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Calculus/Higher Order Derivatives - Wikibooks, open books for an open world
Calculus/Higher Order Derivatives
← Chain Rule Calculus Implicit differentiation →
The second derivative, or second order derivative, is the derivative of the derivative of a function. The derivative of the function
{\displaystyle f(x)}
{\displaystyle f'(x)}
, and its double (or "second") derivative is denoted by
{\displaystyle f''(x)}
. This is read as "
{\displaystyle f}
double prime of
{\displaystyle x}
", or "The second derivative of
{\displaystyle f(x)}
". Because the derivative of function
{\displaystyle f}
is defined as a function representing the slope of function
{\displaystyle f}
, the double derivative is the function representing the slope of the first derivative function.
Furthermore, the third derivative is the derivative of the derivative of the derivative of a function, which can be represented by
{\displaystyle f'''(x)}
{\displaystyle f}
triple prime of
{\displaystyle x}
", or "The third derivative of
{\displaystyle f(x)}
" . This can continue as long as the resulting derivative is itself differentiable, with the fourth derivative, the fifth derivative, and so on. Any derivative beyond the first derivative can be referred to as a higher order derivative.
{\displaystyle f(x)}
be a function in terms of
{\displaystyle x}
. The following are notations for higher order derivatives.
{\displaystyle n}
-th Derivative
{\displaystyle f''(x)}
{\displaystyle f'''(x)}
{\displaystyle f^{(4)}(x)}
{\displaystyle f^{(n)}(x)}
Probably the most common notation.
{\displaystyle {\frac {d^{2}f}{dx^{2}}}}
{\displaystyle {\frac {d^{3}f}{dx^{3}}}}
{\displaystyle {\frac {d^{4}f}{dx^{4}}}}
{\displaystyle {\frac {d^{n}f}{dx^{n}}}}
{\displaystyle {\frac {d^{2}}{dx^{2}}}[f(x)]}
{\displaystyle {\frac {d^{3}}{dx^{3}}}[f(x)]}
{\displaystyle {\frac {d^{4}}{dx^{4}}}[f(x)]}
{\displaystyle {\frac {d^{n}}{dx^{n}}}[f(x)]}
Another form of Leibniz notation
{\displaystyle D^{2}f}
{\displaystyle D^{3}f}
{\displaystyle D^{4}f}
{\displaystyle D^{n}f}
Warning: You should not write
{\displaystyle f^{n}(x)}
to indicate the
{\displaystyle n}
-th derivative, as this is easily confused with the quantity
{\displaystyle f(x)}
all raised to the nth power.
The Leibniz notation, which is useful because of its precision, follows from
{\displaystyle {\frac {d}{dx}}\left({\frac {df}{dx}}\right)={\frac {d^{2}f}{dx^{2}}}}
Newton's dot notation extends to the second derivative,
{\displaystyle {\ddot {y}}}
, but typically no further in the applications where this notation is common.
Find the third derivative of
{\displaystyle f(x)=4x^{5}+6x^{3}+2x+1}
{\displaystyle x}
Repeatedly apply the Power Rule to find the derivatives.
{\displaystyle f'(x)=20x^{4}+18x^{2}+2}
{\displaystyle f''(x)=80x^{3}+36x}
{\displaystyle f'''(x)=240x^{2}+36}
Find the 3rd derivative of
{\displaystyle f(x)=12\sin(x)+{\frac {1}{x+2}}+2x}
{\displaystyle x}
{\displaystyle f'(x)=12\cos(x)-{\frac {1}{(x+2)^{2}}}+2}
{\displaystyle f''(x)=-12\sin(x)+{\frac {2}{(x+2)^{3}}}}
{\displaystyle f'''(x)=-12\cos(x)-{\frac {6}{(x+2)^{4}}}}
For applications of the second derivative in finding a curve's concavity and points of inflection, see "Extrema and Points of Inflection" and "Extreme Value Theorem". For applications of higher order derivatives in physics, see the "Kinematics" section.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Higher_Order_Derivatives&oldid=3744628"
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Let B = left{ begin{bmatrix} 1 -2 end{bmatrix}, begin{bmatrix}2 1 end{bmatrix}
Let B = left{ begin{bmatrix} 1 -2 end{bmatrix}
B=\left\{\left[\begin{array}{c}1\\ -2\end{array}\right],\left[\begin{array}{c}2\\ 1\end{array}\right]\right\}\text{ }and\text{ }C=\left\{\left[\begin{array}{c}2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 3\end{array}\right]\right\}
be bases for
{R}^{2}.
Change-of-coordinate matrix from C to B.
B=\left\{\left[\begin{array}{c}1\\ -2\end{array}\right],\left[\begin{array}{c}2\\ 1\end{array}\right]\right\}\text{ }and\text{ }C=\left\{\left[\begin{array}{c}2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 3\end{array}\right]\right\}
Let x, y are scalare such that
\left[\begin{array}{c}2\\ 1\end{array}\right]=x\left[\begin{array}{c}1\\ -2\end{array}\right]+y\left[\begin{array}{c}2\\ 1\end{array}\right]
x=1,y=0
\left[\begin{array}{c}2\\ 1\end{array}\right]=0\left[\begin{array}{c}1\\ -2\end{array}\right]+1\left[\begin{array}{c}2\\ 1\end{array}\right]\to \left(c\right)
\left[\begin{array}{c}1\\ 3\end{array}\right]=2\left[\begin{array}{c}1\\ -2\end{array}\right]+y\left[\begin{array}{c}2\\ 1\end{array}\right]
=1\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}x+2y\\ -2x+y\end{array}\right]
x+2y=1,-2x+y=3
Solving above equation
\begin{array}{c}2\text{⧸}x+4y=2\\ -\text{⧸}2x+y=3\\ 5y=5⇒y=1\end{array}
-y=1\in x+2y=1
⇒x+2=1
⇒x=-1
\therefore \left[\begin{array}{c}1\\ 3\end{array}\right]=\left(-1\right)\left[\begin{array}{c}1\\ -2\end{array}\right]+1\left[\begin{array}{c}2\\ 1\end{array}\right]\to \left(d\right)
from c, d Transition matrix of C over D
\left[T{\right]}_{C,B}=\left[\begin{array}{cc}0& -1\\ 1& 1\end{array}\right]
Given V = R, and two bases: B and C and the coordinator [v]B a) Find the change of coordinates matrix
PB\to C.
b) Find the coordinator [v]C.
Find the value of x or y so that the line passing through the given points has the given slope. (-9,y),(0,-3),
m=\frac{-7}{9}
Given point
P\left(-2,6,3\right)
B=y{a}_{x}+\left(x+z\right){a}_{y}
, Express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.
\left[\begin{array}{cc}4& \text{ }0\end{array}\right]
Let A and C be the following ordered bases of P3(t):
A=\left(1,1+t,1+t+{t}^{2},1+t+{t}^{2}+{t}^{3}\right)
C=\left(1,-t,{t}^{2}-{t}^{3}\right)
Find tha change of coordinate matrix
{I}_{CA}
How are triple integrals defined in cylindrical and spherical coordinates?
Why might one prefer working in one of these coordinate systems to working in rectangular coordinates?
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Homogeneous Linear Differential Equations | Brilliant Math & Science Wiki
Samir Khan and Sarthak Khattar contributed
A homogeneous linear differential equation is a differential equation in which every term is of the form
y^{(n)}p(x)
i.e. a derivative of
y
times a function of
x
. In general, these are very difficult to work with, but in the case where all the constants are coefficients, they can be solved exactly. In this case, the differential equation looks like
a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,
a_{0}, a_{1} ,a_{n-1} ,\ldots , a_{n}
being real constants, and almost resembles a polynomial. In fact, looking at the roots of this associated polynomial gives solutions to the differential equation.
A reasonable guess for a solution to
a_ny^{(n)}+\cdots+a_0y
y=e^{rx}
- this is called an ansatz. Note that
\dfrac{d^ky}{dx^k} e^{rx} = r^k e^{rx},
e^{rx}
to be a solution, it would have to satisfy
a_{n} r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_{1} r e^{rx} + a_{0} e^{rx} = 0.
e ^ {rx} \neq 0
, it must be the case that
a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0. \qquad (3).
e^{rx}
solves the differential equation when
r
For the differential equation
a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,
the associated characteristic equation is
a_nr^n+\cdots+a_1r+a_0=0
Depending on the nature of the roots of the characteristic polynomial, the differential equation has slightly different solutions.
When the roots are real and distinct, the solutions are just the linear combinations of
e^{rx}
for the different roots
r
n
r_1, r_2, \ldots, r_{n}
y(x) = c_1 e^{r_1x} + c_2e^{r_2x} +\cdots +c_ne^{r_nx}
(1),
c_1, c_2, \ldots, c_n
\begin{array}{c}&y'' + 2y' - 8y = 0, &y(0) = 5, &y'(0) = - 2.\end{array}
r^2 + 2r - 8 = 0.
r = 2, -4
y(x) = c_1 e^{2x} + c_2 e^{-4x}
y'(x) = 2 c_1 e^{2x} - 4 c_2 e^{-4x}
y(0) = c_1 + c_2 = 5, y'(0) = 2c_1 - 4c_2 = -2 \Rightarrow c_1 = 3, c_2=2.
y(x) = 3e^{2x} + 2e^{-4x}. \ _\square
c_1e^{-2x}+c_2e^{-3x}
c_1e^{-2x}+c_2e^{3x}
c_1e^{2x}+c_2e^{3x}
c_1e^{2x}+c_2e^{-3x}
What is the general solution to the differential equation
y''+5y'+6y=0
When the characteristic polynomial has repeated roots, the previous theorem no longer applies.
If the characteristic equation has a repeated real root
r
k,
r
in equation is of the form
(c_1 + c_ {2}x + c_ {3}x^2 + \cdots + c_ {k}x^{k-1}) \cdot e^{rx}.
y^{(4)} + 3 \cdot y^{(3)} + 3 \cdot y'' + y' = 0.
r^4 + 3 \cdot r^3 + 3 \cdot r^2 + r= r \cdot (r+1)^3=0
r_1 = 0,
y_1 = c_1
(k=3)
r_2 = -1,
y_2 = (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}.
y(x) = c_1 + (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}. \ _\square
c_1e^{-2x}
c_1e^{-2x}+c_2xe^{-2x}
(c_1+c_2)e^{-2x}
c_1e^{2x}+c_2xe^{2x}
y''+4y'+4y=0
a \pm bi,
and
b
\sqrt{-1}.
If the characteristic equation has a pair of complex roots not repeated
a \pm bi
, then the relevant part to them of the general solution of equation has the form
e^{ax} \cdot (c_1 \cos bx + c_2 \sin bx)
y'' - 4y' + 5y = 0.
r^2 - 4r + 5 = 0,
2 \pm i.
y(x) = e^{2x} \cdot (c_1 \cos x + c_2 \sin x).
e^{3x}(c_1\cos 3x+c_2\sin 3x)
e^{2x}(c_1\cos 3x+c_2\sin 3x)
e^{2x}(c_1\cos 2x+c_2\sin 3x)
e^{3x}(c_1\cos 2x+c_2\sin 2x)
y''-4y'+13y=0
When there are repeated complex roots, they can be accounted for in the same way as with repeated real roots.
y^{(4)} + 4 \cdot y^{(3)} + 12 \cdot y'' + 16y' + 16y = 0.
(r^2 + 2r + 4)^2 = 0,
-1 \pm i \cdot \sqrt{3}
2.
Therefore, the general solution is
y(x) =e^{-x} \cdot \left(c_1 \cos x\sqrt{3} + d_1 \sin x\sqrt{3}\right) + xe^{-x} \cdot \left(c_2 \cos x\sqrt{3} + d_2 \sin x\sqrt{3}\right).
Cite as: Homogeneous Linear Differential Equations. Brilliant.org. Retrieved from https://brilliant.org/wiki/homogeneous-linear-differential-equations/
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Deven Livingston 2022-04-26 Answered
Given a set {13, x, -1, 11}, for what x would the mean of the set be 3?
occhei042 2022-04-26 Answered
Given a set {-3, x, 3,4,6}, for what x would the mean of the set be -11?
Given a set {-2, x, -3, -4, 1, 15, 22, 2}, for what x would the mean of the set be -12?
g2esebyy7 2022-04-25 Answered
"We need to estimate the conditional mean
{\beta }_{0}+{\beta }_{1}{x}_{0}
{x}_{0}
, so we use
\stackrel{^}{{Y}_{0}}=\stackrel{^}{{\beta }_{0}}+\stackrel{^}{{\beta }_{1}}{x}_{0}
as a natural estimator." here we get
\stackrel{^}{{Y}_{0}}~N\left({\beta }_{0}+{\beta }_{1}{x}_{0},{\sigma }^{2}{h}_{00}\right) \text{ }\text{where }{h}_{00}=\frac{1}{n}+\frac{{\left({x}_{0}-\stackrel{―}{x}\right)}^{2}}{\left(n-1\right){s}_{x}^{2}}
with a confidence interval for
E\left({Y}_{0}\right)={\beta }_{0}+{\beta }_{1}{x}_{0}
\left(\stackrel{^}{{b}_{0}}+\stackrel{^}{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}},\stackrel{^}{{b}_{0}}+\stackrel{^}{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}}\right)
c={t}_{n-2,1-\frac{\alpha }{2}}
Where these results are found by looking at the shape of the distribution and at
E\left(\stackrel{^}{{Y}_{0}}\right)
var\left(\stackrel{^}{{Y}_{0}}\right)
"We want to predict the observation
{Y}_{0}={\beta }_{0}+{\beta }_{1}{x}_{0}+{ϵ}_{0}
{x}_{0}
E\left(\stackrel{^}{{Y}_{0}}-{Y}_{0}\right)=0 \text{ }\text{and }var\left(\stackrel{^}{{Y}_{0}}-{Y}_{0}\right)={\sigma }^{2}\left(1+{h}_{00}\right)
\left(\stackrel{^}{{b}_{0}}+\stackrel{^}{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}+1},\stackrel{^}{{b}_{0}}+\stackrel{^}{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}+1}\right)
Probability vs Confidence
My notes on confidence give this question:
An investigator is interested in the amount of time internet users spend watching TV a week. He assumes
\sigma =3.5
hours and samples
n=50
users and takes the sample mean to estimate the population mean
\mu
n=50
is large we know that
\frac{\stackrel{―}{X}-\mu }{\frac{\sigma }{\sqrt{n}}}
approximates the Standard Normal. So, with probability
\alpha =0.99
, the maximum error of estimate is
E={z}_{\frac{\alpha }{2}}×\frac{\sigma }{\sqrt{n}}\approx 1.27
The investigator collects that data and obtain
\stackrel{―}{X}=11.5
hours. Can he still assert with 99% probability that the error is at most 1.27 hours?
With the answer that:
No he cannot, because the probability describes the method/estimator, not the result. We say that "we conclude with 99% confidence that the error does not exceed 1.27 hours."
I am confused. What is this difference between probability and confidence? Is it related to confidence intervals? Is there an intuitive explanation for the difference?
Given the set: {5,3,6,-2,5,6, -1,x}, for what x would the mean of the set be 0?
Troszokd95 2022-04-25 Answered
Given the set: {5, x, 11, -3, -12, 19, 10}, for what x would the mean of the set be 0?
Given the set: {-1, 6,4,11,x, -7}, for what x would the mean of the set be 1?
Given the set: {-2, 10, -8, -14, x, 7, -6}, for what x would the mean of the set be -8?
Given a set {4, -2, -12, 5, x}, for what x would the mean of the set be -5?
Ricardo Berger 2022-04-25 Answered
Given a set {9,3,4,6,x,6,1,-4}, for what x would the mean of the set be -32?
IQ scores among the general population have a mean of 100 and a standard deviation of 15. A researcher claims that the standard deviation, o, of IQ scores for males is less than 15. A random sample of 19 IQ scores for males had a mean of 101 and a standard deviation of 10. Assuming that IQ scores for males are approximately normally distributed, is there significant evidence (at the 0.05 level of significance) to conclude that the researcher's claim is correct?
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.)
(a) State the null hypothesis
{H}_{0}
, and the alternative hypothesis
{H}_{1}
{H}_{0}
{H}_{1}
(b) Determine the type of test statistic to use.
(c) Find the value of the test statistic. (Round to three or more decimal places.)
(d) Find the critical value. (Round to three or more decimal places.)
(e) Can we support the claim that the standard deviation of IQ scores for males is less than 15?
In your own words, distinguish between measures of association and tests of statistical significance.
In which of the following situations should the chi-square Test for Homogeneity be used?Select the correct answer below:
A professor wants to determine if the students in each of her two sections had the same distribution of grades. She records the grade distribution for each section and compares the two distributions.
A survey company puts out a survey asking people two questions. First, it asks if a person owns a car or not. Second, it asks whether or not the person supports an upcoming environmental protection bill. The data are recorded in a contingency table. The company wants to determine if support for the bill is independent of car ownership.
A professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10. She records the actual distribution of grades and wants to compare it to the normal distribution.
tefoldaavv 2022-04-24 Answered
In the following table, likely voters’ preferences of two candidates are cross-classified by gender.
\begin{array}{|ccc|}\hline & \text{ Male }& \text{ Female }\\ \text{ Candidate A }& 150& 130\\ \text{ Candidate B }& 100& 120\\ \hline\end{array}
For the chi-square test of independence, the value of the test statistic is:
surgerh7a 2022-04-24 Answered
In each of Exercises, we have provided a sample mean, sample size, and population standard deviation. In each case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.
x = 24, n = 15,
\sigma
= 4, H0:
\mu
= 22, Ha:
\mu
Davon Trujillo 2022-04-24 Answered
In a chi-square test for goodness of fit, which of the following is used to compute the degrees of freedom?
a. The number of cells in the table
b. The number of categories for the variable plus 1
c. The number of rows times the number of columns
d. The number of categories for the variable minus 1
acceloq3c 2022-04-24 Answered
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Engineering Acoustics/Moving Coil Loudspeaker - Wikibooks, open books for an open world
Engineering Acoustics/Moving Coil Loudspeaker
1 Moving Coil Transducer
1.1 The Magnet Motor Drive System
1.2 The Loudspeaker Cone System
1.3 The Loudspeaker Suspension
1.4 Modeling the Loudspeaker as a Lumped System
Moving Coil TransducerEdit
The purpose of the acoustic transducer is to convert electrical energy into acoustic energy. Many variations of acoustic transducers exist, although the most common is the moving coil-permanent magnet transducer. The classic loudspeaker is of the moving coil-permanent magnet type.
The classic electrodynamic loudspeaker driver can be divided into three key components:
1) The Magnet Motor Drive System
2) The Loudspeaker Cone System
3) The Loudspeaker Suspension
Figure 1 Cut-away of a moving coil-permanent magnet loudspeaker
The Magnet Motor Drive SystemEdit
The main purpose of the Magnet Motor Drive System is to establish a symmetrical magnetic field in which the voice coil will operate. The Magnet Motor Drive System is comprised of a front focusing plate, permanent magnet, back plate, and a pole piece. In figure 2, the assembled drive system is illustrated. In most cases, the back plate and the pole piece are built into one piece called the yoke. The yoke and the front focusing plate are normally made of a very soft cast iron. Iron is a material that is used in conjunction with magnetic structures because the iron is easily saturated when exposed to a magnetic field. Notice in figure 2, that an air gap was intentionally left between the front focusing plate and the yoke. The magnetic field is coupled through the air gap. The magnetic field strength (B) of the air gap is typically optimized for uniformity across the gap. [1]
Figure 2 Permanent Magnet Structure
When a coil of wire with a current flowing is place inside the permanent magnetic field, a force is produced. B is the magnetic field strength, l is the length of the coil, and I is the current flowing through the coil.
{\displaystyle F=Bli}
Figure 3 Voice Coil Mounted in Permanent Magnetic Structure
The coil is excited with the AC signal that is intended for sound reproduction, when the changing magnetic field of the coil interacts with the permanent magnetic field then the coil moves back and forth in order to reproduce the input signal. The coil of a loudspeaker is known as the voice coil.
Figure 4 Photograph - Voice Coil
The Loudspeaker Cone SystemEdit
On a typical loudspeaker, the cone serves the purpose of creating a larger radiating area allowing more air to be moved when excited by the voice coil. The cone serves a piston that is excited by the voice coil. The cone then displaces air creating a sound wave. In an ideal environment, the cone should be infinitely rigid and have zero mass, but in reality neither is true. Cone materials vary from carbon fiber, paper, bamboo, and just about any other material that can be shaped into a stiff conical shape. The loudspeaker cone is a very critical part of the loudspeaker. Since the cone is not infinitely rigid, it tends to have different types of resonance modes form at different frequencies, which in turn alters and colors the reproduction of the sound waves. The shape of the cone directly influences the directivity and frequency response of the loudspeaker. When the cone is attached to the voice coil, a large gap above the voice coil is left exposed. This could be a problem if foreign particles make their way into the air gap of the voice coil and the permanent magnet structure. The solution to this problem is to place what is known as a dust cap on the cone to cover the air gap. Below a figure of the cone and dust cap are shown.
Figure 6 Cone and Dust Cap attached to Voice Coil
The Loudspeaker SuspensionEdit
Most moving coil loudspeakers have a two piece suspension system, also known as a flexure system. The combination of the two flexures allows the voice coil to maintain linear travel as the voice coil is energized and provide a restoring force for the voice coil system. The two piece system consists of large flexible membrane surrounding the outside edge of the cone, called the surround, and an additional flexure connected directly to the voice coil, called the spider. The surround has another purpose and that is to seal the loudspeaker when mounted in an enclosure. Commonly, the surround is made of a variety of different materials, such as, folded paper, cloth, rubber, and foam. Construction of the spider consists of different woven cloth or synthetic materials that are compressed to form a flexible membrane. The following two figures illustrate where the suspension components are physically at on the loudspeaker and how they function as the loudspeaker operates.
Figure 7 Loudspeaker Suspension System
Figure 8 Moving Loudspeaker
Modeling the Loudspeaker as a Lumped SystemEdit
Before implementing a loudspeaker into a specific application, a series of parameters characterizing the loudspeaker must be extracted. The equivalent circuit of the loudspeaker is key when developing enclosures. The circuit models all aspects of the loudspeaker through an equivalent electrical, mechanical, and acoustical circuit. Figure 9 shows how the three equivalent circuits are connected. The electrical circuit is comprised of the DC resistance of the voice coil, Re, the imaginary part of the voice coil inductance, Le, and the real part of the voice coil inductance, Revc. The mechanical system has electrical components that model different physical parameters of the loudspeaker. In the mechanical circuit, Mm, is the electrical capacitance due to the moving mass, Cm, is the electrical inductance due to the compliance of the moving mass, and Rm, is the electrical resistance due to the suspension system. In the acoustical equivalent circuit, Ma models the air mass and Ra models the radiation impedance[2]. This equivalent circuit allows insight into what parameters change the characteristics of the loudspeaker. Figure 10 shows the electrical input impedance as a function of frequency developed using the equivalent circuit of the loudspeaker.
Figure 9 Loudspeaker Analogous Circuit
Figure 10 Electrical Input Impedance
[1] The Loudspeaker Design Cookbook 5th Edition; Dickason, Vance., Audio Amateur Press, 1997. [2] Beranek, L. L. Acoustics. 2nd ed. Acoustical Society of America, Woodbridge, NY. 1993.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Moving_Coil_Loudspeaker&oldid=3232732"
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Q 3 A beam of light converges towards a point O, 10 cm behind a concave mirror of focal - Science - Light - Reflection and Refraction - 12667539 | Meritnation.com
v=-\frac{20}{3} cm
Now from the mirror equation we can write that ,
1/f = 1/u + 1/v \phantom{\rule{0ex}{0ex}}⇒-1/20 = -1/10 + 1/v \phantom{\rule{0ex}{0ex}}⇒ 1/v = 1/20 \phantom{\rule{0ex}{0ex}}⇒v = 20 cm
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Physics Study Guide/Energy - Wikibooks, open books for an open world
Physics Study Guide/Energy
Kinetic energy is simply the capacity to do work by virtue of motion.
(Translational) kinetic energy is equal to one-half of mass times the square of velocity.
{\displaystyle \mathrm {KE} _{T}={\frac {1}{2}}m\|{\vec {v}}\|^{2}}
(Rotational) kinetic energy is equal to one-half of moment of inertia times the square of angular velocity.
{\displaystyle \mathrm {KE} _{R}={\frac {1}{2}}I\omega ^{2}}
Total kinetic energy is simply the sum of the translational and rotational kinetic energies. In most cases, these energies are separately dealt with. It is easy to remember the rotational kinetic energy if you think of the moment of inertia I as the rotational mass. However, you should note that this substitution is not universal but rather a rule of thumb.
Potential energy is simply the capacity to do work by virtue of position (or arrangement) relative to some zero-energy reference position (or arrangement).
Potential energy due to gravity is equal to the product of mass, acceleration due to gravity, and height (elevation) of the object.
{\displaystyle \mathrm {PE} _{g}=-m{\vec {g}}\cdot {\vec {x}}=mgy}
Note that this is simply the vertical displacement multiplied by the weight of the object. The reference position is usually the level ground but the initial position like the rooftop or treetop can also be used. Potential energy due to spring deformation is equal to one-half the product of the spring constant times the square of the change in length of the spring.
{\displaystyle \mathrm {PE} _{e}={\frac {1}{2}}k\|{\vec {x}}-{\vec {x}}_{e}\|^{2}={\frac {1}{2}}k\|\Delta {\vec {x}}\|^{2}}
The reference point of spring deformation is normally when the spring is "relaxed," i.e. the net force exerted by the spring is zero. It will be easy to remember that the one-half factor is inserted to compensate for finite '"change in length" since one would want to think of the product of force and change in length
{\displaystyle (k\Delta {\vec {x}})\cdot \Delta {\vec {x}}}
directly. Since the force actually varies with
{\displaystyle \Delta {\vec {x}}}
, it is instructive to need a "correction factor" during integration.
K: Kinetic energy (J)
m: mass (kg)
v: velocity (m/s)
I: moment of inertia, (kg·m2)
ω: ("omega") angular momentum (rad/s)
Ug: Potential energy (J)
g: local acceleration due to gravity (on the earth’s surface, 9.8 m/s2)
h: height of elevation (m)
Ue: Potential energy (J)
k: spring constant (N/m)
Δx: change in length of spring (m)
Energy: a theoretically indefinable quantity that describes potential to do work. SI unit for energy is the joule (J). Also common is the calorie (cal).
The joule: defined as the energy needed to push with the force of one newton over the distance of one meter. Equivalent to one newton-meter (N·m) or one watt-second (W·s).
1 joule = 1 J = 1 newton • 1 meter = 1 watt • 1 second
Energy comes in many varieties, including Kinetic energy, Potential energy, and Heat energy.
