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The Demand Curve and Utility | Boundless Economics | Course Hero
The Demand Curve and Utility
Utility is an economic measure of how valuable, or useful, a good or service is to a consumer.
Utility is measured by comparing multiple options.
Utility can be positive and negative.
Ordinal utility ranks a series of preferences without measuring how much more valuable one option is than another. Cardinal utility measures how much more preferable one option is in comparison to another.
Ordinal utility is generally the preferred method of measuring utility.
ordinal: Of a number, indicating position in a sequence.
cardinal: Describing a "natural" number used to indicate quantity (e.g., one, two, three), as opposed to an ordinal number indicating relative position.
Utility is a term used by economists to describe the measurement of "useful-ness" that a consumer obtains from any good or service. Utility may measure how much one enjoys a movie or the sense of security one gets from buying a deadbolt. The utility of any object or circumstance can be considered. Some examples include the utility from eating an apple, from living in a certain house, from voting for a specific candidate, or from having a given wireless phone plan. In fact, every decision that an individual makes in their daily life can be viewed as a comparison between the utility gained from pursuing one option or another.
Apples and Oranges: Utility allows you to compare apples and oranges based on which you prefer.
Utility may be positive or negative with no effect on its interpretation. If one option gives
-15
utility and another gives
-12
, selecting the second is not, as it might seem, the "lesser of two evils," but can only be interpreted as the better option.
Utility can be measured in one of two ways:
Ordinal utility ranks a series of options in order of preference. This ranking does not show how much more valuable one option is than another, only that one option is preferable over another. An example of a statement reflecting ordinal utility is that "I would rather read than watch television." Generally, ordinal utility is the preferred method for gauging utility.
Cardinal utility also ranks a series of options in order of preference, but it also measures the magnitude of the utility differences. An example of a statement reflecting cardinal utility is "I would enjoy reading three times more than watching television." Given how difficult it is to precisely measure preference, cardinal utility is rarely used.
The theory of utility states that, all else equal, a rational person will always choose the option that has the highest utility.
Explain the Theory of Utility
The rationality assumption gives a basis for modeling human behavior and decision making.
Utility includes every element of a decision.
Rationality is dependent on a person's individual preferences. Therefore, what might be a rational decision for one person may not be a rational decision for another.
Rational individual: A person who chooses the option that, all else equal, gives the greatest utility.
The theory of utility is based on the assumption of that individuals are rational. Rationality has a different meaning in economics than it does in common parlance. In economics, an individual is "rational" if that individual maximizes utility in their decisions. Whenever an individual is to choose between a group of options, they are rational if they choose the option that, all else equal, gives the greatest utility. Recalling that utility includes every element of a decision, this assumption is not particularly difficult to accept. If, when everything is taken into account, one decision provides the greatest utility, which is equivalent to meaning that it is the most preferred, then we would expect the individual to take that most preferred option. This should not necessarily be taken to mean that individuals who fail to quantify and measure every decision they make are behaving irrationally. Rather, this means that a rational individual is one who always selects that option that they prefer the most.
Consumer making a decision: When making an economically rational purchasing decision, a consumer must consider all of their personal preferences.
It is important to emphasize how rationality relates to a person's individual preferences. People prioritize different things. For example one person may prioritize flavor while another person may value making healthy choices more. As a result the first person may choose a sugary cereal while the second may choose granola. Based on their preferences, both made the economically rational choice.
The rationality assumption gives a basis for modeling human behavior and decision making. If we could not assume rationality, it would be impossible to say what, when presented with a set of choices, an individual would select. The notion of rationality is therefore central to any understanding of microeconomics.
Marginal utility of a good or service is the gain from an increase or loss from a decrease in the consumption of that good or service.
Marginal utility is measured on a per unit basis.
Since an individual's utility is rarely measured using cardinal means, calculating a product's marginal value for an individual may be difficult.
Instead of trying to calculate a product's marginal value for an individual, economists assign dollar values to products based on their market price. This allows economists to estimate a product's marginal value based on all the consumer's preferences.
The idea of marginal value is an important consideration when making production or purchasing decisions. A person should produce or purchase an additional item when the marginal utility exceeds the marginal cost.
marginal: Of, relating to, or located at or near a margin or edge; also figurative usages of location and margin (edge).
In economic terms, marginal utility of a good or service is the gain from an increase or loss from a decrease in the consumption of that good or service. The idea of marginal value is an important consideration when making production or purchasing decisions. A person should produce or purchase an additional item when the marginal utility exceeds the marginal cost.
Marginal Utility of Housing: The marginal utility of owning a second house is likely less than the marginal utility of owning the first house.
Marginal utility is measured on a per unit basis. When evaluating the marginal utility of any item, it is important to know in what unit utility is measured. The unit is based on the type of activity that you are trying to measure. If you are a consumer of potato chips, you might measure utility based on whether to buy another bag or have another hand full with your lunch. If you are a producer of potato chips, your marginal value might be defined by a pallet of potato chips. In general, marginal value should be measured based on the smallest unit of consumption or production related to the product in question.
It is also important to remember that utility is difficult to quantify since preferences vary based on the individual. Utility is rarely measured in terms of magnitude; utility is normally just about determining which option is the best choice. Since utility is rarely measured using cardinal means, it may seem difficult to determine a product's marginal value. Economists get around this by substituting dollar values. While this may fail to capture a specific individual's preferences and utility, it offers a good approximation based on everyone's collective preferences as defined by the market.
The principle of diminishing marginal utility states that as more of a good or service is consumed, the marginal benefit of the next unit decreases.
Explain diminishing marginal utility
If you consume too much, the marginal utility of a good or service can become negative.
In some circumstances, the marginal utility of producing or consuming an additional unit will increase for a short period of time. Generally there will be a "tipping point" at which marginal utility will then decrease.
Generally these exceptions occur when what is being consumed is a component of a larger whole.
The principle of diminishing marginal utility states that as an individual consumes more of a good, the marginal benefit of each additional unit of that good decreases.
The concept of diminishing marginal utility is easy to understand since there are numerous examples of it in everyday life. Imagine it is a hot summer day and you are hungry, so you get some ice cream. The first bite is great and so is the second. But with each spoonful, your hunger decreases and you become cooler. So while the last bite might still be good, it is probably not as satisfying as the first. This is a simple illustration of diminishing marginal utility.
Total and marginal utility: As you can see in the chart, the more of a good you consume, the further its marginal utility decreases.
While there are some circumstances where there will always be some marginal utility to producing or consuming more of a good, there are also circumstances where marginal utility can become negative. For example, while some antibiotics may be useful in curing diseases. However, if you take too much you can become sick or resistant to the drugs which could lead to future illnesses being incurable. So it is important to remember that "diminishing" does not necessarily mean to zero; you can have too much of a good thing.
This concept suggests a uniform steady decline of marginal utility, but that may not always be the case. There can be situations in which one might gain more utility from consuming a later unit of a good than from earlier consumption. If you are going on a date, for example, getting one ticket to a concert will have some utility but the second arguably has more because it enhances the value of the first.
Generally these exceptions occur when what is being consumed is a component of a larger whole. While utility may increase for a period, there is usually a "tipping point" where afterwards marginal utility decreases. Getting a third ticket for your date will have low marginal utility than the second.
Principles of Economics/Utility. Provided by: Wikibooks. Located at: http://en.wikibooks.org/wiki/Principles_of_Economics/Utility. License: CC BY-SA: Attribution-ShareAlike
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US Navy 060513-N-5174T-045 Lt.nTaylor Forester makes a few last minute decisions before purchasing a gold necklace from a Navy Exchange vendor aboard the Nimitz-class aircraft carrier USS Ronald Reagan (CVN 76). Provided by: Wikipedia. Located at: http://en.wikipedia.org/wiki/File:US_Navy_060513-N-5174T-045_Lt._Taylor_Forester_makes_a_few_last_minute_decisions_before_purchasing_a_gold_necklace_from_a_Navy_Exchange_vendor_aboard_the_Nimitz-class_aircraft_carrier_USS_Ronald_Reagan_(CVN_76).jpg. License: CC BY-SA: Attribution-ShareAlike
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Resolution Scale Calculator
Definition and purpose of resolution scaling
How to scale a resolution using the resolution scale calculator?
How to calculate the resolution scale percentage?
How to manually scale or resize a resolution using the formula?
Standard display resolution scales in games, images and videos
Our resolution scale calculator helps resize digital resolutions and provides other resolutions that maintain the aspect ratio in its dynamic resolution scale chart.
You can easily up-scale or down-scale your resolution by entering your resolution scale as a percentage or by inputting its aspect ratio in the original resolution field and scaling it up.
If you're calculating upscaled resolution for image printing, you can convert your resolution to print size based on its PPI (pixels per inch) and get its framing dimensions using our picture frame calculator.
The act of resizing a resolution to transform how it is displayed on the screen's pixels is known as resolution scaling.
When down-scaled, the display resolution takes up more pixels for each displayed object, providing a zoomed-in display. And when up-scaled, the display resolution takes fewer pixels for each displayed object, providing a zoomed-out display.
Image resolution scaling works in an inverse manner to display resolution scaling. When resizing an image resolution while keeping the resolution of the display constant:
Down-scaling causes the pixels in the image to come closer, sharpening the image and reducing its size.
And when up-scaling the image resolution, its pixels expand, thus blurring or pixelating the image and increasing its size.
💡 Video resolution scaling works similar to image resolution scaling, where up-scaling the video resolution increases its size, but the quality drops, and the details become blurry.
Game resolution scaling
By default, video games run at fullscreen with their objects rendered at 1:1 to match the display resolution. We can down-scale their render resolution to increase the game's performance, but at the cost of visual quality, or up-scale the render resolution to improve the visual quality, but at the expense of performance.
As an example, let's say your native screen resolution is 2560x1440. You can downscale your favorite game's render resolution to 1920x1080 to experience a significant boost in the performance. However, you'll need to increase the sharpening and use anti-aliasing to improve your game's visual experience.
Here's how you can use our calculator to scale your resolution for images or videos:
Enter the original resolution width and height of your object, i.e., 1920 x 1080.
Input the percentage by which you want to scale your object's resolution, i.e., 25%.
If you want the calculator to down-scale your resolution, reduce the percentage.
Increase the percentage if you want the calculator to up-scale your image resolution.
The calculator will then display the scaled values, e.g., when scaling 1920x1080 resolution by 25%, we get 960x540.
🙋 After calculating resolution upscale or downscale values, the calculator gives their aspect ratio and a resolution scale chart (in the advanced mode).
To calculate the resolution scale percentage, enter your scaled resolution after entering the original resolution. The calculator will automatically calculate the scale percentage whilst showing the dynamic resolution scale chart.
For example, if you wish to find the percentage scale of 1440p resolution from 1080p, enter 1920x1080 in the original resolution, then the scaled resolution section to be displayed. Enter 2560x1440 in the scaled resolution. You'll get a percentage scale of 177.8%.
🔎 As an additional feature, you can measure the physical, or print size, of your scaled resolution by entering the PPI (pixels per inch) in the pixel density field, under the digital to physical dimensions section.
We use the following formulas to manually scale the width and height of a resolution.
\text{Width}_{(scaled)} = \sqrt{\frac{\text{pixel}}{\text{W'ratio}}}
\text{Height}_{(scaled)} = \sqrt{\frac{\text{pixel}}{\text{H'ratio}}}
pixel - Total number of pixels in the resolution;
W'ratio - The ratio obtained by dividing the resolution's height by its width; and
H'ratio - The ratio obtained by dividing the resolution's width by its height.
Here's how to use the formula to scale a resolution by percentage:
Find the required pixel percentage by multiplying the original resolution width by its height. Then, divide the result by 100 and multiply it by your scale value.
pixel = width × height × scale / 100
scale - The required percentage value to up-scale or down-scale the resolution.
E.g., if we want to scale 1440p resolution down to 50%, we first find the number of pixels in a 1440p resolution:
pixel = 2560 × 1440 × 50 / 100
pixel = 1,843,200
Divide the resolution's height by its width to find the W'ratio and the width by its height to find the H'ratio.
W'ratio = height / width
H'ratio = width / height
E.g., using the values of a 1440p resolution gives:
W'ratio = 1440 / 2560 = 0.5625
H'ratio = 2560 / 1440 = 1.7778
Divide pixel by the ratios and take a square root of the result to find your scaled resolution.
\text{Width}_{(scaled)} = \sqrt{\frac{\text{pixel}}{\text{W'ratio}}}
\text{Height}_{(scaled)} = \sqrt{\frac{\text{pixel}}{\text{H'ratio}}}
E.g., by using the values of pixel and ratios from Step 1 and 2, we can scale a 1440p resolution down to 50%:
Width(scaled)
= ( 1843200/0.5625 ) ^ (1/2)
Height(scaled)
= ( 1843200/1.7778) ^ (1/2)
Thus, if we down-scale 2560x1440 to 50%, we get a 1810x1018 resolution.
🙋 If we up-scale 1920x1080 resolution, also known as 1080p or FHD, to 177.8%, we get 2560x1440, known as 2K or 1440p. Inversely, if we down-scale the 2K resolution to 56.25%, we get FHD.
Below is a table showing some standard resolution scales for games, images, and videos:
Comparing various prominent resolutions.
Psst. You can also use our pixels to inches converter to find the print size of these resolutions. 😊
How do I calculate image resolution scale?
To calculate image resolution scale:
Divide the width of your image by its height to find the image ratio.
Multiply your preferred height with this ratio to obtain your image width.
If you have a preferred width instead of height, swap them with each other to find the resized image ratio and the new height.
What are standard laptop screen resolutions?
Here are some standard laptop screen resolutions:
16:10 aspect ratio resolutions
16:9 aspect ratio resolutions
PPI stands for pixel per inch.
It is how we understand the sharpness of a display, thus helping us measure its overall quality. PPI tells us the pixel density, i.e., the number of pixels packed in an inch of the display. Higher PPI means there's more room for information on the display.
What is resolution scale?
Visually transforming a resolution across the display pixels is known as resolution scaling, e.g., keeping the resolution ratio the same and resizing an image, video, or display by up-scaling or down-scaling their resolution.
What is the percentage scale of 1080p from 1440p?
If we down-scale 1440p by 56.25%, we get 1080p. And we up-scale 1080p by 177.78%, we get 1440p.
Multiply the width and height of a resolution, i.e., 2560 by 1440 and 1920 by 1080.
Divide both results with each other, i.e., 2,073,600 / 3,686,400.
Multiply the end result by 100, i.e., 1.778 x 100 = 177.78%.
Original resolution (or aspect ratio)
Resizing scale
Resolution width (scaled)
Resolution height (scaled)
Pixel count (scaled)
Digital to physical dimensions
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H. W. Zhang, G. Subhash, X. N. Jing, L. J. Kecskes and R. J. Dowding “Evaluation of hardness–yield strength relationships for bulk metallic glasses“ Philosophical Magazine Letters (21 Aug 2006)
Using the Vickers hardness number calculator
This Vickers hardness number calculator uses the size of indentation to return the hardness of the material. There are different scales to measure hardness, including Brinell hardness, Rockwell hardness, Meyers hardness, and Vickers hardness test.
The Vickers hardness scale, named after the company where it was developed in 1921, is the focus of this article. This test is an alternative to Brinell hardness, and it is relatively easier to use because it does not depend on the cone diameter or size of the indenter. The article also briefly covers how to perform the Vickers hardness test.
Read on to understand how to calculate the Vickers hardness number once you have completed the indentation experiment.
Hardness tests measure the resistance of a material against plastic deformation. To perform these tests (be it Brinell, Rockwell, or Vickers), an indenter is pressed on the surface of the specimen with a specific force
F
for a time interval of, say, 10-15 s. During this interval, the material undergoes plastic deformation, which you will measure after unloading the material specimen.
The indentation could be of different shapes depending on the type of test. For a Vickers hardness test, you'll use a diamond indenter with angle
136\degree
, and it would leave a square pyramid shaped indentation. The diagonals of the indentation are measured and averaged. Once you know the mean diagonal, you can use the Vickers hardness calculation number formula to find its value.
You can find the test conditions, guidelines, and formula for calculating the Vickers hardness number in the international standard document – BS EN ISO 6507-1: Metallic materials — Vickers hardness test.
The Vickers hardness number,
HV
, is a function of normal force in kgf (kilogram-force),
F
, mean diagonal in mm,
d
, and, surface area of the indentation. Mathematically, Vickers hardness number in
\text{kgf/mm}^2
\footnotesize HV = \frac{\text{Force}}{\text{Area}} = \frac{2F \sin{\frac{\alpha}{2}}}{gd^2}
g
– Acceleration due to gravity,
g = 9.80665 \text{ m/s}^2
\alpha
– Angle of indenter, commonly taken as
\alpha = 136°
In order to convert the Vickers hardness number to
\text{MPa}
, you have to multiply the
HV
value by the gravitational acceleration
g
. Such that, hardness
H
\text{GPa}
\footnotesize H = HV \times \frac{9.80665}{1000}
The above parameter
H
is also known as surface area hardness.
You can also estimate the tensile strength from the hardness number by dividing it with a constant
c
\sigma_u = \frac{H}{c}
c
depends on several factors; however, as a rule of thumb,
c
3
for metallic crystalline materials.
Let's calculate the Vickers hardness number for a material specimen having an indentation with a mean diagonal of
0.5 \text{ mm}
under a load of
50 \text{ N}
To calculate Vickers hardness number:
Enter the load value,
F = 50 \text{ N}
Fill in the value of mean diagonal as,
d = 0.5 \text{ mm}
Using the Vickers hardness calculation formula:
\footnotesize \qquad \begin{align*} HV &= \frac{2F \sin{\frac{\alpha}{2}}}{gd^2} \\\\ &= \frac{2 \times 50 \sin{\frac{136}{2}}}{9.80665 \times 0.5^2} \\\\ &= 37.82 \text{ kgf/mm}^2\\ \end{align*}
The calculator also returns the surface area hardness, and tensile strength as
370.9 \text{ MPa}
126.06 \text{ MPa}
Nomenclature of Vickers hardness number
The Vickers hardness number is written as
xxxHVyy
xxxHVyy/zz
xxx
– Hardness number;
yy
– Load value (in kgf); and
zz
– Loading time.
How do I calculate Vickers hardness number?
Divide the angle of indenter by 2.
Multiply force by the sine of the resultant.
Divide the product by gravitational acceleration.
Divide the quotient by square of the mean diagonal.
Multiply the result by 2 to obtain the Vickers hardness number.
What is the Vickers hardness calculation formula?
The following equation gives the Vickers hardness formula: HV = 2 × F × sin(α/2) / (g × d²) for a given loading force F and indenter with angle, α, leaving an indentation with a mean diagonal, d, on the material specimen.
What is the Vickers hardness number for iron?
The Vickers hardness number for iron is 30-80HV5. This implies that if the test is conducted under a load value of 5 kgf, the Vickers hardness number would be 30 to 80 kgf/mm². This value is just an estimate, and the actual value depends on the composition of the iron specimen.
What is the HV value for carbon steel?
The Vickers hardness number for carbon steel is 55-120HV5. If the test is conducted under a load value of 5 kgf, the Vickers hardness number would be in the range of 55 to 120 kgf/mm². As an estimate, the value may vary depending on the composition of carbon steel.
Mean diagonal length on indentation (d)
Material properties — hardness & strength
Surface area hardness (H)
Tensile strength (σᵤ)
Based on BS EN ISO standard:
BS EN ISO 6507-1: Metallic materials — Vickers hardness test
Find the critical load for a column using our buckling calculator.
Buckling Calculator
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Ideal Gas Density Calculator | Ideal gas law
Deriving density from ideal gas law equation
Example: Calculating the density of air ideal gas
Aspects to consider when using ideal gas law to find the density
Other tools similar to our ideal gas density calculator
Calculating the density of an ideal gas is not more complicated than any of its other properties. The only additional task is to express specific volume in terms of density. That's what this calculator does: it's an ideal gas law calculator with density instead of volume as the output.
If you thought the density of all ideal gases was the same, nothing further from reality. Ideal gas densities can be as low as 0.07927 kg/m³ (hydrogen) or as high as 2.281 kg/m³ in the case of butane. In other words, butane is 29 times denser than hydrogen 🙀
Keep reading to learn more about finding density using the ideal gas law!
Using our calculator is the easiest way to find the density using the ideal gas law. Even so, it's essential to know the equations for a better understanding.
Deriving density from the ideal gas law only requires knowing the relationship between specific volume and density and using it in the ideal gas law equation. Let's remember the equation:
Pν = RT
P
— Absolute pressure of the gas;
ν
— Specific volume of the gas;
R
— Gas constant, different for every gas; and
T
— Absolute gas temperature, in kelvin (K).
Specific volume measures how much space a substance occupies per unit mass. It's the inverse of density (
\rho
ν = \frac{1}{\rho}
If we insert this density formula in the ideal gas law:
\frac{P}{\rho} = RT
\rho
\rho = \frac{P}{RT}
Awesome! We derived a density equation using the ideal gas law.
Now, let's calculate density with the ideal gas law for one of the most common fluids: air.
Air density is widely known, but we can calculate it with the previous formula. Follow these steps to do it:
Determine the gas constant. For air, it's
R = 287 \ \text{J/kg⋅K}
Determine the pressure and temperature conditions. Suppose we want to find air density at
T = 15\ \text{°C} = 288.15 \ \text{K}
P = 101325 \ \text{Pa}
Input the values in the equation, taking into account that the density formula for ideal gas law requires temperature in kelvin:
ρ = \frac{P}{RT} = \frac{101325 \ \text{Pa}}{287 \ \text{J/kg⋅K}(288.15 \ \text{K})} = 1.225 \ \text{kg/m³}
That's it. If you search for air density on Google, you'll get this answer.
🔎 We also have a dedicated air density calculator that, besides temperature and pressure, takes into account air moisture properties.
Consider these points when using the ideal gas law:
If you want to find density using the ideal gas law, you need to provide the absolute pressure of the gas, not its manometric pressure.
The ideal gas law is an idealization, and real gases deviate from it. The nearer temperature and pressure to the critical point, the more our gas will differ from this idealization.
We can quantify the deviation from ideal-gas behavior using a correction factor called the "compressibility factor" (you can learn more about it in our compressibility factor calculator).
Now that you know how to calculate density with the ideal gas law, you can take a look at these tools related to our gas density calculator:
At atmospheric pressures below 10 kPa, steam is an ideal gas (or behaves like that), no matter its temperature. The error increases at higher pressures, like the atmospheric pressure, especially near the critical point where it reaches about 100% error.
Yes, we can consider CO2 an ideal gas as long as it keeps far away from its critical pressure (7.39 MPa) or, failing that, far below its critical temperature (31.05 °C). The reduced pressure and temperature quantify how far away it is from those values.
How to determine which gas behaves most ideally?
To determine which gas behaves most ideally:
Determine the reduced temperature (TR) and reduced pressure (PR) of each gas.
With TR and PR, determine the compressibility factor (Z) using a compressibility chart.
The gas with the Z nearer to 1 is the one that behaves most ideally.
Is the density of all ideal gases the same?
No, the density of all ideal gases is not the same. Every substance has a different molar mass that translates into a different density. You can use our gas density calculator and check how different they are!
Absolute presure (P)
J/kg⋅K
Mechanical advantage calculator finds out the force amplification when using levers, pulleys, screws, wedges, ramps, or wheels and axles.
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Wavelength, frequency, and speed: the main components of the sound wave calculator
Sound wavelength formula: sound frequency and wavelength relationship
Exciting aspects about the frequency and wavelength of sound waves
Typical speeds of sound
How to find the wavelength of sound using the sound wavelength calculator
Waves are everywhere, and as sound is a mechanical wave present in many essential aspects of our lives, we created the sound wavelength calculator.
Our sound wave calculator lets you know the wavelength of a sound wave if you know its frequency and speed (or the medium in which it's propagating). You can also use it to calculate the frequency of a wave if you know its wavelength and sound speed.
Keep reading if you want to learn more exciting aspects about sound waves, how to find the wavelength of sound, and how to calculate the speed of sound with frequency and wavelength.
Waves are everywhere and manifest in different ways. Waves occur when there's a disturbance in a system, and the disturbance travels from one place to another.
There are two main kinds of waves: mechanical waves and electromagnetic waves. The main difference is that mechanical waves need a medium to travel (a material), whereas electromagnetic waves can travel through a vacuum space.
Sound is an example of a mechanical wave, and other examples include ripples on the water's surface, seismic shear waves, and water waves. Examples of electromagnetic waves are light, microwaves, and radio waves.
Mechanical waves are classified within three groups, depending on the direction of the periodic motion relative to the movement of the wave:
Longitudinal waves: Each particle moves back and forth in the same direction as the wave.
Transverse waves: The particles move back and forth transversely (at right-angles) to the wave motion.
Combined waves: These are a combination of longitudinal and transversal waves. The most common example of this type is sea waves.
Longitudinal waves are the most relevant in our daily lives as, when a fluid acts as the propagation medium, sound waves are longitudinal. Also, sound waves can behave as longitudinal and transversal when the medium is a solid material.
As you may imagine, the study of sound waves is mainly concerned with how it propagates through that strange fluid called air, as that's how we usually receive sound.
Do check out our modulation calculator for info on how waves carry information.
Pressure disturbance is the cause of sound waves, and we can represent them as sine waves, characterized by three terms:
Speed of sound: Indicates the speed at which the sound wave propagates. It varies depending on the propagation medium.
Wavelength: This is the distance between one point to the corresponding point on the following repetition of the wave shape, and it's what we're mainly concerned about in our sound wavelength calculator.
Frequency: Number of times the wave repeats per unit time, usually measured in Hz. For example, if a wave repeats ten times in a second, it has a frequency of 10 cycles per second or 10 Hz.
Read the following section to know how these variables relate to each other through the sound wavelength formula.
The wavelength formula of sound is the same used for other waves:
\quad\ \ λ=\frac{v}{f}
λ
is the wavelength of the sound wave,
v
its speed (in this case, speed of sound), and
𝑓
its frequency.
There are other ways to express this relationship of wavelength and sound frequency. For example, if we wanted to calculate the frequency of a wave, we would rearrange the equation to obtain the frequency of sound formula:
\quad\ \ f=\frac{v}{λ}
Finally, if you want to know how to calculate the speed of sound with frequency and wavelength, this is the formula:
\quad\ \ v=λf
🔎 Are you further interested in sound speed? Look at our speed of sound calculator. With this tool, you can calculate the speed of sound for air and water, not with frequency and wavelength, but in terms of temperature.
Now that you know the equation of frequency of sound waves and speed of sound, let's look at some exciting aspects about frequency, typical values of the speed of sound, and how to find a sound frequency and wavelength using the calculator.
The frequency of a sound determines how we perceive it. High-pitch sounds are indicative of high frequency, while a low-pitch implies low frequency.
On the other hand, the wavelength is not a dependent quantity, as it depends on the speed of sound (inherent to the environment) and frequency (inherent to the source of the disturbance).
However, we can relate wavelength to the size of the musical instruments. Small size instruments, such as flutes, have a high pitch, and therefore, high frequency and short wavelength. Whereas large instruments, such as trombones, produce long-wavelength sound.
💡 We can say the pitch of a sound (such as the sound produced by musical instruments) is directly related to its frequency. On the other hand, the size of an instrument is related directly to its audio wavelength.
Frequency ranges of infrasound, acoustic, and ultrasound waves
Frequency doesn't only determine how we feel or perceive a sound but also determines if we can sense it. The human ear cannot perceive all sound waves, and we can only perceive sounds with frequencies from 20 Hz to 20,000 Hz. This spectrum of frequencies is known as the human hearing range.
Nevertheless, frequencies outside the human range are present in our daily lives, both in nature and technology.
Source: Wikipedia. Attribution: Derivative of Heath Raftery's own creation.
The infrasound range
Although we believe we can hear all the sounds emitted by elephants, most of the sounds produced by these animals are low-frequency noises below 20 Hz, known as infrasound. They use these signals to communicate over distances up to 10 km.
Some natural phenomena also emit infrasound, such as volcanic eruption (below 20 Hz) and earthquakes (below 10 Hz). Curiously, some animals can perceive this range of sound waves, which is why elephants flee in fear when earthquakes events are about to occur.
The ultrasound range
Have you heard the term "ultrasound imaging" and don't know why it's called that way? Ultrasound imaging uses ultrasound waves to obtain images of the body's internal organs, which can be used, for example, to calculate the volume of your bladder. When a sound wave strikes the targeted object, it bounces back, and with these echoes, physicians construct images of the organs. Ultrasound waves also have applications in therapeutic procedures, and cancer therapy is one of the more promising areas. These applications range mainly from 1 to 3 MHz, but we can even find frequencies up to 7.5 MHz.
The property that describes how much is reflected and the resistance that the ultrasound wave finds through different body tissues is known as acoustic impedance (Z).
In science and technology, we can also use ultrasound for imaging processes in the field of non-destructive testing procedures, such as acoustic microscopy.
Most of the animals are only sensitive to frequencies above the human range. For example, the audible range of dogs 🐶 goes from 67 Hz to 45 kHz, while for cats, from 48 Hz to 85 kHz 🙀
💡 Did you know? Human hearing is sensitive to very low-frequency sound waves compared to most of the frequency emitted by animals. This sensitivity was crucial to the survival of our ancestors, as it allowed us to hear night predators we couldn't see.
The speed of sound in a fluid depends on a measure of resistance to compression called bulk modulus (
B
) and its density (
ρ
\quad\ \ v=\sqrt{\frac{B}{ρ}}
In solids, the speed of sound relationship is similar and depends on a measure of tensile or compressive stiffness called Young's modulus (
E
) and its density:
\quad\ \ v=\sqrt{\frac{E}{ρ}}
In both cases, the more resistant the material is, the greater the speed of sound, and the denser the material is, the lower the speed. Temperature is another crucial factor in the speed of sound in fluids, as it affects bulk modulus and density.
Anyway, you don't have to bother with calculating the speed of sound on your own. Our sound wavelength calculator provides you in advance with the speed of sound in different materials, for example:
Air (20 °C/68 °F): 343 m/s
Water (20 °C/68 °F): 1481 m/s
Let's suppose you want to calculate the wavelength of a woman's voice in the air. If we know the average frequency of the women voice is 210 Hz, these would be the steps:
Establish the medium in which your sound wave propagates. In this case, the medium could be air at 20 °C.
Input 210 Hz in the frequency box.
Now you're done with your sound wave calculation! The result should be 1.6333 m.
You can also select the air as your medium, lock it, and convert our sound wavelength calculator into a sound wavelength in air calculator 😜
This calculator also works the other way, so if you don't know how to find the sound frequency, you only have to follow the same previous steps but input the audio wavelength instead of the frequency in the calculator.
Frequency (𝑓)
Use this tool to calculate the shear strain produced by shear forces, stresses, and twisting couples in circular shafts.
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Parameter advising for multiple sequence alignment | BMC Bioinformatics | Full Text
Dan DeBlasio1 &
John Kececioglu1
While the multiple sequence alignment output by an aligner strongly depends on the parameter values used for its alignment scoring function (i.e. choice of gap penalties and substitution scores), most users rely on the single default parameter setting. A different parameter setting, however, might yield a much higher-quality alignment for a specific set of input sequences. The problem of picking a good choice of parameter values for a given set of input sequences is called parameter advising. A parameter advisor has two ingredients: (i) a set of parameter choices to select from, and (ii) an estimator that estimates the accuracy of a computed alignment; the parameter advisor then picks the parameter choice from the set whose resulting alignment has highest estimated accuracy.
Our estimator Facet (Feature-based Accuracy Estimator) is a linear combination of real-valued feature functions of an alignment. We assume the feature functions are given as well as the universe of parameter choices from which the advisor's set is drawn. For this scenario we define the problem of learning an optimal advisor by finding the best possible parameter set for a collection of training data of reference alignments. Learning optimal advisor sets is NP-complete [1]. For the advisor sets problem, we develop a greedy
\frac{\ell }{k}
-approximation algorithm that finds near optimal sets of size at most k given an optimal solution of size ℓ<k. For the advisor estimator problem, we have an efficient method for finding the coefficients for the estimator that performs well in practice [2, 3].
Parameter advising
We apply parameter advising to boost the true accuracy of the Opal aligner [4, 5], where the advisor is using parameter sets found by the
\frac{\ell }{k}
-approximation algorithm. Figure 1 shows the accuracy of the advisor for a parameter set of size k = 10, where the benchmarks are assigned to bins based on their accuracy using a default parameter choice; the figure also shows the accuracies when using a single default parameter choice, and an oracle. The number of benchmarks per bin is indicated above the columns. An oracle is an advisor that knows the true accuracy of an alignment; its accuracy is shown by the dotted line, which gives the performance of a perfect advisor. Notice that in many cases the performance of the estimator is close to the oracle. This is most clear on the bin which has lowest average accuracy, where advising increases the average accuracy by almost 20% compared to using a single default parameter.
Advising accuracy of Facet within benchmark bins.
Figure 2 shows the average advising accuracy for parameter sets of various cardinalities using as the estimator Facet [3], TCS [6], MOS [7], and PredSP [8], where in the average, benchmark bins contribute equally. The vertical axis is advising accuracy on the testing data, averaged over all benchmarks and all folds using 12-fold cross-validation. The horizontal axis is the cardinality k of the greedy advisor set. Greedy advisor set found by the approximation algorithm are augmented from the exact set of cardinality ℓ = 1 (namely, the best single parameter choice). Notice that Facet (the topmost curve in the plot) continues to increase in advising accuracy up to cardinality k = 6. Notice also that while all of the advisors reach a plateau, for Facet this occurs at a greater cardinality and accuracy than for other estimators.
Average advising accuracy of estimators on sets of varying cardinality.
Our tool Facet (Feature-based Accuracy Estimator) [9] is an easy-to-use, open-source utility for estimating the accuracy of a protein multiple sequence alignment. Facet evaluates the estimated accuracy of a computed alignment as a linear combination of real-valued feature functions. We considered 12 features of which we found an optimal subset of 5 that provide the best performance for alignment advising. Many of the most useful features utilize information about protein secondary structure. We find coefficients by fitting the difference in estimator values to the difference in true accuracy for pairs of examples where the correct alignment is known. This "difference fitting" approach is computationally efficient and yields an estimator that works well for advising.
Facet is open-source software that allows users to estimate accuracy as either (1) a stand alone tool, or (2) a software library that can be integrated into a pre-existing Java application. The implementation provides optimized default coefficients and features. These coefficients may also be specified manually and new features can also be added. Figure 3 shows a simple example of using Facet within a Java application to choose between two alignments of the same set of sequences. The secondary structure predictions are computed on the unaligned sequences and can be reused between the two alignments.
Example of invoking Facet in Java.
The Facet website provides parameter sets that can be used with the Opal aligner (namely substitution matrices and affine gap penalties), as well as scripts for structure prediction.
While the new problem of learning optimal parameter sets for an advisor is NP-complete, in practice our greedy approximation algorithm efficiently learns parameter sets that are remarkably close to optimal. Moreover, these parameter sets significantly boost the accuracy of an aligner compared to a single default parameter choice, when advising using the best accuracy estimators from the literature.
DeBlasio DF, Kececioglu JD: Learning Parameter Sets for Alignment Advising. Proceedings of the 5th ACM Conference on Bioinformatics, Computational Biology and Health Informatics (ACM-BCB). 2014, 10.1145/2649387.2649448.
DeBlasio DF, Wheeler TJ, Kececioglu JD: Estimating the accuracy of multiple alignments and its use in parameter advising. Proceedings of the 16th Conference on Research in Computational Molecular Biology (RECOMB). 2012, 45-59. 10.1007/9 78-3-642-29627- 7_5.
Kececioglu JD, DeBlasio DF: Accuracy Estimation and Parameter Advising for Protein Multiple Sequence Alignment. Journal of Computational Biology. 2013, 20 (4): 259-279. 10.1089/cmb.2013.0007.
Wheeler TJ, Kececioglu JD: Multiple alignment by aligning alignments. Bioinformatics. 2007, 23 (13): 559-68. 10.1093/bioinformatics/btm226.
Wheeler TJ, Kececioglu JD: Opal: multiple sequence alignment software, Version 2.1.0. 2012, [http://opal.cs.arizona.edu]
Chang JM, Tommaso PD, Notredame C: TCS: A new multiple sequence alignment reliability measure to estimate alignment accuracy and improve phylogenetic tree reconstruction. Molecular Biology and Evolution. 2014, 10.1093/molbev/msu117.
Lassmann T, Sonnhammer ELL: Automatic assessment of alignment quality. Nucleic Acids Research. 2005, 33 (22): 7120-7128. 10.1093/nar/gki1020.
Ahola V, Aittokallio T, Vihinen M, Uusipaikka E: Model-based prediction of sequence alignment quality. Bioinformatics. 2008, 24 (19): 2165-2171. 10.1093/bioinformatics/btn414.
DeBlasio DF, Kececioglu JD: Facet: software for accuracy estimation of protein multiple sequence alignments, Version 1.1. 2014, [http://facet.cs.arizona.edu]
Dan DeBlasio & John Kececioglu
Correspondence to Dan DeBlasio.
DeBlasio, D., Kececioglu, J. Parameter advising for multiple sequence alignment. BMC Bioinformatics 16, A3 (2015). https://doi.org/10.1186/1471-2105-16-S2-A3
Optimal Advisor
Protein Multiple Sequence
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Return on Investment (ROI) is a popular profitability metric used to evaluate how well an investment has performed.
ROI is expressed as a percentage and is calculated by dividing an investment's net profit (or loss) by its initial cost or outlay.
ROI can be used to make apples-to-apples comparisons and rank investments in different projects or assets.
The return on investment (ROI) formula is as follows:
\begin{aligned} &\text{ROI} = \dfrac{\text{Current Value of Investment}-\text{Cost of Investment}}{\text{Cost of Investment}}\\ \end{aligned}
ROI=Cost of InvestmentCurrent Value of Investment−Cost of Investment
"Current Value of Investment” refers to the proceeds obtained from the sale of the investment of interest. Because ROI is measured as a percentage, it can be easily compared with returns from other investments, allowing one to measure a variety of types of investments against one another.
ROI is a popular metric because of its versatility and simplicity. Essentially, ROI can be used as a rudimentary gauge of an investment’s profitability. This could be the ROI on a stock investment, the ROI a company expects on expanding a factory, or the ROI generated in a real estate transaction.
The calculation itself is not too complicated, and it is relatively easy to interpret for its wide range of applications. If an investment’s ROI is net positive, it is probably worthwhile. But if other opportunities with higher ROIs are available, these signals can help investors eliminate or select the best options. Likewise, investors should avoid negative ROIs, which imply a net loss.
For example, suppose Jo invested $1,000 in Slice Pizza Corp. in 2017 and sold the shares for a total of $1,200 one year later. To calculate the return on this investment, divide the net profits ($1,200 - $1,000 = $200) by the investment cost ($1,000), for a ROI of $200/$1,000, or 20%.
With this information, one could compare the investment in Slice Pizza with any other projects. Suppose Jo also invested $2,000 in Big-Sale Stores Inc. in 2014 and sold the shares for a total of $2,800 in 2017. The ROI on Jo’s holdings in Big-Sale would be $800/$2,000, or 40%.
Examples like Jo's (above) reveal some limitations of using ROI, particularly when comparing investments. While the ROI of Jo's second investment was twice that of the first investment, the time between Jo’s purchase and the sale was one year for the first investment but three years for the second.
Jo could adjust the ROI of the multi-year investment accordingly. Since the total ROI was 40%, to obtain the average annual ROI, Jo could divide 40% by 3 to yield 13.33% annualized. With this adjustment, it appears that although Jo’s second investment earned more profit, the first investment was actually the more efficient choice.
ROI can be used in conjunction with the rate of return (RoR), which takes into account a project’s time frame. One may also use net present value (NPV), which accounts for differences in the value of money over time, due to inflation. The application of NPV when calculating the RoR is often called the real rate of return.
Recently, certain investors and businesses have taken an interest in the development of a new form of the ROI metric, called "social return on investment," or SROI. SROI was initially developed in the late 1990s and takes into account broader impacts of projects using extra-financial value (i.e., social and environmental metrics not currently reflected in conventional financial accounts).
SROI helps understand the value proposition of certain environmental social and governance (ESG) criteria used in socially responsible investing (SRI) practices. For instance, a company may decide to recycle water in its factories and replace its lighting with all LED bulbs. These undertakings have an immediate cost that may negatively impact traditional ROI—however, the net benefit to society and the environment could lead to a positive SROI.
There are several other new variations of ROI that have been developed for particular purposes. Social media statistics ROI pinpoints the effectiveness of social media campaigns—for example how many clicks or likes are generated for a unit of effort. Similarly, marketing statistics ROI tries to identify the return attributable to advertising or marketing campaigns.
So-called learning ROI relates to the amount of information learned and retained as a return on education or skills training. As the world progresses and the economy changes, several other niche forms of ROI are sure to be developed in the future.
Return on investment (ROI) is calculated by dividing the profit earned on an investment by the cost of that investment. For instance, an investment with a profit of $100 and a cost of $100 would have a ROI of 1, or 100% when expressed as a percentage. Although ROI is a quick and easy way to estimate the success of an investment, it has some serious limitations. For instance, ROI fails to reflect the time value of money, and it can be difficult to meaningfully compare ROIs because some investments will take longer to generate a profit than others. For this reason, professional investors tend to use other metrics, such as net present value (NPV) or the internal rate of return (IRR).
Historically, the average ROI for the S&P 500 has been about 10% per year. Within that, though, there can be considerable variation depending on the industry. For instance, during 2020, many technology companies generated annual returns well above this 10% threshold. Meanwhile, companies in other industries, such as energy companies and utilities, generated much lower ROIs and in some cases faced losses year-over-year. Over time, it is normal for the average ROI of an industry to shift due to factors such as increased competition, technological changes, and shifts in consumer preferences.
World Health Organization. "Social Return on Investment," Pages 2-4.
DQYDJ. "S&P 500 Historical Return Calculator."
Fortune. "The Best Stocks of 2020 Have Made Pandemic Investors Even Richer."
Opportunity cost is the potential loss from a missed opportunity—the result of choosing one alternative and forgoing another.
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Examine the integrals below. Consider the multiple tools available for integrating and use the best strategy. After evaluating each integral, write a short description of your method.
\int _ { 1 } ^ { 4 } \sqrt { 1 + \frac { 9 } { 4 } x } d x
\int\Big(1+w\Big)^{1/2}dw=\frac{2}{3}\Big(1+w\Big)^{3/2}
\int 2 x \operatorname { csc } ( x ^ { 2 } ) \operatorname { cot } ( x ^ { 2 } ) d x
\frac{d}{dx}\csc(x)=-\csc(x)\cot(x)
Use substitution. Let
u = x^2
\int \frac { e ^ { 1 / x } } { 2 x ^ { 2 } } d x
u =\frac{1}{x}
\int _ { - 10 } ^ { 10 } \operatorname { cos } 0 d x
\operatorname{cos}(0)
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What is the effectiveness-NTU method?
How to use the effectiveness-NTU calculator
How do I determine the effectiveness of a heat exchanger?
How do I calculate the NTU of a heat exchanger?
NTU formulas for different types of heat exchangers
Effectiveness formulas for different types of heat exchangers
With Omni's effectiveness-NTU calculator, you'll be able to execute design and performance calculations of heat exchangers. With the design calculations, you can determine the heat transfer area (A) for a given set of conditions. Use the performance calculation to find the outlet temperatures (Tco and Tho) of the fluids and the actual heat transfer rate (q).
If you'd like to learn a bit more about this method, we invite you to keep reading and find out about:
What is the effectiveness-NTU method;
Its advantages over the LMTD method;
How to calculate the effectiveness of a heat exchanger;
The formula for the effectiveness of a heat exchanger;
How to calculate the NTU of a heat exchanger; and
The NTU and ε formulas for different heat exchangers configurations.
Similar to the LMTD (log mean temperature difference), the effectiveness-NTU method is a method used to analyze heat exchangers. This one is preferred when the outlet temperatures of the fluids are unknown, since, in these cases, the LMTD requires a cumbersome iterative solution.
The term effectiveness (ε) is a dimensionless indicator that relates the actual heat transfer rate (q) to the maximum possible heat transfer rate (qmax) that could occur for a particular heat exchanger and a particular set of fluids. The formula for the effectiveness of a heat exchanger is given by the ratio of these heats:
\footnotesize \varepsilon = \dfrac {q}{q_{max}}
The actual heat rate
q
is determined in terms of the associated properties as:
\begin{align*} \footnotesize q = C_{c} \cdot (T_{co} - T_{ci}) \\ \footnotesize = C_{h} \cdot (T_{hi} - T_{ho}) \end{align*}
T_{ci}
T_{co}
- Inlet and outlet temperatures of the cold fluid, respectively;
C_c
- Heat capacity of the cold fluid;
T_{hi}
T_{ho}
- Inlet and outlet temperatures of the hot fluid, respectively; and
C_h
- Heat capacity of the hot fluid.
In order to calculate the maximum possible heat transfer rate (qmax) that could happen in a given heat exchanger, the following expression is typically used:
\footnotesize q_{max} = C_{min} \cdot (T_{hi} - T_{ci})
C_{min}
corresponds to the minimum value of the heat capacities, i.e., whichever is smaller of
C_c
C_h
This heat rate is calculated using the difference between the inlet temperature of the hot fluid,
T_{hi}
, and the inlet temperature of the cold fluid,
T_{ci}
. This represents the maximum temperature difference in the system.
q_{max}
calculated with
C_{min}
and not with
C_{max}
? 🤔 The reason for this is that the fluid with the lowest heat capacity is the one that can experience the maximum temperature change. This could mean receiving or giving heat, whether it's the cold or the hot fluid.
The number of transfer units (NTU) is another dimensionless parameter used in the effectiveness-NTU method. It is defined in terms of the overall heat transfer coefficient,
U
, the area of heat transfer,
A
, and the minimum heat capacity,
C_{min}
\footnotesize NTU = \dfrac {U\cdot A}{C_{min}}
Finally, another dimensionless parameter is used in the effectiveness-NTU method. This one is known as the ratio of the heat capacities, (Cr), which is given by:
\footnotesize C_r = \dfrac {C_{min}}{C_{max}}
With the effectiveness-NTU calculator, you'll be able to perform either design calculations or performance assessments of a specific heat exchanger.
The objective of a design calculation is to find the total area of heat transfer (A) required for a given set of inlet and outlet temperatures. With the performance calculation, you'll be able to find the actual heat transfer rate (q) and the outlet temperatures of both the cold (Tco) and hot (Tho) fluids.
With these distinctions in mind, let's see how to use the effectiveness-NTU calculator for both cases.
Design calculation - determine area (A)
To perform a design calculation, follow these steps:
Select the Design problem option from the Type of calculation field.
On the row below, choose the type of heat exchanger that you'd like to evaluate.
Next, you'll find two blocks, Cold fluid properties and Hot fluid properties, where you'll be able to input the respective properties of the cold and hot fluids. Here, enter the known properties of the fluids, such as mass, inlet and outlet temperatures, and specific heat.
At this point, the calculator will be able to show you the results for the actual heat transfer rate (q), the maximum heat transfer rate (qmax), and the effectiveness (ε) of the chosen device.
Finally, in the Area of heat transfer block, enter the value of the overall heat coefficient (U). Once you've input it, the calculator will be able to determine the NTU value and compute the area of heat transfer (A).
Performance assessment - calculate heat transfer rate (q) and outlet temperatures (Tco and Tho)
On the other hand, if you'd like to execute the performance calculations, this is what you have to do:
On the Type of calculation row, select the Performance calculation option.
On the field below, choose the type of heat exchanger that you'd like to assess.
Next, on the cold and hot fluids properties' blocks Cold fluid properties and Hot fluid properties, enter the respective known properties of the substances: mass, inlet temperature, and specific heat.
Next, on the Area of heat transfer block, input the values for the overall heat coefficient (U) and area of heat transfer (A).
After all these values have been entered, the calculator will show you the results for the outlet temperatures (Tco and Tho) and the actual heat transfer rate (q).
🙋 The effectiveness-NTU calculator also shows the result of other in-between steps, such as the value for the effectiveness (ε), the NTU, and the ratio of the heat capacities (Cr). This way, you can check your results if you're doing the calculations yourself 😉
To determine the effectiveness of a heat exchanger, there're two possibilities:
From fluid's properties, calculate the maximum (qmax) and actual heat transfer (q).
Determine the ratio between the heat capacities of the fluids, Cr = Cmin / Cmáx.
Calculate the effectiveness as the ratio of the heats, ε = q/qmax.
Determine Cr as: Cr = Cmin / Cmáx.
Calculate NTU using the relationship: NTU = U x A / Cmin
With the values of NTU and Cmin, use the ε formula or curves of the respective heat exchanger to calculate ε.
If you don't know the value of the overall heat transfer coefficient (U), you can use our heat transfer coefficient calculator to determine it.
To calculate the NTU of a heat exchanger, there're two possible situations:
From the fluid's properties, calculate the maximum (qmax) and actual heat transfer (q) rates.
Calculate the effectiveness as a ratio of the heats, ε = q/qmax.
With the values of ε and Cr, use the NTU formula or curves of the respective heat exchanger to calculate NTU.
Determine Cmin from the fluids' properties.
Calculate NTU with the relationship: NTU = U x A / Cmin
As you've might have noticed, depending upon the type of calculations that you're interested in performing, the formulas to determine the parameters of effectiveness (ε) and number of transfer units (NTU) are not always the short versions that we indicated earlier in the definitions section of this article. Instead, these are calculated from formulas or charts.
In the following sections, you can find the expressions for the most commonly used configurations of heat exchangers.
To determine the number of transfer units (NTU), you could either use a graph or a formula for a specific type of heat exchanger. In the table below, you can find the expressions to calculate NTU:
\footnotesize NTU = \dfrac{-\ln(1- \varepsilon\cdot(1 + C_r))}{1 + C_r}
\footnotesize C_r <1
\footnotesize NTU = \dfrac{1}{C_r -1} \cdot \ln\bigg (\dfrac{\varepsilon - 1}{\varepsilon C_r - 1}\bigg)
\footnotesize C_r = 1
\footnotesize NTU = \dfrac{\varepsilon}{1- \varepsilon}
Shell and tube (One shell pass and 2, 4, ... tube passes)
\footnotesize A = 1 + {C_r}^2
\footnotesize NTU_1 = -A^{-\frac 12} \cdot \ln \bigg(\dfrac{E - 1}{E+1}\bigg)
\footnotesize E = \dfrac{2/\varepsilon_1 - (1 + C_r)}{(1 + {C_r}^2)^{\frac 12}}
Shell and tube (n shell passes and 2n, 4n, ... tube passes)
\footnotesize \varepsilon_1 = \dfrac{F-1}{F-C_r}
F ={\bigg( \dfrac{\varepsilon C_r-1}{\varepsilon - 1}\bigg)}^{\frac 1n}
\footnotesize NTU = n\cdot NTU_1
Cross flow (Cmax (mixed), Cmin (unmixed))
\footnotesize B = \ln(1- \varepsilon C_r)
\footnotesize NTU = - \ln\bigg(1+\dfrac{B}{C_r}\bigg)
Cross flow (Cmax (unmixed), Cmin (mixed))
\footnotesize D = \ln(1- \varepsilon)
\footnotesize NTU = \dfrac{-1}{C_r}\ln[C_r D + 1]
All exchangers (Cr = 0)
\footnotesize NTU = - \ln(1 - \varepsilon)
Similar to the NTU, in order to calculate the effectiveness (ε), you can use a graph or a formula for a given type of heat exchanger. In the table below, you'll find the expressions to calculate ε:
\footnotesize \varepsilon = \dfrac{1-\exp[-NTU(1+C_r)]}{1+C_r}
\footnotesize C_r <1
\footnotesize G = (1-C_r)
\footnotesize \varepsilon = \dfrac{1-\exp[-NTU\cdot G]}{1-C_r \cdot \exp[-NTU\cdot G]}
\footnotesize C_r = 1
\footnotesize \varepsilon = \dfrac{NTU}{1+NTU}
\footnotesize A = 1 +{C_r}^2
\footnotesize H = 1 +C_r
\footnotesize J = \dfrac{1-\exp[-NTU_1\cdot({A}^{\frac 12})]}{1 +\exp[-NTU_1\cdot({A}^{\frac 12})]}
\footnotesize \varepsilon_1 = 2\cdot(H + A \cdot J)
\footnotesize K = {\bigg(\dfrac{1-\varepsilon_1 C_r}{1-\varepsilon_1}\bigg)}^{n}
\footnotesize \varepsilon = \dfrac{K-1}{K-C_r}
Cross flow (both fluids unmixed)
\footnotesize L = NTU^{0.22}[\exp(-C_r\cdot {NTU}^{0.78})
\footnotesize \varepsilon = 1 - \exp\bigg[ \dfrac{L}{C_r} - 1\bigg]
\footnotesize M = \exp[-C_r(1-\exp(-NTU))]
\footnotesize \varepsilon = \dfrac{1-M}{C_r}
\footnotesize N = 1-\exp(-C_r \cdot NTU)
\footnotesize \varepsilon = 1-\exp \bigg(\dfrac{-N}{C_r}\bigg)
\footnotesize \varepsilon = 1-\exp(-NTU)
What are the advantages of the NTU method over the LMTD method?
The main advantage of the NTU method over the LMTD method is that for performance calculations, i.e., determining heat transfer rate and outlet temperatures, the LMTD requires an iterative solution, while with the NTU, the solution can be obtained directly from the formulas.
Can the effectiveness of a heat exchanger be greater than 1?
No, the effectiveness of a heat exchanger can't be greater than 1. The effectiveness (ε) of a heat exchanger should always be a value between zero and one, 0 < ε < 1. The effectiveness represents the ratio between the actual heat rate (q) and the maximum possible heat transfer rate (qmax) that can occur in a heat exchanger for a given set of fluids' conditions (ε = q/qmax).
Cold fluid properties
Mass flow rate (m꜀)
Specific heat capacity (cₚ꜀)
Heat capacity (C꜀)
Inlet temperature (T꜀ᵢ)
Outlet temperature (T꜀ₒ)
Hot fluid properties
Mass flow rate (mₕ)
Specific heat capacity (cₚₕ)
Heat capacity (Cₕ)
Inlet temperature (Tₕᵢ)
Outlet temperature (Tₕₒ)
Actual heat rate (q)
Area of heat transfer
Overall heat transfer coeff. (U)
BTU/(h⋅ft²⋅°F)
Through the acceleration using force and mass calculator, you can find the acceleration of an object when the force and mass values are known.
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How to calculate density from mass and volume using this calculator
Other related density calculators
If you need to go from the volume of an object to its density, our volume to density calculator is just the tool for you. All you need to know is its mass (or weight, as measured here on Earth).
In this short article, we will explain:
How to calculate density from mass and volume.
How do you find the mass with density and volume?
What is the density of a golf ball?
The density of an object or substance is basically the mass per unit volume of space that it occupies. Given two objects that are the same size, the one that weighs more is denser. Mathematically, the density definition is written as:
\rho = \frac{m}{V}
\rho
– Density;
m
– Mass (same as weight on Earth); and
V
So to go from an object's volume to its density, we just need to know its mass.
Next, we'll explain how to use this volume to density calculator.
The volume to density calculator is straightforward to use by following these instructions:
Enter the volume of the object in your choice of units. We support a wide range of volume units. Click the volume unit drop-down menu to see them.
Optional: If you don't know the object's volume but do know its dimensions, click the advanced mode button to reveal an extra calculator section you can use to calculate the volume.
Input the mass or weight of the object.
Enjoy your object's density result instantly, without pressing a calculate button.
And that's how to calculate density with mass and volume. Actually, the calculator can work in many ways. Input any two values, and the other one is calculated for you.
In need of some more density calculators? Here is a list of related calculators:
1.13 g/cm3. Follow these steps to arrive at this answer:
The mass of a golf ball is 45.9 g, and its volume is 40.7 cm3.
Divide the mass by the volume.
So in this case that is 45.9 g / 40.7 cm3 = 1.13 g/cm3.
How can you use density to identify an unknown substance?
Measuring the density of a substance can tell you a lot about what it is made of by matching it to a list of know densities of different materials. This method allowed Archimedes to tell real gold apart from lead made to look like gold. The density of gold is 1.7 times the density of lead.
Have the energy of a wave or photon, and want to find its wavelength? Look no further than the energy to wavelength calculator!
Use the stress calculator to find the stress, strain and modulus of elasticity of an object.
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Cutoff for random lifts of weighted graphs
January 2022 Cutoff for random lifts of weighted graphs
Guillaume Conchon-Kerjan
Guillaume Conchon-Kerjan1
1LPSM, UMR 8001, Université de Paris
We prove the cutoff phenomenon for the random walk on random n-lifts of finite weighted graphs, even when the random walk on the base graph
\mathcal{G}
of the lift is not reversible. The mixing time is w.h.p.
{\mathit{t}}_{\mathrm{mix}}={\mathit{h}}^{-1}log\mathit{n}
, where h is a constant associated to
\mathcal{G}
, namely the entropy of its universal cover. Moreover, this mixing time is the smallest possible among all n-lifts of
\mathcal{G}
. In the particular case where the base graph is a vertex with
\mathit{d}/2
loops, d even, we obtain a cutoff for a d-regular random graph, as did Lubetzky and Sly in (Duke Math. J. 153 (2010) 475–510) (with a slightly different distribution on d-regular graphs, but the mixing time is the same).
The author’s research is supported by a Ph.D. grant of ED386.
The author thanks Jonathan Hermon and an anonymous referee for many useful comments on a first preprint of this paper.
Guillaume Conchon-Kerjan. "Cutoff for random lifts of weighted graphs." Ann. Probab. 50 (1) 304 - 338, January 2022. https://doi.org/10.1214/21-AOP1534
Received: 1 August 2019; Revised: 1 July 2020; Published: January 2022
Keywords: Cutoff , periodic trees , Random walks
Guillaume Conchon-Kerjan "Cutoff for random lifts of weighted graphs," The Annals of Probability, Ann. Probab. 50(1), 304-338, (January 2022)
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Game mechanics - Tears of Themis Wiki
1.4.1.1 playerAttackValue
1.4.1.2 enemyAttackValue
1.4.2.1 playerDefenceValue
1.4.2.2 enemyDefenceValue
2 Story Mechanics
2.2 Clue Analysis
Debates[edit | edit source | hide | hide all]
Debates are the "battle" stages in Tears of Themis. In debates, there will be an enemy who will have various arguments with HP. The player attacks the enemy by selecting their cards to use against enemy's arguments and lower the enemy's HP. Each card selection is a turn, and the player has a limited amount of turns to complete the stage. Additionally, every time a player selects a card to use, the enemy's "thought cloud bubble" will charge. Once it's fully charged, the enemy will attack and lower the player's HP.
If a player runs out of cards, they can refresh the deck at the cost of 1 turn.
Once all arguments have been defeated, the player wins stage. Players can be defeated though if they:
Run out of turns
The player's HP reaches 0
Attributes[edit | edit source | hide]
Arguments and cards will have different attributes: Intuition, Logic, Empathy, or Null (no typing). The attribute of the card you select will affect how much damage your attack does. Strong attacks will deal 1.5x damage while weak attacks will do 0.5x damage. Attacks that are neutral or against Null will do 1.0x damage.
Empathy Intuition
Logic Empathy
One easy way to remember typings are by color and converting them into elements used in other games. For example, Logic is blue and correlates to Water, which beats Empathy's red (Fire), but weak against Intuition's green (Grass).
See also: Cards
Skills[edit | edit source | hide]
Cards will also have skills. Some cards have skills which passively take effect, and others have skills which take effect when a player selects them.
Cards will affect Influence (Attack stat) or Defense such as boosting your own Influence/Defense stats or lowering the enemy's Influence/Defense stats to boost damage/lower damage you take.
Some cards will even have skills that passively boost the base stats of cards, such as Formidable cards that boost Influence when equipped from the Support deck.
Damage Calculations[edit | edit source | hide]
See also: Deck#Power Calculations, Cards#Power Calculations
Datamined Content Ahead!
The following page has info datamined from the game, which may not be accurate and also subject to change. Content written with datamined info may potentially also be interpreted incorrectly than may actually happen in-game.
{\displaystyle damage = attackValue * (1 - defenceValue) * attributeAddRate * giftSkillAddRate?}
Attack[edit | edit source | hide]
playerAttackValue[edit | edit source | hide]
{\displaystyle playerAttackValue = (cardInfluence * 2 + deckTotalInfluence) / 5 * (1 + cardGroupAddRate? + playerSkillAddRate?) * (1 + buffAddRate)}
enemyAttackValue[edit | edit source | hide]
{\displaystyle enemyAttackValue = enemyInfluence * (1 + buffAddRate)}
Defense[edit | edit source | hide]
playerDefenceValue[edit | edit source | hide]
{\displaystyle defenceValue = deckTotalDefence * (1 + totalSupportDeckDefense) * (1 + playerBuffAddRate)}
{\displaystyle playerDefenceValue = defenceValue / (30000 + defenceValue)}
enemyDefenceValue[edit | edit source | hide]
{\displaystyle defenceValue = bossDefence * (1 + bossBuffAddRate)}
{\displaystyle enemyDefenceValue = defenceValue / (30000 + defenceValue)}
Story Mechanics[edit | edit source | hide]
The story is similar to visual novel where dialogue plays on the screen with character sprites in the background. The story will sometimes present the player with choices. These choices will not have any major effect on the story, will only slightly change the dialogue, and can always be replayed.
In addition to the visual novel portions of the story, the game also has investigative interactive sections of the game as well.
Questioning[edit | edit source | hide]
There are story segments where players must question a character and must find key evidence from the questioning. The questions involve selectable phrases that appear on the screen that will then trigger questioning and dialogue related to that phrase. Some dialogue reveals important information related to the case that will be logged as key evidence. Some phrases can also unlock additional phrases that the player can ask the character.
Clue Analysis[edit | edit source | hide]
Clue Analysis involves inspecting an item, looking for spots of interest, and trying to unlock clues related to the case from the item. The item is in a 3D space, so players can revolve the item in different directions to try to look for clues.
Inspection[edit | edit source | hide]
Inspection consists of analyzing a character for features that stand out about them. Tapping a key section of the character can then either trigger dialogue or show a zoomed-in pop up of the area where you can further inspect the character's. Some zoom-in areas can consist of more than one clue.
If having trouble finding an inspection area, tap the magnifying glass in the top left corner to show the areas of interest.
Investigation[edit | edit source | hide]
During Investigation, a player will be presented with a scene where they will need to look for evidence and items of interest by tapping on them. There can also be items you can in the scene which trigger dialogue but may not count as evidence.
If having trouble finding items, tap the magnifying glass in the top left corner to show key evidence items.
Trials[edit | edit source | hide]
At the end of most episodes, players will then take all the evidence they've gathered to go to court and aim to win for their client.
As the trial progresses, evidence will be asked to be shown to prove certain points. Selecting the wrong evidence will not penalize the player, and they can choose to select an evidence as many times as they want.
There will also be some parts of the trial where the player may be asked to choose between two dialogue options. For most dialogue choices, any of the options is usually fine as you can either re-select or the trial will continue the same with slightly different dialogue for one or two lines.
Episode 1's Trial case is an example of this, where the dialogue differs but the trial result is the same. Episode 1's Trial is a bit of an exception though in that the player's choice will also affect the dialogue for the next chapter as well. The trial can still be replayed though to see the other result.
Retrieved from "https://tot.wiki/w/index.php?title=Game_mechanics&oldid=11932"
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Common parameters - Training parameters | CatBoost
\alpha = 0.1
Depends on the class:
CatBoostClassifier: Logloss if the target_border parameter value differs from None. Otherwise, the default loss function depends on the number of unique target values and is either set to Logloss or MultiClass.
CatBoost and CatBoostRegressor: RMSE
\alpha = 0.1
None (do not output additional metric values)
The default value is defined automatically for Logloss, MultiClass & RMSE loss functions depending on the number of iterations if none of parameters leaf_estimation_iterations, --leaf-estimation-method,l2_leaf_reg is set. In this case, the selected learning rate is printed to stdout and saved in the model.
The default value depends on objective, task_type, bagging_temperature and sampling_unit:
When the objective parameter is QueryCrossEntropy, YetiRankPairwise, PairLogitPairwise and the bagging_temperature parameter is not set: Bernoulli with the subsample parameter set to 0.5.
Neither MultiClass nor MultiClassOneVsAll, task_type = CPU and sampling_unit = Object: MVS with the subsample parameter set to 0.8.
[0; \inf)
w_{i}
o_{i}
w_{j}
g_{j}
o_{i_{j}}
g_{j}
\infty
The minimal number of trees for the best model is not set
GPU — Any integer up to 8 for pairwise modes (YetiRank, PairLogitPairwise, and QueryCrossEntropy), and up to 16 for all other loss functions.
It is assumed that all passed values are feature names if at least one of the passed values can not be converted to a number or a range of numbers. Otherwise, it is assumed that all passed values are feature indices.
For example, use the following construction if features indexed 1, 2, 7, 42, 43, 44, 45, should be ignored: [1,2,7,42,43,44,45]
For example, if training should exclude features with the identifiers 1, 2, 7, 42, 43, 44, 45, use the following construction: 1:2:7:42-45.
False (not used; generates random permutations)
float (0;1]
None (set to 1)
The file is not loaded, the values are generated
Refer to the file format description.
The file is not saved
Default value differs depending on the dataset size and ranges from 1 to 256 inclusively
None (Depends on the training objective)
When leaf_estimation_iterations for the Command-line version is set to n, the leaf estimation iterations are calculated as follows: each iteration is either an addition of the next step to the leaf value, or it's a scaling of the leaf value. Scaling counts as a separate iteration. Thus, it is possible that instead of having n gradient steps, the algorithm makes a single gradient step that is reduced n times, which means that it is divided by
2\cdot n
1+\epsilon
False — Use only а fraction of the fold for calculating the approximated values. The size of the fraction is calculated as follows: \frac{1}{\mbox{X}}, where X is the specified coefficient for changing the length of folds. This mode is faster and in rare cases slightly less accurate
For imbalanced datasets with binary classification the weight multiplier can be set to 1 for class 0 and to
\left(\frac{sum\_negative}{sum\_positive}\right)
for class 1.
