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3Blue1Brown - Vectors, what even are they?
Chapter 1Vectors, what even are they?
Text Adaption by River Way
Interpretations of Vectors
"The introduction of numbers as coordinates is an act of violence."
\qquad
— Hermann Weyl
The fundamental building block for linear algebra is the vector, so it’s worth making sure we’re all on the same page about what exactly a vector is. You see, broadly speaking there are three distinct-but-related interpretations of vectors, which I’ll call the physics student perspective, the computer science perspective, and the mathematician’s perspective.
The physics student perspective is that vectors are arrows pointing in space. What defines a given vector is its length and the direction it’s pointing, but as long as those two facts are the same you can move it around and it’s still the same vector.
Vectors that live in a flat plane are two-dimensional, and those sitting in the broader space that you and I live in are three-dimensional.
The computer science perspective is that vectors are ordered lists of numbers. For example, if you were doing some analytics about house prices, and the only features you cared about were square footage and price, you might model each house as a pair of numbers, the first indicating square footage, and the second indicating price.
Notice that order matters here. In the lingo, you’d be modeling houses as two-dimensional vectors, where “vector” is pretty much a fancy word for list, and what makes it two-dimensional is the fact that its length is two.
Mathematician's Abstraction
The mathematician generalizes both of these views, basically saying that a vector can be anything where there’s a sensible notion of adding two vectors and multiplying a vector by a number, operations that I’ll talk about later in this chapter. The details of this view are rather abstract, and I actually think it’s healthy to ignore it until the last video in this series, favoring a more concrete setting in the interim. The reason I bring it up here is that it hints at the fact that the ideas of vector addition and multiplication by numbers will play an important role throughout these topics.
Thinking About Coordinate Systems
Now, while I’m sure many of you are already familiar with coordinate systems, it’s worth walking through them explicitly since this is where all the important back and forth between the two main perspectives of linear algebra happens. Focusing our attention on two dimensions for the moment, you have a horizontal line, called the x-axis, and a vertical line, called the y-axis. The place where they intersect is the origin, which you should think of as the center of space and the root of all vectors.
After choosing an arbitrary distance to represent a length of
1
, you make tick marks on each axis spaced out by this distance.
When I want to convey the idea of 2d space as a whole, which comes up a lot in this text, I’ll extend these tick marks to make grid lines, like so:
Let’s settle on a specific thought to have in mind when I say the word vector. Given the geometric focus I’m shooting for here, whenever I introduce a new topic involving vectors, I want you to first think about an arrow, and specifically think about an arrow inside a coordinate system, like the
xy
-plane, with its tail sitting at the origin.
The coordinates of a vector are a pair of numbers that basically give instructions for how to get from the tail of that vector at the origin, to its tip. The first number tells you how far to walk along the x-axis, with positive numbers indicating rightward motion and negative numbers indicating leftward motion, and the second number tells you how far to then walk parallel to the y-axis, with positive numbers indicating upward motion, and negative numbers indicating downward motion.
To distinguish vectors from points, the convention is to write this pair of numbers vertically with square brackets around them.
\nwarrow\ =\begin{bmatrix}-2 \\ 3\end{bmatrix}\neq(2,3)
As an important note: every pair of numbers gives you one and only one vector, and every vector is associated with one and only one pair of numbers.
Which vector corresponds with walking
6
units up and then
4
units left?
\begin{bmatrix}6 \\ 4\end{bmatrix}
\begin{bmatrix}4 \\ 6\end{bmatrix}
\begin{bmatrix}-4 \\ 6\end{bmatrix}
\begin{bmatrix}-6 \\ -4\end{bmatrix}
In three-dimensions, you add a third axis, called the z-axis, which is perpendicular to both the x and the y axes. In this case, each vector is associated with an ordered triplet of numbers: the first number tells you how far to move along the x-axis, the second number tells you how far to move parallel to the y-axis, and the third number tells you how far to move parallel to the new z-axis. Every triplet of numbers gives you one unique point in space, and every point in space is associated with exactly one triplet of numbers.
So what about vector addition, and multiplying numbers by vectors? After all, every topic in linear algebra centers around these two operations. Luckily, these are both relatively straight-forward.
Let’s say we have two vectors, one pointing up and a little to the right, and another pointing to the right and a little bit down.
To add these two vectors, move the second vector so that it’s tail sits on the tip of the first one. Then if you draw a new vector from the tail of the first one to where the tip of the second now sits, that new vector is their sum.
Why is this a reasonable thing to do? Why this definition of addition and not some other? Well, the way I like to think about it is that each vector represents a certain movement; a step with a certain distance and direction. If you take a step along the first vector, then take a step in the direction and distance described by the second vector, the overall effect is the same as if it just moved along the sum of those two vectors.
You could think of this as an extension of how we think about adding numbers on a number line. One of the ways we teach kids to think about addition, say
2+5
, is to think of moving
2
steps to the right, followed by another
5
steps to the right. The overall effect is the same as if you just took
7
steps to the right to begin with.
In fact, let’s see how vector addition looks numerically. The first vector here has coordinates
\begin{bmatrix}1\\2\end{bmatrix}
, and the second has coordinates
\begin{bmatrix}3\\-1\end{bmatrix}
. When you take their vector sum using this tip to tail method, you can think of a four step path from the tail of the first to the tip of the second: Walk
1
to the right, then
2
up, then
3
1
Reorganizing these steps so that you first do all the rightward motion, then all the vertical motion, you can read it as saying first move
1+3
to the right, then move
2-1
up. So the new vector has coordinates
1+3
2+(-1)
In general, to add two vectors in the list-of-numbers conception of vectors, match up their terms and add them each together.
\begin{bmatrix} \color{green}{x_1} \\ \color{red}{y_1} \end{bmatrix} + \begin{bmatrix} \color{green}{x_2} \\ \color{red}{y_2} \end{bmatrix} = \begin{bmatrix} \color{green}{x_1+x_2} \\ \color{red}{y_1+y_2} \end{bmatrix}
We have two vectors being added together:
\begin{bmatrix}4\\-2\end{bmatrix}+\begin{bmatrix}6\\2\end{bmatrix}
. Describe how to walk from the origin to their sum.
10
units along the positive
x
4
y
10
x
-axis and don't move on
y
Don't move, the sum is at the origin.
Don't move on the
x
-axis and walk
10
y
The other fundamental vector operation is multiplication by a number. This is best understood by just looking at a few examples.
If you take the number
2
, and multiply it by a given vector, you stretch out that vector so that it’s two times as long as when you started.
If you multiply a vector by
\frac13
, you squish it down so that it is one-third its original length.
If you multiply it by a negative number, like
-1.8
, then the vector gets flipped around, then stretched out by a factor of
1.8
This process of stretching, squishing, and sometimes reversing direction, is called “scaling.” Whenever you catch a number like
2
\frac13
-1.8
acting like this, scaling some vector, you call it a “scalar”. In fact, throughout linear algebra, one of the main things numbers do is scale vectors, so it’s common to use the word scalar interchangeably with the word number. Numerically, stretching out a vector by a factor of
2
corresponds with multiplying each of its coordinates by
2
, so in the conception of vectors as lists of numbers, multiplying a given vector by a scalar means multiplying each one of its components by that scalar.
2\overrightarrow{\mathbf{v}}= 2\cdot \begin{bmatrix} \color{green}{x} \\ \color{red}{y} \end{bmatrix} = \begin{bmatrix} 2\color{green}{x} \\ 2\color{red}{y} \end{bmatrix}
You’ll see in the following chapters what I mean when I say pretty much every linear algebra topic revolves around these two fundamental operations of vector addition and scalar multiplication. I’ll also talk more in the last linear algebra chapter about how and why the mathematician thinks only about these operations, independent and abstracted away from however you choose to represent vectors.
In truth, it doesn’t matter whether you think of vectors as fundamentally being arrows in space that happen to have a nice numerical representation, or fundamentally as lists of numbers that happen to have a nice geometric interpretation. The usefulness of linear algebra has less to do with either one of these views than it does with the ability to translate back and forth between them. It gives the data-analyst a nice way to conceptualize many lists of numbers in a visual way, which can seriously clarify patterns in the data and give a global view of what certain operations do.
On the flip side, it gives people like physicists and computer graphics programmers a language to describe space, and the manipulation of space, using numbers that can be crunched and run through a computer.
When I do mathy animations, for example, I start by thinking about what’s going on in space, then get the computer to represent things numerically, and figure out where to place which pixels on the screen, and doing that often relies on an understanding of linear algebra. In the next lesson, we'll start getting into some neat concepts surrounding vectors, like span, bases and linear dependence.
Linear combinations, span, and basis vectors
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Reactive Intermediates | Brilliant Math & Science Wiki
Sravanth C., Skanda Prasad, Anirudh Chandramouli, and
Thomas Neuschatz
Tylan AVIGA
A reactive intermediate is a molecule that is a product in an intermediate step of a chemical reaction. It is typically very energetic: it usually exists only for a short time, as a product of an earlier step in the reaction, and quickly stabilizes. These reactive intermediates provide a basis for understanding how complex reactions are possible.
Interestingly, identifying the presence of these intermediates is not always simple. Unlike typical reactants and products, reactive intermediates typically cannot be isolated and may only be seen using spectrometry or through using experimentation to infer their existence.
Any organic reaction is characterized by the breaking of one or more covalent bonds and the formation of new ones. The reaction is completed successfully if the bonds formed are stronger than the bonds broken. In other words, the reaction is successful if the products are more stable than the reactants. The breaking of bonds can occur through two different methods. The two types of bond cleavage are thrombolytic bond cleavage and electrolytic bond cleavage.
Homolytic cleavage is the cleavage of a covalent bond between two species A and B such that after the cleavage both A and B acquire an unpaired electron. In organic chemistry, homolytic cleavage of a covalent bond containing carbon in a reaction occurs only under some specific conditions. The conditions are listed below:
1) The two atoms bonded must have a small difference in electronegativity: that is, the bond between the two atoms must be non-polar.
2) The reaction mixture should contain free radical initiators. For example, peroxides, non polar mediums (carbon tetrachloride, etc.), light, and extreme temperature are all free radical initiators.
Note: Homolytic cleavage is sure to happen if you see anyone of the following initiators:
=
Heat, E
=
Electricity, L
=
Light, P
=
Peroxide, R
=
Radicals.
In short, HELPR helps the initiating of a homolytic fission.
Heterolytic cleavage is the cleavage of a covalent bond between two species A and B such that after cleavage one of the atoms acquires positive charge, and the other negative charge. In organic chemistry, heterolytic cleavage of a covalent bond containing carbon in a reaction occurs only under some specific conditions. The conditions are listed below:
1) The two atoms bonded must have large change in electro negativity, that is, the bond between the two atoms must be polar.
2) Factors like polar solvents, low temperature, etc. favor heterolytic bond cleavage.
A series of steps involved in the transformation of reactants into products is called reaction mechanism.
_\square
The organic reactions and their mechanisms are classified into three groups:
Carbon compounds bearing a positive charge on carbon and carrying six electrons in its valence shell are called carbocations.
These are formed by heterolytic cleavage of the covalent bonds.
Types of carbocations: Depending on the nature of the carbon bearing the positive charge, carbocations are classified as primary
(1^\circ)
meaning that the carbon is bonded to just one other carbon, secondary
(2^\circ)
meaning that the carbon is attached to two other carbons, and tertiary
(3^\circ)
meaning that the carbon is attached to three different carbons.
Although there are quaternary carbons, there is no possibility of quaternary carbocations. That is because their octet is filled and if there is cleavage between
\ce{C}-\ce{C}
bond, it will turn into a tertiary carbocation, hence ruling out the possibility of quaternary carbocations.
The reaction intermediate formed due to the heterolytic cleavage of a covalent bond such that the electron pair of the bond is taken by the carbon atom, is called a carbanion.
They are important chiefly as chemical intermediates—that is, as substances used in the preparation of other substances. Important industrial products, including useful plastics, are made using carbanions. The simplest carbanion is the methide ion
(\ce{CH3^{-}})
, which is derived from Methane
(\ce{CH4})
by the loss of a proton. Carbanions exist in a trigonal pyramidal geometry.
Similarly, here too are three types of carbanions, depending on the nature of carbon which bears the negative charge.
They are primary, secondary, and tertiary, named so on the number of other carbons they are bonded to.
There are no quaternary carbanions.
Free radicals are reaction intermediates formed due to the homolytic cleavage of a covalent bond containing carbon such that carbon gets an unpaired electron.
In this process each atom takes away one of the two electrons forming a single covalent bond. It will produce two new species having an unpaired electron. This chemical species with only one unpaired electron are called free radicals. Free radicals actively take part in chemical reactions and are highly reactive.
\huge\ce{A-B->A\cdot \ +\ B^\cdot }
\ce{Cl2}
Since both chlorine atoms in
\text{Cl}_2
have equal electro negativity, they will undergo homolytic fission when energy is applied. The reaction will produce
\ce{2Cl\cdot}
which is highly reactive.
_\square
Usually, when reactions are written, and when there is a possibility of homolytic fission, in other words, formation of free radicals, we draw two queer arrows, also called as fish hook arrow. This represents that the bond between the two atoms in question is going to cleave homolytically. This can be observed in the above example where there is a homolytic fission between
\ce{Cl}-\ce{Cl}
Similarly, there are three types of free radicals. They are primary, secondary, and tertiary.
They have been classified as such depending upon the nature of carbon carrying the single odd/unpaired electron.
Cite as: Reactive Intermediates. Brilliant.org. Retrieved from https://brilliant.org/wiki/carbocations-carbanions-free-radicals/
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group(deprecated)/mulperms - Maple Help
Home : Support : Online Help : group(deprecated)/mulperms
multiply two permutations in disjoint cycle notation
mulperms(perm1, perm2)
perm1, perm2
the permutations in disjoint cycle notation
Important: The group package has been deprecated. Use the superseding command GroupTheory[PermProduct] instead.
The product is expressed in disjoint cycle notation.
The command with(group,mulperms) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{group}\right):
\mathrm{mulperms}\left([[2,3,4],[1,6]],[[4,6]]\right)
[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}]]
The following convention is used for the order of the multiplication: The product is the permutation obtained by first applying perm1, and then perm2. Therefore, the result of the following operation will not be [[2,3]] but [[1,3]].
\mathrm{mulperms}\left([[1,2]],[[1,2,3]]\right)
[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]]
GroupTheory[PermutationRepresentation]
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Engineering Acoustics/Sound Absorbing Structures and Materials - Wikibooks, open books for an open world
Engineering Acoustics/Sound Absorbing Structures and Materials
2 Noise Control Mechanisms
3.1 Sound Absorbing Coefficient
3.2 Non-Porous Absorbers ( Absorbing Resonators )
3.3 Porous Absorbers
4 Physical Characteristic Properties of Porous Absorbers
5 Acoustic modeling of Porous Absorbents
5.1 Wave equation in rigid porous absorbents
5.3 Effective Bulk Module
Noise can be defined as unwanted sound. There are many cases and applications that reducing noise level is of great importance. Loss of hearing is only one of the effects of continuous exposure to excessive noise levels. Noise can interfere with sleep and speech, and cause discomfort and other non-auditory effects. Moreover, high level noise and vibration lead to structural failures as well as reduction in life span in many industrial equipments. As an example in control valves, the vibration caused by flow instability occasionally defects the feedback to the control system and resulting in extreme oscillations. The importance of noise issue could be well understood by looking at regulations that have been passed by governments to restrict noise production in society. Industrial machinery, air/surface transportation and construction activities are assumed to be main contributors in noise production or so called "noise pollution".
Noise Control Mechanisms[edit | edit source]
Modifying and canceling sound field by electro-acoustical approaches is called active noise control. There are two methods for active control. First by utilizing the actuators as an acoustic source to produce completely out of phase signals to eliminate the disturbances. second method is to use flexible and vibro-elastic materials to radiate a sound field interfering with the disturbances and minimize the overall intensity. The latter method is called active structural acoustic control (ASAC).
Passive noise control refers to those methods that aim to suppress the sound by modifying the environment close to the source. Since no input power is required in such methods, Passive noise control is often cheaper than active control, however the performance is limited to mid and high frequencies. active control works well for low frequencies hence, the combination of two methods may be utilized for broadband noise reduction.
Figure 1: Noise Control Mechanisms
Sound Absorption[edit | edit source]
Sound waves striking an arbitrary surface are either reflected, transmitted or absorbed; the amount of energy going into reflection, transmission or absorption depends on acoustic properties of the surface. The reflected sound may be almost completely redirected by large flat surfaces or scattered by a diffused surface. When a considerable amount of the reflected sound is spatially and temporally scattered, this status is called a diffuse reflection, and the surface involved is often termed a diffuser. The absorbed sound may either be transmitted or dissipated. A simple schematic of surface-wave interactions is shown in figure 2.
Figure 2: surface-sound interaction- absorption (left), reflection (middle) and diffusing (right)
Sound energy is dissipated by simultaneous actions of viscous and thermal mechanisms. Sound absorbers are used to dissipate sound energy and to minimize its reflection.[1] The absorption coefficient
{\displaystyle \alpha }
is a common quantity used for measuring the sound absorption of a material and is known to be the function of the frequency of the incident wave. It is defined as the ratio of energy absorbed by a material to the energy incident upon its surface.
Sound Absorbing Coefficient[edit | edit source]
The absorbing coefficient can be mathematically presented as follows:
{\displaystyle \alpha =1-{\frac {I_{R}}{I_{I}}}}
where α,
{\displaystyle I_{R}}
{\displaystyle I_{I}}
are the sound absorption coefficient, one-sided intensity of the reflected sound and the one-sided intensity of the incident sound, respectively.
from the above equation, it can be observed that the absorption coefficient of materials varies from 0 to 1. there are several standard methods to measure sound absorption coefficient. In one of the common approaches,a plane wave impedance tube that is equipped with two microphones is utilized.The experimental setup and dimensions are according to ASTM E1050/ISO 10534-2.[2](Figure 3)The method is done by evaluating the transfer function,
{\displaystyle {\hat {h}}(f)}
between two microphones spaced s apart, and a distance l from sample to get absorption coefficient using the following equations:
{\displaystyle {\hat {h}}={\frac {\hat {p_{1}}}{\hat {p_{2}}}}}
{\displaystyle {\hat {r}}={\frac {{\hat {h}}-e^{-jks}}{e^{jks}+{\hat {h}}}}}
{\displaystyle \alpha =1-|{\hat {r}}|^{2}}
{\displaystyle {\hat {p_{1}}},{\hat {p_{2}}}}
are complex pressure amplitude measured by Mic. 1 and Mic. 2 respectively. k is the wave number, s is the microphones spacing and
{\displaystyle \alpha }
is the absorption coefficent.
According to the standard technique,[2] frequency is limited by microphone spacing as well as tube diameter. It is also recommended that
{\displaystyle 0.05{\frac {c}{s}}<f<0.45{\frac {c}{s}}}
to guarantee the plane wave propagation. The coefficient of commercial absorbing materials is specified in terms of a noise reduction coefficient (NRC) which refers to the average of absorption coefficients at 250 Hz, 500 Hz, 1,000 Hz, and 2,000 Hz. Average values of some acoustic insulating materials that are used in buildings are tabulated in table 1. Based on their construction and material structure, sound absorbers are categorized as non-porous and porous absorbers.
Figure 3: Two-microphone method to obtain sound absorption coefficient
Table 1 - Sound absorbing coefficient of common absorbents [3]
6 mm cork sheet 0.1-0.2
6 mm porous rubber sheet 0.1-0.2
12 mm fiberboard on battens 0.3-0.4
50 mm slag wool or glass silk 0.8-0.9
100 mm mineral wool 0.65
Non-Porous Absorbers ( Absorbing Resonators )[edit | edit source]
There are Two types of non-porous absorbers that are common in industrial applications. Panel (membrane) resonators and Helmholtz resonators. Panel absorbers are light, thin and non-porous sheets or membranes that are tuned to absorb sound waves over a specific frequency range. The structural resistance of the panel to fast shaping leads to sound absorption. Panel absorbers are defined by their geometry and structural vibration properties. Helmholtz Resonators or cavity absorbers are perforated structures containing very small pores; one example is the acoustic liners that are used inside the aircraft engine frame to suppress the noise emission from the compression and combustion stages. Similar structures are applied in fans and ventilators used in ventilation and air-conditioning systems. The size of the opening, the length of the neck, and the volume of the cavity govern the resonant frequency of the resonator and hence the absorption performance.
Porous Absorbers[edit | edit source]
Porous sound absorbers correspond to materials where sound propagation takes place in a network of interconnected pores such that viscous and thermal interaction cause acoustic energy to be dissipated and converted to heat. Absorptive treatment such as mineral wool, glass fiber or high-porosity foams reduces reflection sound. Porous absorbers are in fact thermal materials and usually not effective sound barriers. The need for significant thickness compared to operating sound wavelength makes porous absorbers dramatically inefficient and impractical at low frequencies.
Figure 2: Typical variation of sound absorbing coefficient for different absorbers
Physical Characteristic Properties of Porous Absorbers[edit | edit source]
The propagation of sound in a porous material is a phenomenon that governed by physical characteristics of a porous medium, namely porosity (
{\displaystyle \phi }
) , tortuosity (q), flow resistivity (
{\displaystyle \sigma }
), viscous characteristic length (
{\displaystyle \Lambda }
) and thermal characteristic length(
{\displaystyle \Lambda '}
Defined as the ratio of interconnected void volume (air volume in open pores) and the total volume. Most commercial absorbers have high porosity (greater than 0.95). The higher porosity the easier interaction between solid-fluid phases which leads to more sound attenuation.
{\displaystyle \phi ={\frac {V_{0}}{V_{T}}}}
{\displaystyle V_{0}}
= volume of the void space.
{\displaystyle V_{T}}
= total volume of the porous material.
Tortuosity [1]
the physical characteristics corresponds to the “non-straightness” of the pore network inside the porous material. It shows how well the porous material prevents direct flow through the porous medium. The more complex the path, the more time a wave is in contact with the absorbent and hence the more energy dissipation and more absorbing capability. If the Porous absorber is not conductive, one method to measure it is to saturate the absorbent with an electrically conducting fluid and measure the electrical resistivity of the saturated sample,
{\displaystyle R_{s}}
, and compare to the resistivity of the fluid itself,
{\displaystyle R_{f}}
then the tortuosity can be expressed as follows:
{\displaystyle q=\phi {\frac {R_{s}}{R_{f}}}}
The pressure drop required to drive a unit flow through the material can be related to the viscous losses of the propagating sound waves inside the porous absorber and denoted as flow resistivity . For a wide range of porous materials, the flow resistivity is the major factor for sound absorption. The unit of flow resistance is
{\displaystyle N.s/m^{4}}
{\displaystyle Rayls/m}
and defined as the ratio of static pressure drop
{\displaystyle \Delta P}
to a volume flow (U) for a small sample thickness (d).
{\displaystyle \sigma ={\frac {\Delta P}{Ud}}}
Characteristic lengths [4]
Two more important microstructural properties are the characteristic viscous length
{\displaystyle \Lambda }
and characteristic thermal length
{\displaystyle \Lambda }
’ that contribute viscous and thermal dissipation. The former is related to smaller pores and the latter is related to the larger pores of porous aggregate. The thermal length
{\displaystyle \Lambda '}
is the twice ratio of volume to surface area in connected pores. This is geometric and can be measured directly. The viscous length,
{\displaystyle \Lambda }
, is nearly the same, but each integral is weighted by the square of the fluid velocity v inside the pores and hence, cannot be measured directly.
{\displaystyle \Lambda '=2{\frac {\int dV}{\int dS}}}
{\displaystyle \Lambda =2{\frac {\int v_{fluid}^{2}dV}{\int v_{fluid}^{2}dS}}}
Acoustic modeling of Porous Absorbents[edit | edit source]
Wave equation in rigid porous absorbents[edit | edit source]
the plane wave equation derived from the linearized equations of conservation of mass and momentum should be modified to account for the effects of porosity, tortuosity and flow resistance. The modified wave equation[5] that governs the sound propagation in compressible-gas filled in rigid porous materials is given by:
{\displaystyle {\frac {\partial ^{2}P}{\partial x^{2}}}-({\frac {q\rho _{0}}{k_{eff}}})*{\frac {\partial ^{2}P}{\partial t^{2}}}-({\frac {\sigma \phi }{k_{eff}}})*{\frac {\partial p}{\partial t}}=0}
where, p = sound pressure within the pores of material
{\displaystyle \rho _{0}}
= density of compressible gas
{\displaystyle k_{eff}}
= effective bulk modulus of the gas
q = tortuosity
{\displaystyle \phi }
= porosity
{\displaystyle \sigma }
= flow resistivity
The acoustical behavior of a absorptive porous layer can also be investigated from its basic acoustic quantities: the complex wave number and characteristic impedance. These quantities are obtained as a part of solution of the modified plane wave equation and can be used to determine the absorption coefficient and surface impedance. The most practical and common values for the complex wave number and the surface impedance are based on semi-empirical methods and correlated using the regression analysis. One important correlation is suggested by Delany and Bazely [6]
{\displaystyle k'=\alpha +j\beta ={\frac {\omega }{c}}[1+0.0978({\frac {\rho _{0}f}{\sigma }})^{-0.700}-j0.189({\frac {\rho _{0}f}{\sigma }})^{-0.595}]}
{\displaystyle z'=R+jX=\rho _{0}c[1+0.0571({\frac {\rho _{0}f}{\sigma }})^{-0.754}-j0.087({\frac {\rho _{0}f}{\sigma }})^{-0.732}]}
Where, f = frequency σ = flow resistance
Effective Density[edit | edit source]
By assuming rigid-frame pore network in the absorbent, the solid phase would be completely motionless and the frame bulk modulus is considerably greater than that of compressible gas,hence it can be modeled as an effective fluid using the wave equation for a fluid with complex effective fluid density and complex effective bulk modulus. In this situation the dynamic density accounts for the viscous losses and the dynamic bulk modulus for the thermal losses. The effective density relation as a function of dynamic tortuosity was proposed by Johnson et al. [4]
{\displaystyle \rho _{eff}=q(1+{\frac {\sigma \phi }{j\omega \rho _{0}q}}G(\omega ))\rho _{0}}
{\displaystyle G(\omega )={\sqrt {1+j{\frac {4q^{2}\mu \rho _{0}\omega }{\sigma ^{2}\Lambda ^{2}\phi ^{2}}}}}}
μ = gas viscosity
Effective Bulk Module[edit | edit source]
Another factor that affects the sound propagation in the absorbent is the thermal interaction in material due to the heat exchange between the acoustic wave front traveling in the compressible fluid and the solid phase. Champoux and Allard [7]., have introduced a function
{\displaystyle G'(\omega )}
to evaluate effective bulk module for the gas. as it is observed in the following formula this would be the function of thermal characteristic length (
{\displaystyle \Lambda '}
{\displaystyle k_{eff}={\frac {\gamma p_{0}}{\gamma -(\gamma -1)(1-j{\frac {8\mu }{\Lambda '^{2}Pr^{2}\omega \rho _{0}}}G'(\omega ))\rho _{0}}}}
{\displaystyle G'(\omega )={\sqrt {1+j{\frac {\Lambda '^{2}Pr^{2}\omega \rho _{0}}{16\mu }}}}}
γ = gas specific heat ratio (for air ~ 1.4)
Pr = fluid Prantdl number
↑ a b Cox, T. J. and P. D'antonio, Acoustic Absorbers and Diffusers, SponPress,(2004)
↑ a b ASTM E1050 - 08 Standard Test Method for Impedance and Absorption of Acoustical Materials Using A Tube, Two Microphones and A Digital Frequency Analysis System
↑ Link common absorbing materials, absorbing coefficients.
↑ a b Johnson, D.L., Koplik, J., and Dashen, R.,"Theory of dynamic permeability and tortuosity in fluid-saturated porous media," Journal of Fluid Mechanics, vol. 176, 1987, pp. 379-402
↑ Fahy, F., Foundations of Engineering Acoustics, Academic Press London, (2001).
↑ Delany, M.E., and Bazley, E.N., "Acoustical properties of fibrous absorbent materials" Applied Acoustics, vol. 3, 1970, pp. 105-116.
↑ Champoux, Y., and Allard, J.F., "Dynamic tortuosity and bulk modulus in air saturated porous media," Journal of Applied Physics, vol. 70, no. 4, 1991, pp. 1975-1979.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Sound_Absorbing_Structures_and_Materials&oldid=3825345"
Book:Engineering Acoustics
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The following two-way contingency table gives the breakdown of the population of
The following two-way contingency table gives the breakdown of the population of adults in a town according to their highest level of education and whether or not they regularly take vitamins
\begin{array}{|ccc|}\hline \text{Education}& \text{Use of vitamins takes}& \text{Does not take}\\ \text{No High School Diploma}& 0.03& 0.07\\ \text{High School Diploma}& 0.11& 0.39\\ \text{Undergraduate Degree}& 0.09& 0.27\\ \text{Graduate Degree}& 0.02& 0.02\\ \hline\end{array}
\begin{array}{|cccc|}\hline \text{Education}& \text{Use of vitamins takes}& \text{Does not take}& \text{Total}\\ \text{No High School Diploma}& 0.03& 0.07& 0.10\\ \text{High School Diploma}& 0.11& 0.39& 0.50\\ \text{Undergraduate Degree}& 0.09& 0.27& 0.36\\ \text{Graduate Degree}& 0.02& 0.02& 0.04\\ \text{Total}& 0.25& 0.75& 1\\ \hline\end{array}
The probability that the person does not take vitamins regularly is obtained as follows:
P\left(\text{Does not take vitamins regularly}\right)=0.07+0.39+0.27+0.002=0.75
Thus, the probability that the person does not take vitamins regularly is 0.75
You randomly survey students in your school about whether they have a pet. You display your results in the two-way table. How many female students took the survey?
\begin{array}{|cc|}\hline & \text{Gender}\\ & \text{Male}& \text{Female}\\ \text{Pet}\\ \text{Yes}& 33& 35\\ \text{No}& 8& 11\\ \hline\end{array}
The following two-way contingency table gives the breakdown of a town's population according to party affiliation (A, B, C, or None) and opinion on a property tax issue:
\begin{array}{|cccc|}\hline \text{Affiliation}& \text{Favors}& \text{Opposes}& \text{Undecided}\\ \text{A}& 0.12& 0.09& 0.07\\ \text{B}& 0.16& 0.12& 0.14\\ \text{C}& 0.04& 0.03& 0.06\\ \text{None}& 0.08& 0.06& 0.03\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}
A person is selected at random. What is the probability that the person is affiliated with parties A or B?
A table of values for f,g, f', and g' is given.
\begin{array}{|ccccc|}\hline x& f\left(x\right)& g\left(x\right)& {f}^{\prime }\left(x\right)& {g}^{\prime }\left(x\right)\\ 1& 3& 2& 4& 6\\ 2& 1& 8& 5& 7\\ 3& 7& 2& 7& 9\\ \hline\end{array}
a) If h(x)=f(g(x)), find h'(3)
b) If H(x)=g(f(x)),find H'(1).
\begin{array}{ccc}& \text{ Dry }& \text{ Wet }\\ \text{ Cats }& 10& 30\\ \text{ Dogs}& 20& 20\end{array}
Please, tell which type of distribution shows the individual percentages out of the total in a two-way table?
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I’ve struggled with insomnia for all of my adult life. It began in college and has waxed and waned in severity ever since, correlating with stress levels but not entirely.
My form of insomnia starts with an active mind some evening. Maybe it’s active because I’m thinking through a challenge at work. Or I’m replaying a lively dinner conversation. Or I’m on a trip and excited. So, not necessarily stressful thoughts, just engaging ones. I go to bed but my mind doesn’t quiet down. 30 minutes later, I realize I’m not asleep and I start to think, “I really need to get to sleep. I have a lot going on tomorrow.” That only exacerbates the problem. My thoughts turn entirely meta and self defeating, “What’s wrong with me? Ugh, tomorrow will be miserable. Ok, let’s try for real now.” Repeat ad nauseam, laying in bed for hours.
Then, as you might imagine, the next day is indeed terrible and the next evening my mind is primed with negative thoughts about my sleep habits. The cycle continues. At its worst, I would go entirely without sleep for a half a dozen nights out of a month. In the steady state, it’d be one or two nights a month. Needless to say, both the peaks and troughs were quite disruptive to my quality of life and work output.
Over the years I had come to just accept it and cope. I figured I’d been dealt a poor hand in the relevant genetic lottery and this was the cost of doing business. Sometimes I’d use Ambien or other aids to get through it but I was always wary of the side effects and addiction potential.
In early 2016, I got fed up. I was blissfully unemployed and thoroughly de-stressed from having traveled extensively after quitting my job. Yet, I was still having issues sleeping — still going almost entirely sleepless for a couple nights a month. This seemed ridiculous, so I resolved to fix it.
I tried to read up on the latest research about insomnia but I found the volume of academic literature overwhelming for a layman. And it was mostly pharmacologically-oriented which I was intent on avoiding. Outside of journals, the online information on sleep is largely terrible — a rehash of common wisdom or an attempt to sell you something.
So, I sought out a Stanford researcher with a private sleep-therapy practice. After disqualifying me of more serious mental health issues, she ran me through a variant of a program known as CBT-i or “Cognitive Behavioral Therapy for Insomnia”1. It turned out to be extremely effective and I now have a playbook for good sleep even as my stress levels have rebounded.
I’m thrilled to have made this breakthrough. But it was time consuming and expensive to get there. And it struck me how it all boiled down to a fairly simple algorithm albeit one that requires a lot of willpower and would be hard to derive on your own.
I wish someone had told me to try this many years earlier. So I write this in hopes that it’ll help someone out there.
Common Wisdom Invariants
If you have trouble sleeping, I’m sure you’ve heard these all before. In my experience they are necessary but not sufficient.
Wind down an hour before bed. 2
Invest in your bed. Keep your bedroom dark and cool.
Unintuitive New Invariants
The gist here is: “don’t force it and don’t hang out in bed.”
Never get in bed and try to sleep because “it’s bedtime”. Only get in bed when you are dying to go to bed.
If you’re in bed for more than 20 minutes and haven’t fallen asleep yet, get out and do something else. 3
Get out of bed right away when you wake up in the morning. Don’t linger and don’t try to sleep in.
Don’t change your day based on your quality of sleep. No matter how bad it was, don’t deviate from your normal routine — don’t cancel meetings, don’t skip a workout, don’t try to sleep earlier the following evening.
The gist here is: “sleep a lot less before you can sleep more.”
Start out by setting a wake up time that leaves you plenty of time in the morning. For me, this was 6 a.m.
Pick a target sleep time that leaves you with about two hours of sleep less than you think you need. For me, this was midnight.
Treat the target sleep time as a goalpost to get past. That is, don’t sleep, nap, or get in bed before it — no matter what. And make sure to strictly abide by all the invariants above.
When you wake up, jot down the following:
How long you spent awake in the middle of the night
When you got out of bed
At the end of every week, calculate the average of your time spent asleep divided by your time spent in bed each night. Like so 4:
\text{efficiency} = \text{average(}\frac{\text{time asleep}}{\text{time in bed}}\text{)} = \frac{1}{7}\sum_{i=1}^{7}\frac{d_i-b_i-c_i}{e_i-a_i}
\text{efficiency} < .8
, push your target sleep time back even later by 20 minutes.
.9 \leq \text{efficiency} < .95
, push your target sleep time up earlier by 20 minutes.
Go back to step #3. Repeat forever.
Well, it turns out that insomnia is largely a form of performance anxiety that accrues over years of episodic poor sleep due to stress, travel, environmental changes, etc. Your mind gradually loses confidence in the act of going to sleep and jumps quickly to counter-productive thoughts while you lay in bed. If you expect to sleep poorly, you will.
Drastically restricting your sleep with this regimen wrestles control away from your mind and puts it back into the rightful hands of your body and its circadian rhythm5. If you stick to the rules, you’re effectively running a search algorithm to find your body’s optimal schedule.
The algorithm starts by finding an aggregate level of sleep low enough that your body’s need for it overpowers your mind’s obstructionism. Finding this lower bound is painful but it definitely exists. Once you’ve found it, you relax the constraint gradually as your mind builds confidence that it can indeed sleep well and that the bed is a relaxing oasis. 6
Eventually the algorithm converges and you have a schedule that works without any of the bookkeeping. You still need to stick to the invariants so as not to trip up the rhythm you’ve locked in. If you do get knocked off track by a spike in stress or long term travel, just restart the algorithm.
There’s a lot of literature about it. The algorithm described later is a subset of CBT-i called “sleep restriction therapy.” ↩
I expected the therapist to berate me for device usage in the evening but she considered the issue entirely overblown relative to the research. ↩
It’s helpful to have a default activity you can jump to quickly and not leave any room for engaging with daytime thoughts. I either read or watch videos about chess. ↩
There are some apps that will help you keep track of everything. I used one made by the VA. ↩
The modern model of sleep describes two related processes of which circadian rhythm is one. This regimen primarily acts by increasing “sleep pressure” in sleep-wake homeostasis, the other one. This is a good resource from Michigan. ↩
I imagine that getting very good at evening-time meditation would also work but this seems easier and more direct. ↩
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Anisotropic Behavior of Cosmological Models with Exponential and Hyperbolic Scale Factors
In this paper, the cosmological models of the universe are constructed in
f\left(R,T\right)
gravity with choice of the functional
f\left(R,T\right)
{f}_{1}\left(R\right)=\lambda R
{f}_{2}\left(T\right)=\lambda T
. The space-time considered here is Bianchi Type I and the energy momentum tensor is in the form of perfect fluid. Two cosmological models are presented using a power form of exponential function and a hyperbolic form. The energy conditions along with the state finder diagnostic pair have been obtained and analyzed.
f\left(R,T\right)
Cosmology, Hyperbolic Scale Factor, Bianchi Type I, Hubble Parameter
f\left(R,T\right)
f\left(R\right)
{T}_{\mu \nu }
f\left(R,T\right)
S=\frac{1}{16\text{π}G}\int \sqrt{-g}f\left(R,T\right){\text{d}}^{4}x+\int \text{ }{L}_{m}\sqrt{-g}{\text{d}}^{4}x
{\theta }_{\mu \nu }=-2{T}_{\mu \nu }-p{g}_{\mu \nu }
f\left(R,T\right)
{f}_{R}{R}_{\mu \nu }-\frac{1}{2}f\left(R,T\right){g}_{\mu \nu }+\left({g}_{\mu \nu }\square -{\nabla }_{\mu }{\nabla }_{\nu }\right){f}_{R}=8\text{π}{T}_{\mu \nu }-{f}_{T}{T}_{\mu \nu }-{f}_{T}{\theta }_{\mu \nu }
f\left(R,T\right)
{f}_{R}
{f}_{T}
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)
f\left(R,T\right)=\lambda R+\lambda T
\lambda
{R}_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }=\left(\frac{\text{8π}}{\lambda }+1\right){T}_{\mu \nu }+\Lambda {g}_{\mu \nu }
\Lambda =\frac{1}{2}\left(\rho -p\right)
\text{d}{s}^{2}=\text{d}{t}^{2}-{A}^{2}\text{d}{x}^{2}-{B}^{2}\left(\text{d}{y}^{2}+\text{d}{z}^{2}\right)
{T}_{\mu \nu }=\left(p+\rho \right){u}_{\mu }{u}_{\nu }-p{g}_{\mu \nu }
{u}^{\mu }
2\frac{\stackrel{¨}{B}}{B}+\frac{{\stackrel{\dot{}}{B}}^{2}}{{B}^{2}}=\frac{1}{2}\rho -\alpha p
\frac{\stackrel{\dot{}}{A}\stackrel{\dot{}}{B}}{AB}+\frac{\stackrel{¨}{B}}{B}+\frac{\stackrel{¨}{A}}{A}=\frac{1}{2}\rho -\alpha p
2\frac{\stackrel{˙}{A}\stackrel{˙}{B}}{AB}+\frac{{\stackrel{˙}{B}}^{2}}{{B}^{2}}=\alpha \rho -\frac{1}{2}p
\alpha =\frac{\text{8π}}{\lambda }+\frac{3}{2}
{H}_{x}=\frac{\stackrel{˙}{A}}{A}
{H}_{y}=\frac{\stackrel{˙}{B}}{B}=\frac{\stackrel{˙}{C}}{C}
2{\stackrel{˙}{H}}_{y}+3{H}_{y}^{2}=\frac{1}{2}\rho -\alpha p
{H}_{x}{H}_{y}+{\stackrel{˙}{H}}_{y}+{\stackrel{˙}{H}}_{x}+{H}_{x}^{2}+{H}_{y}^{2}=\frac{1}{2}\rho -\alpha p
2{H}_{x}{H}_{y}+{H}_{y}^{2}=\alpha \rho -\frac{1}{2}p
p=\frac{1}{\frac{1}{4}-{\alpha }^{2}}\left[\left(3\alpha -\frac{1}{2}\right){H}_{y}^{2}+2\alpha {\stackrel{˙}{H}}_{y}-{H}_{x}{H}_{y}\right]
\rho =\frac{1}{\frac{1}{4}-{\alpha }^{2}}\left[\left(\frac{3}{2}-\alpha \right){H}_{y}^{2}+{\stackrel{˙}{H}}_{y}-2\alpha {H}_{x}{H}_{y}\right]
\omega =\frac{p}{\rho }=\frac{\left(6\alpha -1\right){H}_{y}^{2}+4\alpha {\stackrel{˙}{H}}_{y}-2{H}_{x}{H}_{y}}{\left(3-2\alpha \right){H}_{y}^{2}+2{\stackrel{˙}{H}}_{y}-4\alpha {H}_{x}{H}_{y}}
\Lambda =\frac{\rho -p}{2}=\frac{2\left(2{H}_{y}^{2}+{\stackrel{˙}{H}}_{y}+{H}_{x}{H}_{y}\right)}{\left(1+2\alpha \right)}
\zeta tanh\left(\eta t\right)
{\left({\text{e}}^{n\zeta t}-1\right)}^{\frac{1}{n}}
\eta
\zeta
\mathcal{R}=\zeta tanh\left(\eta t\right)
H=\frac{\eta }{\mathrm{cosh}\left(\eta t\right)\mathrm{sinh}\left(\eta t\right)}
q=2{\mathrm{sinh}}^{2}\left(\eta t\right)
{H}_{x}=k{H}_{y}
p=\frac{36{\eta }^{2}}{\left(1-4{\alpha }^{2}\right)\left(k+2\right){\mathrm{cosh}}^{2}\left(\eta t\right){\mathrm{sinh}}^{2}\left(\eta t\right)}\left[3\alpha -\frac{1}{2}-\frac{2}{3}\alpha \mathrm{cosh}\left(2\eta t\right)\left(k+2\right)-k\right]
\rho =\frac{36{\eta }^{2}}{\left(1-4{\alpha }^{2}\right)\left(k+2\right){\mathrm{cosh}}^{2}\left(\eta t\right){\mathrm{sinh}}^{2}\left(\eta t\right)}\left[\frac{3}{2}-\alpha -\frac{1}{3}\mathrm{cosh}\left(2\eta t\right)\left(k+2\right)-2\alpha k\right]
\lambda
\left(\eta =0.01,k=0.45,\lambda =0.5\right).
\left(\eta =0.01,k=0.45,\lambda =0.5\right).
w=\frac{p}{\rho }=\frac{3\alpha -\frac{1}{2}-\frac{2}{3}\alpha \mathrm{cosh}\left(2\eta t\right)\left(k+2\right)-k}{\frac{3}{2}-\alpha -\frac{1}{3}\mathrm{cosh}\left(2\eta t\right)\left(k+2\right)-2\alpha k}
\Lambda =\frac{18{\eta }^{2}}{\left(1-4{\alpha }^{2}\right)\left(k+2\right){\mathrm{cosh}}^{2}\left(\eta t\right){\mathrm{sinh}}^{2}\left(\eta t\right)}\left[\left(1-2\alpha \right)+\left(\frac{2}{3}\alpha -3\right)\mathrm{cosh}\left(2\eta t\right)\right]
\approx -0.715
\left(\eta =0.01,k=0.45,\lambda =0.5\right).
\left(\eta =0.01,k=0.45,\lambda =0.5\right).
R={\left({\text{e}}^{n\zeta t}-1\right)}^{\frac{1}{n}}
H=\frac{3n\zeta {\text{e}}^{n\zeta t}}{{\text{e}}^{n\zeta t}-1}
q\left(t\right)=-1+\frac{n}{{\text{e}}^{n\zeta t}}
p=\left[\frac{1}{\frac{1}{4}-{\alpha }^{2}}\right]\left[\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}}{{\left(k+2\right)}^{2}{\left({\text{e}}^{n\zeta t}-1\right)}^{2}}\right]\left[3\alpha -\frac{1}{2}-\frac{2n\alpha }{3}\left(k+2\right){\text{e}}^{-n\zeta t}-k\right]
\rho =\left[\frac{1}{\frac{1}{4}-{\alpha }^{2}}\right]\left[\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}}{{\left(k+2\right)}^{2}{\left({\text{e}}^{n\zeta t}-1\right)}^{2}}\right]\left[\frac{3}{2}-\alpha -\frac{n}{3}\left(k+2\right){\text{e}}^{-n\zeta t}-2\alpha k\right]
\approx -0.8
w=\frac{3\alpha -\frac{1}{2}-\frac{2n\alpha }{3}\left(k+2\right){\text{e}}^{-n\zeta t}-k}{\frac{3}{2}-\alpha -\frac{n}{3}\left(k+2\right){\text{e}}^{-n\zeta t}-2\alpha k}
\Lambda =\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}\left(1-2\alpha \right)\left(1-\frac{n{\text{e}}^{-n\zeta t}}{3}\right)}{2\left(\frac{1}{4}-{\alpha }^{2}\right){\left({\text{e}}^{n\zeta t}-1\right)}^{2}\left(k+2\right)}
\approx -0.95
\left(\zeta =-0.02,n=0.5,k=0.62,\lambda =0.68\right).
\left(\zeta =-0.02,n=0.5,k=0.62,\lambda =0.68\right).
\left(\zeta =-0.02,n=0.5,k=0.62,\lambda =0.68\right).
\left(\zeta =-0.02,n=0.5,k=0.62,\lambda =0.68\right).
\left(r,s\right)
r={\mathrm{sinh}}^{2}\left(\eta t\right)\left(1-2{\mathrm{sinh}}^{2}\left(\eta t\right)\right)
s=\frac{{\mathrm{sinh}}^{2}\left(\eta t\right)\left(1-2{\mathrm{sinh}}^{2}\left(\eta t\right)\right)-1}{6\mathrm{sinh}\left(\eta t\right)-\frac{3}{2}}
r=\frac{{\text{e}}^{n\zeta t}+\left(n-3\right)n+{n}^{2}{\text{e}}^{-n\zeta t}}{{\text{e}}^{2\zeta nt}}
s=\frac{{\text{e}}^{n\zeta t}\left(1+{n}^{2}\right)+{n}^{2}-3n-1}{3\left(\frac{n-{\text{e}}^{n\zeta t}}{{\text{e}}^{n\zeta t}}-\frac{1}{2}\right)}
\left(r,s\right)\approx \left(1,0\right)
\rho +p\ge 0
\rho +p\ge 0
\rho +3p\ge 0
\rho \pm p\ge 0
\begin{array}{c}p+\rho =\frac{9{\eta }^{2}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\mathrm{sinh}}^{2}\left(\eta t\right){\mathrm{cosh}}^{2}\left(\eta t\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \left[\left(1+2\alpha \right)\left(1-k\right)-\left(k+2\right)\left(3+\frac{2}{3}\alpha \right)\mathrm{cosh}\left(2\eta t\right)\right]\end{array}
\begin{array}{c}\rho +3p=\frac{9{\eta }^{2}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\mathrm{sinh}}^{2}\left(\eta t\right){\mathrm{cosh}}^{2}\left(\eta t\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \left[8\alpha -k\left(3+2\alpha \right)-cosh\left(2\eta t\right)\left(k+2\right)\left(3+2\alpha \right)\right]\end{array}
\begin{array}{c}\rho -p=\frac{9{\eta }^{2}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\mathrm{sinh}}^{2}\left(\eta t\right){\mathrm{cosh}}^{2}\left(\eta t\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \left[\left(k+2\right)\left(1-2\alpha -\mathrm{cosh}\left(2\eta t\right)\left(3+\frac{2}{3}\alpha \right)\right)\right]\end{array}
\rho +p=\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\left({\text{e}}^{n\zeta t}-1\right)}^{2}}\left(1+2\alpha \right)\left[1-k-\frac{n}{3}\left(k+2\right){\text{e}}^{-n\zeta t}\right]
\rho +3p=\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\left({\text{e}}^{n\zeta t}-1\right)}^{2}}\left[8\alpha -k\left(3+2\alpha \right)-n\left(k+2\right){\text{e}}^{-n\zeta t}\left(\frac{1}{3}+2\alpha \right)\right]
\rho -p=\frac{9{\zeta }^{2}{\text{e}}^{2n\zeta t}}{\left(\frac{1}{4}-{\alpha }^{2}\right){\left(k+2\right)}^{2}{\left({\text{e}}^{n\zeta t}-1\right)}^{2}}\left(1-2\alpha \right)\left[1-\frac{n}{3}{\text{e}}^{-n\zeta t}\right]
\eta =0.01
\lambda =0.5
n=0.5
\lambda =0.68
\zeta =-0.02
\lambda
\lambda
\theta =\frac{3\eta }{\mathrm{cosh}\left(\eta t\right)\mathrm{sinh}\left(\eta t\right)}
\theta =\frac{3n\zeta {\text{e}}^{n\zeta t}}{{\text{e}}^{n\zeta t}-1}
t\to 0,\theta \to \infty
t\to \infty ,\theta \to 0
\mathcal{A}=4{\left(\frac{k-1}{k+2}\right)}^{2}
\mathcal{A}=2{\left(\frac{k-1}{k+2}\right)}^{2}
k=1
Esmaeili, F.M. (2018) Anisotropic Behavior of Cosmological Models with Exponential and Hyperbolic Scale Factors. Journal of High Energy Physics, Gravitation and Cosmology, 4, 223-235. https://doi.org/10.4236/jhepgc.2018.42017
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Estimate frequency response and spectrum using spectral analysis with frequency-dependent resolution - MATLAB spafdr - MathWorks Australia
y\left(t\right)=G\left(q\right)u\left(t\right)+v\left(t\right)
G\left({e}^{i\omega }\right)
\frac{2\pi }{N{T}_{s}}
\frac{\pi }{{T}_{s}}
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A Processing Workflow for Ocean Bottom Cable Dual-Sensor Acquisition Data: A Case of 3D Seismic Data in Bohai Bay
A Processing Workflow for Ocean Bottom Cable Dual-Sensor Acquisition Data: A Case of 3D Seismic Data in Bohai Bay
The technique of ocean bottom cable (OBC) dual-sensor acquisition is an effective method to suppress the ghost wave and the reverberation at the receiver. With the advent of this technique, the processing method has become the key to the effective use of the OBC dual-sensor data. This paper has developed a new set of processing workflow based on the principle of combining the hydrophone and geophone data. This new process was applied to the OBC data acquired in Bohai area. The actual processing results show that the ghost and the reverberation are attenuated effectively. The frequency energy of the first notch point of the hydrophone data increased from −22 dB to −13 dB, and the frequency energy of the first notch point of the geophone data increased from −18 dB to −10 dB. The spectral characteristics of the dual-sensor data are more reasonable. The frequency spectrum is broadened and richer, and the resolution of the stack profile is improved greatly.
Ocean Bottom Cable (OBC), Ghost, Frequency Spectrum Notch, Reverberation, Sea Bottom Reflection Coefficient, Summing
In seismic exploration, 3D seismic data describe reservoirs more accurately and guide production more effectively. For a long time, the towed streamer acquisition is a main method in 3D marine seismic exploration. As marine exploration improves day by day, there are many obstacles such as production-related platforms, marine traffic, and so on. These obstacles lead to huge gaps which reduce the number of coverage and ultimately affect the imaging results in towed streamer seismic surveys. In this case, the OBC method which employs a stationary array of receiver stations on the ocean bottom and a marine vessel towing only a seismic energy source becomes a natural choice.
There are some problems in the seismic data acquired by ocean bottom cable. It is well known that the sea surface and the seabed are two strong reflection interfaces. So when the seismic wave travels from the ground up to the sea, it continues to reflect downwards to the bottom of the sea, this process repeats. These interferences which disturb effective signal and generate frequency notches are called receiver-side free-surface multiples. These multiples, or “ghosts”, have a worse effect in the shallow sea. As an example, if the water depth is 20 m and the wave speed is 1480 m/s, the wave field recorded by hydrophone receives a first frequency notch at 37 Hz (the first frequency notch of the hydrophone data is equal to the water sonic speed divided by 2 times the water depth) [1] [2] [3] [4] . The notch means that the seismic energy at certain frequency is cancelled by the interference of the multiple. In shallow water, the frequency components of these notches are within the effective frequency band. These interferences significantly reduce the bandwidth and resolution of the data. Obviously, eliminating “ghost” is the key to use the OBC data effectively. In seismic processing, predictive deconvolution is a common method to suppress the marine multiples. Unfortunately, the ghost filter recorded by the hydrophone is mixed phase, and ghost filter recorded by the geophone is minimum phase. Therefore, predictive deconvolution cannot be used to eliminate the ghost.
Based on a simple observation that the ghost waves recorded by hydrophone and geophone have an opposite polarity, geophysicists propose an OBC dual-sensor acquisition technique to eliminate ghost waves. Barr and Sanders (1989) propose that properly combining the hydrophone and geophone data sets can eliminate not only the receiver ghost, but also all water-column reverberations (at the receiver) [5] . Ralph, Sanders, and Starr (1993) prove that the characters of the OBC dual-sensor data are consistent with the characters of streamer data. Basically, the quality of the OBC dual-sensor data is better than that of the streamer data [6] . So far, the OBC dual-sensor acquisition technique has been widely used in marine exploration.
In the OBC dual-sensor processing, Barr and Sanders (1989) places special calibration shots to record the direct arrivals and then analysis the scale factor at each detection point [7] [8] . This method obtains accurate scale factors, but it will add more cost of a survey. Bill Dragoset and Fred J. Barr (1994) derive calibration scale factors from the data directly. This method based on the criterion that the proper scale factors are those that best whiten the summed data [9] . Hoffe (2000) described the wave field characteristics of OBC dual-sensor seismic data and revealed the physical nature of dual-sensor data summing [10] ; Soudani (2006) proposed a three-dimensional OBC combining processing techniques [11] ; Hugonnet (2011) presented a three-step OBC dual-sensor combining processing technique by using the cross ghost wave operator [12] . Zhang (2015) studied the method of OBC seismic data with low SNR [13] . The above research has studied the method of OBC dual-sensor combining in theory or from a specific step, and does not list detailed processing workflow for actual data.
This paper proposes a multi-domain processing workflow for the OBC dual-sensor acquisition data. In order to obtain the best scale factor, this workflow is more careful and detailed in noise attenuation, and the noise is suppressed in multi-domains. We applied the workflow to an OBC dual-sensor data acquired in 2011 from Bohai Bay. The results show that this processing workflow greatly improves the quality of seismic data.
In OBC dual-sensor survey, the mechanism of receiving signals by hydrophone and geophone is different. The signal recorded by the hydrophone is a pressure scalar, and the polarity of a scalar response is independent of the direction of propagation. The geophone detects the velocity of the particles which is a vector response. The polarity of a vector response is related to the direction of propagation. When the seismic wave propagates from the sea surface to the sea floor, it is a down wave signal; when the seismic wave is reflected from the ground layer to the sea bottom, it is an up wave signal. Figure 1 shows that the hydrophone and the geophone receive the same polarity of up-going wave and opposite polarity down-going wave. Summing the dual-sensor recording can only eliminate the down-going wave, some up-going reverberation still exist in the result.
Suppose that the depth of the receiver and the refection coefficient of the seabed are known, after some mathematical operations, the OBC dual-sensor combination can suppress all the reverberation at the receiver (Figure 2). When the primary wave reaches the bottom of the sea from underground, assuming that the refection coefficient of the seabed is r, the primary wave is
x\left(t\right)
, the two-way travel of seismic wave is
\tau
, the wave field recorded by hydrophone is
G\left(t\right)
, the wave field recorded by geophone is
H\left(t\right)
. The seismic wave field recorded by hydrophone and geophone can be expressed as the following formula:
The wave field of hydrophone
\begin{array}{c}H\left(t\right)=\left(1+r\right)x\left(t\right)-{\left(1+r\right)}^{\text{2}}x\left(t-\tau \right)+{\left(1+r\right)}^{2}rx\left(t-2\tau \right)+\cdots \\ =x\left(t\right)-\left(1+r\right)\underset{i=1}{\overset{\infty }{\sum }}{\left(-r\right)}^{i-1}x\left(t-i\tau \right)\end{array}
Figure 1. The propagation schematic of up-going and down-going wave.
Figure 2. The principle of eliminating ghost by combining of the hydrophone and geophone data.
The wave field of geophone
\begin{array}{c}G\left(t\right)=\left(1-r\right)x\left(t\right)+{\left(1-r\right)}^{\text{2}}x\left(t-\tau \right)-{\left(1-r\right)}^{2}rx\left(t-2\tau \right)+\cdots \\ =x\left(t\right)+\left(1-r\right)\underset{i=1}{\overset{\infty }{\sum }}{\left(-r\right)}^{i-1}x\left(t-i\tau \right)\end{array}
If we multiply the geophone by a proportional coefficient
k=\frac{1+r}{1-r}
0<r<1
and add it with the hydrophone data (Formula (3)), we can get a seismic record in which the reverberation at the receiver is eliminated completely.
Sum=H\left(t\right)+\frac{1+r}{1-r}G\left(t\right)=\frac{2}{1-r}x\left(t\right)
The key of summing is the calculation of the accurate reflection coefficient of seabed or the scale factors. Our method of extracting scale factors from seismic data is based on finding scalar that best whiten the summed data. This method needs minimizing the impact of noise as much as possible. Therefore, the noise in OBC data must be suppressed properly before combination processing.
In fact, hydrophone and geophone have different sensitivities to the various kinds of noises that can be present in the ocean-bottom environment. The appropriate method is taken to attenuate the noise between hydrophone and geophone data in the shot domain, common middle point (CMP) domain, common reflection point (CRP) domain, etc. The processing workflow for OBC dual-sensor acquisition data is shown as Figure 3.
This processing workflow was applied to 3D OBC dual-sensor seismic data acquired in the Bohai Bay Region. In this survey, the depth of the water is 2 - 20 m. The source-line spacing and shot-point interval are 100 × 50 m, with receiver-line spacing and station interval 200 × 50 m. The recording time is 8 s, the sample interval is 1 ms.
Figure 4 shows the noise attenuation processing of geophone data in shot domain. Figure 4(a) is a raw shot gather. After noise attenuation, the noise was suppressed greatly and the signal to noise was improved obviously (Figure 4(b)). Figure 4(c) is the noise, from which we can see that effective waves are
Figure 3. The processing workflow for OBC dual-sensor acquisition data.
Figure 4. Geophone shot gathers. Gather before noise attenuation (a); gather after noise attenuation (b); and the noise removed (c); Noise attenuation of hydrophone shot gathers ((d)-(f)).
well protected. Figures 4(d)-(f) shows the noise attenuation processing of hydrophone data in shot domain. The results are as good as that of the geophone data.
Figure 5(a) shows the significant linear interference in the CRP domain which appears as random noise in other domains. This interference makes the calculation of the scale factor unreliable. After noise attenuation (Figure 5(b)), the quality profile has been greatly improved, and the effective waves are clearly visible. Figure 5(c) is the noise eliminated.
Figure 6 shows the comparison of CMP stack between before and after dual-sensor combination. After dual-sensor combination, the event is clearer.
Figure 5. Geophone CRP gathers. Gather before noise attenuation (a); gather after noise attenuation (b); and the noise removed (c).