Kinetic energy (K): The energy that an object has due to its motion. Half of velocity squared times mass. Units: joules (J)
Potential energy due to gravity (UG): The energy that an object has stored in it by elevation from a mass, such as raised above the surface of the earth. This energy is released when the object becomes free to move. Mass times height time acceleration due to gravity. Units: joules (J)
Potential energy due to spring compression (UE): Energy stored in spring when it is compressed. Units: joules (J)
Heat energy (Q): Units: joules (J)
Spring compression (Dx): The difference in length between the spring at rest and the spring when stretched or compressed. Units: meters (m)
Spring constant (k): a constant specific to each spring, which describes its “springiness”, or how much work is needed to compress the spring. Units: newtons per meter (N/m)
Change in spring length (Δx): The distance between the at-rest length of the spring minus the compressed or extended length of the spring. Units: meters (m)
Moment of inertia (I): Describes mass and its distribution. (kg•m2)
Angular momentum (ω): Angular velocity times mass (inertia). (rad/s)
Retrieved from "https://en.wikibooks.org/w/index.php?title=Physics_Study_Guide/Energy&oldid=3825413"
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Classification Problems | Brilliant Math & Science Wiki
Karleigh Moore and Christopher Williams contributed
Classification is a central topic in machine learning that has to do with teaching machines how to group together data by particular criteria. Classification is the process where computers group data together based on predetermined characteristics — this is called supervised learning. There is an unsupervised version of classification, called clustering where computers find shared characteristics by which to group data when categories are not specified.
A common example of classification comes with detecting spam emails. To write a program to filter out spam emails, a computer programmer can train a machine learning algorithm with a set of spam-like emails labelled as spam and regular emails labelled as not-spam. The idea is to make an algorithm that can learn characteristics of spam emails from this training set so that it can filter out spam emails when it encounters new emails.
Classification is an important tool in today’s world, where big data is used to make all kinds of decisions in government, economics, medicine, and more. Researchers have access to huge amounts of data, and classification is one tool that helps them to make sense of the data and find patterns.
While classification in machine learning requires the use of (sometimes) complex algorithms, classification is something that humans do naturally everyday. Classification is simply grouping things together according to similar features and attributes. When you go to a grocery store, you can fairly accurately group the foods by food group (grains, fruit, vegetables, meat, etc.) In machine learning, classification is all about teaching computers to do the same.
Here are a few examples of situations where classification is useful:
Let’s say you are trying to write a machine learning program that will be able to detect cancerous tumors in lungs. It takes in images of lung x-rays as input and determines if there is a tumor. If there is a tumor, we’d like the computer to output “yes” and if there is not a tumor, we’d like the computer to output “no.” We’d like the computer to output the correct answer as much as possible.
Say the training set for this algorithm consists of several images of x-rays, half of the images contain tumors and are labelled “yes” and the other half do not contain tumors and are labelled “no.”
If the algorithm learns how to identify tumors with high accuracy, you can see why this might be a useful tool in a medical setting — a computer could save doctors time by analyzing x-ray images quickly.
Let’s say there are linguistics researchers studying grammar structures in languages. They have a bunch of text files of transcribed speeches that they want to analyze. They want to teach a computer to recognize parts of speech, such as adjectives, subject, and verbs, in the sentences.
Let’s say you are writing a program that recommends new music to a user based on their music preferences. The user selects from a list of songs the songs that he or she likes and the program should show him or her new songs that they are likely to enjoy.
In this case, what is the input training data? What are the labels?
The input data would be the songs that the user selected that they liked, this could be labelled as “like” and the songs that the user did not select from the list are labelled as “dislike.”
Say you work in a computer processor factory. As the processors are being prepared to be packaged and shipped, you must conduct a quality check to make sure that none of the processors are damaged. Describe how you might get a computer to do this job for you using machine learning and classification.
Show possible solution
You could connect a computer to a camera that photographs each processor before it is shipped. The computer will run an algorithm that classifies the processor as damaged or not damaged. The training set you could use to teach this algorithm to determine which processors are damaged would be images of defective processors and images of functional processors.
Choosing the right classification algorithm is very important. An algorithm that performs classification is called a classifier. A classifier algorithm should be fast, accurate, and sometimes, minimize the amount of training data that it needs. Generally, the more parameters a set of data has, the larger the training set for an algorithm must be. Different classification algorithms basically have different ways of learning patterns from examples. [1]
More formally, classification algorithms map an observation
v
to a concept/class/label
\omega
Many times, classification algorithms will take in data in the form of a feature vector which is basically a vector containing numeric descriptions of various features related to each data object. For example, if the algorithm deals with sorting images of animals into various classes (based on what type of animal they are, for example), the feature vector might include information about the pixels, colors in the image, etc.
Here are some common classification algorithms and techniques:
A common and simple method for classification is linear regression.
(x_i,y_i),
x_i
y_i
are observations of the two variables which are expected to depend linearly on each other.
The best-fitting linear relationship between the variables
x
y
Which of these lines, H1, H2, and H3, represents the worst classifier algorithm? (The classifier algorithms identify and label data and place them on one side of the line or the other according to the results).
H3 and H2 have no error in this graph — all of the solid circles are to one side and all of the hollow circles are to the other side of the lines. H1, on the other hand, groups solid circles with the hollow circles, this is a poor classifier.
A perceptron is an algorithm used to produce a binary classifier. That is, the algorithm takes binary classified input data, along with their classification and outputs a line that attempts to separate data of one class from data of the other: data points on one side of the line are of one class and data points on the other side are of the other. Binary classified data is data where the label is one thing or another, like "yes" or "no"; 1 or 0; etc.
The perceptron algorithm returns values of
w_0, w_1, ..., w_k
b
such that data points on one side of the line are of one class and data points on the other side are of the other. Mathematically, the values of
\boldsymbol{w}
b
are used by the binary classifier in the following way. If
\boldsymbol{w} \cdot \boldsymbol{x} + b > 0
, the classifier returns 1; otherwise, it returns 0. Note that 1 represents membership of one class and 0 represents membership of the other. This can be seen more clearly with the AND operator, replicated below for convenience.
The AND operation between two numbers. A red dot represents one class (
x_1
x_2 = 0
) and a blue dot represents the other class (
x_1
x_2 = 1
). The line is the result of the perceptron algorithm, which separates all data points of one class from those of the other.
The perceptron algorithm is one of the most commonly used machine learning algorithms for binary classification. Some machine learning tasks that use the perceptron include determining gender, low vs high risk for diseases, and virus detection.
Naive Bayes classifiers are probabilistic classifiers with strong independence assumptions between features. Unlike many other classifiers which assume that, for a given class, there will be some correlation between features, naive Bayes explicitly models the features as conditionally independent given the class.
Because of the independence assumption, naive Bayes classifiers are highly scalable and can quickly learn to use high dimensional (many parameters) features with limited training data. This is useful for many real world datasets where the amount of data is small in comparison with the number of features for each individual piece of data, such as speech, text, and image data.
Another way to do a classification is to use a decision tree.
Say you have the following training data set of basketball players that includes information about what color jersey they have, which position they play, and whether or not they are injured. The training set is labelled according to whether or not a player will be able to play for Team A.
Person Jersey Color Offense or Defense Injured? Will they play for Team A?
John Blue Offense No Yes
Steve Red Offense No No
Sarah Blue Defense No Yes
Rachel Blue Offense Yes No
Richard Red Defense No No
Alex Red Defense Yes No
Lauren Blue Offense No Yes
Carol Blue Defense No Yes
What is the rule for whether or not a player may play for Team A?
They must have a blue jersey and not be injured.
To use a decision tree to classify this data, select a rule to start the tree.
Here we will use “jersey color” as the root node.
Next, we will include a node that will distinguish between injured and uninjured players.
Other types of classification algorithms include support vector machines (SVMs), Bayesian, and logistic regression.
Use of Statistics In Input Data
Classification algorithms often include statistics data.
Let's say a company is trying to write software that will take in as input the full text of a book. From the frequency of certain words, the program will determine which genre the book belongs to. Perhaps if the word "detective" appears very frequently, the program will label the book as a "mystery." If the book contains the word "magic" or "wizard" many times, perhaps the software should label the book as "fantasy." And so forth. (In this example, the language of the books is English).
Let's say that the computer program goes through each book and keeps track of the number of times each word occurs. The algorithm might find that across all genres, the words "the," "is," "and,", "I," and other very common English words occur with about the same frequency. Classifying the novels based on these word frequencies would probably not be very helpful. However, if the algorithm notices that a particular subset of words tend to occur more often in science-fiction novels and fantasy novels than in mystery novels or non-fiction novels, the algorithm can use this information to sort future book instances.
In the basketball team example above, the rules for determining if a player would play for Team A were fairly straightforward with just two binary data points to consider. In book genre example, a historical-fiction novel might contain the word "detective" many times if its topic has to do with a famous unsolved crime. It is possible that the machine learning algorithm would classify this novel as a mystery book. This is called error. Many times, error can be reduced by feeding the algorithm more training examples. However, eliminating error completely is very difficult to do, so in general, a good classifier algorithm will have as low an error rate as possible.
Classification, and its unsupervised learning counterpart, clustering, are central ideas behind many other techniques and topics in machine learning. Being able to classify and recognize certain kinds of data allows computer scientists to expand on knowledge and applications in other machine learning fields such as computer vision, natural language processing, deep learning, building predictive economic, market, and weather models, and more.
Schurmann, J. (1996). Pattern Classification. John Wiley & Sons Inc..
Sewaqu, . Linear regression. Retrieved November 5, 2010, from https://en.wikipedia.org/wiki/Least_squares#/media/File:Linear_regression.svg
, Z. File:Svm separating hyperplanes (SVG).svg. Retrieved July 18, 2016, from https://en.wikipedia.org/wiki/File:Svm_separating_hyperplanes_(SVG).svg
Cite as: Classification Problems. Brilliant.org. Retrieved from https://brilliant.org/wiki/classification/
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Normally distributed and uncorrelated does not imply independent - Wikipedia
In probability theory, although simple examples illustrate that linear uncorrelatedness of two random variables does not in general imply their independence, it is sometimes mistakenly thought that it does imply that when the two random variables are normally distributed. This article demonstrates that assumption of normal distributions does not have that consequence, although the multivariate normal distribution, including the bivariate normal distribution, does.
To say that the pair
{\displaystyle (X,Y)}
of random variables has a bivariate normal distribution means that every linear combination
{\displaystyle aX+bY}
{\displaystyle X}
{\displaystyle Y}
for constant (i.e. not random) coefficients
{\displaystyle a}nd
{\displaystyle b}
(not both equal to zero) has a univariate normal distribution. In that case, if
{\displaystyle X}
{\displaystyle Y}
are uncorrelated then they are independent.[1] However, it is possible for two random variables
{\displaystyle X}
{\displaystyle Y}
to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.
1.1 A symmetric example
1.2 An asymmetric example
1.3 Examples with support almost everywhere in ℝ2
A symmetric example[edit]
Joint range of
{\displaystyle X}
{\displaystyle Y}
. Darker indicates higher value of the density function.
{\displaystyle X}
has a normal distribution with expected value 0 and variance 1. Let
{\displaystyle W}
have the Rademacher distribution, so that
{\displaystyle W=1}
{\displaystyle W=-1}
, each with probability 1/2, and assume
{\displaystyle W}
{\displaystyle X}
{\displaystyle Y=WX}
{\displaystyle X}
{\displaystyle Y}
are uncorrelated;
both have the same normal distribution; and
{\displaystyle X}
{\displaystyle Y}
are not independent.[2][3]
{\displaystyle X}
{\displaystyle Y}
are uncorrelated, one may consider the covariance
{\displaystyle \operatorname {cov} (X,Y)}
: by definition, it is
{\displaystyle \operatorname {cov} (X,Y)=\operatorname {E} (XY)-\operatorname {E} (X)\operatorname {E} (Y).}
Then by definition of the random variables
{\displaystyle X}
{\displaystyle Y}
{\displaystyle W}
, and the independence of
{\displaystyle W}
{\displaystyle X}
{\displaystyle \operatorname {cov} (X,Y)=\operatorname {E} (XY)-0=\operatorname {E} (X^{2}W)=\operatorname {E} (X^{2})\operatorname {E} (W)=\operatorname {E} (X^{2})\cdot 0=0.}
{\displaystyle Y}
has the same normal distribution as
{\displaystyle X}
, consider
{\displaystyle {\begin{aligned}\Pr(Y\leq x)&{}=\operatorname {E} (\Pr(Y\leq x\mid W))\\&{}=\Pr(X\leq x)\Pr(W=1)+\Pr(-X\leq x)\Pr(W=-1)\\&{}=\Phi (x)\cdot {\frac {1}{2}}+\Phi (x)\cdot {\frac {1}{2}}\end{aligned}}}
{\displaystyle X}
{\displaystyle -X}
both have the same normal distribution), where
{\displaystyle \Phi (x)}
is the cumulative distribution function of the Standard normal distribution..
{\displaystyle X}
{\displaystyle Y}
are not independent, observe that
{\displaystyle |Y|=|X|}
{\displaystyle \operatorname {Pr} (Y>1|-1/2<X<1/2)=\operatorname {Pr} (X>1|-1/2<X<1/2)=0}
Finally, the distribution of the simple linear combination
{\displaystyle X+Y}
concentrates positive probability at 0:
{\displaystyle \operatorname {Pr} (X+Y=0)=1/2}
. Therefore, the random variable
{\displaystyle X+Y}
is not normally distributed, and so also
{\displaystyle X}
{\displaystyle Y}
are not jointly normally distributed (by the definition above).
An asymmetric example[edit]
The joint density of
{\displaystyle X}
{\displaystyle Y}
. Darker indicates a higher value of the density.
{\displaystyle X}
{\displaystyle Y=\left\{{\begin{matrix}X&{\text{if }}\left|X\right|\leq c\\-X&{\text{if }}\left|X\right|>c\end{matrix}}\right.}
{\displaystyle c}
is a positive number to be specified below. If
{\displaystyle c}
is very small, then the correlation
{\displaystyle \operatorname {corr} (X,Y)}
is near
{\displaystyle -1}
{\displaystyle c}
is very large, then
{\displaystyle \operatorname {corr} (X,Y)}
is near 1. Since the correlation is a continuous function of
{\displaystyle c}
, the intermediate value theorem implies there is some particular value of
{\displaystyle c}
that makes the correlation 0. That value is approximately 1.54.[note 1] In that case,
{\displaystyle X}
{\displaystyle Y}
are uncorrelated, but they are clearly not independent, since
{\displaystyle X}
completely determines
{\displaystyle Y}
{\displaystyle Y}
is normally distributed—indeed, that its distribution is the same as that of
{\displaystyle X}
—one may compute its cumulative distribution function:
{\displaystyle {\begin{aligned}\Pr(Y\leq x)&=\Pr(\{|X|\leq c{\text{ and }}X\leq x\}{\text{ or }}\{|X|>c{\text{ and }}-X\leq x\})\\&=\Pr(|X|\leq c{\text{ and }}X\leq x)+\Pr(|X|>c{\text{ and }}-X\leq x)\\&=\Pr(|X|\leq c{\text{ and }}X\leq x)+\Pr(|X|>c{\text{ and }}X\leq x)\\&=\Pr(X\leq x),\end{aligned}}}
where the next-to-last equality follows from the symmetry of the distribution of
{\displaystyle X}
and the symmetry of the condition that
{\displaystyle |X|\leq c}
In this example, the difference
{\displaystyle X-Y}
is nowhere near being normally distributed, since it has a substantial probability (about 0.88) of it being equal to 0. By contrast, the normal distribution, being a continuous distribution, has no discrete part—that is, it does not concentrate more than zero probability at any single point. Consequently
{\displaystyle X}
{\displaystyle Y}
are not jointly normally distributed, even though they are separately normally distributed.[4]
Examples with support almost everywhere in ℝ2[edit]
It is well-known that the ratio
{\displaystyle C}
of two independent standard normal random deviates
{\displaystyle X_{i}}
{\displaystyle Y_{i}}
has a Cauchy distribution. One can equally well start with the Cauchy random variable
{\displaystyle C}
and derive the conditional distribution of
{\displaystyle Y_{i}}
to satisfy the requirement that
{\displaystyle X_{i}=CY_{i}}
{\displaystyle X_{i}}
{\displaystyle Y_{i}}
independent and standard normal. Cranking through the math one finds that
{\displaystyle Y_{i}=W_{i}{\sqrt {\frac {\chi _{i}^{2}\left(k=2\right)}{1+C^{2}}}}}
{\displaystyle W_{i}}
is a Rademacher random variable and
{\displaystyle \chi _{i}^{2}\left(k=2\right)}
is a Chi-squared random variable with two degrees of freedom.
Consider two sets of
{\displaystyle \left(X_{i},Y_{i}\right)}
{\displaystyle i\in \left\{1,2\right\}}
{\displaystyle C}
is not indexed by
{\displaystyle i}
– that is, the same Cauchy random variable
{\displaystyle C}
is used in the definition of both
{\displaystyle \left(X_{1},Y_{1}\right)}
{\displaystyle \left(X_{2},Y_{2}\right)}
. This sharing of
{\displaystyle C}
results in dependences across indices: neither
{\displaystyle X_{1}}
{\displaystyle Y_{1}}
{\displaystyle Y_{2}}
. Nevertheless all of the
{\displaystyle X_{i}}
{\displaystyle Y_{i}}
are uncorrelated as the bivariate distributions all have reflection symmetry across the axes.
Non-normal joint distributions with normal marginals.
The figure shows scatterplots of samples drawn from the above distribution. This furnishes two examples of bivariate distributions that are uncorrelated and have normal marginal distributions but are not independent. The left panel shows the joint distribution of
{\displaystyle X_{1}}
{\displaystyle Y_{2}}
; the distribution has support everywhere but at the origin. The right panel shows the joint distribution of
{\displaystyle Y_{1}}
{\displaystyle Y_{2}}
; the distribution has support everywhere except along the axes and has a discontinuity at the origin: the density diverges when the origin is approached along any straight path except along the axes.
^ Hogg, Robert; Tanis, Elliot (2001). "Chapter 5.4 The Bivariate Normal Distribution". Probability and Statistical Inference (6th ed.). pp. 258–259. ISBN 0130272949.
^ UIUC, Lecture 21. The Multivariate Normal Distribution, 21.6:"Individually Gaussian Versus Jointly Gaussian".
^ Rosenthal, Jeffrey S. (2005). "A Rant About Uncorrelated Normal Random Variables".
^ Edward L. Melnick and Aaron Tenenbein, "Misspecifications of the Normal Distribution", The American Statistician, volume 36, number 4 November 1982, pages 372–373
^ More precisely 1.53817..., the square root of the median of a chi-squared distribution with 3 degrees of freedom.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Normally_distributed_and_uncorrelated_does_not_imply_independent&oldid=1086853908"
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why NH3 is strong complexing reagent than cl- ?
Is complex salts or coordination compound found in earth crust??
Please explain how to solve the 27th one:
27. \mathrm{CFSE}\left({∆}_{0}\right) \mathrm{for} \mathrm{metal} \mathrm{ion} \mathrm{in} {\mathrm{d}}^{7} \mathrm{configuration} \mathrm{in} \mathrm{presence} \mathrm{of} \mathrm{strong} \mathrm{ligand} \mathrm{field} \mathrm{is}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(1\right) -0.3 {∆}_{0} \left(2\right) -0.8 {∆}_{0} \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) -1.6 {∆}_{0} \left(4\right) -1.8 {∆}_{0}
How to apply VBT and find the no. of unpaired electrons to find magnetic moment if there is more than one ligand in the complex?
Q. The coordination number and magnetic moment of the complex [Cr(C2O4)2(NH3)2]- respectively is
(1) 6, 3.87 BM (2) 4, 3.87 BM
In the following compound why is Fe written as iron and not ferrate? Also what does the (II) in the middle of Hexamamminenickel(II)chloride represent?
Explain how to solve this question. How to find secondary valancy of these compounds? And how to find coordination number from complex?
Q.7. The spin only magnetic moment value of 5.92 BM. Predict the geometry of the complex ion.
How is oxidation nmbr 3 of cobalt in these compunds.?
when do we add -ate in the IUPAC nomenclature of coordination compounds?
What are the oxidation numbers of some frequently used compounds and metal ions/atoms in coordination chemistry?
iupac name of the given structure
{K}_{2} \left[{O}_{S} c{l}_{5} N\right]
\left[{\left(N{H}_{3}\right)}_{5}Cr—OH—Cr{\left(N{H}_{3}\right)}_{5}\right]Cl
of the complex [Ni(NH3)5Br]Cl] the ionization isomer will give colour with AgNO3......
Please explain the right with detailed explanation
Tetra ammine nickel(ii) hexacyano ferrate (iii) representation?
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(→AMEXP: Exponential-Time AM: fixing a wild HTML escape!)
m (→AM[polylog]: AM With Polylog Rounds: oops, and another)
Same as [[#am|AM]], except that we allow polylog(n) rounds of interaction between Arthur and Merlin instead of a constant number.
Not much is known about AM[polylog] -- for example, whether it sits in [[Complexity Zoo:P#ph|PH]]. However, [[zooref#ss04|[SS04]]] show that if AM[polylog] contains [[Complexity Zoo:C#conp|coNP]], then [[Complexity Zoo:E#eh|EH]] collapses to [[Complexity Zoo:S#s2exppnp|S<sub>2</sub>-EXP•P<sup>NP</sup>]]. ([[zooref#pv04|[PV04]]] improved the collapse to [[#amexp|AM<sub>EXP</sub>]].)
Not much is known about AM[polylog] -- for example, whether it sits in [[Complexity Zoo:P#ph|PH]]. However, [[zooref#ss04|[SS04]]] show that if AM[polylog] contains [[Complexity Zoo:C#conp|coNP]], then [[Complexity Zoo:E#eh|EH]] collapses to [[Complexity Zoo:S#s2exppnp|S<sub>2</sub>-EXP•P<sup>NP</sup>]]. ([[zooref#pv04|[PV04]]] improved the collapse to [[#amexp|AM<sub>EXP</sub>]].)
===== <span id="ampmp" style="color:red">AmpMP</span>: Amplifiable [[Complexity Zoo:M#mp2|MP]] =====
The class of decision problems such that for some [[Complexity Zoo:Symbols#sharpp|#P]] function f(x,0<sup>m</sup>),
{\displaystyle t(n)}
{\displaystyle t(n)}
{\displaystyle \Sigma _{0}=\Delta _{0}=\Pi _{0}}
{\displaystyle \times }
{\displaystyle \phi =\forall i<j\psi }
{\displaystyle \phi =\exists i<j\psi }
{\displaystyle j}
{\displaystyle \phi }
{\displaystyle \Sigma _{i+1}}
{\displaystyle \exists X_{1}\dots \exists X_{n},\psi }
{\displaystyle \psi \in \Delta _{i}}
{\displaystyle \Sigma _{i}}
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Tell whether the function represents exponential growth or exponential decay. Then graph t
Tell whether the function represents exponential growth or exponential decay. Then graph the function. y=e^-3x
y={e}^{-3}x
y={e}^{-3}x
The function is an exponential decay when has the form
y=a{e}^{-}kt
y={e}^{-3}x
Solve the equation. Round your answers to the nearest hundredth.
{8}^{x}=12
{8}^{x}=12
Use the two-way table to create a two-way table that shows the joint and marginal relative frequencies.
\overline{)\begin{array}{cc}\text{Class }& \text{Gender}\\ & \text{Male}& \text{Female}& \text{Total}\\ \text{Sophomore}& 26& 54& 89\\ \text{Junior}& 32& 48& 80\\ \text{Senior}& 50& 55& 105\\ \text{Total}& 108& 157& 265\end{array}}
Determine whether each function represents exponential growth or exponential decay. Identify the percent rate of change.
g\left(t\right)=2{\left(\frac{5}{4}\right)}^{t}
Determine whether each function is an example of exponential growth or decay. Then find the y-intercept.
y=0.6\left(1/10{\right)}^{x}
y={\left(\frac{4}{3}\right)}^{x}
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What is the correlation coefficient and what is its significance? Explain why co
Correlation a measure which indicates the “go-togetherness” of two data sets. It can be denoted as r. The value of correlation coefficient lies between –1 and +1. The positive 1 indicates that the two data sets are perfect and both are in same direction. The negative 1 indicates that the two data sets are perfect and both are in opposite direction. It will be zero when there is no relationship between the two data sets.
Correlation coefficient, r:
The Karl Pearson’s product-moment correlation coefficient or simply, the Pearson’s correlation coefficient is a measure of the strength of a linear association between two variables and is denoted by r or
{r}_{xy}
The coefficient of correlation
{r}_{xy}
between two variables x and y for the bivariate data set
\left({x}_{i},{y}_{i}\right)f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}i=1,2,3\dots N
{r}_{xy}=\frac{n\left(\sum xy\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n\left(\sum {x}^{2}\right)-\left(\sum {x}^{2}\right)\right]×\left[n\left(\sum {y}^{2}\right)-\left(\sum {y}^{2}\right)\right]}}
Scatterplot and correlation:
A scatterplot is a type of data display that shows the relationship between two numerical variables. Each member of the data set gets plotted as a point whose (x, y) coordinates relates to its values for the two variables.
When the y variable tends to increase as the x variable increases, it can be said that there is a positive correlation between the variables. In other words, when the points on the scatterplot produce a lower left to upper right pattern, there is a positive correlation between the variables.
When the y variable tends to decrease as the x variable increases, it can be said that there is a negative correlation between the variables. In other words, when the points on the scatterplot produce an upper left to lower right pattern, there is a positive correlation between the variables.
When all the points on a scatterplot lie on a straight line, it can be said that there is a perfect correlation between the two variables.
A scatterplot in which the points do not have a linear trend (either positive or negative) is called a zero correlation or a near-zero correlation.
Form of the association between variables:
The form of the association describes whether the data points follow a linear pattern or some other complicated curves. For data if it appears that a line would do a reasonable job of summarizing the overall pattern in the data. Then, the association between two variables is linear.
Direction of association:
If the increase in the values of one variable increases the values of another variable, then the direction is positive. If the increase in the values of one variable decreases the values of another variable, then the direction is negative.
Strength of the association:
The association is said to be strong if all the points are close to the straight line. It is said to be weak if all points are far away from the straight line and it is said to be moderate if the data points are moderately close to straight line.
Least squares line fitting:
Regression analysis estimates the relationship among variables. That is, it estimates the relationship between one dependent variable and one or more independent variables.
The general form of first-order regression model is
y-\cap ={\beta }_{0}+{\beta }_{1}x+ϵ
, Where, the variable y is the dependent variable that is to be modelled or predicted, the variable x is the independent variable that is used to predict the dependent variable, and ε is the error term.
The difference between of the observed value of y and predicted value of value of y is called as residual. Hence, the value of residual is represented as
y–\left(y-\cap \right)
If the sum of the squares of the residuals is expressed as smallest sum possible, then the straight line satisfies the least squares property. The regression line of the straight line satisfies the least-squares property then that "best fits the points in a scatterplot.
You were asked about advantages of using box plots and dot plots to describe and compare distributions of scores. Do you think the advantages you found would exist not only for these data, but for numerical data in general? Explain.
Find the regression line using the given points.
The two-way table summarizes data from an experiment comparing the effectiveness of three different diets (A, B, and C) on weight loss. Researchers randomly assigned 300 volunteer subjects to the three diets. The response variable was whether each subject lost weight over a 1-year period.
