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How to use the rug size calculator
Manually measuring rug area and sizes for different surfaces
How to calculate rug pad size
How big is a rug? – normal rug sizes
Using our rug size calculator, you can obtain suitable rug measurements for your room based on the style of the rug and also calculate its cost based on the total size of the rug.
Furthermore, we will tell you about different rug sizes normally available in the market, so you can decide if you want a custom rug or buy a standard size based on your budget.
Rugs are the new make-up for our floors, just as the picture frames are for our walls. They add a contemporary style or vibrance to our rooms, which is aesthetically pleasing. Rugs also offer cushion comfort and warmth for our bare feet in chilly winters.
Rugs come in many styles. Our calculator will help you measure rug sizes for areas of different dimensions in several shapes. Before we proceed, please record the length and width of the area where you'll be placing the rug.
Let us provide the necessary guidelines to help you make the most of our rug size calculator.
Choose your preferred rug style:
Rug style tells our calculator the shape of your rug.
Bare floor perimeter:
Do you want to cover the entire floor or only a portion of it? The default value of 1.5 feet means leaving 1.5 feet around the rug's perimeter.
Enter floor dimensions:
Enter your room width in the first field, e.g., 12 ft.
Enter your room length in the second field, e.g., 16 ft.
The tool will then calculate the square footage and display it in area size, i.e., 192 sq. ft.
Any extrusions in the room? (advanced mode)
Add the extrusion width and length if you have an object against the wall that reduces the available floor space, e.g., a fireplace or a fixed cupboard.
Custom rug size:
Once you've entered your floor dimensions, your custom rug length, rug width, and rug size will be displayed.
Under preferred rug style, the default floor to leave uncovered in our rug size calculator is 1.5 ft. Thus, in which case, our custom rectangle rug size equals 10.5 × 14.5 ft, i.e., 152.2 sq. ft.
🙋 If we want to buy a standard rug from the market, the closest matching rug size standard for a surface of 12 ft × 16 ft, leaving 1.5 ft of surface uncovered, is displayed below the result as 10 ft × 14 ft.
Rug cost calculator:
In cost per unit area, enter the price per area of the rug.
And you will see the Total rug cost based on your custom rug size.
To manually calculate an area rug size for various shapes, we will use the following base formula, followed by an example:
Area _{(r)} = W _{(r)} × L _{(r)}
This can be subdivided divided into:
W_{(r)} = W_{(s)} - W_{(e)} - U_{(s)}
L_{(r)} = L_{(s)} - L_{(e)} - U_{(s)}
Area _{(r)}
– Total size of the rug;
L _{(r)} | W _{(r)}
– Respective length and width of the rug;
L _{(s)} | W _{(s)}
– Respective surface length or width where we want to place the rug;
L _{(e)} | W _{(e)}
– Respective extrusion on the surface length or width; and
U_{(s)}
– Amount of surface we don't wish to cover.
Now, let's calculate some area rug sizes in different shapes.
If you want to calculate a rug size for your dining room of 15 ft × 10 ft, leaving 1 ft fireplace extrusion from the width, and 1.5 ft surface uncovered, place the given values in the formula, to get:
W_{(r)} = 15 - 1 - 1.5 = 12.5
L_{(r)} = 10 - 0 - 1.5 = 8.5
Area _{(r)} = 12.5\ \text{ft} \times 8.5\ \text{ft}
Calculating the area of a rectangle rug for our dining room, we get
106.25\ \text{sq. ft.}
What is the size of an area rug for a living room of 12 ft × 14 ft when leaving 2 ft of surface uncovered from all sides?
The square rugs are calculated the same way as rectangle rugs, except we measure our rug around the shorter side of the surface.
W_{(r)} = 12 - 0 - 2 = 10
L_{(r)} = 12 - 0 - 2 = 10
Area _{(r)} = 10\ \text{ft} \times 10\ \text{ft}
Calculating the area of a square rug for our living room, we get
100\ \text{sq. ft.}
We can also measure this area rug size in inches, where 1 sq. ft. = 144 sq. in., thus:
100\ \text{sq. ft.} = 14,399.5\ \text{sq. in.}
Standard runner rugs generally have a width between 2 and 3 ft, primarily used in hallways and kitchens.
If you want a custom kitchen runner rug, multiply the length of your floor, e.g., 12 ft, with your preferred runner width, e.g., 3 ft, and you'll get the size of your runner rug, i.e., 36 ft.
Area _{(r)} = 3\ \text{ft} \times 12\ \text{ft}
The area of our runner rug is
36\ \text{sq. ft.}
Suppose you have an area of 5 ft x 7 ft, where you want to place a circle rug, and you want to find the size of this area rug in inches.
The circle rugs are calculated the same way as square rugs, where we measure our rug around the shorter side of the surface.
We calculate the area of a circle rug using the following formula:
Area _{(r)} = \pi \left(\frac{W _{(r)}}{2}\right)^2
\pi
– Constant with an approximate value of 3.14159.
Inputting the values:
Area _{(r)} = 3.14159 ×(5/2)^2
(For our rug to fit, we take the smaller side of our area as the diameter.)
Thus the required area of our circle rug is
19.6\ \text{sq. ft.}
Multiplying this result by 144, we get
2822.4\ \text{sq. in.}
How about calculating the dining room rug size of 15 ft × 10 ft, leaving 3 ft of the surface uncovered from all sides.
First calculate the new length and width of the rug:
W_{(r)} = 15 - 0 - 3 = 12
L_{(r)} = 10 - 0 - 3 = 7
Then, we calculate the area of an oval rug using the following formula:
Area _{(r)} = \pi \left(\frac{W _{(r)}}{2} × \frac{L _{(r)}}{2}\right)
Area _{(r)} = \pi × (\frac{12}{2} × \frac{7}{2})
Thus, the area of our oval rug is
66\ \text{sq. ft.}
Let's calculate the rug size of an octagon rug for a room that has the dimensions 6 ft x 6 ft.
We first multiply the shorter side of our surface with 0.4142 to get one of the eight sides of our octagon. Then take its square and multiply it by 4.8284 to get our octagon side.
Area _{(r)}= {(W _{(s)} \times 0.414)}^2 \times 4.83
Inputting the values in the formula, we get:
Area _{(r)}= {(6 \times 0.414)}^2 \times 4.83
Thus, calculating the area of an octagon rug for our desired surface, we get
29.8\ \text{sq. ft.}
Rug pad size is the padding underneath your rug that keeps it from slipping, unlike carpets that are glued to the floor.
To calculate rug pad size, subtract an inch (2.5 cm) from all four sides of your rug.
We hope you now understand how to calculate the pad and rug sizes for different shapes. Next, we'll give you an idea of some standard area rug measurements.
Here are some height × width rug standards to help you understand how big a rug is available in a particular shape and size. All measurements are in feet.
Rectangle & oval rug sizes:
Square, circle & octagon rug sizes:
Runner rug sizes:
8′ × 2.5
10′ × 2.5′
Should a standard rug size not meet your requirement, you can always choose a custom rug size.
How to choose a rug size?
Measure the area length and width where you want to place the rug, leaving 2 feet on both sides.
Then multiply the measured length and width to get your rectangle area rug size.
For example, the perfect rug size for a room of 12 ft x14 ft is 10 ft x 12 ft, i.e., 120 ft2.
How do I find the area rug size for a living room?
To find the size of an area rug for the living room:
First, measure the length and width of your living room.
Remove two feet from both measurements.
Multiply the resulting length and width.
You have your rectangle area rug size for the living room.
For example, if your living room is 10 ft x 12 ft, you need an area rug of 8 ft x 10 ft, i.e., 80 ft2.
What are standard kitchen runner rug sizes?
The standard kitchen runner rug sizes are:
If you have a large kitchen:
And if you have an enormous kitchen:
What is a rug pad size?
Rug pad size is the padding underneath your rug that keeps it from slipping, unlike carpets that glue to the surface. To calculate rug pad size, subtract an inch from all four sides of your rug.
Bare floor perimeter
Total rug cost
Time difference calculator enables convenient time subtraction.
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Other area conversion tools!
Hectares to acres converter is a tool that converts between units of area measurements. The tool has many other unit options as well, other than hectares and acres.
If you want to know more about hectares and acres or how to convert a hectare to an acre, continue reading, and you'll be surprised at how amazing these units are.
You will find that the hectares to acres converter not only converts any area value from hectares to acres and vice versa but also comes in handy for conversions between many other area measuring units.
Hectares and acres are units of area that generally measure land.
Hectare is a non-SI metric unit that is equal to:
\footnotesize 1\text{ Hectare} = 2.471 \text { acres} = 10,000 \text{ m}^2
An acre is an imperial unit that measures land. It is equal to:
\footnotesize 1\text{ Acre} = 0.405 \text { hectares} = 4046.86 \text{ m}^2
Input the area in hectares.
The result is the area in acres.
Yes, it is that simple. But wait, there is more. Each input has a list of units from which you can select any other unit as well.
For instance, you just found out that you inherit
3.6 \text { hectares}
of land in the beautiful countryside of Mahone bay, now you are curious, how much is it in acres? Well, the tool will tell you it is
8.8958\text{ acres}
, almost 9 acres.
💡 Did you know that Tasmania, the largest island in Australia is 6.81 million hectares which is 16,827,878 acres!
The conversion from hectares to acres and the other way round is simple and easy to remember.
If you have an amount in hectares and are looking to convert it to acres, use the formula:
\text {acres} = \text {hectares} \times 2.471
So, if you have a plot with an area of
7 \text { hectares}
2.471
and you have
17.29738 \text { acres}
Now, if you have an amount in acres and are looking to convert it to hectares, use the formula:
\text {hectares} = \text {acres} / 2.471
Say you have a farmhouse over
21 \text { acres}
, in hectares, it is
21
2.471
8.4984 \text { hectares}
Our hectares to acres converter is not the only area conversion tool that we have. Below is a list of all the other amazing tools you didn't know you needed.
Square miles to square km converter;
Acres to hectares converter; and
Square yard converter.
Yes, a hectare is bigger than an acre. One hectare is equal to 2.471054 acres.
Both units measure land. When a hectare is a non-SI metric unit, an acre is an imperial unit of land measurement.
There are 7.68903 hectares in 19 acres.
The formula to convert from acres to hectares is:
ha = ac / 2.471
ha - hectares; and
ac - acres.
So, to convert from acres to hectares:
Divide the amount in acres by 2.471.
The result is your quantity in hectares.
A football field is one acre, approximately. This is a general estimation and can vary from field to field, but it is a known standard.
Since there are 2.471 acres in 1 hectare, it implies 1 hectare can contain 2.471 football fields.
Use our gallons to pounds calculator to calculate the weight of gallons of water or other liquids.
Our kg to liters converter will help you measure the weight and quantity of different liquids you use every day, from kilograms to liters and vice versa.
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Validating the Phenomenological Smoke Model at Different Operating Conditions of DI Diesel Engines | J. Eng. Gas Turbines Power | ASME Digital Collection
, Pune, 411 003 India
, Hosur, Tamilnadu, 635 126 India
, Delhi, 110 016 India
Azeem Uddin,
Aghav, Y. V., Lakshminarayanan, P. A., Babu, M. K. G., Uddin, A., and Dani, A. D. (January 9, 2008). "Validating the Phenomenological Smoke Model at Different Operating Conditions of DI Diesel Engines." ASME. J. Eng. Gas Turbines Power. January 2008; 130(1): 012803. https://doi.org/10.1115/1.2771239
A new phenomenological model that was published in Aghav et al. (2005, “Phenomenology of Smoke From Direct Injection Diesel Engines,” Proceedings of ICEF2005, ASME Paper No. 1350) encompasses the spray and the wall interaction by a simple geometrical consideration. The current study extends this earlier work with investigations made on 16 different engines from six-engine families of widely varying features, applied to off-highway as well as on-road duty. A dimensionless factor was introduced to take care of the nozzle hole manufactured by hydroerosion, as well as the conical shape of the nozzle hole (
k
factor) in the case of valve-closed-orifice type of nozzles. The smoke emitted from the wall spray formed after wall impingement is the major contributor to the total smoke at higher loads. As the fuel spray impinges upon the walls of the combustion chamber, its velocity decreases. This low-velocity jet contributes to the higher rate of the smoke production. Therefore, the combustion bowl geometry along with injection parameters play a significant role in the smoke emissions. The new model is one dimensional and based on the recent phenomenological description of spray combustion in a direct injection diesel engine. The satisfactory comparison of the predicted and observed smoke over the wide range of engine operation demonstrated applicability of the model in simulation study of combustion occurring in direct injection (DI) diesel engines.
diesel engines, nozzles, orifices (mechanical), soot, DI diesel, turbulent structure, free jet, wall jet, Kolmogorov scale
Diesel engines, Engines, Smoke, Soot, Sprays, Nozzles, Fuels, Turbulence, Stress, Emissions
The Cause of Exhaust Smoke and Its Reduction Methods in an HSDI Diesel Engine Under High Speed and High Load Conditions
Study of Nozzle Characteristics on the Performance of a Small Bore High Speed D. I. Engine
The Effect of Swirl on Combustion and Exhaust Emissions
Diffusion-Flame/Wall Interactions in a Heavy-Duty DI Diesel Engine
Extinction Diesel Combustion: An Integrated View Combining Laser Diagnostics, Chemical Kinetics and Empirical Validation
A Theoretical Investigation on the Effects of Combustion Chamber Geometry and Engine Speed on Soot and NOx Emissions
,” ASME-ICE Paper No. 99-ICE-207, Vol.
An Innovative Methodology to Improve the Design and the Performance of Direct Injection Diesel Engine
Identifying a Critical Time for Mixing in a Direct Injection Diesel Engine Through the Study of Increased In-Cylinder Mixing and Its Effect on Emissions Through Study of Increased In-Cylinder Mixing and Its Effects on Emissions
Knox-Kelecy
Application of a Phenomenological Soot Model to Diesel Engine Combustion
Third International Symposium COMODIA
Turbulent Mixing Rate—Its Effect on Smoke and Hydrocarbon Emissions From Diesel Engines
M. K. G.
Phenomenology of Smoke From Direct Injection Diesel Engines
Proceedings of ICEF2005
, ASME Paper No. 1350.
Stiech
A Textbook on Modelling Engine Spray and Combustion Processes
,” 1965, Motor Industry Research Association (MIRA) Report No. 10.
http://www.dieselnet.com/standards/http://www.dieselnet.com/standards/
Phenomenological Diesel Combustion Model Including Smoke, and No Emission
Interferometric Studies of Vapourising and Combusting Sprays
A Text Book on Internal Combustion Engine Fundamentals
AVL Boost 4.1 Users Guide
The Influence of Hydrogrinding at VCO Nozzles on the Mixture Preparation in a DI Diesel Engine
Experimental Study on Effects of Nozzle Hole Geometry on Achieving Low Diesel Engine Emissions
Study of Fuel Temperature Effects on Fuel Injection, Combustion, and Emissions of Direct-Injection Diesel Engines
Study of Fuel Temperature Effects on Fuel Injection, Combustion and Emissions of Direct-Injection Diesel Engines
|
Rule_of_inference Knowpia
{\displaystyle A\to B}
{\displaystyle {\underline {A\quad \quad \quad }}\,\!}
{\displaystyle B\!}
Example: Hilbert systems for two propositional logicsEdit
In a Hilbert system, the premises and conclusion of the inference rules are simply formulae of some language, usually employing metavariables. For graphical compactness of the presentation and to emphasize the distinction between axioms and rules of inference, this section uses the sequent notation (
{\displaystyle \vdash }
) instead of a vertical presentation of rules. In this notation,
{\displaystyle {\begin{array}{c}{\text{Premise }}1\\{\text{Premise }}2\\\hline {\text{Conclusion}}\end{array}}}
{\displaystyle ({\text{Premise }}1),({\text{Premise }}2)\vdash ({\text{Conclusion}})}
Admissibility and derivabilityEdit
In a set of rules, an inference rule could be redundant in the sense that it is admissible or derivable. A derivable rule is one whose conclusion can be derived from its premises using the other rules. An admissible rule is one whose conclusion holds whenever the premises hold. All derivable rules are admissible. To appreciate the difference, consider the following set of rules for defining the natural numbers (the judgment
{\displaystyle n\,\,{\mathsf {nat}}}
asserts the fact that
{\displaystyle n}
{\displaystyle {\begin{matrix}{\begin{array}{c}\\\hline {\mathbf {0} \,\,{\mathsf {nat}}}\end{array}}&{\begin{array}{c}{n\,\,{\mathsf {nat}}}\\\hline {\mathbf {s(} n\mathbf {)} \,\,{\mathsf {nat}}}\end{array}}\end{matrix}}}
{\displaystyle {\begin{array}{c}{n\,\,{\mathsf {nat}}}\\\hline {\mathbf {s(s(} n\mathbf {))} \,\,{\mathsf {nat}}}\end{array}}}
{\displaystyle {\begin{array}{c}{\mathbf {s(} n\mathbf {)} \,\,{\mathsf {nat}}}\\\hline {n\,\,{\mathsf {nat}}}\end{array}}}
This is a true fact of natural numbers, as can be proven by induction. (To prove that this rule is admissible, assume a derivation of the premise and induct on it to produce a derivation of
{\displaystyle n\,\,{\mathsf {nat}}}
{\displaystyle {\begin{array}{c}\\\hline {\mathbf {s(-3)} \,\,{\mathsf {nat}}}\end{array}}}
In this new system, the double-successor rule is still derivable. However, the rule for finding the predecessor is no longer admissible, because there is no way to derive
{\displaystyle \mathbf {-3} \,\,{\mathsf {nat}}}
^ Boolos, George; Burgess, John; Jeffrey, Richard C. (2007). Computability and logic. Cambridge: Cambridge University Press. p. 364. ISBN 0-521-87752-0.
^ John C. Reynolds (2009) [1998]. Theories of Programming Languages. Cambridge University Press. p. 12. ISBN 978-0-521-10697-9.
^ Kosta Dosen (1996). "Logical consequence: a turn in style". In Maria Luisa Dalla Chiara; Kees Doets; Daniele Mundici; Johan van Benthem (eds.). Logic and Scientific Methods: Volume One of the Tenth International Congress of Logic, Methodology and Philosophy of Science, Florence, August 1995. Springer. p. 290. ISBN 978-0-7923-4383-7. preprint (with different pagination)
^ Bergmann, Merrie (2008). An introduction to many-valued and fuzzy logic: semantics, algebras, and derivation systems. Cambridge University Press. p. 100. ISBN 978-0-521-88128-9.
|
Section 4.5: Gradient Vector
f\left(x,y\right)
∇f={f}_{x} \mathbf{i}+{f}_{y} \mathbf{j}
, whereas the gradient of
f\left(x,y,z\right)
∇f={f}_{x} \mathbf{i}+{f}_{y} \mathbf{j}+{f}_{z} \mathbf{k}
. Table 5.4.1 lists the five most important properties of the gradient vector.
The gradients of
f\left(x,y\right)
are orthogonal to the level curves
y=y\left(x\right)
defined implicitly by
f\left(x,y\right)=c
c
f\left(x,y,z\right)
are orthogonal to the level surfaces
z=z\left(x,y\right)
f\left(x,y,z\right)=c
c
At any point where
∇f≠\mathbf{0}
, the gradient
∇f
points in the direction of increasing values of
∇f≠\mathbf{0}
, it necessarily points in the direction of maximal increase in
f
The maximal rate of change in
f
, as measured by the directional derivative, has the value
∥∇f∥
Table 5.4.1 Properties of the gradient vector
Obtaining the Gradient in Maple
The Del (or Nabla) operator
∇ =\mathbf{i} \frac{∂}{∂x}+\mathbf{j} \frac{∂}{∂y}+\mathbf{k} \frac{∂}{∂z}
, applied to a scalar
f\left(x,y,z\right)
, results in the gradient vector
∇f={f}_{x} \mathbf{i}+{f}_{y} \mathbf{j}+{f}_{z} \mathbf{k}
. Maple has ∇, the Nabla symbol, in its Common Symbols palette, but it only works as an operator when one of the VectorCalculus packages is loaded.
With the Student VectorCalculus package loaded, the ∇-operator will correctly compute the gradient in a Cartesian frame for any expression in either two or three names of coordinate variables. Thus,
∇\left(x+y\right)
would correctly return
\mathbf{i}+\mathbf{j}
, but applied to
a+x
, would treat both
and
x
as independent variables, and differentiate with respect to both. The Gradient command itself admits a list of names with respect to which differentiation is to take place, but this list cannot be made know to the ∇-operator without the use of the SetCoordinates command. Finally, note that the gradient is returned as a VectorField, a more complex data structure than a "free vector" such as
〈x,y〉
The Gradient command in the Student MultivariateCalculus package never links to the Nabla symbol, and always requires as an additional parameter, a list of the variables of differentiation. However, this list of names can be set equal to a list of coordinates, so that the gradient is computed at a specific point.
Of course, the gradient of, say,
f\left(x,y\right)
, could also be obtained by the construct
〈\frac{∂}{∂ x} f,\frac{∂}{∂ y} f〉
With the Student MultivariateCalculus package loaded, the Context Panel, launched on an expression in two or three variables, provides interactive access to the Gradient command.
There is also a Gradient command in the Physics:-Vectors package, but it only acts on vectors defined within that package. No use is made of that package, or its structures, in this work.
f\left(x,y\right)=5-2 {x}^{2}-3 {y}^{2}
and let P be the point
\left(1,2\right)
∇f
at P.
z=f\left(x,y\right)
On the same set of axes, graph the level curve through P, and
∇f
At P, show that
∇f
is orthogonal to a vector tangent to the level curve through P.
At P, obtain
\mathrm{ψ}=\left(∇f\right)·\mathbf{u}
, the directional derivative of
\mathbf{u}=\mathrm{cos}\left(t\right) \mathbf{i}+\mathrm{sin}\left(t\right) \mathbf{j}
\mathrm{ψ}
is a maximum when u is along
∇f\left(\mathrm{P}\right)
and that this maximum is
{∥∇f\left(\mathrm{P}\right)∥}^{2}
w={x}^{2}+2 {y}^{2}+3 {z}^{2}
\left(1,1,1\right)
∇w
On the same set of axes, graph the level surface
w=6
∇w
∇w
is orthogonal to the level surface
w=6
. Hint: Show that this gradient is orthogonal to the
x
y
-coordinate curves through P.
Prove Property 1 in Table 4.5.1.
Show both graphically and analytically that the level curves of
u={x}^{2}-{y}^{2}-3 x+2
are orthogonal to the level curves of
v=2xy-3 y
u=\mathrm{sin}\left(x\right)\mathrm{cosh}\left(y\right)
v=\mathrm{cos}\left(x\right)\mathrm{sinh}\left(y\right)
u=\frac{{x}^{4}+2{x}^{2}{y}^{2}+{y}^{4}-1}{\left({x}^{2}+{y}^{2}+2y+1\right)\left({x}^{2}+{y}^{2}-2y+1\right)}
v=\frac{4xy}{\left({x}^{2}+{y}^{2}+2y+1\right)\left({x}^{2}+{y}^{2}-2y+1\right)}
, show both graphically and analytically that their level curves are mutually orthogonal.
\left(2,1\right)
, determine the maximal rate of change and its direction for
f\left(x,y\right)=\frac{x y}{{x}^{2}+{y}^{3}}
\left(1,2,3\right)
f\left(x,y,z\right)=\frac{{x}^{2}-2 y+5 z}{x+y z}
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Calorimetry | Boundless Chemistry | Course Hero
Heat capacity is a measure of the amount of heat energy required to change the temperature of a pure substance by a given amount.
Calculate the change in temperature of a substance given its heat capacity and the energy used to heat it
Heat capacity is the ratio of the amount of heat energy transferred to an object to the resulting increase in its temperature.
Molar heat capacity is a measure of the amount of heat necessary to raise the temperature of one mole of a pure substance by one degree K.
Specific heat capacity is a measure of the amount of heat necessary to raise the temperature of one gram of a pure substance by one degree K.
heat capacity: The capability of a substance to absorb heat energy; the amount of heat required to raise the temperature of one mole or gram of a substance by one degree Celsius without any change of phase.
specific heat capacity: The amount of heat that must be added or removed from a unit mass of a substance to change its temperature by one Kelvin.
Heat capacity is an intrinsic physical property of a substance that measures the amount of heat required to change that substance's temperature by a given amount. In the International System of Units (SI), heat capacity is expressed in units of joules per kelvin
\left(J\cdot K^{-1}\right)
. Heat capacity is an extensive property, meaning that it is dependent upon the size/mass of the sample. For instance, a sample containing twice the amount of substance as another sample would require twice the amount of heat energy (Q) to achieve the same change in temperature (
\Delta T
) as that required to change the temperature of the first sample.
There are two derived quantities that specify heat capacity as an intensive property (i.e., independent of the size of a sample) of a substance. They are:
the molar heat capacity, which is the heat capacity per mole of a pure substance. Molar heat capacity is often designated CP, to denote heat capacity under constant pressure conditions, as well as CV, to denote heat capacity under constant volume conditions. Units of molar heat capacity are
\frac{J}{K\cdot\text{ mol}}
the specific heat capacity, often simply called specific heat, which is the heat capacity per unit mass of a pure substance. This is designated cP and cV and its units are given in
\frac{J}{g\cdot K}
Given the molar heat capacity or the specific heat for a pure substance, it is possible to calculate the amount of heat required to raise/lower that substance's temperature by a given amount. The following two formulas apply:
q=mc_p\Delta T
q=nC_P\Delta T
In these equations, m is the substance's mass in grams (used when calculating with specific heat), and n is the number of moles of substance (used when calculating with molar heat capacity).
The molar heat capacity of water, CP, is
75.2\frac{J}{\text{mol}\cdot K}
. How much heat is required to raise the temperature of 36 grams of water from 300 to 310 K?
We are given the molar heat capacity of water, so we need to convert the given mass of water to moles:
\text{36 grams}\times \frac{\text{1 mol }\text{H}_2\text{O}}{\text{18 g}}=\text{2.0 mol H}_2\text{O}
Now we can plug our values into the formula that relates heat and heat capacity:
q=nC_P\Delta T
q=(2.0\;\text{mol})\left(75.2\;\frac{J}{\text{mol}\cdot K}\right)(10\;K)
q=1504\;J
Interactive: Seeing Specific Heat and Latent Heat: Specific heat capacity is the measure of the heat energy required to raise the temperature of a given quantity of a substance by one kelvin. Latent heat of melting describes tœhe amount of heat required to melt a solid. When a solid is undergoing melting, the temperature basically remains constant until the entire solid is molten. The above simulation demonstrates the specific heat and the latent heat.
Specific heat capacity tutorial: This lesson relates heat to a change in temperature. It discusses how the amount of heat needed for a temperature change is dependent on mass and the substance involved, and that relationship is represented by the specific heat capacity of the substance, C.
Constant-volume calorimeters, such as bomb calorimeters, are used to measure the heat of combustion of a reaction.
Describe how a bomb calorimeter works
A bomb calorimeter is used to measure the change in internal energy,
\Delta U
, of a reaction. At constant volume, this is equal to qV, the heat of reaction.
The calorimeter has its own heat capacity, which must be accounted for when doing calculations.
bomb calorimeter: A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction.
calorie: The amount of energy needed to raise the temperature of 1 gram of water by 1 °C. It is a non-SI unit of energy equivalent to approximately 4.18 Joules. A Calorie (with a capital C) = 1000 calories.
Bomb calorimetry is used to measure the heat that a reaction absorbs or releases, and is practically used to measure the calorie content of food. A bomb calorimeter is a type of constant-volume calorimeter used to measure a particular reaction's heat of combustion. For instance, if we were interested in determining the heat content of a sushi roll, for example, we would be looking to find out the number of calories it contains. In order to do this, we would place the sushi roll in a container referred to as the "bomb", seal it, and then immerse it in the water inside the calorimeter. Then, we would evacuate all the air out of the bomb before pumping in pure oxygen gas (O2). After the oxygen is added, a fuse would ignite the sample causing it to combust, thereby yielding carbon dioxide, gaseous water, and heat. As such, bomb calorimeters are built to withstand the large pressures produced from the gaseous products in these combustion reactions.
Bomb calorimeter: A schematic representation of a bomb calorimeter used for the measurement of heats of combustion. The weighed sample is placed in a crucible, which in turn is placed in the bomb. The sample is burned completely in oxygen under pressure. The sample is ignited by an iron wire ignition coil that glows when heated. The calorimeter is filled with fluid, usually water, and insulated by means of a jacket. The temperature of the water is measured with the thermometer. From the change in temperature, the heat of reaction can be calculated.
Once the sample is completely combusted, the heat released in the reaction transfers to the water and the calorimeter. The temperature change of the water is measured with a thermometer. The total heat given off in the reaction will be equal to the heat gained by the water and the calorimeter:
q_{rxn}=-q_{cal}
Keep in mind that the heat gained by the calorimeter is the sum of the heat gained by the water, as well as the calorimeter itself. This can be expressed as follows:
q_{cal}=m_{\text{water}}C_{\text{water}}\Delta T+C_{cal}\Delta T
where Cwater denotes the specific heat capacity of the water
\left(1 \frac{\text{cal}}{\text{g} ^{\circ}\text{C}}\right)
, and Ccal is the heat capacity of the calorimeter (typically in
\frac{\text{cal}}{^{\circ}\text{C}}
). Therefore, when running bomb calorimetry experiments, it is necessary to calibrate the calorimeter in order to determine Ccal.
Since the volume is constant for a bomb calorimeter, there is no pressure-volume work. As a result:
where ΔU is the change in internal energy, and qV denotes the heat absorbed or released by the reaction measured under conditions of constant volume. (This expression was previously derived in the "Internal Energy and Enthalpy " section.) Thus, the total heat given off by the reaction is related to the change in internal energy (ΔU), not the change in enthalpy (ΔH) which is measured under conditions of constant pressure.
The value produced by such experiments does not completely reflect how our body burns food. For example, we cannot digest fiber, so obtained values have to be corrected to account for such differences between experimental (total) and actual (what the human body can absorb) values.
Discuss how a constant-pressure calorimeter works
A constant- pressure calorimeter measures the change in enthalpy (
\Delta H
) of a reaction occurring in solution, during which the pressure remains constant. Under these conditions, the change in enthalpy of the reaction is equal to the measured heat.
Change in enthalpy can be calculated based on the change in temperature of the solution, its specific heat capacity, and mass.
constant-pressure calorimeter: Measures the change in enthalpy of a reaction occurring in solution, during which the pressure remains constant.
adiabatic: Not allowing any transfer of heat energy; perfectly insulating.
coffee-cup calorimeter: An example of constant-pressure calorimeter.
A constant-pressure calorimeter measures the change in enthalpy of a reaction occurring in a liquid solution. In that case, the gaseous pressure above the solution remains constant, and we say that the reaction is occurring under conditions of constant pressure. The heat transferred to/from the solution in order for the reaction to occur is equal to the change in enthalpy (
\Delta H = q_P
), and a constant-pressure calorimeter thus measures this heat of reaction. In contrast, a bomb calorimeter 's volume is constant, so there is no pressure-volume work and the heat measured relates to the change in internal energy (
\Delta U=q_V
A simple example of a constant-pressure calorimeter is a coffee-cup calorimeter, which is constructed from two nested Styrofoam cups and a lid with two holes, which allows for the insertion of a thermometer and a stirring rod. The inner cup holds a known amount of a liquid, usually water, that absorbs the heat from the reaction. The outer cup is assumed to be perfectly adiabatic, meaning that it does not absorb any heat whatsoever. As such, the outer cup is assumed to be a perfect insulator.
Coffee cup calorimeter: A styrofoam cup with an inserted thermometer can be used as a calorimeter, in order to measure the change in enthalpy/heat of reaction at constant pressure.
Data collected during a constant-pressure calorimetry experiment can be used to calculate the heat capacity of an unknown substance. We already know our equation relating heat (q), specific heat capacity (C), and the change in observed temperature (
\Delta T
q=mC\Delta T
We will now illustrate how to use this equation to calculate the specific heat capacity of a substance.
A student heats a 5.0 g sample of an unknown metal to a temperature of 207
^\circ
C, and then drops the sample into a coffee-cup calorimeter containing 36.0 g of water at 25.0
^\circ
C. After thermal equilibrium has been established, the final temperature of the water in the calorimeter is 26.0
^\circ
C. What is the specific heat of the unknown metal? (The specific heat of water is 4.18
\frac {J} {g^\circ C}
The walls of the coffee-cup calorimeter are assumed to be perfectly adiabatic, so we can assume that all of the heat from the metal was transferred to the water:
-q_{\text{metal}}=q_{\text{water}}
-m_{\text{metal}}C_{\text{metal}} \Delta T_{\text{metal}}=m_{\text{water}}C_{\text{water}}\Delta T_{\text{water}}
Then we can plug in our known values:
-(5.0\text{ g})C_{\text{metal}}(26.0^\circ\text{C}-207^\circ\text{C})=(36.0\text{ g})(4.18\; \frac {J}{\text{g}^\circ\text{C}})(26.0^\circ\text{C}-25.0^\circ\text{C})
C_{\text{metal}}
C_{metal}=0.166\; \frac {J} {g^\circ\text{C}}
The specific heat capacity of the unknown metal is 0.166
\frac {J} {g ^\circ\text{C}}
To determine the standard enthalpy of the reaction H+(aq) + OH–(aq) → H2O(l), equal volumes of 0.1 M solutions of HCl and of NaOH can be combined initially at 25°C.
This process is exothermic and as a result, a certain amount of heat qP will be released into the solution. The number of joules of heat released into each gram of the solution is calculated from the product of the rise in temperature and the specific heat capacity of water (assuming that the solution is dilute enough so that its specific heat capacity is the same as that of pure water's). The total quantity of transferred heat can then be calculated by multiplying the result with the mass of the solution.
\Delta H=q_P = m_{\text{sol'n}}C_{\text{water}} \Delta T_{\text{sol'n}}
Note that ΔH = qP because the process is carried out at constant pressure.
Specific heat. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Specific_heat. License: CC BY-SA: Attribution-ShareAlike
Heat Transfer/Introduction. Provided by: Wikibooks. Located at: http://en.wikibooks.org/wiki/Heat_Transfer/Introduction%23Heat_capacity_or_specific_heat. License: CC BY-SA: Attribution-ShareAlike
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specific heat capacity. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/specific_heat_capacity. License: CC BY-SA: Attribution-ShareAlike
Specific heat capacity tutorial. Located at: http://www.youtube.com/watch?v=Kbn_WHEJsyc. License: Public Domain: No Known Copyright. License terms: Standard YouTube license
Constant-volume calorimeter. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Constant-volume_calorimeter%23Calvet-type_calorimeters. License: CC BY-SA: Attribution-ShareAlike
calorie. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/calorie. License: CC BY-SA: Attribution-ShareAlike
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Multidecompositions of the Balanced Complete Bipartite Graph into Paths and Stars
Hung-Chih Lee, Yen-Po Chu, "Multidecompositions of the Balanced Complete Bipartite Graph into Paths and Stars", International Scholarly Research Notices, vol. 2013, Article ID 398473, 4 pages, 2013. https://doi.org/10.1155/2013/398473
Hung-Chih Lee 1 and Yen-Po Chu1
1Department of Information Technology, Ling Tung University, Taichung 40852, Taiwan
Academic Editor: E. Kiliç
Let and denote a path and a star with edges, respectively. For graphs , , and , a -multidecomposition of is a partition of the edge set of into copies of and copies of with at least one copy of and at least one copy of . In this paper, necessary and sufficient conditions for the existence of the (, )-multidecomposition of the balanced complete bipartite graph are given.
Let , , and be graphs. A -decomposition of is a partition of the edge set of into copies of . If has a -decomposition, we say that is -decomposable and write . A -multidecomposition of is a partition of the edge set of into copies of and copies of with at least one copy of and at least one copy of . If has a -multidecomposition, we say that is -multidecomposable and write .
For positive integers and , denotes the complete bipartite graph with parts of sizes and . A complete bipartite graph is balanced if . A -path, denoted by , is a path with edges. A -star, denoted by , is the complete bipartite graph . A -cycle, denoted by , is a cycle of length .
-decompositions of graphs have been a popular topic of research in graph theory. Articles of interest include [1–11]. The reader can refer to [12] for an excellent survey of this topic. Decompositions of graphs into -stars have also attracted a fair share of interest. Articles of interest include [13–18]. The study of the -multidecomposition was introduced by Abueida and Daven in [19]. Abueida and Daven [20] investigated the problem of the -multidecomposition of the complete graph . Abueida and Daven [21] investigated the problem of the -multidecomposition of several graph products where denotes two vertex disjoint edges. Abueida and O'Neil [22] settled the existence problem of the -multidecomposition of the complete multigraph for , and . In [23], Priyadharsini and Muthusamy gave necessary and sufficient conditions for the existence of the -multidecomposition of where . Furthermore, Shyu [24] investigated the problem of decomposing into -paths and -stars, and gave a necessary and sufficient condition for . In [25], Shyu considered the existence of a decomposition of into -paths and -cycles and established a necessary and sufficient condition for . He also gave criteria for the existence of a decomposition of into -paths and cycles in [26]. Shyu [27] investigated the problem of decomposing into -cycles and -stars and settled the case . Recently, Lee [28] established necessary and sufficient conditions for the existence of the -multidecomposition of a complete bipartite graph.
In this paper, we investigate the problem of the -multidecomposition of the balanced complete bipartite graph and give necessary and sufficient conditions for such a multidecomposition to exist.
For our discussions, some terminologies and notations are needed. Let be a graph. The degree of a vertex of , denoted by , is the number of edges incident with . A graph is -regular if each vertex is of degree . The vertex of degree in is called the center of . Let and be subsets of the vertex set and the edge set of , respectively. We use to denote the subgraph of induced by and to denote the subgraph obtained from by deleting . Suppose that are edge-disjoint-graphs. Then, , or , denotes the graph with vertex set , and edge set . Thus, if a graph can be decomposed into subgraphs , we write , or . Moreover, denotes the smallest integer not less than and denotes the largest integer not greater than . Let denote the -path with edges , and denote the -cycle with edges . Throughout the paper, denotes the bipartition of , where and .
For the edge in , the label of is . For example, in the labels of and are and , respectively. Note that each vertex of is incident with exactly one edge with label for . Let be a subgraph of and a nonnegative integer. We use to denote the graph with vertex set and edge set , where the subscripts of are taken modulo . In particular, .
The following results due to Yamamoto et al. and Parker, respectively, are essential for our discussions.
Proposition 1 (see [18]). Let be integers. Then, is -decomposable if and only if and
Proposition 2 (see [7]). There exists a -decomposition of if and only if , and one of aforementioned (see Table 1) cases occurs.
1 Even Even Even , , both are not equalities
2 Even Even Odd ,
3 Even Odd Even ,
4 Odd Even Even ,
5 Odd Even Odd ,
6 Odd Odd Even ,
7 Odd Odd Odd ,
We first give necessary conditions of the -multidecomposition of .
Lemma 3. Let and be positive integers. If there exists a -multidecomposition of , then and .
Proof. The result follows from the fact that the maximum size of a star in is , the size of each member in the multidecomposition is , and .
We now show that the necessary conditions are also sufficient. Since for , the result holds for by Proposition 1. So it remains to consider the case . The proof is divided into cases , and , which are treated in Lemmas 4, 5, and 6, respectively.
Lemma 4. Let and be positive integers with . If , then is -multidecomposable.
Proof. Let where and are integers with . Then, from the assumption . Note that By Proposition 1, is -decomposable. On the other hand, trivially, , and from the assumption . This implies that is -decomposable by Proposition 2. Hence, is -multidecomposable.
Lemma 5. Let be a positive integer with . Then, is -multidecomposable.
Proof. Note that . Trivially, . On the other hand, . Furthermore, for odd , and for even . Hence, is -decomposable by Proposition 2, and is -multidecomposable.
Lemma 6. Let and be integers with . If , then is -multidecomposable.
Proof. Suppose that . Then, from the assumption . Let , , , and . Let for and . Then, . Note that is isomorphic to , is isomorphic to , and is isomorphic to , which is -decomposable by Proposition 1. Hence, it is sufficient to show that is -multidecomposable.
Let . Since , we have , which implies that is a positive integer Let . Then, is a -cycle in . Let . For odd , define a -path in as follows: where the subscripts of are taken modulo . Since and , we have Thus, , which implies the labels of the edges in are and . Note that for , is a -cycle which consists of all of the edges with labels and in . Thus, and are edge-disjoint in .
Define a subgraph of as follows: Since can be decomposed into copies of and for even as well as for odd , can be decomposed into copies of . Let for even and for odd . Note that for even , , and for odd , Let for . Then for even , , and for odd , with the center at . In the following, we will show that can be decomposed into copies of with centers in , and into copies of with centers in for even , and into copies of with centers in , an with the center at , and copies of with centers in for odd .
We show the required star decomposition of by orienting the edges of . For any vertex of , the outdegree (indegree , resp.) of in an orientation of is the number of arcs incident from (to, resp.) . It is sufficient to show that there exists an orientation of such that where , and for even where , and for odd
We first consider the edges oriented outward from according to the parity of . Let and . If is even, then the edges are all oriented outward from , where . If is odd, then the edges for , and , , as well as for are all oriented outward from . In both cases, the subscripts of are taken modulo in the set of numbers . Note that for even we orient edges from each and for odd we orient at most edges from . By inequality (4), we have , which assures us that there are enough edges for the above orientation.
Finally, the edges which are not oriented yet are all oriented from to . From the construction of the orientation, it is easy to see that (9) and (10) are satisfied, and for all , we have So, we only need to check (8).
Since for , it follows from (11) that for . Note that for even , , and for odd , Thus, Therefore from (12), we have for . This establishes (8). Hence, there exists the required decomposition of . Let be the star with center at in for . Then, is a -star. This completes the proof.
Now, we are ready for the main result. It is obtained form the arguments above, Lemma 4 and Lemmas 3, 4, 5, and 6.
Theorem 7. Let and be positive integers. Then, has a -multidecomposition if and only if and .
The authors are grateful to the referees for the valuable comments.
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Copyright © 2013 Hung-Chih Lee and Yen-Po Chu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Section 4.7: Approximations
For a sufficiently well behaved function
f\left(x,y\right)
, the change
\mathrm{Δ}f\equiv f\left(x+\mathrm{dx},y+\mathrm{dy}\right)-f\left(x,y\right)
is approximated by the differential
\mathrm{df}\equiv {f}_{x}\left(x,y\right) \mathrm{dx}+{f}_{y}\left(x,y\right) \mathrm{dy}
. Such a differential is often called either a total or an exact differential. The generalization to functions of more than two variables is obvious.
Contrary to the logic in single-variable calculus where the derivative is defined first, and the differential is defined in terms of the derivative as the derivative times an increment, for functions of several variables the differential must be defined first. If a function has a differential, then it is called differentiable!
A sufficient condition for the existence of a differential, and hence for the function of several variables to be differentiable, is the continuity of the first partial derivatives.
As for the function of one variable, a differentiable function is necessarily continuous.
Functions of several variables can be approximated by Taylor polynomials. Table 4.7.1 lists Taylor polynomials of degrees one and two for the function
f\left(x,y\right)
. The expansion point is taken as
\left(a,b\right)
f\left(a+h,b+k\right)=f\left(a,b\right)+{f}_{x}\left(a,b\right)h+{f}_{y}\left(a,b\right)k
f\left(a+h,b+k\right)=f\left(a,b\right)+{f}_{x}\left(a,b\right)h+{f}_{y}\left(a,b\right)k +\frac{1}{2}\left({f}_{\mathrm{xx}}\left(a,b\right){h}^{2}+2 {f}_{\mathrm{xy}}\left(a,b\right)h k+{f}_{\mathrm{yy}}\left(a,b\right){k}^{2}\right)
Table 4.7.1 Taylor polynomials of degrees one and two
The equality of the mixed partial derivatives
{f}_{\mathrm{xy}}
{f}_{\mathrm{yx}}
is assumed in Table 4.7.1. Hence, instead of having two separate terms containing the product
h k
, both are combined into twice the one. It is sometimes useful to write the second-degree terms as half the quadratic form
\left[\begin{array}{cc}h& k\end{array}\right] \left[\begin{array}{cc}{f}_{\mathrm{xx}}& {f}_{\mathrm{xy}}\\ {f}_{\mathrm{xy}}& {f}_{\mathrm{yy}}\end{array}\right]\left[\begin{array}{c}h\\ k\end{array}\right]
where the row vector
\left[\begin{array}{cc}h& k\end{array}\right]
is the transpose of the column vector
\left[\begin{array}{c}h\\ k\end{array}\right]
and the intervening symmetric matrix of second partial derivatives is called a Hessian.
A general representation of the
n
th-degree term is
\frac{1}{n!}{\left[{\left(h \frac{∂}{∂x}+k \frac{∂}{∂y}\right)}^{n}f\right]}_{\left(a,b\right)}
, but unfortunately, no form of this expression is a valid Maple operator.
The role existence and continuity of partial derivatives plays in differentiability of a function of several variables is examined at length in Section 4.11.
x
changes from 3 to 3.04 and
y
-2
-2.2
\mathrm{Δ}f
\mathrm{df}
f\left(x,y\right)=3 {x}^{2}+4 x y-5 {y}^{2}+7
\sqrt{4{\left(3.03\right)}^{3}+5{\left(6.2\right)}^{2}}
by using the total differential for some appropriate function
f\left(x,y\right)
\frac{1}{2.03}+\frac{1}{3.02}+\frac{1}{4.97}
f\left(x,y,z\right)
Use the total differential to approximate the value of
f\left(x,y\right)={x}^{2}{e}^{2 {\mathrm{xy}}^{2}}
\left(x,y\right)=\left(1.03,0.96\right)
f\left(x,y,z\right)=x \mathrm{ln}\left(\frac{y+ z}{{x}^{2}+y z}\right)
\left(x,y,z\right)=\left(2.03,3.12,1.48\right)
Use differentials to estimate the maximum error in determining the area of a rectangle measured (in cm) to be
4×5
, if each measurement is accurate to .1 cm.
Use differentials to estimate the maximum error in determining the surface area of a closed rectangular box measured (in inches) to be
15×6×21
, if each measurement is accurate to .2 in.
A closed box is constructed from lumber that is
3/4
in thick. The outside measurements are
9×7×5
. Use differentials to estimate the interior volume of the box.
Show that the plane tangent to
f\left(x,y\right)=3 {x}^{2}+5 x y-7 {y}^{2}+4
\left(x,y\right)=\left(2,3\right)
is the first-degree Taylor polynomial constructed at the point of contact. Show further that approximating
\mathrm{Δ}f
by the total differential
\mathrm{df}
amounts to a tangent-plane approximation.
\left(x,y\right)=\left(4,1\right)
, construct the second-degree Taylor polynomial for
f\left(x,y\right)=\sqrt{3 {x}^{2}+{y}^{3}}
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On the Fekete and Szegö Problem for the Class of Starlike Mappings in Several Complex Variables
Qing-Hua Xu, Tai-Shun Liu, "On the Fekete and Szegö Problem for the Class of Starlike Mappings in Several Complex Variables", Abstract and Applied Analysis, vol. 2014, Article ID 807026, 6 pages, 2014. https://doi.org/10.1155/2014/807026
Qing-Hua Xu1 and Tai-Shun Liu2
Let be the familiar class of normalized univalent functions in the unit disk. Fekete and Szegö proved the well-known result for . We investigate the corresponding problem for the class of starlike mappings defined on the unit ball in a complex Banach space or on the unit polydisk in , which satisfies a certain condition.
Let be the class of functions of the form which are analytic in the open unit disk We denote by the subclass of the normalized analytic function class consisting of all functions which are also univalent in . Let denote the class of starlike functions in .
It is well known that the Fekete and Szegö inequality is an inequality for the coefficients of univalent analytic functions found by Fekete and Szegö [1], related to the Bieberbach conjecture. Finding similar estimates for other classes of functions is called the Fekete and Szegö problem.
The Fekete and Szegö inequality states that if , then for . After that, there were many papers to consider the corresponding problems for various subclasses of the class , and many interesting results were obtained. We choose to recall here the investigations by, for example, Kanas [2] (see also [3–5]).
The coefficient estimate problem for the class , known as the Bieberbach conjecture [6], is settled by de Branges [7], who proved that for a function in the class , then , for .
However, Cartan [8] stated that the Bieberbach conjecture does not hold in several complex variables. Therefore, it is necessary to require some additional properties of mappings of a family in order to obtain some positive results, for instance, the convexity and the starlikeness.
In [9], Gong has posed the following conjecture.
Conjecture A. If is a normalized biholomorphic starlike mapping, where is the open unit polydisk in , then
In contrast, although the coefficient problem for the class had been completely solved, only a few results are known for the inequalities of homogeneous expansions for subclasses of biholomorphic mappings in several complex variables (see, for detail, [9]).
Recently, some best-possible results concerning the coefficient estimates for subclasses of holomorphic mappings in several variables were obtained in work of Graham et al. [10], Graham et al. [11], Hamada et al. [12], Hamada and Honda [13], Kohr [14], X. Liu and T. Liu [15], and Xu and Liu [16].
In [17], Koepf obtained the following result for .
Theorem A. Let . Then The above estimation is sharp.
It is natural to ask whether we can extend Theorem A to higher dimensions.
In this paper, we will establish inequalities between the second and third coefficients of homogeneous expansions for starlike mappings defined on the unit ball in Banach complex spaces and the unit polydisc in , respectively, which are the natural extension of Theorem A to higher dimensions.
Let be a complex Banach space with norm ; let be the dual space of ; let be the unit ball in . Also, let denote the boundary of , and let be the distinguished boundary of .
For each , we define According to the Hahn-Banach theorem, is nonempty.
Let denote the set of all holomorphic mappings from into . It is well known that if , then for all in some neighborhood of , where is the th-Fréchet derivative of at , and, for , Furthermore, is a bounded symmetric -linear mapping from into .
A holomorphic mapping is said to be biholomorphic if the inverse exists and is holomorphic on the open set . A mapping is said to be locally biholomorphic if the Fréchet derivative has a bounded inverse for each . If is a holomorphic mapping, then is said to be normalized if and , where represents the identity operator from into . Let be the set of all normalized biholomorphic mappings on . We say that is starlike if is biholomorphic on and is starlike with respect to the origin. Let be the set of normalized starlike mappings on .
Suppose that is a bounded circular domain. The first Fréchet derivative and the -th Fréchet derivative of a mapping at point are written by , , respectively. The matrix representations are where , .
In order to prove the desired results, we first give some lemmas.
Lemma 1 (see [18]). Let be a normalized locally biholomorphic mapping. Then is a starlike mapping on if and only if
Lemma 2. Let be a normalized locally biholomorphic mapping. Then if and only if where and .
Lemma 3 (see [19]). Let , and , ; then
Lemma 4. Suppose that . Then defined by , where , belongs to if and only if .
Proof. Denote ; since , we have Straightforward calculation yields It is not difficult to check that Hence By using (16), we deduce that Therefore, by Lemma 1, we obtain that if and only if . This completes the proof of Lemma 4.
In this section, we state and prove the main results of our present investigation.
Theorem 1. Suppose and Then The above estimate is sharp.
Proof. Fix and denote . Let be given by where . Then , , and Since , from Lemma 1, we have In view of Lemma 3, we obtain that That is,
On the other hand, since , we have Comparing with the homogeneous expansion of two sides of the above equality, we obtain Equation (27) may be rewritten as follows: Thus, from (18) of Theorem 1, (24), (26), and (28), we deduce that If now , then On the other hand, if , then we use and get The following example shows that the estimation of Theorem 1 is sharp.
Example. If , we consider the following example: By Lemma 4, we obtain that .
It is not difficult to check that the mapping satisfies the condition of Theorem 1. Setting () in (32), we obtain that If , we consider the following example: In view of Lemma 4, we deduce that .
It is not difficult to verify that the mapping satisfies the condition of Theorem 1. Taking () in (34), we have This completes the proof of Theorem 1.
Remark 2. When , , Theorem 1 is equivalent to Theorem A.
Theorem 3. Suppose and for , where , . Then The above estimate is sharp.
Proof. For any , denote . Let be given by where and satisfies . Then , , and Since , from Lemma 2, we deduce that , . Therefore, according to Lemma 3, we have Hence, in view of (26), (28), and (32) of Theorem 3, we obtain that If now , then On the other hand, if , then we use and get Then, by using (42) and (43), we have If , then we have Also since is a holomorphic function on , in view of the maximum modulus theorem of holomorphic function on the unit polydisc, we obtain That is, Hence
Finally, in order to see that the estimation of Theorem 3 is sharp, it suffices to consider the following mappings.
If , we consider the following example:
If , we consider the following example: In view of Problem 6.2.5 of [19], we deduce that the mappings , defined in (50) and (51), are in the class .
It is not difficult to verify that the mappings defined in (50) and (51) satisfy the condition of Theorem 3. Taking () in (50) and (51), respectively, we deduce that the equality in (37) holds true. This completes the proof of Theorem 3.
Remark 4. When , Theorem 3 reduces to Theorem A.
This work was supported by NNSF of China (Grant no. 11261022), the Jiangxi Provincial Natural Science Foundation of China (Grant no. 20132BAB201004), the Natural Science Foundation of Department of Education of Jiangxi Province, China (Grant no. GJJ12177), and the Zhejiang Provincial Natural Science Foundation of China (Grant no. Y6110053).
M. Fekete and G. Szegö, “Eine bemerkunguber ungerade schlichte Funktionen,” Journal London Mathematical Society, vol. 8, pp. 85–89, 1933. View at: Google Scholar
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L. Bieberbach, “Über die Koeffizienten der einigen Potenzreihen welche eine schlichte Abbildung des Einheitskreises vermitten,” S. B. Preuss. Akad. Wiss., vol. 38, pp. 940–955, 1916. View at: Google Scholar
H. Cartan and P. Montel, “Sur la possibilité d’étendre aux fonctions de plusieurs variables complexes la théorie des fonctions univalentes,” in Lecons sur les Fonctions Univalentes ou Multivalentes, Gauthier-Villars, Paris, France, 1933. View at: Google Scholar
S. Gong, The Bieberbach Conjecture, American Mathematical Society, International Press, Providence, RI, USA.
I. Graham, H. Hamada, and G. Kohr, “Parametric representation of univalent mappings in several complex variables,” Canadian Journal of Mathematics, vol. 54, no. 2, pp. 324–351, 2002. View at: Publisher Site | Google Scholar | MathSciNet
I. Graham, G. Kohr, and M. Kohr, “Loewner chains and parametric representation in several complex variables,” Journal of Mathematical Analysis and Applications, vol. 281, no. 2, pp. 425–438, 2003. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
H. Hamada, T. Honda, and G. Kohr, “Growth theorems and coefficients bounds for univalent holomorphic mappings which have parametric representation,” Journal of Mathematical Analysis and Applications, vol. 317, no. 1, pp. 302–319, 2006. View at: Publisher Site | Google Scholar | MathSciNet
H. Hamada and T. Honda, “Sharp growth theorems and coefficient bounds for starlike mappings in several complex variables,” Chinese Annals of Mathematics B, vol. 29, no. 4, pp. 353–368, 2008. View at: Publisher Site | Google Scholar | MathSciNet
G. Kohr, “On some best bounds for coefficients of several subclasses of biholomorphic mappings in Cn,” Complex Variables, vol. 36, pp. 261–284, 1998. View at: Google Scholar
X. Liu and T. Liu, “The sharp estimates of all homogeneous expansions for a class of quasi-convex mappings on the unit polydisk in
{\mathbb{C}}^{n}
,” Chinese Annals of Mathematics B, vol. 32, no. 2, pp. 241–252, 2011. View at: Publisher Site | Google Scholar | MathSciNet
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Copyright © 2014 Qing-Hua Xu and Tai-Shun Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Palindrome Date Finder
How to use this palindrome date finder?
Why do we care about palindromes?
Palindromic numbers – the beauty in math
How can you celebrate palindrome day?
Have you noticed something odd in your calendar this February? Don't worry; you didn't miss a leap year. There is, however, something quite unique happening this month. Look to the second half, and you might notice it… that's right, it's the ultimate palindrome day, Tuesday 2/22/22!
Palindromic dates in the MM/DD/YY format are quite a feat. It's not often that the number of months can match the number of years and give such unique results as the upcoming "Twosday".
Even more amazing is that "Twosday" is also a palindrome in the DD/MM/YYYY format: 22/02/2022. According to Aziz Inan, University of Portland's professor of engineering, "Eight-digit palindrome dates are very rare and are clustered in the first three or so centuries at the beginning of a millennial, and then don't show up for 600 to 700 years, until they appear as a cluster in the next millennium".
Although we may not believe in such things as lucky numbers, we genuinely think these dates are special and worth celebrating. They make great occasions to take a moment to ponder on the beauty of numbers and notice all the interesting patterns that appear around us in our seemingly chaotic lives. That's why we decided to celebrate "Twosday", and we want to help you notice all the other special days like it - so we created this palindrome date finder.
Upon hearing the word “palindrome”, most people think of word plays and puns based on the repetition of letters. That is, in most part, correct, but it’s by no means all that the world of palindromes has to offer! In its most broad meaning, a palindrome is a repeating, symmetrical pattern of signs that doesn’t change no matter what side we look at it from – not limited only to the letters of the alphabet!
There are many types of palindromes, and only some of them consist of words and letters. You can also find these repeating patterns in dates, numbers, even music! The most common palindrome types include:
Character-unit palindromes: words that read the same backward and forward. Examples in English include such words as kayak, level, and civic.
💡 You might have heard about the most notorious example of a character-unit palindrome, as it was quite popular around the Internet not too long ago. We are, of course, talking about the word aibohphobia – which, quite ironically, is the official name for an irrational fear of palindromes.
Name palindromes: actually a subset of character-unit palindromes. These are simply names that read the same no matter the direction, such as Bob, Hannah, or Otto.
Musical palindromes that consist of notes in a palindromic pattern. For instance, if you were to play a melody that goes A, B, E, B, A, that would be a palindrome. The musical world has many examples of great works of art based around this principle, such as Joseph Haydn's Symphony No. 47.
Sentence palindromes: complete phrases that consist of the same words read backward and forward. The sentence "Go hang a salami, I'm a lasagna hog" is an example of such a palindrome.
Palindromic numbers: numbers whose digits are the same read backward and forward. For example, 10101 or 22322. Interestingly, in the mathematical view, all single digits are palindromes - after all, they start and end with the same sign! Although usually considered in the decimal system, palindromicity can be applied to the natural numbers in all numerical systems. You can read more about palindromic numbers further down.
Palindrome dates: – the star of this tool – are dates whose digits form a palindrome. For example, did you know that February 2nd, 2020, is officially known as the Universal Palindrome Day? That is because no matter the date format, on that day, the digits form the palindrome: 02/02/2020!
Our palindrome date finder is a concise tool that you’re bound to find easy to use! All you need to do is fill it in with some information, and the calculator will do the rest for you.
The first batch of information you need to give the finder is:
The date format – choose whether you state the day or the month first.
The year format – you can pick from four and two digits.
What you want the finder to do – give you all palindrome dates between two certain days, or show you any number (between 1 and 100) of the upcoming palindrome dates.
Depending on your choice, you will be asked to either choose the start and end date for your search or give the number of palindrome dates you want to see. Once that is done, voila! You will find your results below the calculator.
This question could be answered in three short words: humans love patterns.
The word "love", however, is a bit of a misnomer. It only takes a little bit of research to realize that finding patterns all around us is more of a biological necessity than a simple pleasure. You see, our brains evolved to find them in order to ensure our survival. In the words of science popularizer Neil deGrasse Tyson:
"Over centuries of evolution, humans' pattern recognition skills determined natural selection. Hunters skilled at spotting prey and predator and telling poisonous plants from healthy ones offered them a better chance of survival than those blind to the patterns. It enabled the survivors to pass on those pattern-friendly genes to future generations."
Indeed, once we start thinking about it, it's easy to notice our tendency to see patterns manifesting themselves in many different forms. For example, have you ever looked at a random object and realized that… it’s apparently looking back at you?!
This is an example of a phenomenon known as face pareidolia – the human brain's deeply ingrained tendency to recognize the patterns that define a face even in the most unexpected of places. If you've ever wondered why Jesus appears on so many slices of toast these days – now you know why 🍞😉
Our tendency to see and, above all, enjoy palindromes most likely stems from the same root. We, as human beings, are inherently inclined to not only recognize patterns but also to assign them with meaning. And since most of us no longer have to utilize these tendencies to ensure survival, we have the luxury of making them curiosities and fun, interesting instances of order and repetitiveness in the otherwise quite random world that we live in. Or, in the case of palindrome dates, we even treat them as mnemonic devices for important events in our lives!
The beauty of repetitive patterns in numbers makes palindrome dates stand out in the calendar – and makes them easy to remember. You might be surprised by just how many people decide to celebrate this beauty by choosing palindrome dates for the most important days of their lives. For example, in Las Vegas – a spot known for being a popular wedding destination - almost 2,700 couples got married on October 10th, 2010 (10/10/10)! And even more couples are being drawn by the upcoming "Twosday", February 22nd, with American chapels expecting it to be the busiest wedding day ever.
Palindromes in mathematics are ubiquitous. For one, all the single digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) are technically palindromes – they do look the same no matter which "side" you consider them from. They are just a drop in the sea of interesting examples that showcase the beauty of numbers and the relationships between them.
It’s important to note that palindromic numbers can be seen in practically any numerical system one can think of. It’s not necessary to venture outside most people’s comfort zone of the decimal system to find fascinating instances, though. An excellent example of palindromes occurring in mathematics is the case of perfect powers.
There exist countless palindromic perfect powers nk, where:
n is a natural number (
n \in \mathbb{N}
k = 2
k = 3
k = 4
Examples of such palindromic squares include 121, 484, 676, 10201, 12321, 14641; palindromic perfect cubes are, among others, 1331, 1030301, 1367631; and you can even find palindromic fourth powers such as 14641, 10406040 or 1004006004001.
You don’t have to wander into the realm of powers to find palindromic numbers, though. In fact, there’s quite a lot to come across if you simply enumerate… As we’ve already established, in the decimal system, there are exactly ten single-digit palindromes, but on top of that, we have:
9 palindromic numbers with two digits (11, 22, 33, 44, 55, 66, 77, 88, 99),
90 palindromic numbers with three digits (for example 101, 111, 121, and so on to 999), and
90 palindromic numbers with four digits (1001, 1111, 1221, …, 9779, 9889, 9999).
💡 Did you know you can create palindromes from non-palindromic numbers? The operation is known as the Lychrel process, and it works by "reversing" the digits of any number and then adding them. After doing it several times, you will get a palindrome at some point! Let's go through an example, starting with the number 273.
And voila! Using the Lychrel procedure, we easily arrived at the palindromic number 5115 starting from a non-palindrome.
Palindromes can be a great way to teach children about several things – from math, through new vocabulary, to even music! And what better occasion is there to show your students the beauty of palindromes than the upcoming Twosday? If you're a teacher, we highly encourage you to take this opportunity and get ready with some special lesson plans. Hopefully, this shortlist will give you some inspiration!
Time capsules. If you're teaching first grade, you're presented with a truly unique opportunity of giving your students a gift they will cherish for years after they finally graduate. Celebrate 22/02/2022 by creating a time capsule that your kids will open on the next big palindrome of their life, (0)3/(0)3/(20)33 – coincidentally, their senior year of high school!
Palindrome writing game. You can engage your students with a bit of realization. Have them write down as many 3, 4, or 5-letter-long palindromes as they can!
Encourage creativity by making the students write poems using mostly palindromes. You might be surprised by how wild these poems can get!
Finally, have your students engage in a guessing game – let them estimate how many palindrome dates there will be in the next few years. You can use this palindrome date finder to check who is the closest!
Do you have a fancy for beautiful dates ✨ with symmetry? Use our Palindrome Date Finder to find the upcoming palindrome days 📅 and also a list of all palindromes between two dates! Who knows, maybe one of them will turn out to be your dream day? 😉
What are your preferences? ✍️
Month-Day-Year (M-D-Y)
4-digit (YYYY)
find all palindrome dates between...
Enter the date interval 🗓️
🎊 Fun Fact
8-digit Palindrome Dates are super rare - they occur just 12 times in the 21st century! 😲
Watch or watch not; there is no try: with our Star Wars marathon calculator, you will learn the best way to enjoy all the Star Wars movies and series!
Star Wars Marathon Calculator
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Estimate frequency response and spectrum using spectral analysis with frequency-dependent resolution - MATLAB spafdr - MathWorks América Latina
y\left(t\right)=G\left(q\right)u\left(t\right)+v\left(t\right)
G\left({e}^{i\omega }\right)
\frac{2\pi }{N{T}_{s}}
\frac{\pi }{{T}_{s}}
|
Fit generalized additive model (GAM) for binary classification - MATLAB fitcgam - MathWorks América Latina
{\left({\stackrel{^}{y}}_{k}-{\stackrel{^}{y}}_{k-1}\right)}^{\prime }\left({\stackrel{^}{y}}_{k}-{\stackrel{^}{y}}_{k-1}\right)/{\stackrel{^}{y}}_{k}{}^{\prime }{\stackrel{^}{y}}_{k}
{\stackrel{^}{y}}_{k}
\begin{array}{l}y~Binomial\left(n,\mu \right)\\ g\left(\mu \right)=\mathrm{log}\frac{\mu }{1-\mu }=c+\text{}{f}_{1}\left({x}_{1}\right)+\text{}{f}_{2}\left({x}_{2}\right)+\cdots +{f}_{p}\left({x}_{p}\right),\end{array}
g\left(\mu \right)=c+\text{}{f}_{1}\left({x}_{1}\right)+\text{}{f}_{2}\left({x}_{2}\right)+\cdots +{f}_{p}\left({x}_{p}\right)+\sum _{i,j\in \left\{1,2,\cdots ,p\right\}}{f}_{ij}\left({x}_{i}{x}_{j}\right),
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Reaction Quotient Calculator
How to find the reaction quotient?
Relation between the reaction quotient and the equilibrium constant
An example of how to calculate the reaction quotient
How to use our reaction quotient calculator?
Chemical reactions tend to a state of equilibrium — use our reaction quotient calculator to know in which direction your reaction is moving.
What the reaction quotient is;
How to find the reaction quotient using the reaction quotient equation; and
Similarities with the equilibrium constant equation;
We will end with a short step-by-step tutorial explaining how to calculate the reaction quotient.
The reaction quotient (Q) is a function of the concentrations or pressures of the chemical compounds present in a chemical reaction at a particular point in time. The reaction quotient equation is used to determine the direction of the reaction: from a starting point, every reaction (or almost every: there are cycling reactions too) tends to reach an equilibrium state, where the two rates of transformation (reagents into products and vice versa) are equal.
Both the reaction quotient and the equilibrium constant can be calculated only for reversible reactions: this means that the reaction proceeds in both directions, constantly. This is in contrast to irreversible reactions, like combustion, where it's impossible to obtain the reagents back.
Let us consider a general chemical equation:
aA + bB \rightleftharpoons cC + dD
the lowercase letters indicate the stoichiometric coefficients; and
the uppercase letters indicate the activities of the substances.
Notice that you can add as many chemical species as you need, as long as you maintain the same notation.
To find the reaction quotient
Q
, multiply the activities for the species of the products and divide by the activities of the reagents, raising each one of these values to the power of the corresponding stoichiometric coefficient.
Here's the reaction quotient equation for the reaction given by the equation above:
Q = \frac{\left[C\right]^c\left[D\right]^d}{\left[A\right]^a\left[B\right]^b}
Activities of the chemical species should be used in the calculations, but are usually discarded in favor of concentrations for diluted liquid compounds or partial pressures for gaseous compounds. In most cases, this substitution is fine and won't cause you any trouble, but remember to check if your problem explicitly asks for activities!
If any solid or pure liquid appears in your reaction, either as reagents and/or as products, remember that their activity is 1.
The reaction quotient equation can be used at any point in time during a chemical reaction. There's a point, however, where its quantity equals the equilibrium constant of that reaction, most often denoted by
K
. At that point, the reaction as attained chemical equilibrium. Chemical equilibrium is the dynamic state where the composition of the chemical system does not change with time (even if the transformation of species is allowed, this does not affect the overall concentrations).
The equilibrium constant equation is entirely similar to the reaction quotient:
K = \frac{\left[C\right]^c\left[D\right]^d}{\left[A\right]^a\left[B\right]^b} \ \Bigg|_{\text{equilibrium}}
Just remember that all of the quantities used in the equation for
K
are at their equilibrium value. The equilibrium constant quantifies toward which side of the reaction the equilibrium leans. In particular:
K > 1
, then the equilibrium favors the reagents.
K < 1
, then the equilibrium favors the products.
The thermodynamical equilibrium is a stable state of the reaction and the Le Chatelier principles state that a change in conditions is reflected by a change in the equilibrium constant that counteracts and balances the affecting variations.
The equilibrium constant is constant for a given chemical reaction, with a definite value for each temperature. You can find these values in tables, usually for standardized temperature values — for IUPAC, it is 25°C. Its value equals the reaction quotient at the equilibrium:
Q\big|_{\text{equilibrium}} = K
The state of the reaction when the equilibrium is not yet attained is given by the reaction quotient:
Q < K
, the reaction will proceed from reagents to products:
aA + bB → cC + dD
Q > K
, the reaction will proceed from products to reagents:
cC + dD → aA + bB
When working on acid and bases, the concept of the equilibrium constant is usually discarded in favor of the characteristic dissociation and association constants.