For example, class_weights=[0.1, 4]multiplies the weights of objects from class 0 by 0.1 and the weights of objects from class 1 by 4.
If class labels are not standard consecutive integers [0, 1 ... class_count-1], use the dict or collections.OrderedDict type with label to weight mapping.
For example, class_weights={'a': 1.0, 'b': 0.5, 'c': 2.0} multiplies the weights of objects with class label a by 1.0, the weights of objects with class label b by 0.5 and the weights of objects with class label c by 2.0.
The dictionary form can also be used with standard consecutive integers class labels for additional readability. For example: class_weights={0: 1.0, 1: 0.5, 2: 2.0}.
Class labels are extracted from dictionary keys for the following types of class_weights:
collections.OrderedDict (when the order of classes in the model is important)
The class_names parameter can be skipped when using these types.
Do not use this parameter with auto_class_weights and scale_pos_weight.
For example, class_weights <- c(0.1, 4) multiplies the weights of objects from class 0 by 0.1 and the weights of objects from class 1 by 4.
The quantity of class weights must match the quantity of class names specified in the --class-names parameter and the number of classes specified in the --classes-count parameter.
\left(\frac{sum\_negative}{sum\_positive}\right)
<value for class 1>,..,<values for class N>
CW_k=\displaystyle\frac{max_{c=1}^K(\sum_{t_{i}=c}{w_i})}{\sum_{t_{i}=k}{w_{i}}}
CW_k=\sqrt{\displaystyle\frac{max_{c=1}^K(\sum_{t_i=c}{w_i})}{\sum_{t_i=k}{w_i}}}
Do not use this parameter with class_weights and scale_pos_weight.
For imbalanced datasets, the weight multiplier can be set to
\left(\frac{sum\_negative}{sum\_positive}\right)
Do not use this parameter with auto_class_weights and class_weights.
Depends on the processing unit type, the number of objects in the training dataset and the selected learning mode
Specifies options:
Langevin: true,
DiffusionTemperature: objects in learn pool count,
ModelShrinkRate: 1 / (2. * objects in learn pool count).
Zero constraints for features at the end of the list may be dropped.
In monotone_constraints = "(1,0,-1)"an increasing constraint is set on the first feature and a decreasing one on the third. Constraints are disabled for all other features.
Set constraints individually for each required feature as an array or a dictionary (the number of features is n).
[<constraint_0>, <constraint_2>, .., <constraint_n-1>]
{"<feature index or name>":<constraint>, .., "<feature index or name>":<constraint>}
monotone_constraints = [1, 0, -1]
These dictionary examples
monotone_constraints = {"Feature2":1,"Feature4":-1}
monotone_constraints = {"2":1, "4":-1}
Set the multiplication weight individually for each required feature as an array or a dictionary (the number of features is n).
[<feature-weight_0>, <feature-weight_2>, .., <feature-weight_n-1>]
{"<feature index or name>":<weight>, .., "<feature index or name>":<weight>}
feature_weights = [0.1, 1, 3]
feature_weights = {"Feature2":1.1,"Feature4":0.3}
feature_weights = {"2":1.1, "4":0.3}
These examples first_feature_use_penalties parameter:
Set the penalty individually for each required feature as an array or a dictionary (the number of features is n).
[<feature-penalty_0>, <feature-penalty_2>, .., <feature-penalty_n-1>]
{"<feature index or name>":<penalty>, .., "<feature index or name>":<penalty>}
Array examples.
first_feature_use_penalties = [0.1, 1, 3]
per_object_feature_penalties = [0.1, 1, 3]
first_feature_use_penalties = {"Feature2":1.1,"Feature4":0.1}
first_feature_use_penalties = {"2":1.1, "4":0.1}
per_object_feature_penalties = {"Feature2":1.1,"Feature4":0.1}
per_object_feature_penalties = {"2":1.1, "4":0.1}
--model-shrink-mode for the Command-line version
--monotone-constraints for the Command-line version
1 - model\_shrink\_rate \cdot learning\_rate {,}
model\_shrink\_rate
learning\_rate
is the value of the --learning-ratefor the Command-line version parameter
1 - \frac{model\_shrink\_rate}{i} {,}
model\_shrink\_rate
is the identifier of the iteration.
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Binocular_disparity Knowpia
A similar disparity can be used in rangefinding by a coincidence rangefinder to determine distance and/or altitude to a target. In astronomy, the disparity between different locations on the Earth can be used to determine various celestial parallax, and Earth's orbit can be used for stellar parallax.
Figure 1. Definition of binocular disparity (far and near).
Human eyes are horizontally separated by about 50–75 mm (interpupillary distance) depending on each individual. Thus, each eye has a slightly different view of the world around. This can be easily seen when alternately closing one eye while looking at a vertical edge. The binocular disparity can be observed from apparent horizontal shift of the vertical edge between both views.
At any given moment, the line of sight of the two eyes meet at a point in space. This point in space projects to the same location (i.e. the center) on the retinae of the two eyes. Because of the different viewpoints observed by the left and right eye however, many other points in space do not fall on corresponding retinal locations. Visual binocular disparity is defined as the difference between the point of projection in the two eyes and is usually expressed in degrees as the visual angle.[1]
The term "binocular disparity" refers to geometric measurements made external to the eye. The disparity of the images on the actual retina depends on factors internal to the eye, especially the location of the nodal points, even if the cross section of the retina is a perfect circle. Disparity on retina conforms to binocular disparity when measured as degrees, while much different if measured as distance due to the complicated structure inside eye.
Figure 1: The full black circle is the point of fixation. The blue object lies nearer to the observer. Therefore, it has a "near" disparity dn. Objects lying more far away (green) correspondingly have a "far" disparity df. Binocular disparity is the angle between two lines of projection . One of which is the real projection from the object to the actual point of projection. The other one is the imaginary projection running through the nodal point of the fixation point.
In computer vision, binocular disparity is calculated from stereo images taken from a set of stereo cameras. The variable distance between these cameras, called the baseline, can affect the disparity of a specific point on their respective image plane. As the baseline increases, the disparity increases due to the greater angle needed to align the sight on the point. However, in computer vision, binocular disparity is referenced as coordinate differences of the point between the right and left images instead of a visual angle. The units are usually measured in pixels.
Tricking neurons with 2D imagesEdit
Figure 2. Simulation of disparity from depth in the plane. (relates to Figure 1)
Brain cells (neurons) in a part of the brain responsible for processing visual information coming from the retinae (primary visual cortex) can detect the existence of disparity in their input from the eyes. Specifically, these neurons will be active, if an object with "their" special disparity lies within the part of the visual field to which they have access (receptive field).[2]
Researchers investigating precise properties of these neurons with respect to disparity present visual stimuli with different disparities to the cells and look whether they are active or not. One possibility to present stimuli with different disparities is to place objects in varying depth in front of the eyes. However, the drawback to this method may not be precise enough for objects placed further away as they possess smaller disparities while objects closer will have greater disparities. Instead, neuroscientists use an alternate method as schematised in Figure 2.
Figure 2: The disparity of an object with different depth than the fixation point can alternatively be produced by presenting an image of the object to one eye and a laterally shifted version of the same image to the other eye. The full black circle is the point of fixation. Objects in varying depths are placed along the line of fixation of the left eye. The same disparity produced from a shift in depth of an object (filled coloured circles) can also be produced by laterally shifting the object in constant depth in the picture one eye sees (black circles with coloured margin). Note that for near disparities the lateral shift has to be larger to correspond to the same depth compared with far disparities. This is what neuroscientists usually do with random dot stimuli to study disparity selectivity of neurons since the lateral distance required to test disparities is less than the distances required using depth tests. This principle has also been applied in autostereogram illusions.
Computing disparity using digital stereo imagesEdit
The disparity of features between two stereo images are usually computed as a shift to the left of an image feature when viewed in the right image.[3] For example, a single point that appears at the x coordinate t (measured in pixels) in the left image may be present at the x coordinate t − 3 in the right image. In this case, the disparity at that location in the right image would be 3 pixels.
Stereo images may not always be correctly aligned to allow for quick disparity calculation. For example, the set of cameras may be slightly rotated off level. Through a process known as image rectification, both images are rotated to allow for disparities in only the horizontal direction (i.e. there is no disparity in the y image coordinates).[3] This is a property that can also be achieved by precise alignment of the stereo cameras before image capture.
Computer algorithmEdit
After rectification, the correspondence problem can be solved using an algorithm that scans both the left and right images for matching image features. A common approach to this problem is to form a smaller image patch around every pixel in the left image. These image patches are compared to all possible disparities in the right image by comparing their corresponding image patches. For example, for a disparity of 1, the patch in the left image would be compared to a similar-sized patch in the right, shifted to the left by one pixel. The comparison between these two patches can be made by attaining a computational measure from one of the following equations that compares each of the pixels in the patches. For all of the following equations, L and R refer to the left and right columns while r and c refer to the current row and column of either images being examined. d refers to the disparity of the right image.
Normalized correlation:
{\displaystyle {\frac {\sum {\sum {L(r,c)\cdot R(r,c-d)}}}{\sqrt {(\sum {\sum {L(r,c)^{2}}})\cdot (\sum {\sum {R(r,c-d)^{2}}})}}}}
Sum of squared differences:
{\displaystyle \sum {\sum {(L(r,c)-R(r,c-d))^{2}}}}
Sum of absolute differences:
{\displaystyle \sum {\sum {\left|L(r,c)-R(r,c-d)\right\vert }}}
The disparity with the lowest computed value using one of the above methods is considered the disparity for the image feature. This lowest score indicates that the algorithm has found the best match of corresponding features in both images.
The method described above is a brute-force search algorithm. With large patch and/or image sizes, this technique can be very time consuming as pixels are constantly being re-examined to find the lowest correlation score. However, this technique also involves unnecessary repetition as many pixels overlap. A more efficient algorithm involves remembering all values from the previous pixel. An even more efficient algorithm involves remembering column sums from the previous row (in addition to remembering all values from the previous pixel). Techniques that save previous information can greatly increase the algorithmic efficiency of this image analyzing process.
Uses of disparity from imagesEdit
Knowledge of disparity can be used in further extraction of information from stereo images. One case that disparity is most useful is for depth/distance calculation. Disparity and distance from the cameras are inversely related. As the distance from the cameras increases, the disparity decreases. This allows for depth perception in stereo images. Using geometry and algebra, the points that appear in the 2D stereo images can be mapped as coordinates in 3D space.
This concept is particularly useful for navigation. For example, the Mars Exploration Rover uses a similar method for scanning the terrain for obstacles.[4] The rover captures a pair of images with its stereoscopic navigation cameras and disparity calculations are performed in order to detect elevated objects (such as boulders).[5] Additionally, location and speed data can be extracted from subsequent stereo images by measuring the displacement of objects relative to the rover. In some cases, this is the best source of this type of information as the encoder sensors in the wheels may be inaccurate due to tire slippage.
Cyclodisparity
^ Qian, N., Binocular Disparity and the Perception of Depth, Neuron, 18, 359–368, 1997.
^ Gonzalez, F. and Perez, R., Neural mechanisms underlying stereoscopic vision, Prog Neurobiol, 55(3), 191–224, 1998.
^ a b Linda G. Shapiro and George C. Stockman (2001). Computer Vision. Prentice Hall, 371–409. ISBN 0-13-030796-3.
^ "The Computer Vision Laboratory ." JPL.NASA.GOV. JPL/NASA, n.d. Web. 5 Jun 2011. <[1]>.
^ "Spacecraft: Surface Operations: Rover ." JPL.NASA.GOV. JPL/NASA, n.d. Web. 5 Jun 2011. http://marsrovers.jpl.nasa.gov/mission/spacecraft_rover_eyes.html.
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Watts to Heat Calculator
How to calculate the watts to heat a substance?
Specific heat at constant pressure (cₚ) vs. constant volume (cᵥ).
Example: Calculating how many watts to heat water during a time interval
Thermal energy is present everywhere, and calculating the watts to heat a substance is essential to know how much resources we'll spend.
We use heat for cooking our food, keeping us warm, drying different objects, and more. For that reason, it's so relevant in our lives.
How to calculate the watts to heat any substance.
How much to run a 1500 watt heater (cost per hour, day, or month).
Before digging into how to calculate the watts to heat something, let's look at the specific heat formula. Specific heat (aka specific heat capacity) is the property of a material that specifies the amount of energy needed to increase its temperature in one unit per unit mass. This formula defines it:
c = \frac{Q}{m \times ΔT}
c – Specific heat;
Q – Energy added (usually in the form of heat) to increase the temperature;
ΔT – Temperature change; and
m – Mass of the object.
If you know the heat capacity and mass of some material, you can predict the energy needed to cause any temperature change to it:
Q = c \times m \times ΔT
If we divide both sides of the equation by time, we get the power necessary (Ẇ) to cause a specific temperature change during a certain time interval (Δt):
Ẇ = \frac{Q}{Δt} = \frac{c × m × ΔT}{Δt}
Causing a temperature change can require different amounts of heat, depending on how we execute the process. Suppose we do it at constant pressure and, therefore, the substance is allowed to expand as we transfer the heat. In that case, we require more heat than if it were at constant volume. That occurs because, at constant pressure, we need additional energy to cause the expansion.
For that reason, in thermodynamics, we define two kinds of specific heats: 1. specific heat at constant pressure (cₚ), and 2. specific heat at constant volume (cᵥ).
As they are considered almost incompressible, and their volume doesn't significantly change, for liquids and solids, cₚ and cᵥ are equal (cₚ = cᵥ). But for gases, it's essential to make the differentiation.
💡 This calculator has c values predefined for some common substances, including the differentiation between cₚ and cᵥ for gases.
Suppose you're interested in how many watts are needed to heat 1 kg of water and increase its temperature by ΔT = 40°C = 40 K. The time to accomplish this task is 10 min, and you found on the internet that the specific heat of the water is 4181.3 J/kg·K. To know the required watts to heat that amount of water, follow these steps:
Input 40 °C or 40 K in the "Change of temperature (ΔT)" box.
Input 1 kg in the "Mass (m)" box.
Select a custom substance and type 4181.3 J/kg·K in the "Specific heat capacity (c)" box.
Input 1 min or 600 s the "Time to heat (t)" box.
That's it. The required watts to heat the water in that time should be 278.75.
You can check the results with the formula:
Ẇ = Q/Δt = (4181.3 J/kg·K × 1kg × 40K)/600s = 278.75 W
💡 You can click on the advanced mode of the calculator to define the temperature change based on an initial and final temperature. But be careful, check that your substance doesn't go through a phase change during this temperature change, as it would require additional energy.
What is the difference between work and power?
The difference between work and power is:
Work means energy transfer associated with a force acting through a distance.
Power is how fast work is applied.
If we exert a force to raise an object, we're applying work to increase its potential energy. The faster we lift it, the higher the power.
If an electromotive force moves electrons in a wire, that's an example of electrical work. A more rapid electron transport implies a higher electric power.
How to calculate the cost of an electric heater?
To calculate an electric heater cost, follow these steps:
Determine your heater's power consumption (i.e., 1.5 kW).
Figure out your local electricity cost (i.e., $0.1563 per kW⋅h).
Multiply the power consumption by the electricity cost, and you'll get the hourly consumption (i.e., 1.5 kW × $0.1563/kW⋅h = $0.23445 per hour.).
To calculate the daily cost, multiply the hourly cost by the number of hours you use the heater a day.
For the monthly cost, multiply the daily cost by the number of days you use the heater a month.
The cost to run a 1500 watt heater per hour is $0.1563, which equals $3.7512 per day and $113 per month for a 24h usage. This answer assumes an average electricity cost of 10.42 cents per kW⋅h.
How to calculate heat from watts?
To calculate heat (actually, temperature change) from watts applied to a substance, use the formula: ΔT = (Δt × Ẇ)/(c × m), where:
ΔT – Temperature change experienced by the substance;
Δt – Time during which we apply the heat.
Ẇ – Power in Watts with which we heat the substance.
c – Specific heat of the substance; and
Time to heat (t)
Power (Ẇ)
A negative power means that we're lowering the temperature of the substance 🥶. We're not providing heat to that substance, but it's transferring heat 🔥 to our environment (heat loss).
This tool calculates the power needed to heat a substance in a specific amount of time. If you're searching for the energy to heat something, independently of the time, you can check our sensible heat calculator.
The mean free path calculator lets you find the mean free path of any particle in an ideal gas.
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Astronomical nutation - Wikipedia
For broader coverage of this topic, see Nutation.
Astronomical nutation is a phenomenon which causes the orientation of the axis of rotation of a spinning astronomical object to vary over time. It is caused by the gravitational forces of other nearby bodies acting upon the spinning object. Although they are caused by the same effect operating over different timescales, astronomers usually make a distinction between precession, which is a steady long-term change in the axis of rotation, and nutation, which is the combined effect of similar shorter-term variations.[1]
An example of precession and nutation is the variation over time of the orientation of the axis of rotation of the Earth. This is important because the most commonly used frame of reference for measurement of the positions of astronomical objects is the Earth's equator — the so-called equatorial coordinate system. The effect of precession and nutation causes this frame of reference itself to change over time, relative to an arbitrary fixed frame.
Nutation is one of the corrections which must be applied to obtain the apparent place of an astronomical object. When calculating the position of an object, it is initially expressed relative to the mean equinox and equator — defined by the orientation of the Earth's axis at a specified date, taking into account the long-term effect of precession, but not the shorter-term effects of nutation. It is then necessary to apply a further correction to take into account the effect of nutation, after which the position relative to the true equinox and equator is obtained.
Because the dynamic motions of the planets are so well known, their nutations can be calculated to within arcseconds over periods of many decades. There is another disturbance of the Earth's rotation called polar motion that can be estimated for only a few months into the future because it is influenced by rapidly and unpredictably varying things such as ocean currents, wind systems, and hypothesised motions in the liquid nickel-iron outer core of the Earth.
1 Earth's nutation
1.1 Effect on position of astronomical objects
Earth's nutationEdit
It has been suggested that this section be split out into another article titled Earth's nutation. (Discuss) (October 2020)
Precession and nutation are caused principally by the gravitational forces of the Moon and Sun acting upon the non-spherical figure of the Earth. Precession is the effect of these forces averaged over a very long period of time, and a time-varying moment of inertia (If an object is asymmetric about its principal axis of rotation, the moment of inertia with respect to each coordinate direction will change with time, while preserving angular momentum), and has a timescale of about 26,000 years. Nutation occurs because the forces are not constant, and vary as the Earth revolves around the Sun, and the Moon revolves around the Earth. Basically, there are also torques from other planets that cause planetary precession which contributes to about 2% of the total precession. Because periodic variations in the torques from the sun and the moon, the wobbling (nutation) comes into place. You can think of precession as the average and nutation as the instantaneous.
The largest contributor to nutation is the inclination of the orbit of the Moon around the Earth, at slightly over 5° to the plane of the ecliptic. The orientation of this orbital plane varies over a period of about 18.6 years. Because the Earth's equator is itself inclined at an angle of about 23.4° to the ecliptic (the obliquity of the ecliptic,
{\displaystyle \epsilon }
), these effects combine to vary the inclination of the Moon's orbit to the equator by between 18.4° and 28.6° over the 18.6 year period. This causes the orientation of the Earth's axis to vary over the same period, with the true position of the celestial poles describing a small ellipse around their mean position. The maximum radius of this ellipse is the constant of nutation, approximately 9.2 arcseconds.
Smaller effects also contribute to nutation. These are caused by the monthly motion of the Moon around the Earth and its orbital eccentricity, and similar terms caused by the annual motion of the Earth around the Sun.
Effect on position of astronomical objectsEdit
Because nutation causes a change to the frame of reference, rather than a change in position of an observed object itself, it applies equally to all objects. Its magnitude at any point in time is usually expressed in terms of ecliptic coordinates, as nutation in longitude (
{\displaystyle \Delta \psi }
) and nutation in obliquity (
{\displaystyle \Delta \epsilon }
). The largest term in nutation is expressed numerically (in arcseconds) as follows:
{\displaystyle {\begin{aligned}\Delta \psi &=-17.2\sin \Omega \\\Delta \epsilon &=9.2\cos \Omega \end{aligned}}}
{\displaystyle \Omega }
is the ecliptic longitude of the ascending node of the Moon's orbit. By way of reference, the sum of the absolute value of all the remaining terms is 1.4 arcseconds for longitude and 0.9 arcseconds for obliquity.[2]
Spherical trigonometry can then be used on any given object to convert these quantities into an adjustment in the object's right ascension (
{\displaystyle \alpha }
{\displaystyle \delta }
) For objects that are not close to a celestial pole, nutation in right ascension (
{\displaystyle \Delta \alpha }
{\displaystyle \Delta \delta }
) can be calculated approximately as follows:[3]
{\displaystyle {\begin{aligned}\Delta \alpha &=(\cos \epsilon +\sin \epsilon \sin \alpha \tan \delta )\Delta \psi -\cos \alpha \tan \delta \Delta \epsilon \\\Delta \delta &=\cos \alpha \sin \epsilon \Delta \psi +\sin \alpha \Delta \epsilon \end{aligned}}}
Nutation was discovered by James Bradley from a series of observations of stars conducted between 1727 and 1747. These observations were originally intended to demonstrate conclusively the existence of the annual aberration of light, a phenomenon that Bradley had unexpectedly discovered in 1725-6. However, there were some residual discrepancies in the stars' positions that were not explained by aberration, and Bradley suspected that they were caused by nutation taking place over the 18.6 year period of the revolution of the nodes of the Moon's orbit. This was confirmed by his 20-year series of observations, in which he discovered that the celestial pole moved in a slightly flattened ellipse of 18 by 16 arcseconds about its mean position.[4]
Although Bradley's observations proved the existence of nutation and he intuitively understood that it was caused by the action of the Moon on the rotating Earth, it was left to later mathematicians, d'Alembert and Euler, to develop a more detailed theoretical explanation of the phenomenon.[5]
^ Seidelmann, P. Kenneth, ed. (1992). Explanatory Supplement to the Astronomical Almanac. University Science Books. pp. 99–120. ISBN 0-935702-68-7.
^ "NeoProgrammics - Science Computations".
^ Berry, Arthur (1898). A Short History of Astronomy. John Murray. pp. 265–269.
^ Robert E. Bradley. "The Nodding Sphere and the Bird's Beak: D'Alembert's Dispute with Euler". The MAA Mathematical Sciences Digital Library. Mathematical Association of America. Retrieved 21 April 2014.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Astronomical_nutation&oldid=1043135027"
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Zero - Simple English Wikipedia, the free encyclopedia
Zero (0) is a special number.[1] If there are zero things, then there is nothing at all. For example, if a person has zero hats, that means they do not have any hats.
0, zero, "oh" /ˈoʊ/, nought, naught, nil, o, 〇
(zeroth, noughth)
(except itself)
2 Math with zero
3 History of zero
4 The place of zero as a number
7 The numerical digit zero
8 Telling zero and the letter O apart
9 Zeroes of a function
The symbol for the number zero is "0". It is the additive identity of common numbers.[2] This means that if a number is added to 0, then that number would remain unchanged.[3]
Math with zeroEdit
Adding a number to zero results in that number. For example, adding zero to three gives three. In symbols:
Subtracting zero from a number always gives that number. For example, subtracting zero from three gives three. In symbols:
Subtracting a positive number from zero always makes that number negative (or, if a negative number is subtracted from zero, it makes the number positive). In symbols:
Multiplying a number by zero always gives zero. For example, multiplying forty-three by zero gives zero. In symbols:
Dividing zero by a number always gives zero. For example, dividing zero by forty-three gives zero. In symbols:
Any number divided by zero has no answer. In symbols:
43 ÷ 0 has an undefined answer.
In particular, zero divided by zero has no answer. In symbols:
0 ÷ 0 has no answer.
The following table includes all of the above examples along with other operations in a condensed, generalized form (where x represents any number).
Division 0 ÷ x = 0, when x ≠ 0 0 ÷ 5 = 0
Exponentiation 0 x = 0, when x ≠ 0 05 = 0
x 0 = 1, when x ≠ 0 50 = 1
History of zeroEdit
The idea of zero was first thought about in Babylon, India and in Central America at different times. Some places and countries did not know about zero, which may have made it harder for those people to do mathematics. For example, the year after 1 BC is AD 1 (there is no year zero). In India, zero was theorized in the seventh century by the mathematician Brahmagupta.
Over hundreds of years, the idea of zero was passed from country to country, from India and Babylon to other places, like Greece, Persia and the Arab world. The Europeans learned about zero from the Arabs, and stopped using Roman math. This is why numbers are called "Arabic numerals".
The place of zero as a numberEdit
Zero is almost never used as a place number (ordinal number). This means that it is not used like 1, 2, or 3 to indicate the order, or place, of something, like 1st, 2nd, or 3rd. An exception to this is seen in many programming languages.
Some other things about zero:[4]
The number zero is a whole number (counting number).
The number zero is not a positive number.
The number zero is not a negative number, either.
The number zero is a neutral number.
Any number divided by itself equals one, except if that number is zero. In symbols:
0 ÷ 0 = "not a number."
In time, zero means "now". For example, when a person is counting down the time to the start of something, such as a foot race or when a rocket takes off, the count is: "three, two, one, zero (or go)". Zero is the exact time of the start of the race or when the rocket takes off into the sky.
0 as a numberEdit
0 is the integer that precedes the positive 1, and follows −1. In most (if not all) numerical systems, 0 was identified before the idea of 'negative integers' was accepted. It means "courageous one" in hieroglyphics.
Zero is a number which means an amount of null size; that is, if the number of brothers is zero, that means the same thing as having no brothers, and if something has a weight of zero, it has no weight. If the difference between the numbers of pieces in two piles is zero, it means the two piles have an equal number of pieces. Before counting starts, the result can be assumed to be zero; that is the number of items counted before one counts the first item, and counting the first item brings the result to one. And if there are no items to be counted, zero remains the final result.
While mathematicians all accept zero as a number, some non-mathematicians would say that zero is not a number, arguing that one cannot have zero of something. Others say that if one has a bank balance of zero, one has a specific quantity of money in that account, namely none. It is that latter view which is accepted by mathematicians and most others.
Normally speaking, there was no year zero between 1 BC and 1 AD. More specifically, almost all historians leave out the year zero from the proleptic Gregorian and Julian calendars (that is, from the normal calendar used in English-speaking countries), but astronomers include it in these same calendars. However, the phrase Year Zero may be used to describe any event considered so important, that someone might want to start counting years all over again from zero.
0 as a numeralEdit
On the seven-segment displays of calculators, watches, etc., 0 is usually written with six line segments, though on some historical calculator models, it was written with four line segments. The four-segment 0 is not common.
The number zero (as in the "zero brothers" example above) is not the same as the numeral or digit zero, used in numeral systems using positional notation. Successive positions of digits have higher values, so the digit zero is used to skip a position and give appropriate value to the preceding and following digits. A zero digit is not always necessary in a different positional number system. Something called bijective numeration is a possible example of a system without zeroes.
The numerical digit zeroEdit
0 (zero) is also used as a numerical digit used to represent that number in numerals. It is used to hold the place of that digit, because correct placing of digits affects a numeral's value.
In the numeral 10, which stands for one times ten and zero units (or ones).
In the numeral 100, which stands for one times a hundred plus zero tens plus zero units.
Telling zero and the letter O apartEdit
The number 0 and the letter O are both round, though of different widths. The difference is important on a computer. For one thing, a computer will not do arithmetic with the letter O, because it does not know that it should have been a zero.
The oval-shaped zero and circular letter O came into use together on modern character displays. The zero with a dot in the centre seems to have begun as a choice on IBM 3270 controllers (this has the problem that it looks like the Greek letter theta). The slashed zero, looking like the letter O with a diagonal line drawn inside it, is used in old-style ASCII graphic sets that came from the default typewheel on the well-known ASR-33 Teletype. This format causes problems because it looks like the symbol
{\displaystyle \emptyset }
, representing the empty set,[5] as well as for certain Scandinavian languages which use Ø as a letter.
The rule which has the letter O with a slash and the zero without was used at IBM and a few other early mainframe makers; this is even more of a problem for Scandinavians, because it looks like two of their letters at the same time. Some Burroughs/Unisys computers display a zero with a backwards slash. And yet another convention common on early line printers left zero without any extra dots or slashes but added a tail or hook to the letter O so that it resembled an inverted Q or cursive capital letter O.
A German licence plate showing zeroes
The letters used on some European number plates for cars make the two symbols look different. This is done by making the zero rather egg-shaped and the O more circular, but most of all by cutting open the zero on the upper right side, so the circle is not closed any more (as in German plates). The style of letters chosen is called fälschungserschwerende Schrift (abbr.: FE Schrift), meaning "script which is harder to falsify". But those used in the United Kingdom do not make the letter o and the number 0 look different from each other, because there can never be any mistake if the letters are correctly spaced.
In paper writing you do not have to make the 0 (zero) and O (letter O) look different at all. Or you may add a slash across the zero in order to show the difference.
Zeroes of a functionEdit
Functions are explained in the Function (mathematics) article. If the function f(x) = 0, then x is called a zero (or root) of the function f.[6] For example, if the function f(x) is x2 − 1, then the zeroes of the function are +1 and −1, because f(+1) = (+1)2 − 1 = 0, and f(−1) = (−1)2 − 1 = 0.
Zeroes of a function are used because they are another way to talk about solving an equation, which is a main goal in algebra. If we want to solve an equation like x2 = 1, then we can subtract the right-hand side of the equation from both sides, in this case 1. Whatever we get on the left-hand side, in this case x2 − 1, can be called a function f(x). The right-hand side has to be zero, because we subtracted it from itself. So f(x) = 0. Finding the zeroes of this function is the same as solving this equation. In the paragraph before, the zeroes of this function are +1 and −1, so they are the solutions of this equation. We got this equation by subtracting the same thing from both sides, so we also have solutions to the equation we started with, in this case x2 = 1. More generally, if we could find zeroes of functions, we could solve any equation.
↑ Russell, Bertrand (1942). Principles of mathematics (2 ed.). Forgotten Books. p. 125. ISBN 1-4400-5416-9. , Chapter 14, page 125
↑ "Zero". www.mathsisfun.com. Retrieved 2020-09-22.
↑ Weisstein, Eric W. "Zero". mathworld.wolfram.com. Retrieved 2020-09-22.
↑ Weisstein, Eric W. "Root". mathworld.wolfram.com. Retrieved 2020-09-22.
Tapan Kumar Das Gupta: "Der Ursprung des neuzeitlichen Zahlensystems - Entstehung und Verbreitung." Norderstedt 2013. ISBN 978-3-7322-4809-4.
A History of Zero Archived 2008-12-04 at the Wayback Machine
The Discovery of the Zero Archived 2009-05-26 at the Wayback Machine
The History of Algebra Archived 2014-10-09 at the Wayback Machine
Why numbering should start at zero by Edsger Dijkstra
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Zero&oldid=8181258"
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Total Harmonic Distortion - MapleSim Help
Home : Support : Online Help : MapleSim : MapleSim Component Library : Signal Blocks : Mathematical : Functions : Total Harmonic Distortion
The Total Harmonic Distortion component determines the total harmonic distortion (THD) over the given period
\frac{1}{f}
Consider that the input
u
consists of harmonic RMS components
{U}_{1}
{U}_{2}>
{U}_{3}
, etc. The total RMS component is then determined by:
The calculation of the total harmonic distortion is based on the parameter
\mathrm{useFirstHarmonic}
. The default value
\mathrm{useFirstHarmonic}=\mathrm{true}
represents the standard THD calculation used in electrical engineering http://www.electropedia.org/iev/iev.nsf/display?openform&ievref=551-20-13 . The non-default value
\mathrm{useFirstHarmonic}=\mathrm{false}
calculates the THD typically used for the assessment of audio signals. If
\mathrm{useFirstHarmonic}=\mathrm{true}
, the total higher harmonic content (harmonic order numbers > 1) refers to the RMS value of the fundamental wave:
\mathrm{useFirstHarmonic}=\mathrm{false}
, the total higher harmonic content (harmonic order numbers > 1) refers to the total RMS:
In case of a zero input signal or within the first period of calculation, the boolean output signal
\mathrm{valid}
\mathrm{false}
to indicate that the calculation result is not valid. Valid calculations are indicated by
\mathrm{valid}=\mathrm{true}
u
y
\mathrm{valid}
Boolean output signal; true, if output y is valid
f
\mathrm{Hz}
use first harmonic
\mathrm{true}
THD with respect to first harmonic, if true; otherwise with respect to total RMS
useFirstHarmonic
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Interactive Autograder
Below is the autograder. When entering your answers, make sure that they are in all caps and there is a comma between each question. If you skipped the question, it would look like ",,". Do not add any spaces.
Small example: "A,B,CD,,E,E,E"
Autograder is WIP and currently does not work.
The score is based off of two variables: time and accuracy. The time makes up 20% of the overall score, with accuracy the other 80%. The formula can be written as follows:
S = 0.20 \cdot f(t) + 0.80 \cdot P
S
is score,
f(t)
is points from time, and
P
is points from the exam, all in percent.
The score in time is a piecewise function, meant to allot full points to those who use half the max time or less. Afterwards, it linearly decreases until it hits zero points at the max time. The formula is shown as follows:
f(x) = \left\{ \begin{array}{ll} 1 & \quad t \leq \dfrac{t_{MAX}}{2} \\ 1 + \dfrac{t - 0.5 \cdot t_{MAX}}{0.5 \cdot t_{MAX} - t_{MAX}} & \quad t \geq \dfrac{t_{MAX}}{2} \end{array} \right.
t_{MAX}
is the maximum time alloted for the exam. This ensures that "rushing" the exam will award no extra points compared to spending half the alloted time.
The accuracy function is based on a Danish paper, with modifications made to allow for multiple key (correct options) questions. Overall, it results in a value between -1 and 4 being awarded for each question. To get a score, you simply add all the points together and divide by the number of questions times four. The function is as follows:
P(k, c_a, w_a, c) = \left\{ \begin{array}{ll} 0 & \quad \textrm{if } (c_a + w_a) = k \textrm{ or } (c_a + w_a) = 0 \\ \\ \dfrac{P_c - P_w}{P_{tot}} & \quad \textrm{otherwise} \end{array} \right.
For the inputs:
k
is total number of options,
c_a
is number of correctly chosen answers,
w_a
is number of wrongly chosen answers, and
c
is number of keys (correct options). This function awards zero points if all or none of the options are chosen, as they are equivalent. If only some options are chosen, the second equation is used, where
P_c
is the number of points awarded,
P_w
is number of points subtracted, and
P_{tot}
is used to scale the points. This will always be -1 if only wrong options are chosen, and +4 if only right options are chosen. Examples will be covered in a later section.
P_w
P_c(k, w_a) = \left\{ \begin{array}{ll} 0 & \quad \textrm{if } w_a = 0 \\ \\ \dfrac{w}{k - w}\ln\left(\dfrac{k}{w}\right) & \quad \textrm{if } w_a \ge 0 \end{array} \right.
P_c
P_c(k, c_a, c) = \left\{ \begin{array}{ll} 0 & \quad \textrm{if } c_a = 0 \\ \\ \ln\left(\dfrac{k}{c}\right) & \quad \textrm{if } c_a \ge 0 \textrm{ and } c = 1 \\ \\ \ln\left(\dfrac{k}{3 - c_a}\right) & \quad \textrm{if } c_a \ge 0 \textrm{ and } c = 2 \end{array} \right.
Similarly, the equation for
P_{tot}
P_c(k, a, c) = \left\{ \begin{array}{ll} 0 & \quad \textrm{if } c_a = 0 \\ \\ \dfrac{c_a + w_a}{k - (c_a + wa)}\ln\left(\dfrac{c_a + w_a}{k - (c_a + wa)}\right) & \quad \textrm{if } c_a \ge 0 \textrm{ and } c = 1 \\ \\ \dfrac{3 - c_a + w_a}{k - (3 - c_a + wa)}\ln\left(\dfrac{3 - c_a + w_a}{k - (3 - c_a + wa)}\right) & \quad \textrm{if } c_a \ge 0 \textrm{ and } c = 2 \end{array} \right.
The zero removes undefined behavior of the natural log and fraction. The second function defines points when there is a single key. It was provided by the Danish paper on grading multiple choice questions. The third function, if present, flips the funciton by the horizontal axis. It follows the rule
c + 1 - c_a
and can be extended to have more than two keys. However, it may not be reliable for more than two keys.
There's not a lot of cases to cover for this, but I will give a couple examples. Assume that the exam is 60 minutes long, meaning you get full credit if you turn it in within 30 minutes.
Someone submitted the exam after a minute (somehow): 100%
Someone submitted the exam in 30 minutes: 100%
Someone submitted the exam in 31 minutes: 96.67%
Someone submitted the exam in 59 minutes: 3.33%
Accuracy is a lot more interesting. For all examples, assume that there are five options.
First, we will cover the basic cases: none equivalent, all wrong, and all correct. The points awarded is capped at -1 for all wrong answers and +4 for all correct answers.
No options were chosen: 0 points
All options were chosen: 0 points
1 wrong option was chosen: -1 point
2 wrong answers were chosen: -1 point
1 correct options were chosen: +4 points
If both right and wrong answers are chosen, less points are awarded. The idea is that the student got it right, but was not as sure as an all-correct student. This tapers off from the maximum +4 down to 0. For these examples, assume that there is only one correct option:
1 correct and 1 wrong: +2 points
1 correct and 2 wrong: +1.3 points
This gets more complex if there are two correct options. Picking only one of the two options will award less points, then choosing both options. This is compounded by choosing wrong options. For these examples, assume there are two correct options:
1 correct only: +1.5 points
1 correct and 1 wrong: +0.1
If both correct options are chosen and wrong options are additionally picked, it simply follows the "one correct option" examples.
For an interactive view, check out the Desmos.
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Encode input message using orthogonal space-time block code (OSTBC) - Simulink - MathWorks Italia
OSTBC Encoding Algorithms
Encode input message using orthogonal space-time block code (OSTBC)
The OSTBC Encoder block encodes an input symbol sequence using orthogonal space-time block code (OSTBC). The block maps the input symbols block-wise and concatenates the output codeword matrices in the time domain. For more information, see the OSTBC Encoding Algorithms section of this help page.
The block supports time and spatial domains for OSTBC transmission. It also supports an optional dimension, over which the encoding calculation is independent. This dimension can be thought of as the frequency domain. The following illustration indicates the supported dimensions for the inputs and output of the OSTBC Encoder block.
The following table describes the variables.
F The additional dimension; typically the frequency domain. The encoding does not depend on this dimension.
T Input symbol sequence length for the time domain.
On the output, T/R is the symbol sequence length in time domain.
F can be any positive integer. N can be 2, 3 or 4, indicated by Number of transmit antennas. For N = 2, R must be 1. For N = 3 or 4, R can be 3/4 or 1/2, indicated by Rate. The time domain length T must be a multiple of the number of symbols in each codeword matrix. Specifically, for N = 2 or R = 1/2, T must be a multiple of 2 and when R = 3/4, T must be a multiple of 3.
To understand the block’s dimension propagation, refer to the following table.
F = 1 Column vector 2-D
F > 1 2-D 3-D
For information about the data types each block port supports, see the Supported Data Type table on this page. The output signal inherits the data type from the input signal. For fixed-point signals, the complex conjugation may cause overflows which the fixed-point parameter Saturate on integer overflow must handle.
The output signal inherits frame type from the input signal. A column vector input requires either frame-based or sample-based input; otherwise, the input must be sample-based.
The OSTBC Encoder block supports five different OSTBC encoding algorithms. Depending on the selection for Rate and Number of transmit antennas, the block implements one of the algorithms in the following table:
OSTBC Codeword Matrix
\left(\begin{array}{cc}{s}_{1}& {s}_{2}\\ -{s}_{2}^{*}& {s}_{1}^{*}\end{array}\right)
\left(\begin{array}{ccc}{s}_{1}& {s}_{2}& 0\\ -{s}_{2}^{*}& {s}_{1}^{*}& 0\\ 0& 0& {s}_{1}\\ 0& 0& -{s}_{2}^{*}\end{array}\right)
\left(\begin{array}{ccc}{s}_{1}& {s}_{2}& {s}_{3}\\ -{s}_{2}^{*}& {s}_{1}^{*}& 0\\ {s}_{3}^{*}& 0& -{s}_{1}^{*}\\ 0& {s}_{3}^{*}& -{s}_{2}^{*}\end{array}\right)
\left(\begin{array}{cccc}{s}_{1}& {s}_{2}& 0& 0\\ -{s}_{2}^{*}& {s}_{1}^{*}& 0& 0\\ 0& 0& {s}_{1}& {s}_{2}\\ 0& 0& -{s}_{2}^{*}& {s}_{1}^{*}\end{array}\right)
\left(\begin{array}{cccc}{s}_{1}& {s}_{2}& {s}_{3}& 0\\ -{s}_{2}^{*}& {s}_{1}^{*}& 0& {s}_{3}\\ {s}_{3}^{*}& 0& -{s}_{1}^{*}& {s}_{2}\\ 0& {s}_{3}^{*}& -{s}_{2}^{*}& -{s}_{1}\end{array}\right)
In each matrix, its (l, i) entry indicates the symbol transmitted from the ith antenna in the lth time slot of the block. The value of i can range from 1 to N (the number of transmit antennas). The value of l can range from 1 to the codeword block length.
Sets the number of antennas at the transmitter side. The block supports 2, 3, or 4 transmit antennas. The value defaults to 2.
Sets the symbol rate of the code. You can specify either 3/4 or 1/2. This field only appears when using more than 2 transmit antennas. This field defaults to
\frac{3}{4}
for more than 2 transmit antennas. For 2 transmit antennas, there is no rate option and the rate defaults to 1.
For an example of this block in use, see OSTBC Over 3-by-2 Rayleigh Fading Channel . The model shows the use of a rate ¾ OSTBC for 3 transmit and 2 receive antennas with BPSK modulation using independent fading links and AWGN
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Exact_solutions_in_general_relativity Knowpia
Background and definitionEdit
These tensor fields should obey any relevant physical laws (for example, any electromagnetic field must satisfy Maxwell's equations). Following a standard recipe[which?] which is widely used in mathematical physics, these tensor fields should also give rise to specific contributions to the stress–energy tensor
{\displaystyle T^{\alpha \beta }}
.[1] (A field is described by a Lagrangian, varying with respect to the field should give the field equations and varying with respect to the metric should give the stress-energy contribution due to the field.)
{\displaystyle G^{\alpha \beta }=8\pi \,T^{\alpha \beta }.}
In the above field equations,
{\displaystyle G^{\alpha \beta }}
is the Einstein tensor, computed uniquely from the metric tensor which is part of the definition of a Lorentzian manifold. Since giving the Einstein tensor does not fully determine the Riemann tensor, but leaves the Weyl tensor unspecified (see the Ricci decomposition), the Einstein equation may be considered a kind of compatibility condition: the spacetime geometry must be consistent with the amount and motion of any matter or non-gravitational fields, in the sense that the immediate presence "here and now" of non-gravitational energy–momentum causes a proportional amount of Ricci curvature "here and now". Moreover, taking covariant derivatives of the field equations and applying the Bianchi identities, it is found that a suitably varying amount/motion of non-gravitational energy–momentum can cause ripples in curvature to propagate as gravitational radiation, even across vacuum regions, which contain no matter or non-gravitational fields.
Difficulties with the definitionEdit
take any Lorentzian manifold, compute its Einstein tensor
{\displaystyle G^{\alpha \beta }}
, which is a purely mathematical operation
{\displaystyle 8\pi }
declare the resulting symmetric second rank tensor field to be the stress–energy tensor
{\displaystyle T^{\alpha \beta }}
Types of exact solutionEdit
Vacuum solutions:
{\displaystyle T^{\alpha \beta }=0}
; these describe regions in which no matter or non-gravitational fields are present,
Electrovacuum solutions:
{\displaystyle T^{\alpha \beta }}
must arise entirely from an electromagnetic field which solves the source-free Maxwell equations on the given curved Lorentzian manifold; this means that the only source for the gravitational field is the field energy (and momentum) of the electromagnetic field,
Null dust solutions:
{\displaystyle T^{\alpha \beta }}
must correspond to a stress–energy tensor which can be interpreted as arising from incoherent electromagnetic radiation, without necessarily solving the Maxwell field equations on the given Lorentzian manifold,
Fluid solutions:
{\displaystyle T^{\alpha \beta }}
must arise entirely from the stress–energy tensor of a fluid (often taken to be a perfect fluid); the only source for the gravitational field is the energy, momentum, and stress (pressure and shear stress) of the matter comprising the fluid.
Scalar field solutions:
{\displaystyle T^{\alpha \beta }}
must arise entirely from a scalar field (often a massless scalar field); these can arise in classical field theory treatments of meson beams, or as quintessence,
Lambdavacuum solutions (not a standard term, but a standard concept for which no name yet exists):
{\displaystyle T^{\alpha \beta }}
arises entirely from a nonzero cosmological constant.
non-null electrovacuums have Segre type
{\displaystyle \{\,(1,1)(11)\}}
and isotropy group SO(1,1) x SO(2),
null electrovacuums and null dusts have Segre type
{\displaystyle \{\,(2,11)\}}
and isotropy group E(2),
perfect fluids have Segre type
{\displaystyle \{\,1,(111)\}}
and isotropy group SO(3),
Lambda vacuums have Segre type
{\displaystyle \{\,(1,111)\}}
and isotropy group SO(1,3).
Constructing solutionsEdit
Existence of solutionsEdit
Global stability theoremsEdit
The positive energy theoremEdit
Exact Solutions of Einstein's equations Malcolm A.H. MacCallum Scholarpedia, 8(12):8584. doi:10.4249/scholarpedia.8584
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How to calculate the acceleration due to gravity on Earth
How to use the acceleration due to gravity calculator
Other similar calculators
If you have ever wondered how free-falling objects would behave on different planets, our acceleration due to gravity calculator can help you find out that.
Continue reading to know what acceleration due to gravity is and how to calculate its value on any celestial body. You will also find an example of calculating the magnitude of the acceleration due to gravity using our tool.
As the name suggests, the acceleration due to gravity is the acceleration experienced by a body when it falls freely under the influence of gravity alone. We use the symbol
g
to denote it. The SI unit of
g
is m/s2.
Acceleration due to gravity or
g
is a vector quantity, and it is directed towards the center of the celestial body under consideration.
The formula for determining the magnitude of the acceleration due to gravity
g
at the surface of any celestial body is:
\small g = \frac{G \ M}{R^2}
M
– Mass of the celestial body in kg;
G = 6.674 \times 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}
– Universal gravitational constant; and
R
– Radius of the celestial body in m.
🙋 The acceleration due to gravity experienced by an object is independent of its mass.
As an example, let us calculate the acceleration due to gravity on the surface of Earth. We know that:
Mass of earth,
M = 5.972 \times 10^{24}\ \text{kg}
Radius of Earth,
R = 6. 371 \times 10^6 \ \text{m}
substituting these values in the formula of acceleration due to gravity, we get:
\!\footnotesize \begin{align*} g & = \frac{G \ M}{R^2}\\ \\ & = \frac{6.674 \times 10^{-11}\ \mathrm{m^3 kg^{-1} s^{-2}} \times 5.972 \times 10^{24}\ \text{kg}} {(6. 371 \times 10^6 \ \text{m})^2} \\ \\ & = 9.8\ \mathrm{m/ s^2} \end{align*}
Hence the magnitude of the acceleration due to gravity on the surface of Earth is 9.8 m/s2. In the next section, you will see how to determine the value of
g
using our gravitational acceleration calculator.
Now let us calculate the same problem as above using the acceleration due to gravity calculator:
Type the mass of the Earth as 5.972 e24.
Enter the radius of the Earth as 6371 km or 6371000 m.
The tool will calculate the magnitude of the acceleration due to gravity on the surface of Earth as 9.8 m/s2.
If you liked this gravitational acceleration calculator, do check out our other tools on gravitation, namely:
Gravitational force calculator;
Acceleration using force and mass calculator; and
g-force calculator.
How do I calculate the acceleration due to gravity on Mars?
To calculate the acceleration due to gravity on the surface of Mars, follow the given steps:
Take the square of the radius of Mars (3,390 km or 3.390 × 10⁶ m). You will get 1.1492 × 10¹³ m².
Divide the mass of Mars (6.4185 x 10²³ kg) by the square of its radius in m.
Multiply the result by G, i.e., the universal gravitation constant.
You will get the value of acceleration due to gravity on Mars, i.e., 3.7 m/s².
How do I calculate the acceleration due to gravity on the Moon?
To calculate the acceleration due to gravity on the Moon, proceed as follows:
Divide the mass of the Moon (7.346 x 10²² kg) by the square of its radius (1737.4 km).
Multiply the result from step 1 with 6.674×10⁻¹¹ m³/kg s², i.e., the value of G.
Congrats! You have calculated the acceleration due to gravity on the surface of the Moon as 1.6 m/s².
Check out Wien's law calculator to find the temperature of any hot object, basing on its thermal emission spectrum.
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Density mass and volume equations
How to use this density mass volume calculator
Other density-related calculators
With this density mass volume calculator, you are free to enter any two of these values, and the calculator will instantly calculate the third value. This short article covers:
How to calculate volume with density and mass and other combinations.
How to use this density mass volume calculator.
What is the density of a cork, and how to calculate it.
The density of an object is defined as the mass per unit volume and is mathematically expressed as:
\rho = \frac{m}{V}
\rho
– Density of the substance;
m
– Mass (or weight, if measured on Earth); and
V
– Volume.
We can then rearrange this equation to express the mass in terms of density and volume:
m = \rho V
Finally, by rearranging the equation one final time, we can calculate the volume of an object if the mass and density are known:
V = \frac{m}{\rho}
This density mass volume calculator contains all three of these equations, making it very flexible to your needs.
It couldn't be easier to use this calculator. All you have to do is enter any two values to calculate the third value. Let's go through an example just to make sure it's clear in your mind.
You know the density of the object is 10490 kg/m3, so enter this value in the density field of the calculator. Make sure you select the appropriate unit for each variable before entering the value.
The mass of the object is 6 kg, so go ahead and enter this value into the second row of the calculator.
Oh, look! You've got your answer. Change the volume units to cm3, and the value for the volume is 572 cm3.
Omni has a wide range of density calculators, if you would like to explore this topic further:
Gas density calculator;
Mass to density calculator;
Volume to density calculator;
Density to mass calculator;
Density to weight calculator;
Liquid ethylene density calculator;
Cube density calculator;
Density of a cylinder calculator; and
Sphere density calculator.
How do I calculate volume with density and mass?
To calculate the volume of an object if you know its density and mass:
Weigh the mass of the object.
Lookup the density of the material the object is made from.
Divide the mass by the density.
Enjoy your result for the volume of the object.
Mathematically, volume is given by the equation: volume = mass / density.
What is the density of a cork?
240 kg/m3 or 14.98 lb/ft3. The volume of a champagne cork is around 37.5 cm3 and its weight is 9 grams. To calculate the density, divide the weight by the volume: 9 g / 37.5 cm3 = 0.24 g/cm3 = 240 kg/m3.
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Week by Week of Pregnancy Weight Gain
Pregnancy Weight Gain: Week by Week Breakdown
Pregnancy weight gain breakdown by week
What's the average pregnancy weight gain week by week? Where do all the extra pounds go? Keep on reading to find all the answers!
Your pre-pregnancy body mass index (BMI) determines the amount of weight you should gain during pregnancy. BMI is the simplest way to assess somebody's weight and figure and compare it to the general population. However, one thing might be tricky — BMI and its implications does not apply to pregnant women! That's why we usually record the woman's weight before pregnancy and then monitor its growth rate throughout every week of pregnancy.
Remember, the formula for BMI is:
\text{BMI} = \frac{ \text{weight [kg]} }{ (\text{height [m]})^2 }
Your recommended weight gain depends on your pre-pregnancy BMI:
Underweight before pregnancy (BMI <18.5)
Single pregnancy expected weight gain 25 to 35 pounds (11.3–15.9 kg)
Single pregnancy expected weight gain: 15–25 pounds (6.8–11.3 kg)
Single pregnancy expected weight gain: 11–20 pounds (5–9 kg)
First things firsts: eating for two might not be such a good idea. Your diet should increase by only 300 kcal, starting at the beginning of the second trimester, and grow up to 500 kcal when you begin breastfeeding. 🍼
As you can imagine, all the weight gain during pregnancy must serve a specific purpose. You're not wrong: all that mass goes into the birth weight of the baby, and also allows the gestation to thrive and survive. In a typical healthy pregnancy, an average woman needs to gain around 25–35 pounds (11.3–15.9 kg).
Let's break it down pound by pound!
The baby: An average child's birth weight is 7.7 lb (3.5 kg).
An enlarged uterus weighs around 2 lb (0.9 kg).
Enlarged breasts weigh around 2 lb (0.9 kg).
The placenta adds another 1.5 lb (0.7 kg).
Amniotic fluid is responsible for an additional 2 lb (0.9 kg).
Your blood volume (increased by ~50%) and other bodily fluids: 6.6 lb (3 kg).
The extra fat tissue weighs close to 7.7 lb (3.5 kg).
Achieving a healthy weight gain during pregnancy is not that difficult! Trust your body, exercise, and eat well. Don't get too attached to them sweets; turn to veggies and nuts instead. Keep in mind that the growth rate is different for every pregnancy. There is no single ideal way in which pregnant women gain weight - try to avoid comparing yourself to other future mums!
However, there's one specific thing that should catch your attention: an extremely rapid weight gain exceeding 0.5 kg (1.1 lb) per week. Track your weight, and contact your health care provider if you realize you're gaining pounds way too fast.
Here are the estimates of growth rates for the average singleton pregnancy. Remember: they do not have to fit your case entirely!
<14 weeks — your weight during the first trimester of pregnancy should not change.
14–20 weeks — a steady weight gain appears for a total increase of 2.5 kg.
20–30 weeks — an increase of another 5 kg.
30–36 weeks — another 2.5 kg.
36–40 weeks — at the end of the third trimester, you shouldn't gain weight anymore.
One available study broke it down to even finer pieces:
0 to 10 weeks — 0.065 kg, 2.3 oz, or 0.14 pounds per week.
10 to 20 weeks — 0.335 kg or 0.74 lb per week.
20 to 30 weeks — 0.450 kg or 1 lb per week.
Use the BMI percentile calculator to assess your child's development pattern and learn where they place among their peers.
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What is g force
How to calculate g force
Example: Using the g-force calculator.
This g force calculator helps determine the force due to gravitational acceleration experienced by a moving object. The tool first establishes the acceleration due to gravity and then computes the gravitational force equivalent based on the moving speed of any entity.
Read on to understand how you can calculate the g force from the velocity of an object.
The g force or g-force, otherwise known as the gravitational force equivalent is the force experienced by an object with reference to the acceleration to due to gravity value —
9.81 \text{ m/s}^2
32.17\text{ ft/s}^2
. The force experienced by an object resting on the earth's surface is roughly
1 g
g
is different from the unit of weight grams (
\text{g}
The g force is directly proportional to the object's acceleration, such that a pilot experiences as much as
8 g
-5 g
. When the g value is positive, the pilot's blood rushes towards his feet, leading to blackouts or losing consciousness, whereas, during the negative g value, it causes the blood flow towards the brain and eyes resulting in swelling veins and reddened visions. We can also experience the same rush on roller coaster rides, during which people often lose consciousness. Mathematically, the acceleration due to gravity,
g
for an object is:
\scriptsize g = \frac{GM}{r^2}
G
- Universal gravitational constant,
6.674 \times 10^{-11} \text{ m}^{3} \text{kg}^{-1} \text{s}^{-2}
M
- Mass of the celestial body (e.g. planet Earth); and
r
- Radius of celestial body.
If you know the acceleration due to gravity value, you can further calculate the g force from velocity. Suppose an object has an initial velocity
v_0
and final velocity,
v_1
over a time interval,
t
. The equation gives the g force:
g \text{ Force} = \frac {v_1 - v_0}{t g} = \frac {\Delta v}{t g}
It is interesting to notice that the term
\frac {v_1 - v_0}{t}
is acceleration of an object.
Estimate the g force on a person suddenly stopping his car (in
1 \text{s}
that was traveling at a velocity of
60 \text{km/h}
To calculate the g force:
Enter the initial velocity,
v_0 = 60 \text{ km/h}
16.67 \text{ m/s}
Fill in the final velocity,
v_1 = 0 \text{ km/h}
Set the time,
t = 1\text{ s}
The calculator returns the g value using the equation:
\qquad \scriptsize g \text{ Force} = \frac {16.67 - 0}{1 \times 9.81} = - 1.697 \ g
The driver experiences a negative g force value of
-1.697 \ g
when he suddenly stops his car.
You can explore the similar calculators listed below if you like this one:
Gravitational force calculator; and
How do I calculate g force from velocity?
To calculate g force from velocity:
Subtract initial velocity from final velocity.
Divide the difference by time.
Divide the resultant by the acceleration due to gravity, 9.81 m/s², to obtain the g force value.
What is the g force if I am driving at a speed of 60 km/h?
Considering the initial velocity as 0 km/h, and time as 1 s, the g force value is: 1.697 g. Mathematically, g force = (v1 - v0) / (t × g) = (16.67 - 0) / (9.81) = 1.697 g.
Final velocity (V₁)
Gravitational force equivalent (F)
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What is the definition of damping?
What is the damping coefficient?
What are the three degrees of damping?
What is the critical damping coefficient? The damping coefficient formula
How to use the critical damping calculator
Welcome to our critical damping calculator, which can help you estimate the critical damping coefficient of a damped oscillator.
You know that a simple pendulum's motion is oscillatory and the force describing the motion is proportional to the displacement of the bob from its mean position. However, in real situations, frictional forces like air resistance also influence this motion and oppose the pendulum's movement. Read on to find more about these opposing frictional forces — the definition of damping force, degrees of damping, damping coefficient, and formula for estimating critical damping coefficient.
We also explain underdamped, overdamped, and critically damped oscillations using figures and examples. After reading, you'll never be in doubt about how to calculate any critically damped system.
Damping is the process of dissipating energy from a vibrating structure, resulting in the reduction of the vibration's amplitude. Generally, it involves the conversion of the mechanical energy of the vibrating structure into thermal energy.
We know that a rolling ball comes to a stop because of air resistance and friction due to the surface on which it rolls. This way, any movement of an object is resisted by its surrounding objects. The resistance reduces the energy of the moving object with time, eventually slowing it down to a stop.
Now, let's apply this idea to a vibrating structure. The air surrounding such an object (or any medium in contact with it) tries to resist the vibrational motion and reduce its vibrational energy (mechanical energy). The higher the energy, the higher the vibration's amplitude. Thus, reducing the energy with time reduces the vibration's amplitude too.
Now that we've covered the mechanical theory, let's jump to the next section to learn what is the damping coefficient.
The damping coefficient of the damping agent or damper tells us how effective the damper is in resisting an object's harmonic motion.
We calculate the resistive force exerted by a damper or the damping force as:
F=−c \frac{\text{d}x}{\text{d}t},
c
is the damping coefficient;
x
is the displacement of the oscillating object from its mean position; and
\frac{\text{d}x}{\text{d}t}
denotes the change in displacement (
x
) with time (
t
) — this is also known as the instantaneous velocity.
The damping coefficient
c
helps estimate the energy dissipated due to frictional forces, which slow down oscillation.
Now it's high time to find out what is the critical damping coefficient.
Want to learn more about damping and damping coefficient? Take a look at our damping ratio calculator.
There are three degrees of damping that depend on how quickly the amplitude of the oscillations decreases:
Underdamping (light damping);
Critical damping; and
Overdamping (heavy damping).
In underdamped oscillations, the object comes to rest after a few oscillations with amplitude decreasing exponentially with time. Let's go back to the first oscillating object we have learned — a simple pendulum! It's an underdamped system.
In over damping, the damper exerts a force that is greater than what is required to prevent the oscillations. Hence, the object returns to its equilibrium position without oscillating. Have you noticed that the toilet flush handle returns to its resting position gradually withing oscillating? It's an overdamped system. Another example is an automated door — it's overdamped by design to prevent damage to the glass.
In critical damping, the object reaches the equilibrium point as quickly as possible without oscillating. An automobile shock absorber is a great example of the practical application of critical damping.
The value of the damping coefficient at which critical damping occurs is the critical damping coefficient of a system. It's also equal to the undamped resonant frequency of the oscillator.
In terms of stiffness and mass of the oscillating object, we calculate the critical damping coefficient as:
c_\text{c} = 2\sqrt{k\cdot m},
k
is the stiffness of the oscillated body; and
m
is the mass of the oscillating body.
Select the units for each variable from the drop-down list beside them.
Enter the oscillator's mass in the input box for Mass (m).
Type the oscillator's stiffness in the input for Stiffness (k).
The calculator displays the natural frequency (ωn) and critical damping coefficient (cc) of the oscillator.
✅ You can enter the mass and natural frequency to find out the other two variables, or you can enter the critical damping coefficient and one other variable to estimate the remainders. In short, enter any two variables and obtain the other two variables.
An oscillating body executes simple harmonic motion (SHM) if:
The magnitude of the forces acting on it is directly proportional to the magnitude of its displacement from the mean position; and
The restoring force always acts towards the mean position.
What are the differences between overdamping, underdamping, and critical damping?
There are three damping systems:
An overdamped system returns to the equilibrium position slowly without oscillating.
An underdamped system moves to the equilibrium position quickly and oscillates before coming to rest.
A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.
How do I find the critical damping coefficient using the natural frequency?
The critical damping coefficient (cc) is twice the product of the mass (m) and natural frequency (ωn) of the oscillating object: cc = 2mωn.
So, find the product of the mass and natural frequency and multiply it by 2 to obtain the critical damping coefficient.
How do I calculate the stiffness of an oscillator with mass and natural frequency?
To find the stiffness of an oscillator, equate the two formulas to estimate critical damping coefficient:
cc = 2mωn, where m is the mass and ωn is the oscillating object's natural frequency; and
cc = 2√(k×m), where k is the stiffness.
Thus, stiffness is the product of the oscillator's mass and square of the natural frequency: k = mωn2.
How do I find the critical damping coefficient of an oscillator of mass 2 kg and stiffness 2 N/m?
The formula for calculating critical damping coefficient (cc) using the oscillator's mass (m) and stiffness (k) is:
cc = 2√(k×m).
So, the critical damping coefficient of an oscillator of mass 2 kg and stiffness 2 N/m is 2√(2 × 2) = 4 Ns/m.
How do I calculate the stiffness of an oscillator of mass 5 kg and critical damping coefficient 1 Ns/m?
The formula to calculate stiffness is:
k = cc2/2m,
where k is the stiffness, m is the mass, and ωn is the natural frequency of the oscillating object. Thus, an oscillator of mass 5 kg and critical damping coefficient 1 Ns/m has a stiffness of 12 / (2 × 5) = 0.05 N/m.
Natural circular frequency (ω)
Critical damping coefficient (c꜀)
lbf·sec/in
Given a vehicle's power and weight, you can use this quarter-mile calculator to estimate the elapsed time and final speed over a 1/4-mile distance.
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Equivariant K -theory of semi-infinite flag manifolds and the Pieri–Chevalley formula
15 September 2020 Equivariant
K
-theory of semi-infinite flag manifolds and the Pieri–Chevalley formula
We propose a definition of equivariant (with respect to an Iwahori subgroup)
K
-theory of the formal power series model
{\mathbf{Q}}_{G}
of semi-infinite flag manifolds, and we prove the Pieri–Chevalley formula, which describes the product, in the
K
{\mathbf{Q}}_{G}
, of the structure sheaf of a semi-infinite Schubert variety with a line bundle (associated to a dominant integral weight) over
{\mathbf{Q}}_{G}
. In order to achieve this, we provide a number of fundamental results on
{\mathbf{Q}}_{G}
and its Schubert subvarieties including the Borel–Weil–Bott theory, whose precise shape was conjectured by Braverman and Finkelberg in 2014.
One more ingredient of this article besides the geometric results above is (a combinatorial version of) standard monomial theory for level-zero extremal weight modules over quantum affine algebras, which is described in terms of semi-infinite Lakshmibai–Seshadri paths. In fact, in our Pieri–Chevalley formula, the positivity of structure coefficients is proved by giving an explicit representation-theoretic meaning through semi-infinite Lakshmibai–Seshadri paths.
Syu Kato. Satoshi Naito. Daisuke Sagaki. "Equivariant
K
-theory of semi-infinite flag manifolds and the Pieri–Chevalley formula." Duke Math. J. 169 (13) 2421 - 2500, 15 September 2020. https://doi.org/10.1215/00127094-2020-0015
Received: 23 February 2018; Revised: 14 February 2020; Published: 15 September 2020
Secondary: 14N15 , 33D52 , 81R10
Keywords: K-theory , normality , Pieri–Chevalley formula , semi-infinite flag manifold , semi-infinite Lakshmibai–Seshadri path , standard monomial theory
Syu Kato, Satoshi Naito, Daisuke Sagaki "Equivariant
K
-theory of semi-infinite flag manifolds and the Pieri–Chevalley formula," Duke Mathematical Journal, Duke Math. J. 169(13), 2421-2500, (15 September 2020)
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Where do we use the Hamming distance?
How to calculate the Hamming distance?
How to visualize the Hamming distance?
How to use our Hamming distance calculator?
With our Hamming distance calculator, you will learn why the distance from
01
10
2
9
— at least sometimes!
Are you ready for a short dive into information theory? Here you will learn:
Where do we apply the Hamming distance?
The Hamming distance is a metric used in information theory to measure how much two messages with the same length differ.
For message, we mean a string: whenever you see a number in this article, you should remember that the digits are no more than letters of an alphabet, and you can't perform mathematical operations on them!
🙋 Forget the definition of distance in a physical way: we will travel in different spaces!
The Hamming distance indicates the number of different digits/letters in a pair of messages. Take the words "Hamming" and "humming": what is the Hamming distance?
Let's see. "Hamming" and "humming" — there is one letter of difference, so the Hamming distance is 1.
Increase the Hamming distance to two: farming, humping, camping. As you can see, there are a few examples. If we increase the Hamming distance to three, there is even more: fasting, hosting, hammock or Hamburg.
