ as $Latex formula$ moves is,
$Latex formula$
### Example 13
The diagram shows the graph of the parabola $Latex formula$. The tangent to the parabola at $Latex formula$, $Latex formula$, cuts the $Latex formula$-axis at $Latex formula$. The normal to the parabola at $Latex formula$ cuts the $Latex formula$-axis at $Latex formula$.
a) Derive the equation of the tangent to the parabola at the point $Latex formula$.
b) Show that $Latex formula$ has coordinates $Latex formula$.
c) Let $Latex formula$ be the midpoint of $Latex formula$. Find the Cartesian equation of the locus of $Latex formula$.
### Solution 13
a) We have that
$Latex formula$
$Latex formula$
At the point $Latex formula$,
$Latex formula$
Hence we have the equation of the tangent,
$Latex formula$
$Latex formula$
b) We now need to find the equation of the normal to find the coordinates of the point $Latex formula$. The gradient of the normal at $Latex formula$ is equal to the negative reciprocal of the gradient of the tangent at $Latex formula$. So we have that,
$Latex formula$
$Latex formula$
At the point $Latex formula$ we have that $Latex formula$, since this is the intersection with the $Latex formula$-axis.
$Latex formula$
Hence the point $Latex formula$ is given by $Latex formula$
c) The point $Latex formula$ is given by the intersection of the tangent with the $Latex formula$-axis. So, letting $Latex formula$ gives,
$Latex formula$
Hence the point $Latex formula$ is given by $Latex formula$.
The point $Latex formula$ in Cartesian coordinates, which is the midpoint of $Latex formula$ is given in parametric form by
$Latex formula$
Equating coordinates gives,
$Latex formula$
Thus, substituting for $Latex formula$ gives,
$Latex formula$
Thus, the locus of the point $Latex formula$ is
$Latex formula$
### Example 14
$Latex formula$ and $Latex formula$ are two variable points on a parabola $Latex formula$.
a) If the variable chord $Latex formula$ is always parallel to the line $Latex formula$, show that $Latex formula$.
b) The normals at $Latex formula$ and $Latex formula$ meet at $Latex formula$. Prove that the locus of $Latex formula$ is a straight line.
### Solution 14
Firstly, we draw a diagram to illustrate the information given.
a) The gradient of the chord $Latex formula$, $Latex formula$ is given by,
$Latex formula$
Now, this gradient must b equal to $Latex formula$ as the chord $Latex formula$ is at all times parallel to the lie $Latex formula$.
$Latex formula$
b) From the derivation in the “tangents, normals and chords”, the equation of the normal at the point $Latex formula$ is given by,
$Latex formula$
And the normal at $Latex formula$ is given by,
$Latex formula$
$Latex formula$ gives,
$Latex formula$
$Latex formula$
$Latex formula$
Now, at this point, to find the $Latex formula$ coordinate, we may substitute the $Latex formula$-value into $Latex formula$. However, this is tedious and an easier method is
$Latex formula$
Which gives,
$Latex formula$
Hence the point $Latex formula$ is given by
$Latex formula$
Equating Cartesian and parametric coordinates gives,
$Latex formula$
Now, at this point we perfect the square for the expression for $Latex formula$
$Latex formula$
Now, recall that $Latex formula$. Doing so gives,
$Latex formula$
$Latex formula$
We now eliminate $Latex formula$ from the expressions. Compare this to elimination of a single parameter used in earlier examples. This gives,
$Latex formula$
$Latex formula$
which is of course a straight line.
The technique employed here is an important one and is a vital member of your collection of techniques.
### Example 15
$Latex formula$ and $Latex formula$ are two points with parameters $Latex formula$ and $Latex formula$ respectively on the parabola $Latex formula$. The chord $Latex formula$passes through the fixed point $Latex formula$.
a) Show that $Latex formula$ $Latex formula$.
b) Find the locus of $Latex formula$, the point of intersection of the tangents $Latex formula$ at and $Latex formula$.
### Solution 15
a) Recall that the equation of the chord $Latex formula$ on the chord $Latex formula$ with parameters $Latex formula$ and $Latex formula$ respectively is given by:
$Latex formula$
Now, substituting the point $Latex formula$ into the equation gives,
$Latex formula$
$Latex formula$
b) Recall that the tangent at $Latex formula$ with parameter $Latex formula$ on the parabola $Latex formula$ has equation,
$Latex formula$
Similarly, the tangent at $Latex formula$ with parameter $Latex formula$ has equation,
$Latex formula$
Now, $Latex formula$ gives,
$Latex formula$
$Latex formula$
Substituting back into $Latex formula$ gives,
$Latex formula$
Hence the point of intersection $Latex formula$ has coordinates,
$Latex formula$
Now, equating Cartesian and parametric coordinates gives,
$Latex formula$
Now, recall from (a) that $Latex formula$ $Latex formula$.
So we have that,
$Latex formula$
$Latex formula$
Now, since the value of $Latex formula$ is a constant value regardless of the value of $Latex formula$, it thus follows that the locus of the point $Latex formula$ is the line $Latex formula$, since the values of $Latex formula$ and $Latex formula$ may vary through the real numbers.
A note should be made about the very common mistake of trying to eliminate the parameters without using the information given. Many students will try to find expressions for $Latex formula$ and $Latex formula$ by solving simultaneously. Do not fall into this common pitfall. This method is cumbersome and frivolous, as it leads nowhere. Always use the extra information given in the question. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 480, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9804487228393555, "perplexity": 468.68231724004863}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525133.20/warc/CC-MAIN-20190717081450-20190717103402-00064.warc.gz"} |
https://wwrenderer-staging.libretexts.org/render-api?sourceFilePath=Library/Union/setLimitContinuity/ur_lr_5_6.pg&problemSeed=1234567&courseID=anonymous&userID=anonymous&course_password=anonymous&answersSubmitted=0&showSummary=1&displayMode=MathJax&language=en&outputFormat=nosubmit | Find the value of the constant $b$ that makes the following function continuous on $(-\infty,\infty)$.
$b=$
Now draw a graph of $f$. | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9787745475769043, "perplexity": 54.31537602829291}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00232.warc.gz"} |
http://science.sciencemag.org/content/early/2014/07/16/science.1255380 | Report
# A mixture of Bose and Fermi superfluids
See allHide authors and affiliations
Science 17 Jul 2014:
1255380
DOI: 10.1126/science.1255380
## Abstract
Superconductivity and superfluidity of fermionic and bosonic systems are remarkable many-body quantum phenomena. In liquid helium and dilute gases, Bose and Fermi superfluidity has been observed separately, but producing a mixture in which both the fermionic and the bosonic components are superfluid is challenging. Here, we report on the observation of such a mixture with dilute gases of two Lithium isotopes, 6Li and 7Li. We probe the collective dynamics of this system by exciting center-of-mass oscillations that exhibit extremely low damping below a certain critical velocity. Using high precision spectroscopy of these modes we observe coherent energy exchange and measure the coupling between the two superfluids. Our observations can be captured theoretically using a sum-rule approach that we interpret in terms of two coupled oscillators.
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https://www.bartleby.com/solution-answer/chapter-11cr-problem-50e-algebra-and-trigonometry-mindtap-course-list-4th-edition/9781305071742/4752-matrix-equations-solve-the-matrix-equation-for-the-unknown-matrix-x-or-show-that-no/d6a1194d-eb04-4382-aa24-7be68afdb3f3 | Chapter 11.CR, Problem 50E
### Algebra and Trigonometry (MindTap ...
4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Chapter
Section
### Algebra and Trigonometry (MindTap ...
4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem
# 4 7 − 5 2 . Matrix Equations : Solve the matrix equation for the unknown matrix X, or show that no solution exists, where A = [ 2 1 3 2 ] , B = [ 1 − 2 − 2 4 ] , C = [ 0 1 3 − 2 4 0 ] 2 X + C = 5 A
To determine
To find:
The value of unknown matrix X satisfying matrix equation 2X+C=5A, if solution exists, where A=[2132] and C=[013240].
Explanation
Approach:
Two matrices A and B can be multiplied if they satisfy the condition that the number of columns in A is same as the number of rows in B.
Two matrices A and B can be added or subtracted only if they have same dimension.
Calculation:
A=[2132] and C=[013240]
In order to be added to the matrix C, matrix X must be of the dimension 3×2.
Let, X=[x1x2x3x4x5x6]
2X+C=5A2[x1x2x3x4x5x6]+[013240
### Still sussing out bartleby?
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http://science.sciencemag.org/content/296/5576/2095 | # Changes in Tropical Clouds and Radiation
See allHide authors and affiliations
Science 21 Jun 2002:
Vol. 296, Issue 5576, pp. 2095
DOI: 10.1126/science.296.5576.2095a
Two potentially important papers by Wielicki et al. (1) and Chen et al. (2) dealt with aspects of how clouds and radiation vary and change, and whether climate models simulate the changes correctly. There is ample prior evidence suggesting that models have difficulties in correctly simulating clouds, and clouds are regarded as the biggest source of uncertainty in reports by the Intergovernmental Panel on Climate Change (IPCC) (3). However, an alternative interpretation of the disagreements shown between observations and models is that the analyses of the observations may be flawed.
Making observations from space that are stable over time is difficult. Most satellites last only a few years before they are replaced, orbits of satellites decay, and heating extremes are experienced as the satellite platform moves into and out of sunlight. Therefore, calibration and ground truth validation are vital for accurate measurements. In the case of outgoing longwave radiation (OLR), there are no trends observed like those by the Earth Radiation Budget Satellite (ERBS) (1) in the operational NOAA series of satellites (4, 5). It seems likely that the highly unusual OLR observations during the 1997–98 El Niño were real (6) and failure of models to represent this climatic variation suggest their inadequacy. Because values from instruments on different satellites cannot be trusted without overlapping measurements, the reality of the increase in OLR in the 1990s hinges on the continuity of the ERBS measurements. However, there was a three-month hiatus in those measurements in 1993, after which substantial changes in calibration occurred and an offset of 2.5 W m−2 was introduced (7). Without that offset, the decadal increase in OLR would not exist. At the very least, this raises questions about the reality of the decadal variation reported in (1, 2).