The acid dissociation reaction is described by a given chemical equation:
H_aA + H_2O ⇌ aH_3O^+ + A^{a-}
The equilibrium constant is given by the dissociation constant
K_a
K_a= \frac{\left[A^{-}\right]^a\left[H^+\right]^a}{\left[H_aA\right]}
H_2O
doesn't show up in this equation because the activity of pure water is
1
For a base, the protonation reaction is described by a generic chemical equation:
B + H_2O \rightleftharpoons OH^- + BH^+
In analogy with the previously seen acid dissociation constant, one can define the association constant
K_b
K_b = \frac{\left[OH^-\right]\left[BH^+\right]}{\left[B\right]}
Here the activity of water can be omitted as well.
If you know the concentration of the solution, you can calculate the pH of the solution (both of acid and base) by using our pH calculator.
Calculating the reaction quotient is pretty easy — the implications are the interesting part!
Choose your reaction. Let's assume that it is :
Cd^{2+}_{(aq)} + 4Cl^-_{(aq)} \rightleftharpoons CdCl_{4(aq)}^{2-}
At 25°C, this reaction has an equilibrium constant
K = 108
Choose the concentration of the products and reagents. We will use these values:
\left[Cd^{2+}\right] = 1\ \text{M}
\left[Cl^-\right] = 0.5\ \text{M}
\left[CdCl_4^{2-}\right] = 0.25\ \text{M}
Calculate the reaction quotient by using the equation above:
Q= \frac{\left[CdCl_4\right]}{\left[Cd^{2+}\right] \left[Cl^-\right]^4} =\frac{0.25}{1\times 0.5^4}=4
We see that the value of
Q
is smaller than the value of
K
. Hence, the reaction has just set off, and it's highly unbalanced towards the reagents.
Our reaction quotient calculator can take up to 6 reagents and 6 products. At the beginning, you will only see two of each group — more will appear as you fill the preceding fields.
Remember to write the stoichiometric coefficient and the activities/concentrations in the correct fields!
The concentration of the substances can be computed with our concentration calculator. If you want to calculate the concentration of the diluted solution, why not try our solution dilution calculator?
The concepts of equilibrium and reaction quotient are close to other thermodynamic quantities. You can find more about them at our Gibbs free energy calculator!
We also have a lot of tools to help you with solutions and concentrations, such as:
The molarity calculator;
The molality calculator; and
the PPM to molarity calculator and PPMto mg/l converter.
There's many more on our chemistry page. You will surely find what you need there!
What is the reaction quotient definition?
The reaction quotient is a quantity used in chemistry to understand the progress of a chemical reaction with respect to the equilibrium state. In a reversible chemical reaction, the concentrations of the chemical species vary, with reagents transforming into products and vice versa. The reaction quotient measures the relative abundance of a chemical species at any given time.
How do I calculate the reaction quotient?
Using capital letters to indicate the concentrations of the species and cursive letters to indicate the stoichiometric coefficients, the state of the reaction is defined as Q = ([C]ᶜ·[D]ᵈ) / ([A]ᵃ·[B]ᵇ) (with the products on top and the reagents at the bottom). If Q > 1, then the reaction favors the reagents. If Q < 1, the products are dominant, while Q = 1 implies that the reaction is at the equilibrium.
What are Q and K in a chemical reaction?
Q is the reaction quotient, while K is the equilibrium constant. They are both defined as ([C]ᶜ·[D]ᵈ) / ([A]ᵃ·[B]ᵇ): the ratio of the product of the concentrations of the reaction's products to the product of the concentrations of the reagents, each of them raised to the power of their relative stoichiometric coefficients. K is defined only at the equilibrium while Q is defined during the whole reaction. They are equal at the equilibrium.
What is the concentration of water ?
In the calculations for the reaction quotient, the value of the concentration of water is always 1. Water does not participate in a reaction when it's the solvent and its quantity is so big that its variations are negligible, thus it is excluded from the calculations.
Coefficient of reagent 1
Activity of reagent 1
Coefficient of product 1
Activity of product 1
Our radioactive decay calculator will help you understand what radioactive decay is and how it is related to the activity of a substance.
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Section 42.25 (02SN): Intersecting with an invertible sheaf—The Stacks project
Section 42.25: Intersecting with an invertible sheaf (cite)
42.25 Intersecting with an invertible sheaf
In this section we study the following construction.
Definition 42.25.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. We define, for every integer $k$, an operation
\[ c_1(\mathcal{L}) \cap - : Z_{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X) \]
called intersection with the first Chern class of $\mathcal{L}$.
Given an integral closed subscheme $i : W \to X$ with $\dim _\delta (W) = k + 1$ we define
\[ c_1(\mathcal{L}) \cap [W] = i_*(c_1({i^*\mathcal{L}}) \cap [W]) \]
where the right hand side is defined in Definition 42.24.1.
For a general $(k + 1)$-cycle $\alpha = \sum n_ i [W_ i]$ we set
\[ c_1(\mathcal{L}) \cap \alpha = \sum n_ i c_1(\mathcal{L}) \cap [W_ i] \]
Write each $c_1(\mathcal{L}) \cap W_ i = \sum _ j n_{i, j} [Z_{i, j}]$ with $\{ Z_{i, j}\} _ j$ a locally finite sum of integral closed subschemes of $W_ i$. Since $\{ W_ i\} $ is a locally finite collection of integral closed subschemes on $X$, it follows easily that $\{ Z_{i, j}\} _{i, j}$ is a locally finite collection of closed subschemes of $X$. Hence $c_1(\mathcal{L}) \cap \alpha = \sum n_ in_{i, j}[Z_{i, j}]$ is a cycle. Another, more convenient, way to think about this is to observe that the morphism $\coprod W_ i \to X$ is proper. Hence $c_1(\mathcal{L}) \cap \alpha $ can be viewed as the pushforward of a class in $\mathop{\mathrm{CH}}\nolimits _ k(\coprod W_ i) = \prod \mathop{\mathrm{CH}}\nolimits _ k(W_ i)$. This also explains why the result is well defined up to rational equivalence on $X$.
The main goal for the next few sections is to show that intersecting with $c_1(\mathcal{L})$ factors through rational equivalence. This is not a triviality.
Lemma 42.25.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be an invertible sheaves on $X$. Then
\[ c_1(\mathcal{L}) \cap \alpha + c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}) \cap \alpha \]
in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ for every $\alpha \in Z_{k + 1}(X)$. Moreover, $c_1(\mathcal{O}_ X) \cap \alpha = 0$ for all $\alpha $.
Proof. The additivity follows directly from Divisors, Lemma 31.27.5 and the definitions. To see that $c_1(\mathcal{O}_ X) \cap \alpha = 0$ consider the section $1 \in \Gamma (X, \mathcal{O}_ X)$. This restricts to an everywhere nonzero section on any integral closed subscheme $W \subset X$. Hence $c_1(\mathcal{O}_ X) \cap [W] = 0$ as desired. $\square$
Recall that $Z(s) \subset X$ denotes the zero scheme of a global section $s$ of an invertible sheaf on a scheme $X$, see Divisors, Definition 31.14.8.
Lemma 42.25.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $Y$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ Y$-module. Let $s \in \Gamma (Y, \mathcal{L})$. Assume
$\dim _\delta (Y) \leq k + 1$,
$\dim _\delta (Z(s)) \leq k$, and
for every generic point $\xi $ of an irreducible component of $Z(s)$ of $\delta $-dimension $k$ the multiplication by $s$ induces an injection $\mathcal{O}_{Y, \xi } \to \mathcal{L}_\xi $.
Write $[Y]_{k + 1} = \sum n_ i[Y_ i]$ where $Y_ i \subset Y$ are the irreducible components of $Y$ of $\delta $-dimension $k + 1$. Set $s_ i = s|_{Y_ i} \in \Gamma (Y_ i, \mathcal{L}|_{Y_ i})$. Then
\begin{equation} \label{chow-equation-equal-as-cycles} [Z(s)]_ k = \sum n_ i[Z(s_ i)]_ k \end{equation}
as $k$-cycles on $Y$.
Proof. Let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $k$. Let $\xi \in Z$ be its generic point. We want to compare the coefficient $n$ of $[Z]$ in the expression $\sum n_ i[Z(s_ i)]_ k$ with the coefficient $m$ of $[Z]$ in the expression $[Z(s)]_ k$. Choose a generator $s_\xi \in \mathcal{L}_\xi $. Write $A = \mathcal{O}_{Y, \xi }$, $L = \mathcal{L}_\xi $. Then $L = As_\xi $. Write $s = f s_\xi $ for some (unique) $f \in A$. Hypothesis (3) means that $f : A \to A$ is injective. Since $\dim _\delta (Y) \leq k + 1$ and $\dim _\delta (Z) = k$ we have $\dim (A) = 0$ or $1$. We have
\[ m = \text{length}_ A(A/(f)) \]
which is finite in either case.
If $\dim (A) = 0$, then $f : A \to A$ being injective implies that $f \in A^*$. Hence in this case $m$ is zero. Moreover, the condition $\dim (A) = 0$ means that $\xi $ does not lie on any irreducible component of $\delta $-dimension $k + 1$, i.e., $n = 0$ as well.
Now, let $\dim (A) = 1$. Since $A$ is a Noetherian local ring it has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$. These correspond 1-1 with the $Y_ i$ passing through $\xi '$. Moreover $n_ i = \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i})$. Also, the multiplicity of $[Z]$ in $[Z(s_ i)]_ k$ is $\text{length}_ A(A/(f, \mathfrak q_ i))$. Hence the equation to prove in this case is
\[ \text{length}_ A(A/(f)) = \sum \text{length}_{A_{\mathfrak q_ i}}(A_{\mathfrak q_ i}) \text{length}_ A(A/(f, \mathfrak q_ i)) \]
which follows from Lemma 42.3.2. $\square$
The following lemma is a useful result in order to compute the intersection product of the $c_1$ of an invertible sheaf and the cycle associated to a closed subscheme. Recall that $Z(s) \subset X$ denotes the zero scheme of a global section $s$ of an invertible sheaf on a scheme $X$, see Divisors, Definition 31.14.8.
Lemma 42.25.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $Y \subset X$ be a closed subscheme. Let $s \in \Gamma (Y, \mathcal{L}|_ Y)$. Assume
for every generic point $\xi $ of an irreducible component of $Z(s)$ of $\delta $-dimension $k$ the multiplication by $s$ induces an injection $\mathcal{O}_{Y, \xi } \to (\mathcal{L}|_ Y)_\xi $1.
\[ c_1(\mathcal{L}) \cap [Y]_{k + 1} = [Z(s)]_ k \]
\[ [Y]_{k + 1} = \sum n_ i[Y_ i] \]
where $Y_ i \subset Y$ are the irreducible components of $Y$ of $\delta $-dimension $k + 1$ and $n_ i > 0$. By assumption the restriction $s_ i = s|_{Y_ i} \in \Gamma (Y_ i, \mathcal{L}|_{Y_ i})$ is not zero, and hence is a regular section. By Lemma 42.24.2 we see that $[Z(s_ i)]_ k$ represents $c_1(\mathcal{L}|_{Y_ i})$. Hence by definition
\[ c_1(\mathcal{L}) \cap [Y]_{k + 1} = \sum n_ i[Z(s_ i)]_ k \]
Thus the result follows from Lemma 42.25.3. $\square$
[1] For example, this holds if $s$ is a regular section of $\mathcal{L}|_ Y$.
Comment #5374 by Jonathan Ng on July 06, 2020 at 16:29
Hi, not sure if I am right but think you meant
Z_{k+1}(X)
Z_{k-1}(X)
in 02SP.
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The metric thread callout
How to determine metric thread sizes
How to use the metric thread dimensions calculator
This thread calculator will help you find the different nut and bolt thread dimensions under ISO metric standards. In this metric thread calculator, you will learn:
How to determine metric thread sizes;
How to read and understand metric thread callouts;
The formulas for metric external thread dimensions;
The formulas for metric internal thread dimensions; and
What metric thread class means.
Let's start by learning what threads are. 🔩
Threads can mean a lot of things. In this thread calculator, we'll discuss the screw thread made out of grooves and ridges wrapped around a cylindrical or conical metal shaft in a helix pattern. The thread's helix spiral pattern acts like an inclined plane that translates rotational movement to axial movement while gaining some mechanical advantage.
As a result, a screw thread offers a locking feature that would require a good amount of shear stress to break, making it great for fastening objects together. This kind of thread can be an external thread (like on a screw or bolt) or an internal thread (like what we see inside a nut). We call these threaded materials threaded fasteners.
Though we can also see threads in other applications, like bottle caps, bulbs, pipes and connectors, and even worm gears, this thread calculator focuses specifically on metric threads. Metric threads have symmetrical V-shaped threads that form 60º-angle grooves and have basic dimensions shown below:
The major diameter is the largest diameter of an external or internal thread. We also call this diameter the gross diameter or nominal diameter of the thread. On the other hand, the minor diameter is the smallest diameter of an external or internal thread. We also call it the root or core diameter of the thread.
In between these two diameters is the pitch diameter. The pitch diameter is the thread's diameter in which the thickness of the thread and the space between two threads are equal. We can approximate the pitch diameter's value by taking the average of the minor and major diameters of the thread.
Thread pitch, or simply pitch is the distance between a point on one thread to the corresponding point on the thread adjacent to it (for example, the measurement from crest to crest or from root to root).
Understanding your metric threads starts with knowing how to read metric thread callouts. A thread callout is the same as a threaded fastener's label when you see them in a hardware store. Any thread standard has its own thread callout, and here is an example of a metric thread callout:
\text{M10 × 1 × 25}
We read this callout as "M-ten by one by twenty-five" where:
\text{M}
stands for "metric" and means that the threaded fastener follows the ISO Metric Standard;
10
in this case, which is the number affixed to the letter M, represents the basic major diameter of the thread;
1
represents the thread pitch; and
25
is the length of the shaft.
All the measurements in a metric thread callout are in millimeters.
Now that we can read metric thread callouts, we can start calculating our threads' different dimensions. Knowing how to determine metric thread sizes gives us another way to distinguish one threaded fastener from another. As a guide, we are going to use this illustration:
From the illustration above, we can see that we can use fractions of H to determine the different metric thread dimensions given the basic major diameter (
\small{d}
). But what is H?
\small{H}
is the height of the thread's fundamental triangle. The fundamental triangle is the projected equilateral triangle we can form from the thread's cross-section. We can get its height measurement using this formula:
H = P \times \frac{\sqrt{3}}{2}
By analyzing the same diagram above, we can then formulate the following equations to find the other basic diameters of an external thread:
d_1 = d - 2 \times \frac{5}{8} \times H
d_2 = d - 2 \times \frac{3}{8} \times H
d
– Basic major diameter;
H
– Height of fundamental triangle;
d_1
– Basic minor diameter; and
d_2
– Basic pitch diameter.
💡 As you may have noticed, we use the lowercase letter
d
to designate diameters for external threads. For convenience, we use the uppercase letter
D
for metric internal thread dimensions. That is,
d
for external thread major diameter, and
D
for internal thread major diameter. Coincidentally, the measurement for these basic major diameters is equal, such that:
d = D
As a rule of thumb, an internal thread of a given specified size should have the same basic diameters as an external thread of the same specified size. This means that
d_1 = D_1
d_2 = D_2
D_1
D_2
are the internal thread's basic minor diameter and basic pitch diameter, respectively.
From what we have already discussed, the basic diameters of a pair of external and internal threads precisely match each other. However, most of the time we provide some allowances to our threads to have some leeway when fitting them together. This is where metric thread classes come into play.
Metric thread classes indicate a metric thread's tolerance grade and tolerance position. Each tolerance grade and position have corresponding equations that will tell us how much deeper we should cut or cold-roll the threads to achieve those allowances we want.
However, there are limitations as to how much we could deviate from the basic thread dimensions. Therefore, it's good to know our thread's allowable maximum and minimum diameters.
Here are the formulas that we use to find those maximum and minimum diameters:
For external thread major diameter (d):
d_\text{max} = d + es
d_\text{min} = d + es - T_\text{d}
For external thread minor diameter (d₁):
d_\text{1max} = d_1 - es - 2 \times y
d_\text{1min} = d_1 - es - 2 \times z
For external thread pitch diameter (d₂):
d_\text{2max} = d_2 + es
d_\text{2min} = d_2 + es - T_\text{d2}
For internal thread major diameter (D):
D_\text{min} = D + EI
For internal thread minor diameter (D₁):
D_\text{1min} = D_1 + EI
D_\text{1max} = D_1 + EI + T_\text{D1}
For internal thread pitch diameter (D₂):
D_\text{2min} = D_2 + EI
D_\text{2max} = D_2 + EI + T_\text{D2}
es
EI
– Fundamental deviations, upper and lower deviations, respectively;
T
– Tolerances for their corresponding diameters (i.e.,
T_\text{d2}
is the external thread pitch diameter (
d_2
) tolerance); and
y
z
– Adjustment values for maximum minor diameter and mininmum minor diameters, respectively.
Here are the equations we use to find the fundamental deviations depending on what tolerance position the threads need to be:
Fundamental deviations for external threads
\small{es = -(50 + 11 \times P) / 1000}
\small{es = -(30 + 11 \times P) / 1000}
\small{es = -(15 + 11 \times P) / 1000}
\small{es = 0}
Fundamental deviations for internal threads
\small{EI = (15 + 11 \times P) / 1000}
\small{EI = 0}
On the other hand, here are the different general equations that we use to determine the tolerances based on the tolerance grade of the thread:
\small{T_\text{d}(n) = k \times (180 \times P^{2/3} - 3.15 \times P^{-1/2}) / 1000}
\small{T_\text{d2}(n) = k \times (90 \times P^{0.4} \times d^{0.1}) / 1000}
\small{T_\text{D1}(n) = k \times (433 \times P - 190 \times P^{1.22}) / 1000}
(for 0.2 mm ≤ P ≤ 0.8 mm);
\small{T_\text{D1}(n) = k \times (230 \times P^{0.7}) / 1000}
(for P ≥ 1.0 mm);
\small{T_\text{D2}(n) = k \times (90 \times P^{0.4} \times d^{0.1}) / 1000}
In these equations, the value for
k
depends on which tolerance grade the threads use, indicated by
n
. A bolt with a tolerance grade of
n = 6
for its major diameter should use
k = 1.0
T_\text{D1}
. Here are the other values of k in the table below:
for Td(n)
For the minor diameter, you can also see that we've got there two other adjustment variables,
y
z
, incorporated in the equation. Below are the equations we use for that:
\footnotesize \begin{align*} y &= R_\text{min} \times \bigg[1 - \cos\bigg(\frac{\pi}{3} -\\ &\qquad\cos^{-1}\left(1 - \frac{T_\text{d2}}{(4 \times R_\text{min})}\right)\!\!\bigg)\bigg]\\\\ z &= \frac{H}{4} + \frac{T_\text{d2}}{2} - R_\text{min} \end{align*}
R_\text{min}
\frac{P}{8}
For threads with specified thread class details, we can add their codes at the end of the thread callout. For example, an external thread labeled as
\text{M30 × 2 × 40 - 5g6g}
means that it has a pitch diameter tolerance grade of
5
and a major diameter tolerance grade of
6
, both of which follow a
\text{g}
tolerance position.
Usually, we only perform manual calculations to find a thread's basic diameters. As you may have found out, it would take many extra steps to determine the diameter limits. And that is where our metric thread dimensions calculator comes in very handy! Here are steps to follow in using our tool:
Choose which dimensions you want to calculate – metric external thread dimensions, internal thread dimensions, or both.
Select the thread pitch of your threads from the drop-down menu.
Input the basic major diameter of your threaded fastener. You can get this measurement from the thread callout or measure this dimension using a pair of precision calipers.
Finally, from the Tolerance class details section of our metric thread calculator, select your fastener's different tolerance grades and tolerance positions, whichever is known to you. You can also see this detail on your metric thread callout.
Once you've done the steps above, you'll immediately see not just the basic diameters but also the maximum and minimum diameters of your thread.
You can also click on the Advanced mode button below our metric thread calculator if you wish to explore the different preliminary measurements used in the calculations, such as the height of the fundamental triangle, deviations, tolerance values, and adjustments.
How do I calculate the pitch diameter of a thread?
Let's say we have a bolt with a basic major diameter (d) of 10 mm and a thread pitch (P) of 1.5 mm. To find its pitch diameter:
First, calculate the height of its fundamental triangle (H) using H = P × (√3) / 2. H = 1.5 mm × (√3) / 2 = 1.299 mm.
Then, obtain for the pitch diameter (d₂) using d₂ = d - 2 × (3 / 8) × H, d₂ = 10 mm - 2 × (3 / 8) × 1.299 mm = 9.026 mm.
How do I calculate the minor diameter of a thread?
Let's say we have a bolt with a basic major diameter (d) of 20 mm and a thread pitch (P) of 2 mm. To find its minor diameter:
Calculate the height of its fundamental triangle (H) using H = P × (√3) / 2. H = 2 mm × (√3) / 2 = 1.732 mm.
Obtain for the minor diameter (d₁) using d₂ = d - 2 × (5 / 8) × H, d₂ = 20 mm - 2 × (5 / 8) × 1.732 mm = 17.835 mm.
An M6 thread has a basic major diameter of 6 mm. Typically, metric thread callouts also come with the thread pitch, such that an M6 × 0.1 threaded bolt has a basic major diameter of 6 mm and a thread pitch of 0.1 mm. When purchasing a threaded fastener, make sure you have both the diameter and the thread pitch. Without one of those measurements, you could buy the wrong-sized hardware.
What is the nominal diameter of a thread?
The nominal diameter of a thread is the diameter for which a thread is known. We use the nominal diameter for the general identification of threads. For example, a thread with a callout of M10 × 1.5 has a nominal diameter of 10 mm. Generally, the nominal diameter of a thread is the same as the basic major diameter of the thread.
Basic minor diameter
Note: To calculate the basic major diameter, make sure it is blank before entering your basic pitch or minor diameter.
Pitch diameter tolerance grade
Major diameter tolerance grade
External thread limits
Maximum major diameter
Minimum major diameter
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Element distinctness problem - Wikipedia
In computational complexity theory, the element distinctness problem or element uniqueness problem is the problem of determining whether all the elements of a list are distinct.
It is a well studied problem in many different models of computation. The problem may be solved by sorting the list and then checking if there are any consecutive equal elements; it may also be solved in linear expected time by a randomized algorithm that inserts each item into a hash table and compares only those elements that are placed in the same hash table cell.[1]
Several lower bounds in computational complexity are proved by reducing the element distinctness problem to the problem in question, i.e., by demonstrating that the solution of the element uniqueness problem may be quickly found after solving the problem in question.
1 Decision tree complexity
2 Quantum complexity
3 Generalization: Finding repeated elements
Decision tree complexity[edit]
It is known that, for lists of numbers, the problem's time complexity is Θ(n log n), i.e., both the upper and lower bounds on its time complexity are of order of the linearithmic function in the algebraic decision tree model of computation,[2] a model of computation in which the elements may not be used to index the computer's memory (as in the hash table solution) but may only be accessed by computing and comparing simple algebraic functions of their values. In other words, an asymptotically optimal algorithm of linearithmic time complexity is known for this model. The algebraic computation tree model basically means that the allowable algorithms are only the ones that can perform polynomial operations of bounded degree on the input data and comparisons of the results of these computations.
The same lower bound was later proved for the randomized algebraic decision tree model.[3][4]
Quantum complexity[edit]
It is also known that quantum algorithms can solve this problem faster, in
{\displaystyle \Theta \left(n^{2/3}\right)}
queries. The optimal algorithm is by Andris Ambainis.[5] Yaoyun Shi first proved a tight lower bound when the size of the range is sufficiently large.[6] Ambainis[7] and Kutin[8] independently (and via different proofs) extended his work to obtain the lower bound for all functions.
Generalization: Finding repeated elements[edit]
Elements that occur more than n/k times in a multiset of size n may be found in time O(n log k). The element distinctness problem is a special case of k=n. This algorithm is optimal under the decision tree model of computation.[9]
The algorithm is a generalization of the one for a special case of k=2 (the Boyer–Moore majority vote algorithm), which had a rather convoluted history of publication.[10]
The above algorithms rely only on the test of identity of the elements. If sorting is allowed, previously known order statistics finding algorithms may be exploited. For example, for k=2, a median may be found first in linear time, and then it may be easily tested whether there are more than n/2 median elements. However the above algorithms require fewer comparisons than the order statistics algorithms.[10]
^ Gil, J.; Meyer auf der Heide, F.; Wigderson, A. (1990), "Not all keys can be hashed in constant time", Proc. 22nd ACM Symposium on Theory of Computing, pp. 244–253, doi:10.1145/100216.100247 .
^ Ben-Or, Michael (1983), "Lower bounds for algebraic computation trees", Proc. 15th ACM Symposium on Theory of Computing, pp. 80–86, doi:10.1145/800061.808735 .
^ Grigoriev, Dima; Karpinski, Marek; Heide, Friedhelm Meyer; Smolensky, Roman (1996), "A lower bound for randomized algebraic decision trees", Computational Complexity, 6 (4): 357, doi:10.1007/BF01270387 .
^ Grigoriev, Dima (1999), "Complexity lower bounds for randomized computation trees over zero characteristic fields", Computational Complexity, 8 (4): 316–329, doi:10.1007/s000370050002 .
^ Ambainis, Andris (2007), "Quantum walk algorithm for element distinctness", SIAM Journal on Computing, 37 (1): 210–239, arXiv:quant-ph/0311001, doi:10.1137/S0097539705447311
^ Shi, Y. (2002). Quantum lower bounds for the collision and the element distinctness problems. Proceedings of the 43rd Symposium on Foundations of Computer Science. pp. 513–519. arXiv:quant-ph/0112086. doi:10.1109/SFCS.2002.1181975.
^ Misra, J.; Gries, D. (1982), "Finding repeated elements", Science of Computer Programming, 2 (2): 143–152, doi:10.1016/0167-6423(82)90012-0, hdl:1813/6345 .
^ a b Boyer, R. S.; Moore, J S. (1991), "MJRTY - A Fast Majority Vote Algorithm", in Boyer, R. S. (ed.), Automated Reasoning: Essays in Honor of Woody Bledsoe, Automated Reasoning Series, Dordrecht, The Netherlands: Kluwer Academic Publishers, pp. 105–117 .
Retrieved from "https://en.wikipedia.org/w/index.php?title=Element_distinctness_problem&oldid=959034236"
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Section 59.9 (03NB): Presheaves—The Stacks project
Section 59.9: Presheaves (cite)
59.9 Presheaves
A reference for this section is Sites, Section 7.2.
Definition 59.9.1. Let $\mathcal{C}$ be a category. A presheaf of sets (respectively, an abelian presheaf) on $\mathcal{C}$ is a functor $\mathcal{C}^{opp} \to \textit{Sets}$ (resp. $\textit{Ab}$).
Terminology. If $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then elements of $\mathcal{F}(U)$ are called sections of $\mathcal{F}$ over $U$. For $\varphi : V \to U$ in $\mathcal{C}$, the map $\mathcal{F}(\varphi ) : \mathcal{F}(U) \to \mathcal{F}(V)$ is called the restriction map and is often denoted $s \mapsto s|_ V$ or sometimes $s \mapsto \varphi ^*s$. The notation $s|_ V$ is ambiguous since the restriction map depends on $\varphi $, but it is a standard abuse of notation. We also use the notation $\Gamma (U, \mathcal{F}) = \mathcal{F}(U)$.
Saying that $\mathcal{F}$ is a functor means that if $W \to V \to U$ are morphisms in $\mathcal{C}$ and $s \in \Gamma (U, \mathcal{F})$ then $(s|_ V)|_ W = s |_ W$, with the abuse of notation just seen. Moreover, the restriction mappings corresponding to the identity morphisms $\text{id}_ U : U \to U$ are the identity.
The category of presheaves of sets (respectively of abelian presheaves) on $\mathcal{C}$ is denoted $\textit{PSh} (\mathcal{C})$ (resp. $\textit{PAb} (\mathcal{C})$). It is the category of functors from $\mathcal{C}^{opp}$ to $\textit{Sets}$ (resp. $\textit{Ab}$), which is to say that the morphisms of presheaves are natural transformations of functors. We only consider the categories $\textit{PSh}(\mathcal{C})$ and $\textit{PAb}(\mathcal{C})$ when the category $\mathcal{C}$ is small. (Our convention is that a category is small unless otherwise mentioned, and if it isn't small it should be listed in Categories, Remark 4.2.2.)
Example 59.9.2. Given an object $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, we consider the functor
\[ \begin{matrix} h_ X : & \mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\ & U & \longmapsto & h_ X(U) = \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U, X) \\ & V \xrightarrow {\varphi } U & \longmapsto & \varphi \circ - : h_ X(U) \to h_ X(V). \end{matrix} \]
It is a presheaf, called the representable presheaf associated to $X$. It is not true that representable presheaves are sheaves in every topology on every site.
Lemma 59.9.3 (Yoneda). Let $\mathcal{C}$ be a category, and $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. There is a natural bijection
\[ \begin{matrix} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X, Y) & \longrightarrow & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})} (h_ X, h_ Y) \\ \psi & \longmapsto & h_\psi = \psi \circ - : h_ X \to h_ Y. \end{matrix} \]
Proof. See Categories, Lemma 4.3.5. $\square$
In the terminology subsubsection, the
\mathcal{F}(\varphi)
is going the wrong way since contravariant out of
\mathcal{F}
/covariant out of
\mathcal{F}^{\mathrm{opp}}
Above comment should say "...contravariant out of
\mathcal{C}
\mathcal{C}^{\mathrm{opp}}
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Brayton cycle — Wikipedia Republished // WIKI 2
{\displaystyle c=}
{\displaystyle T}
{\displaystyle \partial S}
{\displaystyle N}
{\displaystyle \partial T}
{\displaystyle \beta =-}
{\displaystyle 1}
{\displaystyle \partial V}
{\displaystyle V}
{\displaystyle \partial p}
{\displaystyle \alpha =}
{\displaystyle 1}
{\displaystyle \partial V}
{\displaystyle V}
{\displaystyle \partial T}
{\displaystyle U(S,V)}
{\displaystyle H(S,p)=U+pV}
{\displaystyle A(T,V)=U-TS}
{\displaystyle G(T,p)=H-TS}
The Brayton cycle is a thermodynamic cycle that describes the operation of certain heat engines that have air or some other gas as their working fluid. The original Brayton engines used a piston compressor and piston expander, but modern gas turbine engines and airbreathing jet engines also follow the Brayton cycle. Although the cycle is usually run as an open system (and indeed must be run as such if internal combustion is used), it is conventionally assumed for the purposes of thermodynamic analysis that the exhaust gases are reused in the intake, enabling analysis as a closed system.
The engine cycle is named after George Brayton (1830–1892), the American engineer who developed it originally for use in piston engines, although it was originally proposed and patented by Englishman John Barber in 1791.[1] It is also sometimes known as the Joule cycle. The reversed Joule cycle uses an external heat source and incorporates the use of a regenerator. One type of Brayton cycle is open to the atmosphere and uses an internal combustion chamber; and another type is closed and uses a heat exchanger.
BRAYTON CYCLE | Animation
Thermodynamic Cycles - Brayton Cycle (Part 4 of 4)
Brayton Cycle - Gas Power Cycles - Thermodynamics
2 Early gas turbine history
4 Methods to increase power
5 Methods to improve efficiency
6.1 Closed Brayton cycle
6.2 Solar Brayton cycle
6.3 Reverse Brayton cycle
6.4 Inverted Brayton cycle
In 1872, George Brayton applied for a patent for his "Ready Motor", a reciprocating heat engine operating on a gas power cycle. The engine was a two-stroke and produced power on every revolution. Brayton engines used a separate piston compressor and piston expander, with compressed air heated by internal fire as it entered the expander cylinder. The first versions of the Brayton engine were vapor engines which mixed fuel with air as it entered the compressor by means of a heated-surface carburetor.[2] The fuel / air was contained in a reservoir / tank and then it was admitted to the expansion cylinder and burned. As the fuel/air mixture entered the expansion cylinder, it was ignited by a pilot flame. A screen was used to prevent the fire from entering or returning to the reservoir. In early versions of the engine, this screen sometimes failed and an explosion would occur. In 1874, Brayton solved the explosion problem by adding the fuel just prior to the expander cylinder. The engine now used heavier fuels such as kerosene and fuel oil. Ignition remained a pilot flame.[3] Brayton produced and sold "Ready Motors" to perform a variety of tasks like water pumping, mill operation, running generators, and marine propulsion. The "Ready Motors" were produced from 1872 to sometime in the 1880s; several hundred such motors were likely produced during this time period. Brayton licensed the design to Simone in the UK. Many variations of the layout were used; some were single-acting and some were double-acting. Some had under walking beams; others had overhead walking beams. Both horizontal and vertical models were built. Sizes ranged from less than one to over 40 horsepower. Critics of the time claimed the engines ran smoothly and had a reasonable efficiency.[3]
Brayton-cycle engines were some of the first internal combustion engines used for motive power. In 1875, John Holland used a Brayton engine to power the world's first self-propelled submarine (Holland boat #1). In 1879, a Brayton engine was used to power a second submarine, the Fenian Ram. John Philip Holland's submarines are preserved in the Paterson Museum in the Old Great Falls Historic District of Paterson, New Jersey.[4]
In 1878, George B. Selden patented the first internal combustion automobile.[5] Inspired by the internal combustion engine invented by Brayton displayed at the Centennial Exposition in Philadelphia in 1876, Selden patented a four-wheel car working on a smaller, lighter, multicylinder version. He then filed a series of amendments to his application which stretched out the legal process, resulting in a delay of 16 years before the patent[5] was granted on November 5, 1895. In 1903, Selden sued Ford for patent infringement and Henry Ford fought the Selden patent until 1911. Selden had never actually produced a working car, so during the trial, two machines were constructed according to the patent drawings. Ford argued his cars used the four-stroke Alphonse Beau de Rochas cycle or Otto cycle and not the Brayton-cycle engine used in the Selden auto. Ford won the appeal of the original case.[6]
Rudolf Diesel originally proposed a very high compression, constant-temperature cycle where the heat of compression would exceed the heat of combustion, but after several years of experiments, he realized that the constant-temperature cycle would not work in a piston engine. Early Diesel engines use an air blast system which was pioneered by Brayton in 1890. Consequently, these early engines use the constant-pressure cycle.[9]
Examples of Brayton engines
Modern Brayton engines are almost always a turbine type, although Brayton only made piston engines. In the original 19th-century Brayton engine, ambient air is drawn into a piston compressor, where it is compressed; ideally an isentropic process. The compressed air then runs through a mixing chamber where fuel is added, an isobaric process. The pressurized air and fuel mixture is then ignited in an expansion cylinder and energy is released, causing the heated air and combustion products to expand through a piston/cylinder, another ideally isentropic process. Some of the work extracted by the piston/cylinder is used to drive the compressor through a crankshaft arrangement.
Gas turbines are also Brayton engines, with three components: a gas compressor, a burner (or combustion chamber), and an expansion turbine.
isentropic process – ambient air is drawn into the compressor, where it is pressurized.
isobaric process – the compressed air then runs through a combustion chamber, where fuel is burned, heating that air—a constant-pressure process, since the chamber is open to flow in and out.
adiabatic process – compression
Since neither the compression nor the expansion can be truly isentropic, losses through the compressor and the expander represent sources of inescapable working inefficiencies. In general, increasing the compression ratio is the most direct way to increase the overall power output of a Brayton system.[11]
The efficiency of the ideal Brayton cycle is
{\displaystyle \eta =1-{\frac {T_{1}}{T_{2}}}=1-\left({\frac {P_{1}}{P_{2}}}\right)^{(\gamma -1)/\gamma }}
{\displaystyle \gamma }
is the heat capacity ratio.[12] Figure 1 indicates how the cycle efficiency changes with an increase in pressure ratio. Figure 2 indicates how the specific power output changes with an increase in the gas turbine inlet temperature for two different pressure ratio values.
Reheat, wherein the working fluid—in most cases air—expands through a series of turbines, then is passed through a second combustion chamber before expanding to ambient pressure through a final set of turbines, has the advantage of increasing the power output possible for a given compression ratio without exceeding any metallurgical constraints (typically about 1000 °C). The use of an afterburner for jet aircraft engines can also be referred to as "reheat"; it is a different process in that the reheated air is expanded through a thrust nozzle rather than a turbine. The metallurgical constraints are somewhat alleviated, enabling much higher reheat temperatures (about 2000 °C). Reheat is most often used to improve the specific power (per throughput of air), and is usually associated with a drop in efficiency; this effect is especially pronounced in afterburners due to the extreme amounts of extra fuel used.
Increasing pressure ratio, as Figure 1 above shows, increasing the pressure ratio increases the efficiency of the Brayton cycle. This is analogous to the increase of efficiency seen in the Otto cycle when the compression ratio is increased. However, practical limits occur when it comes to increasing the pressure ratio. First of all, increasing the pressure ratio increases the compressor discharge temperature. This can cause the temperature of the gases leaving the combustor to exceed the metallurgical limits of the turbine. Also, the diameter of the compressor blades becomes progressively smaller in higher pressure stages of the compressor. Because the gap between the blades and the engine casing increases in size as a percentage of the compressor blade height as the blades get smaller in diameter, a greater percentage of the compressed air can leak back past the blades in higher pressure stages. This causes a drop in compressor efficiency, and is most likely to occur in smaller gas turbines (since blades are inherently smaller to begin with). Finally, as can be seen in Figure 1, the efficiency levels off as pressure ratio increases. Hence, little gain is expected by increasing the pressure ratio further if it is already at a high level.
A Brayton engine also forms half of the combined cycle system, which combines with a Rankine engine to further increase overall efficiency. However, although this increases overall efficiency, it does not actually increase the efficiency of the Brayton cycle itself.
w high-temperature heat exchanger
~ mechanical load, e.g. electric generator
A closed Brayton cycle recirculates the working fluid; the air expelled from the turbine is reintroduced into the compressor, this cycle uses a heat exchanger to heat the working fluid instead of an internal combustion chamber. The closed Brayton cycle is used, for example, in closed-cycle gas turbine and space power generation.
A Brayton cycle that is driven in reverse, via net work input, and when air is the working fluid, is the gas refrigeration cycle or Bell Coleman cycle. Its purpose is to move heat, rather than produce work. This air-cooling technique is used widely in jet aircraft for air conditioning systems using bleed air tapped from the engine compressors. It is also used in the LNG industry where the largest reverse Brayton cycle is for subcooling LNG using 86 MW of power from a gas turbine-driven compressor and nitrogen refrigerant.[19]
Main article: Inverted Brayton cycle
Wikimedia Commons has media related to Brayton cycle.
Britalus rotary engine
Heat engine
^ according to Gas Turbine History Archived June 3, 2010, at the Wayback Machine
^ Frank A. Taylor (1939), "Catalog of the Mechanical Collections Of The Division Of Engineering", United States National Museum Bulletin 173, United States Government Printing Office, p. 147
^ a b US 125166, Brayton, George B., "Improvement in gas-engines", published 1872-04-02
^ "Holland Submarines". Paterson Friends of the Great Falls. Archived from the original on 2007-08-12. Retrieved 2007-07-29.
^ a b US 549160, Selden, George B., "Road-engine", published 1895-11-05
^ "Weird & Wonderful Patents - Selden Patent". www.bpmlegal.com.
^ US 432114, Brayton, George B., "Gas and air engine", published 1890-07-15
^ US 432260, Brayton, George B., "Hydrocarbon-engine", published 1890-07-15
^ "Diesel Engines". www.dieselnet.com.
^ NASA/Glenn Research Center (May 5, 2015). "PV and TS Diagrams". www.grc.nasa.gov.
^ "Brayton Thermodynamic Cycle".
^ Solhyco.com Archived 2011-12-29 at the Wayback Machine Retrieved 2012-01-09
^ Solugas.EU Archived 2014-12-25 at the Wayback Machine Retrieved 2014-11-09
Today in Science article on Brayton Engine
Test and evaluation of a solar powered gas turbine system
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1 Global Energy Parliament, Trivandrum, Kerala, India.
2 Atomic Energy Commission, Cadarache, France.
Abstract: Applying the I-Theory, this paper gives a new outlook about the concept of Entropy and Negentropy. Using S∞ particle as 100% repelling energy and A1 particle as the starting point of attraction, we are able to define Entropy and Negentropy on the quantum level. As the I-Theory explains that repulsion force is driven by Weak Force and attraction is driven by Strong Force, we also analyze Entropy and Negentropy in terms of the Fundamental Forces.
Keywords: I-Theory, I-Particle, Entropy, Negentropy, Syntropy, Intelligence, Black Matter, White Matter, Red Matter, Gravitation, Strong Force, Weak Force, Free Energy
\Delta S=\frac{Q}{T}
\Delta S\ge 0
S={k}_{B}\mathrm{ln}\left(\Omega \right)
\Omega =N\ast P
{\Omega }_{2}>{\Omega }_{1}
\left\{\begin{array}{l}{S}_{2}={k}_{B}\mathrm{ln}\left({\Omega }_{2}\right)\\ {S}_{1}={k}_{B}\mathrm{ln}\left({\Omega }_{1}\right)\end{array}
{S}_{2}>{S}_{1}
{S}_{BH}=\frac{{k}_{B}\ast A}{4\ast {l}_{p}^{2}}
{l}_{p}=\sqrt{G\hslash /{c}^{3}}
J={S}_{\mathrm{max}}-S=-\varphi =-{k}_{B}\mathrm{ln}\left(Z\right)
\varphi
{\Omega }_{{S}_{\infty }}=Max\left(\Omega \right)
{S}_{{S}_{\infty }}={S}_{Max}
{\Omega }_{{S}_{\infty }}\gg {\Omega }_{{A}_{1}}
{S}_{{S}_{\infty }}>{S}_{{A}_{1}}
Cite this paper: Isa, H. and Dumas, C. (2020) Entropy and Negentropy Principles in the I-Theory. Journal of High Energy Physics, Gravitation and Cosmology, 6, 259-273. doi: 10.4236/jhepgc.2020.62020.
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Transfer learning - Wikipedia
Applying previously-learned knowledge to new problems
Transfer learning (TL) is a research problem in machine learning (ML) that focuses on storing knowledge gained while solving one problem and applying it to a different but related problem.[1] For example, knowledge gained while learning to recognize cars could apply when trying to recognize trucks. This area of research bears some relation to the long history of psychological literature on transfer of learning, although practical ties between the two fields are limited. From the practical standpoint, reusing or transferring information from previously learned tasks for the learning of new tasks has the potential to significantly improve the sample efficiency of a reinforcement learning agent.[2]
In 1976, Stevo Bozinovski and Ante Fulgosi published a paper explicitly addressing transfer learning in neural networks training.[3][4] The paper gives a mathematical and geometrical model of transfer learning. In 1981, a report was given on the application of transfer learning in training a neural network on a dataset of images representing letters of computer terminals. Both positive and negative transfer learning was experimentally demonstrated.[5]
In 1993, Lorien Pratt published a paper on transfer in machine learning, formulating the discriminability-based transfer (DBT) algorithm.[6]
In 1997, Pratt and Sebastian Thrun guest edited a special issue of Machine Learning devoted to transfer learning,[7] and by 1998, the field had advanced to include multi-task learning,[8] along with a more formal analysis of its theoretical foundations.[9] Learning to Learn,[10] edited by Thrun and Pratt, is a 1998 review of the subject.
Transfer learning has also been applied in cognitive science, with Pratt also guest editing an issue of Connection Science on reuse of neural networks through transfer in 1996.[11]
Andrew Ng said in his NIPS 2016 tutorial [12][13][14] that TL will be the next driver of ML commercial success after supervised learning to highlight the importance of TL.
The definition of transfer learning is given in terms of domains and tasks. A domain
{\displaystyle {\mathcal {D}}}
consists of: a feature space
{\displaystyle {\mathcal {X}}}
and a marginal probability distribution
{\displaystyle P(X)}
{\displaystyle X=\{x_{1},...,x_{n}\}\in {\mathcal {X}}}
. Given a specific domain,
{\displaystyle {\mathcal {D}}=\{{\mathcal {X}},P(X)\}}
, a task consists of two components: a label space
{\displaystyle {\mathcal {Y}}}
and an objective predictive function
{\displaystyle f:{\mathcal {X}}\rightarrow {\mathcal {Y}}}
{\displaystyle f}
is used to predict the corresponding label
{\displaystyle f(x)}
of a new instance
{\displaystyle x}
. This task, denoted by
{\displaystyle {\mathcal {T}}=\{{\mathcal {Y}},f(x)\}}
, is learned from the training data consisting of pairs
{\displaystyle \{x_{i},y_{i}\}}
{\displaystyle x_{i}\in X}
{\displaystyle y_{i}\in {\mathcal {Y}}}
Given a source domain
{\displaystyle {\mathcal {D}}_{S}}
and learning task
{\displaystyle {\mathcal {T}}_{S}}
, a target domain
{\displaystyle {\mathcal {D}}_{T}}
{\displaystyle {\mathcal {T}}_{T}}
{\displaystyle {\mathcal {D}}_{S}\neq {\mathcal {D}}_{T}}
{\displaystyle {\mathcal {T}}_{S}\neq {\mathcal {T}}_{T}}
, transfer learning aims to help improve the learning of the target predictive function
{\displaystyle f_{T}(\cdot )}
{\displaystyle {\mathcal {D}}_{T}}
using the knowledge in
{\displaystyle {\mathcal {D}}_{S}}
{\displaystyle {\mathcal {T}}_{S}}
Algorithms are available for transfer learning in Markov logic networks[16] and Bayesian networks.[17] Transfer learning has also been applied to cancer subtype discovery,[18] building utilization,[19][20] general game playing,[21] text classification,[22][23] digit recognition,[24] medical imaging and spam filtering.[25]
In 2020 it was discovered that, due to their similar physical natures, transfer learning is possible between Electromyographic (EMG) signals from the muscles when classifying the behaviors of Electroencephalographic (EEG) brainwaves from the gesture recognition domain to the mental state recognition domain. It was also noted that this relationship worked vice versa, showing that EEG can likewise be used to classify EMG in addition.[26] The experiments noted that the accuracy of neural networks and convolutional neural networks were improved[27] through transfer learning both at the first epoch (prior to any learning, ie. compared to standard random weight distribution) and at the asymptote (the end of the learning process). That is, algorithms are improved by exposure to another domain. Moreover, the end-user of a pre-trained model can change the structure of fully-connected layers to achieve superior performance.[28]
In the domain of machine learning on code,[29] it has been shown that transfer learning is useful for automatically repairing security vulnerabilities.[30]
^ West, Jeremy; Ventura, Dan; Warnick, Sean (2007). "Spring Research Presentation: A Theoretical Foundation for Inductive Transfer". Brigham Young University, College of Physical and Mathematical Sciences. Archived from the original on 2007-08-01. Retrieved 2007-08-05.
^ George Karimpanal, Thommen; Bouffanais, Roland (2019). "Self-organizing maps for storage and transfer of knowledge in reinforcement learning". Adaptive Behavior. 27 (2): 111–126. arXiv:1811.08318. doi:10.1177/1059712318818568. ISSN 1059-7123. S2CID 53774629.
^ Stevo. Bozinovski and Ante Fulgosi (1976). "The influence of pattern similarity and transfer learning upon the training of a base perceptron B2." (original in Croatian) Proceedings of Symposium Informatica 3-121-5, Bled.
^ Stevo Bozinovski (2020) "Reminder of the first paper on transfer learning in neural networks, 1976". Informatica 44: 291–302.
^ S. Bozinovski (1981). "Teaching space: A representation concept for adaptive pattern classification." COINS Technical Report, the University of Massachusetts at Amherst, No 81-28 [available online: UM-CS-1981-028.pdf]
^ Pratt, L. Y. (1993). "Discriminability-based transfer between neural networks" (PDF). NIPS Conference: Advances in Neural Information Processing Systems 5. Morgan Kaufmann Publishers. pp. 204–211.
^ Pratt, L. Y.; Thrun, Sebastian (July 1997). "Machine Learning - Special Issue on Inductive Transfer". link.springer.com. Springer. Retrieved 2017-08-10.
^ Caruana, R., "Multitask Learning", pp. 95-134 in Thrun & Pratt 2012
^ Baxter, J., "Theoretical Models of Learning to Learn", pp. 71-95 Thrun & Pratt 2012
^ Thrun & Pratt 2012.
^ Pratt, L. (1996). "Special Issue: Reuse of Neural Networks through Transfer". Connection Science. 8 (2). Retrieved 2017-08-10.
^ NIPS 2016 tutorial: "Nuts and bolts of building AI applications using Deep Learning" by Andrew Ng, archived from the original on 2021-12-19, retrieved 2019-12-28
^ "NIPS 2016 Schedule". nips.cc. Retrieved 2019-12-28.
^ Nuts and bolts of building AI applications using Deep Learning, slides
^ a b Lin, Yuan-Pin; Jung, Tzyy-Ping (27 June 2017). "Improving EEG-Based Emotion Classification Using Conditional Transfer Learning". Frontiers in Human Neuroscience. 11: 334. doi:10.3389/fnhum.2017.00334. PMC 5486154. PMID 28701938. Material was copied from this source, which is available under a Creative Commons Attribution 4.0 International License.
^ Mihalkova, Lilyana; Huynh, Tuyen; Mooney, Raymond J. (July 2007), "Mapping and Revising Markov Logic Networks for Transfer" (PDF), Learning Proceedings of the 22nd AAAI Conference on Artificial Intelligence (AAAI-2007), Vancouver, BC, pp. 608–614, retrieved 2007-08-05
^ Niculescu-Mizil, Alexandru; Caruana, Rich (March 21–24, 2007), "Inductive Transfer for Bayesian Network Structure Learning" (PDF), Proceedings of the Eleventh International Conference on Artificial Intelligence and Statistics (AISTATS 2007), retrieved 2007-08-05
^ Hajiramezanali, E. & Dadaneh, S. Z. & Karbalayghareh, A. & Zhou, Z. & Qian, X. Bayesian multi-domain learning for cancer subtype discovery from next-generation sequencing count data. 32nd Conference on Neural Information Processing Systems (NeurIPS 2018), Montréal, Canada. arXiv:1810.09433
^ Arief-Ang, I.B.; Salim, F.D.; Hamilton, M. (2017-11-08). DA-HOC: semi-supervised domain adaptation for room occupancy prediction using CO2 sensor data. 4th ACM International Conference on Systems for Energy-Efficient Built Environments (BuildSys). Delft, Netherlands. pp. 1–10. doi:10.1145/3137133.3137146. ISBN 978-1-4503-5544-5.
^ Arief-Ang, I.B.; Hamilton, M.; Salim, F.D. (2018-12-01). "A Scalable Room Occupancy Prediction with Transferable Time Series Decomposition of CO2 Sensor Data". ACM Transactions on Sensor Networks. 14 (3–4): 21:1–21:28. doi:10.1145/3217214. S2CID 54066723.
^ Banerjee, Bikramjit, and Peter Stone. "General Game Learning Using Knowledge Transfer." IJCAI. 2007.
^ Do, Chuong B.; Ng, Andrew Y. (2005). "Transfer learning for text classification". Neural Information Processing Systems Foundation, NIPS*2005 (PDF). Retrieved 2007-08-05.
^ Rajat, Raina; Ng, Andrew Y.; Koller, Daphne (2006). "Constructing Informative Priors using Transfer Learning". Twenty-third International Conference on Machine Learning (PDF). Retrieved 2007-08-05.
^ Bickel, Steffen (2006). "ECML-PKDD Discovery Challenge 2006 Overview". ECML-PKDD Discovery Challenge Workshop (PDF). Retrieved 2007-08-05.
^ Bird, Jordan J.; Kobylarz, Jhonatan; Faria, Diego R.; Ekart, Aniko; Ribeiro, Eduardo P. (2020). "Cross-Domain MLP and CNN Transfer Learning for Biological Signal Processing: EEG and EMG". IEEE Access. Institute of Electrical and Electronics Engineers (IEEE). 8: 54789–54801. doi:10.1109/access.2020.2979074. ISSN 2169-3536.
^ Maitra, Durjoy Sen; Bhattacharya, Ujjwal; Parui, Swapan K. (August 2015). "CNN based common approach to handwritten character recognition of multiple scripts". 2015 13th International Conference on Document Analysis and Recognition (ICDAR): 1021–1025. doi:10.1109/ICDAR.2015.7333916. ISBN 978-1-4799-1805-8. S2CID 25739012.
^ Kabir, H. M. Dipu; Abdar, Moloud; Jalali, Seyed Mohammad Jafar; Khosravi, Abbas; Atiya, Amir F.; Nahavandi, Saeid; Srinivasan, Dipti (January 7, 2022). "SpinalNet: Deep Neural Network with Gradual Input". arXiv:2007.03347 – via arXiv.org. {{cite journal}}: Cite journal requires |journal= (help)
^ Allamanis, Miltiadis; Barr, Earl T.; Devanbu, Premkumar; Sutton, Charles (2019-07-31). "A Survey of Machine Learning for Big Code and Naturalness". ACM Computing Surveys. 51 (4): 1–37. doi:10.1145/3212695. ISSN 0360-0300.
^ Chen, Zimin; Kommrusch, Steve James; Monperrus, Martin (2022). "Neural Transfer Learning for Repairing Security Vulnerabilities in C Code". IEEE Transactions on Software Engineering: 1–1. doi:10.1109/TSE.2022.3147265. ISSN 1939-3520.
Thrun, Sebastian; Pratt, Lorien (6 December 2012). Learning to Learn. Springer Science & Business Media. ISBN 978-1-4615-5529-2.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Transfer_learning&oldid=1080756337"
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Chemical Formulas | Boundless Chemistry | Course Hero
Molecular formulas are a compact chemical notation that describe the type and number of atoms in a single molecule of a compound.
molecular formula: A formula that describes the exact number and type of atoms in a single molecule of a compound.
empirical formula: A formula that indicates the simplest whole number ratio of all the atoms in a molecule.
structural formula: A formula that indicates not only the number of atoms, but also their arrangement in space.
Examples of Empirical and Molecular Formulas
Structural formula of butane: The chemical structure of butane indicates not only the number of atoms, but also their arrangement in space.
Empirical formulas describe the simplest whole-number ratio of the elements in a compound.
Derive a molecule's empirical formula given its mass composition
Empirical formulas are the simplest form of notation.
The molecular formula for a compound is equal to, or a whole-number multiple of, its empirical formula.
Like molecular formulas, empirical formulas are not unique and can describe a number of different chemical structures or isomers.
To determine an empirical formula, the mass composition of its elements can be used to mathematically determine their ratio.
empirical formula: A notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
Chemists use a variety of notations to describe and summarize the atomic constituents of compounds. These notations, which include empirical, molecular, and structural formulas, use the chemical symbols for the elements along with numeric values to describe atomic composition.
Empirical formulas are the simplest form of notation. They provide the lowest whole-number ratio between the elements in a compound. Unlike molecular formulas, they do not provide information about the absolute number of atoms in a single molecule of a compound. The molecular formula for a compound is equal to, or a whole-number multiple of, its empirical formula.
Structural Formulas v. Empirical Formulas
An empirical formula (like a molecular formula) lacks any structural information about the positioning or bonding of atoms in a molecule. It can therefore describe a number of different structures, or isomers, with varying physical properties. For butane and isobutane, the empirical formula for both molecules is C2H5, and they share the same molecular formula, C4H10. However, one structural representation for butane is CH3CH2CH2CH3, while isobutane can be described using the structural formula (CH3)3CH.
Isobutane: The structural formula of isobutane.
Empirical formulas can be determined using mass composition data. For example, combustion analysis can be used in the following manner:
A CHN analyzer (an instrument that can determine the composition of a molecule) can be used to find the mass fractions of carbon, hydrogen, oxygen, and other atoms for a sample of an unknown organic compound.
Once the relative mass contributions of elements are known, this information can be converted into moles.
The empirical formula is the lowest possible whole-number ratio of the elements.
Suppose you are given a compound such as methyl acetate, a solvent commonly used in paints, inks, and adhesives. When methyl acetate was chemically analyzed, it was discovered to have 48.64% carbon (C), 8.16% hydrogen (H), and 43.20% oxygen (O). For the purposes of determining empirical formulas, we assume that we have 100 g of the compound. If this is the case, the percentages will be equal to the mass of each element in grams.
Step 1: Change each percentage to an expression of the mass of each element in grams. That is, 48.64% C becomes 48.64 g C, 8.16% H becomes 8.16 g H, and 43.20% O becomes 43.20 g O because we assume we have 100 g of the overall compound.
Step 2: Convert the amount of each element in grams to its amount in moles.
\left(\frac{48.64 { g C}}{1}\right)\left(\frac{1 { mol }}{12.01 { g C}}\right) = 4.049\ \text{mol}
\left(\frac{8.16 { g H}}{1}\right)\left(\frac{1 { mol }}{1.008 { g H}}\right) = 8.095\ \text{mol}
\left(\frac{43.20 { g O}}{1}\right)\left(\frac{1 { mol }}{16.00 { g O}}\right) = 2.7\ \text{mol}
Step 3: Divide each of the mole values by the smallest of the mole values.
\frac{4.049 { mol }}{2.7 { mol }} = 1.5
\frac{8.095 { mol }}{2.7 { mol }} = 3
\frac{2.7 { mol }}{2.7 { mol }} = 1
1.5 \times 2 = 3
3 \times 2 = 6
1 \times 2 = 2
Thus, the empirical formula of methyl acetate is C3H6O2.
The empirical formula of decane is C5H11. Its molecular weight is 142.286 g/mol. What is the molecular formula of decane?
Step 1: Calculate the molecular weight of the empirical formula (the molecular weight of C = 12.011 g/mol and H = 1.008 g/mol)
5 (12.0111 g/mol) + 11 (1.008 g/mol) = C5H11
60.055 g/mol + 11.008 g/mol = 71.143 g/mol per C5H11
Step 2: Divide the molecular weight of the molecular formula by the the molecular weight of the empirical formula to find the ratio between the two.
\frac{142.286 \ g/mol}{71.143 \ g/mol} = 2
Since the weight of the molecular formula is twice the weight of the empirical formula, there must be twice as many atoms, but in the same ratio. Therefore, if the empirical formula of decane is C5H11, the molecular formula of decane is twice that, or C10H22.
From the Molecular Formula to the Empirical Formula - YouTube: This video shows how to go from the molecular formula of a compound to its corresponding empirical formula.
An ionic formula must satisfy the octet rule for the constituent atoms and electric neutrality for the whole compound.
polyatomic ion: A set of covalently bonded atoms that have an overall charge, making them an ion.
monatomic ion: An ion made of only one atom, for example Cl-.
Ionic bonds are formed by the transfer of one or more valence electrons between atoms, typically between metals and nonmetals. The transfer of electrons allows the atoms to effectively achieve the much more stable electron configuration of having eight electrons in the outermost valence shell ( octet rule ). When sodium donates a valence electron to fluorine to become sodium fluoride, that is an example of ionic bond formation.
Formation of sodium flouride: The transfer of electrons between two atoms to create two ions that attract each other because they are oppositely charged.
Introduction To Ionic Compounds Video Series by Leah4sci - YouTube: This video explains the basics of ions.
Cation and Anion Formation - Ionic Compounds Part 2 - YouTube: This video shows you how monoatomic ions get their charge, and how to quickly find the charge of ions by looking at the periodic table.
Chemical formula. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Chemical_formula. License: CC BY-SA: Attribution-ShareAlike
Chemical formula. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Chemical_formula. License: Public Domain: No Known Copyright
Empirical formula. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Empirical_formula. License: CC BY-SA: Attribution-ShareAlike
Empirical formula. Provided by: Wikipedia. Located at: https://en.wikipedia.org/wiki/Empirical_formula. License: CC BY-SA: Attribution-ShareAlike
Chemical%252520formula. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Chemical_formula. License: Public Domain: No Known Copyright
From the Molecular Formula to the Empirical Formula - YouTube. License: Public Domain: No Known Copyright. License terms: Standard YouTube license
cation. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/cation. License: CC BY-SA: Attribution-ShareAlike
anion. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/anion. License: CC BY-SA: Attribution-ShareAlike
Introduction To Ionic Compounds Video Series by Leah4sci - YouTube. Located at: http://www.youtube.com/watch?v=aQzHVSUJBO0. License: Public Domain: No Known Copyright. License terms: Standard YouTube license
Cation and Anion Formation - Ionic Compounds Part 2 - YouTube. Located at: http://www.youtube.com/watch?v=0q5rP7wxrDk. License: Public Domain: No Known Copyright. License terms: Standard YouTube license
The symbols of the elements can be used to represe.docx
SCIENCE 115 • De La Salle University
SHS_GENERAL CHEMISTRY 1_Q1_M4_Chemical and Structural Formula.pdf
BIOCHEMIST 6303 • AMA Computer University - Quezon City
LAB-REPORT-Determination-of-Empirical-Formula.pdf
CHE INORGANIC • Batangas State University
1 GENCHEM M4.pdf
MECH 801 • Gandhi Institute of Engineering & Technology
Chemistry IA1
CHEM MISC • Austin Peay State University
CHEM 105L • Morgan State University
TALLER_1_CUÁNTICA _tercer_corte..docx
CHE MISC • Universidad de Córdoba
GENERAL-CHEMISTRY-1-week-1-2.pdf
BIOLOGY 11 • Lyceum of Aparri
Mass Relationship in Chemical Reaction_ Experimental Data and Empirical Formulas.pdf
Chem lab 01.docx
week 5 lesson 2 2nd quarter.pdf
CHE ORGANIC CH • Nueva Ecija High School
GC1-Module-Wk-2.pdf
SCIENCE 456 • Philippine Christian University
Chem-Week-1.pdf
CHEMISTRY 5.60 • Harvard University
Tutorial03aValensi EleMuatanFormalOK1 (1).docx
TEKNIK KIM 521 • Islamic University of Indonesia
STEM-7-CHEMISTRY-Worksheet-1-Week-1.pdf
CHE Chem317L • Saint Louis University, Baguio City Main Campus - Bonifacio St., Baguio City
SCH3U U1L2 hands on - writing chemical formulas worksheet.pdf
CHEMISTRY SCH3U • Cairo University
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Hoop House Calculator
What is a hoop house or greenhouse?
Greenhouse cover cost
Example: Using the hoop house calculator
The hoop house calculator is helpful to estimate the greenhouse surface area and the size of the heater you would need to maintain appropriate temperatures for your crops.
Plants often require a specific set of conditions to grow, and it is critical to saving them from changes in the environment like extreme cold. A hoop house or greenhouse maintains pressure, carbon dioxide, and lighting. It is also used in normal weather conditions for hobby or gardening purposes. A greenhouse heater regulates temperature and humidity depending on the greenhouse's size and shape.
If you know what type of greenhouse you have, you can start by entering the numbers in the calculator or reading to understand what a hoop house is and the associated costs.
A hoop house or greenhouse is a transparent confined space or a cabin-like area often made from polycarbonate or glass or polyethylene that mimics favorable conditions for the plants to grow. It is mostly used to grow crops in places where it is difficult for plants to grow due to the extreme weather or to grow delicate plants that need particular climatic conditions. A greenhouse works on the basic principles of:
Carbon dioxide enrichment.
It is made from a transparent material to receive heat inside by solar radiation and trap the heat inside it. Based on the space or ground area availability, the greenhouse could be of different shapes and sizes. Types include:
Arched roof;
Quonset; and
Different types of greenhouses — (clockwise from top): Arched roof, gable, quonset, and lean-to type greenhouse.
A hoop house often has a concrete foundation dug into the ground and a wooden frame. A transparent covering of acrylic glass, polycarbonate, or polyethylene is attached to the frame. The amount of greenhouse cover depends on the exposed surface area of the structure.
To find the surface area, first, we need to find the area of individual walls and roofs. You can ignore the area of the ground. The walls are triangular, square, or rectangle in shape with an exception in the Quonset type and arch type greenhouse where it is an arc. You can then multiply the surface area by the material unit cost to find the total material cost.
You can also use the surface area to calculate the walls' net heat transfer. This heat loss is helpful when estimating the size of the heater you need to fit in the greenhouse to maintain adequate temperature levels. To start using the calculator:
Select the type of greenhouse.
Enter the dimensions of the greenhouse, such as:
Height of sidewalls;
The greenhouse cover calculator will return the surface area of the hoop house.
Fill in the unit cost of the material.
The hoop house cost calculator returns the material cost.
Choose the type of construction material or directly enter the heat loss value.
Enter the inside temperature.
Fill in the outside temperature.
The hoop house calculator returns the heat transfer through the greenhouse walls.
Insert the heater efficiency to obtain the heater size needed.
Find the area of the gable type hoop house having the following dimensions:
10 \text{ m}
Height of sidewalls =
7.5 \text{ m}
10 \text{ m}
Width =
8 \text{ m}
The greenhouse is made up of 6 mm polyethylene having cost as
\text{\textdollar}0.5 \text{ per m}^2
. The inside and outside temperatures are
30 \text{ °C}
10 \text{ °C}
10 \text{ m}
7.5 \text{ m}
10 \text{ m}
8 \text{ m}
The surface area of the hoop house is returned as
384.34 \text{ m}^2
Fill in the unit cost of the material as
\text{\textdollar} 0.5 \text{ per m}^2
The material cost for the greenhouse is
\text{\textdollar}192.17
Choose the type of construction material as 6 mm polyethylene.
Enter the inside temperature as
30 \text{ °C}
Fill in the outside temperature as
10 \text{ °C}
The heat transfer through the greenhouse walls is
50.16 \text{ kW}
Insert the heater efficiency as 0.8 to obtain the size of input as
62.7\text{ kW}
What do you mean by hoop house?
The hoop house or greenhouse is a cabin-sized space built to grow plants in favorable conditions. It is very common in places of extreme temperature environments such as snowy areas. The hoop house controls the temperature, pressure, humidity, and carbon dioxide levels to ensure healthy plants.
There are different kinds of greenhouses such as — Gable type, Quonset, arched type, and lean-to type built based on the amount of space available and the arrangement of plants. Each design has its limitations, whether space or heat retention or cost.
How do I calculate the area for my hoop house?
To calculate the area of the hoop house:
Find the surface area of each wall and roof by multiplying their respective dimensions.
Add all the areas to obtain the total area of the hoop house.
Note: Do not consider the floor area while adding the areas.
How do I select a greenhouse heater?
To select a greenhouse heater:
Multiply the total area by the temperature difference.
Multiply the product by heat loss coefficient to obtain the heat requirement for the hoop house.
Use the heat requirement to select the capacity or size of the heater.
Total height (H₁ + H₂)
Height of sidewalls (H₁)
Polyethylene (6 mm)
W/(m²⋅K)
Heater input needed
Use this carpet calculator to estimate the costs of your dream carpet.
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"Antennas" redirects here. For other uses of "antenna", see Antenna (disambiguation).
4 Resonant antennas
4.1 Current and voltage distribution
4.3 Arrays and reflectors
5.5 Field regions
5.8.1 Antenna tuning at the antenna
5.8.2 Line matching at the radio
5.8.3 Extreme examples of loaded small antennas
7 Modeling antennas with line equations
Current and voltage distributionEdit
Arrays and reflectorsEdit
The radiant flux as a function of the distance from the transmitting antenna varies according to the inverse-square law, since that describes the geometrical divergence of the transmitted wave. For a given incoming flux, the power acquired by a receiving antenna is proportional to its effective area. This parameter compares the amount of power captured by a receiving antenna in comparison to the flux of an incoming wave (measured in terms of the signal's power density in Watts per square metre). A half-wave dipole has an effective area of
{\displaystyle 0.13\lambda ^{2}}
seen from the broadside direction. If higher gain is needed one cannot simply make the antenna larger. Due to the constraint on the effective area of a receiving antenna detailed below, one sees that for an already-efficient antenna design, the only way to increase gain (effective area) is by reducing the antenna's gain in another direction.
GainEdit
Gain is a parameter which measures the degree of directivity of the antenna's radiation pattern. A high-gain antenna will radiate most of its power in a particular direction, while a low-gain antenna will radiate over a wide angle. The antenna gain, or power gain of an antenna is defined as the ratio of the intensity (power per unit surface area)
{\displaystyle \scriptstyle I}
radiated by the antenna in the direction of its maximum output, at an arbitrary distance, divided by the intensity
{\displaystyle \scriptstyle I_{\text{iso}}}
radiated at the same distance by a hypothetical isotropic antenna which radiates equal power in all directions. This dimensionless ratio is usually expressed logarithmically in decibels, these units are called "decibels-isotropic" (dBi)
{\displaystyle G_{\text{dBi}}=10\log {I \over I_{\text{iso}}}\,}
A second unit used to measure gain is the ratio of the power radiated by the antenna to the power radiated by a half-wave dipole antenna
{\displaystyle \scriptstyle I_{\text{dipole}}}
; these units are called "decibels-dipole" (dBd)
{\displaystyle G_{\text{dBd}}=10\log {I \over I_{\text{dipole}}}\,}
{\displaystyle G_{\text{dBi}}=G_{\text{dBd}}+2.15\,}
Effective area or apertureEdit
{\displaystyle A_{\mathrm {eff} }={\lambda ^{2} \over 4\pi }\,G}
Radiation patternEdit
Impedance matchingEdit
Antenna tuning at the antennaEdit
Line matching at the radioEdit
Extreme examples of loaded small antennasEdit
Effect of groundEdit
The phase of reflection of electromagnetic waves depends on the polarization of the incident wave. Given the larger refractive index of the ground (typically n ≈ 2) compared to air (n = 1), the phase of horizontally polarized radiation is reversed upon reflection (a phase shift of
{\displaystyle \scriptstyle {\pi }}
radians or 180°). On the other hand, the vertical component of the wave's electric field is reflected at grazing angles of incidence approximately in phase. These phase shifts apply as well to a ground modeled as a good electrical conductor.