As you can see, the higher the Hamming distance, the more difficult it is to "see" the original word: that's because we accumulated errors.
The Hamming distance is a fundamental concept in the field of error detection and correction. An error in a message is quantifiable by measuring the number of different bits between the corrupted message and the original one.
🔎 The Hamming distance bears the name of Richard Hamming, a pioneer of the theory of error correction and detection. He introduced both this concept and the framework of Hamming codes in his 1950 work.
The Hamming distance in an error-correcting code defines the capacity of an algorithm to detect and correct an error. The higher the distance, the higher the number of possible messages between two valid codewords, allowing more space to identify the error. You can learn more on this complex topic with our Hamming codes calculator!
In computer science, the Hamming distance is usually relegated to strings of numbers: letters lie at a higher level!
We will give you a hint on how to calculate the Hamming distance in a binary and a decimal message.
Take two binary messages with equal length, and write them down:
\begin{align*} a=011010100101011\\ b=010110110101001 \end{align*}
Now compare the messages "bitwise", and mark on one of them where the values of the bit in a given position differ between the messages:
\begin{align*} a=011010100101011\\ b=01\textcolor{red}{0}\textcolor{red}{1}101\textcolor{red}{1}01010\textcolor{red}{0}1 \end{align*}
Then count the number of bits you found: that is the Hamming distance. For short messages, it doesn't take long to calculate, but on longer ones, it can take a while — that's why we made this calculator!
You can apply the same procedure for any other message, in any alphabet. In our Hamming distance calculator, we implemented the decimal system too. Let's see an example of Hamming distance with ten digits!
\begin{align*} a=12271995\\ b=02071895 \end{align*}
Mark and count where different digits lie in the same position:
\begin{align*} a=12271995\\ b=\textcolor{red}{0}2\textcolor{red}{0}71\textcolor{red}{8}95 \end{align*}
The Hamming distance in this case is
3
. Easy, isn't it?
There is a neat way to visualize the Hamming distance — it uses geometry to represent the various codewords and find a path connecting them.
This representation can get messy pretty quickly, so don't rely on it to calculate the Hamming distance: writing down the string is usually enough!
Let's start with the simplest of the messages: one binary bit. Our bit can assume a value of either
0
1
The visualization of the Hamming distance in a single bit message.
Assuming the presence of a corruption, we can visualize the path (in red) that connects the two states. In this case, the path is made of a single segment: the Hamming distance is
1
We now increase the number of bits to two: how many messages can we build? The answer is four, and we can represent them on a square!
With an increased number of bits, the complexity increases too.
Let's corrupt two bits: from
00
we pass to
11
. To connect them, we need to depart from
00
, visit either
01
10
, and arrive at
11
, thus traveling two line segments: the Hamming distance is
2
We can do it again! Add another bit; which shape will you get?
Now we have a cube!
A cube, exactly! We created two "levels": in each one, the last two bits are the same as the previous example: the Hamming distance can be at most
2
if you stay on one "floor". Adding another "floor" allows us to increase the maximum distance to
3
Follow the example highlighted in red above:
010 \rightarrow 110 \rightarrow 111 \rightarrow 101
We visited two intermediate states and walked three segments. Notice how multiple minimal paths connect the two selected states: in this case, there are
6
of them. Can you find them all?
🔎 What about paths longer than the maximum Hamming distance? They contain detours! It is impossible to have a Hamming distance
4
3
-bits message: you would introduce and then correct an error!
Let's increase the number of bits one last time: now our message is four bits long, and the corresponding shape is a hypercube, a four-dimensional shape that we can represent in our 3D space with two "concentric" cubes. Take a look!
This is the last time, we swear!
Now we can reach a Hamming distance as big as
4
. We chose a single path to show that, but obviously, the number of possible paths is high!
Let's sum the representations:
1
bits: a segment, the maximum Hamming distance is
1
2
bits: a square, the maximum Hamming distance is
2
3
bits: a cube, the maximum Hamming distance is
3
4
bits: a hypercube, the maximum Hamming distance is
4
Isn't it obvious? You can flip only as many bits as the length of the message to create any other codeword!
There is nothing easier than using our Hamming distance calculator! Choose the numerical system you desire, and input the two messages: we will give you the Hamming distance in the blink of an eye!
What is the Hamming distance between 10101 and 01100?
The Hamming distance between 10101 and 01100 is 3. The two messages differ in positions number 1, 2, and 5. A Hamming distance of 3 on a message with length 5 means that more than half of the digits changed: it is hard to see the original message, as the corruption is pretty high!
How do I measure the Hamming distance?
To calculate the Hamming distance, you simply count the number of bits where two same-length messages differ. An example of Hamming distance 1 is the distance between 1101 and 1001. If you increase the distance to 2, we can give as an example 1001 and 1010.
Is the Hamming distance a metric?
The Hamming distance is a metric (in the mathematical sense) used in error correction theory to measure the distance between two codewords. In detail, the Hamming distance measures the number of different bits in two strings of the same length.
In computer science, distance metrics are a way to measure similarity between data. They don't measure a physical distance between two points in space but the distance of two "codewords" in their relative abstract spaces.
Distance metrics are of great importance in error detection and correction and in machine learning.
The aquarium glass thickness calculator finds required aquarium thickness, volume, glass surface area, and weight for any given measurements.
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An Integrated Method of Data Mining and Flow Unit Identification for Typical Low Permeability Reservoir Prediction
An Integrated Method of Data Mining and Flow Unit Identification for Typical Low Permeability Reservoir Prediction ()
1Guangxi Colleges and Universities Key Laboratory of Beibu Gulf Oil and Natural Gas Resource Effective Utilization, Beibu Gulf University, Qinzhou, China.
2College of Petroleum and Chemical Engineering, Beibu Gulf Universtiy, Qinzhou, China.
With the development of oilfield exploration and mining, the research on continental oil and gas reservoirs has been gradually refined, and the exploration target of offshore reservoir has also entered the hot studystage of small sand bodies, small fault blocks, complex structures, low permeability and various heterogeneous geological bodies. Thus, the marine oil and gas development will inevitably enter thecomplicated reservoir stage; meanwhile the corresponding assessment technologies, engineering measures andexploration method should be designed delicately. Studying on hydraulic flow unit of low permeability reservoir of offshore oilfield has practical significance for connectivity degree and remaining oil distribution. An integrated method which contains the data mining and flow unit identification part was used on the flow unit prediction of low permeability reservoir; the predicted results were compared with mature commercial system results for verifying its application. This strategy is successfully applied to increase the accuracy by choosing the outstanding prediction result. Excellent computing system could provide more accurate geological information for reservoir characterization.
Low Permeability Reservoir, Offshore Oilfield, Hydraulic Flow Unit, Flow Unit Identification, Data Mining
Yu, P. (2019) An Integrated Method of Data Mining and Flow Unit Identification for Typical Low Permeability Reservoir Prediction. World Journal of Engineering and Technology, 7, 122-128. doi: 10.4236/wjet.2019.71008.
With the exploration and mining in marine oil and gas formation, there are an increasing number of low permeability reservoirs being found. Generally speaking, this type of reservoir has the serious heterogeneity characteristics which can lead to the larger difference of permeability among reservoirs of equal porosity basically, and several related logging response characteristics are not obvious. In such environment, a reasonable representation of the reservoir has very important practical significance. In order to characterize oil and gas reservoirs more precisely, the concept of reservoir hydraulic flow unit is introduced, which, after continued development and improvement, for the moment generally refers to the basic unit with consistent petrological, geological and hydrodynamic characteristics within a given reservoir that is different from other rocks, and this method is especially effective for the porous medias paces with strong heterogeneity characteristics [1] [2] [3] [4] .
Reservoir quality index and flow unit are two basic concepts used by scholars. Both petroleum geologists and engineers have long acquainted the importance of these concepts in petroleum field as they have applied them as outstanding methods to properly quantify and characterize reservoir formations [5] . Recent studies also have proved the superiority of data mining technology to empirical and statistical approaches in petroleum and geosciences related problems. A growing tendency utilizing mining algorithm for solving problems of onshore reservoir description is observed [6] [7] [8] . The aim of this study is to characterize target offshore reservoir with the integrated method of data mining and flow unit identification, and core samples were taken from Beibu Gulf typical low permeability operating area.
2. Data Analysis Approach
Hydraulic flow unit, has been well known as a part of the reservoir porous media, characterizes lateral and vertical consistency in reservoir rock and fluid properties. Lots of definitions for the unit were brought forward in the last decades by Hearn et al. [9] , Ebanks [10] and Gunter et al. [11] . It was pointed out that the unit is useful in the reservoir characterization as it combines the most important two petrophysical properties: permeability and porosity. These two properties control reservoir quality in terms of reservoir storativity and transmissibility. In 1993, Amaefule et al. [12] presented the mathematical model for the RQI (Equation (1)) as:
RQI=0.0314\sqrt{\frac{k}{{\varphi }_{e}}}
where k is permeability (mD),
{\varphi }_{e}
is effective porosity (fraction), and RQI is Reservoir Quality Index (μm).
Different RQIs could be a signification for the existence of multiple flow units in the porous media which inevitably influences the expected pressure profiles and flow regimes. In following equations
{\varphi }_{z}
(Equation (2)) and FZI (Equation (3)) are normalized porosity index and Flow Zone Indicator (μm) respectively.
{\varphi }_{z}=\frac{{\varphi }_{e}}{1-{\varphi }_{e}}
FZI=\frac{1}{\sqrt{{F}_{s}}\tau {S}_{gv}}=\frac{RQI}{{\varphi }_{z}}
This model system is based on texture and mineralogy which defines similar fluid flow features that is independent of lithofacies. Authors give an equation which is rearranged to isolate the variable that is constant within a unit based on the concept of bundle of capillary tubes [13] .
3. Flow Unit Division and Characterization
3.1. Flow Unit Division
The division and characterization of the hydraulic flow units can be executed either using the static based or the dynamic based methodology. The common static methods include developing relationship between fluid properties and rock media by core and/or log derived data, and the statistical measurement of permeability and the heterogeneous degree of the reservoir. It has been found that the application effects of single parameter methods are poor, loyalties of results are not high. Considering natural distribution characteristics of porous media itself, clustering analysis was chosen to execute early classification with the Flow Zone Indicator and its related parameters; then the final classification was completed according to the clustering pedigree chart of hydraulic unit samples [14] .
Flow Zone Index and the factor which reflects the features of microscopic pore structure have high correlation by analyzing the relationship between flow parameter and displacement pressure (Pd). Besides, geological parameters for low permeability reservoirs were selected as clustering (Ward’s method) variables:
{\varphi }_{e}
{\varphi }_{z}
and RQI. Sum of squares method which could deal with isolated points reasonably was considered preferentially. Based on the homogeneity of similar samples, 4 schemes were built finally (FU#5, FU#6, FU#7, FU#8).
3.2. Data Mining for Flow Unit Characterization
Due to the differentiation between petrophysical and percolation characteristics, there are some differences on logging response for each category of hydraulic flow unit, and based on the differences, a data mining method was also applied for identify the hydraulic unit of uncored intervals in the crossplot of logging bins. Key link of the whole system is probability database (include core and logging data), its construction process is through comparing each sample data and selecting logging parameters associated with the core. The database consisting of probability of occurrence for each unit corresponding to discretized logging data is assigned to all FUs. Basic computation of the database and inference of FUs was executed by self-software. Meanwhile, back evaluation program was performed to test the software’s effect for known categories of cored interval. The software design procedure is presented in Figure 1.
Probability database of cored sample unit categories were calculated after software debugging completed, and named as PFU5-1, PFU5-2, PFU5-3; PFU6-1, PFU6-2, PFU6-3; PFU7-1, PFU7-2, PFU7-3; PFU8-1, PFU8-2, PFU8-3. The result of back evaluation report shows that the accurate rates of PFU5-3 and PFU6-3 are better than remaining database (Table 1).
3.3. Comparative Verification
The predicted results of software were compared with the mature commercial system results of artificial neural network recognition mode for verifying its application. Firstly, the same cored single well and database PHU5-3 which has the higher accurate rate were choose, and then operated software, result of prediction rate was 82.07%. Then selecting 5 typical logging parameters to be neural network input parameters, and the categories of hydraulic flow unit were selected as the parameter of expectation output, hidden layer neurons number range of 4 - 12 and the output type is hydraulic flow unit. In the training process, the network learning rate is 0.05, permissible error is 0.001, and maximum number of iterations is 10000. The number of hidden layer nodes selected by experience formula, network error showed the minimum error when the number of the nodes reach 11, then supplied single well data, we focus on the training error has reached the requirements when the number of iterations up to 5968 times (Table 2). Weights and thresholds of prediction model were created. The true positive rates of receiver operating characteristic curve increase fast, curves bent upward, areas under curve large, classified performance of model is favorable. Calling learning network to carry out predictive instruction for verified well, the final accuracy rate is 82.97%, close to the predictive results with our software.
Table 1. Statistical results of back evaluation report on cored single wells.
Figure 1. Compiling flowchart of FU recognition system.
Table 2. Input parameters and calculated process factors.
In this work, hydraulic flow units are delineated, and predicted from well loggings and validated on the basis of data mining method, reservoir performance, petrophysical properties and lithology. At the same time, the software was written in the prediction process, and the results were compared with the mature commercial system results for verifying its application. This study proves that in the offshore system of low permeability, utilizing the integrated method of data mining and flow unit identification could reach higher accurate rate on the prediction of uncored intervals. The same way could be promoted on several trial blocks with similar regional geological background.
[1] Prasad, M. (2003) Velocity-Permeability Relations within Hydraulic Units. Geophysics, 68, 108-117.
[2] Svirsky, D., Ryazanov, A., Pankov, M., Corbett, P. and Posysoev, A. (2004) Hydraulic Flow Units Resolve Reservoir Description Challenges in a Siberian Oil Field. SPE Asia Pacific Conference on Integrated Modelling for Asset Management, Kuala Lumpur, 29-30 March 2004, 1-15.
[3] Amabeoku, M., Kersey, D., Bin Nasser, R., Al-Waheed, H. and Al-Belowi, A. (2005) Incorporating Hydraulic Units Concepts in Saturation-Height Modeling in a Gas Field. SPE Asia Pacific Oil and Gas Conference and Exhibition, Jakarta, 5-7 April 2005, 1-17.
[4] Crandall, D., Bromhal, G. and Karpyn, Z.T. (2010) Numerical Simulations Examining the Relationship between Wall-Roughness and Fluid Flow in Rock Fractures. International Journal of Rock Mechanics and Mining Sciences, 47, 784-796.
[5] Abbaszadeh, M., Fujii, H. and Fujimoto, F. (1996) Permeability Prediction by Hydraulic Flow Units—Theory and Applications. SPE Formation Evaluation, 11, 263-271.
[6] Aminian, K., Ameri, S., Oyeerokun, A. and Thomas, B. (2003) Prediction of Flow Units and Permeability Using Artificial Neural Networks. SPE Western Regional/AAPG Pacific Section Joint Meeting, Long Beach, 19-24 May 2013, 1-7.
[7] Perez H.H., Datta-Gupta, A. and Mishra, S. (2005) The Role of Electrofacies, Lithofacies, and Hydraulic Flow Units in Permeability Predictions from Well Logs: A Comparative Analysis Using Classification Trees. SSPE Reservoir Evaluation & Engineering, 8, 143-155.
[8] Ali, S.S., Nizamuddin, S., Abdulraheem, A., Hassan R.M. and Hossain, E.M. (2013) Hydraulic Unit Prediction Using Support Vector Machine. Journal of Petroleum Science and Engineering, 110, 243-252.
[9] Hearn, C.L., Ebanks, W.J.J. and Ranganathan V. (1984) Geological Factors Influencing Reservoir Performance of the Hartzog Draw Field Wyoming. Journal of Petroleum Technology, 36, 1335-1344.
[10] Ebanks W.J.J. (1987) Flow Unit Concept: An Integrated Approach to Reservoir Description for Engineering Projects. AAPG Bulletin, 71, 551-552.
[11] Gunter, G.W., Finneran, J.M. and Hartmann, D.J. (1997) Early Determination of Reservoir Flow Units Using an Integrated Petrophysical Method. SPE Annual Technical Conference and Exhibition, San Antonio, 5-8 October 1997, 1-8.
[12] Amafule, J.O., Altunbay, M., Tiab, D., Kersey, D.G. and Keelan, D.K. (1993) Enhanced Reservoir Description: Using Core and Log Data to Identify Hydraulic (Flow) Units and Predict Permeability in Uncored Intervals/Wells. SPE Annual Technical Conference and Exhibition, Houston, 3-6 October 1993, 205-220.
[13] Yarmohammadi, S., Kadkhodaie-Ilkhchi, A., Rahimpour-Bonab, H. and Shirzadi, A. (2014) Seismic Reservoir Characterization of a Deep Water Sandstone Reservoir using Hydraulic and Electrical Flow Units: A Case Study from the Shah Deniz Gas Field, the South Caspian Sea. Journal of Petroleum Science and Engineering, 118, 52-60.
[14] Aguilar, C., Govea, H. and Rincon, G. (2014) Hydraulic Unit Determination and Permeability Prediction Based on Flow Zone Indicator Using Cluster Analysis. SPE Latin American and Caribbean Petroleum Engineering Conference, Maracaibo, 21-23 May 2014, 1-13.
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Visualizing solid Shapes - Practically Study Material
The circle, the square, the rectangle, the quadrilateral and the triangle are examples of plane figure; the cube, the cuboid, the sphere, the cylinder, the cone and the pyramid are the examples of Solid Shapes. Plane figures are of two dimensions (2–D) and the solid shapes are of three-dimensions (3–D). Three dimensional objects is that they all have some length, breadth and height or depth. That is, they all occupy space and have three dimensions. In this chapter, we learn about various three-dimensional objects.
Match the shape with the name
Sol : After matching the shape with their names are as follow
(i) (b) (ii) (d)
(v) (g) (vi) (e)
Match the 2 dimensional figures with the names
Sol : After matching figures with their names are as the follow :
(iii) (e) (iv) (c)
Perspectives (Views) of 3-D Shapes
Let us view a brick from front, side and top. How do they look like ?
For each of the given solid shapes, the three views are given :
10.2. MAPPING SPACE AROUND US
In Geography, you have been asked to locate a particular State, a particular river, a mountain etc., on a map. In History, you might have been asked to locate a particular place where some event had occured long back. You have traced routes of rivers, roads, railway lines, traders and many others. Look at the map of a house whose picture is given alongside.
Now, look at the adjacent map, which has been drawn by seven year old Raghav, as the route from his house to his school :
From this map, can you tell –
(i) how far is Raghav’s school from his house ?
(ii) would every circle in the map depict a round about ?
(iii) whose school is nearer to the house, Raghav’s or his sister’s ?
It is very difficult to answer the above questions on the basis of the given map. Can you tell why ?
The reason is that we do not know if the distances have been drawn properly or whether the circles drawn are roundabouts or represent something else. Now look at another map drawn by his sister, ten year old Meena, to show the route from her house to her school. This map is different from the earlier maps. Here, Meena has used different symbols for different landmarks. Secondly, longer line segments have been drawn for longer distances and shorter line segments have been drawn for shorter distances, i.e., she has drawn the map to a scale.
Now, you can answer the following questions :
1. How far is Raghav’s school from his residence ?a
2. Whose school is nearer to the house, Raghav’s or Meena’s ?
3. Which are the important landmarks on the route ?
Thus we realise that, use of certain symbols and mentioning of distances has helped us read the map easily. Observe that the distances shown on the map are proportional to the actual distances on the ground. This is done by considering a proper scale. While drawing (or reading) a map, one must know, to what scale it has to be drawn (or has been drawn), i.e., how much of actual distance is denoted by 1mm or 1cm in the map. This means, that if one draws a map, he/she has to decide that 1cm of space in that map shows a certain fixed distance of say 1 km or 10 km. This scale can vary from map to map but not within a map.
For instance, look at the map of India alongside the map of Delhi.
You will find that when the maps are drawn of same size, scales and the distances in the two maps will vary. That is 1 cm of space in the map of Delhi will represent smaller distances as compared to the distances in the map of India.
The larger the place and smaller the size of the map drawn, the greater is the distance represented by 1 cm.
Thus, we can summarise that :
1. A map depicts the location of a particular object/place in relation to other objects/places.
2. Symbols are used to depict the different objects/places.
3. There is no reference or perspective in map, i.e., objects that are closer to the observer are shown to be of the same size as those that are farther away. For example, look at the following illustration.
4. Maps use a scale which is fixed for a particular map. It reduces the real distances proportionately to distances on the paper.
Example 1. A map is given showing some landmarks of a city. The landmarks are replaced by symbols, whose meanings are given on right side of the map.
(1) Ankit lives in the house H. How far is Ankit’s house from the school ?
(2) Which is nearby from Ankit’s house, stadium or hospital ?
(1) School is at a distance of 2 + 1 = 3 km from Ankit’s house.
(2) Distance of stadium from Ankit’s house = 1 + 2 + 1 + 2 = 6 km
Distance of hospital from Ankit’s house = 1 + 2 + 1 + 1 = 5 km
Therefore, hospital is nearer than stadium from Ankit’s house.
The given figure shows the map of a compound of a boarding school. Answer the following questions. (i) Which of the following landmarks is nearest to the girls dormitory ?
A. Boys dormitory
B. Academic block
D. Powerhouse
(ii) Which landmark is situated at the northwest corner of the school compound ?
(i) It can be clearly seen in the map that boys dormitory is nearer to girls dormitory as compared to academic block, auditorium, and power house.
(ii) The North West corner of the school compound in the given map is the dining hall.
The given figure shows the map of a locality which is drawn on a centimetre grid paper.
What is the shortest distance between the hospital and the post office ?
Solution. It can be seen in the map that there are two ways of reaching the post office from the hospital and the shortest is the one which is shown by blue arrows.
The scale used in the map is 1 cm = 500 m
Therefore, the shortest distance between hospital and post office
We know that 1000 m = 1 km
\therefore
2500 m =
\frac{2500}{1000}
2\frac{1}{2}
Thus, the shortest distance between the hospital and the post office is
2\frac{1}{2}
10.3. FACES, EDGES AND VERTICES
The corners of a solid shape are called its vertices, the line segments of its skeleton are its edges; and its flat surface are its faces.
The 8 corners of the cube are its vertices. The 12 line segments that form the skeleton of the cube are its edges. The 6 flat square surfaces that are the skin of the cube are its faces.
Example 4. Complete the following table
Solution. Complete table is as follow :
Polyhedron Let us look at the following solid figures.
All of the above solid figures are made up of polygonal regions, lines and points. There is no curved surface in the given figures. Such solids are called polyhedrons.
The polygonal regions in polyhedrons are called the faces. The faces meet to form line segments which are known as edges. The edges meet at the points which are known as vertices.
Hence, a polyhedron can be defined as a geometric object with flat faces and straight edges.
Polyhedra are named according to the number of faces. For example: tetrahedron (4 faces), pentahedron (5 faces), hexahedron (6 faces) and so on. The first figure is the figure of a tetrahedron. The second figure is an octahedron. Now, what can we say about the solids like cylinders, cones, spheres etc.?
These solids have lateral surfaces as well as curved edges. It means that these solids are not formed strictly with only flat surfaces as well as straight edges. Therefore, we can say that the solids like cylinders, cones and spheres are not polyhedrons.
We can classify polyhedrons into different categories. Let us discuss them one by one.
A polyhedron may be a regular or an irregular polyhedron.
A polyhedron is said to be regular if it satisfies two conditions which are given as follows.
(a) Its faces are made up of regular polygons.
(b) The same number of faces meets at each vertex.
If the polyhedron does not satisfy any one or both of the above conditions, then we can say that the polyhedron is irregular. To understand this concept, let us consider two solids, i.e. a cube and a hexahedron, as shown below.
Here, we can see that the faces of the cube are congruent regular polygons (i.e. all the faces are squares of same dimension) and each vertex is formed by the same number of faces i.e. 3 faces. Therefore, a cube is a regular polyhedron.
For the hexahedron, the faces are triangular in shape and they are congruent to each other. It means that the faces of the hexahedron are congruent regular polygons. If we look at the vertex A, we will notice that 3 faces meet at A. On the other hand, at point B, 4 faces meet. Thus, the vertices are not formed by equal number of faces. Therefore, the hexahedron is an irregular polygon.
The polyhedron may be a concave or a convex polyhedron.
It can be understood easily by taking two solids, i.e. a cube and the star shaped polyhedron, as shown below.
For the cube, the line segment AB joining the two points A and B of the polyhedron is contained either in the polyhedron or on the surface which is clearly shown in the figure. Thus, a cube is a convex polyhedron. For the star shaped polyhedron, the line segment AB joining the two points A and B of the polyhedron is neither contained in the polyhedron nor on the surface.
Thus, the star shaped polyhedron is a concave polyhedron.
In this way, we can easily identify a polyhedron and classify it as concave or convex and regular or irregular.
Let us discuss one more example using the concept of polyhedron.
Example 5. How many faces are at least required to make a polyhedron ?
Solution. At least 4 triangular faces are required to make a polyhedron. For the base of the polyhedron, we require at least a three sided closed figure or a triangle (at least three sides are required to form a closed figure). Let us take another point which is not on the previous triangle. If we join the line segments from that point to each of the vertices of the base triangle, then we will have three triangles. In this way, we require 4 triangular faces to make a polyhedron as shown below.
Thus, at least four faces are required to make a polyhedron.
Two important members of polyhedron family around are prisms and pyramids.
We say that a prism is a polyhedron whose base and top are congruent polygons and whose other faces, i.e., lateral faces are parallelograms in shape.
On the other hand, a pyramid is a polyhedron whose base is a polygon (of any number of sides) and whose lateral faces are triangles with a common vertex. (If you join all the corners of a polygon to a point not in its plane, you get a model for pyramid).
A prism or a pyramid is named after its base. Thus a hexagonal prism has a hexagon as its base; and a triangular pyramid has a triangle as its base. What, then, is a rectangular prism? What is a square pyramid? Clearly their bases are rectangle and square respectively.
Example 6. Identify whether the polyhedron shown in the following figure is a prism or a pyramid and name it accordingly.
Solution. The polyhedron shown in the given figure has two identical hexagonal bases and the lateral surfaces are parallelogram in shape. Therefore, the given figure is a hexagonal prism.
Example 7. Is a rectangular prism a cuboid ?
Solution. In a rectangular prism, the top and base surfaces are rectangles. If the lateral surfaces of that rectangular prism are strictly rectangles, then we can say that the rectangular prism is a cuboid, otherwise not. However, the lateral surfaces may not be rectangles as shown below.
Therefore, a rectangular prism is not always a cuboid.
Every polyhedron has a specific number of faces, edges, and vertices (depending upon the type of polyhedron it is). However, is there any relation that can be applied to the number of faces, edges, and vertices of any polyhedron irrespective of the type of polyhedron?
Tabulate the number of faces, edges and vertices for the following polyhedrons:
(Here ‘V’ stands for number of vertices, ‘F’ stands for number of faces and ‘E’ stands for number of edges).
Solid F V E F + V E+2
Cuboid 6 8 12 14 12 + 2
Triangular pyramid 4 4 6 8 6 + 2
Triangular prism 5 6 9 11 9 + 2
Pyramid with square base 5 5 8 10 8 + 2
Prism with square base 5 6 9 11 9 + 2
What do you infer from the last two columns ? In each case, do you find F + V = E + 2,
i.e., F + V – E = 2 ? This relationship is called Euler’s formula.
In fact, this formula is true for any polyhedron.
← Algebraic expressions NCERT QuestionsDirect and Inverse Proportions →
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reciprocation - Maple Help
Home : Support : Online Help : Mathematics : Geometry : 2-D Euclidean : Transformations : reciprocation
find the reciprocation of a point or a line with respect to a circle
reciprocation(Q, P, c)
c=\mathrm{O}\left(r\right)
be a fixed circle and let P be any ordinary point other than the center O. Let P' be the inverse of P in circle
\mathrm{O}\left(r\right)
. Then the line Q through P' and perpendicular to OPP' is called the polar of P for the circle c. Note that when P is a line, then Q will be a point.
Note that this routine in particular, and the geometry package in general, does not encompass the extended plane, i.e., the polar of center O does not exist (though in the extended plane, it is the line at infinity) and the polar of an ideal point P does not exist either (it is the line through the center O perpendicular to the direction OP in the extended plane).
If line Q is the polar point P, then point P is called the pole of line Q.
The pole-polar transformation set up by circle
c=\mathrm{O}\left(r\right)
is called reciprocation in circle c
The command with(geometry,reciprocation) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{geometry}\right):
\mathrm{circle}\left(c,[\mathrm{point}\left(\mathrm{OO},0,0\right),2],[x,y]\right):
\mathrm{point}\left(P,1,0\right):
\mathrm{inversion}\left(\mathrm{PP},P,c\right):
\mathrm{reciprocation}\left(l,P,c\right)
\textcolor[rgb]{0,0,1}{l}
\mathrm{detail}\left(l\right)
\begin{array}{ll}\textcolor[rgb]{0,0,1}{\text{name of the object}}& \textcolor[rgb]{0,0,1}{l}\\ \textcolor[rgb]{0,0,1}{\text{form of the object}}& \textcolor[rgb]{0,0,1}{\mathrm{line2d}}\\ \textcolor[rgb]{0,0,1}{\text{equation of the line}}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\end{array}
\mathrm{draw}\left({c,l,\mathrm{dsegment}\left(\mathrm{dsg},P,\mathrm{PP}\right)},\mathrm{printtext}=\mathrm{true},\mathrm{view}=[-3..5,-3..3],\mathrm{axes}=\mathrm{NONE}\right)
\mathrm{reciprocation}\left(A,l,c\right):
\mathrm{coordinates}\left(A\right)=\mathrm{coordinates}\left(P\right)
[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}]
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Section 47.1 (08XH): Introduction—The Stacks project
In this chapter we discuss dualizing complexes in commutative algebra. A reference is [RD].
We begin with a discussion of essential surjections and essential injections, projective covers, injective hulls, duality for Artinian rings, and study injective hulls of residue fields, leading quickly to a proof of Matlis duality. See Sections 47.2, 47.3, 47.4, 47.5, 47.6, and 47.7 and Proposition 47.7.8.
This is followed by three sections discussing local cohomology in great generality, see Sections 47.8, 47.9, and 47.10. We apply some of this to a discussion of depth in Section 47.11. In another application we show how, given a finitely generated ideal $I$ of a ring $A$, the “$I$-complete” and “$I$-torsion” objects of the derived category of $A$ are equivalent, see Section 47.12. To learn more about local cohomology, for example the finiteness theorem (which relies on local duality – see below) please visit Local Cohomology, Section 51.1.
The bulk of this chapter is devoted to duality for a ring map and dualizing complexes. See Sections 47.13, 47.14, 47.15, 47.16, 47.17, 47.18, 47.19, 47.20, 47.21, 47.22, and 47.23. The key definition is that of a dualizing complex $\omega _ A^\bullet $ over a Noetherian ring $A$ as an object $\omega _ A^\bullet \in D^{+}(A)$ whose cohomology modules $H^ i(\omega _ A^\bullet )$ are finite $A$-modules, which has finite injective dimension, and is such that the map
\[ A \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \]
is a quasi-isomorphism. After establishing some elementary properties of dualizing complexes, we show a dualizing complex gives rise to a dimension function. Next, we prove Grothendieck's local duality theorem. After briefly discussing dualizing modules and Cohen-Macaulay rings, we introduce Gorenstein rings and we show many familiar Noetherian rings have dualizing complexes. In a last section we apply the material to show there is a good theory of Noetherian local rings whose formal fibres are Gorenstein or local complete intersections.
In the last few sections, we describe an algebraic construction of the “upper shriek functors” used in algebraic geometry, for example in the book [RD]. This topic is continued in the chapter on duality for schemes. See Duality for Schemes, Section 48.1.
It seems like the reference [R+D] doesn't work.
Thanks for this and your other comment. If you want to be listed as a contributor, leave your full name in your next comment. See here for corresponding changes.
There is a theory of noncommutative dualizing complexes, developed by Amnon Yekutieli, James Zhang and Michel Van den Bergh.
It gave rise to the theory of rigid commutative dualizing complexes, developed by the same people.
See the book Derived Categories, by Amnon Yekutieli To be published by Cambridge University Press; prepublication version arXiv:1610.09640 https://arxiv.org/abs/1610.09640v4
(another attempt to get the link active)
Derived Categories, by Amnon Yekutieli To be published by Cambridge University Press prepublication version https://arxiv.org/abs/1610.09640v4
There is a very recent theory of rigid dualizing complexes over commutative DG rings. See
Duality and Tilting for Commutative DG Rings, by Amnon Yekutieli https://arxiv.org/abs/1312.6411v4 and The twisted inverse image pseudofunctor over commutative DG rings and perfect base change, by Liran Shaul Advances in Mathematics 320 (2017) 279–328 https://doi.org/10.1016/j.aim.2017.08.041
There are intimate connections to J. Lurie's dualizing complexes
J. Lurie, Derived algebraic geometry XIV: representability theorems http://www.math.harvard.edu/~lurie/papers/DAG-XIV.pdf
and to D. Gaitsgory's
\otimes^!
D. Gaitsgory, Ind-coherent sheaves, Mosc. Math. J. 13 (3) (2013) 399–528.
Yes, there are many, many references we could and should give here. It would be good to write a short section on the literature but I am not the right person to do this. On the other hand, this chapter in some sense is doing the absolute minimum to get started with the dualizing complex, in the spirit of Grothendieck and in spirit very close to Hartshorne's book (as it concerns the dualizing complex and how we work with it).
Typo in the second to last paragraph: It says "After esthablishing..."
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How to calculate the size of a 220 volt wire?
How do you use this 220 volt wire size calculator?
Other useful tools like the 220 volt wire size calculator
The 220 volt wire size calculator will help you determine the appropriate electrical wire for your next house installation. Do you want to install a new bell, maybe a new water pump? Keep reading to find out how to size your cables.
The following information is required before we start:
Wire size material since it affects the electrical resistivity.
Distance between your energy source and the equipment you want to energize.
The amount of current your equipment needs.
Once we have it all, we will use the following formula:
\quad A = \frac{2\times I \varrho L}{V \times \%\text{drop}},
\%\text{drop}
— Allowable voltage drop, in %.
V
— Source voltage, measured in volts.
I
— Maximum current running through the wire, measured in amps;
\varrho
— Resistivity of the conductor material, measured in ohm meters;
L
— Length of the wire, measured in meters; and
A
— Cross-sectional area of the wire measured in square meters.
The result will be expressed in mm² (if using the metric system).
It is common for wires to indicate their size in AWG (American Wire Gauge) units. Our wire gauge calculator relates circular cross area in mm² to AWG, but do not worry; our 220-volt wire size calculator also provides results in AWG.
Verify your connection will be 220 volt alternate current.
Choose your conductor material.
Define the amount of current your equipment need. For example, a 1 HP pump connected to a 220 volts single phase will require 3.39 amperes.
Finally, input the length of your cable (remember it's a one-way distance).
As a result, the 220-volt wire size calculator will indicate the cross area of the cable you need to install.
Since you already know how to size a 220 volts wire, you can take a look at other calculators:
DC wire size;
100 amp wire size;
12 volt wire size;
Amp to wire size; and
24 volt wire size.
What wire size should I use for a 20 amp 220 volt circuit?
10 AWG or 2.38 mm of wire diameter. Try our 220-volt wire size calculator or follow these steps:
Multiply 20 amperes per two times cable distance and then by the material resistivity.
Divide the result by the voltage drop. Here, you get the cross-section area which you can convert to AWG by using internet charts. For the calculation, we assume a maximum wire size length of 39 m (one-way).
What size wire for 50 amp 220 volt circuit?
6 AWG. In our 220-volt wire size calculator:
Make sure you are in AC single-phase mode.
Add 50 amps in the Current section.
Define the one-way distance of your wire, and you will get 6 AWG as a result. For the calculation, we assume a maximum wire size length of 46 m (one-way).
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DocumentTools/Canvas/AppendToCanvas - Maple Help
Home : Support : Online Help : DocumentTools/Canvas/AppendToCanvas
AppendToCanvas(canvas,elements)
XML canvas string
list of canvas elements
The AppendToCanvas command appends new elements to the end of an existing XML canvas string in order to create a new canvas with merged content.
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Canvas}\right):
\mathrm{cv}≔\mathrm{NewCanvas}\left(["Part 1",\mathrm{sqrt}\left(x\right)]\right):
\mathrm{ShowCanvas}\left(\mathrm{cv}\right)
\mathrm{cv2}≔\mathrm{AppendToCanvas}\left(\mathrm{cv},["Part 2",\mathrm{ln}\left(y\right)]\right):
\mathrm{ShowCanvas}\left(\mathrm{cv2}\right)
\mathrm{ShareCanvas}\left(\mathrm{cv2}\right)
The DocumentTools[Canvas][AppendToCanvas] command was introduced in Maple 2021.
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Egyptian Fractions Calculator | Count Like an Egyptian
What are Egyptian fractions?
Applications of the Egyptian fractions
Greedy algorithms: how to calculate an Egyptian fraction expansion
The greedy algorithm in action
Other Egyptian fractions algorithms
How to use our Egyptian fraction calculator
Our Egyptian fractions calculator will teach you how to write fractions in the Egyptian style. It's not hard — well, the math isn't, at least!
How Egyptians did math;
What an Egyptian fraction is;
How to use Egyptian fractions;
How to calculate Egyptian fractions — the greedy algorithm; and
How to use our Egyptian fractions calculator.
Discover the ancient Egyptian fractions with us.
1/3
1/2
1
, go!
Egyptians were good at math! Their millennial civilization saw the development of many innovative concepts, spanning from arithmetic to geometry. Look at the pyramids — we even call them pyramids!
The first evidence of the use of mathematics dates back to 3200 BC. Let this sink in: we are closer to Rome than Romans were to those first additions! In many papyruses, historians found traces of mathematical problems and solutions. Thanks to them, we know that Egyptians knew how to calculate surfaces and volumes of many shapes and had knowledge of mathematical constants like pi.
Egyptians had a particular way to write fractions (today, we call them Egyptian fractions) and they made extensive use of it. Let's take a look!
Our knowledge of the Egyptian fraction expansion comes mainly from a papyrus called the Rhind papyrus, where, out of 87 mathematical problems, as many as 81 involve fractions. Historians think that the Egyptians' interest in fractions stemmed from the particular labor force they used in their construction projects. When managing hundreds of workers, you don't want conflict over an unevenly split piece of bread! An Egyptian fraction is a way to represent a proper fraction (a fraction where the numerator is smaller than the denominator) as a sum of distinct unit fractions.
\frac{11}{14}=\frac{1}{2}+\frac{1}{4}+\frac{1}{28}
It looks impractical, and it is so. Egyptian fractions were still used shortly after the Egyptian civilization, but quickly gave space to vulgar fractions, which we use today.
Egyptians held certain fractions in high regard.
2/3
3/4
didn't have a unit fraction expansion, and together with
1/2
, they each had a special symbol.
Every real number smaller than 1 can be represented using Egyptian fractions. One of the most prominent mathematicians in history proved this result: Fibonacci. Egyptian fractions are nothing but an addition to the list of discoveries by the man: we already covered the Fibonacci sequence and the fields where his intuitions are used, like the Fibonacci retracement and the linear feedback shift registers.
Using a simple recursion, he also laid down the basis for developing the algorithm we will use in our calculator.
When Egyptians developed their fraction system, they mainly applied it in trade and division problems. One of those problems involved dividing loaves of bread among varying numbers of people — it sounds like the type of exercise we'd meet in primary school!
Why use Egyptian fractions today? When you need to divide a certain number of objects into equal parts, and the partition is not straightforward, put it down in terms of those ancient Egyptian fractions. Are you at a party with other 7 people, and there are only 5 pizzas? Let's ask the ancient Egyptians! Well, it looks like they don't know what a pizza is.
Anyway: 8 people, 5 pizzas. This means everyone gets one pizza each and two are left to divide equally:
\frac{5}{8} = \frac{1}{2}+\frac{1}{8}
Each person gets half a pizza and a small slice. The reasoning also applies (and in particular) for counterintuitive divisions. Think of 12 people and 13 pizzas. Instead of giving a whole pizza to everyone and then slicing the remaining one in twelfths or worse, try this expansion:
\frac{13}{12}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
Where the fractions represent the amount of pizza each partygoer gets.
In order to calculate an Egyptian fraction, you can apply many algorithms. The one we judged easiest to learn is the greedy algorithm:
\frac{x}{y}=\frac{1}{\left\lceil\frac{y}{x}\right\rceil}+\frac{(-y)\mod{x}}{y\left\lceil\frac{y}{x}\right\rceil}
Let's check what do those operations mean.
The weird "half" square brackets indicate the ceiling operator, defined as the smallest integer greater than or equal to its argument. In other words, we round up towards the integer. Take a look at this:
\left\lceil\frac{3}{8}\right\rceil=\left\lceil 0.375 \right \rceil = 1
The ceiling function is often taught in association with the floor function,
\lfloor x \rfloor
. In the latter, the result is the closest integer smaller than (or equal to) the argument. We are not talking about the floors and ceilings of construction and carpentry!
The modulus is the other "uncommon" operator, which returns the remainder of the division between the two arguments. You can learn more at our modulo calculator. Now that you know how the ceiling operation works, it's going to be even easier to understand:
\footnotesize y \operatorname{mod}\ {x} = y - \left\lfloor \frac{y}{x} \right\rfloor\cdot x
The same holds for negative numbers; remember to calculate the remainder in the same "direction"! Here's an example:
\footnotesize \begin{split} &\ (-11)\operatorname{mod}\ {5}\\ =&\ -11 -\left\lfloor\frac{-11}{5}\right\rfloor\cdot 5\\ =&\ \ 4 \end{split}
If the second argument has a numerator
>1
, we feed the fraction back to the algorithm until all fractions have denominator
1
. It will happen sooner or later — the finiteness of this algorithm for Egyptian fractions was proved by the man, the myth, the mathematician Fibonacci.
Let's try the greedy algorithm with an example:
6/7
First step: apply the algorithm!
\footnotesize \begin{split} \frac{6}{7} &= \frac{1}{\left\lceil\frac{7}{6}\right\rceil} + \frac{(-7)\operatorname{mod}\ 6}{7\left\lceil\frac{7}{6}\right\rceil} \\[1em] &= \frac{1}{2}+\frac{5}{14} \end{split}
The first fraction is the first unitary fraction of our result. Now, feed the second fraction to the algorithm again:
\footnotesize \begin{split} \frac{5}{14} &= \frac{1}{\left\lceil\frac{14}{5}\right\rceil} + \frac{(-14)\operatorname{mod}\ 5}{14\left\lceil\frac{14}{5}\right\rceil}\\[1em] &=\frac{1}{3}+\frac{1}{42} \end{split}
The numerator of the second fraction is
1
: our greedy algorithm has done its job. We can now write the Egyptian fraction expansion in its entirety:
\frac{6}{7}=\frac{1}{2}+\frac{1}{3}+\frac{1}{42}
The greedy algorithm is not the only way to calculate an Egyptian fraction expansion. Here we will teach you two other algorithms you can use. They are both based on conflict resolution and share the first step: decomposing the fraction
x/y
x
copies of the fraction
1/y
\frac{x}{y}=\frac{1}{y}+\frac{1}{y}+...+\frac{1}{y}
We need to solve those conflicts since there can't be identical terms in an Egyptian fraction. We have a choice: pair or split.
Solving conflicts by splitting
We apply a simple decomposition of a unity fraction — one you can try on a piece of paper!
\frac{1}{y}=\frac{1}{y+1}+\frac{1}{y\cdot (y+1)}
Every time you find a conflict, one of the two identical fractions "surrender" and yield the two "daughter" fractions.
❗ Careful! This method gives an exponentially big number of terms: they follow the value of the numerator
n
n_{\text{terms}}=2^n-1
Solving conflicts by pairing
Instead of a single unit fraction of a pair disappearing (giving three fractions as a result), why not pair the two sides of the conflict?
Consider a pair of identical unit fractions, two identities hold, depending on the parity of the denominator.
If the denominator is even, we can do:
\frac{1}{a}+\frac{1}{a}=\frac{2}{a}=\frac{1}{a/2}
And if the denominator is odd, we can do:
\frac{1}{a}+\frac{1}{a}=\frac{2}{a+1}+\frac{2}{a\cdot(a+1)}
This "peaceful" solution of the conflict gives shorter expansions than the splitting algorithm.
Using our Egyptian fraction calculator is straightforward. Insert the fraction that you want to convert to Egyptian fractions, select the algorithm you prefer, and you are good to go!
⚠️ We limited the number of iterations that our Egyptian fraction calculator can take when using the splitting algorithm. The value of the numerator can be at most
5
You discovered how Egyptians wrote their fractions, and with it, a piece of the history of math. Numbers and calculations have always accompanied humanity, from Mayan numerals to integrals. Discover more math calculators at the math page of omnicalculator.com!
What is an Egyptian fraction?
An Egyptian fraction is the representation of any real number smaller than 1 as a sum of non-repeating unit fractions (fractions with numerator equal to 1).
The expansion was used especially in equal division problems and is of little use today; however, it remains an interesting mathematical topic.
You can still use Egyptian fractions to solve mathematical problems involving the divisions of a certain number of items in equal parts. The result of such problems can be rather interesting — for example, preferring more divisions rather than smaller fractions.
What is the Egyptian fraction of 4/5?
Using the greedy algorithm to find the Egyptian fraction expansion, we can convert 4/5 in Egyptian fraction: 4/5 = 1/2 + 1/4 + 1/120. As you can see, the denominator grew pretty quickly!
Are Egyptian fractions unique?
For many fractions, the Egyptian fraction expansion is not unique, and it is possible to find different decompositions. However, every number has an expansion: this was already proved centuries ago by the Italian mathematician Fibonacci!
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Comparing Quatities - Practically Study Material
8.1. RECALLING RATIOS AND PERCENTAGES
We know, ratio means comparing two quantities.
The comparison can be done by using fractions as,
\frac{5}{20}=\frac{1}{4}
The number of oranges are
\frac{1}{4}
th the number of apples. In terms of ratio, this is 1 : 4, read as, “1 is to 4”
Number of apples to number of oranges
=\frac{20}{5}=\frac{4}{1}
which means, the number of apples are 4 times the number of oranges. This comparison can also be done using percentages.
So percentage of orange is
\frac{5}{25}×\frac{4}{4}=\frac{20}{100}=20%
[Denominator made 100]
Out of 25 fruits number of oranges are 5. So out of 100 fruits, number of oranges
\frac{5}{25}×100=20
Since contains only apples and oranges, So, percentage of apples + percentage of oranges = 100 or percentage of apples + 20 = 100 or percentage of apples = 100 – 20 = 80
8.2. FINDING THE INCREASE OR DECREASE PERCENT
We often come across such information in our daily life as.
(i) 25% off on marked prices
(ii) 10% hike in the price of petrol
Example : The price of a scooter was Rs. 34,000 last year. It has increased by 20% this year. What is the price now ?
First find the increase in the price, which is 20% of Rs. 34,000, and then find the new price.
20% of Rs 34000
=Rs\frac{20}{100}×34000
New price = Old price + Increase
=Rs 34,000+Rs 6,800
20% increase means, Rs 100 increased to Rs 120.
So, Rs 34,000 will increase to ?
Increased price =
\mathrm{Rs}\frac{120}{100}×34000
Similarly, a percentage decrease in price would imply finding the actual decrease followed by its subtraction the from original price.
Price of scooter = Rs. 34000
Reduction = 5% of Rs. 4000
=Rs·\frac{5}{100}×34000=Rs.1700
New price = Old price – Reduction = Rs. 34000 – Rs. 1700 = Rs. 32300
8.3. FINDING DISCOUNTS
Discount is a reduction given on the Marked Price (MP) of the article.
This is generally given to attract customers to buy goods or to promote sales of the goods. You can find the discount by subtracting its sale price from its marked price.
Example : An item marked at Rs. 840 is sold for Rs. 714. What is the discount and discount % ?
= Rs. 840 – Rs. 714
On marked price of Rs. 840, the discount is Rs. 126.
On MP of Rs. 100, how much will the discount be ?
=\frac{126}{840}×100=15%
We can also find discount when discount % is given.
Your bill in a shop is Rs. 577.80 and the shopkeeper gives a discount of 15%. How would you estimate the amount to be paid ?
(i) Round off the bill to the nearest tens of Rs. 577.80, i.e., to Rs. 580.
(ii) Find 10% of this, i.e., Rs.
\frac{10}{100}×5801=\mathrm{Rs}.58
(iii) Take half of this, i.e.,
\frac{1}{2}×58=Rs.29
(iv) Add the amounts in (ii) and (iii) to get Rs. 87.
You could therefore reduce your bill amount by Rs. 87 or by about Rs. 85, which will be Rs. 495 approximately.
1. Try estimating 20% of the same bill amount.
2. Try finding 15% of Rs. 375.
8.4. PRICES RELATED TO BUYING AND SELLING (PROFIT AND LOSS)
For the school fair (mela) I am going to put a stall of lucky dips. I will charge Rs. 10 for one lucky dip but I will buy items which are worth Rs. 5.
So you are making a profit of 100%.
No, I will spend Rs. 3 on paper to wrap the gift and tape. So my expenditure is Rs. 8.
This gives me a profit of Rs. 2, which is,
\frac{2}{8}×100=25%
Sometimes when an article is bought, some additional expenses are made while buying or before selling it. These expenses have to be included in the cost price.
These expenses are sometimes referred to as overhead charges. These may include expenses like amount spent on repairs, labour charges, transportation etc.
Finding cost price/selling price, profit %/loss%
Example : Sohan bought a second hand refrigerator for Rs. 2,500, then spent Rs. 500 on its repairs and sold it for Rs. 3,300. Find his loss or gain percent.
Solution : Cost Price (CP) = Rs. 2500 + Rs. 500 (overhead expenses are added to give CP)
Sale Price (SP) = Rs. 3300
As SP > CP, he made a profit = Rs. 3300 – Rs. 3000 = Rs. 300
His profit on Rs. 3,000, is Rs. 300. How much would be his profit on Rs. 100 ?
\text{Profit }=\frac{300}{3000}×100%=\frac{30}{3}%=10% \overline{)\mathrm{P}%=\frac{\mathrm{P}}{\mathrm{CP}}×100}
8.5. SALES TAX/VALUE ADDED TAX
The teacher showed the class a bill in which the following heads were written.
Bill amount + ST (5%)
ST means Sales Tax, which we pay when we buy items.
This sales tax is charged by the government on the sale of an item.
It is collected by the shopkeeper from the customer and given to the government.
This is, therefore, always on the selling price of an item and is added to the value of the bill.
These days however, the prices include the tax known as Value Added Tax (VAT).
We might have come across statements like “one year interest for FD (fixed deposit) in the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’.
Interest is the extra money paid by institutions like banks or post offices on money deposited (kept) with them. Interest is also paid by people when they borrow money.
We already know how to calculate Simple Interest.
We have some money in the bank. Every year some interest is added to it, which is shown in the passbook. This interest is not the same, each year it increases. Normally, the interest paid or charged is never simple. The interest is calculated on the amount of the previous year.
This is known as interest compounded or Compound Interest (C.I.).
Let us take an example and find the interest year by year. Each year our sum or principal changes.
A sum of Rs. 20,000 is borrowed by Heena for 2 years at an interest of 8% compounded annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of 2 years. Aslam asked the teacher whether this means that they should find the interest year by year. The teacher said ‘yes’, and asked him to use the following steps :
1. Find the Simple Interest (S.I.) for one year.
Let the principal for the first year be
{P}_{1}
{P}_{1}
= Rs. 20,000
S{I}_{1}
= SI at 8% p.a.
for 1st year = Rs.
\frac{20000×8}{100}=Rs.1600
2. Then find the amount which will be paid or received. This becomes principal for the next year.
Amount at the end of 1st year
{P}_{1}
S{I}_{1}
= Rs. 20000 + Rs. 1600
= Rs. 21600 =
{P}_{2}
(Principal for 2nd year)
3. Again find the interest on this sum for another year.
{\mathrm{SI}}_{2}=\mathrm{SI}\text{ at }8%\text{ p.a. for }2\text{nd year }=\text{ Rs. }\frac{21600×8}{100}=\mathrm{Rs}.1728
4. Find the amount which has to be paid or received at the end of second year.
Amount at the end of 2nd year =
{\mathrm{P}}_{2}+{\mathrm{SI}}_{2}
= Rs. 21600 + Rs. 1728 = Rs. 23328
Total interest given = Rs. 1600 + Rs. 1728 = Rs. 3328
Reeta asked whether the amount would be different for simple interest. The teacher told her to find the interest for two years and see for herself.
SI for 2 years = Rs.
\frac{20000×8×2}{100}=\mathrm{Rs} 3200
Reeta said that when compound interest was used Heena would pay Rs. 128 more.
8.7. FORMULA FOR COMPOUND INTEREST
{p}_{1}
is the sum on which interest is compounded annually at a rate of R% per annum.
{p}_{1}
= Rs. 5000 and R = 5% per annum. Then by the steps mentioned above
{\mathrm{SI}}_{1}=Rs·\frac{5000×5×1}{100}
\text{or }{\mathrm{SI}}_{1}=Rs·\frac{{\mathrm{P}}_{1}×\mathrm{R}×1}{100}
\text{so, }{A}_{1}=Rs.5000+\frac{5000×5×1}{100}
\text{or }{A}_{1}={P}_{1}+S{I}_{1}={P}_{1}+\frac{{P}_{1}R}{100}
=Rs.5000\left(1+\frac{5}{100}\right)={\mathrm{P}}_{2}
\text{or }={P}_{1}\left(1+\frac{R}{100}\right)={P}_{2}
{\mathrm{SI}}_{2}=Rs.5000\left(1+\frac{5}{100}\right)×\frac{5×1}{100}
\text{or }{\mathrm{SI}}_{2}=\frac{{\mathrm{P}}_{2}×\mathrm{R}×1}{100}
=Rs.\frac{5000×5}{100}\left(1+\frac{5}{100}\right)
\text{or }={P}_{1}\left(1+\frac{R}{100}\right)×\frac{R}{100}=\frac{{P}_{1}R}{100}\left(1+\frac{R}{100}\right)
{\mathrm{A}}_{2}=\mathrm{Rs}.5000\left(1+\frac{5}{100}\right)+Rs\frac{5000×5}{100}\left(1+\frac{5}{100}\right)
{\mathrm{A}}_{2}={\mathrm{P}}_{2}+{\mathrm{SI}}_{2}
=Rs.5000\left(1+\frac{5}{100}\right)\left(1+\frac{5}{100}\right)={\mathrm{P}}_{1}\left(1+\frac{\mathrm{R}}{100}\right)+{\mathrm{P}}_{1}\frac{\mathrm{R}}{100}\left(1+\frac{\mathrm{R}}{100}\right)
=Rs.5000{\left(1+\frac{5}{100}\right)}^{2}={P}_{3}={P}_{1}\left(1+\frac{R}{100}\right)\left(1+\frac{R}{100}\right)={P}_{1}{\left(1+\frac{R}{100}\right)}^{2}
Proceeding in this way the amount at the end of n years will be
{A}_{n}={P}_{1}{\left(1+\frac{R}{100}\right)}^{n}
Or, we can say
A=P{\left(1+\frac{R}{100}\right)}^{n}
Using this we get only the formula for the amount to be paid at the end of n years, and not the formula for compound interest.
Aruna at once said that we know that CI = A – P, so we can easily find the compound interest too.
8.8. RATE COMPOUNDED ANNUALLY OR HALF YEARLY (SEMI ANNUALLY)
Time period and rate when interest not compounded annually
The time period after which the interest is added each time to form a new principal is called the conversion period. When the interest is compounded half yearly, there are two conversion periods in a year each after 6months. In such situations, the half yearly r ate will be half of the annual rate. What will happen if interest is compounded quarterly? In this case, there are 4 conversion periods in a year and the quarterly rate will be one-fourth of the annual rate.
8.9. APPLICATIONS OF COMPOUND INTEREST FORMULA
There are some situations where we could use the formula for calculation of amount in CI. Here are a few.
(iii) The value of an item, if its price increases or decreases in the intermediate years.
The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
There is 5% increase in population every year, so every new year has new population.
Thus, we can say it is increasing in compounded form.
Population in the beginning of 1998 = 20000 (we treat this as the principal for the
Increase at
5%=\frac{5}{100}×20000=1000
Population in 1999 = 20000 + 1000 = 21000 [Treat as the Principal for the 2nd year.]
5%=\frac{5}{100}×21000=1050
Population in 2000 = 21000+1050 = 22050 [Treat as the Principal for the 3rd year.]
5%=\frac{5}{100}×22050=1102.5
At the end of 2000 the population = 22050 + 1102.5 = 23152.5 or,
Population at the end of 2000
=20000{\left(1+\frac{5}{100}\right)}^{3}
=20000×\frac{21}{20}×\frac{21}{20}×\frac{21}{20}
So, the estimated population = 23153.
← Cubes and Cube RootsAlgebraic Expressions and Identities →
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Without your calculator, evaluate the following limits.
\lim\limits_ { x \rightarrow 3 ^ { + } } \sqrt { x - 3 }
Notice that this is a right-sided limit because
x = 3
is an endpoint on
y=\sqrt{x-3}.
There is no limit from the left of the endpoint; so, consequently, this must be written as a one-sided limit.
\lim\limits_ { x \rightarrow \infty } \frac { x ^ { 2 } - 2 x + 1 } { x ^ { 3 } }
This is a limit to infinity. We are looking for end-behavior. Compare the highest-power term in the numerator and the denominator.
\lim\limits_{x\rightarrow \infty }\frac{x^{2}}{x^{3}}=\lim\limits_{x\rightarrow \infty }\frac{1}{x}= 0
There is a horizontal asymptote of
y = 0
\lim\limits_ { x \rightarrow \pi } \frac { \operatorname { cos } x + 1 } { x - \pi }
This is Ana's Definition of the Derivative:
(x) =\operatorname{cos}x
a = π
So Ana's derivative can be evaluated as:
The limit is equal to
f^\prime(π)
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Section 47.2 (08XI): Essential surjections and injections—The Stacks project
Section 47.2: Essential surjections and injections (cite)
47.2 Essential surjections and injections
We will mostly work in categories of modules, but we may as well make the definition in general.
An injection $A \subset B$ of $\mathcal{A}$ is essential, or we say that $B$ is an essential extension of $A$, if every nonzero subobject $B' \subset B$ has nonzero intersection with $A$.
A surjection $f : A \to B$ of $\mathcal{A}$ is essential if for every proper subobject $A' \subset A$ we have $f(A') \not= B$.
Some lemmas about this notion.
Lemma 47.2.2. Let $\mathcal{A}$ be an abelian category.
If $A \subset B$ and $B \subset C$ are essential extensions, then $A \subset C$ is an essential extension.
If $A \subset B$ is an essential extension and $C \subset B$ is a subobject, then $A \cap C \subset C$ is an essential extension.
If $A \to B$ and $B \to C$ are essential surjections, then $A \to C$ is an essential surjection.
Given an essential surjection $f : A \to B$ and a surjection $A \to C$ with kernel $K$, the morphism $C \to B/f(K)$ is an essential surjection.
Lemma 47.2.3. Let $R$ be a ring. Let $M$ be an $R$-module. Let $E = \mathop{\mathrm{colim}}\nolimits E_ i$ be a filtered colimit of $R$-modules. Suppose given a compatible system of essential injections $M \to E_ i$ of $R$-modules. Then $M \to E$ is an essential injection.
Proof. Immediate from the definitions and the fact that filtered colimits are exact (Algebra, Lemma 10.8.8). $\square$
Lemma 47.2.4. Let $R$ be a ring. Let $M \subset N$ be $R$-modules. The following are equivalent
$M \subset N$ is an essential extension,
for all $x \in N$ nonzero there exists an $f \in R$ such that $fx \in M$ and $fx \not= 0$.
Proof. Assume (1) and let $x \in N$ be a nonzero element. By (1) we have $Rx \cap M \not= 0$. This implies (2).
Assume (2). Let $N' \subset N$ be a nonzero submodule. Pick $x \in N'$ nonzero. By (2) we can find $f \in R$ with $fx \in M$ and $fx \not= 0$. Thus $N' \cap M \not= 0$. $\square$
The statement of 45.2.4 (2) should read: for all
x \in N
nonzero ...
Thanks Dario. Fixed here.
Comment #3442 by Sebastian Bozlee on July 27, 2018 at 20:21
The second-to-last sentence of the proof of Lemma 08XM should read: By (2) we can find
f \in R
fx \in M
fx \neq 0
I also have to ask if a compatible system of essential injections
E_i \to M
yielding an essential injection
E \to M
was intended in Lemma 08XL. Both statements are true, but it feels odd not to invoke the universal property of the colimit to get the map between E and M.
Thanks for the fix. See changes here.
Lemma 47.2.3 is stated as used later, so I think it is fine as is.
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Harmonic Number Calculator
What is a harmonic number? Harmonic number equation
How do you calculate a harmonic number for integers?
How do you find the harmonic number of a non-integer?
Calculating harmonic series sums and harmonic numbers
How to use this harmonic number calculator
Our harmonic number calculator is the perfect solution for anyone seeking to find the
n
-th harmonic number or to calculate the sum of the harmonic series of the first
n
It is common to have misconceptions regarding the harmonic number and harmonic series. So we shall discuss some basic concepts on this subject. Specifically, let's answer these questions:
What is a harmonic number and its equation?
How do you calculate a harmonic number for integers and non-integers?
What is the formula for the sum of harmonic series?
What is the relation between harmonic number and harmonic series?
Whether you're a novice or an advanced reader, we hope that you'll learn something interesting here today. So grab your favorite snacks and explore this topic with us!
We define the
n
-th harmonic number as the sum of the reciprocals of the first
natural numbers:
\small \begin{align*} H_n &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}\\ &= \sum_{k=1}^n\frac{1}{k} \end{align*}
H_n
n
-th harmonic number; and
n
is any natural number.
As a consequence of Bernard's postulate,
H_n
is never an integer unless
n =1
🔎 Notice that the harmonic numbers are a rough approximation of the natural logarithm, which is given by
\ln n = \int_1^n \frac{1}{x}dx
. To learn more about the natural logarithm, use our natural log calculator.
To calculate the harmonic number Hₙ for any integer n, use the following steps:
Divide 1 by the first n natural numbers and gather them in a sequence to get 1/1, 1/2, 1/3, … 1/n.
Add every number in this sequence to get the n-th harmonic number as Hₙ = 1 + 1/2 + 1/3 + … + 1/n.
Verify your answer using our harmonic number calculator.
For example, to calculate the 5th harmonic number
H_5
, we must evaluate the sum:
\small \begin{align*} H_5 &= \sum_{n=1}^5 \frac{1}{n} \\ &= \frac{1}{1} +\frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} \\ &= \frac{137}{60} \approx 2.28333 \end{align*}
Test your understanding of how to find a harmonic number: Calculate the 8th harmonic number
H_8
yourself using the harmonic number equation. You can verify your answer using our calculator or our FAQ section.
Strictly speaking, harmonic numbers are defined only for natural numbers. However, we can use the following expression to interpolate harmonic numbers for non-integers:
\small \psi(n) = H_{n-1} - \gamma
\psi(n)
is the digamma function; and
\gamma
is the Euler–Mascheroni constant, which is roughly
0.577
Let's break this down, starting with the digamma function, which we define as the logarithmic derivative of the gamma function:
\small \psi(x) = \frac{d}{dx}\ln\left(\Gamma(x)\right) = \frac{\Gamma'(x)}{\Gamma(x)}
\ln(x)
is the natural logarithmic function; and
\Gamma(x)
Please don't panic — you won't need to evaluate these complicated functions to find our harmonic number. Still, it helps to learn more about this gamma function using our gamma function calculator.
There are many methods to evaluate this digamma function, but the simplest one we can use is the following expression, which is valid for any complex number
z
whose real component is positive (i.e.,
\Re(z)>0
\small\psi(z) = \sum_{k=0}^{\infty}\frac{z-1}{(k+1)(k+z)}-\gamma
Re-writing this equation for
z = n + 0i
n
is an arbitrary positive non-integer, we would get:
\small\psi(n) = \sum_{k=0}^{\infty}\frac{n-1}{(k+1)(k+n)} -\gamma
Combining this equation with the initial interpolation relation we've introduced (
\psi(n) = H_{n-1} - \gamma
), we can deduce that
\small\begin{align*} H_{n-1} - \gamma &=\sum_{k=0}^{\infty}\frac{\footnotesize n-1}{\footnotesize (k+1)(k+n)} -\gamma\\ H_{n-1} &= \sum_{k=0}^{\infty}\frac{\footnotesize n-1}{\footnotesize (k+1)(k+n)}\\ \therefore H_n &= \sum_{k=0}^{\infty}\frac{\footnotesize n}{\footnotesize (k+1)(k+n+1)} \end{align*}
And voila! An equation to evaluate harmonic numbers for any non-integer
n
is at hand! And as we promised, you needn't calculate any extra functions to find the answer. Also, you need not sum this series to infinity — this series will converge after a finite number of iterations.
Now that you've learned how to calculate harmonic numbers, let's find out how they are related to calculating the sum of harmonic series.
The harmonic series is the sum of the reciprocals of all natural numbers, given by:
\sum_{n=1}^\infty \frac{1}{n}= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dotsb
While its definition is similar to that of a harmonic number, you can see from the formula for the sum of harmonic series it is an infinite series. The harmonic number is considered a partial sum of the harmonic series, where the series contains the reciprocals of the first
natural numbers, instead of all natural numbers.
If you're interested in how harmonic series and music are related, we urge you to check out our harmonic series calculator.
This harmonic number calculator is easy to use and is always here to help you. To calculate the harmonic number of a positive number
n
, or to calculate the sum of harmonic series up to the
n
-th term:
Enter the positive number in the field labeled n.
This calculator will evaluate the
n
-th harmonic number
H_n
and display it below.
n
H_n
has a determinable fraction form, this calculator will provide it alongside its decimal form.