Figure 6. Comparison of CMP stack between before and after dual-sensor combination. The stack of hydrophone data (a); the stack of geophone data (b); the stack of dual-sensor combination (c).
Figure 7 shows the comparison of spectrum between before and after dual-sensor combination. The notches in the spectrum of hydrophone and geophone data are both caused by reverberation. After the processing of combination, the frequency energy of the first notch point of the hydrophone data increased from −22 dB to −13 dB, and the frequency energy of the first notch point of the geophone data increased from −18 dB to −10 dB. The spectral characteristics of the dual-sensor data are more reasonable.
As shown in Figure 8, before dual-sensor combination, the side lobes energy of both hydrophone and geophone data is very obvious. After the combination processing, the reverberation is eliminated, and the side lobe energy that represents multiple waves is greatly suppressed.
About the dual-sensor OBC data processing, we can come to conclusions:
1) The presence of noise affects the accuracy of the scale factor. So it is necessary to perform noise attenuation in multi-domain.
2) In order to ensure that the opposite polarity noise in the dual-sensor data can exactly cancel each other, each processing step that affects multiple waves in the data should be used with caution. Before combination, the deconvolution processing must not be applied.
3) Dual-sensor OBC combination processing can remove notch, broaden frequency band and enrich the frequency spectrum.
Figure 7. Comparison of spectrum between before and after dual-sensor combination. The spectrum of hydrophone data (a); the spectrum of geophone data (b); the spectrum of dual-sensor combination (c).
Figure 8. Comparison of autocorrelation between before and after dual-sensor combination. The autocorrelation of hydrophone data (a); the autocorrelation of geophone data (b); the autocorrelation of dual-sensor combination (c).
Yang, X.M. and Wang, Y.C. (2019) A Processing Workflow for Ocean Bottom Cable Dual-Sensor Acquisition Data: A Case of 3D Seismic Data in Bohai Bay. Open Journal of Geology, 9, 67-74. https://doi.org/10.4236/ojg.2019.91006
1. Wang, D.K., Tong, S.Y., Liu, H.S. and Zhu, W.Q. (2015) Notch Effect and Frequency Compensation of Dual-Sensor OBC Data in Shallow Water. Journal of Earth Science, 26, 508-514. https://doi.org/10.1007/s12583-015-0559-2
2. Soubaras, R. (1996) Ocean Bottom Hydrophone and Geophone Processing. SEG Technical Program Expanded Abstracts, 1996, 24-27.
3. Gong, X.D. (2014) Influence of Water Depth Error of Detection Point on Combined Processing of OBC Double Inspection Data and Countermeasures. Geophysical Prospecting for Petroleum, 53, 324-329.
4. Zhou, B., Gong, X., Gao, M. and Zhang, J. (2015) An Improvement of The Dual-Sensor Summation Technique for Cross-Ghosting from OBC and Its Application. China Offshore Oil and Gas, 27, 49-52.
5. Barr, F.T. and Paffenholz, J. (1996) The Dual-Sensor Ocean-Bottom Cable Method: Comparative Geophysical Attributes, Quantitative Geophone Coupling Analysis and Other Recent Advances. SEG Technical Program Expanded Abstracts, 1996, 21-23.
6. Ralph, J., Sanders, J.I. and Starr, J. (1993) Extending Offshore Limits with Joint Streamer and Ocean Bottom Cable Operations. SEG Technical Program Expanded Abstracts, 1993, 540-543.
7. Barr, F.J. and Joe, I.S. (1989) Attenuation of Water-Column Reverberations Using Pressure and Velocity Detectors in a Water-Bottom Cable. SEG Technical Program Expanded Abstracts, 1989, 653-656.
8. Barr, F.J. (1997) Dual-Sensor OBC Technology. The Leading Edge, 16, 45-52. https://doi.org/10.1190/1.1437427
9. Dragoset, B., Fred, J. and Barr, F.J. (1994) Ocean-Bottom Cable Dual-Sensor Scaling. SEG Technical Program Expanded Abstracts, 1994, 857-860.
10. Hoffe, B.H., Laurence, R., et al. (2000) Applications of OBC Recording. The Leading Edge, 19, 382-391. https://doi.org/10.1190/1.1438616
11. Soudani, M.T.A, Boelle, J.L. and Hugonnet, P. (2006) 3D Methodology for OBC Pre-Processing. The 68th EAGE Conference & Exhibition Incorporating SPE EUROPEC, Vienna, 12-14 June 2006, B044.
12. Hugonnet, P., Boelle, J.L. and Herrmann, P. (2011) PZ Summation of 3D WAZ OBS Receiver Gathers. The 73rd EAGE Conference & Exhibition Incorporating SPE EUROPEC, Vienna, 23-26 May 2011, B002. https://doi.org/10.3997/2214-4609.20148974
13. Zhang, B.Q., Zhou, H.W., Ding, Z.Y., et al. (2015) Integrated Processing Techniques to Low Signal-to-Noise Ratio OBC Dual-Sensor Seismic Data. The 85th SEG Annual International Meeting, New Orleans, 18-23 October 2015, 2180-2184. https://doi.org/10.1190/segam2015-5860831.1
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GDP Gap Definition
What Is a GDP Gap?
A GDP gap is the difference between the actual gross domestic product (GDP) and the potential GDP of an economy as represented by the long-term trend. A negative GDP gap represents the forfeited output of a country's economy resulting from the failure to create sufficient jobs for all those willing to work. A large positive GDP gap, on the other hand, generally signifies that an economy is overheated and at risk of high inflation.
The difference between real GDP and potential GDP is also known as the output gap.
A GDP gap is represented as the difference between an economy's actual GDP and potential GDP.
Negative GDP gaps are common after economic shocks or financial crises and are reflective of an underperforming economy.
A large positive GDP gap may be a sign that the economy is overheated and poses an inflationary risk.
The term GDP gap is also applied more simply to describe the difference in GDP between two national economies.
Understanding a GDP Gap
A GDP gap can be positive or negative and is calculated as:
(Actual GDP - Potential GDP)/Potential GDP
(ActualGDP−PotentialGDP)/PotentialGDP
From a macroeconomic perspective, you want the smallest possible GDP gap, and preferably no gap at all.
A negative gap shows that an economy is operating at less than its full potential. It's underperforming and essentially leaving money on the table from where it should be trend-wise. Here, production and value are irretrievably lost due to a shortage of employment opportunities.
Negative GDP gaps are common after economic shocks or financial crises. The negative GDP gap, in this case, is mostly a reflection of a hesitant business environment. Companies are unwilling to spend or commit to increased production schedules until stronger signs of a recovery are present. This, in turn, leads to less hiring and perhaps even continued layoffs in all sectors.
That said, a positive GDP gap is also problematic. A large positive GDP gap may be a sign that the economy is overheated and heading toward a correction. The larger the positive GDP gap, the more likely it is that an economy is at risk of a period of high inflation at the very least.
Example of a GDP Gap
According to the Bureau of Economic Analysis (BEA), the actual GDP in the United States for the fourth quarter of 2020 was $20.93 trillion. The Federal Reserve Bank of St. Louis has its own real potential GDP in 2012 dollars. Adjusted to 2020 dollars, it projected a potential GDP of $19.41 trillion.
Running this through the formula—($20.93-$19.41)/$19.41—we get a positive GDP gap of about 0.8%. That is near ideal from the perspective of sustainable economic growth. However, this represents just a moment in time. Policymakers watch the GDP gap closely and make adjustments to try and keep growth in line with the long-term trend.
GDP Gaps Between Nations
In recent years, an increasing amount of attention has been paid to the GDP gap between the United States, the world's largest economy in terms of GDP, and China. In 2020, this GDP gap was estimated to be around $5.9 trillion, which while significant still represents a rapid closing in by China over the last decade.
China has been making up ground since the Great Recession with its huge infrastructure investments and also bounced back quicker than the U.S. from the 2020 economic crisis. Current projections anticipate that China could overtake the U.S. economy in GDP terms by 2028. However, other economists are less convinced, arguing that an aging population and growing debt pile could keep China confined to second place.
Bureau of Economic Analysis. "Gross Domestic Product, 4th Quarter and Year 2020 (Advance Estimate)." Accessed June 3, 2021.
The Federal Reserve Bank of St. Louis. "Real Potential Gross Domestic Product (GDPPOT)." Accessed June 3, 2021.
Bloomberg. "China’s Covid Rebound Edges It Closer to Overtaking U.S. Economy." Accessed June 3, 2021.
A production gap is an economic analytical term denoting the difference between actual industrial production from its perceived potential production.
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Why does \sec(\frac{\pi}{4})=\sqrt{2}?
\mathrm{sec}\left(\frac{\pi }{4}\right)=\sqrt{2}
yotaniwc
\mathrm{sec}\left(\frac{\pi }{4}\right)=\frac{1}{\mathrm{cos}\left(\frac{\pi }{4}\right)}
Trig Table of Special arcs gives -->
\mathrm{cos}\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}
\mathrm{sec}\left(\frac{\pi }{4}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}
{\mathrm{tan}}^{2}\theta -{\mathrm{sin}}^{2}\theta ={\mathrm{tan}}^{2}\theta \mathrm{sin}\theta
\mathrm{tan}\left(\frac{\pi }{3}\right)
\left(\mathrm{tan}\right)\frac{17\pi }{12}
\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{7\pi }{6}\right)\right)
\mathrm{csc}\left(\frac{3\pi }{4}\right)
\frac{\mathrm{sin}x+\mathrm{sin}x\mathrm{tan}x}{1+\mathrm{tan}x}=\mathrm{sin}x
f\left(x\right)={x}^{2},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\le 4\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=m+b,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x>4
|
x=1/\sqrt{3}
C
{x}^{2}+{y}^{2}=1
C
\mathbf{N}\left(1/\sqrt{3}\right)
\mathbf{T}\left(1/\sqrt{3}\right)
\mathbf{R}=\left[\begin{array}{c}x\\ \sqrt{1-{x}^{2}}\end{array}\right]
\mathbf{R}\prime =\left[\begin{array}{c}1\\ -\frac{x}{\sqrt{1-{x}^{2}}}\end{array}\right]
\mathrm{ρ}=1/\sqrt{1-{x}^{2}}
\mathbf{T}=\left[\begin{array}{c}\sqrt{1-{x}^{2}}\\ -x\end{array}\right]
\frac{d\mathbf{T}}{\mathrm{ds}}=\left[\begin{array}{c}- x\\ -\sqrt{1-{x}^{2}}\end{array}\right]
\mathrm{κ}=1
\mathbf{N}=\left[\begin{array}{c}-x\\ -\sqrt{1-{x}^{2}}\end{array}\right]
x=1/\sqrt{3}
\mathbf{T}\left(1/\sqrt{3}\right)=\frac{1}{\sqrt{3}}\left[\begin{array}{c}\sqrt{2}\\ -1\end{array}\right]
\mathbf{N}\left(1/\sqrt{3}\right)=-\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ \sqrt{2}\end{array}\right]
Note that N can be obtained from T by interchanging components and negating the second component to place N to the right of T.
\mathrm{P}:\left(1/\sqrt{3},\sqrt{2/3}\right)
\mathbf{R}\left(1/\sqrt{3}\right)+\mathbf{N}\left(1/\sqrt{3}\right)
\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ \sqrt{2}\end{array}\right]- \frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ \sqrt{2}\end{array}\right]
\left[\begin{array}{c}0\\ 0\end{array}\right]
Hence, the center of curvature is to the right of the point P.
\mathbf{T}\left(1/\sqrt{3}\right)
\mathbf{N}\left(1/\sqrt{3}\right)
R:=PositionVector([x,sqrt(1-x^2)]);
p1:=PlotPositionVector(R,x=-1..1, points=[1/sqrt(3)],normal,tangent, curveoptions=[scaling= constrained,labels=[x,y], size=[300,300]], tangentoptions=[width=.05], normaloptions=[width=.05]);
C
\mathbf{T}\left(1/\sqrt{3}\right),\mathbf{N}\left(1/\sqrt{3}\right)
\mathrm{BasisFormat}\left(\mathrm{false}\right):
C
C
〈x,\sqrt{1-{x}^{2}}〉
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\\ \sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{to position Vector}}{\to }
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\\ \sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{R}
\mathbf{T}\left(1/\sqrt{3}\right)
\mathbf{N}\left(1/\sqrt{3}\right)
x
x=1/\sqrt{3}
x
x=1/\sqrt{3}
\mathbf{R}
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\\ \sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{tangent vector}}{\to }
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{x}}{\sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}}\end{array}\right]
\stackrel{\text{2-normalize}}{\to }
\left[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{1}}{\sqrt{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}}}\\ \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{x}}{\sqrt{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}}\end{array}\right]
\stackrel{\text{evaluate at point}}{\to }
\left[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\\ \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}\end{array}\right]
\mathbf{R}
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\\ \sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{principal normal}}{\to }
\left[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{x}}{\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}}}\\ \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\sqrt{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}}\end{array}\right]
\stackrel{\text{2-normalize}}{\to }
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{-}\sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{evaluate at point}}{\to }
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}\\ \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\end{array}\right]
\mathbf{T}\left(1/\sqrt{3}\right)
\mathbf{N}\left(1/\sqrt{3}\right)
x=1/\sqrt{3},y=\sqrt{2/3}
x∈\left[-1,1\right]
C
\left[\begin{array}{c}\frac{1}{3}\sqrt{3}\sqrt{2}\\ -\frac{1}{3}\sqrt{3}\end{array}\right]
\stackrel{\text{plot arrow}}{\to }
\left[\begin{array}{c}-\frac{1}{3}\sqrt{3}\\ -\frac{1}{3}\sqrt{3}\sqrt{2}\end{array}\right]
\stackrel{\text{plot arrow}}{\to }
\mathbf{R}
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\\ \sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\end{array}\right]
\stackrel{\text{to list}}{\to }
\left[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\sqrt{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\right]
\to
\mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right):
\mathrm{BasisFormat}\left(\mathrm{false}\right):
C
\mathbf{R}≔\mathrm{PositionVector}\left(\left[x,\sqrt{1-{x}^{2}}\right]\right):
x=1/\sqrt{3}
\mathbf{N}≔\mathrm{eval}\left(\mathrm{PrincipalNormal}\left(\mathbf{R},\mathrm{normalized}\right),x=1/\sqrt{3}\right):
C
x=1/\sqrt{3}
\mathbf{T}≔\mathrm{eval}\left(\mathrm{TangentVector}\left(\mathbf{R},\mathrm{normalized}\right),x=1/\sqrt{3}\right):
C
x=1/\sqrt{3}
\mathrm{PlotPositionVector}\left(\mathbf{R},x=-1..1,\mathrm{points}=\left[1/\sqrt{3}\right],\mathrm{normal},\mathrm{tangent},\mathrm{curveoptions}=\left[\mathrm{scaling}=\mathrm{constrained},\mathrm{labels}=\left[x,y\right],\mathrm{size}=\left[300,300\right]\right],\mathrm{tangentoptions}=\left[\mathrm{width}=.05\right],\mathrm{normaloptions}=\left[\mathrm{width}=.05\right]\right)
The principal normal indeed points towards the center of curvature. The components of N could be obtained by interchanging the components of T and negating the second component so that N points to the right of T.
|
Thermodynamic Analysis of the Ceria Redox Cycle With Methane-Driven Reduction for Solar Fuel Production | ES | ASME Digital Collection
Peter Krenzke,
Krenzke, P, & Davidson, J. "Thermodynamic Analysis of the Ceria Redox Cycle With Methane-Driven Reduction for Solar Fuel Production." Proceedings of the ASME 2014 8th International Conference on Energy Sustainability collocated with the ASME 2014 12th International Conference on Fuel Cell Science, Engineering and Technology. Volume 1: Combined Energy Cycles, CHP, CCHP, and Smart Grids; Concentrating Solar Power, Solar Thermochemistry and Thermal Energy Storage; Geothermal, Ocean, and Emerging Energy Technologies; Hydrogen Energy Technologies; Low/Zero Emission Power Plants and Carbon Sequestration; Photovoltaics; Wind Energy Systems and Technologies. Boston, Massachusetts, USA. June 30–July 2, 2014. V001T02A004. ASME. https://doi.org/10.1115/ES2014-6332
The nonstoichiometric cerium oxide (ceria) redox cycle is an attractive pathway for storing energy from concentrated sunlight in chemical bonds by splitting water and carbon dioxide. The endothermic reduction reaction
CeO2-δox→CeO2-δed+Δδ2O2
is favored thermodynamically at high temperatures and low oxygen partial pressures, while the CO2 and H2O splitting reactions (R2, R3) are exothermic and favored at lower temperatures and higher oxygen partial pressures. The produced hydrogen and carbon monoxide, referred to collectively as syngas, are important feedstocks used in the synthesis of ammonia and liquid fuels.
CeO2-δed+ΔδCO2→CeO2-δox+ΔδCO
CeO2-δed+ΔδH2O→CeO2-δox+ΔδH2
Cycles, Fuels, Methane, Solar energy, Carbon dioxide, Water, Oxygen, Carbon, Feedstock, High temperature, Hydrogen, Sunlight, Syngas, Temperature
Syngas and Hydrogen Production by Cyclic Redox of ZrO 2 -Supported CeO 2 in a Volumetric Receiver-Reactor
|
Evaluate the following derivatives. \frac{d}{dx}\int_{3}^{e^{x}}\cos t^{2}dt
\frac{d}{dx}{\int }_{3}^{{e}^{x}}{\mathrm{cos}t}^{2}dt
To evaluate the derivative,
\frac{d}{dx}{\int }_{3}^{{e}^{x}}{\mathrm{cos}t}^{2}dt
According to the Lebnitz Rule of differentiation under the sign of integration,
\frac{d}{dx}{\int }_{g\left(x\right)}^{h\left(x\right)}f\left(t\right)dt=f\left(h\left(x\right)\right)\frac{dh\left(x\right)}{dx}-f\left(g\left(x\right)\right)\frac{dg\left(x\right)}{dx}
Let us apply the above rule,
{\int }_{3}^{{e}^{x}}{\mathrm{cos}t}^{2}dt={\mathrm{cos}\left({e}^{x}\right)}^{2}\frac{d\left({e}^{x}\right)}{dx}-{\mathrm{cos}\left(3\right)}^{2}\frac{d\left(3\right)}{dx}
⇒{\int }_{3}^{{e}^{x}}{\mathrm{cos}t}^{2}dt={\mathrm{cos}e}^{2x}\left({e}^{x}\right)-\mathrm{cos}9\left(0\right)
\therefore {\int }_{3}^{{e}^{x}}{\mathrm{cos}t}^{2}dt={e}^{x}\mathrm{cos}\left({e}^{2x}\right)
Derivatives Find and simplify the derivative of the following functions
y=\left(2\sqrt{x}-1\right){\left(4x+1\right)}^{-1}
Take the derivative of
\frac{2+2x+{x}^{2}}{\mathrm{ln}\left(x\right)}
y=\mathrm{ln}\left(\mathrm{ln}x\right)
f\left(x\right)=\frac{{2}^{x}}{{2}^{x}+1}
f\left(x\right)={\mathrm{sin}}^{-1}\left({e}^{\mathrm{sin}x}\right)
|
Energy Level and Transition of Electrons | Brilliant Math & Science Wiki
Transition of an Electron and Spectral Lines
In this section we will discuss the energy level of the electron of a hydrogen atom, and how it changes as the electron undergoes transition. According to Bohr's theory, electrons of an atom revolve around the nucleus on certain orbits, or electron shells. Each orbit has its specific energy level, which is expressed as a negative value. This is because the electrons on the orbit are "captured" by the nucleus via electrostatic forces, and impedes the freedom of the electron. The orbits closer to the nucleus have lower energy levels because they interact more with the nucleus, and vice versa. Bohr named the orbits as
\text{K }(n=1), \text{L }(n=2), \text{M }(n=3), \text{N }(n=4), \text{O }(n=5), \cdots
in order of increasing distance from the nucleus. Note that
n
refers to the principal quantum number.
The energy of the electron of a monoelectronic atom depends only on which shell the electron orbits in. The energy level of the electron of a hydrogen atom is given by the following formula, where
n
denotes the principal quantum number:
E_n=-\frac{1312}{n^2}\text{ kJ/mol}.
For a single electron instead of per mole, the formula in eV (electron volts) is also widely used:
E_n=-\frac{13.6}{n^2}\text{ eV}.
Observe that the energy level is always negative, and increases as
n.
n
can only take on positive integers, the energy level of the electron can only take on specific values such as
E_1=-13.6\text{ eV},
E_2=-3.39\text{ eV},
E_3=-1.51\text{ eV}, \cdots
and so on. Thus, we can say that the energy level of an electron is quantized, rather than continuous. The figure below shows the electron energy level diagram of a hydrogen atom. Observe how the lines become closer as
n
For atoms other than hydrogen, we simply multiply
-\frac{1312}{n^2}\text{ kJ/mol}
-\frac{13.6}{n^2}\text{ eV}
Z_{\text{eff}}^2,
Z_{\text{eff}}
refers to the effective nuclear charge. Keep in mind that this rule can only be applied to monatomic atoms (or ions) such as
\ce{H}, \ce{He+}, \ce{Li}^{2+}.
Find the ionization energy of hydrogen.
Ionization energy is the energy needed to take away an electron from an atom. It is equivalent to the energy needed to excite an electron from
n=1
(ground state) to
n=\infty,
E_{\infty}-E_1=1312\text{ kJ/mol},
E_{\infty}-E_1=13.6\text{ eV}.\ _\square
In chemistry, energy is a measure of how stable a substance is. The lower the energy level of an electron, the more stable the electron is. Thus an electron would be in its most stable state when it is in the K shell
(n=1).
For this reason, we refer to
n=1
as the ground state of the electron. If the electron is in any other shell, we say that the electron is in excited state.
It is quite obvious that an electron at ground state must gain energy in order to become excited. Likewise, an electron at a higher energy level releases energy as it falls down to a lower energy level. Using the formula above, we can calculate how much energy is absorbed/released during the transition of an electron. The energy change during the transition of an electron from
n=n_1
n=n_2
\Delta E=E_{2}-E_{1}=13.6\times\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ eV}.
Obviously, a positive energy change means that the electron absorbs energy, while a negative energy change implies a release of energy from the electron. Note that the formula is the energy per mole, rather than that of a single photon.
During transition, an electron absorbs/releases energy is in the form of light energy. The energy of the photon
E
absorbed/released during the transition is equal to the energy change
\Delta E
of the electron. Using the properties of DeBroglie waves, we can calculate the wavelength and frequency of the following formula:
E=h\nu=h\frac{c}{\lambda},
h=6.63\times10^{-34}\text{ J}\cdot\text{s}
denotes Planck's constant,
\nu
denotes frequency,
\lambda
denotes wavelength, and
c=3.00\times10^8\text{ m/s}
denotes the speed of light. Combining this formula with the
\Delta E
formula above gives the famous Rydberg formula:
\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ m}^{-1},
R=1.097\times10^7\text{ m}^{-1}
is the Rydberg constant. Using the Rydberg formula, we can compute the wavelength of the light the electron absorbs/releases, which ranges from ultraviolet to infrared.
Because the value of
\frac{1}{n^2}
substantially decreases as
n
increases, the value of the energy change or wavelength depends on the smaller between
n_1
n_2.
For this reason, the light emission by the fall of the energy level of an electron can be categorized into several groups. If an electron falls from any
n\ge2
n=1,
then the wavelength calculated using the Rydberg formula gives values ranging from 91 nm to 121 nm, which all fall under the domain of ultraviolet. As this was discovered by a scientist named Theodore Lyman, this kind of electron transition is referred to as the Lyman series. Similarly, any electron transition from
n\ge3
n=2
emits visible light, and is known as the Balmer series. Electron transition from
n\ge4
n=3
gives infrared, and this is referred to as the Paschen series.
Since the energy level of the electron of a hydrogen atom is quantized instead of continuous, the spectrum of the lights emitted by the electron via transition is also quantized. In other words, the wavelength
\lambda
can only take on specific values since
n_1
n_2
are integers. As a result, the electron transition gives spectral lines as shown in the right figure below (showing only visible light, or Balmer series). Note how this differs to the continuous spectrum shown in the left figure below. Running sunlight through a prism would give a continuous spectrum.
When analyzing spectral lines, we must approach them from the right side. This is because the lines become closer and closer as the wavelength decreases within a series, and it is harder to tell them apart. The line with the longest wavelength within a series corresponds to the electron transition with the lowest energy within that series. Hence in the figure above, the red line indicates the transition from
n=3
n=2,
which is the transition with the lowest energy within the Balmer series.
Recall that the energy level of the electron of an atom other than hydrogen was given by
E_n=-\frac{1312}{n^2}\cdot Z_{\text{eff}}^2\text{ kJ/mol}.
Since each element has a unique
Z_{\text{eff}}
value, the spectral lines of each element would be different. Therefore spectral lines can be thought of the "fingerprints" of an element, and be used to identify an element.
The figure above shows the spectrum of Balmer series. Which of the following electron transitions corresponds to the turquoise line
(\lambda\approx485\text{ nm})
in the figure above?
n=2\rightarrow n=1
n=3\rightarrow n=1
n=3\rightarrow n=2
n=4\rightarrow n=2
Observe that the red line has the longest wavelength within the Balmer series. Since a longer wavelength means smaller energy, the red line correspond to the transition which emits the lowest energy within the Balmer series, which is
n=3\rightarrow n=2.
The turquoise line indicates the transition with the second lowest energy within the Balmer series, which is
n=4\rightarrow n=2.
Therefore our answer is (D).
_\square
Cite as: Energy Level and Transition of Electrons. Brilliant.org. Retrieved from https://brilliant.org/wiki/energy-level-and-transition-of-electrons/
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Chaos Temple (hut) - OSRS Wiki
The RuneScape Wiki also has an article on: rsw:Chaos Temple (hut)The RuneScape Classic Wiki also has an article on: classicrsw:Evil Altar
This article is about the temple in level 38 Wilderness. For other uses, see Chaos Temple.
Ice Path ← Chaos Temple (hut) → Lava Maze
The Chaos Temple is a structure in level 38 Wilderness, west of the Lava Maze. It contains a Chaos Altar where players can recharge and train their Prayer using bones on the altar (demonic ashes do not work). Free players may visit the temple, however only members may take advantage of the altar's benefits.
Praying at the Chaos Altar is a requirement for the Easy Wilderness Diary.
3 Chaos Altar
The easiest way to get to the Chaos Altar is to use a burning amulet's teleport to Lava Maze entrance and run south-west. However, the spot can be often crowded by player killers due to its proximity to Revenant Caves and King Black Dragon Lair, so an alternative and likely a safer way to get there is to use Ghorrock Teleport on Ancient Magicks and run south, or Cemetery Teleport on the Arceuus spellbook and run north. Alternatively, players who have an Obelisk in their player-owned house can use it to teleport to level 44 Wilderness and then run south. Otherwise, this can be done from ANY Obelisk if the Hard Wilderness Diary is complete.
The Chaos Temple hut in Level 38 Wilderness.
A wine of Zamorak spawns on a table inside the hut, which can only be obtained by casting Telekinetic Grab. Players who have completed the hard Wilderness Diary will receive them in noted form. The wine respawns approximately every 28 seconds.
Players who kill the Chaos Fanatic may use the Chaos Altar to recharge their Prayer as it is very close.
Due to the altar's use for Prayer training while being in multi-combat, it is frequently targeted by player killers.
There is also the Elder Chaos druid who can unnote any type of bones for a fee of 50 coins per bone.
Members can offer bones on the Chaos Altar, granting 3.5x Prayer experience per bone, the same bonus as a gilded altar with two burners lit. However, every bone offered on the altar has a 50% chance to not be consumed, which can repeat for the same bone repeatedly. Therefore, the average experience per bone is 2x the amount from a player-owned house gilded altar with both incense burners lit, or 7x the normal burying experience.
The 5% Prayer experience bonus of the Zealot's robes stacks multiplicatively with the Chaos Altar's bonus, yielding a 52.5% chance to not consume a bone, meaning they can effectively be buried an average of 2.1 times.
The Elder Chaos druid outside the temple will unnote bones for 50 coins each.
For more information on this method of training Prayer, see here.
Mathematical Proof[edit | edit source]
Geometric series representation of bones gained.
The 50% chance when offering a bone on this altar to not be consumed means they can be offered again, granting additional prayer experience and another 50% chance of not being consumed. This repeating series results in a 100% gain in experience. A proof demonstrating this is below:
{\displaystyle x}
number of bones, we wish to solve for
{\textstyle S}
, the number of bones you can offer to the altar. At a gilded altar,
{\textstyle S=x}
because you can only offer each of the
{\textstyle x}
bones a single time. At this altar however, after offering the first
{\textstyle x}
bones, 50% (or
{\textstyle 1/2}
) will be left remaining to be offered again, which is
{\textstyle {\frac {1}{2}}x}
bones. Those bones can be offered again, which will leave us with
{\textstyle {\frac {1}{2}}({\frac {1}{2}}x)={\frac {1}{4}}x}
remaining. This can be repeated infinitely many times, and the total number of bones offered to the altar is
{\textstyle S(x)=x+{\frac {1}{2}}x+{\frac {1}{4}}x+{\frac {1}{8}}x+{\frac {1}{16}}x...}
. Factoring out the x, we have
{\textstyle S(x)=x(1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}...)}
. The part in the parentheses is a well-known geometric sum that converges to 2, so
{\textstyle S(x)=x\sum _{n=0}^{\infty }2^{-n}=2x}
given an infinite amount of time and ideal circumstances (never dying and losing bones, for instance). For example, if a player offers 1000 bones to this altar and continues until they run out, they should expect to receive as much experience as offering 2000 of the same type of bone to a gilded altar with two burners lit.
An alternative proof observes that the average number of offerings can be thought of as the expected value of a negative binomial distribution formulated as the number of trials until a certain number of successes. Letting
{\displaystyle x}
be the number of bones a player has to offer, the expected value of this distribution is
{\displaystyle {\frac {x}{p}}}
{\displaystyle p=0.5}
is the probability that a bone will be consumed for a given offering. Therefore, the expected number of offerings for
{\displaystyle x}
bones is
{\displaystyle 2x}
offerings, concordant with the previous result.
The 5% Prayer experience bonus of the Zealot's robes stacks multiplicatively with the Chaos Altar's bonus, yielding a 52.5% chance to not consume a bone. Using the above formulas, each bone can be used an average of 2.1 times, meaning only 95% as many bones are needed compared to using the Chaos Altar without the set, and 47.5% as many bones using a normal altar.
The Prayer training benefits were added.
The wine of zamorak spawn was added.
Retrieved from ‘https://oldschool.runescape.wiki/w/Chaos_Temple_(hut)?oldid=14281978’
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Canuto, B. ; Rosset, Edi ; Vessella, S.
We prove a logarithmic stability estimate for a parabolic inverse problem concerning the localization of unknown cavities in a thermic conducting medium
\Omega
{ℝ}^{n}
n\ge 2
, from a single pair of boundary measurements of temperature and thermal flux.
Mots clés : parabolic equations, strong unique continuation, stability, inverse problems
author = {Canuto, B. and Rosset, Edi and Vessella, S.},
title = {A stability result in the localization of cavities in a thermic conducting medium},
AU - Canuto, B.
AU - Rosset, Edi
AU - Vessella, S.
TI - A stability result in the localization of cavities in a thermic conducting medium
Canuto, B.; Rosset, Edi; Vessella, S. A stability result in the localization of cavities in a thermic conducting medium. ESAIM: Control, Optimisation and Calculus of Variations, Tome 7 (2002), pp. 521-565. doi : 10.1051/cocv:2002066. http://www.numdam.org/articles/10.1051/cocv:2002066/
[1] V. Adolfsson and L. Escauriaza,
{C}^{1,\alpha }
domains and unique continuation at the boundary. Comm. Pure Appl. Math. L (1997) 935-969. | MR 1466583 | Zbl 0899.31004
[2] G. Alessandrini and E. Rosset, The inverse conductivity problem with one measurement: Bounds on the size of the unknown object. SIAM J. Appl. Math. 58 (1998) 1060-1071. | MR 1620386 | Zbl 0953.35141
[3] G. Alessandrini and L. Rondi, Optimal stability for the inverse problem of multiple cavities. J. Differential Equations 176 (2001) 356-386. | MR 1866280 | Zbl 0988.35163
[4] G. Alessandrini, E. Beretta, E. Rosset and S. Vessella, Optimal stability for inverse elliptic boundary value problems with unknown boundaries. Ann. Scuola Norm. Sup. Pisa Cl. Sci (4) XXIX (2000) 755-806. | Numdam | MR 1822407 | Zbl 1034.35148
[5] K. Bryan and L.F. Candill Jr., An inverse problem in thermal imaging. SIAM J. Appl. Math. 56 (1996) 715-735. | MR 1389750 | Zbl 0854.35125
[6] K. Bryan and L.F. Candill Jr., Uniqueness for boundary identification problem in thermal imaging, in Differential Equations and Computational Simulations III, edited by J. Graef, R. Shivaji, B. Soni and J. Zhu.
[7] K. Bryan and L.F. Candill Jr., Stability and reconstruction for an inverse problem for the heat equation. Inverse Problems 14 (1998) 1429-1453. | MR 1662444 | Zbl 0914.35152
[8] B. Canuto, E. Rosset and S. Vessella, Quantitative estimates of unique continuation for parabolic equations and inverse initial-boundary value problems with unknown boundaries. Trans. AMS 354 (2002) 491-535. | MR 1862557 | Zbl 0992.35112
[9] R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. 1. Wiley, New York (1953). | Zbl 0051.28802
[10] D. Gilbarg and N.S. Trudinger, Elliptic partial differential equations of second order. Springer, New York (1983). | MR 737190 | Zbl 0562.35001
[11] V. Isakov, Inverse problems for partial differential equations. Springer, New York (1998). | MR 1482521 | Zbl 0908.35134
[12] S. Ito and H. Yamabe, A unique continuation theorem for solutions of a parabolic differential equation. J. Math. Soc. Japan 10 (1958) 314-321. | MR 99519 | Zbl 0088.30403
[13] O.A. Ladyzhenskaja, V.A. Solonnikov and N.N. Ural'Ceva, Linear and quasilinear equations of parabolic type. Amer. Math. Soc., Providende, Math. Monographs 23 (1968). | Zbl 0174.15403
[14] E.M. Landis and O.A. Oleinik, Generalized analyticity and some related properties of solutions of elliptic and parabolic equations. Russ. Math. Surveys 29 (1974) 195-212. | MR 402268 | Zbl 0305.35014
[15] F.H. Lin, A uniqueness theorem for parabolic equations. Comm. Pure Appl. Math. XLIII (1990) 127-136. | MR 1024191 | Zbl 0727.35063
[16] J.L. Lions and E. Magenes, Non-homogeneous boundary value problems and applications II. Springer, New York (1972). | Zbl 0227.35001
[18] S. Vessella, Stability estimates in an inverse problem for a three-dimensional heat equation. SIAM J. Math. Anal. 28 (1997) 1354-1370. | MR 1474218 | Zbl 0888.35130
|
The IsDihedral( G ) command determines whether the finite group G is isomorphic to a dihedral group of some (even) order, without using an (expensive) isomorphism test. It returns the value true if G is isomorphic to
{\mathrm{D}}_{n}
n
\mathrm{with}\left(\mathrm{GroupTheory}\right):
G≔\mathrm{Group}\left([\mathrm{Perm}\left([[1,2,3,4,5,6,7],[8,9,10,11,12,13,14]]\right),\mathrm{Perm}\left([[2,7],[3,6],[4,5],[9,14],[10,13],[11,12]]\right)]\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\right)\left(\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{12}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{13}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{14}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{14}\right)\left(\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{13}\right)\left(\textcolor[rgb]{0,0,1}{11}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{12}\right)〉
\mathrm{GroupOrder}\left(G\right)
\textcolor[rgb]{0,0,1}{14}
\mathrm{IsDihedral}\left(G\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{AreIsomorphic}\left(G,\mathrm{DihedralGroup}\left(7\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{IsDihedral}\left(\mathrm{QuaternionGroup}\left(\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
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Introduction to Forecasting of Dynamic System Response - MATLAB & Simulink - MathWorks í•œêµ
Forecasting the response of a dynamic system is the prediction of future outputs of the system using past output measurements. In other words, given observations y(t) = {y(1), …, y(N)} of the output of a system, forecasting is the prediction of the outputs y(N+1), …, y(N+H) until a future time horizon H.
When you perform forecasting in System Identification Toolbox™ software, you first identify a model that fits past measured data from the system. The model can be a linear time series model such as AR, ARMA, and state-space models, or a nonlinear ARX model. If exogenous inputs influence the outputs of the system, you can perform forecasting using input-output models such as ARX and ARMAX. After identifying the model, you then use the forecast command to compute y(N+1), …, y(N+H). The command computes the forecasted values by:
Suppose that you have collected time series data y(t) = {y(1), …, y(N)} of a stationary random process. Assuming the data is a second-order autoregressive (AR) process, you can describe the dynamics by the following AR model:
y\left(t\right)+{a}_{1}y\left(tâ1\right)+{a}_{2}y\left(tâ2\right)=e\left(t\right)
You can identify the model using the ar command. The software computes the fit coefficients and variance of e(t) by minimizing the 1-step prediction errors between the observations {y(1), …, y(N)} and model response.
\stackrel{^}{y}\left(t\right)
\stackrel{^}{y}\left(t\right)=â{a}_{1}y\left(tâ1\right)â{a}_{2}y\left(tâ2\right)
\begin{array}{l}\stackrel{^}{y}\left(N+1\right)=â{a}_{1}y\left(N\right)â{a}_{2}y\left(Nâ1\right)\\ \stackrel{^}{y}\left(N+2\right)=â{a}_{1}\stackrel{^}{y}\left(N+1\right)â{a}_{2}y\left(N\right)\\ \stackrel{^}{y}\left(N+3\right)=â{a}_{1}\stackrel{^}{y}\left(N+2\right)â{a}_{2}\stackrel{^}{y}\left(N+1\right)\\ \stackrel{^}{y}\left(N+4\right)=â{a}_{1}\stackrel{^}{y}\left(N+3\right)â{a}_{2}\stackrel{^}{y}\left(N+2\right)\\ \stackrel{^}{y}\left(N+5\right)=â{a}_{1}\stackrel{^}{y}\left(N+4\right)â{a}_{2}\stackrel{^}{y}\left(N+3\right)\end{array}
\stackrel{^}{y}\left(N+2\right)
\stackrel{^}{y}\left(N+1\right)
y\left(t\right)=e\left(t\right)+{c}_{1}e\left(tâ1\right)+{c}_{2}e\left(tâ2\right)
\stackrel{^}{y}\left(t\right)={c}_{1}e\left(tâ1\right)+{c}_{2}e\left(tâ2\right)
\stackrel{^}{y}\left(3\right)=0.1e\left(2\right)+0.2e\left(1\right)
\begin{array}{l}e\left(2\right)=y\left(2\right)-\stackrel{^}{y}\left(2\right)=y\left(2\right)-\left[0.1e\left(1\right)+0.2\text{ }e\left(0\right)\right]\\ e\left(1\right)=y\left(1\right)-\stackrel{^}{y}\left(1\right)=y\left(1\right)-\left[0.1e\left(0\right)+0.2\text{ }e\left(-1\right)\right]\end{array}
\begin{array}{l}e\left(1\right)=5-\left(0.1*0+0.2*0\right)=5\\ e\left(2\right)=10-\left(0.1*5+0.2*0\right)=9.5\\ \stackrel{^}{y}\left(3\right)=0.1*9.5+0.2*5=1.95\end{array}
\begin{array}{l}\stackrel{^}{y}\left(4\right)=0.1e\left(3\right)+0.2e\left(2\right)\\ \stackrel{^}{y}\left(5\right)=0.1e\left(4\right)+0.2e\left(3\right)\end{array}
\stackrel{^}{y}\left(4\right)=0.2*e\left(2\right)=0.2*9.5=1.9
\stackrel{^}{y}\left(5\right)=0
V=e{\left(1\right)}^{2}+e{\left(2\right)}^{2}=\text{ (y(1) - [0}\text{.1 e(0) + 0}{\text{.2 e(-1)])}}^{2}\text{ + (y(2) - [0}\text{.1 e(1) + 0}{\text{.2 e(0)])}}^{2}
V\left(a,b\right)={\left(5-\left[0.1a+0.2b\right]\right)}^{2}+{\left(10-\left[0.1\left(5-\left[0.1a+0.2b\right]\right)+0.2a\right]\right)}^{2}
\begin{array}{l}e\left(1\right)=5-\left(0.1*50+0.2*0\right)=0\\ e\left(2\right)=10-\left(0.1*0+0.2*50\right)=0\\ \stackrel{^}{y}\left(3\right)=0\\ \text{y^(4) = 0}\end{array}
yf_zeroIC = 5×1
yf_estimatedIC = 5×1
\begin{array}{l}x\left(t+1\right)=Ax\left(t\right)+Ke\left(t\right)\\ y\left(t\right)=Cx\left(t\right)+e\left(t\right)\end{array}
\begin{array}{l}\stackrel{^}{x}\left(t+1\right)=\left(A-K*C\right)\text{ }\stackrel{^}{x}\left(t\right)+Ky\left(t\right)\\ \stackrel{^}{y}\left(t\right)=C*\stackrel{^}{x}\left(t\right)\end{array}
\stackrel{^}{y}\left(t\right)
\stackrel{^}{x}\left(0\right)={x}_{0}
\stackrel{^}{x}\left(N+1\right)
\begin{array}{l}\stackrel{^}{x}\left(1\right)=\left(A-K*C\right){\text{ x}}_{0}\text{+}Ky\left(0\right)\\ \stackrel{^}{x}\left(2\right)=\left(A-K*C\right)\stackrel{^}{x}\left(1\right)+Ky\left(1\right)\\ ⋮\\ \stackrel{^}{x}\left(N+1\right)=\left(A-K*C\right)\stackrel{^}{x}\left(N\right)+Ky\left(N\right)\end{array}
\stackrel{^}{x}\left(N+1\right)
y\left(t\right)=f\left(y\left(t-1\right),\text{ }y\left(t-2\right),\text{ }...\text{ },\text{ }y\left(t-N\right)\right)+e\left(t\right)
R\left(t\right)={\left[y\left(tâ1\right),y\left(tâ2\right),y{\left(tâ1\right)}^{2},y{\left(tâ2\right)}^{2},y\left(tâ1\right)y\left(tâ2\right)\right]}^{T}
y\left(t\right)={w}_{1}y\left(t-1\right)+{w}_{2}y\left(t-2\right)+{w}_{3}y{\left(t-1\right)}^{2}+{w}_{4}y{\left(t-2\right)}^{2}+{w}_{5}y\left(t-1\right)y\left(t-2\right)+c+e\left(t\right)
\begin{array}{l}\stackrel{^}{y}\left(101\right)={w}_{1}y\left(100\right)+{w}_{2}y\left(99\right)+{w}_{3}y{\left(}^{100}+{w}_{4}y{\left(}^{99}+{w}_{5}y\left(100\right)y\left(99\right)\\ \stackrel{^}{y}\left(102\right)={w}_{1}\stackrel{^}{y}\left(101\right)+{w}_{2}y\left(100\right)+{w}_{3}\stackrel{^}{y}{\left(}^{101}+{w}_{4}y{\left(}^{100}+{w}_{5}\stackrel{^}{y}\left(101\right)y\left(100\right)\\ \stackrel{^}{y}\left(103\right)={w}_{1}\stackrel{^}{y}\left(102\right)+{w}_{2}\stackrel{^}{y}\left(101\right)+{w}_{3}\stackrel{^}{y}{\left(}^{102}+{w}_{4}\stackrel{^}{y}{\left(}^{101}+{w}_{5}\stackrel{^}{y}\left(102\right)\stackrel{^}{y}\left(101\right)\\ \stackrel{^}{y}\left(104\right)={w}_{1}\stackrel{^}{y}\left(103\right)+{w}_{2}\stackrel{^}{y}\left(102\right)+{w}_{3}\stackrel{^}{y}{\left(}^{103}+{w}_{4}\stackrel{^}{y}{\left(}^{102}+{w}_{5}\stackrel{^}{y}\left(103\right)\stackrel{^}{y}\left(102\right)\end{array}
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Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1.
{\int }_{4}^{\mathrm{\infty }}x{e}^{-2x}
{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}xdx
{e}^{-2x}x
, integrate by parts,
\int fdg=fg-\int gdf
f=x,dg={e}^{-2x}dx,df=dx,g=-\frac{1}{2}{e}^{-2x}
=\left(-\frac{1}{2}{e}^{-2x}x\right){|}_{4}^{\mathrm{\infty }}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx
Possible derivation:
\frac{d}{dx}\left(x\right)
Use the power rule,
\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}
, where n = 1.
\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left({x}^{1}\right)={x}^{0}
\left(-\frac{1}{2}{e}^{-2x}x\right){|}_{4}^{\mathrm{\infty }}=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)-\left(-\frac{1}{2}{e}^{-24}4\right)=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)-\left(-\frac{2}{{e}^{8}}\right)
=\left(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b\right)+\frac{2}{{e}^{8}}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx
\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b=0
=\frac{2}{{e}^{8}}+\frac{1}{2}{\int }_{4}^{\mathrm{\infty }}{e}^{-2x}dx
\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2x}x
Hint: Factor a constant multiple out of the limit.
\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}x{e}^{-2x}=-\frac{1}{2}\underset{x\to \mathrm{\infty }}{lim}x{e}^{-2x}
-\frac{\underset{x\to \mathrm{\infty }}{lim}{e}^{-2x}x}{2}
Hint: | Linear functions grow asymptotically slower than exponential functions.
Since the polynomial x grows asymptotically slower than
{e}^{2x}
as x approaches ∞,
{\int }_{0}^{\frac{\pi }{2}}\mathrm{arccos}\left(\mathrm{sin}x\right)dx
-d\frac{{\left\{\pi \right\}}^{2}}{8}
{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{cos}x}{\text{cosh}x}dx
{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}x}{{e}^{x}-1}dx
Some time ago I came across to the following integral:
I={\int }_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\mathrm{ln}x}
What are the hints on how to compute this integral?
Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
\int 37{e}^{74x}\mathrm{arctan}\left({e}^{37x}\right)dx
Inverse Trigonometric Forms (92):
\int u{\mathrm{tan}}^{-1}u\text{ }du=\frac{{u}^{2}+1}{2}{\mathrm{tan}}^{-1}u-\frac{u}{2}+C
Evaluate integral :
\int {\mathrm{cos}}^{4}\left(x\right)dx
\left(1+{e}^{x}\right)ydy-{e}^{x}dx=0
I need help to evaluate the integral:
\int \frac{{x}^{2}-2x+7}{\sqrt[3]{4x-1}}dx
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Short CV - Latest Works
Professor in Mathematical Analysis
Director of the PhD School in Mathematics
Born in Udine on April 12, 1961
Research areas: Calculus of Variations, Gamma-convergence, Homogenization, Discrete Variational Problems, Fracture Mechanics, Image Processing, Geometric Measure Theory
Associate Professor at SISSA, Trieste 1995-2000
Associate Professor at the University of Brescia, 1992-1995
`Ricercatore' at the University of Brescia, 1988-1992
(`Servizio civile', 1986-1988)
Contract Professor at the University of Udine, 1985-1986
`Perfezionando'(Ph D, adviser E. De Giorgi) at Scuola Normale Superiore, Pisa, 1983-1985
`Laurea' in Mathematics at the University of Pisa, 1983
Distinguished visiting positions, invited lectures, honours
1998, Jan-May. Marie-Curie Scholarship at the Max-Planck Institu ̈t, Leipzig 2002. Invited Distinguished Professorship, Paris XIII
2004. Timoshenko Fellowship, Stanford
2013. Plenary Speaker at GAMM Conference, Novi Sad
2013-14. One-year Visiting Professorship at the Mathematical Institute and Fellowship at Mansfield College, Oxford, UK
2014. Invited Sectional Speaker at the International Congress of Mathematicians ICM 2014, Seoul
2015 (Jan-Mar) John Von Neumann Visiting Professorship at Technische Universität München
2015 Tullio Levi Civita Prize
Ph.D. Theses (Adviser)
R. Alicandro. Approximation of Free-Discontinuity Problems (SISSA, 1999)
M.S. Gelli. Variational Limits of Discrete Systems (SISSA, 1999)
N. Ansini. Homogenization Problems for Multi-dimensional and Multi-scale Structures (SISSA, 2000)
M.Cicalese. Multi-scale analysis for variational problems arising from discrete systems (University of Naples `Federico II', 2003)
C.I Zeppieri Multi-scale analysis via Gamma-convergence (Universita' di Roma La Sapienza, 2006)
L. Sigalotti. Asymptotic analysis of discrete systems with complex interfacial interactions (Sapienza Universita' di Roma, 2010)
G. Scilla Variational motion of discrete interfaces (Sapienza Universita' di Roma 2014)
A. Cancedda (co-supervised with V.Chiadò Piat). Spectral analysis and problems with oscillating constraints in the theory of Homogenization (Politecnico di Torino, 2015)
V. Vallocchia. Some asymptotic problems for non-convex discrete systems(Università di Roma Tor Vergata 2018)
L. Kreutz. Some results on ferromagnetic spin systems and related issues (GSSI, L'Aquila, 2018)
A. Tribuzio (Università di Roma "Tor Vergata'', ongoing)
I also had the pleasure of collaborating to the following thesis:
M. Solci A Variational Model for Phase Separation (Pavia 2003, adviser E.Vitali).
Journal de l'Ecole Polytechnique
Course `Local minimization, Variational Evolution and Gamma-convergence'
A handbook of Gamma-convergence
Course `From Discrete Systems to Continuum Problems' (Würzburg 2012) - LECTURE NOTES
A. Braides.
An Introduction to Homogenization and Gamma-convergence.
In: G. Allaire, A. Braides, G. Buttazzo, A. Defranceschi, L.V. Gibiansky.
School on Homogenization. Lecture Notes of the Courses held at ICTP,
Trieste, 4--17 September 1993
A. Braides and A. Defranceschi.
Homogenization of Multiple Integrals
Lecture Notes in Mathematics No. 1694, Springer Verlag, Berlin, 1998.
A. Braides and M.S. Gelli.
From Discrete to Continuum: a Variational Approach.
Lecture Notes. SISSA, Trieste, 2000.(file pdf)
Gamma-convergence for Beginners.
From discrete to continuous variational problems: an introduction .
Lecture Notes. School on Homogenization Techniques and Asymptotic Methods for Problems with Multiple Scales.
appeared as part of
From discrete systems to continuous variational problems: an introduction.
Topics on concentration phenomena and problems with multiple scales, 3--77, Lect. Notes Unione Mat. Ital., 2, Springer, Berlin, 2006.
in Handbook of Differential Equations. Stationary Partial Differential Equations, Volume 3 (M. Chipot and P. Quittner, eds.) Elsevier 2006.
Lecture Notes in Mathematics No. 2094, Springer Verlag, Berlin, 2013
Topics on concentration phenomena and problems with multiple scales.
Edited by Andrea Braides and Valeria Chiadò Piat.
Lecture Notes of the Unione Matematica Italiana, 2.
Springer-Verlag, Berlin, 2006. xii+316
Recent Papers (see also my web page at CVGMT)
A. Braides - M. Cicalese - M. Ruf (Analysis & PDE)Continuum limit and stochastic homogenization of discrete ferromagnetic thin films (2018)
A. Braides - V. Chiadò Piat (Applicable Analysis) Homogenization of networks in domains with oscillating boundaries (2018)
A. Braides - A. Causin - M. Solci (Ann.Mat.Pura Appl) Asymptotic analysis of a ferromagnetic Ising system with "diffuse" interfacial energy (2018)
A. Braides (Book Chapter- Trends in Applications of Mathematics to Mechanics (E.Rocca et al, eds) Springer) Rigidity effects for antiferromagnetic thin films: a prototypical example (2018)
A. Braides - A. Malusa - M. Novaga (Accepted Paper: Ann. Scuola Normale Sup.Pisa) Crystalline evolutions with rapidly oscillating forcing terms (2018)
A. Braides - P. Cermelli - S. Dovetta (Preprint)
\mathrm{\Gamma }
-limit of the cut functional on dense graph sequences (2018)
A. Braides - L. Kreutz (SIAM J. Math. Anal.) An integral-representation result for continuum limits of discrete energies with multi-body interactions (2018)
A. Bach - A. Braides - C. I. Zeppieri (Submitted Paper) Quantitative analysis of finite-difference approximations of free-discontinuity problems (2018)
A. Braides - L. Kreutz (Calc. Var. Partial Diff. Equations) Design of lattice surface energies (2018)
A. Braides - V. Vallocchia (Acta Applicandae Mathematicae) Static, Quasi-static and Dynamic Analysis for Scaled Perona-Malik Functionals (2018)
A. Braides - A. Causin - A. Piatnitski - M. Solci (J. Stat. Phys.) Asymptotic behaviour of ground states for mixtures of ferromagnetic and antiferromagnetic interactions in a dilute regime (2018)
A. Braides - A. Causin - M. Solci (Proceedings A. Roy. Soc. London) A homogenization result for interacting elastic and brittle media (2018)
A. Braides - M. Cicalese (Arch. Ration. Mech. Anal.) Interfaces, modulated phases and textures in lattice systems (2017)
A. Braides - S. Conti - A. Garroni (Calc. Var. Partial Diff. Equations) Density of polyhedral partitions (2017)
N. Ansini - A. Braides - J. Zimmer (Proc. Roy. Soc. Edinburgh A) Minimising movements for oscillating energies: the critical regime (2017)
A. Braides - A. Cancedda - V. Chiadò Piat (ESAIM: Control Optim. Calc. Var.) Homogenization of metrics in oscillating manifolds (2017)
A. Braides - M. S. Gelli (ESAIM: M2AN) Analytical treatment for the asymptotic analysis of microscopic impenetrability constraints for atomistic systems (2017)
A. Braides - B. Cassano - A. Garroni - D. Sarrocco (Annales de l'Institut Henri Poincaré) Quasi-static damage evolution and homogenization: a case study of non-commutability (2016)
A. Braides - M. Colombo - M. Gobbino - M. Solci (C. R. Acad. Sci. Paris, Ser. I) Minimizing movements along a sequence of functionals and curves of maximal slope (2016)
A. Braides - V. Chiadò Piat - M. Solci (Math. Mech. Complex Syst.) Discrete double-porosity models for spin systems (2016)
A. Braides - M. Solci (J. Nonlinear Sci.) Motion of discrete interfaces through mushy layers (2016)
A. Braides - M. S. Gelli (J. Mech. Phys. Solids) Asymptotic analysis of microscopic impenetrability constraints for atomistic systems (2016)
A. Braides - M. Solci (Math. Mech. Solids) Asymptotic analysis of Lennard-Jones systems beyond the nearest-neighbour setting: a one-dimensional prototypical case (2016)
A. Braides - A. Garroni - M. Palombaro (Multiscale Model. Simul.) Interfacial energies of systems of chiral molecules (2016)
A. Braides - L. Kreutz (Atti Accad. Naz. Lincei Rend. Lincei Mat. Appl.) Optimal bounds for periodic mixtures of nearest-neighbour ferromagnetic interactions (2016)
A. Braides - M. Cicalese - N. K. Yip (J. Stat. Phys.) Crystalline Motion of Interfaces Between Patterns (2016)
A. Braides - L. Kreutz (Preprint) Optimal design of mixtures of ferromagnetic interactions (2016)
A. Braides - V. Vallocchia (Submitted Paper) Static, Quasi-static and Dynamic Analysis for Scaled Perona-Malik Functionals (2016)
A. Braides - M. Cicalese - M. Ruf (Preprint) Continuum limit and stochastic homogenization of discrete ferromagnetic thin films (2016)
A. Braides - M. Cicalese - F. Solombrino (SIAM J. Math. Anal.) Q-tensor continuum energies as limits of head-to-tail symmetric spins systems (2015)
A. Braides - V. Chiadò Piat - A. Piatnitski (SIAM J. Math. Anal.) Homogenization of Discrete High-Contrast Energies (2015)
A. Braides - A. Defranceschi - E. Vitali (Netw. Heterog. Media) Variational evolution of one-dimensional Lennard-Jones systems (2014)
A. Braides (J. Stat. Phys.) An example of non-existence of plane-like minimizers for an almost-periodic Ising system (2014)
A. Braides (Proceedings: International Congress of Mathematicians (ICM 2014), Seoul, 2014) Discrete-to-Continuum Variational Methods for Lattice Systems (2014)
A. Braides - A. Piatnitski (J. Funct. Anal.) Homogenization of surface and length energies for spin systems (2013)
A. Braides - M. Solci (Boll. Unione Mat. Ital.) Multi-scale free-discontinuity problems with soft inclusions (2013)
A. Braides (Book: Lecture Notes in Mathematics, Springer)Local minimization, variational evolution and -convergence (2013)
A. Braides - G. Scilla (C. R. Acad. Sci. Paris) Nucleation and backward motion of discrete interfaces (2013)
A. Braides - G. Scilla (Interfaces Free Bound.)Motion of discrete interfaces in periodic media (2013)
A. Braides - G. Scilla Nucleation and backward motion of discrete interfaces C. R. Acad. Sci. Paris, 351 (2013), 803--806
A. Braides, A. Defranceschi and E. Vitali Variational evolution of one-dimensional Lennard-Jones systems Networks Heterog. Media
A. Braides and M. Solci, Multi-scale free-discontinuity problems with soft inclusions. Boll. Unione Mat. Ital. (IX), 6 (2013), 29-51
A. Braides and A. Piatnitski. Variational problems with percolation: dilute spin systems at zero temperature J. Stat. Phys. 149 (2012), 846-864
A. Braides, A. Causin and M. Solci.Interfacial energies on quasicrystals IMA J Appl Math 77 (2012), 816-836.
A. Braides and N.K. Yip.A quantitative description of mesh dependence for the discretization of singularly perturbed non-convex problems. SIAM J. Numer. Anal. 50 (2012), 1883--1898
A. Braides, A. Defranceschi and E. Vitali. A compactness result for a second-order variational discrete model M2AN 46 (2011), 389-410
A. Braides and A. Piatnitski Homogenization of surface and length energies for spin systems. J. Funct. Anal. 264 (2013), 1296--1328
A. Braides and M. Solci. Interfacial energies on Penrose lattices. M3AS, 21 (2011), 1193-1210
A. Braides and C. Larsen.$\Gamma$-convergence for stable states and local minimizers. Ann. SNS Pisa {\bf 10} (2011), 193-206
A. Braides and L.Sigalotti. Models of defects in atomistic systems. Calc. Var. Partial Diff. Eq., 41 (2011), 71-109
A. Braides, M.S. Gelli, and M. Novaga. Motion and pinning of discrete interfaces. Arch. Ration. Mech. Anal., 95 (2010), 469-498.
A. Braides, M. Briane, and J. Casado Diaz. Homogenization of non-uniformly bounded periodic diffusion energies in dimension two. Nonlinearity 22 (2009), 1459-1484
A. Braides, G. Riey, and M. Solci. Homogenization of Penrose tilings. C.R. Acad. Sci., Ser. I 347 (2009), 697-700
A. Braides - C.I. Zeppieri Multiscale analysis of a prototypical model for the interaction between microstructure and surface energy Interfaces Free Bound. 11 (2009), 61--118
A. Braides, M. Maslennikov, and L. Sigalotti.Homogenization by blow-up. Applicable Anal. 87 (2008), 1341--1356.
R. Alicandro - A. Braides - M. Cicalese Continuum limits of discrete thin films with superlinear growth densities Calc. Var. Partial Diff. Eq. 33 (2008), 267--297
A. Braides and L. Sigalotti. Asymptotic analysis of periodically-perforated nonlinear media at and close to the critical exponent. C.R. Acad. Sci. Paris 346 (2008), 363--367.