{\displaystyle \textstyle {\left|E_{V}\right|=2\left|E_{0}\right|\,\left|\cos \left({2\pi h \over \lambda }\sin \theta \right)\right|}}
{\displaystyle \textstyle {\left|E_{H}\right|=2\left|E_{0}\right|\,\left|\sin \left({2\pi h \over \lambda }\sin \theta \right)\right|}}
{\displaystyle \scriptstyle {E_{0}}}
{\displaystyle \scriptstyle {\lambda }}
{\displaystyle \scriptstyle {h}}
{\displaystyle \scriptstyle {\theta =0}}
{\displaystyle \scriptstyle {\theta =0}}
Modeling antennas with line equationsEdit
Mutual impedance and interaction between antennasEdit
Antenna typesEdit
Retrieved from "https://en.wikipedia.org/w/index.php?title=Antenna_(radio)&oldid=1086279766"
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Even More WordPress Plugins for Writers - SitePoint
The life of a writer (whether author or copywriter) is generally quite misunderstood. Friends assume that you’ve landed the dream job. You don’t have to leave your lounge room to get to work. No pesky boss breathing down your neck. And, free from the constraints of the corporate world, you can take a four-hour lunch break every day if it so pleases you.
While all these things apply some of the time, people tend to forget that the life of a writer is often characterised by hour upon hour of staring into a flickering laptop screen. Desperately trying to find the right word, the perfectly poised phrase, the exact tone. For writers, the ability to make ends meet is quite often reliant upon an intangible, un-controllable quality — inspiration. If writers are unable to transform an idea into words, they simply don’t make money.
So, on that rather depressing note, I thought it would be a gesture of solidarity amongst the online writing community to follow up on my article, The Best WordPress Plugins for Writers and Writing and provide details of a few more WordPress Plugins designed to make the life of the online writer just that smidgen easier.
This plugin automatically uploads your old blog posts onto your Twitter feed, saving you from the embarrassment of having to spruik your old copy just to keep it relevant. It also helps to optimize your Twitter feed and save it from dying a slow, painful death because you simply don’t have the time to update it (or got so wrapped up in drafting your novel that you forgot!). With over 80,000 active installs, and regular updates, this plugin could become a key tool in any writer’s social media arsenal.
Unfortunately, writers don’t always get paid for everything they pen, particularly budding new writers looking to enhance their portfolio. Occasionally, writers have to pump out a blog post just to keep our names out there. This WordPress plugin is a fantastic tool that effectively creates a wall between the general public and your lovingly crafted blog. This wall can be quickly and easily broken down, as long as your readers share your post on social media first. It’s completely free and it’s a great way to generate some publicity.
A word of warning on this one though; if you don’t already have loyal readers, then this plugin does have the potential to backfire. Readers may not be willing to share something on social media if they haven’t actually read it themselves first.
This plugin has something for everyone. If you haven’t already installed it, then I recommend you take a look. It’s a completely free plugin. Rather than waffling on any further, I suggest you check out the plugin page for yourself. You’ll be impressed.
This plugin forces you to do what every writer should be doing regularly (but doesn’t!) — backing up their work. UpdraftPlus Backup Restore is an automatic backup system. It even allows you to choose where you store your backup, integrating with the likes of Dropbox, Google Drive, Amazon S3, or just plain old email. According to the marketing spiel, UpdraftPlus is the highest ranking backup plugin in the WordPress repository, with over 800,000 active installs. Check out their introductory YouTube video for further details.
Everyone wants a fast website — visitors and site owners alike. A slow, clunky website is a one-way ticket to online irrelevance. Even if your website is mostly dedicated to writing (and avoids the big files and images that can slow things down), it’s no excuse for a slow-loading site. This plugin optimizes your site by using techniques such as minifying and database caching, reducing download times, increasing server performance, and integrating with content delivery networks. With more than one million active installs, its popularity speaks for itself. Here’s an overview on YouTube.
Everyone hates spam. It’s totally annoying. This handy little plugin blocks spam from the comments section of your site, meaning you only have to moderate and respond to the comments that are directly related to the content you have slaved over. You don’t have to delete fifteen comments by Dr. Michael Michelson who wants you to visit his ‘five-steps to pimple free skin’ page anymore. You can keep and respond to quality comments on your page and avoid the dreaded spam.
This plugin is great for writers who want to embed PDFs and other files into their content and allow them to be used with the Google Docs viewer. It also gives your readers the option of downloading the content. That means you can pepper your site with more comprehensive samples of your work.
Writers often forget that they can use affiliate links to make a little extra cash off their writing. If you find a service or product that you like or want to endorse use the Amazon Affiliate program (and plugin). Or, better yet, if you’ve published a book yourself, be sure to include the details on your own website. You never know, you might just be able to make a little bit of money every time someone clicks-through and purchases.
An obvious way to make money from your blog is to create a member-only section, jam packed with exclusive, high-quality content for which your members pay a fee. This plugin can protect WordPress pages, posts, tags, and entire categories. It extremely easy to configure, with lots of built-in options available. The s2Member Framework even makes it easy to sell subscriptions to your loyal online readership, integrating with PayPal, to help streamline the entire payment process. There are both free, and paid pro versions (one-off fee of either US
89 or US
Writers live and die by their portfolios. You need to provide your audience with a stylish sample of work and the types of projects you have worked on. This is a fantastic plugin that allows you to do just that.
If you’re in the book writing business you need this plugin. It allows you to showcase your books – complete with descriptions, images, reviews and purchase links – on your WordPress site. You can even bypass the middleman and include a PayPal or eJunkie button on your site and sell the books from your own front yard.
This is the commenting system that every WordPress site should have. It works as a comment network, meaning the comments that are logged on your site are also logged on the Disqus community, which helps people discover content they might be interested in.
Once again, it’s important to remember that the more plugins you have installed on your WordPress site, the more it can impact on performance. Make use of the fantastic plugins out there, but only install what you really need.
ChrisBpluginpluginsWordPresswriters
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Flow Characteristics of a Powder Lubricant Sheared Between Parallel Plates | J. Tribol. | ASME Digital Collection
Department of Mechanical Engineering and Energy Process, Southern Illinois University, Carbondale, IL 62901-6603
M. M. Khonsari, Dow Chemical Endowed Chair in Rotating Machinery and Professor,
Department of Mechanical Engineering, 2508 CEBA, Louisiana State University, Baton Rouge, LA 70803
Contributed by the Tribology Division of THE AMERICAN SOCIETY OF MECHANICAL ENGINEERS and presented at the STLE/ASME Tribology Conference, Orlando, FL, October 10–13, 1999. Manuscript received by the Tribology Division January 26, 1999; revised manuscript received June 1, 1999. Paper No. 99-Trib-16. Associate Technical Editor: R. F. Salant.
Zhou , L., and Khonsari , M. M. (June 1, 1999). "Flow Characteristics of a Powder Lubricant Sheared Between Parallel Plates ." ASME. J. Tribol. January 2000; 122(1): 147–155. https://doi.org/10.1115/1.555337
This paper is devoted to the understanding of the flow characteristics of a powder lubricant. An accurate and realistic model is introduced to investigate the mechanism of powder lubrication sheared between two infinitely wide parallel plates. In this theory, the solid volume fraction,
v,
together with flow velocity and granular temperature are directly obtained from a set of complete governing equations (mass, momentum and pseudo-energy conservation equations). The boundary conditions take into account the effect of slip velocity and the wall surface roughness. A set of parametric simulation results is presented. It is shown that viscous dissipation plays an important role in the pseudo-energy equation. [S0742-4787(00)01901-9]
powders, lubrication, granular flow, machine bearings, tribology, shear deformation, slip flow
Boundary-value problems, Flow (Dynamics), Temperature, Plates (structures), Particulate matter, Energy dissipation, Lubricants, Surface roughness
The Quasi-Hydrodynamic Mechanism of Powder Lubrication: Part 2. Lubricant Film Pressure Profile
High-Temperature Solid-Lubricated Bearing Development—Dry Powder-Lubricated Traction Testing
Elrod, H. G., 1988, “Granular Flow As A Tribological Mechanics,” Interface Dynamics, Leeds-Lyon Conference, p. 75.
Elrod, H. G., and Brewe, D. E., 1992, “Numerical Experiments with Flows of Elongated Granules,” NASA TM105567, AVSCOM TR91-C-006.
On the Lubrication Mechanism of Grain Flows
Yu, C. M., Craig, K., and Tichy, J., 1994, “Granular Collision Lubrication,” J. Rheology, pp. 921–936.
Generalized Boundary Interactions for Powder Lubricated Couette Flow
Granular Collisional Lubrication: Effect of Surface Roughness, Particle Size and Solids Fraction
Grain Flow as a Fluid Mechanical Phenomenon
Kinetic Theories for Granular Flows: Inelastic Particles in Couette Flow and Slightly Inelastic Particles in a General Flow Field
Frictional-Collisional Constitutive Relations for Granular Materials with Applications to Plane Shearing
Observations of Rapidly Flowing Granular Fluid Materials
Computer Simulations of Granular Shear Flows
On the Role of Enduring Contact in Powder Lubrication
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Liquidation - Vertex Protocol
As mentioned, liquidations occur when a user's margin fraction nears or dips below the maintenance margin.
Remember: Vertex is a cross-margin trading platform. Your account's equity is spread (used) between all positions. Cross-margin provides traders with more flexibility and can help avoid unnecessary liquidations. However, it also means that all positions are liquidated at once.
There are two types of liquidations on Vertex: partial and full which are determined when liquidation bots call the clearinghouse contract.
Partial liquidations are when a given % of your positions are liquidated.
Full Liquidations mean all positions, including trades and LPs are fully liquidated.
The price at which liquidations occurs is based on market data from the Oracles.
The mechanism is autonomously handled by liquidation bots. The bots monitor a positions margin fraction and call the clearinghouse to activate liquidations. Anyone with an account on Vertex can call the contract liquidation function, given that the liquidated account is eligible for full or partial liquidation.
Our liquidation algorithm follows the equation:
\text{Max. Liq. Fraction} = -1.5 \times \frac{\text{Margin Fraction}}{\text{Maint. Margin}} + 1.75
The maximum liquidation fraction is clamped between 0 and 1. A value of 0 means that you cannot be liquidated, while a value of 1 means full liquidation. With a default maintenance margin of 0.01 on Vertex Forex markets, this implies full liquidation at a margin fraction of 0.005.
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Erratum: “Electrostatic Forces and Stored Energy for Deformable Dielectric Materials” [Journal of Applied Mechanics, 2005, 72(4), pp. 581–590] | J. Appl. Mech. | ASME Digital Collection
Erratum: “Electrostatic Forces and Stored Energy for Deformable Dielectric Materials” [Journal of Applied Mechanics, 2005, 72(4), pp. 581–590]
This is a correction to: Electrostatic Forces and Stored Energy for Deformable Dielectric Materials
McMeeking, R. M., and Landis, C. M. (August 22, 2008). "Erratum: “Electrostatic Forces and Stored Energy for Deformable Dielectric Materials” [Journal of Applied Mechanics, 2005, 72(4), pp. 581–590]." ASME. J. Appl. Mech. November 2008; 75(6): 067002. https://doi.org/10.1115/1.2966272
deformation, dielectric materials, electrostatics
Dielectric materials, Engineering mechanics, Deformation, Electrostatics
In this Erratum we point out and correct an omission of an important term in Eq. (8) of our original paper. This equation was a statement of the conservation of angular momentum and we inadvertently omitted the electrical body couple term that is needed in this momentum balance. In Eq. (9) the electrical body force components
biE
are defined. A similar equation is needed to define the electrical body couple as
ciE=ϵijkσjkM
Then, the correct statement of the balance of angular momentum and its analysis is given as
∫VciEdV+∫Vϵijkxj(bk+bkE)dV+∫Sϵijkxj(Tk+TkE)dS=ddt∫VρϵijkxjvkdV,
∫Vϵijk(σjk+σjkM)dV=0
The result of this analysis appears in Eq. (13) of the original manuscript as
σji+σjiM=σij+σijM
Note that Eq. (13) does not follow from the original Eq. (8), but does follow from this corrected version. Hence, Eq. (13) and all of the remaining equations within the original manuscript are correct.
New CPC Solar Collector for Planar Absorbers Immersed in Dielectrics. Application to the Treatment of Contaminated Water
Material Interaction Effects in the Reliability of High Density Interconnect (HDI) Boards
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Section 1-2 - LimitPrecise - Maple Help
Home : Support : Online Help : Study Guides : Calculus : Chapter 1 - Limits : Section 1-2 - LimitPrecise
Section 1.2: Precise Definition of a Limit
Section 1.1 provided an intuitive introduction to the notion of a mathematical limit. This intuition was captured in an informal definition. In fact, the statement
\underset{x\to a}{lim} f\left(x\right)=L
means the difference between
f\left(x\right)
L
can be made as small as desired for all values of
x
sufficiently close to -- but different from --
a.
The development of general tools for working with and applying limits requires a more precise definition of limit. This section gives a formal definition of the limit, one that provides a framework upon which a wide range of useful techniques will be developed.
The first step is to develop a graphical interpretation of the precise definition. Next, the algebraic manipulations involved in applying the precise definition will be explored for four examples. The section concludes with a more precise definition of one-sided limits first seen in Section 1.1.
Definition 1.2.1 is a precise definition of the limit of the real-valued function
f\left(x\right)
. The limit is a number designated by the letter
L
. The notation for the "limit of
f\left(x\right)
x
a
" is
\underset{x\to a}{lim}f\left(x\right)
If this limit is the number
L
\underset{x\to a}{lim}f\left(x\right)=L
As Definition 1.2.1 is read, keep in mind that all it is trying to say is that for
x
"near"
a
f\left(x\right)
are "near"
L
. But also keep in mind that it took nearly 100 years for the mathematics community to settle on this form of the statement of what "near" should mean.
f\left(x\right)
x
a
L
\underset{x→a}{lim}f\left(x\right)=L
\mathrm{ϵ}>0
\mathrm{δ}>0
with the property that if
0<\left|x-a\right|<\mathrm{δ}
x
is in the domain of f), then
\left|f\left(x\right)-L\right|<\mathrm{ϵ}.
This definition rigorously captures the essence of "nearness." It says that if the values of
f\left(x\right)
get arbitrarily close to the number
L
x
a,
L
is the limit to those function values. The nearness to
L
f\left(x\right)
is measured by
|f\left(x\right)-L|
while the nearness of
x
a
|x-a|
|f\left(x\right)-L|
must be smaller than ε, where ε represents any possible small number. These differences must be smaller than ε for all
x
x=a
. The nearness of
x
a
is dictated by the inequality
|x-a|<\mathrm{\delta }.
Because this definition might be difficult to fathom at first, several examples that interpret this definition graphically are provided.
Graphical Interpretation of the Limit
The precise definition of a limit expressed in Definition 1.2.1 took mathematicians more than a century to formulate. The central idea expressed by it is easier to grasp if it can be seen graphically. The EpsilonDelta maplet is an interactive tool for visualizing the roles of
\mathrm{ϵ}
\mathrm{δ}
in Definition 1.2.1. Its use is illustrated in Examples 1.2.1-4.
Use the EpsilonDelta maplet to verify
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)=5
Use the EpsilonDelta maplet to show that
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)\ne 2
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}^{2}-3x+3\right)=3
\underset{x→2}{lim}\mathrm{ln}\left(x\right)=\mathrm{ln}\left(2\right)
Using Definition 1.2.1 to Verify a Limit
Definition 1.2.1 can be used to prove or disprove that a given number
L
is, or is not, the limit. It cannot be used to find the value of
L
. Methods for finding
L
appear in the next few sections. In each of Examples 1.2.6-9, an animation captures the relationship between the intervals
\left|x-a\right|<\mathrm{δ}
\left|f\left(x\right)-L\right|<\mathrm{ϵ}
. Then, the following "algorithm" is applied to find
\mathrm{δ}\left(\mathrm{ϵ}\right)
, and to verify that it satisfies Definition 1.2.1.
The essence of Definition 1.2.1 is finding a
\mathrm{δ}
-band around
x=a
inside of which function values
f\left(x\right)
remain within an
\mathrm{ϵ}
L
. See the region shaded in yellow in Figure 1.2.2.
Solve the two equations
f\left(a+{\mathrm{δ}}_{R}\right)=L+\mathrm{ϵ}
f\left(a-{\mathrm{δ}}_{L}\right)=L-\mathrm{ϵ}
{\mathrm{δ}}_{L}
{\mathrm{δ}}_{R}
are positive, and generally different. See the dotted red lines in Figure 1.2.2.
\mathrm{δ}\left(\mathrm{ϵ}\right)≤\mathrm{min}\left\{{\mathrm{δ}}_{L},{\mathrm{δ}}_{R}\right\}
. See the black arrows and the region shaded yellow in Figure 1.2.2.
f:=x->x^2-3*x+3:
g:=x->3/2+(1/2)*sqrt(-3+4*x):
p1:=plot(f,2..3.5,color=black):
p2:=plot(3,x=2..3,color=blue):
p3:=plot([[x,4,x=0..g(4)],[x,2,x=0..g(2)]],color=red,linestyle=dot):
p4:=plots:-textplot({[2.3,4.2,typeset(y=L+epsilon)],[2.28,1.8,typeset(y=L-epsilon)],[2.2,3.2,typeset(y=L)],[2.3,3.64,typeset(` f`(a+delta))],[2.3,2.44,typeset(` f`(a-delta))]},align=LEFT,font=[Times,12]):
p5:=plot([[g(2),t*f(g(2)),t=0..1],[g(4),t*f(g(4)),t=0..1],[3,t*f(3),t=0..1]],color=red,linestyle=dot):
p6:=plots:-textplot({[g(2),-.2,typeset(a-delta[L])],[g(4),-.2,typeset(a+delta[R])],[3,-.2,typeset(a)],[(3+XR)/2,.7,typeset(delta)],[(3+XL)/2,.7,typeset(delta)]}):
XL:=2.75:
XR:=3.25:
p7:=plottools:-polygon([[XL,0],[XL,f(XL)],[0,f(XL)],[0,f(XR)],[XR,f(XR)],[XR,0]],color=yellow,transparency=.8):
p8:=plottools:-arrow([3,.5],Vector([XR-3,0]),.01,.1,.1,double_arrow):
p9:=plottools:-arrow([3,.5],Vector([XL-3,0]),.01,.1,.1,double_arrow):
plots:-display(p||(1..9), view=[2..3.5,-.3..5],tickmarks=[[-1,5],[-1,8]],labels=[" ",""]);
Figure 1.2.2 Schematic for Definition 2.1
The final step is to show that
0<\left|x-a\right|<\mathrm{δ}\left(\mathrm{ϵ}\right)
\left|f\left(x\right)-L\right|<\mathrm{ϵ}
, that is, that for all
x
-values in the
\mathrm{δ}
x=a
, all the function values
f\left(x\right)
remain inside the
\mathrm{ϵ}
y=L
. (Having found a candidate for
\mathrm{δ}\left(\mathrm{ϵ}\right)
, it's much easier to establish the validity of the appropriate inequalities than it is to use inequalities to find
\mathrm{δ}\left(\mathrm{ϵ}\right)
in the first place.)
This demonstration is expedited by expressing the
x
\mathrm{δ}
-band
\left|x-a\right|<\mathrm{δ}
x=a+t \mathrm{δ}\left(\mathrm{ϵ}\right)
0<\left|t\right|<1
t
can be both positive and negative, these
x
-values correspond to all those inside the
\mathrm{δ}
-band. Then, to show that
\left|f\left(x\right)-L\right|<\mathrm{ϵ}
for these
x
-values, show instead that
\left|f\left(a+t \mathrm{δ}\left(\mathrm{ϵ}\right)\right)-L\right|<\mathrm{ϵ}
. The following four examples will verify that these steps form an algorithm that can be applied to a variety of functions without having to make radical modifications because of the peculiar properties of the function
f
Note: For a function that decreases in the vicinity of
x=a
, the two equations for
{\mathrm{δ}}_{L}
{\mathrm{δ}}_{R}
would respectively be
f\left(a+{\mathrm{δ}}_{R}\right)=L-\mathrm{ϵ}
f\left(a-{\mathrm{δ}}_{L}\right)=L+\mathrm{ϵ}
The use of Definition 1.2.1 to verify a limit is illustrated in Examples 1.2.5-8.
Use Definition 1.2.1 to verify
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)=5
, the limit explored in Example 1.2.1.
\underset{x\to 3}{lim} \frac{{x}^{2}-9}{x-3}=6
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\sqrt{x}=\sqrt{3}
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}^{2}-3x+3\right)=3
Modifications for One-Sided Limits
For two-sided limits, values of the independent variable on both sides of the limit point must be considered (assuming they are in the domain of the function). One-sided limits only consider values of the independent variable on one side of the limit point. These differences are summarized in Table 1.2.4.
Values of Independent Variable
0<∣x-a∣<\mathrm{δ}
a-\mathrm{δ}<x<a
a<x<a+\mathrm{δ}
one-sided (right)
0<∣x-a∣<\mathrm{δ}
x>a
a<x<a+\mathrm{δ}
one-sided (left)
0<∣x-a∣<\mathrm{δ}
x<a
a-\mathrm{δ}<x<a
Table 1.2.4 Distinctions between two-sided and one-sided limits
Definition 1.2.1 can be applied to a one-sided limit with the same technique used in Examples 1.2.6 - 9. However, the equation determining
\mathrm{δ}\left(\mathrm{ϵ}\right)
must be selected from Table 1.2.5, which takes into account how
x=a
is approached, and whether the function is increasing or decreasing in a neighborhood of
x=a
f
x=a
f
x=a
f\left(a-\mathrm{δ}\right)=L-\mathrm{ϵ}
f\left(a-\mathrm{δ}\right)=L+\mathrm{ϵ}
f\left(a+\mathrm{δ}\right)=L+\mathrm{ϵ}
f\left(a+\mathrm{δ}\right)=L-\mathrm{ϵ}
Table 1.2.5 Applying Definition 2.1 to one-sided limits
Table 1.2.3 summarizes some of the main points of the section.
When using the definition of the limit to show
\underset{x\to a}{lim} f\left(x\right)=L
, a useful picture would contain the following features:
the graph of the function near the limit point,
x=a
a horizontal band centered at
y=L
|y-L|<\mathrm{ϵ}
a vertical band centered at
x=a
|x-a|<\mathrm{δ}
(technically, the vertical line
x=a
should be excluded from this band)
\mathrm{δ}
is consistent with the definition of the limit, then each value of
x
in the vertical band (except, possibly,
x=a
) will produce a function value that is within the horizontal band. If there is even one function value outside this horizontal band then either the value of
\mathrm{δ}
is too large for this
\mathrm{ϵ}
or the limit value (
L
) is incorrect.
Table 1.2.3 Summary of highlights of Section 1.2
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)=5
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)\ne 2
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}^{2}-3x+3\right)=3
\underset{x→2}{lim}\mathrm{ln}\left(x\right)=\mathrm{ln}\left(2\right)
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}(3x-4)=5
\underset{x\to 3}{lim} \frac{{x}^{2}-9}{x-3}=6
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\sqrt{x}=\sqrt{3}
\underset{x\to 3}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}^{2}-3x+3\right)=3
|
Section 59.6 (03N8): A computation—The Stacks project
Section 59.6: A computation (cite)
59.6 A computation
How do we compute the cohomology of $\mathbf{P}^1_\mathbf {C}$ with coefficients $\Lambda = \mathbf{Z}/n\mathbf{Z}$? We use Čech cohomology. A covering of $\mathbf{P}^1_\mathbf {C}$ is given by the two standard opens $U_0, U_1$, which are both isomorphic to $\mathbf{A}^1_\mathbf {C}$, and whose intersection is isomorphic to $\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} = \mathbf{G}_{m, \mathbf{C}}$. It turns out that the Mayer-Vietoris sequence holds in étale cohomology. This gives an exact sequence
\[ H_{\acute{e}tale}^{i-1}(U_0\cap U_1, \Lambda ) \to H_{\acute{e}tale}^ i(\mathbf{P}^1_ C, \Lambda ) \to H_{\acute{e}tale}^ i(U_0, \Lambda ) \oplus H_{\acute{e}tale}^ i(U_1, \Lambda ) \to H_{\acute{e}tale}^ i(U_0\cap U_1, \Lambda ). \]
To get the answer we expect, we would need to show that the direct sum in the third term vanishes. In fact, it is true that, as for the usual topology,
\[ H_{\acute{e}tale}^ q (\mathbf{A}^1_\mathbf {C}, \Lambda ) = 0 \quad \text{ for } q \geq 1, \]
\[ H_{\acute{e}tale}^ q (\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} , \Lambda ) = \left\{ \begin{matrix} \Lambda & \text{ if }q = 1\text{, and} \\ 0 & \text{ for }q \geq 2. \end{matrix} \right. \]
These results are already quite hard (what is an elementary proof?). Let us explain how we would compute this once the machinery of étale cohomology is at our disposal.
Higher cohomology. This is taken care of by the following general fact: if $X$ is an affine curve over $\mathbf{C}$, then
\[ H_{\acute{e}tale}^ q (X, \mathbf{Z}/n\mathbf{Z}) = 0 \quad \text{ for } q \geq 2. \]
This is proved by considering the generic point of the curve and doing some Galois cohomology. So we only have to worry about the cohomology in degree 1.
Cohomology in degree 1. We use the following identifications:
\begin{eqnarray*} H_{\acute{e}tale}^1 (X, \mathbf{Z}/n\mathbf{Z}) = \left\{ \begin{matrix} \text{sheaves of sets }\mathcal{F}\text{ on the étale site }X_{\acute{e}tale}\text{ endowed with an} \\ \text{action }\mathbf{Z}/n\mathbf{Z} \times \mathcal{F} \to \mathcal{F} \text{ such that }\mathcal{F}\text{ is a }\mathbf{Z}/n\mathbf{Z}\text{-torsor.} \end{matrix} \right\} \Big/ \cong \\ = \left\{ \begin{matrix} \text{morphisms }Y \to X\text{ which are finite étale together} \\ \text{ with a free }\mathbf{Z}/n\mathbf{Z}\text{ action such that } X = Y/(\mathbf{Z}/n\mathbf{Z}). \end{matrix} \right\} \Big/ \cong . \end{eqnarray*}
The first identification is very general (it is true for any cohomology theory on a site) and has nothing to do with the étale topology. The second identification is a consequence of descent theory. The last set describes a collection of geometric objects on which we can get our hands.
The curve $\mathbf{A}^1_\mathbf {C}$ has no nontrivial finite étale covering and hence $H_{\acute{e}tale}^1 (\mathbf{A}^1_\mathbf {C}, \mathbf{Z}/n\mathbf{Z}) = 0$. This can be seen either topologically or by using the argument in the next paragraph.
Let us describe the finite étale coverings $\varphi : Y \to \mathbf{A}^1_\mathbf {C} \setminus \{ 0\} $. It suffices to consider the case where $Y$ is connected, which we assume. We are going to find out what $Y$ can be by applying the Riemann-Hurwitz formula (of course this is a bit silly, and you can go ahead and skip the next section if you like). Say that this morphism is $n$ to 1, and consider a projective compactification
\[ \xymatrix{ {Y\ } \ar@{^{(}->}[r] \ar[d]^\varphi & {\bar Y} \ar[d]^{\bar\varphi } \\ {\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} } \ar@{^{(}->}[r] & {\mathbf{P}^1_\mathbf {C}} } \]
Even though $\varphi $ is étale and does not ramify, $\bar{\varphi }$ may ramify at 0 and $\infty $. Say that the preimages of 0 are the points $y_1, \ldots , y_ r$ with indices of ramification $e_1, \ldots e_ r$, and that the preimages of $\infty $ are the points $y_1', \ldots , y_ s'$ with indices of ramification $d_1, \ldots d_ s$. In particular, $\sum e_ i = n = \sum d_ j$. Applying the Riemann-Hurwitz formula, we get
\[ 2 g_ Y - 2 = -2n + \sum (e_ i - 1) + \sum (d_ j - 1) \]
and therefore $g_ Y = 0$, $r = s = 1$ and $e_1 = d_1 = n$. Hence $Y \cong {\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} }$, and it is easy to see that $\varphi (z) = \lambda z^ n$ for some $\lambda \in \mathbf{C}^*$. After reparametrizing $Y$ we may assume $\lambda = 1$. Thus our covering is given by taking the $n$th root of the coordinate on $\mathbf{A}^1_{\mathbf{C}} \setminus \{ 0\} $.
Remember that we need to classify the coverings of ${\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} }$ together with free $\mathbf{Z}/n\mathbf{Z}$-actions on them. In our case any such action corresponds to an automorphism of $Y$ sending $z$ to $\zeta _ n z$, where $\zeta _ n$ is a primitive $n$th root of unity. There are $\phi (n)$ such actions (here $\phi (n)$ means the Euler function). Thus there are exactly $\phi (n)$ connected finite étale coverings with a given free $\mathbf{Z}/n\mathbf{Z}$-action, each corresponding to a primitive $n$th root of unity. We leave it to the reader to see that the disconnected finite étale degree $n$ coverings of $\mathbf{A}^1_{\mathbf{C}} \setminus \{ 0\} $ with a given free $\mathbf{Z}/n\mathbf{Z}$-action correspond one-to-one with $n$th roots of $1$ which are not primitive. In other words, this computation shows that
\[ H_{\acute{e}tale}^1 (\mathbf{A}^1_\mathbf {C} \setminus \{ 0\} , \mathbf{Z}/n\mathbf{Z}) = \mathop{\mathrm{Hom}}\nolimits (\mu _ n(\mathbf{C}), \mathbf{Z}/n\mathbf{Z}) \cong \mathbf{Z}/n\mathbf{Z}. \]
The first identification is canonical, the second isn't, see Remark 59.69.5. Since the proof of Riemann-Hurwitz does not use the computation of cohomology, the above actually constitutes a proof (provided we fill in the details on vanishing, etc).
Comment #564 by Dustin on April 22, 2014 at 09:37
Hey, first of all thanks for this resource.
Now that those niceties are out of the way, of course, I'm here to nitpick. I think there's an error in the discussion of the H^1 of A^1-0. The answer should be dual to what's written: it should be Hom(\mu_n,Z/nZ). The error is in the "We leave it to the reader to see...".
Thanks for your encouragement. As to your nitpick: I was thinking of (some of) these coverings as given by
y^n = x
with action of
1
\mathbf{Z}/n\mathbf{Z}
y \mapsto \zeta y
\zeta \in \mu_n(\mathbf{C})
n
1
. But I guess your point is that one can equally well think of this the other way round, i.e., as the map
\mu_n(\mathbf{C}) \to \mathbf{Z}/n\mathbf{Z}
\zeta
1
. To check which one is right, you'd have to compute the addition of two of these torsors (addition of torsors is given by taking the product of the torsors over the base and then dividing by the anti-diagonal action of the group). This is annoying, but I did it to make sure that you were right.
Anyhoo, maybe a better way of getting at it is that there is an absolutely canonical
\mu_n
\mathbf{G}_m
\mathbf{G}_m \to \mathbf{G}_m
n
th power map. This gives a canonical element
1 \in H^1(\mathbf{G}_{m, \mathbf{Z}}, \mu_n)
whose image over the complex numbers tells us you are correct.
Regarding the opening line "How do we compute the cohomology of
P^1_C
\Lambda=Z/nZ
?", which cohomology are you referring to - etale or singular (of course the answer turns out to be the same, and perhaps the same method works for both - cech cohomology). From the next sentence involving Cech cohomology I thought you were referring to singular cohomology, and being a beginner I don't know yet if Cech cohomology computes etale cohomology for abelian sheaves (as opposed to quasi-coherent ones) - if so can you link to it (I searched but couldn't find it)? But then the remaining sentences convinced me you were referring to etale cohomology of
P^1_C
minor grammatical fix: in line 3 "and which intersection is isomorphic" I think should be whose intersection is isomorphic
OK, I fixed the which and whose confusion, see here. About the other stuff. There is a bit of information about Cech cohomology later in the chapter. But the discussion is a bit loose, because it comes from a lecture I gave a long time ago. If you have a precise suggestion of how to change the text please feel free to put a further comment here.
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What is the Celsius scale?
Conversion between Celsius and other temperature scales
How do I use the Celsius converter?
The Celsius converter helps you convert from the Celsius scale (or the Centigrade scale) to any other temperature scale such as Kelvin, Fahrenheit, and so on. You could also do the reverse, i.e, convert from any of the temperature scales to find the equivalent temperature in the Celsius scale.
Read on to learn more about what the Celsius scale is and the formulas governing the conversion between the Celsius scale and other units of temperature.
The Celsius scale, invented in 1742 by the Swedish astronomer Anders Celsius, is based on the freezing and boiling points of water. The freezing point of water is set as 0° C, and the boiling point of water is set as 100° C. The range of temperature values between these two bounds is divided into 100 degrees, thus giving it the alternative name, centigrade scale.
The conversion between the Celsius scale and other units of temperature is fairly straightforward and is governed by simple mathematical formulae, as shown below:
Note: Here, T(°C) denotes the temperature in degrees Celsius, T(°F) denotes the temperature in Fahrenheit, and so on.
Degrees Celsius to Kelvin
T(K)\ =\ T(°C)\ +\ 273.15
T(°F)\ =\ \frac{9\ ×\ T(°C)}{5}\ +\ 32
T(°R)\ =\ \frac{9\ ×\ T(°C)}{5}\ +\ 491.67
T(°De)\ =\ -1.5\ ×\ T(°C)\ +\ 150
T(°N)\ =\ 0.33\ ×\ T(°C)
Degrees Celsius to degrees Réaumur
T(°Ré)\ =\ 0.8\ ×\ T(°C)
Degrees Celsius to degrees Rømer
T(°Rø)\ =\ 0.525\ ×\ T(°C)\ +\ 7.5
To use the Celsius converter, you need to do the following:
Enter the temperature in Celsius scale.
Tada! You will directly get the equivalent temperatures in other units!
Alternatively, you can also enter the temperature in other units, and get the equivalent temperature in degrees Celsius!
You can convert between the Celsius scale and any of the following scales of temperature:
Kelvin scale;
Fahrenheit scale;
Rankine scale;
Delisle scale;
Newton scale;
Réaumur scale; and
Rømer scale.
To convert between any of the other units of temperature, you may check out some of our following tools:
Kelvin to Celsius converter; and
How do I convert from Celsius to Fahrenheit?
To convert the temperature from Celsius to Fahrenheit, do the following:
Multiply the Celsius temperature by the fraction 9/5.
Add 32 to the resulting number.
Voila! You'll get the equivalent temperature in the Fahrenheit scale!
What is 100° C in the Kelvin scale?
100° C is equivalent to 373 K in the Kelvin scale. You can convert from the Celsius scale to the Kelvin scale by simply adding 273 to the Celsius temperature!
Energy Conversion Calculator is a handy tool which can alter between different energy units.
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Keshik ger - The RuneScape Wiki
When it stood, this would have been a roomy tent.
Keshik gers are excavation hotspots at the Stormguard Citadel - Keshik memorial excavation site that players can excavate after reaching level 76 Archaeology.
They initially appear as aerated sediment, requiring uncovering before being usable. Uncovering the hotspots yields a one-time reward of 175 Archaeology experience.
Avian song-egg player 6,066.7 3 N/A • Wise Old Man (Wise Am the Music Man)
• Lowse (Armadylean I)
• Velucia (Museum - Armadylean I) N/A
Keshik drum 6,066.7 3 N/A • Wise Old Man (Wise Am the Music Man)
Morin khuur 6,066.7 3 N/A • Wise Old Man (Wise Am the Music Man)
Animal furs 1 2/10 5,676 150
Avian song-egg player (damaged) 1 1/3 Not sold Not alchemisable
Keshik drum (damaged) 1 1/3 Not sold Not alchemisable
Morin khuur (damaged) 1 1/3 Not sold Not alchemisable
Mission Report: Everlight Silvthril 1 Uncommon Not sold Not alchemisable
Mission Report: Warforged Bronze 1 Uncommon Not sold Not alchemisable
Mission Report: Hellfire Metal 1 Uncommon Not sold Not alchemisable
{\displaystyle {\frac {L+E}{250{,}000}}}
{\displaystyle L}
{\displaystyle E}
{\displaystyle {\frac {1}{125{,}000}}}
{\displaystyle {\frac {1}{1{,}042}}}
template = Archaeology Hotspot Calculator Lite/Results form = Keshik_gerArchForm result = ArchResult param=name|Hotspot Name|Keshik ger|combobox|Acropolis debris,Administratum debris,Aetherium forge,Amphitheatre debris,Ancient magick munitions,Animal trophies,Armarium debris,Aughra remains,Autopsy table,Bandos sanctum debris,Bibliotheke debris,Big High War God shrine,Byzroth remains,Carcerem debris,Castra debris,Ceramics studio debris,Chthonian trophies,Crucible stands debris,Culinarum debris,Cultist footlocker,Destroyed golem,Dis dungeon debris,Dis overspill,Dominion Games podium,Dragonkin coffin,Dragonkin reliquary,Experiment workbench,Flight research debris,Gladitorial goblin remains,Goblin dorm debris,Goblin trainee remains,Gravitron research debris,Hellfire forge,Howls workshop debris,Icyene weapon rack,Ikovian memorial,Infernal art,Keshik ger,Keshik weapon rack,Kharid-et chapel debris,Kyzaj champions boudoir,Legionary remains,Lodge art storage,Lodge bar storage,Makeshift pie oven,Moksha device,Monoceros remains,Oikos fishing hut remnants,Oikos studio debris,Optimatoi remains,Orcus altar,Pontifex remains,Praesidio remains,Praetorian remains,Prodromoi remains,Sacrificial altar,Saurthen debris,Stadio debris,Stockpiled art,Tailory debris,Tsutsaroth remains,Venator remains,Varanusaur remains,War table debris,Warforge scrap pile,Warforge weapon rack,Weapons research debris,Xolo mine,Xolo remains,Yubiusk animal pen param=method|Method|High intensity|buttonselect|High intensity,Medium intensity,AFK param=mattock|Mattock|Necronium|select|Bronze,Iron,Steel,Mithril,Adamant,Rune,Orikalkum,Dragon,Necronium,Crystal,Bane,Imcando,Elder Rune,Time and Space,Guildmaster Tony
Keshik (or kheshig) were the imperial guard of Mongol royalty in the Mongol Empire. It is Mongolian for "favoured" or "blessed".
Gers (more commonly known as yurts) are round, portable tents commonly used by nomadic cultures in Central Asia. "Ger" (гэр) is Mongolian for "home".
Retrieved from ‘https://runescape.wiki/w/Keshik_ger?oldid=35701294’
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Modular exponentiation definition
How to use this power mod calculator?
Examples of exponentiation modulo
Omni's power mod calculator is here to help whenever you need to compute powers in modular arithmetic. It uses one of the fast modular exponentiation algorithms, so there's no risk of facing the problem of overflow. Should you ever need to perform exponentiation modulo n by hand, we discuss several helpful methods you can use at home, including Fermat's little theorem.
Have you already checked the rest of our exponentiation calculators?
How to divide exponents?
How to calculate fractional exponents?
What is exponential regression?
Modular exponentiation means that we perform exponentiation over a modulo, i.e., for the given integers a,b,n we want to find c such that
c = a^b \operatorname{mod}n
0 \leq c < n
. Computing power in modular arithmetic is linked to modular inverses.
You can perform this calculation manually, but it can be very time-consuming. Alternatively, some mathematical theorems allow you to simplify the problem at hand - see below. There are also fast algorithms, which will give you the result almost immediately. We use one of these algorithms in this power mod calculator.
This power mod calculator is very user-friendly, so you'll have no trouble using it. You just need to:
Input the data for computing the power of xʸ in modular arithmetic:
Exponent y; and
Modulus n.
Your data will be summarized at the bottom of the calc. Verify if everything is all right.
The result of modular exponentiation will appear there as well. That's it!
Our power mod calculator will be your best friend if you're frequently faced with the problem of computing powers in modular arithmetic. Read on if you need to know how to calculate exponentiation modulo n by hand.
Here we will go through several examples of performing exponentiation modulo by hand using different methods.
Example 1. Direct method
Let's calculate 5⁴ mod 3.
We know that 5⁴ = 625, so our problem is in fact 625 mod 3.
Clearly, 625 is not divisible by 3, but 624 is (this is because the sum of its digits is 6+2+4 = 12, which is divisible by 3).
So 625 - 1 is divisible by 3, which means that 5⁴ mod 3 = 625 mod 3 = 1.
Example 2. Smart method
Let's calculate 5⁴⁴ mod 2.
It's gonna be very hard to compute 5⁴⁴, because this number is very, very big. So, we need to be smart. Recall that mod 2 means that we're asking if the number at hand is even or odd: if it is even, then it's equal to 0 mod 2. If it's odd, it's equal to 1 mod 2.
When we're computing consecutive powers of 5, we get 5, 25, 625,.... As you can see, we always have 5 as the last digit. Indeed, if you have a number with the last digit equal to 5, and you multiply this number by 5, then you're bound to get 5 at the last place again. To see this, imagine performing the long multiplication algorithm - you start by multiplying 5 × 5, and so you get 25. So 5 goes to the result row, and 2 gets transferred to the next column. No matter what happens next, you have 5 as the last digit.
A number that has 5 as its last digit, is odd. So 5⁴⁴ mod 2 = 1.
Example 3. Last digit
Let's calculate 5⁴⁴⁴ mod 10.
First, you need to realize that computing mod 10 is the same as computing the number's last digit. We've already established that raising 5 to any positive integer power gives a number that ends with 5 (see above). Hence, 5⁴⁴⁴ ends with 5 as well, so 5⁴⁴⁴ mod 10 = 5.
Example 4. Fermat's little theorem
Let's calculate 162⁶⁰ mod 61.
Fermat's little theorem states that if n is a prime number, then for any integer a we have:
a^n \operatorname{mod} n = a
If additionally a is not divisible by n, then
a^{n-1} \operatorname{mod} n = 1
Hence, since in our case we have n = 61, which is a prime number, and a = 162, which is not divisible by 61, we obtain
162⁶⁰ mod 61 = 1.
Modular exponentiation means that we're calculating powers in modular arithmetic, that is, performing an operation of the form ab mod n, where a, b and n are integers. If b is negative, modular exponentiation is linked to modular multiplication inverses.
How do I calculate exponential modulo?
If the numbers at hand are not very big, you can simply solve the exponent first and then apply the modulo. Otherwise, you either need to apply some smart reasoning, a math theorem (like Fermat's little theorem or Euler's theorem), or a specialized computer algorithm that performs fast modular exponentiation.
How do I reduce exponential power in modulo?
To reduce power in exponentiation modulo, you need to apply the rules of modular arithmetic, or even some advanced math theorems, like Fermat's little theorem or one of its generalizations, e.g., Euler's theorem.
What is Fermat's little theorem?
Fermat's little theorem is one of the most popular math theorems dealing with modular exponentiation. It has many generalizations, which you can evoke in more complicated calculations. We call it 'little' so as to distinguish it from its much more popular sibling, Fermat's last theorem.
y (exponent)
n (divisor)
xy mod n = ?
Our perimeter of a rectangle with given area calculator lets you find the perimeter of a rectangle once you know the area and a side (length or width).
The slant height calculator lets you calculate the slant height for a right circular cone or a right angle pyramid.
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Process Capability Index Calculator | Find Cpk
What is process capability index?
How to calculate process capability index? Process capability index formula
Interpretation of process capability index and the meaning of Cpk statistics
Examples of process capability index calculation
Welcome to Omni's process capability index calculator, aka the Cpk calculator! With the process capability index calculator, you can quickly determine whether the products of your process meet specification limits (target values, usually set by customers, in which products or services are relative to customer requirements). Do you want to determine whether your process is going smoothly, or whether you need to make some improvements for the future? Read on to learn more about:
How to calculate process capability index (a.k.a. process capability ratio);
What the meaning of Cpk is; and
What values of Cpk statistics are satisfactory.
As a bonus, you will learn why people are so excited when their Cpk is 1.33 or larger! 💡
To better explain the process capability index, let's imagine a scenario. You are craving a delicious cappuccino on a Monday morning ☕. You decide to walk to the coffee shop at the corner of your street. That coffee shop is your favorite, because the way they make cappuccino meets your requirements, which have certain specifications. You know that if the cappuccino's milk temperature exceeds your specification limits of 55–65 °C, the fluffy foam will not be stable, and it will most definitely ruin your Monday morning. Unlike other coffee shops that are even closer to your house, the one at the corner of your street is your favorite because you can get a cappuccino with a perfect milk temperature within your specification limits every time.
Now that you have the scenario in mind, let's define the process capability index. Process capability index, process capability ratio, or Cpk, is a measure of the process's capability to produce desired results that meet the customer's specification limits (values within which a product should perform). In the scenario described, the Cpk of preparing milk for cappuccinos would indicate that the coffee shop is consistent with its average performance of producing cappuccinos for customers (i.e., baristas consistently deliver cappuccinos with milk temperature that is 60 °C on average 🥛, and rarely exceeds 55–65 °C).
Before presenting the process capability index (or process capability ratio) formula, it is essential to note that you can apply different process capability indices (Cpk, Cp, Cpm, Cpkm) to different situations.
For example, you can successfully utilize the process capability index formula when you want to observe variation in your process relative to specification limits and determine how the process can perform in the future. Cpk can be used when your process average is not perfectly centered. The formula for Cpk is as follows:
C_{pk}=\min \left( {\text{USL} - \mu \over 3\sigma} , {\mu - \text{LSL} \over 3\sigma}\right)
C_{pk}
is the process capability index;
\text{USL}
is the upper specification limit;
\text{LSL}
is lower specification limit; and
\sigma
If you want to see the potential capability of the process when the process average is centered, you can use the following formula of Cp:
C_p = \frac{\text{USL} - \text{LSL}}{6\sigma }
C_p
is the potential process capability;
USL
LSL
is the lower specification limit; and
\sigma
If you want to see the capability of your process around a specific target, you can use the following formula for Cpm below. However, remember that Cpm does not account for the process average that is off-centered.
C_{pm} = \frac{C_p}{\sqrt{1 + \left( \frac{\mu-T}{\sigma} \right)^2}}
C_{pm}
is the process capability to be relative to the target process mean. This is also called the Taguchi capability index.
C_p
μ
is the mean;
{\displaystyle {{\sigma }}}
is the standard deviation; and
T
is the target process mean.
If you're determining the process capability around a target, and you don't have a process that is centered around the target value, you can use the Cpkm formula below:
C_{pkm} = \frac{C_{pk}}{\sqrt{1 + \left( \frac{\mu-T}{\sigma} \right)^2}}
C_{pkm}
the "third generation" process capability index;
C_{pk}
μ
\sigma
T
🙋 Before calculating process capability indices, check whether you have normally distributed data and a stable process. Also, don't forget our calculator's Advanced mode 🧠 feature, to calculate Cpm and Cpkm!
Now that you're familiar with the process capability index formula, how to calculate the process capability index, and what Cpk is, let's talk about its interpretation. As mentioned before, the higher your Cpk value, the better your process produces. Experts agree that you want your Cpk index to be at least 1.33. Here is the explanation why you want to reach this magic number:
Rare events in your process become detectable when the mean moves off-center by 1.5 sigmas (i.e. when normal distribution shifts by 1.5 standard deviations on either side). When Cpk is 1.33, upper and lower specification limits are four standard deviations from the process mean. In this case, there is some (one standard deviation) room for variability within specification limits, and you can consider the process capable. However, a Cpk of 1.33 is not ideal since you want larger variability before defects are displayed.
The process with Cpk of 2 corresponds to a 6-sigma level process; therefore, it can bear a larger process mean shift while having most process-related output within specifications (has desired room for variability). Therefore, a Cpk of 2 would indicate excellent process capability.
Is it still confusing? Let's discuss an example. Soups at your buffet must be served between 64–80 °C, as per your customer's specifications 🍲. The mean temperature of the soups is usually 72 °C, and the process of reheating has a standard deviation of 1.5 °C. What is the Cpk value of reheating soups?
\begin{split} C_{pk} &= \min \left[\left(\frac{80-72}{3\times1.5}\right), \left(\frac{72-64}{3\times1.5}\right)\right] \\ C_{pk} &= \frac{8}{4.5} \approx 1.778 \end{split}
Since the smallest acceptable Cpk is 1.33, your value of 1.778 is a great result and we can conclude that your process is capable!
Potential process capability (Cₚ)
Process capability index (Cₚₖ)
The median absolute deviation calculator can calculate the median absolute deviation of a data set of up to 50 points.
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Strength of Bases | Boundless Chemistry | Course Hero
Strength of Bases
Strong bases either dissociate completely in solution to yield hydroxide ions, or deprotonate water to yield hydroxide ions.
Recognize strong bases and their chemical properties.
In chemistry, a base is a substance that can either accept hydrogen ions (protons) or, more generally, donate a pair of valence electrons; it can be thought of as the chemical opposite of an acid.
Strong bases are commonly, though not exclusively, formed from the hydroxides of alkali metals and alkaline earth metals.
base: a proton acceptor, or an electron pair donor
solvate: a complex formed from solvent molecules attaching to a solute
As discussed in the previous concepts on bases, a base is a substance that can: donate hydroxide ions in solution (Arrhenius definition); accept H+ ions (protons) (Bronsted-Lowry definition); or donate a pair of valence electrons (Lewis definition). In water, basic solutions have a pH higher than 7.0, indicating a greater concentration of OH- than H+.
A strong Arrhenius base, like a strong acid, is a compound that ionizes completely or near-completely in solution. Therefore, the concentration of hydroxide ions in a strongly basic solution is equal to that of the undissociated base. Common examples of strong Arrhenius bases are the hydroxides of alkali metals and alkaline earth metals such as NaOH and Ca(OH)2. Strong bases are capable of deprotonating weak acids; very strong bases can deprotonate very weakly acidic C–H groups in the absence of water.
Sodium hydroxide pellets: Sodium hydroxide pellets, before being suspended in water to dissociate.
The cations of these strong bases appear in the first and second groups of the periodic table (alkali and earth alkali metals). Generally, the alkali metal bases are stronger than the alkaline earth metal bases, which are less soluble. When writing out the dissociation equation of a strong base, assume that the reverse reaction does not occur, because the conjugate acid of a strong base is very weak.
Group 1 salts of carbanions (such as butyllithium, LiC4H9, which dissociates into Li+ and the carbanion C4H9-), amides (NH2-), and hydrides (H-) tend to be even stronger bases due to the extreme weakness of their conjugate acids—stable hydrocarbons, amines, and hydrogen gas. Usually, these bases are created by adding pure alkali metals in their neutral state, such as sodium, to the conjugate acid. They are called superbases, because it is not possible to keep them in aqueous solution; this is due to the fact they will react completely with water, deprotonating it to the fullest extent possible. For example, the ethoxide ion (conjugate base of ethanol) will undergo this reaction in the presence of water:
Unlike weak bases, which exist in equilibrium with their conjugate acids, the strong base reacts completely with water, and none of the original anion remains after the base is added to solution. Some other superbases include:
Superbases such as the ones listed above are commonly used as reagents in organic laboratories.
In aqueous solution, a weak base reacts incompletely with water to yield hydroxide ions.
Solve acid-base equilibrium problems involving weak bases.
A base is a substance that can accept hydrogen ions (H+) or, more generally, donate a pair of valence electrons; a weak base does not, therefore, fully ionize or completely accept hydrogen ions in an aqueous solution.
Bases increase pH; weak bases have a less dramatic effect on pH.
pOH is occasionally used as an alternative to pH to quantify the relative H+/hydroxide concentration in solution.
A base dissociation constant, Kb, mathematically represents the base's relative strength and is analogous to the acid dissociation constant; weaker bases have smaller Kb values.
Like weak acids, weak bases can be used to make buffer solutions.
weak base: a proton acceptor that does not ionize fully in an aqueous solution
enol: an organic alcohol with an -OH functional group located off a double bond
enolate: a deprotonated enol
A base is a substance that can accept hydrogen ions (H+) or, more generally, donate a pair of valence electrons. A weak base is a chemical base that does not ionize fully in an aqueous solution. As Brønsted-Lowry bases are proton acceptors, a weak base may also be defined as a chemical base with incomplete protonation. A general formula for base behavior is as follows:
\text{B}(\text{aq}) + \text{H}_2\text{O}(\text{aq}) \rightleftharpoons \text{BH}^+(\text{aq}) + \text{OH}^-(\text{aq})
A base can either accept protons from water molecules or donate hydroxide ions to a solution. Both actions raise the pH of the solution by decreasing the concentration of H+ ions. This results in a relatively low pH compared to that of strong bases. The pH of bases in aqueous solution ranges from greater than 7 (the pH of pure water) to 14 (though some bases have pH values greater than 14). The formula for pH is:
Sometimes, however, it is more convenient to focus on the pOH of bases, rather than the pH. The pOH more directly references the [OH-].
Some common weak bases and their corresponding pKb values include:
C6H5NH2 (9.38)
NH3 (4.75)
CH3NH2 (3.36)
CH3CH2NH2 (3.27)
Smaller pKb values indicate higher values of Kb; this also indicates a stronger base.
Like weak acids, weak bases have important applications in biochemical studies, chemistry reactions, and physiological purposes, particularly because of their role in buffer solutions. Weak bases can also be used to catalyze certain reactions, such as enolate formation, as demonstrated in the figure below:
Weak base catalyzing enolate formation: A weak base, symbolized by B:, can catalyze enolate formation by acting as a proton acceptor.
The base dissociation constant (Kb) measures a base's relative strength.
Calculate the Kw (water dissociation constant) using the following equation: Kw = [H+] x [OH−] and manpulate the formula to determine [OH−] = Kw/[H+] or [H+]=Kw/[OH-]
The base dissociation constant KbE measures a base's basicity, or strength.
Kb is related to the acid dissociation constant, Ka, by the simple relationship pKa + pKb = 14, where pKb and pKa are the negative logarithms of Kb and Ka, respectively.
Kb and Ka are also related through the ion constant for water, Kw, by the relationship
\text{K}_\text{W}=\text{K}_\text{a}\times \text{K}_\text{b}
conjugate acid: the species created when a base accepts a proton
In chemistry, a base is a substance that can accept hydrogen ions (protons) or, more generally, donate a pair of valence electrons. The base dissociation constant, Kb, is a measure of basicity—the base's general strength. It is related to the acid dissociation constant, Ka, by the simple relationship pKa + pKb = 14, where pKb and pKa are the negative logarithms of Kb and Ka, respectively. The base dissociation constant can be expressed as follows:
\text{K}_\text{b} = \dfrac{[\text{BH}^+][\text{OH}^-]}{\text{B}}
\text{B}
\text{BH}^+
is its conjugate acid, and
\text{OH}^-
is hydroxide ions.
\text{B}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \leftrightharpoons \text{HB}^+(\text{aq}) + \text{OH}^-(\text{aq})
\text{K}_{\text{b}} = \frac{[\text{OH}^{-}][\text{HB}^{+}]}{[\text{B}]}
Kb is related to Ka for the conjugate acid. Recall that in water, the concentration of the hydroxide ion, [OH−], is related to the concentration of the hydrogen ion by the autoionization constant of water:
\text{K}_\text{W}=[\text{H}^+][\text{OH}^-]
[\text{OH}^{-}] = \frac{\text{K}_{\text{w}}}{[\text{H}^{+}]}
Substituting this expression for [OH−] into the expression for Kb yields:
\text{K}_{\text{b}} = \frac{\text{K}_{\text{w}}[\text{HB}^{+}]}{[\text{B}][\text{H}^{+}]} = \frac{\text{K}_{\text{w}}}{\text{K}_{\text{a}}}
\text{K}_\text{W}=\text{K}_\text{a}\text{K}_\text{b}
\text{pK}_\text{a}+\text{pK}_\text{b}=14
The pH of a weak base in aqueous solution depends on the strength of the base (given by Kb) and the concentration of the base (the molarity, or moles of the base per liter of solution). A convenient way to find the pH for a weak base in solution is to use an ICE table: ICE stands for "Initial," "Change," and"Equilibrium."
Before the reaction starts, the base, B, is present in its initial concentration [B]0, and the concentration of the products is zero. As the reaction reaches equilibrium, the base concentration decreases by x amount; given the reaction's stoichiometry, the two products increase by x amount. At equilibrium, the base's concentration is [B]0 – x, and the two products' concentration is x.
\text{K}_{\text{b}} = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}
\text{K}_{\text{b}} = \frac{\text{x}^2}{[\text{B}]_{0}-\text{x}}
This quadratic equation can be solved for x. However, if the base is weak, then we can assume that x will be insignificant compared to [B]0, and the approximation [B]0– x ≈ [B]0 can be used. The equation simplifies to:
\text{K}_{\text{b}} = \frac{\text{x}^2}{[\text{B}]_{0}}
Since x = [OH]-, we can calculate pOH using the equation pOH = –log[OH]-; we can find the pH using the equation 14 – pOH = pH.
General Chemistry/Properties and Theories of Acids and Bases. Provided by: Wikibooks. Located at: http://en.wikibooks.org/wiki/General_Chemistry/Properties_and_Theories_of_Acids_and_Bases%23Strong_and_Weak_Acids.2FBases. License: CC BY-SA: Attribution-ShareAlike
solvate. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/solvate. License: CC BY-SA: Attribution-ShareAlike
Sodium hydroxide. Provided by: Wikimedia. Located at: http://commons.wikimedia.org/wiki/File:Sodium_hydroxide.jpg. License: CC BY-SA: Attribution-ShareAlike
weak base. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/weak%20base. License: CC BY-SA: Attribution-ShareAlike
enolate. Provided by: Wiktionary. Located at: http://en.wiktionary.org/wiki/enolate. License: CC BY-SA: Attribution-ShareAlike
Weak base enolate formation. Provided by: Wikimedia. Located at: http://commons.wikimedia.org/wiki/File:Weak_base_enolate_formation.svg. License: CC BY-SA: Attribution-ShareAlike
Acid dissociation constant. Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/Acid_dissociation_constant%23Bases. License: CC BY-SA: Attribution-ShareAlike
General Chemistry/Chemical Equilibria/Acid-Base Equilibrium. Provided by: Wikibooks. Located at: http://en.wikibooks.org/wiki/General_Chemistry/Chemical_Equilibria/Acid-Base_Equilibrium%23Base_Dissociation_Constant. License: CC BY-SA: Attribution-ShareAlike
Original figure by Jeanine Batterton. Licensed CC BY-SA 4.0. Provided by: Jeanine Batterton. License: CC BY-SA: Attribution-ShareAlike
ENGLISH 1 • University of New South Wales
CHEMISTRY 1 • University of Central Oklahoma
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CHEMISTRY 2 • De La Salle University
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S2016_Bis2A_Lecture05_reading
Unit 2. Lesson 6—Strength of Acids and Bases.docx
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CHEMISTRY AP Chemist • Andover High, Andover
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quiz prep packet 8 - relative strength of acids and bases_answer key.pdf
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Strength of Acids and Bases.ppt
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CH.18-Sec.2-Strengths_of_Acids_and_Bases_Study_Guide.doc
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Modularity of membrane-bound charge-translocating protein complexes | Biochemical Society Transactions | Portland Press
Modularity of membrane-bound charge-translocating protein complexes
Filipa Calisto;
1Instituto de Tecnologia Química e Biológica - António Xavier, Universidade Nova de Lisboa, Av. da República EAN, 2780-157 Oeiras, Portugal
2Faculty of Sciences, Department of Chemistry and Biochemistry and BioISI - Biosystems & Integrative Sciences Institute, University of Lisboa, Campo Grande, 1749-016 Lisboa, Portugal
Correspondence: Manuela M. Pereira (mmpereira@fc.ul.pt)
Filipa Calisto, Manuela M. Pereira; Modularity of membrane-bound charge-translocating protein complexes. Biochem Soc Trans 17 December 2021; 49 (6): 2669–2685. doi: https://doi.org/10.1042/BST20210462
Energy transduction is the conversion of one form of energy into another; this makes life possible as we know it. Organisms have developed different systems for acquiring energy and storing it in useable forms: the so-called energy currencies. A universal energy currency is the transmembrane difference of electrochemical potential (
Δμ~
). This results from the translocation of charges across a membrane, powered by exergonic reactions. Different reactions may be coupled to charge-translocation and, in the majority of cases, these reactions are catalyzed by modular enzymes that always include a transmembrane subunit. The modular arrangement of these enzymes allows for different catalytic and charge-translocating modules to be combined. Thus, a transmembrane charge-translocating module can be associated with different catalytic subunits to form an energy-transducing complex. Likewise, the same catalytic subunit may be combined with a different membrane charge-translocating module. In this work, we analyze the modular arrangement of energy-transducing membrane complexes and discuss their different combinations, focusing on the charge-translocating module.
charge translocation, energy metabolism, energy transduction, respiratory chains
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Exponent of next higher power of 2 - MATLAB nextpow2 - MathWorks Deutschland
Next Power of 2 of Double Integer Values
Next Power of 2 of Unsigned Integer Values
Optimize FFT with Padding
Exponent of next higher power of 2
P = nextpow2(A)
P = nextpow2(A) returns the exponents for the smallest powers of two that satisfy
{2}^{p}\ge |A|
for each element in A. By convention, nextpow2(0) returns zero.
You can use nextpow2 to pad the signal you pass to fft. Doing so can speed up the computation of the FFT when the signal length is not an exact power of 2.
Define a vector of double integer values and calculate the exponents for the next power of 2 higher than those values.
a = [1 -2 3 -4 5 9 519];
Calculate the positive next powers of 2.
np2 = 2.^p
np2 = 1×7
1 2 4 4 8 16 1024
Preserve the sign of the original input values.
np2.*sign(a)
1 -2 4 -4 8 16 1024
Define a vector of unsigned integers and calculate the exponents for the next power of 2 higher than those values.
a = uint32([1020 4000 32700]);
p = 1x3 uint32 row vector
Calculate the next powers of 2 higher than the values in a.
2.^p
Use the nextpow2 function to increase the performance of fft when the length of a signal is not a power of 2.
Create a 1-D vector containing 8191 sample values.
x = rand([1,8191]);
Calculate the next power of 2 higher than 8191.
p = nextpow2(8191);
n = 2^p
Pass the signal and the next power of 2 to the fft function.
scalar, vector, or array of real numbers
Input values, specified as a scalar, vector, or array of real numbers of any numeric type.
Example: [-15.123 32.456 63.111]
Example: int16([-15 32 63])
fft | log2 | pow2
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Abstract: Present day Quantum Field Theory (QFT) is founded on canonical quantization, which has served quite well but also has led to several issues. The free field describing a free particle (with no interaction term) can suddenly become nonrenormalizable the instant a suitable interaction term appears. For example, using canonical quantization , has been deemed a “free” theory with no difference from a truly free field [1] [2]. Using the same model, affine quantization has led to a truly interacting theory [3]. This fact alone asserts that canonical and affine tools of quantization deserve to be open to their procedures together as a significant enlargement of QFT.
Keywords: Quantum Field Theory, Canonical Quantization, Affine Quantization
1.1. Classical Starting Point
There are different ways to promote classical to quantum expressions that are useful. For the classical, canonical Hamiltonian model, we have
H\left(\pi ,\phi \right)=\int \left\{\frac{1}{2}\left[\pi {\left(x\right)}^{2}+{\left(\stackrel{\to }{\nabla }\phi \left(x\right)\right)}^{2}+{m}^{2}\phi {\left(x\right)}^{2}\right]+g\phi {\left(x\right)}^{r}\right\}{\text{d}}^{s}x,
r\ge 2
s\ge 3
. The ingredients in this expression are the classical field
\phi \left(x\right)
\pi \left(x\right)
. These fields obey the Poisson bracket
\left\{\phi \left(x\right),\pi \left(y\right)\right\}={\delta }^{s}\left(x-y\right)
However, we can describe the same Hamiltonian in a different way. Let us choose the affine field
k\left(x\right)\equiv \pi \left(x\right)\phi \left(x\right)
, instead of the momentum field, but still keep
\phi \left(x\right)
. However, it is necessary to keep
\phi \left(x\right)\ne 0
for otherwise if
\phi \left(x\right)=0
\pi \left(x\right)
means nothing. Let us exam (1), the same classical Hamiltonian, but now in the new coordinates, leading to
{H}^{\prime }\left(\kappa ,\phi \right)=\int \left\{\frac{1}{2}\left[\kappa {\left(x\right)}^{2}\phi {\left(x\right)}^{-2}+{\left(\stackrel{\to }{\nabla }\phi \left(x\right)\right)}^{2}+{m}^{2}\phi {\left(x\right)}^{2}\right]+g\phi {\left(x\right)}^{r}\right\}{\text{d}}^{s}x.
This new set of fields leads to the Poisson bracket
\left\{\phi \left(x\right),\kappa \left(y\right)\right\}={\delta }^{s}\left(x-y\right)\phi \left(x\right)
1.2. Quantum Starting Point
Promotion of the fields
\phi \left(x\right)\to \stackrel{^}{\phi }\left(x\right)
\pi \left(x\right)\to \stackrel{^}{\pi }\left(x\right)
, leads to the traditional quantum expression for our Hamiltonian, which is given by
H\left(\stackrel{^}{\pi },\stackrel{^}{\phi }\right)=\int \left\{\frac{1}{2}\left[\stackrel{^}{\pi }{\left(x\right)}^{2}+{\left(\stackrel{\to }{\nabla }\stackrel{^}{\phi }\left(x\right)\right)}^{2}+{m}^{2}\stackrel{^}{\phi }{\left(x\right)}^{2}\right]+g\stackrel{^}{\phi }{\left(x\right)}^{r}\right\}{\text{d}}^{s}x.
Now, knowing that the classical variables were no longer the canonical choice but rather the affine coordinates, and after the promotion of affine field variables to new quantum field variables, the new quantum Hamiltonian becomes
{H}^{\prime }\left(\stackrel{^}{\kappa },\stackrel{^}{\phi }\right)=\int \left\{\frac{1}{2}\left[\stackrel{^}{\kappa }\left(x\right)\stackrel{^}{\phi }{\left(x\right)}^{-2}\stackrel{^}{\kappa }\left(x\right)+{\left(\stackrel{\to }{\nabla }\stackrel{^}{\phi }\left(x\right)\right)}^{2}+{m}^{2}\stackrel{^}{\phi }{\left(x\right)}^{2}\right]+g\stackrel{^}{\phi }{\left(x\right)}^{r}\right\}{\text{d}}^{s}x.
Do not worry about
\stackrel{^}{\phi }{\left(x\right)}^{-2}
because we have already insisted that
\phi \left(x\right)\ne 0
\stackrel{^}{\phi }\left(x\right)\ne 0
. In a previous usage, which proved itself by modifying
\stackrel{^}{\phi }{\left(x\right)}^{-2}\to {\left[\stackrel{^}{\phi }{\left(x\right)}^{2}+\epsilon \right]}^{-1}
\epsilon ={10}^{-10}
served as a safeguard [4].
\stackrel{^}{\kappa }\left(x\right)\stackrel{^}{\phi }{\left(x\right)}^{-1/2}=0
, which, in Schrödinger’s representation, leads to
\stackrel{^}{\kappa }\left(x\right)=-i\hslash \left[\phi \left(x\right)\left(\partial /\partial \phi \left(x\right)\right)+\left(\partial /\partial \phi \left(x\right)\right)\phi \left(x\right)\right]/2
\stackrel{^}{\phi }\left(x\right)=\phi \left(x\right)
. Following a suitable regularization process [5], this yields the stated result.
1.3. Advantages of an Affine Quantization
Using the results of the previous sections we propose that
\stackrel{^}{\kappa }\left(x\right){\Pi }_{y}\phi {\left(y\right)}^{-1/2}=0
which exposes our choice for general wave functions as given by
\Psi \left(\phi \right)=\int W\left(\phi \left(x\right)\right){\Pi }_{y}\text{ }\phi {\left(y\right)}^{-1/2}\text{ }\text{d}\phi \left(x\right)
. A regularized version, using
x\to k
k=an=a\times {\left\{\cdots \mathrm{,}-\mathrm{2,}-\mathrm{1,0,1,}\cdots \right\}}^{s}
of this expression looks like
\Psi \left(\phi \right)=w\left(\phi \right){\Pi }_{k}{\left(b{a}^{s}\right)}^{1/2}{\left({\phi }_{k}\right)}^{-\left(1-2b{a}^{s}\right)/2}
b{a}^{s}
is dimensionless and
b\sim 1
We now take a Fourier transformation of the absolute square of our regulated wave function that looks like
F\left(f\right)={\Pi }_{k}\int \left\{{\text{e}}^{i{f}_{k}{\phi }_{k}}{|w\left({\phi }_{k}\right)|}^{2}\left(b{a}^{s}\right){|{\phi }_{k}|}^{-\left(1-2b{a}^{s}\right)}\text{ }\text{d}{\phi }_{k}\right\}.
Normalization ensures that if all
{f}_{k}=0
F\left(0\right)=1
F\left(f\right)={\Pi }_{k}\int \left\{1-\int \left(1-{\text{e}}^{i{f}_{k}{\phi }_{k}}\right){|w\left({\phi }_{k}\right)|}^{2}\left(b{a}^{s}\right)\text{d}{\phi }_{k}/{|{\phi }_{k}|}^{\left(1-2b{a}^{s}\right)}\right\}\text{ }.