\text{Diet}\text{ }\text{Lost weight?}
\begin{array}{lcccc}& \mathrm{A}& \mathrm{B}& \mathrm{C}& \text{ Total }\\ \text{ Yes }& & 60& & 180\\ \text{ No }& & 40& & 120\\ \text{ Total }& 90& 100& 110& 300\end{array}
Suppose we randomly select one of the subjects from the experiment. Show that the events "Diet B" and "Lost weight" are independent.
{s}_{xy}
In bivariate data, we sometimes notice that one of the quantities increases (1, 2, 3...) while the other quantity decreases (20, 19, 18...). Which phrase best describes this association? would it be no correlation, a perfect correlation,a positive correlation or a negative correlation?
Suppose you were to collect data for each pair of variables. You want to make a scatterplot. Which variable would you use as the explanatory variable and which as the response variable? Why? What would you expect to see in the scatterplot? Discuss the likely direction, form, and strength. Gasoline: number of miles you drove since filling up, gallons remaining in your tank
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Solve system of differential equations - MATLAB dsolve - MathWorks Australia
\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}
{C}_{1} {\mathrm{e}}^{a t}
\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}=\mathit{ay}
{C}_{1} {\mathrm{e}}^{-\sqrt{a} t}+{C}_{2} {\mathrm{e}}^{\sqrt{a} t}
\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}
y\left(0\right)=5
5 {\mathrm{e}}^{a t}
\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}={\mathit{a}}^{2}\mathit{y}
y\left(0\right)=b
{y}^{\prime }\left(0\right)=1
\frac{{\mathrm{e}}^{a t} \left(a b+1\right)}{2 a}+\frac{{\mathrm{e}}^{-a t} \left(a b-1\right)}{2 a}
\begin{array}{l}\frac{\mathit{dy}}{\mathit{dt}}=\mathit{z}\\ \frac{\mathit{dz}}{\mathit{dt}}=-\mathit{y}.\end{array}
{C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)
{C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)
{C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)
{C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)
\frac{\partial }{\partial t}y\left(t\right)={e}^{-y\left(t\right)}+y\left(t\right)
\frac{\partial }{\partial t}\mathrm{ }y\left(t\right)={\mathrm{e}}^{-y\left(t\right)}+y\left(t\right)
{\mathrm{W}\text{lambertw}}_{0}\left(-1\right)
F\left(y\left(t\right)\right)=g\left(t\right)
\left(\begin{array}{c}\left({\int \frac{{\mathrm{e}}^{y}}{y {\mathrm{e}}^{y}+1}\mathrm{d}y|}_{y=y\left(t\right)}\right)={C}_{1}+t\\ {\mathrm{e}}^{-y\left(t\right)} \left({\mathrm{e}}^{y\left(t\right)} y\left(t\right)+1\right)=0\end{array}\right)
F\left(y\left(x\right)\right)=g\left(x\right)
{\mathrm{e}}^{y\left(x\right)}+\frac{{y\left(x\right)}^{2}}{2}={C}_{1}+{\mathrm{e}}^{-x}+\frac{{x}^{2}}{2}
\frac{\mathit{dy}}{\mathit{dt}}=\frac{\mathit{a}}{\sqrt{\mathit{y}}}+\mathit{y}
y\left(a\right)=1
{\left({\mathrm{e}}^{\frac{3 t}{2}-\frac{3 a}{2}+\mathrm{log}\left(a+1\right)}-a\right)}^{2/3}
a
\begin{array}{l}\left\{\begin{array}{cl}\left\{\begin{array}{cl}\left\{{\sigma }_{1}\right\}& \text{ if }-\frac{\pi }{2}<{\sigma }_{2}\\ \left\{{\sigma }_{1},-{\left(-a+{\mathrm{e}}^{\frac{3 t}{2}-\frac{3 a}{2}+\mathrm{log}\left(a+{\left(-\frac{1}{2}+{\sigma }_{3}\right)}^{3/2}\right)+2 \pi {C}_{2} \mathrm{i}}\right)}^{2/3} \left(\frac{1}{2}+{\sigma }_{3}\right)\right\}& \text{ if }{\sigma }_{2}\le -\frac{\pi }{2}\end{array}& \text{ if }{C}_{2}\in \mathbb{Z}\\ \varnothing & \text{ if }{C}_{2}\notin \mathbb{Z}\end{array}\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}={\left(-a+{\mathrm{e}}^{\frac{3 t}{2}-\frac{3 a}{2}+\mathrm{log}\left(a+1\right)+2 \pi {C}_{2} \mathrm{i}}\right)}^{2/3}\\ \\ \mathrm{ }{\sigma }_{2}=\text{angle}\left({\mathrm{e}}^{\frac{3 {C}_{1}}{2}+\frac{3 t}{2}}-a\right)\\ \\ \mathrm{ }{\sigma }_{3}=\frac{\sqrt{3} \mathrm{i}}{2}\end{array}
\left({x}^{2}-1{\right)}^{2}\frac{{\partial }^{2}}{\partial {x}^{2}}y\left(x\right)+\left(x+1\right)\frac{\partial }{\partial x}y\left(x\right)-y\left(x\right)=0
{C}_{2} \left(x+1\right)+{C}_{1} \left(x+1\right) \int \frac{{\mathrm{e}}^{\frac{1}{2 \left(x-1\right)}} {\left(1-x\right)}^{1/4}}{{\left(x+1\right)}^{9/4}}\mathrm{d}x
x=-1
\left(\begin{array}{c}x+1\\ \frac{1}{{\left(x+1\right)}^{1/4}}-\frac{5 {\left(x+1\right)}^{3/4}}{4}+\frac{5 {\left(x+1\right)}^{7/4}}{48}+\frac{5 {\left(x+1\right)}^{11/4}}{336}+\frac{115 {\left(x+1\right)}^{15/4}}{33792}+\frac{169 {\left(x+1\right)}^{19/4}}{184320}\end{array}\right)
\infty
\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+1\end{array}\right)
\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}-\frac{1}{90 {x}^{5}}-\frac{37}{336 {x}^{6}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+\frac{37}{336 {x}^{6}}+\frac{37}{1680 {x}^{7}}+1\end{array}\right)
\frac{\mathit{dy}}{\mathit{dx}}=\frac{1}{{\mathit{x}}^{2}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}
{C}_{1}+{\mathrm{e}}^{-\frac{1}{x}}
\mathit{y}\left(0\right)=1
{\mathrm{e}}^{-\frac{1}{x}}+1
{\mathit{e}}^{-\frac{1}{\mathit{x}}}
x=0
\underset{\mathit{x}\to {0}^{+}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=0
\underset{\mathit{x}\to {0}^{-}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=\infty
\mathrm{lim}\text{\hspace{0.17em}}\mathit{x}\to {\mathit{x}}_{0}^{+}
O\left({\mathrm{var}}^{n}\right)
O\left({\mathrm{var}}^{-n}\right)
\mathrm{lim}x\to {x}_{0}^{+}\text{ }
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Descibe in words the region of RR^{3} represented by the equation or inequality.
Descibe in words the region of RR^{3} represented by the equation or inequality.y = -2
Descibe in words the region of
{\mathbb{R}}^{\mathbb{3}}
represented by the equation or inequality.
y=-2
y=-2\text{ }in\text{ }{\mathbb{R}}^{\mathbb{3}}
represenrs the vertical plane that is parallel to the xz - plane and it locates 2 units left to the xz - plane.
R{R}^{3}
is the three dimensional coordinate system which contains x, y and z - coordinates.
y=-2\text{ }in\text{ }{\mathbb{R}}^{\mathbb{3}}
\left(x,y,z\right)|y=-2
{\mathbb{R}}^{\mathbb{3}}
whose y - coordinate is -2 and x, z - coordinates are any values.
y=-2\text{ }in\text{ }{\mathbb{R}}^{\mathbb{3}}
is sketched as shown in Figure 1
From Figure 1, the equation
y=-2\text{ }in\text{ }{\mathbb{R}}^{\mathbb{3}}
represents the vertical plane that is parallel to the xz - plane and it locates 2 units left to the xz - plane.
y=-2\text{ }in\text{ }{\mathbb{R}}^{\mathbb{3}}
is described.
The change - of - coordinate matrix from
\mathcal{B}=\left\{\left[\begin{array}{c}3\\ -1\\ 4\end{array}\right]\left[\begin{array}{c}2\\ 0\\ -5\end{array}\right]\left[\begin{array}{c}8\\ -2\\ 7\end{array}\right]\right\}
to the standard basis in
R{R}^{n}.
Plotting points and finding distance in three dimensions. Two points P and
\underset{―}{O}
(a) Plot P and
\underset{―}{O}
(b) Find the distance between P and
\underset{―}{O}.
P\left(3,1,0\right),\underset{―}{O}\left(-1,2,-5\right)
a\right)\stackrel{\to }{A}\left(x,y,z\right)={\stackrel{\to }{e}}_{x}
d\right)\stackrel{\to }{A}\left(\rho ,\varphi ,z\right)={\stackrel{\to }{e}}_{\rho }
g\right)\stackrel{\to }{A}\left(r,\theta ,\varphi \right)={\stackrel{\to }{e}}_{\theta }
j\right)\stackrel{\to }{A}\left(x,y,z\right)=\frac{-y{\stackrel{\to }{e}}_{x}+x{\stackrel{\to }{e}}_{y}}{{x}^{2}+{y}^{2}}
{R}^{3}
x=5
For the polar point
\left(-4,-\frac{3\pi }{4}\right)
a) Give alternate coordinates with
r>0
-2\pi \le \theta <0
b) Give alternate coordinates with
r<0
0\le \theta <2\pi
c) Give alternate coordinates of your choice.
R{R}^{3}
z\ge -1
Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, rounded to one decimal place.
\left\{\begin{array}{l}y\ge x-3\\ y\ge -2x+6\\ y\le 8\end{array}
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Generate Reward Function from a Model Predictive Controller for a Servomotor - MATLAB & Simulink - MathWorks India
Create Model Predictive Controller
This example shows how to automatically generate a reward function from cost and constraint specifications defined in a model predictive controller object. You then use the generated reward function to train a reinforcement learning agent.
You can use the generateRewardFunction to generate a reward function for reinforcement learning, starting from cost and constraints specified in a model predictive controller. The resulting reward signal is a sum of costs (as defined by an objective function) and constraint violation penalties depending on the current state of the environment.
This example is based on theDC Servomotor with Constraint on Unmeasured Output (Model Predictive Control Toolbox) example, in which you design a model predictive controller for a DC servomechanism under voltage and shaft torque constraints. Here, you will convert the cost and constraints specifications defined in the mpc object into a reward function and use it to train an agent to control the servomotor.
Open the Simulink model for this example which is based on the above MPC example but has been modified for reinforcement learning.
open_system('rl_motor')
Create the open-loop dynamic model of the motor, defined in plant and the maximum admissible torque tau using an helper function.
Specify input and output signal types for the MPC controller. The shaft angular position, is measured as first output. The second output, torque, is unmeasurable.
Specify constraints on the manipulated variable, and define a scale factor.
Impose torque constraints during the first three prediction steps, and specify scale factor for both shaft position and torque.
Specify weights for the quadratic cost function to achieve angular position tracking. Set to zero the weight for the torque, thereby allowing it to float within its constraint.
Create an MPC controller for the plant model with a sample time of 0.1 s, a prediction horizon 10 steps, and a control horizon of 2 steps, using the previously defined structures for the weights, manipulated variables, and output variables.
mpcobj = mpc(plant,0.1,10,2,Weights,MV,OV);
Display the controller specifications.
OutputVariables: [0.1000 0]
ECR: 10000
-220 <= MV1 (V) <= 220, MV1/rate (V) is unconstrained, MO1 (rad) is unconstrained
-78.54 <= UO1 (Nm)(t+1) <= 78.54
UO1 (Nm)(t+4) is unconstrained
The controller operates on a plant with 4 states, 1 input (voltage) and 2 output signals (angle and torque) and has the following specifications:
The cost function weights for the manipulated variable, manipulated variable rate and output variables are 0, 0.1 and [0.1 0] respectively.
The manipulated variable is constrained between -220V and 220V.
The manipulated variable rate is unconstrained.
The first output variable (angle) is unconstrained but the second (torque) is constrained between -78.54 Nm and 78.54 Nm in the first three prediction time steps and unconstrained in the fourth step.
Note that for reinforcement learning only the constraints specification from the first prediction time step will be used since the reward is computed for a single time step.
The generated reward function is a starting point for reward design. You can modify the function with different penalty function choices and tune the weights. For this example, make the following change to the generated code:
Scale the original cost weights Qy, Qmv and Qmvrate by a factor of 100.
After you make changes, the cost and penalty specifications should be as follows:
Qy = [10 0];
Qmvrate = 10;
Py = Wy * exteriorPenalty(y,ymin,ymax,'quadratic');
Pmv = Wmv * exteriorPenalty(mv,mvmin,mvmax,'quadratic');
Pmvrate = Wmvrate * exteriorPenalty(mv-lastmv,mvratemin,mvratemax,'quadratic');
For this example, the modified code has been saved in the MATLAB function file rewardFunctionMpc.m. Display the generated reward function.
type rewardFunctionMpc.m
function reward = rewardFunctionMpc(y,refy,mv,refmv,lastmv)
% REWARDFUNCTIONMPC generates rewards from MPC specifications.
% 02-Jun-2021 16:05:41
ymin = [-Inf -78.5398163397448];
ymax = [Inf 78.5398163397448];
mvmin = -220;
mvmax = 220;
Sy = [6.28318530717959 157.07963267949];
Smv = 440;
Qy = [0.1 0];
Wy = [1 1];
% Alternatively, use the hyperbolicPenalty or barrierPenalty function for
To integrate this reward function, open the MATLAB Function block in the Simulink model.
open_system('rl_motor/Reward Function')
r = rewardFunctionMpc(y,refy,mv,refmv,lastmv);
The MATLAB Function block will now execute rewardFunctionMpc.m during simulation.
The environment dynamics are modeled in the Servomechanism subsystem. For this environment,
The observations are the reference and actual output variables (angle and torque) from the last 8 time steps.
The action is the voltage
\mathit{V}
applied to the servomotor.
{\mathit{T}}_{\mathit{s}}=0.1\mathit{s}
The total simulation time is
{\mathit{T}}_{\mathit{f}}=20\mathit{s}
Specify the total simulation time and sample time.
ainfo = rlNumericSpec([numAct 1],'LowerLimit',-220,'UpperLimit',220);
env = rlSimulinkEnv('rl_motor','rl_motor/RL Agent',oinfo,ainfo);
featureInputLayer(numObs,'Normalization','none','Name','State')
fullyConnectedLayer(numAct,'Name','Action')];
Specify the agent options using rlTD3AgentOptions. The agent trains from an experience buffer of maximum capacity 1e6 by randomly selecting mini-batches of size 256. The discount factor of 0.995 favors long-term rewards.
The exploration model in this TD3 agent is Gaussian. The noise model adds a uniform random value to the action during training. Set the standard deviation of the noise to 100. The standard deviation decays at the rate of 1e-5 every agent step until the minimum value of 0.005.
agentOpts.ExplorationModel.StandardDeviationMin = 0.005;
agentOpts.ExplorationModel.StandardDeviation = 100;
Run each training for at most 2000 episodes, with each episode lasting at most ceil(Tf/Ts) time steps.
Stop the training when the agent receives an average cumulative reward greater than -2 over 20 consecutive episodes. At this point, the agent can track the reference signal.
load('rlDCServomotorTD3Agent.mat')
Validate Controller Response
To validate the performance of the trained agent, simulate the model and view the response in the Scope blocks. The reinforcement learning agent is able to track the reference angle while satisfying the constraints on torque and voltage.
sim('rl_motor');
close_system('rl_motor')
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Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Quantities : Support
compute the support set of a random variable
Support(X, t, opts)
Support(X, opts)
algebraic; any expression
(optional) equation of the form output = form, specifying what form the output should have
Given a random variable X with probability density function
f
, the Support function will find a set A of real numbers such that
f\left(x\right)=0
x
not in A. This set can be used to estimate the size of the support of X.
There are three forms of output that Support can return.
If called with the
\mathrm{output}=\mathrm{property}
option, then Support returns a property that points in the set A have, typically of the form RealRange(a, b). Note that only numeric ranges can be represented in this way.
\mathrm{output}=\mathrm{boolean}
option, then Support returns a boolean expression expressing that t is in A, typically of the form
a\le t<b
\mathrm{output}=\mathrm{range}
option, then Support returns a range describing A, of the form
a..b
If called without an
\mathrm{output}
option, then the default is the
\mathrm{property}
format if t is not given and
\mathrm{boolean}
format if t is given.
The second parameter can be any algebraic expression. It is a necessary argument for
\mathrm{output}=\mathrm{boolean}
and ignored otherwise.
\mathrm{with}\left(\mathrm{Statistics}\right):
Compute the support of the normal distribution.
\mathrm{Support}\left(\mathrm{Normal}\left(3,1\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{real}}
\mathrm{Support}\left(\mathrm{Normal}\left(3,1\right),'\mathrm{output}=\mathrm{range}'\right)
\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}
\mathrm{Support}\left(\mathrm{Uniform}\left(3,5\right)\right)
[\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5})
\mathrm{Support}\left(\mathrm{Binomial}\left(10,0.5\right)\right)
[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{10}]
\mathrm{Support}\left(\mathrm{Uniform}\left(3,5\right),t\right)
\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{5}
\mathrm{Support}\left(\mathrm{Uniform}\left(3,5\right),4\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{Support}\left(\mathrm{Uniform}\left(3,5\right),7\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
Try another distribution.
T≔\mathrm{Distribution}\left(\mathrm{`=`}\left(\mathrm{PDF},t↦\mathrm{piecewise}\left(t<-1,0,t<1,\frac{3\cdot {t}^{2}\cdot \left(5-t\right)}{10},0\right)\right)\right):
X≔\mathrm{RandomVariable}\left(T\right):
\mathrm{Support}\left(X\right)
[\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1})
\mathrm{Support}\left(X,t\right)
\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{1}
\mathrm{Support}\left(X,t,'\mathrm{output}=\mathrm{property}'\right)
[\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1})
\mathrm{Support}\left(X,t,'\mathrm{output}=\mathrm{boolean}'\right)
\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{\le }\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{1}
\mathrm{Support}\left(X,t,'\mathrm{output}=\mathrm{range}'\right)
\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{1}
Statistics[Range]
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Consider the elliptical-cylindrical coordinate system (eta, psi, z), defined by
Consider the elliptical-cylindrical coordinate system (eta, psi, z)
x=a\text{ }\mathrm{cos}h\text{ }\eta \mathrm{cos}\psi ,y=a\mathrm{sin}h\text{ }\eta \mathrm{sin}\psi ;z=z,\text{ }\eta \text{ }GE\text{ }0,0\text{ }LE\text{ }\psi LE\text{ }2\pi ,\text{ }zR.In\text{ }PS6
it was shown that this is an orthogonal coordinate system with scale factors
{h}_{1}={h}_{2}=a{\left({\text{cosh}}^{2}\text{ }\eta -{\mathrm{cos}}^{2}\psi \right)}^{\frac{1}{2}}.
Determine the dual bases
\left(E1,E2,E3\right),\left(\eta ,\eta \psi ,z\right).Showt\stackrel{^}{:}f=a\frac{1}{a}\frac{{\left({\text{cosh}}^{2}eat-{\mathrm{cos}}^{s}\psi \right)}^{1}}{2}\left[\frac{f}{\eta }e1+\frac{f}{\psi }e2+\frac{f}{z}e3,\frac{f}{w}here\left(e1,e2,e3\right)
denotes the unit coordinate basis.
Given that magnitude of gradient along x and y direction is
{h}_{1}={h}_{2}=a{\left({\text{cosh}}^{2}\text{ }\eta -{\mathrm{cos}}^{2}\psi \right)}^{\frac{1}{2}}.
and along z direction = h3 =1 Hence delf = vectors in direction of del/magnitude
=\frac{1}{{\left(\text{cosh}\eta -{\mathrm{cos}}^{2}\psi \right)}^{\frac{1}{2}}}\right\}\left[\frac{\frac{\partial f}{\partial \eta }}{e1}+\frac{\frac{\partial f}{\partial \mathrm{\setminus }\psi }}{e2}\right]+e3\frac{\partial f}{\partial z}
Convert between the coordinate systems. Use the conversion formulas and show work.
\left(8,\frac{\pi }{3},\frac{\pi }{6}\right)
Change to cylindrical.
Find the value of x or y so that the line passing through the given points has the given slope. (9, 3), (-6, 7y),
m=3
The equivalent polar coordinates for the given rectangular coordinates:
\left[\begin{array}{cc}5& \text{ }{180}^{\circ }\end{array}\right]
The intercepts on the coordinate axes of the straight line with the given equation
4x-3y=12.
\gamma =\left\{{t}^{2}-t+1,t+1,{t}^{2}+1\right\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\beta =\left\{{t}^{2}+t+4,4{t}^{2}-3t+2,2{t}^{2}+3\right\}be\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}deredbasesf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{P}_{2}\left(R\right).
Find the change of coordinate matrix Q that changes
\beta \text{ coordinates into }\gamma -\text{ coordinates}
{x}^{2}\text{ }+\text{ }{y}^{2}\text{ }+\text{ }8x=0
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This section gives some examples of using catremote interactively. All the arguments are given on the command line. However, if they are omitted then arguments other than the first (the mode) will usually be prompted for.
List the various modes in which catremote may be used. Type either of:
% catremote
% catremote help
List all the databases in the current configuration file:
% catremote list
List all the name servers (that is, databases of server type ‘namesvr’) in the current configuration file:
% catremote list namesvr
Show details of the USNO PMM astrometric catalogue:
% catremote details usno@eso
Find all the objects in the USNO PMM which lie within ten minutes of arc of Right Ascension
12h\phantom{\rule{0.3em}{0ex}}15m\phantom{\rule{0.3em}{0ex}}00\text{}s\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}.\text{}0
and Declination (J2000):
% catremote query usno@eso 12:15:00 30:30:00 10
The objects selected will be saved as a catalogue called usno_eso_121500_303000.tab created in your current directory. This catalogue will be written in the Tab-Separated Table (TST) format.
12h\phantom{\rule{0.3em}{0ex}}15m\phantom{\rule{0.3em}{0ex}}00\text{}s\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}.\text{}0
and Declination (J2000) which also lie in the magnitude range 14 to 16:
% catremote query usno@eso 12:15:00 30:30:00 10 14,16
Find the equatorial coordinates of the galaxy NGC 3379:
% catremote name simbad_ns@eso ngc3379
The coordinates returned are for equinox J2000.
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Discrete mathematics cardinality using Richard Hammack's Elements of Discrete ma
Discrete mathematics cardinality using Richard Hammack's Elements of Discrete mathematics chapter 18 A superset of uncountable set is uncountable. (We say A is a superset of B if B sube A. )
Discrete mathematics cardinality using Richard Hammacks
Raheem Donnelly
Consider A is a superset of B and A is uncountable. The objective to show that B is uncountable. Assume B is countable. Since using the concept that every subset of countable set is countable. So, A is countable. Which is contradiction, as given A is uncountable. Therefore, B is uncountable.
⇒
a superset of uncountable set is uncountable. Hence proved.
\cap \cup
\left(a,b\right)\in R
Using cardinatility of sets in discrete mathematics the value of N is real numbers
Currently using elements of discrete mathematics by Richard Hammack chapter 18
Let A be a collection of sets such that
X\in A
X\subset N
|X|=n
n\in N
|A|=|N|.
Let consider the following algorithm: Polynomial evaluation
This algorithm evaluates the polynomial
P\left(x\right)=\sum _{k=0}^{n}{c}_{k}{x}^{n-k}
At the point t.
Input: The sequence of coefficients
{C}_{0},{C}_{1},\dots ,{C}_{n}
, the value tandn
Output: p(t)
Procedure poly(c, n, t)
\left({c}_{0}\right)
\left(t:poly\left(c,n-1,t\right)+{c}_{n}\right)
{b}_{n}
, be the number of multiplications required to compute p(t).
a) Find the recurrence relation and initial condition for the sequence
\left\{{b}_{n}\right\}
{b}_{1},{b}_{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{3}
c) Solve the recurrence relation.
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Variables and Scope Practice Problems Online | Brilliant
Suppose that one day you and a group of friends decide to come up with a brand new programming language, with syntax similar to Java's. After it is finished you decide to test it out with a program that prints two integers on the same line as shown below:
def int main() {
def void f() {
def void g() {
def void h() {
You and your friends still have not yet agreed on whether the programming language should follow either static scoping or dynamic scoping.
If if follows static scoping it will print out
and
b
If it follows dynamic scoping, it will print out
c
and
a + b + c + d
The prime factors of a number
n
are all the prime numbers that divide it. For example the prime factors of
2015
are 13, 31, and 5 (
2015 = 5 \times 13 \times 31
). The following Python function finds the largest prime factor of a number.
def largest_pf(n):
for divisor_1 in range(1, n + 1):
for divisor_2 in range(2, divisor_1 + 1):
if divisor_1 % divisor_2 == 0:
max_divisor = divisor_1
return max_divisor
The program works well for most inputs but it contains a variable whose scope is not properly defined. Identify the variable.
divisor_2 i max_divisor is_prime
02 namespace MyNamespace
08 std::cout << x << std::endl;
09 std::cout << MyNamespace::x << std::endl;
10 std::cout << ::x << std::endl;
Consider the above algorithm, in c++ the
::
notation is used to refer back to the the variable defined in the global namespace. Let
l_{1},l_{2}
l_{3}
be the output results of line
8
9
10
, respectively then what is the value of
(l_{1}+l_{2})-l_{3}?
for code...
return find + 1
return leave()
print find + bad() + leave()
Consider the fragment of theoretical python code. Predict what will be outputted when it is run once.
Side note: This is an example of very poor function and variable naming. Proper naming in programming is one of the keys to writing good code.
|
p-Norms (Taxicab Norm, Euclidean Norm and infinity-norm) | Sabbiu Shah
23.12.2018 — linear algebra, norms — 1 min read
Norms are a measure of distance. Norm is defined as follows: For
p\geq1
||x||_p \equiv \sqrt[p]{|x_1|^p + |x_2|^p + ... + |x_n|^p}
Taxicab Norm (1-Norm)
p=1
, then the norm is said to be taxicab norm. The distance derived from this norm is called Manhattan distance.
||x||_1 \equiv |x_1| + |x_2| + ... + |x_n|
Euclidean Norm (2-Norm)
It is the most common notion of distance. When
p=2
, then the norm is said to be euclidean norm.
||x||_2 \equiv \sqrt{|x_1|^2 + |x_2|^2 + ... + |x_n|^2}
Infinity norm is defined as,
||x||_\infty \equiv max(|x_1|, |x_2|, ..., |x_n|)
\begin{aligned} ||x||_p & \equiv \sum_{i=1}^n |x_i|^p\ \ \ \ \ \ \ \ \ \ \text{Equation of p-norm}\\ ||x||_p & \equiv m \sum_{i=1}^n \frac{|x_i|^p}{m}\ \ \ \ \ \ \ \ \ \ m=max(|x_i|)\\ \end{aligned}
p
\infty
, only the term
\frac{max|x_i|}{m}
1
, while other terms approaches to
0
\sum_{i=1}^n \frac{|x_i|^p}{m} = 1
\begin{aligned} \therefore\ ||x||_\infty & \equiv max(|x_i|) \end{aligned}
Visualising norms as a unit circle
This section will show visualization when,
||x||_p \equiv 1
. Let us consider for 2 Dimensional case.