Another contributing factor to a flawed analysis could be the way the data were processed. Because the ERBS satellite precesses with about a 36-day period, monthly averaged data alias the diurnal cycle onto 6-month periods (5 × 36 days). Hence, it is likely that the appearance of the semiannual cycle in the record is spurious and a result of interference between the diurnal and annual cycles. It is not surprising that this scenario is not simulated by current climate models.
Chen et al. (2) suggested that the hydrological cycle is changing, that the Hadley and Walker cells are strengthening, and that other variables, including upper-tropospheric humidity and cloudiness, are changing in consistent ways. Changes in satellites, however—specifically, the late 1991 introduction of NOAA-12—have introduced spurious variability into the cloud record. In addition, the challenge is to ensure that the changes in the variables are quantitatively consistent and not transient. A latitude–time series of the zonally averaged mean annual cycle and the anomalies of the 500 millibar (mb) vertical motion [Fig. 1 and (8, 9)] from the National Centers for Environmental Prediction (NCEP) reanalyses (10) for 1979 to 2000 place the Chen et al. period of January 1985 to August 1994 in context. Because a prolonged but modest El Niño event occurred from mid-1990 to mid-1995 (11–13), changes found by (2) are real, but small and short-lived. El Niño and interannual variability are dominant regardless of the analysis source (10), and decadal trends are not significant.
The overall conclusion is that there is great difficulty in obtaining reliable time series from space unless satellite records overlap, the orbits are stabilized, and calibration and validation are given much more attention. The results presented by Wielicki et al. and Chen et al. reveal the shortcomings in the current climate observing system and the need for a new approach to making stable homogeneous climate observations.
## REFERENCES AND NOTES
Response: We have carefully considered Trenberth's concerns regarding our papers (1, 2) and have reached the following conclusions.
First, Trenberth is concerned that there was an ERBS calibration shift while the instrument was powered down for 4 months from July to November 1993, during a spacecraft battery system anomaly. When the instrument resumed operation, the total channel offsets (zero-level instrument reading) used to provide longwave (LW) fluxes had dropped by about 3 W m−2, roughly the magnitude of the decadal tropical mean increase in LW flux. It is to be expected from both the physics of active-cavity instruments and past experience that changes in offsets will occur after extended power-down periods because of the change in thermal state of the instrument (3, 4). The validity of the ERBS offset change in late 1993 was verified using two independent tests. Offsets determined using the onboard blackbody were verified by direct observations of deep space four times between 1984 and 1999. All four cases agreed with blackbody-determined offsets to within 0.3 to 0.7 W m−2, while pre- and post-1993 values agreed within 0.5 W m−2. In addition, 6-month averages of Advanced Very High Resolution Radiometer (AVHRR), High-Resolution Infrared Radiation Sounder (HIRS), and ERBS LW fluxes before and after the period in question agreed to within 0.5 W m−2. For a 6-month period, AVHRR and HIRS orbit and calibration drift are expected to be small. We conclude that there is no evidence that a change in the ERBS calibration after the 4-month shutdown explains the decadal variations. We also note that both HIRS and AVHRR are only indirect measures of broadband LW flux.
Second, Trenberth suggests that the 36-day period required for the ERBS orbit to precess through 12 hours of local time sampling could alias diurnal cycle sampling errors into an apparent semiannual cycle change in shortwave (SW) flux. We tested this by constructing 36-day averages instead of monthly means, and found that Trenberth is correct: this removed roughly two-thirds of the semiannual signal. The reason for this error was an interaction between the monthly data processing and a slow drift in the phase of the precession by 6 hours over the period from 1985 to 1995. The 36-day average ERBS anomaly records are shown in Figs. 1 and 2, which should be compared with figures 3 and 4 of (1). The decadal variation in SW flux is now clearer, without a strong seasonal component. Some evidence for increased SW and net flux variability in the 1990s remains.
Third, we agree that interannual variability dominates, as indicated previously (2). The weaker decadal signal emerges only in the frequency domain. We plotted the time series of anomalies in frequency of occurrence of several parameters (Fig. 3). The El Niño–Southern Oscillation (ENSO) weakens the Walker cell, increasing occurrences of near-zero vertical velocity omega, mid- to high-range cloud amount, and midrange upper-troposphere relative humidity (UTH). Decadal variation is seen instead as shifts from near-zero to extreme high and low omega, and from higher to lower cloud amount and UTH, between the 1980s and 1990s. The early to mid-1990s do not resemble ENSO in frequency space. Our decadal subtropical omega (−7.6 × 10−4 Pa s−1) and LW flux (-2.8 W m−2) changes (2) are each ∼1.5% of their mean values, suggesting that increased subsidence warming balances increased radiative cooling. The observed cloud decrease from 1985–89 to 1996–99, input to a general circulation model (GCM) radiative transfer model, implies a SW flux decrease of 3.0 W m−2, similar to the 2.5 W m−2 decrease observed. Although relationships among the variables in Fig. 3 are complex and require further study, real evidence exists for consistent decadal variation in fluxes and independently observed climate parameters.
Finally, like Trenberth, we also call for more rigorous future observations of climate change. Current global observations are typically designed either for short NASA research missions or for NOAA weather missions at lower accuracy. For example, there is a planned gap in the radiation time series between the end of the NASA Aqua research mission in 2008 and the restart of the National Polar Orbiting Environmental Satellite System (NPOESS) weather system in 2012. Until a continuous climate observing system is established, both climate models and observations will remain uncertain. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8818190693855286, "perplexity": 2614.171394763392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00065.warc.gz"} |
https://socratic.org/questions/what-is-the-slope-of-the-line-normal-to-the-tangent-line-of-f-x-xcotx-2xsin-x-pi-1 | Calculus
Topics
# What is the slope of the line normal to the tangent line of f(x) = xcotx+2xsin(x-pi/3) at x= (15pi)/8 ?
Aug 12, 2017
$- \frac{2}{15 \sqrt{3} \pi + 4}$
#### Explanation:
$\Rightarrow x \cot x + 2 x \sin \left(x - \frac{\pi}{3}\right)$
For reference :-
$\sin \left(\frac{15 \pi}{8}\right) = \frac{1}{2}$
$\cos \left(\frac{15 \pi}{8}\right) = - \frac{\sqrt{3}}{2}$
$\cos e c \left(\frac{15 \pi}{8}\right) = 2$
$\cot \left(\frac{15 \pi}{8}\right) = - \sqrt{3}$
As slope of any line is the derivative of the equation, I will differentiate the equation of tangent.
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - x \cos e {c}^{2} x \cot x + 2 \left[\sin \left(x - \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)\right]$
clearly,
Here I can see the trigonometric formula of $\sin \left(A - B\right) \mathmr{and} \cos \left(A - B\right)$
Solving,
$\sin \left(x - \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)$
by above 2 formulas, I get,
$\left(\sin x + \cos x\right) + \sqrt{3} \left(\sin x - \cos x\right)$
So, at $x = \frac{15 \pi}{8}$,
The value of above equation becomes $2$
Now, at $x = \frac{15 \pi}{8}$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15 \sqrt{3} \pi + 4}{2} = {m}_{\text{1}}$
Now, when a line is normal to the tangent line, it means that the line is perpendicular to the tangent line.
Let the slope of perpendicular line be ${m}_{\text{2}}$
If the lines are perpendicular to each other then,
${m}_{\text{1"m_"2}} = - 1$
So, using above equation the slope of line normal to the tangent comes out to be,
${m}_{\text{2}} = - \frac{2}{15 \sqrt{3} \pi + 4}$
ENJOY MATHS !!!!!!!!
##### Impact of this question
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http://math.ecnu.edu.cn/RCFOA/seminar_template.php?id=271 | Lifting Commutation Relations in Cuntz Algebras
9:30 am to 10:20 am, June 5th, 2015 Science Building A1414
Abstract:
We examine splitting of the quotient map from the full free product $A * B$, or the unital free product $A *_{\mathbb{C}} B$, to the (maximal) tensor product $A \otimes B$, for unital $C^*$-algebras $A$ and $B$. Such a splitting is very rare, but we show there is one if $A$ and $B$ are both the Cuntz algebra $\mathcal{O}_2$ or $\mathcal{O}_{\infty}$, and in a few other cases. The splitting is not explicit (and in principle probably cannot be). We also describe several $K$-theoretic obstructions to a splitting. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8216598033905029, "perplexity": 466.89648725852294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00407.warc.gz"} |
https://www.physicsforums.com/threads/mandl-and-shaw-5-1.235077/ | # Mandl and Shaw 5.1
1. May 14, 2008
### jdstokes
[SOLVED] Mandl and Shaw 5.1
To show
$-\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2 \quad \mathrm{and} \quad -\frac{1}{2}\partial_\nu A_\mu \partial^\nu A^\mu$
represent the same Lagrangian it suffices to show that
$\partial_\nu A_\mu\partial^\mu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu$ is at most a 4-divergence.
The trouble is, I have no idea why this would be the case. Is this a matter of utilizing the product rule in some clever way?
Edit: Yes it is: factor out $\partial_\nu$.
Last edited: May 14, 2008
2. Oct 25, 2010
Re: [SOLVED] Mandl and Shaw 5.1
hi
does anybody have any suggestion to solve 2.4 too?!
3. Jun 23, 2011
### jmlaniel
Re: [SOLVED] Mandl and Shaw 5.1
Can this be done without using the Lorentz gauge ($\partial_\mu A^\mu = 0$) or is it necessary imposed ? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.952134907245636, "perplexity": 1458.2434280981859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00365.warc.gz"} |
http://lablemminglounge.blogspot.com/2008/03/geochemical-quantification-of-double.html | ## Monday, March 31, 2008
### The geochemical quantification of double beta decay
In 1933, Enrico Fermi proposed a theory of beta decay using the neutrino particle suggested by Wolfgang Pauli three years earlier. During this time period, the promising young physicist Maria Goeppert married an American guy named Mayer and started 15 years of two body-related unemployment in the States (In contrast, the gap between the publication of her book, “Elementary Theory of Nuclear Shell Structure” and the awarding of the Nobel prize for that work was only 13 years).