For non-integer values of
n
, we can only provide
H_n
in decimal form.
Note that we round all decimal answers to 5 decimal places. Also, to keep the computational time to a minimum, we've capped the maximum value of an integer
n
10^6
, beyond which the algorithm will cause a perceptible lag in most machines. For the same reason, a non-integer
n
10^3
Don't hesitate to contact us if you want to compute a harmonic number of any number larger than these limits.
What is the 8ᵗʰ harmonic number?
H₈ = 761/280 = 2.71786. To calculate the 8th harmonic number yourself, follow these instructions:
Divide 1 by the first 8 natural numbers and gather them in a sequence to get 1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8.
Add every number in this sequence to get the n-th harmonic number as Hₙ = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 = 761/280 = 2.71786.
Is the harmonic series a p-series?
Yes. The harmonic series is a special case of p-series given by ∑ 1/nᵖ, where p = 1.
Does the harmonic series converge?
No, the harmonic series does not converge. The consecutive numbers in the series do not get small fast enough to enable a convergence.
nᵗʰ Harmonic number (Hₙ)
Area of triangle with coordinates
Use this simple tool to calculate the area of a triangle with its coordinates.
Area of Triangle with Coordinates Calculator
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Thin-film interference and optical path difference
How to calculate minimum anti-reflective coating thickness
How to calculate reflectivity
How to use this thin-film optical coating calculator
This thin-film optical coating calculator will calculate the reflectivity and interference type of light reflected by a thin optical film. We can evaluate the optical path difference and reflectivity by using Fresnel equations, Snell's law, and trigonometry. Then we can discern the type of interference. Alternatively, we can determine the minimum anti-reflection coating thickness required for a particular wavelength of light.
In this article, we will learn more about thin-film interference, optical path difference, and how to calculate reflectivity.
When light hits the interface between two different mediums, a portion of it is reflected back into the first medium, while the remaining light is transmitted through to the second medium. Depending on the material, some light could be absorbed. The transmitted light undergoes refraction if there is a difference in the refractive indices.
If we place a thin-film between these two mediums, something interesting happens – the top and the bottom interfaces will reflect light into the first medium, resulting in interference of light waves. Learn more about the dual nature of light with our De Broglie wavelength calculator.
Interference of reflected light when there is a thin layer of different material between two mediums.
In order to assess the type of interference, we need to calculate the optical path difference, which describes the phase shift the two different light waves undergo by virtue of their different optical path length. It is given by:
\qquad OPD = 2n_2d\cos(\theta_2)
OPD
– Optical path difference between the two reflected light waves;
n_2
– Refractive index of the thin-film;
d
– Thickness of the thin-film; and
\theta_2
– Refraction angle in the thin-film.
Furthermore, if the refractive index of the first medium is less than the second medium, then there will be a
\text{180}\degree
\pi
radian phase change in the reflected light.
The reflected light waves can interfere in the following ways:
Constructive interference: When the reflected light waves' crests and troughs overlap each other and increase the resultant wave amplitude.
Destructive interference: When the reflected light wave's crests overlap with troughs, destroying the resultant wave amplitude.
Intermediate: Overlapping state, resulting in a wave with amplitude somewhere between the maximum and minimum values.
Based on the OPD and phase change, the conditions for different interference types are given in the following table:
Conditions for different types of interference based on phase shift and optical path difference (OPD).
180\degree
At top and bottom interfaces
At top or bottom interface
n_1\lt n_2, n_2\lt n_3
n_1\gt n_2, n_2\lt n_3
n_1\lt n_2, n_2\gt n_3
OPD for constructive interference
m\lambda
\left(m-\frac{1}{2}\right)\lambda
OPD for destructive interference
\left(m-\frac{1}{2}\right)\lambda
m\lambda
n_1
– Refractive index of the first medium;
n_2
n_3
– Refractive index of the final medium or substrate;
m
– Any positive integer; and
\lambda
– Light's wavelength.
n_1>n_2>n_3
, there is no phase shift in the reflected light at both the top and bottom interfaces.
Anti-reflective coating is a thin optical coating over a transparent material (say, glass) used to reduce reflection. We can reduce reflection by ensuring that the reflected light waves interfere destructively and eliminate each other. Multiple layers of such coating are often employed each targeting specific light wavelengths.
The following equations and inequalities summarize the conditions for destructive interference in such a coating:
\ \ \ \begin{align*} &n_{\text{air}}<n_{\text{film}}<n_{\text{glass}}\\ &2n_{\text{film}}d\cos(\theta_2) = \left(m-\frac{1}{2}\right)\lambda\\ \end{align*}
This leads to the following minimum anti-reflective coating thickness:
\qquad d_{\text{min}} = \frac{\lambda}{4n_{\text{film}}}
Reflectance, reflectivity, or power reflection coefficient, refers to the fraction of incident light reflected at the interface. Fresnel equations for s-polarized and p-polarized reflectivity for non-magnetic media are given by:
\quad\begin{align*} R_s = \left|\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\right|^2\\ R_p = \left|\frac{n_1\cos\theta_t-n_2\cos\theta_i}{n_1\cos\theta_t+n_2\cos\theta_i}\right|^2 \end{align*}
R_s
– s-polarized reflectivity;
\theta_i
– Angle of incidence;
\theta_t
– Angle of transmittance or angle of refraction; and
R_p
– p-polarized reflectivity.
Similarly, transmissivity is a measure of light transmitted at the interface. It is given by:
\quad\begin{align*} T_s = 1-R_s\\ T_p = 1-R_p \end{align*}
T_s
– s-polarized transmissivity; and
T_p
– p-polarized transmissivity.
But how do you calculate reflectivity in our case, where you have three mediums involved? One way to do so would be:
Evaluate the reflectivity at the first interface,
R_{s1}
R_{p1}
Calculate the reflectivity at the second interface,
R_{s2}
R_{p2}
Evaluate the transmissivity of this reflected light at the first interface,
T_{s21}
T_{p21}
(the indices '21' represent transmissivity from the second medium to the first).
Calculate the actual fraction of light reflected from the second interface to the first medium as
R_{s21}=R_{s1} \times T_{s21}
R_{p21}=R_{p1} \times T_{p21}
Finally, obtain overall reflectivity as
R_s = R_{s1} + R_{s21}
R_p = R_{p1} + R_{p21}
🔎 A special case of these Fresnel equations yields Brewster's angle. It is the angle at which all reflected light is of one particular polarisation while the refracted/transmitted light is of the other polarisation. Learn more about Brewster's angle with our brewster angle calculator.
This thin-film optical coating calculator has a lot going on under its hood, but it is relatively straightforward to use:
Enter the incident angle and the optical film thickness, along with refractive indices of the three mediums involved. The calculator will calculate reflectivity, optical path difference, and interference type.
Enter the film's wavelength and refractive index, and this thin-film optical coating calculator will evaluate the minimum thickness of the anti-reflective coating.
Incident Angle (θ₁)
Optical film thickness (d)
Wavelength and refractive indices
Refractive index of first medium (n₁)
Refractive index of optical film (n₂)
Refractive index of substrate (n₃)
OPD and reflectivity
s-polarized reflectivity (Rₛ)
p-polarized reflectivity (Rₚ)
Min anti-reflection coating thickness
|
Algebraic Expressions and Identities - Practically Study Material
a) What are expressions
In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are.
Examples of expressions are:
x+3, 2y–5, 3{x}^{2}, 4xy+7
We can form many more expressions. As we know expressions are formed from variables and constants.
The expression 2y – 5 is formed from the variable y and constants 2 and 5. The expression 4xy + 7 is formed from variables x and y and constants 4 and 7.
We know that, the value of y in the expression 2y – 5, may be anything. It can be
2, 5, –3, 0, \frac{5}{2},–\frac{7}{3}
etc.; actually countless different values. The value of an expression changes with the value chosen for the variables it contains. Thus as y takes on different values, the value of 2y – 5 goes on changing. When y = 2, 2y – 5 = 2(2) – 5 = –1; when y = 0, 2y – 5 = 2 × 0 –5 = –5, etc.
Consider the expression x + 5. Let us say the variable x has a position X on the number line;
X may be anywhere on the number line, but it is definite that the value of x + 5 is given by a point P, 5 units to the right of X. Similarly, the value of x – 4 will be 4 units to the left of X and so on.
Let’s see the position of 4x and 4x + 5
The position of 4x will be point C; the distance of C from the origin will be four times the distance of X from the origin. The position D of 4x + 5 will be 5 units to the right of C.
b) Terms, factors and Coefficients
Take the expression 4x + 5. This expression is made up of two terms, 4x and 5. Terms are added to form expressions. Terms themselves can be formed as the product of factors. The term 4x is the product of its factors 4 and x. The term 5 is made up of just one factor, i.e., 5.
The expression 7xy – 5x has two terms 7xy and –5x. The term 7xy is a product of factors 7, x and y. The numerical factor of a term is called its coefficient. The coefficient in the term 7xy is 7 and the coefficient in the term –5x is –5.
c) Monomials, binomials and polynomials
Expression that contains only one term is called a monomial. Expression that contains two terms is called a binomial. An expression containing three terms is a trinomial and so on. In general, an expression containing, one or more terms with non-zero coefficient (with variables having non negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.
Examples of monomials:
4{x}^{2}, 3xy, –7z, 5x{y}^{2}, 10y, –9,82mnp
Examples of binomials:
\mathrm{a}+\mathrm{b}, 4l+5\mathrm{m}, \mathrm{a}+4, 5–3\mathrm{xy}, {\mathrm{z}}^{2}–4{\mathrm{y}}^{2}
Examples of trinomials:
a+b+c, 2x+3y–5, {x}^{2}y–x{y}^{2}+{y}^{2}
Examples of polynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.
d) Like and Unlike terms
Look at the following expressions:
7x,14x,–13x,5{x}^{2},7y,7xy,–9{y}^{2},–9{x}^{2},–5yx
Like terms from these are:
(i) 7x, 14x, –13x are like terms.
5{x}^{2}
–9{x}^{2}
are like terms.
(iii) 7xy and –5yx are like terms.
Unlike terms from these are:
(i) 7x and 7y are unlike terms.
(ii) 7x and 7xy are unlike terms.
(iii) 7x and
5{x}^{2}
are unlike terms.
9.2. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS
Let’s add
7{x}^{2}–4x+5
and 9x – 10, we do
7{x}^{2}–4x+5\phantom{\rule{0ex}{0ex}}+ 9x–10\phantom{\rule{0ex}{0ex}} \overline{)7{x}^{2}+5x–5 }
During addition of algebraic expression, we write each expression to be added in a separate row. While doing so we write like terms one below the other, and add them, as shown above
Thus 5 + (–10) = 5 –10 = –5.
Similarly, – 4x + 9x = (– 4 + 9)x = 5x.
Let us take some more examples.
9.3. MULTIPLICATION OF MONOMIALS WITH ALGEBRAIC EXPRESSIONS
I. Multiplying two monomials
4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Let’s observe the multiplication of few monomials.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y = –15xy
Note: We observe that all the three products of monomials, 3xy, 15xy, –15xy, are also monomials.
(iv) 5x × 4x2 = (5 × 4) × (x × x2)
= 20 × x3 = 20x3
Note: 5 × 4 = 20
i.e., coefficient of product = coefficient of first monomial × coefficient of second monomial; and x × x2 = x3
algebraic factor of product = algebraic factor of first monomial × algebraic factor of second monomial.
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x2 yz
Let see few more examples.
Suppose we have two monomials 4x and 5y. By multiplying them, we get
4x × 5y = (4 × x) × (5 × y)
= (4 × 5) × (x × y) = 20xy
Hence, we can say that 4x × 5y = 20xy
Now, 10x × 5x2z = (10 × x) × (5 × x2 × z)
This becomes (10 × 5) × (x × x2 × z) = 50x3z
II. Multiplying three or more monomials
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x2 y2 × 6x3y3 = (4xy × 5x2y2 ) × 6x3y3
= 20x3y3 × 6x3y3 = 120x3y3 × x3y3
= 120 (x3×x3) × (y3×y3) = 120x6×y6 = 120x6y6
It is clear that we first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.
Example: Complete the table for area of a rectangle with given length and breadth.
3x×5y = 15xy
9y×4y2 = 36y3
4ab×5bc = 20ab2c
2l2m×3lm2 = 6l3m3
III. Multiplication of a monomial with a polynomial
a) Multiplying a monomial by a binomial
Let us multiply the monomial 3x by the binomial
5y + 2, i.e., find 3x × (5y + 2)
Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law,
3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x
7 × 106 = 7 × (100 + 6) = 7 × 100 + 7 × 6
(Here, we used distributive law)
= 7 × 40 – 7 × 2
Similarly, (–3x) × (–5y + 2)
= (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x
and 5xy × (y2 + 3) = (5xy × y2 ) + (5xy × 3)
Multiplication of binomial × monomial?
For example, (5y + 2) × 3x = ?
We may use commutative law as : 7 × 3 = 3 × 7 (or) in
general a × b = b × a
Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x
b) Multiplying a monomial by a trinomial
Consider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law;
3p × (4p2 + 5p + 7) = (3p × 4p2 ) + (3p × 5p) + (3p × 7)
Observe, by using the distributive law, we are able to carry out the multiplication term by term.
Example: Simplify the expressions and evaluate them as directed:
(i) x (x – 3) + 2 for x = 1
(ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2
Solution: (i) x (x – 3) + 2 = x2 – 3x + 2
For x = 1, x2 – 3x + 2 = (1)2 – 3 (1) + 2
(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y2 – 21y – 3y + 12 – 63
= 6y2 – 24y – 51
For y = –2, 6y2 – 24y – 51 = 6 (–2)2 – 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
9.4. MULTIPLICATION OF BINOMIALS WITH BINOMIALS AND TRINOMIALS
I. Multiplication of a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b).
Observe, every term in one binomial multiplies every term in the other binomial.
(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
= 6a2 + 9ab + 8ba + 12b2
(Since ba = ab)
When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be present. But two of these are like terms, which are combined, and hence we get 3 terms.
In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.
In multiplication of polynomials with polynomials, we sholud always look for like terms, if any, and combine them.
II. Multiplication of a binomial with a trinomial
In this multiplication, we shall have to multiply each of the three terms in the trinomial by each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may reduce to 5 or less, if the term by term multiplication results in like terms. Consider
\underset{\text{binomial }}{\underset{⏟}{\left(a+7\right)}}×\underset{\text{trinomial }}{\underset{⏟}{\left({a}^{2}+3a+5\right)}}
= a × (a + 3a + 5) + 7 × (a + 3a + 5)
[using the distributive law]
= a3 + (3a2 + 7a2 ) + (5a + 21a) + 35
Consider the equality (a + 1) (a +2) = a2 + 3a + 2
We shall evaluate both sides of this equality for some value of a, say a = 10.
LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2)
RHS = a2 + 3a + 2 = 102 + 3 × 10 + 2
Thus, the values of the two sides of the equality are equal for a = 10.
Let us now take a = –5
LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2)
= (– 4) × (–3) = 12
RHS = a2 + 3a + 2 = (–5)2 + 3 (–5) + 2
= 25 – 15 + 2 = 10 + 2 = 12
Thus, for a = –5, also LHS = RHS.
We shall find that for any value of a, LHS = RHS Such an equality, true for every value of the variable in it, is called an identity. Thus,
(a + 1) (a + 2) = a2 + 3a + 2 is an identity.
An equation is true for only certain values of the variable in it. It is not true for all values of the variable. For example, consider the equation
a2 + 3a + 2 = 132
It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc.
9.6. STANDARD IDENTITIES
I. Use of the Identities (a + b)2 and (a – b)2
Let us try to find the square of the number 102. The square of a number, as we know, is the product of the number with itself. One way to do this is by writing the numbers one below the other, and then multiplying them as we normally do. The other way is to break the numbers and then apply distributive property. This will make our work much easier.
= 100 (100 + 2) + 2 (100 + 2)
Let us try to find the square of the expression (a + b).
(a + b)2 = (a + b) (a + b) = a (a + b) + b (a + b)
\overline{)\therefore {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}}
This is, in fact, an important identity. It is known as an identity because for any value of a and b, the LHS is always equal to the RHS. The difference between an identity and an equation is that for an equation, its LHS and RHS are equal only for some values of the variable.
On the other hand, as we discussed, for an identity, the LHS equals the RHS for any value of the variable.
There is another important identity for the square of the expression (a – b). Let us try and simplify it.
(a – b)2 = (a – b) (a – b)
= a (a – b) – b (a – b)
= a × a + a × (–b) – b × a – b × (–b)
= a2 – ab – ab + b2
\overline{)\therefore {\left(a–b\right)}^{2}={a}^{2}–2ab+{b}^{2}}
Many a times, these identities help in shortening our calculations.
II. Use of the Identity (x + a) (x + b)
A very important identity that we have to learn is regarding the expression (x + a) (x + b).
We know how to multiply binomials. By finding the product of these binomials, we can find the identity.
Let’s understand this identity from the following examples:
Example: Use the appropriate identity to simplify the following expressions.
(a) (3p + 5q) (3p – 7z)
\left({z}^{2}–\frac{4}{3}\right) \left({z}^{2}–\frac{3}{2}\right)
(a) The given expression is (3p + 5q)(3p – 7z).
This expression is of the form (x + a) (x + b).
Thus, we can use the identity (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 3p, a = 5q, b = –7z
\therefore
(3p + 5q)(3p – 7z)
= (3p)2 + (5q – 7z)(3p) + (5q)(–7z)
= 9p2 + (5q)(3p) – (7z)(3p) – 35qz
= 9p2 + 15pq – 21pz – 35qz
(b)The given expression is
\left({z}^{2}–\frac{4}{3}\right)\left({z}^{2}–\frac{3}{2}\right)
Here, x = z2,
a=–\frac{4}{3}, b=–\frac{3}{2}
\therefore \left({z}^{2}–\frac{4}{3}\right)\left({z}^{2}–\frac{3}{2}\right)
={\left({z}^{2}\right)}^{2}+\left\{\left(–\frac{4}{3}\right)+\left(–\frac{3}{2}\right)\right\}{z}^{2}+\left(–\frac{4}{3}\right)×\left(–\frac{3}{2}\right)
={z}^{4}+\left(\frac{–8–9}{6}\right){z}^{2}+2={z}^{4}–\frac{17}{6}{z}^{2}+2
III. Use of the Identity (a + b) (a – b)
Suppose we need to find the product of the numbers 79 and 81. Instead of multiplying these two numbers, we can use the identity (a + b) (a – b). This identity is very important and is applicable in various situations.
Let us first understand this identity.
(a + b) (a – b) = a (a – b) + b (a – b)
(By distributive property)
= a2 – ab + ba – b2
= a2 – ab + ab – b2 (ab = ba)
\overline{)\therefore \left(a+b\right)\left(a–b\right)={a}^{2}–{b}^{2}}
Now, let us solve some examples in which the above identity can be applied.
Example: Simplify the following expressions.
\left(\frac{3}{7}l+\frac{4}{5}m\right)\left(\frac{3}{7}l–\frac{4}{5}m\right)
(b) (x2 – y3)(x2 + y3) + (y3 – z4)(y3 + z4) + (z4 – x2)(z4 + x2)
(a) The given expression is
\left(\frac{3}{7}l+\frac{4}{5}m\right)\left(\frac{3}{7}l–\frac{4}{5}m\right)
This expression is of the form (a + b) (a – b).
Hence, we can use the identity (a + b) (a – b) = a2 – b2
\therefore \left(\frac{3}{7}l+\frac{4}{5} \mathrm{m}\right)\left(\frac{3}{7}l–\frac{4}{5} \mathrm{m}\right)={\left(\frac{3}{7}l\right)}^{2}–{\left(\frac{4}{5} \mathrm{m}\right)}^{2}
=\frac{9}{49}{l}^{2}–\frac{16}{25}{\mathrm{m}}^{2}
On using the identity (a + b) (a – b) = a2 – b2, we get
\left\{{\left({x}^{2}\right)}^{2}–{\left({y}^{3}\right)}^{2}\right\}+\left\{{\left({y}^{3}\right)}^{2}–{\left({z}^{4}\right)}^{2}\right\}+\left\{{\left({z}^{4}\right)}^{2}–{\left({x}^{2}\right)}^{2}\right\}
={x}^{4}–{y}^{6}+{y}^{6}–{z}^{8}+{z}^{8}–{x}^{4}
=\left({x}^{4}–{x}^{4}\right)+\left(–{y}^{6}+{y}^{6}\right)+\left(–{z}^{8}+{z}^{8}\right)
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Bayesian First Aid: Pearson Correlation Test - Publishable Stuff
Correlation does not imply causation, right but, as Edward Tufte writes, “it sure is a hint.” The Pearson product-moment correlation coefficient is perhaps one of the most common ways of looking for such hints and this post describes the Bayesian First Aid alternative to the classical Pearson correlation test. Except for being based on Bayesian estimation (a good thing in my book) this alternative is more robust to outliers and comes with a pretty nice default plot. :)
The classical test of Pearson product-moment correlation coefficient between two paired variables $x_i$ and $y_i$ assumes bivariate normality. It assumes that the relation is linear and that both $x_i$ and $y_i$ are normally distributed. I’ve already written about a Bayesian alternative to the correlation test here and about how that model can be made more robust here. The Bayesian First Aid alternative is basically the robustified version with slightly different priors.
Instead of a bivariate normal distribution we’ll assume a bivariate t distribution. This is the same trick as in the Bayesian First Aid alternative to the t-test, compared to the normal distribution the wider tails of the t will downweight the influence of stray data points. The model has six parameters: the means of the two marginal distributions ($\mu_x,\mu_y$), the SDs ($\sigma_x,\sigma_y$), the degree-of-freedoms parameter that influences the heaviness of the tails ($\nu$), and finally the correlation ($\rho$). The SDs and $\rho$ are then used to define the covariance matrix of the t distribution as:
The priors on $\mu_x,\mu_y, \sigma_x,\sigma_y$ and $\nu$ are also the same as in the alternative to the two sample t-test, that is, by peeking at the data we set the hyperparameters
M_\mu, S_\mu, L_\sigma
H_\sigma
resulting in a, for all practical purposes, flat prior. When modeling correlations it is common to directly put a prior distribution on the covariance matrix (the Inverse-Wishart distribution for example). Here we instead do as described by Barnard, McCulloch & Meng (2000) and put separate priors on $\sigma_x,\sigma_y$ and $\rho$, where $\rho$ is given a uniform prior. The advantage with separate priors is that it gives you greater flexibility and makes it easy to add prior information into the mix.
So, here is the final model for the Bayesian First Aid alternative to the correlation test:
The bayes.cor.test Function
The bayes.cor.test function can be called exactly as the cor.test function in R. If you just ran cor.test(x, y), prepending bayes. (like bayes.cor.test(x, y)) runs the Bayesian First Aid alternative and prints out a summary of the model result. By saving the output, for example, like fit <- bayes.cor.test(x, y) you can inspect it further using plot(fit), summary(fit) and diagnostics(fit).
To showcase bayes.cor.test I will use data from the article 2D:4D ratios predict hand grip strength (but not hand grip endurance) in men (but not in women) by Hone & McCullough (2012). The ratio here referred to is that between one’s index finger (2D) and ring finger (4D) which is nicely visualized in the Wikipedia entry for digit ratio:
So why is the 2D:4D ratio an interesting measure? It is believed that the 2D:4D ratio is affected by the prenatal exposure to androgens (hormones that control the development of male characteristics) with a lower 2D:4D ratio being indicative of higher androgen exposure. Stated in a sloppy way, the working hypothesis is that the 2D:4D ratio is a proxy variable for prenatal androgen exposure and could therefore be related to a host of other traits related to “manliness” such as aggression, prostate cancer risk, sperm count, etc. (see Wikipedia for a longer list). Hone & McCullough was interested in the relation between the 2D:4D ratio and a typical “manly” attribute, muscle strength. They therefore measured the grip strength (in kg) and the 2D:4D ratio in 222 psychology students (100 male, 122 female) and found that:
2D:4D ratios significantly predicted [grip strength] in men, but not in women. […] We conclude furthermore that the association of prenatal exposure to testosterone (as indexed by 2D:4D ratios) and strength pertains only to men, and not to women.”
Two points regarding their analysis, before moving on:
They support the conclusion above by comparing the regression p-values between the male group (p < 0.001) and the female group (p = 0.09) and notice that while the regression coefficient is less than 0.05 for the male group it is more than 0.05 for female group. It is not clear to me how p = 0.09 is strong evidence that prenatal exposure to testosterone does not influence strength in women at all.
Hone & McCullough tests the null hypothesis that there is no correlation whatsoever between 2D:4D ratio and strength. This question could be answered without any data, of course there is some correlation between 2D:4D ratio and strength (no matter how tiny). In complex systems, such as humans, it would be extremely unlikely that any given trait (such as hair color, shoe size, movie taste, running speed, no of tweets, etc.) does not correlate at all with any other trait. More interesting is to what degree 2D:4D ratio and strength correlates (which Hone & McCullough also estimated but did not base the final conclusion on).
We’ll leave Hone & McCullough’s analysis behind us (their analysis included many more variables, for one thing) and will just look at the strength and digit ratio data. The data was scraped from two scatter plots using the great and free online tool WebPlotDigitizer (so it might differ slightly from the original data, though I was pretty thorough) and can be downloaded here: 2d4d_hone_2012.csv. Here is how it looks:
d <- read.csv("2d4d_hone_2012.csv")
qplot(ratio_2d4d, grip_kg, data = d, shape = I(1)) + facet_grid(sex ~ ., scales = "free")
Let’s first estimate the correlation for the male group using the classical cor.test:
cor.test( ~ ratio_2d4d + grip_kg, data = d[d$sex == "male", ])
## data: ratio_2d4d and grip_kg
## t = -3.612, df = 95, p-value = 0.0004879
And now using the Bayesian First Aid version:
fit_male <- bayes.cor.test( ~ ratio_2d4d + grip_kg, data = d[ d$sex == "male",])
fit_male
## Bayesian First Aid Pearson's Correlation Coefficient Test
## data: ratio_2d4d and grip_kg (n = 97)
## Estimated correlation:
## -0.34
## -0.52 -0.15
## The correlation is more than 0 by a probability of 0.001
Both estimates are very similar and both indicate a slight negative correlation. Let’s look at the Bayesian First Aid plot:
plot(fit_male)
This plot shows lots of useful stuff! At the top we have the posterior distribution for the correlation $\rho$ with a 95% highest density interval. At the bottom we see the original data with superimposed posterior predictive distributions (that is, the distribution we would expect a new data point to have). This is useful when assessing how well the model fits the data. The two ellipses show the 50% (darker blue) and 95% (lighter blue) highest density regions. The red histograms show the marginal distributions of the data with a smatter of marginal densities drawn from the posterior. Looking at this plot we see that the model fits quite well, however, we could be concerned with the right skewness of the ratio_2d4d marginal which is not captured by the model.
Now let’s look at the corresponding plot for the female data:
fit_female <- bayes.cor.test( ~ ratio_2d4d + grip_kg, data = d[ d$sex == "female",])
plot(fit_female)
Indeed, 2D:4D ratio and strength seems to be slightly less correlated than for the male group, but to claim there is evidence for no correlation seems a bit unfounded. We could, however, take a look at the posterior difference in correlation between the male and the female group. To do this we first extract the MCMC samples from the Bayesian First Aid fit object using the as.data.frame function.
female_mcmc <- as.data.frame(fit_female)
male_mcmc <- as.data.frame(fit_male)
head(female_mcmc)
## rho mu1 mu2 sigma1 sigma2 nu
## 1 0.024912 0.9758 22.84 0.03196 6.209 33.14
## 2 -0.002173 0.9718 22.24 0.03336 6.082 57.15
## 4 0.017139 0.9751 22.10 0.03779 5.687 129.93
We’ll then plot the difference in $\rho$ and calculate the probability that $\rho$ is more negative for men:
hist(male_mcmc$rho - female_mcmc$rho, 30, xlim = c(-1, 1), main = "", yaxt = "n")
mean(male_mcmc$rho - female_mcmc$rho < 0)
Even though we can’t claim that 2D:4D ratio and strength is not correlated in women at all, there is some evidence that 2D:4D ratio and strength is more negatively correlated in men than in women. Given that I have a relatively high (less “manly”) 2D:4D ratio of 1.0 does it mean I have an excuse for my subpar arm strength? Nope, I should just go to the gym more often…
bayes.cor.test Model Code
A great thing with Bayesian data analysis is that it is simple to step away from the well trodden model and explore alternatives. The Bayesian First Aid function that makes this easy is model.code, if you just ran fit <- bayes.cor.test(exposure_to_puppies, rated_happiness) a call to model.code(fit) prints out corresponding R and JAGS code that can be directly copy-n-pasted into an R script and modified from there.
## Model code for the Bayesian First Aid alternative to Pearson's correlation test. ##
x <- exposure_to_puppies
y <- rated_happieness
xy[i,1:2] ~ dmt(mu[], prec[ , ], nu)
# JAGS parameterizes the multivariate t using precision (inverse of variance)
# rather than variance, therefore here inverting the covariance matrix.
# Constructing the covariance matrix
sigma[1] ~ dunif(sigmaLow, sigmaHigh)
mu[1] ~ dnorm(mean_mu, precision_mu)
# Initializing the data list and setting parameters for the priors
# that in practice will result in flat priors on mu and sigma.
xy = cbind(x, y),
n = length(x),
mean_mu = mean(c(x, y), trim=0.2) ,
precision_mu = 1 / (max(mad(x), mad(y))^2 * 1000000),
sigmaLow = max(mad(x), mad(y)) / 1000 ,
sigmaHigh = min(mad(x), mad(y)) * 1000)
inits_list = list(mu=c(mean(x, trim=0.2), mean(y, trim=0.2)), rho=cor(x, y, method="spearman"),
sigma = c(mad(x), mad(y)), nuMinusOne = 5)
params <- c("rho", "mu", "sigma", "nu")
samples <- coda.samples(model, params, n.iter=5000)
One thing that we might want to change is the prior on rho which is currently $\text{Uniform}(-1,1)$. As described by Barnard, McCulloch & Meng (2000), the beta distribution stretched to the interval [-1,1] is a flexible alternative that facilitates the construction of a more informative prior. To use this prior we’ll have to replace
rho_half_with ~ dbeta(1, 1)
# shifting and streching rho_half_with from [0,1] to [-1,1]
rho ~ 2 * rho_half_with - 1
and replace the initialization of rho with the initialization of rho_half_width:
inits_list = list(mu = c(mean(x, trim=0.2), mean(y, trim=0.2)),
rho_half_width = (cor(x, y, method="spearman") + 1) / 2,
Now, by changing the parameters of dbeta we can specify a range of different priors:
Here the top left prior is the uniform, representing little prior information on the correlation between measures of puppy exposure and happiness. The top right prior is skeptical of there being a large correlation but is agnostic with respect to the direction. The lower left prior is just slightly in favor of a positive correlation, with the lower right being quite optimistic putting more than 99% of the prior probability over rho > 0.
Posted by Rasmus Bååth Mar 17th, 2014 Bayesian, R, Statistics
« A Hack to Create Matrices in R, Matlab style Jeffreys’ Substitution Posterior for the Median: A Nice Trick to Non-parametrically Estimate the Median »
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Help:Preferences - Wikiquote
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In 2018 OpenAI made a breakthrough in Deep Reinforcement Learning, this was possible only because of solid hardware architecture and using the state of the art's algorithm: Proximal Policy Optimization.
The main idea of Proximal Policy Optimization is to avoid having too large a policy update. To do that, we use a ratio that tells us the difference between our new and old policy and clip this ratio from 0.8 to 1.2. Doing that will ensure that the policy update will not be too large.
This tutorial will dive into understanding the PPO architecture and implement a Proximal Policy Optimization (PPO) agent that learns to play Pong-v0.
However, if you want to understand PPO, you need first to check all my previous tutorials. In this tutorial, as a backbone, I will use the A3C tutorial code.
Problem with Policy Gradient
If you were reading my tutorial about Policy Gradient, we talked about the Policy Loss function:
{L}^{PG}\left(Q\right)={E}_{t}\left[\mathrm{log}{\pi }_{Q}\left({a}_{t}|{s}_{t}\right)\right]*{A}_{t}
- L(Q) - Policy loss; E - Expected; logπ... - log probability of taking that action at that state, A - Advantage.
The idea of this function is doing gradient ascent step (which is equivalent to taking reversed gradient descent); this way, our agent is pushed to take actions that lead to higher rewards and avoid harmful actions.
However, the problem comes from the step size:
Too small, the training process is too slow;
Too high, there is too much variability in training.
That's where PPO is helpful; the idea is that PPO improves the stability of the Actor training by limiting the policy update at each training step.
To do that, PPO introduced a new objective function called "Clipped surrogate objective function" that will constrain policy change in a small range using a clip.
Clipped Surrogate Objective Function
First, as explained in the PPO paper, instead of using log pi to trace the impact of the actions, PPO uses the ratio between the probability of action under current policy divided by the likelihood of the action under the previous policy:
{r}_{t}\left(Q\right)=\frac{{\pi }_{Q}\left({a}_{t}|{s}_{t}\right)}{{\pi }_{{Q}_{old}}\left({a}_{t}|{s}_{t}\right)}
As we can see, rt(Q) is the probability ratio between the new and old policy:
If rt(Q) >1, the action is more probable in the current policy than the old one.
If rt(Q) is between 0 and 1, the action is less likely for the current policy than for the old one.
As a consequence, we can write our new objective function:
{L}^{CPI}\left(Q\right)={\stackrel{^}{E}}_{t}\left[\frac{{\pi }_{Q}\left({a}_{t}|{s}_{t}\right)}{{\pi }_{{Q}_{old}}\left({a}_{t}|{s}_{t}\right)}{\stackrel{^}{A}}_{t}\right]={\stackrel{^}{E}}_{t}\left[{r}_{t}\left(Q\right){\stackrel{^}{A}}_{t}\right]
However, if the action took more probable in our current policy than in our former, this would lead to a giant policy gradient step and an excessive policy update.
Consequently, we need to constrain this objective function by penalizing changes that lead to a ratio (in the paper, it is said that the ratio can only vary from 0.8 to 1.2). To do that, we have to use the PPO clip probability ratio directly in the objective function with its Clipped surrogate objective function. By doing that, we'll ensure not to have too large policy update because the new policy can't be too different from the older one.
{L}^{CLIP}\left(Q\right)={\stackrel{^}{E}}_{t}\left[min\left({r}_{t}\left(Q\right){\stackrel{^}{A}}_{t}, clip\left({r}_{t}\left(Q\right), 1-\in , 1+\in \right){\stackrel{^}{A}}_{t}\right)\right]
With the Clipped Surrogate Objective function, we have two probability ratios, one non-clipped and one clipped in a range (between [1−∈,1+∈], epsilon is a hyperparameter that helps us to define this clip range (in the paper ∈ = 0.2).
If we take the minimum of the clipped and non-clipped objectives, the final aim would be lower (pessimistic bound) than the unclipped goal. Consequently, we have two cases to consider:
When the advantage is > 0:
If advantage > 0, the action is better than the average of all the actions in that state. Therefore, we should encourage our new policy to increase the probability of taking that action at that state.
This means increasing rt because we increase the probability at new policy and the denominator of old policy stay constant:
{L}^{CPI}\left(Q\right)={\stackrel{^}{E}}_{t}\left[\frac{{\pi }_{Q}\left({a}_{t}|{s}_{t}\right)}{{\pi }_{{Q}_{old}}\left({a}_{t}|{s}_{t}\right)}{\stackrel{^}{A}}_{t}\right]={\stackrel{^}{E}}_{t}\left[{r}_{t}\left(Q\right){\stackrel{^}{A}}_{t}\right]
However, because of the clip, rt(Q) will only grow to as much as 1+∈, which means this action can't be a hundred times probable compared to the old policy (because of the clip). This is done because we don't want to update our policy too much. Taking action at a specific state is only one try. It doesn't mean that it will always lead to a super positive reward, so we don't want to be too greedy because it can also lead to bad policy.
In summary, in the case of positive advantage, we want to increase the probability of taking that action at that step, but not too much.
When the advantage is < 0:
If the advantage < 0, the action should be discouraged because of the negative effect of the outcome. Consequently, rt will decrease (because the action is less probable for the current agent policy than the previous one), but rt will only decrease to as little as 1−∈ because of the clip.
Also, we don't want to make a big change in the policy by being too greedy by ultimately reducing the probability of taking that action because it leads to an opposing advantage.
In summary, thanks to this clipped surrogate objective, the range that the new policy can vary from the old one is restricted because the incentive for the probability ratio to move outside of the interval is removed. If the ratio is > 1+e or < 1-e the gradient will be equal to 0 (no slope) because the clip has the effect to a gradient. So, both of these clipping regions do not allow us to become too greedy and try to upgrade too much at once and upgrade outside the region where this example is has a good approximation.
The final Clipped Surrogate Objective Loss:
{L}_{t}^{CLIP+VF+S}\left(Q\right)={\stackrel{^}{E}}_{t}\left[{L}_{t}^{CLIP}\left(Q\right)-{c}_{1}{L}^{V}{F}_{t}\left(Q\right)+{c}_{2}S\left[{\pi }_{Q}\right]\left({s}_{t}\right)\right]
- c1 and c2 - coefficients; S - denotes an entropy bonus, Lv and Ft - squared-error loss.
Implementing a PPO agent in A3C to play Pong:
So now, we're ready to implement a PPO agent in A3C style. This means that I will use A3C code as a backbone, and there will be the same processes explained in the A3C tutorial.
So, from the theory side, it looks that it's challenging to implement, but in practice, we already have prepared code to be adopted to use the PPO loss function; we will need to change our used loss function and defined replay function mainly:
def ppo_loss(y_true, y_pred):
advantages, prediction_picks, actions = y_true[:, :1], y_true[:, 1:1+action_space], y_true[:, 1+action_space:]
ENTROPY_LOSS = 5e-3
prob = y_pred * actions
r = prob/(old_prob + 1e-10)
p1 = r * advantages
p2 = K.clip(r, min_value=1 - LOSS_CLIPPING, max_value=1 + LOSS_CLIPPING) * advantages
loss = -K.mean(K.minimum(p1, p2) + ENTROPY_LOSS * -(prob * K.log(prob + 1e-10)))
Actor.compile(loss=ppo_loss, optimizer=RMSprop(lr=lr))
As you can see, we replace the categorical_crossentropy loss function with our ppo_loss function.
From the act function, we will need prediction:
return action, prediction
Prediction from the above function is used in replay function:
def replay(self, states, actions, rewards, predictions):
discounted_r = np.vstack(self.discount_rewards(rewards))
advantages = discounted_r - values
self.Actor.fit(states, y_true, epochs=self.EPOCHS, verbose=0, shuffle=True, batch_size=len(rewards))
self.Critic.fit(states, discounted_r, epochs=self.EPOCHS, verbose=0, shuffle=True, batch_size=len(rewards))
You may ask, what is this np.hstack used for? Because we use a custom loss function, we must send data in the right shape of data. So we pack all advantages, predictions, actions to y_true, and when they are received in the custom loss function, we unpack them.
The last step to do is adding prediction memory to our run function:
states, actions, rewards, predictions = [], [], [], []
action, prediction = self.act(state)
self.replay(states, actions, rewards, predictions)
We add the same lines to the defined train_threading function. That's all! We have just created an agent that can learn to play any atari game. That's awesome! You'll need about 12 to 24 hours of training on 1 GPU to have a good agent with Pong-v0.
So this was a pretty long and exciting tutorial, and here is the complete code:
value = Dense(1, activation='linear', kernel_initializer='he_uniform')(X)
class PPOAgent:
# PPO Main Optimization Algorithm
self.path = '{}_APPO_{}'.format(self.env_name, self.lr)
action, prediction = agent.act(state)
#env_name = 'Pong-v0'
agent = PPOAgent(env_name)
#agent.run() # use as PPO
agent.train(n_threads=5) # use as APPO
#agent.test('Pong-v0_APPO_0.0001_Actor.h5', 'Pong-v0_APPO_0.0001_Critic.h5')
So same as before, I trained PongDeterministic-v4 and Pong-v0, from all my previous RL tutorials results, were the best. In PongDeterministic-v4, we reached the best score in around 400 steps (while A3C needed only 800 steps):
And here are the results for the Pong-v0 environment :
Results are pretty nice, comparing with other RL algorithms. I am delighted with the results! My goal wasn't to make it the best, to compare it with the same parameters.
After training it, I thought that this algorithm should learn to play Pong with CNN layers, then my agent would be smaller, and I would be able to upload it to GitHub. So, I changed model architecture in the following way:
X = Conv2D(32, 8, strides=(4, 4),padding="valid", activation="elu", data_format="channels_first", input_shape=input_shape)(X_input)
X = Conv2D(64, 4, strides=(2, 2),padding="valid", activation="elu", data_format="channels_first")(X)
And I was right! I trained this agent for 10k steps, and the results were also quite impressive:
Here is the gif, how my agent plays pong:
Don't forget to implement each part of the code by yourself. It's essential to understand how everything works. Try to modify the hyperparameters, use another gym environment. Experimenting is the best way to learn, so have fun!
I hope this tutorial could give you a good idea of the basic PPO algorithm. You can now build upon this by executing multiple environments in parallel to collect more training samples and solve more complicated game scenarios.
For now, this is my last Reinforcement Learning tutorial. This was quite a long journey where I learned a lot. I plan to return to reinforcement learning later this year, and this will be a more practical and more exciting tutorial; we will use it in finance!
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What is relative humidity? – Relative humidity definition
How to calculate relative humidity
How to use the relative humidity calculator
How do we measure relative humidity?
What is an ideal level of relative humidity?
The relative humidity calculator allows you to determine the relative humidity from the air temperature and dew point.
Remember those hot and humid months during summer and how it feels? The warm and muggy feeling can be attributed to high relative humidity that limits the amount of sweat evaporated from our skin. Less sweat evaporation means we feel much hotter than the actual temperature. To know more about relative humidity and how it affects us, continue reading this article. You will find answers to questions like:
How to calculate relative humidity.
How to measure relative humidity.
We have also included an example of calculating relative humidity using our calculator.
The relative humidity is a measure of how moist the air is. We can define relative humidity as the actual amount of water vapor (moisture) present in the air compared to the total amount of water vapor that the air can hold at a given temperature.
To express relative humidity as the ratio of vapor pressure
P_w
to the saturation vapor pressure
P_{ws}
at a given temperature, we can use the equation:
RH = 100 \times \frac{P_w}{P_{ws}}
Note that we always express relative humidity as a percentage.
Another concept that we should be familiar with while talking about relative humidity is absolute humidity. It is the mass of water vapor present in a unit volume of air. We can write the formula for absolute humidity as:
H = \frac{m}{V}
H
– Absolute humidity usually expressed in g/m3;
m
– Mass of water vapor; and
V
– Volume of air and water vapor mixture.
We can express the relation between dew point, air temperature, and relative humidity as:
D_p = \frac{\lambda \times \bigg (\ln \big(\frac{RH}{100} \big ) + \frac{\beta\ T}{\lambda + T} \bigg)}{\beta - \bigg (\ln \big(\frac{RH}{100} \big ) + \frac{\beta\ T}{\lambda + T} \bigg)}
D_p
– Dew point temperature (in °C);
RH
– Relative humidity;
T
– Air temperature; and
\beta = 17.625
\lambda = 243.04 \text{\degree C}
– Revised Magnus coefficients as recommended by Alduchov and Eskridge.
Substituting the values and rearranging the above equation, we can write the formula for relative humidity as:
RH = 100 \times \left [ \frac{e^{\frac{17.625 \times D_p}{243.04 + D_p}}}{e^{\frac{17.625 \times T}{243.04 + T}}} \right ]
If you want to learn how to calculate relative humidity from the actual and saturation mixing ratios of air, we recommend checking our mixing ratio of air calculator.
Now let's see how we can use the relative humidity (RH) calculator to find out the relative humidity when the dew point is 70 °F and the air temperature is 95 °F.
Enter the air temperature as 95 °F (35 °C). The tool has an inbuilt temperature conversion option that allows you to input the temperature in other units.
Type the dew point temperature as 70 °F (~21.1 °C).
The RH calculator will display the relative humidity as 44.48%.
You can also use this tool to calculate air temperature and dew point.
To measure relative humidity, we use hygrometers. Some commonly used hygrometers are:
Psychrometer: The psychrometers measure the relative humidity by measuring the difference between dry bulb temperature and wet bulb temperature.
Hair hygrometer: This device relies on measuring the relationship between the relative humidity and length of hair.
Capacitive type electronic hygrometer: These hygrometers detect the change in a sensor's capacitance when it absorbs moisture and thus measures the humidity.
Chilled-mirror dew point hygrometer: As the name suggests, these hygrometers measure the dew point temperature, which we can then use to calculate relative humidity.
In general, we feel more comfortable when the relative humidity level is between 30-50%. Lower humidity levels can lead to dryness, irritation, and itching.
Levels higher than 50% are uncomfortable as our sweat won't evaporate efficiently and we will "feel" a higher temperature. Higher humidity levels also cause mold growth which can lead to various health issues.
How do I calculate relative humidity with temperature and dew point
To calculate relative humidity with temperature and dew point, follow the given instructions:
Measure the air temperature T, in °C.
Find out the dew point temperature Dp, in °C.
Calculate relative humidity RH using the formula, RH = 100 × {exp[17.625 × Dp/(243.04 + Dp)]/exp[17.625 × T/(243.04 + T)]}.
What does it mean when relative humidity is 100?
When relative humidity is 100%, it means that air is fully saturated with water. Relative humidity measures the saturation level of air with water; hence a relative humidity of 100% implies that the air cannot take any more moisture.
What happens to relative humidity when temperature increases?
If a system's moisture content remains constant (e.g., in a closed room), the relative humidity decreases with an increase in temperature. This outcome is because the higher the temperature, the higher the amount of water vapor the air can hold, and hence the lower the relative humidity.
Determine the speed of fluid flow through the soil using the hydraulic conductivity calculator.
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L-Model Backward module:
In this part, we will implement the backward function for the whole network. Recall that when we implemented the L_model_forward function, we stored a cache containing a cache at each iteration (X, W, b, and z). In the backpropagation module, we will use those variables to compute the gradients. Therefore, in the L_model_forward function, we will iterate through all the hidden layers backward, starting from layer L. In each step, we will use the cached values for layer l to backpropagate through layer l. The figure below shows the backward pass.
To backpropagate through this network, we know that the output is:
{A}^{\left[L\right]}=\sigma \left({Z}^{\left[L\right]}\right)
Our code needs to compute:
dAL=\frac{\partial L}{\partial {A}^{\left[L\right]}}
To do so, we'll use this formula (derived using calculus which you don't need to remember):
# derivative of cost with respect to AL
We can then use this post-activation gradient dAL to keep going backward. As seen in the figure above, we can now feed in dAL into the LINEAR->SIGMOID backward function we implemented (which will use the cached values stored by the L_model_forward function). After that, we will have to use a for loop to iterate through all the other layers using the LINEAR->RELU backward function. We should store each dA, dW, and db in the grads dictionary. To do so, we'll use this formula:
grads\left["dW"+str\left(l\right)\right]=d{W}^{\left[l\right]}
For example, for l=3 this would store dW[l] in grads["dW3"].
AL - probability vector, the output of the forward propagation L_model_forward();
Y - true "label" vector (containing 0 if non-cat, 1 if cat);
caches - list of caches containing:
1. every cache of linear_activation_forward() with "relu" (it's caches[l], for l in range(L-1) i.e l = 0...L-2);
2. the cache of linear_activation_forward() with "sigmoid" (it's caches[L-1]).
grads - A dictionary with the gradients:
# the number of layers
# after this line, Y is the same shape as AL
# Lth layer (SIGMOID -> LINEAR) gradients. Inputs: "dAL, current_cache". Outputs: "grads["dAL-1"], grads["dWL"], grads["dbL"]
grads["dA" + str(L-1)], grads["dW" + str(L)], grads["db" + str(L)] = linear_activation_backward(dAL, current_cache, "sigmoid")
# Loop from l=L-2 to l=0
# Inputs: "grads["dA" + str(l + 1)], current_cache".
# Outputs: "grads["dA" + str(l)] , grads["dW" + str(l + 1)] , grads["db" + str(l + 1)]
dA_prev_temp, dW_temp, db_temp = linear_activation_backward(grads["dA"+str(l+1)], current_cache, "relu")
Update Parameters module:
In this section, we will update the parameters of the model using gradient descent:
{W}^{\left[l\right]}={W}^{\left[l\right]}-\alpha d{W}^{\left[l\right]}\phantom{\rule{0ex}{0ex}}{b}^{\left[l\right]}={b}^{\left[l\right]}-\alpha d{b}^{\left[l\right]}
here α is the learning rate. After computing the updated parameters, we'll store them in the parameters dictionary.
Code for our update_parameters function:
parameters - python dictionary containing our parameters.
Grads - python dictionary containing our gradients, output of L_model_backward.
parameters - python dictionary containing our updated parameters:
parameters["b" + str(l+1)] = parameters["b" + str(l+1)] - learning_rate*grads["db" + str(l+1)]
Congrats on implementing all the functions required for building a deep neural network. It was a long tutorial, but from now on, it will only get better. We'll put all these together to build An L-layer neural network (deep) in the next part. In fact, we'll use these models to classify cat vs. dog images.
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Snell's Law - Maple Help
Snell's Law, also known as the Law of Refraction, is a formula describing the relationship between the angles of incidence and refraction when light (or other electromagnetic radiation) passes between two media:
\frac{\mathrm{sin}\left({\mathrm{θ}}_{1}\right)}{\mathrm{sin}\left({\mathrm{θ}}_{2}\right)}=\frac{{v}_{1}}{{v}_{2}}=\frac{{n}_{2}}{{n}_{1}}=\frac{{\mathrm{λ}}_{1}}{{\mathrm{λ}}_{2}}
{\mathrm{θ}}_{k}
is the angle measured from the normal (the line perpendicular to the surface)
{v}_{k}
is the velocity of light in each medium (m/s)
{n}_{k}
is the refractive index in each medium (no unit)
{\mathrm{λ}}_{k}
is the wavelength of the light in each medium (m)
Snell's Law is valid only as long as
\mathrm{sin}\left({\mathrm{θ}}_{1}\right)
is no greater than
\frac{{n}_{2}}{{n}_{1}}
{n}_{2}< {n}_{1}
, i.e. light travels from a medium of a higher index of refraction to one with a lower refractive index, the angle
{\mathrm{θ}}_{c}
{\mathrm{θ}}_{c} = {\mathrm{sin}}^{-1}\left(\frac{{n}_{2}}{{n}_{1}}\right)
is called the critical angle. This angle is the largest possible incident angle which will create a refracted ray. As the incident angle
{\mathrm{θ}}_{1}
increases towards
{\mathrm{θ}}_{c}
, the proportion of the light refracted through the surface decreases and the proportion reflected increases. When
{\mathrm{θ}}_{1}≥{\mathrm{θ}}_{c}
, all light is reflected, a situation known as Total Internal Reflection.
Click on the graph or slide the slider to change the angle of incidence.
{n}_{1}
{n}_{2}
Angle of incidence = degrees
A list of refractive indices
Heavy Flint Glass
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Growing Annuity Calculator
What is growing annuity?
What is the growing annuity formula? Future and present values of growing annuity
How to use the growing annuity calculator?
Growing annuity calculator disclaimer
Use the growing annuity calculator (or PV of growing annuity calculator) to determine any of the following variables of a specified growing annuity:
Initial deposit or the present value of the growing annuity (PV);
Final balance or the future value of the growing annuity (FV); and
Annuity amount which is the periodic cashflow (deposit or withdrawal).
In addition, you can analyze the result by following the progression of balances in the dynamic chart or the annuity table.
In the following, you can learn the future value of the growing annuity formula (increasing annuity formula) and we also show you some growing annuity examples. You can also read what a growing annuity is at all and find out the types of annuity.
Present value of annuity calculator; and
Future value of annuity calculator.
A growing annuity refers to the series of payments (deposits) or receipts (withdrawals) made over a specified term that increase each period at a constant rate.
In a growing ordinary annuity, payments or receipts happen at the end of each period; in a growing annuity due, payments or receipts are made at the beginning of each period.
The ordinary present value of growing annuity formula can be found the following equation:
PV_{ga} =\frac{P}{(r - g)}\times \bigg(1 - \bigg(\frac{1 + g}{1 + r}\bigg)^n\bigg)
PV_{ga}
- Present value of the growing annuity;
P
- The first payment or receipt subject to constant periodic increase;
r
- Rate of return or interest rate;
g
- Periodic growth rate of the cash flows; and
n
- Number of periods.
And the future value of growing annuity formula:
FV_{ga} =P\times \frac{(1 + r)^n - (1 + g)^n}{(r - g)}
FV_{ga}
- Future value of the growing annuity.
Note, that the above formula is inadequate when the rate of return and the growth rate are equal. In such a case, the following formula is applicable:
FV_{ga} =P\times n\times (1 + r)^{n-1}
You need to multiply the above equations by
(1 + r)
to get the growing annuity due formula.
You need to proceed with the following simple steps to run the tool:
1. Growing annuity configuration
Initial deposit - The present value of growing annuity;
Final balance - The future value of growing annuity; and
Annuity amount - Periodic deposit or withdrawal.
Direction of cash flows:
Payments (deposits); and
Receipts (withdrawals).
Type of annuity - You can choose between an annuity due (beginning of period) or an ordinary annuity (end of period);
Compounding frequency - The frequency interest is added to the principal balance of your annuity, or, in other words, how often the earned return or interest is reinvested; and
First period starts from (advanced mode) - The first day of the annuity.
Length of annuity - The lifespan of the annuity.
Rate of return - The interest rate of the growing annuity.
Annual growth rate - You can set the annual percentage increase of the growing annuity here.
Periodic growth rate - The percentage growth rate of the periodic deposits of withdrawals. Note, that periodic and annual growth rates are linked together: the other will be calculated according to annuity frequency if you set one.
How do I calculate the cash flows of a growing annuity?
Follow the instructions bellow to find the cash flows of the growing annuity:
For the first cash flow, you need to apply the following formula:
P = FV / [((1 + r)n - (1 + g)n) / (r - g)],
FV - Final balance;
P - First payment or receipt;
If the growth rate and the interest rate are equal, you need to apply the following modified simple formula:
P = FV / (n * (1 + r)ⁿ⁻¹).
If you wish to compute the cash flow in period t:
Pt = P * (1 + r)t-1.
How do I calculate the future value of growing annuity?
To compute the future value of growing annuity you need to apply the formula:
FV = P * [((1 + r)ⁿ - (1 + g)ⁿ) / (r - g)],
Are the periodic cash flows equal in growing annuity?
No. In a growing annuity scheme, the reoccurring payments or recipes increase at a constant rate each period.
You should consider the growing annuity calculator as a model for financial approximation. All payment figures, balances, and interest figures are estimates based on the data you provided in the specifications that are, despite our best effort, not exhaustive.
Growing annuity configuration
final balance (FV)
Direction of cash flows
The final balance (future value) of your growing annuity is $91,514.60.
Monthly deposit $1,000.00 initially
Find out if it is economically viable to buy a 3D printer or not, using the 3D Printer – Buy vs Outsourcing Service Calculator.
3D Printer - Buy vs Outsource Calculator
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Earth Orbit Calculator
What are earth's satellites?
How to calculate the orbital speed of a satellite using this earth orbit calculator
How can you estimate the period of rotation for an earth's satellite?
Example on how to use this earth orbit calculator
This Earth orbit calculator helps you determine the orbital speed and orbital period of Earth's satellites at a given height above the average earth sea level. The fundamental laws that control the motion of Earth's satellites around Earth apply to the motion of planets, the moon, the earth's only natural satellite, and other satellites in the skies. Read on to learn about how different attributes of the satellite rotation are determined by its distance from the Earth's surface.
A satellite is a small object that orbits another larger object. The earth is a satellite because it orbits the sun. We consider the moon a satellite because it orbits the earth. However, most people think of an artificial satellite when they say satellite. These spacecraft are launched into an orbit around the earth or any other celestial body. Sputnik 1 was Earth's first artificial satellite. It was launched into an elliptical low Earth orbit at the height of 939 km. It orbited for three weeks before its batteries ran out.
There are tens of thousands of artificial satellites orbiting the earth. Some photograph our globe, while others photograph other planets, the sun, and other celestial bodies. These images aid scientists in studying the earth, the solar system, and the universe. Other satellites broadcast television and make phone calls all around the world.
This Earth orbit calculator has two features to help you calculate the orbital speed and period of rotation of an Earth's satellite launched at a specific height from the surface of the earth.
According to Nicolaus Copernicus, earth and the other planets orbit the Sun in circles. He also discovered that as the distance from the Sun increased, so did the orbital periods. Kepler later found these orbits to be ellipses, but the orbits of most planets in the solar system are roughly circular. The Earth's orbital distance from the Sun fluctuates by only 2%. The eccentric orbit of Mercury, whose orbital distance changes by approximately 40%, is an exception.
Determining a satellite's orbital speed and period is significantly easier in circular orbits. We use this simplification in the following calculations. This tool focuses on objects orbiting the Earth, but you can apply our findings to other situations. Therefore, the orbital speed of an earth's satellite is given as follows:
\small \text{orbital speed} = \sqrt{\frac{G \cdot M_{E}}{(R_{E}+h)}}
G
– Earth's gravitational constant;
M_E
– Earth's mass;
R_E
– Earth's radius; and
h
– Perpendicular distance of the satellite from the surface of the earth.
You can simply calculate this using this Earth orbit calculator by selecting the option Speed of the satellite, entering the height of the object rotating the earth, and voila, you have the orbital speed of the earth's satellite!
You need a certain amount of speed to reach the orbit (and stay there): learn how to calculate it with our delta-
v
calculator!
Similar to the earth's orbit, all of the earth's satellites orbit the earth at a certain height with a constant period of rotation.
Therefore, the orbital period of an earth's satellite is given as follows:
\small \text{orbital period} = 2\pi \sqrt{\frac{(R_{E}+h)^3}{G \cdot M_{E}}}
You can select the option Period of satellite rotation; enter the height of the object rotating the earth; you now know the orbital period of the earth's satellite!
Let's try to determine the orbital speed and period for the International Space Station (ISS)
Since the ISS orbits at a height of 400 km above Earth's surface, the radius at which it orbits is RE + 400 km.
You can just plug in the height of the ISS from the earth's surface as 400 km, and the calculator will compute its orbital period as 1.54 hrs and orbital speed as 7.672 km/s.
How do you calculate the orbital speed of an earth's satellite?
To calculate the orbital speed of an earth's satellite, you need to know the gravitational constant (G), earth's mass (M), earth's radius (R), and the height of rotation of the satellite (h).
The orbital speed is calculated as:
√((G × M) / (R + h))
What determines the orbital period of a satellite?
The orbital period of a satellite depends on the mass of the planet being orbited and the distance of the satellite from the center of the planet.
We can obtain the orbital period of a satellite T from Newton's form of Kepler's third law.
T2 = 4 π a3 / GM
Period of satellite rotation
Height of the satellite from the Earth.
This AC wattage calculator allows you to calculate the AC wattage from volts and amps.
AC Wattage Calculator
Arrow speed calculator finds the real speed of an arrow for a bow with custom parameters.
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How to calculate modulation index by formula
How to use the modulation calculator
Modulation index example
Using our modulation calculator, you can obtain the ratio of the modulation signal against the amplitude of the carrier signal, known as the modulation index.
The following article will discuss phase, frequency, and amplitude modulation and share how we calculate the modulation index by formula.
Modulation is converting data into radio waves by adding information from a low-frequency signal to a high-frequency electronic or optical carrier signal.
This carrier signal is a pure wave of constant frequency that can travel long distances but doesn't carry any useful information. Thus, we include information by imposing our high-frequency input signal wave on this carrier wave.
💡 To learn about the relationship between frequency and waves, check out our frequency to wavelength calculator.
The shape of the carrier wave changes with modulation, i.e., after we encode the sound waves or data information from the input signal to the carrier signal. The base of these changes are the following wave parameters:
Amplitude – The peak of the wave;
Frequency – Number of waves crossing a point in a given second; and
Phase – The offset of a wave from a given point.
Based on these wave parameters, there are three types of modulation techniques for analog carriers, all of which vary according to the unmodulated levels of the message signal:
Amplitude Modulation – Varying amplitude of the carrier signal;
Frequency Modulation – Varying frequency of the carrier signal; and
Phase Modulation – Varying phase shift of the carrier signal.
🔎 Phase and frequency modulations have many commonalities. So we collectively refer to them as angle modulation, where the angle is the modification of the quantity passed to a sine or cosine function.
Calculating the modulation ratio from the message signal to the carrier signal, this numerically expressed degree of modulation is the modulation index.
Different modulation techniques have different signal-to-noise ratios. Check out our signal to noise ratio calculator to learn more about this aspect.
We use the following modulation index equations to find the variation between the message and the carrier signals:
The definition of AM modulation index is the amplitude ratio of the message signal to the carrier signal:
\mu_a = A_m / A_c
Frequency modulation index
It is the ratio of frequency deviation of the message signal to the sinusoidal message signal frequency:
\mu_f = \Delta f / f_m
It is the maximum phase change corresponding to the maximum message signal amplitude:
\mu_p = \Delta \theta
\mu_a, \mu_f, \mu_p
– Modulation indexes for amplitude, frequency, and phase, respectively, also known as modulation factors;
A_m
– Amplitude of the message signal;
A_c
– Amplitude of the carrier signal;
\Delta f
– Peak frequency deviation in the message signal;
f_m
– Highest frequency in the message signal; and
\Delta \theta
– The peak phase deviation.
Below you will find out how simple it is to use the modulation calculator.
Select the wave parameter for calculating its modulation index. For example, by selecting amplitude, we get the following three fields:
Message signal amplitude (Aₘ)
Here you may enter the amplitude of the message signal, e.g., 40 volts.
Carrier signal amplitude (A꜀)
And here, you may enter the respective amplitude of the carrier signal, e.g., 50 volts.
Modulation index (μa)
You'll then get how much the modulated variable varies around its unmodulated levels, i.e., 0.8.
Similarly, if we select frequency as our wave parameter, we have the following fields for calculating the frequency modulation index:
Max frequency deviation (Δf)
Here you enter the peak frequency-deviation of your message signal.
Message signal frequency (fₘ)
And here, you may enter the highest frequency of the message signal.
🙋 At a constant modulating frequency, phase modulation is indistinguishable from frequency modulation. Thus, the frequency modulation index in such an instance equals the phase deviation.
Let's take an FM broadcast station for a modulation index example:
A station has a maximum frequency deviation of 75 kHz, with the highest message signal frequency at 15 kHz. What is its modulation index?
We know that the equation of modulation index for frequency is:
\mu = \Delta f / f_m
\mu = 75 / 15
\mu = 5
Thus, the frequency modulation index of our station is 5.
How do I calculate modulation index?
Divide the message signal amplitude (Aₘ) by the carrier signal amplitude (A꜀) to obtain the amplitude modulation index (μa). Mathematically, its representation is:
μa = Aₘ / A꜀
Or divide the max frequency deviation (Δf) by the message signal frequency (fₘ) to obtain the frequency modulation index (μf). Mathematically, we represent that as:
μf = Δf / fₘ
Amplitude modulation vs frequency modulation
The modulation index is between 0 and 1.
To send data, the carrier wave's amplitude is modified.
Requires low bandwidth.
The modulation index is always greater than 1.
To send data, the carrier wave's frequency is modified.
Requires high bandwidth.
Sound quality is better than AM.
What is the modulation index when the message signal is 20 V at 100 V carrier?
0.2 is the modulation index when the message signal is 20 V with a 100 V carrier.
We obtain this modulation index by dividing the message signal by the carrier signal:
Modulation index = message signal / carrier signal
Altering the amplitude of the carrier signal by combining it with a message signal to transmit information is known as amplitude modulation. It generally requires low bandwidth and is used for broadcasting audio signals.
Calculate modulation index of
Amplitude modulation index (μₐ)
With the thermal energy calculator, you can estimate the kinetic energy of molecules in an ideal gas.
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What is an employee tenure calculator
How to use our work tenure calculator
Formula to calculate average employment tenure
Our tenure calculator is a handy tool for finding the average service duration of employees in your organization.
The following article will discuss what tenure is, why we calculate the work tenure, and how to calculate the employment tenure by hand using the average tenure formula.
Calculating the employee tenure helps us obtain the average duration of service for employees in an organization. This is the average length of time someone is employed for before they resign or move on, which helps to get an overview of:
How satisfied employees are;
Loyalty of employees towards the organization;
How well the organization's retention policies are; and
The organization's revenue per employee.
💡 Revenue per employee is a measure of the total revenue divided by the number of employees.
Now that you know what employment tenure is, let's find more information on how to calculate the average duration of service for an organization's employees.
You can use our employee work tenure calculators as follows:
For average duration of service
Enter the total number of employees you've had over a specified period, e.g., 20 employees.
What is the sum of the years of service for each employee over that duration? For example, input 50 years in the second field of this employment tenure calculator.
Thus, your employee tenure is 2.5 years.
For specific duration of service
Select an individual's employment starting date in the first field.
Select their ending date in the second field.
The result will give you that individual's employee tenure.
In the next section, you will learn about the average tenure formula we've used above.
🙋 We usually calculate employee tenure for mature organizations holding several years of operations and revenue growth.
Now that you know what employment tenure is and how to calculate the average tenure of employees with our tool, let's see the proper formula to do it manually:
\text{Tenure} = \text{Duration} / \text{Employees}
\text{Employees}
- Average number of total employees in the specified period;
\text{Duration}
- Combined years of service of all those employees; and
\text{Tenure}
- Employee tenure at the organization.
The instruction explaining how to calculate the average tenure of employees is below:
Calculate the total number of employees, i.e., those who have left and who are still with you, e.g., 30 employees
🔎 The types of employees we include in this calculation are the ones that have a noticeable impact on the business.
Add up all the time they've worked at your organization, e.g., the total tenure time is 150 years.
Divide the number of years by the number of employees:
\text{Tenure} = 150 / 30 = 5
Thus, this organization's average employee tenure is 5 years.