The equation is given as,
\begin{aligned} & ||x||_1 = |x_1| + |x_2|\\ \implies & 1= |x_1| + |x_2|\\ \end{aligned}
Thus we get the following equations,
x_1\geq0
x_2\geq0
x_2=1-x_1
[First quadrant]
x_1\leq0
x_2\geq0
x_2=1+x_1
[Second quadrant]
x_1\leq0
x_2\leq0
x_2=x_1-1
[Third quadrant]
x_1\geq0
x_2\leq0
x_2=-x_1-1
[Fourth quadrant]
Plotting these equations, we get,
\begin{aligned} & ||x||_2 = \sqrt{|x_1|^2 + |x_2|^2}\\ \implies & 1= x_1^2 + x_2^2\\ \end{aligned}
As this equation represents a unit circle, we get the following graph,
\begin{aligned} & ||x||_\infty = max(|x_1|, |x_2|)\\ \implies & 1 = max(|x_1|, |x_2|)\\ \end{aligned}
This gives the following graph,
Substituting different values of p, these equations can be further visualised in Wolfram Mathematica Demonstaration
excerpt justin solomon, mathematical methods for robotics, vision, and graphics
wolfram mathematica norms demonstration
wikipedia article on lp space
|
Evaluate the following derivatives. \frac{d}{dx}\int_{7}^{x}\sqrt{1+t^{4}+t^{6}
Evaluate the following derivatives. \frac{d}{dx}\int_{7}^{x}\sqrt{1+t^{4}+t^{6}}dt
Wierzycaz 2021-02-22 Answered
Evaluate the following derivatives.
\frac{d}{dx}{\int }_{7}^{x}\sqrt{1+{t}^{4}+{t}^{6}}dt
\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)dt={\int }_{a\left(x\right)}^{b\left(x\right)}\frac{\partial }{\partial x}\left(f\left(x,t\right)dt+{b}^{\prime }\left(x\right)f\left(x,b\right)-{a}^{\prime }\left(x\right)f\left(x,a\right)
\frac{d}{dx}{\int }_{7}^{x}\sqrt{1+{t}^{4}+{t}^{6}}dt=\left({\int }_{7}^{x}0\right)+\sqrt{1+{x}^{4}+{x}^{6}}-0=\sqrt{1+{x}^{4}+{x}^{6}}
g\left(x\right)=\mathrm{ln}\mid 3x-1\mid
y=\left(1+{x}^{2}\right){e}^{{\mathrm{tan}}^{-1}x}
Calculating derivatives Find dy>dx for the following functions.
y=\frac{1-\mathrm{cos}x}{1+\mathrm{cos}x}
Solve the derivatives of the following functions.
f\left(x\right)=10{x}^{8}-4{x}^{5}+2{x}^{3}-18x-9
z=\frac{4-3x}{3{x}^{2}+x}
|
Autobiographical Numbers Practice Problems Online | Brilliant
This quiz is devoted to solving a single problem:
An autobiographical number is a number
N
such that the first digit of
N
counts how many zeroes are in
N,
the second digit counts how many ones are in
N
and so on. What other autobiographical numbers are there? The autobiographical number 1210 is broken down below as an example.
We'll start with the 10-digit case, and then go back to the general problem.
Note if you watched the related TED-Ed video (screenshot below) which continues to the solution, this quiz takes a different solving route (and gives hints as to how to extend it) so it will be helpful even if you've seen the rest of the video before.
N
N
N,
N
Again, our first goal will be to focus on the question: What is the autobiographical number with 10 digits?
As this problem can be difficult to answer outright, it's broken into parts.
What is the sum of the digits in a 10 digit autobiographical number?
Hint: Consider the 4 digit example from the last page.
N
N
N,
N
Consider the last 5 digits of a 10 digit autobiographical number (as marked above). How many 0 digits must there be in that set?
N
N
N,
N
Using the previous answer, we know the latter digits must be all 0s (except for a 1 somewhere; the extra digit can't be more than 1 because that will cause the overall sum of digits to exceed the target of 10).
What about the 3s and 4s digit places (as marked above)?
They must both be 0 They must be 0 and 1 in some order They must be 0 and 2 in some order They must be 1 and 2 in some order
To summarize the previous results (including information from the solutions), we know the sum of the digits must be 10, and the facts from this diagram:
We're close to an answer now!
N
N
N,
N
What is the first digit of the 10 digit autobiographical number? That is, how many digits of the digits in an 10 digit autobiographical number must be 0?
N
N
N,
N
Combine the information from the previous questions, and you should now be able to answer:
What is the autobiographical number with 10 digits?
Now, how to solve the general case?
Our first steps can apply in a general way. For example, when checking 8-digit autobiographical numbers, this diagram arises.
From these steps it's possible to get a general formula for the first digit for when
4 \leq N \leq 10 .
Once you have this information, you have enough constraints that the different cases fall very quickly. Note, as a word of warning:
At least one of the cases of
N
has two numbers that fit rather than one.
N
has no numbers of that length.
|
Transform position and velocity components from Standard Julian Epoch (J2000) to discontinued Standard Besselian Epoch (B1950) - Simulink - MathWorks Switzerland
\left({\overline{r}}_{J2000}\right)
\left({\overline{v}}_{J2000}\right)
\left({\overline{r}}_{B1950}\right)
\left({\overline{v}}_{B1950}\right)
\left[\begin{array}{c}{\overline{r}}_{B1950}\\ {\overline{v}}_{B1950}\end{array}\right]={\left[\begin{array}{cc}{\overline{M}}_{rr}& {\overline{M}}_{vr}\\ {\overline{M}}_{rv}& {\overline{M}}_{vv}\end{array}\right]}^{T}\left[\begin{array}{c}{\overline{r}}_{J2000}\\ {\overline{v}}_{J2000}\end{array}\right],
\left({\overline{M}}_{rr},{\overline{M}}_{vr},{\overline{M}}_{rv},{\overline{M}}_{vv}\right)
{\overline{M}}_{rr}=\left[\begin{array}{ccc}0.9999256782& -0.0111820611& -0.0048579477\\ 0.0111820610& 0.9999374784& -0.0000271765\\ 0.0048579479& -0.0000271474& 0.9999881997\end{array}\right]
{\overline{M}}_{vr}=\left[\begin{array}{ccc}0.00000242395018& -0.00000002710663& -0.00000001177656\\ 0.00000002710663& 0.00000242397878& -0.00000000006587\\ 0.00000001177656& -0.00000000006582& 0.00000242410173\end{array}\right]
{\overline{M}}_{rv}=\left[\begin{array}{ccc}-0.000551& -0.238565& 0.435739\\ 0.238514& -0.002667& -0.008541\\ -0.435623& 0.012254& 0.002117\end{array}\right]
{\overline{M}}_{vv}=\left[\begin{array}{ccc}0.99994704& -0.01118251& -0.00485767\\ 0.01118251& 0.99995883& -0.00002718\\ 0.00485767& -0.00002714& 1.00000956\end{array}\right]
|
Capacitors Practice Problems Online | Brilliant
The two metal plates in the above figure have net charges of
+140.0\text{ pC}
-140.0\text{ pC},
respectively. If the potential difference between them is
10.0\text{ V},
what is the capacitance of the system consisting of the two metal plates?
7.0 \text{ pF}
14.0 \text{ pF}
21.0 \text{ pF}
28.0 \text{ pF}
The capacitor in the above figure has a capacitance of
40.0 \,\mu\text{F}
and is initially uncharged. The battery provides a potential difference of
90.0\text{ V}.
After switch S is closed, how much charge will pass through it?
6.48 \times 10^{-3} \text{ C}
5.04 \times 10^{-3} \text{ C}
2.16 \times 10^{-3} \text{ C}
3.60 \times 10^{-3} \text{ C}
Consider a spherical capacitor which consists of two concentric spherical shells of radii
38.0\text{ mm}
40.0\text{ mm},
respectively. If we want to make a parallel-plate capacitor with the same separation, what must be the area of each plate for the same capacitance?
\displaystyle k=\frac{1}{4\pi\varepsilon_0}=8.99 \times 10^9 \text{ N}\cdot\text{m}^2\text{/C}^2
\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.
152.8 \text{ cm}^2
248.2 \text{ cm}^2
95.5 \text{ cm}^2
191.0 \text{ cm}^2
One can make a homemade capacitor using an aluminum foil and a plastic or a wooden stick (it's important that the stick is an insulator). Make a small ball out of the aluminum foil and wrap it around the stick. Then make a wider foil sphere around the ball so that they don't touch, and your capacitor is ready!
If the radius of the smaller ball is
5~\mbox{cm}
and of the bigger sphere
15~\mbox{cm}
, which capacitance in pF do you expect to obtain?
The ball and the sphere are centered around the same point
If you want to make a parallel-plate capacitor of capacitance
0.9\text{ pF}
with two metal plates with area
1.1\text{ cm}^2,
what must be the separation between the two plates?
\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.
3.24 \text{ mm}
2.16 \text{ mm}
4.32 \text{ mm}
1.08 \text{ mm}
|
Solution of Two-Dimensional Stokes Flow With Elliptical Coordinates and Its Application to Permeability of Porous Media | J. Appl. Mech. | ASME Digital Collection
Mustapha Hellou,
Civil Engineering and Mechanical Engineering Laboratory-EA3913,
University of Rennes, INSA Rennes
Rennes F-35000
Email: mustapha.hellou@insa-rennes.fr
Franck Lominé,
Email: franck.lomine@insa-rennes.fr
Mohamed Khaled Bourbatache,
Email: mohamed-khaled.bourbatache@insa-rennes.fr
Department of Mechanical and Systems Engineering Complexes,
Institut Pprime, UPR CNRS 3346, University of Poitiers
F-86000, Poitiers
Email: mohamed.hajjam@univ-poitiers.fr
Hellou, M., Lominé, F., Bourbatache, M. K., and Hajjam, M. (April 22, 2021). "Solution of Two-Dimensional Stokes Flow With Elliptical Coordinates and Its Application to Permeability of Porous Media." ASME. J. Appl. Mech. June 2021; 88(6): 061009. https://doi.org/10.1115/1.4050687
In this paper, analytical developments of the biharmonic equation representing two-dimensional Stokes flow are performed with elliptical coordinates. It is found that the streamfunction is expressed with series expansions based on Gegenbauer polynomials of first and second kinds with order one
Cn1
Dn1
for n natural integer number. Application to an elliptical fiber enclosed in an elliptical boundary with uniform flow is made. It is found that a particular solution responsible of the drag must be added to the general solution. Following this, the flow through rectangular arrays of elliptical fibers is studied, and the permeability of this medium is determined as a function of porosity.
Stokes flow, biharmonic equation, elliptical coordinates, drag on an elliptical cylinder, Gegenbauer equation, permeability
Creeping flow, Fibers, Flow (Dynamics), Permeability, Polynomials, Porous materials, Porosity, Drag (Fluid dynamics)
Low-Reynolds-Number Hydrodynamics With Special Applications to Particulate Media
Slow Viscous Flows.
Flow Through a Finned Channel Filled With a Porous Medium
Numerical Simulation of Flowing Blood Cells
Transp. Porous Med.
Casadevall i Solvas
Stokes Flow Through An Array of Rectangular Fibers
Flow Field and Drag for Elliptical Filter Fibers
Effects of Microstructure on Flow Properties of Fibrous Porous Media at Moderate Reynolds Number
Bourbatache
Two-Scale Analysis of the Permeability of 3d Periodic Granular and Fibrous Media
On the Flow of a Viscous Fluid Past An Inclined Elliptic Cylinder At Small Reynolds Numbers
Low-Reynolds-Number Flow Past An Elliptic Cylinder
Generalized Expansions for Slow Flow Past a Cylinder
Q. J. Mech. Appl. Maths.
The Forces Experienced by Randomly Distributed Parallel Circular Cylinders Or Spheres in a Viscous Flow At Small Reynolds Number
Low-Reynolds-Number Flow Past Cylindrical Bodies of Arbitrary Cross-Sectional Shape
Two-Dimensional Stokes Flow Driven by Elliptic Paddles
Numerical Simulation of Fluid Flow Through Random Packs of Ellipses
Stokes’Paradox: Creeping Flow Past a Two-Dimensional Cylinder in An Infinite Domain
A Note on the Stokes’paradox
Q. Appl. Maths.
D. Van Nostrand company LTD
Fonctions Sphériques De Legendre Et Fonctions Sphéroïdales, III
Study of Permeability by Periodic and Self-consistent Homogenisation
Permeability-Porosity Relationship in Rtm for Different Fiberglass and Natural Reinforcements
Calculation of the Permeability and Apparent Permeability of Three-Dimensional Porous Media
Analytical Determination of Viscous Permeability of Fibrous Porous Media
Transport Phenomena in Bispherical Coordinates
Etude théorique et expérimentale de l’écoulement engendré par un cylindre en translation uniforme dans un fluide visqueux en régime de stokes
|
How can you use what you know about transformations of functions to graph radical function
How can you use what you know about transformations of functions to graph radical functions?
Bentley Leach
All that we used in transformations of graphs of linear and square functions, can be applyed to the graphs of radical functions.
If we are adding or subtracting a number from the x, than the function will translate left and right.
If we are adding or subtracting a number from the y, than the function will translate up and down.
Same goes for reflecting given functions across x-axis, y-axis and the (0,0) point.
f\left(x\right)=\sqrt{2x-{x}^{2}}
f\left(x\right)={\mathrm{log}}_{2}x
y=\mathrm{sin}x
sketch the function y=
3\cdot \mathrm{sin}\left(\frac{x}{2}\right)
C\mathrm{\infty }\left[a,b\right]
of infinitely differentiable functions on the interval [a,b], consider the derivate transformation D and the definite integral transformation I defined by
D\left(f\right)\left(x\right)=f\prime \left(x\right)D\left(f\right)\left(x\right)=f\prime \left(x\right)\text{ }and\text{ }I\left(f\right)\left(x\right)=\int xaf\left(t\right)dtf\left(t\right)dt
. (a) Compute
\left(DI\right)\left(f\right)=D\left(I\left(f\right)\right)\left(DI\right)\left(f\right)=D\left(I\left(f\right)\right)
. (b) Compute
\left(ID\right)\left(f\right)=I\left(D\left(f\right)\right)\left(ID\right)\left(f\right)=I\left(D\left(f\right)\right)
. (c) Do this transformations commute? That is to say, is it true that
\left(DI\right)\left(f\right)=\left(ID\right)\left(f\right)\left(DI\right)\left(f\right)=\left(ID\right)\left(f\right)
for all vectors f in the space?
|
Store Sales - Time Series Forecasting Challenge | Research @ WDSS
2.4. Aggregate Correlation of Sales
4. The notion of distance for clustering
4.0.1. Dynamic time warping
1.1. Heatmaps
1. SARIMA
2.1. DLM Theory
3.1. LSTM Theory
Created 2022-02-12| Updated 2022-02-12 |Ivan Silajev, Hugo Ng, Amr Abdelmotteleb|EconomicsFinance
This blog post serves as a review and a showcase of the various statistical inference and modelling techniques utilised by the WDSS Machine Learning Competition Team over the Autumn and Winter seasons in 2021.
Throughout the post, each technique is explained in chronological order to accurately portray the thought process of the team tackling the problem.
The purpose of this report is to potentially inspire readers to participate in data science competitions or even take an interest in data science if they have not already.
Before explaining the team’s problem solving process, the problem itself must be formulated.
The competition we entered was the “Store Sales - Time Series Forecasting” competition on Kaggle. The objective was to create 15 day forecasts for the sales figures of 33 different products for 54 stores belonging to the same retailer.
In summary, we had to make accurate 15 day predictions for 1782 individual time series using data scientific methods.
Before proceeding with creating the predictions, it was essential to examine the information, in the form of tabular data sets, given to us and determine the utility of each set.
The training data set contained past sales data, product families and if the products were on promotion. The rest of the datasets contained additional exogenous variables:
The location, type and the daily number of transactions for each store
Ecuadorian holidays
As with any data scientific project, we all carried out an EDA (Exploratory Data Analysis) to determine how to use the data for modelling.
The following is brief list of what modelling and inference techniques we used for this competition:
Metrics for comparing time series;
Methods of clustering time series;
UMAP-assisted K-means clustering
Time series modelling methods;
In the following sections, we give a detailed overview of:
Our EDA and thought process
The decisions we made regarding how we use the data and their justifications
The underlying theory and code for our methods
Aggregate Correlation of Sales
We will investigate whether all sales time series for a given product family are similar enough to be treated identically.
A simple ‘measure’ for the similarity between two time series is the linear correlation coefficient between the values of both series. Letting
a_i
i
th value in the first time series and
b_i
i
th value in the second series, plot the coordinates
(a_i, b_i)
i
. If a linear relationship is clearly visible, then both time series are approximately similar.
Two time series are completely similar if they are both affine tranformations of each other (i.e. it is possible to express time series
a_i
b_i
A + B a_i
A
B
are real values).
It is costly to compare a large number of time series with each other to produce a correlation matrix, especially when the time series contain more than 1.5 thousand values. A less reliable, but quicker solution is to compare each time series with the sum of all time series (the aggregate time series).
The aggregate time series would be approximately similar to every series that forms the largest similarity group. This way, one can detect the existence of series groupings with a given set of series.
This EDA showed that the degree of similarity between series from the same product family can vary drastically. For example, for the BEAUTY family, it ranged from small negative values to 0.85 with a mean of 0.5. Hence, there is convincing evidence that the time series can’t be trivally grouped by the product family they represent.
By grouping similar time series data together, we may be able to reduce the number of models to fit. We focus on grouping stores and product types.
The clustering technique we use is UMAP-assisted k-Means clustering. We focus on two notions of distance: dynamic time warping and Euclidean distance. There is a reliability-speed trade-off among these two distances, as we explain below.
The notion of distance for clustering
We need to standardize our data, such that comparisons are fair. Naturally, more expensive products may fluctuate more, but the underlying structure may be similar to cheaper products, and we want to be be able to make such comparison.
We can choose dynamic time warping (DTW) or Euclidean distance for our clustering. DTW is a reliable algorithm for measuring similarity between two temporal sequences. It is able to match peaks to peaks, troughs to troughs, and account for any stretching, shrinking, and shifting/phase difference. This is helpful, since two ‘close’ time series can exhibit slight differences as described.
A drawback of dynamic time warping is that it is computationally intensive by design. If we use it for clustering on a large dataset, which we have, we are doing a lot of expensive computations. This limits the usefulness of DTW, which could be remedied for example with cloud computing.
Euclidean distance is fast, as naive comparison of two time series vertically is a simple computation. When two time series only differ in magnitude but match in phase, Euclidean distance is good enough.
However, Euclidean distance is not as reliable as DTW. Imagine two time series that look like sin curves, but one is shifted out of phase. Intuitively, these time series are very similar, and should be grouped together. A vertical point-to-point comparison by Euclidean distance gives us a large distance. As a result, distances between two time series may be erroneously larger than it actually is.
Due to computational constraints, we will be proceeding with euclidean distance.
We choose the smallest number of clusters that capture at least 97% of the distortions. In this case, 5 clusters.
umap_3d = UMAP(n_components=3, n_neighbors=15, metric='euclidean' , init='random', random_state=0)
proj_3d = umap_3d.fit_transform(df.to_numpy())
kmeans = KMeans(n_clusters=5, init='k-means++', random_state=0)
clusters = kmeans.fit_predict(proj_3d)
color=clusters.astype('str'),
labels={'color': 'Cluster'},
fig_3d.update_layout(scene = dict(
xaxis_title='Component 1',
yaxis_title='Component 2',
zaxis_title='Component 3'))
By looking at clustermaps, we can see how similarly each cluster behaves.
We analysed clusters from a number of perspectives: the percentage of cluster members belonging to each family, state, city and store number. We concluded that it is only the product’s family accounts for a reasonable amount of difference and that is not the case for other characteristics. However, aggregate correlation showed that even the similarities within product families are limited. Therefore, we will be proceeding with fitting separate models for each family in each stores.
In this section, we go through fitting an ARIMA model on a time series for one of the families of products in one store; to see how well it performs.
We are now going to create a dataset for the sales of Bread/Bakery in store 44 .
One very important assumption of an ARIMA model is for the data to be stationary. We will test this with a hypothesis test called ADF (Augmented Dickey-Fuller).
NOTE: The null hypothesis of the ADF test is that the time series is non-stationary.
We have a p value of 0.214832, So we do not reject the null hypothesis and so the series is non-stationary. In order to still be able to use ARIMA, we are going to difference our data.
Let us now plot the original timeseries, as well as the 1st and 2nd order differenced time series with their ACF (Autocorrelation) plots to determine the order of differencing we are going to use in our model:
The p-value for the first order differenced data is close to 0, so we can proceed with using this transformation. We now move on to identifying the number of AR terms § and the number of MA terms (q) that we are going to include in our ARIMA model. We do this by inspecting the ACF and PACF plots for the data.
We use q = 2 based on the autocorrelation plot and p = 6 based on the partial autocorrelation plot to fit the ARIMA model. The residuals seem to exhibit no significant patterns. We can now predict the sales for the 16 last observations from the original dataset that we kept as a test set. The predictions seem to align well with the actual data.
Seasonality definitely plays a role in the time series of some of the families of products in some of the stores, so using a seasonal ARIMA (SARIMA) could be a further improvement.
We fit a SARIMA model to explore any weekly trends and we automated the process of selecting the model’s parameters. However, using a grid search for selecting the best SARIMA parameters based on root-mean-square error is computationally expensive. Applying it to thousands of families of products would take a significant amount of time. Hence, we decided to explore other models with fewer parameters to specify.
State space models are a generalisation of general linear regression models. Unlike GLMs, which include ARIMA models, SSMs can consider cases where the underlying distribution of the parameters of the model changes over time. A dynamic linear model is an SSM that assumes the data can be modelled as a linear combination of its parameters.
DLM Theory
A standard DLM can be formulated by the following four equations, given we already have data up until time
t-1
x_t = \phi_t \cdot b_t + \epsilon_t : \epsilon_t \sim N(0, v_t)
b_t = F_t \cdot b_{t-1} + \Delta_t : \Delta_t \sim N(0, v_{t-1} \Sigma_t)
b_{t-1} = m_{t-1} + B_{t-1} : B_{t-1} \sim N(0, v_{t-1} B_{t-1})
v_t = \delta v_{t-1}
The first equation is an ordinary linearity assumption for the data, the same one for ordinary linear regression, except the parameters can vary over time.
The second equation is the hidden model and is the key assumption for a linear model to be dynamic.
The third equation is a reiteration of the fact that the model parameters are random as well. The parameters possess an initial prior distribution which updates as more data is collected.
The fourth equation models the change in variance as time progresses.
Each term at time
t
x_t
b_t
is the parameter vector
\epsilon_t
is a normally distributed variable with variance
v_t
m_t
is the expectation of the parameters
B_{t}
\Delta_t
are multivariate normally distributed vectors with covariance matricies
v_t B_t
v_{t-1} \Sigma_t
\phi_t
F_t
are the transition vector and matrix respectively, which define how the model parameters combine linearly
\delta
is the discount factor, a value in the interval of zero to one that models the increasing uncertainty in the distribution of future data
The model reduces to ordinary linear regression if:
F_t
is the identity matrix for all
t
\Sigma_t
is the zero matrix for all
t
\delta
is one, so there is no increasing uncertainty in the distribution of future data
The dynamic linear model offers a high degree of modelling flexibility, as well as offering a convenient method for updating the model parameters over time.
Long short-term memory (LSTM) is a recurrent neural network (RNN) that utilises two different hidden variables rather than one to learn about the behaviour of a sequential process. In theory, LSTMs could capture deeper non-linear underlying patterns in the data.
LSTM Theory
While a simple RNN uses its hidden variable as a short term memory channel, LSTM uses a second one for long term memory, which, theoretically, preserves significant patterns of the system’s behaviour in the model for as long as they persist.
A standard LSTM can be formulated by the following equations, given we already have data up until time
t-1
f_t = \sigma (W_f \cdot [h_{t-1}, x_t] + b_f)
i_t = \sigma (W_i \cdot [h_{t-1}, x_t] + b_i)
\bar{C}_t = \text{tanh} (W_C \cdot [h_{t-1}, x_t] + b_C)
C_t = f_t \times C_{t-1} + i_t \times \bar{C}_t
o_t = \sigma (W_o \cdot [h_{t-1}, x_t] + b_o)
h_t = o_t \times \text{tanh}(C_t)
The first equation represents the forget gate, the perceptron layer responsible for removing any insignificant patterns the long term memory assumed from the last iteration.
The second equation represents the information gate, the layer accountable for recording new potential patterns into the long term memory and strengthen ones that were present before.
The third equation offers a candidate long term memory term, which is corrected by the information gate, as visible in the fourth equation.
The fifth equation is the output gate, the layer for creating the final output that will be used for creating the next short term hidden variable, as shown in the sixth equation.
\times
symbol represents termwise multiplication.
t
x_t
h_t
is the hidden short term memory variable
C_t
is the hidden long term memory variable
\bar{C}_t
is a candidate for
C_t
f_t
is the forget vector
i_t
is the information vector
o_i
is the output vector
W
matricies are the weights that dictate how much of each term in the input contributes to the value of the output
b
vectors are biases that set the default values of the variables they form
\sigma
is the element-wise sigmoid activator function
\text{tanh}
is the element_wise hyperbolic tangent activator function
LSTM allows for modelling arbitrary patterns in time series or any sequential process becuase of its neural-like nature.
We define an LSTM model with 50 units, outputting a hidden variable of length 50. A dense perceptron layer with a rectified linear unit activation function combines the outputs at the end to create a single prediction.
The model uses an Adam optimiser and a mean square loss function.
The LSTM model produced very monotonuous predictions and thus struggled with predicting the variability present in the data. The quality of predictions remained low even after varying different parameters of the network. Even though LSTMs take longer to learn and their setup is more complex, this model couldn’t recreate the success that ARIMA and DLM models offer in terms of time series prediction.
Authors: Ivan Silajev, Hugo Ng, Amr Abdelmotteleb
Editor: Janique Krasnowska
Permalink: https://research.wdss.io/sales-forecasting-challenge/
machine learningpythonclusteringtime seriessales
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Sensors | Free Full-Text | Model of Magnetically Shielded Ferrite-Cored Eddy Current Sensor
Proposal for an IIoT Device Solution According to Industry 4.0 Concept
Impact Damage Detection Using Chirp Ultrasonic Guided Waves for Development of Health Monitoring System for CFRP Mobility Structures
Individualization of Intensity Thresholds on External Workload Demands in Women’s Basketball by K-Means Clustering: Differences Based on the Competitive Level
Nondestructive Determination of Strength of Concrete Incorporating Industrial Wastes as Partial Replacement for Fine Aggregate
Vasić, D.