During the great depression, Goeppert-Mayer decided to up the intellectual profile of housewives by suggesting, in 1935, the possibility of double beta decay. Where an intervening unstable isotope prevented sequential single beta decay, double beta decay would allow transformation to a lower energy Z-2 nucleus by simultaneous double electron-antineutrino emission. Most of the double decay candidates are neutron-rich R-process elements such as 96Zr, 100Mo, or 130Te. The predicted decay product of 130Te is 130Xe, a noble gas. In the late 40’s, an isotopic study of telluride minerals was performed, and a 130Xe excess in Xenon from telluride ores was reported by Inghram & Reynolds (1950).
Subsequent noble gas results verified the existence of 128Te – 128Xe and 82Se – 82Kr systems, all before the first direct detection of a double beta decay was established in Elliott et al. (1987)- almost 40 years after the initial geochemical result, and 15 years after Goeppert-Mayer's death.
Unfortunately, the ability of Tellurium and Selenium minerals to quantitatively retain noble gasses is poor, as these minerals have low closure temperatures, and are easily deformed. They are also difficult to date directly. But as every hard-up geochemist knows, if you’re desperate for a date and you don’t know where to turn, it never hurts to look for a zircon.
The dating of zircon using the uranium-lead decay scheme is arguably the most popular and rigorous geochronological method currently available. And as it just so happens, 96Zr is expected to decay into 96Mo. Since zircon usually doesn’t have much initial Mo, It should be possibly to detect a 96Mo excess, and use the U/Pb age from the same mineral to calculate a decay constant for 96Zr (Technically, this method determines the 238U/96Zr decay constant ratio).
This has been done a few times (e.g. here and here), but the process is complicated by the spontaneous fission of 238U. The fission products of this decay include most heavy isotopes of Mo, so the fissionogenic Mo excess and the double beta 96Mo excess have to be deconvolved. The result is that the precision on 96Zr double beta decay is fairly poor, with Wieser & De Laeter (2001) reporting a value of 9.4 ± 3.2 x 1018.
However, there is another promising double beta decay isotope. 100Mo decays into 100Ru. And molybdenite generally contains Re, so that the mineral can be dated using the Re/Os single beta decay scheme. And moly contains fuck-all uranium. It looks like the NSF thinks this is a promising technique as well, as they’ve awarded a \$225,000 grant to a leading Re/Os lab for support for this and other experiments. And on the other side of the Pacific, Hidaka et al. 2004 have reported a result of 2.1 ± 0.3 x 1018.
In the meantime, the direct counting mob have continued to count decays. According to Barabash (2006), their current best determination of the 100Mo halflife is 7.1 ± 0.4 x 1018. It will be interesting to see if the Denver Re/Os crowd can do better, and if either group can explain why the direct counting gang have halflives that are approximately a factor of 2 higher than the geochemists (counters have 96Zr as 2.0±0.3x1019- also double the geochemical determination). Barabash 2006 does not address this discrepancy, or even reference the more recent geochemical results.
I’m also curious about the physicist’s budget. After all, I have a sneaking suspicion that the direct counting experiments cost a little bit more than a quarter million dollars.
References:
Barabash (2006) (How is one supposed to reference arXiv entries?)
Elliott S R Hahn A A and Moe M K 1987 Phys. Rev. Lett. 59 2020
Goeppert-Mayer M 1935 Phys. Rev. 48 512
Hidaka H Ly C V Siziki K 2004. Physical Review C, 70, id. 025501
Inghram M G and Reynold J H 1950 Phys. Rev. 78 822
Wieser M E De Laeter J R 2001 Phys. Rev. C 64, 024308
Ed said...
How is 100MO obtained? My periodic table lists molybdenum as atomic weight of 95.96 +/- .02. (Atomic number of 42)
Chuck said...
For multi-isotopic elements, the mean atomic weight listed on your table is the weighted mean of all the natural occurring isotopes.
Mo has 7 naturally occurring isotopes- 92, 94, 95, 96, 97, 98, and 100. To get the mean atomic weight, you multiply each by its relative abundance, and add the products:
92*0.148+94*0.0925+...
Although if you want better than 1% accuracy, you need to use the exact mass, not the nominal mass (e.g. 99.907 amu for 100Mo instead of 100).
Because the half-life of 100Mo (and all other double beta decay isotopes) is millions of times longer than the age of the universe, the proportions of these isotopes have only changed a little bit from what they were when the solar system formed.
Molybdenum has 7 stable
Ed said...
Thanks for the explain. Chem is not my strong suit.
Yes I remember that things were pretty much of a mess back then (when the solor system formed) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8395669460296631, "perplexity": 2645.5314784257225}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119653672.23/warc/CC-MAIN-20141024030053-00104-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://owasa.org/results-from-testing-of-drinking-water-for-fluoride-before-and-after-the-fluoride-overfeed-on-february-2-2017 | # Results from testing of drinking water for fluoride before and after the fluoride overfeed on February 2, 2017
As shown below, fluoride was at normal levels in drinking water sampled from pipes outside the Jones Ferry Road Water Treatment Plant (WTP) immediately following the shutdown of the plant at 3:22 PM on February 2, 2017 due to the accidental fluoride overfeed.
No water with elevated level of fluoride was pumped into the pipes serving our customers (the water distribution system). All water with elevated fluoride levels was contained on-site at the WTP.
The US Public Health Service’s recommended fluoride level is 0.7 parts per million (ppm). This level has been OWASA’s target fluoride level since 2012. By Federal and State regulations, the maximum limit for fluoride in distribution water systems is 4.0 ppm. Fluoride is also regulated with a secondary maximum limit of 2.0 ppm in distribution water systems; if levels were above this secondary limit, a utility would be required to notify their customers.
Prior to the fluoride overfeed on the morning of February 2, 2017, OWASA laboratory staff collected water samples both at the WTP and from sites within our service area as part of our routine distribution system sampling. The level of fluoride in these samples ranged from 0.62 to 0.67 ppm.
The fluoride overfeed was discovered at 3:00 PM that afternoon. At 3:10 PM laboratory staff collected a sample of drinking water from a location just before the water enters the pipes serving our customers. The fluoride level at this location was 0.68 ppm, which is within the normal range of OWASA drinking water; fluoride levels in 2016 ranged from 0.55 to 0.90 ppm with an average 0.69 ppm.
The WTP was shut down at 3:22 PM to contain the over-fluoridated water within the facility. As an added precaution, laboratory staff collected samples of drinking water at sites bracketing the WTP from 3:47 to 3:59 PM. These sites were selected because they are served by water pipes directly leaving the WTP. Fluoride levels in these samples were all 0.64 ppm, also within the normal range of OWASA drinking water.
The 1.5-million-gallon tank that contained the water with elevated fluoride levels at the WTP has heavy-duty curtains that force the water to follow a set path (shown in the figure below). There are three hatches where samples can be collected as water flows through this path. Beginning at 4:10 PM, two samples were collected from each of the three hatches, from the surface of the water and approximately mid-depth of the water. These sites are located further away from where treated drinking water leaves the WTP than the samples collected at 9:00 AM and 3:10 PM. At the time samples were collected from this tank, the WTP was shut down, no water was flowing through this tank, and all water in the tank was contained. As is shown in the graphic below, water above the secondary maximum limit (2.0 ppm) had not yet reached the point where water is pumped out of the tank into the drinking water piping system. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8320059180259705, "perplexity": 2795.9636296435237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00555.warc.gz"} |
https://mvtrinh.wordpress.com/2014/10/31/measure-of-angle/ | ## Measure of Angle
Let $ABCD$ be a rectangle with $BC=2AB$, and let $BCE$ be an equilateral triangle. $\overline{BE}$ and $\overline{EC}$ intersecting $\overline{AD}$ (not the extension of $\overline{AD}$) at $F$ and $G$, respectively. If $M$ is the midpoint of $\overline{EC}$, how many degrees are in angle $CMD$?
Source: NCTM Mathematics Teacher 2006
SOLUTION
$m\angle MCD=90-60=30^\circ$
Triangle $CMD$ is isosceles because $CM=CD=BC/2$
$m\angle CMD=(180-30)/2=150/2=75^\circ$
Answer: $75^\circ$
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## About mvtrinh
Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , . Bookmark the permalink. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9923012852668762, "perplexity": 651.0757170223369}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891811655.65/warc/CC-MAIN-20180218042652-20180218062652-00153.warc.gz"} |
http://www.eeng.dcu.ie/local-docs/latex-help/ensuremath.html | # \ensuremath (LaTeX2e)
\ensuremath{Text set in math mode}
The argument of the \ensuremath command is always set in math mode, regardless of the current mode. Note that math mode is not specifically invoked in the argument of the \ensuremath command.
Thus \ensuremath{$\alpha^2+\beta^2$} is not correct.
The correct form is \ensuremath{\alpha^2+\beta^2}. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9941652417182922, "perplexity": 4263.442503324995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825308.77/warc/CC-MAIN-20171022150946-20171022170946-00088.warc.gz"} |
https://math.stackexchange.com/questions/2029163/frac1n-as-a-difference-of-egyptian-fractions-with-all-denominators-n | # $\frac{1}{n}$ as a difference of Egyptian fractions with all denominators $<n$
Is there a good characterization of the set $S$ of positive integers $n$ such that $\frac{1}{n}$ can be represented as a difference of Egyptian fractions with all denominators $< n$? For example, $44 \in S$ because $$\dfrac{1}{44} = \left( \frac{1}{33} + \frac{1}{12}\right) - \frac{1}{11}$$
If I'm not mistaken, the first few members of $S$ are $$6, 12, 15, 18, 20, 21, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45$$ This does not appear to be in the OEIS yet; I intend to submit it soon. [ EDIT: It is now in OEIS as A278638.]
Here are some things I know so far:
1. If $n \in S$, then $mn \in S$ for any positive integer $m$.
2. $mn \in S$ for integers $m,n$ with $n < m < 2 n$, because $$\dfrac{1}{mn} = \dfrac{1}{n(m-n)} - \dfrac{1}{m(m-n)}$$
3. $S$ contains no prime or prime power.