A. Braides - L. Truskinovsky Asymptotic expansions by Gamma-convergence Cont. Mech. Therm. 20 (2008), 21--62
A.Braides - G. Riey A variational model in image processing with focal points ESAIM: M2AN 42 (2008) 729--748
A. Braides and V. Chiado' Piat Non convex homogenization problems for singular structures Network and Heterogenous Media 3 (2008), 489-508
A. Braides and A. Piatnitski. Overall properties of a discrete membrane with randomly distributed defects Arch. Ration. Mech. Anal. 189 (2008), 301--323
A. Braides - A. Gloria Exact bounds on the effective behaviour of a conducting `discrete' polycrystal Multiscale Modeling and Simulation 6 (2007), 1198-1216
A. Braides - M. Solci - E. Vitali A derivation of linear elastic energies from pair-interaction atomistic systems Network and Heterogenous Media 2 (2007), 551-567
A. Braides - A. Chambolle - M. Solci A relaxation result for energies defined on pairs set-function and applications ESAIM:COCV 13 (2007), 717-734
A. Braides - C. I. Zeppieri A note on equi-integrability in dimension reduction problems Calc. Var. Partial Differential Equations 29 (2007), 231-238
A. Braides - M. Briane Homogenization of non-linear variational problems with thin low-conducting layers. Appl. Math. Optim. 55 (2007), no. 1, 1--29
A. Braides and M. Cicalese. Surface energies in nonconvex discrete systems. M3AS 17 (2007) 985-1037
R. Alicandro, A. Braides and M. Cicalese Phase and anti-phase boundaries in binary discrete systems: a variational viewpoint. Netw. Heterog. Media 1 (2006), no. 1, 85--107
N. Ansini - A. Braides - V. Valente Multi-Scale Analysis by G-convergence of a Shell-Membrane Transition. SIAM J. Math. Anal. 38 (2006), no. 3, 944--976
A. Braides and V. Chiado' Piat Another brick in the wall. Variational problems in materials science, 13--24, Progr. Nonlinear Differential Equations Appl., 68, Birkhäuser, Basel, 2006
A. Braides, A.J. Lew and M. Ortiz. Effective cohesive behavior of layers of interatomic planes. Arch. Ration. Mech. Anal. 180 (2006), no. 2, 151--182
A. Braides and R. March Approximation by $\Gamma$-convergence of a curvature-depending functional in Visual Reconstruction. Comm. Pure Appl. Math. 59 (2006), no. 1, 71--121
Complete List of Publications (as of January 2019)
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What is the Kijun Line (Base Line)?
The Kijun Line, also called the Base Line or Kijun-sen, is one of five components that make up the Ichimoku Cloud indicator. The Kijun Line is typically used in conjunction with the Conversion Line (Tenkan-sen) to generate trade signals when they cross. These signals can be further filtered via the other components of the Ichimoku indicator.
The Kijun Line is the mid-point of the high and low price over the last 26 periods.
When the price is above the Kijun Line it indicates the recent price momentum is to the upside. When the price is below the Kijun Line, recent price momentum is to the downside.
The Kijun Line and Tenkan Line are used together to generate trade signals.
The Base Line is the midpoint price of the last 26-periods.
The Kijun Line is one of five components of the Ichimoku indicator.
The Formula for the Kijun Line (Base Line) is
\begin{aligned} &\text{Kijun line (base line)} = \frac{1}{2}*\left( \max_{\left\{ t .. t-26\right\} } {\left[p\right ]} + \min_{\left\{ t .. t-26\right\}}{\left[p\right ]} \right)\\ &\textbf{where:}\\ &\max_{\left\{ t .. t-26\right\} }{\left[p\right ]} = \text{the maximum price over the last 26 periods}\\ &\min_{\left\{ t .. t-26\right\} }{\left[p\right ]} = \text{the minimum price over the last 26 periods}\\ \end{aligned}
Kijun line (base line)=21∗({t..t−26}max[p]+{t..t−26}min[p])where:{t..t−26}max[p]=the maximum price over the last 26 periods{t..t−26}min[p]=the minimum price over the last 26 periods
How to Calculate the Kijun Line (Base Line)
Find the highest price over the last 26 periods.
Find the lowest price over the last 26 periods.
Combine the high and low, then divide by two.
Update the calculation after each period ends.
What Does the Kijun Line Tell You?
The Kijun Line, or Base Line, is part of the Ichimoku Cloud indicator.
The Ichimoku Cloud is a technical indicator that defines support and resistance, measures momentum, and provides buy and sell signals. Its developer, Goichi Hosoda, designed the indicator to be a "one look equilibrium chart".
There are several different lines included in the Ichimoku Cloud indicator.
Tenkan-Sen—Conversion Line
Kijun-Sen—Base Line
Senkou Span A—Leading Span A
Senkou Span B—Leading Span B
Chikou Span—Lagging Span
While the "cloud", made up of Leading Span A and B, is the most prominent feature of Ichimoku Cloud indicator, the Kijun Line generates trading signals when it is crossed by the Tenkan Line. The Tenkan Line is the 9-period price mid-point, therefore it moves quicker than the Kinjun line which looks at 26 periods.
When the Tenkan Line crosses above the Kijun Line it signals that the short-term price momentum is moving to the upside, and may be interpreted as a buy signal.
When the Tenkan Line crosses below the Kijun Line it signals momentum has shifted to the downside and may be interpreted as a sell signal.
Buy or sell signals should be used within the context of the other components of the Ichimoku indicator. For example, a trader may only wish to trade the buy signals if the price is also above the "cloud" or Leading Span A.
When the Tenkan Line and Kijun Line are crossing back and forth the price lacks a trend, or is moving in a choppy fashion, and therefore the crossovers will not produce reliable trade signals.
On its own, the Kijun Line can also be used for analyzing price momentum. With the price is above the Kijun line, it means the price is above 26-period mid-point and therefore has an upward bias. If the price is below the Kijun Line, it is below the mid-point price, and therefore has a downward bias.
Example of a Kijun Line
The following chart shows an example of an Ichimoku Cloud indicator applied to the SPDR S&P 500 ETF (SPY).
In the chart above, the Kijun Line is red and the Tenkan Line is blue. After a brief selloff, the Tenkan moved above the Kijun in early 2016. This was a potential buy signal. The two lines did not cross again until 2018, which would have provided the sell signal. For most of this time, the price stayed above the Kijun Line and the "cloud", helping to confirm the uptrend.
The Difference Between the Kijun Line and a Moving Average
The Kijun Line is a moving mid-point, based on the high and low over a set number of periods. It is calculated by adding the high and low and dividing by two. A moving average (MA) is different. It sums up the closing prices of a set number of periods and then divides that by the number of periods. A 26-period Kijun Line and a 26-period MA will produce different values, and therefore provide traders with different information.
The Limitations of Using the Kijun Line
Unless there is a very strong trend, the Kijun Line will often appear near the price. When the Kijun Line is often intersecting or near the price, it is not as useful for helping to assess the trend direction.
The same goes for crossovers with the Tenkan Line. When the price trends strongly, crossover signals may be quite profitable. Yet many crossovers signals will be unprofitable if the price fails to trend following the crossover.
The Kijun Line is reactionary, in that it shows what price has done in the past. There are no predictive qualities inherent in the indicator's calculation.
The Kijun Line should ideally be used in conjunction with the other elements of the Ichimoku Cloud indicator, along with price action and other technical indicators.
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Use the table below and answer the questions. 1) P
Use the table below and answer the questions. 1) P (> 1 car | 2 to 4 kids) = ? 2) P (< 4 kids or <= 2 cars) = ?
1) P (> 1 car | 2 to 4 kids) = ?
2) P (< 4 kids or
\le
2 cars) = ?
Solution ia given below
The _______ relative frequencies are the sums of each row and column in a two-way table. (joint, marginal, or conditional)
Like Weight Lifting Yes No Total Like Aerobic Yes 71219 Exercise No 14721 Total 211940
How are the smoking habits of students related to their parents
\left(A\cup D\right)
using the following table.
A 2012 Gallup poll surveyed Americans about their employment status and whether or not they have diabetes.
The survey results indicate that 1.5% of the 47,774 employed (full or part time) and 2.5% of the 5,855 unemployed 18-29 year olds have diabetes.
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Suppose the probability of winning the Powerball lottery is 0.1,
Suppose the probability of winning the Powerball lottery is 0.1, while
tugmiddelc0 2021-11-13 Answered
Suppose the probability of winning the Powerball lottery is 0.1, while the probability of being abducted by aliens is 0.6. If these two events are independent, what is the probability of winning Powerball but not being abducted?
H) 0.36
The probability of winning the Powerball lottery is 0.1. That is,
P\left(W\in n\in g\right)=0.1
, and the probability of being abducted by aliens is 0.6, then the probability of not being abducted by aliens is
0.4\left(=1-0.6\right)
P\left(Not\text{ }abducted\right)=0.4
. The two events are independent.
The probability of winning Powerball but not being abducted is,
P\left(Winning\cap Not\text{ }abducted\right)=P\left(Winning\right)P\left(Not\text{ }abducted\right)
=0.1×0.4
=0.04
Thus, the probability of winning Powerball but not being abducted is 0.04.
Correct Answer: Option (C) 0.04.
A restaurant menu lists seven entrée choices. Two of the entrée choices are vegetarian. One member of a couple chooses one entrée at random and then the other member chooses a different entrée at random. Consider the problem of calculating the probability that the couple choose vegetarian entrées.
- Can a binomial distribution be used for the solution of the above problem? Why or why not?
- What kind of probability distribution can be used to solve this problem?
- What is the probability that both members chose vegetarian entrées?
Sue and Ann are taking the same English class but they do not study together, so whether one passes will be Independent of whether the other passes. In other words, “Sue passes and “Ann passes” are assumed to be independent events. The probebiity that Sue passes English is 0.8 and the probablity that Ann passes English is 0.75 Whats the probabilty both girls pass English?
The probability of a day being sunny, rainy and snowy are 0.5, 0.4 and 0.1, respectively. The probability of a huge thunderstorm happening are 0.1, 0.7 and 0.2, given the day is sunny, rainy and snowy, respectively. Suppose a huge thunderstorm has taken place.
a. What is the probability that the day was sunny?
b. What is the probability that the day was rainy?
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Perform Symbolic Computations - MATLAB & Simulink - MathWorks España
Differentiate Symbolic Expressions
Expressions with One Variable
Second Partial and Mixed Derivatives
Integrate Symbolic Expressions
Indefinite Integrals of One-Variable Expressions
Indefinite Integrals of Multivariable Expressions
If MATLAB Cannot Find a Closed Form of an Integral
Solve Algebraic Equations with One Symbolic Variable
Solve Algebraic Equations with Several Symbolic Variables
Solve Systems of Algebraic Equations
Substitute Symbolic Variables with Numbers
Substitute in Multivariate Expressions
Substitute One Symbolic Variable for Another
Substitute a Matrix into a Polynomial
Substitute the Elements of a Symbolic Matrix
Plot Symbolic Functions
Explicit Function Plot
Implicit Function Plot
With the Symbolic Math Toolbox™ software, you can find
Derivatives of single-variable expressions
For in-depth information on taking symbolic derivatives see Differentiation.
To differentiate a symbolic expression, use the diff command. The following example illustrates how to take a first derivative of a symbolic expression:
For multivariable expressions, you can specify the differentiation variable. If you do not specify any variable, MATLAB® chooses a default variable by its proximity to the letter x:
For the complete set of rules MATLAB applies for choosing a default variable, see Find a Default Symbolic Variable.
To differentiate the symbolic expression f with respect to a variable y, enter:
To take a second derivative of the symbolic expression f with respect to a variable y, enter:
You get the same result by taking derivative twice: diff(diff(f, y)). To take mixed derivatives, use two differentiation commands. For example:
You can perform symbolic integration including:
Integration of multivariable expressions
For in-depth information on the int command including integration with real and complex parameters, see Integration.
Suppose you want to integrate a symbolic expression. The first step is to create the symbolic expression:
To find the indefinite integral, enter
If the expression depends on multiple symbolic variables, you can designate a variable of integration. If you do not specify any variable, MATLAB chooses a default variable by the proximity to the letter x:
You also can integrate the expression f = x^n + y^n with respect to y
If the integration variable is n, enter
To find a definite integral, pass the limits of integration as the final two arguments of the int function:
If the int function cannot compute an integral, it returns an unresolved integral:
You can solve different types of symbolic equations including:
Algebraic equations with one symbolic variable
Algebraic equations with several symbolic variables
Systems of algebraic equations
For in-depth information on solving symbolic equations including differential equations, see Equation Solving.
Use the double equal sign (==) to define an equation. Then you can solve the equation by calling the solve function. For example, solve this equation:
If you do not specify the right side of the equation, solve assumes that it is zero:
If an equation contains several symbolic variables, you can specify a variable for which this equation should be solved. For example, solve this multivariable equation with respect to y:
If you do not specify any variable, you get the solution of an equation for the alphabetically closest to x variable. For the complete set of rules MATLAB applies for choosing a default variable see Find a Default Symbolic Variable.
You also can solve systems of equations. For example:
Symbolic Math Toolbox provides a set of simplification functions allowing you to manipulate the output of a symbolic expression. For example, the following polynomial of the golden ratio phi
You can simplify this answer by entering
and get a very short answer:
Symbolic simplification is not always so straightforward. There is no universal simplification function, because the meaning of a simplest representation of a symbolic expression cannot be defined clearly. Different problems require different forms of the same mathematical expression. Knowing what form is more effective for solving your particular problem, you can choose the appropriate simplification function.
For example, to show the order of a polynomial or symbolically differentiate or integrate a polynomial, use the standard polynomial form with all the parentheses multiplied out and all the similar terms summed up. To rewrite a polynomial in the standard form, use the expand function:
The factor simplification function shows the polynomial roots. If a polynomial cannot be factored over the rational numbers, the output of the factor function is the standard polynomial form. For example, to factor the third-order polynomial, enter:
The nested (Horner) representation of a polynomial is the most efficient for numerical evaluations:
For a list of Symbolic Math Toolbox simplification functions, see Choose Function to Rearrange Expression.
You can substitute a symbolic variable with a numeric value by using the subs function. For example, evaluate the symbolic expression f at the point x = 1/3:
The subs function does not change the original expression f:
When your expression contains more than one variable, you can specify the variable for which you want to make the substitution. For example, to substitute the value x = 3 in the symbolic expression
You also can substitute one symbolic variable for another symbolic variable. For example to replace the variable y with the variable x, enter
You can also substitute a matrix into a symbolic polynomial with numeric coefficients. There are two ways to substitute a matrix into a polynomial: element by element and according to matrix multiplication rules.
Element-by-Element Substitution. To substitute a matrix at each element, use the subs command:
You can do element-by-element substitution for rectangular or square matrices.
Substitution in a Matrix Sense. If you want to substitute a matrix into a polynomial using standard matrix multiplication rules, a matrix must be square. For example, you can substitute the magic square A into a polynomial f:
Create the polynomial:
Create the magic square matrix:
Get a row vector containing the numeric coefficients of the polynomial f:
Substitute the magic square matrix A into the polynomial f. Matrix A replaces all occurrences of x in the polynomial. The constant times the identity matrix eye(3) replaces the constant term of f:
The polyvalm command provides an easy way to obtain the same result:
To substitute a set of elements in a symbolic matrix, also use the subs command. Suppose you want to replace some of the elements of a symbolic circulant matrix A
To replace the (2, 1) element of A with beta and the variable b throughout the matrix with variable alpha, enter
The result is the matrix:
For more information, see Substitute Elements in Symbolic Matrices.
Symbolic Math Toolbox provides the plotting functions:
fplot to create 2-D plots of symbolic expressions, equations, or functions in Cartesian coordinates.
fplot3 to create 3-D parametric plots.
ezpolar to create plots in polar coordinates.
fsurf to create surface plots.
fcontour to create contour plots.
fmesh to create mesh plots.
Create a 2-D line plot by using fplot. Plot the expression
{x}^{3}-6{x}^{2}+11x-6
Add labels for the x- and y-axes. Generate the title by using texlabel(f). Show the grid by using grid on. For details, see Add Title and Axis Labels to Chart.
Plot equations and implicit functions using fimplicit.
\left({x}^{2}+{y}^{2}{\right)}^{4}=\left({x}^{2}-{y}^{2}{\right)}^{2}
-1<x<1
Plot 3-D parametric lines by using fplot3.
\begin{array}{c}x={t}^{2}\mathrm{sin}\left(10t\right)\\ y={t}^{2}\mathrm{cos}\left(10t\right)\\ z=t.\end{array}
Create a 3-D surface by using fsurf.
Plot the paraboloid
z={x}^{2}+{y}^{2}
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Higher Dimensional Regression Practice Problems Online | Brilliant
One of the benefits of least squares regression is that it is easy to generalize from its uses on scatter plots to 3D or even higher dimensional data.
Previously, we learned that when least squares regression is used on two-dimensional data the SSE is given by the formula
SSE = \sum_{i=1}^{n} (y_i - mx_i - b)^2.
This gives us a good idea of what a higher dimensional error function will look like.
predictor variables
\{x_1, x_2,\ldots, x_p\}
and one response variable
y,
then a linear equation which outputs
y
will take the form
y = m_1x_1+m_2x_2+\cdots+m_px_p+b.
Given this information, what is a reasonable formula for the error when there is more than one predictor variable?
\sum_{i=1}^{n} (y_i - m_1x_{1i} - m_2x_{2i} - … - m_px_{pi} - b)^2
\sum_{i=1}^{n} (y_i - m_1x_{1i} - m_2x_{2i} - … - m_px_{pi} - b)^p
\sum_{i=1}^{n} (y_i - m_1x_{1i} * m_2x_{2i} * … * m_px_{pi} - b)^2
Remember, the squared error of a single point is the squared difference between the
y
-value and the predicted
y
-value at that point. The SSE for the best-fit function is the sum of the squared errors for each point.
Earlier, we derived a formula for a best-fit line. Now, we will attempt to modify this formula so that it works for higher dimensional linear regression. Instead of outputting a best-fit line, this formula will now output a best-fit hyperplane--a linear equation in higher dimensions.
In the last chapter, we started our derivation by representing our best-fit equation with a vector. We can do so again with
\vec{x} = \begin{bmatrix} m_1 \\ m_2 \\ \vdots \\ m_p \\ b \\ \end{bmatrix}.
Now, we must create a matrix
A
which, when multiplied with
\vec{x},
outputs a vector containing the predicted value of
y
for each data point in the set.
Previously, we did this by making
A
’s first column the
x
-values of all data points and its second column a line of ones. Now, we can achieve the same results for higher dimensions by adding another column to
A
for each additional predictor variable. This is shown below for a data set with
n
predictor variables:
A = \begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1p} & 1 \\ x_{21} & x_{22} & \cdots & x_{2p} & 1 \\ \vdots & ~ & ~ & ~ & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{np} & 1 \\ \end{bmatrix}.
Once again, we will also initialize the vector
\vec{b}
to contain the
y
-values of every data point.
As it turns out, from this point on the derivation is exactly the same as before. We have to find the vector
\vec{x}
A\vec{x}
b,
and once again we can do this by solving the equation
A^T\vec{b} = A^TA\vec{x}.
After that, we have our answer. The elements of
\vec{x}
will give the coefficient values for the best-fit hyperplane.
Alfred is back, and this time he’s remembered there are multiple types of trees. He’s managed to compile a table of the seeds he planted each spring as well as the number of new sprouts each fall. Using this information, identify the matrix
A
which he needs to create in the process of calculating a best-fit linear equation.
\begin{array}{c|c|c} \text{Oak Seeds} & \text{Maple Seeds} & \text{New Growths} \\ \hline 10&5&9 \\ \hline 4 & 8&7\\ \hline 4 & 3& 5 \\ \hline 6 & 2&4\\ \end{array}
A = \begin{bmatrix} 10 & 5 & 1 \\ 4 & 8 & 1 \\ 4 & 3 & 1 \\ 6 & 2 & 1 \\ \end{bmatrix}
A = \begin{bmatrix} 10 & 1 & 5 & 1\\ 4&1&8&1\\ 4 & 1 & 3 & 1 \\ 6 & 1 & 2 & 1 \\ \end{bmatrix}
A = \begin{bmatrix} 10 & 4 & 4 & 6 & 1 \\5 & 8 & 3 & 2 & 1 \\ \end{bmatrix}
A = \begin{bmatrix} 10 & 9 & 1 \\ 4 & 7 & 1 \\ 4 & 5 & 1 \\ 6 & 4 & 1 \\ \end{bmatrix}
At this point, we can find a best-fit hyperplane for any conceivable data set, as long as there are more data points than predictors. But there’s one major problem. What if the points in a data set are very predictable, but not in a linear fashion?
As it turns out, there is a simple way to expand on our previous model. We can just add new, nonlinear terms to our function and update the rest of our math accordingly.
Generally, this is done by adding powers of the predictor variables, in which case this process is known as polynomial regression.
For instance, say we have a simple data set in which there is one predictor variable
x
y
. The only twist is that we suspect
y
to be best represented by a second degree polynomial of
x
Instead of representing the data with a best-fit line
y = mx + b,
we should now represent it with a best-fit polynomial
y = m_1x^2+m_2x+b.
In many ways, this is the same as creating another predictor variable. We have taken each point in our data set and added another value,
x^2.
After this step, we can calculate the coefficients as we normally would in higher dimensional linear regression.
Franklin is in the business of building toy race cars and is analyzing the relationship between the weight and top speed of a car when all else is held equal. So far he’s managed to collect just five data points, but he’s convinced that the relationship should be modeled with a cubic polynomial.
Given the table below, which matrix
A
must he construct in the process of calculating the best-fit curve?
\begin{array}{c|c} x & y \\ \hline 5&30 \\ \hline 4 & 26\\ \hline 6 & 20 \\ \hline 3 & 18\\ \hline 7 & 15 \end{array}
A = \begin{bmatrix} 155 & 1 \\ 84 & 1\\ 258 & 1\\ 39 & 1 \\ 399 & 1 \end{bmatrix} \hspace{1cm}
A = \begin{bmatrix} 5 & 5 & 5 & 1 \\ 4 & 4 & 4 & 1\\ 6 & 6 & 6 & 1\\ 3 & 3 & 3 & 1 \\ 7 & 7 & 7 & 1 \end{bmatrix}
A = \begin{bmatrix} 125 & 25 & 5 & 1 \\ 64 & 16 & 4 & 1\\ 216 & 36 & 6 & 1\\ 27 & 9 & 3 & 1 \\ 343 & 49 & 7 & 1 \end{bmatrix}
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Fundamental Trigonometric Identities: Level 3 Challenges Practice Problems Online | Brilliant
\sin ^{ 2 }{ 1^{\circ} } +\sin ^{ 2 }{ 2^{\circ} } +\sin ^{ 2 }{ 3^{\circ} } + \ldots \\ +\sin ^{ 2 }{ 88^{\circ} } +\sin ^{ 2 }{ 89^{\circ} } +\sin ^{ 2 }{ 90^{\circ} } = \ ?
by Edgar de Asis Jr.
\large \frac { \sin ^{ 2 }{ \theta } }{ 5 } =\frac { \cos ^{ 2 }{ \theta } }{ 6 }
\theta
is a positive acute angle that satisfies the equation above, find
\sin { \theta }
Note: Give your answer to 3 decimal places.
\displaystyle \sum_{k=1}^{50} \Bigg [ \bigg(1 + \tan(k^\circ)+\sec(k^\circ)\bigg)\ \bigg(1+\cot(k^\circ)-\csc(k^\circ) \bigg)\Bigg]=\ ? \
by K. J. W.
\Large \left(\sqrt{2+\sqrt{2}}\right)^{x} + \left(\sqrt{2-\sqrt{2}}\right)^{x} = 2^{x}
Find the sum of all real
x
by Utsav Banerjee
\large\frac { 1 }{ \cos ^{ 2 }{ \theta } } +\frac { 1 }{ 1+\sin ^{ 2 }{ \theta } } +\frac { 2 }{ 1+\sin ^{ 4 }{ \theta } } +\frac { 4 }{ 1+\sin ^{ 8 }{ \theta } }
\large \sin ^{ 16 }{ \theta } = \frac { 1 }{ 5 }
, what is the value of the expression above?
by Fuad Muhammad
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Batteries | Free Full-Text | Impedance Based Temperature Estimation of Lithium Ion Cells Using Artificial Neural Networks
Recent Development of Nickel-Rich and Cobalt-Free Cathode Materials for Lithium-Ion Batteries
Ströbel, M.
Pross-Brakhage, J.
Marco Ströbel
Julia Pross-Brakhage
Electrical Energy Storage Systems, Institute for Photovoltaics, University of Stuttgart, Pfaffenwaldring 47, 70569 Stuttgart, Germany
Academic Editor: Catia Arbizzani
Tracking the cell temperature is critical for battery safety and cell durability. It is not feasible to equip every cell with a temperature sensor in large battery systems such as those in electric vehicles. Apart from this, temperature sensors are usually mounted on the cell surface and do not detect the core temperature, which can mean detecting an offset due to the temperature gradient. Many sensorless methods require great computational effort for solving partial differential equations or require error-prone parameterization. This paper presents a sensorless temperature estimation method for lithium ion cells using data from electrochemical impedance spectroscopy in combination with artificial neural networks (ANNs). By training an ANN with data of 28 cells and estimating the cell temperatures of eight more cells of the same cell type, the neural network (a simple feed forward ANN with only one hidden layer) was able to achieve an estimation accuracy of
\Delta T=
1 K (10
{}^{\circ }
<T<
{}^{\circ }
C) with low computational effort. The temperature estimations were investigated for different cell types at various states of charge (SoCs) with different superimposed direct currents. Our method is easy to use and can be completely automated, since there is no significant offset in monitoring temperature. In addition, the prospect of using the above mentioned approach to estimate additional battery states such as SoC and state of health (SoH) is discussed. View Full-Text
Keywords: lithium-ion batteries; temperature estimation; sensorless temperature measurement; artificial intelligence; artificial neural network lithium-ion batteries; temperature estimation; sensorless temperature measurement; artificial intelligence; artificial neural network
Ströbel, M.; Pross-Brakhage, J.; Kopp, M.; Birke, K.P. Impedance Based Temperature Estimation of Lithium Ion Cells Using Artificial Neural Networks. Batteries 2021, 7, 85. https://doi.org/10.3390/batteries7040085
Ströbel M, Pross-Brakhage J, Kopp M, Birke KP. Impedance Based Temperature Estimation of Lithium Ion Cells Using Artificial Neural Networks. Batteries. 2021; 7(4):85. https://doi.org/10.3390/batteries7040085
Ströbel, Marco, Julia Pross-Brakhage, Mike Kopp, and Kai P. Birke. 2021. "Impedance Based Temperature Estimation of Lithium Ion Cells Using Artificial Neural Networks" Batteries 7, no. 4: 85. https://doi.org/10.3390/batteries7040085
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Паскалов закон — Википедија
Паскалов закон или принцип за пренос на притисокот (наречен и Паскалово начело[1][2][3]) — начело во механиката на флуидите промените на притисокот настануваат насекаде во затворена непритислива течност на начин така што промените во притисокот се случуваат подеднакво насекаде низ садот.[4] Законот е воспоставен од францускиот математичар Блез Паскал.[5]
3 Pascal's barrel
4 Applications of Pascal's law
DefinitionУреди
Pressure in water and air. Pascal's law applies only for fluids.
Pascal's principle is defined as
This principle is stated mathematically as:
{\displaystyle \Delta P=\rho g(\Delta h)\,}
{\displaystyle \Delta P}
is the hydrostatic pressure (given in pascals in the SI system), or the difference in pressure at two points within a fluid column, due to the weight of the fluid;
ρ is the fluid density (in kilograms per cubic meter in the SI system);
g is acceleration due to gravity (normally using the sea level acceleration due to Earth's gravity, in SI in metres per second squared);
{\displaystyle \Delta h}
is the height of fluid above the point of measurement, or the difference in elevation between the two points within the fluid column (in metres in SI).
The intuitive explanation of this formula is that the change in pressure between 2 elevations is due to the weight of the fluid between the elevations. A more correct interpretation, though, is that the pressure change is caused by the change of potential energy per unit volume of the liquid due to the existence of the gravitational field. Note that the variation with height does not depend on any additional pressures. Therefore, Pascal's law can be interpreted as saying that any change in pressure applied at any given point of the fluid is transmitted undiminished throughout the fluid.
If a U-tube is filled with water and pistons are placed at each end, pressure exerted against the left piston will be transmitted throughout the liquid and against the bottom of the right piston. (The pistons are simply "plugs" that can slide freely but snugly inside the tube.) The pressure that the left piston exerts against the water will be exactly equal to the pressure the water exerts against the right piston. Suppose the tube on the right side is made wider and a piston of a larger area is used; for example, the piston on the right has 50 times the area of the piston on the left. If a 1 N load is placed on the left piston, an additional pressure due to the weight of the load is transmitted throughout the liquid and up against the larger piston. The difference between force and pressure is important: the additional pressure is exerted against the entire area of the larger piston. Since there is 50 times the area, 50 times as much force is exerted on the larger piston. Thus, the larger piston will support a 50 N load - fifty times the load on the smaller piston.
Forces can be multiplied using such a device. One newton input produces 50 newtons output. By further increasing the area of the larger piston (or reducing the area of the smaller piston), forces can be multiplied, in principle, by any amount. Pascal's principle underlies the operation of the hydraulic press. The hydraulic press does not violate energy conservation, because a decrease in distance moved compensates for the increase in force. When the small piston is moved downward 100 centimeters, the large piston will be raised only one-fiftieth of this, or 2 centimeters. The input force multiplied by the distance moved by the smaller piston is equal to the output force multiplied by the distance moved by the larger piston; this is one more example of a simple machine operating on the same principle as a mechanical lever.
Pascal's principle applies to all fluids, whether gases or liquids. A typical application of Pascal's principle for gases and liquids is the automobile lift seen in many service stations (the hydraulic jack). Increased air pressure produced by an air compressor is transmitted through the air to the surface of oil in an underground reservoir. The oil, in turn, transmits the pressure to a piston, which lifts the automobile. The relatively low pressure that exerts the lifting force against the piston is about the same as the air pressure in automobile tires. Hydraulics is employed by modern devices ranging from very small to enormous. For example, there are hydraulic pistons in almost all construction machines where heavy loads are involved.
Pascal's barrelУреди
The effects of Pascal's law in the (possibly apocryphal) "Pascal's barrel" experiment.
Pascal's barrel is the name of a hydrostatics experiment allegedly performed by Blaise Pascal in 1646.[6] In the experiment, Pascal inserted a 10-m long (32.8 ft) vertical tube into a barrel filled with water.[7] When water was poured into the vertical tube, Pascal found that the increase in hydrostatic pressure caused the barrel to burst.[6]
The experiment is mentioned nowhere in Pascal's preserved works and it may be apocryphal, attributed to him by 19th-century French authors, among whom the experiment is known as crève-tonneau (approx.: "barrel-buster");[8] nevertheless the experiment remains associated with Pascal in many elementary physics textbooks.[9]
Applications of Pascal's lawУреди
The underlying principle of the hydraulic jack and hydraulic press.
Force amplification in the braking system of most motor vehicles.
Used in artesian wells, water towers, and dams.
Scuba divers must understand this principle. At a depth of 10 meters under water, pressure is twice the atmospheric pressure at sea level, and increases by about 100 kPa for each increase of 10 m depth.[5]
Usually Pascal's rule is applied to confined space (static flow), but due to the continuous flow process, Pascal's principle can be applied to the lift oil mechanism (which can be represented as a U tube with pistons on either end). However, the lift height will be in microns because energy will be drained and pressure will be diminished after each impact with the lifting material, but force exerted will be equal.
Applied force in cylinder P1A1.
The underlying principal of hot isostatic pressing
Pascal's contributions to the physical sciences
↑ http://www.britannica.com/EBchecked/topic/445445/Pascals-principle
↑ https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/Pascals_principle.html
↑ http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html
↑ Bloomfield, Louis (2006). How Things Work: The Physics of Everyday Life (Third Edition). John Wiley & Sons. стр. 153. ISBN 0-471-46886-X.
↑ 5,0 5,1 Acott, Chris (1999). „The diving "Law-ers": A brief resume of their lives“. South Pacific Underwater Medicine Society journal. 29 (1). ISSN 0813-1988. OCLC 16986801. Посетено на 2011-06-14. .
↑ 6,0 6,1 Merriman, Mansfield (1903). Treatise on hydraulics (изд. 8.). J. Wiley. стр. 22.
↑ Wine East. 22-23. L & H Photo Journalism. 1994. стр. 23.
↑ perhaps first in an educational context; the attribution is found under this name in A. Merlette, L'encyclopédie des écoles, journal de l'enseignement primaire et professionnel (1863) p. 284: l'expérience du crève-tonneau réalisée pour la première fois par le célèbre Biaise Pascal. Ernest Menu de Saint-Mesmin, Problèmes de mathématiques et de physique: donnés dans les Facultés des science et notamment à la Sorbonne, avec les solutions raisonnées, L. Hachette (1862), p. 380.
↑ see e.g. E. Canon-Tapia in: Thor Thordarson (ed.) Studies in Volcanology, 2009, ISBN 9781862392809, p. 273.
Преземено од „https://mk.wikipedia.org/w/index.php?title=Паскалов_закон&oldid=4671759“
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Magnetic flux and Faraday's law (qualitative) Practice Problems Online | Brilliant
Suppose that when you move a magnet through a
40
-turn coil at a certain speed, a voltage of
10\text{ V}
is induced across the coil. If you move the same magnet through a
80
-turn coil at the same speed, what voltage will be induced across the coil, assuming both coils are closely packed?
5\text{ V}
10\text{ V}
20\text{ V}
We don't have enough information
Two circular loops lie in the
xy
z
-axis passes through the center of the left loop, as shown in the above figure. By switching on, a counterclockwise current start to flow in the left loop, as viewed from a point on the positive
z
-axis. Then which of the following statements is true of the right loop at that moment?
An induced current moves counterclockwise. An induced current moves clockwise. The current remains zero. None of these.
According to Faraday's law, an ammeter resists a current in the wire loop when the magnet is moving with respect to the loop. Which of the following can induce an emf?
(a) Change the magnitude
B
of the magnetic field within the loop.
(b) Change the total area of the loop that lies within the magnetic field.
(c) Change the angle between the direction of the magnetic field
\vec{B}
and the plane of the loop.
(b) and (c) only (c) only (a) and (b) only (a), (b) and (c)
A wire loop is moving with constant velocity
v
through a uniform magnetic field, as shown in the above diagram. This field may be produced, for example, by a large electromagnet. The dotted rectangle shows the assumed limits of the magnetic field; the fringing of the field at its edges is neglected. The magnetic field is directed into the plane of the screen. Choose the moment at which the current
is induced in the wire loop.
A and D All of them B and C D only
In diagram A above, an equilateral triangle loop is just entering, at time
t_A=0,
a region of uniform magnetic field. In diagram B, C and D at some later times
t_D > t_C > t_B > t_A,
the triangle loop has moved distances
x_1,
x_2
x_3,
respectively, into the magnetic field. If the triangle loop is moving with constant velocity
v,
what is the appropriate order of the magnitudes of the induced current
in the triangle loop?
i_A > i_D > i_C > i_B
i_D > i_C > i_B > i_A
i_A > i_B > i_C > i_D
i_C > i_B > i_A > i_D
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Archimedes, Vitruvius and Leonardo: The Odometer Connection
Massimo Callegari, Stefano Brillarelli, Cecilia Scoccia
Department of Industrial Engineering & Mathematical Sciences, Polytechnic University of Marche, Ancona, Italy.
Abstract: A multimedia exhibition in Fano (I) in 2019 celebrated the tight links between Vitruvius and Leonardo on the 500th year since Leonardo’s death. The authors realized an interactive animation of some machines to let the visitors enjoy an immersive experience of the studies of the great scholars of the past. They also took the opportunity to review the history of the odometer and to study how Leonardo redesigned the concept of Vitruvius. A few questions are still unanswered, but the search drove them back to another great scientist of the past, Archimedes of Syracuse.
Keywords: History of Machines, Mechanisms, Leonardo, Vitruvius, Archimedes, Odometer
1.1. The Odometer
The noun “odometer” derives from ancient Greek ὁδόμετρον, hodómetron, from ὁδός, hodós (“path” or “gateway”) and μέτρον, métron (“measure”) and identifies an instrument that has the purpose of measuring the distance travelled by a vehicle on land or at sea. The large majority of odometers are nowadays electronic or electromechanical but in ancient times they were purely mechanical. Designs have evolved over the years, but the basic function was always the same. They have always been used to measure distance in things like automobiles, see Figure 1, bicycles, and back before these were invented, horse drawn carts.
Possibly the first evidence for the use of an odometer can be found in the works of the ancient Greek Strabo (64 - 63 BC-24 AD) and the ancient Roman
Figure 1. An odometer and a trip meter installed in a modern car.
Pliny (23 - 79 AD). Both authors list the distances of routes traveled by Alexander the Great (356 - 323 BC) as by his bematists Diognetus and Baeton: bematists were trained to measure distance while counting their steps but the high accuracy of the bematists’ measurements rather indicates the use of a mechanical device. In fact from the nine surviving bematists’ measurements in Pliny’s Naturalis Historia eight show a deviation of less than 5% from the actual distance, three of them being within 1%: the overall accuracy of the measurements implies that the bematists already must have used a sophisticated device for measuring distances, although there is no direct mention of such a device.
The first descriptions of this instrument have reached us through the manuscripts of the Roman architect and military engineer Marcus Vitruvius Pollio (80 - 15 BC). Around 30 - 15 BC he wrote the famous “De architectura” dedicated to his patron, the emperor Caesar Augustus (Vitruvius, 1990): it is a ten books treatise meant as a guide for realizing projects such as buildings, military camps, facilities, instruments and machines. The odometer described by Vitruvius will be deeply explained in the next section.
As a matter of fact, the odometer has been studied by many scholars throughout history: Hero of Alexandria (10 - 70 BC) described an odometer set in motion by the rotation of a cart’s wheel and Leon Battista Alberti in the mid-fifteenth century studied this instrument in the treatise “Ludi rerum mathematicarum”. Between 1500 and 1504 Leonardo da Vinci (1452-1519) too was interested in the design of the odometer and drew few sketches on folio 1 of the Atlantic Codex preserved today in the Veneranda Biblioteca Ambrosiana in Milan.
In the end, it must be noticed that the odometer was also independently invented in ancient China (Needham, 1965) possibly by the prolific inventor and early scientist Zhang Heng (78 AD - 139 AD) or by the later Ma Jun (active during 220 - 265 AD); despite such associations, there is evidence to suggest that the invention of the odometer was a gradual process in Han Dynasty China that centered around the huang men court people (i.e. eunuchs, palace officials, attendants and familiars, actors, acrobats, etc.) that would follow the musical procession of the royal “drum-chariot”.
1.2. The Odometer within the Evolution of Design Theory
Mechanisms design today is a process of creation, invention and definition, involving an eventual synthesis of contributory and often conflicting factors. Design rules and concepts were practiced extensively by the engineers of ancient times leading to machine design from machine elements to the design of a machine as a system.
In fact, since ancient times, many scientists and inventors have contributed to the study of mechanisms and machines. Most of them where Hellenic, Byzantines or Arabic. Originally, this kind of machines have been developed for practical requirements and then improved, using precise scientific concepts. For instance, the lever and the wedge have been used in various forms, even centuries before Archimedes was born (Dimarogonas, 1991): levers appeared in 5000 BC as a simple balance scale and in 3000 BC a shaduf in Mesopotamia, using a lever system, allowed an operator to fill a bucket with water.
The first mathematical approaches to mechanisms design date back to the third century BC, developed by great scholars such as Ctesibius, Philo, Hero of Alexandria and Archimedes of Syracuse (Chondros, 2007, 2009). Their interest was generally devoted to war machines, such as ballistae, catapults, burning glasses, steam cannons and giant claws. As is well known, Leonardo da Vinci himself paid much attention to this type of mechanism and the strength of materials. There are many examples of this in the Codex Atlanticus and in his studies based on the De architectura of Vitruvius (Oliveira, 2019). The most iconic are winches, flying machines, machine guns, aerial screws, cranes and revolving bridges. But speaking about Leonardo’s studies, it cannot be skipped his vision about programmable automation shown in the concepts of the Lion, the Knight and the Bell’s Ringer, as reported in the Codex Madrid I and II (Rosheim, 2006).
Interesting surveys on machine designs of the Renaissance period can be found in (Ceccarelli & De Paolis, 2008; Rossi et al., 2011; Cigola & Ceccarelli, 2016): by using drawings reproducing Vitruvius’s machines from Book X of his De Architectura (e.g. the editions by Fra’ Giovanni Giocondo in 1513, Cesare Cesariano in 1521 and Daniele Barbaro in 1584) they analyze machine designs and drawings, both as an interpretation of Roman machines and inspiration for Renaissance designs.
One of the most important “enabling technology” for the development of modern design theory was certainly gearing systems, born during the Hellenistic period: Cicerone in De Re Publica mentioned a cycloidal gearwheel, stating that it had been built by Archimedes. Screw theory seems to be the common thread that has linked many of the studies conducted by Archimedes, Vitruvius and Leonardo da Vinci (in the Modern Era) (Chondros, 2010).
The gearwheel coupling was largely used for war machines, astronomical devices for the estimation of the position of celestial bodies and also odometers. One of the most important examples of the gear coupling is the mechanism of Antikythera, whose distinctive feature was the triangular shape of the teeth (Sorge, 2012). In this mechanism only one tooth pair is in contact at each instant. A series of studies have been performed to understand its operating conditions, e.g. De Solla Price (1974) and later Wright (2005) reconstructed the planetary system.
With triangular teeth, motion transmission is continuous but the speed ratio is variable: because of the higher velocity of the teeth during approach with respect to recess, an impact occurs at the beginning of the meshing phase. The gearing behavior was certainly characterized by a sensible rattling noise due to the teeth collisions and the energy losses were relevant but with a rudimentary lubrication the contact might be good enough.
As shown in the following sections, Sleeswyk (1981) built a prototype of the Vitruvius odometer in which replaced the square-toothed gear designs of da Vinci with the triangular, pointed teeth found in the Antikythera mechanism: with this modification, the Vitruvius odometer could work.
1.3. The Fano Exhibition
2. Vitruvius’ Odometer
Vitruvius describes two versions of the odometer in chapter 9 of book X of “De Architectura”: one for measuring marine distances and the other for land travels, which is the one that will be commented here. Before turning to Vitruvius’ odometer, it is useful to briefly recall the most important measurement units used in ancient Rome: the “foot”, corresponding to 296.4 mm; the “double step”, which was 5 feet long or 1482 mm; the “mile”, corresponding to 1000 double steps (the word mile derives from the Latin expression “millia passuum”), which was therefore equal to 1482 m.
Vitruvius’ odometer, Figure 3, is driven by one of the two rear wheels of a four-wheeled carriage, called “raeda”. The wheels have a diameter of 4 feet (about 1.2 meters) which roughly corresponds to a circumference of 12.5 feet; therefore it requires 400 turns of the wheels to measure the distance of 1 mile.
Figure 2. A visitor exploiting the digital mock-up of the giant crossbow of Leonardo at the image wall.
3. Leonardo’s Odometer
Figure 9. Longitudinal shaft and the two mileage counting systems.
Figure 10. Peg wheel and drum with pebbles.
Figure 11. Use of the worm gear in the odometer of Leonardo da Vinci.
4. Comparison between the Two Designs
A first difference between the two works lies in the value used for the constant π, when calculating the diameter of the cart wheels. According to Vitruvius the wheels must have a diameter of 4 feet and a circumference of 12.5 feet for which a value π = 3.125 was evidently used; Leonardo, on the other hand, uses the value
\text{π}=\left(\text{3}+\text{1}/\text{7}\right)=3.\stackrel{¯}{142857}
closer to the real value. Therefore the odometer of Vitruvius would commit an error of 2654.8 feet (equal to 786.89 m) in the measurement of 100 Roman miles (with an error of about 0.5%) while that of Leonardo, more precise, would overestimate the distance of 100 Florentine miles by 120 arms (equivalent to 70 m) (with an error less than one tenth of the previous one).
5. The Influence of Archimedes
The study of the writings of Vitruvius and Leonardo tells us how the liaisons between the two scholars were strong despite the 1500 years that separate them. Leonardo was fascinated by the work of Vitruvius and has been looking for a copy of “De Architectura” for many years. The exhibition in Fano (July 12-Oct. 13, 2019) has shown how much Vitruvius influenced Leonardo’s work. As for the odometer, Leonardo tried to draw the original concept by Vitruvius in the central drawing of folio 1r-b of the Codex Atlanticus (Figure 6) but probably gave up understanding it was practically impossible to realize. In fact the teeth of gear wheels in past years had the shape of equilateral triangles as shown for example in the Antikythera gearwheels and a 1:400 gear ratio causes a too large diameter or too small teeth, as would happen todays too with current design criteria. Leonardo overcame the problem in his design (see the left of Figure 6) by using a worm gear and a single tooth meshing with a peg wheel.
Cite this paper: Callegari, M. , Brillarelli, S. and Scoccia, C. (2020) Archimedes, Vitruvius and Leonardo: The Odometer Connection. Advances in Historical Studies, 9, 330-343. doi: 10.4236/ahs.2020.95025.
[1] Borgo, F. (2019). Leonardo legge Vitruvio. In Leonardo e Vitruvio: Oltre il cerchio e il quadrato (pp. 23-39). Venice: Marsilio.
[2] Brillarelli, S., Callegari, M., Carbonari, L., & Clini, P. (2020). Digital Experience of the Work of Vitruvius and Leonardo. IOP Conference Series: Materials Science and Engineering, 949, Article ID: 012041.
[3] Callegari, M., Brillarelli, S., & Scoccia, C. (2021). The Odometers of Marcus Vitruvius Pollio and Leonardo da Vinci. Mechanisms and Machine Science, 91, 75-82.
[4] Ceccarelli, M., & De Paolis, P. (2008). A Survey on Roman Engineers and Their Machines. In Proceedings of the III Congreso Internacional de Patrimonio e Historia de la Ingenieria (pp. 29-48). Las Palmas de Gran Canaria: Centro Internacional de Conservacion de Patrimonio.
[5] Chondros, T. G. (2007). Archimedes (287-212 BC). In M. Ceccarelli (Ed.), Distinguished Figures in Mechanism and Machine Science (pp. 1-30). History of Mechanism and Machine Science, Vol. 1, Dordrecht: Springer.
[6] Chondros, T. G. (2009). The Development of Machine Design as a Science from Classical Times to Modern Era. In H. S. Yan, & M. Ceccarelli (Eds.), International Symposium on History of Machines and Mechanisms (pp. 59-68). Dordrecht: Springer.
[7] Chondros, T. G. (2010). Archimedes Life Works and Machines. Mechanism and Machine Theory, 45, 1766-1775.
[8] Cigola, M., & Ceccarelli, M. (2014). Marcus Vitruvius Pollio (Second Half of the Ist Century B.C.). History of Mechanisms and Machine Science, 26, 307-344.
[9] Cigola, M., & Ceccarelli, M. (2016). Machine Designs and Drawings in Renaissance Editions of de Architectura by Marcus Vitruvius Pollio. History of Mechanisms and Machine Science, 31, 291-307.
[10] De Solla Price, D. J. (1974). Gears from the Greeks. The Antikythera Mechanism: A Calendar Computer from ca. 80 B.C. Transactions of American Philosophical Society, New Series, 64, 19.
[11] Dimarogonas, A. D. (1991). The Origins of the Theory of Machines and Mechanisms. Proceedings 40 Years of Modern Kinematics: A Tribute to Ferdinand Freudenstein Conference, Minneapolis.
[12] Needham, J. (1965). Physics and Physical Technology, Part 2, Mechanical Engineering. Science and Civilisation in China (Vol. 4). Cambridge: Cambridge University Press.
[13] Oliveira, A. R. E. (2019). The Mechanical Sciences in Leonardo da Vinci’s Work. Advances in Historical Studies, 8, 215-238.
[14] Rosheim, M. E. (2006). Leonardo’s Lost Robots. Dordrecht: Springer.
[15] Rossi, C., Ceccarelli, M., & Cigola, M. (2011). La Groma, lo Squadro agrimensorio e il corobate. Note di approfondimento su progettazione e funzionalità di antiche strumentazioni, Disegnare Idee Immagini, anno XI n. 42, 22-33.
[16] Scaglia, G. (1992). Francesco Di Giorgio: Checklist and History of Manuscripts and Drawings in Autographs and Copies from Ca. 1470 to 1687 and Renewed Copies. Bethlehem: Lehigh Univ Pr.
[17] Schofield, R. V. (2016). Notes on Leonardo and Vitruvius. In C. Moffatt, & S. Taglialagamba (Eds.), Illuminating Leonardo (Vol. 1, pp. 120-133). Leiden: Brill.
[18] Sleeswyk, A. W. (1981). Vitruvius’ Odometer. Scientific American, 245, 188-200.
[19] Sorge, F. (2012). Coupling Mechanics of Antikythera Gearwheels. Journal of Mechanical Design, 134, Article ID: 061007.
[20] Taddei, M., Zanon, E., & Laurenza, D. (2005). Leonardo’s Machines. Secrets and Inventions in the Da Vinci Codices. Milano: Giunti.
[21] Vitruvius (1990). De Architectura. Translation by Luciano Migotto (Latin and Italian), Rome: Edizioni Studio Tesi.
[22] Wright, M. T. (2005). The Antikythera Mechanism: A New Gearing Scheme. Bulletin of the Scientific Instrument Society, 85, 2-7.
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Force is required to lift a body from the ground to a height h and work is measured as - Physics - Work Energy And Power - 11770987 | Meritnation.com
Force is required to lift a body from the ground to a height h and work is measured
as the product of force and magnitude of displacement.
a) Name the energy possessed by the body at maximum height. Write an equation for it.
b) A man of mass 60 kg carries a stone of mass 20 kg to the top of a multi-storeyed building of height 50m. Calculate the total energy spent by him? (9.8m/s2)
a\right). the energy possesed is gravitational potential enrgy .\phantom{\rule{0ex}{0ex}}eqn for PE for mass m at height h \phantom{\rule{0ex}{0ex}}U=mgh\phantom{\rule{0ex}{0ex}}b\right). total mass M=60+20=80 kg .\phantom{\rule{0ex}{0ex}}h=50m.\phantom{\rule{0ex}{0ex}}work done by the man =mgh=80*9.8*50=39200J\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Regards
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Unrestricted Hartree–Fock - Wikipedia
Unrestricted Hartree–Fock
Unrestricted Hartree–Fock (UHF) theory is the most common molecular orbital method for open shell molecules where the number of electrons of each spin are not equal. While restricted Hartree–Fock theory uses a single molecular orbital twice, one multiplied by the α spin function and the other multiplied by the β spin function in the Slater determinant, unrestricted Hartree–Fock theory uses different molecular orbitals for the α and β electrons. This has been called a different orbitals for different spins (DODS) method. The result is a pair of coupled Roothaan equations, known as the Pople–Nesbet–Berthier equations.[1][2]
{\displaystyle \mathbf {F} ^{\alpha }\ \mathbf {C} ^{\alpha }\ =\mathbf {S} \mathbf {C} ^{\alpha }\ \mathbf {\epsilon } ^{\alpha }\ }
{\displaystyle \mathbf {F} ^{\beta }\ \mathbf {C} ^{\beta }\ =\mathbf {S} \mathbf {C} ^{\beta }\ \mathbf {\epsilon } ^{\beta }\ }
{\displaystyle \mathbf {F} ^{\alpha }\ }
{\displaystyle \mathbf {F} ^{\beta }\ }
are the Fock matrices for the
{\displaystyle \alpha \ }
{\displaystyle \beta \ }
orbitals,
{\displaystyle \mathbf {C} ^{\alpha }\ }
{\displaystyle \mathbf {C} ^{\beta }\ }
are the matrices of coefficients for the
{\displaystyle \alpha \ }
{\displaystyle \beta \ }
{\displaystyle \mathbf {S} }
is the overlap matrix of the basis functions, and
{\displaystyle \mathbf {\epsilon } ^{\alpha }\ }
{\displaystyle \mathbf {\epsilon } ^{\beta }\ }
are the (diagonal, by convention) matrices of orbital energies for the
{\displaystyle \alpha \ }
{\displaystyle \beta \ }
orbitals. The pair of equations are coupled because the Fock matrix elements of one spin contains coefficients of both spin as the orbital has to be optimized in the average field of all other electrons. The final result is a set of molecular orbitals and orbital energies for the α spin electrons and a set of molecular orbitals and orbital energies for the β electrons.
This method has one drawback. A single Slater determinant of different orbitals for different spins is not a satisfactory eigenfunction of the total spin operator -
{\displaystyle \mathbf {S} ^{2}}
. The ground state is contaminated by excited states. If there is one more electron of α spin than β spin, the ground state is a doublet. The average value of
{\displaystyle \mathbf {S} ^{2}}
{\displaystyle \langle \mathbf {S} ^{2}\rangle }
, should be
{\displaystyle {\tfrac {1}{2}}({\tfrac {1}{2}}+1)=0.75}
but will actually be rather more than this value as the doublet state is contaminated by a quadruplet state. A triplet state with two excess α electrons should have
{\displaystyle \langle \mathbf {S} ^{2}\rangle }
= 1 (1 + 1) = 2, but it will be larger as the triplet is contaminated by a quintuplet state. When carrying out unrestricted Hartree–Fock calculations, it is always necessary to check this contamination. For example, with a doublet state, if
{\displaystyle \langle \mathbf {S} ^{2}\rangle }
= 0.8 or less, it is probably satisfactory. If it is 1.0 or so, it is certainly not satisfactory and the calculation should be rejected and a different approach taken. It requires experience to make this judgment. Even singlet states can suffer from spin-contamination, for example the H2 dissociation curve is discontinuous at the point when spin-contamination states (known as the Coulson–Fischer point[3]).
Despite this drawback, the unrestricted Hartree–Fock method is used frequently, and in preference to the restricted open-shell Hartree–Fock (ROHF) method, because UHF is simpler to code, easier to develop post-Hartree–Fock methods with, and returns unique functions unlike ROHF where different Fock operators can give the same final wave function.
Unrestricted Hartree–Fock theory was discovered by Gaston Berthier and subsequently developed by John Pople; it is found in almost all ab initio programs.
^ Berthier, Gaston (1954). "Extension de la methode du champ moleculaire self-consistent a l'etude des couches incompletes" [Extension of the method of molecular self-consistent field to the study of incomplete layers]. Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences (in French). 238: 91–93.
^ Pople, J. A.; Nesbet, R. K. (1954). "Self-Consistent Orbitals for Radicals". The Journal of Chemical Physics. 22 (3): 571. Bibcode:1954JChPh..22..571P. doi:10.1063/1.1740120.
^ Coulson, C.A.; Fischer, I. (1949). "XXXIV. Notes on the molecular orbital treatment of the hydrogen molecule". Philosophical Magazine. Series 7. 40 (303): 386–393. doi:10.1080/14786444908521726.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Unrestricted_Hartree–Fock&oldid=1027009850"
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The determinant of a square matrix is a value determined by the elements of the matrix. In the case of a
2 \times 2
matrix, the determinant is calculated by
\text{det}\begin{pmatrix}a & b \\ c & d \end{pmatrix} = ad-bc,
\text{det}
from the set of square matrices to the set of real numbers, that satisfies 3 important properties:
\text{det}(I) = 1
\text{det}
is linear in the rows of the matrix.
M
\det(M)=0
(0,2,3)
2 \cdot (0,1,0) + 3 \cdot (0,0,1)
\text{det}\begin{pmatrix}1&0&0\\0&2&3\\0&0&1\end{pmatrix} = 2 \cdot \text{det}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+3 \cdot \text{det}\begin{pmatrix}1&0&0\\0&0&1\\0&0&1\end{pmatrix}=2.
\text{det}(AB)=\text{det}(A)\text{det}(B)
A'
\text{det}(A)=\text{det}(A')
\text{det}(A) = \text{det}(A^T)
A'
\text{det}(A')=-\text{det}(A)
{\begin{cases} a^2 - b^2 &=& 5 \\ c^2 + d^2 &=& 74 \\ (ac)^2 - (bd)^2 &=& 341 \\ (ad)^2 - (bc)^2 &=& ? \end{cases} }
The determinant by minors method calculates the determinant using recursion. The base case is simple: the determinant of a
1 \times 1
a
a
\text{det}\begin{pmatrix}a\end{pmatrix} = a \cdot \text{det}\begin{pmatrix}1\end{pmatrix}=a
\text{det}\begin{pmatrix}1\end{pmatrix} = I
A_{ij}
i^\text{th}
j^\text{th}
A = \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} \implies A_{11} = \begin{pmatrix}5&6\\8&9\end{pmatrix}.
n \times n
A
\text{det}(A) = \sum_{i=1}^n (-1)^{i+1}a_{1,i}\text{det}(A_{1i}) = a_{1,1}\text{det}A_{11}-a_{1,2}\text{det}A_{12}+\cdots.
\begin{pmatrix}a&b\\c&d\end{pmatrix}?
\text{det}\begin{pmatrix}a&b\\c&d\end{pmatrix} = a ~\text{det}\begin{pmatrix}d\end{pmatrix} - b ~\text{det}\begin{pmatrix}c\end{pmatrix} = ad-bc.\ _\square
\det\left(\begin{array}{cc}1& a\\2& b \end{array}\right)=4
\det\left(\begin{array}{cc}1& b\\2& a \end{array}\right)=1,
a^2+b^2?
3 \times 3
An alternate method, determinant by permutations, calculates the determinant using permutations of the matrix's elements. Let
\sigma
\{1, 2, 3, \ldots, n\}
S
n \times n
A
\sum_{\sigma \in S}\left(\text{sgn}(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}\right).
n
A
-1
if the permutation has the odd sign. The determinant is the sum over all choices of these
n
\left(\begin{array}{cc}1&0&-1&9&11\\0&-6&-1&9&11\\0&0&\frac{1}{3}&-80&\frac{1}{3}\\0&0&0&9&7\\0&0&0&0&-5 \end{array}\right).
\begin{pmatrix}a&b\\c&d\end{pmatrix}?
\{1,2\}
\{1,2\}
\{2,1\}
. The first has positive sign (as it has 0 transpositions) and the second has negative sign (as it has 1 transposition), so the determinant is
\text{det}(A) = \sum_{\sigma \in S}\left(\text{sgn}(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}\right) = 1 \cdot a_{1,1}a_{2,2} + (-1) \cdot a_{1,2}a_{2,1} = ad-bc.
_\square
\det\left(\begin{array}{cc}2&6&4\\-3&1&5\\9&3&7 \end{array}\right).
X=\text{det}\begin{vmatrix} a & b & c & d \\ 0 & f & g & h \\ 0 & 0 & k & l \\ 0 & 0 & 0 & p \end{vmatrix}=a\times f\times k\times p.
Lower triangular determinant (elements which are above the main diagonal are zero):
X=\text{det}\begin{vmatrix} a & 0 & 0 & 0 \\ e & f & 0 & 0 \\ i & j & k & 0 \\ m & n & o & p \end{vmatrix}=a\times f\times k\times p.
Diagonal determinant (elements which are under and above the main diagonal are zero):
X=\text{det}\begin{vmatrix} a & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & k & 0 \\ 0 & 0 & 0 & p \end{vmatrix}=a\times f\times k\times p.