Rep, I.
Špikić, D.
Kekelj, M.
Model of Magnetically Shielded Ferrite-Cored Eddy Current Sensor
Dorijan Špikić
Matija Kekelj
INETEC Institute for Nuclear Technology, 10250 Zagreb, Croatia
Academic Editors: Leandro Maio, Vittorio Memmolo and Marco Laracca
(This article belongs to the Special Issue Structural Health Monitoring and Non-destructive Testing for Engineering Applications: Advances in Sensor and Technologies)
Computationally fast electromagnetic models of eddy current sensors are required in model-based measurements, machine interpretation approaches or in the sensor design phase. If a sensor geometry allows it, the analytical approach to the modeling has significant advantages in comparison to numerical methods, most notably less demanding implementation and faster computation. In this paper, we studied an eddy current sensor consisting of a transmitter coil with a finitely long I ferrite core, which was screened with a finitely thick magnetic shield. The sensor was placed above a conductive and magnetic half-layer. We used vector magnetic potential formulation of the problem with a truncated region eigenfunction expansion, and obtained expressions for the transmitter coil impedance and magnetic potential in all subdomains. The modeling results are in excellent agreement with the results using the finite element method. The model was also compared with the impedance measurement in the frequency range from 5 kHz to 100 kHz and the agreement is within
3%
for the resistance change due to the presence of the half-layer and
1%
for the inductance change. The presented model can be used for measurement of properties of metallic objects, sensor lift-off or nonconductive coating thickness. View Full-Text
Keywords: ferrite core; magnetic shield; eddy current; probe; coil; truncated domain; eigenfunction expansion ferrite core; magnetic shield; eddy current; probe; coil; truncated domain; eigenfunction expansion
Vasić, D.; Rep, I.; Špikić, D.; Kekelj, M. Model of Magnetically Shielded Ferrite-Cored Eddy Current Sensor. Sensors 2022, 22, 326. https://doi.org/10.3390/s22010326
Vasić D, Rep I, Špikić D, Kekelj M. Model of Magnetically Shielded Ferrite-Cored Eddy Current Sensor. Sensors. 2022; 22(1):326. https://doi.org/10.3390/s22010326
Vasić, Darko, Ivan Rep, Dorijan Špikić, and Matija Kekelj. 2022. "Model of Magnetically Shielded Ferrite-Cored Eddy Current Sensor" Sensors 22, no. 1: 326. https://doi.org/10.3390/s22010326
Darko Vasić received a Ph.D. degree in electrical engineering from the University of Zagreb, Croatia, in 2010. He is currently an associate professor with the Department of Electronic Systems and Information Processing at the Faculty of Electrical Engineering and Computing, University of Zagreb. Darko Vasić has over 15 years of research, educational and industrial experience in sensors, instrumentation and electronic systems. His specific research interests are in the field of computational electromagnetic sensing and non-destructive testing. He has coauthored more than 50 journal and conference papers.
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If an angular acceleration of
28.0\text{ rad/s}^2
is generated by a
44.0\text{ N}\cdot\text{m}
torque on a wheel, then what is the wheel's approximate rotational inertia?
1.9\text{ kg}\cdot\text{m}^2
1.8\text{ kg}\cdot\text{m}^2
1.7\text{ kg}\cdot\text{m}^2
1.6\text{ kg}\cdot\text{m}^2
Consider a diver who is launching from a diving board. When he launches from the board, his angular speed about his center of mass is changing from
0
5.35\text{ rad/s}
190\text{ ms}.
If the rotational inertia about his center of mass is
12.0\text{ kg}\cdot\text{m}^2,
what is the magnitude of average external torque on him from the board during the launch?
3.26 \times 10^2 \text{ N}\cdot\text{m}
3.38 \times 10^2 \text{ N}\cdot\text{m}
3.14 \times 10^2 \text{ N}\cdot\text{m}
2.90 \times 10^2 \text{ N}\cdot\text{m}
A cylinder is pinned at its center and two forces
F_1=110\text{ N}
F_2=70\text{ N}
are acting on it, as shown above. If the mass of the cylinder is
100\text{ kg},
what is the rotational acceleration produced by the two forces?
12.7\text{ rad/s}^2
10.6\text{ rad/s}^2
11.6\text{ rad/s}^2
13.7\text{ rad/s}^2
Two blocks of masses
m_1=430\text{ g}
m_2=500\text{ g}
are hanging on a pulley, as shown in the above figure. The pulley is fixed on a horizontal axle with negligible friction, and its radius is
R=5.00\text{ cm}.
If they are released from rest, block
m_2
48.0\text{ cm}
4.00\text{ s}.
If there is no slippage between the cord and the pulley, what is the rotational inertia of the pulley?
g=9.8\text{ m/s}^2.
2.36 \times 10^{-2} \text{ kg}\cdot\text{m}^2
3.15 \times 10^{-2} \text{ kg}\cdot\text{m}^2
3.41 \times 10^{-2} \text{ kg}\cdot\text{m}^2
2.62 \times 10^{-2} \text{ kg}\cdot\text{m}^2
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LMIs in Control/pages/LMI for Nonrotating Missiles Attitude Control yaw roll channel - Wikibooks, open books for an open world
LMIs in Control/pages/LMI for Nonrotating Missiles Attitude Control yaw roll channel
Deriving the exact dynamic modeling of a missile is a very complicated procedure. Thus, a simplified model is used to model the missile dynamics. To do so, we consider a simplified attitude system model for the yaw/roll channel of the system. We aim to achieve a non-rotating motion of missiles. Note that the attitude control design for the yaw/roll channel and the pitch channel can be solved exactly in the same way except for different representing matrices of the system.
4 The LMI: LMI for non-rotating missle attitude control
The state-space representation for the yaw/roll channel can be written as follows:
{\displaystyle {\begin{aligned}{\dot {x}}(t)&=A(t)x(t)+B_{1}(t)u(t)+B_{2}(t)d(t)\\y(t)&=C(t)x(t)+D_{1}(t)u(t)+D_{2}(t)d(t)\end{aligned}}}
{\displaystyle x=[\beta \quad w_{y}\quad w_{x}\quad \delta _{x}\quad \delta _{y}]^{\text{T}}}
{\displaystyle u=[\delta _{xc}\quad \delta _{yc}]^{\text{T}}}
{\displaystyle y=[n_{z}\quad w_{x}]^{\text{T}}}
{\displaystyle d=\delta _{z}}
are the state variable, control input, output, and disturbance vectors, respectively. The paprameters
{\displaystyle \alpha }
{\displaystyle w_{z}}
{\displaystyle \delta _{z}}
{\displaystyle \delta _{zc}}
{\displaystyle n_{y}}
{\displaystyle \beta }
{\displaystyle w_{y}}
stand for the attack angle, pitch angular velocity, the elevator deflection, the input actuator deflection, the overload on the side direction, the sideslip angle, and the yaw angular velocity, respectively.
In the aforementioned yaw/roll channel system, the matrices
{\displaystyle A(t),B_{1}(t),B_{2}(t),C(t),D_{1}(t),}
{\displaystyle D_{2}(t)}
{\displaystyle {\begin{aligned}A(t)={\begin{bmatrix}A_{11}(t)&A_{12}(t)\\0&A_{22}(t)\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}A_{11}(t)={\begin{bmatrix}-b_{4}(t)&1&{\frac {\alpha (t)}{57.3}}\\-{\acute {b}}_{1}(t)b_{4}(t)-b_{2}(t)&-{\acute {b}}_{1}(t)-{\acute {b}}_{1}(t)&{\frac {J_{y}-J_{z}}{57.3J_{x}}}{w}_{z}(t)-{\frac {{\acute {b}}_{1}\alpha (t)}{57.3}}\\c_{2}(t)&{\frac {J_{y}-J_{z}}{57.3J_{x}}}{w}_{z}(t)&-c_{1}(t)\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}A_{12}(t)={\begin{bmatrix}0&-b_{5}(t)\\0&-{\acute {b}}_{1}(t){\acute {b}}_{5}-b_{3}(t)\\-c_{3}(t)&c_{4}(t)\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}A_{22}(t)=-{\frac {1}{\tau _{x}\tau _{y}}}{\begin{bmatrix}\tau _{y}&0\\0&\tau _{x}\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}B_{1}(t)={\begin{bmatrix}0&0\\0&0\\0&0\\{\frac {1}{\tau _{x}}}&0\\0&{\frac {1}{\tau _{y}}}\end{bmatrix}},\quad B_{2}(t)={\begin{bmatrix}a_{6}(t)\\-{\acute {b}}_{1}(t)a_{6}(t)\\0\\0\\0\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}C(t)=-{\frac {1}{57.3g}}{\begin{bmatrix}V(t)b_{4}(t)&0&0&0&V(t)b_{5}(t)\\0&0&-57.3g&0&0\end{bmatrix}}\end{aligned}}}
{\displaystyle {\begin{aligned}D_{1}(t)=0,\quad D_{2}(t)=-{\frac {V(t)}{57.3g}}{\begin{bmatrix}b_{6}(t)\\0\end{bmatrix}}\end{aligned}}}
{\displaystyle a_{1}(t)\sim a_{6}(t),\quad b_{1}(t)\sim b_{7}(t),{\acute {a}}_{1}(t),{\acute {b}}_{1}(t)}
{\displaystyle c_{1}(t)\sim c_{4}(t)}
are the system parameters. Moreover,
{\displaystyle V}
is the speed of the missle and
{\displaystyle J_{x}}
{\displaystyle J_{y}}
{\displaystyle J_{z}}
are the rotary inertia of the missle corresponding to the body coordinates.
{\displaystyle u=Kx}
{\displaystyle {\begin{aligned}{\dot {x}}&=(A+B_{1}K)x+B_{2}d\\z&=(C+D_{1}K)x+D_{2}d\end{aligned}}}
{\displaystyle H_{\infty }}
{\displaystyle G_{zd}(s)=(C+D_{1}K)(sI-(A+B_{1}K))^{-1}B_{2}+D_{2}}
{\displaystyle \gamma }
{\displaystyle ||G_{zd}(s)||_{\infty }<\gamma }
The LMI: LMI for non-rotating missle attitude controlEdit
{\displaystyle {\begin{aligned}&{\text{min}}\quad \gamma \\&{\text{s.t.}}\quad X>0\\&{\begin{bmatrix}(AX+B_{1}W)^{T}+AX+B_{1}W&B_{2}&(CX+D_{1}W)^{T}\\B_{2}^{T}&-\gamma I&D_{2}^{T}\\CX+D_{1}W&D_{2}&-\gamma I\end{bmatrix}}<0\end{aligned}}}
{\displaystyle \gamma }
is the disturbance attenuation level. When the matrices
{\displaystyle W}
{\displaystyle X}
{\displaystyle K=WX^{-1}}
https://github.com/asalimil/LMI-for-Attitude-Control-Nonrotating-Missle-Yaw-Roll-Channel
Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/pages/LMI_for_Nonrotating_Missiles_Attitude_Control_yaw_roll_channel&oldid=3794628"
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Pacinian corpuscles found in the skin are sensitive to
Pacinian corpuscles are found in the skin, joints, tendons, muscles and gut area and consist of the endings of a single neuron surrounded by many layers of connective tissue. These receptors respond to pressure.
A. Hyperthyroidism 1. Myxedema
B. Hyposecretion of glucocorticoids 2. Addison's disease
C. Hypothyroidism- I 3. Increased metabolic rate
A- 3; B- 2; C- 1; D- 5
A. Hyperthyroidism 3. Increased metabolic rate
C. Hypothyroidism- I 1. Myxedema
D. Hyperparathyroidism 5. Kidney stone
Choose the correct statements with regard to human respiration.
I. Inspiration is facilitated by the contraction of phrenic and internal intercostal muscles.
II. Healthy human can inhale or exhale about 2000 to 3000 mL of air per minute.
III. Functional residual capacity represent the volume of left out air in lungs after expiration.
IV. Total lung capacity is the total volume of air that can be accomodated in the lung after forced inspiration.
Statement III and IV are correct and statement I and II can be corrected as:-
I. Inspiration is facilitated by the contraction of diaphragm and contraction of external intercostal muscles.
II. Healthy humans can inhale 6000 to 8000 mL of air per minute.
Which one from those given below represents the position of nitrogen in the 9-membered double rings of purines?
Purines are 9-membered double rings with nitrogen at 1, 3, 7 and 9 positions. DNA has two types of purines namely Adenine (A) and Guanine (G).
When a red blood cell loses water by osmosis the shrinking of the cell volume leads to crinkling of the plasma membrane, called
Flacidity
When the external water potential is more negative than that of the cell, the cell lose water by osmosis. Such phenomenon in red blood cell is called crenation.
Flaccid cell in botany, is a cell in which the plasma membrane is not pressed tightly against the cell wall.
Bursting of cell occurs due to an osmotic imbalance that has caused excess water to diffuse into the cell.
Tonicity is a measure of the effective osmotic pressure gradient.
The most common type of ovule found in angiosperms is
Atropous ovule
Hemitropous ovule
Anatropous ovule is a completely inverted ovule turned back 180 degrees on its stalk. It is the most common type of ovule in angiosperms.
Campylotropous ovule refers to the body of the ovule that is curved or bent round so that the micropyle and chalaza so not lie in the same straight line.
Hemitropous ovule refers to the body of the ovule is placed transversely at right angles to the funicle. The micropyle and chalaza lie in one straight line.
are abundant in nervous system of non- vertebrates
consists largely of lipids
has a low electric resistance
allows the continous conduction of impulse
Myelin sheaths consists largely of lipids and has a high electrical resistance. These are abundant in vertebrates and allows the saltatory conduction of impulse.
Pick the reaction from the following, where a water molecule is removed and reduction of NAD+ does not occur in the reactions of respiration.
I. Succinic acid
\to
II. Malic acid
\to
Oxaloacetic acid
III. 2-phosphoglycerate
\to
phosphoenol pyruvic acid
IV. Pyruvic acid
\to
Acetyl Co-A
I and III reactions in which water is removed, but NAD+ is not reduced.
(I) Succinic acid
\to
\to
FADH+ + H2O
(III) 2- Phosphoglycerate
\to
2- Phosphoenol pyruvic acid + H2O
The II and IV reaction occur in the following way:
(II) Malic acid
\to
NAD+ + H2O + NADH
(IV) Pyruvic acid
\to
Acetyl Co- A
\to
NAD+ + CO2 + H2
Taenidia is the vital component of which system in human body?
Taenidia is the vital component respiratory system in human body. It is also known as intima. It is the inner spiral cuticular thickening of the tracheae. It prevents the tracheae from collapsing.
The largest type of nematocysts in Hydra is
Holotrichous isorhizas
Atrichous isorhizas
Desmonemes
Stenoteles
Stenoteles or penetrants are the largest type of nematocysts in Hydra. The thread tube is open at the end. When discharged, it release thread tube by which a poisnous fluid, hypotoxin is injected paralysing the prey.
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Does a square root have two values? | Brilliant Math & Science Wiki
John Rhoades, Ivan Koswara, Zandra Vinegar, and
\large \sqrt{x^2} = \pm x
Why some people say it's true: That's exactly what I was taught in school: when you take a square root, the answer is always "plus or minus" some value.
Why some people say it's false: When you square
x,
it becomes positive no matter what it was before; then, when you take the square root, it's still positive. Therefore, the answer is just
|x|
\pm x
\color{#20A900}{\text{Reveal the Correct Answer:}}
\color{#D61F06}{\textbf{false}}
On its standard domain of non-negative real numbers,
\sqrt{x}
is defined as "the non-negative real number which, when squared, equals
x.
" For example,
\sqrt{25} = 5
\pm 5
\sqrt{x}
is defined this way so that it is a function.
A function is a relation or map between a set of input values (the domain) and a set of output values (the range) which has the property that every accepted input corresponds with exactly one output. This property is commonly known as "passing the vertical line test."
\sqrt{x}
to be a function, its evaluation on any input must be a single, well-defined output. That is why it's defined as it is. Below is a graph of the square-root function. Note that it's only a single arm reaching out from the origin, not a rotation of a full parabola.
The graph of a
90^\circ
rotation of the whole
x^2
parabola would not pass the vertical line test and would not be a function.
Extending the Domain to All Real Numbers and the Range to Complex Numbers:
The behavior of the square root function when extended to the domain of all real numbers (positive reals, negative reals, and 0) precisely mirrors the argument made above. The square root of a negative number is a complex number. But, even with its domain extended,
\sqrt{x}
is similarly defined so that it is still a function. For example,
\sqrt{-25} = 5i
\pm 5i
\color{#3D99F6}{\text{See Common Rebuttals:}}
Rebuttal: But
x^2=y
almost always has two solutions. For example, when
y=25
x=5
x=-5
are both solutions. And this is only one case of the general rule that I learned in school: when you're solving an equation and take the square root of both sides in the process,
x^2
\pm x
x^2=25
x=\pm 5
y=x^2
does have two solutions, i.e. both
x=5
x=-5
are correct, but you're merging two steps of the process you were taught in school when you think about the rule this way. In school, you were likely shown this technique for solving simple quadratic equations in two steps:
x^2=25
x=\pm\sqrt{25}
\pm\sqrt{25} = \pm 5
\pm
is applied this way precisely because the square root is limited in its range to only positive values, whereas the function
x^2
has a domain of both positive and negative values. But the evaluation of the square root itself is simply 5.
The actual rule is that, when you want to invert the
x^2
function, likely because you're solving for a variable or expression that has been squared, you have to put a
\pm
in front of the square root of the side that was not squared initially so that you can include the domain of all negative numbers as potential inputs to the
x^2
The misconception here stems from this whole process frequently being referred to as "taking the square root of both sides." That's a misleading way to describe what you're doing since, in actuality, you are manually applying an actual inversion of the
x^2
function to both sides of the equation.
Rebuttal: The square root is the inverse function of
x^2
. That means that they just cancel out. So,
\sqrt{x^2}
x
, no matter what
x
Reply: The square root function is not the inverse of
x^2
for all real numbers. Instead, it is an inverse of
x^2
only on the interval
[0, \infty)
Here is a graph of the square-root function. Note that it's only a single arm reaching out from the origin, not a rotation of a full parabola.
\large \sqrt{x^2} = \pm x
is not a function. It is an equation. Restrictions on the singular output of a function would only apply when the square root is found a function's value determination.
2 1 0 and 2 0
\frac{\sqrt{x^2}}{|x|} + 1,
x
is a non-zero real number.
Cite as: Does a square root have two values?. Brilliant.org. Retrieved from https://brilliant.org/wiki/plus-or-minus-square-roots/
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Non-Standard Induction | Brilliant Math & Science Wiki
Non-Standard Induction
Mursalin Habib, Mahindra Jain, Gene Keun Chung, and
In standard induction, the most important part is the inductive step where you show that if a statement is true for some natural number
k
k+1
. The base case, albeit important, is obvious most of the time. In other words, proving something by induction requires linking
P(k)
P(k+1)
in some way. But what if you can't do that? What if you somehow showed that
P(k)\Rightarrow P(k+m)
. Can we use that to prove something?
The answer is yes. The induction argument is very flexible and it doesn't have to be restricted to the base case
=1
P(k)\implies P(k+1)
. Non-standard induction refers to forms of induction that don't follow the 'regular' or 'standard' format.
Proof of Non-standard Induction
Here's a way to prove a statement true for all positive integers, if you've managed to show that
P(k)\Rightarrow P(k+m)
Proof by Non-Standard Induction
1
: Prove the base case:
This is where you verify that the statement is true for
1
2
3
\cdots m
2
: The inductive step:
Now you show that
P(k)\Rightarrow P(k+m)
. In other words, you show that if
P(k)
were true, so would be
P(k+m)
If you manage to do both of these things, you'll have ensured that the statement is true for all positive integers
n
. Sometimes it is easier to do step 2 first, as playing around with the problem allows us to naturally discover a suitable
m
This approach is often useful when you see a cyclic term, like
(-1)^n
\sin ( \frac{ n \pi } { 5 })
n^2 \pmod{5}
, or when we spot a certain pattern in the sequence. This often makes the induction step simpler to show, though it means that we will have to consider multiple base cases.
Show that every integer
n > 7
3 a + 5 b
for some non-negative integers
and
b
If we know that the integer
k
3a + 5b
, then it is clear that
k + 3 = 3 ( a+1) + 5b
can also be written in the required format. As such, this tells us that we should try to apply non-standard induction with
m = 3
To prove the base case, observe that:
\begin{aligned} 8 &= 3 \times 1 + 5 \times 1 \\ 9 &= 3 \times 3 + 5 \times 0 \\ 10 &=3 \times 0 + 5 \times 2. \end{aligned}
Hence, by non-standard induction, the statement is true for all integers
n > 7
_\square
Note: This is a special case of the Chicken Mcnugget Theorem.
Prove that for any integer
n>5
, every square can be divided into
n
smaller squares.
A good approach to this problem is noticing that any square can be divided into
4
This means if a square can be divided into
k
smaller squares, then the square can be divided into
k+3
squares. As such, this tells us that we should try to apply non-standard induction with
m = 3
Now if we want to prove the statement for any integer
n>5
, we need to show that the statement holds for
n=6, 7, 8
. If we can do that, we're done. (Why?)
This proves to be actually harder than showing the inductive step. But if you get your hands dirty a little bit, you'll see this.
That covers our base cases and completes the proof.
_\square
[Erdos & Suranyi] Prove that every positive integer can be written in the form
\pm 1^2\pm 2^2\pm 3^2\pm \cdots \pm l^2
l
and the appropriate choices of plus and minus signs.
First, we make the (clever) observation that
(a+1) ^2 - ( a + 2)^2 - (a+3) ^2 + (a+4) ^2 = 4
. This can be arrived at by trying to get rid of the quadratic and linear terms, in order to arrive at a constant.
k=±1^2 ± 2^2±3^2\pm ... ± l^2
k+4=±1^2 ± 2^2±3^2\pm ... ± l^2 +(l+1)^2- (l+2)^2 - (l+3)^2 + (l+4)^2
. As such, this tells us that we should try to apply non-standard induction with
m = 4
Now the only thing left to do is show that such representations exist for integers
1
4
. After playing around for a while we see that
\begin{aligned} 1 & =1^2\\ 2& =-1^2-2^2-3^2+4^2 \\ 3 &=-1^2+2^2 \\ 4 & =-1^2-2^2+3^2. \end{aligned}
This means the statement holds for all positive integers.
_\square
Note: The expression that we found is not necessarily the shortest. For example, we have
5 = 1^2 + 2 ^2
, though our induction step will give us
5 = 1^2 + 2^2 - 3^2 - 4^2 + 5^2
. In this question, we are merely interested in the existence for a representation, and not concerned about finding the most compact representation.
This proof is very similar to the proof of standard induction. Can you spot the difference?
Fix an integer
m > 1
S
\quad
(1) The integers
1, 2, \ldots , m
belong to the set.
\quad
(2) Whenever an integer
k
S
, the integer
k+m
S
S
_\square
T
S
T
is non-empty. Hence it must contain a smallest element, which we will denote by
\alpha
. Then by (1),
0 < \alpha- m < \alpha
\alpha
T
\alpha-m \not\in T \Rightarrow(\alpha-m)\in S
S
(\alpha-m)+m=\alpha
\alpha\subset
T
Hence set
T
is empty, and set
S
_\square
1) Prove the following using non-standard induction with
m = 5
n > 7
3 a + 5 b
a, b
\omega
is a complex third root of unity, what is the value of
\omega^{2n} + \omega^n + 1
\displaystyle \sum_{k=0 } ^ { \lfloor \frac{n}{3} \rfloor } { n \choose 3k }
Cite as: Non-Standard Induction. Brilliant.org. Retrieved from https://brilliant.org/wiki/non-standard-induction/
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Q.10. Find the mode of the following data:
Ashutosh sir didnt understand your explanation, you didnt tell what deviation is and didnt explain clearly.
Please explain again!
The sum of deviation of a set of values {x}_{1},{x}_{2},........,{x}_{n},measured from 50 is -10 and the sum of \phantom{\rule{0ex}{0ex}}deviation of the values from 46 is 70.Find the value of n and the mean.
Q.4. The mean of 'n' observation is
\overline{x}
, if the first term is increased by 1, second by 2 and so on. What will be the new mean?
Please explain what is sum of deviations and solve the sum.... In an easy manner no links plz.. Take this question a bit personally. Thnx
Q1. The sum of deviations of a set of values x1 , x2, xy.............. , yAP measured from 50 is – 10 and the sum of deviations of the values from 46 is 70. Find the value of n and the mean.
21. An incomplete distribution is given as follow, the median is 35 and the sum of all the frequencies is 170. Find missing frequencies
Can someone just give me answer for this.
Q.22. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given below:
No. of seats 100-104 104-108 108-112 112-116 116-120
Determine the mean number of seats occupied over the flights
Q.44. If mean of 1, 2, 3, ...., n is
\frac{16n}{11}
, then value of n is _______
Q.22. The following table show marks of 85 students of a class X in a school. Find the modal marks of the distrbution :
Q). Find the median of the following data.
Question in the pic...
No links pls.... Thnx
{x}_{i}
's are the mid-points of the class intervals of grouped data,
{f}_{i}
's are corresponding frequencies and
\overline{x}
is the mean, find the value of
\sum _{}\left({f}_{i}{x}_{i}-\overline{x}\right)
Can plss any one answer this question
Q). Find the median of the following data by converting into a grouped frequency distribution.
4. In calculating the mean of grouped data,group in classes of equal width. we may use the formula,\phantom{\rule{0ex}{0ex}} \overline{)x}=a+\frac{\sum _{}{f}_{i}{f}_{i}}{\underset{}{\sum {f}_{i}}}\phantom{\rule{0ex}{0ex}}Where ,a is the assumed mean a must be one of the mid points of the classes. Is the last statement correct? Justify\phantom{\rule{0ex}{0ex}}your answer
ques4
Q.27. Compute the median for the following cumulative frequency distribution.
Pls answer 30th que part 1:
30. The arithmetic mean of the given frequency distribution is Rs 472 and the total number of employees is 100. Find the value of p and q.
Daily income (in Rs) 200-300 300-400 400-500 500-600 600-700 700-800 800-900
No. of employees 5 p 24 q 9 6 4
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We have to make a wall of size 16m by 12m How many bricks will be required to construct this - Maths - Mensuration - 8956359 | Meritnation.com
We have to make a wall of size 16m by 12m.How many bricks will be required to construct this wall of size 20cm by 8cm...Kindly answer by today or tomorrow....