4. There are no members of the form $2p^k$ where $p$ is a prime $> 3$.
5. There are no members of the form $3p^k$ where $p$ is a prime $> 11$.
• If you want to describe the solutions of this equation, using this formula. math.stackexchange.com/questions/419766/… Or use a different approach. math.stackexchange.com/questions/450280/… – individ Nov 28 '16 at 5:39
• An interesting subset are made the numbers n where $1/n$ is given by only two terms. I believe for the OP’s sequence from 6 to 45 that is the case for most numbers except 21, 33 and the PO’s example 44. Examples for 21 and 33 are: $1/21= 1/7+1/14 – (1/10+1/15)$, and $1/33= 1/15+1/10-(1/22+1/11)$. – Mikael Jensen Jan 10 '18 at 22:27
• One thing to note: Given property 1 it is useful to consider the set $S'$ such that for all $n\in S'$, there is no proper divisor of $n$ in $S'$ (then $S=\cup_{n\in\mathbb{N}}nS'$). Then the proper generalization of properties 4-5 is there are no members of $S'$ of the form $np^k$ where $p$ is a prime $>H_n \text{lcm}(1,\dots, n)$ where $H_n$ is the $n$-th harmonic number. – Will Fisher Jun 18 '18 at 16:24
• Can someone explain the meaning of 'characterisation' in this context. As, it seems to me that the question already lists a plethora of properties of $\mathbb{S}$. – Devashish Kaushik Aug 23 '18 at 7:56
• @DevashishKaushik characterization = necessary and sufficient condition (ideally one that can be easily tested). – Robert Israel Aug 23 '18 at 14:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9486596584320068, "perplexity": 170.86255164918543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573801.14/warc/CC-MAIN-20190920005656-20190920031656-00525.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/finite-math-and-applied-calculus-6th-edition/chapter-0-section-0-3-multiplying-and-factoring-algebraic-expressions-exercises-page-21/12 | ## Finite Math and Applied Calculus (6th Edition)
$y^{2} + \frac{1}{y^{2}} -2$
$(y -\frac{1}{y})^{2}$ 1. Square the first term. $(y)^{2} = y^{2}$ 2. Multiply the first term and second term together. Then multiply this answer by 2. So $(y)(\frac{-1}{y})(2) = -2$ 3. Square the last term. $(\frac{-1}{y})^{2} = \frac{1}{y^{2}}$ 4. Add your answers from steps 1-3 to get $y^{2} + \frac{1}{y^{2}} -2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.98046875, "perplexity": 862.4505450426279}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998708.41/warc/CC-MAIN-20190618083336-20190618105336-00112.warc.gz"} |
http://mathoverflow.net/questions/68308/algebraic-numbers-and-selberg-class | # Algebraic numbers and Selberg class
Hello,
I would like to know whether, given an algebraic number $\alpha$ of degree $d$, the Dedekind Zeta function $\zeta_{\mathbb{Q}(\alpha)}$ is always a function of the Selberg class of degree $d$ of not. I know that it is true when $\mathbb{Q}(\alpha)$ is an abelian extension of $\mathbb{Q}$, but what about the non abelian case? Thank you in advance.
-
Perhaps it is worthwhile to remark that if $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}(n)$ over a number field with unitary central character, and $\pi$ satisfies the Ramanujan conjecture, then $L(s,\pi)$ is in the Selberg class. Moreover, it is expected that any element in the Selberg class is of that form. – GH from MO Jun 22 '11 at 4:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8999995589256287, "perplexity": 74.05420698883286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153323.32/warc/CC-MAIN-20160205193913-00237-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/165499/would-a-directional-graviton-emitter-violate-any-known-laws-of-physics | # Would a directional “graviton” emitter violate any known laws of physics?
Setting aside that we don't known what the mediating particle in quantum gravity looks like and have no way to manipulate it, what would the implications of a directional graviton source be? Would it allow a "reaction-less" drive without creating other problems (e.g. that a violation of conservation of momentum has corresponding implications with respect to translational invariance)?
Also, presuming that such a device (unlike mass) can be made to turn on and off, would that imply (via conservation of momentum and the finite propagation rate of gravity), that a graviton must have (negative?) momentum?
By "directional", I'm referring to a gravitational effect that, on average and over an arbitrarily long time, will attract particles more strongly in one hemisphere than the other.
• Well, the lowest multipole that is occupied is the quadrupole, so your direction emitter is going to have a fiarly complicated design. – dmckee Feb 17 '15 at 0:30
• @dmckee Why would it inherently be any more complicated than, for example, a Yagi-antenna? What would require a graviton to be harder to manipulate than a photon? -- Besides, I'm specifically ignoring the practicality to look at the practicability and the implications of it being even possible. – BCS Feb 17 '15 at 1:30
• Electromagnetism admits dipole radiators, so the simple direction EM antenna is simpler than the simplest directional GR antenna. And you can expect the same to hold true for any given level of directionality. Yagis are a little complicated but get pretty good focus, to do as well with gravitational waves will take more elements. – dmckee Feb 17 '15 at 1:35
• @dmckee I didn't think gravitational waves (or at least any kind we know how to make) were any more directional than a simple dipole antenna would be. (See edit) – BCS Feb 17 '15 at 2:00
Beyond the answer depending on a quantum theory of gravity, there is an important difference that need to be made. I will take QED as an example to express this point, as directional photon emitter do exists. In this simple analogy, mass would be charge and photons, gravitons. The point I want to illustrate is related to the so-called "virtual" particles.
When two charged particles interact, this interaction is often pictured as relying on the exchange of virtual photons. These corresponds to the inner edges of the Feynman diagrams associated to the interaction. These virtual particles are best seen, in my humble opinion, as possible interaction channels. This way, we see the two charged particle as interacting via the photon field. There can also be freely propagating photons. These are the ones that are associated to the external edges of Feynman diagrams.
Likewise, for the case of gravity, we would expect that the gravitational interaction of two masses is mediated via the exchange of "virtual" gravitons. I suppose these are not the type you are referring about since they are not freely propagating. Indeed, freely propagating gravitons would be more related to gravitational waves. A classical gravitational wave would be expected to be some kind of superposition of gravitons, much the same way classical electromagnetic waves can be seen as a superposition of photons.
Pushing the analogy, lasers are directional sources of electromagnetic waves and the photons in the beam are in a coherent state. I would expect your directional graviton source to be some kind of "gravitational wave laser". In fact, this is something that probably could exist in principle.
Now, regarding your point on a "reaction-less" drive, a "gravity laser" would not be properly such a drive. If what you mean by reaction-less drive is just a drive that would not require propellant mass, than these already exists in the form of directional light sources. No violation of physical principles is involved as photons can carry momentum. Likewise, it may be possible to base a drive on a directional gravitational wave source, but I would not see the point, because there would be no inherent advantage to using a directional light source (since both gravitons and photons are massless, there would be no increase in the energy efficiency), but the complexity of the device would probably be orders of magnitude bigger as user dmckee suggested in his comments to your question.
Moreover, this would suggest that the gravitons must have negative momentum only if you define positive momentum in the direction the spaceship using such a drive is accelerating. In general, gravitons would not have "negative momentum" as this is essentially meaningless, the momentum being a vector quantity the sign of which depend on an inconsequent choice of space orientation if at all defined.
If you rather meant "negative momentum" as meaning that the if the beam emitted by your drive now collides with matter, it will act as an attractive force, then this is certainly not the case, as this would imply a violation of momentum conservation, as you suggested. In fact, the gravitational waves would probably diffract on the mass it "collides" with. A way to phrase it would be to say that some part of the wave would go right through and another part would be absorbed. The part that is absorbed could then be partly re-emitted. At the end the fraction of the momentum of the wave that is absorbed would be transferred to the mass. On the level of freely propagating gravitons, I suggest this would depend on the cross section of the graviton with the mass. With some probability, it will not interact and go right through and with some other probability it will be absorbed and maybe re-emitted, transferring momentum to the mass it collides with so that the total momentum is conserved, overall having a repulsive or no effect at all. An attractive force would rather involve virtual gravitons, unlike the one in the output of your device.
Finally, concerning your last comment, I am not sure of what you meant. It seemed you confused the "virtual" gravitons with the freely propagating ones. If particles are attracted more by one hemisphere, it would mean there is more mass in that hemisphere. It wouldn't have anything to do with the directionality of the emission of the freely propagating gravitons.
• @JDługosz Thank you for the edit. I tried to be consistent but I guess I slipped. To everyone: We should always say gravitational waves in this case instead of gravity waves as these are a fluid phenomenon. – G. Bergeron Dec 8 '16 at 6:53
• I noticed gravity waves used correctly in the most recent SETI seminar talk, on planetary formation. Gravity waves in the disk leading to periodic features and/or instability. – JDługosz Dec 8 '16 at 7:14
• I added a discussion on the interaction of the output beam of your supposed device with matter. – G. Bergeron Dec 8 '16 at 7:52
• @JDługosz Is the answer now satisfactory? – G. Bergeron Dec 11 '16 at 2:54
This is what happens when a gravitational wave passes . It does come from the direction of the source, but note the classical framework of general relativity.
Gravitons are the province of quantum mechanics and one can only speculate , until a definitive quantization of gravity is obtained.
Nevertheless, it is known that the classical wave will be composed by innumerable gravitons, the way classical electromagnetic waves emerge from innumerable photons. One can safely predict that conservation laws in the flat space of a limited environment that you imply all conservation laws will hold and there will only be a negative ! momentum if the coordinate system is designed that way. A machine emitting gravitons will have equal and opposite momentum to the gravitons leaving it.
One can make gravitational quadrupoles easily but generating enough energy is the problem, with the very weak couplings of gravity it is only with LIGO they detected gravitational waves, and individual gravitons have not been seen , due to the very small coupling constant.
So even if in principle one could design a graviton source, one needs astrophysical dimensions to be able to get enough energy in the wave generated .