💡 The greater the employee tenure is in a nation, the lower the unemployment rate.
🙋 To calculate an individual's employment tenure, subtract their end date from their start date.
What is work tenure?
Employee work tenure is the average duration of time an employee works for a particular organization before moving onto another.
To calculate work tenure, we divide the number of combined years of service by the number of employees, i.e.,
Tenure = Duration / Employee
To calculate employee tenure:
Count your total number of employees.
Combine all the years they've worked for you.
Divide the number of years by the number of employees, i.e.,
What is the tenure of 20 employees with 120 years of service?
6 years is the average tenure of 20 employees with 120 combined years of service to an organization.
We obtain this by dividing the combined duration of service by the number of employees, i.e.,
How do I calculate the employee tenure in months or years?
To calculate the tenure in months:
Count the number of months your employees have worked.
Divide this number by the number of employees.
You will have your tenure in months, i.e.,
Divide this result by 12 to calculate the tenure in years.
Combined duration of service
Specific duration of service
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Reciprocal condition number - MATLAB rcond - MathWorks Deutschland
Sensitivity of Badly Conditioned Matrix
Find Condition of Identity Matrix
Reciprocal condition number
C = rcond(A) returns an estimate for the reciprocal condition of A in 1-norm. If A is well conditioned, rcond(A) is near 1.0. If A is badly conditioned, rcond(A) is near 0.
Examine the sensitivity of a badly conditioned matrix.
A notable matrix that is symmetric and positive definite, but badly conditioned, is the Hilbert matrix. The elements of the Hilbert matrix are
H\left(i,j\right)=1/\left(i+j-1\right)
Create a 10-by-10 Hilbert matrix.
Find the reciprocal condition number of the matrix.
The reciprocal condition number is small, so A is badly conditioned.
The condition of A has an effect on the solutions of similar linear systems of equations. To see this, compare the solution of
Ax=b
to that of the perturbed system,
Ax=b+0.01
Create a column vector of ones and solve
Ax=b
Now change
by 0.01 and solve the perturbed system.
Compare the solutions, x and x1.
Since A is badly conditioned, a small change in b produces a very large change (on the order of 1e5) in the solution to x = A\b. The system is sensitive to perturbations.
Examine why the reciprocal condition number is a more accurate measure of singularity than the determinant.
Create a 5-by-5 multiple of the identity matrix.
This matrix is full rank and has five equal singular values, which you can confirm by calculating svd(A).
Although the determinant of the matrix is close to zero, A is actually very well conditioned and not close to being singular.
Calculate the reciprocal condition number of A.
The matrix has a reciprocal condition number of 1 and is, therefore, very well conditioned. Use rcond(A) or cond(A) rather than det(A) to confirm singularity of a matrix.
C — Reciprocal condition number
Reciprocal condition number, returned as a scalar. The data type of C is the same as A.
The reciprocal condition number is a scale-invariant measure of how close a given matrix is to the set of singular matrices.
If C is near 0, the matrix is nearly singular and badly conditioned.
If C is near 1.0, the matrix is well conditioned.
rcond is a more efficient but less reliable method of estimating the condition of a matrix compared to the condition number, cond.
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Hereditarily finite set - Wikipedia
Finite sets whose elements are all hereditarily finite sets
2.1 Ackermann's bijection
3 Axiomatizations
3.1 Theories of finite sets
4 Graph models
{\displaystyle \{\{\},\{\{\{\}\}\}\}}
{\displaystyle \emptyset =\{\}}
{\displaystyle \{7,{\mathbb {N} },\pi \}}
{\displaystyle \{3,\{{\mathbb {N} }\}\}}
{\displaystyle {\mathbb {N} }=\{0,1,2,\dots \}}
{\displaystyle H_{\aleph _{0}}}
{\displaystyle \aleph _{0}}
{\displaystyle H_{\aleph _{0}}}
Ackermann's bijection[edit]
{\displaystyle H_{\aleph _{0}}}
{\displaystyle H_{\aleph _{0}}}
{\displaystyle f(2^{a}+2^{b}+\cdots )=\{f(a),f(b),\ldots \}}
{\displaystyle f(5)=f(4+1)=f(2^{2}+2^{0})=\{f(2),f(0)\}=\{f(2^{1}),\{\}\}=\{\{f(1)\},\{\}\}=\{\{f(2^{0})\},\{\}\}=\{\{\{f(0)\}\},\{\}\}=\{\{\{\{\}\}\},\{\}\}}
{\displaystyle \{\}}
{\displaystyle \emptyset }
{\displaystyle \{\{\}\}}
{\displaystyle \{\emptyset \}}
{\displaystyle \{0\}}
{\displaystyle \{\{\{\}\}\}}
{\displaystyle \{\{\{\{\}\}\}\}}
{\displaystyle \{\{\},\{\{\}\}\}}
{\displaystyle \{0,1\}}
{\displaystyle \{\{\{\{\{\}\}\}\}\}}
{\displaystyle \{\{\{\},\{\{\}\}\}\}}
{\displaystyle \{\{\},\{\{\{\}\}\}\}}
{\displaystyle 6}
{\displaystyle \{\{\{\{\{\{\}\}\}\}\}\}}
{\displaystyle 7}
{\displaystyle \{\{\{\{\{\{\{\}\}\}\}\}\}\}}
{\displaystyle 8}
{\displaystyle \{\{\{\{\{\{\{\{\}\}\}\}\}\}\}\}}
{\displaystyle \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}}
{\displaystyle \{0,1,2\}}
{\displaystyle n}
{\displaystyle 1,1,1,2,3,6,12,25,52,113,247,548,1226,2770,6299,14426,\dots }
Axiomatizations[edit]
Theories of finite sets[edit]
{\displaystyle \emptyset }
{\displaystyle 0}
{\displaystyle H_{\aleph _{0}}}
{\displaystyle Z^{-}}
{\displaystyle H_{\aleph _{0}}}
ZF[edit]
{\displaystyle ~V_{4}~}
{\displaystyle V_{\omega }=\bigcup _{k=0}^{\infty }V_{k}}
Graph models[edit]
{\displaystyle H_{\aleph _{0}}}
{\displaystyle \{\dots \}}
{\displaystyle \{t,t,s\}=\{t,s\}}
{\displaystyle t}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Hereditarily_finite_set&oldid=1054672990"
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Alpha_compositing Knowpia
In computer graphics, alpha compositing or alpha blending is the process of combining one image with a background to create the appearance of partial or full transparency.[1] It is often useful to render picture elements (pixels) in separate passes or layers and then combine the resulting 2D images into a single, final image called the composite. Compositing is used extensively in film when combining computer-rendered image elements with live footage. Alpha blending is also used in 2D computer graphics to put rasterized foreground elements over a background.
This color spectrum image's alpha channel falls off to zero at its base, where it is blended with the background color.
In a 2D image a color combination is stored for each picture element (pixel), often a combination of red, green and blue (RGB). When alpha compositing is in use, each pixel has an additional numeric value stored in its alpha channel, with a value ranging from 0 to 1. A value of 0 means that the pixel is fully transparent and the color in the pixel beneath will show through. A value of 1 means that the pixel is fully opaque.
With the existence of an alpha channel, it is possible to express compositing image operations using a compositing algebra. For example, given two images A and B, the most common compositing operation is to combine the images so that A appears in the foreground and B appears in the background. This can be expressed as A over B. In addition to over, Porter and Duff defined the compositing operators in, held out by (the phrase refers to holdout matting and is usually abbreviated out), atop, and xor (and the reverse operators rover, rin, rout, and ratop) from a consideration of choices in blending the colors of two pixels when their coverage is, conceptually, overlaid orthogonally:
As an example, the over operator can be accomplished by applying the following formula to each pixel:
{\displaystyle \alpha _{o}=\alpha _{a}+\alpha _{b}(1-\alpha _{a})}
{\displaystyle C_{o}={\frac {C_{a}\alpha _{a}+C_{b}\alpha _{b}(1-\alpha _{a})}{\alpha _{o}}}}
{\displaystyle C_{o}}
{\displaystyle C_{a}}
{\displaystyle C_{b}}
stand for the color components of the pixels in the result, image A and image B respectively, applied to each color channel (red/green/blue) individually, whereas
{\displaystyle \alpha _{o}}
{\displaystyle \alpha _{a}}
{\displaystyle \alpha _{b}}
are the alpha values of the respective pixels.
The over operator is, in effect, the normal painting operation (see Painter's algorithm). Bruce A. Wallace derived the over operator based on a physical reflectance/transmittance model, as opposed to Duff's geometrical approach.[2] The in and out operators are the alpha compositing equivalent of clipping. The two use only the alpha channel of the second image and ignore the color components.
Straight versus premultipliedEdit
If an alpha channel is used in an image, there are two common representations that are available: straight (unassociated) alpha and premultiplied (associated) alpha.
With straight alpha, the RGB components represent the color of the object or pixel, disregarding its opacity.
With premultiplied alpha, the RGB components represent the emission of the object or pixel, and the alpha represents the occlusion. The over operator then becomes:
{\displaystyle C_{o}=C_{a}+C_{b}(1-\alpha _{a})}
{\displaystyle \alpha _{o}=\alpha _{a}+\alpha _{b}(1-\alpha _{a})}
A more obvious advantage of this is that, in certain situations, it can save a subsequent multiplication (e.g. if the image is used many times during later compositing). However, the most significant advantages of using premultiplied alpha are for correctness and simplicity rather than performance: premultiplied alpha allows correct filtering and blending. In addition, premultiplied alpha allows regions of regular alpha blending and regions with additive blending mode to be encoded within the same image.[3]
Assuming that the pixel color is expressed using straight (non-premultiplied) RGBA tuples, a pixel value of (0, 0.7, 0, 0.5) implies a pixel that has 70% of the maximum green intensity and 50% opacity. If the color were fully green, its RGBA would be (0, 1, 0, 0.5).
However, if this pixel uses premultiplied alpha, all of the RGB values (0, 0.7, 0) are multiplied, or scaled for occlusion, by the alpha value 0.5, which is appended to yield (0, 0.35, 0, 0.5). In this case, the 0.35 value for the G channel actually indicates 70% green emission intensity (with 50% occlusion). A pure green emission would be encoded as (0, 0.5, 0, 0.5). Knowing whether a file uses straight or premultiplied alpha is essential to correctly process or composite it, as a different calculation is required. It is also entirely acceptable to have an RGBA triplet express emission with no occlusion, such as (0.4, 0.3, 0.2, 0.0). Fires and flames, glows, flares, and other such phenomena can only be represented using associated / premultiplied alpha.
The only important difference is in the dynamic range of the color representation in finite precision numerical calculations (which is in all applications): premultiplied alpha has a unique representation for transparent pixels, avoiding the need to choose a "clear color" or resultant artifacts such as edge fringes (see the next paragraphs). In an associated / premultiplied alpha image, the RGB represents the emission amount, while the alpha is occlusion. Premultiplied alpha has some practical advantages over normal alpha blending because interpolation and filtering give correct results.[4]
Ordinary interpolation without premultiplied alpha leads to RGB information leaking out of fully transparent (A=0) regions, even though this RGB information is ideally invisible. When interpolating or filtering images with abrupt borders between transparent and opaque regions, this can result in borders of colors that were not visible in the original image. Errors also occur in areas of semitransparency because the RGB components are not correctly weighted, giving incorrectly high weighting to the color of the more transparent (lower alpha) pixels.
Premultiplication can reduce the available relative precision in the RGB values when using integer or fixed-point representation for the color components, which may cause a noticeable loss of quality if the color information is later brightened or if the alpha channel is removed. In practice, this is not usually noticeable because during typical composition operations, such as OVER, the influence of the low-precision color information in low-alpha areas on the final output image (after composition) is correspondingly reduced. This loss of precision also makes premultiplied images easier to compress using certain compression schemes, as they do not record the color variations hidden inside transparent regions, and can allocate fewer bits to encode low-alpha areas. The same “limitations” of lower quantisation bit depths such as 8 bit per channel are also present in imagery without alpha, and this argument is problematic as a result.
Gamma correctionEdit
Alpha blending, not taking into account gamma correction
Alpha blending, taking into account gamma correction.
The RGB values of typical digital images do not directly correspond to the physical light intensities, but are rather compressed by a gamma correction function:
{\displaystyle C_{\text{encoded}}=C_{\text{linear}}^{\gamma }}
This transformation better utilizes the limited number of bits in the encoded image by choosing
{\displaystyle \gamma }
that better matches the non-linear human perception of luminance.
Accordingly, computer programs that deal with such images must decode the RGB values into a linear space (by undoing the gamma-compression), blend the linear light intensities, and re-apply the gamma compression to the result:[5][6]
{\displaystyle C_{o}=\left({\frac {C_{a}^{1/\gamma }\alpha _{a}+C_{b}^{1/\gamma }\alpha _{b}(1-\alpha _{a})}{\alpha _{o}}}\right)^{\gamma }}
When combined with premultiplied alpha, pre-multiplication is done in linear space, prior to gamma compression.[7] This results in the following formula:
{\displaystyle C_{o}=\left(C_{a}^{1/\gamma }+C_{b}^{1/\gamma }(1-\alpha _{a})\right)^{\gamma }}
Note that only the color components undergo gamma-correction; the alpha channel is always linear.
Other transparency methodsEdit
Although used for similar purposes, transparent colors and image masks do not permit the smooth blending of the superimposed image pixels with those of the background (only whole image pixels or whole background pixels allowed).
A similar effect can be achieved with a 1-bit alpha channel, as found in the 16-bit RGBA high color mode of the Truevision TGA image file format and related TARGA and AT-Vista/NU-Vista display adapters' high color graphic mode. This mode devotes 5 bits for every primary RGB color (15-bit RGB) plus a remaining bit as the "alpha channel".
Screendoor transparency can be used to simulate partial occlusion where only 1-bit alpha is available.
For some applications, a single alpha channel is not sufficient: a stained-glass window, for instance, requires a separate transparency channel for each RGB channel to model the red, green and blue transparency separately. More alpha channels can be added for accurate spectral color filtration applications.
The concept of an alpha channel was introduced by Alvy Ray Smith and Ed Catmull in the late 1970s at the New York Institute of Technology Computer Graphics Lab, and fully developed in a 1984 paper by Thomas Porter and Tom Duff.[8]
The use of the term alpha is explained by Smith as follows: "We called it that because of the classic linear interpolation formula
{\displaystyle \alpha A+(1-\alpha )B}
that uses the Greek letter
{\displaystyle \alpha }
(alpha) to control the amount of interpolation between, in this case, two images A and B".[9] That is, when compositing image A atop image B, the value of
{\displaystyle \alpha }
in the formula is taken directly from A's alpha channel.
Transparent color in palettes
^ "Definition of alpha blending". PCMAG. Retrieved 2021-08-07.
^ Wallace, Bruce A. (1981). "Merging and transformation of raster images for cartoon animation". SIGGRAPH Computer Graphics. New York City, New York: ACM Press. 15 (3): 253–262. CiteSeerX 10.1.1.141.7875. doi:10.1145/800224.806813. ISBN 0-89791-045-1. S2CID 1147910.
^ "TomF's Tech Blog - It's only pretending to be a wiki". tomforsyth1000.github.io. Archived from the original on 12 December 2017. Retrieved 8 May 2018.
^ "ALPHA COMPOSITING – Animationmet". animationmet.com. Archived from the original on 2019-09-25. Retrieved 2019-09-25.
^ Minute Physics (March 20, 2015). "Computer Color is Broken". YouTube. Archived from the original on 2021-11-22.
^ Novak, John (September 21, 2016). "What every coder should know about gamma".
^ "Gamma Correction vs. Premultiplied Pixels – Søren Sandmann Pedersen". ssp.impulsetrain.com.
^ Porter, Thomas; Duff, Tom (July 1984). "Compositing Digital Images" (PDF). SIGGRAPH Computer Graphics. New York City, New York: ACM Press. 18 (3): 253–259. doi:10.1145/800031.808606. ISBN 9780897911382. S2CID 18663039. Archived (PDF) from the original on 2011-04-29. Retrieved 2019-03-11.
^ Alvy Ray Smith (1995-08-15). "Alpha and the History of Digital Compositing" (PDF). alvyray.com. p. 6. Archived from the original (PDF) on 2021-10-25.
Compositing Digital Images - Thomas Porter and Tom Duff (Original Paper)
Alpha Matting and Premultiplication
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Multiclassification - Objectives and metrics | CatBoost
Multiclassification: objectives and metrics
\displaystyle\frac{\sum\limits_{i=1}^{N}w_{i}\log\left(\displaystyle\frac{e^{a_{it_{i}}}}{ \sum\limits_{j=0}^{M - 1}e^{a_{ij}}} \right)}{\sum\limits_{i=1}^{N}w_{i}} { ,}
t \in \{0, ..., M - 1\}
\displaystyle\frac{\frac{1}{M}\sum\limits_{i = 1}^N w_i \sum\limits_{j = 0}^{M - 1} [j = t_i] \log(p_{ij}) + [j \neq t_i] \log(1 - p_{ij})}{\sum\limits_{i = 1}^N w_i} { ,}
t \in \{0, ..., M - 1\}
\frac{TP}{TP + FP}
\frac{TP}{TP+FN}
(1 + \beta^2) \cdot \frac{Precision * Recall}{(\beta^2 \cdot Precision) + Recall}
\beta
(0; +\infty)
2 \frac{Precision * Recall}{Precision + Recall}
The formula depends on the value of the average parameter:
\frac{\sum\limits_{i=1}^{M} w_{i} F1_{i}}{\sum\limits_{i=1}^{M}w_{i}} {, where}
w_{i}
is the sum of the weights of the documents which correspond to the i-th class. If document weights are not specified
w_{i}
stands for the number of times the i-th class is found among the label values.
\displaystyle\frac{\sum\limits_{i=1}^{M}F1_{i}}{M}
TotalF1 = \displaystyle\frac{2 \cdot TP}{2 \cdot TP + FP + FN} {, where}
TP = \sum\limits_{i=1}^{M} TP_{i}
FP = \sum\limits_{i=1}^{M} FP_{i}
FN = \sum\limits_{i=1}^{M} FN_{i}
The method for averaging the value of the metric that is initially individually calculated for each class.
Default: Weighted.
Possible values: Weighted, Macro, Micro.
This functions is defined in terms of a
k \times k
C
(where k is the number of classes):
\displaystyle\frac{\sum\limits_{k}\sum\limits_{l}\sum\limits_{m} C_{kk} C_{lm} - C_{kl}C_{mk}}{\sqrt{\sum\limits_{k} \left(\sum\limits_{l} C_{kl}\right) \left(\sum\limits_{k' | k' \neq k} \sum\limits_{l'} C_{k'l'}\right)}\sqrt{\sum\limits_{k} \left(\sum\limits_{l} C_{lk}\right) \left(\sum\limits_{k' | k' \neq k} \sum\limits_{l'} C_{l' k'}\right)}}
\displaystyle\frac{\sum\limits_{i=1}^{N}w_{i}[argmax_{j=0,...,M - 1}(a_{ij})==t_{i}]}{\sum\limits_{i=1}^{N}w_{i}} { , }
t \in \{0, ..., M - 1\}
\displaystyle\frac{\sum\limits_{i = 1}^{N} w_{i} [argmax_{j=0,...,M - 1}(a_{ij})\neq t_{i}]}{\sum\limits_{i = 1}^{N} w_{i}}
1 - Accuracy
1 - \displaystyle\frac{1 - Accuracy}{1 - RAccuracy}
RAccuracy = \displaystyle\frac{\sum\limits_{k=0}^{M - 1} n_{k_{a}}n_{k_{t}}}{(\sum\limits_{i=1}^{N}w_{i})^{2}}
k_{a}
is the weighted number of times class k is predicted by the model
k_{t}
is the weighted number of times class k is set as the label for input objects
Refer to the A Performance Metric for Multi-Class Machine Learning Models paper for calculation principles
The value is calculated separately for each class k numbered from 0 to M–1 according to the binary classification calculation principles. The objects of class k are considered positive, while all others are considered negative.
Possible values: Mu, OneVsAll.
Examples: AUC:type=Mu, AUC:type=OneVsAll.
misclass_cost_matrix
The matrix M with misclassification cost values.
M[i,j]
in this matrix is the cost of classifying an object as a member of the class i when its' actual class is j. Applicable only if the used type of AUC is Mu.
Format for a matrix of size C:
<Value for M[0,0]>, <Value for M[0,1]>, ..., <Value for M[0,C-1]>, <Value for M[1,0]>, ..., <Value for M[C-1,0]>, ..., <Value for M[C-1,C-1]>
All diagonal elements
M[i, j]
(such that i=j) must be equal to 0.
The type parameter is optional and is assumed to be set to Mu if the parameter is explicitly specified.
Default: All non-diagonal matrix elements are set to 1. All diagonal elements
М[i, j]
(such that i = j) are set to 0.
Examples: Three classes — AUC:misclass_cost_matrix=0/0.5/2/1/0/1/0/0.5/0, Two classes — AUC:type=Mu;misclass_cost_matrix=0/0.5/1/0.
MultiClass + +
MultiClassOneVsAll + +
TotalF1 - +
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Differential pressure calculation – Calculate flow rate from pressure differential
Example: Using the differential pressure calculator
The differential pressure calculator will assist you in computing the change in pressure across a valve or a fluid flow device in a fluid system. From a pitot tube in an aircraft to measure airspeed to a pneumatic valve in a robotic arm, and all things in between, differential pressure in a fluid flow system play a vital role in their operations.
Pressure switches act on the pre-set values of change in pressure to trigger circuit breaking or make any other adjustments like opening an extra valve, redirecting the fluid flow, or detecting a blockage in the pipes or airways. Such pressure switches operate along with pressure sensors or gauges that monitor the pressure in a fluid system.
The concept is useful for fluid flows in a pipe regardless of the pipe's location. It could be a respiratory medical device, flow measurement in aircraft, heating, ventilation, air conditioning(HVAC) systems, or any other hydraulic or pneumatically operating system. The pressure differential is a function of discharge, flow coefficient, and the specific gravity of the fluid.
This article will explain differential pressure and how to calculate flow rate using differential pressure.
Consider fluid flow through an orifice, a venturi, or a similar constriction having a pressure gauge on either side. The change in pressure between the two gauges is known as differential pressure. It is helpful to find any blockages or leaks in the system as it will cause a change in differential pressure. Mathematically, it is related to the flow factor or flow coefficient, volumetric flow rate, and the fluid's specific gravity:
\quad \begin{align*} K_v &= Q\sqrt{\frac{S}{\Delta P}} \\\\ \Delta P &= \frac{Q^2}{K_v^2} S \end{align*}
K_v
– Flow factor;
S
– Specific gravity of the fluid;
Q
– Volumetric flow rate; and
\Delta P
– Differential pressure.
The flow factor is the efficiency of the fluid flow through a pipeline. It is also expressed in
\text{m}^3/\text{hr}
To calculate differential pressure:
Enter the specific gravity of the fluid or select from the list of fluids.
Fill in the volumetric flow rate for the flow.
Insert the flow factor.
The calculator will return the differential pressure.
To calculate the flow rate, you can leave the field empty and enter the pressure difference instead.
Find the pressure difference if the water has a volumetric flow rate of 10 l/min and a flow factor of 5 l/min.
Enter the specific gravity of water as 1.
Fill in the volumetric flow rate for the flow as
10 \text{ l/min}
Insert the flow factor as
5 \text{ l/min}
Using the differential pressure formula:
\qquad \scriptsize \Delta P = \frac{Q^2}{K_v^2} S = \frac{10^2}{5^2}\times 1 = 4 \text{ bar}
What do you mean by differential pressure?
The change in pressure when a fluid passes through a constriction like a venturi or an orifice is the differential pressure. It has the same units as pressure, i.e., bars or Pa. It is a pressure measurement method and is very common in fluid flow systems.
How do you calculate differential pressure?
Find the square of volumetric flow rate.
Find the square of flow factor.
How do I calculate flow rate using differential pressure?
To calculate flow rate using differential pressure:
Divide the pressure difference (ΔP) by the specific gravity (S) of the fluid.
Find the square root of the resultant division.
Multiply the root by the flow factor (Kv) to obtain the flow rate (Q) for the fluid system. Mathematically, that's:
Q = √(Kv × (ΔP/S))
What are the applications of differential pressure?
The differential pressure is a measurement method that is used in:
Detecting blockages and leaks in the system;
Activating pressure switches and valves;
Determining the proper functioning of fluid flow systems;
Measuring liquid levels and flow rates; and
Respiratory flow in medical devices.
Discharge or volumetric flow rate
Use our centrifugal force calculator to determine the force acting on a rotating object.
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Mauchly’s Test of Sphericity - MATLAB & Simulink - MathWorks España
Mauchly’s Test of Sphericity
The regular p-value calculations in the repeated measures anova (ranova) are accurate if the theoretical distribution of the response variables have compound symmetry. This means that all response variables have the same variance, and each pair of response variables share a common correlation. That is,
\mathrm{Σ}={\mathrm{Ï}}^{2}\left(\begin{array}{cccc}1& \mathrm{Ï}& \cdots & \mathrm{Ï}\\ \mathrm{Ï}& 1& \cdots & \mathrm{Ï}\\ ⋮& ⋮& â±& ⋮\\ \mathrm{Ï}& \mathrm{Ï}& \cdots & 1\end{array}\right).
If the compound symmetry assumption is false, then the degrees of freedom for the repeated measures anova test must be adjusted by a factor ε, and the p-value must be computed using the adjusted values.
Compound symmetry implies sphericity.
For a repeated measures model with responses y1, y2, ..., sphericity means that all pair-wise differences y1 – y2, y1 – y3, ... have the same theoretical variance. Mauchly’s test is the most accepted test for sphericity.
Mauchly’s W statistic is
W=\frac{|T|}{{\left(trace\left(T\right)/p\right)}^{d}},
T=M\text{'}\stackrel{^}{â}M.
M is a p-by-d orthogonal contrast matrix, Σ is the covariance matrix, p is the number of variables, and d = p – 1.
A chi-square test statistic assesses the significance of W. If n is the number of rows in the design matrix, and r is the rank of the design matrix, then the chi-square statistic is
C=â\left(nâr\right)\mathrm{log}\left(W\right)D,
D=1â\frac{2{d}^{2}+d+2}{6d\left(nâr\right)}.
The C test statistic has a chi-square distribution with (p(p – 1)/2) – 1 degrees of freedom. A small p-value for the Mauchly’s test indicates that the sphericity assumption does not hold.
The rmanova method computes the p-values for the repeated measures anova based on the results of the Mauchly’s test and each epsilon value.
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Created by Oghenekaro Elem and Purnima Singh, PhD
Black F., Scholes M. “The pricing of options and corporate liabilities“ The Journal of Political Economy, Vol. 81, No. 3 (1973)
What is Black Scholes?
How to calculate Black Scholes model – Black Scholes formula
How to use the Black Scholes options calculator?
Assumptions and limitations of the Black Scholes Model
This Black Scholes calculator is an important tool for options traders to set a rational price for stock options.
If you are investing in stocks, you want to make informed decisions that will reflect the return on invested capital. Without a mathematical framework as a guide, it will be no different from gambling. Black Scholes removes the guesswork involved in predicting stock price movement so that arbitrage opportunities are minimal across markets.
How do you calculate stock options value using the Black Scholes formula?
You will also find an example of using the Black Scholes model calculator.
Black Scholes is a mathematical model that helps options traders determine a stock option’s fair market price. The Black Scholes model, also known as Black-Scholes-Merton (BSM), was first developed in 1973 by Fisher Black and Myron Scholes; Robert Merton was the first to expand the mathematical understanding of the options pricing model.
The Black-Scholes options pricing model serves as a guide for making rational trading decisions as traders seek to buy options below the calculated value of the Black Scholes formula and sell at a price above the calculated value.
An option is a contract that gives the owner a right to buy or sell an asset for a specific price (also known as the strike price) on or before a specific date (also known as the expiration date).
Although most options traders rarely exercise their option rights before the expiration date, you may exercise an American option at any time before the option expires. But you can only exercise a European option on the expiration date.
Financial market traders buy and sell options as protection (or hedge) against uncertainty in the financial market. There are two types of options:
Call option – gives the owner the right to buy the asset at the strike price; and
Put option – gives the owner the right to sell the asset at the strike price.
For example, assuming you bought 100 shares of Tesla (TSLA) stocks at $500 per share today (present value = 100 × 500 = $50,000). You believe the price of the shares will increase quickly to $600 per share by next month, so you can sell it (future value = 100 × 600 = $60,000) at a profit of $10,000.
But you can’t be too sure, things can go differently, and the price of a TSLA share can fall below the current $500 price, losing you money. Or it can rise even higher than $600, and you’ll be wishing you’d bought 200 shares.
For a fair price, you can buy a put option contract to protect yourself or a call option to take advantage of the opportunity if the price indeed rises higher.
The put option contract will state the strike price, e.g., $550, the number of stocks you will be selling, e.g., 100 shares, and the exact date the contract expires.
If TSLA share price falls below $550 per share by the specified expiry date on the contract, you can choose to exercise your right to sell the stocks for the strike price of $550 to the option writer you bought the put option from. So, assuming the price went as low as $250, exercising your put option means protecting yourself from the market downturn and selling at a profit of $50 earning per share instead of being at a loss. But if TSLA share price rises over $550 or to the expected price of $600, you can decide to let your put option expire since you will earn more money selling at a higher price in the market.
Conversely, if you had bought a call option with the strike price of $550, when the price rises over $550 or to the expected price of $600, you can decide to exercise your call option and buy the shares at a lower price of $550 to sell at a profit in the market. If the price falls below the strike price, you won’t need to exercise your call option.
To further explore the relationship between a call and put option, check out our put call parity calculator.
How do you know what is a fair price to pay for these options contract?
That’s what the Black Scholes calculator helps you determine. The model uses a partial differential equation to predict a stock’s price movement in the financial market and arrive at the price you should buy the call or put option. Hence, to use the Black Scholes formula, you need to provide the following information:
Current price of the stock, also known as its spot price;
Time to the expiration of the options contract;
Risk-free interest rate, or the rate specified in the option for a given stable asset or short-dated government bonds such as US Treasury bills;
Expected volatility or unpredictability of the stock is expressed as the standard deviation of the stock price; and
The Black Scholes equation is a complex mathematical formula. Gladly, you don’t have to go through all the processes involved when using our Black Scholes option pricing calculator. However, the Black Scholes equations have been summarized as follows:
\!\small \begin{align*} C &= S_0e^{-qt}N(d_1) - X e^{-rt}N(d_2) \\\\ P &= X e^{-rt}N(-d_2) - S_0e^{-qt}N(-d_1)\\\\ d_1 &=\frac {\ln (\frac{S_0}{X} ) +(r - d + (v^2/2))t} {v \sqrt{t}}\\\\ d_2 &=\frac {\ln (\frac{S_0}{X} ) +(r - d - (v^2/2))t} {v \sqrt{t}} \end{align*}
C
– Call option price;
P
– Put option price;
S_0
– Current stock price;
X
– Strike price;
N(d_1)
N(d_2)
– Cumulative standard normal distribution functions of
d_1
d_2
T
– Term of the option;
r
– Risk-free interest rate;
q
– Dividend yield percentage; and
v
– Annualized volatility of the stock.
To use the Black Scholes calculator and get the values of a call and put option, you only have to provide details of six main variables. As an example, for the given input data:
Provide the current price of the stock, i.e., $400.
Input the strike price, i.e., $350.
Enter the option contract term or expiration date, i.e., 1 year.
Type the risk-free interest rate in percentage, i.e., 3%.
State the expected volatility of the stock, i.e., 20%.
Input the expected dividend yield as 1%.
The Black Scholes option calculator will give you the call option price and the put option price as $65.67 and $9.30, respectively.
Like all models, it is essential to accept the Black Scholes model's results as estimations that should guide your decision-making, not as absolutes. There are several modifications to the Black Scholes model today that try to fix the model's limitations, but theoretical approximations are not accurate predictors of reality. Therefore, here are some shortcomings to consider about using the Black Scholes option pricing model, and by extension, this Black Scholes model calculator:
The Black Scholes model is most suited to European options because it assumes the option lasts its entire life span until the expiration date. If you are interested in calculating how much you can gain/lose by executing an option contract before expiration, you can check out our call option calculator.
Assumes that the markets are entirely efficient, and we can't predict their movements.
Assumes that volatility is constant.
Assumes the risk-free interest rate remains constant until the expiration date. But it is not known, and it is not constant in reality.
Doesn't account for transactional costs, such as fees and taxes, involved in pricing and trading options.
How does the Black Scholes model work?
The Black Scholes model works by using a stock's volatility, price and strike price, expected dividend yield, and risk-free interest rate for a stable asset to determine the price of a stock option. The model assumes that the stock price follows a lognormal distribution path throughout the life of the stock option.
The Black Scholes model is used by options traders for the valuation of stock options. The model helps determine the fair market price for a stock option using a set of six variables:
Price of the asset;
Risk-free interest rate of return;
Dividend yield; and
What interest rate is used in Black-Scholes?
The Black-Scholes model assumes that the interest rates are constant and known until the option contract's expiration. Hence, it uses the risk-free one-year interest rates to represent this assumption.
Oghenekaro Elem and Purnima Singh, PhD
Put option price
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How to calculate the Graham number
Graham number – Real life examples
Shortcomings of the Graham number
Our smart Graham number calculator helps you calculate a stock's fair value by using the most recent earnings per share and book value per share. In this article, we will cover:
What the Graham number is in investing.
How to calculate the Graham number.
How to use our Graham number calculator.
What is a good Graham number?
As a bonus, we will explore a real-life successful investing example.
Graham number is an investing metric that puts together the current earnings per share (EPS) and the book value per share to obtain a stock price value.
Benjamin Graham suggested that if investors trade a stock at a price under its Graham number, then the stock is undervalued. Contrarily, if the stock trades over its Graham number, it is overvalued.
Consequently, we can consider the Graham number as the intrinsic value which other more sophisticated investors calculate through the discounted cash flow method.
It is essential to notice that the fair value of a stock will rise if the EPS grows. If the EPS grows, the more the retained earnings will be for a company; thus, shareholders' equity will increase too. But hey, everything starts with the revenue growth. Do not forget!
In the chart below, you can see how the intrinsic value (which you can obtain without the Graham number calculator) is between a range of several stock prices, making the stock overvalued or undervalued.
As we now have a rough idea of what the Graham number is, let's discuss how it is computed in more detail. Graham number formula is relatively easy:
\text{GN} = \sqrt{22.5 \times \text{EPS} \times \text{BVPS}}
\text{GN}
– Graham number;
\text{EPS}
– Earnings per share; and
\text{BVPS}
– Book value per share.
It is important to verify the main two requirements for using the Graham number formula:
Price earnings (PE) ratio shall be below or equal to 15, and price to book (PB) ratio shall be below or equal to 1.5;
PE ratio × PB ratio shall be below or equal to 25.
Furthermore, our Graham number calculator also helps you determine the per-share value in case you don't have them. Click in the advanced mode button and fill in the following values:
Common outstanding shares: You will find it in the income statement.
Common shareholders equity: Available in the balance sheet.
TTM net income to common stockholders: You will find it in the income statement too.
You will see how our tool automatically calculates the per-share values.
We are going to cover TD Synnex corporation for this example. Before the 2020, 2nd quarter earnings report, Synnex was trading at a:
PB ratio = 1.3
Before the report, the trailing twelve months (TTM) available information was:
Earnings per share TTM = 10.47 USD/share
Book value per share = 75.82 USD/share
Because the PE ratio was under 15 and the PB ratio was under 1.5, we can use the Graham number formula to calculate the fair value of the stock.
\begin{align*} \text{GN} &= \sqrt{22.5 \times 10.47 \times 75.82}\\ &= 133.65 \end{align*}
Our Graham number calculator indicates that the fair value for TD Synnex corporation at that moment was 133.65 USD per share. Was it undervalued or overvalued?
On June 19th, the stock price was 101.80 USD per share, meaning a potential upside of 31%. The stock eventually spun off, releasing its subsidiary Concentrix and making it a public traded company. Shareholders who had Synnex stocks received shares of Concentrix. Up to date (December 14th, 2021), an investment in Synnex, when the stock was trading at 101.8 USD, has returned 172%.
Graham number does not consider essential investing metrics like free cash flow or EBITDA margin, and, as you could have noted, it only serves with companies of positive net income.
Whenever this doesn't apply, we suggest you explore another valuation method:
EBITDA multiple – You could use this method whenever the company has positive EBITDA or for companies in which depreciation and amortization greatly affect their net income results.
EV to sales – We recommend using it when companies do not generate positive operating income or cash flows yet.
So, if you bought a stock at a higher price than the Graham number, you can consider the other valuation methods we have, as well as its price/earnings to growth ratio. If you are still uncomfortable with the price, you can sell a fraction of your investment to reduce risk or buy more and reduce your cost basis.
How do I calculate Graham number for stocks?
To calculate Graham number for stocks, follow these steps:
Calculate the earnings per share (EPS). It is the result of the net income divided by the outstanding shares.
Calculate the book value per share (BVPS). You will need to divide common shareholders' equity by the outstanding shares.
Get the square root of the result of 22.5 × BVPS × EPS. Also, you can use our handy Graham number calculator to make your life more simple.
A good Graham number is the one that is above the current stock price because it signals a possible profit. Just keep in mind we need to fulfill two main conditions for using the Graham number formula: First, price to earnings (PE) ratio below 15, and price to book (PB) ratio below 1.5; or the result of PE × PB under 22.5.
What is a bad Graham number?
Value investing requires paying a fair price for the stock, which would minimize your drawdown risk in case of a bear market. Our Graham number calculator relates the current earnings per share and the book value to recommend a stock price. Consequently, there cannot be a wrong Graham number, only bad acquisition prices.
Is Graham number still useful today?
Benjamin Graham developed the Graham number concept when interest rates were higher; thus, stock valuations were tighter. At that time, it was easier to find companies with PE and PB ratios below 15 and 1.5, respectively. Nowadays, it is harder to find companies that trade around those metrics, but the example above is still not impossible.
Is it underpriced or overpriced?
Our optimal hedge ratio calculator helps you to calculate the hedge ratio that will optimize your portfolio returns.
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Ptolemy's_inequality Knowpia
In Euclidean geometry, Ptolemy's inequality relates the six distances determined by four points in the plane or in a higher-dimensional space. It states that, for any four points A, B, C, and D, the following inequality holds:
Four points and their six distances. The points are not co-circular, so Ptolemy's inequality is strict for these points.
{\displaystyle {\overline {AB}}\cdot {\overline {CD}}+{\overline {BC}}\cdot {\overline {DA}}\geq {\overline {AC}}\cdot {\overline {BD}}.}
It is named after the Greek astronomer and mathematician Ptolemy.
The four points can be ordered in any of three distinct ways (counting reversals as not distinct) to form three different quadrilaterals, for each of which the sum of the products of opposite sides is at least as large as the product of the diagonals. Thus the three product terms in the inequality can be additively permuted to put any one of them on the right side of the inequality, so the three products of opposite sides or of diagonals of any one of the quadrilaterals must obey the triangle inequality.[1]
As a special case, Ptolemy's theorem states that the inequality becomes an equality when the four points lie in cyclic order on a circle. The other case of equality occurs when the four points are collinear in order. The inequality does not generalize from Euclidean spaces to arbitrary metric spaces. The spaces where it remains valid are called the Ptolemaic spaces; they include the inner product spaces, Hadamard spaces, and shortest path distances on Ptolemaic graphs.
Assumptions and derivationEdit
Ptolemy's inequality is often stated for a special case, in which the four points are the vertices of a convex quadrilateral, given in cyclic order.[2][3] However, the theorem applies more generally to any four points; it is not required that the quadrilateral they form be convex, simple, or even planar.
For points in the plane, Ptolemy's inequality can be derived from the triangle inequality by an inversion centered at one of the four points.[4][5] Alternatively, it can be derived by interpreting the four points as complex numbers, using the complex number identity
{\displaystyle (A-B)(C-D)+(A-D)(B-C)=(A-C)(B-D)}
to construct a triangle whose side lengths are the products of sides of the given quadrilateral, and applying the triangle inequality to this triangle.[6] One can also view the points as belonging to the complex projective line, express the inequality in the form that the absolute values of two cross-ratios of the points sum to at least one, and deduce this from the fact that the cross-ratios themselves add to exactly one.[7]
A proof of the inequality for points in three-dimensional space can be reduced to the planar case, by observing that for any non-planar quadrilateral, it is possible to rotate one of the points around the diagonal until the quadrilateral becomes planar, increasing the other diagonal's length and keeping the other five distances constant.[6] In spaces of higher dimension than three, any four points lie in a three-dimensional subspace, and the same three-dimensional proof can be used.
Four concyclic pointsEdit
For four points in order around a circle, Ptolemy's inequality becomes an equality, known as Ptolemy's theorem:
{\displaystyle {\overline {AB}}\cdot {\overline {CD}}+{\overline {AD}}\cdot {\overline {BC}}={\overline {AC}}\cdot {\overline {BD}}.}
In the inversion-based proof of Ptolemy's inequality, transforming four co-circular points by an inversion centered at one of them causes the other three to become collinear, so the triangle equality for these three points (from which Ptolemy's inequality may be derived) also becomes an equality.[5] For any other four points, Ptolemy's inequality is strict.
A cycle graph in which the distances disobey Ptolemy's inequality
Ptolemy's inequality holds more generally in any inner product space,[1][8] and whenever it is true for a real normed vector space, that space must be an inner product space.[8][9]
For other types of metric space, the inequality may or may not be valid. A space in which it holds is called Ptolemaic. For instance, consider the four-vertex cycle graph, shown in the figure, with all edge lengths equal to 1. The sum of the products of opposite sides is 2. However, diagonally opposite vertices are at distance 2 from each other, so the product of the diagonals is 4, bigger than the sum of products of sides. Therefore, the shortest path distances in this graph are not Ptolemaic. The graphs in which the distances obey Ptolemy's inequality are called the Ptolemaic graphs and have a restricted structure compared to arbitrary graphs; in particular, they disallow induced cycles of length greater than three, such as the one shown.[10]
The Ptolemaic spaces include all CAT(0) spaces and in particular all Hadamard spaces. If a complete Riemannian manifold is Ptolemaic, it is necessarily a Hadamard space.[11]
Inner product spacesEdit
{\displaystyle \|\cdot \|}
is a norm on a vector space
{\displaystyle X.}
Then this norm satisfies Ptolemy's inequality:
{\displaystyle \|x-y\|\,\|z\|~+~\|y-z\|\,\|x\|~\geq ~\|x-z\|\,\|y\|\qquad {\text{ for all vectors }}x,y,z.}
if and only if there exists an inner product
{\displaystyle \langle \cdot ,\cdot \rangle }
{\displaystyle X}
{\displaystyle \|x\|^{2}=\langle x,\ x\rangle }
{\displaystyle x\in X.}
[12] Another necessary and sufficient condition for there to exist such an inner product is for the norm to satisfy the parallelogram law:
{\displaystyle \|x+y\|^{2}~+~\|x-y\|^{2}~=~2\|x\|^{2}+2\|y\|^{2}\qquad {\text{ for all vectors }}x,y.}
If this is the case then this inner product will be unique and it can be defined in terms of the norm by using the polarization identity.
Greek mathematics – Mathematics of Ancient Greeks
Parallelogram law – The sum of the squares of the 4 sides of a parallelogram equals that of the 2 diagonals
Polarization identity – Formula relating the norm and the inner product in a inner product space
Ptolemy – 2nd-century writer and astronomer
Ptolemy's table of chords – 2nd century AD trigonometric table
^ a b Schoenberg, I. J. (1940), "On metric arcs of vanishing Menger curvature", Annals of Mathematics, Second Series, 41: 715–726, doi:10.2307/1968849, MR 0002903 .
^ Steele, J. Michael (2004), "Exercise 4.6 (Ptolemy's Inequality)", The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities, MAA problem books, Cambridge University Press, p. 69, ISBN 9780521546775 .
^ Alsina, Claudi; Nelsen, Roger B. (2009), "6.1 Ptolemy's inequality", When Less is More: Visualizing Basic Inequalities, Dolciani Mathematical Expositions, vol. 36, Mathematical Association of America, pp. 82–83, ISBN 9780883853429 .
^ Apostol (1967) attributes the inversion-based proof to textbooks by R. A. Johnson (1929) and Howard Eves (1963).
^ a b Stankova, Zvezdelina; Rike, Tom, eds. (2008), "Problem 7 (Ptolemy's Inequality)", A Decade of the Berkeley Math Circle: The American Experience, MSRI Mathematical Circles Library, vol. 1, American Mathematical Society, p. 18, ISBN 9780821846834 .
^ a b Apostol 1967.
^ Silvester, John R. (2001), "Proposition 9.10 (Ptolemy's theorem)", Geometry: Ancient and Modern, Oxford University Press, p. 229, ISBN 9780198508250 .
^ a b Giles, J. R. (2000), "Exercise 12", Introduction to the Analysis of Normed Linear Spaces, Australian Mathematical Society lecture series, vol. 13, Cambridge University Press, p. 47, ISBN 9780521653756 .
^ Schoenberg, I. J. (1952), "A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal", Proceedings of the American Mathematical Society, 3: 961–964, doi:10.2307/2031742, MR 0052035 .
^ Howorka, Edward (1981), "A characterization of Ptolemaic graphs", Journal of Graph Theory, 5 (3): 323–331, doi:10.1002/jgt.3190050314, MR 0625074 .
^ Buckley, S. M.; Falk, K.; Wraith, D. J. (2009), "Ptolemaic spaces and CAT(0)", Glasgow Mathematical Journal, 51 (2): 301–314, doi:10.1017/S0017089509004984, MR 2500753 .
^ Apostol, Tom M. (1967). "Ptolemy's Inequality and the Chordal Metric". Mathematics Magazine. 40 (5): 233–235. doi:10.2307/2688275. JSTOR 2688275. MR 0225213.
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How to find the volume of a rectangular pyramid - formulas
How to use the rectangular pyramid volume calculator?
Other pyramid calculators
The rectangular pyramid volume calculator can help you find the volume and surface area of a pyramid with a rectangular base.
What a rectangular pyramid is; and
How to find the volume of a rectangular-based pyramid?
You will also find an example of using the rectangular pyramid volume calculator.
A rectangular pyramid is a polyhedron (three-dimensional shape) with a rectangular base and triangular lateral faces (see figure 1). Some famous examples of rectangular pyramids are the Egyptian pyramids and the Louvre pyramid.
Figure 1: A right rectangular pyramid.
To calculate the volume or capacity of a rectangular pyramid, we will use the formula:
\quad V = \frac{a b H}{3}
a
- Length of the rectangular base;
b
- width of the rectangular base; and
H
- Height of the pyramid.
The formula for calculating the surface area of the pyramid is:
A = ab + a \sqrt {\Big(\frac{b}{2}\Big)^2 + H^2} + b\sqrt {\Big(\frac{a}{2}\Big)^2 + H^2}
Let us see how we can use the rectangular pyramid volume calculator to find the volume of a rectangular pyramid with length and width of base edges as 7 cm and 5 cm, respectively, and height as 10 cm.
Enter the dimensions of the base, i.e., base length = 7 cm and base width = 5 cm.
Type the height of the pyramid, i.e., 10 cm.
The calculator will display the total surface area (160.13 cm2) and volume of the rectangular based pyramid (116.67 cm3).
We hope you enjoyed using our rectangular pyramid volume calculator. Make sure to check out our other tools that deal with the determination of various parameters of a pyramid.
How do I get the volume of a rectangular pyramid?
To get the volume of a rectangular pyramid, follow the given instructions:
Multiply the length and width of the rectangular base to get its area.
Now multiply the base area with the height of the pyramid.
Divide the result from step 2 by three, and you will get the volume of a rectangular pyramid.
How many faces does a rectangular pyramid have?
A rectangular pyramid has five faces and eight edges. Out of these five faces, the base face is rectangular and the other four faces are triangular-shaped.
How many vertices are there in a rectangular pyramid?
There are five vertices in a rectangular pyramid. One vertex is located above the rectangular base of the pyramid. The other four vertices lie at the four corners of the base.
Pyramid parameters
The trapezoid calculator is here to give you all the information about your trapezoid shape - the sides, height, angles, area, and perimeter.
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Introduction to Robotics/Sensors/Lecture/Students - Wikiversity
Introduction to Robotics/Sensors/Lecture/Students
Edit — Print — Printouts
{\displaystyle D=Ct}
D is the distance, measured in meters.
C is the speed of sound in air, measured in meters per second.
t is the amount of time the wave travels, measured in seconds.
At 72°, sound travels at 344.8m/s.
We convert to centimeters because they are more useful, and it helps to remove fractions from our calculation.
{\displaystyle D=Ct\times {\frac {100cm}{1m}}=100Ct}
For the BoeBot
{\displaystyle D={\frac {100CT}{2}}}
Because the sound wave must travel to the object and back.
The BoeBot measures t in "clock cycles", which are each 2μseconds. We must convert the equation to use seconds instead of clock cycles:
{\displaystyle D={\frac {100Ct}{2}}\times {\frac {2}{1,000,000}}={\frac {Ct}{10,000}}=0.03448t}
The BoeBot can do some limited math with fractions.
The "**" operator is used as follows:
{\displaystyle A**B=A\times {\frac {B}{65536}}}
We can use the following code to convert time to distance:
Distance = Conversion ** time
Finding the Conversion Constant
{\displaystyle Conversion=0.03448\times 65536=2260}
Retrieved from "https://en.wikiversity.org/w/index.php?title=Introduction_to_Robotics/Sensors/Lecture/Students&oldid=1608346"
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Cesàro summation - Wikipedia
For the song "Cesaro Summability" by the band Tool, see Ænima.
In mathematical analysis, Cesàro summation (also known as the Cesàro mean[1][2]) assigns values to some infinite sums that are not necessarily convergent in the usual sense. The Cesàro sum is defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series.
This special case of a matrix summability method is named for the Italian analyst Ernesto Cesàro (1859–1906).
The term summation can be misleading, as some statements and proofs regarding Cesàro summation can be said to implicate the Eilenberg–Mazur swindle. For example, it is commonly applied to Grandi's series with the conclusion that the sum of that series is 1/2.
3 (C, α) summation
4 Cesàro summability of an integral
{\displaystyle (a_{n})_{n=1}^{\infty }}
be a sequence, and let
{\displaystyle s_{k}=a_{1}+\cdots +a_{k}=\sum _{n=1}^{k}a_{n}}
be its kth partial sum.
The sequence (an) is called Cesàro summable, with Cesàro sum A ∈
{\displaystyle \mathbb {R} }
, if, as n tends to infinity, the arithmetic mean of its first n partial sums s1, s2, ..., sn tends to A:
{\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}=A.}
The value of the resulting limit is called the Cesàro sum of the series
{\displaystyle \textstyle \sum _{n=1}^{\infty }a_{n}.}
If this series is convergent, then it is Cesàro summable and its Cesàro sum is the usual sum.
Let an = (−1)n for n ≥ 0. That is,
{\displaystyle (a_{n})_{n=0}^{\infty }}
{\displaystyle (1,-1,1,-1,\ldots ).}
Let G denote the series
{\displaystyle G=\sum _{n=0}^{\infty }a_{n}=1-1+1-1+1-\cdots }
The series G is known as Grandi's series.
{\displaystyle (s_{k})_{k=0}^{\infty }}
denote the sequence of partial sums of G:
{\displaystyle {\begin{aligned}s_{k}&=\sum _{n=0}^{k}a_{n}\\(s_{k})&=(1,0,1,0,\ldots ).\end{aligned}}}
This sequence of partial sums does not converge, so the series G is divergent. However, G is Cesàro summable. Let
{\displaystyle (t_{n})_{n=1}^{\infty }}
be the sequence of arithmetic means of the first n partial sums:
{\displaystyle {\begin{aligned}t_{n}&={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}\\(t_{n})&=\left({\frac {1}{1}},{\frac {1}{2}},{\frac {2}{3}},{\frac {2}{4}},{\frac {3}{5}},{\frac {3}{6}},{\frac {4}{7}},{\frac {4}{8}},\ldots \right).\end{aligned}}}
{\displaystyle \lim _{n\to \infty }t_{n}=1/2,}
and therefore, the Cesàro sum of the series G is 1/2.
As another example, let an = n for n ≥ 1. That is,
{\displaystyle (a_{n})_{n=1}^{\infty }}
{\displaystyle (1,2,3,4,\ldots ).}
Let G now denote the series
{\displaystyle G=\sum _{n=1}^{\infty }a_{n}=1+2+3+4+\cdots }
Then the sequence of partial sums
{\displaystyle (s_{k})_{k=1}^{\infty }}
{\displaystyle (1,3,6,10,\ldots ).}
Since the sequence of partial sums grows without bound, the series G diverges to infinity. The sequence (tn) of means of partial sums of G is
{\displaystyle \left({\frac {1}{1}},{\frac {4}{2}},{\frac {10}{3}},{\frac {20}{4}},\ldots \right).}
This sequence diverges to infinity as well, so G is not Cesàro summable. In fact, for any sequence which diverges to (positive or negative) infinity, the Cesàro method also leads to a sequence that diverges likewise, and hence such a series is not Cesàro summable.
(C, α) summationEdit
In 1890, Ernesto Cesàro stated a broader family of summation methods which have since been called (C, α) for non-negative integers α. The (C, 0) method is just ordinary summation, and (C, 1) is Cesàro summation as described above.
The higher-order methods can be described as follows: given a series Σan, define the quantities
{\displaystyle {\begin{aligned}A_{n}^{-1}&=a_{n}\\A_{n}^{\alpha }&=\sum _{k=0}^{n}A_{k}^{\alpha -1}\end{aligned}}}
(where the upper indices do not denote exponents) and define Eα
n to be Aα
n for the series 1 + 0 + 0 + 0 + .... Then the (C, α) sum of Σan is denoted by (C, α)-Σan and has the value
{\displaystyle (\mathrm {C} ,\alpha ){\text{-}}\sum _{j=0}^{\infty }a_{j}=\lim _{n\to \infty }{\frac {A_{n}^{\alpha }}{E_{n}^{\alpha }}}}
if it exists (Shawyer & Watson 1994, pp.16-17). This description represents an α-times iterated application of the initial summation method and can be restated as
{\displaystyle (\mathrm {C} ,\alpha ){\text{-}}\sum _{j=0}^{\infty }a_{j}=\lim _{n\to \infty }\sum _{j=0}^{n}{\frac {\binom {n}{j}}{\binom {n+\alpha }{j}}}a_{j}.}
Even more generally, for α ∈
{\displaystyle \mathbb {R} }
{\displaystyle \mathbb {Z} }
−, let Aα
n be implicitly given by the coefficients of the series
{\displaystyle \sum _{n=0}^{\infty }A_{n}^{\alpha }x^{n}={\frac {\displaystyle {\sum _{n=0}^{\infty }a_{n}x^{n}}}{(1-x)^{1+\alpha }}},}
and Eα
n as above. In particular, Eα
n are the binomial coefficients of power −1 − α. Then the (C, α) sum of Σan is defined as above.
If Σan has a (C, α) sum, then it also has a (C, β) sum for every β > α, and the sums agree; furthermore we have an = o(nα) if α > −1 (see little-o notation).
Cesàro summability of an integralEdit
Let α ≥ 0. The integral
{\displaystyle \textstyle \int _{0}^{\infty }f(x)\,dx}
is (C, α) summable if
{\displaystyle \lim _{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }f(x)\,dx}
exists and is finite (Titchmarsh 1948, §1.15) harv error: no target: CITEREFTitchmarsh1948 (help). The value of this limit, should it exist, is the (C, α) sum of the integral. Analogously to the case of the sum of a series, if α = 0, the result is convergence of the improper integral. In the case α = 1, (C, 1) convergence is equivalent to the existence of the limit
{\displaystyle \lim _{\lambda \to \infty }{\frac {1}{\lambda }}\int _{0}^{\lambda }\int _{0}^{x}f(y)\,dy\,dx}
which is the limit of means of the partial integrals.
As is the case with series, if an integral is (C, α) summable for some value of α ≥ 0, then it is also (C, β) summable for all β > α, and the value of the resulting limit is the same.
Almost convergent sequence
^ Hardy, G. H. (1992). Divergent Series. Providence: American Mathematical Society. ISBN 978-0-8218-2649-2.
^ Katznelson, Yitzhak (1976). An Introduction to Harmonic Analysis. New York: Dover Publications. ISBN 978-0-486-63331-2.
Shawyer, Bruce; Watson, Bruce (1994), Borel's Methods of Summability: Theory and Applications, Oxford University Press, ISBN 0-19-853585-6
Titchmarsh, E. C. (1986) [1948], Introduction to the theory of Fourier integrals (2nd ed.), New York, NY: Chelsea Publishing, ISBN 978-0-8284-0324-5
Volkov, I. I. (2001) [1994], "Cesàro summation methods", Encyclopedia of Mathematics, EMS Press
Zygmund, Antoni (1988) [1968], Trigonometric Series (2nd ed.), Cambridge University Press, ISBN 978-0-521-35885-9
Retrieved from "https://en.wikipedia.org/w/index.php?title=Cesàro_summation&oldid=1082911759"
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Environmental Science Text Compilatioon Committee, Seoul (Korea, Republic of)
[en] This book gives descriptions of environmental pollution such as water and soil pollution, harmful chemicals substances and radiation, nature protection on wild animals, wild plants, and nature park, environmental assessment, and environmental management. It deals with the earth environment on change and the cause of the earth environment, ozone layer, global warming and acid fallout, plan for the earth control and environment information and information system.
Feb 2001; 223 p; Academy seojeok press; Seoul (Korea, Republic of); ISBN 89-7010-219-6; ; 201 refs
CONTROL, ENVIRONMENT, GREENHOUSE EFFECT, INFORMATION SYSTEMS, PLANNING, POLLUTION, RADIATIONS, SAFETY
The temperature map used Landsat data for avoiding chilling injury of spring cabbage in Choshi
Yoshida, S.; Yamaki, A.; Aoki, T.; Anzai, T.; Udagawa, Y.
[en] The detailed temperature(T) map was made by using land sat TM heat band data for avoiding chillinginjury of cabbage in choshi Area, Chiba Prefecture. The accuracy was verified with the radiation and air Ts observed on surface. 1. The daily minimum leaf T of cabbage head in Togawadai-cho was always lower than that in Takagaminishi-cho. The positive correlation of daily minimum leaf T of cabbage head was observed between Togawadai-cho and Takagaminishi-cho. 2. The daily minimum leaf T of cabbage head had positive correlation with the daily minimum air T in both Togawadai-cho and Takagaminishi-cho. The daily minimum leaf T of cabbage head in Takagaminisih-cho became much low with decreasing the daily minimum air T compared with that in Togawadai-cho. 3. CCT map made from land sat TM heat band date was transformed into radiation and air T maps using radiation T and air T of simultaneous surface observation, respectively. 4. Both radiation and air T maps were geometrically corrected, and were piled up on topographical map. It was observed that high T spot was located at low, vacancy or lawn place, and low T spot was located at grove, streets, playground or slope. 5. The surface T and air Tin both radiation and air T maps were accurately corresponded to radiation and air T of surface observation in cabbage field at same time, respectively
Bulletin of the Chiba Prefectural Agricultural Experiment Station; ISSN 0577-6880; ; (no.42); p. 15-22
ACCURACY, BRASSICA, DATA PROCESSING, HEAT, INJURIES, JAPAN, LEAVES, METEOROLOGY, REMOTE SENSING
ASIA, DEVELOPED COUNTRIES, DISEASES, ENERGY, FOOD, MAGNOLIOPHYTA, MAGNOLIOPSIDA, PLANTS, PROCESSING, VEGETABLES
Hormone replacement therapy use dramatically increases breast oestrogen receptor expression in obese postmenopausal women
Lawson, James S; Field, Andrew S; Tran, Dinh D; Houssami, Nehmat
[en] It is known that use of hormone replacement therapy (HRT) by postmenopausal women increases the risk of breast cancer. In this study, oestrogen receptor (ER)-α expression is examined using standard immunoperoxidase technique. Normal breast samples of 11 Australian postmenopausal women have been included in the ER-α study; the result showed a strong correlation (r"2 = 0.80) between ER-α expression in normal breast epithelial cells and body mass index (BMI) in normal women who currently use HRT. This finding confirms that the possibility of increased risk of breast cancer associated with increased ER-α expression in normal breast epithelial cells, in turn associated with high BMI and the use of HRT
Available from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC57804; PMCID: PMC57804; PUBLISHER-ID: BCR-3-5-342; PMID: 11597325; OAI: oai:pubmedcentral.nih.gov:57804; Copyright (c) 2001 Lawson et al, licensee BioMed Central Ltd; Country of input: International Atomic Energy Agency (IAEA)
HAZARDS, HORMONES, MAMMARY GLANDS, NEOPLASMS, RECEPTORS, THERAPY, WOMEN
ANIMALS, BODY, DISEASES, FEMALES, GLANDS, MAMMALS, MAN, MEDICINE, MEMBRANE PROTEINS, ORGANIC COMPOUNDS, ORGANS, PRIMATES, PROTEINS, VERTEBRATES
Breast imaging technology: Application of magnetic resonance imaging to early detection of breast cancer
Schnall, Mitchell D
[en] Since its first introduction approximately 10 years ago, there has been extensive progress in the application of magnetic resonance imaging (MRI) to the detection and diagnosis of breast cancer. Contrast-enhanced MRI has been shown to have value in the diagnostic work-up of women who present with mammogram or clinical abnormalities. In addition, it has been demonstrated that MRI can detect mammogram occult multifocal cancer in patients who present with unifocal disease. Advances in risk stratification and limitations in mammography have stimulated interest in the use of MRI to screen high-risk women for cancer. Several studies of MRI high-risk screening are ongoing. Preliminary results are encouraging
Available from http://dx.doi.org/10.1186/bcr265; Available from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC138672; PMCID: PMC138672; PUBLISHER-ID: bcr265; PMID: 11300101; OAI: oai:pubmedcentral.nih.gov:138672; Copyright (c) 2000 BioMed Central Ltd on behalf of the copyright holder; Country of input: International Atomic Energy Agency (IAEA)
BIOMEDICAL RADIOGRAPHY, MAMMARY GLANDS, NEOPLASMS, NMR IMAGING, WOMEN
ANIMALS, BODY, DIAGNOSTIC TECHNIQUES, DISEASES, FEMALES, GLANDS, MAMMALS, MAN, MEDICINE, NUCLEAR MEDICINE, ORGANS, PRIMATES, RADIOLOGY, VERTEBRATES
http://dx.doi.org/10.1186/bcr265, http://www.ncbi.nlm.nih.gov/pmc/articles/PMC138672
Breast imaging technology: Application of magnetic resonance imaging to angiogenesis in breast cancer
[en] Magnetic resonance imaging (MRI) techniques enable vascular function to be mapped with high spatial resolution. Current methods for imaging in breast cancer are described, and a review of recent studies that compared dynamic contrast-enhanced MRI with histopathological indicators of tumour vascular status is provided. These studies show correlation between in vivo dynamic contrast measurements and in vitro histopathology. Dynamic contrast enhanced MRI is also being applied to assessment of the response of breast tumours to treatment
ANGIOGENESIS, BIOMEDICAL RADIOGRAPHY, CHARGES, IN VITRO, IN VIVO, MAMMARY GLANDS, NEOPLASMS, NMR IMAGING, PERMEABILITY, SPATIAL RESOLUTION
BODY, DIAGNOSTIC TECHNIQUES, DISEASES, GLANDS, MEDICINE, NUCLEAR MEDICINE, ORGANS, PHYSICAL PROPERTIES, RADIOLOGY, RESOLUTION
Hypoxia and oxidative stress in breast cancer: Hypoxia signalling pathways
Pugh, Christopher W; Gleadle, Jonathan; Maxwell, Patrick H
[en] Hypoxia-inducible factor-1 (HIF), which is centrally involved in physiological oxygen homeostasis, is also activated in the majority of tumours. Activation of HIF can occur through genetic mechanisms or as a result of hypoxia within the tumour microenvironment. In some cases HIF activation appears to be intimately linked to the proliferative stimulus itself. HIF affects patterns of gene expression and tumour growth, although precise effects vary between tumour types. Modulation of HIF activity, if correctly applied, may be therapeutically beneficial in tumour therapy
Available from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC138694; PMCID: PMC138694; PUBLISHER-ID: bcr313; PMID: 11597320; OAI: oai:pubmedcentral.nih.gov:138694; Copyright (c) 2001 BioMed Central Ltd; Country of input: International Atomic Energy Agency (IAEA)
ANGIOGENESIS, MAMMARY GLANDS, NEOPLASMS, OXYGEN, STRESSES
BODY, DISEASES, ELEMENTS, GLANDS, NONMETALS, ORGANS
Huiping, Chen; Kristjansdottir, Sigrun; Jonasson, Jon G; Magnusson, Jonas; Egilsson, Valgardur; Ingvarsson, Sigurdur
[en] The E-cadherin-catenin complex plays a crucial role in epithelial cell-cell adhesion and in the maintenance of tissue architecture. Perturbation in the expression or function of this complex results in loss of intercellular adhesion, with possible consequent cell transformation and tumour progression. We studied the alterations of E-cadherin and β-catenin in a set of 50 primary gastric tumours by using loss of heterozygosity (LOH) analysis, gene mutation screening, detection of aberrant transcripts and immunohistochemistry (IHC). A high frequency (75%) of LOH was detected at 16q22.1 containing E-cadherin locus. Three cases (6%) showed the identical missense mutation, A592T. This mutation is not likely to contribute strongly to the carcinogenesis of gastric cancer, because a low frequency (1.6%) of this mutation was also found in 187 normal individuals. We also detected a low frequency (0.36%, 0%) of this mutation in 280 breast tumours and 444 other tumours, including colon and rectum, lung, endometrium, ovary, testis, kidney, thyroid carcinomas and sarcomas, respectively. We also analyzed the aberrant E-cadherin mRNAs in the gastric tumours and found that 7 tumours (18%) had aberrant mRNAs in addition to the normal mRNA. These aberrant mRNAs may produce abnormal E-cadherin molecules, resulting in weak cell-cell adhesion and invasive behaviour of carcinoma cells. Reduced expression of E-cadherin and β-catenin was identified at the frequency of 42% and 28%, respectively. Specially, 11 tumours (22%) exhibited positive cytoplasmic staining for β-catenin IHC. An association was found between reduced expression of E-cadherin and β-catenin. Moreover, an association was detected between reduced expression of E-cadherin and diffuse histotype. Our results support the hypothesis that alterations of E-cadherin and β-catenin play a role in the initiation and progression of gastric cancer
Available from http://dx.doi.org/10.1186/1471-2407-1-16; Available from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC60969; PMCID: PMC60969; PUBLISHER-ID: 1471-2407-1-16; PMID: 11747475; OAI: oai:pubmedcentral.nih.gov:60969; Copyright (c) 2001 Huiping et al; licensee BioMed Central Ltd. This is an Open Access article: verbatim copying and redistribution of this article are permitted in all media for any purpose, provided this notice is preserved along with the article's original URL.; Country of input: International Atomic Energy Agency (IAEA)
ADHESION, CARCINOGENESIS, CARCINOMAS, DISTURBANCES, GENE MUTATIONS, KIDNEYS, LUNGS, NEOPLASMS, RECTUM, SARCOMAS, SCREENING, THYROID, UTERUS
BODY, DIGESTIVE SYSTEM, DISEASES, ENDOCRINE GLANDS, FEMALE GENITALS, GASTROINTESTINAL TRACT, GLANDS, INTESTINES, LARGE INTESTINE, MUTATIONS, NEOPLASMS, ORGANS, PATHOGENESIS, RESPIRATORY SYSTEM
http://dx.doi.org/10.1186/1471-2407-1-16, http://www.ncbi.nlm.nih.gov/pmc/articles/PMC60969
Fermi surfaces of LaRhSi3 and CeRhSi3
Kimura, N.; Umeda, Y.; Asai, T.; Terashima, T.; Aoki, H., E-mail: nkimura@mail.cc.tohoku.ac.jp
[en] CeRhSi3 has the large electronic specific-heat coefficient
\text{γ=120}\phantom{\rule{2px}{0ex}}
mJ/mol K2 and the high Kondo temperature TK∼100 K, although its magnetic susceptibility follows the Curie–Weiss law and the specific heat indicates magnetic ordering at 1.8 K. In order to elucidate the 4f-electron nature in CeRhSi3, we have grown single crystals of CeRhSi3 and of its reference material LaRhSi3 and measured the electrical resistivity, the magnetic susceptibility and the de Haas–van Alphen (dHvA) effect. Comparison of the angular dependence of the dHvA frequency in CeRhSi3 to that in LaRhSi3, indicates that the 4f-electrons of CeRhSi3 are itinerant.