X=\begin{vmatrix} 1 & 2 & 2 & 1 \\ 1 & 2 & 4 & 2 \\ 2 & 7 & 5 & 2 \\ -1 & 4 & -6 & 3 \end{vmatrix}.
\begin{aligned} [X]=&\begin{bmatrix} 1 & 2 & 2 & 1 \\ 1 & 2 & 4 & 2 \\ 2&7&5&2 \\ -1&4&-6&3 \end{bmatrix} \\\\\\ \begin{matrix} \text{row}_1 \rightarrow \text{row}_1 \\ \text{row}_2 - 2\text{row}_1 \rightarrow \text{row}_2 \\ \text{row}_3 - 2\text{row}_1 \rightarrow \text{row}_3 \\ \text{row}_4 - 3\text{row}_1 \rightarrow \text{row}_4 \end {matrix} \Rightarrow &\begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -4&-2&-12&0 \end{bmatrix} \\\\\\ \begin{matrix} \text{row}_1 \rightarrow \text{row}_1 \\ \text{row}_2 \rightarrow \text{row}_2 \\ \text{row}_3 \rightarrow \text{row}_3 \\ \text{row}_4 +12\text{row}_3 \rightarrow \text{row}_4 \end {matrix} \Rightarrow &\begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -4&34&0&0 \end{bmatrix} \\\\\\ \begin{matrix} \text{row}_1 \rightarrow \text{row}_1 \\ \text{row}_2 \rightarrow \text{row}_2 \\ \text{row}_3 \rightarrow \text{row}_3 \\ \text{row}_4 +17\text{row}_2 \rightarrow \text{row}_4 \end {matrix} \Rightarrow &\begin{bmatrix} 1&2&2&1\\ -1&-2&0&0\\ 0&3&1&0\\ -21&0&0&0 \end{bmatrix} \\\\\\ \begin{matrix} \text{row}_4 \rightarrow \text{row}_1 \\ \text{row}_2 \rightarrow \text{row}_2 \\ \text{row}_3 \rightarrow \text{row}_3 \\ \text{row}_1 \rightarrow \text{row}_4 \end {matrix} \Rightarrow - &\begin{bmatrix} -21&0&0&0\\ -1&-2&0&0\\ 0&3&1&0\\ 1&2&2&1 \end{bmatrix}. \end{aligned}
\det {[X]} = X = -(-21)(-2)(1)(1) = -42.\ _\square
Sarrus' rule is a shortcut for calculating the determinant of a
3 \times 3
Rewrite the first two rows while occupying hypothetical fourth and fifth rows, respectively:
\left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right| \Rightarrow \left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right| \\ \quad \quad \quad \quad \quad \quad \ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}
Multiply the diagonal elements:
\begin{matrix} \left| \begin {matrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{matrix}\right| \\ \begin{matrix} 1 & 2 & 3 \\ 4& 5 & 6 \end{matrix}\end{matrix}= 1 \cdot 5 \cdot 9+4 \cdot 8\cdot 3+7\cdot 2 \cdot 6 -3\cdot 5 \cdot 7 -6 \cdot 8 \cdot 1 - 9 \cdot 2 \cdot 4 = 0.
The descending diagonal from left to right has a
+
sign , while the descending diagonal from right to left has a
-\text{}
\left| \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \\ 5 & 7 & 5 \end{matrix} \right|.
\left| \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \\ 5 & 7 & 5 \end{matrix} \right| \Rightarrow \left| \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \\ 5 & 7 & 5 \end{matrix} \right| \\\quad \quad \quad\quad \quad \quad \begin{matrix} 0 & 1 & 2 \\ 3 & 5 & 5 \end{matrix}
(0\times 5\times 5)+(3\times 7\times 2)+(5\times 1\times 5)-(2\times 5\times 5)-(5\times 7\times 0)-(5\times 1\times 3)=2.\ _\square
|
Compare generalized linear mixed-effects models - MATLAB - MathWorks Deutschland
altglme
Compare Mixed-Effects Models
Compare generalized linear mixed-effects models
results = compare(glme,altglme)
results = compare(glme,altglme,Name,Value)
results = compare(glme,altglme) returns the results of a likelihood ratio test that compares the generalized linear mixed-effects models glme and altglme. To conduct a valid likelihood ratio test, both models must use the same response vector in the fit, and glme must be nested in altglme. Always input the smaller model first, and the larger model second.
H0: Observed response vector is generated by glme.
H1: Observed response vector is generated by model altglme.
results = compare(glme,altglme,Name,Value) returns the results of a likelihood ratio test using additional options specified by one or more Name,Value pair arguments. For example, you can check if the first input model, glme, is nested in the second input model, altglme.
You can create a GeneralizedLinearMixedModel object by fitting a generalized linear mixed-effects model to your sample data using fitglme. To conduct a valid likelihood ratio test on two models that have response distributions other than normal, you must fit both models using the 'ApproximateLaplace' or 'Laplace' fit method. Models with response distributions other than normal that are fitted using 'MPL' or 'REMPL' cannot be compared using a likelihood ratio test.
altglme — Alternative generalized linear mixed-effects model
Alternative generalized linear mixed-effects model, specified as a GeneralizedLinearMixedModel object. altglme be must fit to the same response vector as glme, but with different model specifications. glme must be nested in altglme, such that you can obtain glme from altglme by setting some of the model parameters of altglme to fixed values such as 0.
Indicator to check nesting between two models, specified as the comma-separated pair consisting of 'CheckNesting' and either true or false. If 'CheckNesting' is true, then compare checks if the smaller model glme is nested in the larger model altglme. If the nesting requirements are not satisfied, then compare returns an error. If 'CheckNesting' is false, then compare does not perform this check.
results — Results of likelihood ratio test
Results of the likelihood ratio test, returned as a table with two rows. The first row is for glme, and the second row is for altglme. The columns of results contain the following.
LRStat Likelihood ratio test statistic for comparing altglme and glme
deltaDF DF for altglme minus DF for glme
Fit a fixed-effects-only model using newprocess, time_dev, temp_dev, and supplier as fixed-effects predictors. Specify the response distribution as Poisson, the link function as log, and the fit method as Laplace. Specify the dummy variable encoding as 'effects', so the dummy variable coefficients sum to 0.
FEglme = fitglme(mfr,'defects ~ 1 + newprocess + time_dev + temp_dev + supplier','Distribution','Poisson','Link','log','FitMethod','Laplace','DummyVarCoding','effects');
Fit a second model that uses the same fixed-effects predictors, response distribution, link function, and fit method. This time, include a random-effects intercept grouped by factory, to account for quality differences that might exist due to factory-specific variations.
{\text{defects}}_{ij}\sim \text{Poisson}\left({\mu }_{ij}\right)
\mathrm{log}\left({\mu }_{ij}\right)={\beta }_{0}+{\beta }_{1}{\text{newprocess}}_{ij}+{\beta }_{2}{\text{time}\text{_}\text{dev}}_{ij}+{\beta }_{3}{\text{temp}\text{_}\text{dev}}_{ij}+{\beta }_{4}{\text{supplier}\text{_}\text{C}}_{ij}+{\beta }_{5}{\text{supplier}\text{_}\text{B}}_{ij}+{b}_{i},
{\text{defects}}_{ij}
i
j
{\mu }_{ij}
i
i=1,2,...,20
j
j=1,2,...,5
{\text{newprocess}}_{ij}
{\text{time}\text{_}\text{dev}}_{ij}
{\text{temp}\text{_}\text{dev}}_{ij}
i
j
{\text{newprocess}}_{ij}
i
j
{\text{supplier}\text{_}\text{C}}_{ij}
{\text{supplier}\text{_}\text{B}}_{ij}
i
j
{b}_{i}\sim N\left(0,{\sigma }_{b}^{2}\right)
i
Compare the two models using a theoretical likelihood ratio test. Specify 'CheckNesting' as true, so compare returns a warning if the nesting requirements are not satisfied.
Since compare did not return an error, the nesting requirements are satisfied. The small
p
-value indicates that compare rejects the null hypothesis that the observed response vector is generated by the model FEglme, and instead accepts the alternate model glme. The smaller AIC and BIC values for glme also support the conclusion that glme provides a better fitting model for the response.
A likelihood ratio test compares the specifications of two nested models by assessing the significance of restrictions to an extended model with unrestricted parameters. Under the null hypothesis H0, the likelihood ratio test statistic has an approximate chi-squared reference distribution with degrees of freedom deltaDF.
When comparing two models, compare computes the p-value for the likelihood ratio test by comparing the observed likelihood ratio test statistic with this chi-squared reference distribution. A small p-value leads to a rejection of H0 in favor of H1, and acceptance of the alternate model altglme. On the other hand, a large p-value indicates that we cannot reject H0, and reflects insufficient evidence to accept the model altglme.
The p-values obtained using the likelihood ratio test can be conservative when testing for the presence or absence of random-effects terms, and anti-conservative when testing for the presence or absence of fixed-effects terms. Instead, use the fixedEffects or coefTest methods to test for fixed effects.
To conduct a valid likelihood ratio test on GLME models, both models must be fitted using a Laplace or approximate Laplace fit method. Models fitted using a maximum pseudo likelihood (MPL) or restricted maximum pseudo likelihood (REMPL) method cannot be compared using a likelihood ratio test. When comparing models fitted using MPL, the maximized log likelihood of the pseudodata from the final pseudo likelihood iteration is used in the likelihood ratio test. If you compare models with non-normal distributions fitted using MPL, then compare gives a warning that the likelihood ratio test is using maximized log likelihood of pseudodata from the final pseudo likelihood iteration. To use the true maximized log likelihood in the likelihood ratio test, fit both glme and altglme using approximate Laplace or Laplace prior to model comparison.
To conduct a valid likelihood ratio test, glme must be nested in altglme. The 'CheckNesting',true name-value pair argument checks the following requirements, and returns an error if any are not satisfied:
You must fit both models (glme and altglme) using the 'ApproximateLaplace' or 'Laplace' fit method. You cannot compare GLME models fitted using 'MPL' or 'REMPL' using a likelihood ratio test.
You must fit both models using the same response vector, response distribution, and link function.
The smaller model (glme) must be nested within the larger model (altglme), such that you can obtain glme from altglme by setting some of the model parameters of altglme to fixed values such as 0.
The maximized log likelihood of the larger model (altglme) must be greater than or equal to the maximized log likelihood of the smaller model (glme).
The weight vectors used to fit glme and altglme must be identical.
The random-effects design matrix of the larger model (altglme) must contain the random-effects design matrix of the smaller model (glme).
The fixed-effects design matrix of the larger model (altglme) must contain the fixed-effects design matrix of the smaller model (glme).
GeneralizedLinearMixedModel | covarianceParameters | fixedEffects | randomEffects
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Convert magnitude to decibels - MATLAB mag2db - MathWorks Deutschland
Magnitude Response of a Highpass Filter
Convert magnitude to decibels
ydb = mag2db(y) expresses in decibels (dB) the magnitude measurements specified in y. The relationship between magnitude and decibels is ydb = 20 log10(y).
Design a 3rd-order highpass Butterworth filter having a normalized 3-dB frequency of
0.5\pi
rad/sample. Compute its frequency response. Express the magnitude response in decibels and plot it.
Repeat the computation using fvtool.
Input array, specified as a scalar, vector, matrix, or N-D array. When y is nonscalar, mag2db is an element-wise operation.
ydb — Magnitude measurements in decibels
Magnitude measurements in decibels, returned as a scalar, vector, matrix, or N-D array of the same size as y.
|
Some Basic Principles of Organic Chemistry Previous Year Questions - Advanced Chemistry
Basic Principles of Organic Chemistry Question Bank
Some Basic Principles of Organic Chemistry Previous Year Questions
December 23, 2019 Mudassar Husain 0 Comments
NEET, JEE Main Previous Year Questions Solutions for Class 11th Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques / Special Cases JEE MAINS – NEET
NEET Previous Year Questions of Organic Chemistry – Some Basic Principles and Techniques
The students who are targeting NEET 2021 exam should prepare for all the important chapters of NEET syllabus to secure highest marks in NEET 2021 and NEET 2022. This questions for NEET 2020 designed by analyzing the previous years question paper and considering the weightage.
According to the NEET Chapter Wise Weightage Organic Chemistry – Some basic Principle and techniques chapter contributes for around 3% of the total number of questions asked in the last 32 years.
Last 7 Years NEET/AIPMT Sub-Topic wise Weightage Analysis for Organic Chemistry – Some basic Principle and techniques:
Now that you are aware of the topic & sub-topic analysis, the next step would be to prepare yourself for the questions that have already been asked in the previous papers for these topics.
NEET 2020 F 1: A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following ?
(1) – R effect of – CH3 groups (2) Hyperconjugation (3) – I effect of – CH3 groups (4) + R effect of – CH3 groups
5. The correct statement regarding electrophile is NEET- 2017
(a) Electrophile is a negatively charge species and can form a bond by accepting a pair of electrons from a nucleophile
(b) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
(c) Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
(d) Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Concept Questions :- Nucleophile and Electrophile
6. With respect to the conformers of ethane, which of the following statements is true? NEET- 2017
(a) Bond angle remains same but bond length changes
(b) Bond angle changes but bond length remains same
(c) Both bond angle and bond length change
(d) Both bond angles and bond length remain same
Concept Questions :- Conformational Isomers
7. The IUPAC name of the compound NEET- 2017
1. 3-keto-2-methylhex-4-enal
2. 5-formylhex-2-en-3-one
3. 5-methyl-4-oxohex-2-en-5-al
8. Which one is the correct order of acidity? NEET- 2017
(a) CH2=CH2 > CH3-CH=CH2> CH3CECH > CH≡CH
(b) CH≡CH > CH3-C≡CH > CH2=CH2> CH3-CH3
(c) CH≡CH > CH2=CH2 > CH3-C≡CH > CH3-CH3
(d) CH3-CH3> CH2=CH2 > CH3-C≡CH > CH≡CH
Concept Questions :- Acidic and Basic Character
9. The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is NEET-2017
(a) sublimation (b) chromatography (c) crystallisation (d) steam distillation
Concept Questions :- Purification of Organic Compounds, Cromatography and Distillation
Q 10. NEET 2016
13. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is NEET- 2016
(a) The eclipsed conformation of ethane is more stable than staggered conformation. because eclipsed conformation has no torsional strain
(b) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
(c) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain
(d) The staggered conformation of ethane is less stable than eclipsed conformation. because staggered conformation has torsional strain
14. In which of the following compounds, the C- CI bond ionisation shall give most stable carbonium ion ? NEET- 2015
15. consider the following compounds NEET- 2015
hyperconjugation occurs in (a) I only (b) II only (c) III only (d) I and Ill
Concept Questions :- Hyperconjugation Effects
16. The enolic form of ethyl acetoacelate as below has:- NEET 2015
(a) 18 sigma bonds and 2 pi-bonds (b) 16 sigma bonds and 1 pi-bond (c) 9 sigma bonds and 2 pi-bonds (d) 9 sigma bonds and 1 pi-bond
Concept Questions :- Shape, Hybridisation and Structure of Carbon Compounds
17. Which of the given compounds can exhibit tautomerism? NEET- 2015
(a) I and II (b) I and III (c) II and III (d) I, II and III
Concept Questions :- Structural Isomer
18. Which of the following statements is not correct for a nucleophile? NEET 2015
(a) Nucleophile is a Lewis acid (b) Ammonia is a nucleophile (c) Nucleophiles attack low electrons density sites (d) Nucleophiles are not electron seeking
19. In the Kjeldahl’s method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralised 10 mL of 1 M H2SO4. The percentage of nitrogen in the soil is- NEET 2014
Concept Questions :- Quantitative Analysis of Organic Compounds
20. Which one is most reactive towards nucleophilic addition reaction? NEET 2014
21. Structure of the compound whose IUPAC name is 3-ethy1-2-hydroxy-4-methylhex-3-en-5-ynoic acid is ? NEET 2013
22. The structure of isobutyl group in an organic compound is NEET 2013
23. Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? NEET 2013
(a) -C≡N (b) -SO3H (c) -COOH (d) -NO2
24. The radical
is aromatic because it has NEET 2013
(a) 6 p-orbitals and 6 unpaired electrons (b) 7 p-orbitals and 6 unpaired electrons (c) 7 p-orbitals and 7 unpaired electrons (d) 6 p-orbitals and 7 unpaired electrons
25. The order of stability of the following tautomeric compound is NEET 2013
a) 1>2>3
b) 3>2>1
d) 2>3>1
26. Among the following compounds the one that is most reactive towards electrophilic nitration is NEET 2013
(a) benzoic acid (b) nitrobenzene (c) toluene (d) benzene
27. The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is NEET 2012
(a) B>A>D>C (b) B>D>C>A (c) A>B>C>D (d) A>C>B>D
28. Which nomenclature is not according to IUPAC system? NEET 2012
29. Which of the following does not exhibit optical isomerism? NEET 2012
30. The correct IUPAC name of the compound NEET 2011
(a) 3-ethyI-4-ethenyIheptane
(b) 3-ethyI-4-propyIhex-5-ene
(c) 3-(1-ethyl propyl) hex-I -ene
(d) 4-ethyI-3-propyIhex-I -ene
31. Which one of the following is most reactive towards electrophilic reagent? NEET 2011
32. In Duma’s method of estimation of nitrogen 0.35 g of an organic compound gave 55 ml of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K, 15 mm) NEET 2011
33. The Lassaigne’s extract is boiled with con. HN03 while testing for halogens. By doing so it. NEET 2011
1. helps in the precipitation of AgCl
2. increases the solubility product of AgCl
3. increases the concentration of N03- ions
4. decomposes Nas and NaCN, if formed.
34. Which one of the following has the most acidic nature? NEET 2010
Ans =b
35. The correct order of increasing reactivity of C-X bond towards nucleophile in the following compounds is NEET 2010
(a) I < II < IV < III (b) II < III < I < IV (c) IV < III < I < IV (d) III < II < I < IV
36. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is NEET 2010
3. CH3COOCOCH3
4. CH3COCl
37. Which one is most reactive towards electrophilic reagent? NEET 2010
38. The II-JPAC name of the compound having the formula CH≡C-CH=CH2 is NEET 2019
(a) 3-butene-l -yne
(b) I-butyn-3-ene
(c) but-I-yne-3-ene
(d) I-butene-3-yne
39. The relative reactivates of acyl compounds towards nucleophilic substitution are in the order of NEET 2008
(a) Acyl chloride > Acid anhydride > Ester > Amide
(b) Ester > Acyl Chloride > Amide > Acid anhydride
(c) Acid anhydride > amide > Ester > Acyl chloride
(d) Acyl chloride > Ester > Acid anhydride > Amide
40. Which one of the following is most reactive towards electrophillic attack? NEET 2008
A) (a) C6H5−CH2OH
B) (b) C6H5−NO2
C) (c) C6H5−OH
D) (d) C6H5−Cl
41.Base strength of NEET 2008
(1) H3CCH2 (2) H2C=CH and (3) H-C=C
(a) (2) > (1) > (3) (c) (1) > (3) > (2) (b) (3) > (2) > (1) (d) (1) > (2) > (3)
Concept Questions :- Electron Displacement Effects
42. A strong base can abstract an a-hydrogen from NEET 2008
(d) alkane
43.The stability of carbanions in the following NEET 2008
is in the order of (a) (1)>(2)> (3)>(4) (b) (2)>(3)>(4)>(1) (c) (4)>(2)>(3)>(1) (d) (1)>(3)>(2)>(4)
Concept Questions :- Reaction Intermediates
44. Which of the following represents the correct order of the acidity in the given compounds ? NEET 2007
(a) CH3COOH > BrCH2COOH > CICH2COOH > FCH2COOH
(b) FCH2COOH > CH3COOH > BrCH2COOH > CICH2COOH
(c) BrCH2COOH > CICH2COOH > FCH2COOH > CH3COOH
(d) FCH2COOH > CICH2COOH > BrCH2COOH > CH3COOH
45. If there is no rotation of plane polarized light by a compound in a specific solvent, thought to be chiral, it may mean that: NEET 2007
(a) the compound is certainly a chiral
(c) there is no compound in the solvent
(d) the compound may be a racemic mixture
Concept questions: optical isomerism for neet
46. For the following : (i) I– (ii) CI– (iii) Br– NEET 2007
the increasing order of nucleophilicity would be : (a) I–<Br–<CI–(b) CI– < Br– < I– (c) I– <CI– < Br– (d) Br– < CI–< I–
The IUPAC name of NEET 2006
1. 3, 4-dimethylpentanoyl chloride
2. 1-chloro-1-oxo-2, 3-dimethylpentane
3. 2-ethy1-3-methylbutanoyl chloride
JEE Previous Year Questions of Organic Chemistry – Some Basic Principles and Techniques
(I) Shows hydrogen bonding from –OH group only.
(II) Shows strongest hydrogen bonding from both sides of –OH group as well as –NO2 group.
(III) Shows stronger hydrogen from both side of –OH group as well as –NH2 group.
(IV) Shows stronger hydrogen from one side –OH group and another side of –OCH3 group shows only dipole-dipole interaction.
Hence correct order of boiling point is :
I < IV < III < II
Spent-lye and glycerol are separated by distillation under reduced pressure.
Under the reduced pressure the liquid boil at low temperature and the temperature of decomposition will not reach. e.g. glycerol boils at
{290}^{\circ }
C
with decomposition but at reduced pressure it boils at
{180}^{\circ }C
without decomposition.
HCl
=20×0.1×{10}^{-3}=2×{10}^{-3}
HCl
neutralised by
NaOH
=15×0.1×{10}^{-3}=1.5×{10}^{-3}
HCl
neutralised by ammonia
\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=2×{10}^{-3}-1.5×{10}^{-3}
\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.5×{10}^{-3}
%
of nitrogen
=\frac{1.4×N×V}{w.t.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}of\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}Subs\mathrm{tan}ce}×100
NOTE : Among isomeric alkanes, the straight chain isomer has higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. Further due to the presence of
\pi
electrons, these molecules are slightly polar and hence have higher boiling points than the corresponding alkanes.
B.pt.
>
>
alkanes (straight chain) > branched chain alkanes.
In aromatic acids presence of electron withdrawing substituent e.g.
-N{O}_{2}
disperses the negative charge of the anion and stablises it and hence increases the acidity of the parent benzoic acid.
O
– isomer will have higher acidity than corresponding
m
p
isomers. Since nitro group at
p
-position have more pronounced electron withdrawing than
-N{O}_{2}
group at
m
-position hence the correct order is the one given above.
P Block Elements Group 15 important MCQs →
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Blender 3D: Noob to Pro/3D Geometry - Wikibooks, open books for an open world
Blender 3D: Noob to Pro/3D Geometry
1 Coordinates And Coordinate Systems
1.2 Handedness Of Coordinate Systems
If you haven't previously studied 3D graphics, technical drawing, or analytic geometry, you are about to learn a new way of visualizing the world, an ability that's fundamental to working with Blender or any 3D modeling tool.
3D modeling is based on geometry, the branch of mathematics concerned with spatial relationships, specifically analytical geometry, which expresses these relationships in terms of algebraic formulas. If you have studied geometry, some of the terminology will be familiar.
Coordinates And Coordinate SystemsEdit
Now imagine there’s a fly buzzing around the room. The fly is moving in three-dimensional space. In mathematical terms, that means its position within the room at any given moment, can be expressed in terms of a unique combination of three numbers.
There are an infinite number of ways —coordinate systems— in which we could come up with a convention for defining and measuring these numbers, i.e. the coordinates. Each convention will yield different values even if the fly is in the same position. Coordinates only make sense with reference to a specific coordinate system! To narrow down the possibilities (in a purely arbitrary fashion), let us label the walls of the room with the points of the compass: in a clockwise direction, North, East, South and West. (If you know which way really is north, feel free to use that to label the walls of your room. Otherwise, choose any wall you like as north.)
Consider the point at floor level in the south-west corner of the room. We will call this (arbitrary) point the origin of our coordinate system, and the three numbers at this point will be
{\displaystyle (0,0,0)}
. The first of the three numbers will be the distance (in some suitable units, let’s say meters) eastwards from the west wall, the second number will be the distance north from the south wall, and the third number will be the height above the floor.
Each of these directions is called an axis (plural: axes), and they are conventionally labelled X, Y and Z, in that order. With a little bit of thought, you should be able to convince yourself that every point within the space of your room corresponds to exactly one set of
{\displaystyle (x,y,z)}
values, and that every possible combination of
{\displaystyle (x,y,z)}
values, with
{\displaystyle 0\leq x\leq W}
{\displaystyle 0\leq y\leq L}
{\displaystyle 0\leq z\leq H}
{\displaystyle W}
is the east-west dimension of your room,
{\displaystyle L}
is its north-south dimension, and
{\displaystyle H}
is the height between ceiling and floor) corresponds to a point in the room.
The following diagram illustrates how the coordinates are built up, using the same colour codes that Blender uses to label its axes: red for X, green for Y and blue for Z (an easy way to remember this if you're familiar with RGB is the order -- Red X, Green Y, Blue Z). In the second picture, the x value defines a plane parallel to the west wall of the room. In the third picture, the y value defines a plane parallel to the south wall, and in the fourth picture, the z value defines a plane parallel to the floor. Put the planes together in the fifth picture, and they intersect at a unique point.
Another simple way to understand what the coordinates of a point say (x,y,z) means is, if one starts from origin and moves x, y, and z units of distance parallel to x, y, and z axes respectively, in any sequence, one will reach that point. Thus, for example, a coordinate of (3,4,5) means the point which is reached when one moves, starting from origin, 3 units of distance along x-axis, 4 units of distance along y-axis and 5 units of distance along z-axis.
There are other ways to define coordinate systems, for example by substituting direction angles in place of one or two of the distance measurements. These can be useful in certain situations, but usually all coordinate systems in Blender are Cartesian. However, in Blender, switching between these coordinate systems is simple and easy to do.
Negative CoordinatesEdit
Can coordinate values be negative? Depending on the situation, yes. Here we are only considering points within our room. But suppose instead of placing our origin in the bottom southwest corner, we put it in the middle of the room, halfway between the floor and ceiling. (After all, it is an arbitrary point, we can place it wherever we like, as long as we agree on its location.) If the X-coordinate is the distance east from the origin, how do we define a point west of the origin? We simply give it a negative X-coordinate. Similarly, points north of the origin have a positive Y-coordinate, those south of it, have negative Y-coordinates. Points above the origin have a positive Z-coordinate, those below it, a negative Z-coordinate.
Handedness Of Coordinate SystemsEdit
It is conventional for most Cartesian coordinate systems to be right-handed. To understand this, hold the thumb, index finger and middle finger of your right hand perpendicular to each other:
Now orient your hand so your thumb points along the X-axis in the positive direction (direction of increasing coordinate numbers), your index finger along the positive Y-axis, and your middle finger along the positive Z-axis. Another way of looking at it is, if you placed your eye at the origin, and you could see the three arrows pointing in the directions of positive X, positive Y and positive Z as in Figure 1, the order X, Y, Z would go counter clockwise.
Another way to visualize this is to make a fist with your right hand, with your curled fingers towards you. Stick out your thumb directly to the right (X). Now aim your pointer finger straight up (Y). Finally, make your middle finger point toward yourself (Z). This is the view from directly above the origin.
Axes Of RotationEdit
Consider a spinning sphere. Every point on it is moving, except the ones along the axis. These form a motionless line around which the rest of the sphere spins. This line is called the axis of rotation.
More precisely, the axis of rotation is a point or a line connecting points that do not change position while that object rotates, drawn when the observer assumes he/she does not change position relative to that object over time.
Conventionally, the direction of the axis of rotation is such that if you look in that direction, the rotation appears clockwise, as illustrated below, where the yellow arrow shows the rotational movement, while the purple one shows the rotation axis:
To remember this convention, hold your right hand in a thumbs-up gesture:
If the rotation follows the direction of your curled fingers, then the direction of the axis of rotation is considered to be the same as the direction which the thumb is pointing in.
This gesture is a different form of the right-hand rule and is sometimes called the right-hand grip rule, the corkscrew-rule or the right-hand thumb rule. From now on we will refer to it as 'the right-hand grip rule'.
When describing the direction of a rotating object, do not say that it rotates left-to-right/clockwise, or right-to-left/counterclockwise. Each of these on their own are meaningless, because they're relative to the observer. Instead of saying this, find the direction of the axis of rotation and draw an arrow to represent it. Those who know the right-hand grip rule will be able to figure out what the direction of rotation of the object is, by using the rule when interpreting your drawing.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Blender_3D:_Noob_to_Pro/3D_Geometry&oldid=3698514"
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Determine which method (AA\cong AA, SAS Theorem of Similarity, SSS
Determine which method (AA\cong AA, SAS Theorem of Similarity, SSS Theorem of Similarity) proves that the following triangles are similar.
\mathrm{\forall }\stackrel{\sim }{=}\mathrm{\forall }
Similarity of traingles.
\mathrm{△}KLM
\mathrm{△}PQR
\frac{KM}{RP}=\frac{ML}{PQ}=\frac{KL}{QR}=\frac{5}{6}
⇒\mathrm{△}KLM\sim \mathrm{△}PQR
(by SSS similarity).
y=2x+3
If hearts are geometrically similar and the volume of blood pumped in one beat is proportional to the volume of the heart, how much more blood will a heart 5 cm wide pump in one beat than a heart that is 3cm wide?
\mathrm{△}FGH\sim \mathrm{△}KJH
To determine:To prove:The congruency of
\mathrm{\angle }PBC\stackrel{\sim }{=}\mathrm{\angle }PBD
\mathrm{\angle }BPC=\mathrm{\angle }BPD={90}^{\circ }
\mathrm{\angle }C=\mathrm{\angle }D
There is not enough information to determine whether the triangles are similar or not
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Bernoulli Distribution | Brilliant Math & Science Wiki
Ognjen Vukadin, Scott Lee, Christopher Williams, and
The Bernoulli distribution essentially models a single trial of flipping a weighted coin. It is the probability distribution of a random variable taking on only two values,
1
("success") and
0
("failure") with complementary probabilities
p
1-p,
respectively. The Bernoulli distribution therefore describes events having exactly two outcomes, which are ubiquitous in real life. Some examples of such events are as follows: a team will win a championship or not, a student will pass or fail an exam, and a rolled dice will either show a 6 or any other number.
What is the probability of getting an even number when a fair die is thrown once?
Hint: A fair die has 6 faces.
The Bernoulli distribution serves as a building block for discrete distributions which model Bernoulli trials, such as binomial distribution and geometric distribution.
The Bernoulli distribution is the probability distribution of a random variable
X
having the probability density function
\text{Pr}(X=x) = \begin{cases} p && x = 1 \\ 1-p && x = 0 \\ \end{cases}
0<p<1
Intuitively, it describes a single experiment having two outcomes: success ("1") occurring with probability
p,
and failure ("0") occurring with probability
1-p.
It describes a single trial of a Bernoulli experiment.
A closed form of the probability density function of Bernoulli distribution is
P(x) = p^{x}(1-p)^{1-x}
One can represent the Bernoulli distribution graphically as follows:
p=0.3
A fair coin is flipped once. The outcome of the experiment is modeled by the Bernoulli distribution with
p=0.5
The expected value of a Bernoulli distribution is
E(X) = 0\times (1-p) + 1\times p = p.
The variance of a Bernoulli distribution is calculated as
Var(X) = E(X^2) - E(X)^2 = 1^2 \times p + 0^2 \times (1-p) - p^2 = p - p^2 = p(1-p).
The mode, the value with the highest probability of occurring, of a Bernoulli distribution is
1
p>0.5
0
p<0.5
p=0.5
, success and failure are equally likely and both
0
1
are modes. This is intuitively clear: since there are only two outcomes with complementary probabilities,
p>0.5
implies that the probability of success is higher than the probability of failure.
Basic properties of Bernoulli distribution can be calculated by taking
n=1
in the binomial distribution.
Using properties such as linearity of expectation and rules for calculating the variance, Bernoulli distribution is used in the calculation of the properties of distributions based on the Bernoulli experiment, such as the binomial distribution.
Bernoulli distribution models the following situations:
A newborn child is either male or female. (Here the probability of a child being a male is roughly 0.5.)
You either pass or fail an exam.
A tennis player either wins or loses a match.
A dart thrown at a circular dartboard lands randomly over its area (example). The dart will either land closer to the center than to the edge or not (in the second case it is either closer to the edge or equally distant from the center and the edge). In this case
p=0.25
n\in \{1,\ldots, 999999 \}
is chosen randomly. We consider three variables,
X_1,X_2,
X_3
X_1
assumes the value
1
if the sum of the digits of
n
9
0
X_2
1
n
can be expressed as a sum of four squares of integers and
0
X_3
assumes values
0,1
2
n
0,1
2
3
The sum of digits of a positive integer
n
9
9
n
. The probability that a randomly chosen integer in
\{1,\ldots, 999999 \}
9
\frac{1}{9}.
X_1
is a Bernoulli distributed random variable with
p=\frac{1}{9}
Every positive integer can be expressed as a sum of four squares, so the variable
X_2
is not a random variable and is not Bernoulli distributed. In the definition of the Bernoulli distribution the restriction
0<p<1
excludes the case
p=1
X_3
models an experiment with more than two outcomes, and hence it is not Bernoulli distributed.
a + b
independent Bernoulli
(p)
trials. Let
N_a
be the number of successes in the first
a
of these trials, and
N_b
be the number of successes in the last
b
of these trials. Using properties of the Bernoulli distribution, we can then say the following:
N_a
~ Binomial
(a, p)
because we have
a
(p)
trials. We don't care about the last
b
N_b
(b, p)
b
(p)
trials. We don't care about the first
a
N_a + N_b
(a + b, p)
because all of the Bernoulli trials are independent, and we can treat them as i.i.d.
N_a
N_b
are independent because the two groups of trials (the first
a
trials and the last
b
trials) are independent.
Cite as: Bernoulli Distribution. Brilliant.org. Retrieved from https://brilliant.org/wiki/bernoulli-distribution/
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The RuneScape Wiki also has an article on: rsw:Recruitment Drive
Recruitment Drive (#83)
2.3 Sir Kuam Ferentse
2.4 Sir Spishyus
2.5 Lady Table
2.6 Miss Cheevers
2.6.1 For the first door
2.6.2 For the second door
2.7 Sir Ren Itchood
2.8 Ms. Hynn Terprett
2.9 Sir Tinley
Talk to Sir Amik Varze in the White Knights' Castle in Falador.
The Temple Knights of Saradomin, a secret organisation founded many centuries ago by Saradomin himself, are currently looking to expand their ranks with some new blood. After the successful thwarting of the Black Knights' plans to take over Asgarnia, and with the personal recommendation of Sir Amik, you have now been offered the chance to apply for membership in this organisation... but are you up to the challenge?
3,000 coins if you are a male character. (You will be reimbursed in the form of a Makeover voucher)
Sir Leye (level 20)
You will lose your Hardcore ironman status if you die to Sir Leye.
You will lose your Hardcore ironman status if you die to Sir Kuam Ferentse.
This quest requires an empty inventory and no items worn.
{"requirements":"*Completion of the following quests:\n**[[Black Knights' Fortress]]\n**[[Druidic Ritual]]","desc":"The Temple Knights of Saradomin, a secret organisation founded many centuries ago by Saradomin himself, are currently looking to expand their ranks with some new blood. After the successful thwarting of the Black Knights' plans to take over Asgarnia, and with the personal recommendation of Sir Amik, you have now been offered the chance to apply for membership in this organisation... but are you up to the challenge?","difficulty":"Novice","kills":"[[Sir Leye]] (level 20)","name":"Recruitment Drive","start":"Talk to [[Sir Amik Varze]] in the [[White Knights' Castle]] in [[Falador]].","length":"Short","ironman":"[[File:Hardcore ironman chat badge.png]] You will lose your Hardcore ironman status if you die to Sir Leye.<br/>\n[[File:Hardcore ironman chat badge.png]] You will lose your Hardcore ironman status if you die to Sir Kuam Ferentse.\n\n[[File:Ultimate ironman chat badge.png]] This quest requires an empty inventory and no items worn.","items":"* 3,000 [[coins]] if you are a male character. ''(You will be reimbursed in the form of a [[Makeover voucher]])''"}
The Make-over mage's house, location just west of Falador.
To start the quest, speak to Sir Amik Varze who is located on the 2nd floor[UK]3rd floor[US] of the western part of the White Knights' Castle. He will tell you to talk to Sir Tiffy Cashien in Falador Park to be tested. In order to be able to be tested, you need to have an empty inventory and no items equipped. At this point, it is safe to change your gender to female but you will not be refunded unless you speak to Sir Amik first.
Now, head to Falador Park, and talk to Sir Tiffy Cashien. He will tell you that you must go through a mental test. You need to be female for one of the tests so visit the Make-over Mage before starting if you aren't already. You must start the quest as male to receive the make-over voucher. If you switch to female before you talk to Sir Amik Varze, you will not receive a voucher or a refund.
Sir Tiffy Cashien will take you to the testing grounds. There are seven different testing rooms. In each room, there is a yellow portal at the beginning of the room (takes you back to Falador) and a yellow portal at the end of the room, which you may use to travel to the next room after you pass the test. You need to complete five in a row. Which five you'll get, as well as the order that they are in, is completely random. However, you will always be required to defeat Sir Leye in one of the five tests. If you fail a test, you have no choice but to return to Falador and start over.
Sir Kuam Ferentse[edit | edit source]
He tells you that you must defeat Sir Leye, who is level 20. Sir Leye has been blessed by Saradomin so that no man may defeat him. So, unless you have a female character, you will not be able to hit the final blow on him. You must defeat him without wearing any armour, weapon, food or potions to pass the test. If you die, you may return and finish the quest after one final trial. Note: Skillers can trap Leye in the corners of the room and flinch him by waiting for his HP bar to go away and then hitting and running. Skillers may find this part impractical because Sir Leye is very difficult to get into the corner positions and he will take damage very slowly, to the point where he may heal faster than the damage being done. One of the corners where flinching is possible is on the south-west corner. Sir Leye should be on the south tile with you on the west tile. Wait until his HP bar disappears, then attack and move back to the west tile. Setting this up is tricky and requires patience.
Sir Spishyus[edit | edit source]
Right next to him, you'll see a 5 kg chicken, a 5 kg bag of grain, and a 5 kg fox. You must get all three of them across a bridge that only supports five kilograms at a time. The tricky part is that if you leave the fox and chicken alone, the fox will eat the chicken. The chicken will also eat the bag of grain if the two of them are left alone.
Take the chicken to the other side, since the fox and grain are the only pair that actually get along.
Drop the chicken, then come back and grab the grain.
Take it to the other side then grab the chicken and drop the grain. Then take the chicken back to the start.
Grab the fox, drop the chicken, and take the fox across the bridge.
Drop the fox, then go back to the beginning for the chicken, take it across and drop the chicken.
Drop the chicken, then come back and grab the fox.
Take it to the other side then grab the chicken and drop the fox. Then take the chicken back to the start.
Grab the grain, drop the chicken, and take the grain across the bridge.
Drop the grain, then go back to the beginning for the chicken and take it across.
Once you complete the puzzle, the door now unlocks and you can proceed to the next puzzle via the portal.
Lady Table[edit | edit source]
The bronze, silver, and gold statues used in the memory test.
Important: Make a screenshot of the statues as you enter the room (or remember the statues if you have a good memory). You only have a few seconds after you enter the room! If you talk to Lady Table before the room changes, the room won't change until you exit the chat dialog, so you can use this time to study the room layout.
Lady Table will test your memory. She will have 11 statues of a knight in front of her. There are supposed to be 12, but one has been taken away. You will have a few seconds to look at the statues, then the missing one will be returned. You must then touch the one that was missing.
The trick to passing this one is to analyse the types of statues. There should be four statues of each colour: bronze, silver, and gold. First, see which colour only has three statues. After you figure out the colour, figure out the weapon. There are four weapons: sword, halberd, greataxe, and mace. Figure out which weapon is missing. After you have figured it out, touch the statue.
Miss Cheevers[edit | edit source]
There are many bookcases and crates with items. Search everything (including the chest) and take everything. There are two doors that you must get through. The first door is missing a handle, while the second one is locked.
For the first door[edit | edit source]
Using the vial of liquid on the new handle.
Take the metal spade, and use it on the Bunsen burner to remove the wood.
Use the metal spade (without handle) on the stone door.
Use the Cupric Sulfate (not to be confused with the other orange cupric powder!) on the door with the metal spade in it
Use a vial of liquid on the door.
The metal spade will expand and be jammed in the hole, and you can now open the door.
For the second door[edit | edit source]
Use a vial of liquid on the tin (which looks like a cake tin);
Use gypsum on the tin. Do not use the liquids on each other.
Use the tin with the lumpy white liquid inside of it on the key (on the ground south of the portal you came from) to get an imprint of the key.
Use tin ore powder on the tin.
Use cupric ore powder on the tin.
Use the tin on the Bunsen burner to make a duplicate bronze key.
Use a knife, chisel or bronze wire on the tin to obtain the bronze key; you can now use the bronze key on the door to open it.
Sir Ren Itchood[edit | edit source]
When you talk to Sir Ren Itchood, he will speak in riddles, the answer to which is a four letter word, made up of the first letter of each line of his riddle. Use the answer to unlock the combination lock on the next door. The possible solutions are: BITE, FISH, LAST, MEAT, RAIN or TIME.
Ms. Hynn Terprett[edit | edit source]
She will give you a multiple choice riddle. There are many to choose from - they are as follows:
Riddle: If you were sentenced to death, what would you rather choose - being drowned in a lake of acid, burned on a fire, thrown to a pack of wolves that have not been fed in over a month or thrown from the walls of a castle, many hundreds of feet high.
Answer: Being fed to the wolves - wolves cannot survive for 30 days without food, thus they would all be dead.
Riddle: I have both a husband and daughter. My husband is four times older than my daughter. In twenty years time, he will be twice as old as my daughter. How old is my daughter now?
Answer: Her daughter is 10 years old because
{\displaystyle 10*4=40}
{\displaystyle {\frac {(40+20)}{2}}=10+20}
Riddle: I dropped four identical stones, into four identical buckets, each containing an identical amount of water. The first bucket was at 32 degrees Fahrenheit, the second was at 33 degrees, the third was at 34 and the fourth was at 35 degrees. Which bucket's stone dropped to the bottom of the bucket last?
Answer: The first bucket (Bucket A). At 32 Fahrenheit (0 Celcius) the bucket of water is frozen, so the stone never reaches the bottom
Riddle: Counting the creatures and humans in RuneScape, you get about a million inhabitants. If you multiply the fingers on everything's left hand by zero, how many would you get?
Answer: Zero. You are multiplying by zero, which yields zero.
Riddle: Which of the following is true? The number of false statements here is one. The number of false statements here is two. The number of false statements here is three. The number of false statements here is four. How many false statements are there?
Answer: Logic dictates that the number of false statements must be three. There are four possible answers, so for one of them to be true, all others must be false.
Riddle: What would be the number you would get if you multiply the number of fingers on everythings left hand to the nearest million
Answer: Zero. If even one creature has no fingers on their left hand, you are multiplying by zero, which yields zero.
Sir Tinley[edit | edit source]
Talk to Sir Tinley, click on "continue". Stand still for a few moments, do not click on anything after clicking continue, or you will fail this task.
After completing all five tests subsequently, you will automatically be teleported back into Falador Park where Sir Tiffy Cashien will congratulate you on a job well done.
1,000.5 Prayer experience
1,000.5 Herblore experience
Access to initiate armour. You will be given a sallet for free, and you can buy subsequent sallets from Sir Tiffy Cashien for 6,000 coins, the cuisse for 8,000 coins, the hauberk for 10,000 coins, or the full set for 20,000 coins. It is aesthetically similar to white armour, with a gold trim.
The Gaze of Saradomin - When you die, you will have the option of returning to Falador instead of Lumbridge. Talk to Sir Tiffy Cashien to change your spawn point at any time.
If you were a male at the start of the quest, you'll be given your 3,000 coins back, as well as a free Makeover voucher to change back.
Completion of Recruitment Drive is required for the following:
Recruitment Drive is one of five quests not to play the standard quest completion music. While Mountain Daughter plays Asleif's singing, Sins of the Father plays a melancholy tune, and Monkey Madness II plays monkey chattering, Regicide and Recruitment Drive play nothing.
When asking Sir Tiffy Cashien about "Testing...?" near the beginning of the quest, the player mentions fetching common items and delivering them across the country, a reference to the quest One Small Favour.
All of the knights' names are puns:
Sir Tiffy Cashien - Certification
Sir Kuam Ferentse - Circumference
Sir Leye - Surly
Sir Spishyus - Suspicious
Miss Hinn Terprett - Misinterpret
Sir Tinley - Certainly
Sir Ren Itchood - Serenitude
Lady Table - Lay the Table
Miss Cheevers - Mischievous
Sir Vey - Survey
Sir Amik Varze - Ceramic Vase
Sir Vyvin - Surviving
Sir Renitee - Serenity
King Vallance was presumably once Sir Vallance - Surveillance
Sir Rebral - Cerebral
Retrieved from ‘https://oldschool.runescape.wiki/w/Recruitment_Drive?oldid=14270020’
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How do I integrate the following: \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}dx
How do I integrate the following:
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx
Stuart Rountree
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx
u=x-\frac{1}{x}
du=\left(1+\frac{1}{{x}^{2}}\right)dx
\frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}=-\frac{{x}^{2}\left(1+\frac{1}{{x}^{2}}\right)}{x\left(x-\frac{1}{x}\right)\sqrt{{x}^{2}\left({x}^{2}+\frac{1}{{x}^{2}}\right)}}=-\frac{1+\frac{1}{{x}^{2}}}{\left(x-\frac{1}{x}\right)\sqrt{{\left(x-\frac{1}{x}\right)}^{2}+2}}
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx
=-\int \frac{du}{u\sqrt{{u}^{2}+2}}
Somewhat inspired by Morons
Without loss of generality we may assume that 1>x>0. Put
x:=\sqrt{y},\text{ }1>y>0
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx=\int \frac{1+y}{2\left(1-y\right)\sqrt{1+{y}^{2}}\sqrt{y}}dy
Introduce the new variable
t:=\frac{1+y}{1-y},\text{ }1<t<\mathrm{\infty }
y=\frac{-1+t}{1+t}
y=\frac{2}{\left(1=t{\right)}^{2}}dt
Substituting back we obtain
\int \frac{1+y}{2\left(1-y\right)\sqrt{1+{y}^{2}}\sqrt{y}}dy=\int \frac{t}{2\sqrt{1+\left(\frac{-1+t}{1+t}{\right)}^{2}\sqrt{\frac{-1+t}{1+t}}}}\frac{2}{\left(1+t{\right)}^{2}}dt
=\frac{1}{\sqrt{2}}\int \frac{t}{\sqrt{{t}^{4}-1}}dt
=\frac{1}{2\sqrt{2}}\mathrm{ln}\left({t}^{2}+\sqrt{{t}^{4}-1}\right)+C
Putting back everything we obtain
\frac{1}{2\sqrt{2}}\mathrm{ln}\left(\frac{\left(1+{x}^{2}{\right)}^{2}+2\sqrt{2}x\sqrt{1+{x}^{4}}}{\left(1-{x}^{2}{\right)}^{2}}\right)+C
{\int }_{0}^{\mathrm{\infty }}\frac{x\mathrm{cos}ax}{\text{sinh}x}dx\frac{{\pi }^{2}}{4}{\text{sech}}^{2}\left(\frac{a\pi }{2}\right)
Can someone simply explain to me how to calculate linear integral linke below?
{\int }_{L}5ydL
where L is line segment from (0;0) to (0,2;0,2)
\int \mathrm{sin}\left({x}^{3}\right)dx
{\int }_{0}^{1}\frac{\left(x-4\right)}{\left({x}^{2}-5x+6dx\right)}
I={\int }_{0}^{\mathrm{\infty }}\frac{x-1}{\sqrt{{2}^{x}-1}\mathrm{ln}\left({2}^{x}-1\right)}dx
How is the integral
\frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\mathrm{cos}\left(nx\right)dx=\frac{4{\left(-1\right)}^{n}}{{n}^{2}}
I thought it would be this :
\frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\mathrm{cos}\left(nx\right)dx=\frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}{\left(-1\right)}^{n}=\frac{2}{\pi }{\left(-1\right)}^{n}{\int }_{0}^{\pi }{x}^{2}=\frac{2}{\pi {\left(-1\right)}^{n}}{\left[\frac{{x}^{3}}{3}\right]}_{0}^{\pi }=\frac{2{\left(-1\right)}^{n}}{3{\pi }^{3}}
But it is actually
\frac{2}{\pi }{\int }_{0}^{\pi }{x}^{2}\mathrm{cos}\left(nx\right)dx=\frac{4{\left(-1\right)}^{n}}{{n}^{2}}
Find the area of the part of the plane
5x+4y+z=20
that lies in the first octant.
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Inclusion_map Knowpia
In mathematics, if
{\displaystyle A}
{\displaystyle B,}
then the inclusion map (also inclusion function, insertion,[1] or canonical injection) is the function
{\displaystyle \iota }
that sends each element
{\displaystyle x}
{\displaystyle A}
{\displaystyle x,}
treated as an element of
{\displaystyle B:}
{\displaystyle A}
{\displaystyle B,}
{\displaystyle B}
{\displaystyle A.}
{\displaystyle \iota :A\rightarrow B,\qquad \iota (x)=x.}
A "hooked arrow" (U+21AA ↪ RIGHTWARDS ARROW WITH HOOK)[2] is sometimes used in place of the function arrow above to denote an inclusion map; thus:
{\displaystyle \iota :A\hookrightarrow B.}
(However, some authors use this hooked arrow for any embedding.)
This and other analogous injective functions[3] from substructures are sometimes called natural injections.
Given any morphism
{\displaystyle f}
between objects
{\displaystyle X}
{\displaystyle Y,}
if there is an inclusion map into the domain
{\displaystyle \iota :A\to X,}
then one can form the restriction
{\displaystyle f\,\iota }
{\displaystyle f.}
In many instances, one can also construct a canonical inclusion into the codomain
{\displaystyle R\to Y}
known as the range of
{\displaystyle f.}
Applications of inclusion mapsEdit
Inclusion maps tend to be homomorphisms of algebraic structures; thus, such inclusion maps are embeddings. More precisely, given a substructure closed under some operations, the inclusion map will be an embedding for tautological reasons. For example, for some binary operation
{\displaystyle \star ,}
{\displaystyle \iota (x\star y)=\iota (x)\star \iota (y)}
is simply to say that
{\displaystyle \star }
is consistently computed in the sub-structure and the large structure. The case of a unary operation is similar; but one should also look at nullary operations, which pick out a constant element. Here the point is that closure means such constants must already be given in the substructure.
Inclusion maps are seen in algebraic topology where if
{\displaystyle A}
is a strong deformation retract of
{\displaystyle X,}
the inclusion map yields an isomorphism between all homotopy groups (that is, it is a homotopy equivalence).
Inclusion maps in geometry come in different kinds: for example embeddings of submanifolds. Contravariant objects (which is to say, objects that have pullbacks; these are called covariant in an older and unrelated terminology) such as differential forms restrict to submanifolds, giving a mapping in the other direction. Another example, more sophisticated, is that of affine schemes, for which the inclusions
{\displaystyle \operatorname {Spec} \left(R/I\right)\to \operatorname {Spec} (R)}
{\displaystyle \operatorname {Spec} \left(R/I^{2}\right)\to \operatorname {Spec} (R)}
may be different morphisms, where
{\displaystyle R}
{\displaystyle R}
{\displaystyle R.}
Identity function – In mathematics, a function that always returns the same value that was used as its argument
^ MacLane, S.; Birkhoff, G. (1967). Algebra. Providence, RI: AMS Chelsea Publishing. p. 5. ISBN 0-8218-1646-2. Note that “insertion” is a function S → U and "inclusion" a relation S ⊂ U; every inclusion relation gives rise to an insertion function.
^ "Arrows – Unicode" (PDF). Unicode Consortium. Retrieved 2017-02-07.
^ Chevalley, C. (1956). Fundamental Concepts of Algebra. New York, NY: Academic Press. p. 1. ISBN 0-12-172050-0.
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Propagate orbit of one or more spacecraft - Simulink - MathWorks í•œêµ
φθψ
αδW
φθψ — Moon libration angles
αδW — Right ascension, declination, and rotation angle
To specify libration angles (φ θ ψ) for Moon orientation, select this check box.
When Propagation method is Numerical (high precision) and Central Body is Custom, the fixed-frame coordinate system is defined by the poles of rotation and prime meridian defined by the block input α, δ, W, or the spin axis properties.
This propagation method is always performed in the ICRF intertial coordinate system with origin at the center of the central body. Given initial intertial position r0 and velocity v0 at time t0, first find orbital energy, ξ, and the reciprocal of the semi-major axis, α:
\begin{array}{l}\mathrm{ξ}=\frac{{v}_{0}{}^{2}}{2}â\frac{\mathrm{μ}}{{r}_{0}}\\ \mathrm{α}=\frac{â2\mathrm{ξ}}{\mathrm{μ}},\end{array}
where μ is the standard gravitation parameter of the central body. Next, determine the orbit type from the sign of α.
α>0 => Circular or elliptical
α<0 => Hyperbolic
α≈0 => Parabolic
To initialize the Newton-Raphson iteration, select an initial guess for χ based on the orbit type:
{\mathrm{Ï}}_{0}â\sqrt{\mathrm{μ}}\left(\mathrm{Î}t\right)\mathrm{α},
where Δt is the propagation step size (simulation time step). If Δt exceeds the orbital period
T=2\mathrm{Ï}\sqrt{\frac{{a}^{3}}{\mathrm{μ}}}
, wrap Δt.
{\mathrm{Ï}}_{0}â\sqrt{p}2\mathrm{cot}\left(2w\right),
\begin{array}{l}\stackrel{â}{h}=\stackrel{â}{{r}_{0}}Ã\stackrel{â}{{v}_{0}}\\ p=\frac{hâ
h}{\mathrm{μ}}\\ \mathrm{cot}\left(2s\right)=3\sqrt{\frac{\mathrm{μ}}{{p}^{3}}}\left(\mathrm{Î}t\right)\\ {\mathrm{tan}}^{3}\left(w\right)=\mathrm{tan}\left(s\right).\end{array}
{\mathrm{Ï}}_{0}â\text{sign}\left(\mathrm{Î}t\right)\sqrt{â\frac{1}{\mathrm{α}}}\mathrm{ln}\left(\frac{â2\mathrm{μ}\mathrm{α}\left(\mathrm{Î}t\right)}{\stackrel{â}{{r}_{0}}â
\stackrel{â}{{v}_{0}}+\text{sign}\left(\mathrm{Î}t\right)\sqrt{â\frac{\mathrm{μ}}{\mathrm{α}}}\left(1â{r}_{0}\mathrm{α}\right)}\right).
\begin{array}{l}{\mathrm{Ï}}_{n+\text{1}}={\mathrm{Ï}}_{n}+\frac{\sqrt{\mathrm{μ}}\left(\mathrm{Î}t\right)â{\mathrm{Ï}}_{n}{}^{3}{c}_{3}â\frac{\stackrel{â}{{r}_{0}}â
\stackrel{â}{{v}_{0}}}{\sqrt{\mathrm{μ}}}{\mathrm{Ï}}_{n}{}^{2}{c}_{2}â{r}_{0}\text{â}{\mathrm{Ï}}_{n}\left(1â\mathrm{Ï}{c}_{3}\right)}{{\mathrm{Ï}}_{n}{}^{2}{c}_{2}+\frac{\stackrel{â}{{r}_{0}}â
\stackrel{â}{{v}_{0}}}{\sqrt{\mathrm{μ}}}{\mathrm{Ï}}_{n}\left(1â\mathrm{Ï}{c}_{3}\right)+{r}_{0}\left(1â\mathrm{Ï}{c}_{2}\right)}\\ {\mathrm{Ï}}_{n}â{\mathrm{Ï}}_{n+\text{1}},\end{array}
\mathrm{Ï}={\mathrm{Ï}}_{n}{}^{2}\mathrm{α}.
(if ψ>0),
\begin{array}{l}{c}_{2}=\frac{1â\mathrm{cos}\left(\sqrt{\mathrm{Ï}}\right)}{\mathrm{Ï}}\\ {c}_{3}=\frac{\sqrt{\mathrm{Ï}}â\mathrm{sin}\left(\sqrt{\mathrm{Ï}}\right)}{\sqrt{{\mathrm{Ï}}^{3}}}.\end{array}
(if ψ<0),
\begin{array}{l}{c}_{2}=\frac{1â\mathrm{cosh}\left(\sqrt{â\mathrm{Ï}}\right)}{\mathrm{Ï}}\\ {c}_{3}=\frac{\mathrm{sinh}\left(\sqrt{â\mathrm{Ï}}\right)â\sqrt{â\mathrm{Ï}}}{\sqrt{{\left(â\mathrm{Ï}\right)}^{3}}}.\end{array}
(if ψ≈0),
\begin{array}{l}{c}_{2}=\frac{1}{2}\\ {c}_{3}=\frac{1}{6}.\end{array}
f
\stackrel{Ë}{f}
\text{g}
\stackrel{Ë}{g}
\begin{array}{l}f=1â\frac{{\mathrm{Ï}}_{n}{}^{2}}{{r}_{0}}{c}_{2}\\ \stackrel{Ë}{f}=\frac{\sqrt{\mathrm{μ}}}{r{r}_{0}}{\mathrm{Ï}}_{n}\left(\mathrm{Ï}{c}_{3}â1\right)\\ g=\left(\mathrm{Î}t\right)â\frac{{\mathrm{Ï}}_{n}{}^{3}}{\sqrt{\mathrm{μ}}}{c}_{3}\\ \stackrel{Ë}{g}=1â\frac{{\mathrm{Ï}}_{n}{}^{2}}{r}{c}_{2}.\end{array}
\begin{array}{l}\stackrel{â}{{r}_{\text{icrf}}}=f\stackrel{â}{{r}_{0}}+g\stackrel{â}{{v}_{0}}\\ \stackrel{â}{{v}_{\text{icrf}}}=\stackrel{Ë}{f}\stackrel{â}{{r}_{0}}+\stackrel{Ë}{g}\stackrel{â}{{v}_{0}}.\end{array}
This option uses the Simulink® solver to integrate position and velocity from central body gravitational acceleration at each simulation timestep (Δt). The method for computing central body acceleration depends on the current setting for parameter Gravitational potential model. You can also include custom acceleration components in to the propagation algorithm using the block Aicrf (applied acceleration) input port. For gravity models that include nonspherical acceleration terms, the block computes nonspherical gravity in a fixed-frame coordinate system (ITRF, in the case of Earth). Numerical integration, however, is always performed in the inertial ICRF coordinate system. Therefore, at each timestep, the block:
\begin{array}{l}\stackrel{â}{{a}_{\text{icrf}}}={\stackrel{â}{a}}_{\text{central}\text{â}\text{body}\text{â}\text{gravity}}+{\stackrel{â}{a}}_{\text{applied}}\\ \stackrel{â}{{a}_{\text{icrf}}}\underset{\text{integrate}}{â}\stackrel{â}{{r}_{\text{icrf}}},\stackrel{â}{{v}_{\text{icrf}}}\end{array}
{\stackrel{â}{a}}_{\text{centralbodygravity}}=â\frac{\mathrm{μ}}{{r}^{2}}\frac{\stackrel{â}{{r}_{\text{icrf}}}}{r},
where μ is the standard gravitation parameter of the central body.