• Sure emitting gravitational radiation is a rocket (as G. Bergeron already noted). But look at the closing paragraph in the question: will that beam have any affects on matter is passes through, to act as a tractor or repulser? – JDługosz Dec 8 '16 at 7:34 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989644050598145, "perplexity": 518.7791702839914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00557.warc.gz"} |
https://www.transtutors.com/questions/now-assume-that-bon-temps-is-expected-to-experience-nonconstant-253454.htm | # Now assume that Bon Temps is expected to experience nonconstant
Now assume that Bon Temps is expected to experience nonconstant growth of 30% for the next 3 years, then return to its long-run constant growth rate of 6%. What is the stock’s value under these conditions? What are its expected dividend and capital gains yields in Year 1? Year 4? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8985263705253601, "perplexity": 1862.2068660638438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744561.78/warc/CC-MAIN-20181118180446-20181118201614-00027.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/14%3A_Organohalogen__Organometallic_Compounds/14.07%3A_Aryl_Halides | # 14.7: Aryl Halides
Aryl halides have a halogen directly bonded to a carbon of an aromatic ring. Examples are bromobenzene, fluorobenzene, and 2,4-dichloromethylbenzene:
Some of the methods by which alkyl halides are prepared do not work for aryl halides because it is difficult to form $$\ce{C}$$-halogen bonds at aromatic ring carbons by nucleophilic displacement reactions. The most common ways of forming $$\ce{C}_\text{aryl}$$-halogen bonds are by substitution of $$\ce{C}_\text{aryl} \ce{-H}$$ by electrophilic halogenating agents (e.g., $$\ce{Br_2}$$ or $$\ce{Cl_2}$$),
and by replacement of $$\ce{C-NH_2}$$ by $$\ce{C}-$$halogen. These reactions are listed in Table 14-5 and will be discussed in more detail in Chapters 22 and 23.
## Nucleophilic Aromatic Displacement Reactions
The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides (see Table 4-6). The simple aryl halides generally are resistant to attack by nucleophiles in either $$S_\text{N}1$$ or $$S_\text{N}2$$ reactions (Table 14-6). However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid (a) when the aryl halide is activated by substitution with strongly electron-attracting groups such as $$\ce{NO_2}$$, and (b) when very strongly basic nucleophilic reagents are used.
## Addition-Elimination Mechanism of Nucleophilic Substitution
Although the simple aryl halides are inert to the usual nucleophilic reagents, considerable activation is produced by strongly electron-attracting substituents provided these are located in either the ortho or para positions, or both. For example, the displacement of chloride ion from 1-chloro-2,4-dinitrobenzene by dimethylamine occurs readily in ethanol solution at room temperature. Under the same conditions chlorobenzene completely fails to react; thus the activating influence of the two nitro groups amounts to a factor of at least $$10^8$$:
A related reaction is that of 2,4-dinitrofluorobenzene with the amino groups of peptides and proteins, and this reaction provides a means for analysis of the N-terminal amino acids in polypeptide chains. (See Section 25-7B.)
In general, the reactions of activated aryl halides closely resemble the $$S_\text{N}2$$-displacement reactions of aliphatic halides. The same nucleophilic reagents are effective (e.g., $$\ce{CH_3O}^\ominus$$, $$\ce{HO}^\ominus$$, and $$\ce{RNH_2}$$); the reactions are second order overall (first order in halide and first order in nucleophile); and for a given halide the more nucleophilic the attacking reagent, the faster the reaction. However, there must be more than a subtle difference in mechanism because an aryl halide is unable to pass through the same type of transition state as an alkyl halide in $$S_\text{N}2$$ displacements.
The generally accepted mechanism of nucleophilic aromatic substitution of aryl halides carrying activating groups involves two steps that are closely analogous to those briefly described in Section 14-4 for alkenyl and alkynyl halides. The first step involves attack of the nucleophile $$\ce{Y}^\ominus$$ at the carbon bearing the halogen substituent to form an intermediate carbanion $$4$$ (Equation 14-3). The aromatic system is destroyed on forming the anion, and the carbon at the reaction site changes from planar ($$sp^2$$ bonds) to tetrahedral ($$sp^3$$ bonds).
In the second step, loss of an anion, $$\ce{X}^ominus$$ or $$\ce{Y}^\ominus$$, regenerates an aromatic system, and, if $$\ce{X}^\ominus$$ is lost, the overall reaction is nucleophilic displacement of $$\ce{X}$$ by $$\ce{Y}$$ (Equation 14-4).
In the case of a neutral nucleophilic reagent, $$\ce{Y}$$ or $$\ce{HY}$$, the reaction sequence would be the same except for the necessary adjustments in the charge of the intermediate:
Why is this reaction pathway generally unfavorable for the simple aryl halides? The answer is that the intermediate $$4$$, which we can express as a hybrid of the valence-bond structures $$4a$$-$$4c$$, is too high in energy to be formed at any practical rate. Not only has $$4$$ lost the aromatic stabilization of the benzene ring, but its formation results in transfer of negative charge to the ring carbons, which themselves are not very electronegative:
However, when strongly electron-attracting groups are located on the ring at the ortho-para positions, the intermediate anion is stabilized by delocalization of electrons from the ring carbons to more favorable locations on the substituent groups. As an example, consider the displacement of bromine by $$\ce{OCH_3}$$ in the reaction of 4-bromonitrobenzene and methoxide ion:
The anionic intermediate formed by addition of methoxide ion to the aryl halide can be described by the valence-bond structures $$5a$$-$$5d$$. Of these structures $$5d$$ is especially important because in it the charge is transferred from the ring carbons to the oxygen of the nitro substituent:
Substituents in the meta positions have much less effect on the reactivity of an aryl halide because delocalization of electrons to the substituent is not possible. No formulas can be written analogous to $$5c$$ and $$5d$$ in which the negative charges are both on atoms next to positive nitrogen, $$\overset{\ominus}{\ce{C}} \overset{\oplus}{\ce{-N}-} \overset{\ominus}{\ce{O}}$$ and $$\overset{\ominus}{\ce{O}} \overset{\oplus}{\ce{-N}-} \overset{\ominus}{\ce{O}}$$,
In a few instances, stable compounds resembling the postulated reaction intermediate have been isolated. One classic example is the complex $$7$$ (isolated by J. Meisenheimer), which is the product of the reaction of either the methyl aryl ether 6$$6$$ with potassium ethoxide, or the ethyl aryl ether $$8$$ and potassium methoxide:
## Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution. Arynes
The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around $$340^\text{o}$$ and this reaction was once an important commercial process for the production of benzenol (phenol):
In addition, aryl chlorides, bromides, and iodides can be converted to areneamines $$\ce{ArNH_2}$$ by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as $$-33^\text{o}$$ with liquid ammonia as solvent:
However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, the entering group does not always occupy the same position on the ring as that vacated by the halogen substituent. For example, the hydrolysis of 4-chloromethylbenzene at $$340^\text{o}$$ gives an equimolar mixture of 3- and 4-methylbenzenols:
Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change (Section 1-1H):
The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed elimination of hydrogen halide $$\left( \ce{HX} \right)$$ from the aryl halide - as illustrated below for the amination of bromobenzene:
Elimination
The product of the elimination reaction is a highly reactive intermediate $$9$$ called benzyne, or dehydrobenzene, which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product:
The rearrangements in these reactions result from the attack of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, $$10$$, from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols:
In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in Section 14-10C and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as $$11$$, in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without:
When $$11$$ decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself:
## Uses for Aryl Halogen Compounds
As with most organic halides, aryl halides most often are synthetic intermediates for the production of other useful substances. For example, chlorobenzene is the starting aryl halide for the synthesis of DDT; it also is a source of benzenol (phenol, Section 14-6C) which, in turn, has many uses (Section 26-1).
Several aromatic chloro compounds are used extensively as insecticides, herbicides, fungicides, and bactericides. They also have acquired much notoriety because in some instances their indiscriminate usage has led to serious problems. For example, hexachlorophene is an external bactericide that until recently was used in cosmetic preparations such as soaps, deodorants, and so on. Its use has been discontinued because of compelling evidence that it can be absorbed through the skin in amounts that are dangerous, if not lethal, for infants and small children.
Other pesticides, notably DDT and the herbicides 2,4-D and 2,4,5-T have been partially banned for different reasons. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8186822533607483, "perplexity": 3362.7035921834404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00047.warc.gz"} |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Characteristic_impedance | # Characteristic impedance
The characteristic impedance or surge impedance (usually written Z0) of a uniform transmission line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that is, a wave travelling in one direction in the absence of reflections in the other direction. Alternatively and equivalently it can be defined as the input impedance of a transmission line when its length is infinite. Characteristic impedance is determined by the geometry and materials of the transmission line and, for a uniform line, is not dependent on its length. The SI unit of characteristic impedance is the ohm.
The characteristic impedance of a lossless transmission line is purely real, with no reactive component. Energy supplied by a source at one end of such a line is transmitted through the line without being dissipated in the line itself. A transmission line of finite length (lossless or lossy) that is terminated at one end with an impedance equal to the characteristic impedance appears to the source like an infinitely long transmission line and produces no reflections.
## Transmission line model
The characteristic impedance ${\displaystyle Z(\omega )}$ of an infinite transmission line at a given angular frequency ${\displaystyle \omega }$ is the ratio of the voltage and current of a pure sinusoidal wave of the same frequency travelling along the line. This definition extends to DC by letting ${\displaystyle \omega }$ tend to 0, and subsists for finite transmission lines until the wave reaches the end of the line. In this case, there will be in general a reflected wave which travels back along the line in the opposite direction. When this wave reaches the source, it adds to the transmitted wave and the ratio of the voltage and current at the input to the line will no longer be the characteristic impedance. This new ratio is called the input impedance. The input impedance of an infinite line is equal to the characteristic impedance since the transmitted wave is never reflected back from the end. It can be shown that an equivalent definition is: the characteristic impedance of a line is that impedance which, when terminating an arbitrary length of line at its output, produces an input impedance of equal value. This is so because there is no reflection on a line terminated in its own characteristic impedance.