S0921452600006591; Available from http://dx.doi.org/10.1016/S0921-4526(00)00659-1; Copyright (c) 2017 Elsevier Science B.V., Amsterdam, The Netherlands, All rights reserved.; Country of input: International Atomic Energy Agency (IAEA)
Physica. B, Condensed Matter; ISSN 0921-4526; ; CODEN PHYBE3; v. 294-295; p. 280-283
CERIUM COMPOUNDS, CURIE-WEISS LAW, DE HAAS-VAN ALPHEN EFFECT, ELECTRIC CONDUCTIVITY, ELECTRONIC SPECIFIC HEAT, ELECTRONS, FERMI LEVEL, KONDO EFFECT, MAGNETIC SUSCEPTIBILITY, MAGNETIZATION, MONOCRYSTALS
CRYSTALS, ELECTRICAL PROPERTIES, ELEMENTARY PARTICLES, ENERGY LEVELS, FERMIONS, LEPTONS, MAGNETIC PROPERTIES, PHYSICAL PROPERTIES, RARE EARTH COMPOUNDS, SPECIFIC HEAT, THERMODYNAMIC PROPERTIES
Deep level defects in sublimation-grown 6H silicon carbide investigated by DLTS and EPR
Irmscher, K.; Pintilie, I.; Pintilie, L.; Schulz, D., E-mail: irmscher@ikz-berlin.de
[en] 6H-SiC bulk single crystals grown by physical vapor transport (PVT) were investigated by deep-level transient spectroscopy (DLTS) and electron paramagnetic resonance (EPR). One of the observed deep level defects was identified as isolated tungsten on Si sites by EPR. The electron spin of
\text{1}\text{2}
could be explained by W5+ (5d1). This is equivalent to the single positive charge state of a double donor when taking into account the Fermi level position in the n-type samples. The interpretation is also consistent with the DLTS detection of a W related deep level which showed a behavior of the capture of electrons and holes that hints at a double donor. In addition a tantalum related deep level is tentatively discussed. W and Ta were incorporated on electrically active sites in 6H-SiC only in low concentrations (2–4×1014 cm−3) during crystal growth by PVT.
CHARGE STATES, CONCENTRATION RATIO, CRYSTAL GROWTH, DEEP LEVEL TRANSIENT SPECTROSCOPY, DEFECTS, ELECTRON SPIN RESONANCE, ELECTRONS, FERMI LEVEL, HOLES, MONOCRYSTALS, SILICON, SILICON CARBIDES, SUBLIMATION, TANTALUM, TUNGSTEN, TUNGSTEN IONS, VAPORS
CARBIDES, CARBON COMPOUNDS, CHARGED PARTICLES, CRYSTALS, DIMENSIONLESS NUMBERS, ELEMENTARY PARTICLES, ELEMENTS, ENERGY LEVELS, EVAPORATION, FERMIONS, FLUIDS, GASES, IONS, LEPTONS, MAGNETIC RESONANCE, METALS, PHASE TRANSFORMATIONS, REFRACTORY METALS, RESONANCE, SEMIMETALS, SILICON COMPOUNDS, SPECTROSCOPY, TRANSITION ELEMENTS
The electric field effect in g3 model of the layer superconductors
Yang, Tzong-Jer; Lee, Wen-Der, E-mail: yangtj@cc.nctu.edu.tw
[en] High-Tc superconductor is considered as alternately stacked metal and insulating layers. The electric field is applied along the stacking direction and investigated in the frame of Ginzburg–Landau theory with the coupling g3 between metallic layers for the transition temperature Tc. Then the system of equations for Tc can be solved by the transfer matrix method. The shift of the transition temperature is analytically derived and we find the criterion for the enhancement of Tc while the electric field is applied.
Physica. C, Superconductivity; ISSN 0921-4534; ; CODEN PHYCE6; v. 364-365; p. 166-169
ELECTRIC FIELDS, GINZBURG-LANDAU THEORY, HIGH-TC SUPERCONDUCTORS, METALS, TRANSFER MATRIX METHOD, TRANSITION TEMPERATURE
CALCULATION METHODS, ELEMENTS, PHYSICAL PROPERTIES, SUPERCONDUCTORS, THERMODYNAMIC PROPERTIES, TYPE-II SUPERCONDUCTORS
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Compare accuracies of two classification models using new data - MATLAB compareHoldout
{\stackrel{^}{\pi }}_{ijk}
{\stackrel{^}{\pi }}_{ijk}
\sum _{i,j,k}{n}_{ijk}={n}_{test}.
\sum _{i,j,k}{\pi }_{ijk}=\sum _{i,j,k}{\stackrel{^}{\pi }}_{ijk}=1.
\delta =\sum _{i=1}^{K}\sum _{j=1}^{K}\sum _{k=1}^{K}\left({c}_{ki}-{c}_{kj}\right){\pi }_{ijk}.
\begin{array}{c}{H}_{0}:\delta =0\\ {H}_{1}:\delta \ne 0\end{array}.
{t}_{{\chi }^{2}}^{\ast }=\sum _{i\ne j}\sum _{k}\frac{{\left({n}_{ijk}+1-\left({n}_{test}+{K}^{3}\right){\stackrel{^}{\pi }}_{ijk}^{\left(1\right)}\right)}^{2}}{{n}_{ijk}+1}.
1-{F}_{{\chi }^{2}}\left({t}_{{\chi }^{2}}^{\ast };1\right)<\alpha
{\stackrel{^}{\pi }}_{ijk}^{\left(1\right)}
{t}_{{\chi }^{2}}^{\ast }
{F}_{{\chi }^{2}}\left(x;1\right)
{t}_{LRT}^{\ast }=2\mathrm{log}\left[\frac{P\left(\underset{i,j,k}{\cap }{N}_{ijk}={n}_{ijk};{n}_{test},{\stackrel{^}{\pi }}_{ijk}={\stackrel{^}{\pi }}_{ijk}^{\left(2\right)}\right)}{P\left(\underset{i,j,k}{\cap }{N}_{ijk}={n}_{ijk};{n}_{test},{\stackrel{^}{\pi }}_{ijk}={\stackrel{^}{\pi }}_{ijk}^{\left(3\right)}\right)}\right].
1-{F}_{{\chi }^{2}}\left({t}_{LRT}^{\ast };1\right)<\alpha ,
{\stackrel{^}{\pi }}_{ijk}^{\left(2\right)}=\frac{{n}_{ijk}}{{n}_{test}}
{\stackrel{^}{\pi }}_{ijk}^{\left(3\right)}=\frac{{n}_{ijk}}{{n}_{test}+\lambda \left({c}_{ki}-{c}_{kj}\right)}
\sum _{i,j,k}\frac{{n}_{ijk}\left({c}_{ki}-{c}_{kj}\right)}{{n}_{test}+\lambda \left({c}_{ki}-{c}_{kj}\right)}=0.
{F}_{{\chi }^{2}}\left(x;1\right)
{\stackrel{^}{\pi }}_{2•}={n}_{2•}/n
{\stackrel{^}{\pi }}_{•2}={n}_{•2}/n
\begin{array}{c}{H}_{0}:{\pi }_{•2}={\pi }_{2•}\\ {H}_{1}:{\pi }_{•2}\ne {\pi }_{2•}\end{array}.
{H}_{0}:{\pi }_{12}={\pi }_{21}.
{t}_{a1}^{\ast }=\frac{{n}_{12}-{n}_{21}}{\sqrt{{n}_{12}+{n}_{21}}}.
1-\Phi \left(|{t}_{1}^{\ast }|\right)<\alpha ,
{t}_{a2}^{\ast }=\frac{{\left({n}_{12}-{n}_{21}\right)}^{2}}{{n}_{12}+{n}_{21}}.
1-{F}_{{\chi }^{2}}\left({t}_{2}^{\ast };m\right)<\alpha
{F}_{{\chi }^{2}}\left(x;m\right)
{n}_{d}={n}_{12}+{n}_{21}
{t}_{1}^{\ast }={n}_{12}.
{F}_{\text{Bin}}\left({t}_{1}^{\ast };{n}_{d},0.5\right)<\alpha
{F}_{\text{Bin}}\left(x;n,p\right)
{t}_{2}^{\ast }=\mathrm{min}\left({n}_{12},{n}_{21}\right).
{F}_{\text{Bin}}\left({t}_{2}^{\ast };{n}_{d},0.5\right)<\alpha /2
{t}_{1}^{\ast }={n}_{12}.
{F}_{\text{Bin}}\left({t}_{1}^{\ast }-1;{n}_{12}+{n}_{21},0.5\right)+0.5{f}_{\text{Bin}}\left({t}_{1}^{\ast };{n}_{12}+{n}_{21},0.5\right)<\alpha
{F}_{\text{Bin}}\left(x;n,p\right)
{f}_{\text{Bin}}\left(x;n,p\right)
{t}_{2}^{\ast }=\mathrm{min}\left({n}_{12},{n}_{21}\right).
{F}_{\text{Bin}}\left({t}_{2}^{\ast }-1;{n}_{12}+{n}_{21}-1,0.5\right)+0.5{f}_{\text{Bin}}\left({t}_{2}^{\ast };{n}_{12}+{n}_{21},0.5\right)<\alpha /2
{\stackrel{^}{\pi }}_{ijk}=\frac{{n}_{ijk}}{{n}_{test}}.
\sum _{i,j,k}{n}_{ijk}={n}_{test}.
{e}_{1}=\sum _{j=1}^{K}\sum _{k=1}^{K}\sum _{i\ne k}^{}{\stackrel{^}{\pi }}_{ijk}.
{e}_{1}=\sum _{j=1}^{K}\sum _{k=1}^{K}\sum _{i\ne k}^{}{\stackrel{^}{\pi }}_{ijk}{c}_{ki},
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Through the looking gl... the eyepiece
What is the field of view of a telescope?
How to calculate the field of view of a telescope
How to use our field of view of a telescope calculator
Telescopes open our eyes to the marvel of the cosmos: the telescope field of view calculator will tell you exactly how much!
The basics of telescopes and eyepieces;
What are the mysterious numbers on an eyepiece?;
What is the field of view of a telescope?; and
How to calculate the field of view of a telescope.
And much more. Travel the cosmo from your backyard with Omni Calculator! 💫
Our eyes work pretty okay, but they are absolutely useless when it comes to distant things. It makes sense: we evolved to see predators a hundred meters away, not stars on the other side of the galaxy. That's where telescopes really give us an edge.
A telescope is an optical device that collects light, thanks to a well-thought set of lenses, which focuses and magnifies an image on our retina.
There are two main types of telescopes:
Refractors; and
The difference lies in how they focus the light. Refractors use a lens called an objective lens, while reflectors use a mirror placed at the bottom of the telescope (in this case, the collecting end is called the aperture). Either way, the light gets focused at a certain distance (the focal length), from where it starts diverging again.
It's at this point that the eyepieces enter the scene.
🙋 If you are interested in other optics calculators, check our thin lens equation calculator and our lens maker equation calculator! Aren't lenses enough? What about our mirror equation calculator?
The eyepiece is another lens (or group of lenses) that collects the diverging light rays coming out of the telescope's focal point, and makes them parallel (again: that's how they entered the telescope), thus recreating the image. Two things happened:
The object got magnified: you can learn more at our telescope magnification calculator; and
The perceived brightness increased since the light collected from the whole opening of the telescope is now concentrated in a small spot (that's why the bigger the telescopes, the better they are: as they collect more light).
A telescope collects light and focuses it on your eyes. The focal lengths of the optic tube and the eyepiece determine the magnification.
But this is not where it ends. Our knowledge of the universe would be way different if we could explore it only through a pinhole. Luckily, telescope makers learned how to control and tune the telescope's field of view.
The night sky is so distant we can assume it is infinitely so. Even when talking of objects like our neighboring planets and galaxies billions of years away, we perceive them as placed on the same, unreachable plane.
That gives us difficulty specifying distances: can we say, "look three meters on the left of Jupiter"? Clearly, the answer is "no". That's why astronomers measure distances using angles.
What happened? Don't worry. We just moved for a second into deep space! Now, the night sky surrounds you all around. Please take a second to relax, and now, follow us.
We told you that everything in space looks infinitely distant: this is akin to saying, "every object lies on a sphere with radius
r=\infty
". And guess what? You are at the center of that sphere. Do you feel important? You are not, but this is fine.
Now, it's straightforward to think in terms of angles. A sphere is a perfect place to say "two degrees on the left" since every point on it corresponds to a radius converging on the center.
We can use two angles varying from
0\degree
360\degree
to identify any point on the sphere's surface. And every object has an angular dimension too. The Sun has an approximate angular diameter of
32'
(the apostrophe indicates the minutes of arc). The Moon's diameter oscillates between
29'
34'
(this remarkable coincidence makes it possible for humans to witness eclipses). The Andromeda galaxy, M31, has angular dimensions
3\degree 10'
1\degree
, while Neptune is just
2''
(seconds of arc).
🔎 Proxima Centauri, the second-closest star to Earth (do you know which one is the closest?), has an angular diameter equal to
0.001''
. Take a pin, place it somewhere, and get
410
kilometers away from it. The pinhead now looks as big as that star!
Back to telescopes: the combination of telescope and eyepiece allows you to see a particular portion of the sky, which has an angular diameter called field of view. Is it better to choose, for your telescope, a wide field of view or a narrow one? And what's the definition of telescope field of view? Keep reading!
The field of view, along with the magnification and the opening, is one of the most critical parameters of a telescope.
The larger the field of view, the more you can see at once: at high magnification, this makes the difference between — for example — looking at a single crater or a striking lunar landscape.
Every eyepiece has a specified apparent field of view
\text{fov}_a
. This quantity defines the angular diameter of the portion of the sky you'd see by looking through the eyepiece alone. The values of
\text{fov}_a
30\degree
110\degree
To compute the field of view, we take the value of
\text{fov}_a
, and we divide it by the magnification
m
of the combination telescope-eyepiece:
\text{fov}_c=\frac{\text{fov}_a}{m}
The magnification
m
is calculated with the ratio of the focal lengths of the telescope
f_t
and the eyepiece
f_e
; thus, we can rewrite the formula as:
\text{fov}_c=\frac{\text{fov}_a}{{f_t}/{f_e}}
🔎 Remember that the field of view of a telescope doesn't depend on the magnification!
The views in the images correspond to eyepiece with equal focal length but different apparent fields of view. Which one do you prefer?
As you can see, the field of view of a telescope depends on both the optic tube and the eyepiece. The telescope alone doesn't have a field of view, though there are limits to the portion of the sky it collects light from. The eyepiece is the part defining the width of the field of view. Usually, the field of view doesn't exceed
10\degree
. To give your telescope a wide field of view, mount an eyepiece with an apparent field of view on the "higher-end" of the spectrum and vice versa.
You can find the value of the field of view of your telescope in a few steps with our telescope field of view calculator.
The most basic way to use our calculator is to provide the values of the magnification and the apparent field of view. However, you may not know the magnification: you can either go to our magnification calculator or click on advanced mode.
There are too many types of telescopes and eyepieces out there to create a comprehensive database of their focal lengths. We made two lists with some selected values (the most common in the field); if you can't find the specifications of your set-up, select custom in the drop-down lists.
🙋 Feel free to use our calculators in reverse: insert the desired field of view value and find the correct apparent field of view, or find the best magnification for any pair of apparent and "true" fields of view.
Tip: Don't know (or can't find) the value of the apparent field of view of the eyepiece you are using? Measure the diameter
d
of the lens you can see at the bottom of your eyepiece. Multiply it by
57.3
and use the result to approximate the apparent field of view. You can perform the operation directly in the field of our calculator!
What about an example? Let's take a good set-up, with a magnification of
560\times
(enough to see the Cassini division in the Saturn's rings), for example, given by a telescope with focal length
f_t = 2800\ \text{mm}
and an eyepiece with
f_e=5\ \text{mm}
. The chosen eyepiece has a
58\degree
apparent field of view: what is the true field of view?
\begin{align*} \text{fov}_c&=\frac{\text{fov}_a}{m} = \frac{58\degree}{560}\\\\ &= 0.104\degree = 6.24' \end{align*}
Saturn has, on average, an angular diameter of
15''
. You can fit around 25 Saturns in your field of view!
Omni tip: at least once in your life, you should see Saturn with your own eyes. We promise it's quite an experience!
Have you ever seen Saturn through a telescope? Here it is, floating with its rings in your field of view (well, an artist's impression anyway).
We hope our telescope field of view calculator helped you understand the concept of field of view and what those numbers on your instrument mean. Are you looking for something other than telescopes? We covered binoculars, too, in our binoculars range calculator!
If you were looking for the concept of field of view in photography, visit our camera field of view calculator!
The field of view of a telescope measures the size of the portion of the sky associated with a particular set-up. Wide fields of view give you a better view of the surroundings of an object and are ideal for observations of open clusters, nebulae, and the Moon. A smaller field of view helps you focus on the details.
What is the field of view of the Hubble space telescope in square degrees?
The WFC3 (wide-field camera 3) of the Hubble space telescope has a field of view (in square degrees) equal to 1.98 × 10⁻³ sq. deg., which translates into 25600 square seconds of arc.
The images beamed down to Earth by Hubble when using WFC3 are squares with sides equal to 160", about one-tenth of the diameter of the Moon, from the Earth.
How do I calculate the field of view of a telescope?
To calculate the field of view of a telescope, you need to find some characteristic values of the set-up and an equation:
Locate the apparent field of view of the eyepiece.
Find the magnification m of the ensemble telescope-eyepiece, or, alternatively:
The focal length of the optic tube fₜ; and
The focal length of the eyepiece fₑ.
Then, use the formula:
fov = fovₐ/m = fovₐ/(fₜ/fₑ)
The result is the diameter of the field of view.
Where do I find the apparent field of view of my eyepiece?
The apparent field of view should be marked on the eyepiece itself. Look for a number followed by the degree symbol, °. Most eyepieces come with an apparent field of view between 30° and 110°. If you really can't find the value, you can approximate it with the formula 57.3 × d, where d is the diameter of the entrance lens of the eyepiece.
The apparent field of view value varies with the magnification on zoom eyepieces.
What is the field of view of a telescope magnifying 80×, using an eyepiece with an apparent field of view of 75°?
0.94° or 56.4'. We find this value with the formula for the field of view of a telescope:
fov = fovₐ/m = 75°/80 = 0.94°
This value is almost a full degree, which is enough to contain two times the Moon, or the Nebula Carina, the brightest nebula in the night sky.
Apparent field of view (fovₐ)
Area of the field of view
The skin depth calculator allows you to determine the skin effect in conductors for any signal frequency.
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Training neural network requires specifying an initial value of the weights. You'll see that a well-chosen initialization method could improve learning and accuracy
Improving Deep Neural Networks - Initialization:
Training neural network requires specifying an initial value of the weights. A well-chosen initialization method would improve learning and accuracy.
If you were reading my previous tutorial pars, you probably followed my instructions for weight initialization, and it has worked out so far. But how do we choose the initialization for another neural network? In this part, I will show you that different initialization methods can lead to different results.
A well-chosen initialization can speed up the convergence of gradient descent and increase the odds of gradient descent converging to a lower training error.
I will use code from my last tutorial, where we used to classify circles. Before we used random weights initialization, now we'll try "He Initialization", this is named for the first author of He et al., 2015. (If you have heard of "Xavier initialization", this is similar, except Xavier initialization uses a scaling factor for the weights W[l] of:
\sqrt{\frac{1}{layer_dim{s}^{\left[l-1\right]}}}, \text{where He initialization would use} \sqrt{\frac{2}{layer_dim{s}^{\left[l-1\right]}}}
Before, we used the following code:
The only difference is that instead of multiplying np.random.randn(..,..) by 0.01, we will multiply it by sqrt(2/dimensions of the previous layer), which is what "He" initialization recommends for layers with a ReLU activation:
def initialize_parameters_he(layer_dimension):
parameters["W" + str(l)] = np.random.randn(layer_dimension[l], layer_dimension[l-1]) * np.sqrt(2./layers_dims[l-1])
At first, let's test our results with the random initialization, that we could compare the difference:
From this above plot, we can see that cost drops only after 8000 training steps.
Classification results are quite fine for us.
From this above plot, we can see that cost drops almost instantly; we don't need to wait for 8000 steps to see improvement.
Classification results are 1% better than using random initialization.
We should remember that different parameters initialization methods lead to different results. Random initialization is used to break symmetry and to make sure different hidden units can learn different things.
Don't initialize to values that are too large, but initializing with overly large random numbers you'll slow down the optimization. As you can see, "He" initialization works well for networks with ReLU activations.
This is the last deep learning tutorial. To get better results, you can optimize this deep neural network by implementing L2 regularization, dropout, and gradient checking. You can try to optimize using momentum, adam optimizer or implement training with mini-batches.
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Now when we have initialized our parameters, we will do the forward propagation module by implementing functions that we'll use when implementing the model.
Now when we have initialized our parameters, we will do the forward propagation module. We will start by implementing some basic functions that we will use later when implementing the model. We will complete three functions in this order:
LINEAR -> ACTIVATION where ACTIVATION will be either ReLU or Sigmoid;
[LINEAR -> RELU] × (L-1) -> LINEAR -> SIGMOID (whole model).
I could write all these functions in one block, but then it's harder to understand code to leave it for learning purposes. I'll remind you that the linear forward module (vectorized over all the examples) computes the following equation:
{Z}^{\left[l\right]}={W}^{\left[l\right]}{A}^{\left[l-1\right]}+{b}^{\left[l\right]}
where A[0]=X
Code for our linear_forward function:
A - activations from previous layer (or input data): (size of the previous layer, number of examples);
W - weights matrix: NumPy array of shape (size of current layer, size of the previous layer);
b - bias vector, NumPy array of shape (size of the current layer, 1).
Z - the input of the activation function, also called a pre-activation parameter;
cache - a python dictionary containing "A", "W", and "b"; stored for computing the backward pass efficiently.
Z = np.dot(W,A)+b
cache = (A, W, b)
Linear-Activation Forward function:
In this tutorial, we will test two activation functions, Sigmoid and ReLu:
\sigma \left(Z\right)=\sigma \left(WA+b\right)=\frac{1}{1+{e}^{-\left(WA+b\right)}}
I will write you the sigmoid function. This function returns two items: the activation value "a" and a "cache" that contains "Z" (it's what we will feed into the corresponding backward function). To use it, we'll call:
Numpy sigmoid activation implementation
Z - numpy array of any shape
A - output of sigmoid(z), same shape as Z
cache - returns Z as well, useful during backpropagation
A = 1/(1+np.exp(-Z))
cache = Z
ReLu: The mathematical formula for ReLu is A=RELU(Z)=max(0, Z). I will write you the ReLu function. This function returns two items: the activation value "A" and a "cache" that contains "Z" (it's what we will feed into the corresponding backward function). To use it, we'll call:
A, activation_cache = relu(Z)
ReLu function:
Numpy Relu activation implementation
Z - Output of the linear layer, of any shape
A - Post-activation parameter, of the same shape as Z
cache - a python dictionary containing "A"; stored for computing the backward pass efficiently
A = np.maximum(0,Z)
For more convenience, we are going to group two functions (Linear and Activation) into one function (LINEAR->ACTIVATION). Hence, we will implement a function that does the LINEAR forward step followed by an ACTIVATION forward step.
Code for our linear_activation_forward function:
A_prev - activations from previous layer (or input data): (size of the previous layer, number of examples);
b - bias vector, NumPy array of shape (size of the current layer, 1);
activation - the activation to be used in this layer, stored as a text string: "sigmoid" or "relu".
A - the output of the activation function, also called the post-activation value;
cache - a python dictionary containing "linear_cache" and "activation_cache" stored efficiently for computing the backward pass.
# Inputs: "A_prev, W, b". Outputs: "A, activation_cache".
Z, linear_cache = linear_forward(A_prev,W,b)
L-Layer Model implementation:
For more convenience when implementing the L-layer Neural Network, we will need a function that replicates the above (linear_activation_forward with RELU) L−1 times and then follows that with one linear_activation_forward SIGMOID.
So writing our code, we'll use the functions we had previously written. In the code below, the variable AL will denote:
{A}^{\left[L\right]}=\sigma \left({Z}^{\left[L\right]}\right)=\sigma \left({W}^{\left[L\right]}{A}^{\left[L-1\right]}+{b}^{\left[L\right]}\right),\phantom{\rule{0ex}{0ex}}\text{This is sometimes is called }\mathbf{\text{Yhat}}\text{, i. e., this is }\stackrel{^}{Y}\text{.}\phantom{\rule{0ex}{0ex}}
Code for our L_model_forward function:
X - data, numpy array of shape (input size, number of examples);
parameters - output of initialize_parameters_deep() function.
AL - last post-activation value;
caches - list of caches containing every cache of linear_activation_forward() (there are L-1 of them, indexed from 0 to L-1)
# Using a for loop to replicate [LINEAR->RELU] (L-1) times
# Implementation of LINEAR -> RELU.
A, cache = linear_activation_forward(A_prev, parameters['W' + str(l)], parameters['b' + str(l)], activation = "relu")
# Adding "cache" to the "caches" list.
# Implementation of LINEAR -> SIGMOID.
AL, cache = linear_activation_forward(A, parameters['W' + str(L)], parameters['b' + str(L)], activation = "sigmoid")
Now we can test the functions we implemented in this tutorial with random numbers to see if they work well. I'll initialize the neural network with 2 deep layers, with 4 inputs and one output. For inputs, I'll use a command np.random.randn(input size, number of examples). Then I'll call the L_model_forward function:
layers_dims = [4,3,2,1]
print("X.shape = ",X.shape)
print("AL =",AL)
print("Length of caches list =",len(caches))
print("parameters:",parameters)
I received such results; you may receive them a little different:
X.shape = (4, 10)
AL = [[0.5 0.5 0.5 0.5 0.50000402 0.5 0.50000157 0.5 0.5 0.50000136]]
Length of caches list = 3
'W1': array([[ 0.00315373, -0.00545479, 0.00453286, -0.00320905],
[-0.00219829, -0.00134337, 0.0017775 , 0.01365787],
'b1': array([[0.],
'W2': array([[ 0.00094041, -0.01439561, -0.0117556 ],
[-0.00528372, -0.00807826, 0.02711167]]),
'W3': array([[0.00180661, 0.01540206]]),
'b3': array([[0.]])}
cache -- returns Z as well, useful during backpropagation
def initialize_parameters_deep(layer_dimension):
L = len(layer_dimension)
parameters["W" + str(l)] = np.random.randn(layer_dimension[l], layer_dimension[l-1]) * 0.01
parameters["b" + str(l)] = np.zeros((layer_dimension[l], 1))
layer_dims = [4,3,2,2,1]
parameters = initialize_parameters_deep(layer_dims)
print("X.shape =", X.shape)
print("AL =", AL)
print("Lenght of caches list = ", len(caches))
print("parameters:", parameters)
Now we have a full forward propagation that takes the input X and outputs a row vector A[L] containing our predictions. It also records all intermediate values in "caches". Using A[L] we can compute the cost of our predictions.
So this was a more difficult tutorial part to understand, but don't worry. You can read it few more times, print values out to get a better understanding.
We'll build a cost function in the next tutorial, and we'll start building backpropagation functions.
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DocumentTools/Canvas/Script/Annotate - Maple Help
Home : Support : Online Help : DocumentTools/Canvas/Script/Annotate
script action to annotate math on a canvas
Annotate(script,caption)
Annotate(script,caption,source)
(optional) Math record
The Annotate command adds message text next to math on a canvas.
The source option is an element of the array returned by GetMath. It can be omitted if SetActive is used to specify the active math container prior to calling Annotate.
In this example, we will insert a canvas, get the math from the canvas, and manually add annotations to the math input.
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Canvas}\right):
create the canvas:
\mathrm{cv}≔\mathrm{NewCanvas}\left(["Sample Canvas",x+y]\right):
\mathrm{ShowCanvas}\left(\mathrm{cv},'\mathrm{input}'=[\mathrm{box}[A],\mathrm{box}[B]]\right)
get references to the math data in the canvas:
\mathrm{data}≔\mathrm{GetMath}\left(\right):
look at the math property of the first element:
\mathrm{data}[1]:-\mathrm{math}
create a script object and add some annotations
\mathrm{sc}≔\mathrm{Script}\left(\right):
\mathrm{Annotate}\left(\mathrm{sc},"this is a sum",\mathrm{data}[1]\right)
\mathrm{Annotate}\left(\mathrm{sc},"this is box A",\mathrm{data}[2]\right)
loop through all of the math and annotate each element with its type
\mathbf{for}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{elem}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{data}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{Annotate}\left(\mathrm{sc},["This contains something of type",\mathrm{whattype}\left(\mathrm{elem}[\mathrm{math}]\right)],\mathrm{elem}\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{end}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}:
In this example, we will define an application that allows user input and provides feedback. The FindPrimes procedure will run when a button is pressed. It will find all the math in the canvas area, and add annotations to the inputs.
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Canvas}\right):
\mathrm{cv}≔\mathrm{NewCanvas}\left(["Write Some Prime Numbers Anywhere",\mathrm{ScriptButton}\left("Check Your Page",\mathrm{FindPrimes},\mathrm{position}=[500,50]\right)]\right):
\mathrm{ShowCanvas}\left(\mathrm{cv}\right)
\mathrm{ShareCanvas}\left(\mathrm{cv}\right)
The DocumentTools[Canvas][Script][Annotate] command was introduced in Maple 2021.
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The Barn-Pole Paradox - Maple Help
Home : Support : Online Help : Math Apps : Natural Sciences : Physics : The Barn-Pole Paradox
The Barn-Pole Paradox
\textcolor[rgb]{0,0.329411764705882,0.501960784313725}{}
The barn-pole paradox is a thought experiment that asks what appears to be a simple yes or no question, namely, whether or not a certain pole can fit into a certain barn; but apparently this paradox does not have a simple answer. In particular, no matter what the answer may be, if the pole is at rest relative to the barn, that answer can change if the pole is moving fast enough relative to the barn, due to relativistic length contraction and an appropriate choice of reference frame.
Suppose there is a pole of length 20 m and a barn of width 10 m. According to Einstein's theory of special relativity, if the pole is moving fast enough from the barn's perspective, it will be length-contracted and thus fit inside the barn. In contrast, from the pole's perspective the barn is moving and is length-contracted so there is no way the pole will fit!
The paradox results from the mistaken notion of absolute simultaneity. If both ends of the pole are in the barn simultaneously, then the pole fits in the barn. In special relativity, having two events be simultaneous in one reference frame does not imply that they are simultaneous in another frame.
Therefore, since simultaneity is relative to each observer, both observers were correct! From the barns perspective the pole fit in the barn, and from the pole's perspective, it did not fit in the barn.
Einstein's theory of special relativity describes the relationship between space and time; in particular, they are not independent quantities, but rather two sides of the same coin, namely spacetime. There are two postulates in special relativity:
The laws of physics remain the same in all inertial (non-accelerating) frames of reference.
The speed of light is constant and does not depend on the speed of the light source relative to the observer.
These postulates are in direct contradiction with the traditional notions of absolute space and time that provided the background for pre-20th century physics, but experiments have proven them to be correct.
The Lorentz transformation, time dilation and length contraction
The coordinates defining the position of an event in spacetime depend on the choice of inertial reference frame. The two postulates mentioned earlier imply that these coordinates must transform in a certain way known as the Lorentz transformation when switching reference frames. Suppose that
S
S'
are two reference frames with coordinates
\left(x,y,z,t\right)
\left(x', y', z', t'\right)
, such that the origin of both frames corresponds to the same event, and
S'
is moving at speed
v
x
-direction relative to
S
. The coordinates are related via:
t' = \mathrm{\gamma } \left(t-\frac{v x}{{c}^{2}}\right)
x' = \mathrm{γ} \left(x-v t\right)
y' = y
z' =z
\mathrm{γ} = \frac{c}{\sqrt{{c}^{2}- {v}^{2}}}>1
In particular, if an object accelerates to a speed
v
close to the speed of light, relative to an observer, then that observer will see that the object has shrunk by a factor of
\mathrm{γ}
relative to its rest length. Similarly, it will appear to be aging more slowly.
As mentioned earlier, suppose there is a pole of length 20 m and a barn of width 10 m.
Exactly how fast would the pole have to be going for it to fit inside the barn from the barn's perspective?
Let the common origin
\left(x,t\right) = \left(x',t'\right) = \left(0,0\right)
in both frames be the event when the leading end of the pole gets to the back door of the barn (assuming the pole is moving through the barn from front to back). We hypothesize that at the same time in the barn's frame of reference, the trailing edge of the pole coincides with (has the same
x
-coordinate as) the front door of the barn. The coordinates for the second event in the barn's frame are
\left(x,t\right) =\left(-10, 0\right)
, and in the pole's frame of reference they are
\left(x',t'\right)= \left(-20, t'\right)
. Plugging these values into the second equation of the Lorentz transformation yields:
-20 \mathrm{m} = \mathrm{γ} \left(-10 \mathrm{m}-v \cdot 0\right)
which requires
\mathrm{γ}=2
v = \frac{\sqrt{3}}{2}c=0.866 c
From the pole's point of view, what time is it when the end of the pole finally makes it into the barn?
The answer is just the value of
t'
stated earlier. To find it, use the first equation in the Lorentz transformation:
t' = 2 \left(0-\frac{\frac{\sqrt{3}}{2}c \cdot \left(-10 \mathrm{m}\right)}{{c}^{2}}\right)=\frac{5 \sqrt{3}\mathrm{m}}{c}= 1.67 ×{10}^{-8}\mathrm{s}
Admittedly, this is a really short time after
t'=0
, when the front of the pole was already hitting the end of the barn, but given the speed involved it is still an important difference, and it proves that the pole did not quite fit into the barn (in the pole's reference frame).
From the pole's perspective, how much of the pole had still not fit into the barn when the front of the pole hit the back of the barn?
We need to find the
x'
coordinate at time
t'=0
, so we use the second equation of the inverse Lorentz transformation, noting that the
x
coordinate for this event is -10 m:
x = \mathrm{γ} \left(x'+v t'\right)
-10 \mathrm{m}= 2 \left(x'+0\right)
x' = -5 \mathrm{m}
In other words, from the pole's perspective, only one quarter of the pole can fit into the barn!
Adjust the speed of the pole relative to the barn and manually move the pole to see if it fits into the barn. Then switch between reference frames to see how much difference it makes! Also, try starting with a pole with rest length less than 10 m and try adjusting the parameters so that the pole no longer fits.
Rest length of pole l (m)
\frac{v}{c}
x
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What is a pooled variance? - The DO Loop
s_1^2
s_2^2
n_1
n_2
s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2- 2}
n_1 = n_2
s_p^2 = (s_1^2 + s_2^2)/2
k
s_i^2
i
n_i
s_p^2 = \Sigma_{i=1}^k (n_i-1)s_i^2 / \Sigma_{i=1}^k (n_i - 1)
\Sigma_{i=1}^k s_i^2 / k
k
s_p^2 = (21*4.69 + 14*2.89 + 17*1.32) / (21+14+17) = 3.10
Hoon Park on October 12, 2021 10:41 am
Hi, I am a graduate student in South Korea. Thank you for posting such a nice lecture. It was very helpful.
I have a question, though. When you get the pooled variance with three samples for example, you calculated the pooled variance like,
Spool^2 = ((n1-1)*s1^2 + (n2-1) *s2^2 + (n3-1)*s3^2 )/df (where df= n1+n2+n3-3). This.
But when you get the Standard Error to get t-value using the pooled variance, how would you do it? my friend told me to do like SE= ((Spool^2) /(n1-1) + (Spool^2)/(n2-1))^(1/2) this. But I do not understand why you do not consider all three samples. I mean, it seems like my friend only took into account 2 samples according to the above equation.
I don't know if my question is understandable. I will looking forward to your reply. Thank you.
Rick Wicklin on October 12, 2021 11:36 am
I understand your question. I suggest you discuss it with your advisor.
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DocumentTools/Canvas/Script/Cursor - Maple Help
Home : Support : Online Help : DocumentTools/Canvas/Script/Cursor
Cursor(script,position)
The Cursor command allows you to record cursor movement script commands intended for use in playback within the Maple Learn environment, and are invisible to operation within Maple.
Some cursor script commands are as follows:
(where n is an integer) Move to position n in the current horizontal run (starting at zero).
Cursor down (May move between math containers)
Cursor up (May move between math containers)
Move to the very end of the expression
Move to the very start of the expression
Move to the end of current horizontal run. This is not necessarily the end of the expression. With the cursor on the + in (a+b)/(c-d), ENDPOS would move to after the b
decrement position in current horizontal run. Does not obey traversal rules like Left would. Stays in current horizontal run.
increment position in current horizontal run. Does not obey traversal rules like Right would. Stays in current horizontal run.
Go up one level of nesting. E.g. from inside a sqrt sign to the surrounding expression or from the denominator of a fraction to the surrounding expression. New cursor position within the outer expression is undefined, so this should be followed by POS or similar
Find first occurrence of id in current horizontal run, starting from the left. Position cursor before that occurrence. An id is specified as an id on the MathML tag, on input. I.e. <math><mfrac><mn id=numerator>1</mn><mi>x</mi></mfrac></math>
IDEND:id
Find first occurrence of id in current horizontal run, starting from the right. Position cursor behind that occurrence
\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Canvas}\right):
\mathrm{sc}≔\mathrm{Script}\left('\mathrm{mode}'='\mathrm{Learn}'\right):
\mathrm{SetActive}\left(\mathrm{sc},"0:0"\right)
\mathrm{Cursor}\left(\mathrm{sc},"End"\right)
\mathrm{cv}≔\mathrm{NewCanvas}\left(["Title"],\mathrm{sc}\right):
\mathrm{ShareCanvas}\left(\mathrm{cv}\right)
The DocumentTools[Canvas][Script][Cursor] command was introduced in Maple 2021.
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Denmark Chemistry Olympiad
The Danish Chemistry Olympiad has existed since 1982 and consists of 5 rounds with three selections (preliminary, semifinal and final). The entire competition season starts in mid-November and ends May where the national team is determined. The national team continues with training to Nordic Chemistry Olympiad (NChO) and the International Chemistry Olympiad (IChO) which are held in July or August. Figure 1 shows an overview of the structure and rounds of the competition.
Figure 1: An overview of the structure and rounds of the Danish Chemistry Olympiad. In the first round, everyone can participate. The 60 highest scorers are sent on to the semifinals. Here, the 14-16 best are selected for the finals, where the four best participants are found over three rounds to form the national team. The national team then participates in two international competitions, NChO and IChO
For the Danish Chemistry Olympiad, there are two rules for participation. The first is that all high school students can participate in the 1st and 2nd rounds regardless of age. If the student is over 20 years old on the first of July of the given year, he or she can not participate in the final rounds, NChO and IChO. The age limit is an international rule set by IChO. The second rule is that there can be a maximum of four students from each school for the 2nd round, three students for the final rounds and a maximum of two students among the four who make up the national team.
Preliminary Round (1st Round)
The preliminary round is held every year in mid-November locally at schools, and everyone can participate. One of your school’s chemistry teachers will hold the exam. If your school does not regularly participate in the Chemistry Olympiad, please contact your chemistry teacher. The first round consists of a 120 minute theoretical test of approximately seven tasks divided into around 30 total parts. The test allows limited aids in the exam: Formula Collection Chemistry A (both 1st and 2nd edition), DATA book physics & chemistry and a calculator or PC / MAC with CAS tool (Maple). Using books and notes during the exam is therefore prohibited. Participants are not expected to do all the tasks correctly or respond to all tasks. A general rule of thumb is that a score over 45% qualifies for the 2nd round.
The syllabus is everything from Chemistry A, including spectroscopy. Typical topics are stoichiometry, inorganic chemistry, organic chemistry, chemical equilibrium, acid and base, reaction kinetics, thermodynamics and spectroscopy.
Semifinals (2nd round)
The 60 highest scorers from round 1 get an invite around December to attend the 2nd round held the last weekend in January, from Friday to Sunday. A typical program consists of: the theoretical exam Friday, lectures followed by the practical exam and a tour of the Copenhagen University (KU) Chemistry Institute on Saturday and a ceremony on Sunday, concluding the round.
The theoretical exam consists of two parts in 120 minutes. Part 1 is 25 multiple choice questions and makes up 35% of the total theoretical score. Part 2 consists of a series of larger tasks in the same style as in the 1st round. The syllabus is the same as the 1st round but has 3-5 multiple choice questions in topics from the final rounds (3rd-5th round). The exam’s overall level is a bit higher than that of the 1st round. Permitted aids are: Formula Collection Chemistry A (both 1st and 2nd edition), DATA book physics & chemistry (Danish: Formelsamling Kemi A (1. and 2. edition), DATAbog fysik & kemi) and a calculator or PC/MAC with CAS tool (Maple).
The practical exam consists of 2 parts. The participant has 2 hours for each part with a break in between without any aids. Part 1 is a titration exercise where the participant can for example determine the concentration of a solution. Common titration types:
Redox titration Permanganate or Iodometric titration.
EDTA titration (complexometric) with EDTA and an indicator like Eriochrome-T
Part 2 of the practical exam consists of a qualitative inorganic analysis sometimes referred to as “Ionjagt” eller “dryp-dryp kemi” in Danish (Eng. “ion hunting” or “drip-drip chemistry”), during which you are given 8 solutions and a list with 8 inorganic compounds they contain. The task is to determine which solution contains which inorganic compound by systematically mixing them and noting the result.
The total 2nd round score is 60% of the theoretical and 40% of the practical score.
The ceremony on Sunday will reveal which approx. 16 out of the approx. 60 participants will advance to the final rounds. During the event, diplomas, books, etc. are given out.
The Final Rounds (3rd - 5th round)
The following 3 rounds constitute the final rounds, and the results, with and that of the 2nd round, decide who will be on the year's national team. The chemistry level in the finals corresponds to the 1st and 2nd year of Denmark’s bachelor in chemistry. Each of the rounds is held over an entire weekend, spaced approximately 1 month apart. The round structures are very similar to that of the 2nd round.
The order of the rounds may change, but in the last few years they have been in the following order:
Round 3 (Aarhus University, AU round)
Round 3, also known as the AU round, is held around the end of February or beginning of March. It is all held over a weekend from Friday to Sunday at Aarhus University (AU) with accommodation. What is unique about this round is that it is themed and the participants are mainly tested in organic chemistry and a new topic. This topic is based on the IChO preparation problems published every year around the beginning of February. The last few years (2015-) this topic has been crystallography, but polymer chemistry has also emerged once.
In preparation, participants will be given a syllabi of organic chemistry based on the book McMurry’s Organic Chemistry by John E. McMurry, and related assignments that will set the exam's framework. For the new topic, participants will receive lectures followed by a problem-solving session. The typical program: Friday is spent on reviewing organic chemistry homework followed by a lecture with a problem-solving session on the new topic. Saturday consists of the practical exam, a 4-hour exam consisting of organic synthesis with TLC analysis and a titration determination of an unknown acid. After the exam, the participants have a guest lecture and tour of AU. Sunday is then a 4 hour theoretical test without any aids. Only paper, a pencil/pen and a calculator (without graph function) are allowed. The test consists of 20 multiple choice questions and contains topics from the previous rounds and 5 problems with sub-parts (example of distribution: 3 organic, 1 spectroscopy and 1 crystallography/polymer chemistry). The participant’s total score for the 3rd round is divided between 60% of the theoretical and 40% of the practical score.
Recommended preparation: Ch. 1-8. Fundamentals of Organic Chemistry, International Edition by John Mcmurry, 7th edition
Round 4 (Technical University of Denmark, DTU round)
Round 4, also known as the DTU round, is held around the beginning of April. It is held over a weekend from Friday to Sunday at the Technical University of Denmark (DTU) with accommodation. This round is also themed, and the participants will mainly be tested on physical and coordination chemistry.
In preparation, the participants receive a booklet with assignments and short theory sections that set the exam framework. A typical program: Friday consists of lectures and problem-solving sessions in physical and coordination chemistry. Saturday is spent on a 4 hour practical exam consisting of inorganic synthesis (synthesis of a complex) and titration analysis on the synthesis product. Sunday wraps up round 4 with a 4-hour theoretical exam without any aids. Only paper, pencil/pen and calculator (without graph function) are allowed. The test consists of 10 multiple choice questions containing topics from the previous rounds and 8 problems with sub-parts. The total score for the 4th round is divided between 60% of the theoretical and 40% of the practical score.
Recommended preparation: The booklet for Chemistry Olympiad Round 4 (given by the Danish Chemistry Olympiad Committee)
Round 5 (University of Copenhagen, KU round)
Round 5 is the lastround of the Danish Chemistry Olympiad.
The typical program: Friday, reviewing inorganic, organic, and expanded chemical equilibrium theory (mass and charge balances) followed by problem-solving sessions. Saturday, lecture in inorganic chemistry followed by problem-solving sessions. After inorganic chemistry, the participants will take the practical exam. Here the participants get approx. 4 hours to do a titration and a qualitative inorganic analysis just like in the 2nd round. Sunday is spent on the theoretical test of 5 hours without any aids. Only paper, a pencil/pen and a calculator (without graph function) are allowed. The exam consists of approx. 10 multiple choice questions and approx. 10 problems with sub-parts based on all topics. The participant’s total score for the 5th round is 60% of the theoretical and 40% of the practical score.
Recommended preparation: Ch. 9-12. Fundamentals of Organic Chemistry, International Edition by John Mcmurry, 7th edition and Basic Analytical Chemistry and the Chemistry of The Elements in Aqueous Solution by Peter Andersen and Ole Mønsted 2010 edition (Danish: Grundlæggende Analytisk Kemi og Grundstoffernes Kemi i Vandig Opløsning )
The Danish Chemistry Olympiad ends Monday with a ceremony where the national team consisting of the 4 highest scoring participants of the approx. 16 will be announced. The final score is calculated as follows:
P_{tot} = 0.2 \cdot P_{\%}(R2) + 0.2 \cdot P_{\%}(R3) + 0.3 \cdot P_{\%}(R4) + 0.3 \cdot P_{\%}(R5)
P_{\%}
is one's score in percent and
R\#
represents teh round with its associated number. This ceremony is in conjunction with all the other science olympiads (mathematics, physics, biology, computer science, and geography).
After this, national team participants receive further information about training for the Nordic Chemistry Olympiad (NChO) and the International Chemistry Olympiad (IChO).
The Danish Chemistry Olympiad website: https://www.kemiolympiade.dk/
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Deep Neural Networks Final Predict Model
In this tutorial, we will use the functions we had implemented in the previous parts to build a deep network and apply it to cat vs dog classification
Deep Neural Network for Image Classification:
This tutorial will use the functions we had implemented in the previous parts to build a Deep Neural Network and apply it to cat vs. dog classification. Hopefully, we will see an improvement in accuracy relative to our previous logistic regression implementation. After this part, we will build and apply a Deep Neural Network to supervised learning using only the NumPy library.
Let's first import all the packages that we will need during this part. We will use the same "Cat vs. Dog" dataset as in the Logistic Regression as a Neural Network tutorial. The model we had built had 60% test accuracy on classifying cats vs. dogs images. Hopefully, our new model will perform better!
I'll import code we already wrote from the "Logistic Regression" tutorial series:
L-layer deep neural network:
It is hard to represent a deep neural network with a figure. However, here is a simplified network representation:
The model can be summarized as: [LINEAR -> RELU] × (L-1) -> LINEAR -> SIGMOID. Detailed Architecture of above figure:
The input is a (64,64,3) image that is flattened to a vector of size (12288,1);
The corresponding vector:
{\left[{x}_{0}, {x}_{1}, ..., {x}_{12287}\right]}^{T}
is then multiplied by the weight matrix W[1], and then you add the intercept b[1]. The result is called the linear unit;
Next, you take the ReLu of the linear unit. This process could be repeated several times for each (W[l], b[l]) depending on the model architecture;
As usual, we will follow the Deep Learning methodology to build the model:
1. Initialize parameters / Define hyperparameters.
Forward propagation;
Compute cost function;
Backward propagation;
Update parameters (using parameters and grads from backprop).
3. Use trained parameters to predict labels.
L-layer Neural Network:
We'll use the helper functions we have implemented previously to build an L-layer neural network with the following structure: [LINEAR -> RELU]×(L-1) -> LINEAR -> SIGMOID.
X - data, NumPy array of shape (number of examples, ROWS * COLS * CHANNELS );
Y - true "label" vector (containing 0 if cat, 1 if dog), of shape (1, number of examples);
layers_dims - a list containing the input size and each layer size, of length (number of layers + 1);
print_cost - if True, it prints the cost every 100 steps.
Parameters -- parameters learned by the model. They can then be used to predict.
# keep track of cost
Before training our model, we need to write a predict function that will be used to predict the results of an L-layer neural network:
X - a data set of examples you would like to label.
Parameters - parameters of the trained model.
p - predictions for the given dataset X.
# number of layers in the neural network
n = len(parameters) // 2
probas, caches = L_model_forward(X, parameters)
We are now ready to train our deep neural networks model. So we can run the cell below to train our model. The cost should decrease on every iteration. In my case, with 6000 images, it may take more than a day on my CPU to run 10000 iterations. Here is the code:
layers_dims = [12288, 800, 10, 1] # 4-layer model
parameters = L_layer_model(train_set_x, train_set_y, layers_dims, learning_rate = 0.1, num_iterations = 10000, print_cost = True)
print("train accuracy: {} %".format(100 - np.mean(np.abs(predict(train_set_x, parameters) - train_set_y)) * 100))
print("test accuracy: {} %".format(100 - np.mean(np.abs(predict(test_set_x, parameters) - test_set_y)) * 100))
My updated Cat vs. Dog training results:
And here is the training cost plot:
From our test accuracy, you can see that it's almost not better in classifying cats versus dogs. It's because simple neural networks can't do it very well. We need to use Convolutional Neural Networks to get better results (using them in future tutorials).
We will test our deep neural network code with the below code. So there are two functions: to create data, which we will use to train our model. Another function will be used to plot decision boundary in our plot, which was predicted by our neural network; not getting deeper into these functions, here they are:
def load_dataset(DataNoise = 0.05, Visualize = False):
train_X, train_Y = sklearn.datasets.make_circles(n_samples=300, noise=DataNoise)
test_X, test_Y = sklearn.datasets.make_circles(n_samples=100, noise=DataNoise)
train_X = train_X.T
train_Y = train_Y.reshape((1, train_Y.shape[0]))
test_X = test_X.T
test_Y = test_Y.reshape((1, test_Y.shape[0]))
if Visualize == True:
plt.scatter(train_X[0, :], train_X[1, :], c=train_Y[0], s=40, cmap=plt.cm.Spectral)
plt.scatter(X[0, :], X[1, :], c=y[0], cmap=plt.cm.Spectral)
Let's visualize our training data with the following line:
train_X, train_Y, test_X, test_Y = load_dataset(DataNoise = 0.15, Visualize = True)
Let's plot our training data:
At first, we'll try to use a neural network with one hidden layer, where input is 2, 4 hidden layers and output equal to 1:
layers_dims = [2, 4, 1]
parameters = L_layer_model(train_X, train_Y, layers_dims, learning_rate = 0.2, num_iterations = 15000, print_cost = True)
print("train accuracy: {} %".format(100 - np.mean(np.abs(predict(train_X, parameters) - train_Y)) * 100))
plot_decision_boundary(lambda x: predict(x.T, parameters), train_X, train_Y)
With 15000 iterations and a learning rate of 0.2, we received an accuracy of 61%:
Cost after iteration 14600: 0.654744
train accuracy: 61.333333333333336 %
And here is the classification plot:
It doesn't look nice so let's try the same 15000 iterations and learning rate of 0.2 with 10 hidden layers:
Now we received 73% accuracy. That's nice; let's plot our result:
And now, let's test our hard work and try a deeper network with layers_dims = [2, 128, 4, 1].
Now we received 74% accuracy. That's nice, but as you can see, it's getting harder to get better results. Let's plot our result:
Congrats! It seems that our Deep Neural Network has better performance (74%) than our 2-layer neural network (73%) on the same data-set. This is quite a good performance for this task. Nice job!
This was quite a long tutorial. Now when we know how to build deep neural networks, we can learn how to optimize them. We will learn how to obtain even higher accuracy in the next tutorials by systematically searching for better hyperparameters (learning_rate, layers_dims, num_iterations, and others). See you in the next tutorial.
Deep Neural Networks step by step initialize
Deep Neural Networks forward propagation
Deep Neural Networks backward propagation
Deep Neural Networks Backward module
Deep Neural Networks Final Model parameters
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Cell response distribution to autoinducer induction in the stochastic model. Cell response probability after 10 hours (top: A, B) and 100 hours (middle: C, D) of induction at different autoinducer concentrations for the lux01 (left: A, C) and lux02 (right: B, D) operons in the stochastic model. The distribution reveals the coexistence of two subpopulations with low and high GFP expression when the cells are induced at intermediate autoinducer concentrations. The region of bistability (precision) is defined by the range of
{c}_{A}^{\ast }
for which the response is bimodal according to the following criterion: the lower/upper limit of the bistable region (orange lines) is defined by the value of
{c}_{A}^{\ast }
for which 90% of the cells are in the low/high state. The black line stands for the concentration of GFP (normalized) as a function of
{c}_{A}^{\ast }
in the deterministic model at the steady state. After 10 hours of induction (top: A, B) most cells are still in a transient state if
{c}_{A}^{\ast }<70\text{nM}
. After 100 hours of induction (middle: C, D), the bimodality region shrinks and the precision increases. The population average curves of the induction and dilution experiments in the stochastic model (bottom: E, F, dashed lines) show that the intrinsic noise allows cells to jump to the high state inside the deterministic bistable region. On the other hand, the transition from high to low follows the deterministic path thus indicating that the switching rate in this case is close to zero.
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Topic: Rocketry
You are looking at all articles with the topic "Rocketry". We found 10 matches.
United States Aviation Spaceflight Aviation/aircraft project Rocketry
The Lockheed Martin X-33 was an uncrewed, sub-scale technology demonstrator suborbital spaceplane developed in the 1990s under the U.S. government–funded Space Launch Initiative program. The X-33 was a technology demonstrator for the VentureStar orbital spaceplane, which was planned to be a next-generation, commercially operated reusable launch vehicle. The X-33 would flight-test a range of technologies that NASA believed it needed for single-stage-to-orbit reusable launch vehicles (SSTO RLVs), such as metallic thermal protection systems, composite cryogenic fuel tanks for liquid hydrogen, the aerospike engine, autonomous (uncrewed) flight control, rapid flight turn-around times through streamlined operations, and its lifting body aerodynamics.
Failures of its 21-meter wingspan and multi-lobed, composite-material fuel tank during pressure testing ultimately led to the withdrawal of federal support for the program in early 2001. Lockheed Martin has conducted unrelated testing, and has had a single success after a string of failures as recently as 2009 using a 2-meter scale model.
A rolleron is a type of aileron used for rockets, placed at the trailing end of each fin, and used for passive stabilization against rotation. Inherent to the rolleron is a metal wheel with notches along the circumference. On one side, the notches protrude into the airflow. During flight, this will spin the wheels up to a substantial speed. The wheels then act as gyroscopes. Any tendency of the rocket to rotate around its major axis will be counteracted by the rollerons: the gyroscopic precession acts to move the rolleron in the opposite direction to the rotation.
Rollerons were first used in the AIM-9 Sidewinder missile.
"Rolleron" | 2013-12-19 | 11 Upvotes 2 Comments
Companies Aviation Spaceflight Aviation/aircraft project Rocketry
Rotary Rocket Company was a rocketry company that developed the Roton concept in the late 1990s as a fully reusable single-stage-to-orbit (SSTO) crewed spacecraft. The design was initially conceived by Bevin McKinney, who shared it with Gary Hudson. In 1996, Rotary Rocket Company was formed to commercialize the concept. The Roton was intended to reduce costs of launching payloads into low earth orbit by a factor of ten.
The company gathered considerable venture capital from angel investors and opened a factory headquartered in a 45,000-square-foot (4,200 m2) facility at Mojave Air and Space Port in Mojave, California. The fuselage for their vehicles was made by Scaled Composites, at the same airport, while the company developed the novel engine design and helicopter-like landing system. A full-scale test vehicle made three hover flights in 1999, but the company exhausted its funds and closed its doors in early 2001.
"Rotary Rocket" | 2015-04-15 | 30 Upvotes 13 Comments
United States Aviation Spaceflight Rocketry
The Space Transportation System (STS), also known internally to NASA as the Integrated Program Plan (IPP), was a proposed system of reusable manned space vehicles envisioned in 1969 to support extended operations beyond the Apollo program. (NASA appropriated the name for its Space Shuttle Program, the only component of the proposal to survive Congressional funding approval). The purpose of the system was twofold: to reduce the cost of spaceflight by replacing the current method of launching capsules on expendable rockets with reusable spacecraft; and to support ambitious follow-on programs including permanent orbiting space stations around the Earth and Moon, and a human landing mission to Mars.
In February 1969, President Richard Nixon appointed a Space Task Group headed by Vice President Spiro Agnew to recommend human space projects beyond Apollo. The group responded in September with the outline of the STS, and three different program levels of effort culminating with a human Mars landing by 1983 at the earliest, and by the end of the twentieth century at the latest. The system's major components consisted of:
A permanent space station module designed for 6 to 12 occupants, in a 270-nautical-mile (500 km) low Earth orbit, and as a permanent lunar orbit station. Modules could be combined in Earth orbit to create a 50 to 100 person permanent station.
A chemically fueled Earth-to-station shuttle.
A chemically fueled space tug to move crew and equipment between Earth orbits as high as geosynchronous orbit, which could be adapted as a lunar orbit-to-surface shuttle.
A nuclear-powered ferry using the NERVA engine, to move crew, spacecraft and supplies between low Earth orbit and lunar orbit, geosynchronous orbit, or to other planets in the solar system.
The tug and ferry vehicles would be of a modular design, allowing them to be clustered and/or staged for large payloads or interplanetary missions. The system would be supported by permanent Earth and lunar orbital propellant depots. The Saturn V might still have been used as a heavy lift launch vehicle for the nuclear ferry and space station modules. A special "Mars Excursion Module" would be the only remaining vehicle necessary for a human Mars landing.
As Apollo accomplished its objective of landing the first men on the Moon, political support for further manned space activities began to wane, which was reflected in unwillingness of the Congress to provide funding for most of these extended activities. Based on this, Nixon rejected all parts of the program except the Space Shuttle, which inherited the STS name. As funded, the Shuttle was greatly scaled back from its planned degree of reusability, and deferred in time. The Shuttle first flew in 1981, and was retired in 2011.
A second part of the system, Space Station Freedom, was approved in the early 1980s and announced in 1984 by president Ronald Reagan. However, this also became politically unviable by 1993, and was replaced with the International Space Station (ISS), with substantial contribution by Russia. The ISS was completed in 2010.
"Space Transportation System" | 2018-10-21 | 87 Upvotes 60 Comments
Spaceflight Physics Rocketry
Direct Fusion Drive (DFD) is a conceptual low radioactivity, nuclear-fusion rocket engine designed to produce both thrust and electric power for interplanetary spacecraft. The concept is based on the Princeton field-reversed configuration reactor invented in 2002 by Samuel A. Cohen, and is being modeled and experimentally tested at Princeton Plasma Physics Laboratory, a US Department of Energy facility, and modeled and evaluated by Princeton Satellite Systems. As of 2018, the concept has moved on to Phase II to further advance the design.
"Direct Fusion Drive" | 2020-09-17 | 23 Upvotes 2 Comments
{\displaystyle \Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}=I_{\text{sp}}g_{0}\ln {\frac {m_{0}}{m_{f}}}}
{\displaystyle \Delta v\ }
{\displaystyle m_{0}}
is the initial total mass, including propellant, also known as wet mass.
{\displaystyle m_{f}}
is the final total mass without propellant, also known as dry mass.
{\displaystyle v_{\text{e}}=I_{\text{sp}}g_{0}}
{\displaystyle I_{\text{sp}}}
{\displaystyle g_{0}}
{\displaystyle \ln }
"Tsiolkovsky Rocket Equation" | 2020-09-21 | 17 Upvotes 1 Comments
Aviation History Spaceflight Military history Military history/Military science, technology, and theory Spaceflight/Timeline of spaceflight working group Physics Lists Military history/World War II Military history/Cold War Rocketry Military history/European military history Military history/British military history
Spaceflight as a practical endeavor began during World War II with the development of operational liquid-fueled rockets. Beginning life as a weapon, the V-2 was pressed into peaceful service after the war at the United States' White Sands Missile Range as well as the Soviet Union's Kapustin Yar. This led to a flourishing of missile designs setting the stage for the exploration of space. The small American WAC Corporal rocket was evolved into the Aerobee, a much more powerful sounding rocket. Exploration of space began in earnest in 1947 with the flight of the first Aerobee, 46 of which had flown by the end of 1950. These and other rockets, both Soviet and American, returned the first direct data on air density, temperature, charged particles and magnetic fields in the Earth's upper atmosphere.
By 1948, the United States Navy had evolved the V-2 design into the Viking capable of more than 100 miles (160 km) in altitude. The first Viking to accomplish this feat, number four, did so 10 May 1950. The Soviet Union developed a virtual copy of the V-2 called the R-1, which first flew in 1948. Its longer-ranged successor, the R-2, entered military service in 1950. This event marked the entry of both superpowers into the post-V-2 rocketry era.
"Spaceflight Before 1951" | 2021-09-06 | 63 Upvotes 31 Comments
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A Stan case study, sort of: The probability my son will be stung by a bumblebee - Publishable Stuff
The Stan project for statistical computation has a great collection of curated case studies which anybody can contribute to, maybe even me, I was thinking. But I don’t have time to worry about that right now because I’m on vacation, being on the yearly visit to my old family home in the north of Sweden.
What I do worry about is that my son will be stung by a bumblebee. His name is Torsten, he’s almost two years old, and he loves running around barefoot on the big lawn. Which has its fair share of bumblebees. Maybe I should put shoes on him so he wont step on one, but what are the chances, really.
Well, what are the chances? I guess if I only had
Data on the bumblebee density of the lawn.
Data on the size of Torsten’s feet and how many steps he takes when running around.
A reasonable Bayesian model, maybe implemented in Stan.
I could figure that out. “How hard can it be?”, I thought. And so I made an attempt.
To get some data on bumblebee density I marked out a 1 m² square on a representative part of the lawn. During the course of the day, now and then, I counted up how many bumblebees sat in the square.
Most of the time I saw zero bumblebees, but 1 m² is not that large of an area. Let’s put the data into R:
bumblebees <- c(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
During the same day I kept an eye on Torsten, and sometimes when he was engaged in active play I started a 60 s. timer and counted how many steps he took while running around. Here are the counts:
Finally I needed to know the area of Torsten’s feet. At a moment when he was not running around I put his foot on a piece of graph paper and traced its edge, which made him giggle. I then calculated how many of the 0.5 cm² squares were fully covered and how many were partially covered.
To estimate of the area of his foot I took the average of the number of full squares and the number partial squares, and then converted to m²:
foot_m2
Turns out my son’s feet are about 0.0053 m² each.
The idea I had here was that if you know the average number of steps Torsten takes per minute while playing, and if you know the average number of bumblebees per m², then you could calculate the average area covered by Torsten’s tiny feet while running around, and then finally you could calculate the probability of a bumblebee being in that area. This would give you a probability for how likely Torsten is to be stung by a Bumblebee. However, while we have data on all of the above, we don’t know any of the above for certain.
So, to capture some of this uncertainty I thought I would whip together a small Bayesian model and, even though it’s a bit overkill in this case, why not fit it using Stan? (If you are unfamiliar with Bayesian data analysis and Stan I’ve made a short video tutorial you can find here.)