{\stackrel{â}{a}}_{centralbodygravity}=â\frac{\mathrm{μ}}{{r}^{2}}\frac{\stackrel{â}{{r}_{\text{icrf}}}}{r}+fixed2inertial\left({\stackrel{â}{a}}_{nonspherical}\right),
\begin{array}{l}{\stackrel{â}{a}}_{\text{nonspherical}}=\\ \left\{\left[\frac{1}{r}\frac{â}{âr}Uâ\frac{{r}_{\text{ff}}{}_{{}_{k}}}{{r}^{2}\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}\frac{â}{â\mathrm{Ï}}U\right]{r}_{\text{ff}}{}_{{}_{i}}\right\}i\\ +\left\{\left[\frac{1}{r}\frac{â}{âr}U+\frac{{r}_{\text{ff}}{}_{{}_{k}}}{{r}^{2}\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}\frac{â}{â\mathrm{Ï}}U\right]{r}_{\text{ff}}{}_{{}_{j}}\right\}j\\ +\left\{\frac{1}{r}\left(\frac{â}{âr}U\right){r}_{k}+\frac{\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}{{r}^{2}}\frac{â}{â\mathrm{Ï}}U\right\}k,\end{array}
\begin{array}{l}\frac{â}{âr}U=\frac{3\mathrm{μ}}{{r}^{2}}{\left(}^{\frac{{R}_{\text{cb}}}{r}}{P}_{2,0}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]{J}_{2}\\ \frac{â}{â\mathrm{Ï}}U=â\frac{\mathrm{μ}}{r}{\left(}^{\frac{{R}_{\text{cb}}}{r}}{P}_{2,1}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]{J}_{2}\end{array}
Ï• and λ — Satellite geocentric latitude and longitude.
μ — Standard gravitation parameter of the central body.
{\stackrel{â}{a}}_{centralbodygravity}=â\frac{\mathrm{μ}}{{r}^{2}}\frac{\stackrel{â}{{r}_{\text{icrf}}}}{r}+fixed2inertial\left({\stackrel{â}{a}}_{nonspherical}\right),
\begin{array}{l}{\stackrel{â}{a}}_{nonspherical}=\\ \left\{\left[\frac{1}{r}\frac{â}{âr}Uâ\frac{{r}_{\text{ff}}{}_{{}_{k}}}{{r}^{2}\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}\frac{â}{â\mathrm{Ï}}U\right]{r}_{\text{ff}}{}_{{}_{i}}â\left[\frac{1}{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}\frac{â}{â\mathrm{λ}}U\right]{r}_{\text{ff}}{}_{{}_{j}}\right\}i\\ +\left\{\left[\frac{1}{r}\frac{â}{âr}U+\frac{{r}_{\text{ff}}{}_{{}_{k}}}{{r}^{2}\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}\frac{â}{â\mathrm{Ï}}U\right]{r}_{\text{ff}}{}_{{}_{j}}+\left[\frac{1}{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}\frac{â}{â\mathrm{λ}}U\right]{r}_{\text{ff}}{}_{{}_{i\text{â}}}\right\}j\\ +\left\{\frac{1}{r}\left(\frac{â}{âr}U\right){r}_{\text{ff}}{}_{{}_{k}}+\frac{\sqrt{{r}_{\text{ff}}{{}_{{}_{i}}}^{2}+{r}_{\text{ff}}{{}_{{}_{j}}}^{2}}}{{r}^{2}}\frac{â}{â\mathrm{Ï}}U\right\}k,\end{array}
\begin{array}{l}\frac{â}{âr}U=â\frac{\mathrm{μ}}{{r}^{2}}\underset{l=2}{\overset{{l}_{\mathrm{max}}}{{â}^{\text{â}}}}\underset{m=0}{\overset{l}{{â}^{\text{â}}}}{\left(}^{\frac{{R}_{\text{cb}}}{r}}\left(l+1\right){P}_{l,m}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]\left\{{C}_{l,m}\mathrm{cos}\left(m\mathrm{λ}\right)+{S}_{l,m}\mathrm{sin}\left(m\mathrm{λ}\right)\right\}\\ \frac{â}{â\mathrm{Ï}}U=\frac{\mathrm{μ}}{r}\underset{l=2}{\overset{{l}_{\mathrm{max}}}{{â}^{\text{â}}}}\underset{m=0}{\overset{l}{{â}^{\text{â}}}}{\left(}^{\frac{{R}_{\text{cb}}}{r}}\left\{{P}_{l,m+1}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]â\left(m\right)\mathrm{tan}\left(\mathrm{Ï}\right)\text{â}{P}_{l,m}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]\right\}\left\{{C}_{l,m}\mathrm{cos}\left(m\mathrm{λ}\right)+{S}_{l,m}\mathrm{sin}\left(m\mathrm{λ}\right)\right\}\\ \frac{â}{â\mathrm{λ}}U=\frac{\mathrm{μ}}{r}\underset{l=2}{\overset{{l}_{\mathrm{max}}}{{â}^{\text{â}}}}\underset{m=0}{\overset{l}{{â}^{\text{â}}}}{\left(}^{\frac{{R}_{\text{cb}}}{r}}\left(m\right){P}_{l,m}\left[\mathrm{sin}\left(\mathrm{Ï}\right)\right]\left\{{S}_{l,m}\mathrm{cos}\left(m\mathrm{λ}\right)â{C}_{l,m}\mathrm{sin}\left(m\mathrm{λ}\right)\right\},\end{array}
[5] Seidelmann, P.K., Archinal, B.A., A’hearn, M.F. et al. "Report of the IAU/IAG Working Group on cartographic coordinates and rotational elements: 2006." Celestial Mech Dyn Astr 98, 155–180 (2007).
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Electrical conductivity - New World Encyclopedia
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Next (Electrical conductor)
Electrical conductivity or specific conductivity is a measure of a material's ability to conduct an electric current. When an electrical potential difference is placed across a conductor, its movable charges flow, giving rise to an electric current. The conductivity σ is defined as the ratio of the current density
{\displaystyle \mathbf {J} }
to the electric field strength
{\displaystyle \mathbf {E} }
{\displaystyle \mathbf {J} =\sigma \mathbf {E} }
1.1 Understanding conductors and insulators
Conductivity is the reciprocal (inverse) of electrical resistivity and has the SI units of siemens per meter (S•m-1) i.e. if the electrical conductance between opposite faces of a one-meter cube of material is one Siemens then the material's electrical conductivity is one Siemens per meter. Electrical conductivity is commonly represented by the Greek letter σ, but κ or γ are also occasionally used.
Understanding conductors and insulators
All conductors contain electric charges which will move when an electric potential difference (measured in volts) is applied across separate points on the material. This flow of charge (measured in amperes) is what is meant by electric current. In most materials, the rate of current is proportional to the voltage (Ohm's law), provided the temperature remains constant and the material remains in the same shape and state. The ratio between the voltage and the current is called the resistance (measured in ohms) of the object between the points where the voltage was applied. The resistance across a standard mass (and shape) of a material at a given temperature is called the resistivity of the material. The inverse of resistance and resistivity is conductance and conductivity. Some good examples of conductors are metal.
Most familiar conductors are metallic. Copper is the most common material for electrical wiring, (silver is the best but expensive), and gold for high-quality surface-to-surface contacts. However, there are also many non-metallic conductors, including graphite, solutions of salts, and all plasmas.
Non-conducting materials lack mobile charges, and so resist the flow of electric current, generating heat. In fact, all materials offer some resistance and warm up when a current flows. Thus, proper design of an electrical conductor takes into account the temperature that the conductor needs to be able to endure without damage, as well as the quantity of electrical current. The motion of charges also creates an electromagnetic field around the conductor that exerts a mechanical radial squeezing force on the conductor. A conductor of a given material and volume (length x cross-sectional area) has no real limit to the current it can carry without being destroyed as long as the heat generated by the resistive loss is removed and the conductor can withstand the radial forces. This effect is especially critical in printed circuits, where conductors are relatively small and close together, and inside an enclosure: the heat produced, if not properly removed, can cause fusing (melting) of the tracks.
Since all conductors have some resistance, and all insulators will carry some current, there is no theoretical dividing line between conductors and insulators. However, there is a large gap between the conductance of materials that will carry a useful current at working voltages and those that will carry a negligible current for the purpose in hand, so the categories of insulator and conductor do have practical utility.
(S•m-1)
Annealed Copper 58.0 × 106 20 Referred to as 100 percent IACS or International Annealed Copper Standard. The unit for expressing the conductivity of nonmagnetic materials by testing using the eddy-current method. Generally used for temper and alloy verification of Aluminum.
Gold 45.2 × 106 20 Gold is commonly used in electrical contacts
Aluminum 37.8 × 106 20
Seawater 5 23 Refer to Kaye and Laby for more detail as there are many variations and significant variables for seawater.
5(S•m-1) would be for an average salinity of 35 g/kg at about 23(°C) Copyright on the linked material can be found here.
deionized water 5.5 × 10-6[1] changes to 1.2 × 10-4 in water with no gas present[1]
To analyze the conductivity of materials exposed to alternating electric fields, it is necessary to treat conductivity as a complex number (or as a matrix of complex numbers, in the case of anisotropic materials mentioned above) called the admittivity. This method is used in applications such as electrical impedance tomography, a type of industrial and medical imaging. Admittivity is the sum of a real component called the conductivity and an imaginary component called the susceptivity.[2]
{\displaystyle \sigma _{T'}={\sigma _{T} \over 1+\alpha (T-T')}}
The temperature compensation slope for most naturally occurring waters is about two %/°C, however it can range between (one to three) %/°C. This slope is influenced by the geochemistry, and can be easily determined in a laboratory.
At extremely low temperatures (not far from absolute zero K), a few materials have been found to exhibit very high electrical conductivity in a phenomenon called superconductivity.
↑ 1.0 1.1 See J. Phys. Chem. B 2005, 109, 1231-1238 In particular page 1235. Note that values in this paper are given in S/cm, not S/m, which differs by a factor of 100. Retrieved September 25, 2008.
↑ Otto H. Schmitt, Mutual Impedivity Spectrometry and the Feasibility of its Incorporation into Tissue-Diagnostic Anatomical Reconstruction and Multivariate Time-Coherent Physiological Measurements University of Minnesota. Retrieved September 25, 2008.
Giancoli, Douglas. Physics for Scientists and Engineers, with Modern Physics (Chapters 1-37), 4th ed. Mastering Physics Series. Upper Saddle River, NJ: Prentice Hall, 2007. ISBN 978-0136139263
Maini, A.K. Electronics and Communications Simplified, 9th ed. New Delhi: Khanna Publishers, 1997.
Plonus, Martin. Electronics and Communications for Scientists and Engineers. San Diego: Harcourt/Academic Press, 2001. ISBN 0125330847
Tipler, Paul Allen, and Gene Mosca. Physics for Scientists and Engineers, Volume 2: Electricity and Magnetism, Light, Modern Physics, 5th ed. New York: W.H. Freeman, 2004. ISBN 0716708108
Electrical_conductivity history
Electrical_conductor history
History of "Electrical conductivity"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Electrical_conductivity&oldid=1062082
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Automatic segmentation of the left ventricle (LV) of a living human heart in a magnetic resonance (MR) image (2D+t) allows to measure some clinical significant indices like the regional wall thicknesses (RWT), cavity dimensions, cavity and myocardium areas, and cardiac phase. Here, we propose a novel framework made of a sequence of two fully convolutional networks (FCN). The first is a modified temporal-like VGG16 (the "localization network") and is used to localize roughly the LV (filled-in) epicardium position in each MR volume. The second FCN is a modified temporal-like VGG16 too, but devoted to segment the LV myocardium and cavity (the "segmentation network"). We evaluate the proposed method with 5-fold-cross-validation on the MICCAI 2019 LV Full Quantification Challenge dataset. For the network used to localize the epicardium, we obtain an average dice index of 0.8953 on validation set. For the segmentation network, we obtain an average dice index of 0.8664 on validation set (there, data augmentation is used). The mean absolute error (MAE) of average cavity and myocardium areas, dimensions, RWT are
{\displaystyle 114.77~{\text{mm}}^{2}}
; 0.9220~mm; 0.9185~mm respectively. The computation time of the pipeline is less than 2~s for an entire 3D volume. The error rate of phase classification is 7.6364%, which indicates that the proposed approach has a promising performance to estimate all these parameters.
RWT are $114.77~\text{mm}^2$; 0.9220~mm; 0.9185~mm
respectively. The computation time of the pipeline is less
than 2~s for an entire 3D volume. The error rate of phase
classification is 7.6364\%, which indicates that the
proposed approach has a promising performance to estimate
all these parameters.}
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The AGI Landscape - AGI University
AGI University
The AGI Landscape
Forget-me-not-Process
Thermodynamics as a theory of decision-making with informationprocessing costs
Reinforcement learning: An introduction 1e
Reactive bandits with attitude
Data clustering by markovian relaxation and the information bottleneck method
Bounded Rationality, Abstraction, and Hierarchical Decision-Making: An Information-Theoretic Optimal
Risk sensitive path integral control
Hysteresis effects of changing the parameters of noncooperative games
An algorithm with nearly optimal pseudo-regret for both stochastic and adversarial bandits
New criteria and a new algorithm for learning in multi-agent systems
On the likelihood that one unknown probability exceeds another in view of the evidence of two sample
\Omega
is going to push the boundary of artificial general intelligence.
\mathbf{\Omega} = \underset{\theta}{\arg\max}\ \mathcal{AGI}(\theta)
Kolmogorov complexity
https://github.com/deepmind/pysc2
https://pythonprogramming.net/starcraft-ii-ai-python-sc2-tutorial/
Important Papers
Universal Transformers
The Forget-me-not Process
AGI Safety Literature Review : summary of general safety research in agi
Out-of-sample extension of graph adjacency spectral embedding: consider the problem of obtaining an out-of-sample extension for the adjacency spectral embedding, a procedure for embedding the vertices of a graph into Euclidean space.
Alignment for Advanced Machine Learning Systems
Measuring and avoiding side effects using relative reachability: introduces a general definition of side effects, based on relative reachability of states compared to a default state, that avoids these undesirable incentives.
The Importance of Sampling in Meta-Reinforcement Learning
Inequity aversion improves cooperation in intertemporal social dilemmas
R. Durrett Probability: Theory and Examples (4th edition).
P. Billingsley Probability and Measure (3rd Edition). Chapters 1-30 contain a more careful and detailed treatment of some of the topics of this semester, in particular the measure-theory background. Recommended for students who have not done measure theory.
R. Leadbetter et al A Basic Course in Measure and Probability: Theory for Applications is a new book giving a careful treatment of the measure-theory background.
D. Khoshnevisan Probability is a well-written concise account of the key topics in 205AB.
R. Bhattacharya and E. C. Waymire A Basic Course in Probability Theory is another well-written account, mostly on the 205A topics.
K.L. Chung A Course in Probability Theory covers many of the topics of 205A: more leisurely than Durrett and more focused than Billingsley.
D. Williams Probability with Martingales has a uniquely enthusiastic style; concise treatment emphasizes usefulness of martingales.
Y.S. Chow and H. Teicher Probability Theory: Independence, Interchangeability, Martingales . Uninspired exposition, but has useful variations on technical topics such as inequalities for sums and for martingales.
R.M. Dudley Real Analysis and Probability. Best account of the functional analysis and metric space background relevant for research in theoretical probability.
B. Fristedt and L. Gray A Modern Approach to Probability Theory. 700 pages allow coverage of broad range of topics in probability and stochastic processes.
L. Breiman Probability. Classical; concise and broad coverage.
O. Kallenberg Foundations of Modern Probability. Quoting an amazon.com reviewer: ``.... a compendium of all the relevant results of probability ..... similar in breadth and depth to Loeve's classical text of the mid 70's. It is not suited as a textbook, as it lacks the many examples that are needed to absorb the theory at a first pass. It works best as a reference book or a "second pass" textbook."
John B. Walsh Knowing the Odds: An Introduction to Probability. New in 2012. Looks very nice -- concise treatment with quite challenging exercises developing part of theory.
George Roussas An Introduction to Measure-Theoretic Probability. Recent treatment of classical content.
Santosh Venkatesh The Theory of Probability: Explorations and Applications. Unique new book, intertwining a broad range of undergraduate and graduate-level topics for an applied audience.
I. Florescu Probability and Stochastic Processes. Very clearly written, and with 550 pages gives a broad coverage of topics including intro to SDEs.
Jim Pitman has his very useful lecture notes linked to the Durrett text; these notes cover more ground than my course will! Also some lecture notes by Amir Dembo for the Stanford courses equivalent to our 205AB.
The Books: https://www.stat.berkeley.edu/~aldous/205B/index.html, by Professor David Aldous from UC Berkeley.
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Gauss: The Prince of Mathematics | Brilliant Math & Science Wiki
Andrew Ellinor, Frank Aiello, Peter Taylor, and
As you progress further into college math and physics, no matter where you turn, you will repeatedly run into the name Gauss. Johann Carl Friedrich Gauss is one of the most influential mathematicians in history. Gauss was born on April 30, 1777 in a small German city north of the Harz mountains named Braunschweig. The son of peasant parents (both were illiterate), he developed a staggering number of important ideas and had many more named after him. Many have referred to him as the princeps mathematicorum, or the “prince of mathematics.”
As part of his doctoral dissertation (at the age of 21), Gauss was one of the first to prove the fundamental theorem of algebra. He went on to publish seminal works in many fields of mathematics including number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy, optics, etc. Number theory was Gauss’s favorite and he referred to number theory as the “queen of mathematics.”
One of the reasons why Gauss was able to contribute so much math over his lifetime was that he got a very early start. There are many tales of his childhood precociousness. The most famous anecdote of young Gauss is the time he found the shortcut for calculating the sum of an arithmetic progression at the tender age of 10.
The anecdote involves his schoolteacher who wanted to take a rest and asked the students to sum the integers from 1 to 100 as busy work. After a few seconds, the teacher saw Gauss sitting idle. When asked why he was not frantically doing addition, Gauss quickly replied that the sum was 5050. His classmates and teacher were astonished, and Gauss ended up being the only pupil to calculate the correct answer.
The story may be apocryphal, and is told different ways in different sources. Nobody is sure which method of summing an arithmetic sequence Gauss figured out as a child. Though there are several ways young Gauss might have solved it, one of them has a concise, intuitive, and elegant visual representation.
1+2+3+ \cdots +(n-1)+n = ?
Fig 1: Arithmetic Progression
Consider two sets of marbles as shown in the Figure 1. The left pile has
n
rows of blue marbles, where the
j^\text{th}
row contains
j
marbles. The right pile has
n
rows of red marbles, where the
j^\text{th}
n+1-j
The total number of blue marbles is given by
1+2 +3+ \cdots + (n-1)+n,
while the total number of red marbles is given by
n+(n-1)+(n-2) + \cdots + 2 + 1,
and clearly both contain the same number of marbles. Now if we were to add these piles together as shown in Figure 2, we would then get a stack with
n
rows, where each row contains
n + 1
marbles:
The total number of marbles in the added pile would be
n(n + 1)
. Since both the red pile and the blue pile have an equal number of marbles, each pile must have contributed
\frac{n(n + 1)}{2}
marbles. Hence, we obtain
1+2+3+...+(n-1)+n= \frac{n(n+1)}{2}.
To sum all the numbers from 1 to 100, Gauss simply calculated
\frac{100\times (100+1)}{2}=5050
, which is immensely easier than adding all the numbers from 1 to 100. Note that
1+2+3 + \cdots +(n-1)+n
must always be a positive integer. Even though the above formula divides by 2, the result will always be a positive integer. This is because the numerator will always be conveniently even due to the multiplication properties of parity. For example,
n
could either be even or odd. If
n
n+1
is odd and hence
n \times (n + 1) = even \times odd = even.
n
n + 1
is even and hence
n \times (n + 1) = odd \times even = even.
Therefore, the numerator is always even and
\frac{n(n + 1)}2
\frac{n(n + 1)}{2}
are called triangular numbers, for reasons well illustrated in the above figures. The first few triangular numbers are
1,3,6,10,15,21,28,36, \ldots.
It is commonplace to encounter an application of summing an arithmetic sequence, both in classroom problems, and in describing the broader world. It is less common to meet 10 year olds who figure out the tricks of arithmetic progression for themselves. It is even less common for a precocious 10 year old to grow up to be nearly as prolific as Gauss.
Proved the law of quadratic reciprocity. The law of quadratic reciprocity is a theorem about quadratic residues modulo an odd prime. It states:
\large \left(\dfrac{p}{q}\right)\left(\dfrac{q}{p}\right)=(-1)^{\frac{p-1}{2} \frac{q-1}{2} },
p
q
are odd prime numbers, and
\left(\dfrac{p}{q}\right)
Formulated Gauss' lemma, which gives a condition for an integer to be a quadratic residue. Gauss' lemma states:
\frac{a}{p}
= 1 iff an an even number of least positive residues of
{a, 2a, 3a, ..., ((p-1)/2)a}
exceed
\frac{p}{2}
Proved that every positive integer is the sum of at most 3 triangular numbers. The proof follows from the result that every positive integer
\equiv
3 (mod 8) can be written as a sum of three squares.
Proved the Theorema Egregium, a major theorem in the differential geometry of curved surfaces. This theorem states that the Gaussian curvature is unchanged when the surface is bent without stretching.
Made important contributions to statistics and probability theory. The Gaussian probability distribution is named after Gauss.
Cite as: Gauss: The Prince of Mathematics. Brilliant.org. Retrieved from https://brilliant.org/wiki/gauss-the-prince-of-mathematics/
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Multistage decimator design - MATLAB designMultistageDecimator - MathWorks España
designMultistageDecimator
Design Efficient Decimator
Compare Multistage Decimator Designs
Determining Best Multistage Decimator Design
Multistage decimator design
C = designMultistageDecimator(M)
C = designMultistageDecimator(M,Fs,TW)
C = designMultistageDecimator(M,Fs,TW,Astop)
C = designMultistageDecimator(___,Name,Value)
C = designMultistageDecimator(M) designs a multistage decimator that has an overall decimation factor of M. In order for C to be multistage, M must not be a prime number. For details, see Algorithms. The design process can take a while if M has many factors.
C = designMultistageDecimator(M,Fs,TW) designs a multistage decimator with a sampling rate of Fs and a transition width of TW. Sampling rate in this case refers to the input sampling rate of the signal before the multistage decimator.
The multistage decimator has a cutoff frequency of Fs/(2M).
C = designMultistageDecimator(M,Fs,TW,Astop) specifies a minimum attenuation of Astop dB for the resulting design.
C = designMultistageDecimator(___,Name,Value) specifies additional design parameters using one or more name-value pair arguments.
Example: C = designMultistageDecimator(48,48000,200,80,'NumStages','auto') designs a multistage decimator with the least number of multiplications per input sample (MPIS).
Design a single-stage decimator using the designMultirateFIR function and a multistage decimator using the designMultistageDecimator function. Determine the efficiency of the two designs using the cost function. The implementation efficiency is characterized by two cost metrics - NumCoefficients and MultiplicationsPerInputSample.
Choose a decimation factor of 48, input sample rate of
30.72×48\text{\hspace{0.17em}}\mathrm{MHz}
, one-sided bandwidth of 10 MHz, and a stopband attenution of 90 dB.
Fin = 30.72e6*M;
Designing the decimation filter using the designMultirateFIR function yields a single-stage design. Set the half-polyphase length to a finite integer, in this case 8.
b = designMultirateFIR(1,M,HalfPolyLength,Astop);
d = dsp.FIRDecimator(M,b)
Numerator: [0 -5.7242e-08 -1.2617e-07 -2.0736e-07 -3.0130e-07 ... ]
Compute the cost of implementing the decimator. The decimation filter requires 753 coefficients and 720 states. The number of multiplications per input sample and additions per input sample are 15.6875 and 15.6667, respectively.
Using the designMultistageDecimator Function
Design a multistage decimator with the same filter specifications as the single-stage design. Compute the transition width using the following relationship:
Fc = Fin/(2*M);
c = designMultistageDecimator(M,Fin,TW,Astop)
Calling the info function on c shows that the filter is implemented as a cascade of four dsp.FIRDecimator objects, with decimation factors of 3, 2, 2, and 4, respectively.
Compute the cost of implementing the decimator.
The NumCoefficients and the MultiplicationsPerInputSample parameters are lower for the four-stage filter designed by the designMultistageDecimator function, making it more efficient.
Compare the magnitude response of both the designs.
Using the 'design' Option in the designMultistageDecimator Function
The filter can be made even more efficient by setting the 'CostMethod' argument of the designMultistageDecimator function to 'design'. By default, this argument is set to 'estimate'.
cOptimal = designMultistageDecimator(M,Fin,TW,Astop,'CostMethod','design')
Design a decimator with an overall decimation factor of 24 using the designMultistageDecimator function. Design the filter in two configurations:
Choose a decimation factor of 24, input sample rate of 6 kHz, stopband attenuation of 90 dB, and a transition width of
0.03×\frac{6000}{2}
Design the two filters using the designMultistageDecimator function.
cAuto = designMultistageDecimator(M,Fs,TW,Astop,'NumStages','Auto')
cTwo = designMultistageDecimator(M,Fs,TW,Astop,'NumStages',2)
View the filter information using the info function. The 'Auto' configuration designs a cascade of three FIR decimators with decimation factors 2, 3, and 4, respectively. The two-stage configuration designs a cascade of two FIR decimators with decimation factors 4 and 6, respectively.
The 'Auto' configuration decimation filter yields a three-stage design that out-performs the two-stage design on all cost metrics.
Compare the Magnitude Response
Comparing the magnitude response of the two filters, both the filters have the same transition-band behavior and follow the design specifications.
hfvt = fvtool(cAuto,cTwo,Analysis="magnitude");
legend(hfvt,"Auto multistage","Two-stage")
However, to understand where the computational savings are coming from in the three-stage design, look at the magnitude response of the three stages individually.
autoSt1 = cAuto.Stage1;
hfvt = fvtool(autoSt1, autoSt2, autoSt3,Analysis="magnitude");
legend(hfvt,"Stage 1","Stage 2","Stage 3")
The third stage provides the narrow transition width required for the overall design (0.03
×
Fs/2). However, the third stage operates at 1.5 kHz and has spectral replicas centered at that frequency and its harmonics.
The first stage removes such replicas. This first and second stages operate at a faster rate but can afford a wide transition width. The result is a decimate-by-2 first-stage filter with only 7 nonzero coefficients and a decimate-by-3 second-stage filter with only 19 nonzero coefficients. The third stage requires 47 coefficients. Overall, there are 73 nonzero coefficients for the three-stage design and 100 nonzero coefficients for the two-stage design.
The combined decimation must equal the overall decimation required.
For an overall decimation factor of 48, there are several combinations of individual stages.
cMinCoeffs = designMultistageDecimator(M,Fs,TW,Astop,'MinTotalCoeffs',true)
To obtain the design with the least number of multiplications per input sample, set 'NumStages' to 'auto'.
cMinMulti = designMultistageDecimator(M,Fs,TW,Astop,'NumStages','auto')
Compare the magnitude response of both filters using fvtool. Both filters have the same transition-band behavior and a stopband attenuation that is below 80 dB.
hvft = fvtool(cMinCoeffs,cMinMulti)
hvft =
Overall decimation factor, specified as a positive integer greater than one. In order for C to be multistage, M must not be a prime number. For details, see Algorithms.
Input sampling rate prior to the multistage decimator, specified as a positive real scalar. If not specified, Fs defaults to 48,000 Hz. The multistage decimator has a cutoff frequency of Fs/(2M).
0.2×Fs/M (default) | positive real scalar
Transition width, specified as a positive real scalar less than Fs/M. If not specified, TW defaults to 0.2×Fs/M. Transition width must be less than Fs/M.
Example: C = designMultistageDecimator(48,48000,200,80,'NumStages','auto') designs a multistage decimator with the lowest number of multiplications per input sample.
NumStages — Number of decimator stages
Number of decimator stages, specified as a positive integer. If set to 'auto', the design algorithm determines the number of stages that result in the lowest number of multiplications per input sample. If specified as a positive integer, N, the overall decimation factor, M, must be able to factor into at least N factors, not counting 1 or M as factors.
'design' –– The function designs each stage and computes the filter order. This method leads to an optimal overall design.
Tolerance, specified as a positive scalar. The tolerance is used to determine the multistage configuration with the least MPIS. When multiple configurations result in the same lowest MPIS within the tolerance specified, the configuration that yields the lowest number of coefficients overall is chosen. To view the total number of coefficients and MPIS for a specific filter, use the cost function.
The overall decimation factor is split into smaller factors with each factor being the decimation factor of the corresponding individual stage. The combined decimation of all the individual stages must equal the overall decimation. The combined response must meet or exceed the given design specifications.
The function determines the number of decimator stages through the 'NumStages' argument. The sequence of stages is determined based on the implementation cost. By default, 'NumStages' is set to 'auto', resulting in a sequence that gives the lowest number of MPIS. When multiple configurations result in the same lowest MPIS within the tolerance specified, the configuration that yields the lowest number of coefficients overall is chosen. If 'MinTotalCoeffs' is set to true, the function determines the sequence that requires the lowest number of total coefficients.
By default, the 'CostMethod' is set to 'estimate'. In this mode, the function estimates the filter order required for each stage and designs the filter based on the estimate. This method is faster than 'design', but can lead to suboptimal designs. For an optimal design, set 'CostMethod' to 'design'. In this mode, the function designs each stage and computes the filter order.
dsp.FilterCascade | dsp.FIRDecimator
designMultirateFIR | designMultistageInterpolator | info | cost
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Using the derivative theorem and, find the Laplace transform of
Using the derivative theorem and, find the Laplace transform of the following functions. r \cos h(t)
Using the derivative theorem and, find the Laplace transform of the following functions
r\mathrm{cos}h\left(t\right)
Laplase transform of
t\mathrm{cos}\left(h\right)t
L\left(t\mathrm{cos}\left(h\right)t\right)
derivative of Laplase transform of
\mathrm{cos}\left(h\right)t
\therefore L\left(\mathrm{cos}ht\right)=\frac{s}{{s}^{2}-1}
\therefore L\left(t\mathrm{cos}ht\right)=\frac{d}{ds}\frac{s}{{s}^{2}-1}=\frac{{s}^{2}-1-2{s}^{2}}{{\left({s}^{2}-1\right)}^{2}}{\left(-1\right)}^{2}
=\frac{-{s}^{2}-1}{{\left({s}^{2}-1\right)}^{2}}\left(-1\right)
=\frac{{s}^{2}+1}{{\left({s}^{2}-1\right)}^{2}}
F\left(s\right)=\frac{3{e}^{-2s}}{s\left(s+3\right)}
F\left(s\right)=\frac{{e}^{-2s}}{s\left(s+1\right)}
F\left(s\right)=\frac{{e}^{-2s}-{e}^{-3s}}{2}
L\left\{\mathrm{sin}\left(t-k\right)\cdot H\left(t-k\right)\right\}
4{t}^{2}+5{t}^{3}
\frac{dy}{dx}-\frac{dx}{dy}=\frac{y}{x}-\frac{x}{y}
y\left(t\right)+2{\int }_{0}^{t}{e}^{-2\left(t-\tau \right)}y\left(\tau \right)d\tau ={e}^{-2t}
What is the inverse laplace transform of
F\left(s\right)=\frac{1}{2}\mathrm{ln}\left(\frac{{s}^{{}^{\left\{2\right\}}}+{b}^{2}}{{s}^{{}^{\left\{2\right\}}}+{a}^{2}}\right)
a,b\text{ }ϵ\text{ }\mathbb{R}
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SDK FAQs | RudderStack Docs
How does RudderStack handle anonymousId ?
The following are the different ways in which RudderStack handles anonymousId across different SDKs:
For the JavaScript SDK
The RudderStack JavaScript SDK automatically generates one unique anonymousId to identify a user uniquely. It then stores it in a cookie named rl_anonymous_id and attaches it to every subsequent event. This helps in identifying the users from other sites that are hosted under a sub-domain.
If anonymousId is explicitly provided by the user using the setAnonymousId method, the user-specified anonymousId overrides the SDK-generated one.
For more information on how RudderStack handles overriding anonymousId, please refer to our docs.
For the Android SDK
RudderStack captures your deviceId and uses that as anonymousId for identifying the user. It is used to track the users across the application installation. To attach more information to the user, you can use the identify method.
You can use the setAnonymousId method to override and use your own anonymousId with the SDK.
On the Android devices, the deviceId is assigned during the first boot. It remains consistent across the applications and installs. It changes only after a factory reset.
For more information on how RudderStack handles anonymousId in the iOS SDK, please refer to our docs.
For the iOS SDK
RudderStack captures deviceId and uses that as anonymousId for identifying the user. To attach more information to the user, you can use the identify method.
According to the Apple documentation, if the device has multiple apps from the same vendor, all those apps will be assigned the same deviceId. If all these apps are uninstalled, then on the next install, the apps will be assigned a new deviceId.
For more information on how RudderStack handlesanonymousId in the iOS SDK, please refer to our docs.
How do I identify anonymous users across client-side and server-side?
To identify anonymous users across both client-side and server-side, it is advisable to use a separate, new cookie at your end.
During the user's first visit, your server generates a new anonymousId to make the event calls using the server-side SDKs and sends the set_cookie response to the browser to set the visitor_id cookie.
If the RudderStack JavaScript SDK is not blocked, you can use the setAnonymousId method to set the same value as the visitor_id.
In case the RudderStack JavaScript SDK is blocked, still the next requests to the server will have the visitor_id cookie which can be used by the server-side events for anonymousId.
The RudderStack JavaScript SDK generates a unique anonymousId for every unique user visit. It then stores this value in a cookie named rl_anonymous_id and attaches it to every subsequent event.
Users sometimes try to directly use the browser APIs to get or set the value for this cookie. However, this is not advisable since the RudderStack cookies are encrypted, and the cookie may not be present altogether (if the SDK is blocked).
It is, therefore, always advisable to use RudderStack's getAnonymousId and setAnonymousId methods to update the cookie value.
To set anonymousId, use the setAnonymousId call after the SDK snippet as below:
rudderanalytics.setAnonymousId("my-anon-id");
To get the anonymousId stored in a RudderStack cookie, use the getAnonymousId call inside the ready callback - this ensures that the method is available and returns the previously set anonymousId value.
rudderanalytics.ready(
var anonId = window.rudderanalytics.getAnonymousId();
console.log(anonId);
What is the RudderStack retry and backoff logic after the connection fails?
When the dataplane gets disconnected from the SDK and events are no longer able to be sent to Rudder Server, then some of the SDK's will store events and retry sending them to Rudder Server with a certain backoff logic.
NOTE: The retry of failed events is not supported by all SDKs. Please see table below for support.
General Support and Logic
JavaScript SDK Yes 100 events in Local Storage 10 times
Android SDK Yes 10k events in sqlite db Infinity
iOS SDK Yes 10k events in sqlite db Infinity
Node SDK Yes 20k events in-memory 10 times
All Other SDKs No N/A N/A
This SDK can be configured to match your requirements for retry and backoff logic. By default, if the dataplane goes down and the JS SDK cannot send events to the Rudder Server, up to 100 events will be stored. While still disconnected from the dataplane, the JS SDK will try to resend the stored events to the Rudder Server. However, for each retry, the delay duration will grow. The equation to get the duration of delay is as follows
dt = md * (F^n)
dt
is the delay time in ms,
md
is the minRetryDelay (configurable; default is 1000 ms),
F
is the backoffFactor (configurable; default is 2), and
n
is the current retry attempt. The SDK will retry until the attempts surpass the maxAttempts value. This is by default set to 10 attempts but is configurable. Each retry attempt, the delay time grows exponentially. However, it will max out at whatever the maxRetryDelay is. By default, this value is set at 360000 ms, but it is configurable.
iOS SDK and Android SDK
Both the iOS and Android SDKs share similar retry and backoff logic for when the dataplane connection fails. If the dataplane goes down, up to 10,000 events will be stored. There is no limit to how many times the SDK will try to send failed events. However, the delay duration in between the attempts will grow by 1 second after each retry. For example, after the first failed attempt, there will be a delay of 1 second. After the second failed attempt, the SDK will wait 2 seconds before it retries. The third failed attempt will cause a delay of 3 seconds, and this behavior will repeat until the connection is re-established.
Currently the Node SDK is the only server-side SDK that supports event retry and backoff logic. The logic is quite similar to the JavaScript SDK. If the connection fails, up to 20,000 events will be stored. However, this is in-memory storage and can result in data loss. The SDK will retry a maximum of 10 times, by default. For each retry the delay duration between retries will grow and can be calculated using the following equation.
dt = 1000 * (2^n)
dt
is the delay time in ms and
n
is the current retry attempt. The SDK will retry until the attempts surpass the maxAttempts value which is set to 10 attempts. With each retry attempt, the delay time will grow exponentially. However, it will never be greater than the maximum delay duration which is 30 seconds.
The Node SDK does have a feature to persist the event data in Redis for more event storage and better guarantees of failed event delivery. Instructions on how to configure the Redis solution can be found here.
Can I filter and selectively send the event data to certain destinations?
Yes, you can use RudderStack's Client-side Event Filtering feature to specify which events should be discarded or allowed to flow through - by whitelisting or blacklisting them in the RudderStack dashboard while setting up your destination. This method is useful if you are sending the events via the device mode.
For more information on the RudderStack SDKs, you can contact us. You can also talk to us in our Slack community - we will be happy to help you!
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Wightman_axioms Knowpia
W0 (assumptions of relativistic quantum mechanics)Edit
Quantum mechanics is described according to von Neumann; in particular, the pure states are given by the rays, i.e. the one-dimensional subspaces, of some separable complex Hilbert space. In the following, the scalar product of Hilbert space vectors Ψ and Φ is denoted by
{\displaystyle \langle \Psi ,\Phi \rangle }
, and the norm of Ψ is denoted by
{\displaystyle \lVert \Psi \rVert }
. The transition probability between two pure states [Ψ] and [Φ] can be defined in terms of non-zero vector representatives Ψ and Φ to be
{\displaystyle P{\big (}[\Psi ],[\Phi ]{\big )}={\frac {|\langle \Psi ,\Phi \rangle |^{2}}{\lVert \Psi \rVert ^{2}\lVert \Phi \rVert ^{2}}}}
{\displaystyle \langle \Psi (a,L),\Phi (a,L)\rangle =\langle \Psi ,\Phi \rangle .}
{\displaystyle U(a,L)U(b,M)=\pm U{\big (}(a,L)\cdot (b,M){\big )},}
i.e. the phase is a multiple of
{\displaystyle \pi }
. For particles of integer spin (pions, photons, gravitons, ...) one can remove the ± sign by further phase changes, but for representations of half-odd-spin, we cannot, and the sign changes discontinuously as we go round any axis by an angle of 2π. We can, however, construct a representation of the covering group of the Poincare group, called the inhomogeneous SL(2, C); this has elements (a, A), where as before, a is a four-vector, but now A is a complex 2 × 2 matrix with unit determinant. We denote the unitary operators we get by U(a, A), and these give us a continuous, unitary and true representation in that the collection of U(a, A) obey the group law of the inhomogeneous SL(2, C).
An ensemble corresponding to U(a, L)|v⟩ is to be interpreted with respect to the coordinates
{\displaystyle x'=L^{-1}(x-a)}
in exactly the same way as an ensemble corresponding to |v⟩ is interpreted with respect to the coordinates x; and similarly for the odd subspaces.
The group of spacetime translations is commutative, and so the operators can be simultaneously diagonalised. The generators of these groups give us four self-adjoint operators
{\displaystyle P_{0},P_{j},\ j=1,2,3,}
which transform under the homogeneous group as a four-vector, called the energy–momentum four-vector.
{\displaystyle P_{0}\geq 0,\quad P_{0}^{2}-P_{j}P_{j}\geq 0.}
W1 (assumptions on the domain and continuity of the field)Edit
For each test function f,[clarification needed] there exists a set of operators
{\displaystyle A_{1}(f),\ldots ,A_{n}(f)}
which, together with their adjoints, are defined on a dense subset of the Hilbert state space, containing the vacuum. The fields A are operator-valued tempered distributions. The Hilbert state space is spanned by the field polynomials acting on the vacuum (cyclicity condition).
W2 (transformation law of the field)Edit
{\displaystyle U(a,L)^{\dagger }A(x)U(a,L)=S(L)A{\big (}L^{-1}(x-a){\big )}.}
W3 (local commutativity or microscopic causality)Edit
Cyclicity of a vacuum and uniqueness of a vacuum are sometimes considered separately. Also, there is property of asymptotic completeness – that Hilbert state space is spanned by the asymptotic spaces
{\displaystyle H^{\text{in}}}
{\displaystyle H^{\text{out}}}
, appearing in the collision S matrix. The other important property of field theory is mass gap, which is not required by the axioms – that energy–momentum spectrum has a gap between zero and some positive number.
Consequences of the axiomsEdit
Relation to other frameworks and concepts in quantum field theoryEdit
Existence of theories that satisfy the axiomsEdit
Osterwalder–Schrader reconstruction theoremEdit
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Rajdhani Express leave Delhi for Jaipur at a speed of 120km/h After 90 minutes,another trian Delhi Express Leaves - Maths - Playing with Numbers - 7873829 | Meritnation.com
Assuming that, the two trains A and B are travelling in opposite directions.
Speed of train A travelling from Delhi to Jaipur = 120 km/hr
Speed of train B travelling from Jaipur to Delhi = 100 km/hr
Now, it is given that the train B starts after 90 min than that of the train A.
Suppose the train A reaches the point P after 90 minutes.
Distance covered by the train A in 90 minutes =
120\times \frac{90}{60}=120\times \frac{3}{2}=180\quad \text{km}
Let the total distance between Delhi and Jaipur = (x + 180) km
Here x = distance between point P and Jaipur.
After the train A has reached point P, let after t hours both the trains meet at point Q.
As the trains are moving towards each other so the relative velocity of train A with respect to train B = 120 + 100 = 220 km/h
\mathrm{Now},\quad \mathrm{t}\quad =\quad \frac{\mathrm{x}}{220}\quad \mathrm{hrs}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\quad \mathrm{distance}\quad \mathrm{travelled}\quad \mathrm{by}\quad \mathrm{train}\quad A\quad \mathrm{in}\quad \mathrm{t}\quad \mathrm{hrs}\quad =\quad \frac{6\mathrm{x}}{11}\quad \mathrm{km}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\quad \mathrm{distance}\quad \mathrm{travelled}\quad \mathrm{by}\quad \mathrm{train}\quad B\quad \mathrm{in}\quad \mathrm{t}\quad \mathrm{hrs}\quad =\quad \mathrm{x}\quad -\quad \frac{6\mathrm{x}}{11}\quad =\quad \frac{5\mathrm{x}}{11}\quad \mathrm{km}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\quad \mathrm{when}\quad \mathrm{the}\quad \mathrm{two}\quad \mathrm{trains}\quad \mathrm{cross}\quad \mathrm{each}\quad \mathrm{other},\quad \mathrm{then}\quad \phantom{\rule{0ex}{0ex}}\mathrm{distance}\quad \mathrm{travelled}\quad \mathrm{by}\quad \mathrm{train}\quad A\quad =\left[\quad 180\quad +\quad \frac{6\mathrm{x}}{11}\right]\quad \mathrm{km}\phantom{\rule{0ex}{0ex}}\mathrm{distance}\quad \mathrm{travelled}\quad \mathrm{by}\quad \mathrm{train}\quad B\quad =\quad \frac{5\mathrm{x}}{11}\quad \mathrm{km}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}\quad \mathrm{the}\quad \mathrm{train}\quad A\quad \mathrm{is}\quad \mathrm{travelling}\quad \mathrm{towards}\quad \mathrm{Jaipur}\quad \mathrm{and}\quad \mathrm{train}\quad B\quad \mathrm{is}\quad \mathrm{travelling}\quad \mathrm{towards}\quad \mathrm{Delhi},\phantom{\rule{0ex}{0ex}}\mathrm{so},\quad \mathrm{it}\quad \mathrm{is}\quad \mathrm{clear}\quad \mathrm{that}\quad \mathrm{when}\quad \mathrm{the}\quad 2\quad \mathrm{trains}\quad \text{meet }\mathrm{each}\quad \mathrm{other},\text{ they are from same distance Jaipur}.\phantom{\rule{0ex}{0ex}}\text{After crossing each other, the train A will be closer to Jaipur as it is heading towards Jaipur.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}
I think some data is missing from your question, please recheck your question once again, and do get back to us.
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A Korovkin Type Approximation Theorem and Its Applications
Malik Saad Al-Muhja, "A Korovkin Type Approximation Theorem and Its Applications", Abstract and Applied Analysis, vol. 2014, Article ID 859696, 6 pages, 2014. https://doi.org/10.1155/2014/859696
Malik Saad Al-Muhja1
1Department of Mathematics and Computer Application, College of Sciences, University of Al-Muthanna, Samawa, Iraq
We present a Korovkin type approximation theorem for a sequence of positive linear operators defined on the space of all real valued continuous and periodic functions via A-statistical approximation, for the rate of the third order Ditzian-Totik modulus of smoothness. Finally, we obtain an interleave between Riesz's representation theory and Lebesgue-Stieltjes integral-i, for Riesz's functional supremum formula via statistical limit.
Some will accept the notes and definitions used in this paper. The concept of -statistical approximation for regular summability matrix (see [1, 2]). Let , , be an infinite summability matrix. For a given sequence , the -transform of , denoted by , is given by , provided that the series converges for each . is said to be regular if , whenever . Then , for all . In [3], Dzyubenko and Gilewicz have given the notion.
is nonnegative regular summability matrix. Then is -statistically convergent to , if, for every , .
We denote by the space of all -periodic and continuous functions on . Endowed with the norm , this space is a Banach space, where . Now, recall that, in [4], the th order Ditzian-Totik modulus of smoothness in the uniform metric is given by where is the symmetric th difference. We have to recall the Korovkin type theorem.
Theorem 1 (see [2]). Let be a sequence of infinite nonnegative real matrices such that and let be a sequence of positive linear operators mapping into . Then, for all , we have uniformly in , if and only if (), uniformly in , where , , and , for all .
It is worth noting that the statistical analog of Theorem 1 has been studied by Radu [2], as follows.
Theorem 2. Let be a sequence of nonnegative regular summability matrices and let be a sequence of positive linear operators mapping into . Then, for all , we have , uniformly in , if and only if (), uniformly in , where , , and , for all .
The following notations are used this paper (see [5, 6]).
Let be fixed and sufficiently large. If and , then it is convenient to denote and therefore , for . Recall that is the sign of on .
Now, let us introduce our theorems as follows.
Theorem 3. Let be a sequence of infinite nonnegative real matrices such that and let be a sequence of positive linear operators mapping into . Then, for all , we have uniformly in , if and only if uniformly in , where , , and , for all . And the constant does not depend on .
Theorem 4. Let be a sequence of nonnegative regular summability matrices and let be a sequence of positive linear operators mapping into . Then, if there exists , we have uniformly in , if and only if uniformly in , where , , and , for all .
Proof of Theorem 3. Since () belong to , implications (5) (6) are obvious. Now, assume that (6) holds. Let , and, be a closed subinterval of length of . And let be defined by and also where and are chosen so that
In [5] Kopotun, we have and , where is the Lagrange polynomial of degree , which interpolates at , , and . Inequality (11) is an analog of Whitney's inequality for Ditzian-Totik moduli. Using (11) and the above presentations of and , we write, for , Taking supremum over and , we obtain Suppose , let us write sets as follows: Consequently, we get and implies
Proof of Theorem 4. Since () belong to , implications (8) (7) are obvious. Assume that the condition (7) is satisfied. Let and be a closed subinterval of length of ; we have
Now, given , choose , where implied , and define the following set: Thus, where polynomial and . Since is -statistically convergent, we can easily show that implies .
Now, let , and using (7) implies This is a complete proof.
3. Application to Functional Approximation
In this section we give some applications which satisfy our theorems, but it's not the classical Korovkin theorem. It has been treated with the Weierstrass second approximation theorem via -statistical convergence (see [6–8]). If , then there is a sequence of polynomials and -statistically uniformly convergent to on (not uniformly convergent). Observe that Fejer operators may be written in the form of We now consider the linear operator defined by where is a matrix of real numbers and also and are Fourier coefficients. Now, let be a nonnegative regular summability matrice. Assume that the following statements are satisfied: (i);(ii). We get where is the sequence of linear operators given by (21).
In [9], Sakaoğlu and Ünver proved the following theorem by using and denoted the space of all functions defined on , for which , . In this case, the norm of a function in , denoted by , is given by .
Theorem 5 (see [9]). Let be a nonnegative regular summability matrix and let be an -statistically uniformly bounded sequence of positive linear operators from into and . Then, for any function, if and only if , where , , , and .
The theory of the Lebesgue integral can be developed in several distinct ways (see [10, 11]). Only one of these methods will be discussed here.
Now, let us introduce our definition as follows.
Definition 6 (Lebesgue-Stieltjes integral-). Let be measurable set, be a bounded function, and be nondecreasing function for . For Lebesgue partition of , put and such that measurable function of ; , , and . Also, , , , and , where and . If , where . Then is integral according to for .
Now, we can provide our theorem as follows as a case which is an illustrative application of approximation theory in functional analysis using functional supremum to limit convergence that acts as support and reinforcement of the concept of Riesz's representation.
Theorem 7. If a sequence is positive linear functional and bounded on , is bounded measurable function to . Then, there exists nondecreasing function to such that .
Proof. Assume that functional supremum is as follows: where converges to ; that is, let be Lebesgue partition such that where .
Since positive linear functional and bounded on , then also, respect between sum and Lebesgue-Stieltjes integral- are , we have as ; hence satisfies Lebesgue-Stieltjes integral- of .
Now, since is functional supremum and satisfies Lebesgue-Stieltjes integral-, and us Definition 6, we have Note that effect sum on measurable function by using Lebesgue partition let , choose , and define the following sets: Then , which gives we obtain that implies .
Now, in this paper we have proved Riesz's representation theory with Lebesgue-Stieltjes integral-, by using Korovkin type approximation which is one of the threads in the development of Riesz's theorem to support the definition of Lebesgue integral, Rudin [10]. This integration toxicity ratio for the world on behalf of the French Lebesgue, who came in his thesis for a doctorate in 1902.
The author is grateful for hospitality at the University of Kufa. He thanks his fellows for the fruitful discussions while preparing this paper. He was partially supported by University of Al-Muthanna.
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-summability and approximation of continuous periodic functions,” Studia Universitatis, Babeş-Bolyai: Mathematica, vol. 52, no. 4, pp. 155–161, 2007. View at: Google Scholar | MathSciNet
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P. P. Korovkin, Linear Operators and Approximation Theory, Gordon and Breach, Delhi, India, 1960. View at: MathSciNet
R. J. Serfling, Approximation Theorems of Mathematical Statistics, John Wiley & Sons, 1980. View at: MathSciNet
I. Sakaoğlu and M. Ünver, “Statistical approximation for multivariable integrable functions,” Miskolc Mathematical Notes, vol. 13, no. 2, pp. 485–491, 2012. View at: Google Scholar | MathSciNet
R. G. Bartle, The Elements of Real Analysis, John Wiley & Sons, 3rd edition, 1976. View at: MathSciNet
Copyright © 2014 Malik Saad Al-Muhja. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Remote Sensing | Free Full-Text | Mesoscale Temporal Wind Variability Biases Global Air–Sea Gas Transfer Velocity of CO2 and Other Slightly Soluble Gases
Automated Mapping of Ms 7.0 Jiuzhaigou Earthquake (China) Post-Disaster Landslides Based on High-Resolution UAV Imagery
Cassar, N.
Division of Earth and Ocean Sciences, Nicholas School of the Environment, Duke University, Durham, NC 27708, USA
Nicholas School of the Environment, Box 90328, Duke University, Durham, NC 27708, USA
Department of Civil and Environmental Engineering, Duke University, Durham, NC 27708, USA
CNRS, Univ Brest, IRD, Ifremer, LEMAR, F-29280 Plouzané, France
Academic Editor: Peter Minnett
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\mathrm{U}
⟨\mathrm{U}⟩,
⟨.⟩
\mathrm{U}
\mathrm{U}
{\mathsf{\sigma }}_{\mathrm{U}}
⟨\mathrm{U}⟩
{\mathrm{I}}_{\mathrm{u}}
{\mathsf{\sigma }}_{\mathrm{U}}
⟨\mathrm{U}⟩
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{\mathrm{I}}_{\mathrm{u}}^{2}
{\mathrm{I}}_{\mathrm{u}}^{2}
to the time-averaging interval (spanning from 6 h to a month) is presented to enable other sub-monthly averaging periods to be used. While the focus here is on CO2, the theoretical tactic employed can be applied to other slightly soluble gases. As monthly and climatological wind data are often used in climate models for gas transfer estimates, the proposed approach provides a robust scheme that can be readily implemented in current climate models. View Full-Text
Keywords: carbon dioxide; gas transfer velocity; time-averaging; wind speeds carbon dioxide; gas transfer velocity; time-averaging; wind speeds
Gu, Y.; Katul, G.G.; Cassar, N. Mesoscale Temporal Wind Variability Biases Global Air–Sea Gas Transfer Velocity of CO2 and Other Slightly Soluble Gases. Remote Sens. 2021, 13, 1328. https://doi.org/10.3390/rs13071328
Gu Y, Katul GG, Cassar N. Mesoscale Temporal Wind Variability Biases Global Air–Sea Gas Transfer Velocity of CO2 and Other Slightly Soluble Gases. Remote Sensing. 2021; 13(7):1328. https://doi.org/10.3390/rs13071328
Gu, Yuanyuan, Gabriel G. Katul, and Nicolas Cassar. 2021. "Mesoscale Temporal Wind Variability Biases Global Air–Sea Gas Transfer Velocity of CO2 and Other Slightly Soluble Gases" Remote Sensing 13, no. 7: 1328. https://doi.org/10.3390/rs13071328
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the mean and standard deviation of 100 observations were calculated as 40 and 5 1 resp by a student - Maths - Statistics - 7256457 | Meritnation.com
\overline{x}
= 40 ,
\sigma
\overline{x}=\frac{{\displaystyle \sum _{x}}x}{n}So,\phantom{\rule{0ex}{0ex}}\sum _{x}x\quad =\quad n\overline{x}\quad =100\times 40=4000Now\quad This\quad sum\quad is\quad incorrect,So\quad for\quad correct\quad sum\quad we\quad will\quad subtract\quad old\quad vlaue\quad and\quad add\quad new\quad valueSo,\quad New\quad (Correct)sum=4000-50+40=4000-10=3990So,\quad Now\quad means\quad i.e.\quad correct\quad \overline{x}=\frac{3990}{100}=39.9Now\quad given\quad \sigma =5.1According\quad to\quad formula\sigma
2=∑xx2n-x¯2 ⇒5.12=∑xx2100-402 ⇒∑xx2100=5.12+402 =1626.01 So, ∑xx2 =1626.01×100=162601 But calculated ∑xx2 is incorrect, so for correct sum of square we subtract square of 50 and ass square of 40 So, new ∑xx2=162601-502+402=161701 So, correct σ2=correct ∑xx2n-correct x¯2 correct σ2 =161701100-39.92=1617.01-1592.01=25 So, σ =25=5 Hence new mean is 39.9 and standard deviation is 5
Ayusman Sahu answered this
refer to NCERT t.b example...
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Forecast univariate autoregressive integrated moving average (ARIMA) model responses or conditional variances - MATLAB forecast - MathWorks Benelux
{y}_{t}=1+0.3{y}_{t-1}+2{x}_{t}+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
{\mathit{x}}_{\mathit{t}}
{\mathit{y}}_{0}
E\left({y}_{t}\right)=\frac{1+2\left(1\right)}{1-0.3}.
\left(1,0,0\right){\left(1,1,0\right)}_{4}
\left(1,0,0\right){\left(1,1,0\right)}_{4}
\left(1-0.5L\right)\left(1-0.2{L}^{4}\right)\left(1-{L}^{4}\right){y}_{t}=1+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
\mathit{t}
\begin{array}{l}{y}_{t}=0.073+0.138{y}_{t-1}+{\epsilon }_{t}\\ {\sigma }_{t}^{2}=0.022+0.873{\sigma }_{t-1}^{2}+0.119{\epsilon }_{t-1},\end{array}
{\epsilon }_{\mathit{t}}
{\left[\begin{array}{cc}{y}_{T-K-1}& {y}_{T-K}\end{array}\right]}^{\prime }
\left[\begin{array}{ccc}{x}_{1,\left(T-K+1\right):T}& {x}_{2,\left(T-K+1\right):T}& {x}_{3,\left(T-K+1\right):T}\end{array}\right]
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Solve the following differential equation by the laplace transform method.
Solve the following differential equation by the laplace transform method. 6y^{(4)}+7y'''+21y"+28y'-12y=t(\cos(2t)+te^{\frac{-3t}{2}}).
Solve the following differential equation by the laplace transform method
6{y}^{\left(4\right)}+7y{}^{‴}+21y28{y}^{\prime }-12y=t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)
y\left(0\right)=0,{y}^{\prime }\left(0\right)=1,y"\left(0\right)=0,{y}^{‴}\left(0\right)=-1
6{y}^{\left(4\right)}+7y{}^{‴}+21y28{y}^{\prime }-12y=t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)
Apply Laplace transform on both sides
L\left\{6{y}^{\left(4\right)}+7{y}^{‴}+21y"+28{y}^{\prime }-12y\right\}=L\left\{t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)\right\}
6L\left\{{y}^{‴}\left(t\right)\right\}+7L\left\{{y}^{‴}\left(t\right)\right\}+21L\left\{y"\left(t\right)\right\}+28L\left\{{y}^{\prime }\left(t\right)\right\}+12L\left\{y\right\}=L\left\{t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)\right\}
6\left({s}^{4}L\left\{y\right\}-{s}^{3}y\left(0\right)-{s}^{2}{y}^{\prime }\left(0\right)-sy"\left(0\right)-{y}^{‴}\left(0\right)\right)+7\left({s}^{3}L\left\{y\right\}-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)\right)+21\left({s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+28L\left(sL\left\{y\right\}-y\left(0\right)\right)-12L\left\{y\right\}=-\left(-\frac{\left(s+2\right)\left(s-2\right)}{\left({s}^{2}+4{\right)}^{2}}-\frac{16}{\left(2s+3{\right)}^{3}}\right)
Plug the initial conditions ,
y\left(0\right)=0,{y}^{\prime }\left(0\right)=1,y0\right)=0,y{}^{‴}\left(0\right)=-1
6\left({s}^{4}L\left\{y\right\}-{s}^{3}y\left(0\right)-{s}^{2}{y}^{\prime }\left(0\right)-sy"\left(0\right)-{y}^{‴}\left(0\right)\right)+7\left({s}^{3}L\left\{y\right\}-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)\right)+21\left({s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+28\left(sL\left\{y\right\}-y\left(0\right)\right)-12L\left\{y\right\}=-\left(-\frac{\left(s+2\right)\left(s-2\right)}{\left({s}^{2}+4{\right)}^{2}}-\frac{16}{\left(2s+3{\right)}^{3}}\right)
y{}^{″}-\frac{{y}^{\prime }}{x}=4{x}^{2}y
\left(2x+{e}^{y}\right)dx+x{e}^{y}dy=0
yy=\mathrm{sin}x
x={e}^{t}+{e}^{-t},y=5-2t
0\le t\le 3
{y}^{″}+6{y}^{\prime }+5y=t\cdot U\left(t-2\right)
y\left(0\right)=1
{y}^{\prime }\left(0\right)=0
Use Laplace transforms to solve the following initial value problem.
{x}^{″}+x=3\mathrm{cos}3t,x\left(0\right)=1,{x}^{\prime }\left(0\right)=0
x\left(t\right)=?
Use a Laplace transform to determine the solution of the following systems with differential equations
{x}^{\prime }+4x+3y=0\text{ with }x\left(0\right)=0
{y}^{\prime }+3x+4y=2{e}^{t},y\left(0\right)=0
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Correction of measurement, state, and state estimation error covariance - MATLAB - MathWorks United Kingdom
vision.KalmanFilter
Track Location of An Object
Correction of measurement, state, and state estimation error covariance
The Kalman filter object is designed for tracking. You can use it to predict a physical object's future location, to reduce noise in the detected location, or to help associate multiple physical objects with their corresponding tracks. A Kalman filter object can be configured for each physical object for multiple object tracking. To use the Kalman filter, the object must be moving at constant velocity or constant acceleration.
The Kalman filter algorithm involves two steps, prediction and correction (also known as the update step). The first step uses previous states to predict the current state. The second step uses the current measurement, such as object location, to correct the state. The Kalman filter implements a discrete time, linear State-Space System.