Applying the transmission line model based on the telegrapher's equations as derived below,[1][2] the general expression for the characteristic impedance of a transmission line is:
${\displaystyle Z_{0}={\sqrt {\frac {R+j\omega L}{G+j\omega C}}}}$
where
${\displaystyle R}$ is the resistance per unit length, considering the two conductors to be in series,
${\displaystyle L}$ is the inductance per unit length,
${\displaystyle G}$ is the conductance of the dielectric per unit length,
${\displaystyle C}$ is the capacitance per unit length,
${\displaystyle j}$ is the imaginary unit, and
${\displaystyle \omega }$ is the angular frequency.
Although an infinite line is assumed, since all quantities are per unit length, the characteristic impedance is independent of the length of the transmission line.
The voltage and current phasors on the line are related by the characteristic impedance as:
${\displaystyle {\frac {V^{+}}{I^{+}}}=Z_{0}=-{\frac {V^{-}}{I^{-}}}}$
where the superscripts ${\displaystyle +}$ and ${\displaystyle -}$ represent forward- and backward-traveling waves, respectively. A surge of energy on a finite transmission line will see an impedance of Z0 prior to any reflections arriving, hence surge impedance is an alternative name for characteristic impedance.
## Derivation
### Using telegrapher's equation
The differential equations describing the dependence of the voltage and current on time and space are linear, so that a linear combination of solutions is again a solution. This means that we can consider solutions with a time dependence ${\displaystyle e^{j\omega t}}$, and the time dependence will factor out, leaving an ordinary differential equation for the coefficients, which will be phasors depending on space only. Moreover, the parameters can be generalized to be frequency-dependent.[1]
Let
${\displaystyle V(x,t)\equiv V(x)\ e^{+j\omega t}}$
and
${\displaystyle I(x,t)\equiv I(x)\ e^{+j\omega t}}$
Take the positive direction for ${\displaystyle V}$ and ${\displaystyle I}$ in the loop to be clockwise.
We find that
${\displaystyle {\text{d}}V=-(R+j\omega L)\ I\ dx=-Z\ I\ {\text{d}}x}$
and
${\displaystyle {\text{d}}I=-(G+j\omega C)\ V\ {\text{d}}x=-Y\ V\ {\text{d}}x}$
or
${\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I}$
and
${\displaystyle {\frac {{\text{d}}I}{{\text{d}}x}}=-Y\ V}$
These two first-order equations are easily uncoupled by a second differentiation, with the results:
${\displaystyle {\frac {{\text{d}}^{2}V}{{\text{d}}x^{2}}}=ZY\ V}$
and
${\displaystyle {\frac {{\text{d}}^{2}I}{{\text{d}}x^{2}}}=ZY\ I}$
Notice that both ${\displaystyle V}$ and ${\displaystyle I}$ satisfy the same equation.
Since ${\displaystyle ZY}$ is independent of ${\displaystyle x}$ and ${\displaystyle t}$, it can be represented by a single constant ${\displaystyle -k^{2}}$. That is:
${\displaystyle -k^{2}\equiv Z\ Y\ }$
so
${\displaystyle j\ k=\pm {\sqrt {Z\ Y\ }}}$
The minus sign is included for later convenience. Because of it, we can write the above equation as
${\displaystyle k=\pm \omega {\sqrt {(L-jR/\omega )(C-jG/\omega )\ }}}$
which is correct for all transmission lines. And for typical transmission lines, that are built to make wire resistance loss ${\displaystyle R}$ small and insulation leakage conductance ${\displaystyle G}$ low, the constant ${\displaystyle k}$ is very close to being a real number:
${\displaystyle k\approx \pm \omega {\sqrt {LC\ }}.}$
Further, with this definition of ${\displaystyle k}$ the position- or ${\displaystyle x}$-dependent part will appear as ${\displaystyle \ \pm j\ k\ x\ }$ in the exponential solutions of the equation, similar to the time-dependent part ${\displaystyle \ +j\ \omega \ t\ }$, so the solution reads
${\displaystyle V(x)=v^{+}\ e^{-jkx}+v^{-}e^{+jkx}}$
where ${\displaystyle v^{+}}$ and ${\displaystyle v^{-}}$ are the constants of integration. When we recombine the time-dependent part we obtain the full solution:
${\displaystyle V(x,t)\quad =\quad V(x)\ e^{+j\omega t}\quad =\quad v^{+}\ e^{-jkx+j\omega t}+v^{-}e^{+jkx+j\omega t}}$
Since the equation for ${\displaystyle I}$ is the same form, it has a solution of the same form:
${\displaystyle I(x)=i^{+}\ e^{-jkx}+i^{-}e^{+jkx}}$
where ${\displaystyle i^{+}}$ and ${\displaystyle i^{-}}$ are again constants of integration.
The above equations are the wave solution for ${\displaystyle V}$ and ${\displaystyle I}$. In order to be compatible, they must still satisfy the original differential equations, one of which is
${\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I}$
Substituting the solutions for ${\displaystyle V}$ and ${\displaystyle I}$ into the above equation, we get
${\displaystyle {\frac {\text{d}}{{\text{d}}x}}\left[v^{+}\ e^{-jkx}+v^{-}\ e^{+jkx}\right]=-(R+j\omega L)\left[\ i^{+}\ e^{-jkx}+i^{-}\ e^{+jkx}\right]}$
or
${\displaystyle -jk\ v^{+}\ e^{-jkx}+jk\ v^{-}\ e^{+jkx}=-(R+j\omega L)\ i^{+}\ e^{-jkx}-(R+j\omega L)\ i^{-}\ e^{+jkx}}$
Isolating distinct powers of ${\displaystyle e}$ and combining identical powers, we see that in order for the above equation to hold for all possible values of ${\displaystyle x}$ we must have:
For the co-efficients of ${\displaystyle e^{-jkx}\quad {\text{ : }}\quad -j\ k\ v^{+}=-(R+j\omega \ L)\ i^{+}}$
For the co-efficients of ${\displaystyle e^{+jkx}\quad {\text{ : }}\quad +j\ k\ v^{-}=-(R+j\omega L)\ i^{-}}$
Since ${\displaystyle jk={\sqrt {(R+j\omega L)(G+j\omega C)\ }}}$
${\displaystyle +{\frac {v^{+}}{i^{+}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}$
${\displaystyle -{\frac {v^{-}}{i^{-}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}$
hence, for valid solutions require
${\displaystyle v^{+}=+Z_{\text{o}}\ i^{+}\quad {\text{ and }}\quad v^{-}=-Z_{\text{o}}\ i^{-}}$
It can be seen that the constant ${\displaystyle Z_{\text{o}}}$, defined in the above equations has the dimensions of impedance (ratio of voltage to current) and is a function of primary constants of the line and operating frequency. It is called the “characteristic impedance” of the transmission line, and conventionally denoted by ${\displaystyle Z_{\text{o}}}$.[2]
${\displaystyle Z_{\text{o}}\quad =\quad {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\quad =\quad {\sqrt {{\frac {L-jR/\omega }{C-jG/\omega }}\ }}}$
for any transmission line, and for well-functioning transmission lines, with ${\displaystyle R}$ and ${\displaystyle G}$ both very small, or ${\displaystyle \omega }$ very high, or all of the above, we get
${\displaystyle Z_{\text{o}}\approx {\sqrt {{\frac {L}{C}}\ }}}$
hence the characteristic impedance is typically very close to being a real number (see also the Heaviside condition.)
### Alternative approach
We follow an approach posted by Tim Healy.[3] The line is modeled by a series of differential segments with differential series ${\displaystyle (R{\text{d}}x,L{\text{d}}x)}$ and shunt ${\displaystyle (C{\text{d}}x,G{\text{d}}x)}$ elements (as shown in the figure above). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance ${\displaystyle Z_{\text{o}}}$. That is, the impedance looking into the line on the left is ${\displaystyle Z_{\text{o}}}$. But, of course, if we go down the line one differential length ${\displaystyle {\text{d}}x}$, the impedance into the line is still ${\displaystyle Z_{\text{o}}}$. Hence we can say that the impedance looking into the line on the far left is equal to ${\displaystyle Z_{\text{o}}}$ in parallel with ${\displaystyle C{\text{d}}x}$ and ${\displaystyle G{\text{d}}x}$, all of which is in series with ${\displaystyle R{\text{d}}x}$ and ${\displaystyle L{\text{d}}x}$. Hence:
${\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {1}{\ (G+j\omega C){\text{d}}x+{\frac {1}{Z_{\text{o}}}}\ }}}$
${\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {\ Z_{\text{o}}\ }{Z_{\text{o}}(G+j\omega C){\text{d}}x+1\ }}}$
${\displaystyle Z_{\text{o}}+Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C){\text{d}}x(R+j\omega L){\text{d}}x+Z_{\text{o}}}$
The ${\displaystyle Z_{\text{o}}}$ terms cancel, leaving
${\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C)(R+j\omega L)({\text{d}}x)^{2}}$
The first-power ${\displaystyle {\text{d}}x}$ terms are the highest remaining order. In comparison to ${\displaystyle {\text{d}}x}$, the term with the factor ${\displaystyle ({\text{d}}x)^{2}}$ may be discarded, since it is infinitesimal in comparison, leading to:
${\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x}$
and hence
${\displaystyle Z_{\text{o}}=\pm {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}}$
Reversing the sign on the square root has the effect of changing the direction of the flow of current.
## Lossless line
The analysis of lossless lines provides an accurate approximation for real transmission lines that simplifies the mathematics considered in modeling transmission lines. A lossless line is defined as a transmission line that has no line resistance and no dielectric loss. This would imply that the conductors act like perfect conductors and the dielectric acts like a perfect dielectric. For a lossless line, R and G are both zero, so the equation for characteristic impedance derived above reduces to:
${\displaystyle Z_{0}={\sqrt {\frac {L}{C}}}.}$
In particular, ${\displaystyle Z_{0}}$ does not depend any more upon the frequency. The above expression is wholly real, since the imaginary term j has canceled out, implying that Z0 is purely resistive. For a lossless line terminated in Z0, there is no loss of current across the line, and so the voltage remains the same along the line. The lossless line model is a useful approximation for many practical cases, such as low-loss transmission lines and transmission lines with high frequency. For both of these cases, R and G are much smaller than ωL and ωC, respectively, and can thus be ignored.