Given : Size of wall = 16 m by 12 m
We know 1 m = 100 cm , So
Size of wall = 1600 cm by 1200 cm
Area of wall = 1600
×
1200 = 1920000 cm2
Size of brick = 20 cm by 8 cm
Area of a brick = 20
×
8 = 160 cm2
Total number of bricks required for wall =
\frac{1920000}{160} = \mathbf{12000}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{ }\mathbit{A}\mathbit{n}\mathbit{s}\mathbf{ }\mathbf{\right)}
Dimensions of the wall=16 m by 12 m/1600 cm by 1200 cm
Dimensions of the brick=20 cm by 8 cm
Bricks needed to construct the wall =1600*1200
=12000 bricks
Ans)12000 bricks will be needed to construct a wall of size 16 m by 12 m where each brick measures 20 cm by 8 cm.
Manya Agrawal answered this
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Client hints - ImageKit.io Docs
ImageKit, with its support of client hints, can automatically deliver responsive images without you changing the image URL. Learn the different ways you can use client hints.
Client hints are the hints provided by the client device to the server along with the request itself. These hints allow the server to fulfill a particular request with the most optimal resource.
You can learn more about client hints from the responsive image guide.
Enable client hints before using them Not every request has these HTTP headers. You will have to explicitly tell the browser to include these client hints using a meta tag:
ImageKit supports Client hints
ImageKit supports the following client hints:
DPR
Save-Data
WebP conversion using Accept header is enabled by default and part of the automatic format conversion feature.
To allow ImageKit to read values from the client hint request headers (DPR and Width), you have to pass the transformation parameters dpr and width with their values set to auto. For example, when the browser requests:
GET: https://ik.imagekit.io/your_imagekitid/tr:w-100,dpr-auto/image_name.jpg
In this case, an extrinsic width of 100 pixels is required. To calculate the extrinsic width of the to-be-delivered image, ImageKit reads the client hint header DPR value and multiplies it with the specified extrinsic width. Therefore, the final actual width of the delivered image is 200
100 * 2 = 200 px
If the browser requests:
https://ik.imagekit.io/your_imagekitid/tr:w-auto,dpr-auto/image_name.jpg
In this case, an intrinsic width of 600 pixels is required. The browser sends Width request header and also considers the DPR of the user-device while calculating the value of Width header. Therefore, ImageKit ignores the DPR value and delivers an image of width 600. ImageKit will return Content-DPR response header so that browser can scale the image correctly.
The Content-DPR Header
ImageKit rounds the intrinsic size of the image to the next smallest 100. If the Widthheader indicates a width of 150 px, then ImageKit will deliver an image with a width of 200 px. Now, if the DPR of the device is 2, then the device will end up rendering an image of width 100 px (200 / 2), which is the incorrect width. The correct intended width to be displayed is 75px (150 / 2). To rectify this miscalculation due to the rounding to the next 100, the Content-DPR header is used.
Content-DPR is a response header and indicates the actual DPR of the response image. It is calculated as follows:
Content-DPR = [Selected Image Size] / (Width / DPR)
Let's learn this with a few examples:
GET: https://ik.imagekit.io/your_imagekitid/tr:w-auto,dpr-auto/image_name.jpg
ImageKit rounds off 212 to 300, and an image of width 300 pixels is delivered. Now, the Content-DPR header is calculated as follows:
Content-DPR = 300/ (212 / 2) = 2.83
Hence, ImageKit responds with a 300 pixels wide image and Content-DPR response header with a value 2.83
Content-DPR: 2.83
When the browser receives the image and the header, it scales it down as 300 / 2.83 = 106 px, which was is intended final width of the rendered image. If there were no Content-DPR header received, the browser would scale down the image as 300 / 2 = 150 px, which might break your layout.
You should be aware of the different terms. Here are a few important definitions from Google developer docs.
The actual dimensions of a media resource. For example, if you open an image in Photoshop, the dimensions shown in the image size dialogue describe its intrinsic size.
Extrinsic size:
The size of a media resource after CSS and other layout factors (such as width and height attributes) have been applied to it. Let’s say you have an <img> element that loads an image with a density-corrected intrinsic size of 320x240, but it also has CSS width and height properties with values of 256px and 192px applied to it, respectively. In this example, the extrinsic size of that <img> element becomes 256x192.
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Two oppositely charged but otherwise identical conducting plates of area 2.50 sq
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick,
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is
1.20×{10}^{6}
volts per meter.
Compute the magnitude of the charge per unit area
\sigma
on the conducting plate.
\sigma =\frac{c}{{m}^{2}}
{\sigma }_{1}
on the surfaces of the dielectric.
{\sigma }_{1}=\frac{c}{{m}^{2}}
Find the total electric-field energy U stored in the capacitor.
u=J
a) the magnitude of the charge per unit area
\sigma
on the conducting plate:
E effective=
\frac{\sigma }{k{ϵ}_{0}}
\sigma =3.6\cdot 8.854{e}^{-12}\cdot 1.20{e}^{6}=3.82\cdot {e}^{-5}\frac{C}{{m}^{2}}
b) The magnitude of the charge per unit area σ1 on the surfaces of the dielectric.
\sigma
(on the dielectric)
=\sigma \left(\left(\frac{1}{k}\right)-1\right)
=-2.75{e}^{-5}\text{ }\frac{C}{{m}^{2}}
4.50\cdot {10}^{4}
\theta
{13.0}^{\circ }
\theta
A block of mass m=2.20 kg slides down a 30 degree incline which is 3.60 m high. At the bottom, it strikes a block of mass=7.00 kg which is at rest on a horizontal surface in the picture. If the collision is elastic, and friction can be ignored, determine (a)the speeds of the two blocks after the collision, and (b) how far back up the incline the smaller mass will go.
Change from rectangular to spherical coordinates. (0, -2, 0)
Suppose unpolarized light of intensity
150\frac{W}{{m}^{2}}
falls on the polarizer at an angle 30. What is the light intensity reaching the photocell?
Anovice pilot sets a plane's controls, thinking the plane will fly at
2.50×{10}^{2}\frac{km}{h}
to the north. If the wind blows at
75\frac{km}{h}
toward the southeast, what is the plane’s resultant velocity? Use graphical techniques.
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Rates Module - ABM Protocol
Jug - Detailed Documentation
The ABM Protocol's Rate Accumulation Mechanism
Module Name: Rates Module
Type/Category: Rates
Contract Sources:
Jug
Pot
A fundamental feature of the MCD system is to accumulate stability fees on Vault debt balances, as well as interest on dotBTC Savings Rate (DSR) deposits.
The mechanism used to perform these accumulation functions is subject to an important constraint: accumulation must be a constant-time operation with respect to the number of Vaults and the number of DSR deposits. Otherwise, accumulation events would be very gas-inefficient (and might even exceed block gas limits).
For both stability fees and the DSR, the solution is similar: store and update a global "cumulative rate" value (per-collateral for stability fees), which can then be multiplied by a normalized debt or deposit amount to give the total debt or deposit amount when needed.
This can be described more concretely with mathematical notation:
Discretize time in 1-second intervals, starting from t_0;
Let the (per-second) stability fee at time t have value F_i (this generally takes the form 1+x, where x is small)
Let the initial value of the cumulative rate be denoted by R_0
Let a Vault be created at time t_0 with debt D_0 drawn immediately; the normalized debt A (which the system stores on a per-Vault basis) is calculated as D_0/R_0
Then the cumulative rate R at time T is given by:
R(t) \equiv R_0 \prod_{i=t_0 + 1}^{t} F_i = R_0 \cdot F_{t_0 + 1} \cdot F_{t_0 + 2} \cdots F_{t-1} \cdot F_t
And the total debt of the Vault at time t would be:
D(t) \equiv A \cdot R(t) = D_0 \prod_{t=1}^{T} F_i
In the actual system, R is not necessarily updated with every block, and thus actual R values within the system may not have the exact value that they should in theory. The difference in practice, however, should be minor, given a sufficiently large and active ecosystem.
Detailed explanations of the two accumulation mechanisms may be found below.
Stability Fee Accumulation
Stability fee accumulation in MCD is largely an interplay between two contracts: the Vat (the system's central accounting ledger) and the Jug (a specialized module for updating the cumulative rate), with the Vow involved only as the address to which the accumulated fees are credited.
The Vat stores, for each collateral type, an Ilk struct that contains the cumulative rate (rate) and the total normalized debt associated with that collateral type (Art). The Jug stores the per-second rate for each collateral type as a combination of a base value that applies to all collateral types, and a duty value per collateral. The per-second rate for a given collateral type is the sum of its particular duty and the global base.
Calling Jug.drip(bytes32 ilk) computes an update to the ilk's rate based on duty, base, and the time since drip was last called for the given ilk (rho). Then the Jug invokes Vat.fold(bytes32 ilk, address vow, int rate_change) which:
adds rate_change to rate for the specified ilk
increases the Vow's surplus by Art*rate_change
increases the system's total debt (i.e. issued dotBTC ) by Art*rate_change.
Each individual Vault (represented by an Urn struct in the Vat) stores a "normalized debt" parameter called art. Any time it is needed by the code, the Vault's total debt, including stability fees, can be calculated as art*rate (where rate corresponds to that of the appropriate collateral type). Thus an update to Ilk.rate via Jug.drip(bytes32 ilk) effectively updates the debt for all Vaults collateralized with ilk tokens.
Example With Visualizations
Suppose at time 0, a Vault is opened and 20 dotBTC is drawn from it. Assume that rate is 1; this implies that the stored art in the Vault's Urn is also 20. Let the base and duty be set such that after 12 years, art*rate = 30 (this corresponds to an annual stability of roughly 3.4366%). Equivalently, rate = 1.5 after 12 years. Assuming that base + duty does not change, the growth of the effective debt can be graphed as follows:
Now suppose that at 12 years, an additional 10 dotBTC is drawn. The debt vs time graph would change to look like:
What art would be stored in the Vat to reflect this change? (hint: not 30!) Recall that art is defined from the requirement that art * rate = Vault debt. Since the Vault's debt is known to be 40 and rate is known to be 1.5, we can solve for art: 40/1.5 ~ 26.67.
The art can be thought of as "debt at time 0", or "the amount of dotBTC that if drawn at time zero would result in the present total debt". The graph below demonstrates this visually; the length of the green bar extending upwards from t = 0 is the post-draw art value.
Some consequences of the mechanism that are good to keep in mind:
There is no stored history of draws or wipes of Vault debt
There is no stored history of stability fee changes, only the cumulative effective rate
The rate value for each collateral perpetually increases (unless the fee becomes negative at some point)
Who calls drip?
The system relies on market participants to call drip rather than, say, automatically calling it upon Vault manipulations. The following entities are motivated to call drip:
Keepers seeking to liquidate Vaults (since the accumulation of stability fees can push a Vault's collateralization ratio into unsafe territory, allowing Keepers to liquidate it and profit in the resulting collateral auction)
Vault owners wishing to draw dotBTC (if they don't call drip prior to drawing from their Vault, they will be charged fees on the drawn dotBTC going back to the last time drip was called—unless no one calls drip before they repay their Vault, see below)
ABM holders (they have a vested interest in seeing the system work well, and the collection of surplus in particular is critical to the ebb and flow of ABM in existence)
Despite the variety of incentivized actors, calls to drip are likely to be intermittent due to gas costs and tragedy of the commons until a certain scale can be achieved. Thus the value of the rate parameter for a given collateral type may display the following time behavior:
Debt drawn and wiped between rate updates (i.e. between drip calls) would have no stability fees assessed on it. Also, depending on the timing of updates to the stability fee, there may be small discrepancies between the actual value of rate and its ideal value (the value if drip were called in every block). To demonstrate this, consider the following:
at t = 0, assume the following values:
\text{rate} = 1 \text{ ; total fee} = f
in a block with t = 28, drip is called—now:
\text{rate} = f^{28}
in a block with t = 56, the fee is updated to a new, different value:
\text{totalfee} \xrightarrow{} g
in a block with t = 70, drip is called again; the actual value of rate that obtains is:
\text{rate} = f^{28} g^{42}
however, the "ideal" rate (if drip were called at the start of every block) would be:
\text{rate}_{ideal} = f^{56}g^{14}
Depending on whether f > g or g > f, the net value of fees accrued will be either too small or too large. It is assumed that drip calls will be frequent enough such inaccuracies will be minor, at least after an initial growth period. Governance can mitigate this behavior by calling drip immediately prior to fee changes. The code in fact enforces that drip must be called prior to a duty update, but does not enforce a similar restriction for base (due to the inefficiency of iterating over all collateral types).
dotBTC Savings Rate Accumulation
DSR accumulation is very similar to stability fee accumulation. It is implemented via the Pot, which interacts with the Vat (and again the Vow's address is used for accounting for the dotBTC created). The Pot tracks normalized deposits on a per-user basis (pie[usr]) and maintains a cumulative interest rate parameter (chi). A drip function analogous to that of Jug is called intermittently by economic actors to trigger savings accumulation.
The per-second (or "instantaneous") savings rate is stored in the dsr parameter (analogous to base+duty in the stability fee case). The chi parameter as a function of time is thus (in the ideal case of drip being called every block) given by:
\text{chi}(t) \equiv \text{chi}0 \prod{i=t_0 + 1}^{t} \text{dsr}_i
where chi_0 is simply chi(t_0).
Suppose a user joins N dotBTC into the Pot at time t_0. Then, their internal savings dotBTC balance is set to:
\text{pie[usr]} = N / \text{chi}_0
The total dotBTC the user can withdraw from the Pot at time t is:
\text{pie[usr]} \cdot \text{chi}(t) = N \prod_{i=t_0 + 1}^{t} \text{dsr}_i
Thus we see that updates to chi effectively increase all Pot balances at once, without having to iterate over all of them.
After updating chi, Pot.drip then calls Vat.suck with arguments such that the additional dotBTC created from this savings accumulation is credited to the Pot contract while the Vow's sin (unbacked debt) is increased by the same amount (the global debt and unbacked debt tallies are increased as well). To accomplish this efficiently, the Pot keeps track of a the total sum of all individual pie[usr] values in a variable called Pie.
The following points are useful to keep in mind when reasoning about savings accumulation (all have analogs in the fee accumulation mechanism):
if drip is called only infrequently, the instantaneously value of chi may differ from the ideal
the code requires that drip be called prior to dsr changes, which eliminates deviations of chi from its ideal value due to such changes not coinciding with drip calls
chi is a monotonically increasing value unless the effective savings rate becomes negative (dsr < ONE)
There is no stored record of depositing or withdrawing dotBTC from the Pot
There is no stored record of changes to the dsr
The following economic actors are incentivized (or forced) to call Pot.drip:
any user withdrawing dotBTC from the Pot (otherwise they lose money!)
any user putting dotBTC into the Pot—this is not economically rational, but is instead forced by smart contract logic that requires drip to be called in the same block as new dotBTC is added to the Pot (otherwise, an economic exploit that drains system surplus is possible)
any actor with a motive to increase the system debt, for example a Keeper hoping to trigger flop (debt) auctions
A Note On Setting Rates
Let's see how to set a rate value in practice. Suppose it is desired to set the DSR to 0.5% annually. Assume the real rate will track the ideal rate. Then, we need a per-second rate value r such that (denoting the number of seconds in a year by N):
r^N = 1.005
An arbitrary precision calculator can be used to take the N-th root of the right-hand side (with N = 31536000 = 3652460*60), to obtain:
r = 1.000000000158153903837946258002097...
The dsr parameter in the Pot implementation is interpreted as a ray, i.e. a 27 decimal digit fixed-point number. Thus we multiply by 10^27 and drop anything after the decimal point:
\text{dsr} = 1000000000158153903837946258
The dsr could then be set to 0.5% annually by calling:
Pot.file("dsr", 1000000000158153903837946258)
Spell - Detailed Documentation
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Home : Support : Online Help : Education : Student Packages : Linear Algebra : Computation : Solvers : LeastSquares
compute a least squares solution to a set of equations
LeastSquares(A, B, options)
Matrix or set
column Vector or set of variables
(optional) parameters; for a complete list, see LinearAlgebra[LeastSquares]
For Matrix A and Vector B, the LeastSquares(A, B) command returns a Vector that best satisfies the condition
A·x
is approximately equal to B, in the least squares sense. The result that is returned is the Vector x which minimizes Norm(A . x - B, 2).
Parameter A can also be a set of equations that describe the linear least squares problem. In this case, B is the set of variables in which the equations in A occur.
\mathrm{with}\left(\mathrm{Student}[\mathrm{LinearAlgebra}]\right):
A≔〈〈3,0,4〉|〈-2,3,4〉〉:
b≔〈1,2,4〉:
X≔\mathrm{LeastSquares}\left(A,b\right)
\textcolor[rgb]{0,0,1}{X}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{351}}{\textcolor[rgb]{0,0,1}{625}}\\ \frac{\textcolor[rgb]{0,0,1}{62}}{\textcolor[rgb]{0,0,1}{125}}\end{array}]
\mathrm{Norm}\left(A·X-b\right)
\frac{\textcolor[rgb]{0,0,1}{16}}{\textcolor[rgb]{0,0,1}{25}}
E≔{2x-y+2=0,mx+ny-3=0}:
V≔{x,y}:
\mathrm{LeastSquares}\left(E,V\right)
{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{n}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\right)}{\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{n}}}
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State whether the investigation in question is an observational study or a desig
State whether the investigation in question is an observational study or a designed experiment. Justify your answer in each case. The Salk Vaccine. In
State whether the investigation in question is an observational study or a designed experiment. Justify your answer in each case.
The Salk Vaccine. In the 1940s and early 1950s, the public was greatly concerned about polio. In an attempt to prevent this disease, Jonas Salk of the University of Pittsburgh developed a polio vaccine. In a test of the vaccine’s efficacy, involving nearly 2 million grade-school children, half of the children received the Salk vaccine, the other half received a placebo, in this case an injection of salt dissolved in water. Neither the children nor the doctors performing the diagnoses knew which children belonged to which group, but an evaluation center did. The center found that the incidence of polio was far less among the children inoculated with the Salk vaccine. From that information, the researchers concluded that the vaccine would be effective in preventing polio for all U.S. school children, consequently, it was made available for general use.
The investigators may simply observe a predefined set of individuals and collect the data from the observations, without influencing any factors involved. This is an “observational study”.
The investigator may design and perform their own experiments my controlling and manipulating factors of their choice and collect the resulting data. This is the “designed experiment”.
The investigators knowingly assigned the treatments- Salk vaccine or placebo, to the children and collect data on the effects of the treatments. The investigators can not only observe, but also manipulate the treatment that results in the outcomes.
Thus, the investigation is a designed experiment.
Whether the given study depicts an observational study or a designed experiment.
The most appropriate study design depends, among other things, on the distribution of
1.A) The risk factor in the population of interest
2.B) The participants
3.C) The outcome in the population of interest
4.D)
A\text{ }\mathrm{&}\text{ }C
Many people believe that students gain weight as freshmen. Suppose we plan to conduct a study to see if this is true.
a) Describe a study design that would require a matched- pairs t procedure to analyze the results.
b) Describe a study design that would require a two- sample t procedure to analyze the results.
An observational study is retrospective if it considers only existing data. It is prospective if the study design calls for data to be collected as time goes on. Tell which of the following observational studies are retrospective and which are prospective. A sample of moose in a national park are given ear tags so that naturalists can track their growth, movement, and physical condition during a ten-year study.
Describe the Process design phase?
Describe a study that would provide more useful information.
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The figure above depicts the developmental figure of a cylinder whose base radius is
r
and height is
h.
The surface area is equal to the sum of the areas of the two circular bases and the rectangular side. The area of each base is
\pi r^2.
Since the width of the rectangular side must be equal to the circumference of the base, the area of the rectangular side is
2\pi rh.
Therefore the total surface area is
2\pi r^2+2\pi rh=2\pi r(r+h).
2 \pi \times 3 \times 4 +2 \pi \times 3^2 = 42 \pi
_\square
Suppose that the sum of the areas of 2 identical circular ends in a cylinder is the same as the area of the curved side of the cylinder. If the radii of of the flat circular ends are each
r,
what is the height of the cylinder?
The sum of the areas of the 2 identical flat circular ends in the cylinder is
2 \pi r^2.
The area of the curved side of the cylinder is
2 \pi r h,
h
is the height of the cylinder. Equating these two gives
2 \pi r^2=2 \pi r h \Rightarrow h=r.
r.
_\square
Suppose the surface area of a circular cylinder with height
h
r
is half the surface area of a circular cylinder with height
5h
r.
r:h?
2 \pi r h +2 \pi r^2
for the surface area of a circular cylinder, we have the following relation between the two surface areas of interest:
2 \pi r h +2 \pi r^2=\frac{1}{2}\times\left(2 \pi r \cdot (5h) +2 \pi r^2\right).
\pi r
\begin{aligned} 2h+2r&=5h+r\\ r&=3h\\ r:h&=3:1. \ _\square \end{aligned}
Suppose that the surface area of a circular cylinder is
20\pi.
If both the radius
r
h
of the cylinder are integers and
r>1,
r+h?
2 \pi r h +2 \pi r^2
for the surface area of a circular cylinder, we have
\begin{aligned} 2 \pi r h +2 \pi r^2&=20 \pi \\ r(h+r)&=10. \qquad (1) \end{aligned}
r>1
by assumption, if
r=2,
h=3.
Then no other integer value of
r>2
(1).
r+h=2+3=5.
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Introducing Binary Practice Problems Online | Brilliant
In our base-10 system, the positions (from right to left) correspond to "number of 1s", "number of 10s", "number of 100s", and so on, where the value of each position is 10 times the previous one.
Binary, or base 2, uses only the digits 0 and 1. Each position corresponds to "number of 1s", "number of 2s", "number of 4s", "number of 8s", and so on. The value of each position is 2 times the previous one.
The number above expressed in base 10 is
8 + 1 = 9 ,
since there is one 8 and one 1.
What is the value (in base 10) of the binary number 10000?
Introducing Binary
Which binary number is larger?
If a binary number ends in 1 (like 1001 or 111011), then it is
\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}.
always odd always even either odd or even, depending on circumstances
What is the result (in binary) of multiplying the binary number 1100 by the binary number 10?
Here are the binary numbers listed starting from 2. Consider the second from the last digit.
10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, ...
Notice the pattern 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, ...
Does this pattern continue for all positive integers?
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ProjectiveGeneralUnitaryGroup - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : ProjectiveGeneralUnitaryGroup
construct a permutation group isomorphic to a projective general unitary group
ProjectiveGeneralUnitaryGroup(n, q)
PGU(n, q)
a positive integer greater than 1
The projective general unitary group
PGU\left(n,q\right)
is the quotient of the general unitary group
GU\left(n,q\right)
The ProjectiveGeneralUnitaryGroup( n, q ) command returns a permutation group isomorphic to the projective general unitary group of degree
n
over the field with
{q}^{2}
elements. In general, this is not a transitive representation.
n=2
PGU\left(n,q\right)
PGL\left(n,q\right)
are isomorphic, so the latter is returned in this case.
The ranges for n and q which are implemented are as follows:
n=2
q\le 100
n=3
q\le 5
n=4
q\le 4
n=5,6
q=2
If either or both of the arguments n and q are non-numeric, then a symbolic group representing the projective general unitary group is returned.
The command PGU( n, q ) is provided as an abbreviation.
\mathrm{with}\left(\mathrm{GroupTheory}\right):
\mathrm{ProjectiveGeneralUnitaryGroup}\left(2,13\right)
\textcolor[rgb]{0,0,1}{\mathbf{PGU}}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{13}\right)
G≔\mathrm{PGU}\left(4,4\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathbf{PGU}}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)
\mathrm{GroupOrder}\left(G\right)
\textcolor[rgb]{0,0,1}{1018368000}
\mathrm{IsTransitive}\left(\mathrm{PGU}\left(3,3\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{IsPrimitive}\left(\mathrm{PGU}\left(3,3\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{GroupOrder}\left(\mathrm{PGU}\left(4,q\right)\right)
{\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
The GroupTheory[ProjectiveGeneralUnitaryGroup] command was introduced in Maple 17.
The GroupTheory[ProjectiveGeneralUnitaryGroup] command was updated in Maple 2020.
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Overview - DeFiner.org
We have four main contracts in our Protocol, and they are SavingAccount, Accounts, Bank, TokenRegistry, and GlobalConfig.
The first three contracts are upgradable contracts, and they contain the main business logic of our Protocol. The overall structure of our Protocol is monolithic, which means that SavingAccount, Accounts, and Bank contracts work closely with each other. We decoupled this into three contracts only for code size limit reasons. The naming here maybe a little counter intuitive. Normally when we want to deposit money to a bank, we first go to the bank and open a saving account, then we deposit money to this account. Here, SavingAccount contract behaves like a traditional bank, which is the entry point for you to deposit or borrow money from DeFiner. The Bank contract records the state of the pool of DeFiner's protocol, like the total deposit, total borrow, and the current rate. The Accounts contract records the deposit and borrow balances for each specific user.
The workflow of a transaction to DeFiner works as the following diagram. The user sends transactions by calling functions on SavingAccount contract, then the SavingAccount contract will call the Bank contract to create a rate index checkpoint, which is used to compute the accumulated interests for all the user. Then the Bank contract will call the functions in Account contract to compute the balance change in the user account.
The workflow of a transaction
The TokenRegistry and GlobalConfig contracts are used to save data that will be used among the three main contracts. Like all three contracts' addresses, so in these three contract, they only need to keep global config's address in contract's memory space.
This is the interface contract that our users will mainly interact with. Users will send transactions to this contract to do deposit, withdraw, borrow, repay, withdrawAll, and liquidate functions to this contract to interact with our protocol. For each operation, this contract is the contract that receives and sends out the tokens. Whenever there is a transaction sent to SavingAccount contract, it will emit an event to log this transaction.
The borrowing power of a user is related to the collateral that the user deposited into our system. The price is obtained from Chainlink's oracle. If the LTV of a user is too large then it is at the risk of being liquidated. Currently, the liquidation threshold is set to around 0.8.
This contract is used to create rate indexes whenever there is an interaction with our SavingAccount contract. We are using rate indexes to compute the borrow and deposit interests. Given the different indexes, we compute the user balances here. Whenever an index is created, this contract will emit an event.
i and j here represent two different blocks.
RateIndex_i = RateIndex_j * (1 + RatePerBlock_{ij} * (i - j))
Balance_i = Balance_j \times RateIndex_j \div RateIndex_i
Borrow balances and deposit balances both use this method to compute.
This contract is used to record the balances of each user account for different tokens. This contract also contains a BitMap which will quickly show whether a user has depositings/borrowings for each token while costs less gas.
This contract is used to store all the contract's addresses, so all other contracts can only keep GlobalConfig's address to call functions of other contracts. It also used to config the reserve ratio and community fund ratio of the protocol.
This contract records the meta-data about each contract supported token. We can add new supported tokens using this contract.
Current Supported Tokens
Token Address: 0x6b175474e89094c44da98b954eedeac495271d0f
cToken Address: 0x5d3a536e4d6dbd6114cc1ead35777bab948e3643
cToken Address: 0x39aa39c021dfbae8fac545936693ac917d5e7563
Token Address: 0xdac17f958d2ee523a2206206994597c13d831ec7
cToken Address: 0xf650c3d88d12db855b8bf7d11be6c55a4e07dcc9
Token Address: 0x0000000000085d4780B73119b644AE5ecd22b376
cToken Address: Not Compound supported.