Now, at last, we can let
a\to 0
to fix the Fourier transformation1
F\left(f\right)=\mathrm{exp}\left\{-b\int {\text{d}}^{s}x\left(1-{\text{e}}^{if\left(x\right)\text{ }\phi \left(x\right)}\right){|w\left(\phi \left(x\right)\right)|}^{2}\text{d}\phi \left(x\right)/|\phi \left(x\right)|\right\}.
Observe that the affine quantization his led to a Poisson distribution, which is the only other term, besides a Gaussian expression, as dictated by The Central Limit Theorem [6]. Nevertheless, the same expression as in (7) could have arisen when
g=0
, or even when
g↗\text{ }↘0
, asserting that our final result is definitely not a Gaussian! Of significant is the fact that if the coupling g, or even the mass m, are smoothly changed, there are only continuous changes within
w\left(\phi \right)
. Also, the fact that
\stackrel{^}{\kappa }\left(x\right)\phi {\left(x\right)}^{-1/2}=0
, which is a dramatic change from canonical theory’s equivalent relation, i.e.,
\stackrel{^}{\pi }\left(x\right)11=0
, makes a big difference; indeed, the factor
{|{\phi }_{k}|}^{-\left(1-2b{a}^{s}\right)}
in (5) is the key to avoiding a Gaussian result. Apparently, this behavior of affine quantization adopts the least final domain at the outset, which overcomes any threat of nonrenormalizability.2
We have obtained a continuous, fully regularized, expression that implicitly involves a large sample of quantum field models. The application of affine quantization, but not canonical quantization, has offered us a treasure of interest that presently rests in the Fourier representation space. Understanding the physics needed to clarify our results requires a second Fourier transformation back into the original space of the classical field, here given by
\phi \left(x\right)
. That issue is purely a mathematical task, and the implications of such an effort are certainly of great interest!
1Any change of
w\left(\phi \right)
a\to 0
is left implicit.
2For those who wish to learn more about affine quantization see [5]. For beginners, canonical quantization deals with the harmonic oscillator, but the half-harmonic oscillator requires affine quantization [7].
Cite this paper: Klauder, J. (2021) Evidence for Expanding Quantum Field Theory. Journal of High Energy Physics, Gravitation and Cosmology, 7, 1157-1160. doi: 10.4236/jhepgc.2021.73067.
[1] Aizenman, M. (1981) Proof of the Triviality of Field Theory and Some Mean-Field Features of Ising Models for d > 4. Physical Review Letters, 47, 1-4, E-886.
[2] Fröhlich, J. (1982) On the Triviality of λφd4 Theories and the Approach to the Critical Point in d ≥ 4 Dimensions. Nuclear Physics B, 200, 281-296.
[3] Fantoni, R. and Klauder, J.R. (2021) Affine Quantization of φ44 Succeeds, While Canonical Quantization Fails. Physical Review D, 103, Article ID: 076013.
[4] Fantoni, R. (year) Monte Carlo Evaluation of the Continuum Limit of φ312.
[6] Central Limit Theorem. Wikipedia.
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Bowling Average Calculator (Cricket)
What is bowling average?
How to calculate bowling average in cricket with this calculator?
Common bowling average values
Some other helpful cricket statistics and tools
With this bowling average calculator, you'll evaluate your cricket performance or your favorite player's bowling skills. The main objective of the fielding team in this sport is to minimize the number of runs scored by the batting team. To achieve this goal, bowlers are a fundamental part. Due to the importance of bowling in this sport, some cricket statistical measures of how well bowlers perform exist. Bowling average is one the most important ones.
⚠️ Using only the term "bowling average" may be confusing. In this article, we're talking about the statistic used in cricket🏏 and not in bowling sport🎳. We'll mainly use "bowling average" instead of "cricket bowling average" for practical purposes.
Do you have an interest in knowing how to calculate the bowling average in cricket on your own? Do you want to have an idea of what is a good bowling average?
We believe our bowling average calculator can help you to answer these questions and calculate this metric quickly and reliably, so you get a sense of the performance of yourself or your favorite player.
Bowling average is a measure of the performance of a cricket bowler. It compares the runs conceded by a bowler to the wickets taken by himself.
The lower this number is, the better the performance of the player.
To use this calculator, all you need to do is:
Take the total number of runs the bowler conceded.
The total number of wickets the bowler took.
Input these numbers in the calculator, and our tool will assess the bowler's performance in a flash using the following formula:
\quad\ \ \footnotesize \text{bowling average}=\frac{\text{runs conceded}}{\text{wickets taken}}
The bowling average of a cricket bowler is usually between 20 and 50. The following is a frequent interpretation of the findings that can give you an idea of what is a good bowling average:
Under 20: extraordinary bowling skills
20 to 25: excellent bowling skills
25 to 30: good bowling skills
30 to 35: average bowling skills
35 to 40: below average bowling skills
Above 40: minimal bowling skills
A higher bowling average doesn't mean we're in front of a bad player, but it suggests that this cricketer is selected more probably for batting roles.
The best test cricket bowling average 🧐🧐🧐
To have an idea of how extraordinary a bowling average under 20 is, the lowest test cricket bowling average (for players who bowled at least 600 balls in the test cricket modality) is 10.74. It belongs to the English cricketer George Lohman.
There are more cricket statistics beyond bowling average. We have developed calculators we think can be valuable for you, no matter you're a professional or amateur cricket player (or simply a fan):
Batting Average: this is the equivalent statistic in batting. It compares the total number of runs the batsman scored to the overall number of times the batsman was out. Contrary to what we'd encounter with the bowling average calculator, a higher batting average indicates better performance.
Net Run Rate: another cricket statistic used in performance assessment. Unlike bowling and batting averages, net run rate measures team's performance instead of individual performance.
Duckworth Lewis: this calculator will help you decide which cricket team won a match if it can't be completed. It can also help the batting team set a target score if the match is interrupted and a second part is left.
Cricket Follow-On: a calculator used for strategy purposes that will help you determine whether you should force your opponent to bat again or keep the traditional sequence. "Follow-on" is a critical decision assigned to the captain, so it would be a good idea you make efforts to become the leader of your team and have the opportunity to use this great calculator.
How to calculate bowling average in cricket?
Divide the number of runs the bowler conceded by the number of wickets he took. With this procedure, calculating the bowling average in cricket becomes an easy task.
They usually consider that a bowling average under 30 is a good result in cricket.
What is the importance of bowling average and other cricket statistics?
These cricket statistics objectively try to predict the performance of a player or team in a future game. Although they help predict others' performance, you can also use them to assess your cricket skills and improve them in time.
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Gauss–Hermite quadrature - Wikipedia
Form of Gaussian quadrature
Weights versus xi for four choices of n
In numerical analysis, Gauss–Hermite quadrature is a form of Gaussian quadrature for approximating the value of integrals of the following kind:
{\displaystyle \int _{-\infty }^{+\infty }e^{-x^{2}}f(x)\,dx.}
{\displaystyle \int _{-\infty }^{+\infty }e^{-x^{2}}f(x)\,dx\approx \sum _{i=1}^{n}w_{i}f(x_{i})}
where n is the number of sample points used. The xi are the roots of the physicists' version of the Hermite polynomial Hn(x) (i = 1,2,...,n), and the associated weights wi are given by [1]
{\displaystyle w_{i}={\frac {2^{n-1}n!{\sqrt {\pi }}}{n^{2}[H_{n-1}(x_{i})]^{2}}}.}
Example with change of variable[edit]
Consider a function h(y), where the variable y is Normally distributed:
{\displaystyle y\sim {\mathcal {N}}(\mu ,\sigma ^{2})}
. The expectation of h corresponds to the following integral:
{\displaystyle E[h(y)]=\int _{-\infty }^{+\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {(y-\mu )^{2}}{2\sigma ^{2}}}\right)h(y)dy}
As this doesn't exactly correspond to the Hermite polynomial, we need to change variables:
{\displaystyle x={\frac {y-\mu }{{\sqrt {2}}\sigma }}\Leftrightarrow y={\sqrt {2}}\sigma x+\mu }
Coupled with the integration by substitution, we obtain:
{\displaystyle E[h(y)]=\int _{-\infty }^{+\infty }{\frac {1}{\sqrt {\pi }}}\exp(-x^{2})h({\sqrt {2}}\sigma x+\mu )dx}
{\displaystyle E[h(y)]\approx {\frac {1}{\sqrt {\pi }}}\sum _{i=1}^{n}w_{i}h({\sqrt {2}}\sigma x_{i}+\mu )}
^ Abramowitz, M & Stegun, I A, Handbook of Mathematical Functions, 10th printing with corrections (1972), Dover, ISBN 978-0-486-61272-0. Equation 25.4.46.
Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., eds. (2010), "Quadrature: Gauss–Hermite Formula", NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0-521-19225-5, MR 2723248
Shao, T. S.; Chen, T. C.; Frank, R. M. (1964). "Tables of zeros and Gaussian weights of certain associated Laguerre polynomials and the related generalized Hermite polynomials". Math. Comp. 18 (88): 598–616. doi:10.1090/S0025-5718-1964-0166397-1. MR 0166397.
Steen, N. M.; Byrne, G. D.; Gelbard, E. M. (1969). "Gaussian quadratures for the integrals
{\displaystyle \textstyle \int _{0}^{\infty }e^{-x^{2}}f(x)dx}
{\displaystyle \textstyle \int _{0}^{b}e^{-x^{2}}f(x)dx}
". Math. Comp. 23 (107): 661–671. doi:10.1090/S0025-5718-1969-0247744-3. MR 0247744.
Shizgal, B. (1981). "A Gaussian quadrature procedure for use in the solution of the Boltzmann equation and related problems". J. Comput. Phys. 41: 309–328. doi:10.1016/0021-9991(81)90099-1.
For tables of Gauss-Hermite abscissae and weights up to order n = 32 see http://www.efunda.com/math/num_integration/findgausshermite.cfm.
Generalized Gauss–Hermite quadrature, free software in C++, Fortran, and Matlab
Retrieved from "https://en.wikipedia.org/w/index.php?title=Gauss–Hermite_quadrature&oldid=1035185973"
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Truncated 5-cell - WikiMili, The Best Wikipedia Reader
Schlegel diagrams centered on [3,3] (cells at opposite at [3,3])
In geometry, a truncated 5-cell is a uniform 4-polytope (4-dimensional uniform polytope) formed as the truncation of the regular 5-cell.
Related regular skew polyhedron
Disphenoidal 30-cell
There are two degrees of truncations, including a bitruncation.
Cells 10 5 (3.3.3)
Equilateral-triangular pyramid
Symmetry group A4, [3,3,3], order 120
Properties convex, isogonal
Uniform index 2 3 4
The truncated 5-cell, truncated pentachoron or truncated 4-simplex is bounded by 10 cells: 5 tetrahedra, and 5 truncated tetrahedra. Each vertex is surrounded by 3 truncated tetrahedra and one tetrahedron; the vertex figure is an elongated tetrahedron.
The truncated 5-cell may be constructed from the 5-cell by truncating its vertices at 1/3 of its edge length. This transforms the 5 tetrahedral cells into truncated tetrahedra, and introduces 5 new tetrahedral cells positioned near the original vertices.
The truncated tetrahedra are joined to each other at their hexagonal faces, and to the tetrahedra at their triangular faces.
Seen in a configuration matrix, all incidence counts between elements are shown. The diagonal f-vector numbers are derived through the Wythoff construction, dividing the full group order of a subgroup order by removing one mirror at a time. [1]
k-figure
20 1 3 3 3 3 1 {3}v( ) A4/A2 = 5!/3! = 20
2 10 * 3 0 3 0 {3} A4/A2A1 = 5!/3!/2 = 10
A1A1 2 * 30 1 2 2 1 { }v( ) A4/A1A1 = 5!/2/2 = 30
A2A1 t{3}
6 3 3 10 * 2 0 { } A4/A2A1 = 5!/3!/2 = 10
A2 {3} 3 0 3 * 20 1 1 A4/A2 = 5!/3! = 20
A3 t{3,3}
12 6 12 4 4 5 * ( ) A4/A3 = 5!/4! = 5
{3,3} 4 0 6 0 4 * 5
The truncated tetrahedron-first Schlegel diagram projection of the truncated 5-cell into 3-dimensional space has the following structure:
The projection envelope is a truncated tetrahedron.
One of the truncated tetrahedral cells project onto the entire envelope.
One of the tetrahedral cells project onto a tetrahedron lying at the center of the envelope.
Four flattened tetrahedra are joined to the triangular faces of the envelope, and connected to the central tetrahedron via 4 radial edges. These are the images of the remaining 4 tetrahedral cells.
Between the central tetrahedron and the 4 hexagonal faces of the envelope are 4 irregular truncated tetrahedral volumes, which are the images of the 4 remaining truncated tetrahedral cells.
This layout of cells in projection is analogous to the layout of faces in the face-first projection of the truncated tetrahedron into 2-dimensional space. The truncated 5-cell is the 4-dimensional analogue of the truncated tetrahedron.
Truncated pentatope
Truncated pentachoron (Acronym: tip) (Jonathan Bowers)
The Cartesian coordinates for the vertices of an origin-centered truncated 5-cell having edge length 2 are:
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\sqrt {3 \over 2}},\ \pm {\sqrt {3}},\ \pm 1\right)}
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\sqrt {3 \over 2}},\ 0,\ \pm 2\right)}
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\frac {-1}{\sqrt {6}}},\ {\frac {2}{\sqrt {3}}},\ \pm 2\right)}
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\frac {-1}{\sqrt {6}}},\ {\frac {-4}{\sqrt {3}}},\ 0\right)}
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\frac {-5}{\sqrt {6}}},\ {\frac {1}{\sqrt {3}}},\ \pm 1\right)}
{\displaystyle \left({\frac {3}{\sqrt {10}}},\ {\frac {-5}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ 0\right)}
{\displaystyle \left(-{\sqrt {2 \over 5}},\ {\sqrt {2 \over 3}},\ {\frac {2}{\sqrt {3}}},\ \pm 2\right)}
{\displaystyle \left(-{\sqrt {2 \over 5}},\ {\sqrt {2 \over 3}},\ {\frac {-4}{\sqrt {3}}},\ 0\right)}
{\displaystyle \left(-{\sqrt {2 \over 5}},\ -{\sqrt {6}},\ 0,\ 0\right)}
{\displaystyle \left({\frac {-7}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {1}{\sqrt {3}}},\ \pm 1\right)}
{\displaystyle \left({\frac {-7}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ 0\right)}
{\displaystyle \left({\frac {-7}{\sqrt {10}}},\ -{\sqrt {3 \over 2}},\ 0,\ 0\right)}
More simply, the vertices of the truncated 5-cell can be constructed on a hyperplane in 5-space as permutations of (0,0,0,1,2) or of (0,1,2,2,2). These coordinates come from positive orthant facets of the truncated pentacross and bitruncated penteract respectively.
The convex hull of the truncated 5-cell and its dual (assuming that they are congruent) is a nonuniform polychoron composed of 60 cells: 10 tetrahedra, 20 octahedra (as triangular antiprisms), 30 tetrahedra (as tetragonal disphenoids), and 40 vertices. Its vertex figure is a hexakis triangular cupola.
Schlegel diagram with alternate cells hidden.
Cells 10 (3.6.6)
({ }v{ })
dual polytope Disphenoidal 30-cell
Symmetry group Aut(A4), [[3,3,3]], order 240
Properties convex, isogonal, isotoxal, isochoric
The bitruncated 5-cell (also called a bitruncated pentachoron, decachoron and 10-cell) is a 4-dimensional polytope, or 4-polytope, composed of 10 cells in the shape of truncated tetrahedra.
Topologically, under its highest symmetry, [[3,3,3]], there is only one geometrical form, containing 10 uniform truncated tetrahedra. The hexagons are always regular because of the polychoron's inversion symmetry, of which the regular hexagon is the only such case among ditrigons (an isogonal hexagon with 3-fold symmetry).
E. L. Elte identified it in 1912 as a semiregular polytope.
Each hexagonal face of the truncated tetrahedra is joined in complementary orientation to the neighboring truncated tetrahedron. Each edge is shared by two hexagons and one triangle. Each vertex is surrounded by 4 truncated tetrahedral cells in a tetragonal disphenoid vertex figure.
The bitruncated 5-cell is the intersection of two pentachora in dual configuration. As such, it is also the intersection of a penteract with the hyperplane that bisects the penteract's long diagonal orthogonally. In this sense it is a 4-dimensional analog of the regular octahedron (intersection of regular tetrahedra in dual configuration / tesseract bisection on long diagonal) and the regular hexagon (equilateral triangles / cube). The 5-dimensional analog is the birectified 5-simplex, and the
{\displaystyle n}
-dimensional analog is the polytope whose Coxeter–Dynkin diagram is linear with rings on the middle one or two nodes.
The bitruncated 5-cell is one of the two non-regular uniform 4-polytopes which are cell-transitive. The other is the bitruncated 24-cell, which is composed of 48 truncated cubes.
This 4-polytope has a higher extended pentachoric symmetry (2×A4, [[3,3,3]]), doubled to order 240, because the element corresponding to any element of the underlying 5-cell can be exchanged with one of those corresponding to an element of its dual.
Bitruncated 5-cell (Norman W. Johnson)
10-cell as a cell-transitive 4-polytope
Bitruncated pentachoron
Bitruncated pentatope
Decachoron (Acronym: deca) (Jonathan Bowers)
[[5]] = [10] [4] [[3]] = [6]
stereographic projection of spherical 4-polytope
(centred on a hexagon face)
Net (polytope)
The Cartesian coordinates of an origin-centered bitruncated 5-cell having edge length 2 are:
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ {\frac {5}{\sqrt {6}}},\ {\frac {2}{\sqrt {3}}},\ 0\right)}
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ {\frac {5}{\sqrt {6}}},\ {\frac {-1}{\sqrt {3}}},\ \pm 1\right)}
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {4}{\sqrt {3}}},\ 0\right)}
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ \pm 2\right)}
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ -{\sqrt {\frac {3}{2}}},\ \pm {\sqrt {3}},\ \pm 1\right)}
{\displaystyle \pm \left({\sqrt {\frac {5}{2}}},\ -{\sqrt {\frac {3}{2}}},\ 0,\ \pm 2\right)}
{\displaystyle \pm \left(0,\ 2{\sqrt {\frac {2}{3}}},\ {\frac {4}{\sqrt {3}}},\ 0\right)}
{\displaystyle \pm \left(0,\ 2{\sqrt {\frac {2}{3}}},\ {\frac {-2}{\sqrt {3}}},\ \pm 2\right)}
More simply, the vertices of the bitruncated 5-cell can be constructed on a hyperplane in 5-space as permutations of (0,0,1,2,2). These represent positive orthant facets of the bitruncated pentacross. Another 5-space construction, centered on the origin are all 20 permutations of (-1,-1,0,1,1).
The bitruncated 5-cell can be seen as the intersection of two regular 5-cells in dual positions. = ∩ .
Isotopic uniform truncated simplices
t{3} = {6} Octahedron
r{3,3} = {31,1} = {3,4}
{\displaystyle \left\{{\begin{array}{l}3\\3\end{array}}\right\}}
Decachoron
2t{33} Dodecateron
2r{34} = {32,2}
{\displaystyle \left\{{\begin{array}{l}3,3\\3,3\end{array}}\right\}}
Tetradecapeton
3t{35} Hexadecaexon
{\displaystyle \left\{{\begin{array}{l}3,3,3\\3,3,3\end{array}}\right\}}
Octadecazetton
4t{37}
{ }×{ }
{3}v{3} {3,3}x{3,3}
{3,3}v{3,3}
{3} t{3,3} r{3,3,3} 2t{3,3,3,3} 2r{3,3,3,3,3} 3t{3,3,3,3,3,3}
A 3D net for {6,4|3}, with pairs of yellow triangles folded together into 4D and removed
The regular skew polyhedron, {6,4|3}, exists in 4-space with 4 hexagonal around each vertex, in a zig-zagging nonplanar vertex figure. These hexagonal faces can be seen on the bitruncated 5-cell, using all 60 edges and 30 vertices. The 20 triangular faces of the bitruncated 5-cell can be seen as removed. The dual regular skew polyhedron, {4,6|3}, is similarly related to the square faces of the runcinated 5-cell.
Type perfect [2] polychoron
Symbol f1,2A4 [2]
Cells 30 congruent tetragonal disphenoids
Faces 60 congruent isosceles
(2 short edges)
Edges 40 20 of length
{\displaystyle \scriptstyle 1}
20 of length
{\displaystyle \scriptstyle {\sqrt {3/5}}}
Dual Bitruncated 5-cell
Coxeter group Aut(A4), [[3,3,3]], order 240
Orbit vector (1, 2, 1, 1)
Properties convex, isochoric
The disphenoidal 30-cell is the dual of the bitruncated 5-cell. It is a 4-dimensional polytope (or polychoron) derived from the 5-cell. It is the convex hull of two 5-cells in opposite orientations.
Being the dual of a uniform polychoron, it is cell-transitive, consisting of 30 congruent tetragonal disphenoids. In addition, it is vertex-transitive under the group Aut(A4).
These polytope are from a set of 9 uniform 4-polytope constructed from the [3,3,3] Coxeter group.
3r{3,3,3}
r2r{3,3,3}
t2r{3,3,3}
A3 Coxeter plane
A cuboctahedron is a polyhedron with 8 triangular faces and 6 square faces. A cuboctahedron has 12 identical vertices, with 2 triangles and 2 squares meeting at each, and 24 identical edges, each separating a triangle from a square. As such, it is a quasiregular polyhedron, i.e. an Archimedean solid that is not only vertex-transitive but also edge-transitive. It is radially equilateral.
In geometry, the truncated tetrahedron is an Archimedean solid. It has 4 regular hexagonal faces, 4 equilateral triangle faces, 12 vertices and 18 edges. It can be constructed by truncating all 4 vertices of a regular tetrahedron at one third of the original edge length.
In geometry, the 5-cell is the convex 4-polytope with Schläfli symbol {3,3,3}. It is a 5-vertex four-dimensional object bounded by five tetrahedral cells. It is also known as a C5, pentachoron, pentatope, pentahedroid, or tetrahedral pyramid. It is the 4-simplex (Coxeter's polytope), the simplest possible convex 4-polytope, and is analogous to the tetrahedron in three dimensions and the triangle in two dimensions. The 5-cell is a 4-dimensional pyramid with a tetrahedral base and four tetrahedral sides.
The cubic honeycomb or cubic cellulation is the only proper regular space-filling tessellation in Euclidean 3-space made up of cubic cells. It has 4 cubes around every edge, and 8 cubes around each vertex. Its vertex figure is a regular octahedron. It is a self-dual tessellation with Schläfli symbol {4,3,4}. John Horton Conway called this honeycomb a cubille.
The icosahedral honeycomb is one of four compact regular space-filling tessellations in hyperbolic 3-space. With Schläfli symbol {3,5,3}, there are three icosahedra around each edge, and 12 icosahedra around each vertex, in a regular dodecahedral vertex figure.
In geometry, a truncated 120-cell is a uniform 4-polytope formed as the truncation of the regular 120-cell.
In geometry, a rhombicuboctahedral prism is a convex uniform polychoron.
In the field of hyperbolic geometry, the hexagonal tiling honeycomb is one of 11 regular paracompact honeycombs in 3-dimensional hyperbolic space. It is paracompact because it has cells composed of an infinite number of faces. Each cell is a hexagonal tiling whose vertices lie on a horosphere, a surface in hyperbolic space that approaches a single ideal point at infinity.
In hyperbolic 3-space, the order-6 tetrahedral honeycomb is a paracompact regular space-filling tessellation. It is paracompact because it has vertex figures composed of an infinite number of faces, and has all vertices as ideal points at infinity. With Schläfli symbol {3,3,6}, the order-6 tetrahedral honeycomb has six ideal tetrahedra around each edge. All vertices are ideal, with infinitely many tetrahedra existing around each vertex in a triangular tiling vertex figure.
The order-6 cubic honeycomb is a paracompact regular space-filling tessellation in hyperbolic 3-space. It is paracompact because it has vertex figures composed of an infinite number of facets, with all vertices as ideal points at infinity. With Schläfli symbol {4,3,6}, the honeycomb has six ideal cubes meeting along each edge. Its vertex figure is an infinite triangular tiling. Its dual is the order-4 hexagonal tiling honeycomb.
The triangular tiling honeycomb is one of 11 paracompact regular space-filling tessellations in hyperbolic 3-space. It is called paracompact because it has infinite cells and vertex figures, with all vertices as ideal points at infinity. It has Schläfli symbol {3,6,3}, being composed of triangular tiling cells. Each edge of the honeycomb is surrounded by three cells, and each vertex is ideal with infinitely many cells meeting there. Its vertex figure is a hexagonal tiling.
In geometry of 4 dimensions, a 4-6 duoprism, a duoprism and 4-polytope resulting from the Cartesian product of a square and a hexagon.
Coxeter, The Beauty of Geometry: Twelve Essays, Dover Publications, 1999, ISBN 0-486-40919-8 p. 88 (Chapter 5: Regular Skew Polyhedra in three and four dimensions and their topological analogues, Proceedings of the London Mathematics Society, Ser. 2, Vol 43, 1937.)
Coxeter, H. S. M. Regular Skew Polyhedra in Three and Four Dimensions. Proc. London Math. Soc. 43, 33-62, 1937.
1. Convex uniform polychora based on the pentachoron - Model 3 , George Olshevsky.
Klitzing, Richard. "4D uniform polytopes (polychora)". x3x3o3o - tip, o3x3x3o - deca
↑ Klitzing, Richard. "x3x4o3o - tip".
1 2 On Perfect 4-Polytopes Gabor Gévay Contributions to Algebra and Geometry Volume 43 (2002), No. 1, 243-259 ] Table 2, page 252
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How to use the metric thread pitch diameter calculator?
Different pitch diameter formulas
How to calculate thread pitch diameter example
This pitch diameter calculator will help you find one of the essential dimensions of threaded fasteners – the pitch diameter. In this calculator, you will learn:
What pitch diameter is;
How to use the metric thread pitch diameter calculator;
The different pitch diameter formulas; and
How to calculate thread pitch diameter.
Ready to learn about pitch diameters? Then keep on reading 🙂.
Threads on threaded fasteners have three different diameters essential in determining the proper fit of bolts in nuts or bolts in threaded holes. These three diameters are the major diameter, minor diameter, and pitch diameter, as shown in the diagram below:
We can also see in the diagram above that the thickness of the thread and the space between two threads are equal along the pitch diameter. At the same time, the sum of these distances also equates to the thread's pitch, hence the name pitch diameter.
Before we learn how to find a pitch diameter's value, let us first discuss how to use this pitch diameter calculator in the next section of this text.
🙋 Please note that this tool only works as an ISO Metric Thread pitch diameter calculator. That means all the calculations this tool performs are for 60º-angle threads under ISO Metric Standards only.
Let's say we want to calculate the pitch diameter of an
\text{M25 × 1.5 - 6g}
bolt. Since this is a callout for a metric bolt thread, our first step is to:
Pick external thread dimensions for the option of what we want to calculate.
Next, we select
\small{1.5\ \text{mm}}
for the thread pitch and enter
\small{25\ \text{mm}}
for the Basic major diameter.
We then select
\small{6}
\small{\text{g}}
for the tolerance grade and tolerance position, respectively.
Upon completing these steps, we should already get the values for the basic pitch diameter of
\small{24.026\ \text{mm}}
, the maximum pitch diameter of
\small{23.994\ \text{mm}}
, and the minimum pitch diameter of
\small{23.850\ \text{mm}}
If you want to see the preliminary values used to obtain these pitch diameter values, go to the advanced mode of the calculator. Doing so will display the values for the fundamental deviation and tolerances, upper deviation (
\small{es}
) and pitch diameter tolerance (
\small{T_\text{d2}}
), and the fundamental triangle height (
\small{H}
) of the thread.
In the next section of this text, we'll discuss the different formulas we use in our pitch diameter calculator.
From our discussion of how to use the pitch diameter calculator, we know that we need the values of the basic major diameter and the pitch of the thread we are investigating. Knowing those values, we can easily find the basic pitch diameter using this formula:
d_2 = d - \left(3\times \sqrt{3} \times \frac{P}{8}\right)
d_2
– Basic pitch diameter;
d
– Basic major diameter; and
P
– Thread pitch.
We derive this formula by substituting the value of the thread's fundamental triangle height,
H
\small{(\sqrt{3}) \times \frac{P}{2}}
into this equation
\small{d_2 = d - (2 \times \frac{3}{8}}) \times H
, where it means that
\small{\frac{3}{8}\ \text{of}\ H}
gets deducted from both sides of the basic major diameter, as illustrated below:
💡 An alternate equation to find pitch diameter is:
\small{d_2 = d - 0.6495 \times P}
\small{3 \times \frac{\sqrt{3}}{8}}
part of the previously given formula is equal to 0.6495190528 or approximately 0.6495.
Now that we know how to find the pitch diameter's base value, let us now figure out the maximum and minimum limits of the pitch diameter. Here are the equations that we use to find those values:
For external thread (bolt or screw) pitch diameter (d₂):
d_\text{2max} = d_2 + es
d_\text{2min} = d_2 + es - T_\text{d2}
For internal thread (nut or threaded hole) pitch diameter (D₂):
D_\text{2max} = D_{2} + EI + T_\text{D2}
D_\text{2min} = D_{2} + EI
es
EI
– fundamental deviations, upper and lower deviations, respectively; and
T_\text{d2}
T_\text{D2}
– thread pitch diameter tolerance for external and internal threads, respectively.
The fundamental deviations and tolerances are the minute allowances that we apply on the threads to give the threaded fasteners some wiggle room when they are threaded together. We can either have them fit very tightly or have more wiggle room, but we cannot file down threads such that they become very loose even at their recommended bolt torques.
We can calculate the fundamental deviations,
es
EI
, depending on the thread's tolerance position:
For external threads:
For e position:
\small{es = -(50 + 11 \times P) / 1000}
For f position:
\small{es = -(30 + 11 \times P) / 1000}
For g position:
\small{es = -(15 + 11 \times P) / 1000}
For h position:
\small{es = 0}
For internal threads:
\small{EI = (15 + 11 \times P) / 1000}
\small{EI = 0}
On the other hand, we determine the tolerance values using these equations for external and internal threads, respectively:
\small{T_\text{d2}(n) = k \times (90 \times P^{0.4} \times d^{0.1} / 1000)}
\small{T_\text{D2}(n) = k \times (90 \times P^{0.4} \times d^{0.1} / 1000)}
They are rather the same, but their multipliers,
\small{k}
, depend on the tolerance grade of the thread indicated by the value of
\small{n}
. For a tolerance grade of 6 for an external thread, we should use
\small{1.0}
for the value of
\small{k}
. We can see the other values of
\small{k}
in the table below:
for Td2(n)
Let's say we have an M30×2-6h bolt, and we want to determine its screw pitch diameter. We know that our bolt has a 30-mm basic major diameter and a 2-mm thread pitch given the thread callout. From there, we can already calculate our bolt's basic pitch diameter, as shown below:
\scriptsize \begin{align*} d_2 &= d - \left(3\times \sqrt{3} \times \frac{P}{8}\right)\\ &= 30\ \text{mm} - \left(3\times \sqrt{3} \times \frac{2\ \text{mm}}{8}\right)\\ &= 28.701\ \text{mm} \end{align*}
Since our bolt has a tolerance class of 6h, we can use the table in the previous section to find the multiplier
\small{k}
to be equal to 1, and from the list of fundamental deviation formulas, we know that
\small{es = 0\ \text{mm}}
. We can then obtain the tolerance,
\small{T_{d2}}
with this equation:
\scriptsize \begin{align*} T_\text{d2}(n) &= k \times \frac{(90 \times P^{0.4} \times d^{0.1})}{1000}\\\\ T_\text{d2}(6) &= 1 \times \frac{(90 \times 2\ \text{mm}^{0.4} \times 30\ \text{mm}^{0.1})}{1000}\\\\ &= (90 \times 1.31950 \times 1.40511) / 1000\\ &= 0.16687\ \text{mm} \end{align*}
Finally, we can solve for the maximum pitch diameter, as follows:
\scriptsize \begin{align*} d_\text{2max} &= d_2 + es\\ &= 28.701\ \text{mm} + 0\ \text{mm}\\ &= 28.701\ \text{mm} \end{align*}
And the minimum pitch diameter:
\scriptsize \begin{align*} d_\text{2min} &= d_2 + es - T_{d2}\\ &= 28.701\ \text{mm} + 0\ \text{mm} - 0.16687\ \text{mm}\\ &= 28.534\ \text{mm} \end{align*}
If you found this calculator useful, perhaps you'll also find our bolt circle calculator informational. That tool will help you estimate the location of holes in a circular pattern, especially when drilling bolt holes on a flange.
What is the pitch diameter of a gear?
The pitch diameter of a gear is the diameter of the gear's pitch circle, which has a radius equal to the distance from the gear's center to its pitch point. The pitch point is the point wherein two gears interact with each other. Unlike in threaded fasteners, where external and internal threads should have the same thread pitch and pitch diameter to mate perfectly, gears can mate together even when they don't have the same pitch diameters as long as they have the same gear pitch.
How do I find the pitch diameter?
Let's say you have a metric thread bolt with an outside diameter (or major diameter) of 20 mm and has threads that are 1.5 mm apart.
Multiply 1.5 mm by 0.6495 to get 0.9743 mm.
Subtract 0.9743 mm from the 20-mm outside diameter to obtain the pitch diameter of 19.0257 mm.
How do I find the pitch diameter of an internal thread?
You can use the formula pitch diameter = major diameter - 0.6495 × thread pitch if you know the major diameter of the internal thread in question. On the other hand, you can also measure the minor diameter of an internal thread using a pair of calipers and then use this equation: pitch diameter = minor diameter + 0.4330 × thread pitch.
What is the pitch diameter limits?
Pitch diameter limits are the maximum and minimum pitch diameters that a threaded fastener can have. Setting limits for thread diameters provides standard allowances for every manufactured thread depending on preferred tolerance. Exceeding these limits can lead to bolts and nuts of supposedly the same measurements to either loosen when fit together or not fit at all.
external thread dimensions
Tolerance class details
External thread pitch diameters
Minimum pitch diameter
The roofing calculator lets you estimate the area and cost of your roof.
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Definition of the Bivariate Limit
It is first necessary to distinguish between interior and boundary points of the domain of a function
f
A point P is an interior point of the domain of a function
f
if P is contained in a neighborhood that lies completely in the domain of
A point P is a boundary point of the domain of a function
f
if every neighborhood of P contains points that are in the domain and points that are not in the domain.
Definition 3.2.1 formalizes the meaning of a limit in the plane, called in this guide, the bivariate limit.
Definition 3.2.1: The Bivariate Limit
If P:
\left(a,b\right)
is an interior point of the domain of
L
is the bivariate limit of
f\left(x,y\right)
at P, that is,
\underset{\left(x,y\right)→P}{lim} f\left(x,y\right)=L
, when, for every number
\mathrm{ε}>0
there is a corresponding number
\mathrm{δ}
0<\sqrt{{\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}}<\mathrm{δ}
\left|f\left(x,y\right)-L\right|<\mathrm{ε}
\left(a,b\right)
is a boundary point of the domain of
\underset{\left(x,y\right)→P}{lim} f\left(x,y\right)=L
\mathrm{ε}>0
\mathrm{δ}
0<\sqrt{{\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}}<\mathrm{δ}
\left(x,y\right)
is in the domain of
\left|f\left(x,y\right)-L\right|<\mathrm{ε}
The points satisfying
0<\sqrt{{\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}}<\mathrm{\delta }
are said to lie in a deleted neighborhood of
\left(a,b\right)
. This deleted neighborhood is actually the interior of an annulus that is an open disk of radius
\mathrm{δ}
with the center point
\left(a,b\right)
removed.
Conceptually, Definition 3.2.1 is the generalization of the formal definition of a limit along the real line: the values of
L
by the expedient of taking
\left(x,y\right)
sufficiently close to P. Thus,
L
f\left(x,y\right)
at P if all the points in the deleted neighborhood of P produce function values that are close to
L
The distinction between interior and boundary points amounts to this: The limit is taken over the domain of the function.
L
It is generally difficult to prove that
L
is the limiting value of
f\left(x,y\right)
because the requisite estimates demand a facility with manipulating inequalities.
Table 3.2.1 lists several inequalities that are useful for proving that a real number
L
is indeed the bivariate limit of
f\left(x,y\right)
. Inequality 3 is the "triangle" inequality. Inequalities 4, 5, 6, and 7 should be self-evident. A proof of Inequality 2 uses Inequality 1, which itself is proved in Example 3.2.28.
\left|x y\right|
≤\left({x}^{2}+{y}^{2}\right)/2
|x|+\left|y\right|
≤\sqrt{2} \sqrt{{x}^{2}+{y}^{2}}
|x+y|≤\left|x\right|+\left|y\right|
\left|x\right|≤\sqrt{{x}^{2}+{y}^{2}}
\left|y\right|≤\sqrt{{x}^{2}+{y}^{2}}
{x}^{2}≤{x}^{2}+{y}^{2}
{y}^{2}≤{x}^{2}+{y}^{2}
Table 3.2.1 Useful inequalities
The Bivariate Limit in Maple
Maple's limit command can determine the bivariate limit of rational functions in two variables. Access to this functionality is provided through the Context Panel in the option Limit (Bivariate).
Showing the Limit Does Not Exist
It is far easier to show that
f\left(x,y\right)
does not have a limit at
\left(a,b\right)
. Recall that for a limit on the line to exist, both left-hand and right-hand limits must exist, and be equal. The same idea holds for the bivariate limit. If the limits taken along two different paths are not equal, then the (bivariate) limit cannot exist. Consequently, to show a bivariate limit does not exist at
\left(a,b\right)
, it suffices to show that along two different paths through
\left(a,b\right)
the limits differ, or that one such limit does not exist.
Typical trial paths are the axes, lines
y=m x
, parabolas
y={x}^{2}
x={y}^{2}
, and on rare occasions, the curves
y={x}^{3/2}
y={x}^{2/3}
Just as with functions of a single variable, continuity is defined in terms of the limit. Essentially, a function is continuous at a point if it is defined at that point, and its defined value equals its bivariate limit at that point. If the limit point is interior to the domain of the function, the first part of Definition 3.2.1 applies; if a boundary point, the second.
Definition 3.2.2: Continuity at a Point
f
is continuous at P:
\left(a,b\right)
\underset{\left(x,y\right)→P}{lim} f =f\left(a,b\right)
If
is continuous at every point in its domain, then
f
is said to be a continuous function on that domain.
Just as with functions of a single variable, composition of continuous functions results in a continuous function. This is formalized in Theorem 3.2.1.
Theorem 3.2.1: Composition of Continuous Functions
f\left(x,y\right)
\left(a,b\right)
g\left(x\right)
f\left(a,b\right)
h=g∘f
h\left(x,y\right)=g\left(f\left(x,y\right)\right)
h
is continuous at P
Let P be the generic point
\left(x,y\right)
and O, the origin,
\left(0,0\right)
f
in Examples 3.2.(1-10), show that
\underset{\mathrm{P}→\mathrm{O}}{\mathrm{lim}}f\left(x,y\right)
, the bivariate limit at the origin, does not exist.
f=\frac{{x}^{2}}{{x}^{2}+{y}^{2}}
f=\frac{x-y}{x+y}
f=\frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}
f=\frac{x y}{{x}^{4}+{y}^{4}}
f=\frac{x-y}{{x}^{2}+{y}^{2}}
f=\frac{{x}^{3}{y}^{2}}{{x}^{6}+{y}^{4}}
f=\frac{x {y}^{2}}{{x}^{2}+{y}^{4}}
f=\frac{{x}^{4}+2 {x}^{2}{y}^{2}+3 x {y}^{3}}{{\left({x}^{2}+{y}^{2}\right)}^{2}}
f=\frac{{x}^{2}+y}{\sqrt{{x}^{2}+{y}^{2}}}
f=\frac{{x}^{4}{y}^{4}}{{\left({x}^{2}+{y}^{4}\right)}^{3}}
f=\frac{x y}{{x}^{2}+{y}^{2}}
the bivariate limit at the origin does not exist, but the iterated limits
\underset{y→0}{lim}\left(\underset{x→0}{lim}f\right)
\underset{x→0}{lim}\left(\underset{y→0}{lim}f\right)
are both zero.
f=\frac{x y}{{x}^{3}+{y}^{2}}
\underset{y→0}{lim}\left(\underset{x→0}{lim}f\right)
\underset{x→0}{lim}\left(\underset{y→0}{lim}f\right)
Prove that the bivariate limit at the origin for
f=\frac{{x}^{3}}{{x}^{2}+{y}^{2}}
f=\frac{2 {x}^{3}-{y}^{3}}{{x}^{2}+{y}^{2}}
f=x y \frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}
f=\frac{{x}^{2}{y}^{2}}{{x}^{2}+{y}^{2}}
f=\frac{{x}^{4}+{y}^{4}}{{x}^{2}+{y}^{2}}
f=\frac{2 {x}^{5}+2 {y}^{3}\left(2 {x}^{2}-{y}^{2}\right)}{{\left({x}^{2}+{y}^{2}\right)}^{2}}
f\left(x,y\right)=2 {x}^{2}-6 x y+5 {y}^{2},
\sqrt{{x}^{2}+{y}^{2}}<\sqrt{\mathrm{ε}/8}=\mathrm{δ}
|f\left(x,y\right)|<\mathrm{ε}
Prove the inequality
\left|{x}^{3}-{y}^{3}\right|≤{\left({x}^{2}+{y}^{2}\right)}^{3/2}
f=x \mathrm{sin}\left(1/y\right)+y \mathrm{sin}\left(1/x\right)
the bivariate limit at the origin is zero, but both of the iterated limits
\underset{y→0}{lim}\left(\underset{x→0}{lim}f\right)
\underset{x→0}{lim}\left(\underset{y→0}{lim}f\right)
fail to exist.
Hint: Show
\left|f\right|
≤\left|x\right|+\left|y\right|≤2\sqrt{{x}^{2}+{y}^{2}}
f=\frac{x y}{{x}^{2}+{y}^{2}}+x \mathrm{sin}\left(1/y\right)
the bivariate limit at the origin and the iterated limit
\underset{x→0}{lim}\left(\underset{y→0}{lim}f\right)
both fail to exist, but the iterated limit
\underset{y→0}{lim}\left(\underset{x→0}{lim}f\right)
Show that the bivariate limit at the origin for
f={\begin{array}{cc}\frac{\left|x\right|}{{y}^{2}}{e}^{-|x|/{y}^{2}}& y≠0\\ 0& y=0\end{array}
f=\frac{{x}^{2}y\mathrm{cos}\left(xy\right)}{{x}^{2}+{y}^{2}}
g\left(x,y\right)
that is continuous at the origin.
f=\frac{\mathrm{tan}\left(xy\right)}{\mathrm{tan}\left(x\right)\mathrm{tan}\left(y\right)}
g\left(x,y\right)
f=\frac{x\mathrm{sin}\left(xy\right)}{2-\mathrm{cos}\left(x\right)-\mathrm{cos}\left(y\right)}
g\left(x,y\right)
f=\frac{{x}^{4}+{x}^{2}+x {y}^{2}+{y}^{2}}{{x}^{2}-{x}^{2}y+{y}^{2}}
g\left(x,y\right)
Prove Inequalities 1 and 2 in Table 3.2.1.
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Other units of capacitance
How to use the capacitance converter
Omni's capacitance converter allows you to quickly convert between different units of capacitance. So whether you want the conversion between µF to nF or nF to pF, you can use our tool.
If you want to learn more about the different units of capacitance, keep scrolling. You will also find:
Dimensional formula for capacitance; and
An example of using our calculator for capacitor conversion.
Let's start with the SI unit of capacitance.
As we know, to express any physical quantity, we also need to specify its unit. In the international system of units, i.e., SI units, we represent electrical capacitance in farad (or F).
Before going any further, let us first try to understand what a farad is?
We define the capacitance
C
of a capacitor as the ratio:
\quad C = \frac{Q}{V}
Q
– Maximum charge that can be stored in the capacitor; and
V
– Voltage applied across the plates of a capacitor.
Q = 1\ \rm C
V=1\ \rm V
, we can write one farad as:
\quad 1 \ \rm F = \frac{1\ C}{1\ V}
Hence, we can define 1 F as the capacitance of a capacitor that can store 1 C of charge when we apply a potential difference of 1 V across its plate.
You may not realize, but 1 coulomb is actually an enormous amount of charge. If we use Coulomb's law to calculate the force between two charges of 1 C separated by a distance of 1 m, we get:
\quad \begin{align*} F & = k_e \frac{q_1q_2}{r^2}\\ F &= 9 \times 10^9 \times \frac{1\ \rm C \times 1\ C}{1 \ \rm m}\\ F & = 9 \times 10^9 \ \rm N \end{align*}
Using our force converter, you can see that this force is about
2.0 \times 10^{9} \ \rm lbf
To get an idea about the magnitude of the force mentioned above, you can compare it with the maximum thrust of
2.8 \times 10^{6} \ \rm lbf
exerted by a space shuttle solid rocket booster during liftoff!
Therefore, one farad is also a massive value for capacitance and is not very commonly used. The capacitance values that we use typically range from picofarads (pF) to millifarads (mF).
The different units of capacitance are related as:
1\ \rm pF\ (picofarad) = 10^{-12}\ F
1\ \rm nF\ (nanofarad) = 10^{-9}\ F
1\ \rm \mu F\ (microfarad) = 10^{-6}\ F
1\ \rm mF\ (millifarad)= 10^{-3}\ F
Hence, if we want to convert between different units, say, from F to nF, we must multiply the capacitance in farads by
10^9
. So, a capacitance of
10\ \rm F
is equivalent to a capacitance of
10 \times 10^9\ \rm nF
In the next section, we will see how to use the capacitance unit conversion calculator to change between different units.
To use the capacitor conversion calculator to perform the same calculations, follow these instructions:
Enter the value of capacitance against the given unit field. Say we want to convert
10\ \rm F
, enter the value 10 in the first row.
The calculator will convert the capacitance to different units and display it in the respective rows.
So, now you know that a capacitance of
10\ \rm F
10,000\ \rm mF
10^7\ \rm \mu F
10^{10}\ \rm nF
10^{13}\ \rm pF
The SI unit of capacitance is the farad (denoted by the symbol F). The unit is named after the English scientist Michael Faraday for his significant contributions to the field of electromagnetism.
[M⁻¹ L⁻² T⁴ I²]. Capacitance is the charge per unit voltage (Q/V), and voltage is work done per unit charge (W/Q). Hence, we can write capacitance as Q²/W.
The dimensional formula for the charge is [I T], and work is [M¹ L² T⁻² ]. Consequently, the dimensional formula for capacitance is [M⁻¹ L⁻² T⁴ I²].
How do I convert pF to nF?
We know that, 1 nF (nanoFarad) = 1000 pF (picoFarad). Hence, to convert a given capacitance in pF to nF, divide the capacitance by 1000. Whereas to convert nF to pF, multiply the capacitance by 1000.
How do I convert µF to F?
To convert capacitance in µF (microFarad) to F (Farad), proceed as follows:
Using the information, 1 µF = 10⁻⁶ F, divide the capacitance in µF by 1,000,000 to get the capacitance in farads.
To convert from F to µF, multiply the capacitance by 1,000,000.
Enter the Capacitance value
Our Kelvin to Celsius converter will help you convert the temperature from Kelvin to Celsius scales and vice versa.
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Section 59.22 (03OY): Cohomology of quasi-coherent sheaves—The Stacks project
Section 59.22: Cohomology of quasi-coherent sheaves (cite)
59.22 Cohomology of quasi-coherent sheaves
We start with a simple lemma (which holds in greater generality than stated). It says that the Čech complex of a standard covering is equal to the Čech complex of an fpqc covering of the form $\{ \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)\} $ with $A \to B$ faithfully flat.
Lemma 59.22.1. Let $\tau \in \{ fppf, syntomic, smooth, {\acute{e}tale}, Zariski\} $. Let $S$ be a scheme. Let $\mathcal{F}$ be an abelian sheaf on $(\mathit{Sch}/S)_\tau $, or on $S_\tau $ in case $\tau = {\acute{e}tale}$, and let $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a standard $\tau $-covering of this site. Let $V = \coprod _{i \in I} U_ i$. Then
$V$ is an affine scheme,
$\mathcal{V} = \{ V \to U\} $ is an fpqc covering and also a $\tau $-covering unless $\tau = Zariski$,
the Čech complexes $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ and $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F})$ agree.
Proof. The defintion of a standard $\tau $-covering is given in Topologies, Definition 34.3.4, 34.4.5, 34.5.5, 34.6.5, and 34.7.5. By definition each of the schemes $U_ i$ is affine and $I$ is a finite set. Hence $V$ is an affine scheme. It is clear that $V \to U$ is flat and surjective, hence $\mathcal{V}$ is an fpqc covering, see Example 59.15.3. Excepting the Zariski case, the covering $\mathcal{V}$ is also a $\tau $-covering, see Topologies, Definition 34.4.1, 34.5.1, 34.6.1, and 34.7.1.
Note that $\mathcal{U}$ is a refinement of $\mathcal{V}$ and hence there is a map of Čech complexes $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$, see Cohomology on Sites, Equation (21.8.2.1). Next, we observe that if $T = \coprod _{j \in J} T_ j$ is a disjoint union of schemes in the site on which $\mathcal{F}$ is defined then the family of morphisms with fixed target $\{ T_ j \to T\} _{j \in J}$ is a Zariski covering, and so
\begin{equation} \label{etale-cohomology-equation-sheaf-coprod} \mathcal{F}(T) = \mathcal{F}(\coprod \nolimits _{j \in J} T_ j) = \prod \nolimits _{j \in J} \mathcal{F}(T_ j) \end{equation}
by the sheaf condition of $\mathcal{F}$. This implies the map of Čech complexes above is an isomorphism in each degree because
\[ V \times _ U \ldots \times _ U V = \coprod \nolimits _{i_0, \ldots i_ p} U_{i_0} \times _ U \ldots \times _ U U_{i_ p} \]
as schemes. $\square$
Note that Equality (59.22.1.1) is false for a general presheaf. Even for sheaves it does not hold on any site, since coproducts may not lead to coverings, and may not be disjoint. But it does for all the usual ones (at least all the ones we will study).
Remark 59.22.2. In the statement of Lemma 59.22.1 the covering $\mathcal{U}$ is a refinement of $\mathcal{V}$ but not the other way around. Coverings of the form $\{ V \to U\} $ do not form an initial subcategory of the category of all coverings of $U$. Yet it is still true that we can compute Čech cohomology $\check H^ n(U, \mathcal{F})$ (which is defined as the colimit over the opposite of the category of coverings $\mathcal{U}$ of $U$ of the Čech cohomology groups of $\mathcal{F}$ with respect to $\mathcal{U}$) in terms of the coverings $\{ V \to U\} $. We will formulate a precise lemma (it only works for sheaves) and add it here if we ever need it.
Lemma 59.22.3 (Locality of cohomology). Let $\mathcal{C}$ be a site, $\mathcal{F}$ an abelian sheaf on $\mathcal{C}$, $U$ an object of $\mathcal{C}$, $p > 0$ an integer and $\xi \in H^ p(U, \mathcal{F})$. Then there exists a covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ of $U$ in $\mathcal{C}$ such that $\xi |_{U_ i} = 0$ for all $i \in I$.
Proof. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $. Then $\xi $ is represented by a cocycle $\tilde{\xi } \in \mathcal{I}^ p(U)$ with $d^ p(\tilde{\xi }) = 0$. By assumption, the sequence $\mathcal{I}^{p - 1} \to \mathcal{I}^ p \to \mathcal{I}^{p + 1}$ in exact in $\textit{Ab}(\mathcal{C})$, which means that there exists a covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ such that $\tilde{\xi }|_{U_ i} = d^{p - 1}(\xi _ i)$ for some $\xi _ i \in \mathcal{I}^{p-1}(U_ i)$. Since the cohomology class $\xi |_{U_ i}$ is represented by the cocycle $\tilde{\xi }|_{U_ i}$ which is a coboundary, it vanishes. For more details see Cohomology on Sites, Lemma 21.7.3. $\square$
Theorem 59.22.4. Let $S$ be a scheme and $\mathcal{F}$ a quasi-coherent $\mathcal{O}_ S$-module. Let $\mathcal{C}$ be either $(\mathit{Sch}/S)_\tau $ for $\tau \in \{ fppf, syntomic, smooth, {\acute{e}tale}, Zariski\} $ or $S_{\acute{e}tale}$. Then
\[ H^ p(S, \mathcal{F}) = H^ p_\tau (S, \mathcal{F}^ a) \]
for all $p \geq 0$ where
the left hand side indicates the usual cohomology of the sheaf $\mathcal{F}$ on the underlying topological space of the scheme $S$, and
the right hand side indicates cohomology of the abelian sheaf $\mathcal{F}^ a$ (see Proposition 59.17.1) on the site $\mathcal{C}$.
Proof. We are going to show that $H^ p(U, f^*\mathcal{F}) = H^ p_\tau (U, \mathcal{F}^ a)$ for any object $f : U \to S$ of the site $\mathcal{C}$. The result is true for $p = 0$ by the sheaf property.
Assume that $U$ is affine. Then we want to prove that $H^ p_\tau (U, \mathcal{F}^ a) = 0$ for all $p > 0$. We use induction on $p$.
Pick $\xi \in H^1_\tau (U, \mathcal{F}^ a)$. By Lemma 59.22.3, there exists an fpqc covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ such that $\xi |_{U_ i} = 0$ for all $i \in I$. Up to refining $\mathcal{U}$, we may assume that $\mathcal{U}$ is a standard $\tau $-covering. Applying the spectral sequence of Theorem 59.19.2, we see that $\xi $ comes from a cohomology class $\check\xi \in \check H^1(\mathcal{U}, \mathcal{F}^ a)$. Consider the covering $\mathcal{V} = \{ \coprod _{i\in I} U_ i \to U\} $. By Lemma 59.22.1, $\check H^\bullet (\mathcal{U}, \mathcal{F}^ a) = \check H^\bullet (\mathcal{V}, \mathcal{F}^ a)$. On the other hand, since $\mathcal{V}$ is a covering of the form $\{ \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)\} $ and $f^*\mathcal{F} = \widetilde{M}$ for some $A$-module $M$, we see the Čech complex $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F})$ is none other than the complex $(B/A)_\bullet \otimes _ A M$. Now by Lemma 59.16.4, $H^ p((B/A)_\bullet \otimes _ A M) = 0$ for $p > 0$, hence $\check\xi = 0$ and so $\xi = 0$.
Pick $\xi \in H^ p_\tau (U, \mathcal{F}^ a)$. By Lemma 59.22.3, there exists an fpqc covering $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ such that $\xi |_{U_ i} = 0$ for all $i \in I$. Up to refining $\mathcal{U}$, we may assume that $\mathcal{U}$ is a standard $\tau $-covering. We apply the spectral sequence of Theorem 59.19.2. Observe that the intersections $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ are affine, so that by induction hypothesis the cohomology groups
\[ E_2^{p, q} = \check H^ p(\mathcal{U}, \underline{H}^ q(\mathcal{F}^ a)) \]
vanish for all $0 < q < p$. We see that $\xi $ must come from a $\check\xi \in \check H^ p(\mathcal{U}, \mathcal{F}^ a)$. Replacing $\mathcal{U}$ with the covering $\mathcal{V}$ containing only one morphism and using Lemma 59.16.4 again, we see that the Čech cohomology class $\check\xi $ must be zero, hence $\xi = 0$.
Next, assume that $U$ is separated. Choose an affine open covering $U = \bigcup _{i \in I} U_ i$ of $U$. The family $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ is then an fpqc covering, and all the intersections $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ are affine since $U$ is separated. So all rows of the spectral sequence of Theorem 59.19.2 are zero, except the zeroth row. Therefore
\[ H^ p_\tau (U, \mathcal{F}^ a) = \check H^ p(\mathcal{U}, \mathcal{F}^ a) = \check H^ p(\mathcal{U}, \mathcal{F}) = H^ p(U, \mathcal{F}) \]
where the last equality results from standard scheme theory, see Cohomology of Schemes, Lemma 30.2.6.
The general case is technical and (to extend the proof as given here) requires a discussion about maps of spectral sequences, so we won't treat it. It follows from Descent, Proposition 35.9.3 (whose proof takes a slightly different approach) combined with Cohomology on Sites, Lemma 21.7.1. $\square$
Remark 59.22.5. Comment on Theorem 59.22.4. Since $S$ is a final object in the category $\mathcal{C}$, the cohomology groups on the right-hand side are merely the right derived functors of the global sections functor. In fact the proof shows that $H^ p(U, f^*\mathcal{F}) = H^ p_\tau (U, \mathcal{F}^ a)$ for any object $f : U \to S$ of the site $\mathcal{C}$.
Comment #4587 by Shiro on October 08, 2019 at 09:00
In lemma 56.22.1, for the Zariski covering case, the covering
{V \rightarrow W}
may not be an open immersion. I wonder what's the precise meaning of covering in a big Zariski topos?
Dear Shiro, you are very right and thanks very much for pointing this out. The statement of the lemma has to be modified for the case of the Zariski topology because
V \to U
isn't a Zariski covering in that case (and only in that case -- the other cases are fine). All the definitions and conventions regarding Zariski coverings are given in Section 34.3. I will fix this the next time I go through all the comments.
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SVM - Understanding the math - Part 1 - The margin - SVM Tutorial
What is a separating hyperplane?
Just by looking at the plot, we can see that it is possible to separate the data. For instance, we could trace a line and then all the data points representing men will be above the line, and all the data points representing women will be below the line.
Such a line is called a separating hyperplane and is depicted below:
If it is just a line, why do we call it an hyperplane ?
Even though we use a very simple example with data points laying in
R^2
the support vector machine can work with any number of dimensions !
A hyperplane is a generalization of a plane.
in one dimension, a hyperplane is called a point
in two dimensions, it is a line
in three dimensions, it is a plane
in more dimensions you can call it an hyperplane
The point L is a separating hyperplane in one dimension
What is the optimal separating hyperplane?
The fact that you can find a separating hyperplane, does not mean it is the best one ! In the example below there is several separating hyperplanes. Each of them is valid as it successfully separates our data set with men on one side and women on the other side.
There can be a lot of separating hyperplanes
Suppose we select the green hyperplane and use it to classify on real life data.
This hyperplane does not generalize well
This time, it makes some mistakes as it wrongly classify three women. Intuitively, we can see that if we select an hyperplane which is close to the data points of one class, then it might not generalize well.
So we will try to select an hyperplane as far as possible from data points from each category:
This one looks better. When we use it with real life data, we can see it still make perfect classification.
The black hyperplane classifies more accurately than the green one
That's why the objective of a SVM is to find the optimal separating hyperplane:
because it correctly classifies the training data
and because it is the one which will generalize better with unseen data
What is the margin and how does it help choosing the optimal hyperplane?
The margin of our optimal hyperplane
Given a particular hyperplane, we can compute the distance between the hyperplane and the closest data point. Once we have this value, if we double it we will get what is called the margin.
Basically the margin is a no man's land. There will never be any data point inside the margin. (Note: this can cause some problems when data is noisy, and this is why soft margin classifier will be introduced later)
For another hyperplane, the margin will look like this :
As you can see, Margin B is smaller than Margin A.
If an hyperplane is very close to a data point, its margin will be small.
The further an hyperplane is from a data point, the larger its margin will be.
This means that the optimal hyperplane will be the one with the biggest margin.
That is why the objective of the SVM is to find the optimal separating hyperplane which maximizes the margin of the training data.
This concludes this introductory post about the math behind SVM. There was not a lot of formula, but in the next article we will put on some numbers and try to get the mathematical view of this using geometry and vectors.
If you want to learn more read it now :
SVM - Understanding the math - Part 2 : Calculate the margin
Categories Mathematics, SVM Tutorial Tags hyperplane, math Post navigation
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User:Bernejay - Wikipedia
Bernejay
This user believes in the right of every human being to have access to Wikipedia.
This user remembers Microsoft Windows 3.1x
This user remembers
This user believes that Windows and Linux both have strengths and weaknesses and people should not argue one is better than the other
This user thinks that every user SHOULD UPGRADE to Windows 10 as it's the last version of Windows.
This user once used DOS back in the day.
This user knows that math is
{\displaystyle \int e^{xy}\,}
This user likes to create and draw graphs of functions that barely make sense during his spare time.
Pi ≈ 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679 82148 08651 32823 06647 09384 46095 50582 23172 53594 08128 48111 74502 84102 70193 85211 05559 64462 29489 54930 38196 44288 10975 66593 34461 28475 64823 37867 83165 27120 19091 45648 56692 34603 86104 54326 64821 33936 07260 24914 12737 24587 00660 63155 81748 81520 92096 28292 54091 71536 43678 92590 36001 13305 30548 82046 65213 84146 95194 15116 09...
This user's deviation is standard.
{\displaystyle \lim _{t\to \infty }(Wikipedia[t])=}
CBN This user watches ABS-CBN programs and is "proud to be a Kapamilya."
I am a solid Kapamilya.
This user likes the BBC.
This user watches BBC World News.
CNN This user watches CNN.
This user dislikes commercials that happen during TV shows and movies and finds them annoying
to protect his computer.
This user is a Football fan who enjoys the game. Get in there!
This user is a Madridista.
This user's favorite football player is Kaká
This user is an educator.
This user is an aspiring academic.
This user teaches at a university or other institution of higher education.
This user works in retail.
This user is a Digital Marketing Expert.
It is approximately 6:18 AM where this user lives (Davao)
This user is diurnal.
This user sleeps, but not for predictable amounts of time.
Confusing This user wants to use the UTC time zone but it is too confusing for him to understand.
Taas noong nag-aaral / nakapagtapos mula sa Unibersidad ng Pilipinas.
BCS This user has a Bachelor of Computer Science degree.
This user is a MSc student at university.
PhD This user aspires to be a doctoral student.
This user's favorite subject is computing.
CHI This user's favourite subject is Chinese.
MS This user's favourite subject is Media Studies.
My Nations
Tambayan Philippines.
This user is an Austronesian.
This user lives on the Ring of Fire.
This user is interested in England.
This user is interested in Wales.
This user is interested in Northern Ireland.
This user is an Anglophile.
This user sometimes wishes that he/she were British.
This user is not British, but shares a common heritage and respects a parity of esteem between the many people of "these islands".
sax-en This brooker wishes to cleanse English of needless fremd words.
This user is not British, but can speak with a British accent.
あ This user has full understanding of the Katakana or Hiragana.
漢字-3 This user has an advanced understanding of the Kanji script.
This user thinks that the beauty of Chinese characters is ruined by simplifying them.
This user likes Java.
?> This user will stop using PHP short tags when you pry them from his cold, dead hands.
vb-1 This user is a beginner Visual Basic programmer.
AYB This user knows that
first-person shooters.
This user is still alive.
Scrabble This user enjoys playing Scrabble.
AB This user plays Angry Birds games.
This user wants to become a true Christian.
This user prays.
? This user follows his own political ideals.
This user believes that movie stars and celebrities should stick to what they're good at, not become politicians.
MadayawEdit
Davao City-born Bernejay is a graduate of the Bachelor of Science in Computer Science degree program in the University of the Philippines Mindanao. He works as an IT administrator and trainer of a non-profit organization based in Cebu City.
He is currently taking his Master's degree in Computer Education, previously taking a Master's degree in Information Technology.
He used to be a teacher in a tourism and hospitality institution in Davao City. Prior to this, he used to work as a Virtual Assistant of a small BPO company in Davao City. He then became a Technical Assistant of a shopping mall in the said city.
He was formerly a Website Administrator at Himati, the official student publication of UP Mindanao.
This user spearheaded the expansion of the University of the Philippines Mindanao page in 2007. He edits pages of his interest such as the Philippines national football team, also known as the "Azkals," the United Football League, Vallacar Transit Incorporated (a local bus conglomerate), TV programs, and many others.
Retrieved from "https://en.wikipedia.org/w/index.php?title=User:Bernejay&oldid=1073763717"
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Regressions with Multiple Fixed Effects – Comparing Stata and R | R-bloggers
Regressions with Multiple Fixed Effects – Comparing Stata and R
Posted on April 5, 2014 by Thiemo Fetzer in R bloggers | 0 Comments
In my paper on the impact of the recent fracking boom on local economic outcomes, I am estimating models with multiple fixed effects. These fixed effects are useful, because they take out, e.g. industry specific heterogeneity at the county level – or state specific time shocks.
The models can take the form:
y_{cist} = \alpha_{ci} + b_{st} + \gamma_{it}+ X_{cist}'\beta + \epsilon_{cist}
\alpha_{ci}
is a set of county-industry,
b_{ci}
a set of state-time and
\gamma_{it}
is a set of industry-time fixed effects.
Such a specification takes out arbitrary state-specific time shocks and industry specific time shocks, which are particularly important in my research context as the recession hit tradable industries more than non-tradable sectors, as is suggested in Mian, A., & Sufi, A. (2011). What Explains High Unemployment ? The Aggregate Demand Channel.
How can we estimate such a specification?
Running such a regression in R with the lm or reg in stata will not make you happy, as you will need to invert a huge matrix. An alternative in Stata is to absorb one of the fixed-effects by using xtreg or areg. However, this still leaves you with a huge matrix to invert, as the time-fixed effects are huge; inverting this matrix will still take ages.
However, there is a way around this by applying the Frisch-Waugh Lovell theorem iteratively (remember your Econometrics course?); this basically means you iteratively take out each of the fixed effects in turn by demeaning the data by that fixed effect. The iterative procedure is described in detail in Gaure (2013), but also appears in Guimaraes and Portugal(2010).
Simen Gaure has developed an R-package called lfe, which performs the demeaning for you and also provides the possibility to run instrumental variables regressions; it theoretically supports any dimensionality of fixed effects. The key benefit of Simen Gaure’s implementation is the flexibility, the use of C in the background for some of the computing and its support for multicore processing, which speeds up the demeaning process dramatically, especially the larger your samples get..
In Stata there is a package called reg2hdfe and reg3hdfe which has been developed by Guimaraes and Portugal (2010). As the name indicates, these support only fixed effects up to two or three dimensions.
Lets see how – on the same dataset – the runtimes of reg2hdfe and lfe compare.
Comparing Performance of Stata and R
I am estimating the following specification
y_{cist} = \alpha_{ci} + b_{sit} + \gamma_{it}+ X_{cist}'\beta + \epsilon_{cist}
\alpha_{ci}
are county industry fixed effects and
b_{sit}
are state-time-industry fixed effects. There are about 3000 counties in the dataset and 22 industries. Furthermore, there are 50 states and the time period is also about 50 quarters. This means – in total – there are 3000 x 22 = 66,000 county-industry fixed effects to be estimated and 22 x 50 x 50 = 55,000 time fixed effects to be estimated. The sample I work with has sufficient degrees of freedom to allow the estimation of such a specification – I work with roughly 3.7 million observations.
I have about 10 covariates that are in
X_{cist}
, i.e. these are control variables that vary within county x industry over state x industry x time.
Performance in Stata
In order to time the length of a stata run, you need to run
set rmsg on, which turns on a timer for each command that is run.
The command I run in stata is
reg2hdfe logy x1-x10, id1(sitq ) id2(id) cluster(STATE_FIPS )
You should go get a coffee, because this run is going to take quite a bit of time. In my case, it took t=1575.31, or just about 26 minutes.
In order to make the runs of reg2hdfe and lfe, we need to set the tolerance level of the convergence criterion to be the same in both. The standard tolerance in Stata is set at
1e^{-6}
, while for lfe package it is set at
1e^{-8}
. In order to make the runs comparable you can set the options in the R package lfe options explicitly:
options(lfe.eps=1e-6)
The second change we need to make is to disallow lfe to use multiple cores, since reg2hdfe uses only a single thread. We can do this by setting:
options(lfe.threads=1)
Now lets run this in R using:
system.time(summary(felm(log(y) ~ x1 + x2 +x3 +x4 + x5 + x6 + x7 +x8 + x9 + x10 + G(id)+G(sitq), data=EMP, cluster=c("STATE_FIPS"))))
The procedure converges in a lot quicker than Stata…
user system elapsed 208.450 23.817 236.831
It took a mere 4 minutes. Now suppose I run this in four separate threads…
Running this on four threads saves about one minute in processing time; not bad, but not too much gained; the gains from multi-threading increase, the more fixed-effects are added and the larger the samples are.
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Level_(logarithmic_quantity) Knowpia
The type of level and choice of units indicate the scaling of the logarithm of the ratio between the quantity and its reference value, though a logarithm may be considered to be a dimensionless quantity.[1][2][3] The reference values for each type of quantity are often specified by international standards.
Power and field levels are used in electronic engineering, telecommunications, acoustics and related disciplines. Power levels are used for signal power, noise power, sound power, sound exposure, etc. Field levels are used for voltage, current, sound pressure.[4][clarification needed]
Power levelEdit
Level of a power quantity, denoted LP, is defined by
{\displaystyle L_{P}={\frac {1}{2}}\log _{\mathrm {e} }\!\left({\frac {P}{P_{0}}}\right)\!~\mathrm {Np} =\log _{10}\!\left({\frac {P}{P_{0}}}\right)\!~\mathrm {B} =10\log _{10}\!\left({\frac {P}{P_{0}}}\right)\!~\mathrm {dB} .}
P is the power quantity;
P0 is the reference value of P.