To make configuring a Kalman filter easier, you can use the configureKalmanFilter object to configure a Kalman filter. It sets up the filter for tracking a physical object in a Cartesian coordinate system, moving with constant velocity or constant acceleration. The statistics are the same along all dimensions. If you need to configure a Kalman filter with different assumptions, do not use the function, use this object directly.
In the state space system, the state transition model, A, and the measurement model, H, are set as follows:
A [1 1 0 0; 0 1 0 0; 0 0 1 1; 0 0 0 1]
H [1 0 0 0; 0 0 1 0]
kalmanFilter = vision.KalmanFilter
kalmanFilter = vision.KalmanFilter(StateTransitionModel,MeasurementModel)
kalmanFilter = vision.KalmanFilter(StateTransitionModel,MeasurementModel,ControlModel,Name,Value)
kalmanFilter = vision.KalmanFilter returns a kalman filter for a discrete time, constant velocity system.
kalmanFilter = vision.KalmanFilter(StateTransitionModel,MeasurementModel) additionally configures the control model, B.
kalmanFilter = vision.KalmanFilter(StateTransitionModel,MeasurementModel,ControlModel,Name,Value) configures the Kalman filter object properties, specified as one or more Name,Value pair arguments. Unspecified properties have default values.
StateTransitionModel — Model describing state transition between time steps (A)
[1 1 0 0; 0 1 0 0; 0 0 1 1; 0 0 0 1] (default) | M-by-M matrix
Model describing state transition between time steps (A), specified as an M-by-M matrix. After the object is constructed, this property cannot be changed. This property relates to the A variable in the state-space model.
MeasurementModel — Model describing state to measurement transformation (H)
[1 0 0 0; 0 0 1 0] (default) | N-by-M matrix
Model describing state to measurement transformation (H) , specified as an N-by-M matrix. After the object is constructed, this property cannot be changed. This property relates to the H variable in the state-space model.
ControlModel — Model describing control input to state transformation (B)
[] (default) | M-by-L matrix
Model describing control input to state transformation (B) , specified as an M-by-L matrix. After the object is constructed, this property cannot be changed. This property relates to the B variable in thestate-space model.
State — State (x)
[0] (default) | scalar | M-element vector.
State (x), specified as a scalar or an M-element vector. If you specify State as a scalar, it will be extended to an M-element vector. This property relates to the x variable in the state-space model.
StateCovariance — State estimation error covariance (P)
[1] (default) | scalar | M-by-M matrix
State estimation error covariance (P), specified as a scalar or an M-by-M matrix. If you specify StateCovariance as a scalar it will be extended to an M-by-M diagonal matrix. This property relates to the P variable in the state-space system.
ProcessNoise — Process noise covariance (Q)
Process noise covariance (Q) , specified as a scalar or an M-by-M matrix. If you specify ProcessNoise as a scalar it will be extended to an M-by-M diagonal matrix. This property relates to the Q variable in the state-space model.
MeasurementNoise — Measurement noise covariance (R)
[1] (default) | scalar | N-by-N matrix
Measurement noise covariance (R) , specified as a scalar or an N-by-N matrix. If you specify MeasurementNoise as a scalar it will be extended to an N-by-N diagonal matrix. This property relates to the R variable in the state-space model.
Use the predict and correct functions based on detection results. Use the distance function to find the best matches.
When the tracked object is detected, use the predict and correct functions with the Kalman filter object and the detection measurement. Call the functions in the following order:
[...] = predict(kalmanFilter);
[...] = correct(kalmanFilter,measurement);
When the tracked object is not detected, call the predict function, but not the correct function. When the tracked object is missing or occluded, no measurement is available. Set the functions up with the following logic:
If measurement exists
If the tracked object becomes available after missing for the past t-1 contiguous time steps, you can call the predict function t times. This syntax is particularly useful to process asynchronous video.. For example,
[...] = correct(kalmanFilter,measurement)
correct Correction of measurement, state, and state estimation error covariance
predict Prediction of measurement
distance Confidence value of measurement
Track the location of a physical object moving in one direction.
Generate synthetic data which mimics the 1-D location of a physical object moving at a constant speed.
detectedLocations = num2cell(2*randn(1,40) + (1:40));
Simulate missing detections by setting some elements to empty.
detectedLocations{1} = [];
for idx = 16: 25
detectedLocations{idx} = [];
Create a figure to show the location of detections and the results of using the Kalman filter for tracking.
ylabel('Location');
xlim([0,length(detectedLocations)]);
Create a 1-D, constant speed Kalman filter when the physical object is first detected. Predict the location of the object based on previous states. If the object is detected at the current time step, use its location to correct the states.
kalman = [];
for idx = 1: length(detectedLocations)
location = detectedLocations{idx};
if isempty(kalman)
if ~isempty(location)
stateModel = [1 1;0 1];
measurementModel = [1 0];
kalman = vision.KalmanFilter(stateModel,measurementModel,'ProcessNoise',1e-4,'MeasurementNoise',4);
kalman.State = [location, 0];
trackedLocation = predict(kalman);
plot(idx, location,'k+');
d = distance(kalman,location);
title(sprintf('Distance:%f', d));
trackedLocation = correct(kalman,location);
title('Missing detection');
plot(idx,trackedLocation,'ro');
legend('Detected locations','Predicted/corrected locations');
Use Kalman filter to remove noise from a random signal corrupted by a zero-mean Gaussian noise.
Synthesize a random signal that has value of 1 and is corrupted by a zero-mean Gaussian noise with standard deviation of 0.1.
z = x + 0.1 * randn(1,len);
Remove noise from the signal by using a Kalman filter. The state is expected to be constant, and the measurement is the same as state.
stateTransitionModel = 1;
measurementModel = 1;
obj = vision.KalmanFilter(stateTransitionModel,measurementModel,'StateCovariance',1,'ProcessNoise',1e-5,'MeasurementNoise',1e-2);
z_corr = zeros(1,len);
for idx = 1: len
z_corr(idx) = correct(obj,z(idx));
figure, plot(x * ones(1,len),'g-');
plot(1:len,z,'b+',1:len,z_corr,'r-');
legend('Original signal','Noisy signal','Filtered signal');
This object implements a discrete time, linear state-space system, described by the following equations.
x\left(k\right)=Ax\left(k-1\right)+Bu\left(k-1\right)+w\left(k-1\right)
z\left(k\right)=Hx\left(k\right)+v\left(k\right)
k Time. Scalar
x State. Gaussian vector with covariance P. [
x~N\left(\overline{x},P\right)
] M-element vector
P State estimation error covariance. M-by-M matrix
A State transition model. M-by-M matrix
B Control model. M-by-L matrix
u Control input. L-element vector
w Process noise; Gaussian vector with zero mean and covariance Q. [
w~N\left(0,Q\right)
Q Process noise covariance. M-by-M matrix
z Measurement. For example, location of detected object. N-element vector
H Measurement model. N-by-M matrix
v Measurement noise; Gaussian vector with zero mean and covariance R. [
v~N\left(0,R\right)
] N-element vector
R Measurement noise covariance. N-by-N matrix
[1] Welch, Greg, and Gary Bishop, An Introduction to the Kalman Filter, TR 95–041. University of North Carolina at Chapel Hill, Department of Computer Science.
[2] Blackman, S. Multiple-Target Tracking with Radar Applications. Artech House, Inc., pp. 93, 1986.
configureKalmanFilter | assignDetectionsToTracks
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Multinomial Theorem | Brilliant Math & Science Wiki
Sandeep Bhardwaj, Lino Demasi, Patrick Corn, and
The multinomial theorem describes how to expand the power of a sum of more than two terms. It is a generalization of the binomial theorem to polynomials with any number of terms. It expresses a power
(x_1 + x_2 + \cdots + x_k)^n
as a weighted sum of monomials of the form
x_1^{b_1} x_2^{b_2} \cdots x_k^{b_k},
where the weights are given by generalizations of binomial coefficients called multinomial coefficients.
Multinomial Theorem Statement
Proof of Multinomial Theorem
Multinomial Theorem Examples - Number of Terms
Multinomial Theorem Examples - Specific Terms
The first important definition is the multinomial coefficient:
b_1, b_2, \ldots, b_k
\displaystyle \sum_{i=1}^{k} b_i = n,
the multinomial coefficient is
\binom{n}{b_1, b_2,\ldots , \ b_k} = \frac{n!}{b_1!b_2!\cdots b_k!}.
k = 2,
this is the binomial coefficient:
\binom{n}{b_1,b_2} = \frac{n!}{b_1!b_2!} = \frac{n!}{b_1!(n-b_1)!} = \binom{n}{b_1}.
Now the multinomial theorem can be stated as follows:
k
and a non-negative integer
n
\left(x_1 + x_2 + \cdots + x_k\right)^{n} = \sum_{b_1 + b_2 + \cdots +b_k = n} \binom{n}{b_1, b_2, b_3, \ldots, b_k} \prod_{j=1}^{k} x_j^{b_j}.
The number of terms of this sum are given by a stars and bars argument: it is
\binom{n+k-1}{k}
There are two proofs of the multinomial theorem, an algebraic proof by induction and a combinatorial proof by counting. The algebraic proof is presented first.
Proceed by induction on
m.
k = 1
the result is true, and when
k = 2
the result is the binomial theorem. Assume that
k \geq 3
and that the result is true for
k = p.
k = p+1,
\left(x_1 + x_2 + \cdots + x_p+x_{p+1}\right)^{n} = \left(x_1 + x_2 + \cdots +x_{p-1} + (x_p + x_{p+1})\right)^{n}.
x_p + x_{p+1}
as a single term and using the induction hypothesis,
\sum_{b_1 + b_2 + \cdots +b_{p-1} + B = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \prod_{j=1}^{p-1} x_j^{b_j} \times (x_p + x_{p+1})^B.
By the binomial theorem, this becomes
\sum_{b_1 + b_2 + \cdots +b_{p-1} + B = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \prod_{j=1}^{p-1} x_j^{b_j} \times \sum_{b_p + b_{p+1} = B}\binom{B}{b_p} x_p^{b_p} x_{p+1}^{b_{p+1}}.
\binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \binom{B}{b_p} = \binom{n}{b_1, b_2, b_3, \ldots, b_{p+1}},
\sum_{b_1 + b_2 + \cdots +b_{p+1} = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p+1}} \prod_{j=1}^{k} x_j^{b_j}.\ _\square
Here is the combinatorial proof, which relies on a fact from the Multinomial Coefficients wiki.
Consider a term in the expansion of
\left(x_1 + x_2 + \cdots + x_k\right)^{n}
. It must be of the form
\displaystyle \alpha\prod_{i=1}^{k} x_i^{\beta_i}
\alpha
\beta_i.
Since each term must come from choosing one summand from
x_1 + x_2 + \cdots +x_k,
\beta_1 + \beta_2 + \cdots + \beta_k = n.
The number of different ways that we can get this term will be the number of ways to choose
\beta_1
x_1
\beta_2
x_2
, and so on. By a result from the Multinomial Coefficients wiki, this is exactly
\alpha = \binom{n}{\beta_1, \beta_2, \ldots, \beta_k}.\ _\square
How many terms are in the expansion of
(x_1+x_2+x_3)^4?
There is one term for each ordered triple
(b_1,b_2,b_3)
b_1+b_2+b_3= 4
. One way to count these triples is to represent them as collections of
2
bars and
4
stars; for instance,
*|***|
represents the triple
(1,3,0)
. The number of such collections is
\binom{4+2}{4} = 15
15
terms in the expansion.
_\square
n
n+1
2n
2n+1
\big(1+2x+x^2\big)^n
How many terms are there in the expansion of the above trinomial, when expanded in descending powers of
x?
How many total distinct terms are there in the expansion of
(x+y+z+t)^{10}?
Determine the coefficient of
a^2b^4d
in the expansion of the polynomial
(3a + 5b -2c +d)^7.
A general term in the expansion of
(3a + 5b -2c +d)^7
will be of the form
\binom{7}{b_1,b_2,b_3,b_4}(3a)^{b_1}(5b)^{b_2}(-2c)^{b_3}(d)^{b_4}.
To have the term with
a^2b^4d,
b_1 = 2, b_2 = 4, b_3 = 0, b_4 = 1.
\begin{aligned}\binom{7}{2,4,1}(3a)^{2}(5b)^{4}(d)^{1} &= \frac{7!}{4!2!1!}\big(9a^2\big)\big(625b^4\big)(d) \\ &= 105(5625)a^2b^4d \\ &= 590625a^2b^4d,\end{aligned}
implying the answer is 590625.
_\square
t^{20}
\left(t^3 - 3t^2 + 7t +1\right)^{11}.
A general term of the expansion has the form
\binom{11}{b_1, b_2, b_3, b_4}\left(t^3\right)^{b_1}\left(-3t^2\right)^{b_2}(7t)^{b_3}(1)^{b_4}.
In order to have a coefficient of
t^{20},
b_1 + b_2 + b_3 +b_4 = 11
3b_1 + 2b_2 + b_3 = 20.
b_3 = 20 - 2b_2 - 3b_1
b_4 = 11 - b_1 - b_2 - b_3 = 2b_1 + b_2 - 9.
Thus, the coefficient will be the sum
\displaystyle\sum_{b_1,b_2 \geq 0} \binom{11}{b_1,b_2,20-2b_2-3b_1,2b_1+b_2-9}
2b_2+3b_1 \leq 20
2b_1+b_2 \geq 9.
We can evaluate this as
-7643472342.
_\square
If the coefficient of
a^8b^4c^9d^9
(abc+abd+acd+bcd)^{10}
N
, then what is the sum of the digits of
N
x^7
\big(1+x+2x^3\big)^{10}?
Find the term that does not contain the variable
x
in the complete expansion of
\left(x^2+x+\frac{1}{x}\right)^{10}.
To be clear, if an expression is completely expanded, then all like terms have been combined together, leaving unlike terms in the final answer. For example,
(a+1)^4=a^4+4a^3+6a^2+4a+1.
Find coefficient of
x^6
\big(1-x+2x^2\big)^{10}.
t^8
\big(1+2t^2-t^3\big)^9.
Cite as: Multinomial Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/multinomial-theorem/
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Table shuffleboard is a game (usually found in higher quality bars) in which a heavy puck is glided down a long table toward an end zone. Points are scored by getting the puck to come to rest as close to the edge as possible, the closer the puck, the higher the reward.
If the puck slides off the end of the table, however, the player gets zero points. In the table shown below, three points are awarded for the far zone, one for the near zone, and zero if the puck doesn't cross the first line.
Crudely, the difficulty
D
of making it into each zone is proportional to
1/\Delta s
, the range of allowable speeds between which the puck still stop inside the given zone. For example, if the puck is released with speed
s_3^\textrm{max}
, it will stop just before falling off the end of the table, and if released with speed
s_3^\textrm{min}
, it will stop just inside the three point zone, and
\Delta s_3 = s_3^\textrm{max} - s_3^\textrm{min}
Suppose the puck and the table have a coefficient of kinetic friction given by
\mu_k
, and that the puck is pulled down by the acceleration due to gravity,
g
. How much more difficult is it to get the puck to stop in the three point zone than in the one point zone? In other words, find
\frac{D_3}{D_1} = \frac{\Delta s_1}{\Delta s_3}
d_0=15
l=5
Four identical metal balls are hanging from a light string of length
5l
at equally placed points as shown in figure. The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. Calculate
\theta_2
in degrees if
\theta_1=\tan^{-1}2
l
\lambda
lies in a heap on a floor. You grab one end of the rope and pull upward with a non-impulsive constant force
{ F }_{ o }
. Find the time period require to completely lift off the rope.
Assume that the rope is greased.
The answer is in the form
\displaystyle{{ T=\cfrac { a }{ g } (\sqrt { \cfrac { { F }_{ o } }{ \lambda } } -\sqrt { \cfrac { { F }_{ o } }{ \lambda } -\cfrac { bgl }{ c } } })}
a+b+c
I got this question from Deepanshu.
A right prism that has an equilateral triangular base with length
a=50cm
is placed in a horizontal slit between two tables, so that one of the side faces is vertical. How small can the width
d
of the slit be made in
meters
before the prism falls out of the slit.
There is no friction between the prism and the tables and the prism is made of a homogeneous material.
The edges of the slit are parallel
This was a past IPHO problem.
When tearing a sheet of paper most people exert a force perpendicular to the sheet, rather than grabbing the sheet by the edges and trying to pull it apart. The physics of this can be understood with a simple model.
Consider a row of three springs along the x-axis with the same spring constant and length, connected end to end. Initially, they are all at their natural length. If any of them stretch by more than one percent they will break. This is our paper. Tearing apart is equivalent to putting forces N1 and N2 (each with magnitude N) at the two points where two springs join, in opposite directions and perpendicular to the springs. Pulling apart is equivalent to putting equal and opposite forces F1 and F2 (magnitude F) at the same points but now parallel to the springs. Find the ratio
F/N
when the middle spring breaks.
Hint 1: Do NOT assume that the ends of the springs move only vertically.
Hint 2: tear the paper slowly and think about force balances.
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Robust Stability and Stabilization of Interval Uncertain Descriptor Fractional-Order Systems with the Fractional-Order : The Case
Yuanhua Li, Heng Liu, Hongxing Wang, "Robust Stability and Stabilization of Interval Uncertain Descriptor Fractional-Order Systems with the Fractional-Order : The Case", Mathematical Problems in Engineering, vol. 2015, Article ID 606048, 8 pages, 2015. https://doi.org/10.1155/2015/606048
Yuanhua Li,1 Heng Liu ,1 and Hongxing Wang1
1School of Finance, Huainan Normal University, Huainan 232038, China
Stability and stabilization of fractional-order interval system is investigated. By adding parameters to linear matrix inequalities, necessary and sufficient conditions for stability and stabilization of the system are obtained. The results on stability check for uncertain FO-LTI systems with interval coefficients of dimension n only need to solve one 4n-by-4n LMI. Numerical examples are presented to shown the effectiveness of our results.
During the last two decades, the study of fractional-order control systems has received more and more attention. As a generalization of the traditional calculus, the fractional calculus have found many applications in viscoelastic systems, robotics, finance, and so on ([1–4], etc.). Studying on fractional-order calculus has become an active research field.
Stability analysis is a basic problem in control theory. For Caputo fractional derivative-based linear systems, the stability results were formulated with the fractional-order belonging to and , in [5–7] and so on. In [8], the stability issue of interval fractional-order linear time-invariant (FO-LTI) systems was first presented and discussed. The stability of single-input single-output FO-LTI systems was further discussed based on an experimentally verified Kharitonov-like procedure [9].
Robust stability analysis was carried out for FO-LTI interval system with fractional commensurate order based on the maximum eigenvalue of a Hermitian matrix by applying Lyapunov inequality in [10] and then further discussed in [7].
The uncertain FO-LTI systems have been wildly studied. In [11], the robust stability problem was discussed based on the ranges of the corresponding interval eigenvalues by applying the matrix perturbation theory. In [12], a new and effectively robust stability checking method was first proposed for FO-LTI interval uncertain systems in terms of LMIs, and an analytical design of the stabilizing controllers for fractional-order dynamic interval systems was given. Note that the above-mentioned results on stability check for uncertain FO-LTI systems with interval coefficients of dimension need to solve one LMI. Therefore, it is valuable to seek some simple necessary and sufficient conditions for checking robust stability of uncertain FO-LTI systems with interval coefficients. With the above motivation, based on the results of [12], the robust stability and stabilization problems of uncertain FO-LTI interval systems with the fractional-order belonging to are further investigated in this paper.
This paper is organized as follows: in Section 2, we present some preliminaries results on the fractional derivative, the linear algebra and the matrix theory. In Section 3, we study the problems of the stability and stabilization of uncertain FO-LTI systems with interval coefficients in terms of LMIs. In Section 4, numerical examples are presented to illustrate our proposed results. Finally, Section 5 concludes this work.
Notations. Throughout this paper, stands for the set of by matrices with real entries. The symbols , , and stand for the transpose of , the expression , and the identity matrix of order , respectively. The symbol is used to denote the row vector with the th element being , ; that is, The symbol is the Kronecker product of two matrices and The symbol will be used in some matrix expressions to indicate a symmetric structure; that is, if matrices and were given, thenLet ; consider the symbol in which , .
Throughout the paper, only the Caputo definition is used. The following Caputo definition is adopted for fractional derivatives of order of function [13]: with , , where is the Gamma function:
Consider the following FO-LTI interval system: where is the fractional commensurate order, and stand for the state vector and control input, respectively, and the system matrices and are interval uncertain in the sense that where , , , and are given matrices.
To take into account the stability [14], we introduce the following definition.
Definition 1. The fractional-order interval system (7) is said to be asymptotically stabilizable via linear state-feedback control if there exists a state-feedback controller such that the closed-loop system is asymptotically stable.
Denote To handle the interval uncertain, the following notations are introduced:
Lemma 2. Let ,: Then , .
Proof. Since , , , and are all diagonal, it is easy to check that It follows that Thus, .
Noting that the above proof is reversible, it is easy to know that . Therefore, .
In the same way, we have .
Lemma 3 (see [7, 15]). Let be a deterministic real matrix without uncertainty. Then, a necessary and sufficient condition for the asymptotical stability of is where is the spectrum of all eigenvalues of .
Lemma 4 (see [16]). Let be a real matrix. Then where , if and only if there exists such that where .
Lemma 5 (see [17]). For any matrices and with appropriate dimensions, we have
Lemma 6 (see [18]). Let , , and be real matrices of suitable dimensions. Then, for any ,
Lemma 7 (see [18]). Let , , and be symmetric matrices such that, , and . Furthermore, assume that for all nonzero . Then, there exists a constant such that
In this section, by adding parameters into linear matrix inequalities, necessary and sufficient conditions for stability and stabilization of the system are obtained. Those results are generalization of the main theorems in [12].
Theorem 8. Let . The uncertain FO-LTI interval system (7) with controller is asymptotically stable if and only if there exist some symmetric positive definite matrix and a real scalar constant such that where
Proof. It is easy to check that Denote ; then we see that where is an arbitrary positive define matrix.
By applying (13), we havein which Sufficiency. Suppose that there exists a symmetric positive definite matrix such that (22) holds. By applying (26), (27), and Lemma 5, we have By using the Schur complement of (22), one obtains It follows from Lemma 4 that . Therefore, by Lemma 3, the uncertain FO-LTI interval system (7) is asymptotically stable.
Necessity. Suppose that the uncertain FO-LTI interval system (7) is asymptotically stable. Then, . It follows from Lemma 4 that there exists a symmetric positive definite matrix such that By using Lemma 2 and after some calculations, one can obtain from (27) that Therefore, for all , , that is, Consequently, given any and , we have Applying Lemma 6, we obtain It follows by Lemma 7 that there exists a constant such that So we derive that Applying the well-known Schur complement yields (22).
Next, let us establish a stabilization result.
Theorem 9. Let . The uncertain FO-LTI interval system (7) is asymptotically stable if and only if there are a matrix , a symmetric positive definite matrix , and two real scalars , such that where Moreover, the robustly asymptotically stabilizing state-feedback gain matrix is given by
Remark 10. When one takes and , it is easy to obtain equivalence of Theorem 8 and [12, Theorem 1] and Theorem 9 and [12, Theorem 2], respectively.
Example 1 (see [12]). Consider the robust stability of the following uncertain FO-LTI interval system: where and : Taking , a feasible solution of(22) is as follows:
Letand let the initial conditions be , , and . Time response of the state variables is depicted in Figure 1.
Time response of the state variables.
In the following example, we have shown the effectiveness of our results by choosing different parameters.
Example 2 (see [12]). Consider the robust stability of the following uncertain FO-LTI interval system: where and , with (I)Taking and , a feasible solution of (39) is as follows: Finally, the asymptotically stabilizing state-feedback gain matrix is obtained as (II)Taking and , we have (III)Taking and , we have
Let and let the initial conditions be , , and , and let be as in (49), (51), and (53), respectively. Time response of the state variables is depicted in Figures 2–4, respectively.
Time response of the state variables when is as in (49).
Remark 11. Applying [12, Theorem 2 (28)] to Example 2 gives the same results as in Example 2 (II).
In this paper, the robust asymptotical stability of fractional-order interval systems with the fractional-order belonging to has been studied. The results on stability check for uncertain FO-LTI systems with interval coefficients of dimension only need to solve one -by- LMI. LMI stability conditions for fractional systems are proposed. Numerical examples have shown the effectiveness of our results. To the best of our knowledge, the idea of introducing free parameters is used for the first time to derive an analytical design of the stabilizing controllers for fractional-order dynamic interval systems. Working towards relaxing the requirements for the knowledge of system uncertainties and applying the proposed control methods to fractional-order nonlinear systems while maintaining the simplicity of the controller design are our further investigation directions.
The authors would like to thank the referees for their helpful comments and suggestions. The work of authors was supported in part by the Natural Science Foundation of China (Grant nos. 11401243 and 61403157) and the Science Foundation of Huainan Normal University (no. 2014xj45).
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Copyright © 2015 Yuanhua Li et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Multinomial Distribution - MATLAB & Simulink - MathWorks í•œêµ
Multinomial distribution models the probability of each combination of successes in a series of independent trials. Use this distribution when there are more than two possible mutually exclusive outcomes for each trial, and each outcome has a fixed probability of success.
Multinomial distribution uses the following parameter.
0â¤\text{Probabilities}\left(i\right)â¤1\text{â};\text{â}\underset{\text{all}\left(i\right)}{â}\text{Probabilities}\left(i\right)=1
The multinomial pdf is
f\left(x|n,p\right)=\frac{n!}{{x}_{1}!\cdots {x}_{k}!}{p}_{1}^{{x}_{1}}\cdots {p}_{k}^{{x}_{k}},
where k is the number of possible mutually exclusive outcomes for each trial, and n is the total number of trials. The vector x = (x1...xk) is the number of observations of each k outcome, and contains nonnegative integer components that sum to n. The vector p = (p1...pk) is the fixed probability of each k outcome, and contains nonnegative scalar components that sum to 1.
The expected number of observations of outcome i in n trials is
\text{E}\left\{{x}_{i}\right\}=n{p}_{i}\text{â},
where pi is the fixed probability of outcome i.
The variance is of outcome i is
\text{var}\left({x}_{i}\right)=n{p}_{i}\left(1â{p}_{i}\right)\text{â}.
The covariance of outcomes i and j is
\mathrm{cov}\left({x}_{i},{x}_{j}\right)=ân{p}_{i}{p}_{j}\text{â},\text{â}iâ j.
The multinomial distribution is a generalization of the binomial distribution. While the binomial distribution gives the probability of the number of “successes†in n independent trials of a two-outcome process, the multinomial distribution gives the probability of each combination of outcomes in n independent trials of a k-outcome process. The probability of each outcome in any one trial is given by the fixed probabilities p1,..., pk.
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Two-way analysis of variance - MATLAB anova2 - MathWorks Nordic
anova2 performs two-way analysis of variance (ANOVA) with balanced designs. To perform two-way ANOVA with unbalanced designs, see anovan.
p = anova2(y,reps) returns the p-values for a balanced two-way ANOVA for comparing the means of two or more columns and two or more rows of the observations in y.
reps is the number of replicates for each combination of factor groups, which must be constant, indicating a balanced design. For unbalanced designs, use anovan. The anova2 function tests the main effects for column and row factors and their interaction effect. To test the interaction effect, reps must be greater than 1.
anova2 also displays the standard ANOVA table.
p = anova2(y,reps,displayopt) enables the ANOVA table display when displayopt is 'on' (default) and suppresses the display when displayopt is 'off'.
[p,tbl] = anova2(___) returns the ANOVA table (including column and row labels) in cell array tbl. To copy a text version of the ANOVA table to the clipboard, select Edit > Copy Text menu.
[p,tbl,stats] = anova2(___) returns a stats structure, which you can use to perform a multiple comparison test. A multiple comparison test enables you to determine which pairs of group means are significantly different. To perform this test, use multcompare, providing the stats structure as input.
The data is from a study of popcorn brands and popper types (Hogg 1987). The columns of the matrix popcorn are brands, Gourmet, National, and Generic, respectively. The rows are popper types, oil and air. In the study, researchers popped a batch of each brand three times with each popper, that is, the number of replications is 3. The first three rows correspond to the oil popper, and the last three rows correspond to the air popper. The response values are the yield in cups of popped popcorn.
Perform a two-way ANOVA. Save the ANOVA table in the cell array tbl for easy access to results.
The column Prob>F shows the p-values for the three brands of popcorn (0.0000), the two popper types (0.0001), and the interaction between brand and popper type (0.7462). These values indicate that popcorn brand and popper type affect the yield of popcorn, but there is no evidence of an interaction effect of the two.
Display the cell array containing the ANOVA table.
Store the F-statistic for the factors and factor interaction in separate variables.
Sample data, specified as a matrix. The columns correspond to groups of one factor, and the rows correspond to the groups of the other factor and the replications. Replications are the measurements or observations for each combination of groups (levels) of the row and column factor. For example, in the following data the row factor A has three levels, column factor B has two levels, and there are two replications (reps = 2). The subscripts indicate row, column, and replication, respectively.
\begin{array}{c}\begin{array}{cc}B=1& B=2\end{array}\\ \left[\begin{array}{cc}{y}_{111}& {y}_{121}\\ {y}_{112}& {y}_{122}\\ {y}_{211}& {y}_{221}\\ {y}_{212}& {y}_{222}\\ {y}_{311}& {y}_{321}\\ {y}_{312}& {y}_{322}\end{array}\right]\end{array}\begin{array}{c}\\ \begin{array}{c}\begin{array}{c}\\ \end{array}\right\}A=1\\ \begin{array}{c}\\ \end{array}\right\}A=2\\ \begin{array}{c}\\ \end{array}\right\}A=3\end{array}\end{array}
reps — Number of replications
1 (default) | an integer number
Number of replications for each combination of groups, specified as an integer number. For example, the following data has two replications (reps = 2) for each group combination of row factor A and column factor B.
\begin{array}{c}\begin{array}{cc}B=1& B=2\end{array}\\ \left[\begin{array}{cc}{y}_{111}& {y}_{121}\\ {y}_{112}& {y}_{122}\\ {y}_{211}& {y}_{221}\\ {y}_{212}& {y}_{222}\\ {y}_{311}& {y}_{321}\\ {y}_{312}& {y}_{322}\end{array}\right]\end{array}\begin{array}{c}\\ \begin{array}{c}\begin{array}{c}\\ \end{array}\right\}A=1\\ \begin{array}{c}\\ \end{array}\right\}A=2\\ \begin{array}{c}\\ \end{array}\right\}A=3\end{array}\end{array}
When reps is 1 (default), anova2 returns two p-values in vector p:
The p-value for the null hypothesis that all samples from factor B (i.e., all column samples in y) are drawn from the same population.
The p-value for the null hypothesis, that all samples from factor A (i.e., all row samples in y) are drawn from the same population.
When reps is greater than 1, anova2 also returns the p-value for the null hypothesis that factors A and B have no interaction (i.e., the effects due to factors A and B are additive).
Example: p = anova(y,3) specifies that each combination of groups (levels) has three replications.
displayopt — Indicator to display the ANOVA table
Indicator to display the ANOVA table as a figure, specified as 'on' or 'off'.
p-value for the F-test, returned as a scalar value. A small p-value indicates that the results are statistically significant. Common significance levels are 0.05 or 0.01. For example:
A sufficiently small p-value for the null hypothesis for group means of row factor A suggests that at least one row-sample mean is significantly different from the other row-sample means; i.e., there is a main effect due to factor A
A sufficiently small p-value for the null hypothesis for group (level) means of column factor B suggests that at least one column-sample mean is significantly different from the other column-sample means; i.e., there is a main effect due to factor B.
A sufficiently small p-value for combinations of groups (levels) of factors A and B suggests that there is an interaction between factors A and B.
The rows of the ANOVA table show the variability in the data, divided by the source into three or four parts, depending on the value of reps.
Columns Variability due to the differences among the column means
Rows Variability due to the differences among the row means
Variability due to the interaction between rows and columns (if reps is greater than its default value of 1)
Error Remaining variability not explained by any systematic source
stats — Statistics for multiple comparison test
Statistics for multiple comparisons tests, returned as a structure. Use multcompare to perform multiple comparison tests, supplying stats as an input argument. stats has nine fields.
sigmasq Mean squared error
colmeans Estimated values of the column means
coln Number of observations for each group in columns
rowmeans Estimated values of the row means
rown Number of observations for each group in rows
inter Number of interactions
pval p-value for the interaction term
df Error degrees of freedom (reps — 1)*r*c where reps is the number of replications and c and r are the number of groups in factors, respectively.
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Low-Velocity Impact Response Characterization of a Hybrid Titanium Composite Laminate | J. Eng. Mater. Technol. | ASME Digital Collection
S. Bernhardt,
A. S. Kobayashi
Bernhardt, S., Ramulu, M., and Kobayashi, A. S. (July 13, 2006). "Low-Velocity Impact Response Characterization of a Hybrid Titanium Composite Laminate." ASME. J. Eng. Mater. Technol. April 2007; 129(2): 220–226. https://doi.org/10.1115/1.2400272
The low-velocity impact response of a hybrid titanium composite laminate, known as
TiGr
, was compared to that of graphite/epoxy composite. The
TiGr
material comprised of two outer plies of titanium foil surrounding a composite core. The composite core was PIXA-M (a high temperature thermoplastic) reinforced by IM-6 graphite fibers and consolidated by an induction heating process. The impact response of
TiGr
was characterized by two modes of failure which differed by failure or nonfailure in tension of the bottom titanium ply. The ductility of titanium caused buckling by yielding whereas the brittle adjacent composite ply lead to fracture. The maximum failure force of the material correlated well with the previously reported static flexural data, and the material outperformed the commonly used graphite/epoxy.
titanium, carbon fibre reinforced composites, laminates, foils, induction heating, ductility, buckling, yield stress, brittleness, fracture
Buckling, Composite materials, Failure, Fracture (Materials), Fracture (Process), Laminates, Titanium, Failure mechanisms, Displacement, Brittleness, Electromagnetic induction, Heating, Damage, Delamination, Ductility, Fibers, Tension
M. O. W.
Low-Velocity Impact Loading on Fibre Reinforced Aluminium Laminates (ARALL) and Other Aircraft Sheet Materials
The Impact Properties of Metal-Composite Laminates
Proceedings of the ICCM & ECCM
Impact Loading on Fibre Metal Laminates
Del Luongo
Low-Velocity Impact Behaviour of Fiberglass-Aluminium Laminates
Modeling of Facesheet Crack Growth in Titanium-Graphite Hybrid Laminates. Part II: Experimental results
Delamination Growth From Face Sheet Seams in Cross-Ply Titanium/Graphite Hybrid Laminates
Manufacturing of GLARE Parts and Structures
Net Shape Manufacturing and the Performance of Polymer Composites under Dynamic Loads
Damage Tolerance of Graphite/Epoxy Composites
Model 208C04 Spec Sheet
,” http://www.pcb.com/spec-sheet.asp?model=208C04&item-id=3715http://www.pcb.com/spec-sheet.asp?model=208C04&item-id=3715
KD-2300 Specifications
,” and Kaman Measuring Systems, “KD-2300 Sensors,” http://www.kamaninstrumentation.com/html/products/datasheets/kd2300-specs.htmhttp://www.kamaninstrumentation.com/html/products/datasheets/kd2300-specs.htm
Impact Response Characterization of a Hybrid Titanium Composite Laminate
, MS thesis, University of Washington, Seattle, Washington.
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Why does \cot(2 \arctan (Ax))=\frac{1-(Ax)^2}{2Ax}
\mathrm{cot}\left(2\mathrm{arctan}\left(Ax\right)\right)=\frac{1-{\left(Ax\right)}^{2}}{2Ax}
We have the following fundamental trigonometric identity:
\mathrm{tan}2t=\frac{2\mathrm{tan}t}{1-{\mathrm{tan}}^{2}t}
Take reciprocal on both sides:
\frac{1}{\mathrm{tan}2t}=\frac{1-{\mathrm{tan}}^{2}t}{2\mathrm{tan}t}
t=\mathrm{arctan}Ax
to get the identity in the question.
\left(\mathrm{cot}x\equiv \frac{1}{\mathrm{tan}x}\right)
\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }
And the double angle formula for tangent:
\mathrm{tan}2\theta =\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }
Lastly, before I confuse myself lets
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
Solve the following equation for all radian solutions and if
0\le x\le 2\pi
. Give all answers as exact values in radians.
0\le x\le 2\pi
x=\frac{7\pi }{6},\frac{\pi }{2},\frac{11\pi }{6}\text{ }rad
I calculated
{\mathrm{sin}75}^{\circ }
\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}
, but the answer is
\frac{\sqrt{2}+\sqrt{6}}{4}
. What went wrong?
I calculated the exact value of
{\mathrm{sin}75}^{\circ }
{\mathrm{sin}75}^{\circ }=\mathrm{sin}\left({30}^{\circ }+{45}^{\circ }\right)
=\mathrm{sin}30°\mathrm{cos}45°+\mathrm{cos}30°\mathrm{sin}45°
=\frac{1}{2}·\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}
=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}
\frac{\sqrt{2}+\sqrt{6}}{4}
{\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)
{\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{12}}{4}\right)+{\mathrm{csc}}^{-1}\left(\sqrt{2}\right)\right)
T={\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{12}}{4}\right)+{\mathrm{csc}}^{-1}\left(\sqrt{2}\right)\right)
{\mathrm{csc}}^{-1}\sqrt{2}={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4};\text{ }{\mathrm{cos}}^{-1}\left(\frac{\sqrt{12}}{4}\right)={\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}
T={\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+\frac{\pi }{4}+\frac{\pi }{6}\right)
Proving that an
{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{k}^{k+1}\mathrm{sin}\left(\mathrm{exp}\left(x\right)\right)dxdk
integral converges
\mathrm{cot}\left(\theta \right)=7
\mathrm{sin}\left(\theta \right),\mathrm{cos}\left(\theta \right),\mathrm{sec}\left(\theta \right)
2\pi
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Fractions - Infinity Learn
MathsFractions
A fraction shows part of a whole. This whole can be a region or a collection. The word fraction is derived from the Latin word “fractio” which means ‘to break’. The Egyptians, being the earliest civilization to study fractions, used fractions to resolve their mathematical problems, which included the division of food, supplies, and the absence of a bullion currency.
A mixed fraction is a mixture of a whole number and a proper fraction. For example,
5\frac{1}{3, where 5 is the whole number and 1/3 is the proper fraction, or,}
2\frac{2}{5,}
7\frac{9}{11}
equi. fraction are the fractions that represent the same value after they are simplified. To get equivalent fractions of any given fraction:
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The probability it will drill a well in field A
The probability it will drill a well in field A is 40%. If it does, the probability the well will be successful is 45% Calculate
An oil company is bidding for the rights to drill a well in field A and awell in field B. The probability it will drill a well in field A is 40%. Ifit does, the probability the well will be successful is 45%, The probability it will drill a well in field B is 30% Ifit
Correct answer is on photo
Flaws occur in telephone cable at the average rate of 4.4 flaws per km of cable.
what is the probability of 1 flaw in 100m of cable?
3,645 rolls of landscape fabric are manufactured during one day. After being inspected, 121 of these rolls are rejected as imperfect. What percent of the rolls is rejected? (Round to the nearest whole percent.)
Suppose that the random variable x, shown below, represents the number times . P(x) represents the probability of a randomly selected person having received that number of speeding tickets during that period. Use the probability distribution table shown below to answer the following questions.
\begin{array}{|cc|}\hline x& P\left(x\right)\\ 0& 0.3176\\ 1& 0.2705\\ 2& 0.2469\\ 3& 0.0832\\ 4& 0.0485\\ 5& 0.0333\\ 6+& 0.0000\\ \hline\end{array}
a) What is the probability that a randomly selected person has received four tickets in a three-year period?
P\left(x=4\right)=P\left(x=4\right)=
b) What is the probability that a randomly selected person has received three tickets in a three-year period?
P\left(x=3\right)=P\left(x=3\right)=
c) What is the probability that that a randomly selected person has received more than one tickets in a three-year period?
P\left(x>1\right)=P\left(x>1\right)=
d) What is the probability that that a randomly selected person has received two or less tickets in a three-year period?
P\left(x\le 2\right)=P\left(x\le 2\right)=
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The Following are the scores of 20 students who took
The Following are the scores of 20 students who took a calculus exam. Find the mean and standard deviation of the given data. Round to the nearest tenth, if necessary.
\begin{array}{|ccccccc|}\hline Score& 76& 82& 85& 90& 92& 95\\ Frequency& 5& 2& 3& 4& 3& 3\\ \hline\end{array}
=\frac{\sum {x}_{i}{f}_{i}}{\sum {f}_{i}}
=\frac{76\left(5\right)+82\left(2\right)+85\left(3\right)+90\left(4\right)+92\left(3\right)+95\left(3\right)}{20}
=86
For standard deviation, we make a table
\begin{array}{|ccccc|}\hline data& data-mean& \left(data-mean{\right)}^{2}& frequency& \left(data-mean{\right)}^{2}frequency\\ 76& -10& 100& 5& 500\\ 82& -4& 16& 2& 32\\ 85& -1& 1& 3& 3\\ 90& 4& 16& 4& 64\\ 92& 6& 36& 3& 108\\ 95& 9& 81& 3& 243\\ \hline\end{array}
Then we add the values form the last column
\sum {\left({x}_{i}-\stackrel{―}{x}\right)}^{2}{f}_{i}=950
Then we use it to find the standard deviation
\sqrt{\frac{\sum {\left({x}_{i}-\stackrel{―}{X}\right)}^{2}{f}_{i}}{n}}=\sqrt{\frac{950}{20}}=6.9
Mean=86
Standard deviatiion
=6.9
If an experiment involves: a) a random sample of size n is selected without replacement from N items, and b) of the N items, k may be classified as successes and
N-k
are classified as failures, this falls under:
a. negative binomial
d. hypergeometric
(True/False) The central limit theorem implies that:
1.a All variables have bell-shaped sample data distributions if a random sample contains at least 30 observations.
2.b. Population distributions are normal whenever the population size is large.
3.c. For large random samples, the sampling distribution of
\stackrel{―}{y}
is approximately normal, regardless of the shape of the population distribution.
3.d. The sampling distribution looks more like the population distribution as the sample size increases.
Find the standard deviation for the set of grouped sample data.
\begin{array}{|cc|}\hline Interval& Frequency\\ 0.5-3.5& 3\\ 3.5-6.5& 5\\ 6.5-9.5& 4\\ 9.5-12.5& 5\\ \hline\end{array}
s=?
The honey farm has installed a filling machine for honey jars that hold 500 g of honey. Doris is calibrating the new machine. She sets the machine to a mean of 500 g and performs a test run of 48 jars. The table attached to the quiz post in the Google Classroom shows the results. Assume it is normally distributed.
Determine the mean of the data
Determine the standard deviation of the data.
What is the probability that a jar contains at least 504 g of honey?
In an orchard, harvested apples are randomly distributed in packets, each containing 7 fruits.
The orchard had a total of 25 apples collected, and 10 fruits were considered of excellent quality. Let X be the number of optimal apples allocated to first package, which of the probability distributions would be adequate to describe the random variable X?
(a) Geometric
\left(\frac{7}{25}\right)
(b) Binomial
\left(7,\frac{3}{25}\right)
(c) Poisson (3)
(d) Geometric
\left(\frac{3}{25}\right)
(e) Hypergeometric (25,10,7)
What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample?
Construct all random samples consisting three observations from the given data. Arrange the observations in ascending order without replacement and repetition.
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Derivatives of constant multiples of functions Evaluate the following derivatives. \frac{d}{dt}(\frac{3}{8}\sqrt{t})
Miguel Reynolds 2022-01-15 Answered
Derivatives of constant multiples of functions Evaluate the following derivatives.
\frac{d}{dt}\left(\frac{3}{8}\sqrt{t}\right)
\frac{d}{dt}\left(\frac{3}{8}\sqrt{t}\right)
\frac{d}{dt}\left(\frac{3}{8}\sqrt{t}\right)
Take constant out:
=\frac{3}{8}\frac{d}{dt}\left({t}^{\frac{1}{2}}\right)\text{ }\text{ }\text{ }\left(\because \sqrt[n]{a}={a}^{\frac{1}{n}}\right)
Apply power rule:
\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}
=\frac{3}{8}\left(\frac{1}{2}{t}^{\frac{1}{2}-1}\right)
=\frac{3}{16}\left({t}^{-\frac{1}{2}}\right)
=\frac{3}{16{t}^{\frac{1}{2}}}\text{ }\text{ }\text{ }\left(\because {a}^{-n}=\frac{1}{{a}^{n}}\right)
=\frac{3}{16\sqrt{t}}\text{ }\text{ }\text{ }\left(\because \sqrt[n]{a}={a}^{\frac{1}{n}}\right)
What is the instantaneous rate of change of f(x)=3x+5 at x=1?
How do you find the slope of a tangent line to the graph of the function
\sqrt{x}
at (4,2)?
Find the derivative of this problem.
h\left(x\right)=\mathrm{sin}2x\mathrm{cos}2x
f\left(x\right)={\left(5+4x\right)}^{2}
y={x}^{2}-4x
How do you find f'(x) using the definition of a derivative for
f\left(x\right)={e}^{x}
Find the equation of the tangent plane to the graph of
f\left(x,y\right)=8{x}^{2}-2x{y}^{2}
A)z=23x-15y+42
B)z=48x-80y+120
C)0=48x-80y+120
D)0=23x-15y+42
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Overview - Idle Finance
Products > Best Yield > Overview
Best Yield strategy constantly monitors interest rates on various DeFi yield sources to ensure the current allocation is yielding the best aggregate interest rate available on the market.
Users' funds are pooled together and programmatically deposited into one or more of the available lending protocols.
By analyzing supply rate functions across integrated platforms and total funds in the pool, the strategy is able to constantly rebalance capital across any number of protocols to earn the highest interest rate possible with very high precision.
When a user deposits funds, an equivalent amount (value-wise) of strategy tokens (idleTokens) is minted to his or her account. When a user withdraws funds from the strategy, the equivalent amount of value in the strategy token is burned from his or her account.
As soon as the user deposits, he or she starts earning yield. Essentially, idleTokens holdings and yield are split up across the pool token holders proportionally to their balances. For example, users will get IdleDAI if they deposit into the DAI BY pool.
Best Yield strategy maximises the current aggregated interest rate, this can be modelled as follows:
max\ q(x)= \sum_{i=0}^{n} \frac{x_i}{tot} * nextRate_i(x_i)
where n is the number of lending protocols used, x_i is the amount (in underlying) allocated in protocol i , nextRate(x_i) is a function that returns the new APR for protocol i after supplying x_i amount of underlying and is tot the total
tot=\sum_{i=0}^{n} x_i
Protocols and assets
Currently, the Best Yield strategy is available on Ethereum and Polygon blockchains. For each network, there is a different basket of assets available in the pools.
Currently, there are three protocols integrated into Idle:
Idle DAO has established a series of Integration Standard Requirements required to implement a new yield source or an asset in the Best Yield strategy.
SUSD
TUSD
Benefits of using Best Yield
A superior optimisation algorithm for automatic management of users funds;
Gas fees savings for funds rebalance (which the user would have to pay to deposit funds/interact from one platform to another);
Participating in $IDLE liquidity mining program and leveraging all the advantages linked to its multiple use-cases;
By depositing into BY pools users can get several other underlying governance tokens (e.g. COMP or AAVE);
For integrators, no need to stitch together disparate protocols or spend months integrating and updating yield functionality.
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Berkson's paradox - Wikipedia
Tendency to misinterpret statistical experiments involving conditional probabilities
An example of Berkson's paradox:
In figure 1, assume that talent and attractiveness are uncorrelated in the population.
In figure 2, someone sampling the population using celebrities may wrongly infer that talent is negatively correlated with attractiveness, as people who are neither talented nor attractive do not typically become celebrities.
Berkson's paradox, also known as Berkson's bias, collider bias, or Berkson's fallacy, is a result in conditional probability and statistics which is often found to be counterintuitive, and hence a veridical paradox. It is a complicating factor arising in statistical tests of proportions. Specifically, it arises when there is an ascertainment bias inherent in a study design. The effect is related to the explaining away phenomenon in Bayesian networks, and conditioning on a collider in graphical models.
1.2 Original illustration
1.3 Ellenberg example
1.4 Quantitative example
An illustration of Berkson's Paradox. The top graph represents the actual distribution, in which a positive correlation between quality of burgers and fries is observed. However, an individual who does not eat at any location where both are bad observes only the distribution on the bottom graph, which appears to show a negative correlation.
The most common example of Berkson's paradox is a false observation of a negative correlation between two desirable traits, i.e., that members of a population which have some desirable trait tend to lack a second. Berkson's paradox occurs when this observation appears true when in reality the two properties are unrelated—or even positively correlated—because members of the population where both are absent are not equally observed. For example, a person may observe from their experience that fast food restaurants in their area which serve good hamburgers tend to serve bad fries and vice versa; but because they would likely not eat anywhere where both were bad, they fail to allow for the large number of restaurants in this category which would weaken or even flip the correlation.
Original illustration[edit]
Berkson's original illustration involves a retrospective study examining a risk factor for a disease in a statistical sample from a hospital in-patient population. Because samples are taken from a hospital in-patient population, rather than from the general public, this can result in a spurious negative association between the disease and the risk factor. For example, if the risk factor is diabetes and the disease is cholecystitis, a hospital patient without diabetes is more likely to have cholecystitis than a member of the general population, since the patient must have had some non-diabetes (possibly cholecystitis-causing) reason to enter the hospital in the first place. That result will be obtained regardless of whether there is any association between diabetes and cholecystitis in the general population.
Ellenberg example[edit]
An example presented by Jordan Ellenberg: Suppose Alex will only date a man if his niceness plus his handsomeness exceeds some threshold. Then nicer men do not have to be as handsome to qualify for Alex's dating pool. So, among the men that Alex dates, Alex may observe that the nicer ones are less handsome on average (and vice versa), even if these traits are uncorrelated in the general population. Note that this does not mean that men in the dating pool compare unfavorably with men in the population. On the contrary, Alex's selection criterion means that Alex has high standards. The average nice man that Alex dates is actually more handsome than the average man in the population (since even among nice men, the ugliest portion of the population is skipped). Berkson's negative correlation is an effect that arises within the dating pool: the rude men that Alex dates must have been even more handsome to qualify.
Quantitative example[edit]
As a quantitative example, suppose a collector has 1000 postage stamps, of which 300 are pretty and 100 are rare, with 30 being both pretty and rare. 10% of all his stamps are rare and 10% of his pretty stamps are rare, so prettiness tells nothing about rarity. He puts the 370 stamps which are pretty or rare on display. Just over 27% of the stamps on display are rare (100/370), but still only 10% of the pretty stamps are rare (and 100% of the 70 not-pretty stamps on display are rare). If an observer only considers stamps on display, they will observe a spurious negative relationship between prettiness and rarity as a result of the selection bias (that is, not-prettiness strongly indicates rarity in the display, but not in the total collection).
Two independent events become conditionally dependent given that at least one of them occurs. Symbolically:
{\displaystyle 0<P(A)<1}
{\displaystyle 0<P(B)<1}
{\displaystyle P(A|B)=P(A)}
{\displaystyle P(A|B,A\cup B)=P(A)}
{\displaystyle P(A|A\cup B)>P(A)}
{\displaystyle A}
and event
{\displaystyle B}
may or may not occur
{\displaystyle P(A|B)}
, a conditional probability, is the probability of observing event
{\displaystyle A}
{\displaystyle B}
Explanation: Event
{\displaystyle A}
{\displaystyle B}
are independent of each other
{\displaystyle P(A|B,A\cup B)}
is the probability of observing event
{\displaystyle A}
{\displaystyle B}
{\displaystyle A}
{\displaystyle B}
) occurs. This can also be written as
{\displaystyle P(A|B\cap (A\cup B))}
Explanation: The probability of
{\displaystyle A}
{\displaystyle B}
{\displaystyle A}
{\displaystyle B}
) is smaller than the probability of
{\displaystyle A}
given (
{\displaystyle A}
{\displaystyle B}
In other words, given two independent events, if you consider only outcomes where at least one occurs, then they become conditionally dependent, as shown above.
The cause is that the conditional probability of event
{\displaystyle A}
occurring, given that it or
{\displaystyle B}
occurs, is inflated: it is higher than the unconditional probability, because we have excluded cases where neither occur.
{\displaystyle P(A|A\cup B)>P(A)}
conditional probability inflated relative to unconditional
One can see this in tabular form as follows: the yellow regions are the outcomes where at least one event occurs (and ~A means "not A").
A & B ~A & B
A & ~B ~A & ~B
For instance, if one has a sample of
{\displaystyle 100}
, and both
{\displaystyle A}
{\displaystyle B}
occur independently half the time (
{\displaystyle P(A)=P(B)=1/2}
), one obtains:
So in
{\displaystyle 75}
outcomes, either
{\displaystyle A}
{\displaystyle B}
occurs, of which
{\displaystyle 50}
{\displaystyle A}
occurring. By comparing the conditional probability of
{\displaystyle A}
to the unconditional probability of
{\displaystyle A}
{\displaystyle P(A|A\cup B)=50/75=2/3>P(A)=50/100=1/2}
We see that the probability of
{\displaystyle A}
is higher (
{\displaystyle 2/3}
) in the subset of outcomes where (
{\displaystyle A}
{\displaystyle B}
) occurs, than in the overall population (
{\displaystyle 1/2}
). On the other hand, the probability of
{\displaystyle A}
{\displaystyle B}
{\displaystyle A}
{\displaystyle B}
) is simply the unconditional probability of
{\displaystyle A}
{\displaystyle P(A)}
{\displaystyle A}
{\displaystyle B}
. In the numerical example, we have conditioned on being in the top row:
Here the probability of
{\displaystyle A}
{\displaystyle 25/50=1/2}
Berkson's paradox arises because the conditional probability of
{\displaystyle A}
{\displaystyle B}
within the three-cell subset equals the conditional probability in the overall population, but the unconditional probability within the subset is inflated relative to the unconditional probability in the overall population, hence, within the subset, the presence of
{\displaystyle B}
decreases the conditional probability of
{\displaystyle A}
(back to its overall unconditional probability):
{\displaystyle P(A|B,A\cup B)=P(A|B)=P(A)}
{\displaystyle P(A|A\cup B)>P(A)}
Berkson, Joseph (June 1946). "Limitations of the Application of Fourfold Table Analysis to Hospital Data". Biometrics Bulletin. 2 (3): 47–53. doi:10.2307/3002000. JSTOR 3002000. PMID 21001024. (The paper is frequently miscited as Berkson, J. (1949) Biological Bulletin 2, 47–53.)
Jordan Ellenberg, "Why are handsome men such jerks?"
Numberphile: Does Hollywood ruin books? – An education video on Berkson's paradox in popular culture
Retrieved from "https://en.wikipedia.org/w/index.php?title=Berkson%27s_paradox&oldid=1084824766"
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Growth and Characterization of 8-Hydroxy Quinoline Nitrobenzoate
Growth and Characterization of 8-Hydroxy Quinoline Nitrobenzoate
Department of Physics, Sri Ramakrishna Institute of Technology, Coimbatore, India
In this work we report the newly formed crystal structure of 8-Hydroxy quinoline nitro benzoate. The systematic investigation has been carried out on the growth and characterizations with a view of using this organic material in semiconductor devices apart from its various biological applications. Single crystals of 8-Hydroxy Quinoline Nitro Benzoate (8-HQNB) were grown successfully by slow evaporation solution growth technique. The new formation of the crystal with molecular formula C16H14N2O6 is confirmed by single crystal X-ray diffraction analysis. The crystallographic data has been deposited in Cambridge Crystallographic Data Centre [CCDC NO. 1005192]. Fourier Transform Infra Red (FTIR) spectroscopic analysis confirms the functional groups of the newly confirmed crystal. The UV-Vis-NIR studies reveal that there is no remarkable absorption in the visible region which proves its suitability for optical applications.
Crystal Growth, Organic, FTIR
The organic materials have been well known for their wide applications in superconductors [1] , semiconductors [2] , nonlinear optical devices [3] and photonic devices [4] [5] . 8-Hydroxy quinoline is a derivative of the heterocycle quinoline by placement of an OH group on carbon number 8. It is a monoprotic bidentate chelating agent. It exhibits antiseptic, disinfectant and pesticide properties [6] . It once was of interest as an anti-cancer drug and its solution in alcohol is used in liquid bandages [7] . Growth, structural, thermal and optical properties of 8-hydroxy quinoline by slow evaporation solution growth technique were already reported [8] [9] . In the present investigation an attempt has been made to grow 8-hydroxy quinoline nitro benzoate single crystals. And the newly formed crystal structure has been reported. The newly formed crystal has been confirmed by the single crystal X-ray diffraction analysis and the data has been deposited in Cambridge crystallographic data centre (CCDC).
Analar grade samples of 8-hydroxy quinoline, 2-nitro benzoic acid acid and acetone were obtained from Sigma Aldrich for the growth of single crystals.
The scheme of preparation was done in the following way. 8-hydroxy quinoline and 2-nitro benzoic acid were taken into 1:1 molar ratio and was dissolved in acetone. They are stirred well for about 20 minutes using magnetic stirrer in order to achieve homogeneous mixing of the solvent. The saturated solution was filtered using Whatmann filter paper and transferred into crystallizing vessel and kept undisturbed for crystallization. After a period of one week, good quality crystals were harvested from the solution and are shown in Figure 1.
Single crystal X-ray diffraction analysis proves the newly formed crystal with molecular formula C16H14N2O6 and lattice parameters a = 7.693(5) Å, b = 23.203(5) Å and c = 8.654(5) Å. It crystallizes in monoclinic crystal system under centrosymmetric P21/n space group. FT-IR spectrum was recorded to confirm the presence of functional groups. The optical transmittance spectrum for the newly formed crystal shows its wide transparency in the visible region.
The grown crystals were subjected to single crystal X-ray diffraction analysis using Enraf Nonius-CAD 4 diffractometer. Crystallographic data of HQNB has been deposited with the Cambridge Crystallographic Data Centre [CCDC No. 1005192]. The crystallographic data are presented in Table 1.
The crystal structure of the grown sample is shown in Figure 2.
Figure 1. Photograph of 8-HQNB single crystals.
Table 1. Crystal data.
Figure 2. Crystal structure of 8-HQNB.
FT-IR spectrum was recorded to confirm the presence of functional groups using Bruker: RFS 27 spectrometer. The recorded spectrum of 8-HQNB crystal is depicted in Figure 3.
The O−H stretching vibration appears at 3385 cm−1 and the aromatic C−H stretching appears at 3076 cm−1. The absorption band at 2889 cm−1 is assigned due to symmetric C−H stretching vibration of −CH2 group. The carbonyl stretching vibrations band at 1716 cm−1 as well as anti symmetric and symmetric stretching vibration bands at 1534 cm−1 and 1375 cm−1 of COO− group are obtained. The peak observed at 1601 cm−1 is assigned to C=N ring stretching vibration.
Figure 3. F-T-IR spectrum of 8-HQNB.
The NO2 stretching vibration of 2-nitrobenzoic acid moiety in the complex salt appears at 1534 cm−1. The band observed at 875 cm−1 is assigned to out of plane bending of C−O deformation. The C-C in plane and out plane bending vibration was observed at 816 cm−1. Also the vibrations around 746 - 550 cm−1 are assigned to in-plane bending vibration of quinoline. The observed frequencies with their assignments are summarized in Table 2.
3.3. UV-Vis-NIR Studies
The optical transmittance spectrum for the grown crystal was recorded using a Perkin-Elmer Lambda 35 Spectrophotometer in the wavelength range from 200 - 1200 nm. The recorded spectrum is shown in Figure 4(a).