Token Address: 0x9f8f72aa9304c8b593d555f12ef6589cc3a579a2
Token Address: 0x0d8775f648430679a709e98d2b0cb6250d2887ef
cToken Address: 0x6c8c6b02e7b2be14d4fa6022dfd6d75921d90e4e
Token Address: 0xe41d2489571d322189246dafa5ebde1f4699f498
cToken Address: 0xb3319f5d18bc0d84dd1b4825dcde5d5f7266d407
Token Address: 0x1985365e9f78359a9B6AD760e32412f4a445E862
cToken Address: 0x158079ee67fce2f58472a96584a73c7ab9ac95c1
Token Address: 0x2260fac5e5542a773aa44fbcfedf7c193bc2c599
cToken Address: 0xc11b1268c1a384e55c48c2391d8d480264a3a7f4
Token Address: 0x054f76beED60AB6dBEb23502178C52d6C5dEbE40
cToken Address: Not Compound supported
FIN LPToken
For SavingAccount, Accounts, and Bank contracts, we are using OpenZeppelin proxies to conduct the upgrading process. So other than the contracts, we also have an OpenZeppelin proxy contract for each of these three main contracts. The proxy contract addresses can never be changed, but the contract of the underlying implementation is changeable.
The workflow of a proxy contract.
For the GlobalConfig and TokenRegistry contracts, we can deploy a new contract and call the setter function in three main contracts to point to the new contracts, since they don't save any transactional data.
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Engineering Acoustics/Flow-induced oscillations of a Helmholtz resonator and applications - Wikibooks, open books for an open world
Engineering Acoustics/Flow-induced oscillations of a Helmholtz resonator and applications
2 Feedback loop analysis
3 Acoustical characteristics of the resonator
3.1 Lumped parameter model
3.2 Production of self-sustained oscillations
4 Applications to Sunroof Buffeting
4.1 How are vortices formed during buffeting?
4.2 How to identify buffeting
The importance of flow excited acoustic resonance lies in the large number of applications in which it occurs. Sound production in organ pipes, compressors, transonic wind tunnels, and open sunroofs are only a few examples of the many applications in which flow excited resonance of Helmholtz resonators can be found.[4] An instability of the fluid motion coupled with an acoustic resonance of the cavity produce large pressure fluctuations that are felt as increased sound pressure levels. Passengers of road vehicles with open sunroofs often experience discomfort, fatigue, and dizziness from self-sustained oscillations inside the car cabin. This phenomenon is caused by the coupling of acoustic and hydrodynamic flow inside a cavity which creates strong pressure oscillations in the passenger compartment in the 10 to 50 Hz frequency range. Some effects experienced by vehicles with open sunroofs when buffeting include: dizziness, temporary hearing reduction, discomfort, driver fatigue, and in extreme cases nausea. The importance of reducing interior noise levels inside the car cabin relies primarily in reducing driver fatigue and improving sound transmission from entertainment and communication devices. This Wikibook page aims to theoretically and graphically explain the mechanisms involved in the flow-excited acoustic resonance of Helmholtz resonators. The interaction between fluid motion and acoustic resonance will be explained to provide a thorough explanation of the behavior of self-oscillatory Helmholtz resonator systems. As an application example, a description of the mechanisms involved in sunroof buffeting phenomena will be developed at the end of the page.
Feedback loop analysisEdit
As mentioned before, the self-sustained oscillations of a Helmholtz resonator in many cases is a continuous interaction of hydrodynamic and acoustic mechanisms. In the frequency domain, the flow excitation and the acoustic behavior can be represented as transfer functions. The flow can be decomposed into two volume velocities.
qr: flow associated with acoustic response of cavity
qo: flow associated with excitation
Figure 1 shows the feedback loop of these two volume velocities.
Acoustical characteristics of the resonatorEdit
Lumped parameter modelEdit
The lumped parameter model of a Helmholtz resonator consists of a rigid-walled volume open to the environment through a small opening at one end. The dimensions of the resonator in this model are much less than the acoustic wavelength, in this way allowing us to model the system as a lumped system.
where re is the equivalent radius of the orifice.
Figure 2 shows a sketch of a Helmholtz resonator on the left, the mechanical analog on the middle section, and the electric-circuit analog on the right hand side. As shown in the Helmholtz resonator drawing, the air mass flowing through an inflow of volume velocity includes the mass inside the neck (Mo) and an end-correction mass (Mend). Viscous losses at the edges of the neck length are included as well as the radiation resistance of the tube. The electric-circuit analog shows the resonator modeled as a forced harmonic oscillator. [1] [2][3]
V: cavity volume
{\displaystyle \rho }
: ambient density
c: speed of sound
S: cross-section area of orifice
{\displaystyle M_{a}}
: acoustic mass
{\displaystyle C_{a}}
: acoustic compliance
The equivalent stiffness K is related to the potential energy of the flow compressed inside the cavity. For a rigid wall cavity it is approximately:
{\displaystyle K=\left({\frac {\rho c^{2}}{V}}\right)S^{2}}
The equation that describes the Helmholtz resonator is the following:
{\displaystyle S{\hat {P}}_{e}={\frac {{\hat {q}}_{e}}{j\omega S}}(-\omega ^{2}M+j\omega R+K)}
{\displaystyle {\hat {P}}_{e}}
: excitation pressure
M: total mass (mass inside neck Mo plus end correction, Mend)
R: total resistance (radiation loss plus viscous loss)
From the electrical-circuit we know the following:
{\displaystyle M_{a}={\frac {L\rho }{S}}}
{\displaystyle C_{a}={\frac {\pi V}{\rho c^{2}}}}
{\displaystyle L'=\ L+\ 1.7\ re}
The main cavity resonance parameters are resonance frequency and quality factor which can be estimated using the parameters explained above (assuming free field radiation, no viscous losses and leaks, and negligible wall compliance effects)
{\displaystyle \omega _{r}^{2}={\frac {1}{M_{a}C_{a}}}}
{\displaystyle f_{r}=c2\pi {\sqrt {\frac {S}{L'V}}}}
The sharpness of the resonance peak is measured by the quality factor Q of the Helmholtz resonator as follows:
{\displaystyle Q=2\pi {\sqrt {V\left({\frac {L'}{S}}\right)^{3}}}}
{\displaystyle f_{r}}
: resonance frequency in Hz
{\displaystyle \omega _{r}}
: resonance frequency in radians
L: length of neck
L': corrected length of neck
From the equations above, the following can be deduced:
-The greater the volume of the resonator, the lower the resonance frequencies.
-If the length of the neck is increased, the resonance frequency decreases.
Production of self-sustained oscillationsEdit
The acoustic field interacts with the unstable hydrodynamic flow above the open section of the cavity, where the grazing flow is continuous. The flow in this section separates from the wall at a point where the acoustic and hydrodynamic flows are strongly coupled. [5]
The separation of the boundary layer at the leading edge of the cavity (front part of opening from incoming flow) produces strong vortices in the main stream. As observed in Figure 3, a shear layer crosses the cavity orifice and vortices start to form due to instabilities in the layer at the leading edge.
From Figure 3, L is the length of the inner cavity region, d denotes the diameter or length of the cavity length, D represents the height of the cavity, and
{\displaystyle \delta }
describes the gradient length in the grazing velocity profile (boundary layer thickness).
The velocity in this region is characterized to be unsteady and the perturbations in this region will lead to self-sustained oscillations inside the cavity. Vortices will continually form in the opening region due to the instability of the shear layer at the leading edge of the opening.
Applications to Sunroof BuffetingEdit
How are vortices formed during buffeting?Edit
In order to understand the generation and convection of vortices from the shear layer along the sunroof opening, the animation below has been developed. At a certain range of flow velocities, self-sustained oscillations inside the open cavity (sunroof) will be predominant. During this period of time, vortices are shed at the trailing edge of the opening and continue to be convected along the length of the cavity opening as pressure inside the cabin decreases and increases. Flow visualization experimentation is one method that helps obtain a qualitative understanding of vortex formation and conduction.
The animation below, shows in the middle, a side view of a car cabin with the sunroof open. As the air starts to flow at a certain mean velocity Uo, air mass will enter and leave the cabin as the pressure decreases and increases again. At the right hand side of the animation, a legend shows a range of colors to determine the pressure magnitude inside the car cabin. At the top of the animation, a plot of circulation and acoustic cavity pressure versus time for one period of oscillation is shown. The symbol x moving along the acoustic cavity pressure plot is synchronized with pressure fluctuations inside the car cabin and with the legend on the right. For example, whenever the x symbol is located at the point where t=0 (when the acoustic cavity pressure is minimum) the color of the car cabin will match that of the minimum pressure in the legend (blue).
The perturbations in the shear layer propagate with a velocity of the order of 1/2Uo which is half the mean inflow velocity. [5] After the pressure inside the cavity reaches a minimum (blue color) the air mass position in the neck of the cavity reaches its maximum outward position. At this point, a vortex is shed at the leading edge of the sunroof opening (front part of sunroof in the direction of inflow velocity). As the pressure inside the cavity increases (progressively to red color) and the air mass at the cavity entrance is moved inwards, the vortex is displaced into the neck of the cavity. The maximum downward displacement of the vortex is achieved when the pressure inside the cabin is also maximum and the air mass in the neck of the Helmholtz resonator (sunroof opening) reaches its maximum downward displacement. For the rest of the remaining half cycle, the pressure cavity falls and the air below the neck of the resonator is moved upwards. The vortex continues displacing towards the downstream edge of the sunroof where it is convected upwards and outside the neck of the resonator. At this point the air below the neck reaches its maximum upwards displacement.[4] And the process starts once again.
How to identify buffetingEdit
Flow induced tests performed over a range of flow velocities are helpful to determine the change in sound pressure levels (SPL) inside the car cabin as inflow velocity is increased. The following animation shows typical auto spectra results from a car cabin with the sunroof open at various inflow velocities. At the top right hand corner of the animation, it is possible to see the inflow velocity and resonance frequency corresponding to the plot shown at that instant of time.
It is observed in the animation that the SPL increases gradually with increasing inflow velocity. Initially, the levels are below 80 dB and no major peaks are observed. As velocity is increased, the SPL increases throughout the frequency range until a definite peak is observed around a 100 Hz and 120 dB of amplitude. This is the resonance frequency of the cavity at which buffeting occurs. As it is observed in the animation, as velocity is further increased, the peak decreases and disappears. In this way, sound pressure level plots versus frequency are helpful in determining increased sound pressure levels inside the car cabin to find ways to minimize them. Some of the methods used to minimize the increased SPL levels achieved by buffeting include: notched deflectors, mass injection, and spoilers.
Useful WebsitesEdit
This link: [1] takes you to the website of EXA Corporation, a developer of PowerFlow for Computational Fluid Dynamics (CFD) analysis.
This link: [2] is a small news article about the current use of(CFD) software to model sunroof buffeting.
This link: [3] is a small industry brochure that shows the current use of CFD for sunroof buffeting.
[1] Acoustics: An introduction to its Physical Principles and Applications ; Pierce, Allan D., Acoustical Society of America, 1989.
[2] Prediction and Control of the Interior Pressure Fluctuations in a Flow-excited Helmholtz resonator ; Mongeau, Luc, and Hyungseok Kook., Ray W. Herrick Laboratories, Purdue University, 1997.
[3]Influence of leakage on the flow-induced response of vehicles with open sunroofs ; Mongeau, Luc, and Jin-Seok Hong., Ray W. Herrick Laboratories, Purdue University.
[4]Fluid dynamics of a flow excited resonance, part I: Experiment ; P.A. Nelson, Halliwell and Doak.; 1991.
[5]An Introduction to Acoustics ; Rienstra, S.W., A. Hirschberg., Report IWDE 99-02, Eindhoven University of Technology, 1999.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Flow-induced_oscillations_of_a_Helmholtz_resonator_and_applications&oldid=3232722"
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Will the refractive index table , which is there in NCERT come in the exams ?
why u is always negative in lens formula where u is equal to object which should be in plus
What change do you expect in the focal length of a concave mirror and a convex lens when immersed in water. Justify your answer.
Pls answer que 51 and 52
51. the focal length of the concave mirror in an experimental setup shown below, is
(b) 11.0cm
(c) 11.4cm
52. Four students, Ameeta, Zahira, Ravi and David performed the experiment of determination of focal length of a concave mirror separately. They measured the distance between the screen and the mirror as shown in the following diagram. Which one of there student is likely to get the correct value of local length of the convave mirror?
(a) Ameeta, (b) Zehira (c) Ravi (d) David.
14) With the help of a ray diagram explain the phenomenon of refraction of a ray through glass prism. Mark on the diagram - angle of prism, angle of incidence, angles of refraction, emergent ray and angle of deviation.
Solve this please...
Q. A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under
(ii) Where will the image be formed if he shifts the candle towards the lens
Ankit Gangopadhyay
Sahil Haque
Q 13. Draw ray diagram showing the image formation by a convex lens when an object is placed at twice the focal length of the lens.
answer 4th question
4. Light is incident at an angle of
°
°
, on the same face of a given rectangular slab. If the angles of refraction, at this face are r1 and r2 in the two cases. Obtain the relation in these angles?
Shreyas A K
If f1 = the focal length of a concave lens ; f2 = the focal length of a convex lens ; x = distance between the two lenses ; and F = total focal length (combined nature of the lenses) .Prove that 1/F = 1/f1 + 1/f2 - x/(f1*f2)
If an optically denser medium need not have greater mass density then why speed is reduced in that medium?? What causes the speed to slow down?? Jaldi batao!!
Other than plane mirrors what other types of mirrors and lenses provide lateral inversion
Sample Problem 3. The rear view mirror of a car is plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear view mirror, the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be:
Q). Draw a diagram to trace the path of a ray light incident at angle of 40
°
on a rectangular glass slab. Write the measure of the angle of refraction and angle of emergence.
A student focuses the image of a candle name, placed at about 2 from a convex lens of focal length 10 cm, on a screen. After that he gradually moves the flame towards the lens and each time focuses its image on the screen.
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Aidan knows that the observation deck on the Vancouver Lookout is 130 m above the ground.
Aidan knows that the observation deck on the Vancouver Lookout is 130 m above the ground. He measures the angle between the ground and his line of sig
{77}^{\circ }
Since x is the side adjacent the
{77}^{\circ }
angle and we are given the length of the side opposite the angle, then we can use the tangent ratio to write
\mathrm{tan}=\frac{\text{opposite leg}}{\text{adjacent leg}}
\mathrm{tan}{77}^{\circ }=\frac{130}{x}
x=\frac{130}{\mathrm{tan}{77}^{\circ }}
x\approx 30m
y=5x+3
\mathrm{tan}\left(\mathrm{arcsin}\left(\frac{x}{8}\right)\right)
\mathrm{cos}\left(ar\mathrm{sin}\left(\frac{x}{8}\right)\right)
\left(\frac{1}{2}\right)\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{x}{8}\right)\right)
\mathrm{sin}\left(\mathrm{arctan}\left(\frac{x}{8}\right)\right)
\mathrm{cos}\left(\mathrm{arctan}\left(\frac{x}{8}\right)\right)
\mathrm{sin}Y=\mathrm{cos}\left(Y+20°\right)
SL,2SL,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}SL\sqrt{2}
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Experimental Verification of the Roles of Intrinsic Matrix Viscoelasticity and Tension-Compression Nonlinearity in the Biphasic Response of Cartilage | J. Biomech Eng. | ASME Digital Collection
Chun-Yuh Huang,
Departments of Mechanical Engineering and Biomedical Engineering, Columbia University, 500 West 120th St., New York, NY 10027
Monika Kopacz,
Contributed by the Bioengineering Division for publication in the JOURNAL OF BIOMECHANICAL ENGINEERING. Manuscript received Jun. 2001; revised manuscript received Jun. 2002. Associate Editor: M. S. Sacks.
Huang , C., Soltz , M. A., Kopacz , M., Mow , V. C., and Ateshian, G. A. (February 14, 2003). "Experimental Verification of the Roles of Intrinsic Matrix Viscoelasticity and Tension-Compression Nonlinearity in the Biphasic Response of Cartilage ." ASME. J Biomech Eng. February 2003; 125(1): 84–93. https://doi.org/10.1115/1.1531656
A biphasic-CLE-QLV model proposed in our recent study [2001, J. Biomech. Eng., 123, pp. 410–417] extended the biphasic theory of Mow et al. [1980, J. Biomech. Eng., 102, pp. 73–84] to include both tension-compression nonlinearity and intrinsic viscoelasticity of the cartilage solid matrix by incorporating it with the conewise linear elasticity (CLE) model [1995, J. Elasticity, 37, pp. 1–38] and the quasi-linear viscoelasticity (QLV) model [Biomechanics: Its foundations and objectives, Prentice Hall, Englewood Cliffs, 1972]. This model demonstrates that a simultaneous prediction of compression and tension experiments of articular cartilage, under stress-relaxation and dynamic loading, can be achieved when properly taking into account both flow-dependent and flow-independent viscoelastic effects, as well as tension-compression nonlinearity. The objective of this study is to directly test this biphasic-CLE-QLV model against experimental data from unconfined compression stress-relaxation tests at slow and fast strain rates as well as dynamic loading. Twelve full-thickness cartilage cylindrical plugs were harvested from six bovine glenohumeral joints and multiple confined and unconfined compression stress-relaxation tests were performed on each specimen. The material properties of specimens were determined by curve-fitting the experimental results from the confined and unconfined compression stress relaxation tests. The findings of this study demonstrate that the biphasic-CLE-QLV model is able to describe the strain-rate-dependent mechanical behaviors of articular cartilage in unconfined compression as attested by good agreements between experimental and theoretical curvefits
(r2=0.966±0.032
for testing at slow strain rate;
r2=0.998±0.002
for testing at fast strain rate) and predictions of the dynamic response
r2=0.91±0.06.
This experimental study also provides supporting evidence for the hypothesis that both tension-compression nonlinearity and intrinsic viscoelasticity of the solid matrix of cartilage are necessary for modeling the transient and equilibrium responses of this tissue in tension and compression. Furthermore, the biphasic-CLE-QLV model can produce better predictions of the dynamic modulus of cartilage in unconfined dynamic compression than the biphasic-CLE and biphasic poroviscoelastic models, indicating that intrinsic viscoelasticity and tension-compression nonlinearity of articular cartilage may play important roles in the load-support mechanism of cartilage under physiologic loading.
biological tissues, biomechanics, viscoelasticity, torque, physiological models
Biological tissues, Cartilage, Compression, Relaxation (Physics), Stress, Tension, Viscoelasticity, Equilibrium (Physics), Curve fitting, Materials properties, Transients (Dynamics), Physiology
A Fibril-Network Reinforced Model of Cartilage in Unconfined Compression
DiSilvestro, M. R., Zhu, Q., and Suh, J.-K., 1999, “Biphasic Poroviscoelastic Theory Predicts the Strain Rate Dependent Viscoelastic Behavior of Articular Cartilage,” Proc. 1999 Bioeng. Conf., ASME BED-42, pp. 105–106.
A Finite Deformation Theory for Cartilage and Other Soft Hydrated Connective Tissues—I. Equilibrium Results
The Nonlinear Characteristics of Soft Eels and Hydrated Connective Tissues in Ultrafiltration
Fung, Y.-C. B., 1972, “Stress-Strain History Relation of Soft Tissues in Simple Elongation,” In Biomechanics: Its Foundations and Objectives, Prentice Hall, Englewood Cliffs, 1972.
A Mixture Theory for Charged Hydrated Soft Tissues Containing Multi-electrolytes: Passive Transport and Swelling Behaviors
A Microstructural Model for the Tensile Constitutive and Failure Behavior of Soft Skeletal Connective Tissues
Experimental Verification and Theoretical Prediction of Cartilage Interstitial Fluid Pressurization At An Impermeable Contact Interface in Confined Compression
Confined and Unconfined Stress Relaxation of Cartilage: Appropriateness of a Transversely Isotropic Analysis
An Analysis of Unconfined Compression of Articular Cartilage
Biphasic Indentation of Articular Cartilage—I. Theoretical Analysis
Biphasic Indentation of Articular Cartilage-Part II. A Numerical Algorithm and an Experimental Study
Interspecies Comparisons of in situ Intrinsic Mechanical Properties of Distal Femoral Cartilage
A New Method To Determine the Tensile Properties of Articular Cartilage Using the Indentation Test
Trans. Annu. Meet. — Orthop. Res. Soc.
Biphasic Poroviscoelastic Behavior of Articular Cartilage in Creep Indentation Test
The Role of Flow-independent Viscoelasticity In The Tensile Response of Biphasic Articular Cartilage
Compressive Behavior of Articular Cartilage Is Not Completely Explained by Proteoglycan Osmotic Pressure
Cautionary Note About R2,
Numerical Conversion of Transient to Harmonic Response Functions for Linear Viscoelastic Materials
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Applying the Perfect Square Identity | Brilliant Math & Science Wiki
Ashley Toh, Sathvik Acharya, Calvin Lin, and
The perfect square forms
(a+b)^2
(a-b)^2
come up a lot in algebra. We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often.
(a+b)^2
\begin{aligned} (a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ & = a^2 + ab + ba +b^2 \\ &= a^2 + 2ab +b^2.\ _\square \end{aligned}
(a-b)^2
\begin{aligned} (a-b)^2 &= (a-b)(a-b) \\ &= a(a-b) - b(a-b) \\ & = a^2 - ab - ba +b^2 \\ &= a^2 - 2ab +b^2.\ _\square \end{aligned}
(x+2)^2
\begin{aligned} (x+2)^2 &= (x+2)(x+2) \\ &= x(x+2) + 2(x+2) \\ & = x^2 + 2x + 2x +4 \\ &= x^2 + 4x +4.\ _\square \end{aligned}
r
is a number such that
{ r }^{ 2 }-6r+5=0
{ \left( r-3 \right) }^{ 2 }?
For these problems, you will need to recognize the perfect square form, in order to quickly solve it.
73 ^2 + 2 \times 27 \times 73 + 27 ^2
Observe that with
a = 73, b = 27
a^2 + 2 \times b \times a + b^2 = ( a + b) ^2 = 100 ^2 = 10000
_ \square
n^4 + 4
This isn't immediately a perfect square as yet. If we attempt to complete the square, we see that we need the term
4 n^2
. So, let's add and subtract this to obtain
n^4 + 4n^2 + 4 - 4n^2 = ( n^2 + 2) ^2 - 4 n^ 2.
We now apply the difference of two squares identity, to conclude that
( n^2 + 2) ^2 - 4 n^ 2 = ( n^2 + 2)^2 - (2n) ^2 = ( n^2 + 2 + 2n) ( n^2 + 2 - 2n ).\ _\square
The above factorization is known as the Sophie-Germain identity:
n^4 + 4 = ( n^2 + 2n + 2 ) ( n^2 - 2n + 2 ).
\large{a^2+b^2+c^2+d^2=ab+bc+cd+da}
Do there exist distinct reals
a,b,c,d
which satisfy the above equation?
\begin{aligned} a+ b & = & c + 6 \\ ab - ac & = & bc - 1 \\ \end{aligned}
Given the equations above, what is
a^2 + b^2 + c^2?
Cite as: Applying the Perfect Square Identity. Brilliant.org. Retrieved from https://brilliant.org/wiki/applying-the-perfect-square-identity/
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Canonical map - Wikipedia
See also Natural transformation, a related concept in category theory.
For the canonical map of an algebraic variety into projective space, see Canonical bundle § Canonical maps.
In mathematics, a canonical map, also called a natural map, is a map or morphism between objects that arises naturally from the definition or the construction of the objects. In general, it is the map which preserves the widest amount of structure, and it tends to be unique. In the rare cases where latitude in choices remains, the map is either conventionally agreed upon to be the most useful for further analysis, or sometimes the most elegant map known to date.
A standard form of canonical map involves some function mapping a set
{\displaystyle X}
{\displaystyle X/R}
{\displaystyle X}
{\displaystyle R}
{\displaystyle R}
{\displaystyle X}
.[1] A closely related notion is a structure map or structure morphism; the map or morphism that comes with the given structure on the object. These are also sometimes called canonical maps.
A canonical isomorphism is a canonical map that is also an isomorphism (i.e., invertible). In some contexts, it might be necessary to address an issue of choices of canonical maps or canonical isomorphisms; for a typical example, see prestack.
If N is a normal subgroup of a group G, then there is a canonical surjective group homomorphism from G to the quotient group G/N, that sends an element g to the coset determined by g.
If I is an ideal of a ring R, then there is a canonical surjective ring homomorphism from R onto the quotient ring R/I, that sends an element r to its coset I+r.
If V is a vector space, then there is a canonical map from V to the second dual space of V, that sends a vector v to the linear functional fv defined by fv(λ) = λ(v).
If f: R → S is a homomorphism between commutative rings, then S can be viewed as an algebra over R. The ring homomorphism f is then called the structure map (for the algebra structure). The corresponding map on the prime spectra f*: Spec(S) → Spec(R) is also called the structure map.
If E is a vector bundle over a topological space X, then the projection map from E to X is the structure map.
In topology, a canonical map is a function f mapping a set X → X/R (X modulo R), where R is an equivalence relation on X, that takes each x in X to the equivalence class [x] modulo R.[2]
^ Weisstein, Eric W. "Canonical Map". mathworld.wolfram.com. Retrieved 2019-11-20.
^ Vialar, Thierry (2016-12-07). Handbook of Mathematics. BoD - Books on Demand. p. 274. ISBN 9782955199008.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Canonical_map&oldid=1053093378"
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Least-squares reverse time migration in the presence of density variationsLSRTM with density variations | Geophysics | GeoScienceWorld
Least-squares reverse time migration in the presence of density variations
. E-mail: hongyuyjzh@hotmail.com; liuyuzhu@tongji.edu.cn; dlg@tongji.edu.cn.
Jizhong Yang, Yuzhu Liu, Liangguo Dong; Least-squares reverse time migration in the presence of density variations. Geophysics 2016;; 81 (6): S497–S509. doi: https://doi.org/10.1190/geo2016-0075.1
Least-squares migration (LSM) is commonly regarded as an amplitude-preserving or true amplitude migration algorithm that, compared with conventional migration, can provide migrated images with reduced migration artifacts, balanced amplitudes, and enhanced spatial resolution. Most applications of LSM are based on the constant-density assumption, which is not the case in the real earth. Consequently, the amplitude performance of LSM is not appropriate. To partially remedy this problem, we have developed a least-squares reverse time migration (LSRTM) scheme suitable for density variations in the acoustic approximation. An improved scattering-integral approach is adopted for implementation of LSRTM in the frequency domain. LSRTM images associated with velocity and density perturbations are simultaneously used to generate the simulated data, which better matches the recorded data in amplitudes. Summation of these two images provides a reflectivity model related to impedance perturbation that is in better accordance with the true one, than are the velocity and density images separately. Numerical examples based on a two-layer model and a small part of the Sigsbee2A model verify the effectiveness of our method.