Field (or root-power) levelEdit
The level of a root-power quantity (also known as a field quantity), denoted LF, is defined by[5]
{\displaystyle L_{F}=\log _{\mathrm {e} }\!\left({\frac {F}{F_{0}}}\right)\!~\mathrm {Np} =2\log _{10}\!\left({\frac {F}{F_{0}}}\right)\!~\mathrm {B} =20\log _{10}\!\left({\frac {F}{F_{0}}}\right)\!~\mathrm {dB} .}
F is the root-power quantity, proportional to the square root of power quantity;
F0 is the reference value of F.
If the power quantity P is proportional to F2, and if the reference value of the power quantity, P0, is in the same proportion to F02, the levels LF and LP are equal.
The neper, bel, and decibel (one tenth of a bel) are units of level that are often applied to such quantities as power, intensity, or gain.[6] The neper, bel, and decibel are related by
1 B = 1/2 loge10 Np;
1 dB = 0.1 B = 1/20 loge10 Np.
Level and its units are defined in ISO 80000-3.
The ISO standard defines each of the quantities power level and field level to be dimensionless, with 1 Np = 1. This is motivated by simplifying the expressions involved, as in systems of natural units.
Logarithmic ratio quantityEdit
Power and field quantities are part of a larger class, logarithmic ratio quantities.
ANSI/ASA S1.1-2013 defines a class of quantities it calls levels. It defines a level of a quantity Q, denoted LQ, as[7]
{\displaystyle L_{Q}=\log _{r}\!\left({\frac {Q}{Q_{0}}}\right)\!,}
r is the base of the logarithm;
Q is the quantity;
Q0 is the reference value of Q.
For the level of a root-power quantity, the base of the logarithm is r = e. For the level of a power quantity, the base of the logarithm is r = e2.[8]
Frequency levelEdit
Frequency level of a frequency f is the logarithm of a ratio of the frequency f to a reference frequency f0. The reference frequency is C0, four octaves below middle C. [9]
In electronics, the octave (oct) is used as a unit with logarithm base 2, and the decade (dec) is used as a unit with logarithm base 10:
{\displaystyle L_{f}=\log _{2}\!\left({\frac {f}{f_{0}}}\right)~{\text{oct}}=\log _{10}\!\left({\frac {f}{f_{0}}}\right)~{\text{dec}}.}
In music theory, the octave is a unit used with logarithm base 2 (called interval).[10] A semitone is one twelfth of an octave. A cent is one hundredth of a semitone.
Decibel § Definition
Sound level (disambiguation)
Leveling (tapered floating point)
Level-index arithmetic (LI) and symmetric level-index arithmetic (SLI)
^ IEEE/ASTM SI 10 2016, pp. 26–27.
^ ISO 80000-3 2006.
^ D'Amore 2015.
^ ANSI/ASA S1.1 2013, entry 3.01.
^ Ainslie 2015.
^ ANSI/ASA S1.1 2013.
Fletcher, H (1934), "Loudness, pitch and the timbre of musical tones and their relation to the intensity, the frequency and the overtone structure", Journal of the Acoustical Society of America, 6 (2): 59, Bibcode:1934ASAJ....6...59F, doi:10.1121/1.1915704
Taylor, Barry (1995), Guide for the Use of the International System of Units (SI): The Metric System, Diane Publishing Co., p. 28, ISBN 9780788125799
ISO 80000-3 (2006), Quantities and units, vol. Part 3: Space and Time, International Organization for Standardization
Carey, W. M. (2006), "Sound Sources and Levels in the Ocean", IEEE Journal of Oceanic Engineering, 31 (1): 61, Bibcode:2006IJOE...31...61C, doi:10.1109/JOE.2006.872214, S2CID 30674485
ISO 80000-8 (2007), Quantities and units, vol. Part 8: Acoustics, International Organization for Standardization
ANSI/ASA S1.1 (2013), Acoustical Terminology, vol. ANSI/ASA S1.1-2013, Acoustical Society of America
Ainslie, M. A. (2015), "A Century of Sonar: Planetary Oceanography, Underwater Noise Monitoring, and the Terminology of Underwater Sound", Acoustics Today, 11 (1)
D'Amore, F. (2015), Effect of moisturizer and lubricant on the finger‒surface sliding contact: tribological and dynamical analysis
IEEE/ASTM SI 10 (2016), American National Standard for Metric Practice, IEEE Standards Association
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(Redirected from Twenty-four)
(twenty-fourth)
tetravigesimal
The SI prefix for 1024 is yotta (Y), and for 10−24 (i.e., the reciprocal of 1024) yocto (y). These numbers are the largest and smallest number to receive an SI prefix to date.
24 is an even composite number, with 2 and 3 as its distinct prime factors. It is first number of the form 23q, where q is an odd prime. It is the smallest number with exactly eight positive divisors: 1, 2, 3, 4, 6, 8, 12, and 24; thus it is a highly composite number, having more divisors than any smaller number.[1] Furthermore, it is an abundant number, since the sum of its proper divisors (36) is greater than itself, as well as a superabundant number.
In number theory, and algebra:
24 is the smallest 5-hemiperfect number, as it has a half-integer abundancy index:
1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60 = 5/2 × 24
24 is a semiperfect number, since adding up all the proper divisors of 24 except 4 and 8 gives 24.[2]
24 is a practical number, since all smaller positive integers than 24 can be represented as sums of distinct divisors of 24.
24 is a Harshad number, since it is divisible by the sum of its digits in base 10. [3]
24 is a highly totient number, as there are 10 solutions to the equation φ(x) = 24, which is more than any integer below 24. 144, which is the square of 12, and 576, which is the square of 24, are also highly totient.[4]
24 is a polite number, an amenable number, an idoneal number, and a tribonacci number.
24 forms a Ruth-Aaron pair with 25, since the sums of distinct prime factors of each are equal (5).
24 is a compositorial, as it is the product of composite numbers up to 6.
24 is a pernicious number, since its Hamming weight in its binary representation (11000) is prime (2).
24 is the third nonagonal number.[5]
24 is a congruent number, as 24 is the area of a right triangle with a rational number of sides.
24 is a semi-meandric number, where an order-6 semi-meander intersects an oriented ray in R2 at 24 points.
Subtracting 1 from any of its divisors (except 1 and 2, but including itself) yields a prime number; 24 is the largest number with this property.
24 is the largest integer that is divisible by all natural numbers no larger than its square root.
The product of any four consecutive numbers is divisible by 24. This is because among any four consecutive numbers there must be two even numbers, one of which is a multiple of four, and there must be at least one multiple of three.
24 = 4!, the factorial of 4. It is the largest factorial that does not contain a trailing zero at the end of its digits. It represents the number of ways to order 4 distinct items:
(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (2,1,3,4), (2,1,4,3), (2,3,1,4), (2,3,4,1), (2,4,1,3), (2,4,3,1), (3,1,2,4), (3,1,4,2), (3,2,1,4), (3,2,4,1), (3,4,1,2), (3,4,2,1), (4,1,2,3), (4,1,3,2), (4,2,1,3), (4,2,3,1), (4,3,1,2), (4,3,2,1).
24 is the only nontrivial solution to the cannonball problem, that is: 12 + 22 + 32 + … + 242 is a perfect square (702).[6]
24 is the only number whose divisors — 1, 2, 3, 4, 6, 8, 12, 24 — are exactly those numbers n for which every invertible element of the commutative ring Z/nZ is a square root of 1. It follows that the multiplicative group of invertible elements (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine.
It follows that any number n relatively prime to 24 (that is, any number of the form 6K ± 1), and in particular any prime n greater than 3, has the property that n2 – 1 is divisible by 24.
The modular discriminant Δ(τ) is proportional to the 24th power of the Dedekind eta function η(τ): Δ(τ) = (2π)12η(τ)24.
24 degrees is the measure of the central angle and external angle of a pentadecagon.
An icositetragon is a regular polygon with 24 sides and Dih24 symmetry of order 48. It can fill a plane-vertex alongside a triangle and octagon.
24 is the Euler characteristic of a K3 surface: a general elliptic K3 surface has exactly 24 singular fibers.
24 is the order of the octahedral group — the group of rotations of the regular octahedron and the group of rotations of the cube. The binary octahedral group is a subgroup of the 3-sphere S3 consisting of the 24 elements {±1, ±i, ±j, ±k, (±1±i±j±k)/2} of the binary tetrahedral group and the 24 elements contained in its coset {(±1±i)/√2, (±1±j)/√2, (±1±k)/√2, (±i±j)/√2, (±i±k)/√2, (±j±k)/√2}. These two cosets each form the vertices of two self-dual 24-cells. (See point below on 24-cell).
24 is the count of different elements in various uniform polyhedron solids. Within the family of Archimedean and Catalan solids, there are 24 edges in a cuboctahedron and rhombic dodecahedron, 24 vertices in a rhombicuboctahedron, truncated cube, truncated octahedron, and snub cube, as well as 24 faces in a deltoidal icositetrahedron, tetrakis hexahedron, triakis octahedron, and pentagonal icositetrahedron. The cube-octahedron compound, with a rhombic dodecahedral convex hull and the first stellation of the cuboctahedron, has 24 edges.
There are 12 non-prismatic uniform polyhedron compounds (UC01, UC03, UC08, UC10, UC12, UC30, UC42, UC46, UC48, UC50, UC52, and UC54) and 12 uniform star polyhedra (U03, U13, U14, U15, U17, U18, U19, U21, U36, U37, U41, and U58) with a vertex, edge, or face count of 24. The great disnub dirhombidodecahedron, also called Skilling's figure, is a degenerate uniform star polyhedron with a Euler characteristic of 24, when pairs of coinciding edges are considered to be single edges.
Finally, 6 Johnson solids (J17, J27, J37, J45, J61, and J90) also have vertex, edge, or face counts of 24. The pseudo great rhombicuboctahedron, one of two known pseudo-uniform polyhedra alongside the elongated square gyrobicupola (J37), has 24 vertices.
The tesseract has 24 two-dimensional faces (which are all squares). Its dual polychoron, the 16-cell, has 24 edges.
The 24-cell, with 24 octahedral cells and 24 vertices, is a self-dual convex regular 4-polytope. It possesses 576 (24×24) rotational symmetries and 1152 isometries altogether. It tiles 4-dimensional space in a 24-cell honeycomb, in which each 24-cell is surrounded by 24 24-cells.
The vertices of the 24-cell honeycomb can be chosen so that in 4-dimensional space, identified with the ring of quaternions, they are precisely the elements of the subring generated by the binary tetrahedral group as represented by the set of 24 quaternions
{\displaystyle \{\pm 1,\pm i,\pm j,\pm k,{\tfrac {1}{2}}(\pm 1\pm i\pm j\pm k)\}}
in the F4 lattice. Known as the ring of Hurwitz quaternions, this set of 24 quaternions forms the set of vertices of a single 24-cell, all lying on the sphere S3 of radius one centered as the origin. S3 is the Lie group of unit quaternions (isomorphic to the Lie groups SU(2) and Spin(3)), and so the binary tetrahedral group — of order 24 — is a subgroup of S3.
The 24 vertices of the 24-cell are contained in the regular complex polygon 4{3}4, or of symmetry order 1152, as well as 24 4-edges of 24 octahedral cells (of 48). Its representation in the F4 Coxeter plane contains two rings of 12 vertices each.[7]
Truncations, runcinations, and omnitruncations of the 24-cell yield polychora whose Petrie polygons are 24-sided icositetragons; i.e. within the truncated 24-cell, runcinated 24-cell, and omnitruncated 24-cell, amongst others.
24 is the kissing number in 4-dimensional space: the maximum number of unit spheres that can all touch another unit sphere without overlapping. (The centers of 24 such spheres form the vertices of a 24-cell).
The Barnes–Wall lattice contains 24 lattices.
In 24 dimensions there are 24 even positive definite unimodular lattices, called the Niemeier lattices. One of these is the exceptional Leech lattice which has many surprising properties; due to its existence, the answers to many problems such as the kissing number problem and densest lattice sphere-packing problem are known in 24 dimensions but not in many lower dimensions. The Leech lattice is closely related to the equally nice length-24 binary Golay code and the Steiner system S(5,8,24) and the Mathieu group M24. (One construction of the Leech lattice is possible because 12 + 22 + 32 + ... + 242 = 702).
24 is the order of the cyclic group equal to the stable 3-stem in homotopy groups of spheres: πn+3(Sn) = Z/24Z for all n ≥ 5.
The atomic number of chromium[8]
The number of hours in a mean solar day
24! is an approximation (exceeding by just over 3%) of the Avogadro constant.
The number of books in the Tanakh.
In Christian apocalyptic literature it represents the complete Church, being the sum of the 12 tribes of Israel and the 12 Apostles of the Lamb of God. For example, in The Book of Revelation: "Surrounding the throne were twenty-four other thrones, and seated on them were twenty-four elders. They were dressed in white and had crowns of gold on their heads."[9]
Number of Tirthankaras in Jainism.
Number of spokes in the Ashok Chakra.
There are a total of 24 major and minor keys in Western tonal music, not counting enharmonic equivalents. Therefore, for collections of pieces written in each key, the number of pieces in such a collection; e.g., Chopin's 24 Preludes.
Four-and-Twenty was an American racehorse.
The FIFA World Cup final tournament featured 24 men's national teams from 1982 to 1994.
The FIFA Women's World Cup final tournament featured 24 national teams in 2015 and 2019.
In the NBA, the time on a shot clock is 24 seconds.
The number of bits a computer needs to represent 24-bit color images (for a maximum of 16,777,216 colours — but greater numbers of bits provide more accurate colors)
The number of karats representing 100% pure gold.[10]
The number of cycles in the Chinese solar year.
The number of years from the start of the Cold War until the signing of the Seabed Arms Control Treaty, which banned the placing of nuclear weapons on the ocean floor within certain coastal distances.
The number of frames per second at which motion picture film is usually projected, as this is sufficient to allow for persistence of vision.
The number of letters in both the modern and classical Greek alphabet.[11] For the latter reason, also the number of chapters or "books" into which Homer's Odyssey and Iliad came to be divided.
The number of runes in the Elder Futhark.
The number of points on a backgammon board.[12]
A children's mathematical game involving the use of any of the four standard operations on four numbers on a card to get 24 (see 24 Game)
The maximum number of Knight Companions in the Order of the Garter.
The number of the French department Dordogne.
Four and twenty is the number of blackbirds baked in a pie in the traditional English nursery rhyme "Sing a Song of Sixpence".
^ "Sloane's A097942 : Highly totient numbers". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-31.
^ Weisstein, Eric W. "Cannonball Problem". mathworld.wolfram.com. Retrieved 2020-08-19.
^ Coxeter, H.S.M. (1991), Regular Complex Polytopes (2nd ed.), Cambridge: Cambridge University Press
^ Meija, Juris; Coplen, Tyler B.; Berglund, Michael; Brand, Willi A.; Bièvre, Paul De; Gröning, Manfred; Holden, Norman E.; Irrgeher, Johanna; Loss, Robert D.; Walczyk, Thomas; Prohaska, Thomas (2016-03-01). "Atomic weights of the elements 2013 (IUPAC Technical Report)". Pure and Applied Chemistry. 88 (3): 265–291. doi:10.1515/pac-2015-0305. ISSN 0033-4545. S2CID 101719914.
^ "Revelation 4:4, New International Version (1984)". Bible.cc. Retrieved 2013-05-03.
^ "Is 24K gold pure?". Scientific American. Retrieved 2020-08-12.
^ "Greek alphabet | History, Definition, & Facts". Encyclopedia Britannica. Retrieved 2020-08-12.
^ "GammonSite - Rules of backgammon". www.gammonsite.com. Retrieved 2020-08-12.
My Favorite Numbers: 24, John C. Baez
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Weighting function with monotonic gain profile - MATLAB makeweight
\begin{array}{c}W\left(0\right)=\text{dcgain}\\ W\left(\text{Inf}\right)=\text{hfgain}\\ |W\left(j\cdot \text{freq}\right)|=\text{mag}\text{.}\end{array}
\begin{array}{c}W\left(1\right)=\text{dcgain}\\ W\left(-1\right)=\text{hfgain}\\ |W\left({e}^{j\cdot \text{freq}\cdot \text{Ts}}\right)|=\text{mag}\text{.}\end{array}
\begin{array}{c}W\left(0\right)=\text{dcgain}\\ W\left(\text{Inf}\right)=\text{hfgain}\\ |W\left(j\cdot \text{freq}\right)|=\text{mag}\text{.}\end{array}
\begin{array}{c}W\left(1\right)=\text{dcgain}\\ W\left(-1\right)=\text{hfgain}\\ |W\left({e}^{j\cdot \text{freq}\cdot \text{Ts}}\right)|=\text{mag}\text{.}\end{array}
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Section 1-7 - LimitIntermediateValue - Maple Help
Home : Support : Online Help : Study Guides : Calculus : Chapter 1 - Limits : Section 1-7 - LimitIntermediateValue
Section 1.7: Intermediate Value Theorem
In essence, the Intermediate Value theorem states that a continuous function
f
takes on every value between
f\left(a\right)
f\left(b\right)
. This property of a continuous function is called the "Darboux" property. Hence, every continuous function has the Darboux property, but there are functions that have the Darboux property that are not continuous. As a counterexample, take the function whose rule is
\mathrm{sin}\left(1/x\right)
x≠0
, and has the value 0 at
x=0
. This function takes on every value between
-1
and 1, yet is not continuous.
As intuitive as the Intermediate Value theorem (IVT) might seem, it is equivalent to the far deeper completeness property of the real numbers, which essentially states that there are no gaps in the real number line. If there were, then the IVT might not hold because a crucial number could be missing from the range of
. So, asserting that
f
takes on every possible value between
f\left(a\right)
f\left(b\right)
is an assertion that there are no gaps in the real line between these two values, that is, the real numbers are "complete."
For more than a century after Newton and Leibniz articulated the tools of the calculus, these tools were applied to a host of problems in physics and engineering, with spectacular success. It wasn't until much later that mathematicians began to consider such foundational issues as the IVT and its relation to completeness of the real numbers. However, theorems like the IVT are behind interesting observations even today. Click here for an article discussing the use of the IVT in proving that a "wobbly table" can be stabilized by a rotation about its center.
A formal statement of the IVT is the following.
Theorem 1.7.1: The Intermediate Value Theorem
Let f be a continuous function defined on a closed interval
[a,b]
z
f\left(a\right)
f\left(b\right)
f\left(c\right)=z
has at least one solution
c
(a,b)
For an alternate statement of the Intermediate Value theorem, consider
A continuous function on a closed interval must attain every value between the function values at the endpoints.
Stated this way, the IVT might sound "obvious," but either way, it is actually a deep theorem because it is equivalent to the completeness of the real numbers, itself a deep characterization of the reals.
The IVT is the theoretical basis for the Bisection method, a linearly converging numeric technique for solving algebraic equations. This computational method is detailed in Example 1.7.1.
The IVT is also the key in Examples 1.7.2 and 1.7.3.
Use bisection to approximate the zeros of
f\left(x\right)=\frac{{x}^{5}-4{x}^{2}+2}{x-1}
Suppose the temperature at midnight is
55°
F, increases to a high of
85°
F, then returns to
55°
F at midnight the next day. Assuming that the temperature throughout the day is a continuous function of the time of day, show that there is at least one time in the morning when the temperature is the same as the temperature exactly 12 hours later.
A simple closed path is walked in rugged terrain. Suppose the path is parametrized by
s
0≤s≤1
. and suppose further that the heights along the path are given by
h\left(s\right)
. Prove that there are two points along the path where
h\left(a+1/2\right)=h\left(a\right)
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Operator of proper-time-derivative - Wikiversity
{\displaystyle ~{\frac {D}{D\tau }}=u^{\mu }\nabla _{\mu }}
{\displaystyle ~D}
– the symbol of differential in curved spacetime,
{\displaystyle ~\tau }
– proper time, which is measured by a clock moving with test particle,
{\displaystyle ~u^{\mu }}
– 4-velocity of test particle or local volume of matter,
{\displaystyle ~\nabla _{\mu }}
– covariant derivative.
{\displaystyle ~{\frac {d}{d\tau }}=u^{\mu }\partial _{\mu }}
To prove this expression it can be applied to an arbitrary 4-vector
{\displaystyle ~A^{\nu }}
{\displaystyle ~u^{\mu }\partial _{\mu }A^{\nu }={\frac {c{}dt}{d\tau }}{\frac {\partial A^{\nu }}{c{}\partial t}}+{\frac {dx}{d\tau }}{\frac {\partial A^{\nu }}{\partial x}}+{\frac {dy}{d\tau }}{\frac {\partial A^{\nu }}{\partial y}}+{\frac {dz}{d\tau }}{\frac {\partial A^{\nu }}{\partial z}}=}
{\displaystyle ~={\frac {dt}{d\tau }}\left({\frac {\partial A^{\nu }}{\partial t}}+{\frac {dx}{dt}}{\frac {\partial A^{\nu }}{\partial x}}+{\frac {dy}{dt}}{\frac {\partial A^{\nu }}{\partial y}}+{\frac {dz}{dt}}{\frac {\partial A^{\nu }}{\partial z}}\right)={\frac {dt}{d\tau }}{\frac {dA^{\nu }}{dt}}={\frac {dA^{\nu }}{d\tau }}}
Above was used material derivative in operator equation for an arbitrary function
{\displaystyle ~F}
{\displaystyle ~{\frac {dF}{dt}}={\frac {\partial F}{\partial t}}+\mathbf {V} \cdot \nabla F}
{\displaystyle ~\mathbf {V} }
is the velocity of local volume of matter,
{\displaystyle ~\nabla }
– nabla operator.
In turn, the material derivative follows from the representation of differential function
{\displaystyle ~F}
of spatial coordinates and time:
{\displaystyle ~dF(t,x,y,z)={\frac {\partial F}{\partial t}}dt+{\frac {\partial F}{\partial x}}dx+{\frac {\partial F}{\partial y}}dy+{\frac {\partial F}{\partial z}}dz}
Operator of proper-time-derivative is applied to different four-dimensional objects – to scalar functions, 4-vectors and 4-tensors. One exception is 4-position (4-radius), which in four-Cartesian coordinates has the form
{\displaystyle ~x^{\mu }=(ct,x,y,z)=(ct,\mathbf {r} )}
because 4-position is not a 4-vector in curved space-time, but its differential (displacement)
{\displaystyle ~dx^{\mu }=(c{}dt,dx,dy,dz)=(cdt,d\mathbf {r} )}
is. Effect of the left side of operator of proper-time-derivative on the 4-position specifies the 4-velocity:
{\displaystyle ~{\frac {Dx^{\mu }}{D\tau }}=u^{\mu }}
, but the right side of the operator does not so:
{\displaystyle ~u^{\nu }\nabla _{\nu }x^{\mu }\not =u^{\mu }}
In covariant theory of gravitation operator of proper-time-derivative is used to determine the density of 4-force acting on a solid point particle in curved spacetime:[2]
{\displaystyle ~f^{\nu }={\frac {DJ^{\nu }}{D\tau }}=u^{\mu }\nabla _{\mu }J^{\nu }={\frac {dJ^{\nu }}{d\tau }}+\Gamma _{\mu \lambda }^{\nu }u^{\mu }J^{\lambda }}
{\displaystyle ~J^{\nu }=\rho _{0}u^{\nu }}
is 4-vector momentum density of matter,
{\displaystyle ~\rho _{0}}
– density of matter in its rest system,
{\displaystyle ~\Gamma _{\mu \lambda }^{\nu }}
– Christoffel symbol.
{\displaystyle ~f_{\alpha }=\nabla _{\beta }{B_{\alpha }}^{\beta }=-u_{\alpha k}J^{k}=\rho _{0}{\frac {DU_{\alpha }}{D\tau }}-J^{k}\nabla _{\alpha }U_{k}=\rho _{0}{\frac {dU_{\alpha }}{d\tau }}-J^{k}\partial _{\alpha }U_{k},}
{\displaystyle ~{B_{\alpha }}^{\beta }}
is the acceleration stress-energy tensor with the mixed indices,
{\displaystyle ~u_{\alpha k}}
is the acceleration tensor, and the 4-potential of acceleration field is expressed in terms of the scalar
{\displaystyle ~\vartheta }
{\displaystyle ~\mathbf {U} }
{\displaystyle ~U_{\alpha }=\left({\frac {\vartheta }{c}},-\mathbf {U} \right).}
{\displaystyle ~a^{\nu }={\frac {Du^{\nu }}{D\tau }}=u^{\mu }\nabla _{\mu }u^{\nu }={\frac {du^{\nu }}{d\tau }}+\Gamma _{\mu \lambda }^{\nu }u^{\mu }u^{\lambda }=0}
{\displaystyle ~ds=cd\tau }
, then equation of motion of the body along a geodesic in general relativity can be rewritten in equivalent form:
{\displaystyle ~{\frac {d}{ds}}\left({\frac {dx^{\nu }}{ds}}\right)+\Gamma _{\mu \lambda }^{\nu }{\frac {dx^{\mu }}{ds}}{\frac {dx^{\lambda }}{ds}}=0.}
If, instead of the proper time to use a parameter
{\displaystyle ~p}
, and equation of a curve set by the expression
{\displaystyle ~x^{\mu }(p)}
, then there is the operator of derivative on the parameter along the curve:[5]
{\displaystyle ~{\frac {D}{Dp}}={\frac {dx^{\mu }}{dp}}\nabla _{\mu }}
↑ Fedosin S.G. Fizicheskie teorii i beskonechnaia vlozhennost’ materii. – Perm, 2009-2011, 858 pages, Tabl. 21, Pic. 41, Ref. 293. ISBN 978-5-9901951-1-0. (in Russian).
↑ Fedosin S.G. The General Theory of Relativity, Metric Theory of Relativity and Covariant Theory of Gravitation: Axiomatization and Critical Analysis. International Journal of Theoretical and Applied Physics, Vol. 4, No. 1, pp. 9-26 (2014). http://dx.doi.org/10.5281/zenodo.890781.
↑ Fock, V. A. (1964). "The Theory of Space, Time and Gravitation". Macmillan.
↑ Carroll, Sean M. (2004), Spacetime and Geometry, Addison Wesley, ISBN 0-8053-8732-3
Retrieved from "https://en.wikiversity.org/w/index.php?title=Operator_of_proper-time-derivative&oldid=2065311"
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Inverse function rule - Wikipedia @ WordDisk
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse o{\displaystyle f}
{\displaystyle f^{-1}}
{\displaystyle f^{-1}(y)=x}
{\displaystyle f(x)=y}
{\displaystyle \left[f^{-1}\right]'(a)={\frac {1}{f'\left(f^{-1}(a)\right)}}}
Calculus identity
{\displaystyle {\color {CornflowerBlue}{f'}}(x)={\frac {1}{{\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x))}}}
{\displaystyle x_{0}\approx 5.8}
{\displaystyle {\color {CornflowerBlue}{f'}}(x_{0})={\frac {1}{4}}}
{\displaystyle {\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x_{0}))=4~}
{\displaystyle f}
{\displaystyle f}
{\displaystyle f^{-1}(a)}
{\displaystyle \in I}
{\displaystyle f'(f^{-1}(a))\neq 0}
{\displaystyle {\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},}
{\displaystyle {\mathcal {D}}}
{\displaystyle \circ }
{\displaystyle y=x}
{\displaystyle f}
{\displaystyle x}
{\displaystyle x}
{\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.}
{\displaystyle f^{-1}(y)=x}
{\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}={\frac {dx}{dx}}}
This article uses material from the Wikipedia article Inverse function rule, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.
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Complement of an open subset
This article is about the complement of an open set. For a set closed under an operation, see closure (mathematics). For other uses, see Closed (disambiguation).
In geometry, topology, and related branches of mathematics, a closed set is a set whose complement is an open set.[1][2] In a topological space, a closed set can be defined as a set which contains all its limit points. In a complete metric space, a closed set is a set which is closed under the limit operation. This should not be confused with a closed manifold.
1 Equivalent definitions of a closed set
2 More about closed sets
3 Properties of closed sets
4 Examples of closed sets
Equivalent definitions of a closed set[edit]
By definition, a subset
{\displaystyle A}
{\displaystyle (X,\tau )}
is called closed if its complement
{\displaystyle X\setminus A}
{\displaystyle (X,\tau )}
{\displaystyle X\setminus A\in \tau .}
A set is closed in
{\displaystyle X}
if and only if it is equal to its closure in
{\displaystyle X.}
Equivalently, a set is closed if and only if it contains all of its limit points. Yet another equivalent definition is that a set is closed if and only if it contains all of its boundary points. Every subset
{\displaystyle A\subseteq X}
is always contained in its (topological) closure in
{\displaystyle X,}
{\displaystyle \operatorname {cl} _{X}A;}
{\displaystyle A\subseteq X}
{\displaystyle A\subseteq \operatorname {cl} _{X}A.}
{\displaystyle A}
{\displaystyle X}
{\displaystyle A=\operatorname {cl} _{X}A.}
An alternative characterization of closed sets is available via sequences and nets. A subset
{\displaystyle A}
{\displaystyle X}
{\displaystyle X}
if and only if every limit of every net of elements of
{\displaystyle A}
{\displaystyle A.}
In a first-countable space (such as a metric space), it is enough to consider only convergent sequences, instead of all nets. One value of this characterization is that it may be used as a definition in the context of convergence spaces, which are more general than topological spaces. Notice that this characterization also depends on the surrounding space
{\displaystyle X,}
because whether or not a sequence or net converges in
{\displaystyle X}
depends on what points are present in
{\displaystyle X.}
{\displaystyle x}
{\displaystyle X}
is said to be close to a subset
{\displaystyle A\subseteq X}
{\displaystyle x\in \operatorname {cl} _{X}A}
(or equivalently, if
{\displaystyle x}
{\displaystyle A}
in the topological subspace
{\displaystyle A\cup \{x\},}
{\displaystyle x\in \operatorname {cl} _{A\cup \{x\}}A}
{\displaystyle A\cup \{x\}}
is endowed with the subspace topology induced on it by
{\displaystyle X}
[note 1]). Because the closure of
{\displaystyle A}
{\displaystyle X}
is thus the set of all points in
{\displaystyle X}
that are close to
{\displaystyle A,}
this terminology allows for a plain English description of closed subsets:
a subset is closed if and only if it contains every point that is close to it.
In terms of net convergence, a point
{\displaystyle x\in X}
{\displaystyle A}
if and only if there exists some net (valued) in
{\displaystyle A}
{\displaystyle x.}
{\displaystyle X}
is a topological subspace of some other topological space
{\displaystyle Y,}
{\displaystyle Y}
is called a topological super-space of
{\displaystyle X,}
then there might exist some point in
{\displaystyle Y\setminus X}
that is close to
{\displaystyle A}
(although not an element of
{\displaystyle X}
), which is how it is possible for a subset
{\displaystyle A\subseteq X}
to be closed in
{\displaystyle X}
but to not be closed in the "larger" surrounding super-space
{\displaystyle Y.}
{\displaystyle A\subseteq X}
{\displaystyle Y}
is any topological super-space of
{\displaystyle X}
{\displaystyle A}
is always a (potentially proper) subset of
{\displaystyle \operatorname {cl} _{Y}A,}
which denotes the closure of
{\displaystyle A}
{\displaystyle Y;}
indeed, even if
{\displaystyle A}
{\displaystyle X}
(which happens if and only if
{\displaystyle A=\operatorname {cl} _{X}A}
), it is nevertheless still possible for
{\displaystyle A}
to be a proper subset of
{\displaystyle \operatorname {cl} _{Y}A.}
{\displaystyle A}
{\displaystyle X}
{\displaystyle A=X\cap \operatorname {cl} _{Y}A}
for some (or equivalently, for every) topological super-space
{\displaystyle Y}
{\displaystyle X.}
Closed sets can also be used to characterize continuous functions: a map
{\displaystyle f:X\to Y}
{\displaystyle f\left(\operatorname {cl} _{X}A\right)\subseteq \operatorname {cl} _{Y}(f(A))}
{\displaystyle A\subseteq X}
; this can be reworded in plain English as:
{\displaystyle f}
{\displaystyle A\subseteq X,}
{\displaystyle f}
{\displaystyle A}
{\displaystyle f(A).}
{\displaystyle f}
{\displaystyle x\in X}
{\displaystyle x}
{\displaystyle A\subseteq X,}
{\displaystyle f(x)}
{\displaystyle f(A).}
More about closed sets[edit]
Whether a set is closed depends on the space in which it is embedded. However, the compact Hausdorff spaces are "absolutely closed", in the sense that, if you embed a compact Hausdorff space
{\displaystyle D}
in an arbitrary Hausdorff space
{\displaystyle X,}
{\displaystyle D}
will always be a closed subset of
{\displaystyle X}
; the "surrounding space" does not matter here. Stone–Čech compactification, a process that turns a completely regular Hausdorff space into a compact Hausdorff space, may be described as adjoining limits of certain nonconvergent nets to the space.
Closed sets also give a useful characterization of compactness: a topological space
{\displaystyle X}
is compact if and only if every collection of nonempty closed subsets of
{\displaystyle X}
with empty intersection admits a finite subcollection with empty intersection.
{\displaystyle X}
is disconnected if there exist disjoint, nonempty, open subsets
{\displaystyle A}
{\displaystyle B}
{\displaystyle X}
{\displaystyle X.}
{\displaystyle X}
is totally disconnected if it has an open basis consisting of closed sets.
Properties of closed sets[edit]
See also: Kuratowski closure axioms
A closed set contains its own boundary. In other words, if you are "outside" a closed set, you may move a small amount in any direction and still stay outside the set. Note that this is also true if the boundary is the empty set, e.g. in the metric space of rational numbers, for the set of numbers of which the square is less than
{\displaystyle 2.}
Any intersection of any family of closed sets is closed (this includes intersections of infinitely many closed sets)
The union of finitely many closed sets is closed.
The empty set is closed.
The whole set is closed.
In fact, if given a set
{\displaystyle X}
and a collection
{\displaystyle \mathbb {F} \neq \varnothing }
{\displaystyle X}
{\displaystyle \mathbb {F} }
have the properties listed above, then there exists a unique topology
{\displaystyle \tau }
{\displaystyle X}
such that the closed subsets of
{\displaystyle (X,\tau )}
are exactly those sets that belong to
{\displaystyle \mathbb {F} .}
The intersection property also allows one to define the closure of a set
{\displaystyle A}
{\displaystyle X,}
which is defined as the smallest closed subset of
{\displaystyle X}
that is a superset of
{\displaystyle A.}
Specifically, the closure of
{\displaystyle X}
can be constructed as the intersection of all of these closed supersets.
Sets that can be constructed as the union of countably many closed sets are denoted Fσ sets. These sets need not be closed.
Examples of closed sets[edit]
The closed interval
{\displaystyle [a,b]}
of real numbers is closed. (See Interval (mathematics) for an explanation of the bracket and parenthesis set notation.)
The unit interval
{\displaystyle [0,1]}
is closed in the metric space of real numbers, and the set
{\displaystyle [0,1]\cap \mathbb {Q} }
of rational numbers between
{\displaystyle 0}
{\displaystyle 1}
(inclusive) is closed in the space of rational numbers, but
{\displaystyle [0,1]\cap \mathbb {Q} }
is not closed in the real numbers.
Some sets are neither open nor closed, for instance the half-open interval
{\displaystyle [0,1)}
in the real numbers.
Some sets are both open and closed and are called clopen sets.
{\displaystyle [1,+\infty )}
The Cantor set is an unusual closed set in the sense that it consists entirely of boundary points and is nowhere dense.
Singleton points (and thus finite sets) are closed in T1 spaces and Hausdorff spaces.
{\displaystyle \mathbb {Z} }
is an infinite and unbounded closed set in the real numbers.
{\displaystyle f:X\to Y}
is a function between topological spaces then
{\displaystyle f}
is a continuous if and only if preimages of closed sets in
{\displaystyle Y}
{\displaystyle X.}
Clopen set – Subset which is both open and closed
Open set – Basic subset of a topological space
Neighbourhood – Open set containing a given point
Regular closed set
^ In particular, whether or not
{\displaystyle x}
{\displaystyle A}
depends only on the subspace
{\displaystyle A\cup \{x\}}
and not on the whole surrounding space (e.g.
{\displaystyle X,}
or any other space containing
{\displaystyle A\cup \{x\}}
as a topological subspace).
^ Rudin, Walter (1976). Principles of Mathematical Analysis. McGraw-Hill. ISBN 0-07-054235-X.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Closed_set&oldid=1080618190"
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math - SVM Tutorial
SVM - Understanding the math - the optimal hyperplane
January 12, 2021 June 8, 2015 by Alexandre KOWALCZYK
This is the Part 3 of my series of tutorials about the math behind Support Vector Machine.
If you did not read the previous articles, you might want to start the serie at the beginning by reading this article: an overview of Support Vector Machine.
The main focus of this article is to show you the reasoning allowing us to select the optimal hyperplane.
Here is a quick summary of what we will see:
How can we find the optimal hyperplane ?
How do we calculate the distance between two hyperplanes ?
What is the SVM optimization problem ?
How to find the optimal hyperplane ?
At the end of Part 2 we computed the distance
\|p\|
between a point
A
and a hyperplane. We then computed the margin which was equal to
2 \|p\|
However, even if it did quite a good job at separating the data it was not the optimal hyperplane.
Figure 1: The margin we calculated in Part 2 is shown as M1
Categories Mathematics, SVM Tutorial Tags hyperplane, math, vectors
July 26, 2020 November 9, 2014 by Alexandre KOWALCZYK
This is Part 2 of my series of tutorial about the math behind Support Vector Machines.
If you did not read the previous article, you might want to start the serie at the beginning by reading this article: an overview of Support Vector Machine.
In the first part, we saw what is the aim of the SVM. Its goal is to find the hyperplane which maximizes the margin.
But how do we calculate this margin?
SVM = Support VECTOR Machine
In Support Vector Machine, there is the word vector.
That means it is important to understand vector well and how to use them.
Here a short sum-up of what we will see today:
its norm
How to add and subtract vectors ?
What is the dot product ?
How to project a vector onto another ?
Once we have all these tools in our toolbox, we will then see:
What is the equation of the hyperplane?
How to compute the margin?
SVM - Understanding the math - Part 1 - The margin
January 31, 2021 November 2, 2014 by Alexandre KOWALCZYK
This is the first article from a series of articles I will be writing about the math behind SVM. There is a lot to talk about and a lot of mathematical backgrounds is often necessary. However, I will try to keep a slow pace and to give in-depth explanations, so that everything is crystal clear, even for beginners.
If you are new and wish to know a little bit more about SVMs before diving into the math, you can read the article: an overview of Support Vector Machine.
What is the goal of the Support Vector Machine (SVM)?
The goal of a support vector machine is to find the optimal separating hyperplane which maximizes the margin of the training data.
The first thing we can see from this definition, is that a SVM needs training data. Which means it is a supervised learning algorithm.
It is also important to know that SVM is a classification algorithm. Which means we will use it to predict if something belongs to a particular class.
For instance, we can have the training data below:
We have plotted the size and weight of several people, and there is also a way to distinguish between men and women.
With such data, using a SVM will allow us to answer the following question:
Given a particular data point (weight and size), is the person a man or a woman ?
For instance: if someone measures 175 cm and weights 80 kg, is it a man of a woman?
Categories Mathematics, SVM Tutorial Tags hyperplane, math
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Mathematics - SVM Tutorial
SVM - Understanding the math - Duality and Lagrange multipliers
August 19, 2021 September 11, 2016 by Alexandre KOWALCZYK
This is the Part 6 of my series of tutorials about the math behind Support Vector Machines. Today we will learn about duality, optimization problems and Lagrange multipliers. If you did not read the previous articles, you might want to start the serie at the beginning by reading this article: an overview of Support Vector Machine. ... Read more
Categories Mathematics, SVM Tutorial Tags duality, Lagrange, Lagrange multiplier, lower bound, optimization
SVM - Understanding the math - Convex functions
July 26, 2020 September 11, 2016 by Alexandre KOWALCZYK
This is the Part 5 of my series of tutorials about the math behind Support Vector Machines. Today we are going to study convex functions. If you did not read the previous articles, you might want to start the serie at the beginning by reading this article: an overview of Support Vector Machine. How can we ... Read more
Categories Mathematics, SVM Tutorial Tags convex, convexity, function
This is the Part 4 of my series of tutorials about the math behind Support Vector Machines. Today we are going to learn how to solve an unconstrained minimization problem. If you did not read the previous articles, you might want to start the serie at the beginning by reading this article: an overview of Support ... Read more
Categories Mathematics, SVM Tutorial Tags Hessian, matrix, minimization, unconstrained
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2 \|p\|
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How to use the square feet (sq. ft) to cubic yards calculator
If you want to convert between square feet and cubic yards, our square feet to cubic yards calculator is the only tool you need. Just enter the area in square feet and the height/depth, and you will be able to calculate volume in cubic yards in no time.
To calculate volume in cubic yards from the base dimensions, i.e., length, width, and height, check out our cubic yard calculator.
Continue reading this article to know what a cubic yard is and how to convert square feet to cubic yards. You will also learn how to use our sq ft to cubic yards tool.
Before going any further, let us first try to understand what are square feet and cubic yards?
Square feet or
\rm ft^2
is a unit of area, whereas, cubic yards or
\rm yd^3
is a unit of volume.
Often we need to calculate the volume for construction or landscape material, for example, gravel, cement, soil, mulch, etc., and we use cubic yards to express this volume. For example, if you are preparing a flower bed and want to know how much mulch and soil you need.
Let us consider the given figure to understand the difference and relation between square feet and cubic yards. The figure shows a rectangular plot area that is 9 feet long and 6 feet wide. We can calculate the area of this rectangular plot as:
\begin{align*} Area &= length \times width\\ & = 9\ \rm ft \times 6\ ft\\ & = 54\ \rm ft^2 \end {align*}
If we dig up the whole plot to 6-inch (or 0.5 ft) depth, the volume of the resulting bed would be:
\begin{align*} Volume &= Area \times depth\\ & = 54\ \rm ft^2 \times 0.5\ ft\\ & = 27\ \rm ft^3 \end {align*}
1\ \rm yd = 3\ ft
1\ \rm yd^3 = 27\ ft^3
, we can write the above equation as:
\begin{align*} Volume &= 27\ \rm ft^3\\ & = 1\ \rm yd^3 \end {align*}
Since square feet is a measure of area, and cubic yards is a measure of volume, to convert square feet to cubic yards, we need to follow the given steps:
Determine the area by using the appropriate formula and converting the unit of area to square feet.
Measure the depth/height of the area and convert it to feet as well.
Multiply the area and depth/height, and you will get volume in cubic feet.
Divide the volume in cubic feet by 27 to get the volume in cubic yards.
You can also use our square feet to cubic yards calculator for this purpose!
Now let's see how we can use the square feet (sq. ft) to cubic yards calculator:
Enter the area of the surface in square feet. You can use our area calculator for this purpose.
Input the depth/height of the area. You can choose different units using the drop-down menu.
The sq. ft to cubic yards calculator will give you the volume in cubic yards.
You can also use this tool as cubic yards to square feet calculator.
A cubic yard is a unit of volume. We can define a cubic yard as the volume of a cube with length, width, and height of one yard.
To convert square feet to cubic yards, follow the given instructions:
Determine the length and breadth of the region of interest and calculate its area in sq. feet.
Find out the region's depth or height in feet.
Multiply results from steps 1 and 2 and divide the result by 27.
Congratulations! You have converted square feet to cubic yards.
To convert cubic yards to square feet, proceed as follows:
Multiply the volume in cubic yards with 27 to get the volume in cubic feet.
Divide the volume by depth or height measured in feet.
You have calculated the area in square feet.
27 feet. One yard is equal to 3 feet. Hence to determine the number of cubic feet in a cubic yard, take the cube of both sides, and you will get 1 cubic yard = 27 cubic feet.
This mortar calculator is a tool that will come handy during any wall construction project. Use it to calculate the number of mortar bags needed.
Need to estimate how much water you'll need to fill a pond and what size of pond liner to get for your pond? Try our pond calculator now to answer those questions.
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Cleve’s Corner, Blogs Edition » Cleve’s Corner: Cleve Moler on Mathematics and Computing - MATLAB & Simulink
Cleve’s Corner: Cleve Moler on Mathematics and Computing
Stuart’s MATLAB Videos
MATLAB ユーザーコミュニティー
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< Fibonacci Matrices < Previous
Biorhythms >Next >
Cleve’s Corner, Blogs Edition
This is the debut of the MATLAB Central Blogs edition of Cleve's Corner. For years, I have been writing a column named Cleve's Corner in MathWorks News and Notes. The News and Notes Edition is in an electronic magazine format that is published once or twice a year and that has very specific length and word count constraints. The Blogs Edition can be any length, and I hope to have enough time and interesting material to post a couple of times per month. In fact, I may revive some old newsletter Corners, bring them up to date, and post them here.
I am pleased to be able to use MATLAB and its publish command as my Web authoring environmnent. I have been writing MATLAB code for even longer than I have been writing newsletter columns. Now I can write a blog without learning a new word processor.
I am also pleased to be able to use MathJax to display typeset mathematics. MathJax is, among other things, a LaTex interpreter written in JavaScript that displays mathematics properly in modern Web browsers. Here's an example. My next blog posting will involve a discussion of
\phi
, the Golden Ratio. Now look again at that Greek letter
\phi
. It should have the correct font size and, most importantly, have the same baseline as the surrounding text. This is how would have looked before MathJax. The old is a .png image. Some browsers place it far above the baseline of rest of the sentence. If you zoom in on the page, or print the page, the new
\phi
should scale and print nicely, but the old will show the pixelation inherent in an sampled image.
MathJax has been developed by Davide Cervone of Union College and at Design Science, the makers of the MathType equation editor. MathJax has the support of SIAM (Society for Industrial and Applied Mathematics), the AMS (American Mathematical Society), the AIP (American Institute of Physics), a few other professional societies, and several publishers. For more about MathJax, see the recent article by Cervone in the AMS Notices.
Discover…
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Data type whose definition depends on a value
In computer science and logic, a dependent type is a type whose definition depends on a value. It is an overlapping feature of type theory and type systems. In intuitionistic type theory, dependent types are used to encode logic's quantifiers like "for all" and "there exists". In functional programming languages like Agda, ATS, Coq, F*, Epigram, and Idris, dependent types help reduce bugs by enabling the programmer to assign types that further restrain the set of possible implementations.
Two common examples of dependent types are dependent functions and dependent pairs. The return type of a dependent function may depend on the value (not just type) of one of its arguments. For instance, a function that takes a positive integer
{\displaystyle n}
may return an array of length
{\displaystyle n}
, where the array length is part of the type of the array. (Note that this is different from polymorphism and generic programming, both of which include the type as an argument.) A dependent pair may have a second value of which the type depends on the first value. Sticking with the array example, a dependent pair may be used to pair an array with its length in a type-safe way.
2.1 Π type
2.2 Σ type
2.2.1 Example as existential quantification
3 Systems of the lambda cube
3.1 First order dependent type theory
3.2 Second order dependent type theory
3.3 Higher order dependently typed polymorphic lambda calculus
4 Simultaneous programming language and logic
5 Comparison of languages with dependent types
(Because of this and other work by Howard, propositions-as-types is known as the Curry–Howard correspondence.)
Π type[edit]
Loosely speaking, dependent types are similar to the type of an indexed family of sets. More formally, given a type
{\displaystyle A:{\mathcal {U}}}
in a universe of types
{\displaystyle {\mathcal {U}}}
, one may have a family of types
{\displaystyle B:A\to {\mathcal {U}}}
, which assigns to each term
{\displaystyle a:A}
{\displaystyle B(a):{\mathcal {U}}}
. We say that the type B(a) varies with a.
A function whose type of return value varies with its argument (i.e. there is no fixed codomain) is a dependent function and the type of this function is called dependent product type, pi-type or dependent function type.[3] From a family of types
{\displaystyle B:A\to {\mathcal {U}}}
we may construct the type of dependent functions
{\textstyle \prod _{x:A}B(x)}
, whose terms are functions which take a term
{\displaystyle a:A}
and return a term in
{\displaystyle B(a)}
. For this example, the dependent function type is typically written as
{\displaystyle \prod _{x:A}B(x),}
{\textstyle \prod _{x:A}B(x),}
{\textstyle \prod (x:A),B(x).}
{\displaystyle B:A\to {\mathcal {U}}}
is a constant function, the corresponding dependent product type is equivalent to an ordinary function type. That is,
{\textstyle \prod _{x:A}B}
is judgmentally equal to
{\displaystyle A\to B}
when B does not depend on x.
The name 'pi-type' comes from the idea that these may be viewed as a Cartesian product of types. Pi-types can also be understood as models of universal quantifiers.
For example, if we write
{\displaystyle \operatorname {Vec} (\mathbb {R} ,n)}
for n-tuples of real numbers, then
{\textstyle \prod _{n:\mathbb {N} }\operatorname {Vec} (\mathbb {R} ,n)}
would be the type of a function which, given a natural number n, returns a tuple of real numbers of size n. The usual function space arises as a special case when the range type does not actually depend on the input. E.g.
{\textstyle \prod _{n:\mathbb {N} }{\mathbb {R} }}
is the type of functions from natural numbers to the real numbers, which is written as
{\displaystyle \mathbb {N} \to \mathbb {R} }
in typed lambda calculus.
For a more concrete example, taking A to be equal to the family of unsigned integers from 0 to 255, (the ones you can fit into 8 bits or 1 byte) and B(a) = Xa for 256 arbitrary Xa's, then
{\textstyle \prod _{x:A}B(x)}
devolves into the product of X0 × X1 × X2 × ... × X253 × X254 × X255 precisely because the finite set of integers from 0 to 255 would ultimately stop at the bounds just mentioned, resulting in an finite codomain of the dependent function.
Σ type[edit]
The dual of the dependent product type is the dependent pair type, dependent sum type, sigma-type, or (confusingly) dependent product type.[3] Sigma-types can also be understood as existential quantifiers. Continuing the above example, if, in the universe of types
{\displaystyle {\mathcal {U}}}
, there is a type
{\displaystyle A:{\mathcal {U}}}
and a family of types
{\displaystyle B:A\to {\mathcal {U}}}
, then there is a dependent pair type
{\textstyle \sum _{x:A}B(x)}
. (The alternate notations are similar to that of Π types.)
The dependent pair type captures the idea of an ordered pair where the type of the second term is dependent on the value of the first. If
{\textstyle (a,b):\sum _{x:A}B(x),}
{\displaystyle a:A}
{\displaystyle b:B(a)}
. If B is a constant function, then the dependent pair type becomes (is judgementally equal to) the product type, that is, an ordinary Cartesian product
{\displaystyle A\times B}
For a more concrete example, taking A to again be equal to the family of unsigned integers from 0 to 255, and B(a) to again be equal to Xa for 256 more arbitrary Xa's, then
{\textstyle \sum _{x:A}B(x)}
devolves into the sum X0 + X1 + X2 + ... + X253 + X254 + X255 for the same reasons as to what happened to the codomain of the dependent function.
Example as existential quantification[edit]
{\displaystyle A:{\mathcal {U}}}
be some type, and let
{\displaystyle B:A\to {\mathcal {U}}}
. By the Curry–Howard correspondence, B can be interpreted as a logical predicate on terms of A. For a given
{\displaystyle a:A}
, whether the type B(a) is inhabited indicates whether a satisfies this predicate. The correspondence can be extended to existential quantification and dependent pairs: the proposition
{\displaystyle \exists {a}{\in }A\,B(a)}
is true if and only if the type
{\textstyle \sum _{a:A}B(a)}
{\displaystyle m:\mathbb {N} }
{\displaystyle n:\mathbb {N} }
if and only if there exists another natural number
{\displaystyle k:\mathbb {N} }
such that m + k = n. In logic, this statement is codified by existential quantification:
{\displaystyle m\leq n\iff \exists {k}{\in }\mathbb {N} \,m+k=n.}
This proposition corresponds to the dependent pair type:
{\displaystyle \sum _{k:\mathbb {N} }m+k=n.}
That is, a proof of the statement that m is less than or equal to n is a pair that contains both a non-negative number k, which is the difference between m and n, and a proof of the equality m + k = n.
Systems of the lambda cube[edit]
Henk Barendregt developed the lambda cube as a means of classifying type systems along three axes. The eight corners of the resulting cube-shaped diagram each correspond to a type system, with simply typed lambda calculus in the least expressive corner, and calculus of constructions in the most expressive. The three axes of the cube correspond to three different augmentations of the simply typed lambda calculus: the addition of dependent types, the addition of polymorphism, and the addition of higher kinded type constructors (functions from types to types, for example). The lambda cube is generalized further by pure type systems.
First order dependent type theory[edit]
{\displaystyle \lambda \Pi }
of pure first order dependent types, corresponding to the logical framework LF, is obtained by generalising the function space type of the simply typed lambda calculus to the dependent product type.
Second order dependent type theory[edit]
{\displaystyle \lambda \Pi 2}
of second order dependent types is obtained from
{\displaystyle \lambda \Pi }
by allowing quantification over type constructors. In this theory the dependent product operator subsumes both the
{\displaystyle \to }
operator of simply typed lambda calculus and the
{\displaystyle \forall }
binder of System F.
Higher order dependently typed polymorphic lambda calculus[edit]
The higher order system
{\displaystyle \lambda \Pi \omega }
{\displaystyle \lambda \Pi 2}
to all four forms of abstraction from the lambda cube: functions from terms to terms, types to types, terms to types and types to terms. The system corresponds to the calculus of constructions whose derivative, the calculus of inductive constructions is the underlying system of the Coq proof assistant.
Simultaneous programming language and logic[edit]
The Curry–Howard correspondence implies that types can be constructed that express arbitrarily complex mathematical properties. If the user can supply a constructive proof that a type is inhabited (i.e., that a value of that type exists) then a compiler can check the proof and convert it into executable computer code that computes the value by carrying out the construction. The proof checking feature makes dependently typed languages closely related to proof assistants. The code-generation aspect provides a powerful approach to formal program verification and proof-carrying code, since the code is derived directly from a mechanically verified mathematical proof.
Comparison of languages with dependent types[edit]
See also: Proof assistant § Comparison
Paradigm[a]
Types can depend on[b]
Extraction erases irrelevant terms
Ada 2012 Yes[4] Imperative Yes[5] Yes (optional)[6] ? Any term[c] ? ? Ada ?
Agda Yes[7] Purely functional Few/limited[d] Yes Yes (optional) Any term Yes (optional)[e] Proof-irrelevant arguments[9] Proof-irrelevant propositions[10] Haskell, JavaScript Yes[9]
ATS Yes[11] Functional / imperative No[12] Yes Yes Static terms[13] ? Yes Yes Yes
Cayenne No Purely functional No Yes No Any term No No ? ?
(Coq) Yes[14] Purely functional Yes Yes Yes Any term Yes[f] Yes[15] Haskell, Scheme and OCaml Yes
Dependent ML No[g] ? ? Yes ? Natural numbers ? ? ? ?
F* Yes[16] Functional and imperative Yes[17] Yes Yes (optional) Any pure term Yes Yes OCaml, F#, and C Yes
Guru No[18] Purely functional[19] hypjoin[20] Yes[19] Yes Any term No Yes Carraway Yes
Idris Yes[21] Purely functional[22] Yes[23] Yes Yes (optional) Any term Yes No Yes Yes, aggressively[23]
Lean Yes Purely functional Yes Yes Yes Any term Yes Yes Yes Yes
Matita Yes[24] Purely functional Yes Yes Yes Any term Yes Yes OCaml Yes
NuPRL Yes Purely functional Yes Yes Yes Any term Yes ? Yes ?
PVS Yes ? Yes ? ? ? ? ? ? ?
Sage No[h] Purely functional No No No ? No ? ? ?
Twelf Yes Logic programming ? Yes Yes (optional) Any (LF) term No No ? ?
^ This refers to the core language, not to any tactic (theorem proving procedure) or code generation sublanguage.
^ Subject to semantic constraints, such as universe constraints
^ Static_Predicate for restricted terms, Dynamic_Predicate for Assert-like checking of any term in type cast
^ Ring solver[8]
^ Optional universes, optional universe polymorphism, and optional explicitly specified universes
^ Universes, automatically inferred universe constraints (not the same as Agda's universe polymorphism) and optional explicit printing of universe constraints
^ Has been superseded by ATS
^ Last Sage paper and last code snapshot are both dated 2006
^ Sørensen, Morten Heine B.; Pawel Urzyczyn (1998). "Lectures on the Curry-Howard Isomorphism". CiteSeerX 10.1.1.17.7385. {{cite journal}}: Cite journal requires |journal= (help)
^ Bove, Ana; Peter Dybjer (2008). "Dependent Types at Work" (PDF). {{cite journal}}: Cite journal requires |journal= (help)
^ a b "ΠΣ: Dependent Types without the Sugar" (PDF).
^ "GNAT Community download page".
^ "RM3.2.4 Subtype Predicates".
^ SPARK is a provable subset of Ada
^ "Agda download page".
^ "Agda Ring Solver".
^ a b "Announce: Agda 2.2.8". Archived from the original on 2011-07-18. Retrieved 2010-09-28.
^ "Agda 2.6.0 changelog".
^ "ATS2 downloads".
^ "email from ATS inventor Hongwei Xi".
^ "Applied Type System: An Approach to Practical Programming with Theorem-Proving" (PDF).
^ "Coq CHANGES in Subversion repository".
^ "Introduction of SProp in Coq 8.10".
^ "F* changes on GitHub". GitHub.
^ "F* v0.9.5.0 release notes on GitHub". GitHub.
^ "Guru SVN".
^ a b Aaron Stump (6 April 2009). "Verified Programming in Guru" (PDF). Archived from the original (PDF) on 29 December 2009. Retrieved 28 September 2010.
^ Adam Petcher (1 April 2008). "Deciding Joinability Modulo Ground Equations in Operational Type Theory" (PDF). Retrieved 14 October 2010.
^ "Idris git repository". GitHub. 17 May 2022.
^ "Idris, a language with dependent types - extended abstract" (PDF). Archived from the original (PDF) on 2011-07-16.
^ a b Edwin Brady. "How does Idris compare to other dependently-typed programming languages?".
^ "Matita SVN". Archived from the original on 2006-05-08. Retrieved 2010-09-29.
Martin-Löf, Per (1984). Intuitionistic Type Theory (PDF). Bibliopolis.
Nordström, Bengt; Petersson, Kent; Smith, Jan M. (1990). Programming in Martin-Löf's Type Theory: An Introduction. Oxford University Press. ISBN 9780198538141.
Barendregt, H. (1992). "Lambda calculi with types". In Abramsky, S.; Gabbay, D.; Maibaum, T. (eds.). Handbook of Logic in Computer Science. Oxford Science Publications.
McBride, Conor; McKinna, James (January 2004). "The view from the left". Journal of Functional Programming. 14 (1): 69–111. doi:10.1017/s0956796803004829. S2CID 6232997.
Altenkirch, Thorsten; McBride, Conor; McKinna, James (2006). "Why dependent types matter" (PDF). Proceedings of the 33rd ACM SIGPLAN-SIGACT Symposium on Principles of Programming Languages, POPL 2006, Charleston, South Carolina, USA, January 11-13. ISBN 1-59593-027-2.
Norell, Ulf (September 2007). Towards a practical programming language based on dependent type theory (PDF) (PhD). Göteborg, Sweden: Department of Computer Science and Engineering, Chalmers University of Technology. ISBN 978-91-7291-996-9.
Oury, Nicolas; Swierstra, Wouter (2008). "The Power of Pi" (PDF). ICFP '08: Proceedings of the 13th ACM SIGPLAN international conference on Functional programming. pp. 39–50. doi:10.1145/1411204.1411213. ISBN 9781595939197. S2CID 16176901.
Norell, Ulf (2009). "Dependently Typed Programming in Agda" (PDF). In Koopman, P.; Plasmeijer, R.; Swierstra, D. (eds.). Advanced Functional Programming. AFP 2008. Lecture Notes in Computer Science. Vol. 5832. Springer. pp. 230–266. doi:10.1007/978-3-642-04652-0_5. ISBN 978-3-642-04651-3.
Sitnikovski, Boro (2018). Gentle Introduction to Dependent Types with Idris. Lean Publishing. ISBN 978-1723139413.
McBride, Conor; Nordvall-Forsberg, Fredrik (2022). "Type systems for programs respecting dimensions" (PDF). Advanced Mathematical and Computational Tools in Metrology and Testing XII. Advances in Mathematics for Applied Sciences. World Scientific. pp. 331–345. doi:10.1142/9789811242380_0020. ISBN 9789811242380. S2CID 243831207.
Dependently Typed Programming 2008
"Dependent type" at the Haskell Wiki
dependent type theory in nLab
dependent type in nLab
dependent product type in nLab
dependent sum type in nLab
dependent product in nLab
dependent sum in nLab
Retrieved from "https://en.wikipedia.org/w/index.php?title=Dependent_type&oldid=1089182693"
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d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
\frac{\partial u}{\partial t}=\Delta u
u=0
u\left(0\right)=1
{x}^{2}+{y}^{2}\le 0.{4}^{2}
u\left(0\right)=0
\frac{\partial u}{\partial t}=\Delta u
u=0
u\left(0\right)=1
{x}^{2}+{y}^{2}\le 0.{4}^{2}
u\left(0\right)=0
{u}_{0}=1{0}^{-3}xyz.
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f.
coefficient value of 1. The values of the other coefficients in this problem are
d=1
a=0
f=0
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
m\frac{{\partial }^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla ·\left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{in }\Omega
\rho C\frac{\partial u}{\partial t}-\nabla \text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(k\nabla u\right)+h\left(u-{u}_{\infty }\right)=f
u\left(x,t\right)=\sum _{i}{U}_{i}\left(t\right){\varphi }_{i}\left(x\right)
\begin{array}{l}\sum _{i}\underset{\Omega }{\int }d{\varphi }_{j}{\varphi }_{i}\text{\hspace{0.17em}}\frac{d{U}_{i}\left(t\right)}{dt}\text{\hspace{0.17em}}dx+\sum _{i}\left(\underset{\Omega }{\int }\left(\nabla {\varphi }_{j}\cdot \left(c\nabla {\varphi }_{i}\right)+a{\varphi }_{j}{\varphi }_{i}\right)\text{\hspace{0.17em}}dx+\underset{\partial \Omega }{\int }q{\varphi }_{j}{\varphi }_{i}\text{\hspace{0.17em}}ds\right){U}_{i}\left(t\right)\\ \text{ }\text{ }\text{ }\text{ }=\underset{\Omega }{\int }f{\varphi }_{j}\text{\hspace{0.17em}}dx+\underset{\partial \Omega }{\int }g{\varphi }_{j}\text{\hspace{0.17em}}ds\text{ }\forall j\end{array}
M\frac{dU}{dt}+KU=F
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Divergence theorem - Wikiversity
A novice might find a proof easier to follow if we greatly restrict the conditions of the theorem, but carefully explain each step. For that reason, we prove the divergence theorem for a rectangular box, using a vector field that depends on only one variable.
Fig. 1: A region V bounded by the surface S = ∂V with the surface normal n
Fig. 2: Using only the fundamental theorem of calculus in one dimension, students can verify the divergence theorem by direct integration.
See Wikipedia:Divergence theorem for more rigorous proofs.
See the subpage, Divergence theorem/Proof for another proof.
The Divergence (Gauss-Ostrogradsky) theorem relates the integral over a volume,
{\displaystyle {\mathsf {V}}}
, of the divergence of a vector function,
{\displaystyle \mathbf {\vec {F}} }
, and the integral of that same function over the the volume's surface:
{\displaystyle \int _{\mathsf {V}}\underbrace {\left(\mathbf {\vec {\nabla }} \cdot \mathbf {\vec {F}} \right)\,dV} _{\text{triple integral}}=\oint _{\mathsf {S}}\underbrace {\left(\mathbf {\vec {F}} \cdot \mathbf {\hat {n}} \right)\,dS} _{\text{double integral}}.}
Special case of a field that depends on only one variable:[edit | edit source]
Figure 2 shows a vector field using red arrows to denote the field's direction and magnitude. The field occupies all three dimensions, but is only directed in the x or y directions. And, the field varies only along the x-axis. In other words,
{\displaystyle \mathbf {\vec {F}} }
is of the form,
{\displaystyle \mathbf {\vec {F}} =F_{x}(x)\mathbf {\hat {i}} +F_{y}(x)\mathbf {\hat {j}} }
{\displaystyle F_{x}}
{\displaystyle F_{y}}
are arbitrary "smooth" functions of only the variable
{\displaystyle x.}
A z-component,
{\displaystyle F_{z}(x)}
, could be included in (2) without adding any complexity to the proof, except that the third component could not easily be depicted in Figure 2.
Volume integral[edit | edit source]
Our first task is to use the fundamental theorem of calculus to perform this integral over the volume of the box. Despite the fact that the vector field points in two directions, the divergence operator
{\displaystyle \mathbf {\vec {\nabla }} }
removes any involvement of the y-component, since the field does not depend on
{\displaystyle y}
{\displaystyle \mathbf {{\vec {\nabla }}\cdot {\vec {F}}} ={\frac {\partial F_{x}}{\partial x}}={\text{ a function of }}x{\text{ only.}}}
Integrals over three-dimensional boxes are easy to perform:
{\displaystyle \int \left(\mathbf {{\vec {\nabla }}\cdot {\vec {F}}} \right)\,dV=\int _{x_{1}}^{x_{2}}dx\int _{y_{1}}^{y_{2}}dy\int _{z_{1}}^{z_{2}}dz\left({\frac {\partial F_{x}}{\partial x}}\right)}
{\displaystyle \partial F_{x}/\partial x}
{\displaystyle y}
{\displaystyle z,}
the partial derivative may be treated as a constant with regard to those to variables. The integrations in (4) are structured so that they may be performed in any order, which permits us to write:
{\displaystyle \int \left(\mathbf {{\vec {\nabla }}\cdot {\vec {F}}} \right)\,dV=}
{\displaystyle \underbrace {\int _{x_{1}}^{x_{2}}dx\left({\frac {\partial F_{x}}{\partial x}}\right)} _{F_{x}(x_{2})-F_{x}(x_{1})}\;\underbrace {\int _{y_{1}}^{y_{2}}dy} _{y_{2}-y_{1}}\;\underbrace {\int _{z_{1}}^{z_{2}}dz} _{z_{2}-z_{1}}\,}
{\displaystyle =\,\left[F_{x}(x_{2})-F_{x}(x_{1})\right]\cdot A}
{\displaystyle A=(y_{2}-y_{1})(z_{2}-z_{1})}
is the area of the two sides at
{\displaystyle x=x_{1}}
{\displaystyle x=x_{2}}
Surface integrals[edit | edit source]
Now we calculate the surface integral and verify that it yields the same result as (5). The surface integral over the box involves six rectangles. One way to define a two-dimensional surface in three-dimensions is to state one equation involving the three variables.[1] The six surfaces of the box in Figure 2 can be labeled by the set of six (infinite) planes:
{\displaystyle \{x=x_{1},x=x_{2},y=y_{1},y=y_{2},z=z_{1},z=z_{2}\}}
. In contrast with one-dimensional integrals, surface and volume integrals require much more than simply stating the two endpoints associated with a one dimensional integral. Each of the six surfaces must be integrated separately, and the integral over the entire surface equals the sum of the six individual integrals. A common shorthand for surface integrals is,
{\displaystyle {\hat {\mathbf {n} }}\,d\mathbf {S} =d{\vec {A}}}
{\displaystyle {\hat {\mathbf {n} }}}
the outward unit normal, and
{\displaystyle d\mathbf {S} =\left|d{\vec {A}}\right|.}
{\displaystyle \int \underbrace {(\,.\,.\,.)} _{\text{6 sides}}=\overbrace {\int \underbrace {(\,.\,.\,.)} _{x=x_{1}}+\int \underbrace {(\,.\,.\,.)} _{x=x_{2}}} ^{\text{x-pair}}+\overbrace {\int \underbrace {(\,.\,.\,.)} _{y=y_{1}}+\int \underbrace {(\,.\,.\,.)} _{y=y_{2}}} ^{\text{y-pair}}+\overbrace {\int \underbrace {(\,.\,.\,.)} _{z=z_{1}}+\int \underbrace {(\,.\,.\,.)} _{z=z_{2}}} ^{\text{z-pair}}}
Integration over x-pair surfaces[edit | edit source]
The two most important surfaces are the pair already discussed, with area
{\displaystyle A}
, and outward unit normals in the x-direction, shown as the left and right ends of the box shown in Figure 2. Since
{\displaystyle x}
is constant over each of these two surfaces, the the surface integral of anything that depends only on
{\displaystyle \mathbf {\vec {F}} (x)}
is trivial, and depends only on the area, which is the same area,
{\displaystyle A}
, shown in Figure 2 and also at (5). The algebra for the surface at
{\displaystyle x=x_{1}}
{\displaystyle \underbrace {\int \left(\mathbf {\vec {F}} \cdot \mathbf {\hat {n}} \right)dS} _{x=x_{1}}=\underbrace {\left(\mathbf {\vec {F}} \cdot \mathbf {\hat {n}} \right)} _{-F_{x}(x_{1})}\cdot \underbrace {\int \,dS} _{A}=-F_{x}(x_{1})\cdot A}
To understand how
{\displaystyle \mathbf {\vec {F}} \cdot \mathbf {\hat {n}} }
was evaluated in (7), recall that the outward unit normal points in the negative x direction on that side. Also, when a vector is dotted with a unit vector, only one component of the vector survives:[2]
{\displaystyle \underbrace {\mathbf {\vec {F}} \cdot \mathbf {\hat {n}} } _{x=x_{1}}=\left(F_{x}\mathbf {\hat {i}} +F_{y}\mathbf {\hat {j}} \right)\cdot \left(-\mathbf {\hat {i}} \right)=-F_{x}}
{\displaystyle \mathbf {\hat {n}} =-\mathbf {\hat {i}} }
{\displaystyle x=x_{1}.}
{\displaystyle \mathbf {\hat {n}} }
points in the opposite direction, so that combining (5) with (7) establishes that (1) is true, but only if we can establish that the other four surface integrals vanish.