The solutions to the long line transmission equations include incident and reflected portions of the voltage and current:
${\displaystyle V={\frac {V_{r}+I_{r}Z_{c}}{2}}\varepsilon ^{\gamma x}+{\frac {V_{r}-I_{r}Z_{c}}{2}}\varepsilon ^{-\gamma x}}$
${\displaystyle I={\frac {V_{r}/Z_{c}+I_{r}}{2}}\varepsilon ^{\gamma x}-{\frac {V_{r}/Z_{c}-I_{r}}{2}}\varepsilon ^{-\gamma x}}$
When the line is terminated with its characteristic impedance, the reflected portions of these equations are reduced to 0 and the solutions to the voltage and current along the transmission line are wholly incident. Without a reflection of the wave, the load that is being supplied by the line effectively blends into the line making it appear to be an infinite line. In a lossless line this implies that the voltage and current remain the same everywhere along the transmission line. Their magnitudes remain constant along the length of the line and are only rotated by a phase angle.
In electric power transmission, the characteristic impedance of a transmission line is expressed in terms of the surge impedance loading (SIL), or natural loading, being the power loading at which reactive power is neither produced nor absorbed:
${\displaystyle {\mathit {SIL}}={\frac {{V_{\mathrm {LL} }}^{2}}{Z_{0}}}}$
in which ${\displaystyle V_{\mathrm {LL} }}$ is the line-to-line voltage in volts.
Loaded below its SIL, a line supplies reactive power to the system, tending to raise system voltages. Above it, the line absorbs reactive power, tending to depress the voltage. The Ferranti effect describes the voltage gain towards the remote end of a very lightly loaded (or open ended) transmission line. Underground cables normally have a very low characteristic impedance, resulting in an SIL that is typically in excess of the thermal limit of the cable. Hence a cable is almost always a source of reactive power.
## Practical examples
Standard Impedance (Ω) Tolerance
Ethernet Cat.5 100±5 Ω[4]
USB 90±15%[5]
HDMI 95±15%[6]
IEEE 1394 108+3
−2
%[7]
VGA 75±5%[8]
DisplayPort 100±20%[6]
DVI 95±15%[6]
PCIe 85±15%[6]
### Coaxial cable
The characteristic impedance of coaxial cables (coax) is commonly chosen to be 50 Ω for RF and microwave applications. Coax for video applications is usually 75 Ω for its lower loss.
## References
1. "The Telegrapher's Equation". mysite.du.edu. Retrieved 2018-09-09.
2. "Derivation of Characteristic Impedance of Transmission line". GATE ECE 2018. 2016-04-16. Retrieved 2018-09-09.
3. "Characteristic Impedance". www.ee.scu.edu. Retrieved 2018-09-09.
4. "SuperCat OUTDOOR CAT 5e U/UTP" (PDF). Archived from the original (PDF) on 2012-03-16.
5. "USB in a NutShell—Chapter 2—Hardware". Beyond Logic.org. Retrieved 2007-08-25.
6. https://www.nxp.com/documents/application_note/AN10798.pdf (PDF) modified 2011-07-04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 98, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9109994173049927, "perplexity": 421.0697372927744}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00477.warc.gz"} |
https://www.snapxam.com/topic/multiplication-of-integers | # Multiplication of integers
## Definition
The result of multiplying two integers is another integer.
### Struggling with math?
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http://zbmath.org/?q=an:0721.30035 | # zbMATH — the first resource for mathematics
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Conjugacy invariants of Möbius transformations. (English) Zbl 0721.30035
This paper is devoted to generalizing the trace-squared (conjugation) invariant of Möbius transformations with complex entries to elements of a certain set of Clifford matrices ${𝒞}^{n}$. These were first considered by K. Th. Vahlen [Math. Ann. 55, 585-593 (1902)] and later by L. V. Ahlfors [see e.g. Ann. Acad. Sci. Fenn., Ser. A I 105, 15-27 (1985; Zbl 0586.30045)]. The Clifford matrices considered here are slightly different from those obtained by Ahlfors and provide a generalization to orientation-reversing maps. The main theorem is that, for each $k=0,1,2,···,n+2$, the ${T}_{k}$ function (the definition is too complicated to give here)
${T}_{k}:M\left(2,{𝒞}^{n}\right)\to ℝ$
is a conjugation invariant. Here ${𝒞}^{n}$ is the algebra of Clifford numbers generated by n roots of -1. The conjugation is by Clifford matrices. Among the properties of the invariant ${T}_{k}$ are the following:
(i) ${T}_{k}\left(-g\right)=T\left(g\right)$ for all $g\in {𝒞}^{n}$. Thus ${T}_{k}$ is well-defined on the projectivization of ${𝒞}^{n}·$
(ii) ${T}_{k}\left(g\right)=0$ for all k odd (resp. even) if g is orientation preserving (resp. reversing)
$\sum _{0}^{n+2}{T}_{k}\left(g\right)=0·$
(iii) $g\in {𝒞}^{n}$ is hyperbolic if and only if ${T}_{k}\left(g\right)<0$ for the largest k so that ${T}_{k}\left(g\right)\ne 0·$
(iv) $g\in {𝒞}^{n}$ is elliptic if and only if the quadratic form $N\left(z\right):=z\overline{z}$, when restricted to the kernel of Id-r(g) is not positive semidefinite. Here r takes g from the isometries of the ball model of hyperbolic space to the hyperboloid model.
##### MSC:
30G35 Functions of hypercomplex variables and generalized variables | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8521421551704407, "perplexity": 3084.1372672321872}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164758033/warc/CC-MAIN-20131204134558-00039-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://edoc.unibas.ch/661/ | # Nullforms, polarization and tensorpowers
Bürgin, Matthias. Nullforms, polarization and tensorpowers. 2006, Doctoral Thesis, University of Basel, Faculty of Science.
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Official URL: http://edoc.unibas.ch/diss/DissB_7697
## Abstract
Part I: Singular Spaces of the Nullcone:
Given a complex reductive group G and a complex representation V , one of the main
goals of invariant theory is to describe - in terms of generators and relations - the
ring of invariant polynomial functions, denoted by O(V )G. However, for most pairs
G and V , finding explicitly all generators of O(V )G is very difficult. An important
step in this search is to find homogeneous invariants whose zero set is the nullcone
Nv ⊂ V , i.e. the zero set of all homogeneous non-constant invariant functions on
V . Such invariants are strongly related to O(V )G as Hilbert proved the following
result: If f1, . . . , fr are homogeneous invariants whose zero set is equal to Nv then
O(V )G is a finitely generated module over the subalgebra C[f1, . . . , fr].
Given some invariants fi " O(V )G as above one can apply the so called polarization
process to obtain a set of functions lying in O(V ⊕k)G. Our main interest in
this work is to analyze whether the set of functions obtained in this manner defines
the nullcone NV !k . Due to an observation of Kraft and Wallach, this is equivalent
to the question whether for every linear subspace H ⊂ Nv of dimension at most
k there exists a one-parameter subgroup : C* G such that limt0 (t) ·H = 0.
For example, for G = SL2 and V = Vn, the binary forms of degree n, this
amounts to the question whether every subspace H that consists of forms having
a root of multiplicity greater than n/2. This is indeed the case, as we will see. Furthermore
we settle the question for G = SLn and V = S2(Cn)* (symmetric bilinear forms),
V = 2(Cn)* (skew-symmetric bilinear forms) and G = SL3 and V = S3(C3)*
(ternary cubics).
Part II: Multiplicities in Tensor Monomials:
There exist a lot of formulas to decompose a tensor product of representations
V ⊕ W into a direct sum of irreducible representations with respect to an algebraic
group G. However these formulas usually involve summing over the Weyl-group,
which makes explicit calculations often tedious. When considering multiple tensor
products, i.e. tensor monomials V1⊕n1 ⊕ V2⊕n2 · · · Vr⊕nr, then, even with the
use of descent computers, an explicit decomposition is mostly impossible because
of the complexity that arises. For this reason problems involving tensor monomials
remain challenging. The starting point of this work was the following question
asked by Finkelberg: For which (d1, d2, . . . , dn−1) ∈ Nn−1 does the tensor
monomial Cn⊕d1 ⊕ 2Cn⊕d2 ⊕ 3Cn⊕d3 ⊕· · · ⊕ n−1Cn⊕dn−1 , considered as SLn-representation, contain the trivial representation exactly once? We solve this problem
and some related generalizations. However, representations occuring with multiplicity
one in the decomposition of a tensor monomial V1⊕n1 ⊕ V2⊕n2 · · · Vr⊕nr
are rather rare as we prove that multiplicities of subrepresentations of tensor monomials
grow exponentially with respect to ∑ ni. More precisely, we prove, that if G
is a simple complex group and V1, . . . , Vr and W irreducible non-trivial representations
then there is a constant N and a real number > 1 such that if ∑ ni N
then mult(W, V1⊕n1 ⊕ V2⊕n2 · · · Vr⊕nr) ∑ni unless it is zero.
In its current form, this part is a preprint which evolved from my diploma thesis,
where I solved special cases of the two main results Theorem A and Theorem C.
Part III: The Hilbert Nullcone on Tuples of Matrices and Bilinear Forms:
In this joint work with Jan Draisma we explicitly determine the irreducible components
of the nullcone of the representation of G on M!p, where either G =
SL(W) x SL(V) and M = Hom(V,W) (linear maps), or G = SL(V) and M is one
of the representations S2(V*) (symmetric bilinear forms), 2(V*) (skew bilinear
forms), or V * ⊕ V * (arbitrary bilinear forms). Here V and W are vector spaces over
an algebraically closed field K of characteristic zero. We also answer the question
of when the nullcone in M⊕p is defined by the polarisations of the invariants on M;
typically, this is only the case if either dimV or p is small. A fundamental tool in
our proofs is the Hilbert-Mumford criterion for nilpotency.