The grown crystal has a wider transparency range in the visible and NIR region. From the figure it is observed that the crystal has UV cut-off wavelength at 453 nm. A band observed at 225 nm is attributed to the excitation of aromatic ring (π to π*). High transmittance of grown crystal in the visible region facilitates it to be a potential crystal for optoelectronics applications. The higher optical transmission in solution grown 8-HQNB crystal may be due to lesser defects and absence of inclusions, which in turn reduced scattering in HQNB crystals and increases the output intensity. The optical absorption coefficient (α) was calculated from the transmittance using the given formula
\alpha =\frac{2.3036\mathrm{log}\left(1/T\right)}{d}
where T is the measured crystal transmittance and d is the thickness of the sample. Assuming parabolic trends, the relation between α and Eg is given by
\alpha =\frac{A{\left(h\nu -{E}_{g}\right)}^{n}}{h\nu }
Table 2. FT-IR spectral data.
Figure 4. (a) Optical transmittance spectrum of 8-HQNB; (b) Plot of (αhν)2 versus photon energy of 8-HQNB.
where A is a constant,
{E}_{g}
is the optical band gap of the material, υ is the frequency of the incident photon and h is the Planck’s constant.
Figure 4(b) shows the plot of hυ and (αhυ)2. Optical band gap of the material was obtained by extrapolating the linear portion of the plots of (αhυ)2 and hυ. The optical band gap was found to be 2.75 eV. Crystals with wide band gap expected to possess high laser damage threshold and large transmittance in the visible region.
Single crystals of 8-hydroxy quinoline nitro benzoate (C16H14N2O6) were grown by slow evaporation technique. The single crystal X-ray diffraction analysis shows that the grown crystal crystallizes under monoclinic crystal system with lattice parameters a = 7.693 Å, b = 23.203 Å and c = 8.654 Å. It belongs to centrosymmetric P21/n space group with volume 1484.4 Å3 and density 1.478 mg/m3. The crystallographic data has been deposited in Cambridge Crystallographic Data Centre [CCDC No. 1005192]. The presence of functional groups was confirmed by FT-IR spectroscopic analysis. The UV-Vis-NIR studies reveal that there is no remarkable absorption in the visible region. The lower cut off wavelength of 8-HQNB is found to be 453 nm which proves its suitability for optical applications. The optical band gap energy of the grown crystal was calculated as 2.75 eV. Hence all the above attributes 8-hydroxy quinoline nitrobenzoate to a novel material for device applications.
The author expresses her sincere thanks to SAIF IITM for rendering characterization facilities.
Damodharan, J. (2019) Growth and Characterization of 8-Hydroxy Quinoline Nitrobenzoate. Journal of Minerals and Materials Characterization and Engineering, 7, 64-70. https://doi.org/10.4236/jmmce.2019.72005
1. Farges, J.P. (1994) Organic Conductors: Fundamentals and Applications. Marcel Dekker, New York.
2. Ishiguro, T. and Yamaji, K. (1990) Organic Superconductors. Springer-Verlag, New York. https://doi.org/10.1007/978-3-642-97190-7
3. Dou, S.X., Josse, D. and Zyss, J. (1993) Near-Infrared Pulsed Optical Parametric Oscillation in N-(4-Nitrophenyl)-L-Prolinol at the 1-ns Time Scale. Journal of the Optical Society of America B, 10, 1708-1715. https://doi.org/10.1364/JOSAB.10.001708
4. Saleh, B.E.A. and Teich, M.C. (1991) Fundamentals of Photonics. John Wiley and Sons, New York. https://doi.org/10.1002/0471213748
5. Penn, B.G., Cardelino, B.H., Moore, C.E., Shields, A.W. and Fraizer, D.O. (1991) Growth of Bulk Single Crystals of Organic Materials for Nonlinear Optical Devices: An Overview. Progress in Crystal Growth and Characterization of Materials, 22, 19-51. https://doi.org/10.1016/0960-8974(91)90024-7
6. Phillips, J.P. (1956) The Reactions of 8-Quinolinol. Chemical Reviews, 56, 271-297. https://doi.org/10.1021/cr50008a003
7. Shen, A.Y., Wu, S.N. and Chiu, C.T. (1999) Synthesis and Cytotoxicity Evaluation of Some 8-Hydroxyquinoline Derivatives. Journal of Pharmacy and Pharmacology, 51, 543-548. https://doi.org/10.1211/0022357991772826
8. Vijayan, N., Bhagavannarayana, G., Maurya, K.K., Datta, S.N., Gopalakrishnan, R. and Ramasamy, P. (2007) Studies on the Structural, Thermal and Optical Behavior of Solution Grown Organic NLO Material: 8-Hydroxyquinoline. Crystal Research and Technology, 42, 195-200. https://doi.org/10.1002/crat.200610796
9. Krishna Kumar, V., Nagalakshmi, R. and Janaki, P. (2005) Growth and Spectroscopic Characterization of a New Organic Nonlinear Optical Crystal 8-Hydroxyquinoline. Spectrochimica Acta Part A, 61, 1097-1103. https://doi.org/10.1016/j.saa.2004.06.029
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Prime Zeta Function | Brilliant Math & Science Wiki
Patrick Corn, Ishan Singh, A Former Brilliant Member, and
The prime zeta function is an expression similar to the Riemann zeta function. It has interesting properties that are related to the properties of the Riemann zeta function, as well as a connection to Artin's conjecture about primitive roots.
The prime zeta function
\zeta_{\mathbb{P}}(s)
s
is a complex number, is defined by the series
\zeta_{\mathbb{P}}(s) = \sum_{p \in \mathbb{P}} \dfrac{1}{p^{s}}
\mathbb{P}
is the set of prime numbers.
Divergence of
\zeta_{\mathbb{P}}(1)
Expression in terms of the Riemann Zeta Function
Connection with the Twin Prime Constant
Connection with Artin's Conjecture
Relation with other Zeta Functions
\zeta_{\mathbb{P}}(1)
\zeta_{\mathbb{P}}(1)
is the sum of the reciprocals of the primes. This series diverges, but very slowly:
\sum_{\stackrel{p \in \mathbb{P}}{p < n}} \frac1{p} - \log\big(\log(n)\big) \to M\ \text{ as }\ n\to \infty,
M = 0.261497\ldots
is a constant (called the Meissel-Mertens constant). This is reminiscent of the definition of the Euler-Mascheroni constant
\gamma
Euler asserted that the sum of the reciprocals of the primes diverged, and derived a correct estimate for it, but his proof involved manipulations of divergent infinite series and products (for instance, the Euler product expansion for
\zeta(s)
s=1
). As is often the case with Euler's arguments, it can be made rigorous with some extra work.
The Euler product for
\zeta(s)
\big(
\text{Re}(s)>1\big)
\zeta(s) = \prod_{p \text{ prime}} \left( 1-\frac1{p^s} \right)^{-1}
and taking the natural log of both sides gives
\begin{aligned} \log\big(\zeta(s)\big) &= - \sum_{p \text{ prime}} \log\left( 1-\frac1{p^s} \right) \\ &= \sum_{p \text{ prime}} \sum_{k=1}^{\infty} \frac{\hspace{1.5mm} \frac{1}{p^{ks}}\hspace{1.5mm} }{k}, \end{aligned}
using the Maclaurin series
-\log(1-x) = x+\frac{x^2}2 + \frac{x^3}3 + \cdots.
Now switching the two sums gives
\begin{aligned} \log\big(\zeta(s)\big) &= \sum_{k=1}^{\infty} \sum_{p \text{ prime}} \frac{\hspace{1.5mm} \frac{1}{p^{ks}}\hspace{1.5mm} }{k} \\ &= \sum_{k=1}^{\infty} \frac1{k} \sum_{p \text{ prime}} \frac1{p^{ks}} \\ &= \sum_{k=1}^{\infty} \frac{\zeta_{\mathbb{P}}(ks)}{k}. \end{aligned}
A generalization of Möbius inversion says that
f(x) = \sum_{k=1}^{\infty} \frac{g(kx)}{k} \Leftrightarrow g(x) = \sum_{k=1}^{\infty} \mu(k)\frac{f(kx)}{k},
\mu
is the Möbius function, as long as the sums are absolutely convergent (the proof is straightforward). Applying this gives
\zeta_{\mathbb{P}}(s) = \sum_{k=1}^{\infty} \mu(k) \frac{\log\big(\zeta(ks)\big)}{k}.
Note that this gives an idea of why
\zeta_{\mathbb{P}}(1)
diverges at the same speed as
\log\big(\log(n)\big)
k=1
term is the only undefined term at
s= 1,
\zeta(1)
diverges like
\log(n)
. In fact, expanding near
s=1
(
x > 0)
\zeta_{\mathbb{P}}(1+x) = -\log(x)+(M-\gamma) + O(x),
\gamma
is the Euler-Mascheroni constant. The
O(x)
terms go to
0
x \to 0^+ .
The twin prime conjecture is that there are infinitely many primes
p
p+2
is also prime. While this is still open, heuristics suggest that it is true and that in fact the function
\pi_2(x)
that counts twin primes
\le x
\pi_2(x) \sim 2\Pi_2 \int_2^x \frac{dx}{\big(\log(x)\big)^2},
\Pi_2 =0.66016\ldots
is the twin prime constant
\sum_{p \text{ odd prime}} \left( 1-\frac1{(p-1)^2}\right).
A computation involving Taylor expansions of logarithms (similar to the above one) shows that the constant
\Pi_2
is related to the values of
P(s)
\log(\Pi_2) = - \sum_{k=2}^{\infty} \frac{2^k-2}{k} \big(\zeta_{\mathbb{P}}(k)-2^{-k}\big).
The point is that the prime zeta function comes up in evaluating constants involving products over all primes.
Artin's conjecture states that if
a
is an integer that is not a perfect square or
-1
, then it is a primitive root modulo
p
p.
As usual, there is a heuristic estimate for the probability that
a
is a primitive root mod a given prime
p.
a
is squarefree and not congruent to
1
4
, this probability is conjectured to be Artin's constant
C = \prod_{p \text{ prime}} \left( 1-\frac1{p(p-1)} \right) = 0.37395\ldots.
(For other values of
a
, there is also a conjectural probability that is a rational multiple of Artin's constant.)
A calculation involving taking logs and expanding gives
\log(C) = \sum_{n=2}^{\infty} (1-L_n) \frac{\zeta_{\mathbb{P}}(n)}{n},
L_n
n^\text{th}
\big(
L_1 = 1, L_2 = 3, L_n = L_{n-1}+L_{n-2}\big).
An interesting generalization can be made by summing over inverse of positive integers raised to power
s
which are a product of
k
primes (not necessarily distinct).
k
\zeta_{\mathbb{P}}(k,s)
s
k
is a non-negative integer, is defined by the series
\zeta_{\mathbb{P}}(k,s) = \sum_{n : \Omega(n) = k} \dfrac{1}{n^{s}},
\Omega
denotes the number of prime factors.
Try proving these interesting identities involving generalized prime zeta function:
\begin{aligned} \sum_{k=0}^{\infty}\zeta_{\mathbb{P}}(k,s) &= \zeta (s) \\\\ \lim_{s \to 1} (s-1) \zeta_{\mathbb{P}}(k,s) &= 0 \ \forall \ k \in \mathbb N \\\\ \zeta_{\mathbb{P}}(2,s) &= \dfrac{\zeta_{\mathbb{P}}(s)^2 + \zeta_{\mathbb{P}}(2s)}{2}. \end{aligned}
Cite as: Prime Zeta Function. Brilliant.org. Retrieved from https://brilliant.org/wiki/prime-zeta-function/
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Free answers to college discrete math questions
Discrete math questions and answers
Recent questions in Discrete math
qtbabe9876a9 2022-05-22 Answered
Can someone help me to expand the below equation for
N=2
N=3
{f}_{i}=\sum \left\{{e}_{i}:j\le i\le k\right\}
1\le j\le k\le N
{e}_{i}
is the unit vector of coordinate i in an N-dimensional space.
Alani Conner 2022-05-22 Answered
The following cards are dealt with 3 people at random so that every one of them gets the same number of cards:
{R}_{1},{R}_{2},{R}_{3},{B}_{1},{B}_{2},{B}_{3},{Y}_{1},{Y}_{2},{Y}_{3}
where R, B and Y denote red, blue and yellow, respectively. Find the probability that everyone gets a red card.
Richardtb 2022-05-22 Answered
How to inductively prove a graph property?
I'm stuck on Part 1, I can't find it inductively. The distance from
\left[k,k+1\right]
is going to be a non-zero positive value with a vertex. Therefore, there is going to be a positive edge weight from w(uv) that goes from
\left[k,k+1\right].
. Therefore, there exists a v on the set of V such at
A\left[u,k+1\right]=A\left[v,k\right]+w\left(uv\right).
Given an undirected graph
G=\left(V,E\right)
with positive edge weights w(e) for each edge
e\in E
, we want to find a dynamic programming algorithm to compute the longest path in G from a given source s that contains at most n edges.
To do this first define A[v,k] as the weight for the longest path from node s to node v of at most k edges.
1. First we need to prove an optimal sub-structure by induction. Show that if A[v,k] is the weight of the longest path, then for all
u\in V
v\in V
A\left[u,k+1\right]=A\left[v,k\right]+w\left(uv\right)
2. Describe a dynamic programming algorithm that finds the optimal length using part 1. Specifically: describe (1) the OPT recurrence (2) the running time of the iterative solution for computing the OPT table.
skottyrottenmf 2022-05-21 Answered
Proving two integers are relatively prime using Bezout's Theorem.
\mathrm{gcd}\left(a,b\right)=1,
, a and b are relatively prime and also if gcd(a,b) is equal to 1 there should be 2 integers k and m such that
ak+bm=1.
If we can find such integers k and m, is it a proof that a and b are relatively prime ?
What you think about that proof? Is it correct way?
Marianna Stone 2022-05-21 Answered
A+{A}^{\prime }B+{A}^{\prime }{B}^{\prime }C+{A}^{\prime }{B}^{\prime }{C}^{\prime }D=A+B+C+D
Prove the above relationship by using the Boolean definition. I tried
A+{A}^{\prime }B=A+B
, but end up with
A+B+{A}^{\prime }{B}^{\prime }\left(C+D\right)
, how can I go next?
Liberty Mack 2022-05-21 Answered
Does LCM have distributive property of multiplication?
Is statement
lcm\left(ax,ay\right)=a\ast lcm\left(x,y\right)
true? (assuming all variables are positive integers)
Brooke Webb 2022-05-21 Answered
Why does "Some student has asked every faculty member a question" translate to
\mathrm{\forall }y\left(F\left(y\right)\to \mathrm{\exists }x\left(S\left(x\right)\vee A\left(x,y\right)\right)\right)
Context: I'm in undergrad discrete math, this is a textbook question from Discrete Mathematics and its Applications 7th edition
Let S(x) be the predicate "x is a student," F(x) the predicate "x is a faculty member," and A(x, y) the predicate "x has asked y a question," where the domain consists of all people associated with your school. Use quantifiers to express the following statement.
Some student has asked every faculty member a question.
This is what the textbook says is the correct answer:
\mathrm{\forall }y\left(F\left(y\right)\to \mathrm{\exists }x\left(S\left(x\right)\vee A\left(x,y\right)\right)\right)
I mostly understand how this is correct, but shouldn't it be
\wedge
\vee
hughy46u 2022-05-21 Answered
Bounding the order of tournaments without transitive subtournaments of certain size.
A tournament of order N is a directed graph on [N] obtained by assigning a direction to each edge of
{K}_{N}
. A tournament D is transitive if for every triple
a,b,c\in N
\left(a,b\right),\left(b,c\right)\in E\left(D\right)
\left(a,c\right)\in E\left(D\right)
n\in \mathbb{N}
let f(n) be the maximum integer such that there exists a tournament of order f(n) without a transitive sub-tournament of size n. Show that
f\left(n\right)>\left(1+o\left(1\right)\right){2}^{\frac{n-1}{2}}
Erick Clay 2022-05-21 Answered
The random variable X has homogeneous distrigution
I have a problem with exercise:
The random variable X has a homogeneous distribution in the interval [0,1]. Random variable
Y=max\left(X,1/2\right)
. Please find the expected value of a random variable Y. I need to do this for discrete variables and also for continuous variable
Aiden Barry 2022-05-21 Answered
There is a subset
T\subset S
|T|=k+1
a,b\in T
{a}^{2}-{b}^{2}
k\ge 1
If S is a set of positive integers with
|S|=N
, then there is a subset
T\subset S
|T|=k+1
a,b\in T
{a}^{2}-{b}^{2}
What is the smallest value of N as a function of k so that the above statement is true?
I have observed that perfect squares end with 0,1,4,5,6 and 9. If we have two perfect squares that end with the same as one of 0,1,4,5,6 and 9, then we are done.
I think by PHP we should have
⌈\frac{N}{k}⌉=6
Elisha Kelly 2022-05-21 Answered
Discrete Mathematics what exactly does from A to B mean?
Like if there's some relation R from a set A to a set B what exactly does this mean?
I know it's just a subset of the Cartesian product
A×B,
, but what exactly does from A to B mean..? Like obviously from B to A is different so can anyone explain what the FROM really means here?
Simone Werner 2022-05-21 Answered
⌊-x⌋=-⌈x⌉
I wanted to ask if this kind of reasoning for proving the result in the title could be considered correct:
⌈x⌉=n
n-1<x\le n
-⌈x⌉=-n
-n-1<x\le -n
Then multiplying by -1 the formula
-n-1<x\le -n
n+1>-x\ge n
, inverting the sign.
n+1>-x\ge n
n\le -x<n+1
⌊x⌋=n
n\le x<n+1
n\le -x<n+1
we can infer that
⌊-x⌋=
n=-⌈x⌉
f:E\to E
f\circ f=f
. Prove that if f is surjective or injective then
f=I{d}_{E}
Aidyn Cox 2022-05-21 Answered
Find all complex numbers z such as z and 2/z have both real and imaginary part integers
I am really struggling to solve this one. I feel like I am missing the key part of the solution, so I would like to see how it's done.
Find all complex numbers
z=x+yi
such as z and
2{z}^{-1}
have both real and imaginary part integers
2{z}^{-1}=\frac{2}{z}=\frac{2}{x+yi}=\frac{2}{x+yi}\cdot \frac{x-yi}{x-yi}=\frac{2x-2yi}{{x}^{2}+{y}^{2}}.
2{z}^{-1}
have its imaginary part
\in \mathbb{Z}
, we should equal 2y to 0
2y=0⇒y=0
yi=0
x must also be an integer. We simply assume
x\in \mathbb{Z}
(no matter what value x has, as long as it's an integer, we are good).
We do the same for z and find out the same values
yi=0
x\in \mathbb{Z}
A=\left\{\left(x,y\right)\in \mathbb{C}|x\in \mathbb{Z}\text{ and }y=0\right\}
is the set of all complex numbers whose real and imaginary part are integers.
Carlie Fernandez 2022-05-20 Answered
Let's say that we get a table with zeros and ones. We need to get it into disjunctive normal form or conjuctive normal form. We also have discrete variables
{x}_{1},..,{x}_{n}
that are either 1 or 0. How do you determine where to put negation and where not to put it.
for instance: we have a row:
p=0,q=1,r=0,\phantom{\rule{1em}{0ex}}\text{table row result = 1}
Should I write this as:
...\vee \left(\mathrm{¬}p\wedge q\wedge \mathrm{¬}r\right)\vee ...
...\wedge \left(\mathrm{¬}p\vee q\vee \mathrm{¬}r\right)\wedge ...
What is the correct way ? What if the table row result would be zero?
Or the other way with negations? So my question is how do we know where the negations are?
madridomot 2022-05-20 Answered
How to find the probability and generating function for this problem?
We toss a coin k times which is having probability p of landing on heads and a probability
q=1-p
of landing on tails. The probability of getting exactly n heads is denoted by
{a}_{n}
{a}_{n}
and the generating function of the sequence
\left({a}_{n}\right)
Given a rearrangement of the numbers from 1 to n, each pair of consecutive elements a and b of the sequence can be either increasing or decreasing. How many rearrangements have of the numbers from 1 to n have exactly two increasing pairs?
Gael Gardner 2022-05-19 Answered
How to solve the recurrence relation
{a}_{n}=\sum _{i=1}^{n-1}{a}_{i}{a}_{n-i}
The relation is:
{a}_{n}=\sum _{i=1}^{n-1}{a}_{i}{a}_{n-i}\phantom{\rule{0ex}{0ex}}n>1,{a}_{1}=2
I know I probably should create a generating function for
{a}_{n}
G\left(x\right)=f\left(x\right)\ast g\left(x\right)
, but I'm not sure what f(x) and g(x) exactly should be, are they just
\sum {a}_{i}{x}^{i}
\sum {a}_{n-i}{x}^{n-i}
? And are there any other methods to solve it? Any hint would be useful to me.
ownerweneuf 2022-05-19 Answered
When proving divisibility by induction, how does
f\left(k+1\right)-f\left(k\right)
help us to prove it?
osmane5e 2022-05-19 Answered
\forall x\forall y\left(P\left(x,y\right)\to ¬P\left(y,x\right)\right),\forall x\exists yP\left(x,y\right)
be deduced to
¬\exists v\forall zP\left(z,v\right)
Dealing with discrete Math is an interesting subject because discrete Math equations can be encountered basically anywhere from scheduling of sports games and live shows to education where each person is examined online. It is a reason why discrete math questions that we have collected for you are aimed at solutions that go beyond equations to provide you with the answers that will help you understand the concept. Still, discrete Math equations are explained as well by turning to problems in computer science, programming, software, and cryptography among other interesting subjects like software and mobile apps development.
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How can you evaluate \int_0^{\pi/2}\log\cos(x)dx ?
{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\mathrm{cos}\left(x\right)dx
For the sake of simplicity, all the integral variables I use are x even there are a lot of substitutions. Because lots of variables could make one confused.
Let I denote the integral value. By substitute x for
\frac{\pi }{2}-x
I={\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\mathrm{cos}\left(x\right)dx={\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\mathrm{sin}\left(x\right)dx
I={\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(2\mathrm{cos}\left(\frac{x}{2}\right)\mathrm{sin}\left(\frac{x}{2}\right)\right)dx
=\frac{\pi }{2}\mathrm{log}2+{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\mathrm{cos}\left(\frac{x}{2}\right)dx+{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\mathrm{sin}\left(\frac{x}{2}\right)dx
=\frac{\pi }{2}+2{\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\mathrm{cos}\left(x\right)dx+2{\int }_{0}^{\frac{\pi }{4}}+2{\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\mathrm{sin}\left(x\right)dx
=\frac{\pi }{2}\mathrm{log}2+{I}_{1}+{I}_{2}
In the second step from bottom, I use the substitution that
x=\frac{x}{2}
{I}_{1}
, use the substitution that
x=\frac{\pi }{2}-x
{I}_{1}=2{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\mathrm{log}\mathrm{sin}\left(x\right)dx
It gives that
{I}_{1}+{I}_{2}=2I
I=\frac{\pi }{2}\mathrm{log}2+2I
I=-\frac{\pi }{2}\mathrm{log}2
{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\mathrm{cos}xdx=I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\mathrm{sin}xdx
By symmetry we have
\mathrm{ln}\mathrm{cos}x=\mathrm{ln}\mathrm{sin}x
\left[0,\pi .2\right]
. This is true for any even/odd function on this interval, as is an exercise in Demidovich-Problems in Analysis. Thus we have
2I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\mathrm{cos}xdx+{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\mathrm{sin}xdx={\int }_{0}^{\frac{\pi }{2}\mathrm{ln}\left(\mathrm{sin}x\mathrm{cos}x\right)dx={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\frac{1}{2}\cdot \mathrm{sin}\left(2x\right)\right)dx}
All I used was
\mathrm{ln}\left(a\cdot b\right)=\mathrm{ln}\left(a\right)+\mathrm{ln}\left(b\right)
2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}\left(2x\right)
. Now we split the integral back up to obtain
-{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(2\right)dx+{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{sin}\left(2x\right)\right)dx=2I
But the integral of
\mathrm{ln}\mathrm{sin}u
2I
-\frac{\pi \mathrm{ln}\left(2\right)}{2}+I=2I\to I=-\frac{\pi \mathrm{ln}\left(2\right)}{2}
We have a well-known identity:
\prod _{k=1}^{n-1}\mathrm{sin}\left(\frac{\pi k}{n}\right)=\frac{2n}{{2}^{n}}
\mathrm{log}\mathrm{sin}x
is an improperly Riemann-integrable function over
\left(0,\pi \right)
{\int }_{0}^{\pi }\mathrm{log}\mathrm{sin}\theta d\theta =\underset{n\to +\mathrm{\infty }}{lim}\frac{\pi }{n}\sum _{k=1}^{n-1}\mathrm{log}\mathrm{sin}\left(\frac{\pi k}{n}\right)=-\pi \mathrm{log}2
{\int }_{0}^{\pi /2}\mathrm{log}\mathrm{sin}\theta d\theta ={\int }_{0}^{\pi /2}\mathrm{log}\mathrm{sin}\theta d\theta =-\frac{\pi }{2}\mathrm{log}2
\frac{7}{8}{e}^{2t}{t}^{2}
calculate this integral:
{\int }_{0}^{\frac{\pi }{2}}\mathrm{arccos}\left(\mathrm{sin}x\right)dx
-d\frac{{\left\{\pi \right\}}^{2}}{8}
\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx
{\int }_{0}^{1}\frac{\mathrm{log}\left(x\right)}{1-x}dx
Use the reduction formulas in a table of integrals to evaluate the following integrals.
\int {x}^{3}{e}^{2x}dx
Use the method of reduction of order to find the general of the differential equation
t\left(t+3\right)y{}^{″}-3\left(t+2\right){y}^{\prime }+3y=0
\int y3\text{ }ds,\text{ }C÷x=t3,\text{ }y=t,\text{ }0?\text{ }t?\text{ }3
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Represent the plane curve by a vector-valued function y=(x-2)^{2}
y={\left(x-2\right)}^{2}
Egreane61
y={\left(x-2\right)}^{2}
x\left(t\right)=t
x\left(t\right)=t
y={\left(x-2\right)}^{2}
y={\left(t-2\right)}^{2}
r\left(t\right)=x\left(t\right)i+y\left(t\right)j
⇒r\left(t\right)=ti+{\left(t-2\right)}^{2}j
\left(x,\text{ }y\right)=
\stackrel{\to }{A}
is 3.00 units in length and points along the positive x-axis. Vector
\stackrel{\to }{B}
is 4.00 units in tength and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the vectors
\stackrel{\to }{A}+\stackrel{\to }{B}
Determine the line integral along the curve C from A to B. Find the parametric form of the curve C
Use the vector field:
\stackrel{\to }{F}=\left[-bx\text{ }cy\right]
Use the following values:
a=5,\text{ }b=2,\text{ }c=3
Find parametric equations for the line of intersection of the planes
3x-2y+z=1
2x+y-3z=3
\left(-1,1\right)
from Cartesian coordinates to polar coordinates
Find the point(s) of intersection of the following two parametric curves, by first eliminating the parameter, then solving the system of equations.
x=t+5,y={t}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=\frac{1}{16}t,y=t
\left(2\sqrt{3},\text{ }2,\text{ }-1\right)
from rectangular to cylindrical coordinates?
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Modular Arithmetic Operations Warmup Practice Problems Online | Brilliant
The pattern above is blue-orange-blue-orange-blue-orange ... on to infinity.
If all integers are colored in the same pattern as the numbers in this grid, what color is 1234?
The pattern above is blue-orange-blue-orange-blue-orange ... on to infinity. If you multiply a blue number by an orange number, what color will the product be?
Note: Assume that all integers are colored according to the same pattern as the numbers in the grid.
Blue Orange It depends on the numbers you choose
The pattern is purple-blue-green-yellow-orange-purple-.... on to infinity. If you add a blue number to a yellow number, what color will the sum be?
Purple Blue Green Yellow Orange It depends on the numbers you choose
The pattern is purple-blue-green-yellow-orange-purple-.... on to infinity. If you multiply a green number by a yellow number, what color will the product be?
The pattern is blue-green-orange-.... on to infinity. What color is
8 \times 17 \, ?
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The joint density of X and Y is given by, f_{XY}(x,y)=\frac{6}{7}(x^{2}+\frac{xy}{2})\
The joint density of X and Y is given by,f_{XY}(x,y)=\frac{6}{7}(x^{2}+\frac{xy}{2})\
The joint density of X and Y is given by,
{f}_{XY}\left(x,y\right)=\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)\text{ }0<x<1,0<y<2
P\left(X<1,Y>1\right)
b) Find the marginal probability distributions of X and Y.
c) Find the conditional probability density function of Y given
X=0.5
P\left(Y<1/X=0.5\right)
joint density of X and Y
{f}_{xy}\left(x,y\right)=\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)\text{ }0<x<1,0<y<2
P\left(x<1,y>1\right)
P\left(x<1,y>1\right)={\int }_{x=0}^{1}{\int }_{y=1}^{2}\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)dydx
P\left(x<1,y>1\right)=\frac{6}{7}{\int }_{0}^{1}{\left[{x}^{2}y+\frac{x}{2}\frac{{y}^{2}}{2}\right]}_{1}^{2}dx
P\left(x<1,y>1\right)=\frac{6}{7}{\int }_{0}^{1}\left({x}^{2}+\frac{3x}{4}\right)dx=\frac{6}{7}{\left[\frac{{x}^{3}}{3}+\frac{3}{4}\frac{{x}^{2}}{2}\right]}_{0}^{1}=\frac{6}{7}×\frac{17}{24}=\frac{102}{168}
=0.6071
P\left(x<1,y>1\right)=0.6071
b) Marginal probability density of x and y.
Marginal probability density of x
{f}_{x}\left(x\right)={\int }_{y=0}^{2}\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)dy=\frac{6}{7}{\left[{x}^{2}y+\frac{x}{2}\frac{{y}^{2}}{2}\right]}_{0}^{2}=\frac{6}{7}\left(2{x}^{2}+x\right)
{f}_{x}\left(x\right)=\frac{6}{7}\left(2{x}^{2}+x\right)
now, Marginal probability density of y
\theta
\left[X\right]
if X has a density function given by
f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{4}x{e}^{-x/2}& x>0\\ 0& otherwise\end{array}
f\left(x\right)=\left\{\begin{array}{ll}c\left(1-{x}^{2}\right)& -1<x<1\\ 0& otherwise\end{array}
f\left(x\right)=\left\{\begin{array}{ll}\frac{5}{{x}^{2}}& x>5\\ 0& x\le 5\end{array}
A proton is placed in a uniform electric field of
2.75×{10}^{3}\frac{N}{C}
a) the magnitude of the electric force felt by the proton
b) the proton’s acceleration;
c) the proton’s speed after
1.00\mu s
in the field, assuming it starts from rest.
Let A be a 3x5 matrix
(a) Give all possible values for the rank of A
(b) If the rank of A is 3, what is the dimension of its columnspace
(c) If the rank of A is 3, what is the dimension of the solutionspace of the homogeneous system Ax=0
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Cross-entropy loss for classification tasks - MATLAB crossentropy - MathWorks Nordic
{\text{loss}}_{j}=-\left({T}_{j}\text{ln}{Y}_{j}+\left(1-{T}_{j}\right)\text{ln}\left(1-{Y}_{j}\right)\right),
\text{loss}=\frac{1}{N}\sum _{j}{m}_{j}{w}_{j}{\text{loss}}_{j},
{\text{loss}}_{j}^{*}={m}_{j}{w}_{j}{\text{loss}}_{j}
\text{loss}=-\frac{1}{N}\sum _{n=1}^{N}\sum _{i=1}^{K}{T}_{ni}\text{ln}{Y}_{ni},
\text{loss}=-\frac{1}{N}\sum _{n=1}^{N}\sum _{i=1}^{K}\left({T}_{ni}\mathrm{ln}\left({Y}_{ni}\right)+\left(1-{T}_{ni}\right)\mathrm{ln}\left(1-{Y}_{ni}\right)\right),
\text{loss}=-\frac{1}{N}\sum _{n=1}^{N}\sum _{i=1}^{K}{w}_{i}{T}_{ni}\text{ln}{Y}_{ni},
\text{loss}=-\frac{1}{N}\sum _{n=1}^{N}\sum _{t=1}^{S}{m}_{nt}\sum _{i=1}^{K}{T}_{nti}\text{ln}{Y}_{nti},
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Frictional brake with pressure-applying cylinder and pads with faulting - MATLAB - MathWorks Nordic
Mean pad radius
Coulomb friction coefficient vector
Brake pressure when faulted
Frictional brake with pressure-applying cylinder and pads with faulting
The Disc Brake block represents a brake arranged as a cylinder applying pressure to one or more pads that can contact the shaft rotor. Pressure from the cylinder causes the pads to exert friction torque on the shaft. The friction torque resists shaft rotation.
You can also enable faulting. When faulting occurs, the brake will exert a user-specified pressure. Faults can occur at a specified time or due to an external trigger at port T.
This figure shows the side and front views of a disc brake.
The equation that the block uses to calculate brake torque, depends on the wheel speed, Ω, such that when
\Omega \ne 0
T=\frac{{\mu }_{k}P\pi {D}_{b}{}^{2}{R}_{m}N}{4}.
However when
\Omega =0,
the torque applied by the brake is equal to the torque that is applied externally for wheel rotation. The maximum value of the torque that the brake can apply when
\Omega =0,
T=\frac{{\mu }_{s}P\pi {D}_{b}{}^{2}{R}_{m}N}{4}.
Rm=\frac{Ro+Ri}{2}
T is the brake torque.
P is the applied brake pressure.
Ω is the wheel speed.
N is the number of brake pads in disc brake assembly.
μs is the disc pad-rotor coefficient of static friction.
μk is the disc pad-rotor coefficient of kinetic friction.
Db is the brake actuator bore diameter.
Rm is the mean radius of brake pad force application on brake rotor.
Ro is the outer radius of brake pad.
Ri is the inner radius of brake pad.
The block default models a dry brake. You can model fluid friction in a wet brake by setting the Viscous friction coefficient, kv, to a nonzero value. The torque on the wheel in a wet brake system is:
{T}_{wet}=T+{k}_{v}\Omega .
When faults are enabled, a brake pressure is applied in response to one or both of these triggers:
If a fault trigger occurs, the input pressure is replaced by the Brake pressure when faulted value for the remainder of the simulation. A value of 0 implies that no braking will occur. A relatively large value implies that the brake is stuck.
P — Cylinder pressure, bars
Physical signal port associated with cylinder pressure.
S — Shaft
Rotational mechanical conserving port associated with the shaft.
Exposing this port makes related parameters and variables visible.
Mean pad radius — Pad radius
150 mm (default) | positive scaler
Mean radius of the friction pads.
Cylinder bore — Bore diameter
10 mm (default) | positive scaler
Diameter of the piston.
Number of brake pads — Pad quantity
Number of friction pads.
When this parameter is set to Model, the thermal port and related parameter and variables are visible.
Static friction coefficient — Static friction
Coefficient of static friction. The value that you specify for this parameter must be greater than the value that you specify for the Coulomb friction coefficient parameter.
Static friction coefficient vector — Static friction
[.9, .8, .7] | vector
Coefficient of static friction, such that:
The values must be greater than the corresponding value that you specify for the Coulomb friction coefficient vector parameter.
Coulomb friction coefficient — Contact friction
Coulomb friction coefficient vector — Contact friction
[.8, .7, .6] (default) | increasing vector of positive values
Breakaway friction velocity — Friction threshold
0.1 rad/s (default) | scalar
Angular speed at which friction switches from static to kinetic.
For a wet brake, the viscous friction represents the energy loss to the cooling/lubricating fluid between the clutch plates. To model a wet brake, specify a nonzero value for the coefficient of viscous friction. The default value represents a dry brake.
Enable externally or temporally triggered faults. When faulting occurs, the brake pressure normally received at port P will be set to the value specified in the Brake pressure when faulted parameter.
Brake pressure when faulted — Set faulted brake pressure
Set faulted brake pressure. When faulting occurs, the brake pressure normally received at port P will be set to the value specified in the Brake pressure when faulted parameter. A value of 0 implies that braking does not occur. A relatively large value implies that the brake is stuck.
Enables fault triggering at a specified time. When the Simulation time for fault event is reached, the brake pressure normally received at port P will be set to the value specified in the Brake pressure when faulted parameter.
When the Simulation time for fault event is reached, the brake pressure normally received at port P will be set to the value specified in the Brake pressure when faulted parameter.
This parameter is only visible when in the Friction settings, the Thermal Port parameter is set to Model.
Band Brake | Double-Shoe Brake | Loaded-Contact Rotational Friction | Rotational Detent
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Compiler Design Dynamic Programming (DP)
Code generation with dynamic programming.
Contiguous evaluation.
We first discuss an algorithm to generate code from a labeled expression tree that works for machines whereby computations are done in registers and instructions consist of an operator that is applied to one or two registers and a memory location then proceed into the dynamic programming technique works for a broad class of register machines with a complex instruction set for which optimal code is generated in linear time.
An example - Generating code from a labeled expression tree.
Input: A labeled tree with each operand appearing once and a number of registers r ≥ 2.
Output: An optimal sequence of machine instructions to evaluate the root into a register using no more than r registers, we assume they are R1, R2, ..., Rr.
Method: We apply a recursive algorithm, starting from the root of the tree with the base b = 1 and work on each side of the trees separately then store the result in a large subtree, this is for interior nodes with label k > r.
This result is brought back from memory before node N is evaluated and the final step takes place in registers
{\mathrm{R}}_{\mathrm{r-1}}
{\mathrm{R}}_{\mathrm{r}}
Node N has at least one child with label r ≥ 2 therefore we pick the larger child as the big child and let the other child be little child
We recursively generate code for the selected big child using base b = 1 and store this result in register
{\mathrm{R}}_{\mathrm{r}}
Next, we generate the machine instruction ST
{\mathrm{t}}_{\mathrm{k}}
{\mathrm{R}}_{\mathrm{r}}
{\mathrm{t}}_{\mathrm{k}}
is a temporary variable that stores temporary results that assists in the evaluation of label k.
Now to generate code for the little child, if it has a label r or greater, pick base = 1, if the label of the child is j < r, pick b = r - j and recursively apply this algorithm to the little child, the result will appear in
{\mathrm{R}}_{\mathrm{r}}
Generate the instruction LD
{\mathrm{R}}_{\mathrm{r-1}}
{\mathrm{t}}_{\mathrm{k}}
If the big child is the right child of N, generate the instruction OP
{\mathrm{R}}_{\mathrm{r}}
{\mathrm{R}}_{\mathrm{r}}
{\mathrm{R}}_{\mathrm{r-1}}
. If the big child is left child generate OP
{\mathrm{R}}_{\mathrm{r}}
{\mathrm{R}}_{\mathrm{r-1}}
{\mathrm{R}}_{\mathrm{r}}
The above algorithm produces optimal code from an expression tree using time which is a linear function of the size of the tree.
This works for machines whereby all computations are done in registers whereby instructions consist of operators applied in two registers or a register and memory location.
A dynamic programming algorithm is used to extend the class of machines for which optimal code can be generated from expressions trees in linear time. This algorithm works for a broad class of register machines with complex instruction sets.
Such an algorithm is used to generate code for machines with r interchangeable registers R0, R1, ...,
{\mathrm{R}}_{\mathrm{r-1}}
and load, store and operation instructions.
Throughout this article, we will assume that every instruction has a cost of one unit however a dynamic programming algorithm works even with instructions having their own costs.
The algorithm divides the problem of generating optimal code for an expression into subproblems of generating optimal code for subexpressions of the given expression.
For example consider an expression E in the form of
{\mathrm{E}}_{\mathrm{1}}
{\mathrm{E}}_{\mathrm{2}}
To obtain an optimal program of E, we combine optimal programs for
{\mathrm{E}}_{\mathrm{1}}
{\mathrm{E}}_{\mathrm{2}}
followed by the code to evaluate the + operator.
Subproblems for generating code for
{\mathrm{E}}_{\mathrm{1}}
{\mathrm{E}}_{\mathrm{2}}
are also solved in a similar manner.
The program produced by the dynamic programming algorithm has the following property, it evaluates expression E =
{\mathrm{E}}_{\mathrm{1}}
{\mathrm{E}}_{\mathrm{2}}
"contiguously".
Let's look at the syntax tree T for E to have a better idea:
{\mathrm{T}}_{\mathrm{1}}
{\mathrm{T}}_{\mathrm{2}}
are trees for
{\mathrm{E}}_{\mathrm{1}}
{\mathrm{E}}_{\mathrm{2}}
We say that a program P evaluates tree T contiguously if it first, evaluates the subtrees of T that need to be computed in memory followed by evaluating the remainder of T by using previously computed values from memory.
Let's look at an example of a noncontiguous evaluation:
P first evaluates part of
{\mathrm{T}}_{\mathrm{1}}
leaving the value in a register, next it evaluates
{\mathrm{T}}_{\mathrm{2}}
then it returns to evaluate T1.
For the register machine in this article, we prove that for any given machine-language program P that evaluates an expression tree T, there is an equivalent program P' such that;
It's cost is less than that of P.
It uses fewer registers than P.
P' evaluates the tree contiguously.
The above implies that every expression tree can be evaluated optimally by a contiguous program.
The contiguous evaluation property defined here above ensures for any expression tree T, there exists an optimal program consisting of optimal programs for subtrees of the root followed by an instruction to evaluate the root.
Such a property will allow us to use a dynamic programming algorithm to generate an optimal program for T.
The algorithm has three phases, we assume the machine has three registers;
First we compute bottom-up for each node n of the expression tree T an array C of costs whereby its ith component C[i] is the optimal cost of computing the subtree S rooted at n into a register, assuming i registers are available for the computation for 1 ≤ i ≤ r.
Traverse T using the cost vectors to determine subtrees of T that must be computed into memory.
Traverse each tree using the cost vectors and associate instructions to generate the target code. The code for the subtrees that are computed into memory locations is generated first.
Each phase is implemented to run in time linearly proportional to the size of the tree.
The cost of computing a node n includes loads and stores used to evaluate S in the given number of registers and the cost of computing the operator at the root of S.
The 0th component of the cost vector is the optimal cost of computing the subtree S into memory.
The contiguous evaluation property ensures that an optimal program for S is generated by considering combinations of optimal programs only for subtrees of the root of S.
This restriction reduces the number of cases that need to be considered.
To compute the costs C[i] at node n, we look at the instructions as tree rewriting rules.
Now we consider each template E matching the input tree at node n and by examining the cost vectors at the corresponding descendants of n we determine the cost of evaluating the operands at the leaves of E.
For those operands, we consider all possible orders in which the corresponding subtrees of T can be evaluated into registers.
For each ordering, the first subtree that corresponds to a register operand is evaluated using i available registers, the second is evaluated using i-1 registers, and so on.
To account for node n, we add the cost of the instruction associated with template E.
As mentioned previously, the cost vectors for the whole tree are computed bottom-up in time linearly proportional to the number of nodes in T.
It is therefore convenient to store the instruction used to achieve the best cost for C[i] for each i at each node.
The smallest cost in the vector for the root of T yields the minimum cost of evaluating T.
We have a machine with two registers R0 and R1 and the following instructions each of unit cost.
In the instructions Ri is either R0 or R1 and Mj is a memory location.
op represents any arithmetic operator.
We also have the following syntax tree.
Now we apply a dynamic programming algorithm to generate optimal code for the above tree.
The first phase involves computing the cost vectors at each node.
We illustrate this cost computation by considering the cost vector at leaf a.
C[0], is the cost of computing a into memory, its value is 0 since it is already there.
C[1] is the cost of computing a into a register, its value is 1 since we can load it into a register with the instruction ld Ro, a.
C[2] is the cost of loading a into a register with two registers available, it is similar to the one with a single available register.
The cost vector at leaf a is (0, 1, 1) as can be seen from the tree.
Now we consider the cost vector at the root. First, we determine the minimum cost of computing the root with one and two registers available.
The instructions ADD RO, RO, M match the root since it is labeled with the + operator.
Using this instruction, the minimum cost of evaluating the root with a single register is the minimum cost of computing its right subtree into memory plus the minimum cost of computing its left subtree into the register plus 1 for the instruction itself.
The cost vectors at the left and right children of the root show that the minimum cost of computing the root with one register available is 5 + 2 + 1 = 8.
Looking at the minimum cost of evaluating the root with two registers available, we have three cases depending on the instructions used to compute the root and the order in which the left and right subtrees are evaluated.
The first case is to compute the left subtree with two registers available into register RO, compute the right subtree with one register available into register R1 the use the instructions ADD RO, RO, RO to compute the root. The cost is 2 + 5 + 1 = 8.
The second case is to compute the right subtree with two registers available into R1, compute the left subtree with a single register available into RO the use the instruction ADD RO, RO, RO. The cost is 4 + 2 + 1 = 7.
The third case is to compute the right subtree into a memory location M, compute the left subtree with two registers into register RO and use the instruction ADD RO, RO, M. This has a cost of 5 + 2 + 1 = 8.
The minimum cost of computing the root into memory is determined by adding 1 to the minimum cost of computing the root with all registers available, i.e, we compute the root into a register then store the result. The cost vector is at the root is (8, 8, 7).
From the cost vector, we construct the code sequence by traversing the tree and we have the following optimal code sequence.
Code generation is the final phase of a compiler .
DP has been used in various compilers e.g PCC2, its techniques facilitate re-targeting since its applicable to a broad class of machines.
A re-targetable compiler can generate code for multiple instruction sets.
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Building syntax trees.
The structure of a type.
The main application of Syntax-Directed Translation is in the construction of syntax trees. Compilers use syntax trees as an intermediate representation, using a common form of Syntax-Directed Definitions, the input string is converted into a tree.
The compiler then traverses the tree using rules that are in effect an SDD on the syntax tree rather than the parse tree.
We will look at two SDDs used to construct syntax trees for expressions. The first is S-Attributed and is suitable in cases we perform bottom-up parsing and the second is L-Attributed and is suitable during top-down parsing. The third is L-Attributed dealing with basic array types.
Every node in a syntax tree represents a construct and the children of the node represent meaningful components of the construct.
Given a syntax tree that represents an expression E1 + E2. We have the label + and two children that represent the subexpressions.
To implement the nodes of a syntax tree we use objects with several fields. Every object will have an op field which will act as its label.
In addition to the op field, the object will also have a field that holds a lexical value for a leaf node. That is if the node is at all a leaf nod
Also, if the node is an interior node, the number of fields will be proportional to the number of children it has in the syntax tree. Here we use a constructor function Node that takes two or more arguments then creates an object with the first field op and k additional fields for the k children c1, ..., ck.
An example - S-Attributed.
We have the following S-Attributed definition. It constructs syntax trees for a simple expression grammar involving binary operators.
These operators have the same precedence level and are jointly left-associative. Non-terminals have a synthesized attribute node that represents a node in the syntax tree.
Each time the production E -> E1 + T is used, its corresponding rule creates a node with + for op and two children these are, E1.node, and T.node for the subexpression. The second production will also have a similar rule
For the third production, no node is created. This is because E.node is similar t T.node. Similarly, no node is created for the fourth production.
The value of T.node is similar to E.node and since we only use parenthesis to group them, they influence the structure of the parse tree and syntax tree.
The last final two productions have a single terminal to the right. The constructor Leaf is used to create a suitable node that will become the value of T.node.
The following is a syntax tree for an input a - 4 + x;
Steps involved in the construction of the above syntax tree - image 3
Nodes in the syntax tree are shown as records, they have the op field as their first field. Edges of the syntax tree are solid lines.
A parse tree is shown with dotted edges even though its construction is unnecessary.
We also have dashed lines, these represent the values of E.node and T.node. Each line points to the appropriate node in the syntax tree.
We also see leaves a, 4, and c constructed by Leaf. We assume that the lexical value of id.entry points to the symbol table, and the lexical value of num.val is a numerical value of a constant. Leaves or pointers to them become the value of T.node at the parse tree nodes that are labeled T, this is according to the fifth and sixth rules.
By the third rule, the pointer to the leaf for a is also the value of E.node for the leftmost E in the tree.
By the second rule, we create a node with op equal to the minus sign and pointers to the first two leaves. And, by the first rule, we produce a root node of the syntax tree by combining the node for the minus sign with the third leaf.
If rules are evaluated during a postorder traversal of the parse tree or with reductions during a bottom-up parse, then the sequence of steps from image 3 results int p5 pointing to the root of the built syntax tree.
Given a grammar that was designed for top-down parsing, a similar syntax tree is constructed using a similar sequence of steps even if the structure of the parse tree significantly differs from syntax trees.
An example. L-Attributed
We have the following L-attributed definition that performs a similar translation as the S-Attributed definition.
The idea is to build a syntax tree for x + y by passing x as an inherited attribute, this is because x and +y live in different subtrees. A non-terminal E' is the counterpart of a non-terminal T' as we shall come to see.
Non-terminal E' has an inherited attribute inh and a synthesized one syn. E'.inh represents the partial syntax tree that is constructed so far.
It represents the root of the tree for the prefix of the input string to the left of the subtree for E'.
Looking at node 5 in the graph above, E'.inh denotes the root of the partial syntax tree for the identifier a, that is, the leaf for a.
At node 6, E'.inh denotes the root for the partial syntax tree for an input a - 4.
At node 9, E'.inh points to the root of the whole syntax tree.
syn attributes pass this value up the parse tree until it becomes the value of E.node. In other words, the attribute value at node 10 is defined by rule E'.syn = E'.inh that is associated with the production E'
\to ϵ
The value of the attribute at node 11 is defined by the rule E'.syn =
{\mathrm{E}}_{1}^{\text{'}}
.syn that is associated with production 2 from IMAHE HERE (5.13)
Also note that similar rule also define the values of attributes at nodes 12 and 13.
When the structure of a parse tree is different from the abstract syntax of the input, inherited attributes are useful since they can carry information from one part of the tree to another.
In the following example, we see how a mismatch in a structure can be caused by the language design and not constraints on the parsing method.
In the C-programming language, type int[2][3] can be read as 'array of 2 arrays of 3 integers'. The type expression array(2 array(3, integer)), int[2][3] is shown below; - image 6
The array operator takes two parameters, the first a number and the second a type. If types are represented by trees, then this operator will return a tree node that is labeled with two children for a number and type.
We have the following SDD;
Non-terminal T generates a basic type or an array type. Non-terminal B generates one of the basic types int and float. T generates a basic type when it derives BC and derives
ϵ
. Otherwise C generates array components that consist of a sequence of integers each surrounded by a pair of brackets.
Non-terminals B and T synthesize attribute t that represents a type. Non-terminal C has two attributes, the first an inherited attribute b and a synthesized attribute t. Inherited b attributes pass a basic type down the tree and the synthesized t attributes accumulate results.
We have the following annotated syntax tree for an input string int[2][3]
The corresponding type expression in *image 6 *is constructed by passing the type integer from B down the Cs chain through the attributes of t.
In detail at the root of T
\to
BC, nonterminal C inherits type from B using an inherited attribute C.b. At the rightmost node for C, the production is C
\to ϵ
therefore, C.t is equal to C.b.
Semantic rules for C
\to
[num]C1 for C.t by applying the operator array to the operands num.val and C1.t.
Compilers use syntax trees as intermediate representations, using a common form of Syntax-Directed Definitions, the input string is converted into a tree.
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Which of the following r chemical change Why 1 Biogas is produced by decomposition of plant and animal waste - Science - Physical and Chemical Changes - 9498401 | Meritnation.com
Which of the following r chemical change?Why?
1.Biogas is produced by decomposition of plant and animal waste by anaerobic bacteria
2. Biogas is burned as fuel
\mathrm{Dear} \mathrm{user},\phantom{\rule{0ex}{0ex}}\mathrm{Chemical} \mathrm{Change} :\phantom{\rule{0ex}{0ex}}\mathrm{Chemical} \mathrm{change} \mathrm{is},\mathrm{as} \mathrm{the} \mathrm{name} \mathrm{suggests} \mathrm{change} \mathrm{in} \mathrm{the} \mathrm{chemical} \mathrm{state}\phantom{\rule{0ex}{0ex}}\mathrm{or} \mathrm{change} \mathrm{in} \mathrm{chemical} \mathrm{composition} \mathrm{or} \mathrm{formation} \mathrm{of} \mathrm{a} \mathrm{totally} \mathrm{new}\phantom{\rule{0ex}{0ex}}\mathrm{substance} \mathrm{from} \mathrm{an} \mathrm{existing} \mathrm{compound}.\mathrm{Talking} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{molecula},\mathrm{it} \mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{actually} \mathrm{formation} \mathrm{or} \mathrm{breakdown} \mathrm{of} \mathrm{bonds} \mathrm{between} \mathrm{atoms}. \phantom{\rule{0ex}{0ex}}\mathrm{For} \mathrm{example}:\mathrm{rusting} \mathrm{of} \mathrm{iron}\phantom{\rule{0ex}{0ex}} \mathrm{gasoline} \mathrm{burning} \left(\mathrm{water} \mathrm{vapor} \mathrm{and} \mathrm{carbon} \mathrm{dioxide} \mathrm{form}\right),\mathrm{biogas}\phantom{\rule{0ex}{0ex}}\left(\mathrm{breakdown} \mathrm{of} \mathrm{biodegradable} \mathrm{waste}\right).\phantom{\rule{0ex}{0ex}}\mathrm{Therefore} \mathrm{both} \mathrm{are} \mathrm{chemical} \mathrm{change}.
Nitin Kumar answered this
IS THE CHEMICAL CHANGE
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Evaluate the following integrals. \int e^{3-4x}dx
\int {e}^{3-4x}dx
Bubich13
Given integral:
\int {e}^{3-4x}dx
Substitute t=3-4x and differentiate it w.r.t "x"
\frac{dt}{dx}=\frac{d}{dx}\left(3-4x\right)
\frac{dt}{dx}=-4
\frac{-dt}{4}=dx
Therefore, the given integral becomes,
\frac{-1}{4}\int {e}^{t}dt
\int {e}^{x}dx={e}^{x}+c
Where, "c" is integration constant.
⇒\frac{-1}{4}{e}^{t}+c
Substitute t=3-4x in above equation, we get
⇒\frac{-1}{4}{e}^{\left(3-4x\right)}+c
⇒\frac{-{e}^{\left(3-4x\right)}}{4}+c
\int {e}^{3-4x}dx
\int -\frac{{e}^{t}}{4}dt
-\frac{1}{4}\cdot \int {e}^{t}dt
-\frac{1}{4}{e}^{t}
-\frac{1}{4}{e}^{3-4x}
-\frac{{e}^{3-4x}}{4}
-\frac{{e}^{3-4x}}{4}+C
\begin{array}{}\int {e}^{3-4x}dx\\ =-\frac{1}{4}\int {e}^{u}du\\ \int {e}^{u}du\\ ={e}^{u}\\ -\frac{1}{4}\int {e}^{u}du\\ =-\frac{{e}^{u}}{4}\\ u=3-4x:\\ =-\frac{{e}^{3-4x}}{4}\\ Solution:\\ =-\frac{{e}^{3-4x}}{4}+C\end{array}
{\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx
\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz
\left({x}^{2}\right)\left({e}^{y}\right)\frac{dy}{dx}=4
How do you solve for xy'-y=3xy given y(1)=0?
In terms of integration, how would you obtain the "exact-value" of
{\int }_{0}^{\frac{\pi }{4}}\left(\mathrm{sec}x-x\right)\left(\mathrm{sec}x+x\right)dx.
1+{\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x
What is a general solution to the differential equation y'-3y=5?
\frac{dy}{dx}=\frac{2x}{{\left(y+{x}^{2}y\right)}^{2}}
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Existence of Solutions to a Viscous Thin Film Equation
Yue Qiu1,2, Bo Liang3
1Foundation Building, 765 Brownlow Hill, University of Liverpool, Liverpool, UK
2Foundation Building, Xi’an Jiaotong-Liverpool University, Suzhou, China
3School of Science, Dalian Jiaotong University, Dalian, China
A fourth-order degenerate parabolic equation with a viscous term:
\{\begin{array}{l}{u}_{t}-{\left(m\left(u\right){w}_{x}\right)}_{x}=0\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\left(-1,1\right)\times \left(0,T\right),\\ w=-{u}_{xx}+\nu {u}_{t}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\left(-1,1\right))\times \left(0,T\right)\end{array}
is studied with the initial-boundary conditions
{u}_{x}={w}_{x}=0
\left\{-1,1\right\}\times \left(0,T\right)
u\left(x,0\right)={u}_{0}\left(x\right)
\left(-1,1\right)
. It can be taken as a thin film equation or a Cahn-Hilliard equation with a degenerate mobility. The entropy functional method is introduced to overcome the difficulties that arise from the degenerate mobility
m\left(u\right)
and the viscosity term. The existence of nonnegative weak solution is obtained.
Fourth-Order Degenerate Parabolic, Thin Film Equation, Cahn-Hilliard Equation, Entropy Functional
In recent years, the research of nonlinear fourth-order degenerate parabolic equations has become an interesting topic. The typical examples include the Cahn-Hilliard equation and the thin film equation. The Cahn-Hilliard equation can describe the evolution of a conserved concentration field during phase separation. It (see [1] ) has the form
{u}_{t}+\nabla \cdot \left(k\nabla \left({\epsilon }^{2}\Delta u+{A}^{\prime }\left(u\right)\right)\right)=0
where the constants k, A,
{\epsilon }^{2}
denote the atomic mobility, the free energy, the parameter proportional to the interface energy respectively and
-\left({\epsilon }^{2}\Delta u+{A}^{\prime }\left(u\right)\right)
is a kind of chemical potential. For the existence and the properties of solutions, Elliott, Zheng and Garcke (see [2] [3] ) have studied this equation with a linear and a degenerate mobility respectively. Xu, Zhou, Liang and Zheng (see [4] [5] [6] ) have applied the semi-discrete method to obtain the existence and stability results to this model with a gradient mobility.
The thin film equation can analyze the motion of a very thin layer of viscous incompressible fluids along an inclined plane or model the fluid flows such as draining of foams and the movement of contact lenses. The thin film equation belongs to a class of fourth order degenerate parabolic equations (see [7] ) and the first mathematic result, the existence and nonnegativity of weak solutions, are given by Bernis and Friedman [8] to the equation
{u}_{t}+{\left({u}^{n}{u}_{xxx}\right)}_{x}=0
. The thin film equation with a second-order diffusion term was studied by Bertozzi and Pugh [9] . Moreover, for a generalized thin-film equation with period boundary in multidimensional space, Boutat et al. [10] obtained its existence. For other results, the readers may refer to the papers [11] [12] .
In this paper, we study the following initial and boundary value problems for the viscous thin film equation:
\left\{\begin{array}{l}{u}_{t}-{\left(m\left(u\right){w}_{x}\right)}_{x}=0\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{ }{Q}_{T},\\ w=-{u}_{xx}+\nu {u}_{t}\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{ }{Q}_{T},\\ {u}_{x}={w}_{x}=0\text{ }\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{ }\Gamma ,\\ u\left(x,0\right)={u}_{0}\left(x\right),\end{array}
T>0
m\left(u\right)=u
\Omega =\left(-1,1\right)
{Q}_{T}=\Omega \times \left(0,T\right)
\Gamma =\partial \Omega \times \left(0,T\right)
Formally, if we substitute the second equation into the first one, we can get another form for this question:
\left\{\begin{array}{l}{u}_{t}+{\left(m\left(u\right){\left({u}_{xx}-\nu {u}_{t}\right)}_{x}\right)}_{x}=0\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{ }{Q}_{T},\\ {u}_{x}={u}_{xxx}=0\text{ }\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{ }\Gamma ,\\ u\left(x,0\right)={u}_{0}\left(x\right).\end{array}
Our main result is the following theorems.