Integration over the y-pair and z-pair surfaces[edit | edit source]
We begin with the y-pair of surfaces, i.e., those that occupy a portion of the
{\displaystyle y=y_{1}}
{\displaystyle y=y_{2}}
planes. Since the vector field
{\displaystyle \mathbf {\vec {F}} (x,y,z)}
{\displaystyle x}
-variable, the integral is not trivial. By the reasoning used at (7) and (8), we only need integrate the
{\displaystyle y}
-component,
{\displaystyle F_{y}}
, because the unit vector points in the
{\displaystyle \pm \mathbf {\hat {j}} }
direction. Fortunately, the absence of a
{\displaystyle y}
-dependence of our vector field ensures that the following two integrals are equal:
{\displaystyle \underbrace {\iint F_{y}(x)dS} _{y=y_{1}}=\underbrace {\iint F_{y}(x)dS} _{y=y_{2}}=\int _{z_{1}}^{z_{2}}dz\int _{x_{1}}^{x_{2}}F_{y}(x)dx=(z_{2}-z_{1})\int _{x_{1}}^{x_{2}}F_{y}(x)dx}
Since the outward unit normals,
{\displaystyle \mathbf {\hat {n}} =\pm \mathbf {\hat {j}} ,}
are equal and opposite on both sides, the pair of surface integrals at (9) cancel. The same argument can be the
{\displaystyle z}
-pair of surfaces.
Although we have proven the divergence theorem on a rectangular box for a small subset of all possible differentiable vector fields
{\displaystyle {\vec {F}}({\vec {r}})}
, we have established the essential role played by the fundamental theorem of calculus in one-dimension. Essentially, we have placed the two endpoints of segment along the x-axis by the six surfaces that form the boundary of a three-dimensional box. Also, the reader who can grasp this proof is probably capable of developing a proof for any vector field
{\displaystyle {\vec {F}}({\vec {r}})}
for which the divergence is well defined. It just takes a lot of algebra.
↑ For example,
{\displaystyle x^{2}+y^{2}+z^{3}=1}
is the surface of a unit sphere.
↑ Here we adopt the notation that
{\displaystyle (\mathbf {\hat {i}} ,\mathbf {\hat {j}} ,\mathbf {\hat {k}} )}
refer to unit vectors in the
{\displaystyle (x,y,z)}
Retrieved from "https://en.wikiversity.org/w/index.php?title=Divergence_theorem&oldid=2269955"
|
The average relative expression ratios (
\overline{\mathbit{R}}
). (A) The
\overline{R}
values of the total genes of the mutant and wt viruses plotted against time. The average level of the transcripts of the mutant virus is lower in the early stage, but higher in the late stage of infection. (B) The
\overline{R}
values of the different kinetic classes of mutant and wt viruses. In the mutant virus, the L genes are repressed in the early stage of infection, whereas the E gene expressions are enhanced in the late stage of infection.
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Section 42.3 (0EAA): Calculation of some multiplicities—The Stacks project
Section 42.3: Calculation of some multiplicities (cite)
42.3 Calculation of some multiplicities
To prove equality of certain cycles later on we need to compute some multiplicities. Our main tool, besides the elementary lemmas on multiplicities given in the previous section, will be Algebra, Lemma 10.121.7.
Lemma 42.3.1. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Let $x \in R$. Assume that
$\dim (\text{Supp}(M)) \leq 1$, and
$\dim (\text{Supp}(M/xM)) \leq 0$.
Write $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ t\} $. Then
\[ e_ R(M, 0, x) = \sum \nolimits _{i = 1, \ldots , t} \text{ord}_{R/\mathfrak q_ i}(x) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}). \]
Proof. We first make some preparatory remarks. The result of the lemma holds if $M$ has finite length, i.e., if $t = 0$, because both the left hand side and the right hand side are zero in this case, see Lemma 42.2.4. Also, if we have a short exact sequence $0 \to M \to M' \to M'' \to 0$ of modules satisfying (1) and (2), then lemma for 2 out of 3 of these implies the lemma for the third by the additivity of length (Algebra, Lemma 10.52.3) and additivty of multiplicities (Lemma 42.2.3).
Denote $M_ i$ the image of $M$ in $M_{\mathfrak q_ i}$, so $\text{Supp}(M_ i) = \{ \mathfrak m, \mathfrak q_ i\} $. The kernel and cokernel of the map $M \to \bigoplus M_ i$ have support $\{ \mathfrak m\} $ and hence have finite length. By our preparatory remarks, it follows that it suffices to prove the lemma for each $M_ i$. Thus we may assume that $\text{Supp}(M) = \{ \mathfrak m, \mathfrak q\} $. In this case we have a finite filtration $M \supset \mathfrak qM \supset \mathfrak q^2M \supset \ldots \supset \mathfrak q^ nM = 0$ by Algebra, Lemma 10.62.4. Again additivity shows that it suffices to prove the lemma in the case $M$ is annihilated by $\mathfrak q$. In this case we can view $M$ as a $R/\mathfrak q$-module, i.e., we may assume that $R$ is a Noetherian local domain of dimension $1$ with fraction field $K$. Dividing by the torsion submodule, i.e., by the kernel of $M \to M \otimes _ R K = V$ (the torsion has finite length hence is handled by our preliminary remarks) we may assume that $M \subset V$ is a lattice (Algebra, Definition 10.121.3). Then $x : M \to M$ is injective and $\text{length}_ R(M/xM) = d(M, xM)$ (Algebra, Definition 10.121.5). Since $\text{length}_ K(V) = \dim _ K(V)$ we see that $\det (x : V \to V) = x^{\dim _ K(V)}$ and $\text{ord}_ R(\det (x : V \to V)) = \dim _ K(V) \text{ord}_ R(x)$. Thus the desired equality follows from Algebra, Lemma 10.121.7 in this case. $\square$
Lemma 42.3.2. Let $R$ be a Noetherian local ring. Let $x \in R$. If $M$ is a finite Cohen-Macaulay module over $R$ with $\dim (\text{Supp}(M)) = 1$ and $\dim (\text{Supp}(M/xM)) = 0$, then
\[ \text{length}_ R(M/xM) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}). \]
where $\mathfrak q_1, \ldots , \mathfrak q_ t$ are the minimal primes of the support of $M$. If $I \subset R$ is an ideal such that $x$ is a nonzerodivisor on $R/I$ and $\dim (R/I) = 1$, then
\[ \text{length}_ R(R/(x, I)) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}((R/I)_{\mathfrak q_ i}) \]
where $\mathfrak q_1, \ldots , \mathfrak q_ n$ are the minimal primes over $I$.
Proof. These are special cases of Lemma 42.3.1. $\square$
Here is another case where we can determine the value of a multiplicity.
Lemma 42.3.3. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\varphi : M \to M$ be an endomorphism and $n > 0$ such that $\varphi ^ n = 0$ and such that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\varphi ^{n - 1})$ has finite length as an $R$-module. Then
\[ e_ R(M, \varphi ^ i, \varphi ^{n - i}) = 0 \]
for $i = 0, \ldots , n$.
Proof. The cases $i = 0, n$ are trivial as $\varphi ^0 = \text{id}_ M$ by convention. Let us think of $M$ as an $R[t]$-module where multiplication by $t$ is given by $\varphi $. Let us write $K_ i = \mathop{\mathrm{Ker}}(t^ i : M \to M)$ and
\[ a_ i = \text{length}_ R(K_ i/t^{n - i}M),\quad b_ i = \text{length}_ R(K_ i/tK_{i + 1}),\quad c_ i = \text{length}_ R(K_1/t^ iK_{i + 1}) \]
Boundary values are $a_0 = a_ n = b_0 = c_0 = 0$. The $c_ i$ are integers for $i < n$ as $K_1/t^ iK_{i + 1}$ is a quotient of $K_1/t^{n - 1}M$ which is assumed to have finite length. We will use frequently that $K_ i \cap t^ jM = t^ jK_{i + j}$. For $0 < i < n - 1$ we have an exact sequence
\[ 0 \to K_1/t^{n - i - 1}K_{n - i} \to K_{i + 1}/t^{n - i - 1}M \xrightarrow {t} K_ i/t^{n - i}M \to K_ i/tK_{i + 1} \to 0 \]
By induction on $i$ we conclude that $a_ i$ and $b_ i$ are integers for $i < n$ and that
\[ c_{n - i - 1} - a_{i + 1} + a_ i - b_ i = 0 \]
For $0 < i < n - 1$ there is a short exact sequence
\[ 0 \to K_ i/tK_{i + 1} \to K_{i + 1}/tK_{i + 2} \xrightarrow {t^ i} K_1/t^{i + 1}K_{i + 2} \to K_1/t^ iK_{i + 1} \to 0 \]
\[ b_ i - b_{i + 1} + c_{i + 1} - c_ i = 0 \]
Since $b_0 = c_0$ we conclude that $b_ i = c_ i$ for $i < n$. Then we see that
\[ a_2 = a_1 + b_{n - 2} - b_1,\quad a_3 = a_2 + b_{n - 3} - b_2,\quad \ldots \]
It is straighforward to see that this implies $a_ i = a_{n - i}$ as desired. $\square$
Lemma 42.3.4. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$ with $M$ finite and with cohomology groups of finite length over $R$. Let $x \in R$ be such that $\dim (\text{Supp}(M/xM)) \leq 0$. Then
\[ e_ R(M, x\varphi , \psi ) = e_ R(M, \varphi , \psi ) - e_ R(\mathop{\mathrm{Im}}(\varphi ), 0, x) \]
\[ e_ R(M, \varphi , x\psi ) = e_ R(M, \varphi , \psi ) + e_ R(\mathop{\mathrm{Im}}(\psi ), 0, x) \]
Proof. We will only prove the first formula as the second is proved in exactly the same manner. Let $M' = M[x^\infty ]$ be the $x$-power torsion submodule of $M$. Consider the short exact sequence $0 \to M' \to M \to M'' \to 0$. Then $M''$ is $x$-power torsion free (More on Algebra, Lemma 15.88.4). Since $\varphi $, $\psi $ map $M'$ into $M'$ we obtain a short exact sequence
\[ 0 \to (M', \varphi ', \psi ') \to (M, \varphi , \psi ) \to (M'', \varphi '', \psi '') \to 0 \]
of $(2, 1)$-periodic complexes. Also, we get a short exact sequence $0 \to M' \cap \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Im}}(\varphi '') \to 0$. We have $e_ R(M', \varphi , \psi ) = e_ R(M', x\varphi , \psi ) = e_ R(M' \cap \mathop{\mathrm{Im}}(\varphi ), 0, x) = 0$ by Lemma 42.2.5. By additivity (Lemma 42.2.3) we see that it suffices to prove the lemma for $(M'', \varphi '', \psi '')$. This reduces us to the case discussed in the next paragraph.
Assume $x : M \to M$ is injective. In this case $\mathop{\mathrm{Ker}}(x\varphi ) = \mathop{\mathrm{Ker}}(\varphi )$. On the other hand we have a short exact sequence
\[ 0 \to \mathop{\mathrm{Im}}(\varphi )/x\mathop{\mathrm{Im}}(\varphi ) \to \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(x\varphi ) \to \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi ) \to 0 \]
This together with (42.2.2.1) proves the formula. $\square$
Comment #6725 by jok on November 14, 2021 at 16:21
In lemma 42. 3. 1, the second sentence, what is t=0?
@#6725. The integer
t
denotes the number of dimension
1
primes in the support of
M
t = 0
M
0
or has support of dimension
0
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The Multiplication Process - MATLAB & Simulink - MathWorks France
The multiplication of an n-bit binary number with an m-bit binary number results in a product that is up to m + n bits in length for both signed and unsigned words.
Suppose you want to multiply three numbers. Each of these numbers is represented by a 5-bit word, and each has a different binary-point-only scaling. Additionally, the output is restricted to a 10-bit word with binary-point-only scaling of 2-4. The multiplication is shown in the following model for the input values 5.75, 2.375, and 1.8125.
Applying the rules from the previous section, the multiplication follows these steps:
The first two numbers (5.75 and 2.375) are multiplied:
\begin{array}{c}{Q}_{RawProduct}=\text{ 1}01.11\\ \text{ }\frac{×\text{\hspace{0.17em}}10.011}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}01.11\cdot {2}^{-3}}\\ \text{ }\frac{\begin{array}{l}\text{ 1}01.11\cdot {2}^{-2}\\ +\text{\hspace{0.17em}}\text{1}01.11\cdot {2}^{1}\end{array}}{\text{011}01.10101}\begin{array}{c}\\ =13.65625.\end{array}\end{array}
Note that the binary point of the product is given by the sum of the binary points of the multiplied numbers.
The result of step 1 is converted to the output data type:
\begin{array}{c}{Q}_{Temp}=convert\left({Q}_{RawProduct}\right)\\ =001101.1010=13.6250.\end{array}
Signal Conversions discusses conversions. Note that a loss in precision of one bit occurs, with the resulting value of QTemp determined by the rounding mode. For this example, round-to-floor is used. Furthermore, overflow did not occur but is possible for this operation.
The result of step 2 and the third number (1.8125) are multiplied:
\begin{array}{c}{Q}_{RawProduct}=\text{ 011}01.1010\\ \text{ }\frac{\text{ }×\text{\hspace{0.17em}}1.1101}{\text{\hspace{0.17em}}\text{ 11}01.1010\cdot {2}^{-4}}\\ \text{ }\frac{\begin{array}{l}\text{ 11}01.1010\cdot {2}^{-2}\\ \text{ 11}01.1010\cdot {2}^{-1}\\ +\text{\hspace{0.17em}}\text{11}01.1010\cdot {2}^{0}\end{array}}{\text{0011000}.10110010}\begin{array}{c}\\ =24.6953125.\end{array}\end{array}
The product is converted to the output data type:
\begin{array}{c}{Q}_{a}=convert\left({Q}_{RawProduct}\right)\\ =011000.1011=24.6875.\end{array}
Signal Conversions discusses conversions. Note that a loss in precision of 4 bits occurred, with the resulting value of QTemp determined by the rounding mode. For this example, round-to-floor is used. Furthermore, overflow did not occur but is possible for this operation.
Blocks that perform multiplication include the Product, Discrete FIR Filter, and Gain blocks.
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Combustion Analysis Calculator
How to find the empirical formula from combustion analysis?
How to find the molecular formula?
How to use the combustion analysis calculator
Empirical and molecular formula of C, H, O compounds - An example
Empirical and molecular formula of hydrocarbons — an example
Welcome to Omni's combustion analysis calculator that will determine the empirical and molecular formulas of C, H, O organic compounds from combustion data 🔥. We invite you to read on and learn about:
How to find the empirical formula from combustion analysis; and
How to find the molecular formula yourself.
In chemistry, combustion analysis is a quantitative analysis used to determine the empirical formula of an unknown organic compound containing carbon (C), hydrogen (H), and oxygen (O).
The unknown substance, initially weighted, undergoes a combustion process on a combustion apparatus that collects the combustion products carbon dioxide (CO2) and water (H2O), which are weighed afterwards. Then, the empirical formula and the molar masses of C, H, and O are obtained with this information.
Let's take a look at how to find the empirical formula of a C, H, O organic compound. The process can be divided into three steps:
Calculate the masses of each element;
Determine each's moles; and
Obtain the empirical formula.
Let's see each of these steps in detail 🔎
When calculating the masses, we assume that the organic substance is undergoing complete combustion — that is, the only products of the reaction are carbon dioxide (CO2) and water vapor (H2O), as can be seen in the combustion reaction equation:
\small \text C_\alpha \text H_\beta \text O_\gamma + a \text O_2 \longrightarrow b \text C \text O_2 + c \text H_2 \text O
From here, we can tell that all the carbon (C) initially present in the C, H, O compound is now in the dioxide carbon (CO2) and all the hydrogen (H) is contained in the water vapor (H2O) molecule. With these assumptions, we can calculate the masses of carbon
m_\text{C}
and hydrogen
m_\text{H}
\begin{align*} \footnotesize m_\text{C} & \footnotesize = m_{\text{CO}_2}\cdot \dfrac{M_\text{C}}{M_{\text{CO}_2}} \\[1em] \footnotesize m_\text{H} & \footnotesize = m_{\text{H}_2\text{O}}\cdot \dfrac{2 M_\text{H}}{M_{\text{H}_2\text{O}}} \end{align*}
m_{\text{CO}_2}
m_{\text{H}_2\text{O}}
are the masses of carbon dioxide and water;
M_\text{C}
M_\text{H}
are the molar masses of carbon and hydrogen; and
M_{\text{CO}_2}
M_{\text{H}_2\text{O}}
are the molecular masses of dioxide carbon and water.
The mass of oxygen
m_\text{O}
is obtained as the difference of carbon and hydrogen masses from the sample mass
m_\text{sample}
\footnotesize m_\text{O} = m_\text{sample} - m_\text{C} - m_\text{H}
Once the values of the masses are known, we can determine the moles of each element. For this, we divide each element's mass by its molar mass:
\footnotesize \text{mol}_\text{C}=\dfrac {m_\text{C}}{M_\text{C}} \\[1em] \footnotesize \text{mol}_\text{H}=\dfrac {m_\text{H}}{M_\text{H}} \\[1em]\footnotesize \text{mol}_\text{O}=\dfrac{m_\text{O}}{M_\text{O}}
Finally, to obtain the empirical formula, divide each molar mass by the smallest molar value to get the proportion between the atoms of each element.
Not sure about the difference between molecular weight and molar mass? Check out our molecular weight calculator!
Now that you know how to find the empirical formula of an organic substance, maybe you'd like to know as well how to find its molecular formula. You'll see this is even simpler, all we need is:
The empirical formula of a given substance; and
Its molecular mass.
With these known, we can divide the general procedure to get the molecular formula into three steps:
Step 1. From the empirical formula, calculate the empirical molar mass
\text{EFM}
\begin{align*} \footnotesize \text{EFM} & \footnotesize = \text{mol}_\text{C} \cdot M_\text{C} + \text{mol}_\text{H} \cdot M_\text{H} \\ & \footnotesize + \text{mol}_\text{O}\cdot M_\text{O} \end{align*}
\text{mol}_\text{C}
\text{mol}_\text{H}
\text{mol}_\text{O}
are the moles of carbon, hydrogen and oxygen from the empirical formula; and
M_\text{C}
M_\text{H}
M_\text{O}
are their molar masses.
Step 2. Determine
n
as the ratio between the molar mass and the empirical molar mass of the substance:
\footnotesize n = \dfrac {\text{Molar mass}}{\text{Empirical formula mass}}
Step 3. Finally, multiply the moles of each element in the empirical formula by
n
to get the molecular formula. And that's it! 😀
The combustion analysis calculator will help you find the empirical and molecular formula of C, H, O compound or for a hydrocarbon:
Choose the type of substance that you'd like to study.
Input the molar mass, sample mass, CO2 mass, and H2O mass from the combustion analysis. For hydrocarbons, the sample mass is not required.
The calculator will display your substance's empirical formula, empirical mass, and molecular formula.
If you'd like to know the masses of C, H, and O of the sample, select Yes from the drop-down menu on the last row.
💡 If you don't need the molecular formula, it's not necessary to input the substance's molar mass. The combustion analysis calculator will still give you the empirical formula.
Let's see how to get the empirical and molecular formulas of a C, H, O compound with a numerical example!
Consider that from a combustion analysis report, we get that after burning a sample of 12.915 g of a C, H, O compound, 18.942 g CO2 and 7.749 g of H2O are formed. The molar mass is 90.0779 g/mol. What are the empirical and molecular formulas of the substance? 🤔
To solve the problem, we divide the solution process into two phases. We begin by obtaining the empirical formula, then we obtain the molecular formula.
First, determine the masses of C, H, and O that are present in the sample:
\begin{align*} \footnotesize m_\text{C} & \footnotesize = 18.942\: \text{g} \: \text{CO}_2 \cdot\dfrac{12.011\: \tfrac{\text{g}}{\text{mol}} \ \text C}{44.010 \: \tfrac{\text{g}}{\text{mol}} \: \text C \text O_2} \\ & \footnotesize = 5.1694 \: \text{g} \ \text C \\ \footnotesize m_\text{H} & \footnotesize = 7.749 \: \text{g} \: \text H_2 \text O \cdot\dfrac{2 \cdot 1.00797 \: \tfrac{\text{g}}{\text{mol}} \ \text H}{18.0153 \: \tfrac{\text{g}}{\text{mol}} \: \text H_2 \text O} \\ & \footnotesize = 0.8669 \: \text{g} \ \text H \\ \footnotesize m_\text{O} & \footnotesize = 12.915 \: \text{g} - 5.1694 \: \text{g}\ \text C - 0.8669 \: \text{g}\ \text H \\ & \footnotesize = 6.879 \: \text{g} \ \text O \end{align*}
Once we know the values of the masses, next we calculate the number of moles of each element:
\begin{align*} \footnotesize \text{mol}_\text{C} & \footnotesize =\dfrac {5.1694 \: \text{g} \: \text C }{12.011 \: \tfrac{\text{g}}{\text{mol}} \: \text C}= 0.43039 \: \text{mol} \: \text C \\ \footnotesize \text{mol}_\text{H} & \footnotesize =\dfrac {0.8669 \: \text{g}\: \text H }{1.00797 \: \tfrac{\text{g}}{\text{mol}} \: \text H} = 0.8600 \: \text{mol} \: \text H \\ \footnotesize \text{mol}_\text{O} & \footnotesize =\dfrac {6.879 \: \text{g}\: \text O }{15.9994 \: \tfrac{\text{g}}{\text{mol}} \: \text O} = 0.42995 \: \text{mol}\ \text O \end{align*}
Finally, to obtain the empirical formula, we divide the molar masses by the smallest value of them. This way, we can obtain the proportion between the three elements.
In our example, the smallest value of moles corresponds to oxygen:
\begin{align*} \footnotesize \text{mol}_\text{C} & \footnotesize= \dfrac {0.43039 \: \text{mol} \ \text C}{0.42995} = 1.0010 \approx 1 \ \text{mol}\ \text C \\ \footnotesize \text{mol}_\text{H} & \footnotesize= \dfrac {0.8600 \: \text{mol} \ \text H }{0.42995} = 2.0002 \approx 2 \ \text{mol}\ \text H \\ \footnotesize \text{mol}_\text{O} & \footnotesize= \dfrac {0.42995 \: \text{mol} \ \text O}{0.42995} = 1\: \text{mol}\ \text O \end{align*}
From here, we get the empirical formula for our unknown substance:
\text C \text H_2 \text O
💡 Notice that we approximate the number of moles to the closest integer when calculating the proportion between the elements.
To find the molecular formula, we start by calculating the empirical molar mass
\text{EFM}
\begin{align*} \footnotesize \text{EFM} & \footnotesize = \bigg(\dfrac{1 \: \text{mol}\ \text C}{\text{mol} \: \text{substance}} \cdot \dfrac{12.011 \: \text{g}}{\text{mol} \ \text C}\bigg) \\ & \footnotesize +\bigg(\dfrac{2 \: \text{mol}\ \text{H}}{\text{mol} \: \text{substance}}\cdot \dfrac{1.00797 \: \text{g}}{\text{mol}\ \text H}\bigg) \\ & \footnotesize+\bigg(\dfrac{1 \: \text{mol}\ \text O} {\text{mol} \: \text{substance}}\cdot \dfrac {15.9994 \: \text{g}}{\text{mol}\ \text O}\bigg) \\ & \footnotesize = 30.031 \: \dfrac{\text{g}}{\text{mol}} \end{align*}
Next, we calculate the ratio
n
between the molar masses of the molar and empirical formulas:
\footnotesize n = \dfrac {90.0779 \ \tfrac{\text{g}}{\text{mol}}}{30.031 \ \tfrac{\text{g}}{\text{mol}}}\approx 3
Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio n:
(\text C \text H_2 \text O)_3
\text C_3 \text H_6 \text O_3
The method to determine the empirical formula of a hydrocarbon by combustion analysis is similar to the one we studied for C, H, O compounds. Again, to make this procedure clear and illustrate the differences between the first one, we’ll check a numerical example.
Suppose that from a combustion analysis, we get the following information: after burning a sample of 12.501 g of a hydrocarbon, we see that 33.057 g CO2 and 10.816 g of H2O have formed. The molar mass is 204.35 g/mol. What are the empirical and molecular formulas of the hydrocarbon? 🤔
Again, we'll separate the solution onto two stages:
Empirical formula obtention; and
Following the steps explained before, first we calculate the masses of C and H that are present in the sample compound:
\begin{align*} \footnotesize m_\text{C} &\footnotesize= 33.057 \: \text{g} \: \text C\text O_2 \cdot \dfrac{12.011\: \tfrac{\text{g}}{\text{mol}}\: \text C}{44.010\: \tfrac{\text{g}}{\text{mol}}\: \text C\text O_2} \\ & \footnotesize = 9.0218\: \text{g}\: \text C \\ \footnotesize m_\text{H} &\footnotesize= 10.816\: \text{g}\: \text H_2 \text O \cdot \dfrac{2 \cdot 1.00797 \: \tfrac{\text{g}}{\text{mol}} \: \text H}{18.0153\: \tfrac{\text{g}}{\text{mol}}\: \text H_2\text O} \\ & \footnotesize = 1.2103\: \text{g}\: \text H \end{align*}
💡 Notice this time, we didn't use the sample mass value in our calculation! This amount is used to find the mass of oxygen in the case of a C, H, O substance.
With these values known, next we calculate the number of moles of each element:
\begin{align*} \footnotesize \text{mol}_\text{C} &\footnotesize=\dfrac {9.0218 \ \text{g} \ \text C }{12.011 \ \tfrac{\text{g}}{\text{mol}} \ \text C}\\ & \footnotesize = 0.75112 \ \text{mol}\ \text C \\[1em] \footnotesize \text{mol}_\text{H}&\footnotesize=\dfrac {1.2103 \ \text{g} \ \text H }{1.00797 \ \tfrac{\text{g}}{\text{mol}} \ \text H}\\ & \footnotesize= 1.20076 \ \text{mol} \ \text H \end{align*}
Finally, to obtain the empirical formula, we divide each of the amounts of moles by the smallest of them. In this example, the smallest value of moles corresponds to hydrogen:
\begin{align*} \footnotesize \text{mol}_\text{C} &\footnotesize= \dfrac {0.75112\: \text{mol}\: \text C} {0.75112} \\ &\footnotesize= 1\: \text{mol}\: \text C \\[1em] \footnotesize \text{mol}_\text{H} &\footnotesize= \dfrac {1.20076 \: \text{mol}\: \text H }{0.75112} \\ &\footnotesize= 1.5968\: \text{mol}\: \text H \end{align*}
Note this time we aren't rounding the moles of hydrogen to 2. Doing that will yield an incorrect proportion between the elements. Instead, we express the decimal value 1.5968 into the fraction
\tfrac{8}{5}
— then the ratio between the moles of carbon and hydrogen is 5 mol C : 8 mol H.
\text C_5 \text H_8
To get the molecular formula, first we calculate the empirical molar mass
\text{EFM}
\begin{align*} \footnotesize \text{EFM} &\footnotesize= \bigg(\dfrac{5 \ \text{mol}\ \text C}{\text{mol substance}}\cdot \dfrac{12.011 \ \text{g}}{ \text{mol}\ \text C}\bigg) \\ &\footnotesize + \bigg(\dfrac{8 \ \text{mol}\ \text H}{\text{mol substance}}\cdot \dfrac{1.00797 \ \text{g}}{ \text{mol}\ \text H}\bigg) \\ &\footnotesize = 68.119\: \dfrac{\text{g}}{\text{mol}} \end{align*}
n
\footnotesize n = \dfrac {204.35\: \tfrac{\text{g}}{\text{mol}}}{68.119 \: \tfrac{\text{g}}{\text{mol}}}\approx 3
Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio
n
(\text C_5 \text H_8)3
\text C_{15} \text H_{24}
Hydrocarbons (C-H)
Sample's molar mass
Carbon dioxide (CO₂) mass
Water (H₂O) mass
Show atom masses from sample
Find the interplanar distance for a cubic crystal system using our Miller indices calculator.
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\mathrm{LerchPhi}\left(z,a,v\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{z}^{n}}{{\left(v+n\right)}^{a}}
|z|<1
|z|=1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}1<\mathrm{ℜ}\left(a\right)
z
and
v
v
are positive integers, LerchPhi(z, a, v) has a branch cut in the
z
z=1
z=1
a
z
1-a
z=1
1<\mathrm{ℜ}\left(a\right)
\mathrm{ℜ}\left(a\right)\le 1
0\le \mathrm{ℜ}\left(a\right)
a
\mathrm{LerchPhi}\left(3,4,1\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{polylog}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)}{\textcolor[rgb]{0,0,1}{3}}
\mathrm{LerchPhi}\left(0,7,4\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{16384}}
\mathrm{LerchPhi}\left(4,0,3\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}
\mathrm{LerchPhi}\left(z,a,1\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{polylog}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\right)}{\textcolor[rgb]{0,0,1}{z}}
\mathrm{LerchPhi}\left(1,z,1\right)
\textcolor[rgb]{0,0,1}{\mathrm{\zeta }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)
\mathrm{diff}\left(\mathrm{LerchPhi}\left(z,3,4\right),z\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{LerchPhi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)}{\textcolor[rgb]{0,0,1}{z}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{LerchPhi}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)}{\textcolor[rgb]{0,0,1}{z}}
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Dictionary:Kirchhoff’s equation - SEG Wiki
Revision as of 03:32, 23 February 2015 by Chrisdinneen (talk | contribs) (Fixed Equations.)
{\displaystyle \psi _{p}}
{\displaystyle \tau =(t-r/V)}
{\displaystyle \tau }
{\displaystyle \psi _{p}}
{\displaystyle \psi }
{\displaystyle (t-r/V)}
{\displaystyle \psi _{p}={\frac {1}{4\pi }}\iint \left\{[\psi ]{\frac {\partial {\frac {1}{r}}}{\partial n}}-{\frac {1}{Vr}}{\frac {\partial r}{\partial n}}\left[{\frac {\partial \psi }{\partial t}}\right]-{\frac {1}{r}}\left[{\frac {\partial \psi }{\partial n}}\right]\right\}ds}
{\displaystyle \tau =(t-r/V)}
{\displaystyle \psi (x,z,t)={\frac {z}{\pi }}\int \left[1/r^{3}-(2/Vr^{2})\times \left({\frac {\partial }{\partial t}}\right)\psi (x',0,t+T)\right]dx'}
{\displaystyle \tau }
{\displaystyle \psi _{p}(x,T,t)=-{\frac {2T}{\pi V^{2}}}\int {\frac {1}{T^{2}}}{\frac {\partial }{\partial t}}\psi (x',0,t+T)dx'}
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The Dynamic Compressive Response of an Open-Cell Foam Impregnated With a Non-Newtonian Fluid | J. Appl. Mech. | ASME Digital Collection
G. H. McKinley,
Dawson, M. A., McKinley, G. H., and Gibson, L. J. (July 23, 2009). "The Dynamic Compressive Response of an Open-Cell Foam Impregnated With a Non-Newtonian Fluid." ASME. J. Appl. Mech. November 2009; 76(6): 061011. https://doi.org/10.1115/1.3130825
The response of a reticulated, elastomeric foam filled with colloidal silica under dynamic compression is studied. Under compression beyond local strain rates on the order of
1 s−1
, the non-Newtonian, colloidal silica-based fluid undergoes dramatic shear thickening and then proceeds to shear thinning. In this regime, the viscosity of the fluid is large enough that the contribution of the foam and the fluid-structure interaction to the stress response of the fluid-filled foam can be neglected. An analytically tractable lubrication model for the stress-strain response of a non-Newtonian fluid-filled, reticulated, elastomeric foam under dynamic compression between two parallel plates at varying instantaneous strain rates is developed. The resulting lubrication model is applicable when the dimension of the foam in the direction of fluid flow (radial) is much greater than that in the direction of loading (axial). The model is found to describe experimental data well for a range of radius to height ratios
(∼1–4)
and instantaneous strain rates of the foam (
1 s−1
4×102 s−1
). The applicability of this model is discussed and the range of instantaneous strain rates of the foam over which it is valid is presented. Furthermore, the utility of this model is discussed with respect to the design and development of energy absorption and blast wave protection equipment.
colloids, elastomers, flow through porous media, foams, lubrication, non-Newtonian fluids, plates (structures), shear flow, silicon compounds, stress-strain relations, viscosity, foam, lubrication approximation, non-Newtonian, porous media, shear-thickening fluid
Fluids, Non-Newtonian fluids, Shear (Mechanics), Stress, Viscosity, Lubrication, Compression, Fluid dynamics, Plates (structures)
High Deformation Rate Behavior of Polymeric Foams Filled With Concentrated Silica Suspensions
Society of Rheology 77th Annual Meeting
Observations on the Impact Behaviour of Polyurethane Foams; II. The Effect of Fluid Flow
Experimentally Separating Fluid and Matrix Contributions to Polymeric Foam Behavior
Modeling Air Flow in Impacted Polyurethane Foam
A Multi-Field Approach to Modeling the Dynamic Response of Cellular Materials
Brunjail
Liquid-Solid Mass Transfer in Packed Beds of Variously Shaped Particles at Low Reynolds Numbers: Experiments and Model
A New Model for Determining Mean Structure Parameters of Fixed Beds From Pressure Drop Experiments: Application to Packed Beds With Parallelepipedal Particles
Pressure Drop in Non-Newtonian Purely Viscous Fluid Flow Through Porous Media
Discontinuous and Dilatant Viscosity Behavior in Concentrated Suspensions. II Theory and Experimental Tests
Shear-Thickening (‘Dilatancy’) in Suspensions of Nonaggregating Solid Particles Dispersed in Newtonian Liquids
The Rheology of Brownian Suspensions
Kinetic Theory of Jamming in Hardsphere Startup Flows
Structure Diffusion and Rheology of Brownian Suspensions by Stokesian Dynamics Simulation
Shear Thickening and Order-Disorder Effects in Concentrated Colloids at High Shear Rates
Wall Slip Corrections for Couette and Parallel Disk Viscometer
Fatigue Testing of High Performance Flexible Polyurethane Foam
Rheological Behavior and Stability of Concentrated Silica Suspensions
Explanations for the Cause of Shear Thickening in Concentrated Colloidal Suspensions
Rheology of Extremely Shear Thickening Polymer Dispersions
Hadjistamov
Proceedings of the Ninth International Congress on Rheology
Garcfa-Rejon
The Rheology of Charged Stabilized Silica Suspensions
The Rheology and Microstructure of Acicular Precipitated Calcium Carbonate Colloidal Suspensions Through the Shear Thickening Transition
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Section 3.3: Quadric Surfaces
Table 3.3.1 lists the quadric surfaces, surfaces described by equations quadratic in the three variables
x,y,z
, that is, equations of the form
{\mathrm{α}}_{1} {x}^{2}+{\mathrm{α}}_{2} {y}^{2}+{\mathrm{α}}_{3} {z}^{2}+{\mathrm{β}}_{1} x+{\mathrm{β}}_{2} y+{\mathrm{β}}_{3} z+\mathrm{γ}=0
\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}+\frac{{\left(z-{z}_{0}\right)}^{2}}{{c}^{2}}=1
Hyperboloid of 1 sheet
\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}-\frac{{\left(z-{z}_{0}\right)}^{2}}{{c}^{2}}=1
\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}-\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}-\frac{{\left(z-{z}_{0}\right)}^{2}}{{c}^{2}}=1
\frac{{\left(z-{z}_{0}\right)}^{2}}{{c}^{2}}=\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}
\frac{z-{z}_{0}}{c}=\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}
\frac{z-{z}_{0}}{c}=\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{a}^{2}}
\frac{z-{z}_{0}}{c}=\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}-\frac{{\left(y-{y}_{0}\right)}^{2}}{{a}^{2}}
{\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}={r}^{2}
\frac{{\left(x-{x}_{0}\right)}^{2}}{{a}^{2}}+\frac{{\left(y-{y}_{0}\right)}^{2}}{{b}^{2}}=1
y-{y}_{0}=a {\left(x-{x}_{0}\right)}^{2}
Table 3.3.1 Quadric surfaces
Standard form is typically obtained by completing the square in each of the three variables, as applicable. The point
\left({x}_{0},{y}_{0},{z}_{0}\right)
is generally a central point for the quadric surface.
Table 3.3.2 provides additional details for some of the quadric surfaces listed in Table 3.3.1.
The hyperboloid of one sheet is characterized by a single minus sign in front of one of the squared terms.
The orientation of the central axis for this surface is determined by the variable before which there is the minus sign.
The hyperboloid of two sheets is characterized by two minus signs in front of two of the squared terms.
The orientation of the central axis for this surface is determined by the variable before which there is the single plus sign.
The orientation of the central axis for the cone is determined by the variable on the left: this variable can be any one of the three, with the other two on the right having plus signs before them.
As for the cone, the orientation of the central axis of a paraboloid is determined by the variable on the left: this variable can be any one of the three, with the other two on the right having plus signs before them. The difference between the cone and the paraboloid is that the variable isolated on the left is not squared for the paraboloid.
The central axis for a cylinder is determined by the variable that is "missing" from the equation. Consequently, any contour in the plane can be "extruded" along an axis perpendicular to the plane of the curve, thereby forming a cylinder whose cross section has the shape of the planar curve.
Table 3.3.2 Notes pertinent to the quadrics listed in Table 3.3.1
In Examples 3.3.(1-12), put the given equation into standard form for a quadric surface, identify the surface, draw its graph, and discuss the nature of the level curves and plane sections.
4{x}^{2}+4{y}^{2}-{z}^{2}-8 x+8 y+4 z+4=0
2 {x}^{2}-{y}^{2}-4 x-2 y-z+2=0
{z}^{2}-4 {x}^{2}-4{y}^{2}-4 z+8 x-8 y-8=0
3 {x}^{2}- 6 x- y+ 1=0
16{x}^{2}+9{y}^{2}+36{z}^{2}-32 x+18 y-144 z+25=0
{x}^{2}+{y}^{2}-2 x+2 y-z+3=0
9{x}^{2}+4{y}^{2}-18 x+16 y-11=0
{z}^{2}-4 {x}^{2}-2{y}^{2}-4 z+8 x-4 y-6=0
2 {x}^{2}+{y}^{2}-4 x+2 y-z+4=0
4{x}^{2}+4{y}^{2}-{z}^{2}-8 x+8 y+4 z=0
4{x}^{2}+2{y}^{2}-{z}^{2}-8 x+4 y+4 z+2=0
4{x}^{2}+2{y}^{2}-{z}^{2}-8 x+4 y+4 z-2=0
Give an equation for the elliptic cylinder whose central axis is the line through
\left(1,2,3\right)
, parallel to the
y
-axis. The cross sections perpendicular to the central axis should be ellipses with a vertical semi-major axis of length 3 and a horizontal semi-minor axis of length 1. Graph this cylinder for
y∈\left[1,5\right]
z∈\left[0,3\right]
, draw a cylinder whose axis is parallel to the
z
-axis, and whose cross sections are the shape of the curve defined by the equation
{x}^{2}+{y}^{2}+x=\sqrt{{x}^{2}+{y}^{2}}
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Section 50.20 (0FW3): Poincaré duality—The Stacks project
Section 50.20: Poincaré duality (cite)
50.20 Poincaré duality
In this section we prove Poincar'e duality for the de Rham cohomology of a proper smooth scheme over a field. Let us first explain how this works for Hodge cohomology.
Lemma 50.20.1. Let $k$ be a field. Let $X$ be a nonempty smooth proper scheme over $k$ equidimensional of dimension $d$. There exists a $k$-linear map
\[ t : H^ d(X, \Omega ^ d_{X/k}) \longrightarrow k \]
unique up to precomposing by multiplication by a unit of $H^0(X, \mathcal{O}_ X)$ with the following property: for all $p, q$ the pairing
\[ H^ q(X, \Omega ^ p_{X/k}) \times H^{d - q}(X, \Omega ^{d - p}_{X/k}) \longrightarrow k, \quad (\xi , \xi ') \longmapsto t(\xi \cup \xi ') \]
Proof. By Duality for Schemes, Lemma 48.27.1 we have $\omega _ X^\bullet = \Omega ^ d_{X/k}[d]$. Since $\Omega _{X/k}$ is locally free of rank $d$ (Morphisms, Lemma 29.34.12) we have
\[ \Omega ^ d_{X/k} \otimes _{\mathcal{O}_ X} (\Omega ^ p_{X/k})^\vee \cong \Omega ^{d - p}_{X/k} \]
Thus we obtain a $k$-linear map $t : H^ d(X, \Omega ^ d_{X/k}) \to k$ such that the statement is true by Duality for Schemes, Lemma 48.27.4. In particular the pairing $H^0(X, \mathcal{O}_ X) \times H^ d(X, \Omega ^ d_{X/k}) \to k$ is perfect, which implies that any $k$-linear map $t' : H^ d(X, \Omega ^ d_{X/k}) \to k$ is of the form $\xi \mapsto t(g\xi )$ for some $g \in H^0(X, \mathcal{O}_ X)$. Of course, in order for $t'$ to still produce a duality between $H^0(X, \mathcal{O}_ X)$ and $H^ d(X, \Omega ^ d_{X/k})$ we need $g$ to be a unit. Denote $\langle -, - \rangle _{p, q}$ the pairing constructed using $t$ and denote $\langle -, - \rangle '_{p, q}$ the pairing constructed using $t'$. Clearly we have
\[ \langle \xi , \xi ' \rangle '_{p, q} = \langle g\xi , \xi ' \rangle _{p, q} \]
for $\xi \in H^ q(X, \Omega ^ p_{X/k})$ and $\xi ' \in H^{d - q}(X, \Omega ^{d - p}_{X/k})$. Since $g$ is a unit, i.e., invertible, we see that using $t'$ instead of $t$ we still get perfect pairings for all $p, q$. $\square$
Lemma 50.20.2. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$. The map
\[ \text{d} : H^0(X, \mathcal{O}_ X) \to H^0(X, \Omega ^1_{X/k}) \]
Proof. Since $X$ is smooth over $k$ it is geometrically reduced over $k$, see Varieties, Lemma 33.25.4. Hence $H^0(X, \mathcal{O}_ X) = \prod k_ i$ is a finite product of finite separable field extensions $k_ i/k$, see Varieties, Lemma 33.9.3. It follows that $\Omega _{H^0(X, \mathcal{O}_ X)/k} = \prod \Omega _{k_ i/k} = 0$ (see for example Algebra, Lemma 10.158.1). Since the map of the lemma factors as
\[ H^0(X, \mathcal{O}_ X) \to \Omega _{H^0(X, \mathcal{O}_ X)/k} \to H^0(X, \Omega _{X/k}) \]
by functoriality of the de Rham complex (see Section 50.2), we conclude. $\square$
Lemma 50.20.3. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$ equidimensional of dimension $d$. The map
\[ \text{d} : H^ d(X, \Omega ^{d - 1}_{X/k}) \to H^ d(X, \Omega ^ d_{X/k}) \]
Proof. It is tempting to think this follows from a combination of Lemmas 50.20.2 and 50.20.1. However this doesn't work because the maps $\mathcal{O}_ X \to \Omega ^1_{X/k}$ and $\Omega ^{d - 1}_{X/k} \to \Omega ^ d_{X/k}$ are not $\mathcal{O}_ X$-linear and hence we cannot use the functoriality discussed in Duality for Schemes, Remark 48.27.3 to conclude the map in Lemma 50.20.2 is dual to the one in this lemma.
We may replace $X$ by a connected component of $X$. Hence we may assume $X$ is irreducible. By Varieties, Lemmas 33.25.4 and 33.9.3 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a finite separable extension $k'/k$. Since $\Omega _{k'/k} = 0$ (see for example Algebra, Lemma 10.158.1) we see that $\Omega _{X/k} = \Omega _{X/k'}$ (see Morphisms, Lemma 29.32.9). Thus we may replace $k$ by $k'$ and assume that $H^0(X, \mathcal{O}_ X) = k$.
Assume $H^0(X, \mathcal{O}_ X) = k$. We conclude that $\dim H^ d(X, \Omega ^ d_{X/k}) = 1$ by Lemma 50.20.1. Assume first that the characteristic of $k$ is a prime number $p$. Denote $F_{X/k} : X \to X^{(p)}$ the relative Frobenius of $X$ over $k$; please keep in mind the facts proved about this morphism in Lemma 50.19.5. Consider the commutative diagram
\[ \xymatrix{ H^ d(X, \Omega ^{d - 1}_{X/k}) \ar[d] \ar[r] & H^ d(X^{(p)}, F_{X/k, *}\Omega ^{d - 1}_{X/k}) \ar[d] \ar[r]_{\Theta ^{d - 1}} & H^ d(X^{(p)}, \Omega ^{d - 1}_{X^{(p)}/k}) \ar[d] \\ H^ d(X, \Omega ^ d_{X/k}) \ar[r] & H^ d(X^{(p)}, F_{X/k, *}\Omega ^ d_{X/k}) \ar[r]^{\Theta ^ d} & H^ d(X^{(p)}, \Omega ^ d_{X^{(p)}/k}) } \]
The left two horizontal arrows are isomorphisms as $F_{X/k}$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The right square commutes as $\Theta _{X^{(p)}/X}$ is a morphism of complexes and $\Theta ^{d - 1}$ is zero. Thus it suffices to show that $\Theta ^ d$ is nonzero (because the dimension of the source of the map $\Theta ^ d$ is $1$ by the discussion above). However, we know that
\[ \Theta ^ d : F_{X/k, *}\Omega ^ d_{X/k} \to \Omega ^ d_{X^{(p)}/k} \]
is surjective and hence surjective after applying the right exact functor $H^ d(X^{(p)}, -)$ (right exactness by the vanishing of cohomology beyond $d$ as follows from Cohomology, Proposition 20.20.7). Finally, $H^ d(X^{(d)}, \Omega ^ d_{X^{(d)}/k})$ is nonzero for example because it is dual to $H^0(X^{(d)}, \mathcal{O}_{X^{(p)}})$ by Lemma 50.20.1 applied to $X^{(p)}$ over $k$. This finishes the proof in this case.
Finally, assume the characteristic of $k$ is $0$. We can write $k$ as the filtered colimit of its finite type $\mathbf{Z}$-subalgebras $R$. For one of these we can find a cartesian diagram of schemes
\[ \xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & \mathop{\mathrm{Spec}}(R) } \]
such that $Y \to \mathop{\mathrm{Spec}}(R)$ is smooth of relative dimension $d$ and proper. See Limits, Lemmas 32.10.1, 32.8.9, 32.17.3, and 32.13.1. The modules $M^{i, j} = H^ j(Y, \Omega ^ i_{Y/R})$ are finite $R$-modules, see Cohomology of Schemes, Lemma 30.19.2. Thus after replacing $R$ by a localization we may assume all of these modules are finite free. We have $M^{i, j} \otimes _ R k = H^ j(X, \Omega ^ i_{X/k})$ by flat base change (Cohomology of Schemes, Lemma 30.5.2). Thus it suffices to show that $M^{d - 1, d} \to M^{d, d}$ is zero. This is a map of finite free modules over a domain, hence it suffices to find a dense set of primes $\mathfrak p \subset R$ such that after tensoring with $\kappa (\mathfrak p)$ we get zero. Since $R$ is of finite type over $\mathbf{Z}$, we can take the collection of primes $\mathfrak p$ whose residue field has positive characteristic (details omitted). Observe that
\[ M^{d - 1, d} \otimes _ R \kappa (\mathfrak p) = H^ d(Y_{\kappa (\mathfrak p)}, \Omega ^{d - 1}_{Y_{\kappa (\mathfrak p)}/\kappa (\mathfrak p)}) \]
for example by Limits, Lemma 32.18.2. Similarly for $M^{d, d}$. Thus we see that $M^{d - 1, d} \otimes _ R \kappa (\mathfrak p) \to M^{d, d} \otimes _ R \kappa (\mathfrak p)$ is zero by the case of positive characteristic handled above. $\square$
Proposition 50.20.4. Let $k$ be a field. Let $X$ be a nonempty smooth proper scheme over $k$ equidimensional of dimension $d$. There exists a $k$-linear map
\[ t : H^{2d}_{dR}(X/k) \longrightarrow k \]
unique up to precomposing by multiplication by a unit of $H^0(X, \mathcal{O}_ X)$ with the following property: for all $i$ the pairing
\[ H^ i_{dR}(X/k) \times H_{dR}^{2d - i}(X/k) \longrightarrow k, \quad (\xi , \xi ') \longmapsto t(\xi \cup \xi ') \]
Proof. By the Hodge-to-de Rham spectral sequence (Section 50.6), the vanishing of $\Omega ^ i_{X/k}$ for $i > d$, the vanishing in Cohomology, Proposition 20.20.7 and the results of Lemmas 50.20.2 and 50.20.3 we see that $H^0_{dR}(X/k) = H^0(X, \mathcal{O}_ X)$ and $H^ d(X, \Omega ^ d_{X/k}) = H_{dR}^{2d}(X/k)$. More precisesly, these identifications come from the maps of complexes
\[ \Omega ^\bullet _{X/k} \to \mathcal{O}_ X[0] \quad \text{and}\quad \Omega ^ d_{X/k}[-d] \to \Omega ^\bullet _{X/k} \]
Let us choose $t : H_{dR}^{2d}(X/k) \to k$ which via this identification corresponds to a $t$ as in Lemma 50.20.1. Then in any case we see that the pairing displayed in the lemma is perfect for $i = 0$.
Denote $\underline{k}$ the constant sheaf with value $k$ on $X$. Let us abbreviate $\Omega ^\bullet = \Omega ^\bullet _{X/k}$. Consider the map (50.4.0.1) which in our situation reads
\[ \wedge : \text{Tot}(\Omega ^\bullet \otimes _{\underline{k}} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]
For every integer $p = 0, 1, \ldots , d$ this map annihilates the subcomplex $\text{Tot}(\sigma _{> p} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet )$ for degree reasons. Hence we find that the restriction of $\wedge $ to the subcomplex $\text{Tot}(\Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p}\Omega ^\bullet )$ factors through a map of complexes
\[ \gamma _ p : \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]
Using the same procedure as in Section 50.4 we obtain cup products
\[ H^ i(X, \sigma _{\leq p} \Omega ^\bullet ) \times H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \longrightarrow H_{dR}^{2d}(X, \Omega ^\bullet ) \]
We will prove by induction on $p$ that these cup products via $t$ induce perfect pairings between $H^ i(X, \sigma _{\leq p} \Omega ^\bullet )$ and $H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet )$. For $p = d$ this is the assertion of the proposition.
The base case is $p = 0$. In this case we simply obtain the pairing between $H^ i(X, \mathcal{O}_ X)$ and $H^{d - i}(X, \Omega ^ d)$ of Lemma 50.20.1 and the result is true.
Induction step. Say we know the result is true for $p$. Then we consider the distinguished triangle
\[ \Omega ^{p + 1}[-p - 1] \to \sigma _{\leq p + 1}\Omega ^\bullet \to \sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p] \]
and the distinguished triangle
\[ \sigma _{\geq d - p}\Omega ^\bullet \to \sigma _{\geq d - p - 1}\Omega ^\bullet \to \Omega ^{d - p - 1}[-d + p + 1] \to (\sigma _{\geq d - p}\Omega ^\bullet )[1] \]
Observe that both are distinguished triangles in the homotopy category of complexes of sheaves of $\underline{k}$-modules; in particular the maps $\sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p]$ and $\Omega ^{d - p - 1}[-d + p + 1] \to (\sigma _{\geq d - p}\Omega ^\bullet )[1]$ are given by actual maps of complexes, namely using the differential $\Omega ^ p \to \Omega ^{p + 1}$ and the differential $\Omega ^{d - p - 1} \to \Omega ^{d - p}$. Consider the long exact cohomology sequences associated to these distinguished triangles
\[ \xymatrix{ H^{i - 1}(X, \sigma _{\leq p}\Omega ^\bullet ) \ar[d]_ a \\ H^ i(X, \Omega ^{p + 1}[-p - 1]) \ar[d]_ b \\ H^ i(X, \sigma _{\leq p + 1}\Omega ^\bullet ) \ar[d]_ c \\ H^ i(X, \sigma _{\leq p}\Omega ^\bullet ) \ar[d]_ d \\ H^{i + 1}(X, \Omega ^{p + 1}[-p - 1]) } \quad \quad \xymatrix{ H^{2d - i + 1}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \\ H^{2d - i}(X, \Omega ^{d - p - 1}[-d + p + 1]) \ar[u]_{a'} \\ H^{2d - i}(X, \sigma _{\geq d - p - 1}\Omega ^\bullet ) \ar[u]_{b'} \\ H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \ar[u]_{c'} \\ H^{2d - i - 1}(X, \Omega ^{d - p - 1}[-d + p + 1]) \ar[u]_{d'} } \]
By induction and Lemma 50.20.1 we know that the pairings constructed above between the $k$-vectorspaces on the first, second, fourth, and fifth rows are perfect. By the $5$-lemma, in order to show that the pairing between the cohomology groups in the middle row is perfect, it suffices to show that the pairs $(a, a')$, $(b, b')$, $(c, c')$, and $(d, d')$ are compatible with the given pairings (see below).
Let us prove this for the pair $(c, c')$. Here we observe simply that we have a commutative diagram
\[ \xymatrix{ \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \ar[d]_{\gamma _ p} & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[l]_-{\gamma _{p + 1}} } \]
Hence if we have $\alpha \in H^ i(X, \sigma _{\leq p + 1}\Omega ^\bullet )$ and $\beta \in H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet )$ then we get $\gamma _ p(\alpha \cup c'(\beta )) = \gamma _{p + 1}(c(\alpha ) \cup \beta )$ by functoriality of the cup product.
Similarly for the pair $(b, b')$ we use the commutative diagram
\[ \xymatrix{ \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[d]_{\gamma _{p + 1}} & \text{Tot}(\Omega ^{p + 1}[-p - 1] \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \Omega ^{p + 1}[-p - 1] \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p + 1] \ar[l]_-\wedge } \]
and argue in the same manner.
For the pair $(d, d')$ we use the commutative diagram
\[ \xymatrix{ \Omega ^{p + 1}[-p] \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p] \ar[d] & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p]) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p}\Omega ^\bullet ) \ar[l] } \]
and we look at cohomology classes in $H^ i(X, \sigma _{\leq p}\Omega ^\bullet )$ and $H^{2d - i}(X, \Omega ^{d - p - 1}[-d + p])$. Changing $i$ to $i - 1$ we get the result for the pair $(a, a')$ thereby finishing the proof that our pairings are perfect.
We omit the argument showing the uniqueness of $t$ up to precomposing by multiplication by a unit in $H^0(X, \mathcal{O}_ X)$. $\square$
Comment #5056 by 鷺宮しおり on April 26, 2020 at 10:37
A typo: In proposition 0FW7, it should be
t: H_\mathrm{dR}^{2d}(X/k) \to k
t: H_\mathrm{dR}^d(X/k) \to k
Dear Saginomiya, thanks very much and fixed here.
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Vector calculus - Wikiversity
Material from Vectors was moved[1] here.
Used by Physics equations
Here we extend the concept of vector to that of the vector field. A familiar example of a vector field is wind velocity: It has direction and magnitude, which makes it a vector. But it also depends on position (and ultimately on time). Wind velocity is a function of (x,y,z) at any given time, equivalently we can say that wind velocity is a time-dependent field:
{\displaystyle {\vec {V}}_{wind}={\vec {V}}({\vec {r}},t)}
1 Derivative of a vector valued function
3 Gradient of a scalar field
4 Divergence of a vector field
5 Curl of a vector field
6 Laplacian of a scalar or vector field
7 Identities in vector calculus
8 Fundamental theorems of vector calculus
8.1 Gradient theorem
Derivative of a vector valued functionEdit
{\displaystyle \mathbf {a} (x)\,}
be a vector function that can be represented as
{\displaystyle \mathbf {a} (x)=a_{1}(x)\mathbf {e} _{1}+a_{2}(x)\mathbf {e} _{2}+a_{3}(x)\mathbf {e} _{3}\,}
{\displaystyle x\,}
{\displaystyle \mathbf {a} (x)\,}
{\displaystyle x\,}
{\displaystyle {\cfrac {d\mathbf {a} (x)}{dx}}=\lim _{\Delta x\rightarrow 0}{\cfrac {\mathbf {a} (x+\Delta x)-\mathbf {a} (x)}{\Delta x}}={\cfrac {da_{1}(x)}{dx}}\mathbf {e} _{1}+{\cfrac {da_{2}(x)}{dx}}\mathbf {e} _{2}+{\cfrac {da_{3}(x)}{dx}}\mathbf {e} _{3}~.}
Note: In the above equation, the unit vectors
{\displaystyle \mathbf {e} _{i}}
(i=1,2,3) are assumed constant.
{\displaystyle \mathbf {a} (x)\,}
{\displaystyle \mathbf {b} (x)\,}
are two vector functions, then from the chain rule we get
{\displaystyle {\begin{aligned}{\cfrac {d({\mathbf {a} }\cdot {\mathbf {b} })}{x}}&={\mathbf {a} }\cdot {\cfrac {d\mathbf {b} }{dx}}+{\cfrac {d\mathbf {a} }{dx}}\cdot {\mathbf {b} }\\{\cfrac {d({\mathbf {a} }\times {\mathbf {b} })}{dx}}&={\mathbf {a} }\times {\cfrac {d\mathbf {b} }{dx}}+{\cfrac {d\mathbf {a} }{dx}}\times {\mathbf {b} }\\{\cfrac {d[{\mathbf {a} }\cdot {({\mathbf {a} }\times {\mathbf {b} })}]}{dt}}&={\cfrac {d\mathbf {a} }{dt}}\cdot {({\mathbf {b} }\times {\mathbf {c} })}+{\mathbf {a} }\cdot {\left({\cfrac {d\mathbf {b} }{dt}}\times {\mathbf {c} }\right)}+{\mathbf {a} }\cdot {\left({\mathbf {b} }\times {\cfrac {d\mathbf {c} }{dt}}\right)}\end{aligned}}}
Scalar and vector fieldsEdit
{\displaystyle \mathbf {x} \,}
be the position vector of any point in space. Suppose that there is a scalar function (
{\displaystyle g\,}
) that assigns a value to each point in space. Then
{\displaystyle g=g(\mathbf {x} )\,}
represents a scalar field. An example of a scalar field is the temperature. See Figure4(a).
Figure 4: Scalar and vector fields.
If there is a vector function (
{\displaystyle \mathbf {a} \,}
) that assigns a vector to each point in space, then
{\displaystyle \mathbf {a} =\mathbf {a} (\mathbf {x} )\,}
represents a vector field. An example is the displacement field. See Figure 4(b).
Gradient of a scalar fieldEdit
{\displaystyle \varphi (\mathbf {x} )\,}
be a scalar function. Assume that the partial derivatives of the function are continuous in some region of space. If the point
{\displaystyle \mathbf {x} \,}
has coordinates (
{\displaystyle x_{1},x_{2},x_{3}\,}
) with respect to the basis (
{\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\,}
), the gradient of
{\displaystyle \varphi \,}
{\displaystyle {\boldsymbol {\nabla }}{\varphi }={\frac {\partial \varphi }{\partial x_{1}}}~\mathbf {e} _{1}+{\frac {\partial \varphi }{\partial x_{2}}}~\mathbf {e} _{2}+{\frac {\partial \varphi }{\partial x_{3}}}~\mathbf {e} _{3}~.}
In index notation,
{\displaystyle {\boldsymbol {\nabla }}{\varphi }\equiv \varphi _{,i}~\mathbf {e} _{i}~.}
The gradient is obviously a vector and has a direction. We can think of the gradient at a point being the vector perpendicular to the level contour at that point.
It is often useful to think of the symbol
{\displaystyle {\boldsymbol {\nabla }}{}}
as an operator of the form
{\displaystyle {\boldsymbol {\nabla }}{}={\frac {\partial }{\partial x_{1}}}~\mathbf {e} _{1}+{\frac {\partial }{\partial x_{2}}}~\mathbf {e} _{2}+{\frac {\partial }{\partial x_{3}}}~\mathbf {e} _{3}~.}
Divergence of a vector fieldEdit
If we form a scalar product of a vector field
{\displaystyle \mathbf {u} (\mathbf {x} )\,}
{\displaystyle {\boldsymbol {\nabla }}{}}
operator, we get a scalar quantity called the divergence of the vector field. Thus,
{\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {u} ={\frac {\partial u_{1}}{\partial x_{1}}}+{\frac {\partial u_{2}}{\partial x_{2}}}+{\frac {\partial u_{3}}{\partial x_{3}}}~.}
{\displaystyle {\boldsymbol {\nabla }}\mathbf {u} \equiv u_{i,i}~.}
{\displaystyle {\boldsymbol {\nabla }}\mathbf {u} =0}
{\displaystyle \mathbf {u} \,}
is called a divergence-free field.
The physical significance of the divergence of a vector field is the rate at which some density exits a given region of space. In the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or out of the region.
Curl of a vector fieldEdit
The curl of a vector field
{\displaystyle \mathbf {u} (\mathbf {x} )\,}
is a vector whose expression can be obtained with
{\displaystyle {\boldsymbol {\nabla }}\times {\mathbf {u} }={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\{\frac {\partial }{\partial x_{1}}}&{\frac {\partial }{\partial x_{2}}}&{\frac {\partial }{\partial x_{3}}}\\u_{1}&u_{2}&u_{3}\\\end{vmatrix}}}
The physical significance of the curl of a vector field is the amount of rotation or angular momentum of the contents of a region of space.
Laplacian of a scalar or vector fieldEdit
The Laplacian of a scalar field
{\displaystyle \varphi (\mathbf {x} )\,}
is a scalar defined as
{\displaystyle \nabla ^{2}{\varphi }:={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}{\varphi })={\frac {\partial ^{2}\varphi }{\partial x_{1}}}+{\frac {\partial ^{2}\varphi }{\partial x_{2}}}+{\frac {\partial ^{2}\varphi }{\partial x_{3}}}~.}
The Laplacian of a vector field
{\displaystyle \mathbf {u} (\mathbf {x} )\,}
is a vector defined as
{\displaystyle \nabla ^{2}{\mathbf {u} }:=(\nabla ^{2}{u_{1}})\mathbf {e} _{1}+(\nabla ^{2}{u_{2}})\mathbf {e} _{2}+(\nabla ^{2}{u_{3}})\mathbf {e} _{3}~.}
Identities in vector calculusEdit
Some frequently used identities from vector calculus are listed below.
{\displaystyle {\boldsymbol {\nabla }}(\mathbf {a} +\mathbf {b} )={\boldsymbol {\nabla }}\cdot {\mathbf {a} }+{\boldsymbol {\nabla }}\cdot {\mathbf {b} }}
{\displaystyle {\boldsymbol {\nabla }}\times {(\mathbf {a} +\mathbf {b} )}={\boldsymbol {\nabla }}\times {\mathbf {a} }+{\boldsymbol {\nabla }}\times {\mathbf {b} }}
{\displaystyle {\boldsymbol {\nabla }}(\varphi \mathbf {a} )=\cdot {({\boldsymbol {\nabla }}{\varphi })}{\mathbf {a} }+\varphi ({\boldsymbol {\nabla }}\cdot {\mathbf {a} })}
{\displaystyle {\boldsymbol {\nabla }}\times {(\varphi \mathbf {a} )}={({\boldsymbol {\nabla }}{\varphi })}\times {\mathbf {a} }+\varphi ({\boldsymbol {\nabla }}\times {\mathbf {a} })}
{\displaystyle {\boldsymbol {\nabla }}({\mathbf {a} }\times {\mathbf {b} })={\mathbf {b} }\cdot {({\boldsymbol {\nabla }}\times {\mathbf {a} })}-{\mathbf {a} }\cdot {({\boldsymbol {\nabla }}\times {\mathbf {b} })}}
Fundamental theorems of vector calculusEdit
One version of the fundamental theorem of one-dimensional calculus is
{\displaystyle \int _{a}^{b}f\,'(x)dx=f(b)-f(a)}
This is a theorem about a function,
{\displaystyle f(x)}
, its first derivative, and a line segment. Two notations used to denote this line segment are [a,b] and the inequality, a<x<b. In the field of topology,
{\displaystyle \partial }
denotes boundary. If we let the symbol
{\displaystyle {\mathcal {L}}}
denote the infinite number of points in the line segment [a,b], then the symbol
{\displaystyle \partial {\mathcal {L}}}
denotes the two endpoints (at x = a and x = b ) of the line segment
{\displaystyle {\mathcal {L}}}
. These endpoints form the boundary of
{\displaystyle {\mathcal {L}}}
Gradient theoremEdit
The gradient theorem is a direct generalization of the fundamental theorem of calculus:
{\displaystyle \int _{\ell [{\vec {p}}\to {\vec {q}}]\subset \mathbb {R} ^{n}}{\vec {\nabla }}f\cdot d{\vec {\ell }}=f\left({\vec {q}}\right)-f\left({\vec {p}}\right)}
The subscript,
{\displaystyle \ell [{\vec {p}}\to {\vec {q}}]\subset \mathbb {R} ^{n}}
informs this is an integral over the over a one-dimensinal curve (or 'path') line integral
{\displaystyle \ell }
{\displaystyle {\vec {r}}={\vec {p}}}
{\displaystyle {\vec {r}}={\vec {q}}}
. The function,
{\displaystyle f=f({\vec {r}})}
is any scalar field that is differentiable. The expression
{\displaystyle \subset \mathbb {R} ^{n}}
informs us that
{\displaystyle {\vec {r}}}
can be a member of an n-dimensional space. (In other words the theorem is easily generalized to more than three dimensions.) A consequence of this theorem is that
{\displaystyle \int {\vec {\nabla }}f\cdot d{\vec {\ell }}=0}
for any "closed curve" The figure shows the closed curve A, as well as the "open curve", B. Two endpoints form the "boundary" of curve B.
Stokes' theorem states:
{\displaystyle \int _{\Sigma \subset \mathbb {R} ^{3}}{\vec {\nabla }}\times {\vec {F}}\cdot {\vec {dA}}=\oint _{\partial \Sigma }{\vec {F}}\cdot d{\hat {\ell }}}
The integral subscript,
{\displaystyle \Sigma \subset \mathbb {R} ^{3}}
informs us that this theorem is valid only in a three-dimensional vector space. The integral is over a two-dimensional surface,Σ ,with
{\displaystyle {\vec {dA}}={\hat {n}}dA}
{\displaystyle {\hat {n}}}
is normal to the surface. The integral over the surface, Σ, is nonzero only if its boundary, ∂Σ, exists. Surfaces with such boundaries are called open surfaces, and the boundary, ∂Σ, is a curve in 3-space that goes along the "edge" of the surface. This curve is integrated in the direction of positive orientation, meaning that
{\displaystyle {\vec {d\ell }}}
and the surface normal follow
{\displaystyle {\hat {n}}dA}
follow the right-hand rule.
Footnote: According to Wikipedia[2], this form of the theorem was first discovered by Lord Kelvin, who communicated it to George Stokes in a letter dated July 2, 1850. Stokes set the theorem as a question on the 1854 Smith's Prize exam, which led to the result bearing his name.
closed | open: Ω & ∂Ω
The divergence theorem states:
{\displaystyle \int _{\Omega }{\vec {\nabla }}\cdot {\vec {F}}\;dV=\oint _{\partial \Omega \subset \mathbb {R} ^{3}}{\vec {F}}\cdot d{\vec {A}}}
{\displaystyle \Omega \subset \mathbb {R} ^{n}}
, informs us that this theorem is valid in an (arbitrary) n-dimensional vector space. The n-dimensional volume is Ω, and ∂Ω is its boundary. If n =3 dimensions, ∂Ω is a surface. Since this surface encloses a volume, it has no boundary of its own, and is therefore called a closed surface. The figure shows six surfaces. The three on the left have no boundary and are therefore closed; the ones to the right have a boundary (shown in red) and are therefore open. Note that the closed surfaces to the left are themselves boundaries volumes which are defined as what is "inside" the surface.
Footnote: In index notation, the gradient theorem can be written as
{\displaystyle \int _{\Omega }u_{i,i}~dV=\int _{\partial \Omega }n_{i}u_{i}~dA}
↑ https://en.wikiversity.org/w/index.php?title=Vectors&oldid=1194637
↑ https://en.wikipedia.org/w/index.php?title=Stokes%27_theorem&oldid=608564969
Retrieved from "https://en.wikiversity.org/w/index.php?title=Vector_calculus&oldid=2254812"
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Comparative Study Using the 2-Hydrological Models with the Global Weather in a Small Watershed, a Case Study in the Upper Tha Chin River Basin, Thailand
Comparative Study Using the 2-Hydrological Models with the Global Weather in a Small Watershed, a Case Study in the Upper Tha Chin River Basin, Thailand ()
Sombat Chuenchooklin1*, Udomporn Pangnakorn2, Puripus Soonthornnonda3
1Department of Civil Engineering, Water Resources Research Center, Faculty of Engineering, Naresuan University, Phitsanulok, Thailand.
2Department of Agricultural Science, Faculty of Agriculture Natural Resources and Environment, Naresuan University, Phitsanulok, Thailand.