This preprint has already been accepted for publication in the Mathematische
Zeitschrift. I mainly contributed to the first problem we solved: counting and describing
the components of the nullcone of the symmetric bilinear forms. Most other cases evolved from this one, however.
Advisors: Kraft, Hanspeter Schwarz, Gerald W. 05 Faculty of Science > Departement Mathematik und Informatik > Ehemalige Einheiten Mathematik & Informatik > Algebra (Kraft) Kraft, Hanspeter Thesis Doctoral Thesis 7697 Complete 1 English doi: 10.5451/unibas-004102036urn: urn:nbn:ch:bel-bau-diss76974 10.5451/unibas-004102036 22 Jan 2018 15:50 13 Feb 2009 16:14
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https://hbfs.wordpress.com/2016/09/27/weird-binomial-coefficients/ | ## Weird binomial coefficients
The binomial coefficients find great many uses in combinatorics, but also in calculus. The usual way we understand the binomial coefficients is
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$,
where $n$ and $k$ are integers. But what do you do with $\displaystyle \binom{\frac{1}{2}}{k}$?! Is it even defined?
While this is weird-looking, it is well-defined. If we write
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$,
which is strictly equivalent to the usual definition, but now depends more on $k$ than on $n$, it allows us to write…
$\displaystyle \binom{\frac{1}{2}}{n}=\frac{\frac{1}{2}( \frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-k+1)}{k!}$.
The first few $\displaystyle \binom{\frac{1}{2}}{k}$ are
$\displaystyle 1, \frac{1}{2}, -\frac{1}{8}, \frac{1}{16}, -\frac{5}{128}, \frac{7}{256}, -\frac{21}{1024}, \frac{33}{2048}, \ldots$.
### One Response to Weird binomial coefficients
1. […] expansion. The coefficients seems cumbersome: until we remember that we’ve already saw them before. This allows us to […] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.944391667842865, "perplexity": 997.0280670121551}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424090.53/warc/CC-MAIN-20170722182647-20170722202647-00697.warc.gz"} |
https://www.mathcity.org/msc/notes/linear-algebra-important-definitions-and-results | # Linear Algebra: Important Definitions and Results
These notes are made and shared by Mr. Akhtar Abbas. We are really very thankful to him for providing these notes and appreciates his efforts to publish these notes on MathCity.org. These notes contains important definitions with examples and related theorem, which might be helpful to prepare interviews or any other written test after graduation like PPSC, FPSC or etc.
Name Linear Algebra: Important Definitions and Results Mr. Akhtar Abbas 22 pages PDF (see Software section for PDF Reader) 560 kB
• If a vector space $V$ has a basis with $n$ elements, then $n$ is called dimension of $V$.
• If $W$ is a subspace of $V$, then dim(W) = dim(V).
• The span of column of a matrix $A$ is called column space of $A$.
• The span of rows of a matrix $A$ is called row space of $A$.
Please click on View Online to see inside the PDF.
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https://en.wikipedia.org/wiki/Unitarity_%28physics%29 | # Unitarity (physics)
Jump to navigation Jump to search
In quantum physics, unitarity is the condition that the time evolution of a quantum state according to the Schrödinger equation is mathematically represented by a unitary operator. This is typically taken as an axiom or basic postulate of quantum mechanics, while generalizations of or departures from unitarity are part of speculations about theories that may go beyond quantum mechanics.[1] A unitarity bound is any inequality that follows from the unitarity of the evolution operator, i.e. from the statement that time evolution preserves inner products in Hilbert space.
## Hamiltonian evolution and scattering matrix
Time evolution described by a time-independent Hamiltonian is represented by a one-parameter family of unitary operators, for which the Hamiltonian is a generator: ${\displaystyle U(t)=e^{-i{\hat {H}}t/\hbar }}$. The expectation value of the Hamiltonian is conserved under the time evolution that the Hamiltonian generates.[2] If the Hamiltonian itself has an intrinsic time dependence, as occurs when interaction strengths or other parameters vary over time, then computing the family of unitary operators becomes more complicated (see Dyson series). In the Schrödinger picture, the unitary operators are taken to act upon the system's quantum state, whereas in the Heisenberg picture, the time dependence is incorporated into the observables instead.[3]
Similarly, the S-matrix that describes how the physical system changes in a scattering process must be a unitary operator as well; this implies the optical theorem.
## Optical theorem
Unitarity of the S-matrix implies,[why?] among other things, the optical theorem. The optical theorem in particular implies that unphysical particles must not appear as virtual particles in intermediate states. The mathematical machinery which is used to ensure this includes gauge symmetry and sometimes also Faddeev–Popov ghosts.
According to the optical theorem, the imaginary part of a probability amplitude Im(M) of a 2-body forward scattering is related to the total cross section, up to some numerical factors. Because ${\displaystyle |M|^{2}}$ for the forward scattering process is one of the terms that contributes to the total cross section, it cannot exceed the total cross section i.e. Im(M). The inequality
${\displaystyle |M|^{2}\leq {\mbox{Im}}(M)}$
implies that the complex number M must belong to a certain disk in the complex plane. Similar unitarity bounds imply that the amplitudes and cross section cannot increase too much with energy or they must decrease as quickly as a certain formula[which?] dictates.
## References
1. ^ Ouellette, Jennifer. "Alice and Bob Meet the Wall of Fire". Quanta Magazine. Retrieved 8 July 2016.
2. ^ Barnum, Howard; Müller, Markus P.; Ududec, Cozmin (2014). "Higher-order interference and single-system postulates characterizing quantum theory". New Journal of Physics. 16: 123029. arXiv:1403.4147.
3. ^ "Lecture 5: Time evolution" (PDF). 22.51 Quantum Theory of Radiation Interactions. MIT OpenCourseWare. Retrieved 2019-08-21. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9588721990585327, "perplexity": 640.1133646132292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00018.warc.gz"} |
http://www.zora.uzh.ch/id/eprint/24346/ | # Efficient radiative transfer calculation and sensor performance requirements for the aerosol retrieval by airborne imaging spectroscopy
Seidel, F; Schläpfer, D; Itten, K I (2009). Efficient radiative transfer calculation and sensor performance requirements for the aerosol retrieval by airborne imaging spectroscopy. In: 6th EARSeL Imaging Spectroscopy SIG workshop: proceedings , Tel Aviv, Israel, 16 March 2009 - 19 March 2009, online.
## Abstract
Detailed aerosol measurements in time and space are crucial to address open questions in climate research. Earth observation is a key instrument for that matter but it is biased by large uncertainties. Using airborne imaging spectroscopy, such as ESA's upcoming airborne Earth observing instrument APEX, allows determining the widely used aerosol optical depth (AOD) with unprecedented accuracy
thanks to its high spatial and spectral resolution, optimal calibration and high signal-to-noise ratios (SNR).
This study was carried out within the overall aim of developing such a tropospheric aerosol retrieval
algorithm. Basic and efficient radiative transfer equations were applied to determine the sensor performance
requirement and a sensitivity analysis in context of the aerosol retrieval. The AOD retrieval sensitivity requirement was chosen according to the demands of atmospheric correction processes. Therefore, a novel parameterization of the diffuse path-radiance was developed to simulate the atmospheric and surface effects on the signal at the sensor level. It was found for typical remote sensing conditions and a surface albedo of less than 30% that a SNR of circa 300 is sufficient to meet the AOD retrieval sensitivity requirement at 550nm. A surface albedo
around 50% requires much more SNR, which makes the AOD retrieval very difficult. The retrieval performance is further analyzed throughout the visual spectral range for a changing solar geometry and different aerosol characteristics. As expected, the blue spectral region above dark surfaces and high aerosol loadings will provide the most accurate retrieval results. In general, the AOD
retrieval feasibility could be proven for the analyzed cases for APEX under realistic simulated conditions.
## Abstract
Detailed aerosol measurements in time and space are crucial to address open questions in climate research. Earth observation is a key instrument for that matter but it is biased by large uncertainties. Using airborne imaging spectroscopy, such as ESA's upcoming airborne Earth observing instrument APEX, allows determining the widely used aerosol optical depth (AOD) with unprecedented accuracy
thanks to its high spatial and spectral resolution, optimal calibration and high signal-to-noise ratios (SNR).
This study was carried out within the overall aim of developing such a tropospheric aerosol retrieval
algorithm. Basic and efficient radiative transfer equations were applied to determine the sensor performance
requirement and a sensitivity analysis in context of the aerosol retrieval. The AOD retrieval sensitivity requirement was chosen according to the demands of atmospheric correction processes. Therefore, a novel parameterization of the diffuse path-radiance was developed to simulate the atmospheric and surface effects on the signal at the sensor level. It was found for typical remote sensing conditions and a surface albedo of less than 30% that a SNR of circa 300 is sufficient to meet the AOD retrieval sensitivity requirement at 550nm. A surface albedo
around 50% requires much more SNR, which makes the AOD retrieval very difficult. The retrieval performance is further analyzed throughout the visual spectral range for a changing solar geometry and different aerosol characteristics. As expected, the blue spectral region above dark surfaces and high aerosol loadings will provide the most accurate retrieval results. In general, the AOD
retrieval feasibility could be proven for the analyzed cases for APEX under realistic simulated conditions. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8495827317237854, "perplexity": 2773.395850611465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00372.warc.gz"} |
http://proceedings.mlr.press/v28/agarwal13.html | # Selective sampling algorithms for cost-sensitive multiclass prediction
Alekh Agarwal ;
Proceedings of the 30th International Conference on Machine Learning, PMLR 28(3):1220-1228, 2013.
#### Abstract
In this paper, we study the problem of active learning for cost-sensitive multiclass classification. We propose selective sampling algorithms, which process the data in a streaming fashion, querying only a subset of the labels. For these algorithms, we analyze the regret and label complexity when the labels are generated according to a generalized linear model. We establish that the gains of active learning over passive learning can range from none to exponentially large, based on a natural notion of margin. We also present a safety guarantee to guard against model mismatch. Numerical simulations show that our algorithms indeed obtain a low regret with a small number of queries. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.861916720867157, "perplexity": 680.8513261810093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00230.warc.gz"} |
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