{u}_{0}\in {L}^{2}\left(\Omega \right)
\nu >0
. Then there exists at least one pair
\left(u,w\right)
of (1) satisfying
u\in {L}^{\infty }\left(0,T;{H}^{1}\left(\Omega \right)\right)\cap {L}^{2}\left(0,T;{H}^{2}\left(\Omega \right)\right)\cap C\left(\left[0,T\right];{L}^{2}\left(\Omega \right)\right)
w\in {L}^{2}\left(0,T;{H}^{1}\left(\Omega \right)\right)
{u}_{t}\in {L}^{2}\left({Q}_{T}\right)
2) For any test function
\varphi \in {L}^{2}\left(0,T;{H}^{1}\left(\Omega \right)\right)
, it has
{\iint }_{{Q}_{T}}{u}_{t}\varphi \text{d}x\text{d}t+{\iint }_{{Q}_{T}}u{w}_{x}{\varphi }_{x}\text{d}x\text{d}t=0,
{\iint }_{{Q}_{T}}w\varphi \text{d}x\text{d}t=-{\iint }_{{Q}_{T}}{u}_{xx}\varphi \text{d}x\text{d}t+\nu {\iint }_{{Q}_{T}}{u}_{t}\varphi \text{d}x\text{d}t.
u\left(x,0\right)={u}_{0}\left(x\right)
{u}_{0}\in {L}^{2}\left(\Omega \right)
\nu >0
\left(u,w\right)
u\in {L}^{\infty }\left(0,T;{H}^{1}\left(\Omega \right)\right)\cap {L}^{2}\left(0,T;{H}^{2}\left(\Omega \right)\right)\cap C\left(\left[0,T\right];{L}^{2}\left(\Omega \right)\right)
{u}_{t}\in {L}^{2}\left({Q}_{T}\right)
\varphi \in {L}^{2}\left(0,T;{H}^{2}\left(\Omega \right)\right)
{\varphi }_{x}\left(-1,t\right)={\varphi }_{x}\left(1,t\right)=0
{\iint }_{{Q}_{T}}{u}_{t}\varphi \text{d}x\text{d}t+{\iint }_{{Q}_{T}}{u}_{xx}{u}_{x}{\varphi }_{x}\text{d}x\text{d}t-\nu {\iint }_{{Q}_{T}}u{u}_{t}{\varphi }_{xx}\text{d}x\text{d}t=0.
u\left(x,0\right)={u}_{0}\left(x\right)
The following lemmas are needed in the paper:
Lemma 1. (Aubin-Lions, see [13] ) Let X, B and Y be Banach spaces and assume
X\to B\to Y
with compact imbedding
X\to B
\mathfrak{F}
be bounded in
{L}^{p}\left(0,T;X\right)
1\le p<\infty
\frac{\partial \mathfrak{F}}{\partial t}=\left\{\frac{\partial f}{\partial t}:f\in \mathfrak{F}\right\}
{L}^{1}\left(0,T;Y\right)
\mathfrak{F}
{L}^{p}\left(0,T;B\right)
\mathfrak{F}
{L}^{\infty }\left(0,T;X\right)
\frac{\partial \mathfrak{F}}{\partial t}=\left\{\frac{\partial f}{\partial t}:f\in \mathfrak{F}\right\}
{L}^{r}\left(0,T;Y\right)
r>1
\mathfrak{F}
C\left(\left[0,T\right];B\right)
Lemma 2. (see [14] or [15] ) Let V be a real, separable, reflexive Banach space and H is a real, separable, Hilbert space.
V\to H
is continuous and V is dense in H. Then
\left\{u\in {L}^{2}\left(0,T;V\right)|{u}_{t}\in {L}^{2}\left(0,T;{V}^{\prime }\right)\right\}
is continuously imbedded in
C\left(\left[0,T\right];H\right)
In this paper, C is denoted as a positive constant and may change from line to line. The paper is arranged as follows. The existence of solutions to the approximate problem will be proved in Section 2. In Section 3, we will take the limit for small parameters
\delta \to 0
2. Approximate Problem
0<\delta <1
, we consider the following approximate problem. In order to apply existence theory better, we transform (1) into a system:
\left\{\begin{array}{l}{u}_{\delta t}-{\left({m}_{\delta }\left({u}_{\delta }\right){w}_{\delta x}\right)}_{x}=0\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{ }{Q}_{T},\\ {w}_{\delta }=-{u}_{\delta xx}+\nu {u}_{\delta t}\text{ }\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{ }{Q}_{T},\\ {u}_{\delta x}={w}_{\delta x}=0\text{ }\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{ }\Gamma ,\\ {u}_{\delta }\left(x,0\right)={u}_{\delta 0}\left(x\right)\end{array}
{u}_{\delta 0}\left(x\right)={u}_{0}\left(x\right)+\delta
{m}_{\delta }\left({u}_{\delta }\right)={u}_{\delta +}+\delta
{u}_{\delta +}=\mathrm{max}\left\{{u}_{\delta },0\right\}
Lemma 3. There exists at least one solution
{u}_{\delta }
to (3) satisfying
{w}_{\delta }\in {L}^{2}\left(0,T;{H}^{1}\left(\Omega \right)\right)
{u}_{\delta }\in {L}^{2}\left(0,T;{H}^{2}\left(\Omega \right)\right)\cap {L}^{\infty }\left(0,T;{H}^{1}\left(\Omega \right)\right)\cap C\left(\left[0,T\right];{L}^{2}\left(\Omega \right)\right)
{u}_{\delta t}\in {L}^{2}\left({Q}_{T}\right)
{u}_{\delta }\left(x,0\right)={u}_{\delta 0}
\varphi \in {L}^{2}\left(0,T;{H}^{1}\left(\Omega \right)\right)
{\iint }_{{Q}_{T}}{u}_{\delta t}\varphi \text{d}x\text{d}t+{\iint }_{{Q}_{T}}{m}_{\delta }\left({u}_{\delta }\right){w}_{\delta x}{\varphi }_{x}\text{d}x\text{d}t=0,
{\iint }_{{Q}_{T}}{w}_{\delta }\varphi \text{d}x\text{d}t=-{\iint }_{{Q}_{T}}{u}_{\delta xx}\varphi \text{d}x\text{d}t+\nu {\iint }_{{Q}_{T}}{u}_{\delta t}\varphi \text{d}x\text{d}t.
Proof. We apply the Galerkin method to prove this Lemma and so we choose
{\left\{{\varphi }_{i}\right\}}_{i=1,2,3,\cdots }
as the eigenfunctions of the Laplace operator with Neumann boundary value conditions such that
-{\varphi }_{ixx}={\lambda }_{i}{\varphi }_{i}
. Moreover, we can suppose that the eigenfunctions are orthogonal in the H1 and L2 spaces. We use
\left(\cdot ,\cdot \right)
to denote the scalar product in L2 space and we can normalize
{\varphi }_{i}
\left({\varphi }_{i},{\varphi }_{j}\right)={\delta }_{ij}=\left\{\begin{array}{l}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{ }i=j,\\ 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }i\ne j.\end{array}
Besides, we can choose
{\lambda }_{1}=0
{\varphi }_{1}=1
For any positive integer M, we define
{u}_{\delta }^{M}\left(x,t\right)=\sum _{i=1}^{M}\text{ }{c}_{i}\left(t\right){\varphi }_{i}\left(x\right)
{u}_{\delta }^{M}\left(x,0\right)=\sum _{i=1}^{M}\left({u}_{0},{\varphi }_{i}\right){\varphi }_{i}
{w}_{\delta }^{M}\left(x,t\right)=\sum _{i=1}^{M}\text{ }{d}_{i}\left(t\right){\varphi }_{i}\left(x\right)
. Now we consider the following ordinary differential equations system:
\frac{\text{d}}{\text{d}t}\left({u}_{\delta }^{M},{\varphi }_{j}\right)=-\left({m}_{\delta }\left({u}_{\delta }^{M}\right){w}_{\delta x}^{M},{\varphi }_{jx}\right),
\left({w}_{\delta }^{M},{\varphi }_{j}\right)=-\left({u}_{\delta xx}^{M},{\varphi }_{j}\right)+\nu \frac{\text{d}}{\text{d}t}\left({u}_{\delta }^{M},{\varphi }_{j}\right),
for , which yields an initial value problem for the ordinary differential equations:
with . A standard argument can show that this ODE has a local solution by Peano existence theorem since the matrix is positive definite. In order to get the global solvability, we need establish more energy estimates. Multiply (4) by to get
Taking as the test function in (5), we have
Therefore, for any , it has
Since by (4) with , we can apply Poincaré’s inequality to obtain the following estimates:
By taking as the test function in (5), we have
By integrating over and applying the Höler's inequality, we have
There exists a subsequence of and a pair such that, as ,
where the last estimate is from Lemma 1. By (13)-(17), we can perform the limit in a standard fashion and the strong convergence in implies .
In the section, we will perform the limit to the solutions from Lemma 3. For the purpose of the existence, we need establish some uniform estimates independent of . Thus, we define a convex function as following (see [10] ):
Moreover, the function satisfies , , .
By applying this function, we can get the following estimates.
Lemma 4. There exist some constants C independent of such that
Proof. Taking as a test function in the first equation of (3), we have
Thus, it yields the results 1 - 3. We can prove 4 and 5 from (8). By choosing as a test function in the second equation of (3), we get
We have completed the proof of this lemma.
Lemma 5. There exists a pair such that, as ,
3) and a.e. in ;
5) a.e. in .
Proof. By Lemma 4, we can get the results 1 - 2 and 4 directly. Lemma 1 yields 3. By applying the definition of and (18), we get
Letting , we obtain which completes the proof of 6.
Proof of Theorem 1 and Theorem 2. Taking as a test function in Lemma 3, we have
which yields Theorem 1.
On the other hand, by integrating by parts, it implies
Thus, it has
It gives Theorem 2.
Through this paper, two forms of a viscous thin film equation are studied (see the Equations (1) and (2)) and we give the corresponding existence theorems of weak solutions (see Theorem 1 and Theorem 2). For any test function , we have proved that the weak solutions satisfy the equalities:
Since the thin film equation is a degenerate parabolic equation, it is hard to give the existence of strong solutions. On the another hand, the viscous term affects the regularity of solutions and we have shown that .
We can expect that we can show that the existence results would be true with some conditions in high-dimensional space.
The work was supported by the Education Department Science Foundation of Liaoning Province of China (No. JDL2016029) and the Natural Science Fund of Liaoning Province of China (No. 20170540136).
Qiu, Y. and Liang, B. (2018) Existence of Solutions to a Viscous Thin Film Equation. Journal of Applied Mathematics and Physics, 6, 2119-2126. https://doi.org/10.4236/jamp.2018.610178
1. Cahn, J.M. and Hilliard, J.E. (1958) Free Energy of a Non-Uniform System I. Interfacial Free Energy. The Journal of Chemical Physics, 28, 258-367. https://doi.org/10.1063/1.1744102
2. Elliott, C.M. and Zheng, S. (1986) On the Cahn Hilliard Equation. Archive for Rational Mechanics and Analysis, 96, 339-357. https://doi.org/10.1007/BF00251803
3. Elliott, C.M. and Garcke, H. (1996) On the Cahn Hilliard Equation with Degenerate Mobility. SIAM Journal on Mathematical Analysis, 27, 404-423. https://doi.org/10.1137/S0036141094267662
4. Xu, M. and Zhou, S. (2005) Existence and Uniqueness of Weak Solutions for a Generalized Thin Film Equation. Nonlinear Analysis: Theory, Methods & Applications, 60, 755-774. https://doi.org/10.1016/j.na.2004.01.013
5. Xu, M. and Zhou, S. (2008) Stability and Regularity of Weak Solutions for a Generalized Thin Film Equation. Journal of Mathematical Analysis and Applications, 337, 49-60. https://doi.org/10.1016/j.jmaa.2007.03.075
6. Liang, B. and Zheng, S. (2008) Existence and Asymptotic Behavior of Solutions to a Nonlinear Parabolic Equation of Fourth Order. Journal of Mathematical Analysis and Applications, 348, 234-243. https://doi.org/10.1016/j.jmaa.2008.07.022
7. Myers, T.G. (1998) Thin Films with High Surface Tension. SIAM Reviews, 40, 441-462. https://doi.org/10.1137/S003614459529284X
8. Bernis, F. and Friedman, A. (1990) Higher Order Nonlinear Degenerate Parabolic Equations. Journal of Differential Equations, 83, 179-206. https://doi.org/10.1016/0022-0396(90)90074-Y
9. Bertozzi, A.L. and Pugh, M. (1994) The Lubrication Approximation for Thin Viscous Films. Nonlinearity, 7, 1535-1564. https://doi.org/10.1088/0951-7715/7/6/002
10. Boutat, M., Hilout, S., Rakotoson, J.E., et al. (2008) A Generalized Thin-Film Equation in Multidimensional Space. Nonlinear Analysis, 69, 1268-1286. https://doi.org/10.1016/j.na.2007.06.028
11. Ansini, L. and Giacomelli, L. (2004) Doubly Nonlinear Thin-Film Equations in One Space Dimension. Archive for Rational Mechanics and Analysis, 173, 89-131. https://doi.org/10.1007/s00205-004-0313-x
12. Beretta, E., Bertsch, M. and Dal Passo, R. (1995) Nonnegative Solutions of a Fourth-Order Nonlinear Degenerate Parabolic Equation. Archive for Rational Mechanics and Analysis, 129, 175-200. https://doi.org/10.1007/BF00379920
13. Simon, J. (1987) Compact Sets in the Space . Annali di Matematica Pura ed Applicata, 146, 65-96. https://doi.org/10.1007/BF01762360
14. Grün, G. (1995) Degenerate Parabolic Differential Equations of Fourth Order and a Plasticity Model with Nonlocal Hardening. Zeitschrift für Analysis und Ihre Anwendungen, 14, 541-574. https://doi.org/10.4171/ZAA/639
15. Zeidler, E. (1997) Nonlinear Functional Analysis and Its Applications. Springer, New York.
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Algebra/Linear Equations and Functions - Wikibooks, open books for an open world
Algebra/Linear Equations and Functions
← The Coordinate (Cartesian) Plane Linear Equations and Functions Intercepts →
1 What are Linear Equations?
2 '"`UNIQ--postMath-0000000E-QINU`"'
3 '"`UNIQ--postMath-0000000F-QINU`"'
What are Linear Equations?[edit | edit source]
In the functions section we talked about how a function is like a box that takes an independent input value and uses a rule defined mathematically to create a unique output value. The value for the output is dependent on the value that is put in the box. We call the values that are going into the box the independent values or the domain. We call the values coming out of the box the dependent values or the range.
Unless we specify differently, on Cartesian graphs the domain is the real numbers. In the Cartesian Plane section we saw how running different values through a function to identify the points on the Cartesian plane by picking the first (x) value of the point the domain and the second (y) value from the range. To restate this: by convention the two variables for a function on the Cartesian Plane are x for the domain, the independent variable, and y for the range, the dependent variable. The variable y is the same as writing f(x). Mathematicians recognize this equivalence but generally prefer to write y because its shorter. Because a function definition has an input and an output it must also contain an equal sign. The section in this book solving equations showed the various operations we can perform on both sides of an equal sign and still maintain the notion of equivalence. In this section we plug different values into the independent variable and solve to find the associated dependent variable. For instance if we start with the equation:
{\displaystyle y=x}
We can add a -x to both sides to get the equation
{\displaystyle y-x=x-x}
which we then simplify to
{\displaystyle y-x=0}
Or we can add a -y to both sides to get the equation
{\displaystyle y-y=x-y}
{\displaystyle 0=x-y}
{\displaystyle y-x=0\equiv 0=x-y}
{\displaystyle y-x=0=x-y}
Using the transitive property
{\displaystyle y-x=x-y}
adding x + y to both sides gives us
{\displaystyle (y-x)+(y+x)=(x-y)+(y+x)}
using the associative property we change this to
{\displaystyle (y-y)+(x+x)=(x-x)+(y+y)}
{\displaystyle 2x=2y}
And divide both sides by 2
{\displaystyle 2x/2=2y/2}
To simply reverse the order and show
{\displaystyle x=y}
We have not really proved anything mathematically above, but these operations allow us to manipulate equations to get the dependent variable by itself on one side of the equals sign. Then we can plug numbers into the independent variable to discover the function values for those numbers. Then we can draw these values as points on the Cartesian plane and get a feel for what the function would look like if we could see all the points defined by the function at once.
{\displaystyle y=0x}
Equations of the form y = C2 are linear functions of the general form y = m x + b where slope m = 0 and the constant C2 is the y-intercept b (in the general form). The graph of this zero-slope function is a straight horizontal line, intercepting the y-axis at C2, including zero and extends infinitely in the positive and negative directions for all R values of x (see the following diagram).
The domain for such functions is R covering all real numbers (unless otherwise specified), but the range is just the set { c }. The equation y = 0 is the x-axis.
{\displaystyle 0y=x}
Equation x = C1, x is one single value C1 and y, being unrestricted, is every R number. The graph of x = C1 is a straight vertical line where x = C1, covering all positive, negative and zero values of y (see the following diagram).
Its domain is set { C1 } and range is set R (unless otherwise specified). x = C1 is technically not a function (there is more than one value of y for each value of x), but it's a relation. Vertical lines have no slope (m = divide by zero, undefined, plus and minus infinity). These are the only types of linear equations of the general form shown previously which are not linear functions. The equation x = 0 is exactly the y-axis. Lines with steepness approaching vertical have very large-magnitude slopes but are still functions.
We are going to start by looking at simple functions called linear equations. When none of the instances of x and y in the algebraic expression defining the function rule have exponents then all the instances of x and y can be combined into just two occurrences. the graph of the expression can be represented as a straight line. The equation that expresses the function is considered a linear equation with two variables. The following equation is a simple example of such a linear equation:
{\displaystyle y-x=2\,}
Since y is the dependent variable it is standing in for the function. We can re-write the expression as f(x) - x = 2. If we add an x to both sides the equality property holds and we get the expression f(x) - x + x = x + 2. Simplifying we get f(x) = x + 2. In the following table we'll pick 3 values for x, and then calculate the dependent (y) values from f(x).
y value (abscissa)
where x and y are variables to be plotted in a two-dimensional Cartesian coordinate graph as shown here:
This function is equivalent to the previous example of a linear equation, y - x = 2. The arrows at each end of the line indicate that the line extends infinitely in both directions. All linear functions of a single input variable have or can be algebraically arranged to have the general form:
{\displaystyle y=f(x)=mx+b\,}
where x and y are variables, f(x) is the function of x, m is a constant called the slope of the line, and b is a constant which is the ordinate of the y-intercept (i. e. the value of y where the function line crosses the y-axis). The slope indicates the steepness of the line. In the previous example where y = x + 2, the slope m = 1 and the y-intercept ordinate b = 2. The y = mx+b form of a linear function is called the slope-intercept form.
Unless a domain for x is otherwise stated, the domain for linear functions will be assumed to be all real numbers and so the lines in graphs of all linear functions extend infinitely in both directions. Also in linear functions with all real number domains, the range of a linear function will cover the entire set of real numbers for y, unless the slope m = 0 and the function equals a constant. In such cases, the range is simply the constant.
Conversely, if a function has the general form y = mx + b or if it can be arranged to have that form, the function is linear. A linear equation with two variables has or can be algebraically rearranged to have the general form1:
{\displaystyle Ax+By=C\,}
where x and y represent the linear variables, and the letters A, B, and C can represent any real constants, either positive or negative. Conversely, an equation with two variables x and y having that general form, or being able to be arranged in that form, would be linear as long as A and B are not both equal to 0. In the preceding equation, capital letters are to avoid confusion with other constants in this chapter and for consistency with Reference 1.
If one divides the preceding equation by B (when B is not 0) and solves for y, the following form can be obtained:
{\displaystyle y=(-A/B)x+(C/B)\,}
If one equates -A/B to the slope m and C/B to the y-intercept ordinate b, it can be seen that the general form for a linear equation and the slope-intercept form for a linear function are practically interconvertible except for the fact that, in a linear function, the B constant in the linear equation form cannot equal 0.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Algebra/Linear_Equations_and_Functions&oldid=3251294"
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PUCCH format 2 DRS resource element indices - MATLAB ltePUCCH2DRSIndices - MathWorks United Kingdom
ind = ltePUCCH2DRSIndices(ue,chs) returns a matrix of resource element indices for the demodulation reference signal (DRS) associated with the PUCCH format 2 transmission given structures containing the UE-specific settings, and the channel transmission configuration.
Generate the PUCCH format 2 DM-RS indices for four transmit antenna paths, and display the output information structure.
Because there are four antennas, the DM-RS indices are output as a four-column vector and the info output structure contains four elements. View ind and the size of info to confirm.
{n}_{PUCCH}^{\left(2\right)}
ltePUCCH2 | ltePUCCH2Decode | ltePUCCH2Indices | ltePUCCH2DRS | ltePUCCH2DRSDecode | ltePUCCH2PRBS | ltePUCCH1DRSIndices | ltePUCCH3DRSIndices
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Momentum Warmup Practice Problems Online | Brilliant
If a car and a bus are moving with the same momentum, which has greater kinetic energy?
The car has a much smaller mass than the bus.
Bus Car Both have equal kinetic energy
A person is falling freely under the gravity. He has a heavy briefcase in his hand. There is swimming pool just beside his vertical line of fall. If he wants to land in the swimming pool, what should he do?
He should throw his briefcase opposite to the swimming pool He should throw his briefcase vertically downwards He should throw his briefcase towards the swimming pool He should keep his briefcase with him
Rain drops fall on the ground with a speed and a momentum. They hit different objects and create sound. Suppose, 100 drops are hitting a horizontal surface of unit area per second. The speed of rain drops 15 m/s in vertically downward direction. The mass of each drop is 0.05 grams. If the rain drops come to rest immediately after hitting the surface then, find the force experienced by the horizontal surface?
0.075 N 75 N 0.75 N 7.5 N
When two objects collide, they apply forces to each other called impulse forces. Such forces act for a very small amount of time, yet produce large changes in the momentum of the colliding objects.
Let's consider a case of collision between a ball and a bat. The ball of mass
m
is approaches the batsman with a speed of
v
m/s and the bat hits the ball so that it recoils with the same speed, along the same line, but in the opposite direction.
What will be the total impulse of the bat, upon the ball?
0
\frac{mv}{2}
mv
2mv
When a ball is dropped from a height, after hitting the ground, the ball rebounds. The rebounded height is less than the original height of the drop. What can be said about the collision between the ball and the ground?
The collision is perfectly inelastic The collision is perfectly elastic The collision is inelastic
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Draw a graph between simple interest verses time when principal is Rs 4000 and the rate of interest is - Maths - Introduction to Graphs - 11075365 | Meritnation.com
Draw a graph between simple interest verses time when principal is Rs. 4000 and the rate of interest is 3%per annum
EXPERT plz solve this question in detail and plz don't send any link
\mathrm{S}.\mathrm{I}=\frac{\mathrm{P}×\mathrm{R}×\mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\mathrm{S}.\mathrm{I}=\frac{4000×3×\mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\mathrm{S}.\mathrm{I}=120 \mathrm{T} \phantom{\rule{0ex}{0ex}}\mathrm{Take} \mathrm{time} \mathrm{on} \mathrm{X}-\mathrm{axis} \mathrm{and} \mathrm{simple} \mathrm{interest} \mathrm{on} \mathrm{y}-\mathrm{axis} .\phantom{\rule{0ex}{0ex}}\mathrm{When} \mathrm{T}=0 , \mathrm{S}.\mathrm{I}=0\phantom{\rule{0ex}{0ex}}\mathrm{When} \mathrm{T}=1 , \mathrm{S}.\mathrm{I}=120\phantom{\rule{0ex}{0ex}}\mathrm{When} \mathrm{T}=2 , \mathrm{S}.\mathrm{I}=240.\phantom{\rule{0ex}{0ex}}\mathrm{Plot} \mathrm{these} \mathrm{points} \mathrm{on} \mathrm{graph}.
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In the given equation as follows, use partial fractions to
In the given equation as follows, use partial fractions to find the indefinite i
\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx
We have to evaluate the following integral using partial fractions:
\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx
=\int \left(\frac{{x}^{2}-2x+1+2x-1}{{x}^{2}-2x+1}\right)dx
=\int \left(\frac{{x}^{2}-2x+1}{{x}^{2}-2x+1}\right)dx+\int \left(\frac{2x}{{x}^{2}-2x+1}\right)dx-\int \left(\frac{1}{{x}^{2}-2x+1}\right)dx
=\int dx+\int \frac{2xdx}{{\left(x-1\right)}^{2}}-\int \frac{dx}{{\left(x-1\right)}^{2}}
=\int dx+\int \left[\frac{\left(x-1\right)+\left(x-1\right)}{{\left(x-1\right)}^{2}}\right]dx+\int \frac{2dx}{{\left(x-1\right)}^{2}}-\int \frac{dx}{{\left(x-1\right)}^{2}}
=\int dx+\int \frac{dx}{x-1}+\int \frac{dx}{x-1}+\int \frac{dx}{{\left(x-1\right)}^{2}}
=\int dx+2\int \frac{dx}{x-1}+\int \frac{dx}{{\left(x-1\right)}^{2}}
=x+2\mathrm{ln}|x-1|-\frac{1}{x-1}+C
\frac{4}{{x}^{2}-3x+2};\frac{x}{{x}^{2}-4}
\frac{x}{4}=2+\frac{x-3}{3}
can be solved by multiplying both sides by the__________ of the fractions, which is________
Among all pairs of numbers (x,y)
3x+y=8
find the pair for which the sum of squares,
{x}^{2}+{y}^{2}
is minimum. Write your answers as fractions reduced to lowest terms
Find two fractions between
\frac{1}{2}\text{ }\text{ and }\text{ }\frac{6}{7}
What is the simplified value of
\frac{3}{12}-\frac{1}{8}+\frac{5}{6}
How Do I Simplify
1-\frac{1}{{2}^{m-1}}+\frac{1}{{2}^{m}}
Someone help. Somehow my textbook says that it simplifies to
1-\frac{2}{{2}^{m}}+\frac{1}{{2}^{m}}
I don't see this at all.
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Represent the plane curve by a vector-valued function. x^{2}+y^{2}=25
inlays85k5 2021-11-23 Answered
{x}^{2}+{y}^{2}=25
memomzungup4
{x}^{2}+{y}^{2}=25
x=5\mathrm{cos}t
y=\mathrm{sin}t
{\left(5\mathrm{cos}t\right)}^{2}+{\left(5\mathrm{sin}t\right)}^{2}=25\left({\mathrm{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)\right)
=25
The function of the form
r\left(t\right)=x\left(t\right)i+y\left(t\right)j
is a vector valued function.
Substitute x and y in the above equation, to get
r\left(t\right)=\left(5\mathrm{cos}t\right)i+\left(5\mathrm{sin}t\right)j
\left(x,\text{ }y\right)=
\left(12,\frac{\pi }{8}\right),\left(-8,\frac{5\pi }{8}\right)
Find parametric equations for the line that crosses the x-axis where
x=2
and the z-axis where
z=-4
r\left(t\right)=\sqrt{{\left(4-t\right)}^{2}}i+{t}^{2}j-6t\text{ }k
Find the maximum and minimum values attained by the function f along the path
c\left(t\right)
\left(a\right)f\left(x,y\right)=xy;c\left(t\right)=\left(\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)\right);0\le t\le 2\pi
maximum value__________
minimum value__________
f\left(x,y\right)={x}^{2}+{y}^{2};c\left(t\right)=\left(\mathrm{cos}\left(t\right),8\mathrm{sin}\left(t\right)\right);0\le t\le 2\pi
Consider the following planes:
5x-4y+z=1
4x+y-5z=5
a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)
b) Find the angle between the planes. (Round your answer to one decimal place.)
\frac{{d}^{3}y}{{dx}^{3}}
for the following parametric equation:
x={t}^{2}-3
y={t}^{8}
t=2
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Show that \int_0^{\infty} \frac{\sin(t)}{t}dt=\frac{\pi}{2} by using Laplace Transform method. I know that L\{\sin(t)\}=\int_0^{\infty}
Show that NSK PSK\int_0^{\infty} \frac{\sin(t)}{t}dt=\frac{\pi}{2}ZSK NSK by using Laplace Transform method. I know that NSK PSKL\{\sin(t)\}=\int_0^{\infty}
{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(t\right)}{t}dt=\frac{\pi }{2}
L\left\{\mathrm{sin}\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{sin}\left(t\right)dt=\frac{1}{{s}^{2}+1}
L\left(\frac{\mathrm{sin}t}{t}\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}{e}^{-st}dt=\mathrm{arctan}\frac{1}{s}
EDIT: To arrive at this result, note that
I={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}{e}^{-st}dt
-\frac{\partial I}{\partial s}={\int }_{0}^{\mathrm{\infty }}\mathrm{sin}t{e}^{-st}dt
=\frac{1}{1+{s}^{2}}
{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}t}{t}dt
={\int }_{0}^{\mathrm{\infty }}{e}^{-0\cdot t}\frac{\mathrm{sin}t}{t}dt
=\underset{s⇒0}{lim}{\int }_{s}^{\mathrm{\infty }}L\left[\mathrm{sin}t\right]ds
=\underset{s⇒0}{lim}{\int }_{s}^{\mathrm{\infty }}\frac{1}{{s}^{2}+1}ds
=\underset{s⇒0}{lim}{\mathrm{tan}}^{-1}\left(s\right){\mid }_{s}^{\mathrm{\infty }}
=\underset{s⇒0}{lim}\frac{\pi }{2}-{\mathrm{tan}}^{-1}\left(s\right)
=\frac{\pi }{2}
y"+{\omega }^{2}y=\mathrm{sin}\gamma t,y\left(0\right)=0,{y}^{\prime }\left(0\right)=0
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\omega }^{2}{\right)}^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{{s}^{2}+{\omega }^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\gamma }^{2}{\right)}^{2}}\right)
y\left(t\right)={L}^{-1}\left(\frac{\gamma }{\left({s}^{2}+{\gamma }^{2}\right)\left({s}^{2}+{\omega }^{2}\right)}\right)
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
L\left\{{e}^{3t}\mathrm{sin}\left(4t\right)-{t}^{4}+{e}^{t}\right\}
\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)=\frac{1-a\mathrm{cos}x}{1-2a\mathrm{cos}x+{a}^{2}},|a|<1
can be derived by using the fact that
\sum _{k=0}^{\mathrm{\infty }}{a}^{k}\mathrm{cos}\left(kx\right)=Re\sum _{k=0}^{\mathrm{\infty }}{\left(a{e}^{ix}\right)}^{k}
But can it be derived without using complex variables?
L\left\{10{e}^{-5t}\mathrm{sin}\frac{t}{2}\right\}
L\left\{{\int }_{0}^{t}\frac{1-{e}^{u}}{u}du\right\}
f\left(t\right)=t{u}_{2}\left(t\right)
F\left(s\right)=\left(\frac{1}{{s}^{2}}+\frac{2}{s}\right){e}^{-2s}
Determine the Laplace transforms of these functions:
a\right)\text{ }f\left(t\right)=\left(t-4\right)u\left(t-2\right)
b\right)\text{ }g\left(t\right)=2{e}^{-4t}u\left(t-1\right)
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A company with a fleet of 150 cars found that the emissions systems of 7 out of the 22 they tested failed to meet pollution control guidelines.
A company with a fleet of 150 cars found that the emissions systems of 7 out of the 22 they tested failed
A company with a fleet of 150 cars found that the emissions systems of 7 out of the 22 they tested failed to meet pollution control guidelines. Is this strong evidence that more than 20% of the fleet might be out of compliance? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.
{H}_{0}:p=0.20
{H}_{a}:p>0.20
2. Assumed SRS. sample is less than 10% of population . (0.20)(22)<10 and (0.80)(22)>10
The probability that an automobile being filled with gasoline will also need an oil change is 0.37. the probability that it needs a new oil filter is 0.45. and the probability that both the oil and filter need changing is 0.20.
If a new oil filter is needed, what is the probability that the oil has to be changed?
There are a total of 24 people in an application pool, that are equally qualified for a job. The pool consists of;
10 people from California
9 people from Nevada.
5 people from Arizonia.
There are 3 job openings for three different people. Type your answers as a decimal to 4 decimal places.
What is the probability that all three people are from California?
Find the two positive numbers that satisfy the given requirements. - The sum of the first number squared and the second number is 54 and the product is a maximum.
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Icosane - Simple English Wikipedia, the free encyclopedia
Icosane, also commonly spelled eicosane, is an alkaline hydrocarbon with the chemical formula H20C42.[2][3] It has 366,319 structural isomers. Its high flash point makes it a very inefficient fuel, so it is not much use in the petrochemical industry.[4] However, an isomer of icosane, n-Icosane (the straight-chain structural isomer of icosane) is the shortest compound found in paraffin waxes (CnH2n+2, where
{\displaystyle n=\mathbb {Z} ={[20,40]}}
) used to form candles. Icosane's phase transition at a moderate temperature makes it a candidate for PCM, which is used to store thermal energy and control temperature.
Icosane[1]
MeSH eicosane
3AYA9KEC48 Y
Melting point 36 to 38 °C; 97 to 100 °F; 309 to 311 K
Specific heat capacity, C 602.5 J K−1 mol−1 (at 6.0 °C)
Flash point > 113 °C (235 °F; 386 K)
Icosane is a non-polar molecule: quite unreactive except when it burns (see the NFPA Diamond in the infobox). It is also way less dense than insoluble in water. This also means it shares properties with its smaller alkaline counterparts.
Icosane can also be detected in the body odor of people diagnosed with Parkinson's Disease.[4]
The compound is found in the highest concentrations in plants such as Mexican ageratum, licorice, and the Bayrum tree.[5]
↑ "eicosane - Compound Summary". PubChem Compound. USA: National Center for Biotechnology Information. 16 September 2004. Identification and Related Records. Retrieved 4 January 2012.
↑ PubChem. "Eicosane". pubchem.ncbi.nlm.nih.gov. Retrieved 2021-05-20.
↑ "PDBeChem: Ligand Dictionary (PDB Ligand Chemistry - chemical component dictionary)". www.ebi.ac.uk. Retrieved 2021-06-03.
↑ 4.0 4.1 "icosane (CHEBI:43619)". www.ebi.ac.uk. Retrieved 2021-05-20.
↑ "Activities of a Specific Chemical Query". web.archive.org. 2015-09-23. Retrieved 2021-05-21.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Icosane&oldid=7644562"
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(a) Solve the following inequation and write down the solution set:
11𝑥 − 4 < 15x + 4 ≤ 13𝑥 + 14, 𝑥𝜖W
(b) A man invests Rs 4500 in shares of a company which is paying 7.5% dividend. If Rs 100 shares are available at a discount of 10%. Find: (i) Number of shares he purchases.
(c) In a class of 40 students, marks obtained by the students in a class test (out of 10) are given below:
Marks 1 2 3 4 5 6 7 8 9 10 Number of students 1 2 3 3 6 10 5 4 3 3 Calculate the following for the given distribution: (i) Median
(ii) Mode VIEW SOLUTION
(a) Using the factor theorem, show that (x – 2) is a factor of 𝑥3 + 𝑥2 − 4𝑥 − 4. Hence factorize the polynomial completely.
(cosec𝜃 − sin𝜃)(sec𝜃 − cos𝜃)(tan𝜃 + cot𝜃) = 1
(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the: (i) first term (ii) common difference (iii) sum of the first 20 terms.
(a) Simplify
\mathrm{sin}A\left[\begin{array}{cc}\mathrm{sin}A& -\mathrm{cos}A\\ \mathrm{cos}A& \mathrm{sin}A\end{array}\right]+\mathrm{cos}A\left[\begin{array}{cc}\mathrm{cos}A& \mathrm{sin}A\\ -\mathrm{sin}A& \mathrm{cos}A\end{array}\right]
(b) M and N are two points on the X axis and Y axis respectively. P (3, 2) divides the line segment MN in the ratio 2 : 3. Find: (i) the coordinates of M and N (ii) slope of the line MN.
(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the: (i) radius of the cylinder (ii) curved surface area of the cylinder. VIEW SOLUTION
(a) The following numbers,
K+3, K+2, 3K-7 \mathrm{and} 2K-3
are in proportion. Find
K
(b) Solve for x the quadratic equation
{x}^{2}-4x-8=0
. Give your answer correct to three significant figures.
(c) Use ruler and compass only for answering this question. Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre. Construct two tangents to the circle from the external point P. Measure and write down the length of any one tangent. VIEW SOLUTION
(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is: (i) an odd number (ii) divisible by 2 and 3 both. (iii) a number less than 16.
(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives Rs 441 as interest at the time of maturity. Find the amount Rekha deposited each month.
(c) Use a graph sheet for this question. Take 1 cm = 1 unit along both x and y axis. (i) Plot the following points: A(0,5), B(3,0), C(1,0) and D(1,–5) (ii) Reflect the points B, C and D on the y axis and name them as B', C' and D' respectively. (iii) Write down the coordinates of B', C' and D'. (iv) Join the points A, B, C, D, D', C', B', A in order and give a name to the closed figure ABCDD'C'B VIEW SOLUTION
(a) In the given figure,
\angle \mathrm{PQR}=\angle \mathrm{PST}=90°
, PQ = 5 cm and PS = 2 cm.
(i) Prove that
△\mathrm{PQR}~△\mathrm{PST}
(ii) Find Area of
△\mathrm{PQR}
: Area of quadrilateral
\mathrm{SRQT}
(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:
The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to the nearest whole number. Take π =22 7 .
(a) In the given figure AC is a tangent to the circle with centre O. If
\angle \mathrm{ADB}=55°
, find x and y. Give reasons for your answers.
(b) The model of a building is constructed with the scale factor 1 : 30.
(i) If the height of the model is 80 cm, find the actual height of the building in meters.
(ii) If the actual volume of a tank at the top of the building is 27 m3, find the volume of the tank on the top of the model.
(c) Given
\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]M=6I\phantom{\rule{0ex}{0ex}}
, where M is a matrix and I is unit matrix of order
2×2
(ii) Find the matrix M. VIEW SOLUTION
The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference.
The vertices of a
△\mathrm{ABC}
are A(3, 8), B(–1, 2) and C(6, –6). Find:
Using ruler and a compass only construct a semi-circle with diameter BC = 7cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down.
The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method.
Using properties of proportion solve for x, given
\frac{\sqrt{5x}+\sqrt{2x-6}}{\sqrt{5x}-\sqrt{2x-6}}=4
Sachin invests Rs 8500 in 10%, Rs 100 shares at Rs 170. He sells the shares when the price of each share rises by Rs 30. He invests the proceeds in 12% Rs 100 shares at Rs 125. Find:
(ii) the number of Rs 125 shares he buys.
Use graph paper for this question.The marks obtained by 120 students in an English test are given below:
No. of students 5 9 16 22 26 18 11 6 4 3
A man observes the angle of elevation of the top of the tower to be
45°
. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to
60°
. Find the height of the tower correct to 2 significant figures.
Using the Remainder Theorem find the remainders obtained when
{x}^{3}+\left(kx+8\right)x+k
is divided by x + 1 and x – 2.
Hence find k if the sum of the two remainders is 1.
The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers.
In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC//AE. If ∠BAC = 50°, find giving reasons:
Hence prove that BE is also a diameter
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Large Eddy Simulation of a Flow Past a Free Surface Piercing Circular Cylinder | J. Fluids Eng. | ASME Digital Collection
International Research Centre for Computational Hydrodynamics (ICCH), Agern Alle´ 5, 2970 Hørsholm, Denmark
A. Garapon,
Contributed by the Fluids Engineering Division for publication in the JOURNAL OF FLUIDS ENGINEERING. Manuscript received by the Fluids Engineering Division May 22, 2000; revised manuscript received August 24, 2001. Associate Editor: P. W. Bearman.
J. Fluids Eng. Mar 2002, 124(1): 91-101 (11 pages)
Kawamura, T., Mayer , S., Garapon , A., and Sørensen, L. (August 24, 2001). "Large Eddy Simulation of a Flow Past a Free Surface Piercing Circular Cylinder ." ASME. J. Fluids Eng. March 2002; 124(1): 91–101. https://doi.org/10.1115/1.1431545
Interactions between surface waves and underlying viscous wake are investigated for a turbulent flow past a free surface piercing circular cylinder at Reynolds number
Re=2.7×104
using large eddy simulation (LES). The computations have been performed for three Froude numbers
Fr=0.2,
0.5 and 0.8 in order to examine the influence of the Froude number. A second-order finite volume method coupled with a fractional step method is used for solving the grid-filtered incompressible Navier-Stokes equations. The computational results are found to be in good agreement with the available experimental data. At low Froude numbers
Fr=0.2
and 0.5, the amplitude of generated surface wave is small and the influence on the wake is not evident. On the other hand, strong wave-wake interactions are present at
Fr=0.8,
when the generated free surface wave is very steep. It is shown that structures of the underlying vortical flow correlate closely with the configuration of the free surface. Computational results show presence of a recirculation zone starting at the point where the surface slope changes discontinuously. Above this zone the surface elevation fluctuates intensively. The computed intensity of the surface fluctuation is in good agreement with the measurements. It is also shown that the periodic vortex shedding is attenuated near the free surface at a high Froude number. The region in which the periodic vortex shedding is hampered extends to about one diameter from the mean water level. It is qualitatively shown that the separated shear layers are inclined outward near the free surface due to the generation of the surface waves. This change in the relation between two shear layers is suggested to be responsible for the attenuation of the periodic vortex shedding.
surface waves (fluid), wakes, flow simulation, Navier-Stokes equations, vortices, fluctuations
Computation, Flow (Dynamics), Large eddy simulation, Surface waves (Fluid), Surface-piercing cylinders, Turbulence, Vortex shedding, Wakes, Waves, Cylinders, Reynolds number, Fluctuations (Physics), Navier-Stokes equations, Shear (Mechanics), Vortices, Water, Computer simulation, Fluid-dynamic forces
Experimental and numerical study of viscous flow field around an advancing vertical circular cylinder piercing a free-surface
J. Kansai Soc. Naval Archit. of Japan
Interaction of two-dimensional separated flows with a free surface at low Froude numbers
Chiba, S., and Kuwahara, K., 1989, “Numerical analysis for free surface flow around a vertical circular cylinder,” Proceedings of Third Symposium on Computational Fluid Dynamics, Tokyo, Japan, pp. 295–299.
, “Numerical experiments on the flow past a circular cylinder at sub-critical Reynolds number,” Technical Report TF-62, Department of Mechanical Engineering, Stanford University.
Suitability of upwind-biased finite-difference schemes for large-eddy simulation of turbulent flows
Breuer., M., 1997, “Numerical and modeling influences on large eddy simulations for the flow past a circular cylinder,” Proceedings of 11th Symposium on Turbulent Shear Flows, 8–10 Sept. Grenoble, France, pp. 26–27.
Application of large-eddy simulation to open-channel flow
Salvetti, M., Zang, Y., Street, R., and Banerjee, S., 1996, “Large-eddy simulation of decaying free-surface turbulence with dynamic mixed subgrid-scale models,” Proceedings of Twenty-First Symposium on Naval Hydrodynamics, Trondheim, Norway, National Academy Press, pp. 1018–1032.
General circulation experiments with the primitive equations I. The basic experiment
Model consistency in large eddy simulation of turbulent channel flows
Dommermuth, D., and Novikov, E. A., “Direct-numerical and large-eddy simulation of turbulent free surface flows,” Proceedings of the Sixth International Conference on Numerical Ship Hydrodynamics, Iowa City.
Kawamura, T., 1998, “Numerical simulation of 3D turbulent free-surface flows,” PhD thesis, Department of Naval Architecture and Ocean Engineering, School of Engineering, University of Tokyo.
Nonlinear interaction of shear flow with a free surface
Instabilities of a horizontal shear flow with a free surface
Batchelor, G. K., 1967, Introduction to Fluid Dynamics, Cambridge University Press.
Neuere Festellungen u¨ber die Gesetze des Flu¨ssigkeits-und Luftwiderstands
J. Soc. Naval Archit. Japan
Large Eddy Simulation of Flow Past Free Surface Piercing Circular Cylinders
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Solve fully implicit differential equations — variable order method - MATLAB ode15i
f\left(t,y,y\text{'}\right)=0
{\mathrm{ty}}^{2}{\left({\mathit{y}}^{\prime }\right)}^{3}-{\mathit{y}}^{3}{\left({\mathit{y}}^{\prime }\right)}^{2}+\mathit{t}\left({\mathit{t}}^{2}+1\right){\mathit{y}}^{\prime }-{\mathit{t}}^{2}\mathit{y}=0
\mathit{f}\left(\mathit{t},\mathit{y},{\mathit{y}}^{\prime }\right)=0
\mathit{t}
\mathit{y}
{\mathit{y}}^{\prime }
\mathit{f}\left({\mathit{t}}_{0},\mathit{y},{\mathit{y}}^{\prime }\right)=0
\mathit{y}\left({\mathit{t}}_{0}\right)=\sqrt{\frac{3}{2}}
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)=0
\left[1\text{\hspace{0.17em}}10\right]
\mathit{y}\left(\mathit{t}\right)=\sqrt{{\mathit{t}}^{2}+\frac{1}{2}}
f\left(t,y,y\text{'}\right)
y\text{'}-y=0
\begin{array}{l}y{\text{'}}_{1}-{y}_{2}=0\\ y{\text{'}}_{2}+1=0\text{\hspace{0.17em}},\end{array}
f\left({t}_{0},{y}_{0},y{\text{'}}_{0}\right)=0
f\left({t}_{0},{y}_{0},y{\text{'}}_{0}\right)=0
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Hypothesis Test Assumptions - MATLAB & Simulink - MathWorks í•œêµ
Different hypothesis tests make different assumptions about the distribution of the random variable being sampled in the data. These assumptions must be considered when choosing a test and when interpreting the results.
For example, the z-test (ztest) and the t-test (ttest) both assume that the data are independently sampled from a normal distribution. Statistics and Machine Learning Toolbox™ functions are available for testing this assumption, such as chi2gof, jbtest, lillietest, and normplot.
Both the z-test and the t-test are relatively robust with respect to departures from this assumption, so long as the sample size n is large enough. Both tests compute a sample mean
\stackrel{¯}{x}
, which, by the Central Limit Theorem, has an approximately normal sampling distribution with mean equal to the population mean μ, regardless of the population distribution being sampled.
The difference between the z-test and the t-test is in the assumption of the standard deviation σ of the underlying normal distribution. A z-test assumes that σ is known; a t-test does not. As a result, a t-test must compute an estimate s of the standard deviation from the sample.
Test statistics for the z-test and the t-test are, respectively,
\begin{array}{l}z=\frac{\stackrel{¯}{x}â\mathrm{μ}}{\mathrm{Ï}/\sqrt{n}}\\ t=\frac{\stackrel{¯}{x}â\mathrm{μ}}{s/\sqrt{n}}\end{array}
Under the null hypothesis that the population is distributed with mean μ, the z-statistic has a standard normal distribution, N(0,1). Under the same null hypothesis, the t-statistic has Student's t distribution with n – 1 degrees of freedom. For small sample sizes, Student's t distribution is flatter and wider than N(0,1), compensating for the decreased confidence in the estimate s. As sample size increases, however, Student's t distribution approaches the standard normal distribution, and the two tests become essentially equivalent.
Knowing the distribution of the test statistic under the null hypothesis allows for accurate calculation of p-values. Interpreting p-values in the context of the test assumptions allows for critical analysis of test results.
Assumptions underlying Statistics and Machine Learning Toolbox hypothesis tests are given in the reference pages for implementing functions.
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Interpreting Continuous Wavelet Coefficients - MATLAB & Simulink - MathWorks India
Because the CWT is a redundant transform and the CWT coefficients depend on the wavelet, it can be challenging to interpret the results.
To help you in interpreting CWT coefficients, it is best to start with a simple signal to analyze and an analyzing wavelet with a simple structure.
A signal feature that wavelets are very good at detecting is a discontinuity, or singularity. Abrupt transitions in signals result in wavelet coefficients with large absolute values.
For the signal create a shifted impulse. The impulse occurs at point 500.
For the wavelet, pick the Haar wavelet.
[~,psi,xval] = wavefun('haar',10);
plot(xval,psi); axis([0 1 -1.5 1.5]);
title('Haar Wavelet');
To compute the CWT using the Haar wavelet at scales 1 to 128, enter:
CWTcoeffs = cwt(x,1:128,'haar');
CWTcoeffs is a 128-by-1000 matrix. Each row of the matrix contains the CWT coefficients for one scale. There are 128 rows because the SCALES input to cwt is 1:128. The column dimension of the matrix matches the length of the input signal.
Recall that the CWT of a 1D signal is a function of the scale and position parameters. To produce a plot of the CWT coefficients, plot position along the x-axis, scale along the y-axis, and encode the magnitude, or size of the CWT coefficients as color at each point in the x-y, or time-scale plane.
You can produce this plot using cwt with the optional input argument 'plot'.
cwt(x,1:128,'haar','plot');
The preceding figure was modified with text labels to explicitly show which colors indicate large and small CWT coefficients.
You can also plot the size of the CWT coefficients in 3D with
cwt(x,1:64,'haar','3Dplot'); colormap jet;
where the number of scales has been reduced to aid in visualization.
Examining the CWT of the shifted impulse signal, you can see that the set of large CWT coefficients is concentrated in a narrow region in the time-scale plane at small scales centered around point 500. As the scale increases, the set of large CWT coefficients becomes wider, but remains centered around point 500. If you trace the border of this region, it resembles the following figure.
This region is referred to as the cone of influence of the point t=500 for the Haar wavelet. For a given point, the cone of influence shows you which CWT coefficients are affected by the signal value at that point.
To understand the cone of influence, assume that you have a wavelet supported on [-C, C]. Shifting the wavelet by b and scaling by a results in a wavelet supported on [-Ca+b, Ca+b]. For the simple case of a shifted impulse,
\delta \left(t-\tau \right)
, the CWT coefficients are only nonzero in an interval around τ equal to the support of the wavelet at each scale. You can see this by considering the formal expression of the CWT of the shifted impulse.
C\left(a,b;\delta \left(t-\tau \right),\psi \left(t\right)\right)={\int }_{-\infty }^{\infty }\delta \left(t-\tau \right)\frac{1}{\sqrt{a}}{\psi }^{*}\left(\frac{t-b}{a}\right)dt=\frac{1}{\sqrt{a}}{\psi }^{*}\left(\frac{\tau -b}{a}\right)
For the impulse, the CWT coefficients are equal to the conjugated, time-reversed, and scaled wavelet as a function of the shift parameter, b. You can see this by plotting the CWT coefficients for a select few scales.
plot(CWTcoeffs(10,:)); title('Scale 10');
The cone of influence depends on the wavelet. You can find and plot the cone of influence for a specific wavelet with conofinf.
The next example features the superposition of two shifted impulses,
\delta \left(t-300\right)+\delta \left(t-500\right)
. In this case, use the Daubechies' extremal phase wavelet with four vanishing moments, db4. The following figure shows the cone of influence for the points 300 and 500 using the db4 wavelet.
Look at point 400 for scale 20. At that scale, you can see that neither cone of influence overlaps the point 400. Therefore, you can expect that the CWT coefficient will be zero at that point and scale. The signal is only nonzero at two values, 300 and 500, and neither cone of influence for those values includes the point 400 at scale 20. You can confirm this by entering:
CWTcoeffs = cwt(x,1:128,'db4');
plot(CWTcoeffs(20,:)); grid on;
Next, look at the point 400 at scale 80. At scale 80, the cones of influence for both points 300 and 500 include the point 400. Even though the signal is zero at point 400, you obtain a nonzero CWT coefficient at that scale. The CWT coefficient is nonzero because the support of the wavelet has become sufficiently large at that scale to allow signal values 100 points above and below to affect the CWT coefficient. You can confirm this by entering:
plot(CWTcoeffs(80,:));
In the preceding example, the CWT coefficients became large in the vicinity of an abrupt change in the signal. This ability to detect discontinuities is a strength of the wavelet transform. The preceding example also demonstrated that the CWT coefficients localize the discontinuity best at small scales. At small scales, the small support of the wavelet ensures that the singularity only affects a small set of wavelet coefficients.
To demonstrate why the wavelet transform is so adept at detecting abrupt changes in the signal, consider a shifted Heaviside, or unit step signal.
x = [zeros(500,1); ones(500,1)];
CWTcoeffs = cwt(x,1:64,'haar','plot'); colormap jet;
Similar to the shifted impulse example, the abrupt transition in the shifted step function results in large CWT coefficients at the discontinuity. The following figure illustrates why this occurs.
In the preceding figure, the red function is the shifted unit step function. The black functions labeled A, B, and C depict Haar wavelets at the same scale but different positions. You can see that the CWT coefficients around position A are zero. The signal is zero in that neighborhood and therefore the wavelet transform is also zero because any wavelet integrates to zero.
Note the Haar wavelet centered around position B. The negative part of the Haar wavelet overlaps with a region of the step function that is equal to 1. The CWT coefficients are negative because the product of the Haar wavelet and the unit step is a negative constant. Integrating over that area yields a negative number.
Note the Haar wavelet centered around position C. Here the CWT coefficients are zero. The step function is equal to one. The product of the wavelet with the step function is equal to the wavelet. Integrating any wavelet over its support is zero. This is the zero moment property of wavelets.
At position B, the Haar wavelet has already shifted into the nonzero portion of the step function by 1/2 of its support. As soon as the support of the wavelet intersects with the unity portion of the step function, the CWT coefficients are nonzero. In fact, the situation illustrated in the previous figure coincides with the CWT coefficients achieving their largest absolute value. This is because the entire negative deflection of the wavelet oscillation overlaps with the unity portion of the unit step while none of the positive deflection of the wavelet does. Once the wavelet shifts to the point that the positive deflection overlaps with the unit step, there will be some positive contribution to the integral. The wavelet coefficients are still negative (the negative portion of the integral is larger in area), but they are smaller in absolute value than those obtained at position B.
The following figure illustrates two other positions where the wavelet intersects the unity portion of the unit step.
In the top figure, the wavelet has just begun to overlap with the unity portion of the unit step. In this case, the CWT coefficients are negative, but not as large in absolute value as those obtained at position B. In the bottom figure, the wavelet has shifted past position B and the positive deflection of the wavelet begins to contribute to the integral. The CWT coefficients are still negative, but not as large in absolute value as those obtained at position B.
You can now visualize how the wavelet transform is able to detect discontinuities. You can also visualize in this simple example exactly why the CWT coefficients are negative in the CWT of the shifted unit step using the Haar wavelet. Note that this behavior differs for other wavelets.
% plot a few scales for visualization
plot(CWTcoeffs(5,:)); title('Scale 5');
Next consider how the CWT represents smooth signals. Because sinusoidal oscillations are a common phenomenon, this section examines how sinusoidal oscillations in the signal affect the CWT coefficients. To begin, consider the sym4 wavelet at a specific scale superimposed on a sine wave.
Recall that the CWT coefficients are obtained by computing the product of the signal with the shifted and scaled analyzing wavelet and integrating the result. The following figure shows the product of the wavelet and the sinusoid from the preceding figure.
You can see that integrating over this product produces a positive CWT coefficient. That results because the oscillation in the wavelet approximately matches a period of the sine wave. The wavelet is in phase with the sine wave. The negative deflections of the wavelet approximately match the negative deflections of the sine wave. The same is true of the positive deflections of both the wavelet and sinusoid.
The following figure shifts the wavelet 1/2 of the period of the sine wave.
Examine the product of the shifted wavelet and the sinusoid.
You can see that integrating over this product produces a negative CWT coefficient. That results because the wavelet is 1/2 cycle out of phase with the sine wave. The negative deflections of the wavelet approximately match the positive deflections of the sine wave. The positive deflections of the wavelet approximately match the negative deflections of the sinusoid.
Finally, shift the wavelet approximately one quarter cycle of the sine wave.
The following figure shows the product of the shifted wavelet and the sinusoid.
Integrating over this product produces a CWT coefficient much smaller in absolute value than either of the two previous examples. That results because the negative deflection of the wavelet approximately aligns with a positive deflection of the sine wave. Also, the main positive deflection of the wavelet approximately aligns with a positive deflection of the sine wave. The resulting product looks much more like a wavelet than the other two products. If it looked exactly like a wavelet, the integral would be zero.
At scales where the oscillation in the wavelet occurs on either a much larger or smaller scale than the period of the sine wave, you obtain CWT coefficients near zero. The following figure illustrates the case where the wavelet oscillates on a much smaller scale than the sinusoid.
The product shown in the bottom pane closely resembles the analyzing wavelet. Integrating this product results in a CWT coefficient near zero.
The following example constructs a 60-Hz sine wave and obtains the CWT using the sym8 wavelet.
CWTcoeffs = cwt(x,1:64,'sym8','plot'); colormap jet;
Note that the CWT coefficients are large in absolute value around scales 9 to 21. You can find the pseudo-frequencies corresponding to these scales using the command:
freq = scal2frq(9:21,'sym8',1/1000);
Note that the CWT coefficients are large at scales near the frequency of the sine wave. You can clearly see the sinusoidal pattern in the CWT coefficients at these scales with the following code.
surf(CWTcoeffs); colormap jet;
shading('interp'); view(-60,12);
The final example constructs a signal consisting of both abrupt transitions and smooth oscillations. The signal is a 2-Hz sinusoid with two introduced discontinuities.
plot(t,x); xlabel('t'); ylabel('x');
Note the discontinuities near t=0.3 and t=0.7.
Obtain and plot the CWT using the sym4 wavelet.
CWTcoeffs = cwt(x,1:180,'sym4');
imagesc(t,1:180,abs(CWTcoeffs));
colormap jet; axis xy;
xlabel('t'); ylabel('Scales');
Note that the CWT detects both the abrupt transitions and oscillations in the signal. The abrupt transitions affect the CWT coefficients at all scales and clearly separate themselves from smoother signal features at small scales. On the other hand, the maxima and minima of the 2–Hz sinusoid are evident in the CWT coefficients at large scales and not apparent at small scales.
The following general principles are important to keep in mind when interpreting CWT coefficients.
Cone of influence— Depending on the scale, the CWT coefficient at a point can be affected by signal values at points far removed. You have to take into account the support of the wavelet at specific scales. Use conofinf to determine the cone of influence. Not all wavelets are equal in their support. For example, the Haar wavelet has smaller support at all scales than the sym4 wavelet.
Detecting abrupt transitions— Wavelets are very useful for detecting abrupt changes in a signal. Abrupt changes in a signal produce relatively large wavelet coefficients (in absolute value) centered around the discontinuity at all scales. Because of the support of the wavelet, the set of CWT coefficients affected by the singularity increases with increasing scale. Recall this is the definition of the cone of influence. The most precise localization of the discontinuity based on the CWT coefficients is obtained at the smallest scales.
Detecting smooth signal features— Smooth signal features produce relatively large wavelet coefficients at scales where the oscillation in the wavelet correlates best with the signal feature. For sinusoidal oscillations, the CWT coefficients display an oscillatory pattern at scales where the oscillation in the wavelet approximates the period of the sine wave.
The basic algorithm for the continuous wavelet transform (CWT) is:
Take a wavelet and compare it to a section at the start of the original signal.
Calculate a number, C, that represents how closely correlated the wavelet is with this section of the signal. The larger the number C is in absolute value, the more the similarity. This follows from the fact the CWT coefficients are calculated with an inner product. See Inner Products for more information on how inner products measure similarity. If the signal energy and the wavelet energy are equal to one, C may be interpreted as a correlation coefficient. Note that, in general, the signal energy does not equal one and the CWT coefficients are not directly interpretable as correlation coefficients.
As described in Continuous and Discrete Wavelet Transforms, the CWT coefficients explicitly depend on the analyzing wavelet. Therefore, the CWT coefficients are different when you compute the CWT for the same signal using different wavelets.
Shift the wavelet to the right and repeat steps 1 and 2 until you've covered the whole signal.
Scale (stretch) the wavelet and repeat steps 1 through 3.
Repeat steps 1 through 4 for all scales.
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