3Department of Civil Engineering, Faculty of Engineering, Naresuan University, Phitsanulok, Thailand.
The hydrological study in the upstream of the Huai Khot Wang Man diversion canal in Huai Khun Kaew watershed of the Upper Tha Chin River Basin in Uthai Thani Province, Thailand was studied. The soil and water assessment tool (SWAT) and the integrated flood analysis systems (IFAS) applied to the analysis of flow at the outlet. The global weather data provided automatically by the models including land use covers and soil types. The climate forecast system reanalysis (CFSR) and the near real-time precipitation (GSMaP_NRT) used in SWAT and IFAS, respectively. The model sensitivity with Nash and Sutcliffe efficiency (NSE), correlation (R2), and root mean square error (RSME) were applied. The monthly calibrated results from SWAT fitted to the observed data in 2007-2010 with 0.77, 0.88, and 9.08 m3/s, and verified in 2011 with 0.25, 0.61, and 14.30 m3/s, respectively. The daily results from IFAS during a flood period in 2011 fitted to the observed data with 0.21, 0.39, and 34.32 m3/s. Both models showed applicable for efficient gate operation of the diversion canal from this watershed to the Nong Mamong District in Chai Nat Province.
Hydrological Model, SWAT, CFSR, IFAS, PUB
Chuenchooklin, S. , Pangnakorn, U. and Soonthornnonda, P. (2019) Comparative Study Using the 2-Hydrological Models with the Global Weather in a Small Watershed, a Case Study in the Upper Tha Chin River Basin, Thailand. World Journal of Engineering and Technology, 7, 21-26. doi: 10.4236/wjet.2019.72B003.
The ungauged small or medium basin is generally difficult in flood estimation as Prediction in Ungauged Basin (PUB) [1]. Accurate estimates of stream runoff and other hydrologic quantities are needed for numerous purposes of water resources planning and management. The way of obtaining such estimates by modeling methods e.g. the rational and index-flood method, which can be found such as the Hydrologic Modeling System (HMS) [2]. The Soil and Water Assessment Tool (SWAT) [3], is a river basin scale model developed to quantify the impact of land management practices on water, sediment and agricultural chemical yields in large, complex watersheds with various soil, land use and management conditions over long periods of time. SWAT main components include weather, surface runoff, return flow, percolation, evapotranspiration, transmission losses, pond, and reservoir storage, crop growth, and irrigation, groundwater flow, reach routing, nutrient and pesticide loading, and water transfer. It has proven to be an effective tool for assessing water resource and non-point source pollution problems for a wide range of scales and environmental conditions across the globe with a wide range of other water use and water quality applications through the standalone or any geographic information system (GIS) including QGIS as QSWAT [4]. Moreover, an open source distributed rainfall-runoff hydrological model as the Integrated Flood Analysis System (IFAS) developed by the Public Works Research Institute of Japan (PWRI) [5] which can be applied to the very large watershed, e.g. the Chao Phraya Basin, the Indus River Basin, the SEA countries [6], Pakistan [7], Taiwan [8], Japan, and others. Since the proper selection of grid sizing in the IFAS should be undertaken otherwise the longer computation time will occur [9].
Poor water management of existing diversion structures without any storage dam as a head-water source in the central plain of Thailand, including the Tha Chin and Sakae Krang River Basins was reported that it is not realized in the estimation of the upstream discharges to those diversion structures as difficulty in gate management with full potential [10] [11]. Thus, this study aimed to analyze the stream flow discharge by using the QSWAT and IFAS to the Huai Khun Kaew Watershed at the upstream of the Huai Khot-Wang Man diversion point in the Nong Mamong District, Chai Nat Province. The study was a part of the overall research on the Development of Supporting Mechanisms for Budget Planning of Water Resources and Agriculture based on the Application of Information Technological Linkages in the Chai Nat Province which was sponsored by the Thailand Research Fund (TRF) in 2016 [12].
2.1. Modeling Setting in the Study Area
The study area, Huai Khun Kaew watershed, with a drainage area of 1066 km2 locates in the Upper Tha Chin River Basin in Uthai Thani and Chai Nat Provinces, Thailand. The Huai Khun Kaew is a major stream and contains with the largest tributary: Huai Khot with the flow direction from east to west. The hydrological observation station, C.51, has been installed at the outlet by the Royal Irrigation Department (RID). The Huai Khot-Wang Man diversion canal was built to convey some part of a flood from existing watershed to the Nong Mamong District in Chai Nat Province.
This study applied the two different hydrological models as for the comparative study in hydrography products throughout the outlet of the study area base on daily and monthly computation. The models were SWAT via QGIS platform as QSWAT [4] and IFAS [5] which applied to the Huai Khun Kaew Watershed in the Upper Tha Chin River Basin as shown in Figure 1.
Firstly, the IFAS model set applied with the 2-layers and 3-tanks with surface water, groundwater and river course tanks [13]. The sub-basins, river networks, and other components were applied by using the square grid-cell size of 300 m based on direct access to the global data sets e.g. digital elevation model (DEM): GTOPO30, land use covers: GLCC, and near real-time satellite mapping of precipitation: GSMap_NRT [14] [15].
Secondly, SWAT’s water balance conceptual model, as well as all variables and parameters [16] [17], was applied in the study area based on the ASTER Global DEM (GDEM) with 30 m resolution [18]. The land use covers, soil data, and slope were generated to the 16-hydrologic response units (HRUs). The weather data were obtained from SWAT editor based on the Climate Forecast System Reanalysis (CFSR) [19].
2.2. Sensitivity of the Models
The Nash-Sutcliffe efficiency (NSE) is a normalized statistic that determines the relative magnitude of the residual variance as noise compared to the measured data variance as information. NSE is computed as shown in “Equation (1)” [20] where Yiobs is the ith observation for the constituent being evaluated, Yisim is the ith simulated value for the constituent being evaluated, Ymean is the mean of observed data for the constituent being evaluated, and n is the total number of observations.
NSE=1-\left[\frac{{\sum }_{i=1}^{n}{\left({Y}_{i}^{obs}-{Y}_{i}^{sim}\right)}^{2}}{{\sum }_{i=1}^{n}{\left({Y}_{i}^{obs}-{Y}^{mean}\right)}^{2}}\right]
The root mean square error (RSME) incorporates the benefits of error between simulated result and observed data is computed as shown in “Equation (2)”.
RSME=\sqrt{\frac{{\sum }_{i=1}^{n}{\left({Y}_{i}^{obs}-{Y}_{i}^{sim}\right)}^{2}}{n}}
The use of different DEM from each source e.g. GTOPO30 with grid sizing of 300 m and GDEM with the grid sizing of 30 m were applied in the IFAS and SWAT, respectively. The streamlines and sub-basins using the watershed delineation module produced by each model were close to existing GIS.
Figure 1. Map of the Huai Khun Kaew watershed and stream layouts the Huai Khot-Wang Man diversion canal, and stream gauging stations in the Upper Tha Chin River Basin, Chai Nat and Uthai Thani Provinces.
Table 1. The model sensitivity from SWAT during the calibration and verification.
Table 2. Model sensitivity verfied of IFAS and SWAT based on daily during a flood period 2011.
The monthly calibration results from SWAT fitted to the observed in 2007-2010 with NSE of 0.77, R2 of 0.88, and RMSE of 9.08 m3/s, respectively as shown in Table 1. It had verified in 2011 with 0.25, 0.61, and 14.30 m3/s, respectively.
Since IFAS was verified during a flood period in 2011 based on daily flow using the range of those parameters which depends on the classification of tank’s characteristics [21] [22] fitted to the observed with NSE of 0.21, R2 of 0.39, RMSE of 34.32 m3/s, respectively. It was compared to SWAT result during the verification in 2011 as shown in Table 2.
The calibrated and validated results from SWAT and IFAS fitted to the observed in 2007-2011. Both models showed applicable for further efficient gate operation of the diversion canal from this watershed to the Nong Mamong District in Chai Nat Province. The applicable of IFAS is better to simulate daily flows to any point in the basin with the short period e.g. during a flood. However, it takes longer computation time if the model applied with time series for more than 1 year. The SWAT model seems to be very applicable and results are realized to the observed data. However, the complication of the calibration parameters of SWAT is more difficult than IFAS with less parameter. The inspection on models sensitivity should be further carried out and compared to ground-based observation data.
[1] Blöschl, G. (2016) Predictions in Ungauged Basins—Where Do We Stand? Proceedings of the International Association of Hydrological Sciences, Vol. 373, 57-60. https://doi.org/10.5194/piahs-373-57-2016
[2] US Army Corps of Engineers (2017) (2016) Hydrological Modeling System HEC-HMS’s User Manual Version 4.2. https://www.hec.usace.army.mil/software/hec-hms/documentation/HEC-HMS_Users_Manual_4.2.pdf
[3] Arnold, J.G. and Fohrer, N. (2005). SWAT2000: Current Capabilities and Research Opportunities in Applied Watershed Modeling. Hydrological Process, 19, 563-572. https://doi.org/10.1002/hyp.5611
[5] Public Works Research Institute (2017) Integrated Flood Analysis Systems: IFAS Version 1.2 User’s Manual. http://www.icharm.pwri.go.jp/research/ifas/.
[6] Shrestha, B.B., Okazumi, T., Miyamoto, M., Nabesaka, S., Tanaka, S. and Sugiura, A. (2014) Fundamental Analysis for Flood Risk Management in the Selected River Basins of Southeast Asia. J. of Disaster Research, 9, 858-869. https://doi.org/10.20965/jdr.2014.p0858
[7] Aziz, A. and Tanaka, S. (2011) Regional Parameterization and Applicability of Integrated Flood Analysis System (IFAS) for Flood Forecasting of Upper-Middle Indus River. Pak. J. Meteorol., 8, 21-38
[8] Kimura, N., Chiang, S., Wei, H.P., Su, Y.F., Chu, J.L., Cheng, C. and Lin, L. (2014) Tsengwen Reservoir Water-shed Hydrological Flood Simulation under Global Climate Change Using the 20 km Mesh Meteorological Research Institute Atmospheric General Circulation Model (MRIAGCM). Terrestrial. Atmospheric and Oceanic Sciences, 25, 449-461. https://doi.org/10.3319/TAO.2014.01.02.01(Hy)
[9] Sugiura, T., Fukami, K. and Inomata, H. (2008) Development of Integrated Flood Analysis System (IFAS) and Its Applications. World Environmental and Water Resources Congress, 1-10. https://doi.org/10.1061/40976(316)279
[10] Hungspreug, S., Khaouppatum, W. and Thanopanuwat, S. (2000) Operational Flood Forecasting for Chao Phraya River Basin. Proceedings of the International Conference on the Chao Phraya Delta: Historical Development, Dynamics and Challenges of Thailand’s Rice Bowl, Bangkok, 12-15 December 2000.
[11] Japan International Cooperation Agency (1999) The Study on Integrated Plan for Flood Mitigation in Chao Phraya River Basin Summary and Main Report. Royal Irrigation Department Kingdom of Thailand.
[12] Thailand Research Fund (2018) Final Report: The Development of Supporting Mechanisms for Budget Planning of Water Resources and Agriculture Based on the Application of Information Technological Linkages (ITLs) in Chai Nat Province. Water Resource Research Center of Naresuan University.
[13] Sayama, T., Ozawa, G., Kawakami, T., Nabesaka, S. and Fukami, K. (2012) Rainfall-Runoff-Inundation Analysis of the 2010 Pakistan Flood in the Kabul River Basin. Hydrological Sci. J., 57, 298-312. https://doi.org/10.1080/02626667.2011.644245
[14] Shrestha, B.B., Okazumi, T., Miyamoto, M., Nabesaka, S., Tanaka, S. and Sugiura, A. (2014) Fundamental Analysis for Flood Risk Management in the Selected River Basins of Southeast Asia. J. of Disaster Research, 9, 858-869. https://doi.org/10.20965/jdr.2014.p0858
[15] Shiraishi, S.Y., Fukami, K. and Inomata, H. (2009) The Proposal of Correction Method Using the Movement of Rainfall Area on Satellite-Based Rainfall Information by Analysis in the Yoshino River Basin. Annual Journal of Hydraulic Engineering JSCE, 53, 385-390.
[16] SWAT’s Official Website. http://swat.tamu.edu/
[17] Dile, Y., Srinivasan, R. and George, C. (2018) QGIS Interface for SWAT (QSWAT Version 1.5). https://swat.tamu.edu/media/115806/qswat-manual_v15.pdf
[18] USGS (2017) ASTER Global Digital Elevation Model (GDEM), Land Cover Products—Global Land Cover Characterization (GLCC). https://earthexplorer.usgs.gov/
[19] Fuka, D.R., Walter, M.T., MacAlister, C., Degaetano, A.T., Steenhuis, T.S. and Easton, Z.M. (2013) Using the Climate Forecast System Reanalysis as Weather Input Data for Watershed Models. Hydrological Processes. https://doi.org/10.1002/hyp.10073
[20] Nash, J.E. and Sutcliffe.J.V. (1970) River Flow Forecasting through Conceptual Models: Part 1. A Discussion of Principles. J. Hydrology, 10, 282-290. https://doi.org/10.1016/0022-1694(70)90255-6
[21] Fukami, K., Sugiura, Y., Magome, J. and Kawakami, T. (2014) Integrated Flood Analysis System: IFAS Ver. 2.0 User’s Manual. Japan PWRI-Technical Note Inc., 238-259.
[22] Sugiura, A., Fujioka, S., Nabesaka, S., Tsuda, M. and Iwami, I. (2014) Development of a Flood Forecasting System on Upper Indus Catchment Using IFAS. Proceedings of the 6th International Conference on Flood Management: ICFM, 1-12.
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Annealing Temperature Calculator | PCR Method
Allawi H. T., SantaLucia Jr J. “Thermodynamics and NMR of internal G.T mismatches in DNA“ Biochemistry, 36(34), 10581–10594 (1997 Aug 26)See 2 more sources
Wallace R. B. et al. “Hybridization of synthetic oligodeoxyribonucleotides to phi chi 174 DNA: the effect of single base pair mismatch“ Nucleic acids research, 6(11), 3543–3557. (1979 Aug 10)Rychlik W., Spencer W. J., Rhoads R. E. “Optimization of the annealing temperature for DNA amplification in vitro“ Nucleic Acids Res 18(21):6409–6412 (1990)
An introduction to the Polymerase Chain Reaction
A brief overview of DNA structure
A really brief explanation of DNA replication
The ingredients of PCR
The steps of a PCR cycle
What is the PCR annealing temperature?
How to use our PCR annealing temperature calculator
The polymerase chain reaction, or PCR, is nothing short of a miracle in the field of biotechnologies: learn how to optimize this beautiful technique with our PCR annealing temperature calculator!
Here you will learn everything you need to know about PCR, the materials needed, and the process itself. We will answer the following questions:
What is the annealing in PCR?
How do I find the annealing temperature for the best results?
How do I use our primer annealing temperature calculator? (don't worry, it's extremely simple!)
The polymerase chain reaction is an unexpected gift from biology. It's a laboratory technique that allows multiplying a small strand of DNA into an enormous amount of identical copies, only by providing the right tools to an enzyme and the DNA's building blocks.
The process uses the natural mechanism of DNA replication (evolved over millions and billions of years) to its advantage: harnessed and carefully controlled, this allows us to target a single specific fragment of genetic code and greatly amplify its signal when analyzed.
To understand how PCR works, we need to discover DNA first!
🧬 DNA — deoxyribonucleic acid — is the manual of life on Earth. It's a simple molecule that contains the instructions to build a perfectly functional organism, from bacteria 🦠 to raccoons 🦝. Its structure is the famous double helix, where two strands of a biopolymer link and twist. The structure is similar to the single helix, one of the other nucleic acids fundamental to life, the RNA (ribonucleic acid).
You can picture a DNA molecule as a twisted ladder. Each step is made of a pair of molecules called nucleotides, each one in turn composed of a nucleoside and a phosphate. The steps are linked through the latter one. There is more: each nucleoside contains a nitrogenous base and a five-carbon sugar. Before the next step, let's check it out again:
DNA = sequence of paired nucleotides;
Nucleotide = nucleoside + phosphate; and
Nucleoside = nitrogenous base + five-carbon sugar.
The pairing in each step happens at the connection between the bases. There are four types of bases, which pair in the following couples:
Adenine, A + T, Thymine; and
Guanine, G + C, Cytosine.
To be used, each strand of DNA must be "read" in the correct direction. Taking a look at the structure of a nucleotide, biologists decided to create a notation to aid in understanding which direction we are looking at every moment.
The enumeration of the five-carbon sugar makes it possible to identify two carbon atoms (one per nucleotide) connecting the phosphate groups in a DNA strand. Take a look at the image below:
This is the chemical diagram for the Deoxyadenosine monophosphate, one of the four building blocks of DNA. We numbered the carbon atoms on the sugar's ring. The nucleotides connect through the phosphate groups, in correspondence of the carbon number 3.
The two carbons are named 3' and 5', and it is then possible to define a direction like
3'\rightarrow5'
, or its opposite,
5'\rightarrow3'
The image shows a couple of connected nucleotides (the lower one is guanine, while the upper one is a general "R" group). The connection through the phosphate occurs between carbons 3' and 5'of the two sugars, hence creating a "direction", marked in red in the diagram.
Now that you know how DNA is structured, let's learn how it replicates.
The cells in your body and every cell on this planet (apart from some exceptions) constantly follow a reproduction cycle (we have a tool to help you with counting them — the cell doubling time calculator). The process involves a copy of the genetic code (the DNA) to create two cells with a complete set of genes, the instructions needed to "build" an individual. The process itself is complicated, to say the least: we will be able to explain it only briefly to introduce the actors of the PCR.
At the beginning of the replication process, the DNA double helix is "flattened" and split into two single-strands (ssDNA). Each strand exposes its bases, allowing a complex group of enzymes to recreate the complementary strands.
🔎 When we talk of actions performed on DNA like "flattening" and "splitting", we're talking of chemical reactions — that's how cells work. In these reactions, the presence of enzymes is fundamental. Knowing their name is not essential here, but remember that almost every process in your body is somehow helped by them, acting as catalyzers!
The first thing that happens once the double helix opens is the priming of the strand: the enzymes have some limitations and can't start "copying" directly on DNA. A short segment of RNA called a primer (complementary to a specific sequence of bases) is attached to the target strand. From there, an enzyme called DNA polymerase starts attaching new nucleotides, pairing them to the strand's ones, and connecting them via the phosphate bridge.
The DNA polymerase works exclusively in the
5'\rightarrow3'
direction, and therefore it can operate directly on one strand of the separated double helix only (it starts the replication from the opening point on both ssDNA). The copy on the "wrong" strand proceeds in small batches, joined afterward.
Many other processes happen; we've described only the fundamentals. The replication goes on, with the DNA's double helix progressively open until it unravels entirely, and we obtain two daughter filaments, both of them carrying half of the parent molecule.
PCR, the polymerase chain reaction, is the most ground-breaking simple technique in biology. It takes a short sample of DNA, and exponentially replicates it millions of times with the help of the polymerase enzyme. The entire process is driven by a thermal cycle, making it extremely simple and convenient to perform.
🔎 Biochemist Kary Mullis devised the PCR technique in 1983, not in a lab or a university, but in his car on the way for a weekend retreat! The eccentric biochemist imagined this replication mechanism and admittedly stopped the car twice to write it down. He later received the Nobel Prize in Chemistry, and on another occasion, met a talking fluorescent raccoon. But that's another story.
The principle of PCR is to reproduce the replication process of DNA in a vial, specifically on the target fragment, and to allow this process to repeat multiple times. From a single-strand, you would get two daughters, from two, four, and so on — the only limit is set by the available quantity of reagents.
This process allows making small fragments, impossible to find in a genetic code by search (it would be even worse than looking for the proverbial needle in the haystack), but easily detectable after their exponential amplification. This technique became the best tool in the shed of every scientist working with scarce DNA: archeologists and paleontologists, forensic chemists, and geneticists. Everyone got their slice of cake!
🔎 PCR itself got its amplification during the coronavirus pandemic. The most accurate test for the virus employs this technique. After you receive your swab, technicians add the specific primer for the viral genetic code to the sample from the back of your nose. If the primer matches the viral gene, the amplification begins, and you get a positive result! 😷
PCR requires a surprisingly small amount of "ingredients" to be carried out successfully. Let's take a look at them.
The target fragment, or DNA template, is the specific portion of genetic code that will undergo amplification.
The DNA primers are short single-strand fragments, usually given in two types that correspond to the target's specific side of each end (the
3'
). Only when the genetic sequence of the target is known they can be synthesized.
The polymerase enzyme, the real star of the process. It allows amplification by attaching new nucleotides to the parent strands. Currently, laboratories use a heat-resistant enzyme called Taq-polymerase.
The nucleotides needed to create the copies. They are in the form of deoxynucleotide triphosphates: the two "excess" phosphate groups provide energy to the enzyme.
The buffer solution, the environment where the reaction takes place, stabilized by divalent cations like
\text{Mg}^{2+}
Now that we know the ingredients, let's go to the recipe — this is almost a biochemist's food blog!
A temperature cycle drives PCR since each of its steps (chemical reactions) happen at very strict and distinct temperature ranges. It is possible to "activate" and "deactivate" a step by heating or cooling the solution where PCR is happening. This mechanism allowed for the easy automation of PCR in thermal cyclers.
Let's look at the steps of the process in detail. First, pour all of your ingredients into the vial. Turn on your thermal cycler, and let the biology happen!
Denaturation. In this step the temperature is raised to 94–98 °C for 20–30 seconds. The target DNA separates in two single-strands as the hydrogen bonds between bases break.
Annealing. The temperature is lowered to 50–65 °C for another 20–40 seconds. The primers anneal (attach to) the target DNA in this phase. The temperature of this step is crucial since only the correct value allows for a perfect match between target and primer - which otherwise may join the wrong sequence.
Elongation phase. During this step, the temperature is raised to 75–80 °C, precisely at the optimum temperature for the polymerase enzyme. The free nucleotides are paired with the target DNA, effectively creating a new copy of the original fragment. The duration depends on the length of the target and the chosen enzyme.
Once we reach the third step, we start again with the first one, thus opening all of the copied target fragments and having four of the ssDNA available.
When the nucleotides are depleted, the process ends by itself. According to the number of "fully" performed cycles
n
(where there still was an excess of bases), it is possible to estimate the number of fragments
N_\text{DNA}
now present in the solution. The formula is so simple it looks wrong:
N_\text{DNA}=2^n
So, if the thermal cycler performs ten cycles, we end up with
2^{10}=1,\!024
fragments. But another ten cycles (thus twenty in total) would bring the number of copies to
2^{20}=1,\!048,\!756
. The increment is purely exponential, and since it all begins from a single fragment, it is a pure chain reaction.
In the next section, we will learn how to find the primer annealing temperature in PCR!
The PCR annealing temperature is the temperature of the annealing step in a PCR thermal cycle. Its value depends on the denaturation temperatures of both the (less stable) primer and the target DNA.
The empirical formula used to determine the optimal annealing temperature
T_{a}^*
\footnotesize T_{a}^* = 0.3\cdot T_{m}^{p} + 0.7\cdot T_{m}^{t} - 14.9
T_{m}^{p}
is the denaturation (melting) temperature of the most unstable primer, and
T_{m}^{t}
the melting temperature for the target.
Their values are determined with other formulas, depending on the composition and length of the filaments.
⚠️ The constant
14.9
works only when you use temperatures measured in Celsius. For Fahrenheit, the constant is
58.82
, and for Kelvin, it's
288.05
The annealing temperature controls how the primer fragments join the target strands. A higher temperature means that the bonds between the primer and the target may not form at all. On the other hand, lower temperatures allow for bonding on the incorrect sequence of bases.
After primer and target bond, the polymerase enzyme joins the party, connecting to the
3'
extremity of the primer, and waits for the temperature to be optimal for elongation.
We will quickly teach you how to use our optimal primer annealing temperature calculator!
Let's go for an example. We chose a strand of DNA from a cat gene from an online repository. The target strand we looked for is
\thicksim 2\ \text{kb}
\text{kb}
means kilobases, or a thousand nucleotides. We are NOT going to write it here!
We set up the right experimental conditions (the concentrations of ions and other reagents), and using the formula by Wallace we obtained a melting temperature
T_m^t=88.6\ \degree\textrm{C}
Now we need to design the primers. It's an art by itself: for this example, we simply copied the last 30 bases on both terminations of the target DNA. Using the formula by Allawi and SantaLucia, we determined the melting temperature for the primers. Here are the two filaments:
GGGGGATCTTTCTCTATAGGAAACAATTAA with
T_m^{p_1}=65.5\ \degree\textrm{C}
CACAAGCACACATGCGCACATTTGCACACA with
T_m^{p_2}=74.6\ \degree\textrm{C}
From here on, you can help us! Take the melting temperature of the target and the melting temperature of the less stable primer, and plug them into the calculator. The result comes from the formula:
\begin{align*}\footnotesize T_a^* & = \footnotesize 0.3\cdot T_m^p +0.7\cdot T_m^t-14.9 \\ & \footnotesize =66.77\ \degree\textrm{C} \end{align*}
This temperature is slightly above the suggested range for annealing (60–65 °C), but we weren't designing a real experiment, it's still good!
You learned everything from DNA structure (but we guess you already knew something!) to how to find the annealing temperature in a PCR cycle. We hope our primer annealing temperature calculator helped you design your experiment or just satisfied your curiosity.
Try other biology 👩⚕️ calculators at the biology page at omnicalculator.com, like the ligation calculator or the trihybrid cross calculator!
The annealing step in PCR is the second stage of a thermal cycle where the target DNA has its terminations paired with the two 5' → 3' primers, from which the enzyme polymerase can start the replication.
What is the primer annealing temperature in PCR?
The annealing step must happen at a specific temperature to ensure a good matching between the two strands. The annealing temperature lies in the range between 55 and 60 degrees Celsius.
How do I find the annealing temperature in PCR?
To calculate the annealing temperature of the primers in a PCR cycle, you need the melting temperature of the less stable primer Tₘᵖ and the melting temperature of the target DNA Tₘᵗ, both in Celsius.
Then, apply the formula:
Tₐ = 0.3 × Tₘᵖ + 0.7 × Tₘᵗ − 14.9
Or visit the annealing temperature calculator at omnicalculator.com!
What are the effects of a wrong PCR annealing temperature?
A temperature lower than the optimum may cause incorrect matches between primer and target. This error happens because of the energetic advantage of the bonding that facilitates the reaction. On the other hand, a high temperature gives the molecules enough energy to remain unbonded. That's why it is essential to know how to find the annealing temperature in PCR.
Target melting temperature
Annealing temperature (Tₐ)
Allele frequencyDihybrid cross Punnet squareDNA concentration… 3 more
The corn yield calculator estimates the amount of corn produced by a field, based on a 1/1000 acre sample.
The lawn fertilizer calculator tells you how much fertilizer you need to achieve a desired rate of N, P or K, and how many bags to purchase.
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Combination with repetition formula
Example of combination with repetition
Other Omni tools similar to the combinations with repetition calculator
In this article, we deal with how to calculate combinations with repetition.
Combinations are present in many everyday life situations, such as when wanting to know the chances to win the main prize in a lottery (check more about it in our lottery calculator). For that reason, we've created this and other similar calculators.
The easiest way to calculate combinations with repetition is using our calculator, but knowing the equation behind the results is helpful for a deeper understanding. For that reason, in the next section, we present the formula for combinations with repetition and an example of calculation.
🙋 This tool doesn't only calculate the number of combinations with repetitions but also serves as a combination generator with repetition, showing you the list of all those possible combinations.
Don't confuse combinations with permutations.
This formula gives the number of combinations with repetition:
\footnotesize C'(n,r) = \frac{(r+n-1)!}{r!(n-1)!},
r
— The sample size;
n
— The total number of objects; and
!
— The factorial of a number.
Suppose you want to know how many combinations of 5 numbers with repetition are possible. Follow these steps to get the answer:
Identify the total number of objects (n) and the sample size (r). There exist ten digits in the numeric system (from 0 to 9), and we want five-numbers combinations. Therefore,
n = 10
r = 5
Input the values in the combination with repetition formula:
C'(n,r) = \frac{(r+n-1)!}{r!(n-1)!}) = \frac{(5+10-1)!}{5!(10-1)!} = \frac{14!}{5!×9!} = \frac{87,178,291,200}{120×362,880} = 2002
That's it. The number of combinations with repetition with five numbers is 2002. You can check these results in our combinations with repetition calculator.
You can generate combinations with repetition and many other measures using these calculators:
Combinations without repetition;
Permutation and combination; and
How many combinations with 5 numbers with repetition are possible?
There are 2,002 possible combinations with repetitions and 100,000 permutations with repetitions of arranging the digits 0-9 to form a five-digit number.
How to calculate combinations with repetition?
To calculate combinations with repetition C'(n,r):
Identify the total number of objects (n)
Establish or identify the sample size (r)
C'(n,r) = (r+n-1)! / (r!(n-1)!).
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On Wednesdays at Tara's Taquería four tacos are the same price as three burritos. Last Wednesday the Lunch Bunch ordered five tacos and six burritos, and their total bill was
\$8.58
(with no tax or drinks included). Nobody in the Lunch Bunch can remember the cost of one of Tara's tacos. Help them figure it out.
Define two variables: Let
t=
the cost of a taco and
b=
the cost of a burrito.
Write two equations to describe the situation:
4t=3b
5t+6b=8.58
Solve the system of equations and answer the question. Check your work!
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Wheel Offset Calculator - Wheel Positioning
What is wheel offset — What does wheel offset mean?
How to calculate wheel offset for wheel change
Example: Using the wheel offset calculator
This wheel offset calculator will help you determine the wheel offset for your vehicle. When you change the wheels of your truck or car to make customizations, they are often not the same width or size as the ones coming from the original manufacturer. Such wheels or tires are aftermarket wheels. Picking a tire in a larger or smaller diameter can cause errors in the speedometer.
Changing the wheel with a different width might provide you with a better grip but consume a bit more fuel. It is also necessary to check for the contacts with the fenders and the brake pads that might need adjustments by offsetting to avoid interferences.
If you are looking to rig your ride with a new set of aftermarket wheels, you must first understand what wheel offset means. You can use this tool for wheel offset comparison, and the article below will help you know how wheel offset works and how to measure it.
Before we go into wheel offset, let's understand the basic terms concerning the setup.
Declared width or rated width — Width of the bead seat where the tire sits on the wheel. The rims have lips or walls on either side. The width is the inner distance between those walls.
Overall wheel width — Complete width of the wheel, including the walls or flanges on either side of the wheel.
Centreline — Line at the center of the wheel. In other words, the line dividing the cross-section of the wheel into two halves.
Schematics of a wheel-suspension assembly.
The wheel has a mounting pad that is aligned and mounted on the hub of the car. The wheel offset is the distance between the mounting pad and the centerline of the rim. You can use the width and offset of two wheels to find the change in clearance and the position. Mathematically the clearance change is:
\footnotesize \text{Clearance} = \frac{W_c - W_n}{2} + (O_c- O_n)
W_c
– Width of current wheel;
W_n
– Width of new wheel;
O_c
– Offset for current wheel; and
O_n
– Offset for new wheel.
Similarly, the change in position is:
\footnotesize \text{Position} = \frac{W_c - W_n}{2} - (O_c - O_n)
A positive offset means the wheel is more outward and away from the suspension, whereas the negative offset means the wheel is closer to the suspension.
Offset (OF) and backspacing (BS) distance in wheels.
To calculate wheel offset for wheel change:
Enter the width of the current wheel.
Fill in the offset for the current wheel.
Insert the width of the new wheel.
Enter the offset for the new wheel.
The wheel offset calculator will return the change in clearance and the position of the wheel.
Calculate the wheel offset for the new arrangement using the following data:
Enter the width of the current wheel as 7 inches or 177.8 mm.
Fill in the offset for the current wheel as 42 mm.
Insert the width of the new wheel as 8 inches or 203.2 mm.
Enter the offset for the new wheel as 32 mm.
The change in clearance is:
\scriptsize \quad \ \ \ \begin{align*} \text{Clearance} &= \frac{177.8 - 203.2}{2} + (42-32)\\\\ &= -2.7\ \text{mm} \end{align*}
The change in position is:
\scriptsize \qquad \ \begin{align*} \text{Position} &= \frac{177.8 - 203.2}{2} - (42-32) \\\\ &= -22.7\ \text{mm} \end{align*}
Since the position and clearance are negative, the new wheel is closer to the suspension by 22.7 mm, i.e., it is has reduced the clearance (< 0) by 2.7 mm. The position change and clearance change tell the user how much the new wheel will affect the existing values, and therefore, they can make the adjustments as such.
What do you mean by wheel offset?
The wheel offset is the distance between the centerline of the wheel and the mounting surface or pad of the wheel. The offset can be positive, negative, or zero based on the position of the mounting pad on either side of the centerline or at the centerline.
How do I calculate clearance change?
To calculate wheel offset:
Subtract the wheel width for the new wheel from the current wheel.
Divide the difference by two to find the difference between the centerlines.
Find the difference between the offset of both wheels.
Add the resultant to the difference between centerlines to obtain the change in clearance.
What do you mean by wheel backspacing?
The wheel backspace refers to the distance between the wheel's mounting pad or mounting surface and the inner lip or flange. It depicts the location of the mounting pad with respect to the wheel width and changes with increase or decrease in the wheel width.
What are the type of offsets?
A wheel offset could be either positive, negative, or zero. A zero offset means the mounting hub and the wheel's centerline are aligned or have zero distance between them. The negative and positive offset means the mounting hub of the wheel is towards or away from the suspension.
What is the clearance for wheels width and offsets having a difference of 2 mm?
For a difference of 2 mm between the wheel widths, and offsets, the clearance change is 3 mm. Mathematically, the given wheel change is given by the equation:
Clearance = 2/2 + 1 = 3 mm.
Clearance & position
Clearance change
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Whitney umbrella - Wikipedia
Section of the surface
In mathematics, the Whitney umbrella (or Whitney's umbrella, named after American mathematician Hassler Whitney, and sometimes called a Cayley umbrella) is a specific self-intersecting surface placed in three dimensions. It is the union of all straight lines that pass through points of a fixed parabola and are perpendicular to a fixed straight line, parallel to the axis of the parabola and lying on its perpendicular bisecting plane.
Whitney's umbrella can be given by the parametric equations in Cartesian coordinates
{\displaystyle \left\{{\begin{aligned}x(u,v)&=uv,\\y(u,v)&=u,\\z(u,v)&=v^{2},\end{aligned}}\right.}
where the parameters u and v range over the real numbers. It is also given by the implicit equation
{\displaystyle x^{2}-y^{2}z=0.}
This formula also includes the negative z axis (which is called the handle of the umbrella).
Whitney umbrella as a ruled surface, generated by a moving straight line
Whitney umbrella made with a single string inside a plastic cube
Whitney's umbrella is a ruled surface and a right conoid. It is important in the field of singularity theory, as a simple local model of a pinch point singularity. The pinch point and the fold singularity are the only stable local singularities of maps from R2 to R3.
It is named after the American mathematician Hassler Whitney.
In string theory, a Whitney brane is a D7-brane wrapping a variety whose singularities are locally modeled by the Whitney umbrella. Whitney branes appear naturally when taking Sen's weak coupling limit of F-theory.
"Whitney's Umbrella". The Topological Zoo. The Geometry Center. Retrieved 2006-03-08. (Images and movies of the Whitney umbrella.)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Whitney_umbrella&oldid=1088737875"
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Effect of Doppler Shifts on Photon and Particle Flux of Beams
Tribco Inc., Cleveland, OH, USA.
Abstract: Reasons are given for a Doppler shift on the number of observed (sensed) photons in a light beam not just a shift in frequency, also a similar (non-relativistic) effect on the number of observed (sensed) particles (with non-zero rest mass) in a particle beam. Optics texts have neither effect.
Keywords: Photon Flux, Beams, Intensity, Irradiance, Particle Flux
1. Observation of Objects on a Conveyor Belt [1]
Every 1/ω seconds, an object placed on a conveyor belt moving with velocity c (a vector) I is energy flux (intensity of the light) seen by an observer, and Io is I, at v = 0, where KA = any Doppler Shift of the cross-sectional area (velocity distortion of area) being observed moving in observer’s (sensor’s) reference frame, if the example in question has an area distortion. A ward-looking detector moving above the belt with a vector velocity v relative to the same stationary observer. The positive direction is a velocity moving from the source to the observer. Here, c is used because the belt speed has the same effect as the group velocity of light in the optical Doppler shift. θ is the angle between the two velocities. The closing velocity (the sum of the components of c and v moving the items toward the detector) is
\omega \left(c-v\mathrm{cos}\left(\theta \right)\right)
As a result, in one second,
\left[\omega \left(c-v\mathrm{cos}\theta \right)/c\right]
items are detected. That is the same equation as the axial Doppler shift of the frequency for waves moving at speed c and an observer with a velocity v at an angle θ to c. By changing c to speed of light one gets Doppler shift on particles of light photons.
Also it should be obvious from the definition of frequency as 1/(time period) that anything that changes observed frequency (like Doppler shifts) changes observed time periods including the period between photons. Since it has not been obvious for 100 years, [2] gives direct proof of that the axial Doppler changes observed time periods, based on the behavior of Fourier series of modulated signals. When the frequencies (of all harmonics of the series) are multiplied by the same factor K as in Doppler shifts (both axial and transverse). The value of any point in time of any of the harmonics occurs at t/K seconds in the shifted harmonic if it occurred at t seconds in the original harmonic term. Therefore, by superposition the value of every point at time t in the original modulated signal is not at time t/K. The observed time period time has been changed from t to t/K. The same would happen for the time period between observed photons.
2. For Beams Not Belts
The point being that the formula is universal and applies to all moving objects not just waves. In the case of particles in an electron beam or the photons in a light beam it applies to their flux (particles/(second × area)). This is besides the Doppler effect of frequency of the photon or energy of each particle. That means for light the Doppler effect on energy flux is:
I/{I}_{o}={\left[1-\left(v/c\right)\right]}^{2}/KA
I is energy flux (intensity of the light) seen by an observer, and Io is I, at v = 0, where KA = any Doppler Shift of the cross-sectional area (velocity distortion of area) being observed moving in observer’s (sensor’s) reference frame, if the example in question has an area distortion.
Also it should be obvious from the definition of frequency as 1/(time period) that anything that changes observed frequency (like Doppler shifts) changes observed time periods including the period between photons. Since it has not been obvious for 100 years, [2] gives direct proof of that the axial Doppler changes observed time periods, based on the behavior of Fourier series of modulated signals when the frequency (of all harmonics of the series) are multiplied by the same factor as in Doppler shifts (both axial and transverse).
For a particle beam’s E:
E/{E}_{o}=\left(m/{m}_{o}\right)\left[1-\left(v/c\right)\mathrm{cos}\theta \right]/KA
Here unlike in light, c = the group velocity of the particles in the beam (which is belt velocity the last section) and (vcosθ) is the closing velocity between the observer (sensor) and the observed beam. Eo and mo is E and m at v = 0. Note there is an observation angle θ, which becomes a dimension of everything that is distorted by velocity.
To simplify the above gave just only the axial shift for K, but transverse shift exists for light. Therefore the resultant K of the axial and transverse shift should be used in place of the axial shift alone when calculating things in the real world. There is no transverse shift for particle beams with a non-zero rest mass. To have a transverse shift the observed time of travel must be same for observation angles (an Einstein postulate). For light beams, people generally just multiply the axial and transverse shifts to get the resultant that is a few percent off, see [1] for a derivation of the resultant and other impacts of the axial Doppler shift.
Section 1 gave the thought experiment of objects on a conveyor belt being sensed by some electronic device such as a photo switch or a magnetic switch which is moving at other velocity and direction. The analysis showed the rate of observation of the objects is where v = the velocity the sensor is moving, c = the velocity the belt is moving, θ = the angel between the belt’s motion and a line of sight of the sensor, Ro = the rate the objects would be sensed if v = 0. That formula is exactly that of the axial Doppler shift. Instead of objects on a conveyor belt, it could be a stream of machinegun bullets or rain drops. [2] gives a mathematical proof that the axial Doppler shift like the transverse does change observed time periods (based on the behavior Fourier series of modulated signals).
The point being that formula is universal and applies to all moving objects, not just waves. In the case of particles in an electron beam or the photons in a light beam it applies to their flux (particles/(second × area)). This is besides the Doppler effect of frequency of the photon or energy of each particle. That means for light, the Doppler effect on energy flux is:
I/{I}_{o}={\left[1-\left(v/c\right)\right]}^{2}/KA
I is energy flux (intensity of the light) seen by an observer, and Io is I, at v = 0.
But for a beam of particles (with a non-zero rest mass), c = group velocity of the particles (not speed of light). It is based on the motion of the sensor (observer) relative to the beam. The sensor in addition sees a smaller relativistic time change which must be added.
E/{E}_{o}=\left(m/{m}_{o}\right)\left[1-\left(v/c\right)\mathrm{cos}\theta \right]/KA
where KA is the velocity distortion of the cross-sectional area of the beam if any, m is mass of a particle and mo is ma at v = 0.
To simplify the above, only gave the axial shift, but transverse shift exists for light. Therefore, [K] and [KA] should be the resultant of the axial and transverse shift not just the axial shift alone for light. Reference [1] gives a slightly better resultant than multiplying the two shifts.
Cite this paper: Lewis Reich, S. (2019) Effect of Doppler Shifts on Photon and Particle Flux of Beams. Journal of High Energy Physics, Gravitation and Cosmology, 5, 995-998. doi: 10.4236/jhepgc.2019.54055.
[2] Reich, S.L. (2019) A Fourier Series Proof that All Optical Doppler Shifts Impacts Rate Information Is Transmitted and Its Probable Implication for Time. Reseachgate.net.
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This problem is a checkpoint for multiplying polynomials. It will be referred to as Checkpoint 5A.
(x+1)(2x^2−3)
(x+1)(x^2−2x+3)
2(x+3)^2
(x+1)(2x−3)^2
Answers and extra practice for the Checkpoint problems are located in the back of your printed textbook or in the Reference Tab of your eBook. If you have an eBook for CCA2, login and then click the following link: Checkpoint 5A: Multiplying Polynomials
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Legendary reward book - The RuneScape Wiki
Legendary reward book
Item JSON: {"edible":"no","disassembly":"no","stackable":"no","stacksinbank":"yes","death":"never","name":"Legendary reward book","bankable":"no","gemw":false,"equipable":"no","members":"yes","id":"35785","release_date":"26 October 2015","release_update_post":"New Auras - Gauntlet of Souls","lendable":"no","destroy":"Your item will be destroyed permanently and it may not be reclaimable.","highalch":false,"weight":0.001,"lowalch":false,"tradeable":"no","examine":"Your reward for multi-skilling.","noteable":"no"}
The legendary reward book is a gift from Xuan once you have completed the Legendary jack of trades aura's regular skilling. Once the aura depletes on its own or the owner removes the aura after he has trained in 25 different skills, they may get the book, but will not be able to receive another book until they use the first one. This book cannot be banked but can be received as a free player after membership expires. In this case the book cannot be used until membership is resumed.
This book works similar to an experience lamp or Penguin Points in the sense that players specify a single skill in which to receive their bonus. The formula for the experience gained is
{\displaystyle y=3\left(x^{2}-2x+100\right)}
where y = Experience earned. It gives 150% more experience than the reward book from the regular jack of trades aura.
param = type|Jack of trade|Legendary|select|Normal,Master,Supreme,Legendary
Retrieved from ‘https://runescape.wiki/w/Legendary_reward_book?oldid=35650605’
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x^2+8x
Look for a greatest common factor.
x^2y^2−81z^2
This is the difference of squares:
(xy)^2 - (9z)^2
If you need more help, look up difference of squares in the glossary or find your class work from Lesson 3.1.2.
2x^2+14x−16
Always start by looking for a greatest common factor.
Can you factor the polynomial further?
3x^2−11x−4
Use a generic rectangle. Remember that the products of the diagonals must be equal.
. Image of 2 by 2 rectangle.
Start by filling in the first and last terms.
Added to interior labeled, top left, 3, x squared, bottom right, negative 4.
What two numbers multiply to
−12
−11
? Added to rectangle, interior labels, top right, 1, x, bottom left, negative 12, x.
Find the greatest common factor for each row and column.Added to rectangle, labels, left edge, top & bottom, each with question mark, bottom edge, left, 3, x, right, + 1.
3x^2-11x-4=(3x+1)(?+?)
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Using the sequences in the previous problem, suppose we define a new sequence,
s(n)
s(n)=q(t(n))
, a compostion of two sequences. Do you think the new sequence will be arithmetic? Geometric? Neither? Explain. Make a table of values. Does the table support your hypothesis, or do you want to change your guess? Explain.
(n)=(50−7n)^2−6(50−7n)+17
Is this arithmetic, geometric, or quadratic?
\left. \begin{array} { | c | c | c | c | c | c | } \hline n & { 1 } & { 2 } & { 3 } & { 4 } & { 5 } \\ \hline s ( n ) & { 1608 } & { 1097 } & { } & { } \\ \hline \end{array} \right.
Is there a constant difference or a constant multiplier?
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Possible Combinations Calculator/Generator
Example: Calculating how many possible combinations of 5 letters exist
Other tools similar to our possible combinations calculator
If you've ever wondered how to calculate combination possibilities, our possible combinations calculator is what you need.
This tool generates the number of all possible combinations, given a total number of objects and sample size. It also generates all the possible combination elements in a list.
As an example, to better understand the concept, we'll solve the common problem of knowing how many combinations of 5 letters are possible.
🙋 This tool focuses on how to generate all possible combinations, not permutations. The following section discusses the difference between these usually confused quantities. Look at our permutation calculator for a tool that allows permutations calculations.
A combination is the number of ways you can choose r elements out of a set containing n distinct objects, no matter the order. If we must consider the order in which we obtain those elements, then we're dealing with permutations.
We know these definitions can sound complicated, and as a picture is worth a thousand words, we created the following graphic, which describes perfectly these concepts and their differences.
In the example of the image above, there are n = 4 balls of different colors, and the objective is to know how many distinct sets we can get by randomly sampling r = 3 elements.
As you can see, the combination process only considers sets with a different combination of elements each, no matter the order. In contrast, permutation additionally considers the various orders of a specific group of elements.
Also, there are two types of combinations and permutations: without repetition vs. with repetition. "With repetition" means we can have elements repeated within a group of r elements, while "without repetition" implies the opposite.
The number of possible combinations without repetition is given by the formula:
C (n,r) = \frac{n!}{r!(n-r)!},
C (n,r)
— The number of possible combinations without repetition;
n
— The total number of objects;
r
— Sample size; and
!
— The factorial of the number to the left.
And if you're interested in how to calculate possible combinations with repetitions, this is the formula:
C' (n,r) = \frac{(r+n-1)!}{r!(n-r)!}
Now that you've learned the concepts let's see how to calculate how many combinations are possible through an example.
If you're interested in how to find all possible combinations, the easiest way is to use our calculator. Let's see an example.
Suppose you want to calculate how many combinations of 5 letters are possible. Follow these steps to discover it:
Identify the total number of objects (n) and the sample size (r).
Assuming we're dealing with the English alphabet (and not the Russian or the Chinese ones 😳), n = 26.
A combination of 5 letters means r = 5.
Go to the possible combinations calculator and input 26 in the "Total number of objects" box and 5 in the "Sample size" box.
That's it. The amount of possible combinations of 5 letters is 65,780 and 142,506 without and with repetition, respectively. Now you know how to calculate combination possibilities.
Now that you know how to generate all possible combinations with our calculator, you can take a look at these other tools:
Combinations with repetition; and
Combinations without repetition.
How many combinations of 4 digits are possible?
These are the possible combinations and permutations of forming a four-digit number from the 0 to 9 digits:
Without repetitions: 210
With repetitions: 715
Without repetitions: 5,040
With repetitions: 10,000
How many combinations with 12 numbers are possible?
The possible combinations and permutations of arranging twelve numbers (i.e., the 1-12 number list) are:
Without repetitions: 1
With repetitions: 1,352,078
Without repetitions: 479,001,600
With repetitions: 8,916,100,448,256
The number of possible combinations without repetitions C(n,r) equals one because the total number of objects n (twelve digits) equals our sample size r (the twelve digits we'll arrange). In other words, if n = r, then C(n,r) = 1.
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Free cash flow per share (FCF) is a measure of a company's financial flexibility that is determined by dividing free cash flow by the total number of shares outstanding. This measure serves as a proxy for measuring changes in earnings per share.
Ideally, a business will generate more cash flow than is required for operational expenses and capital expenditures. When they do, the free cash flow per share metric below will increase, as the numerator grows holding shares outstanding constant. Increasing free cash flow to outstanding shares value is a positive, as a company is regarded as improving prospects and more financial & operational flexibility.
Free cash flow per share is also called: Free cash flow for [to] the firm. In this case, it is notated as FCFF. The selection of a name is often a matter of preference. It is very common to see it describes as FCF in the newspaper and FCFF in an analyst research note, although they're speaking to the same value.
\text{Free Cash Flow per Share}\ =\ \frac{\text{Free Cash Flow}}{\# \text{ Shares Outstanding}}
Free Cash Flow per Share = # Shares OutstandingFree Cash Flow
BREAKING DOWN Free Cash Flow Per Share
This measure signals a company's ability to pay debt, pay dividends, buy back stock and facilitate the growth of the business. Also, the free cash flow per share can be used to give a preliminary prediction concerning future share prices. For example, when a firm's share price is low and free cash flow is on the rise, the odds are good that earnings and share value will soon be on the up because a high cash flow per share value means that earnings per share should potentially be high as well.
Of the popular financial condition ratios, Free Cash Flow per Share is the most comprehensive, as it's the cash flow available to be distributed to both debt and equity shareholders. An alternative but similar ratio is Free Cash Flow to Equity (FCFE). Free cash flow to equity begins with free cash flow to the firm, but strips out interest expenses on debt-related instruments, as they're senior in the capital structure. This leaves the free cash flow available to equity shareholders, who are at the bottom of the capital structure.
Another key element of free cash flow measures is the exclusion of non-cash related items found on income and cash flow statements. Principally, depreciation and amortization. Although depreciation is reported for tax and other purposes, it is a non-cash item. And free cash flow measures are only interested in cash related items.
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Three-Dimensional Heat Transfer of a Confined Circular Impinging Jet With Buoyancy Effects | J. Heat Transfer | ASME Digital Collection
Koichi Ichimiya,
Koichi Ichimiya
Department of Mechanical Systems Engineering, Yamanashi University, Takeda-4, Kofu, Yamanashi 400-8511, Japan
e-mail: ichimiya@ccn.yamanashi.ac.jp
Sanshin Industry Co. Ltd., Shinbashimachi-1400, Hamamatsu, Shizuoka 432-8528, Japan
Contributed by the Heat Transfer Division for publication in the JOURNAL OF HEAT TRANSFER. Manuscript received by the Heat Transfer Division December 10, 2001; revision received April 15, 2002. Associate Editor: C. Amon.
Ichimiya, K., and Yamada, Y. (March 21, 2003). "Three-Dimensional Heat Transfer of a Confined Circular Impinging Jet With Buoyancy Effects ." ASME. J. Heat Transfer. April 2003; 125(2): 250–256. https://doi.org/10.1115/1.1527901
This paper describes the heat transfer and flow characteristics of a single circular laminar impinging jet including buoyancy force in a comparatively narrow space with a confined wall. Temperature distribution and velocity vectors in the space were obtained numerically by solving three-dimensional governing equations for the Reynolds number
Re=umD/ν=400-2000
and the dimensionless space,
H=h/D=0.25-1.0.
After impingement, heat transfer behavior on the impingement surface is divided into a forced convection region, a mixed convection region, and a natural convection region in the radial direction. The local heat flux corresponding to these three regions was visualized using a thermosensitive liquid crystal. Moreover, with the increase in Reynolds number, Re, and dimensionless space, H, the recirculation flow on the impingement surface moves downstream and its volume increases correspondingly. The Nusselt number averaged from r=0 to the minimum point of peripherally averaged Nusselt number, Num, was evaluated as a function of Re and H.
jets, confined flow, heat transfer, temperature distribution, convection, laminar flow, Analytical, Heat Transfer, Impingement, Jets, Mixed Convection, Visualization
Buoyancy, Flow (Dynamics), Heat transfer, Nozzles, Reynolds number, Temperature distribution, Visualization, Mixed convection, Jets, Liquid crystals, Heat flux, Fluids, Forced convection
Impingement Cooling of Electronics
Hamadah, T. T., 1989, “Impingement Cooling of a Simulated Electronics Package with a Square Array of Round Air Jets,” Proceedings of the National Heat Transfer Conference, HTD-Vol. 111, pp. 107–112.
Loo, E., and Mujumdar, A. S., 1984, “A Simulation Model for Combined Impingement and Through Drying Using Superheated Steam as the Drying Medium,” Drying ’84, pp. 264–280.
Mass Transfer Due to a Confined Laminar Impinging Two-dimensional Jet
Popiel, Cz. O., and Boguslawski, L., 1986, “Mass or Heat Transfer in Impinging Single, Round Jets Emitted by a Bell-shaped Nozzle and Sharp-ended Orifice,” Proceedings of 8th International Heat Transfer Conference (San Francisco), 3, pp. 1187–1192.
Heat Transfer Characteristics of Impinging Two-Dimensional Air Jets
The Effect of Thermal Entrainment on Jet Impingement Heat Transfer
Effect of Surface Renewal Due to Large-Scale Eddies on Jet Impingement Heat Transfer
Confined Multiple Impinging Slot Jets without Crossflow Effects
Effect of Flow Pulsations on the Cooling Effectiveness of an Impinging Jet
Local and Instantaneous Heat Transfer Characteristics of Arrays of Pulsating Jets
Arjocu
Influence of Shear Layer Dynamics on Impingement Heat Transfer
Impingement Cooling of a Confined Circular Air Jet
El-Glenk
An Experimental Investigation of the Effect of the Diameter of Impinging Air Jets on the Local and Average Heat Transfer
Numerical Study of Turbulent Heat Transfer in Confined and Unconfined Impinging Jets
Martin, H., 1977, “Heat and Mass Transfer Between Impinging Gas Jets and Solid Surfaces,” Advances in Heat Transfer, Hartnett, J. P., and Irvine, T. F., eds., Academic Press, London, pp. 1–60.
Jamubunathan
Webb, B. W., and Ma, C. F., 1995, “Single-Phase Liquid Jet Impingement Heat Transfer,” Advances in Heat Transfer, Hartnett, J. P., and Irvine, T. F., eds., Academic Press, London, pp. 105–217.
Patankar, S. V., 1980, Numerical Heat Transfer and Fluid Flow, Hemisphere Publishing Corp., London.
Leonard, B. P., 1980, “The QUICK Algorithm,” Computer Methods in Fluids, Pentech Press, Plymouth.
Fletcher, C. A. J., 1988, Computational Techniques for Fluid Dynamics 1, Springer Series in Computational Physics, Springer, Berlin.
Akino, N., and Kubo, S., 1995, “Heat Flux Visualization Using Liquid Crystal Sheet,” Proceedings of 23rd Visualization Symposium of Japan, 15(1), pp. 67–70.
Mixed Convection Flow and Heat Transfer in the Entry Region of a Horizontal Rectangular Duct
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130 Elektra - Simple English Wikipedia, the free encyclopedia
130 Elektra is a very big farther main belt asteroid. It was found by C. H. F. Peters on February 17, 1873 and named after Electra, an avenger in Greek mythology.
Its spectrum is of the G type, so it is probably made up like Ceres. Spectral signatures of organic compounds have been seen on Elektra's surface [1]
Recent optical sightings have found a moon (see below). Using its orbit, Elektra's mass can be found more correctly. The value of 1.3×1019 kg indicates an unusually high density (for asteroids) of 3.8 ± 0.3 g/cm³. Optical sightings have also determined that Elektra's shape is quite non-spherical, as well as giving indications of albedo differences of 5-15% on its surface.[2]
Moon (S/2003 (130) 1)[change | change source]
In 2003, a small moon of Elektra was detected using the Keck II telescope. The diameter of the moon is 4 km and it orbits at a distance of about 1170 km. The moon has been given the provisional designation S/2003 (130) 1. Due to only a few sightings to date, its orbit is still not well known [3]
W. J. Merline, P. M. Tamblyn,
C. Dumas, L. M. Close,
C. R. Chapman, and F. Menard
unknown, likely small
6 ± 2 km [2]
~4×1014 kg [5]
215×155 ± 12 km[8][9][10][11]
1.28±0.10×1019 kg [12][13]
0.07 m/s²[14]
0.230103 d (5.52247 h)[15]
-88°[2]
0.076 ± 0.011 [16]
↑ D.P. Cruikshank and R.H. Brown (1987). "Organic Matter on Asteroid 130 Elektra". Science. 238 (4824): 183–184. Bibcode:1987Sci...238..183C. doi:10.1126/science.238.4824.183. PMID 17800458. S2CID 46168765. {{cite journal}}: CS1 maint: uses authors parameter (link)
↑ 2.0 2.1 2.2 2.3 F. Marchis; et al. (2006). "Shape, size and multiplicity of main-belt asteroids I. Keck Adaptive Optics survey". Icarus. 185 (1): 39–63. Bibcode:2006Icar..185...39M. doi:10.1016/j.icarus.2006.06.001. PMC 2600456. PMID 19081813.
↑ 3.0 3.1 3.2 130 Elektra and S/2003 (130) 1, orbit data website maintained by F. Marchis. Archived 2010-01-17 at WebCite
↑ "IAUC 8183". Archived from the original on 2006-05-01. Retrieved 2007-12-25.
↑ Discovery Circumstances: Numbered Minor Planets, Minor Planet Centre Archived 2007-06-21 at WebCite
↑ ASTORB[permanent dead link] orbital elements database, Lowell Observatory
↑ Based in IRAS mean diameter of 182±12 km, a/b ratio of 1.4 as per the following references
↑ PDS node spin vector database(in particular, the synthetic compiled value of a/b=1.4). Archived 2010-01-17 at WebCite
↑ (130) Elektra and S/2003 (130) 1, at Johnston's archive (maintained by W. R. Johnston). Archived 2011-05-26 at WebCite
↑ 130 Elektra and S/2003 (130) 1, F. Marchis Archived 2010-01-17 at WebCite
{\displaystyle M\propto a^{3}/P^{2}}
and given errors in a and P. See propagation of uncertainty.
↑ 14.0 14.1 On the extremities of the long axis.
↑ F. Marchis; et al. (2006). "Shape, size and multiplicity of main-belt asteroids I. Keck Adaptive Optics survey". Icarus. 185 (1): 39–63. Bibcode:2006Icar..185...39M. doi:10.1016/j.icarus.2006.06.001. PMC 2600456. PMID 19081813.
↑ Supplemental IRAS minor planet survey Archived 2010-01-17 at WebCite
↑ PDS node taxonomy database Archived 2010-01-17 at WebCite
130 Elektra and S/2003 (130) 1, orbit data website maintained by F. Marchis. Includes adaptive optics image of primary, and satellite orbit diagram.
Data on (130) Elektra from Johnston's archive (maintained by W. R. Johnston)
129 Antigone | 130 Elektra | 131 Vala
Retrieved from "https://simple.wikipedia.org/w/index.php?title=130_Elektra&oldid=8051511"
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What is a molar ratio? What is molar ratio formula?
How to calculate molar ratio?
Significance of a molar ratio and its equation
How do you find molar ratio using this online molar ratio calculator?
Our molar ratio calculator will help you determine the molar ratio between the different chemicals reacting and the different chemicals produced during the reaction. It can also help you determine the mass or the number of moles of each chemical required to carry out the reaction completely. Equipped with that knowledge, you can find out the limiting agent in the reaction or which chemical is available in excess.
Are you wondering what is a molar ratio and how to calculate it? Do you what to learn the importance of a molar ratio? Join us in this article as we discuss the definition of a molar ratio, how to determine it using the molar ratio formula, and how to use it to get more information from your balanced chemical equation.
A molar ratio is between the number of moles (or molecules) of reactants consumed and the number of moles (or molecules) of products generated in a chemical reaction. You can also express it as the ratio of the number of moles (or molecules) of one reactant required to completely react with another reactant or one product produced to another product.
\begin{align*} \frac{\text{10 mol N}_2}{\text{30 mol H}_2} = \frac{1}{3}\\\\ \frac{\text{20 mol NH}_3}{\text{30 mol H}_2} = \frac{2}{3}\\\\ \frac{\text{20 mol NH}_3}{\text{10 mol N}_2} = \frac{2}{1}\\ \end{align*}
Obtain the coefficients of the corresponding elements or compounds in the balanced equation.
Calculate the ratio between these coefficients.
In other words, the formula for molar ratio between any two elements or compounds is as simple as:
\text{A} : \text{B} = \frac{\text{Coefficient of A}}{\text{Coefficient of B}}
\text{A}
– Any element or compound reacting or produced in the reaction;
\text{B}
– Any other element or compound reacting or produced in the reaction;
\text{Coefficient of A}
– Coefficient of
\text{A}
in the balanced chemical reaction; and
\text{Coefficient of B}
\text{B}
in the balanced chemical reaction.
The coefficient in a balanced reaction is useful to determine the molar ratio because the coefficient represents the number of molecules of that element (or compound) needed for the reaction to happen without leaving behind any excess reactants.
For example, consider the following balanced chemical equation:
\text{2H}_2 + \text{O}_2 \rightarrow \text{2H}_2\text{O}
Reaction between hydrogen and oxygen to produce water.
From the coefficients in this balanced chemical reaction, we can derive the following information:
For every two molecules of
\text{H}_2
, it takes one molecule of
\text{O}_2
to react completely;
\text{H}_2
reacting, two molecules of
\text{H}_2\text{O}
are formed; and
For every molecule of
\text{O}_2
\text{H}_2\text{O}
are formed.
This result also means that if we had twenty molecules of hydrogen, we would need ten molecules of oxygen to completely react with it to produce twenty molecules of water.
In every balanced reaction, this proportion between the number of molecules stands between different reactants and products. One mole of a substance is simply the aggregation of
6.02214076 \times 10^{23}
number of atoms or molecules. Use our avogadro's number calculator to understand this further.
Now that you understand how to determine a chemical reaction's molar ratio, let's go on to learn its importance.
Once you have the molar ratio, you can figure out how many moles of each item you need for the reaction to take place completely. For example, if the molar ratio between two reactants is 2:3, and you have
12 \text{ mol}
of the first reactant, then the number of moles of the second reactant required is given by:
\scriptsize \begin{align*} \frac{\text{12 mol 1}^{\text{st}} \text{reactant}}{\text{n mol 2}^{\text{nd}} \text{reactant}} &= \frac{2}{3}\\\\ \text{n mol 2}^{\text{nd}} \text{reactant}&= \frac{3}{2}\times \text{12 mol } \\\\ \text{moles of 2}^{\text{nd}} \text{reactant}&= 18 \text{ mol} \end{align*}
You can use the molar ratio and molecular weight of the molecules to determine the mass of the elements (or compounds) required to complete the reaction. Use our grams to moles calculator calculator and mole calculator to understand how molecular weight and molar mass of a compound are related and how to convert the number of moles to the required compound's mass. Hence, once you know how to find the molar ratio and number the moles required, you can "convert" the molar ratio to grams or "mass" ratio:
\scriptsize \begin{align*} \frac{\text{n mol 1}^{\text{st}} \text{reactant}}{\text{m mol 2}^{\text{nd}} \text{reactant}} &= \frac{ \frac{ \text{mass of 1}^{\text{st}} \text{reactant} }{ \text{molar mass of 1}^{\text{st}} \text{reactant} } }{\frac{ \text{mass of 2}^{\text{nd}} \text{reactant} }{ \text{molar mass of 2}^{\text{nd}} \text{reactant} } } \\ \end{align*}
Since the molar ratio helps determine how much of each substance is required to complete the reaction, you can also use it to find out which substance is the limiting reagent and which substance is in excess. For example, we've seen that the molar ratio equation between hydrogen and oxygen in water production is 2:1. If we had 12 mol of hydrogen for 10 mol of oxygen, we immediately know hydrogen is the limiting reagent because the reaction demands that we have 20 mol of hydrogen. We can also phrase that oxygen is in excess by 4 mol since only 6 mol can react with 12 mol of hydrogen.
\scriptsize \begin{align*} \underbrace{\text{12 mol H}_2}_{\text{limiting reagent}} + \text{10 mol O}_2 \rightarrow & \\ \text{12 mol H}_2\text{O} &+ \underbrace{\text{4 mol O}_2}_{\text{excess}} \end{align*}
Oxidation of methane to form carbon dioxide and water.
To calculate the molar ratio from a balanced reaction is the most straightforward method. In this molar ratio calculator, first set Select calculation type to Calculate molar ratio. Then carefully enter the coefficients of each reactant and product in the corresponding Coefficient field. You will get the resultant molar ratio at the very bottom as a table.
Once you have the molar ratio, you can determine the number of moles of each reactant and product required to complete the reaction. Conversely, if you know the number of moles needed, you can compute the molar ratio. In this calculator, first set Select calculation type to Calculate molar ratio and moles and carefully enter the coefficients or the number of moles in the corresponding fields. You will get the resultant molar ratio at the very bottom as a table.
It is possible to calculate the mass of each reactant and product required if you know the molar ratio, number of moles required, and molecular weight of each reactant and product. In this calculator, first set Select calculation type to Calculate molar ratio, moles, and mass. Then enter the coefficients, the number of moles, and the mass in the corresponding fields.
To evaluate the molecular weight, we need to know each element in the compound along with the number of atoms per molecule. For example, for
\text{H}_2\text{O}
2
in the first Atoms per molecule field and select
\text{H}
as its unit, and enter
1
in the second Atoms per molecule field and select
\text{O}
as its unit. This calculator can support up to five elements per reactant or product.
We recommend you utilize the Atoms per molecule field to calculate the molecular weight, but you're free to manually enter any value in the molecular weight field. The calculator may continue to prompt you with a message "Number of atoms per molecule has to be a positive integer." This is because this calculator uses atomic masses as listed in the IUPAC atomic weight data, and the values you're using may be slightly different.
How to calculate molar ratio from molecular weight?
To determine the molar ratio from the molecular weight, you need the mass of each substance.
Convert the molecular weight of the first substance into its molar mass.
Divide the mass of the first substance by its molar mass to obtain the number of moles used (or produced) in the reaction.
Repeat steps one and two to the second substance to obtain the number of moles used (or produced) in the reaction.
Calculate the ratio between the number of moles of the two substances to obtain their molar ratio.
What is the molar ratio between sodium and chlorine in the formation of table salt?
2:1. For every two moles of sodium, one mole of chlorine is required to form two moles of sodium chloride, commonly known as table salt.
How do I convert molar ratio to volume ratio?
The molar ratio and volume ratio are equal if all the gases reacting and produced are at the same temperature and pressure.
Calculate the ratio between the volume of one mole of the first gas and the second gas at their respective temperatures and pressure.
Multiply this ratio with the molar ratio to obtain the volume ratio.
Be proud of yourself for having solved a complex problem.
How to calculate molar ratio from grams?
To calculate the molar ratio from the mass of the reactants (or products), follow these simple steps:
Calculate the molar mass of each substance by evaluating their molecular weight.
Repeat step one for the second substance to obtain the number of moles used (or produced) in the reaction.
First reactant
Coefficient in balanced reaction
Our buffer pH calculator will help you painlessly compute the pH of a buffer based on an acid or a base.
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Reef_knot Knowpia
The reef knot, or square knot, is an ancient and simple binding knot used to secure a rope or line around an object. It is sometimes also referred to as a Hercules knot. The knot is formed by tying a left-handed overhand knot between two ends, instead of around one end, and then a right-handed overhand knot via the same procedure, or vice versa. A common mnemonic for this procedure is "right over left; left over right", which is often appended with the rhyming suffix "... makes a knot both tidy and tight". Two consecutive overhands tied as described above of the same handedness will make a granny knot. The working ends of the reef knot must emerge both at the top or both at the bottom, otherwise a thief knot results.
Reef knot, Square knot, Hercules knot, Double knot, brother hood knot
Thief knot, Granny knot, Grief knot, Surgeon's knot, Shoelace knot
Joining two ends of a single line to bind around an object.
Not secure as a bend unless secured by additional knots(ex: overhand). Spills easily if one of the free ends is pulled outward. Does not hold well if the two lines are not the same thickness.
#74, #75, #460, #1204, #1402, #2096, #2573, #2574, #2577, #2580
The reef knot or square knot consists of two half knots, one left and one right, one being tied on top of the other, and either being tied first...The reef knot is unique in that it may be tied and tightened with both ends. It is universally used for parcels, rolls and bundles. At sea it is always employed in reefing and furling sails and stopping clothes for drying. But under no circumstances should it ever be tied as a bend, for if tied with two ends of unequal size, or if one end is stiffer or smoother than the other, the knot is almost bound to spill. Except for its true purpose of binding it is a knot to be shunned.
The reef knot is not recommended for tying two ropes together, because of the potential instability of the knot when not stabilized; something that has resulted in many deaths (see Misuse as a bend).