Let’s start our Stan program by declaring what data we have:
Now we need to decide on a model for the number of bumblebees in a m² and a model for the number of toddler_steps in 60 s. For the bees I’m going with a simple Poisson model, that is, the model assumes there is an average number of bees per m² (
\mu_\text{bees}
) and that this is all there is to know about bees.
For the number of steps I’m going to step it up a notch with a negative binomial model. The negative binomial distribution can be viewed in a number of ways, but one particularly useful way is as an extension of the Poisson distribution where the mean number of steps for each outcome is not fixed, but instead is varying, where this is controlled by the precision parameter
\phi_\text{steps}
. The larger
\phi_\text{steps}
is the less the mean number of steps for each outcome is going to vary around the overall mean
\mu_\text{steps}
\phi_\text{steps} \rightarrow \infty
then the negative binomial becomes a Poisson distribution. The point with using the negative binomial is to capture that Torsten’s activity level when playing on the lawn is definitely not constant. Sometimes he can run full speed for minutes, but sometimes he spends a long time being fascinated by the same little piece of rubbish and then he’s not taking many steps.
So we almost have a Bayesian model for our data, we just need to put priors on the parameters
\mu_\text{bees}
\mu_\text{steps}
\phi_\text{steps}
. All three parameters are positive and a quick n’ dirty prior when you have positive parameters is a half-Cauchy distribution centered at zero: It’s just like half a normal distribution centered at zero, but with a much fatter tail. It has one free parameter, the scale, which also happens to be the median of the half-Cauchy. Set the scale/median to a good guess for the parameter in question and you are ready to go! If your guess is good then this prior will give the parameter the slightest nudge in the right direction, if your guess is bad then, as long as you have enough data, the fat tail of the half-Cauchy distribution will help you save face.
My guess for
\mu_\text{bees}
is (4.0 + 4.0) / (10 × 10) = 0.08. I base this on that while crossing the lawn I usually scan an area of about 10×10 m², I then usually see around four bees, and I assume I’m missing half of the bees. So eight bees on a 100 m² lawn gives 0.08 bees per m². My guess for the number of steps per minute is going to be 60. For the precision parameter
\phi_\text{steps}
I have no clue, but it shouldn’t be too large nor to small, so my guess is going to be 1.0. Here are the priors:
With the priors specified we now have all the required parts and here is the resulting Bayesian model:
And here is the Stan code implementing the model above:
# Since we have contrained the parameters to be positive we get implicit
# half-cauchy distributions even if we declare them to be 'full'-cauchy.
mu_bees ~ cauchy(0, (4.0 + 4.0) / (10.0 * 10.0) );
The final step is to predict how many bumblebees Torsten will step on during, say, one hour of active play. We do this in the generated quantities code block in Stan. The code below will step through 60 minutes (a.k.a. one hour) and for each minute: (1) Sample pred_steps from the negative binomial, (2) calculate the area (m²) covered by these steps, and (3) sample the number of bees in this area from the Poisson giving a predicted stings_by_minute. Finally we sum these 60 minutes worth of strings into stings_by_hour.
pred_steps = neg_binomial_2_rng(mu_steps, precision_steps);
We have data, we have a model, and now all we need to do is fit it. After we’ve put the whole Stan model into bee_model_code as a string we do it like this using the rstan interface:
# First we put all the data into a list
# Then we fit the model which description is in bee_model_code .
With the model fitted, let’s take a look at the posterior probability distribution of our parameters.
## mean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat
## mu_bees 0.12 0.00 0.05 0.04 0.08 0.11 0.15 0.23 3551 1
## mu_steps 50.88 0.22 11.65 32.56 43.09 49.34 57.13 78.00 2801 1
## precision_steps 1.80 0.01 0.68 0.78 1.32 1.70 2.17 3.41 2706 1
This seems reasonable, but note that the distributions are pretty wide, which means there is a lot of uncertainty! For example, looking at mu_bees it’s credible that there is everything from 0.04 bees/m² to 0.2 bees/m².
Anyway, let’s take a look at what we are really interested in, the predictive distribution over how many stings per hour Torsten will suffer during active play:
## stings_by_hour
Ok, it seems like it is most probable that Torsten will receive one sting per hour, but we should not be surprised if it’s two or even three stings. I’d better put some shoes on him! The problem is that after a couple of days full of active barefoot play, my son Torsten’s feet look like this:
As you can see his feet are not swollen from all the bee stings they should receive according to the model. Actually, even after a week, he has not gotten a single bee sting! Which is good, I suppose, in a way, but, well, it means that my model is likely pretty crap. How can this be? Well,
I should maybe have gotten more and better data. Perhaps the square I monitored for bumblebees didn’t yield data that was really representative of the bumblebee density in general.
The assumption that bumblebees always sting when stepped upon might be wrong. Or maybe Torsten is so smart so that he actively avoids them…
Maybe the model was too simplistic. I really should have made a hierarchical model incorporating data from multiple squares and several days of running around. To factor in the effect of the weather, the flower density, and the diet of Torsent also couldn’t hurt.
I guess I could spend some time improving the model. And I guess there is a lesson to be learned here about that it is hard to predict the predictive performance of a model. But all that has to wait. I am on vacation after all.
The full data and code behind this blog post can be found here.
« Video Introduction to Bayesian Data Analysis, Part 3: How to do Bayes? My introductory course on Bayesian statistics »
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Home : Support : Online Help : Mathematics : Geometry : 2-D Euclidean : Triangle Geometry : IsRightTriangle
IsRightTriangle(ABC, cond )
This routine tests if the given triangle ABC is a right triangle. It returns true if ABC is a right triangle; false if it is not; and FAIL if it is unable to reach a conclusion.
\mathrm{expr}=0
&\mathrm{or}\left(\mathrm{expr_1}=0,\mathrm{expr_2}=0,...,\mathrm{expr_n}\right)
The command with(geometry,IsRightTriangle) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{geometry}\right):
\mathrm{triangle}\left(T,[2,2,3]\right)
\textcolor[rgb]{0,0,1}{T}
\mathrm{IsRightTriangle}\left(T\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{triangle}\left(\mathrm{ABC},[\mathrm{point}\left(A,0,0\right),\mathrm{point}\left(B,2,0\right),\mathrm{point}\left(C,0,2\right)]\right)
\textcolor[rgb]{0,0,1}{\mathrm{ABC}}
\mathrm{IsRightTriangle}\left(\mathrm{ABC}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{point}\left(B,1,b\right):
\mathrm{IsRightTriangle}\left(\mathrm{ABC},'\mathrm{cond}'\right)
IsRightTriangle: "hint: one of the following conditions must be satisfied: {-4*b = 0, 4*b-8 = 0, -2*b^2+4*b-2 = 0}"
\textcolor[rgb]{0,0,1}{\mathrm{FAIL}}
\mathrm{cond}
\textcolor[rgb]{0,0,1}{\mathrm{&or}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\right)
\mathrm{assume}\left(\mathrm{op}\left(1,\mathrm{cond}\right)\right)
\mathrm{IsRightTriangle}\left(\mathrm{ABC}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
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Just like with the forward propagation, we'll implement helper functions for backpropagation and calculate the gradient of the loss function with respect to the parameters
So up to this point, we have initialized our deep parameters and wrote forward propagation module. Now we will implement the cost function and backward propagation module. Same as before, we need to compute the cost because we want to check if our model is actually learning.
To compute the cross-entropy cost J, we'll be using the formula we already saw before:
-\frac{1}{m}\sum _{i=1}^{m}\left({y}^{\left(i\right)}\mathrm{log}\left({a}^{\left[L\right]\left(i\right)}\right)-\left(1-{y}^{\left(i\right)}\right)\mathrm{log}\left(1-{a}^{\left[L\right]\left(i\right)}\right)\right)
Code for our compute_cost function:
AL - probability vector corresponding to your label predictions, shape (1, number of examples).
Y - true "label" vector (for example: containing 0 if a dog, 1 if a cat), shape (1, number of examples).
cost - cross-entropy cost.
cost = -1./m * np.sum(Y*np.log(AL)+(1-Y)*np.log(1-AL))
# To make sure our cost's shape is what we expect (e.g. this turns [[23]] into 23).
This is only one line of code, so we won't test it, as it was tested in my previous tutorials.
Backward propagation function:
Just like with the forward propagation, we will implement helper functions for backpropagation. We know that propagation is used to calculate the gradient of the loss function for the parameters. We need to write Forward and Backward propagation for LINEAR->RELU->LINEAR->SIGMOID model. This will look like this:
Similar to the forward propagation, we are going to build the backward propagation in three steps:
LINEAR backward;
LINEAR -> ACTIVATION backward where ACTIVATION computes the derivative of either the ReLU or sigmoid activation;
Linear backward:
For layer l, the linear part is:
{Z}^{\left[l\right]}={W}^{\left[l\right]}{A}^{\left[l-1\right]}+{b}^{\left[l\right]}
I suppose we have already calculated the derivative:
d{Z}^{\left[l\right]}=\frac{\partial L}{\partial {Z}^{\left[l\right]}} \text{and we want to get} \left(d{W}^{\left[l\right]}, d{A}^{\left[l-1\right]}, d{b}^{\left[l\right]}\right)
Three outputs (dW[l], db[l], dA[l]) will be computed using the input dZ[l]. Here are the formulas we need from our two-layer neural networks tutorial series:
d{W}^{\left[l\right]}=\frac{\partial L}{\partial {b}^{\left[l\right]}}=\frac{1}{m}d{Z}^{\left[l\right]}{A}^{\left[l-1\right]T}\phantom{\rule{0ex}{0ex}}d{b}^{\left[l\right]}=\frac{\partial L}{\partial {b}^{\left[l\right]}}=\frac{1}{m}\sum _{i=1}^{m}d{Z}^{\left[l\right]\left(i\right)}\phantom{\rule{0ex}{0ex}}d{A}^{\left[l-1\right]}=\frac{\partial L}{\partial {A}^{\left[l-1\right]}}={W}^{\left[l\right]T}d{Z}^{\left[l\right]}
Code for our linear_backward function:
dZ - Gradient of the cost concerning the linear output (of current layer l);
cache - tuple of values (A_prev, W, b) coming from the forward propagation in the current layer.
dA_prev - Gradient of the cost concerning the activation (of the previous layer l-1), the same shape as A_prev;
dW - Gradient of the cost to W (current layer l), the same shape as W;
db - Gradient of the cost to b (current layer l), the same shape as b.
A_prev, W, b = cache
dW = 1./m * np.dot(dZ, A_prev.T)
db = 1./m * np.sum(dZ, axis=1, keepdims=True)
Linear-Activation backward function:
Next, we will create a function that merges the two helper functions: linear_backward and the backward step for the activation linear_activation_backward.
To implement linear_activation_backward, we will write two backward functions:
sigmoid_backward: the backward propagation for SIGMOID unit:
def sigmoid_backward(dA, cache):
The backward propagation for a single SIGMOID unit.
dA - post-activation gradient, of any shape
cache - 'Z' where we store for computing backward propagation efficiently
dZ - Gradient of the cost with respect to Z
Z = cache
dZ = dA * s * (1-s)
relu_backward: the backward propagation for RELU unit:
The backward propagation for a single RELU unit.
# just converting dz to a correct object.
# When z <= 0, we should set dz to 0 as well.
If g(.) is the activation function, sigmoid_backward and relu_backward compute:
d{Z}^{\left[l\right]}=d{A}^{\left[l\right]}·g\text{'}\left({Z}^{\left[l\right]}\right)
Code for our linear_activation_backward function:
dA - post-activation gradient for current layer l;
cache - tuple of values (linear_cache, activation_cache) we store for computing backward propagation efficiently;
dA_prev - Gradient of the cost to the activation (of the previous layer l-1), the same shape as A_prev;
dZ = relu_backward(dA, activation_cache)
dA_prev, dW, db = linear_backward(dZ, linear_cache)
dZ = sigmoid_backward(dA, activation_cache)
dZ = np.array(dA, copy=True) # just converting dz to a correct object.
# When z <= 0, you should set dz to 0 as well.
cost = -np.sum(Y*np.log(AL)+(1-Y)*np.log(1-AL))/m
dZ = relu_backward(dA, cache[1])
dA_prev, dW, db = linear_backward(dZ, cache[0])
dZ = sigmoid_backward(dA, cache[1])
Y = np.array([[1, 1, 0, 0]])
print("cost = ", compute_cost(AL, Y))
This tutorial took a while. I will complete our backward function in the next tutorial. In the next tutorial, we'll write our backward final module and function to update our parameters. After that, we'll finally train our model.
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This article is about the estimation technique. For Fermi's question about extraterrestrial life, see Fermi paradox.
Find sources: "Fermi problem" – news · newspapers · books · scholar · JSTOR (July 2015) (Learn how and when to remove this template message)
In physics or engineering education, a Fermi problem, Fermi quiz, Fermi question, Fermi estimate, order-of-magnitude problem, order-of-magnitude estimate, or order estimation is an estimation problem designed to teach dimensional analysis or approximation of extreme scientific calculations, and such a problem is usually a back-of-the-envelope calculation. The estimation technique is named after physicist Enrico Fermi as he was known for his ability to make good approximate calculations with little or no actual data. Fermi problems typically involve making justified guesses about quantities and their variance or lower and upper bounds. In some cases, order-of-magnitude estimates can also be derived using dimensional analysis.
3 Advantages and scope
An example is Enrico Fermi's estimate of the strength of the atomic bomb that detonated at the Trinity test, based on the distance traveled by pieces of paper he dropped from his hand during the blast. Fermi's estimate of 10 kilotons of TNT was well within an order of magnitude of the now-accepted value of 21 kilotons.[1][2]
Examples of Fermi Questions are often extreme in nature, and cannot usually be solved using common mathematical or scientific information.
Example questions given by the official Fermi Competition:
"If the mass of one teaspoon of water could be converted entirely into energy in the form of heat, what volume of water, initially at room temperature, could it bring to a boil? (litres)."
"How much does the Thames River heat up in going over the Fanshawe Dam? (Celsius degrees)."
"What is the mass of all the automobiles scrapped in North America this month? (kilograms)." [3][4]
Possibly the most famous Fermi Question is the Drake equation, which seeks to estimate the number of intelligent civilizations in the galaxy. The basic question of why, if there were a significant number of such civilizations, ours has never encountered any others is called the Fermi paradox.[5]
Advantages and scopeEdit
Scientists often look for Fermi estimates of the answer to a problem before turning to more sophisticated methods to calculate a precise answer. This provides a useful check on the results. While the estimate is almost certainly incorrect, it is also a simple calculation that allows for easy error checking, and to find faulty assumptions if the figure produced is far beyond what we might reasonably expect. By contrast, precise calculations can be extremely complex but with the expectation that the answer they produce is correct. The far larger number of factors and operations involved can obscure a very significant error, either in mathematical process or in the assumptions the equation is based on, but the result may still be assumed to be right because it has been derived from a precise formula that is expected to yield good results. Without a reasonable frame of reference to work from it is seldom clear if a result is acceptably precise or is many degrees of magnitude (tens or hundreds of times) too big or too small. The Fermi estimation gives a quick, simple way to obtain this frame of reference for what might reasonably be expected to be the answer.
As long as the initial assumptions in the estimate are reasonable quantities, the result obtained will give an answer within the same scale as the correct result, and if not gives a base for understanding why this is the case. For example, suppose you were asked to determine the number of piano tuners in Chicago. If your initial estimate told you there should be a hundred or so, but the precise answer tells you there are many thousands, then you know you need to find out why there is this divergence from the expected result. First looking for errors, then for factors the estimation didn't take account of – Does Chicago have a number of music schools or other places with a disproportionately high ratio of pianos to people? Whether close or very far from the observed results, the context the estimation provides gives useful information both about the process of calculation and the assumptions that have been used to look at problems.
Fermi estimates are also useful in approaching problems where the optimal choice of calculation method depends on the expected size of the answer. For instance, a Fermi estimate might indicate whether the internal stresses of a structure are low enough that it can be accurately described by linear elasticity; or if the estimate already bears significant relationship in scale relative to some other value, for example, if a structure will be over-engineered to withstand loads several times greater than the estimate.[citation needed]
Although Fermi calculations are often not accurate, as there may be many problems with their assumptions, this sort of analysis does tell us what to look for to get a better answer. For the above example, we might try to find a better estimate of the number of pianos tuned by a piano tuner in a typical day, or look up an accurate number for the population of Chicago. It also gives us a rough estimate that may be good enough for some purposes: if we want to start a store in Chicago that sells piano tuning equipment, and we calculate that we need 10,000 potential customers to stay in business, we can reasonably assume that the above estimate is far enough below 10,000 that we should consider a different business plan (and, with a little more work, we could compute a rough upper bound on the number of piano tuners by considering the most extreme reasonable values that could appear in each of our assumptions).
Fermi estimates generally work because the estimations of the individual terms are often close to correct, and overestimates and underestimates help cancel each other out. That is, if there is no consistent bias, a Fermi calculation that involves the multiplication of several estimated factors (such as the number of piano tuners in Chicago) will probably be more accurate than might be first supposed.
In detail, multiplying estimates corresponds to adding their logarithms; thus one obtains a sort of Wiener process or random walk on the logarithmic scale, which diffuses as
{\displaystyle {\sqrt {n}}}
(in number of terms n). In discrete terms, the number of overestimates minus underestimates will have a binomial distribution. In continuous terms, if one makes a Fermi estimate of n steps, with standard deviation σ units on the log scale from the actual value, then the overall estimate will have standard deviation
{\displaystyle \sigma ^{\sqrt {n}}}
, since the standard deviation of a sum scales as
{\displaystyle {\sqrt {n}}}
in the number of summands.
For instance, if one makes a 9-step Fermi estimate, at each step overestimating or underestimating the correct number by a factor of 2 (or with a standard deviation 2), then after 9 steps the standard error will have grown by a logarithmic factor of
{\displaystyle {\sqrt {9}}=3}
, so 23 = 8. Thus one will expect to be within 1⁄8 to 8 times the correct value – within an order of magnitude, and much less than the worst case of erring by a factor of 29 = 512 (about 2.71 orders of magnitude). If one has a shorter chain or estimates more accurately, the overall estimate will be correspondingly better.
^ Von Baeyer, Hans Christian (September 1988). "How Fermi Would Have Fixed It". The Sciences. 28 (5): 2–4. doi:10.1002/j.2326-1951.1988.tb03037.x. This essay was cited in von Baeyer's 1989 AAAS Kavli Science Journalism Award; see "Hans Christian von Baeyer". American Association for the Advancement of Science. Retitled, this essay was included as the first in von Baeyer's collection of essays first published in 1993, and later reprinted; see Von Baeyer, Hans Christian (2001). "The Fermi Solution". The Fermi Solution: Essays on Science. Dover Publications. pp. 3–12. ISBN 9780486417073. OCLC 775985788.
^ "Eyewitnesses to Trinity" (PDF). Nuclear Weapons Journal. Los Alamos National Laboratory. 2005. p. 45. Retrieved 18 February 2014.
^ Fermi Questions. 2012. Prof. L.B. Weinstein, Old Dominion University.
^ Fermi Questions. Richard K Curtis. 2001.
^ The Great Silence: Science and Philosophy of Fermi's Paradox By Milan M. Ćirković
The following books contain many examples of Fermi problems with solutions:
John Harte, Consider a Spherical Cow: A Course in Environmental Problem Solving University Science Books. 1988. ISBN 0-935702-58-X.
John Harte, Consider a Cylindrical Cow: More Adventures in Environmental Problem Solving University Science Books. 2001. ISBN 1-891389-17-3.
Clifford Swartz, Back-of-the-Envelope Physics Johns Hopkins University Press. 2003. ISBN 0-8018-7263-4. ISBN 978-0801872631.
Sanjoy Mahajan, Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic Problem Solving MIT Press. 2010. ISBN 026251429X. ISBN 978-0262514293.
Göran Grimvall, Quantify! A Crash Course in Smart Thinking Johns Hopkins University Press. 2010. ISBN 0-8018-9717-3. ISBN 978-0-8018-9717-7.
Lawrence Weinstein, Guesstimation 2.0: Solving Today's Problems on the Back of a Napkin Princeton University Press. 2012. ISBN 978-0-691-15080-2.
Sanjoy Mahajan, The Art of Insight in Science and Engineering MIT Press. 2014. ISBN 9780262526548.
Dmitry Budker, Alexander O. Sushkov, Physics on your feet. Berkeley Graduate Exam Questions Oxford University Press. 2015. ISBN 978-0199681655.
Rob Eastaway, Maths on the Back of an Envelope: Clever ways to (roughly) calculate anything HarperCollins. 2019. ISBN 978-0008324582.
There are a number of university-level courses devoted to estimation and the solution of Fermi problems. The materials for these courses are a good source for additional Fermi problem examples and material about solution strategies:
6.055J / 2.038J The Art of Approximation in Science and Engineering taught by Sanjoy Mahajan at the Massachusetts Institute of Technology (MIT).
Physics on the Back of an Envelope taught by Lawrence Weinstein at Old Dominion University.
Order of Magnitude Physics taught by Sterl Phinney at the California Institute of Technology.
Order of Magnitude Estimation taught by Patrick Chuang at the University of California, Santa Cruz.
Order of Magnitude Problem Solving taught by Linda Strubbe at the University of Toronto.
Order of Magnitude Physics taught by Eugene Chiang at the University of California, Berkeley.
Chapter 2: Discoveries on the Back of an Envelope from Frontiers of Science: Scientific Habits of Mind taught by David Helfand at Columbia University.
Fermi Questions: A Guide for Teachers, Students, and Event Supervisors by Lloyd Abrams.
An example of a Fermi Problem relating to total gasoline consumed by cars since the invention of cars and comparison to the output of the energy released by the sun.
"How should mathematics be taught to non-mathematicians?" by Timothy Gowers.
"What if? Paint the Earth" from the book What if? Serious Scientific Answers to Absurd Hypothetical Questions by Randall Munroe.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Fermi_problem&oldid=1077642620"
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Electric Circuit Analysis/Kirchhoff's Current Law - Wikiversity
< Electric Circuit Analysis(Redirected from Kirchhoff's Current Law)
Lesson Review 5:
What you need to remember from Kirchhoff's Voltage Law. If you ever feel lost, do not be shy to go back to the previous lesson & go through it again. You can learn by repitition.
{\displaystyle \sum _{n}v_{n}=v_{4}+v_{1}+v_{2}+v_{3}=0}
This Lesson is about Kirchhoff's Current Law. The student/User is expected to understand the following at the end of the lesson.
Define Kirchhoff's Current Law ( word-by-word ).
Kirchhoff's Current Law:
{\displaystyle \sum _{n}i_{n}=i_{1}+i_{2}+i_{3}+i_{4}=0}
Part 1: Kirchhoff's Current Law
Kirchhoff's Current Law states:
The sum of the currents entering a particular point must be zero.
We can now define the electrical point physically connecting two or more electric circuit components, as a NODE. Note that a positive current leaving a point is considered to be a negative current entering that point.
Mathematically, Kirchhoff's Current Law is given by
{\displaystyle \sum _{n}i_{n}=0}
For reference, this law is sometimes called Kirchhoff's first law, Kirchhoff's point rule, Kirchhoff's junction rule, and Kirchhoff's first rule.
Part 2:Kirchhoff's Current Law (Cont...)
{\displaystyle i_{1}+i_{2}+i_{3}+i_{4}=0}
We observe four currents "entering" the junction depicted as the bold black dot in Figure 6.1. Of course, two currents are actually exiting the junction , but for the purposes of circuit analysis it is generally less restrictive to consider what are in actuality positive currents flowing out of a junction to be negative currents flowing into that junction (mathematically the same thing). Doing so allows us to write Kirchhoff's law for this example as:
{\displaystyle \sum _{n}i_{n}=i_{1}+i_{2}+i_{3}+i_{4}=0}
It may not be clear at this point why we insist on thinking of negative currents flowing into a junction instead of positive currents flowing out. But note that Figure 6.1 provides us with more information that we generally can expect to get when analysing circuits, namely the helpful arrows indicating the direction of current flow. If we don't have such assistance, we generally should not pass judgment on the direction of current flow (i.e., placing a negative sign before our current variable) until we calculate it, lest we confuse ourselves and make mistakes.
Nevertheless in this case we have the extra information of directional arrows in Figure 6.1, so we should take advantage of it. We know that currents i2 and i3 flow into the junction and the currents i1 and i4 flow out. Thus we can write
{\displaystyle i_{1}+i_{4}=i_{2}+i_{3}}
Kirchhoff's Current Law as written is only applicable to steady-state current flow (i.e., no alternating current, no signal transmission). It can be extended to include time-dependent current flow, but that is beyond the scope of this section.
Kirchhoff's Current Law is used in a method of circuit analysis referred to as nodal analysis to be discussed in Lecture 7. A node is a section of a circuit where there is no change in voltage (where there are no components, wire is often assumed to be perfectly conductive).
Each node is used to form an equation, and the equations are then solved simultaneously, giving the voltages at each node.
{\displaystyle V_{1}=15V}
{\displaystyle V_{2}=7V}
{\displaystyle R_{1}=20\Omega }
{\displaystyle R_{2}=5\Omega }
{\displaystyle R_{3}=10\Omega }
{\displaystyle R_{3}}
using Kirchhoff's Current Law.
Figure 6.3: Voltages at nodes
This is the same example we solved in Exercise 5.
Figure 6.3 shows Voltages at Nodes a, b, c and d.
We use node a as common node ( ground if you like ). thus
{\displaystyle V_{a}=0V}
From Node b we get:
{\displaystyle V_{b}=-V_{1}=-15V}
From Node d we get:
{\displaystyle V_{d}=V_{2}=7V}
It is clear that we must solve Vc, in order to complete Voltage definitions at all nodes. Vc will be found by applying KCL at node c and solving the resulting equations as follows:
{\displaystyle i_{1}=i_{2}+i_{3}}
{\displaystyle {\frac {V_{b}-V_{c}}{R_{1}}}={\frac {V_{c}}{R_{3}}}+{\frac {V_{c}-V_{d}}{R_{2}}}}
{\displaystyle {\frac {V_{b}}{R_{1}}}-{\frac {V_{c}}{R_{1}}}-{\frac {V_{c}}{R_{3}}}-{\frac {V_{c}}{R_{2}}}+{\frac {V_{d}}{R_{2}}}=0}
We can group the like terms to get the following equation:
{\displaystyle V_{c}\left({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{3}}}\right)={\frac {V_{b}}{R_{1}}}+{\frac {V_{d}}{R_{2}}}}
Substitute values into previous equations you get:
{\displaystyle V_{c}\left({\frac {1}{20\Omega }}+{\frac {1}{5\Omega }}+{\frac {1}{10\Omega }}\right)={\frac {-15V}{20\Omega }}+{\frac {7V}{5\Omega }}}
{\displaystyle V_{c}(0.35)=0.65}
{\displaystyle V_{c}=1.857V}
{\displaystyle R_{3}}
{\displaystyle {\begin{matrix}\ I_{R3}&=&{\frac {V_{c}}{R_{3}}}\\\ \\\ &=&{\frac {1.857}{10}}\\\ \\\ &=&0.186A\end{matrix}}}
Just as we expected! Note that the current here is simplified because we were following voltage definitions and current paths in Figure 6.3.
This method becomes tedious as the complexity of the circuit is increased.
{\displaystyle V_{1}=15V}
{\displaystyle V_{2}=7V}
{\displaystyle R_{1}=20\Omega }
{\displaystyle R_{2}=5\Omega }
{\displaystyle R_{3}=10\Omega }
{\displaystyle R_{3}}
Part 8: Completion List
Retrieved from "https://en.wikiversity.org/w/index.php?title=Electric_Circuit_Analysis/Kirchhoff%27s_Current_Law&oldid=1985578"
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Egg Freezing Calculator | Live Births
How to use the egg freezing calculator?
Why do we freeze eggs in fertility treatment?
When can we freeze eggs?
How to calculate egg freezing success rate?
How many eggs should I freeze if I'm 36 years old?
The egg freezing calculator helps you estimate how many oocytes you need to freeze to ensure a good probability of future pregnancies and live births.
Our egg count calculator will explain the process of egg freezing that's used in assisted reproductive technology (ART). Read on to find out how many eggs you should freeze to increase family-building success changes, depending on your age and the number of the desired offspring.
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Our tool will show you the expected probabilities of giving birth to one, two, or three children. Adjust the number of eggs to find a probability that best suits your needs.
We may freeze eggs to preserve our fertility, postpone becoming parents, and keep the ability to build a family with our genes. As we all know, our fertility decreases with age, and this is unfortunately more prominent in women.
A fantastic thing about women is that a part of you already existed in your grandmother's body. But how is that possible? Let's break it step by step.
Every woman is born with a complete set of unfertilized eggs, called oocytes. The number of eggs decreases with age and with every ovulation, each one resulting in either a period or pregnancy.
The eggs are present in the woman's body since the first moments of her existence, meaning all the eggs in our body are exactly as old as we are. 🥚
When your grandmother was pregnant with your mother, the eggs in your mother's ovaries already existed. One of these eggs was destined to become you one day!
Hungry for more? We have plenty of pregnancy-related calculators to dig into, such as:
The VTE risk score in pregnancy calculator;
The estimated date of delivery calculator;
The due date calculator; and
The pregnancy weight gain calculator.
The best time to freeze eggs is when you're between 24 and 35 years old — this is the time when your fertility is relatively stable and at the highest peak of its possibilities.
Whether your reason is unexpected sickness, a demanding treatment, male infertility, or a simple desire to postpone becoming a parent, freezing your eggs can be a wonderful way to secure your chances of becoming a mother in the future.
The how many eggs should I freeze calculator on your left will help you estimate how many frozen eggs should be enough.
❗ Remember, no method will ever guarantee you a 100% success rate! ART, IVF, ISCI and all the available calculations can increase the possibility of taking home your own baby, but it cannot ensure it.
The ISCI and IVF success rate calculator uses the following formula for calculating egg freezing success:
P_{\geq 1} = 1 - (1 - 0.6 \cdot P_E \cdot P_B)^{n}
P_{\geq 1}
is the probability of at least one live birth;
P_E
is the probability of a euploid blastocyst;
P_B
is the probability of an oocyte maturing into a blastocyst; and
n
is the number of retrieved frozen mature eggs.
While the number of eggs retrieved and the age of a woman pose no problems, the two last variables require a bit of an explanation:
The probability of an oocyte maturing into a blastocyst (
P_B
). A blastocyst is a structure that develops a few days after ovulation and successful conception — it's an initial stage of human embryo development.
The probability of a euploid blastocyst (
P_E
). A euploid blastocyst is a blastocyst with a normal number of chromosomes — genetic material also gets old, and the number of mistakes increases along with age.
💡 All of the calculations used in our egg freezing success rate calculator were taken from the research by R.H. Goldman et al.
First, you need to decide what probability of giving birth to one, two, or three children do you find satisfying? The likelihood of a successful procedure can be predicted for a given number of frozen eggs. 🤰
1 live birth probability
No. of frozen mature eggs
Probability of at least one live birth:
Probability of at least two live births:
Probability of at least three live births:
The Air Force PT calculator allows you to estimate the number of points and performance level achieved during a physical fitness test.
The iron deficiency calculator uses the patient's weight and difference in hemoglobin levels (target Hb - actual Hb) to determine their total iron deficit expressed in miligrams (mg).
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MaximalPalindromicSubstring - Maple Help
Home : Support : Online Help : Programming : Names and Strings : StringTools Package : Combinatorics on Words : MaximalPalindromicSubstring
find a maximal palindromic substring of a string
MaximalPalindromicSubstring( s )
The MaximalPalindromicSubstring command computes a maximal palindromic substring of the string s. A string
t
is a palindrome if it is equal to itself reversed, that is,
t=\mathrm{Reverse}\left(t\right)
If s is nonempty and contains no substrings that are palindromes, the first character of s is the maximal palindromic substring. If s is empty, the empty string ("") is the maximal palindromic substring.
The maximal palindromic substring of s is indicated by returning a sequence of two non-negative integers:
The first is the index of the beginning of the palindromic substring in the string s.
The second is the length of the palindromic substring.
8
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{MaximalPalindromicSubstring}\left(""\right)
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}
\mathrm{MaximalPalindromicSubstring}\left("abcde"\right)
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
\mathrm{pos},\mathrm{len}≔\mathrm{MaximalPalindromicSubstring}\left("abcbde"\right)
\textcolor[rgb]{0,0,1}{\mathrm{pos}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{len}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}
"abcbde"[\mathrm{pos}..\mathrm{pos}+\mathrm{len}-1]
\textcolor[rgb]{0,0,1}{"bcb"}
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Cells | Free Full-Text | Acrylonitrile and Pullulan Based Nanofiber Mats as Easily Accessible Scaffolds for 3D Skin Cell Models Containing Primary Cells
PROX1, a Key Mediator of the Anti-Proliferative Effect of Rapamycin on Hepatocellular Carcinoma Cells
Post-mortem Findings of Inflammatory Cells and the Association of 4-Hydroxynonenal with Systemic Vascular and Oxidative Stress in Lethal COVID-19
Innovative Platform for the Advanced Online Monitoring of Three-Dimensional Cells and Tissue Cultures
Rimann, M.
Moeschlin, N.
Calcagni, M.
Wuertz-Kozak, K.
Dudli, S.
Distler, O.
Adlhart, C.
Acrylonitrile and Pullulan Based Nanofiber Mats as Easily Accessible Scaffolds for 3D Skin Cell Models Containing Primary Cells
Nicole Moeschlin
Institute of Chemistry and Biotechnology (ICBT), ZHAW Zurich University of Applied Sciences, 8820 Wädenswil, Switzerland
Center of Experimental Rheumatology, University Hospital Zurich and Balgrist University Hospital, University of Zurich, 8091 Zurich, Switzerland
Department of Plastic Surgery and Hand Surgery, University Hospital Zurich, University of Zurich, 8091 Zurich, Switzerland
Department of Biomedical Engineering, Rochester Institute of Technology (RIT), Rochester, NY 14623, USA
Schön Clinic Munich Harlaching, Spine Center, Academic Teaching Hospital and Spine Research Institute of the Paracelsus Medical University Salzburg (Austria), 81547 Munich, Germany
Department of Physical Medicine and Rheumatology, Balgrist University Hospital, University of Zurich, 8008 Zurich, Switzerland
Academic Editors: Friedrich Jung, Michael Raghunath and Anna Blocki
(This article belongs to the Collection Advances in Cell Culture and Tissue Engineering)
(1) Background: Three-dimensional (3D) collagen I-based skin models are commonly used in drug development and substance testing but have major drawbacks such as batch-to-batch variations and ethical concerns. Recently, synthetic nanofibrous scaffolds created by electrospinning have received increasing interest as potential alternatives due to their morphological similarities to native collagen fibrils in size and orientation. The overall objective of this proof-of-concept study was to demonstrate the suitability of two synthetic polymers in creating electrospun scaffolds for 3D skin cell models. (2) Methods: Electrospun nanofiber mats were produced with (i) poly(acrylonitrile-co-methyl acrylate) (P(AN-MA)) and (ii) a blend of pullulan (Pul), poly(vinyl alcohol) (PVA) and poly(acrylic acid) (PAA) (Pul/PVA/PAA) and characterized by scanning electron microscopy (SEM) and diffuse reflectance infrared Fourier transform (DRIFT) spectra. Primary skin fibroblasts and keratinocytes were seeded onto the nanofiber mats and analyzed for phenotypic characteristics (phalloidin staining), viability (Presto Blue HS assay), proliferation (Ki-67 staining), distribution (H/E staining), responsiveness to biological stimuli (qPCR), and formation of skin-like structures (H/E staining). (3) Results: P(AN-MA) mats were more loosely packed than the Pul/PVA/PAA mats, concomitant with larger fiber diameter (340 nm ± 120 nm vs. 250 nm ± 120 nm, p < 0.0001). After sterilization and exposure to cell culture media for 28 days, P(AN-MA) mats showed significant adsorption of fetal calf serum (FCS) from the media into the fibers (DRIFT spectra) and increased fiber diameter (590 nm ± 290 nm, p < 0.0001). Skin fibroblasts were viable over time on both nanofiber mats, but suitable cell infiltration only occurred in the P(AN-MA) nanofiber mats. On P(AN-MA) mats, fibroblasts showed their characteristic spindle-like shape, produced a dermis-like structure, and responded well to TGF
\mathsf{\beta }
stimulation, with a significant increase in the mRNA expression of PAI1, COL1A1, and αSMA (all p < 0.05). Primary keratinocytes seeded on top of the dermis equivalent proliferated and formed a stratified epidermis-like structure. (4) Conclusion: P(AN-MA) and Pul/PVA/PAA are both biocompatible materials suitable for nanofiber mat production. P(AN-MA) mats hold greater potential as future 3D skin models due to enhanced cell compatibility (i.e., adsorption of FCS proteins), cell infiltration (i.e., increased pore size due to swelling behavior), and cell phenotype preservation. Thus, our proof-of-concept study shows an easy and robust process of producing electrospun scaffolds for 3D skin cell models made of P(AN-MA) nanofibers without the need for bioactive molecule attachments. View Full-Text
Keywords: 3D cell culture; microenvironment; tissue engineering; biomaterial; alternative methods 3D cell culture; microenvironment; tissue engineering; biomaterial; alternative methods
Rimann, M.; Jüngel, A.; Mousavi, S.; Moeschlin, N.; Calcagni, M.; Wuertz-Kozak, K.; Brunner, F.; Dudli, S.; Distler, O.; Adlhart, C. Acrylonitrile and Pullulan Based Nanofiber Mats as Easily Accessible Scaffolds for 3D Skin Cell Models Containing Primary Cells. Cells 2022, 11, 445. https://doi.org/10.3390/cells11030445
Rimann M, Jüngel A, Mousavi S, Moeschlin N, Calcagni M, Wuertz-Kozak K, Brunner F, Dudli S, Distler O, Adlhart C. Acrylonitrile and Pullulan Based Nanofiber Mats as Easily Accessible Scaffolds for 3D Skin Cell Models Containing Primary Cells. Cells. 2022; 11(3):445. https://doi.org/10.3390/cells11030445
Rimann, Markus, Astrid Jüngel, Sara Mousavi, Nicole Moeschlin, Maurizio Calcagni, Karin Wuertz-Kozak, Florian Brunner, Stefan Dudli, Oliver Distler, and Christian Adlhart. 2022. "Acrylonitrile and Pullulan Based Nanofiber Mats as Easily Accessible Scaffolds for 3D Skin Cell Models Containing Primary Cells" Cells 11, no. 3: 445. https://doi.org/10.3390/cells11030445
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Lidar and Camera Calibration - MATLAB & Simulink - MathWorks Switzerland
Detect Checkerboard Corners
Detect Checkerboard Plane
Calibrate Lidar and Camera
Calibration on Real Data
This example shows you how to estimate a rigid transformation between a 3-D lidar sensor and a camera, then use the rigid transformation matrix to fuse the lidar and camera data.
Lidar sensors and cameras are commonly used together in autonomous driving applications because a lidar sensor collects 3-D spatial information while a camera captures the appearance and texture of that space in 2-D images. You can fuse the data from these sensors to improve your object detection and classification. Lidar-camera calibration estimates a transformation matrix that gives the relative rotation and translation between the two sensors. You use this matrix when performing lidar-camera data fusion.
This diagram illustrates the workflow for the lidar and camera calibration (LCC) process, where we use checkerboard as a calibration object. We extract the checkerboard corners and planes from lidar and camera data, then establish a geometrical relationship between their coordinate systems to perform calibration. For more information on lidar-camera calibration process, see What Is Lidar-Camera Calibration?
This example uses data from two different lidar sensors, a
VelodyneLiDAR®
HDL-64 sensor and a VLP-16 sensor. For the HDL-64 sensor, use data collected from a Gazebo environment.
The HDL-64 sensor captures data as a set of PNG images and corresponding PCD point clouds. This example assumes that you already know the intrinsic parameters of the camera. For more information on extracting camera intrinsic parameters, see Evaluating the Accuracy of Single Camera Calibration.
Load the Velodyne HDL-64 sensor data from Gazebo.
imagePath = fullfile(toolboxdir('lidar'),'lidardata','lcc','HDL64','images');
ptCloudPath = fullfile(toolboxdir('lidar'),'lidardata','lcc','HDL64','pointCloud');
cameraParamsPath = fullfile(imagePath,'calibration.mat');
% Load camera intrinsics.
intrinsic = load(cameraParamsPath);
% Load images using imageDatastore.
imds = imageDatastore(imagePath);
imageFileNames = imds.Files;
% Load point cloud files.
pcds = fileDatastore(ptCloudPath,'ReadFcn',@pcread);
ptCloudFileNames = pcds.Files;
% Square size of the checkerboard.
This example uses a checkerboard pattern for calibration. First, estimate the checkerboard edges from the camera data. Use the estimateCheckerboardCorners3d function to calculate the coordinates of the checkerboard corners and size of the actual checkerboard in millimeters. The function estimates corners as 3-D coordinates in world coordinate system.
[imageCorners3d,checkerboardDimension,dataUsed] = ...
estimateCheckerboardCorners3d(imageFileNames,intrinsic.cameraParams,squareSize);
% Remove image files that are not used.
imageFileNames = imageFileNames(dataUsed);
Visualize the results by using the helperShowImageCorners helper function.
% Display checkerboard corners.
helperShowImageCorners(imageCorners3d,imageFileNames,intrinsic.cameraParams)
Next, use the detectRectangularPlanePoints function to detect the checkerboard plane in the lidar data. The function detects the checkerboard using the board dimensions calculated by the estimateCheckerboardCorners3d function.
% Extract the checkerboard ROI from the detected checkerboard image corners.
roi = helperComputeROI(imageCorners3d,5);
% Filter the point cloud files that are not used for detection.
ptCloudFileNames = ptCloudFileNames(dataUsed);
[lidarCheckerboardPlanes,framesUsed,indices] = ...
detectRectangularPlanePoints(ptCloudFileNames,checkerboardDimension,ROI=roi);
% Remove ptCloud files that are not used.
ptCloudFileNames = ptCloudFileNames(framesUsed);
% Remove image files.
imageFileNames = imageFileNames(framesUsed);
% Remove 3-D corners from images.
imageCorners3d = imageCorners3d(:,:,framesUsed);
Visualize the detected checkerboard by using the helperShowCheckerboardPlanes function.
helperShowCheckerboardPlanes(ptCloudFileNames,indices)
Use the estimateLidarCameraTransform function to estimate the rigid transformation matrix between the lidar sensor and the camera.
[tform,errors] = estimateLidarCameraTransform(lidarCheckerboardPlanes, ...
imageCorners3d,intrinsic.cameraParams);
After calibration, you can use this transformation matrix to:
Project lidar point clouds on images, using the projectLidarPointsOnImage function.
Enhance lidar point clouds using color information from images, using the fuseCameraToLidar function.
Use the helperFuseLidarCamera function to visualize the lidar and the image data fused together.
helperFuseLidarCamera(imageFileNames,ptCloudFileNames,indices, ...
intrinsic.cameraParams,tform);
You can estimate the calibration accuracy using these types of errors.
Translation Error — The difference between the centroid coordinates of the checkerboard planes in the point clouds and those in the corresponding images, in meters.
Rotation Error — The difference between the normal angles defined by the checkerboard planes in the point clouds and those in the corresponding images, in radians.
Reprojection Error — The difference between the projected (transformed) centroid coordinates of the checkerboard planes from the point clouds and those in the corresponding images, in pixels.
Plot the estimated error values by using the helperShowError function.
helperShowError(errors)
Test the LCC workflow on actual VLP-16 lidar data to evaluate its performance.
imagePath = fullfile(toolboxdir('lidar'),'lidardata','lcc','vlp16','images');
ptCloudPath = fullfile(toolboxdir('lidar'),'lidardata','lcc','vlp16','pointCloud');
% Load camera intrinscs.
% Square size of the checkerboard in mm.
% Extract checkerboard corners from the images.
% Remove the unused image files.
% Extract ROI from the detected checkerboard image corners.
% Extract checkerboard plane from point cloud data.
[lidarCheckerboardPlanes,framesUsed,indices] = detectRectangularPlanePoints( ...
ptCloudFileNames,checkerboardDimension,RemoveGround=true,ROI=roi);
% Plot the estimated error values.
helperShowError(errors);
This example provides an overview of the lidar-camera calibration workflow and shows you how to use a rigid transformation matrix to fuse lidar and camera data.
[2] Arun, K. S., T. S. Huang, and S. D. Blostein. “Least-Squares Fitting of Two 3-D Point Sets.” IEEE Transactions on Pattern Analysis and Machine Intelligence PAMI-9, no. 5 (September 1987): 698–700. https://doi.org/10.1109/TPAMI.1987.4767965.
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Reciprocating combustion engine with variable number of pistons - MATLAB - MathWorks Nordic
Offset angle vector
Base and follower bearing viscous friction coefficients
Initial angular deflection
Fuel consumption per piston revolution
Reciprocating combustion engine with variable number of pistons
The Piston Engine block represents a reciprocating combustion engine with multiple cylinders. The piston model accounts for the instantaneous torque transmitted to the engine drive shaft. The instantaneous torque enables you to model vibrations in the drivetrain due to piston revolution. To model just the piston mechanism of a combustion engine, use the Piston block.
Port B represents the translating piston and port F the rotating crankshaft. The piston force follows from the cylinder pressure and cross-sectional area. The block obtains the combustion pressure from a lookup table parameterized in terms of the crank angle and, optionally, the crank angular velocity and engine throttle level.
The crank torque follows from the piston force and crank angle as well as the crank and connecting rod lengths. In terms of these inputs, the ratio of the piston force and crank torque is
\frac{{T}_{\text{F}}}{{F}_{\text{B}}}=-\text{c}\left(sin\left(\theta \right)+\frac{sin\left(2\theta \right)}{2\sqrt{{\left(\frac{\text{r}}{\text{c}}\right)}^{2}-{\mathrm{sin}}^{2}\left(\theta \right)}}\right),
Physical signal port T lets you specify the engine throttle level as a fraction between 0 and 1. This fraction corresponds to the percentage of full power generated. The block uses the physical signal input whenever the pressure lookup table in the block property inspector is parameterized only in terms of the crank angle.
Fuel consumed by engine.
B — Base port associated with piston
Follower port of engine. The crankshaft transmits the power generated from the combustion process. Typically, this is where you would attach a clutch and transmission.
Number of pistons — Total displacement calculation
Number of pistons in the combustion engine.
Offset angle vector — Top dead center offset for each piston
[0, 180, 360, -180] (default) | vector
Vector of piston offset angles. The offset angle specifies the point in the engine cycle when the piston reaches top dead center. The engine cycle spans in angle from -S*180 to +S*180 degrees, where S is the number of strokes per cycle.
The vector size must be the same as the number of pistons. The default vector corresponds to a four-stroke, four-piston engine.
Inside diameter of the piston cylinder wall. The block uses this measurement to compute the torque table. You must specify a value greater than zero.
Distance from the fully retracted position to the fully extended position of the piston. The block uses this measurement to convert pressure on the piston into torque values. You must specify a value greater than zero.
Pressure parameterization — Determine pressure applied to piston
Shaft dynamics — Enable shaft dynamics modeling parameters
No shaft dynamics - Suitable for HIL simulation (default) | Specify shaft stiffness, damping, and inertia
Option to parameterize the shaft dynamics.
Base and follower bearing viscous friction coefficients — Characterize system friction
[0, 0] deg (default) | vector
Viscous friction coefficients for the base bearing and follower bearing, in that order.
Initial crank angle — Starting crank position
Crank angle at time zero relative to a top dead center position. The engine cycle spans in angle from -S*180 to +S*180 degrees, where S is the number of strokes per cycle.
Translational spring stiffness of engine crankshaft. The spring stiffness accounts for elastic energy storage in the crankshaft due to material compliance.
Stiffness coefficient of the engine crankshaft. This parameter accounts for resistance to shaft deformation.
To enable this parameter, set Shaft dynamics to Specify shaft stiffness, damping, and inertia.
Damping — Tendency to drain energy from the system
1000 N*m/(rad/s) (default) | positive scalar
Translational damping of engine crankshaft. The damping accounts for energy dissipation in the crankshaft due to material compliance.
Inertia — Tendency to resist change in motion
.02 kg*m^2 (default) | positive scalar
Moment of inertia of crankshaft about its rotational axis. This parameter accounts for resistance to sudden changes in motion.
Initial angular deflection — Angular deflection initializing parameter
Deflection angle between the base and follower ends of the crankshaft at time zero. The deflection angle measures the angular deformation of the crankshaft due to torsion.
Initial angular velocity — Angular velocity initializing parameter
Angular velocity of the crankshaft at time zero.
Fuel consumption model — Choose No fuel consumption, Constant per revolution, Fuel consumption by speed and torque, Brake specific fuel consumption by speed and torque, or Brake specific fuel consumption by speed and brake mean effective pressure
Interpolation method — Choose Linear or Smooth
Fuel consumption model based on available data. Select a model for calculating engine-fuel consumption. Model parameterizations are compatible with typical industrial data. Choose from the following options:
No fuel consumption — The default option
If you leave Fuel consumption model set to No fuel consumption, the block does not calculate fuel consumption even when the FC port is connected to another block. Selecting this option increases simulation speed.
Fuel consumption per piston revolution — Constant
To enable this parameter, set Fuel consumption to Fuel consumption by speed and torque, Brake specific fuel consumption by speed and torque, or Brake specific fuel consumption by speed and brake mean effective pressure.
Matrix of fuel consumption values that correspond Enter matrix with fuel consumption rates corresponding to engine speed and torque vectors. The number of rows must equal the number of elements in the Speed vector. The number of columns must equal the number of elements in the Torque vector. The default is [.5, .9, 1.4, 1.6, 1.9, 2.7, 3.4, 4.4; 1, 1.7, 2.7, 3.1, 3.6, 5, 6, 7.4; 1.4, 2.7, 4, 4.8, 5.6, 7.5, 8.5, 10.5; 2, 3.6, 5.8, 6.7, 8, 10.4, 11.7, 13.3; 2.5, 4.8, 7.9, 9.4, 10.8, 14, 16.2, 18.6; 3.1, 6, 10.3, 11.9, 13.8, 18.4, 22, 26.5] g/s.
BMEP=T\cdot \left(\frac{2\pi \cdot {n}_{c}}{{V}_{d}}\right),
For the Brake specific fuel consumption by speed and torque fuel model, enter the matrix with brake specific fuel consumption (BSFC) rates corresponding to engine speed and torque vectors. BSFC is the ratio of the fuel consumption rate to the output power. The number of rows must equal the number of elements in the Speed vector. The number of columns must equal the number of elements in the Torque vector.
For the Brake specific fuel consumption by speed and brake mean effective pressure fuel model, enter the matrix with brake specific fuel consumption (BSFC) rates corresponding to engine speed and brake mean effective pressure (BMEP) vectors. BSFC is the ratio of the fuel consumption rate to the output power. The number of rows must equal the number of elements in the Speed vector. The number of columns must equal the number of elements in the Brake mean effective pressure vector.
For both fuel-consumption models, the default is [410, 380, 300, 280, 270, 290, 320, 380; 410, 370, 290, 270, 260, 270, 285, 320; 415, 380, 290, 275, 265, 270, 270, 300; 420, 390, 310, 290, 285, 280, 280, 285; 430, 410, 340, 320, 310, 300, 310, 320; 450, 430, 370, 340, 330, 330, 350, 380] g/hr/kW.
Generic Engine | Piston
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Complete lattice — Wikipedia Republished // WIKI 2
Partially ordered set in which all subsets have both a supremum and infimum
In mathematics, a complete lattice is a partially ordered set in which all subsets have both a supremum (join) and an infimum (meet). Specifically, every non-empty finite lattice is complete. Complete lattices appear in many applications in mathematics and computer science. Being a special instance of lattices, they are studied both in order theory and universal algebra.
Complete lattices must not be confused with complete partial orders (cpos), which constitute a strictly more general class of partially ordered sets. More specific complete lattices are complete Boolean algebras and complete Heyting algebras (locales).
25. Bounded, Complete and Compliment Lattice - Gate
[Коллоквиум]: Formal Concept Analysis: A Useful Example of Modern Mathematics
Introduction to lattices - Mathematical Morphology Lectures
1.1 Complete semilattices
1.2 Complete sublattices
3 Locally finite complete lattices
4 Morphisms of complete lattices
4.1 Galois connections and adjoints
5 Free construction and completion
5.1 Free "complete semilattices"
5.2 Free complete lattices
7 Further results
A partially ordered set (L, ≤) is a complete lattice if every subset A of L has both a greatest lower bound (the infimum, also called the meet) and a least upper bound (the supremum, also called the join) in (L, ≤).
The meet is denoted by
{\displaystyle \bigwedge A}
, and the join by
{\displaystyle \bigvee A}
In the special case where A is the empty set, the meet of A will be the greatest element of L. Likewise, the join of the empty set yields the least element. Since the definition also assures the existence of binary meets and joins, complete lattices thus form a special class of bounded lattices.
More implications of the above definition are discussed in the article on completeness properties in order theory.
In order theory, arbitrary meets can be expressed in terms of arbitrary joins and vice versa (for details, see completeness (order theory)). In effect, this means that it is sufficient to require the existence of either all meets or all joins to obtain the class of all complete lattices.
As a consequence, some authors use the terms complete meet-semilattice or complete join-semilattice as another way to refer to complete lattices. Though similar on objects, the terms entail different notions of homomorphism, as will be explained in the below section on morphisms.
On the other hand, some authors have no use for this distinction of morphisms (especially since the emerging concepts of "complete semilattice morphisms" can as well be specified in general terms). Consequently, complete meet-semilattices have also been defined as those meet-semilattices that are also complete partial orders. This concept is arguably the "most complete" notion of a meet-semilattice that is not yet a lattice (in fact, only the top element may be missing). This discussion is also found in the article on semilattices.
Complete sublattices
A sublattice M of a complete lattice L is called a complete sublattice of L if for every subset A of M the elements
{\displaystyle \bigwedge A}
{\displaystyle \bigvee A}
, as defined in L, are actually in M.[1]
If the above requirement is lessened to require only non-empty meet and joins to be in L, the sublattice M is called a closed sublattice of M.
Any non-empty finite lattice is trivially complete.
The power set of a given set, ordered by inclusion. The supremum is given by the union and the infimum by the intersection of subsets.
The unit interval [0,1] and the extended real number line, with the familiar total order and the ordinary suprema and infima. Indeed, a totally ordered set (with its order topology) is compact as a topological space if it is complete as a lattice.
The non-negative integers, ordered by divisibility. The least element of this lattice is the number 1, since it divides any other number. Perhaps surprisingly, the greatest element is 0, because it can be divided by any other number. The supremum of finite sets is given by the least common multiple and the infimum by the greatest common divisor. For infinite sets, the supremum will always be 0 while the infimum can well be greater than 1. For example, the set of all even numbers has 2 as the greatest common divisor. If 0 is removed from this structure it remains a lattice but ceases to be complete.
The subgroups of any given group under inclusion. (While the infimum here is the usual set-theoretic intersection, the supremum of a set of subgroups is the subgroup generated by the set-theoretic union of the subgroups, not the set-theoretic union itself.) If e is the identity of G, then the trivial group {e} is the minimum subgroup of G, while the maximum subgroup is the group G itself.
The open sets of a topological space, ordered by inclusion. The supremum is given by the union of open sets and the infimum by the interior of the intersection.
The convex subsets of a real or complex vector space, ordered by inclusion. The infimum is given by the intersection of convex sets and the supremum by the convex hull of the union.
The lattice of all transitive relations on a set.
The lattice of all equivalence relations on a set; the equivalence relation ~ is considered to be smaller (or "finer") than ≈ if x~y always implies x≈y.
The lattice of self-adjoint projections (also known as orthogonal projections) of a von Neumann algebra.
Locally finite complete lattices
A complete lattice L is said to be locally finite if the supremum of any infinite subset is equal to 1, or equivalently, the set
{\displaystyle \{y\in L~|~y\leq x\}\,\!}
is finite for any
{\displaystyle 1\neq x\in L}
. The lattice (N, |) is locally finite. Note that in this lattice, the element generally denoted "0" is actually 1 and vice versa.
{\displaystyle f\left(\bigwedge A\right)=\bigwedge \{f(a)\mid a\in A\}}
{\displaystyle f\left(\bigvee A\right)=\bigvee \{f(a)\mid a\in A\}}
for all subsets A of L. Such functions are automatically monotonic, but the condition of being a complete homomorphism is in fact much more specific. For this reason, it can be useful to consider weaker notions of morphisms, that are only required to preserve all joins (giving a category Sup) or all meets (giving a category Inf), which are indeed inequivalent conditions. This notion may be considered as a homomorphism of complete meet-semilattices or complete join-semilattices, respectively.
Galois connections and adjoints
Furthermore, morphisms that preserve all joins are equivalently characterized as the lower adjoint part of a unique Galois connection. For any pair of preorders P and Q, these are given by pairs of monotone functions f and g such that
{\displaystyle f(x)\leq y\iff x\leq g(y)}
where f is called the lower adjoint and g is called the upper adjoint. By the adjoint functor theorem, a monotone map between any pair of preorders preserves all joins if and only if it is a lower adjoint, and preserves all meets if and only if it is an upper adjoint.
As such, each join-preserving morphism determines a unique upper adjoint in the inverse direction that preserves all meets. Hence, considering complete lattices with complete semilattice morphisms boils down to considering Galois connections as morphisms. This also yields the insight that the introduced morphisms do basically describe just two different categories of complete lattices: one with complete homomorphisms and one with meet-preserving functions (upper adjoints), dual to the one with join-preserving mappings (lower adjoints).
A particularly important special case is for lattices of subsets P(X) and P(Y) and a function from X to Y. In this case, the direct image and inverse image maps between the power sets are upper and lower adjoints to each other, respectively.
As usual, the construction of free objects depends on the chosen class of morphisms. Let us first consider functions that preserve all joins (i.e. lower adjoints of Galois connections), since this case is simpler than the situation for complete homomorphisms. Using the aforementioned terminology, this could be called a free complete join-semilattice.
Using the standard definition from universal algebra, a free complete lattice over a generating set S is a complete lattice L together with a function i:S→L, such that any function f from S to the underlying set of some complete lattice M can be factored uniquely through a morphism f° from L to M. Stated differently, for every element s of S we find that f(s) = f°(i(s)) and that f° is the only morphism with this property. These conditions basically amount to saying that there is a functor from the category of sets and functions to the category of complete lattices and join-preserving functions which is left adjoint to the forgetful functor from complete lattices to their underlying sets.
Free complete lattices in this sense can be constructed very easily: the complete lattice generated by some set S is just the powerset 2S, i.e. the set of all subsets of S, ordered by subset inclusion. The required unit i:S→2S maps any element s of S to the singleton set {s}. Given a mapping f as above, the function f°:2S→M is defined by
{\displaystyle f^{\circ }(X)=\bigvee \{f(s)|s\in X\}}
Then f° transforms unions into suprema and thus preserves joins.
Our considerations also yield a free construction for morphisms that do preserve meets instead of joins (i.e. upper adjoints of Galois connections). In fact, we merely have to dualize what was said above: free objects are given as powersets ordered by reverse inclusion, such that set union provides the meet operation, and the function f° is defined in terms of meets instead of joins. The result of this construction could be called a free complete meet-semilattice. One should also note how these free constructions extend those that are used to obtain free semilattices, where we only need to consider finite sets.
The situation for complete lattices with complete homomorphisms obviously is more intricate. In fact, free complete lattices do generally not exist. Of course, one can formulate a word problem similar to the one for the case of lattices, but the collection of all possible words (or "terms") in this case would be a proper class, because arbitrary meets and joins comprise operations for argument-sets of every cardinality.
Now one might still hope that there are some useful cases where the set of generators is sufficiently small for a free complete lattice to exist. Unfortunately, the size limit is very low and we have the following theorem:
The free complete lattice on three generators does not exist; it is a proper class.
A proof of this statement is given by Johnstone;[2] the original argument is attributed to Alfred W. Hales;[3] see also the article on free lattices.
As long as one considers meet- or join-preserving functions as morphisms, this can easily be achieved through the so-called Dedekind–MacNeille completion. For this process, elements of the poset are mapped to (Dedekind-) cuts, which can then be mapped to the underlying posets of arbitrary complete lattices in much the same way as done for sets and free complete (semi-) lattices above.
Already G. Birkhoff's Lattice Theory book[4] contains a very useful representation method. It associates a complete lattice to any binary relation between two sets by constructing a Galois connection from the relation, which then leads to two dually isomorphic closure systems. Closure systems are intersection-closed families of sets. When ordered by the subset relation ⊆, they are complete lattices.
A special instance of Birkhoff's construction starts from an arbitrary poset (P,≤) and constructs the Galois connection from the order relation ≤ between P and itself. The resulting complete lattice is the Dedekind-MacNeille completion. When this completion is applied to a poset that already is a complete lattice, then the result is isomorphic to the original one. Thus we immediately find that every complete lattice is represented by Birkhoff's method, up to isomorphism.
The construction is utilized in formal concept analysis, where one represents real-word data by binary relations (called formal contexts) and uses the associated complete lattices (called concept lattices) for data analysis. The mathematics behind formal concept analysis therefore is the theory of complete lattices.
Another representation is obtained as follows: A subset of a complete lattice is itself a complete lattice (when ordered with the induced order) if and only if it is the image of an increasing and idempotent (but not necessarily extensive) self-map. The identity mapping obviously has these two properties. Thus all complete lattices occur.
Besides the previous representation results, there are some other statements that can be made about complete lattices, or that take a particularly simple form in this case. An example is the Knaster–Tarski theorem, which states that the set of fixed points of a monotone function on a complete lattice is again a complete lattice. This is easily seen to be a generalization of the above observation about the images of increasing and idempotent functions, since these are instances of the theorem.
lattice (order).
^ Burris, Stanley N., and H.P. Sankappanavar, H. P., 1981. A Course in Universal Algebra. Springer-Verlag. ISBN 3-540-90578-2 (A monograph available free online).
^ P. T. Johnstone, Stone Spaces, Cambridge University Press, 1982; (see paragraph 4.7)
^ A. W. Hales, On the non-existence of free complete Boolean algebras, Fundamenta Mathematicae 54: pp.45-66.
^ Garrett Birkhoff, Lattice Theory, AMS Colloquium Publications Vol. 25, ISBN 978-0821810255
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drinkR: Estimate your Blood Alcohol Concentration using R and Shiny. - Publishable Stuff
Inspired by events that took place at UseR 2014 last month I decided to implement an app that estimates one’s blood alcohol concentration (BAC). Today I present to you drinkR, implemented using R and Shiny, Rstudio’s framework for building web apps using R. So, say that I had a good dinner, drinking a couple of glasses of wine, followed by an evening at a divy karaoke bar, drinking a couple of red needles and a couple of beers. By entering my sex, height and weight and the times when I drank the drinks in the drinkR app I end up with this estimated BAC curve:
(Now I might be totally off with what drinks I had and when but Romain Francois, Karl Broman, Sandy Griffith, Karthik Ram and Hilary Parker can probably fill in the details.) If you want to estimate your current BAC (or a friend’s…) then head over to the drinkr app hosted at ShinyApps.io. If you want to know how the app estimates BAC read on below. The code for drinkR is available on GitHub, any suggestion on how it can be improved is greatly appreciated.
drinkR estimates the BAC according to the formulas given in The estimation of blood alcohol concentration by Posey and Mozayani (2007). I was also helped by reading through Computer simulation analysis of blood alcohol and the Widmark factor (explained below) was calculated according to The calculation of blood ethanol concentrations in males and females. Unfortunately all these articles are behind paywalls, that is how most publicly funded research works these days…
The BAC estimates you get out of drinkR will be as good as the formulas in Posey and Mozayani (2007). I don’t know how good they are and I don’t know how well they’ll fit you. Estimating BAC is of course a prediction problem and what you really would want to have is data so that you could build a predictive model and get an idea of how well it predicts BAC. Unfortunately I haven’t found any data on this so the Posey and Mozayani formulas is as good as I can do.
The three steps of estimating BAC
Estimating the BAC (according to Posey and Mozayani, 2007) after you have drunken, say, a beer requires “simulating” three processes:
Alcohol absorption. Just because you drank a beer doesn’t mean it goes directly into your blood stream, it has to be absorbed by your digestive system first and this takes some time.
Alcohol distribution. Your BAC depends on how much of you the absorbed alcohol will be “diluted” by. This depends on, among other things, your weight, height and sex.
Alcohol elimintation. How drunk you get (and how soon you will get sober) depends on how fast your body eliminates the absorbed alcohol.
Alcohol absorption can be approximated by assuming it is first order, that is, assuming there is an alcohol halflife, a time it takes for half of a drink to be absorbed. When measured this halflife tend to be between 6 min to 18 min, depending on how much you have reacently eaten. If you haven’t eaten for a while your halflife might be closer to 6 min while if you just had a big döner kebab it might be closer to 18 min.
Alcohol distribution depends on the amount of water that the alcohol in your body will be diluted in. It can be estimated by the following equation:
where $C$ is the alcohol concentration, $A$ is the mass of the alcohol, $W$ is your body weight and $r$ is the Widmark factor. This factor can be seen as an adjustment that is necessary because your whole body is not made of water, thus the alcohol is not “diluted by” your whole weight. There are many different formulas for estimating $r$ and drinkR uses the one given by Seidl et al. (2000) which estimates $r$ dependent on sex, height and weight:
r_{\text{female}} = 0.31 - 0.0064 \times \text{weight in kg} + 0.0045 \times \text{height in cm}
r_{\text{male}} = 0.32 - 0.0048 \times \text{weight in kg} + 0.0046 \times \text{height in cm}
These linear equations can give really strange values for $r$, for example, if you weight a lot. Therefore I also bound $r$ to be within the limits found by Seidl et al. (2000): 0.44 to 0.80 in women and 0.60 to 0.87 in men.