The reef knot is at least 4,000 years old. The name "reef knot" dates from at least 1794[2] and originates from its common use to reef sails,[3][4] that is to tie part of the sail down to decrease its effective surface area in strong winds. To release the knot a sailor could collapse it with a pull of one hand; the sail's weight would make the collapsed knot come apart. It is specifically this behavior which makes the knot unsafe for connecting two ropes together.[5]
The name "square knot" is found in Dana's 1841 maritime compendium A Seaman's Friend, which also gives "reef knot" as an alternative name.[6][7]
The name square knot is often used for the unslipped version of reef knot. Reef knot itself then is understood as the single slipped version, while the name shoelace knot is to indicate double slipped version. Sometimes the name bowtie also may be used to indicate a double slipped version, but tying a bowtie is usually performed on flat material, and involves a slip knot of one end holding a bight of the other end i.e. not really a double slipped reef knot. The name "Square knot" is also used for completely different other knots such as the mathematical concept of square knot, or friendship knot; this last one earns the name by being flat and drawing a square on one face (and a cross on the other face).
The reef knot is used to tie the two ends of a single rope together such that they will secure something, for example a bundle of objects, that is unlikely to move much. In addition to being used by sailors for reefing and furling sails, it is also one of the key knots of macrame textiles.[8]
The knot lies flat when made with cloth and has been used for tying bandages for millennia. As a binding knot it was known to the ancient Greeks as the Hercules knot (Herakleotikon hamma) and is still used extensively in medicine.[9] In his Natural History, Pliny relates the belief that wounds heal more quickly when bound with a Hercules knot.[10]
It has also been used since ancient times to tie belts and sashes. A modern use in this manner includes tying the obi (or belt) of a martial arts keikogi.
With both ends tucked (slipped) it becomes a good way to tie shoelaces, whilst the non-slipped version is useful for shoelaces that are excessively short. It is appropriate for tying plastic garbage or trash bags, as the knot forms a handle when tied in two twisted edges of the bag.
The reef knot figures prominently in Scouting worldwide. It is included in the international membership badge[11] and many scouting awards.[12] In the Boy Scouts of America demonstrating the proper tying of the square knot is a requirement for all boys joining the program.[13] In Pioneering (Scouting), it is commonly used as a binding knot to finish off specialized lashing (ropework) and whipping knots.[14] However, it is an insecure knot, unstable when jiggled, and is not suitable for supporting weight.
A surgeon's variation, used where a third hand is unavailable, is made with two or three twists of the ropes on bottom, and sometimes on top, instead of just one.
Detail of Egyptian statue dating from 2350 BC depicting a reef knot securing a belt
Ancient Greek jewelry from Pontika (now in Russia), 300 BC, in the form of a reef knot
Singly slipped reef knot
Diagram of common shoelace bow knot, a doubly slipped reef knot
Weight for weighing gold dust - Knot – MHNT
Misuse as a bendEdit
The reef knot can capsize (spill) when one of the free ends is pulled outward.
The reef knot's familiarity, ease of tying, and visually appealing symmetry conceal its weakness. The International Guild of Knot Tyers warns that this knot should never be used to bend two ropes together.[15] However, modern instruction teaches that it is fine for noncritical applications,[16] especially if stabilized. A proper bend knot, for instance a sheet bend or double fisherman's knot, should be used instead. Knotting authority Clifford Ashley claimed that misused reef knots have caused more deaths and injuries than all other knots combined.[17] Further, it is easily confused with the granny knot, which is a very poor knot.
Physical analysisEdit
An approximate physical analysis[18] predicts that a reef knot will hold if
{\displaystyle 2\mu e^{\mu \pi }\geq 1}
, where μ is the relevant coefficient of friction. This inequality holds if
{\displaystyle \mu \gtrsim 0.24}
. Experiments show that the critical value of μ is actually somewhat lower.[19]
Related knotsEdit
Wikimedia Commons has media related to Square knots.
Square knot (emblem or insignia)
^ Ashley, Clifford W. (1944). The Ashley Book of Knots, p.220. Doubleday. ISBN 0-385-04025-3.
^ David Steel (1794), The Elements and Practice of Rigging and Seamanship, London: David Steel, p. 183
^ Lever, Darcy (1998) [1819], The Young Sea Officer's Sheet Anchor (2nd ed.), Mineola, NY: Dover Publications, p. 83, ISBN 978-0-486-40220-8
^ Cyrus Lawrence Day (1986), The Art of Knotting and Splicing (4th ed.), Annapolis: Naval Institute Press, p. 42
^ Ashley, Clifford W. (1944), The Ashley Book of Knots, New York: Doubleday, p. 258, ISBN 978-0-385-04025-9
^ Richard Henry Dana Jr. (1997) [1879], The Seaman's Friend: A Treatise on Practical Seamanship (14th revised and corrected ed.), Mineola, NY: Dover, p. 49, ISBN 0-486-29918-X
^ Ashley, pp. 399-400.
^ Hage, J. Joris (April 2008), "Heraklas on Knots: Sixteen Surgical Nooses and Knots from the First Century A.D.", World Journal of Surgery, vol. 32, no. 4, pp. 648–655, doi:10.1007/s00268-007-9359-x, PMID 18224483, S2CID 21340612
^ Pliny the Elder, Bostock, John; Riley, H. T. (eds.), The Natural History, p. 28.17, retrieved 2009-08-23
^ See File:World Scout Emblem 1955.svg for an image of the emblem.
^ Square Knots - Meaning and Placement, retrieved 2009-08-17
^ Boy Scouts of America, Boy Scout Joining Requirements, retrieved 2009-08-23
^ "Foolproof Way to ALWAYS Tie a Square Knot Right". www.scoutpioneering.com. 15 June 2013. Retrieved 2013-06-17.
^ International Guild of Knot Tyers, Sea Cadet Knots, retrieved 2016-04-19
^ "How to Tie a Square Knot | Boat Safe | Water Sports, Product Reviews, and Nautical News".
^ Ashley, p. 18.
^ Maddocks, J.H. and Keller, J. B., "Ropes in Equilibrium," SIAM J Appl. Math., 47 (1987), pp. 1185-1200
^ Crowell, "The physics of knots," http://www.lightandmatter.com/article/knots.html
Look up square knot in Wiktionary, the free dictionary.
Ancient symbolism of Hercules knot
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P. Novák; V. Havlíček. “Protein Extraction and Precipitation“ ScienceDirect (April 2016)
What is isoelectric point?
Isoelectric point calculator usage
Isoelectric point in chemistry and biochemistry
Our isoelectric point calculator is here to help you determine the pH value of a molecule, which is free of any net charge.
In chemistry, isoelectric point and "the point of zero charge" are usually used interchangeably. However, there are some exceptions, and we will discuss them along with how to calculate the isoelectric point and what the isoelectric point equation looks like.
The pH value determines whether a substance is acidic or basic by nature. And we know that a pH value lesser than seven indicates acidic and above 7 indicates basic substance.
The isoelectric point is the pH value at which a molecule has zero net electrical charges and is considered neutral.
One important thing to remember is that pH values below zero and above 14 are possible. For instance, concentrated HCl (hydrochloric acid) has a pH of -1, and concentrated NaOH (sodium hydroxide) has 15.
Our isoelectric point calculator is a tool you would love to use on the go. It determines the isoelectric point of molecules based on their pKa and pKb values. What are pKa and pKb values, you might wonder? These are the dissociation constants of acids and bases. The dissociation constant represents the capability of a substance to dissociate into ions in a solution.
Now, coming back to the point at hand, how to use the isoelectric point calculator? Follow the given steps, and that's it:
Input the pKa value of the molecule.
Input the pKb value of the molecule.
The result is that molecule's isoelectric point (pI value).
This value indicates the pH at which your molecule carries no net electric charge.
Let's consider an example. Say you have molecule X, its
pKa = 3.7
pKb = 9.4
. Input these values in the calculator and the result is
6.55
Remember, our tool works in reverse and any other order you want to. As long as you have any two of the three values to input, the calculator will work fine for you.
The isoelectric point formula is fairly simple.
pI = \frac {(pKa + pKb)}{2}
pKa
– Dissociation constant of acid;
pKb
– Dissociation constant of base; and
pI
- Isoelectric point.
Adding the pKa and pKb values and dividing them by two is all you have to do to determine the
pI
value of your molecule. It is just like estimating the mean of any two numbers.
The isoelectric point impacts the solubility of a molecule at a given pH. So, knowing how to calculate the isoelectric point is essential.
You already know how this tool works. Now let's take a look at using the isoelectric point equation:
pI = (pKa + pKb) / 2
According to this equation, you need:
pKa
value of the molecule; and
pKb
value of the molecule.
And the result is the
pI
pKa = 2.2
pKb = 7.5
First, add 2.2 to 7.9, which is 9.7.
Now, divide 9.7 by 2, which results in 4.85.
4.85 is the isoelectric point.
The isoelectric point is an essential topic in chemistry and biochemistry. It tells us when a particular molecule has attained the electrically neutral state, and the pI can affect the solubility of said molecule.
Let's look at the isoelectric points of specific biomolecules.
Isoelectric point of amino acids
Amino acids are zwitterions when they attain their isoelectric point. The acidic amino acids have pI values between a pH of 5.0 to 7.0, while the basic amino acids have a higher pH to attain their isoelectric point.
The knowledge of isoelectric points is significant in protein studies. It helps in methods used for protein separation like:
Ion-exchange chromatography – Chromatographic separation method that separates proteins in a solution/environment based on their net charge; and
Protein electrophoresis – A test that measures the amount of a specific protein, in the blood, based on its net electrical charge.
The algorithms that calculate the isoelectric point of proteins and peptides use the Henderson Hasselbalch equation.
As we mentioned before, the isoelectric point and point of zero charge are interchangeable, but there are situations when they are distinct. They both represent the pH at which there is no electric charge.
The isoelectric point represents only the external surface charges, while the point of zero charge represents the particle's external and internal surface charge in a solution.
How can I calculate the isoelectric point?
The formula to calculate the isoelectric point of a substance is:
pI = (pKa + pKb) / 2
pKa – Dissociation constant of acid;
pKb – Dissociation constant of base; and
pI – Isoelectric point.
So, to compute the isoelectric point:
First, you add the pKa and pKb values.
Then, divide the sum by 2.
The result is the isoelectric point.
What is the isoelectric point of pka = 3.2 and pKb = 8.1?
The isoelectric point is 5.65 for pKa = 3.2 and pKb = 8.1.
This isoelectric point value shows that your acidic molecule is electrically neutral at a pH value of 5.65. Remember, the pH of a molecule is affected by the pH of its surroundings
Can a molecule have more than one pka value?
Yes, a molecule can have more than one pKa value. A molecule with multiple pKa values is an indication of more than one ionizable functional group. It means it has more than one hydrogen atom, which can be transferred or deprotonated in an acid-base reaction.
How does the isoelectric point effect solubility of a molecule?
The isoelectric point is a pH where a molecule is free of net charge and is electrically neutral. If there is no charge on the molecule, it will not interact with its surroundings. In the case of a solute, it will not dissolve in the solvent and instead form precipitates.
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Electrolysis Calculator | First Faraday's Law
How to calculate Faraday's law of electrolysis
How to use our Faraday's law of electrolysis equation calculator
Solve your electrochemical problems with our Faraday's law of electrolysis calculator!
Electrochemistry is the perfect fusion of physics and chemistry: it's not an easy topic, but our calculator will do the job, helping you calculate the electrolysis equation. Here you will learn:
Faraday's law of electrolysis;
How to calculate Faraday's law of electrolysis;
Calculations for the electrolysis of water.
Are you charged up? Let's start!
Electrolysis is a chemical technique that uses electric currents to allow otherwise non-spontaneous chemical reactions.
🔎 An electrical current is defined by the flow of electrons. An electron
\text{e}^-
is a fundamental particle with unitary charge, or charge exactly equal to
1.602176634\cdot 10^{-19}\ \text{C}
\text{C}
stands for coulomb, the unit of the electric charge.
To understand electrolysis, we need to introduce its ingredients:
Two electrodes;
An electric current flowing between them; and
An electrolyte surrounding the electrodes.
The electrolyte is an electric conductor (often a solution with free ions) that carry the charge from one electrode to the other. The excess of electrons at one of the electrodes (and the shortage at the other) drive chemical reactions of either oxidation or reduction.
The chemical reactions cause the transformation of one chemical species into another. Let's analyze a standard copper-zinc cell (a battery).
The electrolyte can be a copper salt (not the normal salt) in an aqueous solution, for example
\text{CuSO}_4
(copper sulphate). The two electrodes are two metal objects of copper and zinc. Without the need to apply an external electric current to the system, a current will flow (of course, only if the two electrodes are connected to complete the circuit). Inside the solution, two reactions happen:
\small \begin{split} \text{Zn}_{\text{(s)}} &\rightarrow \text{Zn}^{2+}_{\text{(aq)}} + 2\text{e}^-\\ \text{Cu}^{2+}_{\text{(aq)}} + 2\text{e}^- &\rightarrow \text{Cu}_{\text{(s)}} \end{split}
(\text{aq})
(\text{s})
tell us the form of the elements on each side of the reaction.
\text{aq}
stands for aqueous, meaning that the element is in solution (for example the copper after the dissolution of the solfate salt). On the other hand,
\text s
means solid: in our electrolytic cell, this describes elements on the electrodes.
Look at the first reaction: the solid zinc from the electrode breaks up in a zinc ion positively charged and two electrons. At the same time, the bottom equation describes the creation of solid copper from the ions dissolved in the electrolyte solution. This process needs two electrons (coming from the zinc electrode: remember that they are connected) and creates free negatively charged
\text{SO}_4^{2-}
ions, which then travel to the zinc electrode, balancing the negative charged that flowed from zinc to copper.
How do you quantify the mass that an electrode gained or lost? Michael Faraday, famous physicist, to the rescue!
Michael Faraday is a well-known name in physics: he was one of the pioneers of electricity. If you want, read more at our Faraday's law calculator. He developed an interest in electrolysis, which led to the two laws of electrolysis bearing his name. We are learning about the first one here!
The first law of electrolysis concerns the accumulation (or removal) of mass on the electrodes. This quantity is directly proportional to the electric current flowing in the electrolyte. Faraday introduced a proportionality constant
Z
(defined as mass per unit of current) called the electrochemical equivalent, which allows us to write Faraday's electrolysis equation:
\small m = Z \cdot Q
m
is the mass; and
Q
is the charge.
🙋 You can also write the charge
Q
as the product of the current
I
t
: the coulomb SI unit is defined as
\text{C}=\text{A} \cdot \text{s}
Faraday's law of electrolysis calculations are straightforward: the only thing you need to know apart from the charge and the mass is the electro-chemical constant. You can easily find tables of its values online, like this one.
Let's use the same example as before: an electrolytic cell with copper and zinc electrodes, but this time in the other direction.
We supply the cell a current of
0.1\ \text{A}
for one minute: what is the mass variation at the electrodes?
First let's check the values of the electro-chemical constant for both copper and zinc. By checking a table we find that:
\small \begin{align*} Z_{\text{Cu}}&=3.295\cdot 10^{-7}\ \text{kg}/\text{C}\\ Z_{\text{Zn}}&=1.58\cdot 10^{-7}\ \text{kg}/\text{C} \end{align*}
Now we have all the elements to calculate
m_{\text{Cu}}
m_{\text{Zn}}
\small \begin{align*} m_{\text{Cu}} &= Z_\text{Cu} \cdot Q \\ & = \left( 3.295\cdot 10^{-7}\ \text{kg}/\text{C} \right) \\ &\quad\times \left(0.1\ \text{A} \cdot 60\ \text{s}\right) \\ & = 1.977\cdot 10^{-6}\ \text{kg} \\ & = 1.977\ \text{mg} \end{align*}
\small \begin{align*} m_{\text{Zn}} & = \left(1.58\cdot 10^{-7}\ \text{kg}/\text{C}\right) \\ &\quad\times \left(0.1\ \text{A}\cdot 60\ \text{s}\right)\\ & =0.948\cdot 10^{-6}\ \text{kg} \\ & = 0.948\ \text{mg} \end{align*}
In the equations above, we leveraged the fact that the charge
Q
has units
\text{A}\cdot\text{s}
and can be calculated by multiplying the current
0.1\ \text{A}
with the time
60\ \text{s}
Now, remember which direction the reactions are going: the copper ions pass from the solution to the electrodes (thus, we add the mass). In contrast, the zinc ions migrate from the electrode to the solution (and we subtract the mass). It is possible to put this idea in mathematical terms by carefully considering the flow of charge in the system and using the appropriate sign.
To use our Faraday's law of electrolysis calculator, simply insert the values you know in the relative fields. Select the element whose
Z
you're using, or insert a customized one by choosing Custom at the bottom of the list.
⚠️ Be careful and always use the appropriate measurement units!
Let's try to calculate the water electrolysis! First, select the value of the electro-chemical constant for hydrogen
\text{H}_2
, which is one of the components of water,
\text{H}_2\text{O}
. Now let's assume that you connected your phone's battery to it. An average battery contains a charge of
4,\!000\ \text{mAh}
(milliampere-hour). Insert this value in the Charge field after selecting the correct unit (
\text{mAh}
). How much hydrogen would we produce?
\small \begin{align*} m_{\text{H}_2} & =1.044\cdot 10^{-7} \cdot 4,\!000\ \text{mAh} \\ & \approx 1.5\ \text{g} \end{align*}
To find the value for the oxygen, change the element in the list of constants.
\small \begin{align*} m_{\text{O}_2} & =8.28\cdot 10^{-8} \cdot 4,\!000\ \text{mAh} \\ & \approx 1.2\ \text{g} \end{align*}
And this was only using the battery of a phone. Imagine how much gas you can produce using a Tesla!
🔎 The electrolysis of water is a technology that, in the future, may help develop new clean engines that use hydrogen as fuel.
What is Faraday's law for electrolysis?
The first Faraday's law for electrolysis is an equation that links the mass of a chemical species added or removed at an electrode and the charge in an electrolytic cell.
The electrolysis equation describes the relation:
m is the mass in kg (kilogram);
Q is the charge in C (coulomb); and
Z is the electrochemical constant of proportionality in kg/C (kilograms per coulomb).
How do I calculate the mass lost at one electrode?
To calculate the mass lost at one electrode, you must know the value of the electrochemical constant Z and the charge — or electric current — flowing towards/from the electrode. Apply the first Faraday's law of electrolysis to calculate the mass, or go to omnicalculator.com to do it with even fewer troubles!
What is the electrolysis constant Z?
Z is the electrochemical constant, a quantity defined for every chemical species used as an electrode. Its dimensions are mass over electric charge, with units usually chosen between grams and milligrams and coulomb.
How do I calculate water electrolysis?
Water electrolysis is a process in which an electric current breaks the water molecule into the gaseous elements hydrogen and oxygen. To calculate the mass of gas produced, you can use Faraday's law of electrolysis, keeping in mind that the electrochemical constant is computed for the diatomic gases H₂ and O₂.
Electrochemical constant (Z)
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What is radioactive decay? Types of radioactive decay
Examples of radioactive decay
Activity in radioactive materials
How to calculate radioactive decay?
What is specific activity? Specific activity formula
How to use our radioactive decay calculator?
Radioactive decay in the real life
Radiation is quite dangerous, and in order to understand it and its risks, scientists introduced many measurements units: our radioactive decay calculator will teach you how the activity of a substance is measured. Here you will learn:
A short history of the discovery of radioactivity;
What the radioactive decay is;
What types of radioactive decay are known;
How to calculate radioactive decay, plus how to measure it; and
What is and how to calculate specific activity.
We will also see a few examples of radioactive decay!
Sometimes, important discoveries happen by chance: this was particularly true at the end of the XIX century when decades of experiments with electricity and photography finally led to the initial understanding of the phenomenon of radioactivity.
While experimenting with cathodic tubes and "cathode rays" at the end of 1895, Wilhelm Röntgen discovered radiation (that is, a mean of transmission of energy either by means of a wave or of a particle) that we now know is made of electrons. He discovered that shielding that radiation was not stopping another kind of ray: X-rays.
The origin of X-rays was not known at the time, and the French physicist Henri Becquerel, was investigating the possibility of using solar-stimulated uranium salts to emit X-rays. He did this by trying to impress a photographic film after the salts were exposure to the sun, and wrapping them in black paper.
🔎 The dangers of radiation were not known at the time, and many experimenters ended up suffering after years of exposition to the ionizing radiations. Maria Skłodowska-Curie's notebooks are still heavily contaminated, and it has to be stored in lead-lined boxes. Those were truly the Age of Exploration in physics.
Here is where two coincidences happened. Speculating on what would have happened if one or both of them didn't happen is pretty much useless: humanity was on the path of discovering the power of radiations anyway.
The first coincidence is an overcast sky. The clouds made it impossible for Becquerel to conduct his experiment, so he put his uranium salt sample in a drawer in proximity of the photographic film.
A few days later, the sun still hiding, Becquerel decided to develop the film anyway: no one knows why. This is our second coincidence. He found out that the film was neatly exposed, even without the presence of the sun: Becquerel quickly understood that the uranium salt themselves emitted rays able to pass through the opaque paper, and exposed the photography.
So, what Becquerel discovered was a spontaneous emission of radiation that the physicist Rutherford correctly identified as a kind of decay coming from the atomic nuclei.
The time has come to discuss the definition of radioactive decay. Massive atomic nuclei or nuclei with a high imbalance of neutron and protons can undergo processes of decay to reach a less energetic state. In these processes, the atoms emit radiation: waves or particles (or both? check our De Broglie wavelength calculator). Many types of radioactive decay are known to scientists:
Alpha decay, where two protons and two neutrons (bound together) are released from the nucleus. It's indistinguishable from a helium nucleus.
Beta decay, where two processes are identified:
Beta minus, where a neutron becomes a proton, with an emission of an electron and an antineutrino; and
Beta plus where a proton turns into a neutron, emitting a positron (the positively charged electron antiparticle) and a neutrino.
Gamma decay, where a high energy photon with the shortest wavelength possible is emitted by the nucleus as a result of one of the previously mentioned decays.
Neutron emission, in which a neutron-rich nucleus emits one or more neutrons to reach a more stable isotope of the same element.
Decays where more particles are released at the same time (cluster decay and nuclear fission).
💡 Other radioactive decay mechanisms are known, but they are of secondary importance: the ones listed here are the most likely to happen — even if we hope you will not stumble upon them!
A few notes are required here. Notice how in the beta processes, the total charge is conserved. This is one of the strongest universal principles, the conservation of charge! And it's worth remembering that two general types of phenomena can be identified: one in which the daughter atom (the one resulting from the decay) is of a different type than the original one (transmutation) and one where the two atoms belong to the same element. The difference lies in whether the process involves a change in the number of protons and neutrons (alpha, beta: transmutation) or not (gamma, neutron emission: no transmutation).
As you're now familiar with the definition of radioactive decay, here we are going to lay down some examples of decay, in the notation used by nuclear physicists!
₈₄²¹⁰Po → ₈₂²⁰⁶Pb + ₂⁴He
An unstable polonium atom decays in a stable lead isotope emitting an alpha particle (a helium nucleus).
₆¹⁴C → ₇¹⁴N + e⁻ + ν*
A radioactive isotope of carbon transmutes in a nitrogen atom. e⁻ is an electron and ν* is an antineutrino.
₆¹¹C → ₅¹¹B + e⁺ + ν
A different isotope of carbon loses a proton, transmuting in an atom of boron, with the emission of a positron (e⁺) and a neutrino.
₂₈⁶⁰Ni → ₂₈⁶⁰Ni + γ
The unstable isotope on the left decays by emitting a high energy photon to the stable state on the right.
₄¹³Be → ₄¹²Be + ₀¹n
Beryllium isotope with a high number of neutrons can emit one of them. The number of protons doesn't change: there is no transmutation.
The decay processes are usually linked in longer chains that start from a radioactive nucleus (radionuclides) and, through various steps, reach a final stable state.
🔎 Radiations are a natural occurrence, and they come from every direction! The granite in the New York Grand Central Station emits enough radiation to exceed the safety regulations for nuclear power plants (but it's safe to pass through it, don't worry). And the sky around us is filled with gamma rays, coming from space, from the ground and... from lightning! The amount of radiation increases with altitude: flying a lot can be dangerous! Check out our flight radiation calculator if you want to know more!
Physicists define the activity of a sample of radioactive material as the number of disintegrations per unit of time: it is a measure of the decay of a certain radionuclide.
🔎 The activity of a sample depends on its size. The bigger the sample, the higher the activity, because more nuclei will decay in the same time period.
Activity is inversely proportional to the quantity called half-life, t½. The half-life is the time required to halve the quantity of a certain radioactive species in a sample. Notice that it doesn't depend on the size of a sample! If you are interested in this topic, check out our half-life calculator!
One can define activity utilizing the following equation:
A = \lambda N
A
is the activity;
N
is the number of radionuclides in a given sample; and
\lambda
(the decay constant) is the probability of one of them decaying.
\lambda
is associated with the half-life through the following formula:
λ = \ln 2 \ /\ t_½
The SI measurement unit of activity is the Becquerel, symbol
\text{Bq}
. A Becquerel equals a decay per second, regardless of the process involved: notice how it corresponds to a unit of frequency.
Another unit widely used for the activity was the Curie,
\text{Ci}
, defined as the activity of a gram of radium. This quantity is extremely large, and the conversion between the two units reads:
1\ \textrm{Ci} = 3.7 \times 10^{10}\ \textrm{Bq} = 37\ \textrm{GBq}
🔎 The Curie is named in honor of Maria Skłodowska-Curie, the famous Polish physicist. When the unit was proposed, someone asked to reduce the mass of radium used to 1 nanogram: that would have reduced its value by a factor
10^{-9}
. Maria Skłodowska opposed the decision: for her, that would diminish the efforts put into the research. We agree with her!
🔎 Did you know that Maria Skłodowska is also known for other scientific discoveries? Check out our Curie constant calculator and Curie's Law calculator. She is one of the few people to win two Nobel Prizes and the only one to win them in different fields — chemistry and physics!
Here, we introduce you to the radioactive decay formula. Namely, the radioactivity of a sample with mass
m
, composed by a chemical species with molar mass
m_a
and half-life
t_½
A = N_A \cdot \frac{m}{m_a} \cdot \frac{\ln2}{t_{½}}
N_A
is Avogadro's number.
Let's take a closer look at the above formula for radioactive decay. The first part of the equation,
N_A \cdot \frac{m}{m_a}
is equal to the number of atoms in a sample. The mass over the molar mass equals the number of moles, which when multiplied by the Avogadro number
N_A
gives the number of atoms of the given species. The second part is the decay rate
\lambda
. In fact, this equation is the same as the more abstract
A = \lambda N
💡 The Avogadro number is defined as the number of atoms in a mole of a certain element. It equals
6.022 \times 10^{23}
, and it's a huge number!
That's it! Knowing how to calculate radioactive decay requires you to know only the weight of the specimen you are studying. All the other quantities are written down in tables.
Specific activity is a useful quantity defined as the activity per quantity of radionuclide, meaning that the units of specific activity are
\text{Bq}/\text{g}
The specific activity formula involves molar mass
m_a
, half-life
t_{½}
, and Avogadro's number
N_A
. It reads:
a = \frac{N_A}{m_a} \cdot \frac{\ln2}{t_{½}}
Note that the above formula for specific activity equals the activity expression divided by the mass of the sample, giving a more specific indication of the radioactivity of a chemical species.
Now that you know what the specific activity is and how to calculate it, it's time to learn when to use it! Specific activity is a fixed quantity for every different radioactive species. You can find tables containing values for many elements and their isotopes.
Using our radioactive decay calculator is extremely easy! We ask you about the weight of your sample, and the molar mass of its component (here is a handy table where you can find these values), and then its half-life (you can find its values for many isotopes on this page).
You can use our radioactive decay calculator in reverse too! Input the data you know, and find out the half-life of an element.
Let's see a couple of examples.
The core of the nuclear bomb Fat Man, the second and last atomic weapon used in any conflict, was composed of two hollow hemispheres of plutonium (²³⁹Pu), weighting a total of 6.19 kg. A shockwave created by high potential explosive was used to compress the core onto an initiator, increasing twofold the density of the metal itself. The reaction was then free to start.
Try to insert the values of the mass of the core, the molecular mass of the plutonium (239.05 g/mol) and its half-life (24,100 years) in the corresponding fields. The result is quite a high number: more than 14 TBq (that's Terabecquerel, a thousand billion Becquerels).
To give a scale of the immensity of that number, let's consider the classic example used in radiation courses, the banana.
A banana contains around 0.5 gram of potassium. The isotope ⁴⁰K is a radioactive element that composes 0.012% of the total amount of potassium in a sample (this quantity is called abundance). Its half-life is 1.248 × 10⁹ years and its molar mass is 39.96 g/mol.
Inserting the quantities in the calculator, we obtain the result 15.91 Bq. This is about 12 times smaller than the amount of radioactivity emitted by the core of a nuclear weapon.
As we saw, a large banana has an activity of about 15 Bq. It means that 15 atoms (usually of an unstable potassium isotope) decay every second.
Only a radioactive gaseous atomic species is known to scientists: radon. It can reach the surface of the Earth and leak in basements, where it can be a major hazard. The highest value ever measured was a staggering 100,000 Bq/m³, measured in the basement of a nuclear power plant worker in the US. It was noticed because he was triggering every radiation alarm in the power plant, but the power plant was still being built. Imagine the confusion!
Half-life and activity play a fundamental role in historical dating: the amount of carbon-14 remaining in a sample allows us to determine its age, since its specific activity and half-life are stable quantities. It works by measuring the amount of carbon-14 in a sample, and then (knowing that going backward in time its value doubles roughly every 5,730 years) establish when the ratio between carbon-14 and carbon-12 (the stable isotope of carbon) was as expected. We have a calculator that will help you with radiocarbon dating. Try it out!
🔎 Nuclear testing after the 1950s affected the ratio of carbon-14 and carbon-12 by changing their relative abundances of the two isotopes. Fortunately, carbon-14 is not used for recent dating, and its spike in the '60s gave a tool for scientists to follow the growth of cells, trees, and other "real-time" phenomena!
Radioactive decay is a process in which unstable nuclei reach more stable states by emitting particles or electromagnetic radiation.
What is the activity of a radioactive substance?
The activity of a radioactive substance is the number of disintegrations per unit of time. This quantity tells you how dangerous a radioactive specimen is: the higher the activity, the more radiation the object is emitting.
Which are the measurement units of radioactivity?
Radioactivity is measured in Becquerels. A Becquerel corresponds to the emission of any kind of radiation in a second. A million Becquerels (megabecquerel) is sometimes called a Rutherford, in honor of the physicist that first gave names to the types of radiations.
In the past, the Curie was used instead, but today it's not accepted in the International System.
How do I calculate specific activity?
Specific activity refers to the activity of a given quantity of a radioactive material. The unit of specific activity is Becquerel over grams.
The formula for specific activity is a = Nₐ/mₐ × ln(2)/tₕ, where Nₐ is Avogadro's number, mₐ the molar mass of the substance, and tₕ the half life.
How do I calculate activity?
In order to calculate activity, use the specific activity of a substance, which is the amount of radiation of a radioactive sample. You can then multiply this quantity by the mass of the sample to get its activity.
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KnownColor - Maple Help
Home : Support : Online Help : Graphics : Packages : ColorTools : Palette Object Commands : KnownColor
check if a string is a valid color name in a palette
KnownColor(P,color)
The KnownColor command returns true if color is the name of a color in the palette P. It is useful for avoiding the errors thrown by the Lookup command for colors not in a palette.
P≔\mathrm{ColorTools}:-\mathrm{GetPalette}\left("CSS"\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}〈\textcolor[rgb]{0,0,1}{Palette CSS:}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,0,0}{$\textcolor[rgb]{1,1,1}{Black}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{1,0,0}{$\textcolor[rgb]{1,1,1}{Red}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0.501960784313725,0,0}{$\textcolor[rgb]{1,1,1}{Maroon}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{1,0.650980392156863,0}{$Orange$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{1,1,0}{$Yellow$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0.501960784313725,0.501960784313725,0}{$Olive$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,1,0}{$Lime$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,0.501960784313725,0}{$\textcolor[rgb]{1,1,1}{Green}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,1,1}{$Aqua$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,0.501960784313725,0.501960784313725}{$\textcolor[rgb]{1,1,1}{Teal}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,0,1}{$\textcolor[rgb]{1,1,1}{Blue}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0,0,0.501960784313725}{$\textcolor[rgb]{1,1,1}{Navy}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{1,0,1}{$Fuchsia$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0.501960784313725,0,0.501960784313725}{$\textcolor[rgb]{1,1,1}{Purple}$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0.501960784313725,0.501960784313725,0.501960784313725}{$Gray$}\textcolor[rgb]{0,0,1}{}\colorbox[rgb]{0.749019607843137,0.749019607843137,0.749019607843137}{$Silver$}\textcolor[rgb]{0,0,1}{}White〉
\mathrm{KnownColor}\left(P,"Grey"\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{KnownColor}\left(P,"Gray"\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{KnownColor}\left(P,"Griy"\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
An error will be thrown if a color is not in a palette
\mathrm{Lookup}\left(P,"Griy"\right)
Error, (in ColorTools:-Palette:-Lookup) invalid color Griy is not defined in this palette
The ColorTools[Palette][KnownColor] command was introduced in Maple 16.
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Ounces to Milliliters Converter
How do you convert ounces to milliliters?
How to use our ounces to milliliters converter
Volume conversion beyond ounces
Ounces are standard measurement units for liquids; however, it may be helpful to know that quantity in milliliters: find it quickly with our ounces to milliliters converter.
The different types of fluid ounces;
How to convert from fluid ounce to milliliters; and
How to use our fluid ounces to milliliters converter.
Keep reading! 🥛
A fluid ounce, despite the name, is not a liquid but a measurement unit. Commonly used in the imperial system, it is particularly useful to indicate everyday quantities of liquids—being good where milliliters and liters are respectively too small or too big.
Ounces are abbreviated in fl oz, or fl. oz.. There are two widespread units with the same name, but they differ in their value:
The British Imperial fluid ounce; and
The United States Customary fluid ounce.
Be careful, and remember to be sure which one you are using!
⚠️ Remember not to mistake the fluid ounce with the ounce, the measurement unit of mass!
An imperial fluid ounce is defined in terms of the imperial gallon:
1\ \text{fl oz} = \frac{1}{160}\ \text{gal}
While the US fluid ounce is defined using the US liquid gallon:
1\ \text{fl oz} = \frac{1}{128}\ \text{gal}
Let's check how to convert from ounces to milliliters: jump with us to the metric system!
We need two conversions for the two different types of fluid ounces.
The conversion for the imperial fluid ounces is:
1\ \text{fl oz} = 28.4\ \text{ml}
While the **US fluid ounce$$ follows a slightly different conversion:
1\ \text{fl oz} = 29.6\ \text{ml}
The difference between the two is slightly more than
4%
: in everyday life, it's not relevant—your cake will be delicious anyway. However, it is fundamental to be sure of the units you are working with when it comes to measurements that require better accuracy.
Misunderstanding on the topic caused, over the years, millions of dollars in damages!
Our fluid ounces to milliliters converter is straightforward to use! Choose which one of the two units you desire to convert from and input the value. In the blink of an eye, we will give you the answer: as easy as drinking a glass of water. Or should we say, "As easy as drinking eight fluid ounces of water"? 😉
We have many other conversion tools here at Omni. Why don't you try the:
How to convert from fluid ounces to milliliters?
Depending on which units you are using, there are two conversions:
For the US fluid ounce, 1 fl oz = 29.6 ml; and
For the British fluid ounce, 1 fl oz = 28.4 ml.
There are 29.6 milliliters in a US fluid ounce, while 28.4 milliliters in a British fluid ounce. The difference dates back to 1824!
How many milliliters are 8 US fluid ounces?
The volume of a glass of water, which is 8 US fluid ounces, correspond to 237 ml. Some says you need 8 glasses of water a day: that's 64 ounces! Even if there is little scientific backing to that claim, remember to keep hydrated!
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Radon Transform - MATLAB & Simulink - MathWorks í•œêµ
Projections can be computed along any angle theta (θ). In general, the Radon transform of f(x,y) is the line integral of f parallel to the y´-axis
{R}_{\mathrm{θ}}\left(xâ²\right)={â«}_{â\infty }^{\infty }f\left(xâ²\mathrm{cos}\mathrm{θ}âyâ²\mathrm{sin}\mathrm{θ},xâ²\mathrm{sin}\mathrm{θ}+yâ²\mathrm{cos}\mathrm{θ}\right)dyâ²
\left[\begin{array}{c}xâ²\\ yâ²\end{array}\right]=\left[\begin{array}{cc}\text{ }\mathrm{cos}\mathrm{θ}& \text{sin}\mathrm{θ}\\ â\mathrm{sin}\mathrm{θ}& \text{cos}\mathrm{θ}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]
The Radon transform for a large number of angles is often displayed as an image. In this example, the Radon transform for the square image is computed at angles from 0° to 180°, in 1° increments.
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What is alligation method?
Alligation ratio definition
Alligation formula
Alligation calculation in pharmacy
Alligation calculator: How to calculate the volume of an alligation ratio
One of the most important concepts while studying solutions is their concentration, and that is where the alligation calculator comes into play.
You might be interested in making alligation calculations in pharmacy studies, where most of the applications for the alligation calculator lie, but we will also attempt to explain what the alligation method is and how to calculate the volume of an alligation ratio.
You will also learn about the alligation formula, along with some practical alligation calculation examples.
The alligation definition says it's the process of calculating the proportions of two solutions to be mixed in order to produce a required solution with a specific concentration.
The number it outputs is the ratio that two solutions of the same type but of different concentrations must be mixed in to create a third solution of an intermediate concentration. That is what makes it different from dilution.
The process of dilution of a solution requires you to prepare a solution of lower concentration from a solution of higher concentration. Dilution is so simple that you can quickly obtain it by mixing a solvent into the higher concentration solution to bring it to the lower concentration.
For instance, say you have a 16 molar brine solution (the molar unit shows the concentration), and you want to obtain 75 ml of 9 molar solution. To find the result, place these values in the dilution formula:
M_1 × V_1 = M_2 × V_2
M_1
- Initial molarity;
V_1
- Initial volume;
M_2
- Final molarity; and
V_2
- Final volume.
After rearranging the formula to calculate the required volume, it looks like this:
V_1 = (M_2 × V_2) / M_1
V_1 = 9 × 75 / 16
V_1 = 42.19\text{ ml}
This means that you should mix 42.19 ml of the higher concentration with 32.81 ml of water to dilute the solution enough to obtain 75 ml of the diluted solution. This is a simple method of dilution.
On the other hand, we use the alligation method formula when you have two different concentrations and are looking to obtain an intermediate concentration. The two concentrations can be of the same or different solutions, but they must be measured in the same units.
A similar "How to calculate alligation?" example might look like this:
How many parts of 12 molar and 5 molar solutions should you mix to obtain a 9 molar solution?
Continue reading to find out the answer to this question!
Alligation is important when you have to change the concentration of a solution, whether it is in liquid or solid form, to make a 3rd concentration. We require the appropriate volumes of solutions with specific concentrations to make a balanced solution.
The alligation calculator estimates the alligation ratio, which tells you what ratio of the two concentrations should you use to obtain the intermediate concentration. The calculator also calculates the volumes of these different ratios.
We can represent the alligation method formula in the following form using different concentrations (conc.):
H = \text{required conc. - lower conc.}
L = \text{higher conc. - required conc.}
H
- Higher ratio; and
L
- Lower ratio.
We usually use the alligation ratio format
H:L
to show the final result. Isn't the alligation method formula a straightforward way to find the mixture ratio? Using our alligation calculator is even faster!
Indeed, by now, you are aware of what the alligation definition is. Let's now focus on alligation and its practical calculation in pharmacy. In pharmaceuticals and medicine, the amount and concentration of solutions and their components need to be extremely precise and accurate.
An example of a common and simple question of alligation would be "In what proportion should a pharmacist mix 20% and 5% zinc oxide ointments to prepare a 10% zinc oxide ointment?" You need to use the alligation mixture formula to find the answer!
In this scenario, the proportion of ointments is a percentage, but since all the concentrations are also in percentages, it will not affect the ratio. This means that the concentrations can be in any required unit, but they all need to be the same.
So, in this example,
20%
is the higher,
5%
is the lower, and
10%
is the required concentration.
The ratio of the higher concentration is obtained by subtracting the lower from the required concentration.
The ratio of the lower concentration is obtained by subtracting the required from the higher concentration.
H = 10 - 5 = 5
L = 20 - 10 = 10
5:10
, but, if you further simplify it, then the alligation ratio is
1:2
It means to get 10% of the ointment, mix one part of the higher concentration ointment (20%) to two parts of the lower concentration ointment (5%).
Using our alligation calculator is very simple. You need to know the solution concentrations you are calculating the alligation ratios for, and the calculator will do the rest for you.
To calculate the alligation ratio, you need to provide three values:
Higher concentration of the solution;
Lower concentration of the solution; and
Required concentration of the solution.
Once you input these three values, the calculator will use the alligation mixture formula to find the alligation ratio for you.
After that, you have the option to determine the volume of your solution if you need in order to calculate the concentrations in a per volume manner. To do that, simply input the volume of any one of the concentrations. The calculator automatically finds the other two volumes required to produce that particular amount.
Let's take an alligation calculation example to clarify things in a better way and get back to the question we left unanswered earlier.
How many parts of 12 molar and 5 molar solution should be mixed in order to obtain a 9 molar solution?
Remember, the units for all three concentrations should be the same.
Then follow the steps to obtain the ratio using the alligation calculator:
Input 12 as the higher concentration.
Input 5 as the lower concentration.
Input 9 as the required concentration.
The result is the alligation ratio, 4:3.
If you want to find the volume of these concentrations, enter the volume of your desired concentration. Suppose you want to add the volume of the higher concentration, let's say 350 ml. The calculator will tell you that you should mix 350 ml of the 12 molar solution with 262.5 ml of the 5 molar solution. This will produce 612.5 ml of 9 molar solution.
Be sure to check our other calculators, which every chemist will find very handy:
Grams to ml; and
How many parts of 22% and 15% solution should be mixed to produce 19% of the solution?
Using the alligation method formula you find that the alligation ratio is 4:3.
The alligation ratio suggests that to obtain a solution with concentration a of 19%, you need to mix 4 parts of the 22% solution of with 3 parts of the 15% solution.
Is alligation the same as dilution?
No, alligation and dilution are not the same. Both methods provide two distinct solutions to the problem of obtaining specific concentrations.
Dilution is a simple method in which you simply reduce the solution's concentration by adding water or any required diluent. Alligation, however, does not use any diluent, but instead gives the ratio of two solutions to be mixed in different concentrations to produce an intermediate concentration.
How many milliliters of 13 M and 4 M solutions do you need create 120 ml of a 8 M solution?
The alligation ratio is 4:5. It means that you should mix four parts of the 13 molar solution with five parts of the 4 molar solution to obtain the 8 molar solution.
The volume of your required concentration is 120 ml, which means you need to mix:
Volume of higher concentration = 53.33 ml, and
Volume of lower concentration = 66.67 ml.
How can I calculate the alligation volumes of the alligation ratio?
Once you find the alligation ratio, the next step is to calculate the volumes of the higher and lower concentrations of the solution using volume or required concentration Vᵣ:
You can calculate the volume of any two solutions as long as you know at least one of the volumes and the higher H and lower L concentration alligation ratios.
The formula to calculate the volume of the higher concentration Vₕ is:
Vₕ = H / (H + L) × Vᵣ
The formula to calculate the volume of the lower concentration Vₗ is:
Vₗ = L / (H + L) × Vᵣ
The concentrations can have any unit, but they all must be same: % , w/v , mg/ml, etc.
Find the corresponding volumes
Higher concentration volume
Lower concentration volume
Required concentration volume
Actual yieldAtomAvogadro's number… 33 more
The isoelectric point calculator will help you determine the pH value of a molecule that carries no charge.
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Greenhouses Gases, Carbonyls, and Volatile Organic Compounds Surface Flux Emissions at Three Final Waste Disposal Sites Located in the Metropolitan Area of Costa Rica ()
The surface flux emissions for volatile organic compounds (VOC’s) (alcohols and aromatic species), priority carbonyls and greenhouse gases, were measured in three different final disposal sites for urban solid waste located in the metropolitan area of Costa Rica, between July and October 2014. The emissions fluxes were determined using the static sampling chamber technique coupled to two different adsorption tubes: active charcoal (Supelco, ORBO 32) to capture BTEX and alcohols; and 2,4-DNPH coated silica gel (SKC, 226-119) for carbonyls. As for the VOCs, the BTEX, Alcohols, and Carbonyls total fluxes were in the range of 3 to 258, 1 to 318 and 0.4 to 8.5 mg/(m2día), respectively. The magnitudes per site were in the following order La Carpio > El Huaso > Rio Azul. Ethanol and BTEX presented a high correlation in all the cases because possibly they are sharing the same sources or formation mechanisms. The emission fluxes spatial distributions among the sites were very variable and dependent on the location of the active cells and their age. Only La Carpio showed a more homogeneous distribution due to its middle age.
GHG, Volatile Organic Compounds, Costa Rica, Landfills, Surface Emission Fluxes
Murillo, J. , Marín, J. , Guerrero, V. and Salazar, J. (2017) Greenhouses Gases, Carbonyls, and Volatile Organic Compounds Surface Flux Emissions at Three Final Waste Disposal Sites Located in the Metropolitan Area of Costa Rica. Open Journal of Air Pollution, 6, 149-164. doi: 10.4236/ojap.2017.64012.
F=\frac{PVM\left(\frac{\text{d}C}{\text{d}t}\right)}{ATR}
\text{Flux}=\frac{C\times V}{A\times t}
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Why are delay and reverb important?
The usefulness of a reverb time calculator
How do I calculate reverb time?
What is pre-delay in reverb?
Our delay and reverb calculator is a handy tool for audio producers/engineers or people just starting in music production looking to improve their creations with reverb and delay in sync with the tempo of their song. Just input the bpm of your song, and the calculator will do the rest!
This calculator functions as a:
Reverb/delay calculator;
Pre-delay calculator; or as a
Reverb decay calculator.
Within a few paragraphs, you will learn how to calculate reverb time by hand and with our delay and reverb calculator. Later, we will answer 'What is pre-delay in reverb?', 'What is reverb decay time'? and other useful questions.
Delay and reverb (reverberation) are two effects added to improve the quality of sound productions.
While they are similar, they differ from one another because reverb aims to simulate the sound being played in different rooms. On the other hand, delay replicates the initial sound multiple times (lowering the volume each time) after being played.
These effects have different use cases, and each (when used correctly) gives an entirely different mood to a song or musical piece.
To make the reverb or delay effect appear and disappear at the correct times, we need to time their duration in sync with our song's tempo or bpm. Otherwise, the delay of a note might still be playing when the following note comes in, reducing sound clarity. The reverb and delay calculator quickly obtains the required time for the effect based on each note's length.
To calculate reverb or delay time:
Obtain your song's bpm (beats per minute) and time signature.
Divide 60,000 by the bpm number.
Write down the result. This is the duration of the beat unit (quarter, eighth, etc.) used in the time signature expressed in ms.
Multiply or divide this result by two to obtain longer or shorter notes' lengths.
For example, let's say we have a 120 bpm song with a 3/4 time signature. This means that in a minute, 120 quarter note beats are played. To know how long each of these quarter notes is in milliseconds, we need to divide the number of milliseconds in a minute by the bpm:
{\huge♩} = \frac{60000\ \text{ms}}{120}= 500\ \text{ms}
Now, if we want to use reverb or delay for longer or shorter notes, we divide or multiply by two this result n times proportionally to the relationship between each note and the beat note, in this case, the quarter note:
Half, quarter, and eighth notes duration.
🙋 Our delay and reverb calculator automatically does this calculation for every note, saving you lots of time ⏱. Just input your bpm and start creating!
Pre-delay is the amount of time (in milliseconds) before the onset of the reverb/delay effect after a note is played.
The best way to obtain a correct pre-delay length is first to know how long the delay effect should last (half, quarter, etc., in ms) and then try different shorter notes values for the pre-delay duration (1/16th, 1/32th, etc.) to test which sounds better. The total delay time will then be:
\quad {\scriptsize \text{Total delay} = \text{pre-delay} + \text{decay time}}
Total delay is the note duration we have chosen to use as delay length.
Pre-delay is a shorter note during which we won't hear the delay effect.
Decay time is the actual time during which the delay effect is applied and heard until vanishing.
For example, let's say we have already figured out each note's duration for a specific song, and we would like to apply a delay effect to some eighths notes lasting 300 ms.
We test multiple times and find that a pre-delay of 1/64th or 37.5 ms fits nicely into our composition. Now, if we want to find the decay time, we need to subtract the pre-delay time from the eighth's duration, resulting in a 262.5 ms decay time.
What is decay in reverb?
Decay time is the amount of time during which the reverb effect is applied and heard before fading. A longer pre-delay will produce a shorter decay time.
How do I calculate bpm by ear?
To calculate bpm by ear:
The result is the bpm number for your song.
How long is a quarter note in 120 bpm?
500 ms. To get a note's duration, you need to divide 60,000 by the bpm number. This math will give you the time signature's beat duration in milliseconds. Then, multiply or divide by two accordingly to get each note's duration for that bpm.
Input your bpm:
Pre-delay and decay times
Total delay/reverb duration:
Use the tip calculator to avoid awkward situations at the restaurant or bar. Know exactly how much you and your friends should tip, every time.
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Stochastically Ultimate Boundedness and Global Attraction of Positive Solution for a Stochastic Competitive System
Shengliang Guo, Zhijun Liu, Huili Xiang, "Stochastically Ultimate Boundedness and Global Attraction of Positive Solution for a Stochastic Competitive System", Journal of Applied Mathematics, vol. 2014, Article ID 963712, 8 pages, 2014. https://doi.org/10.1155/2014/963712
Shengliang Guo,1,2 Zhijun Liu,1,2 and Huili Xiang 1,2
A stochastic competitive system is investigated. We first show that the positive solution of the above system does not explode to infinity in a finite time, and the existence and uniqueness of positive solution are discussed. Later, sufficient conditions for the stochastically ultimate boundedness of positive solution are derived. Also, with the help of Lyapunov function, sufficient conditions for the global attraction of positive solution are established. Finally, numerical simulations are presented to justify our theoretical results.
In recent years, many researches have been done on the dynamics of many types of Lotka-Volterra competitive systems. Owing to their theoretical and practical significance, these competitive systems have been investigated extensively and there exists a large volume of literature relevant to many good results (see [1–7]). Particularly, in [8], Gopalsamy introduced the following autonomous two-species competitive system: where and can be interpreted as the population size of two competing species at time , respectively. All parameters involved with the above model are positive constants and can be interpreted in more detail; and are the intrinsic growth rates of two species; , , , and represent the effects of intraspecific competition; and are the effects of interspecific competition. Recently, Tan et al. [9] have considered the effect of impulsive perturbations and discussed the uniformly asymptotic stability of almost periodic solutions for a corresponding nonautonomous impulsive version of (1). It has also been noticed that the ecological systems, in the real world, are often perturbed by various types of environmental noise. May [10] also pointed out that, due to environmental fluctuation, the birth rate, the death rate, and other parameters usually show random fluctuation to a certain extent. To accurately describe such systems, it is necessary to use stochastic differential equations. Recently, various stochastic dynamical models have been introduced extensively in [11–17] and many interesting and valuable results including extinction, persistence, and stability can be found in [18–20].
Motivated by the above works, in this contribution, we assume that the environmental noise affects mainly the intrinsic growth rate with where are independent white noises, are standard Brownian motions defined on the complete probability space with a filtration satisfying the usual conditions, and represent the intensities of the white noises. Then the stochastically perturbed sytem (1) can be Itô’s equations with the initial values .
In this paper, we will focus on the stochastically ultimate boundedness and global attraction of positive solutions of system (3). To the best of our knowledge, there are few published papers concerning system (3). The rest of this paper is organized as follows. In Section 2, we present some assumptions, definitions, and lemmas. In Section 3, we investigate the existence and uniqueness of positive solution, and then, we discuss the stochastically ultimate boundedness of positive solutions. In Section 4, we discuss the global attraction of positive solutions. Finally, we conclude and present a specific example to justify the analytical results.
Throughout this paper, we give the notation and assumptions. , . For any initial value , there exists such that , .
In the following, let us briefly review several basic definitions and lemmas which will be useful for establishing our main results.
Definition 1. The solution of system (3) is stochastically ultimately bounded a.s. if for arbitrary , there exists a positive constant such that
Definition 2. Let be a positive solution of system (3). Then is said to be globally attractive provided that any other solution of system (3) satisfies
Lemma 3 (see [21]). Let such that where is the family of processes such that Then
Lemma 4 (see [22]). Suppose that are real numbers; then where and
Lemma 5 (see [23]). Assume that an n-dimensional stochastic process on satisfies the condition for positive constants , , and . Then there exists a continuous version of which has the property that, for every , there is a positive random variable such that In other words, almost every sample path of is locally but uniformly Hölder continuous with exponent .
Lemma 6 (see [24]). Let be a nonnegative function on such that is integrable on and is uniformly continuous on . Then .
3. Stochastically Ultimate Boundedness
In this section, we first show, under the assumption , that the positive solution of system (3) will not explode to infinity at any finite time.
Lemma 7. Let hold and the initial value . Then system (3) has a unique solution on , which will remain in with probability one.
Proof. It is easy to see that the coefficients of system (3) satisfy the local Lipschitz condition. Then for any given initial value , there exists a unique local solution on , where is the explosion time. To show that the positive solution is global, we only need to show that . Let be sufficiently large for every component of which remains in the interval . For each integer , one can define the stopping time Clearly, is increasing as . Assign , whence . If we can show that ., then . and . for all . In other words, to complete the proof, we just need to show that .
By reduction to absurdity, we suppose that ; then there exists a pair of constants and such that As a result, there exists an integer such that for all Define a -function as Obviously, is a nonnegative function. If , then using Itô’s formula, one can derive that where A simple calculation shows that It then follows from that the upper bound of , noted by , exists. We therefore have Integrating both sides from to yields that whence taking expectations leads to Set for , and then . Note that arbitrary , there exist some such that equals either or . Then is not less than As a consequence, where is the indicator function of . Let lead to the contradiction So we must have . This completes the proof of Lemma 7.
Lemma 7 is fundamental in this paper. In the following, we will show that the th moment of the positive solution of system (3) is upper bounded and then discuss the stochastically ultimate boundedness.
Theorem 8. Let and hold; then the positive solution of system (3) with initial value is stochastically ultimately bounded.
Proof. From Lemma 7, we can see that system (3) has a unique positive solution under assumption . Assign arbitrarily; then by Itô’s formula we can show that Integrating and taking expectations on both sides yield that We then derive that By Hölder inequality one has and moreover Denote , ; then (31) can be rewritten as It follows from that that is, Meanwhile, it is easy to see that By the standard comparison theorem, we therefore derive that which implies that the th moment of positive solution is upper bounded.
Let us now proceed to discuss the stochastically ultimate boundedness of system (3). Setting , then by the Chebyshev inequality, we obtain that This gives that The proof of Theorem 8 is complete.
In this section, we will establish sufficient conditions for global attraction of system (3).
Lemma 9. Let hold and let be a solution of (3) with initial value ; then almost every sample path of is uniformly continuous for .
Proof. We first prove . Let us consider the following integral equation: where Recalling , (32), and the standard comparison theorem, we can know that Then from Lemma 4 and (41) one derives that Meanwhile, by Lemma 3, we obtain that, for and , Let and then from (42), (43), and Lemma 4, one can derive that Thus, it follows from Lemma 5 that almost every sample path of is locally but uniformly Hölder-continuous with exponent for and therefore almost every sample path of is uniformly continuous on .
By a similar procedure as above, can be proven. Thus, is uniformly continuous on . The proof of Lemma 9 is complete.
Theorem 10. Let , , and hold; then the unique positive solution of system (3) is globally attractive for initial data .
Proof. It follows from that, for initial data , system (3) has a unique solution (see Lemma 7). Assume that is another solution of system (3) with initial values .
Define a Lyapunov function as Using Itô’s formula, a calculation of the right differential along (3) shows that Integrating both sides yields that that is, In view of and , then it follows from (49) that So recalling Lemmas 9 and 6, we can show that This completes the proof of Theorem 10.
In this paper, we derived the sufficient conditions for the existence, uniqueness, stochastically ultimate boundness, and global attraction of positive solutions of system (3). In order to illustrate the above theoretical results, we will perform a specific example. Motivated by the Milsten method mentioned in Higham [25], we can obtain the following discrete version of system (3): where and are Gaussian random variables that follow . Let , , , , and ; , , , , and ; ; and the initial value , . After a calculation, we can see that the assumptions of Theorems 8 and 10 hold. Figures 1 and 2 show that the positive solution of system (52) is stochastically ultimately bounded and globally attractive.
The sample path of in the same coordinate system.
It follows from Theorem 8 that a preliminary result on the stochastically ultimate boundness of system (3) is obtained. We would like to mention here that an interesting but challenging problem associated with the investigation of system (3) should be the stochastic permanence; we leave this for future work.
The authors thank the anonymous referees for their valuable comments. The work is supported by the National Natural Science Foundation of China (no. 11261017), the Foundation of Hubei University for Nationalities (PY201401), the Key Laboratory of Biological Resources Protection and Utilization of Hubei Province (PKLHB1329), and the Key Subject of Hubei Province (Forestry).
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Copyright © 2014 Shengliang Guo et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Raoult's Law Calculator
Raoult's law equation - calculate the total pressure with Raoult's law
How to use our Raoult's law calculator - a Raoult's law example
What is Raoult's law used for?
Raoult's law graph
Applications of Raoult's law
Our Raoult's law calculator is a convenient tool to calculate the vapor pressure of an ideal solution. It not only provides calculation options for the vapor pressure of a solution 🧪 but also for the solvent and mole fraction of the solute as well.
Here you will learn the definition of Raoult's law, what Raoult's law is used for and what its applications are. You will also find answers to your questions of how to find the vapor pressure of a solution and how to calculate mole fraction using Raoult's law.
Are you one of those curious people who have wondered what is Raoult's law in chemistry? Well, Raoult's law bears similarities to the ideal gas law - except that it applies to solutions. Raoult's law states that:
The relative lowering of the vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
Mathematically, we express it using the following formula:
p_{solution} = x_{solvent} × p\degree_{solvent}
p_{solution}
- Vapor pressure of an ideal solution;
x_{solvent}
- Mole fraction of the solvent; and
p\degree_{solvent}
-Partial pressure of the pure solvent.
This Raoult's law formula forms the basis of the calculation performed by this Raoult's law calculator.
🔎 In case you are interested, you might find it useful to take a look at our alligation, molarity and molality calculators.
Our Raoult's law calculator primarily determines the vapor pressure of a solution, but you can also use it to calculate the partial pressure or mole fraction of the solvent.
You can choose from two different formulas: one uses the number of moles while the other uses the mole fractions of the solvent.
Start by selecting how you want to determine the vapor pressure. The first option uses moles. To calculate vapor pressure, follow these steps:
Input the partial pressure of the solvent. It has a list of units to choose from as well.
Input the moles of the solute.
Input moles of the solvent.
As a result, the calculator will compute the vapor pressure of the solution. It will also plot a graph of vapor pressure vs. mole fraction for you, based on the moles of the solute and solvent provided.
The second option finds vapor pressure using the mole fraction:
Input the partial pressure of the solvent.
Input the mole fraction of the solvent.
As a result, the calculator will work out the** vapor pressure of the solution**. A graph of vapor pressure of the solution vs. the mole fraction of the solvent will be plotted for you.
Let us consider an example of a Raoult's law calculation.
Let's say you have a solution where chemical x is the solvent. It has a partial pressure of 12 mmHg and a mole fraction of 0.25. Multiply 0.25 by 12 to get a vapor pressure of 3 mmHg.
Raoult's law applies to solutions containing non-volatile solutes ⚗️, which make ideal solutions.
In certain situations, you may want to do the calculations by hand and not use Raoult's law calculator. For that, the following points may come in handy:
Calculate the total pressure with Raoult's law
To calculate the total pressure with Raoult's law, we will use the formula for Raoult's law mentioned earlier:
p_{solution} = x_{solvent} × p\degree_{solvent}
All you need to do is find the vapor pressure of the solvent and multiply it by the mole fraction of the solvent As a result, you get the vapor pressure of the solution. This method applies to non-volatile solutes, and you may need to use other methods to calculate the total pressure for volatile solutes.
How to find the vapor pressure of a solvent
Vapor pressure is a characteristic of solids and liquid that depends on the type of liquid and its temperature.
The formula for Raoult's law can be modified to calculate the vapor pressure of the solvent:
p\degree_{solvent} = p_{solution} / x_{solvent}
In this case, you divide the vapor pressure of the solution by the mole fraction of the solvent, and as a result, you will get the vapor pressure of the pure solvent.
How to calculate mole fraction using Raoult's law
The Raoult's law calculator can also calculate the mole fraction of a solvent in a solution.
The modified form of the Roault's law formula becomes:
x_{solvent} = p_{solution} / p\degree_{solvent}
To calculate the mole fraction using Raoult's law, you divide the solution's vapor pressure by the solvent's vapor pressure.
There are other methods used to calculate the mole fraction of a solution without using Raoult's law. If you want to know more, check out our mole fraction calculator.
Our tool also allows you to plot a graph showing the vapor pressure and its mole fraction of the solution.
The mole fraction is on the x-axis, and vapor pressure is on the y-axis.
The Raoult's law graph is a straight line, which means that the partial vapor pressure of a substance is proportional to its mole fraction under the condition that the temperature remains the same.
One of the most widely used laws to study non-aqueous mixtures is Raoult's law. It applies to ideal solutions, which means solutions where the solute is non-volatile.
You can find some applications of Raoult's law below:
It helps determine the molecular mass of an unknown solute.
It can also be applied to non-ideal solutions by accounting for certain other factors, such as the intermolecular interactions.
Raoults's law is significant in the calculation of the contributions of each component in a mixture to the total pressure exerted by the system. This is particularly true for mixtures where we know the quantity of each component.
Raoult's law states that:
It gives the value of vapor pressure at which the solvent evaporates from liquid to gas.
What is the equation of Raoult's law?
The equation of Raoul's law is:
p = x × p°
x - Mole fraction of solvent;
p - Vapor pressure of an ideal solution; and
p° - Partial pressure of the pure solvent.
To calculate the vapor pressure of an ideal solution, you have to multiply the mole fraction of the solvent by the partial pressure of the solvent.
For instance, if the mole fraction is 0.3 and partial pressure is 9 mmHg, the vapor pressure would be 2.7 mmHg.
How do I calculate mole fraction using Raoult's law ?
To calculate the mole fraction using Raoult's law:
Find the vapor pressure of an ideal solution.
Determine the partial pressure of the pure solvent.
Divide the first value by the second to obtain the mole fraction.
For instance, 3 mmHg and 5 mmHg are the vapor and partial pressure values, respectively. To calculate the mole fraction, divide 3 by 5 to get 0.6.
What is the vapor pressure of a solution with a mole fraction of 0.86?
Let's assume the partial pressure of the solution is 24.7 mmHg, in which case its vapor pressure is 21.24 mmHg.
This value indicates the pressure at which the solvent molecules in the solution will evaporate. The value is determined by simply multiplying the partial pressure by the mole fraction. So, multiplying 0.86 by 24.7, you get 21.24 mmHg.
Partial pressure of solvent (pº)
Moles of solute (n₁)
Moles of solvent (n₂)
Mole fraction of solvent (x)
Vapor pressure of solution (p)
Discover crystallography with out cubic cell calculator.
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Jigu Suanjing - Wikipedia
Jigu suanjing (Chinese: 緝古算經, Continuation of Ancient Mathematics) was the work of early Tang dynasty calendarist and mathematician Wang Xiaotong, written some time before the year 626, when he presented his work to the Emperor. Jigu Suanjing was included as one of the requisite texts for Imperial examination; the amount of time required for the study of Jigu Suanjing was three years, the same as for The Nine Chapters on the Mathematical Art and Haidao Suanjing.
Facsimile of a Qing dynasty block printed Jigu Suanjing
The book began with presentations to the Emperor, followed by a pursuit problem similar to the one in Jiu Zhang Suan shu,[1] followed by 13 three-dimensional geometry problems based mostly on engineering construction of astronomic observation tower, dike, barn, excavation of a canal bed etc., and 6 problems in right angled triangle plane geometry. Apart from the first problem which was solved by arithmetic, the problems deal with the solution of cubic equations, the first known Chinese work to deal with complete cubic equations, as such, it played important roles in the development for solution of high order polynomial equations in the history of Chinese mathematics. Before his time, The Nine Chapters on the Mathematical Art developed algorithm of solving simple cubic equation
{\displaystyle x^{3}=N}
numerically, often referred to as the "finding the root method".
Wang Xiaotong used an algebraic method to solve three-dimensional geometry problems, and his work is a major advance in Algebra in the history of Chinese mathematics.
Each problem in Jigu Suanjing follows the same format, the question part begins with "suppose we have such and such,... question:...how many are there?"; followed by "answer:", with concrete numbers; then followed by "The algorithm says:...", in which Wang Xiaotong detailed the reasoning and procedure for the construction of equations, with a terse description of the method of solution. The emphasis of the book is on how to solve engineering problems by construction of mathematical equations from geometric properties of the relevant problem.
In Jigu Suanjin, Wang established and solved 25 cubic equations, 23 of them from problem 2 to problem 18 have the form
{\displaystyle x^{3}+px^{2}+qx=N,\,}
The remaining two problems 19, and 20 each has a double quadratic equation:
{\displaystyle x^{4}+px^{2}+q=0}
Problem 3, two cubic equations:
{\displaystyle x^{3}+{\frac {3cd}{b-c}}x^{2}+{\frac {3(a+c)hd^{2}}{(H-h)(b-c)}}x={\frac {6Vd^{2}}{(H-h)(b-c)}}}
{\displaystyle x^{3}+5004x^{2}+1169953{\frac {1}{3}}x=41107188{\frac {1}{3}}}
Problem 4 two cubic equations:
{\displaystyle x^{3}+62x^{2}+696x=38448,\quad x=18;}
{\displaystyle x^{3}+594x^{2}=682803,\quad x=33;}
{\displaystyle x^{3}+15x^{2}+66x-360,\quad x=3}
{\displaystyle x^{3}+(D+G)x^{2}+\left(DG+{\frac {D^{2}}{3}}\right)x=P-{\frac {D^{2}G}{3}}}
{\displaystyle X^{+}3{\frac {hs}{D}}x^{2}+3\left({\frac {hs}{D}}\right)^{2}x={\frac {P'}{3}}{\frac {h^{2}}{D^{2}}}}
{\displaystyle x^{3}+90x^{2}-839808,\quad x=72}
{\displaystyle x^{3}+{\frac {S}{2}}x^{2}-{\frac {P^{2}}{2S}}=0}
{\displaystyle x^{3}+{\frac {5}{2}}Dx^{2}+2D^{2}x={\frac {P^{2}}{2D}}-{\frac {D^{2}}{2}}}
Problem 20:"Suppose the long side of a right angle triangle equals to sixteen and a half, the square of the product of the short side and the hypothenuse equals to one hundred sixty four and 14 parts of 25, question, what is the length of the short side ?"
Answer: "the length of the short side is eight and four fifth."
Algorithm:"Let the square of the square of product as 'shi' (the constant term), and let the square of the long side of right angle triangle be the 'fa' (the coefficient of the y term). Solve by 'finding the root method', then find the square root again."
The algorithm is about setting up a double quadratic equation:
{\displaystyle x^{4}+\left(16{\frac {1}{2}}\right)^{2}x^{2}=\left(164{\frac {14}{15}}\right)^{2}}
where x is the short side.
Wang's work influence later Chinese mathematicians, like Jia Xian and Qin Jiushao of Song dynasty.
During the Tang dynasty there were hand-copied Jigu Suanjing in circulation. During the Song dynasty there were 1084 government-printed edition copies. However, by the Ming dynasty the Tang dynasty hand-copied editions and Song dynasty printed editions were almost all lost; only a single copy of a Southern Song print survived. This copy was later obtained by early Qing dynasty publisher Mao Jin, who made an image hand copy (hand-copied character by character, following the printed form closely) of it. Mao Jin's image copy of Jigu Suanjing later became the source for a printed edition during the Qianlong era and was also incorporated into the Siku Quanshu. The Qianlong era printed edition disappeared, and only Mao Jin's image copy edition of Jigu Suanjing survived at the Forbidden City Museum. The copy in the Siku Quanshu still exists.
During the Qing dynasty, study of Jigu Suangjing was in vogue; half a dozen books devoted to the study of Jigu Suanjing by mathematicians were published, some of which concentrated on filling the gaps left by many missing characters due to age, and some devoted to the detail elaboration of algorithm either from geometry point of view (Li Huang) or from Tian yuan shu (Zhang Dunren).
In 1963, Chinese mathematics historian Qian Baocong published his annotated The Ten Computational Canons, which included Jigu Suanjing.
Jigu Suanjing was introduced to the English speaking world by Alexander Wylie in his book Notes on Chinese Literature.[5]
^ Jean-Claude Martzloff A History of Chinese Mathematics, "Jigu Suanjing", p140, Springer ISBN 3-540-33782-2
^ Bai Shangshu Critical Edition of Wang Xiaotong's Ji gu Suan Jing, P83 ISBN 978-7-303-09242-0
^ Yoshio Mikami, The Development of Mathematics in China and Japan p54, 1913. Chelsea Publishing Company, New York
^ Yoshio Mikami The Development of Mathematics in China and Japan, p55, 1912
^ Alexander Wylie, Notes on Chinese Literature, p115-116, 1902, Shanghai, reprint ISBN 0-548-86642-2
Retrieved from "https://en.wikipedia.org/w/index.php?title=Jigu_Suanjing&oldid=1049807610"
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