Finally, alcohol elimination can be reasonably approximated by a constant elimination rate of the BAC. This rate can vary from around 0.009 % per hour to 0.035 % per hour with 0.018 % per hour being a reasonable average.
drinkR puts these three processes together and estimates your BAC over time given a number of drinks with time stamps. Assuming that you are also interested in how drunk you are right now, drinkR shows an estimate of your current BAC by fetching your computers local time (see this stackoverflow question for how this is done). The estimate given by drinkR might be very missleading so don’t use it for any serious purposes! To get a sense of the uncertainty in the BAC estimate play around with the parameters (especially the alcohol elimination rate) and see how much your BAC curve changes.
If you want to see how different levels of BAC could affect you see the Progressive effects of alcohol chart over at Wikipedia and if you want to try out drinkR live I would recommend one of my favorite drinks: Absinthe mixed with Orange soda (say Fanta orange). It’s better than you think it is! :)
Posey, D., & Mozayani, A. (2007). The estimation of blood alcohol concentration. Forensic Science, Medicine, and Pathology, 3(1), 33-39. Link (Unfortunately behind paywall)
Rockerbie, D. W., & Rockerbie, R. A. (1995). Computer simulation analysis of blood alcohol. Journal of clinical forensic medicine, 2(3), 137-141. Link (Unfortunately behind paywall)
Seidl, S., Jensen, U., & Alt, A. (2000). The calculation of blood ethanol concentrations in males and females. International journal of legal medicine, 114(1-2), 71-77. Link (Unfortunately behind paywall)
Posted by Rasmus Bååth Jul 30th, 2014 R, Statistics
« Chillin' at UseR! 2014 Subjective Rhythmization at ICMPC 2014 in Seoul »
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In the last tutorial series, we wrote 2 layers neural networks model, now it's time to build a deep neural network, where we could have whatever count of layers we want
Welcome to another tutorial. In the last tutorial series, we wrote 2 layers neural networks model. Now it's time to build a deep neural network, where we could have whatever count of layers we want.
So the same as in previous tutorials, we'll first implement all the functions required to build a deep neural network. Then we will use these functions to build a deep neural network for image classification (cats vs. dogs).
In this tutorial series, I will use non-linear units like ReLU to improve our model. Our deep neural network model will be built to easily define deep layers that our model would be easy to use.
This is a continuation tutorial of my one hidden layer neural network tutorial to use the same data set. If you see this tutorial first time, cats vs. dogs data-set you can get from my GitHub page. If you don't know how to use a data set, check my previous tutorials.
First, let's define our model structure:
Same as in previous tutorials, build a deep neural network and implement several "helper functions". These helper functions will be used to build a two-layer neural network and an L-layer neural network. In each small helper function, we will implement, I will try to give you a detailed explanation. So what we'll do in this tutorial series:
Initialize the parameters for a two-layer network and an L-layer neural network;
Implement the forward propagation module (shown in the figure below);
- Complete the LINEAR part of a layer's forward propagation step (resulting in Z[l]);
- We'll write the ACTIVATION function (relu/sigmoid);
- We'll combine the previous two steps into a new [LINEAR->ACTIVATION] forward function;
- Stack the [LINEAR->RELU] forward function L-1 time (for layers 1 through L-1) and add a [LINEAR->SIGMOID] at the end (for the final layer L). This will give us a new L_model_forward function;
Compute the loss;
Implement the backward propagation module (denoted in red in the figure below);
- Complete the LINEAR part of a layer's backward propagation step;
- We will write the gradient of the ACTIVATE function (relu_backward/sigmoid_backward);
- Combine the previous two steps into a new [LINEAR->ACTIVATION] backward function;
- Stack [LINEAR->RELU] backward L-1 times and add [LINEAR->SIGMOID] backward in a new L_model_backward function;
Finally, update the parameters.
You will see that for every forward function; there will be a corresponding backward function. That is why we will be storing some values in a cache at every step of our forward module. The cached values will be useful for computing gradients. In the backpropagation module, we will then use the cache to calculate the gradients. In this tutorial series, I will show you exactly how to carry out each of these steps.
Parameters initialization:
I will write two helper functions that will initialize the parameters for our model. The first function will be used to initialize parameters for a two-layer model. The second one will generalize this initialization process to L layers.
So not to write my code twice, I will copy my initialization function for a two-layer model from my previous tutorial:
def initialize_parameters(input_layer, hidden_layer, output_layer):
# initialize 1st layer output and input with random values
W1 = np.random.randn(hidden_layer, input_layer) * 0.01
# initialize 1st layer output bias
b1 = np.zeros((hidden_layer, 1))
# initialize 2nd layer output and input with random values
W2 = np.random.randn(output_layer, hidden_layer) * 0.01
# initialize 2nd layer output bias
b2 = np.zeros((output_layer,1))
Parameters initialization for a deeper L-layer neural network is more complicated because there are many more weight matrices and bias vectors. When we complete our initialize_parameters_deep function, we should make sure that our dimensions match each layer.
Recall that n[l] is the number of units in layer l. Thus, for example, if the size of our input X is (12288,6002) (with m=6002 examples), then:
\begin{array}{ccccc}& Shape of W& Shape of b& Activation& Shape of Activation\\ Layer 1& \left({n}^{\left[1\right]},12288\right)& \left({n}^{\left[1\right]},1\right)& {Z}^{\left[1\right]}={W}^{\left[1\right]}X+{b}^{\left[1\right]}& \left({n}^{\left[1\right]}, 6002\right)\\ Layer 2& \left({n}^{\left[2\right]}, {n}^{\left[1\right]}\right)& \left({n}^{\left[2\right]},1\right)& {Z}^{\left[2\right]}={W}^{\left[2\right]}{A}^{\left[1\right]}+{b}^{\left[2\right]}& \left({n}^{\left[2\right]}, 6002\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ Layer L-1& \left({n}^{\left[L-1\right]}, {n}^{\left[L-2\right]}\right)& \left({n}^{\left[L-1\right]}, 1\right)& {Z}^{\left[L-1\right]}={W}^{\left[L-1\right]}{A}^{\left[L-2\right]}+{b}^{\left[L-1\right]}& \left({n}^{\left[L-1\right]}, 6002\right)\\ Layer L& \left({n}^{\left[L\right]}, {n}^{\left[L-1\right]}\right)& \left({n}^{\left[L\right]}, {n}^{\left[L-1\right]}\right)& {Z}^{\left[L\right]}={W}^{\left[L\right]}{A}^{\left[L-1\right]}+{b}^{\left[L\right]}& \left({n}^{\left[L\right]}, 6002\right)\end{array}
I should remind you that when we compute WX+b in python, it carries out broadcasting. For example, if:
W=\left[\begin{array}{ccc}j& k& l\\ m& n& o\\ p& q& r\end{array}\right] X=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right] b=\left[\begin{array}{c}s\\ t\\ u\end{array}\right]
Then WX+b will be:
WX+b=\left[\begin{array}{ccc}\text{(ja+kd+lg)+s}& \text{(jb+ke+lh)+s}& \text{(jc+kf+li)+s}\\ \text{(na+nd+og)+t}& \text{(mb+ne+oh)+t}& \text{(mc+nf+oi)+t}\\ \text{(pa+qd+rg)+u}& \text{(pb+qe+rh)+u}& \text{(pc+qf+ri)+u}\end{array}\right]
So what we'll do to implement the initialization function for an L-layer Neural Network:
The model's structure will be: *[LINEAR -> RELU] × (L-1) -> LINEAR -> SIGMOID*. In our example, it will have L−1 layers using a ReLU activation function followed by an output layer with a sigmoid activation function;
We'll use random initialization for the weight matrices: np.random.randn(shape)∗0.01;
We'll use zeros initialization for the biases. Use np.zeros(shape);
We will store n[l], the number of units in different layers, in a variable 'layer_dims'. For example, if we declare 'layer_dims' to be [2,4,1]: There will be two inputs, one hidden layer with 4 hidden units and an output layer with 1 output unit. This means W1's shape was (4,2), b1 was (4,1), W2 was (1,4), and b2 was (1,1). Now we will generalize this to L layers.
Before moving further, we need to overview the whole notation:
Superscript [l] denotes a quantity associated with the lth layer. For example, a[L] is the Lth layer activation. W[L] and b[L] are the Lth layer parameters;
Superscript (i) denotes a quantity associated with the ith example. Example: x(i) is the ith training example;
Lowerscript i denotes the ith entry of a vector. Example: a[i] denotes the ith entry of the lth layer's activations.
Code for our deep parameters initialization:
layer_dimensions - python array (list) containing the dimensions of each layer in our network
parameters - python dictionary containing our parameters "W1", "b1", ..., "WL", "bL":
Wl - weight matrix of shape (layer_dimensions[l], layer_dimensions[l-1]);
bl - bias vector of shape (layer_dimensions[l], 1).
def initialize_parameters_deep(layer_dimensions):
# number of layers in the network
L = len(layer_dimensions)
parameters['W' + str(l)] = np.random.randn(layer_dimensions[l], layer_dimensions[l-1]) * 0.01
parameters['b' + str(l)] = np.zeros((layer_dimensions[l], 1))
So we wrote our function. Let's test it out with random numbers:
parameters = initialize_parameters_deep([4,5,3])
print("W1 = ", parameters["W1"])
print("b1 = ", parameters["b1"])
We'll receive:
W1 = [[ 0.00415369 -0.00965885 0.0009098 -0.00426353]
[-0.00807062 0.01026888 0.00625363 0.0035793 ]
[ 0.00103969 -0.00245844 0.02399076 -0.02490289]]
b1 = [[0.]
W2 = [[-0.00999478 -0.01071253 -0.00471277 0.01213727 -0.00630878]
[-0.005903 0.00163863 -0.0143418 0.00660419 0.00885867]
[ 0.00554906 -0.00170923 -0.00708474 -0.0086883 0.00935947]]
You may receive different values because of random initialization. Let's test deeper initialization:
parameters = initialize_parameters_deep([4,5,3,2])
Then we'll receive something to this:
W1 = [[-0.00787148 0.00351103 0.00031584 0.01036506]
[-0.01367634 -0.00592318 -0.01703005 -0.0008115 ]
[-0.00830651 0.0013722 0.01029079 -0.00819454]]
W2 = [[-0.00646581 0.00884422 0.00472376 0.01447212 -0.00341151]
W3 = [[-0.00515514 -0.01256405 0.00632316]
So in our first deep learning tutorial, we defined our model structure and the steps we need to do. So we finished the first step, to initialize deep network parameters. We tried to initialize the neural network with one hidden layer and two hidden layers; everything works fine. In the next tutorial, we'll start building forward propagation functions.
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Vibration Characteristics of a 75kW Turbo Machine With Air Foil Bearings | J. Eng. Gas Turbines Power | ASME Digital Collection
75kW
Kyeong-Su Kim,
Kyeong-Su Kim
, 373-1 Guseong-dong, Yuseong-gu, Daejeon, 305-701, Korea
e-mail: inlee@asdl.kaist.ac.kr
Kim, K., and Lee, I. (July 15, 2006). "Vibration Characteristics of a
75kW
Turbo Machine With Air Foil Bearings." ASME. J. Eng. Gas Turbines Power. July 2007; 129(3): 843–849. https://doi.org/10.1115/1.2718220
Air foil bearings are very attractive bearing systems for turbomachinery because they have several advantages over conventional bearings in terms of oil-free environment, low power loss, long life, and no maintenance. However, most of the developed machines using air foil bearings are limited to small and high-speed rotors of 60,000–120,000 rpm, since the increase in power of turbomachinery requires lower rotor speed and greater loading in bearings, which makes it difficult to use air foil bearings for large machines. In this paper, a 75 kW turboblower using air foil bearings is introduced, and the vibration characteristics of the machine have been investigated experimentally under a wide range of operating conditions, including compressor surge in the performance test. The machine is designed to be fully air lubricated and air cooled, and its operating speed is 20,000–26,000 rpm with maximum pressure ratio of 1.8. The results show that the air foil bearings offer adequate damping to ensure dynamically stable operation in the whole range.
vibrations, turbomachinery, aerodynamics, machine bearings, compressors, rotors, air foil bearings, turbo blower, environment-friendly system
Bearings, Compressors, Foil bearings, Machinery, Rotors, Surges, Vibration, Turbochargers, Pressure, Turbomachinery
, Orlando, ASME Paper No. 97-GT-347.
Development of Foil Journal Bearing for Small High Speed Cryogenic Turboexpander
Evaluation of Advanced Solid Lubricant Coatings for Foil Air Bearings Operation at 25° and 500°C
Oil-Free Turbocharger Demonstration Paves Way to Gas Turbine Engine Applications
Proc. of TURBOEXPO2000
, Munich, ASME, New York, ASME Paper No. 2000-GT-620.
Kirshmann
High Temperature Foil Air Bearing Development for A Missile/UAV Engine Application
HONDA R and D Technical Review
Development of a 1000F, 10,000 Start/Stop Cycle Foil Journal Bearing for A Gas Turbine Engine
, Paper No. 86-1457, A87-23253.
A Test Stand for Dynamic Characterization of Oil-Free Bearings for Modern Gas Turbine Engines
An OiḻFree Thrust Foil Bearing Facility Design, Calibration, and Operation
Munich, ASME
Load Capacity Estimation of Foil Air Journal Bearings for Oil-Free Turbomachinery Application
,” NASA/TM-2000-209782, NASA, Cleveland.
Design Improvements of Power Conversion System of Gas Turbine High Temperature Reactor (GTHTR300)
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MultivariatePowerSeries/UnivariatePolynomialOverPowerSeries - Maple Help
Home : Support : Online Help : MultivariatePowerSeries/UnivariatePolynomialOverPowerSeries
Create a univariate polynomial over power series
UnivariatePolynomialOverPowerSeries(lp,v)
UnivariatePolynomialOverPowerSeries(p,v)
UnivariatePolynomialOverPowerSeries(ps, v)
UnivariatePolynomialOverPowerSeries(u)
UnivariatePolynomialOverPowerSeries(u, v)
list, Array, or Vector of power series generated by this package
The function call UnivariatePolynomialOverPowerSeries(lp,v) creates a univariate polynomial over power series with main variable v and with coefficients that are power series from lp. The degree of the resulting polynomial is equal to the length of lp minus one. The coefficient of v^(i-1) is the i-th element of lp. In particular, the first element of lp is the constant coefficient. The main variable, v, cannot occur in any of the power series in lp.
The function call UnivariatePolynomialOverPowerSeries(p,v) creates a univariate polynomial with power series coefficients. It returns the same result as UnivariatePolynomialOverPowerSeries(lp,v) where lp := [seq(PowerSeries(coeff(p,v,i)),i=0..degree(p,v))].
The function call UnivariatePolynomialOverPowerSeries(ps, v) creates a univariate polynomial over power series representing ps, with v as its main variable. This is only possible if ps is known to be a polynomial function of v, which is the case if it is independent of v (in which case it is trivially polynomial) or if the analytic expression for ps is known and it is polynomial in v. If neither of the former two cases is true, then an error is raised.
The function call UnivariatePolynomialOverPowerSeries(u, v) copies u. You may omit v in this case, but if you specify it, it must be equal to the main variable of u.
\mathrm{with}\left(\mathrm{MultivariatePowerSeries}\right):
Create a univariate polynomial over power series from a list of power series.
f≔\mathrm{UnivariatePolynomialOverPowerSeries}\left([\mathrm{PowerSeries}\left(1\right),\mathrm{PowerSeries}\left(0\right),\mathrm{PowerSeries}\left(x\right),\mathrm{PowerSeries}\left(y\right),\frac{1}{\mathrm{PowerSeries}\left(1+x+y\right)}],z\right)
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{UnⅈvarⅈatⅇPolynomⅈalOvⅇrPowⅇrSⅇrⅈⅇs:}\left(\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{4}}]
\mathrm{Degree}\left(f\right)
\textcolor[rgb]{0,0,1}{4}
[\mathrm{seq}\left(\mathrm{GetCoefficient}\left(f,i\right),i=0..4\right)]
[[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs:}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs:}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs:}\textcolor[rgb]{0,0,1}{x}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs:}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs of}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{:}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }]]
We compute its linear and quadratic truncation. These are defined in terms of the homogeneous degree of the coefficients, ignoring the degree in the main variable.
\mathrm{Truncate}\left(f,1\right)
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}
\mathrm{Truncate}\left(f,2\right)
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}
Create a univariate polynomial over power series from a polynomial
p≔\left(z-1\right)\left(z-2\right)\left(z-3\right)+x\left({z}^{2}+z\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\right)
\mathrm{UnivariatePolynomialOverPowerSeries}\left(p,z\right)
[\textcolor[rgb]{0,0,1}{UnⅈvarⅈatⅇPolynomⅈalOvⅇrPowⅇrSⅇrⅈⅇs:}\left(\textcolor[rgb]{0,0,1}{-6}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-6}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{3}}]
The following attempt will not work, because Maple cannot determine that d is polynomial in z (though actually it is).
d≔\mathrm{PowerSeries}\left(d↦\mathrm{ifelse}\left(d=0,0,\frac{z\cdot {x}^{d-1}}{\left(d-1\right)!}\right),\mathrm{variables}={x,z}\right)
\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs:}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }]
\mathrm{UnivariatePolynomialOverPowerSeries}\left(d,z\right)
Error, (in MultivariatePowerSeries:-UnivariatePolynomialOverPowerSeries) attempted to convert a power series involving z to a univariate polynomial over power series in z, but it is not known to be polynomial in z
We define e in the same way as d but specify the analytic expression. Then we can successfully convert it to a univariate polynomial over power series.
e≔\mathrm{PowerSeries}\left(d↦\mathrm{ifelse}\left(d=0,0,\frac{z\cdot {x}^{d-1}}{\left(d-1\right)!}\right),\mathrm{analytic}=z\mathrm{exp}\left(x\right)\right)
\textcolor[rgb]{0,0,1}{e}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{PowⅇrSⅇrⅈⅇs of}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{:}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }]
k≔\mathrm{UnivariatePolynomialOverPowerSeries}\left(e,z\right)
\textcolor[rgb]{0,0,1}{k}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{UnⅈvarⅈatⅇPolynomⅈalOvⅇrPowⅇrSⅇrⅈⅇs:}\left(\textcolor[rgb]{0,0,1}{0}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}]
To copy k, we can specify the main variable explicitly or omit it. If we specify it explicitly, it has to be z, otherwise we obtain an error:
\mathrm{k1}≔\mathrm{UnivariatePolynomialOverPowerSeries}\left(k,x\right)
Error, (in MultivariatePowerSeries:-UnivariatePolynomialOverPowerSeries) you specified x as the main variable, but the main variable of the first argument is z
\mathrm{k1}≔\mathrm{UnivariatePolynomialOverPowerSeries}\left(k,z\right)
\textcolor[rgb]{0,0,1}{\mathrm{k1}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{UnⅈvarⅈatⅇPolynomⅈalOvⅇrPowⅇrSⅇrⅈⅇs:}\left(\textcolor[rgb]{0,0,1}{0}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}]
\mathrm{k2}≔\mathrm{UnivariatePolynomialOverPowerSeries}\left(k\right)
\textcolor[rgb]{0,0,1}{\mathrm{k2}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{UnⅈvarⅈatⅇPolynomⅈalOvⅇrPowⅇrSⅇrⅈⅇs:}\left(\textcolor[rgb]{0,0,1}{0}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\dots }\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}]
The MultivariatePowerSeries[UnivariatePolynomialOverPowerSeries] command was introduced in Maple 2021.
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Section 60.2 (07H7): Divided power envelope—The Stacks project
Section 60.2: Divided power envelope (cite)
60.2 Divided power envelope
The construction of the following lemma will be dubbed the divided power envelope. It will play an important role later.
Lemma 60.2.1. Let $(A, I, \gamma )$ be a divided power ring. Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal with $IB \subset J$. There exists a homomorphism of divided power rings
\[ (A, I, \gamma ) \longrightarrow (D, \bar J, \bar\gamma ) \]
\[ \mathop{\mathrm{Hom}}\nolimits _{(A, I, \gamma )}((D, \bar J, \bar\gamma ), (C, K, \delta )) = \mathop{\mathrm{Hom}}\nolimits _{(A, I)}((B, J), (C, K)) \]
functorially in the divided power algebra $(C, K, \delta )$ over $(A, I, \gamma )$. Here the LHS is morphisms of divided power rings over $(A, I, \gamma )$ and the RHS is morphisms of (ring, ideal) pairs over $(A, I)$.
Proof. Denote $\mathcal{C}$ the category of divided power rings $(C, K, \delta )$. Consider the functor $F : \mathcal{C} \longrightarrow \textit{Sets}$ defined by
\[ F(C, K, \delta ) = \left\{ (\varphi , \psi ) \middle | \begin{matrix} \varphi : (A, I, \gamma ) \to (C, K, \delta ) \text{ homomorphism of divided power rings} \\ \psi : (B, J) \to (C, K)\text{ an } A\text{-algebra homomorphism with }\psi (J) \subset K \end{matrix} \right\} \]
We will show that Divided Power Algebra, Lemma 23.3.3 applies to this functor which will prove the lemma. Suppose that $(\varphi , \psi ) \in F(C, K, \delta )$. Let $C' \subset C$ be the subring generated by $\varphi (A)$, $\psi (B)$, and $\delta _ n(\psi (f))$ for all $f \in J$. Let $K' \subset K \cap C'$ be the ideal of $C'$ generated by $\varphi (I)$ and $\delta _ n(\psi (f))$ for $f \in J$. Then $(C', K', \delta |_{K'})$ is a divided power ring and $C'$ has cardinality bounded by the cardinal $\kappa = |A| \otimes |B|^{\aleph _0}$. Moreover, $\varphi $ factors as $A \to C' \to C$ and $\psi $ factors as $B \to C' \to C$. This proves assumption (1) of Divided Power Algebra, Lemma 23.3.3 holds. Assumption (2) is clear as limits in the category of divided power rings commute with the forgetful functor $(C, K, \delta ) \mapsto (C, K)$, see Divided Power Algebra, Lemma 23.3.2 and its proof. $\square$
Definition 60.2.2. Let $(A, I, \gamma )$ be a divided power ring. Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal with $IB \subset J$. The divided power algebra $(D, \bar J, \bar\gamma )$ constructed in Lemma 60.2.1 is called the divided power envelope of $J$ in $B$ relative to $(A, I, \gamma )$ and is denoted $D_ B(J)$ or $D_{B, \gamma }(J)$.
Let $(A, I, \gamma ) \to (C, K, \delta )$ be a homomorphism of divided power rings. The universal property of $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$ is
\[ \begin{matrix} \text{ring maps }B \to C \\ \text{ which map }J\text{ into }K \end{matrix} \longleftrightarrow \begin{matrix} \text{divided power homomorphisms} \\ (D, \bar J, \bar\gamma ) \to (C, K, \delta ) \end{matrix} \]
and the correspondence is given by precomposing with the map $B \to D$ which corresponds to $\text{id}_ D$. Here are some properties of $(D, \bar J, \bar\gamma )$ which follow directly from the universal property. There are $A$-algebra maps
\begin{equation} \label{crystalline-equation-divided-power-envelope} B \longrightarrow D \longrightarrow B/J \end{equation}
The first arrow maps $J$ into $\bar J$ and $\bar J$ is the kernel of the second arrow. The elements $\bar\gamma _ n(x)$ where $n > 0$ and $x$ is an element in the image of $J \to D$ generate $\bar J$ as an ideal in $D$ and generate $D$ as a $B$-algebra.
Lemma 60.2.3. Let $(A, I, \gamma )$ be a divided power ring. Let $\varphi : B' \to B$ be a surjection of $A$-algebras with kernel $K$. Let $IB \subset J \subset B$ be an ideal. Let $J' \subset B'$ be the inverse image of $J$. Write $D_{B', \gamma }(J') = (D', \bar J', \bar\gamma )$. Then $D_{B, \gamma }(J) = (D'/K', \bar J'/K', \bar\gamma )$ where $K'$ is the ideal generated by the elements $\bar\gamma _ n(k)$ for $n \geq 1$ and $k \in K$.
Proof. Write $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$. The universal property of $D'$ gives us a homomorphism $D' \to D$ of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we see that $D' \to D$ is surjective (see remarks above). It is clear that $\bar\gamma _ n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e., we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided power structure on $\bar J'/K' \subset D'/K'$, see Divided Power Algebra, Lemma 23.4.3. Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win. $\square$
In the situation of Definition 60.2.2 we can choose a surjection $P \to B$ where $P$ is a polynomial algebra over $A$ and let $J' \subset P$ be the inverse image of $J$. The previous lemma describes $D_{B, \gamma }(J)$ in terms of $D_{P, \gamma }(J')$. Note that $\gamma $ extends to a divided power structure $\gamma '$ on $IP$ by Divided Power Algebra, Lemma 23.4.2. Hence $D_{P, \gamma }(J') = D_{P, \gamma '}(J')$ is an example of a special case of divided power envelopes we describe in the following lemma.
Lemma 60.2.4. Let $(B, I, \gamma )$ be a divided power algebra. Let $I \subset J \subset B$ be an ideal. Let $(D, \bar J, \bar\gamma )$ be the divided power envelope of $J$ relative to $\gamma $. Choose elements $f_ t \in J$, $t \in T$ such that $J = I + (f_ t)$. Then there exists a surjection
\[ \Psi : B\langle x_ t \rangle \longrightarrow D \]
of divided power rings mapping $x_ t$ to the image of $f_ t$ in $D$. The kernel of $\Psi $ is generated by the elements $x_ t - f_ t$ and all
\[ \delta _ n\left(\sum r_ t x_ t - r_0\right) \]
whenever $\sum r_ t f_ t = r_0$ in $B$ for some $r_ t \in B$, $r_0 \in I$.
Proof. In the statement of the lemma we think of $B\langle x_ t \rangle $ as a divided power ring with ideal $J' = IB\langle x_ t \rangle + B\langle x_ t \rangle _{+}$, see Divided Power Algebra, Remark 23.5.2. The existence of $\Psi $ follows from the universal property of divided power polynomial rings. Surjectivity of $\Psi $ follows from the fact that its image is a divided power subring of $D$, hence equal to $D$ by the universal property of $D$. It is clear that $x_ t - f_ t$ is in the kernel. Set
\[ \mathcal{R} = \{ (r_0, r_ t) \in I \oplus \bigoplus \nolimits _{t \in T} B \mid \sum r_ t f_ t = r_0 \text{ in }B\} \]
If $(r_0, r_ t) \in \mathcal{R}$ then it is clear that $\sum r_ t x_ t - r_0$ is in the kernel. As $\Psi $ is a homomorphism of divided power rings and $\sum r_ tx_ t - r_0 \in J'$ it follows that $\delta _ n(\sum r_ t x_ t - r_0)$ is in the kernel as well. Let $K \subset B\langle x_ t \rangle $ be the ideal generated by $x_ t - f_ t$ and the elements $\delta _ n(\sum r_ t x_ t - r_0)$ for $(r_0, r_ t) \in \mathcal{R}$. To show that $K = \mathop{\mathrm{Ker}}(\Psi )$ it suffices to show that $\delta $ extends to $B\langle x_ t \rangle /K$. Namely, if so the universal property of $D$ gives a map $D \to B\langle x_ t \rangle /K$ inverse to $\Psi $. Hence we have to show that $K \cap J'$ is preserved by $\delta _ n$, see Divided Power Algebra, Lemma 23.4.3. Let $K' \subset B\langle x_ t \rangle $ be the ideal generated by the elements
$\delta _ m(\sum r_ t x_ t - r_0)$ where $m > 0$ and $(r_0, r_ t) \in \mathcal{R}$,
$x_{t'}^{[m]}(x_ t - f_ t)$ where $m > 0$ and $t', t \in I$.
We claim that $K' = K \cap J'$. The claim proves that $K \cap J'$ is preserved by $\delta _ n$, $n > 0$ by the criterion of Divided Power Algebra, Lemma 23.4.3 (2)(c) and a computation of $\delta _ n$ of the elements listed which we leave to the reader. To prove the claim note that $K' \subset K \cap J'$. Conversely, if $h \in K \cap J'$ then, modulo $K'$ we can write
\[ h = \sum r_ t (x_ t - f_ t) \]
for some $r_ t \in B$. As $h \in K \cap J' \subset J'$ we see that $r_0 = \sum r_ t f_ t \in I$. Hence $(r_0, r_ t) \in \mathcal{R}$ and we see that
\[ h = \sum r_ t x_ t - r_0 \]
is in $K'$ as desired. $\square$
Lemma 60.2.5. Let $(A, I, \gamma )$ be a divided power ring. Let $B$ be an $A$-algebra and $IB \subset J \subset B$ an ideal. Let $x_ i$ be a set of variables. Then
\[ D_{B[x_ i], \gamma }(JB[x_ i] + (x_ i)) = D_{B, \gamma }(J) \langle x_ i \rangle \]
Proof. One possible proof is to deduce this from Lemma 60.2.4 as any relation between $x_ i$ in $B[x_ i]$ is trivial. On the other hand, the lemma follows from the universal property of the divided power polynomial algebra and the universal property of divided power envelopes. $\square$
Conditions (1) and (2) of the following lemma hold if $B \to B'$ is flat at all primes of $V(IB') \subset \mathop{\mathrm{Spec}}(B')$ and is very closely related to that condition, see Algebra, Lemma 10.99.8. It in particular says that taking the divided power envelope commutes with localization.
Lemma 60.2.6. Let $(A, I, \gamma )$ be a divided power ring. Let $B \to B'$ be a homomorphism of $A$-algebras. Assume that
$B/IB \to B'/IB'$ is flat, and
$\text{Tor}_1^ B(B', B/IB) = 0$.
Then for any ideal $IB \subset J \subset B$ the canonical map
\[ D_ B(J) \otimes _ B B' \longrightarrow D_{B'}(JB') \]
Proof. Set $D = D_ B(J)$ and denote $\bar J \subset D$ its divided power ideal with divided power structure $\bar\gamma $. The universal property of $D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in the lemma. It suffices to show that the divided powers $\bar\gamma $ extend to $D \otimes _ B B'$ since then the universal property of $D_{B'}(JB')$ will produce a map $D_{B'}(JB') \to D \otimes _ B B'$ inverse to the one in the lemma.
Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$. In particular $B \to P$ is flat, hence $D \to D \otimes _ B P$ is flat by Algebra, Lemma 10.39.7. Then $\bar\gamma $ extends to $D \otimes _ B P$ by Divided Power Algebra, Lemma 23.4.2; we will denote this extension $\bar\gamma $ also. Set $\mathfrak a = \mathop{\mathrm{Ker}}(P \to B')$ so that we have the short exact sequence
\[ 0 \to \mathfrak a \to P \to B' \to 0 \]
Thus $\text{Tor}_1^ B(B', B/IB) = 0$ implies that $\mathfrak a \cap IP = I\mathfrak a$. Now we have the following commutative diagram
\[ \xymatrix{ B/J \otimes _ B \mathfrak a \ar[r]_\beta & B/J \otimes _ B P \ar[r] & B/J \otimes _ B B' \\ D \otimes _ B \mathfrak a \ar[r]^\alpha \ar[u] & D \otimes _ B P \ar[r] \ar[u] & D \otimes _ B B' \ar[u] \\ \bar J \otimes _ B \mathfrak a \ar[r] \ar[u] & \bar J \otimes _ B P \ar[r] \ar[u] & \bar J \otimes _ B B' \ar[u] } \]
This diagram is exact even with $0$'s added at the top and the right. We have to show the divided powers on the ideal $\bar J \otimes _ B P$ preserve the ideal $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$, see Divided Power Algebra, Lemma 23.4.3. Consider the exact sequence
\[ 0 \to \mathfrak a/I\mathfrak a \to P/IP \to B'/IB' \to 0 \]
(which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above). As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after applying $B/J \otimes _{B/IB} -$, see Algebra, Lemma 10.39.12. Hence
\[ \mathop{\mathrm{Ker}}(B/J \otimes _{B/IB} \mathfrak a/I\mathfrak a \to B/J \otimes _{B/IB} P/IP) = \mathop{\mathrm{Ker}}(\mathfrak a/J\mathfrak a \to P/JP) \]
is zero. Thus $\beta $ is injective. It follows that $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$ is the image of $\bar J \otimes \mathfrak a$. Now if $f \in \bar J$ and $a \in \mathfrak a$, then $\bar\gamma _ n(f \otimes a) = \bar\gamma _ n(f) \otimes a^ n$ hence the result is clear. $\square$
The following lemma is a special case of [Proposition 2.1.7, dJ-crystalline] which in turn is a generalization of [Proposition 2.8.2, Berthelot].
Lemma 60.2.7. Let $(B, I, \gamma ) \to (B', I', \gamma ')$ be a homomorphism of divided power rings. Let $I \subset J \subset B$ and $I' \subset J' \subset B'$ be ideals. Assume
$B/I \to B'/I'$ is flat, and
$J' = JB' + I'$.
Then the canonical map
\[ D_{B, \gamma }(J) \otimes _ B B' \longrightarrow D_{B', \gamma '}(J') \]
Proof. Set $D = D_{B, \gamma }(J)$. Choose elements $f_ t \in J$ which generate $J/I$. Set $\mathcal{R} = \{ (r_0, r_ t) \in I \oplus \bigoplus \nolimits _{t \in T} B \mid \sum r_ t f_ t = r_0 \text{ in }B\} $ as in the proof of Lemma 60.2.4. This lemma shows that
\[ D = B\langle x_ t \rangle / K \]
where $K$ is generated by the elements $x_ t - f_ t$ and $\delta _ n(\sum r_ t x_ t - r_0)$ for $(r_0, r_ t) \in \mathcal{R}$. Thus we see that
\begin{equation} \label{crystalline-equation-base-change} D \otimes _ B B' = B'\langle x_ t \rangle /K' \end{equation}
where $K'$ is generated by the images in $B'\langle x_ t \rangle $ of the generators of $K$ listed above. Let $f'_ t \in B'$ be the image of $f_ t$. By assumption (1) we see that the elements $f'_ t \in J'$ generate $J'/I'$ and we see that $x_ t - f'_ t \in K'$. Set
\[ \mathcal{R}' = \{ (r'_0, r'_ t) \in I' \oplus \bigoplus \nolimits _{t \in T} B' \mid \sum r'_ t f'_ t = r'_0 \text{ in }B'\} \]
To finish the proof we have to show that $\delta '_ n(\sum r'_ t x_ t - r'_0) \in K'$ for $(r'_0, r'_ t) \in \mathcal{R}'$, because then the presentation (60.2.7.1) of $D \otimes _ B B'$ is identical to the presentation of $D_{B', \gamma '}(J')$ obtain in Lemma 60.2.4 from the generators $f'_ t$. Suppose that $(r'_0, r'_ t) \in \mathcal{R}'$. Then $\sum r'_ t f'_ t = 0$ in $B'/I'$. As $B/I \to B'/I'$ is flat by assumption (1) we can apply the equational criterion of flatness (Algebra, Lemma 10.39.11) to see that there exist an $m > 0$ and $r_{jt} \in B$ and $c_ j \in B'$, $j = 1, \ldots , m$ such that
\[ r_{j0} = \sum \nolimits _ t r_{jt} f_ t \in I \text{ for } j = 1, \ldots , m \]
\[ i'_ t = r'_ t - \sum \nolimits _ j c_ j r_{jt} \in I' \text{ for all }t \]
Note that this also implies that $r'_0 = \sum _ t i'_ t f_ t + \sum _ j c_ j r_{j0}$. Then we have
\begin{align*} \delta '_ n(\sum \nolimits _ t r'_ t x_ t - r'_0) & = \delta '_ n( \sum \nolimits _ t i'_ t x_ t + \sum \nolimits _{t, j} c_ j r_{jt} x_ t - \sum \nolimits _ t i'_ t f_ t - \sum \nolimits _ j c_ j r_{j0}) \\ & = \delta '_ n( \sum \nolimits _ t i'_ t(x_ t - f_ t) + \sum \nolimits _ j c_ j (\sum \nolimits _ t r_{jt} x_ t - r_{j0})) \end{align*}
Since $\delta _ n(a + b) = \sum _{m = 0, \ldots , n} \delta _ m(a) \delta _{n - m}(b)$ and since $\delta _ m(\sum i'_ t(x_ t - f_ t))$ is in the ideal generated by $x_ t - f_ t \in K'$ for $m > 0$, it suffices to prove that $\delta _ n(\sum c_ j (\sum r_{jt} x_ t - r_{j0}))$ is in $K'$. For this we use
\[ \delta _ n(\sum \nolimits _ j c_ j (\sum \nolimits _ t r_{jt} x_ t - r_{j0})) = \sum c_1^{n_1} \ldots c_ m^{n_ m} \delta _{n_1}(\sum r_{1t} x_ t - r_{10}) \ldots \delta _{n_ m}(\sum r_{mt} x_ t - r_{m0}) \]
where the sum is over $n_1 + \ldots + n_ m = n$. This proves what we want. $\square$
Comment #4034 by BAH on March 05, 2019 at 04:14
Hi. Small typo: in the proof of lemma 55.2.4, second line after introducing
\mathcal{R}
, sign "-" instead of sign "=".
Comment #6512 by Matthias Hutzler on August 21, 2021 at 10:14
The universal property of the devided power envelope says that it is a left adjoint to the forgetful functor from
(A, I, \gamma)
-algebras to
(A, I)
-algebras. Is there a reason why this is not mentioned here?
@#6512. Lack of time?
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Swizzling (computer graphics) - Wikipedia
Swizzling (computer graphics)
Vector computation used in computer graphics
Not to be confused with Pointer swizzling.
In computer graphics, swizzling is the ability to compose vectors by arbitrarily rearranging and combining components of other vectors.[1] For example, if A = {1,2,3,4}, where the components are x, y, z, and w respectively, you could compute B = A.wwxy, whereupon B would equal {4,4,1,2}. Additionally, combining two two-component vectors can create a four-component vector, or any combination of vectors and swizzling. This is common in GPGPU applications[example needed].
In terms of linear algebra, this is equivalent to multiplying by a matrix whose rows are standard basis vectors. If
{\displaystyle A=(1,2,3,4)^{T}}
, then swizzling
{\displaystyle A}
as above looks like
{\displaystyle A.wwxy={\begin{bmatrix}0&0&0&1\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}}{\begin{bmatrix}1\\2\\3\\4\end{bmatrix}}={\begin{bmatrix}4\\4\\1\\2\end{bmatrix}}}
^ Lawlor, Orion. "OpenGL ARB_fragment_program Quick Reference ("Cheat Sheet")". University of Alaska Fairbanks. Retrieved 21 January 2014.
OpenGL Vertex Program documentation
Retrieved from "https://en.wikipedia.org/w/index.php?title=Swizzling_(computer_graphics)&oldid=1081237221"
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Finance - Thalesians Wiki
1.3 The leverage effect
Idealized portrait, artist unknown
Josseph de la Vega's (ca. 1650 – Amsterdam, November 13, 1692) Confusion de Confusiones [V88] was published in 1688. It is the oldest book ever written about stock exchanges. It predates Saeed Amen's Trading Thalesians by 326 years!
Written in Spanish by a member of the Sephardi community of Amsterdam in the form of four dialogues between a philosopher, a merchant, and a shareholder, it focusses on the activities of the Amsterdam bourse in the second half of the 17th century.
The author clearly appreciates the importance of (different kinds of) information in the price formation process:
The price of the shares is now 580, [and let us assume that] it seems to me that they will climb to a much higher price because of the extensive cargoes that are expected from India, because of the good business of the Company, of the reputation of its goods, of the prospective dividends, and of the peace in Europe.
The expectation of an event creates a much deeper impression upon the exchange than the event itself. When large dividends or rich imports are expected, shares will rise in price; but if the expectation becomes a reality, the shares often fall; for the joy over the favourable development and the jubilation over a lucky chance have abated in the meantime.
The term volatility is a relatively recent arrival. It originates from the Latin word volatilitas — "readiness to fly swiftly on" (from volo, "to fly") [C92].
We find an unflattering passage describing Louis Napoléon Bonaparte (1808 – 1873) in an 1863 journal [K63]
He was a buyer and seller of those fractional and volatile interests in trading adventures, which go by the name of "Shares," and since it has chanced that the nature of some of his transactions has been brought to light by the public tribunals, it is probable that the kind of repute in which he is held may be owing in part to those disclosures.
In 1915, Wesley Clair Mitchell (1874 – 1948) studied what he referred to as fluctuations in commodity prices [M15]. He was the first person [M63, H08] to empirically show the existence of fat-tailed distributions and time-varying volatility in price data.
In the context of financial markets, the term volatility reappears in 1935 in Harold M. Gartley's (1899 – 1972) early work [G35] on technical analysis.
The first mathematical model for the volatility (which he called le coefficient d'instabilité) of asset prices was proposed earlier and is due to Louis Jean-Baptiste Alphonse Bachelier (1870 – 1946). In his 1900 thesis [B00], he modelled the asset prices as a simple symmetric random walk and, in the limit, Brownian motion.
{\displaystyle dS_{t}=\sigma _{B}\,dW_{t},\qquad S_{0}=s_{0}.}
Fischer Black and Myron Scholes [BS73] model the asset price as a GBM with constant percentage volatility,
{\displaystyle dS_{t}=\nu S_{t}\,dt+\sigma S_{t}\,dW_{t},\qquad S_{0}=s_{0},}
equivalently, they model the log-price
{\displaystyle X:=\ln S}
as a Wiener process with drift
{\displaystyle \nu -{\frac {1}{2}}\sigma ^{2}}
and infinitesimal variance
{\displaystyle \sigma ^{2}}
{\displaystyle dX_{t}=\mu \,dt+\sigma \,dW_{t},\qquad X_{0}=\ln s_{0}.}
One of the "ideal conditions" assumed by them:
The variance rate of the return on the stock is constant.
As mentioned previously, some early evidence for time-varying volatility was provided by Mitchell in 1915 [M15]. See [H08] for details.
In his 1962/3 article [M63], Benoit B. Mandelbrot (1924 – 2010) observed that
...the movement of prices in periods of tranquillity seem to be smoother than predicted by my process. In other words, large changes tend to be followed by large changes — of either sign — and small changes tend to be followed by small changes...
and quoted the earlier empirical observations by Mitchell [M15] and others.
This stylized fact, known as volatility clustering, is one of several (see [C01]) that have motivated the search for more realistic models for volatility dynamics.
Another stylized fact, highlighted in [BS73] is the relation between stock returns and changes in volatility:
I have believed for a long time that stock returns are related to volatility changes. When stocks go up, volatilities seem to go down; and when stocks go down, volatilities seem to go up. The extreme example of this is the depression of the 30's. Stocks went way down, and volatilities went way up.
The term leverage effect refers to thus observed negative correlation between changes in asset prices and changes in volatility. The term takes its origin in the following explanation suggested by Black [BS73]:
A drop in the value of the firm will cause a negative return on its stock, and will usually increase the leverage of the stock.
Whether or not (and to what extent) the leverage effect is due to leverage is debatable, see e.g. [S89]:
While aggregate leverage is significantly correlated with volatility, it explains a relatively small part of the movements in stock volatility. The amplitude of the fluctuations in aggregate stock volatility is difficult to explain using simple models of stock valuation, especially during the Great Depression.
Wiktionary refers to "buy low, sell high" as
Commonplace investment advice, recommending that a prospective investor purchase an asset at a low cost and sell it later for a high price.
Often used in a humorous manner, since this advice is too trite and vague to be helpful in specific situations.
Thales of Miletus (c. 626/623 BC – c. 548/545 BC) is probably the earliest adopeter of this advice. According to Aristotle, Politics 1.1259a,
Although, ibid., Aristotle draws a different conclusion from this story:
Thales then is reported to have thus displayed his wisdom, but as a matter of fact this device of taking an opportunity to secure a monopoly is a universal principle of business; hence even some states have recourse to this plan as a method of raising revenue when short of funds: they introduce a monopoly of marketable goods.
The earliest reference that we could find to the adage "buy low, sell high" in the literature was in the letter ON PRESERVING CORN IN GRANARIES written by a certain J. K. "To the Editor of the Commercial and Agricultural Magazine" for 1801, Vol. IV, from December to June, inclusive, printed and published by Vaughan Griffiths, Paternoster-Row:
[B00] Louis Jean-Baptiste Alphonse Bachelier. Théorie de la spéculation. Annales Scientifiques de l'École Normale Supérieure, 1900, 3, 21-86.
[B76] Fischer Black. Studies of stock price volatility changes. Proceedings of the Business and Economic Statistics Section, American Statistical Association, 1976.
[BS73] Fischer Black and Myron Scholes. The Pricing of Options and Corporate Liabilities. Journal of Political Economy, 1973, 81(3), 637-654.
[C01] Rama Cont. Empirical properties of asset returns: stylized facts and statistical issues. Quantitative Finance, 2001, 1, 223-236.
[C92] George Crabb. English Synonymes Explained in Alphabetical Order. Harper & Brothers, Publishers, 1892.
[G35] Harold M. Gartley. Profits in the Stock Market. Health Research Books, 1935.
[H08] Espen Gaarder Haug. Derivatives: Models on Models. Wiley, 2008.
[K63] Alexander William Kinglake. Napoléon Bonaparte. The Living Age, April 1863, 983.
[M63] Benoit B. Mandelbrot. The variation of certain speculative prices. Journal of Business, 1963, XXXVI, 392-417.
[M15] Wesley Clair Mitchell. The Making and Using of Index Numbers: Introduction to Index Numbers and Wholesale Prices in the United States and Foreign Countries, Bulletins of the U.S. Bureau of Labor Statistics, U.S. Bureau of Labor Statistics, 1915, 173.
[S89] G. William Schwert. Why Does Stock Market Volatility Change Over Time? The Journal of Finance, 1989, XLIV, 1115-1153.
[V88] Josseph de la Vega. Confusion de Confusiones: Dialogos Curiosos Entre un Philosopho agudo, un Mercaderdiscreto, y un Accionista erudito. Amsterdam, 1688.
Retrieved from "https://wiki.thalesians.com/index.php?title=Finance&oldid=395"
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Attenuator basics and the pi attenuator formula
How to use this pi attenuator calculator?
Using the pi attenuator formula to calculate a 40 dB attenuator circuit
Applications of pi attenuators
Beyond pi pad calculator: other helpful electronics tools
With the pi attenuator calculator, you will save the valuable time required in calculating pi attenuator resistor values. You can do it with a couple of slightly complicated formulas (not so funny 🥱) or using our pi pad calculator 😎.
Attenuators play valuable functions in electronics circuits, and you should calculate the required value of each resistor to achieve a particular function in a specific way.
Keep reading this article if you want to know what is a pi attenuator and how to calculate its resistors (using the pi attenuator formula or our pi attenuator calculator). We'll also explore a practical example of a 40 dB attenuator circuit.
In a broad sense, attenuators are electronic devices that reduce the power of a signal through an amplitude reduction caused by some resistors. Attenuators connect to a source on one side and a load on the other.
Although attenuators are relatively simple devices, we can use them to cause a voltage drop, power dissipation, or improve impedance matching. The resistors inside are the essential components and the architects of these tasks, working as energy dissipators.
Within the family of attenuators, we have the pi attenuator, also known as "pi circuit" or "pi pad". It's called this way because its topology resembles the Greek letter "Π." There are two kinds of attenuators within this family:
Pi attenuator with equal impedances.
Pi attenuator with unequal impedances.
Pi attenuator with equal impedances
The equal impedances attenuator is the simplest pi attenuator, as R₁ and R₃ are in parallel and have the same value. Therefore, the source and load impedances are the same.
With the following formula, you can check the results of our calculator:
\quad\ \ R_1=Z_0\left(\frac{K+1}{K-1}\right)
\quad\ \ R_2=Z_0\left(\frac{K^2-1}{2K}\right)
\quad\ \ K=10^{\frac{\text{atten}}{20}}
R_1
R_2
are the resulting resistance values,
Z_0
represents the impedance required,
K
is called the impedance factor, and
\text{atten}
is the desired attenuation.
Pi attenuator with unequal impedances
Are you looking for impedance matching? In that case, the version with unequal impedances is what you need. In this variation, the three resistors have different values, so the input and output impedances are different.
The formula to calculate resistor values is as follows:
R_1=Z_S\left(\frac{K^2-1}{K^2-2K\sqrt{\frac{Z_S}{Z_L}}+1}\right)
R_2=0.5\sqrt{Z_sZ_L}\left(\frac{K^2-1}{K}\right)
R_3=Z_L\left(\frac{K^2-1}{K^2-\frac{2K}{\sqrt{Z_S/Z_L}}+1}\right)
R_S
is the source impedance,
R_L
is the load impedance, and
R_3
stands for the third resistor resistance.
Although the formula to calculate resistances can seem somewhat complicated, our tool is the opposite. This calculator is so straightforward to use that you only need to type the input parameters, and you'll get your results in a flash.
For example, if you want to calculate an equal impedances 40 dB attenuator connected to a 50 Ω source and load, these are the steps:
Select "Equal impedances" in the "Circuit type" box.
Set the attenuation to 40 dB in the second box of the calculator.
In the Z₀ box, type a value of 50 Ω.
And that's all! With the previous inputs, your pi attenuator resistor values should be 51 and 2500 Ω for R₁ and R₂, respectively.
Now, if we wanted to solve an unequal impedances 40 dB attenuator circuit connected to a 75 Ω source and a 50 Ω load, these would be the steps:
Select "Unequal impedances" in the "Circuit type" box.
In the ZS box, type a value of 75 Ω.
In the ZL box, type a value of 50 Ω.
Now you're done with your resistors! The values should be R₁ = 76.9 Ω, R₂ = 3,062 Ω, and R₃ = 50.8 Ω.
We know confidence in a relationship takes time to build up. So, if you still don't trust our pi pad calculator or us, you can still use the attenuator formula and start to strengthen this bond (or simply use it to double-check the calculator results).
Let's use again the example in which we wanted to calculate an equal impedances 40 dB attenuator circuit. If we input the same proposed values (
\text{atten}
= 40 dB and
Z_0
= 50 Ω), the results are:
\scriptsize\ \ \ \ K=10^{\frac{\text{atten}}{20}}=10^{\frac{40\ \text{dB}}{20}}=100
\scriptsize \ \ \ R_1=Z_0\left(\frac{K+1}{K-1}\right)=50\ Ω\left(\frac{100+1}{100-1}\right)
\scriptsize\qquad =50\ Ω\left(\frac{101}{99}\right)=51\ Ω
\scriptsize\ \ \ R_2=Z_0\left(\frac{K^2-1}{2K}\right)=50\ Ω\left(\frac{100^2-1}{2\times100}\right)
\scriptsize\qquad =50\ Ω\left(\frac{9999}{200}\right)
\scriptsize\qquad =2500\ Ω
The same pi attenuator resistor values we obtained with the pi circuit calculator.
The following are some specific applications of these devices:
Signal generation: Under working conditions, the source of a signal is usually encountered "naturally" in the circuit in which we're working. But for test purposes, flexible signal sources are a way to study a system's behavior under different possible conditions. Signal generators can achieve this by modifying amplitude, frequency, and wave shape. Attenuators are a tool that can accomplish the task of amplitude (and therefore power) modification.
Impedance matching: To minimize signal reflection or maximize power transmission in electronics, source and load impedances must be equal (match). When these impedances don't match, you can place an attenuator between them, whose impedance has to match the impedance of the source. In this way, you connect the source to a device of the same impedance, and power transmission reaches its maximum value. Unequal impedance attenuators are essential in this case.
Isolation between circuit stages: Some devices can have problems if they are directly connected one to another. For example, some amplifiers oscillate if their output directly drives a sharp frequency response filter. You can use attenuators to provide isolation between these devices.
We're conscious that circuit design is a complex process in which other devices intervene. For that reason, we have designed other calculators, beyond the pi circuit calculator, that can be very useful to you in this process:
Impedance matching: Use this tool to know the impedance required in your radiofrequency application. Then, come back here and use it in our pi attenuator calculator with the known impedances.
Voltage standing wave ratio (VSWR): Isolation between circuit stages (mentioned above) is a way to minimize VSWR, and in this way, improve impedance matching. After using our pi pad calculator and installing your attenuator, you can use our VSWR calculator to know how your VSWR has improved.
Cable impedance: To check if there's impedance matching, you first have to know the impedance of the elements, one of them usually a transmission line (i.e., a cable). Remember that the cable impedance calculator requires knowing the inner wire's diameter and the outer shielding.
Equal impedances
Impedance (Z₀)
Resistor 1 (R₁)
Resistor 2 (R₂)
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Steradian - Simple English Wikipedia, the free encyclopedia
The steradian (symbol: sr) or square radian is an SI unit. It is like a radian, but for smooth 3D objects, for example, a sphere. A steradian is about one twelfth the surface area of a sphere, regardless of how big the sphere is.[1]
A steradian on a sphere
Sphere vs Steradian[change | change source]
{\displaystyle 4\pi r^{2}}
The surface area of a steradian is just
{\displaystyle r^{2}}
So a sphere measures 4π steradians, or about 12.57 steradians. Likewise a steradian is 1/12.57, or about 8% of a sphere. And because we measure an angle, it doesn't matter what size the sphere is, it will always measure 4π steradians.[2]
↑ Weisstein, Eric W. "Steradian". mathworld.wolfram.com. Retrieved 2020-03-27.
↑ "Steradian". www.mathsisfun.com. Retrieved 2020-03-27.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Steradian&oldid=7387200"
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Summer and Jackie are on a floating platform on a lake and decide to race to the snack bar. Summer is a better swimmer and Jackie is a better runner. They make a bet that whoever loses the race has to buy the snacks. Summer swims at a rate of
12
feet per second and runs at a rate of
20
feet per second. Jackie swims at a rate of
7
30
feet per second. Assuming each takes the optimal path to reach the snack bar, who is buying the snacks? .
In order to optimize the time it takes to reach the snack bar, both girls will swim part of the way and run part of the way. Notice that the running part is horizontal but the swimming part is diagonal.
d(x) =
(distance swimming)
+
(distance running)
=
(hypotenuse of triangle)
+
(horizontal line)
\text{distance} = \left(\text{rate}\right)\left(\text{time}\right)
\text{time}=\frac{d(x)}{\text{rate}}.
Write specific
t(x)
equations for each girl.
Before you apply Calculus to optimize, write a geometric equation describing the distance,
d(x)
, of a generic journey to the snack bar. Let
x
represent the distance on land that each girl is NOT running.
Convert the distance equation into time equations,
t(x)
t^\prime(x) = 0
will give you the optimal value of
x
for each girl. In other words, how far each girl should swim and run. Use that information to calculate the time it takes each girl to reach the snack bar.
Click on the link to the right to view to full version of the eTool. Calc 6-107 HW eTool
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You are looking at all articles with the topic "Computing". We found 363 matches.
{\displaystyle O(b^{d})}
The Aamber Pegasus is a home computer first produced in New Zealand in 1981 by Technosys Research Labs.
The hardware was designed by Stewart J Holmes. The software was designed by Paul Gillingwater, Nigel Keam and Paul Carter.
It is thought that Apple Computers introduction of the Apple II computer into the New Zealand market, and its subsequent heavy educational discounting was the final nail in the coffin for Technosys and the Aamber Pegasus computer. Total production numbers are unknown, but it is thought "around one hundred" were sold.
"Aamber Pegasus" | 2020-01-22 | 18 Upvotes 3 Comments
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Forensic: the science in the crime scenes
What do we calculate with the angle of impact of blood droplets?
How to calculate the angle of impact of a blood spatter
Pulling the right strings: how to use the angle of impact
How to use our angle of impact calculator
DIY forensic: home's Holmes
Learn one of the fundamentals of forensics with our angle of impact calculator: here, helped by a bit of math, you'll discover how detectives recreate one of the elements of a crime scene.
What is the angle of impact;
How to calculate the angle of impact;
Where we use the angle of impact; and
Forensics at home.
Don your gloves and follow the clues with Omni Calculator.
Trials and procedural justice lacked a solid and reliable scientific base for a long time. The best source of information was usually — and in the best case — a witness. More often than not, this resulted in unfair, wrong, or downright malicious outcomes of criminal procedures.
Then, around the time of the Enlightenment science and reason came into the spotlight. The justice system started introducing elements of pathology, physiology, chemistry, and physics in court, alongside "simple" human witnesses.
In an empowering move, numbers and facts made trials fairer. From then on, it was only going to get better: starting from simple anatomical analysis, we ended a few centuries later with a complex and refined set of tools spanning from DNA analysis to computerized tools that allow building a crime scene back from latent elements.
🙋 The road to this outcome was never easy, and many times forensics risked becoming a laughable pseudo-scientific endeavor: consider one of the spearheads of criminology in the 70s and 80s, bite mark analysis. Not even 30 years later, many scientists question its efficacy, leading to many wrongful convictions.
In case the victim suffered blood loss, droplets are one of the most valuable tools in the shed of criminologists, helping them reconstruct the dynamics of the crime.
Crime scenes are often bloody: forensic scientists managed to turn murder into number and developed a framework able to roughly understand the dynamics of a crime through the analysis of the shape of blood droplets.
Blood droplets travel with an almost spherical shape, and then... splat! Droplets land, creating a specific elongated shape (an ellipse), containing information on the angle at which they impact the final surface.
The calculations of the angle of impact of blood spatter followed — and complete — the analysis of the horizontal direction of the droplets. This first step is pretty straightforward: the splatter's elongated shape suggests to the detectives the direction from which the droplets came.
Tracing a set of lines along the major axis of the ellipses of each blood droplet converges in the surrounding of a point: that's the location of the "spill". However, what do we know about the height of the wound?
The location of a wound tells a story: was the victim lying down, sitting, or standing? Maybe even suspended. The height at which a blood drop starts its quick journey to the floor contains crucial information that often helps in the resolution of a case.
That information is encoded in the angle of impact of blood spatter: we define this quantity as the angle between the impact surface and the trajectory of the splatter itself. The angle of impact can vary from
0\degree
(excluded) to
90\degree
(included). Its value dramatically affects the final shape of blood spatters: look at the figure below!
As you can see, an angle of impact equal to
90\degree
corresponds to a (theoretically) circular shape: the blood dripped and was not traveling far from the body.
The lower the angle, the more elongated the shape becomes: eventually, the angle is so low that the shape is only a long smear on the surface.
Together with assessing the horizontal provenience of the droplets, the calculation of the angle of impact composes the broader bloodstain pattern analysis. Let's proceed with our investigation with the formula for the angle of impact of a blood droplet.
Given the great importance of blood pattern analysis, you'll be pleasantly surprised by the innocence of the formula than tells you how to calculate the angle of impact of a blood spatter.
Once you reach the crime scene, you need to whip out a ruler and start measuring the shape of each blood spatter: we are interested in two quantities:
The major axis
L
The minor axis
W
The two elements are straightforward to identify in the typical elliptical shape of the final form of a blood droplet.
The bloodstain may have a tail, an irregular ending that smears the elliptic shape. Don't worry about it, and ignore it in your measurements: this won't be the last approximation leading to the formula for the angle of impact.
Your colleague, detective Perry Masin is here to help. The sine of the angle of impact equals the ratio between the two axes of the ellipse; to calculate the angle itself, we make use of the arcsine function: an inverse trigonometric function (you can learn more about it at our arcsin calculator).
\alpha = \arcsin{\left(\frac{W}{L}\right)}
We told you that the expression was simple!
Measuring the dimensions of blood spatters can give you a clear indication of the angle of impact.
Now, to some sober reality: the calculations for the angle of impact are almost criminally approximated. Many factors, from gravity (we assume that the blood drops travel in a straight line, point-blank, and not following a parabola) to blood viscosity or air resistance, don't enter the formula. It's hard to quantify them, and — at least in part — they probably cancel each other contributions. Still, those approximations affect the final result, introducing errors as big as
50\%
of the actual value.
We now know that the analysis of the angle of impact is approximate. Is it still helpful? Yes!
After calculating angles of impact and horizontal direction, forensic scientists start drawing lines, both:
Following the horizontal direction departing from the center of the stain; and
Rising from the surfaces with the value calculated with the formula for the angle of impact of a blood droplet.
Eventually, those lines more or less converge in a particular area, called area of convergence appropriately. We can infer that the blood originated from that area.
Professionals call this process stringing: it became popular due to the TV show Dexter, but it has comprehensive and valuable use, unlike many other fictional forensic techniques shown on the screen!
You are on the crime scene, and you can see blood spatters everywhere around you. Let's see if we can reconstruct what happened here.
Take three splatters, and measure them. You found:
W_1 = 2.1\ \text{cm}
L_1=2.3\ \text{cm}
for the first one;
W_2 = 1.4\ \text{cm}
L_2 =1.7\ \text{cm}
for the second one; and
W_3 =1.9\ \text{cm}
L_3= 3.1\ \text{cm}
for the third one.
The horizontal analysis suggests that the blood droplets originated from a distance of, respectively:
33\ \text{cm}
(first droplet);
52\ \text{cm}
(second droplet); and
98\ \text{cm}
(third droplet).
Now we have enough data to start looking for an answer. Take the formula for the angle of impact, and feed it the first splatter's measurements:
\footnotesize \begin{align*} \alpha_1 & = \arcsin{\left(\frac{W_1}{L_1}\right)}=\\ &=\arcsin{\left(\frac{2.1\ \text{cm}}{2.3\ \text{cm}}\right)}\\ & =65.93\degree \end{align*}
Considering the angle and the distance, we can infer that the point of origin lies at about
74\ \text{cm}
from the ground (we used the trigonometric equality
33\tan{(65.93\degree)}=73.91\simeq74
for a right triangle: learn how at the right triangle side and angle calculator).
To seek confirmation, let's calculate the angle of impact and then the height of the other droplets. Use our calculator to find the values: insert the width and length of the spatter, using the correct measurement units.
For the angle of impact, we find:
\footnotesize \begin{align*} \alpha_2 & = \arcsin{\left(\frac{1.4\ \text{cm}}{1.7\ \text{cm}}\right)} \\ & =55.44\degree \\ \alpha_3 & = \arcsin{\left(\frac{1.9\ \text{cm}}{3.1\ \text{cm}}\right)}\\ & =37.80\degree \end{align*}
Corresponding to the height calculated with the tangent of these angles, respectively of
75\ \text{cm}
76\ \text{cm}
A representation of the stringing process. By joining "backward" the blood spatters, we can identify the approximate location of the wound.
As you can see, the results of this "mathematical" stringing lead us to a spot a few centimeters wide, about
75\ \text{cm}
high. The victim was not standing, nor were they on the floor. Maybe kneeling?
You can experiment with the angle of impact at your home. Prepare a fluid with a behavior similar to the one of blood, and try to throw some drops on a floor, measure them and calculate the angle of impact!
🔎 Blood is a non-newtonian fluid: it doesn't follow Newton's law of viscosity. Substances of this kind change their viscosity when subjected to a force, either getting thinner or thicker.
Here's a recipe for fake blood. Feel free to modify it until it feels "right". Mix:
4-5 tablespoons of corn starch;
We skipped the coloring because it would stain a lot! Boil the concoction (slowly) and let it simmer for a few minutes. Wait for it to cool down, and eventually add some dish soap to give it a better texture.
Now, start dripping and measuring. Then check if the science works!
You learned that you could reconstruct a small part of a crime scene with a bit of math and a ruler. Isn't it amazing? However, keep in mind the many simplifications introduced to make this forensic tool easy to use. We hadn't been scared of approximating; from ignoring the viscosity to assuming that the droplets travel in a straight line, we hadn't been scared of approximating!
We are glad we could be your accomplices in this short introduction to forensic science. And trust us, our angle of impact calculator is a truthful witness. The case is closed!
In forensic science, the angle of impact is the angle at which a blood droplet impacts a surface. By definition, it is an acute (or right) angle (varying from 0° to 90°). Knowing the angle of impact, detectives can partially reconstruct a crime scene by drawing "strings" from the spatter to an origin area. The position of the victim and other details arise from the calculations.
How do I calculate the angle of impact?
To calculate the angle of impact of a blood spatter:
Measure the width W and the length L of the spatter (they correspond to the minor and major axis of an ellipse);
Ignore eventual "tails": approximate the shape as much as you can;
Compute the arcsine of the ratio between width and length: α = arcsin(W/L).
What is the angle of impact of a spatter 2 cm long and 1 cm wide?
30°. To calculate the angle of impact of an oval-shaped spatter with:
Major axis L = 2 cm; and
Minor axis W = 1 cm;
compute the arcsine of the ratio 1/2:
α = arcsin(W/L) = arcsin(1/2) = 30°.
What is stringing in forensic science?
Once you have calculated the angle of impact and the direction of origin of each blood spatter at a crime scene, you can proceed with the stringing. Pull wires from the center of each spatter, following the directions you already calculated. They will eventually — almost — converge in a specific area: that's where the blood originated from.
Angle of impact (α)
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COVID-19 Outbreak | Toph
COVID-19 outbreak in Atlantis is so bad that the government is forced to lock down the whole country. Athens, the capital of Atlantis is an overpopulated city. The situation here is very severe. People are having a shortage of food as a result of the lockdown. So, all other cities decided to send food to Athens by trucks.
N cities in Atlantis. They are connected through
N - 1
N−1 bi-directed roads. It is possible to reach from any city to another using exactly one path. Cities are numbered from 1 to
N. The city Athens is numbered by an integer
Z. For each i from 1 to
N there are
X_i
Xi food trucks in city
i.
Due to the lockdown, no road is allowed to pass more than
Y trucks. This limit of
Y differs from road to road. You have to find the maximum total number of food trucks Athens can have, if every city sends food trucks in the optimal way.
First line of input contains two integers
N and
Z (
1 \le Z \le N \le 10^6
1≤Z≤N≤106). Each of the next
N - 1
N−1 lines contains three integers
U,
V (
1 \le U, V \le N
1≤U,V≤N) and
Y (
1 \le Y \le 10^6
1≤Y≤106) meaning there is a road between city
U and
V and the capacity of that road is
Y. Next line contains N integers (
0 \le N_i \le 10^6
0≤Ni≤106). The
i_{th}
ith integer indicates the number of food trucks in the city
i.
Print one integer, the maximum possible number of food trucks that Athens can have.
BFS, DFS, DP
dontquitEarliest, Mar '20
rayhan_labibFastest, 0.4s
user.479611Lightest, 86 MB
Run BFS from source Z. Save distance of each node from Z. For each node, set the flow of this node =...
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