Sigsbee model
Preconditioned acoustic least-squares two-way wave-equation migration with exact adjoint operator
l1/l2
Least-squares migration with primary- and multiple-guided weighting matrices
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m (→Automatising Conversions: bash script for DN > Radiance > ToAR)
The conversion process can be scripted to avoid repeating the same steps for each band separately. '''Note,''' however, in this example script constants, band parameters and acquisition related metadata are hard-coded! Review the code and alter appropriately, i.e. check for the parameters <code>ESD</code>, <code>SEA</code>, <code>PanTDI</code>, <code>K_BAND</code>.
Scripting the conversion process avoids repeating the same steps for each band separately. '''Note,''' however, in the following example, constants, band parameters and acquisition related metadata are hard-coded! Reviewing the code and altering appropriately is required, i.e. checking for the parameters <code>ESD</code>, <code>SEA</code>, <code>PanTDI</code>, <code>K_BAND</code>.
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L(\lambda ,Band)={\frac {K*DN\lambda }{Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
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Tell whether the function represents exponential growth or exponential decay. Then graph the function. y=3e^{2x}
y=3{e}^{2x}
Explain how zeros and end behavior of polynomial functions and their graphs are related to the degree and the factors of the polynomial.
f\left(x\right)={\left(1.5\right)}^{x}
Describe a difference between exponential growth and logistic growth.
y=3{e}^{-0.75x}
In 2005, the population of a district was 28,300, With a continuous annual growth rate of approximately 7% what will the population be in 2020 according to the exponential growth function? Round the answer to the nearest whole number.
In 1995 the population of a certain city was 34,000. Since then, the population has been growing at the rate of 4% per year.
a) Is this an example of linear or exponential growth?
b) Find a function f that computes the population x years after 1995?
c) Find the population in 2002
The value of a home y (in thousands of dollars) can be approximated by the model,
y=192\left(0.96{\right)}^{t}
where t is the number of years since 2010.
1. The model for the value of a home represents exponential _____. (Enter growth or decay in the blank.)
2. The annual percent increase or decrease in the value of the home is ______ %. (Enter the correct number in the blank.)
3. The value of the home will be approximately $161,000 in the year
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Find the function f given that the slope of the tangent line at any point (x, f(
Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point. f '(x)=5(2x − 1)^{4}, (1, 9)
Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point.
f{ }^{\prime }\left(x\right)=5\left(2x-1{\right)}^{4},\left(1,9\right)
Start by finding the slope of f(x). To do this, integrate f'(x). The integral comes out to
\frac{\left(2x-1{\right)}^{5}}{2}+C
. Now solve for C by solving
f\left(1\right)=\frac{\left(2×1-1{\right)}^{5}}{2}+C=9
, which gives us a C value of
\frac{17}{2}
, thus the function
f\left(x\right)=\frac{\left(2x-1{\right)}^{5}}{2}+\frac{17}{2}
f\left(x\right)=-3x+4
f\left(x\right)=-3{x}^{2}+7
f\left(x\right)=\frac{x+1}{x+2}
d\right)f\left(x\right)={x}^{5}+1
f\left(x\right)=\sqrt{1-x}
a=0
\sqrt{0.9}
\sqrt{0.99}
What is the minimum vertical distance between the parabolas
y={x}^{2}+1
y=x-{x}^{2}
f\left(t\right)=\left(\mathrm{sin}t–\mathrm{cos}t\right)2
f\left(x\right)=-{x}^{2}+bx-75
; Maximum value: 25
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From the following groups, select the one which has only secondary metabolites?
Arbrin, cellulose, arginine, tyrosine
Glycine, gums, serine, diterpenes
Carotenoids, phenylalanine, curcumin, rubber
Conclavin-A, morphine, codeine, vinblastin
Secondary metabolites are derivatives of primary metabolites which have no direct function in growth and development of plants. These compounds arc accessory rather than central to the functioning, e.g., arbrin, cellulose, gums, diterpenes, carotenoids, curcumin, rubber etc. Arginine, tyrosine, glycine, serine and phenylalanine are amino acids, which are primary metabolites.
Which of the following is not an invasive species?
Non-native or alien species are often introduced accidently for their economic and other uses. They often become invasive ' and drive away the local species. Therefore, these species are second major cause of extinction.
Lantana camara has replaced many species in forests of Central India.
Parthenium hystero- phorus has pushed out various shrubs and herbs from open places in the plains.
Water hyacinth Eicchornia crassipes is used in India to reduce pollution.
In which of the following sets of organisms, does the external fertilization occur?
Echinodermata and mosses
Hemichordata and ferns
Amphibians and algae
Reptiles and gymnosperms
External fertilization is the fertilization that occurs outside the body of the organism. It requires an external medium such as water for its fertilization.Thus, in most aquatic organisms such as a majority of algae, fishes and amphibians, this type of fertilization occurs.
Stinging capsules (nematocysts) are found in
cactus and Venus flytrap
Stinging cells are used for offence and defence. They have nematocysts, composed of capsule, shaft and thread tube. The thread tube coils around the prey or attaches to it and injects a toxin called hypnotoxin which paralyzes the victim. They are found in cnidarians- sea pen (Pennatula), sea fan (Gorgonia) etc.
Occurrence of triploid (3n) primary endosperm nucleus is a characteristic feature of
Endosperm nourishes embryo in seed plants. In gymnosperms, it represents the female gametophyte and is haploid (n). In angiosperms, it is a special tissue which is formed as a result of fusion of a male gamete with diploid secondary nucleus of the central cell. The fusion product is primary endosperm cell having a triploid (3n) endosperm nucleus.
In a diploid cell, at which stage of cell cycle, the amount of DNA is doubled?
S, G2 and M phase
S- phase or Synthesis phase is the phase where DNA replication takes place. It takes place on the template of the existing DNA and thus, the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C, then it increases to 4C.
India is one of the twelve megadiversity countries with of genetic resources of the world
The total number of species estimated is about 1.74 million. Out of these , species in India is roughly 8.1%. India with about 45000 species of plants and twice as many species of animals is one of the 12 megadiversity countries of the world.
Intercalated discs are characteristic of muscles found in
Cardiac muscle fibres are found in the wall of heart. They have dark intercalated discs at intervals. These are the specialized regions of cell membranes of two adjacent fibres. They permit the wave of muscle contraction to be transmitted from one cardiac fibre to another.
C4 plants have better productivity because
C4 plants absorb more light
C4 plants absorb more CO2
C4 plants does not carry photorespiration
C4 plants have more amount of RuBisCO
Presence of photorespiration is a wasteful and energy consuming process in crop plants which ultimately leads to reduction in final yield of crops. During C3 photosynthesis, upto 50% of the CO2 fixed may have to pass through photorespiratory process, thereby resulting in considerable decrease in photosynthetic productivity. In C3 plants, there is little loss of photosynthetic activity on account of photorespiration which is absent in C4 plants and hence they have better productivity.
Match the source gland with its respective hormone and function and select the correct option.
Anterior pituitary Oxytocin Contraction of uterine muscle
Anterior pituitary Vasopressin Induces reabsorption of water in nephron
Thymus Thymosin Proliferation of T- lymphocytes
\mathrm{\alpha }
- cells of islets of Langerhans Glucagon Uptake of glucose into the cell
Oxytocin is released by posterior pituitary.
Vasopressin decreases the amount of urine by increasing reabsorption of water from DCT and collecting tubules.It stimulates the contraction of walls of blood vessels.
Glucagon stimulates liver to convert stored glycogen into glucose and thus raises the blood sugar level.
Thymus releases thymosin which aids in proliferation of T- lymphocytes.
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Risks Control Model - DeFiner.org
Details of how liquidation works in DeFiner
_liquidator: address of the liquidator.
_borrower: address of the borrower, also the account which will be liquidated
_borrowedToken: address of the borrowed token
_collateralToken: address of the collateral token
By calling this function, the liquidator repays the borrower's loan if the borrower is liquidatable and also ends up buying the borrower's collateral at a 5% discounted price. This also resets the borrower's LTV back to what it was initially. The tokens that the liquidator uses to liquidate the borrower's account should be deposited in DeFiner.
By calling this function, the liquidator repays the borrower's loan if the borrower is liquidatable and also ends up buying the borrower's collateral at a 5% discounted price. This resets the borrower's LTV back to what it was initially. The tokens that the liquidator uses to liquidate the borrower's account should be deposited in DeFiner.
In order to be liquidatable, an account's LTV should be greater than 85%, since there is a high risk that this account will not repay it's debt once the LTV goes above 85%.
If all conditions are met, the liquidate function majorly executed the following operations:
Liquidator deposits collateral tokens equivalent to the payAmount calculated
Withdraws borrowed tokens equivalent to the repayAmount to the liquidator
Withdraws collateral tokens equivalent to the payAmount.
Repays the borrower's loan.
Please refer to the pseudocode given below for more details on how these calculations are performed.
CBB: Current borrow balance = principle + accrued interest
LDR: Liquidation Discount ratio: The discount ratio the liquidator will get when buying other's assets during the liquidation process.
CCV: Current Collateral Value = Collateral price * Collateral Amount
UAAL: User asset at Liquidation: The maximum collateral that can be liquidated or swapped.
BP: Borrow Power of borrower
ILTV: Initial LTV ratio of collateral token: 0.6 currently for most tokens
We have to make sure that:
UAAL= (CBB – BP) *100 / (LDR-ILTV)
Before liquidation:
(Assuming 1 DAI = 1 USDT = $1) USDT Price drops by 65% after user deposits, setting LTV to 92%
(Assuming 1 USDT = 1 DAI = $1) Collateral price drops to 65% after User 1 borrows
UAAL=(60-39)*100/(95-60)=60
Deposits: 100 USDT
Loans: 60 DAI
Collateral price drops to 65%, new collateral value = $65
LTV: 0.92, liquidatable
New borrow power = initial BP * %age price drop = 60 * 0.65 = 39
Deposits 200 DAI
It calls liquidate(user1, DAIAdress, USDTAddress)
The maximum amount of DAI that user2 can transfer to user1 is 200 since it doesn't have any borrows.
Since DAI's price is 1, and 60 < 200, so user2 is able to pay the maximum value, which is 60 DAI. This is called full liquidation.
After Liquidation: User 1:
Deposits: 5 USDT
Borrows: 0
User is not liquidatable
Deposits: 95 USDT
borrow power = initial BP * %age price drop = 60 * 0.65 = 39
Deposits 50 DAI
Here, UAAL is computed the same way as the previous example.
UAAL=(60-39)*100/(95-60)=60
But here user2 only has 50DAI, which worth $50 and 50 < 60 so user2 can't be swapped to the maximum value UAAL. That way, user2 only pays 50 DAI and user1 will pay 50 / 0.95 = 52.6 USDC.
Deposits: 100 - 50/0.95 = 47.4 USDT
Borrows: 60-50 = 10DAI
LTV: 10 / 47.4 = 0.21, not liquidatable
Deposits: 50/0.95 = 52.6 USDC
Notice here although user2 doesn't fully liquidate user1, user1 is not liquidatable after liquidation. This is because there is a gap between the initial borrow LTV and the LTV to become liquidatable.
require(isAccountLiquidatable(_borrower);
if (liquidator has borrows){
require(Liquidator's borrow value < Liquidator's borrow power;
uint tokenBalLiquidator = get deposit balance of Liquidator;
uint tokenBalBorrowedUser = get borrow balance of Borrower;
uint borrowedTokenAmountForLiquidation = tokenBalLiquidator.min(tokenBalBorrowedUser)
uint borrowerCollateralVal = getDepositBalanceCurrent(_collateralToken, _borrower);
uint collateralLTV = 60%;
uint totalBorrowwValBorwer = getBorrowETH(_borrower);
uint borrowPowerBorrower = getBorrowPower(borrower);
uint liquidationDiscountRatio = 95%;
uint limiRepaymentVal = (totalBorrowwValBorwer - borrowPowerBorrower) / (liquidationDiscountRatio - collateralLTV);
uint collateralTokenValueForLiquidation = limiRepaymentVal.min(tokenBalLiquidator);
uint liquidationVal = collateralTokenValueForLiquidation.min(borrowedTokenAmountForLiquidation * borrowedTokenPrice / liquidationDiscountRatio);
uint repaymentAmount = (liquidationVal * borrowTokenDivisor) / borrowTokenPrice;
uint payAmount = (repaymentAmount * liquidateTokenDivisor * borrowTokenPrice) / (borrowTokenDivisor * liquidationDiscountRatio * liquidateTokenPrice);
deposit(_liquidator, _collateralToken, payAmount);
_withdrawLiquidate(_liquidator, _borrowedToken, repaymentAmount);
_withdrawLiquidate(_borrower, _collateralToken, payAmount);
repay(_borrower, _borrowedToken, repaymentAmount);
return (repaymentAmount, payAmount);
address _liquidator,
address _borrower,
address _borrowedToken,
address _collateralToken
) external onlyAuthorized returns (uint256, uint256) {
initCollateralFlag(_liquidator);
initCollateralFlag(_borrower);
require(isAccountLiquidatable(_borrower), "borrower is not liquidatable");
// It is required that the liquidator doesn't exceed it's borrow power.
// if liquidator has any borrows, then only check for borrowPower condition
Account storage liquidateAcc = accounts[_liquidator];
if (liquidateAcc.borrowBitmap > 0) {
require(getBorrowETH(_liquidator) < getBorrowPower(_liquidator), "No extra funds used for liquidation");
LiquidationVars memory vars;
ITokenRegistry tokenRegistry = globalConfig.tokenInfoRegistry();
// _borrowedToken balance of the liquidator (deposit balance)
vars.targetTokenBalance = getDepositBalanceCurrent(_borrowedToken, _liquidator);
require(vars.targetTokenBalance > 0, "amount must be > 0");
// _borrowedToken balance of the borrower (borrow balance)
vars.targetTokenBalanceBorrowed = getBorrowBalanceCurrent(_borrowedToken, _borrower);
require(vars.targetTokenBalanceBorrowed > 0, "borrower not own any debt token");
// _borrowedToken available for liquidation
uint256 borrowedTokenAmountForLiquidation = vars.targetTokenBalance.min(vars.targetTokenBalanceBorrowed);
// _collateralToken balance of the borrower (deposit balance)
vars.liquidateTokenBalance = getDepositBalanceCurrent(_collateralToken, _borrower);
uint256 targetTokenDivisor;
(, targetTokenDivisor, vars.targetTokenPrice, vars.borrowTokenLTV) = tokenRegistry.getTokenInfoFromAddress(
_borrowedToken
uint256 liquidateTokendivisor;
uint256 collateralLTV;
(, liquidateTokendivisor, vars.liquidateTokenPrice, collateralLTV) = tokenRegistry.getTokenInfoFromAddress(
_collateralToken
// _collateralToken to purchase so that borrower's balance matches its borrow power
vars.totalBorrow = getBorrowETH(_borrower);
vars.borrowPower = getBorrowPower(_borrower);
vars.liquidationDiscountRatio = globalConfig.liquidationDiscountRatio();
vars.limitRepaymentValue = vars.totalBorrow.sub(vars.borrowPower).mul(100).div(
vars.liquidationDiscountRatio.sub(collateralLTV)
uint256 collateralTokenValueForLiquidation = vars.limitRepaymentValue.min(
vars.liquidateTokenBalance.mul(vars.liquidateTokenPrice).div(liquidateTokendivisor)
uint256 liquidationValue = collateralTokenValueForLiquidation.min(
borrowedTokenAmountForLiquidation.mul(vars.targetTokenPrice).mul(100).div(targetTokenDivisor).div(
vars.liquidationDiscountRatio
vars.repayAmount = liquidationValue.mul(vars.liquidationDiscountRatio).mul(targetTokenDivisor).div(100).div(
vars.targetTokenPrice
vars.payAmount = vars.repayAmount.mul(liquidateTokendivisor).mul(100).mul(vars.targetTokenPrice);
vars.payAmount = vars.payAmount.div(targetTokenDivisor).div(vars.liquidationDiscountRatio).div(
vars.liquidateTokenPrice
deposit(_liquidator, _collateralToken, vars.payAmount);
_withdrawLiquidate(_liquidator, _borrowedToken, vars.repayAmount);
_withdrawLiquidate(_borrower, _collateralToken, vars.payAmount);
repay(_borrower, _borrowedToken, vars.repayAmount);
return (vars.repayAmount, vars.payAmount);
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Arenes | Brilliant Math & Science Wiki
Kïñshük Sïñgh and Jimin Khim contributed
Arenes are aromatic hydrocarbons. The term "aromatic" originally referred to their pleasant smells, but now implies a particular sort of delocalized bonding.
The simplest of them is benzene itself,
\ce{C6H6}
The next simplest is methylbenzene (oldname: toluene) which has one of the hydrogen atoms attached to the ring replaced by a methyl group:
\ce{C6H5CH3}
Aryl Halides - Preparation
Aryl Halides - Chemical Properties
Aryl Halides - Mechanisms
From dizoniam salts:
2. Halogenation of aromatic systems:
Reaction occurs instantly with tertiary alkyl or benzyl halides, and within five minutes or so with primary and secondary halides, but halobenzenes or vinyl halides react very slowly
\big(
heated for days with
alc.\ce{AgNO3}\big).
Aryl halides undergo nucleophilic substitution reaction only in extreme conditions, when conditions are feasible.
Aryl halides can't be used in Friedel–Crafts alkylation reactions.
The presence of electron withdrawing groups like
\ce{NO2, CF3},
etc. at ortho/para position to the halogen atom makes the aryl halides more susceptible to nucleophilic attack.
\text{S}_{\text{N}}
Ar Mechanism
Nucleophilic aromatic substitution (Addition-Elimination)
Rate determining step (R.D.S.): first step, when nucleophile
(\ce{O{ H }^{ - }})
attacks aryl halide
Nucleophile attacks the carbon having halogen because only that carbon has
{ \partial }^{ + }
charge.
CINE substitution is followed when nucleophile is present on the carbon other than the one in which the leaving group is present.
Presence of electron withdrawing group at ortho/para to the halide group enhances reactivity.
Aryl halides can be distinguished from alkyl halides by decolorisation of bromine water:
Cite as: Arenes. Brilliant.org. Retrieved from https://brilliant.org/wiki/arenes/
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Home : Support : Online Help : Programming : Bits : Split
split number into bit multiples
Split(number, options)
Split(number, sbits, options)
(optional) positive integer number of bits to split on
The Split command breaks the input number into a list of sbits bit length (default is 1) numbers, least significant bits first.
Put simply, the Split command converts a number into a list of the base 2^sbits digits.
The most common usage of this command would be for conversion of a number into its base-2 digits, but it could also be used, for example, to convert a number to octal or hex (see convert/octal or convert/hex).
The String command is closely related, but the output for that command is a string.
The option bits=n tells Split how many bits to consider in the split. Any bits in a higher position are simply ignored.
The reverse operation, converting the digits to a number, is accomplished using the Join command.
\mathrm{with}\left(\mathrm{Bits}\right):
\mathrm{Split}\left(255\right)
[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]
\mathrm{Split}\left(255,2\right)
[\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]
\mathrm{Split}\left(255,3\right)
[\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]
Convert first 6 bits to octal
\mathrm{Split}\left(255,3,\mathrm{bits}=6\right)
[\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}]
Convert first 32 bits to hex
\mathrm{Split}\left(255,4,\mathrm{bits}=32\right)
[\textcolor[rgb]{0,0,1}{15}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{15}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}]
To bits, then back to a number
\mathrm{dig}≔\mathrm{Split}\left(395718860534\right)
\textcolor[rgb]{0,0,1}{\mathrm{dig}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]
\mathrm{num}≔\mathrm{Join}\left(\mathrm{dig}\right)
\textcolor[rgb]{0,0,1}{\mathrm{num}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{395718860534}
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A random sample of 88 U.S. 11th- and 12th-graders was selected. The two-way tabl
A random sample of 88 U.S. 11th- and 12th-graders was selected. The two-way table summarizes the gender of the students and their response to the question "Do you have allergies?"
A random sample of 88 U.S. 11th- and 12th-graders was selected. The two-way table summarizes the gender of the students and their response to the question "Do you have allergies?" Suppose we choose a student from this group at random.
\begin{array}{cccc}& \text{ Female }& \text{ Male }& \text{ Total }\\ \text{ Yes }& 19& 15& 34\\ \text{ No }& 24& 30& 54\\ \text{ Total }& 43& 45& 88\end{array}
What is the probability that the student is female or has allergies?
\left(a\right)\frac{19}{88}
\frac{39}{88}
\frac{58}{88}
\frac{77}{88}
Aniqa O'Neill
P\left({A}^{c}\right)=P\left(\text{not}\text{ }A\right)=1-P\left(A\right)
General addition rule for any two events:
P\left(A\text{ }or\text{ }B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{ and }B\right)
\begin{array}{cccc}& \text{ Female }& \text{ Male }& \text{ Total }\\ \text{ Yes }& 19& 15& 34\\ \text{ No }& 24& 30& 54\\ \text{ Total }& 43& 45& 88\end{array}
We note that the table contains information about 88 peoples (given in the bottom right corner of the table).
Moreover, 34 of the 88 people have allergies, becatise 34 is mentioned in the row Yes” and in the column *Total” of the table.
P\left(Allergies\right)=\frac{\mathrm{#}\text{ of favorable outcomes}}{\mathrm{#}\text{ of possible outcomes}}=\frac{34}{88}
We note that 43 of the 88 people are female, becatise 43 is mentioned in the row * Total” and in the column *Female” of the given table.
P\left(Female\right)=\frac{\mathrm{#}\text{ of favorable outcomes}}{\mathrm{#}\text{ of possible outcomes}}=\frac{43}{88}
We note that 19 of the 88 people are females and have allergies, because 19 is mentioned in the row ” Yes” and in the column *Female” of the given table.
P\left(\text{ Allergies and Female}\right)=\frac{\mathrm{#}\text{ of favorable outcomes}}{\mathrm{#}\text{ of possible outcomes}}=\frac{19}{88}
Use the general addition rule:
P\left(\text{ Allergies and Female}\right)=P\left(\text{Allergies}\right)+P\left(\text{Female}\right)-P\left(\text{ Allergies and Female}\right)
=\frac{34}{88}+\frac{43}{88}-\frac{19}{88}
=\frac{34+43-19}{88}
=\frac{58}{88}
\left(c\right)\frac{58}{88}
The two-way table summarizes data on whether students at a certain high school eat regularly in the school cafeteria by grade level.
\text{Grade}\text{ }\text{Eat in cafeteria}\begin{array}{lrrrrr}& 9\mathrm{th}& 10\mathrm{th}& 11\mathrm{th}& 12\mathrm{th}& \text{ Total }\\ \text{ Yes }& 130& 175& 122& 68& 495\\ \text{ No }& 18& 34& 88& 170& 310\\ \text{ Total }& 148& 209& 210& 238& 805\end{array}
If you choose a student at random who eats regularly in the cafeteria, what is the probability that the student is a 10th-grader?
A group of 125 truck owners were asked what brand of truck they owned and whether or not the truck has four-wheel drive. The results are summarized in the two-way table below. Suppose we randomly select one of these truck owners.
\begin{array}{ccc}& \text{ Four-wheel drive}& \text{ No four-wheel drive }\\ \text{ Ford }& 28& 17\\ \text{ Chevy }& 32& 18\\ \text{ Dodge }& 20& 10\end{array}
What is the probability that the person owns a Dodge or has four-wheel drive?
\left(a\right)\frac{20}{80}
\left(b\right)\frac{20}{125}
\left(c\right)\frac{80}{125}
\left(d\right)\frac{90}{125}
\left(e\right)\frac{110}{125}
A survey of 4826 randomly selected young adults (aged 19 to 25) asked, "What do you think are the chances you will have much more than a middle-class income at age 30?" The two-way table summarizes the responses.
\begin{array}{cccc}& \text{ Female }& \text{ Male }& \text{ Total }\\ \text{ Almost no chance }& 96& 98& 194\\ \text{ Some chance but }\text{ }\text{ probably not }& 426& 286& 712\\ \text{ A 50-50 chance }& 696& 720& 1416\\ \text{ A good chance }& 663& 758& 1421\\ \text{ Almost certain }& 486& 597& 1083\\ \text{ Total }& 2367& 2459& 4826\end{array}
Choose a survey respondent at random. Define events G: a good chance, M: male, and N: almost no chance. Find
P\left(C\mid M\right)
. Interpret this value in context.
A study in Sweden looked at former elite soccer players, people who had played soccer but not at the elite level, and people of the same age who did not play soccer. Here is a two-way table that classifies these individuals by whether or not they had arthritis of the hip or knee by their mid-fifties:
\text{Soccer level}\phantom{\rule{0ex}{0ex}}\begin{array}{llccc}& & \text{ Ellte }& \text{ Non-elite }& \text{ Did not play }\\ \text{ Whether person }& \text{ Yes }& 10& 9& 24\\ \text{ developed arthritis }& \text{ No }& 61& 206& 548\end{array}
Researchers suspected that the more serious soccer players were more likely to develop arthritis later in life. Do the data confirm this suspicion? Calculate appropriate percentages to support your answer.
In 1912 the Titanic struck an iceberg and sank on its first voyage. Some passengers got off the ship in lifeboats, but many died. The following two-way table gives information about adult passengers who survived and who died, by class of travel.
\begin{array}{lc}& \text{Class}\\ \text{Survived}& \begin{array}{cccc}& \text{ First }& \text{ Second }& \text{ Third }\\ \text{ Yes }& 197& 94& 151\\ \text{ No }& 122& 167& 476\end{array}\end{array}
Suppose we randomly select one of the adult passengers who rode on the Titanic. Define event D as getting a person who died and event F as getting a passenger in first class. Find P (not a passenger in first class and survived).
The General Social Survey (GSS) asked a random sample of adults their opinion about whether astrology is very scientific, sort of scientific, or not at all scientific. Here is a two-way table of counts for people in the sample who had three levels of higher education:
\begin{array}{lcccc}& \text{ Associate's }& \text{ Bachelor's }& \text{ Master's }& \text{ Total }\\ \begin{array}{c}\text{ Not al all }\\ \text{ scientific }\end{array}& 169& 256& 114& 539\\ \begin{array}{c}\text{ Very or sort }\\ \text{ of scientific }\end{array}& 65& 65& 18& 148\\ \text{ Total }& 234& 321& 132& 687\end{array}
State appropriate hypotheses for performing a chi-square test for independence in this setting.
1. Who seems to have more variability in their shoe sizes, men or women?
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of
n-1
rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
n-1
provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
\begin{array}{|cc|}\hline \text{Shoe Size (in cm)}& \text{Gender (M of F)}\\ 25.7& M\\ 25.4& F\\ 23.8& F\\ 25.4& F\\ 26.7& M\\ 23.8& F\\ 25.4& F\\ 25.4& F\\ 25.7& M\\ 25.7& F\\ 23.5& F\\ 23.1& F\\ 26& M\\ 23.5& F\\ 26.7& F\\ 26& M\\ 23.1& F\\ 25.1& F\\ 27& M\\ 25.4& F\\ 23.5& F\\ 23.8& F\\ 27& M\\ 25.7& F\\ \hline\end{array}
\begin{array}{|cc|}\hline \text{Shoe Size (in cm)}& \text{Gender (M of F)}\\ 27.6& M\\ 26.9& F\\ 26& F\\ 28.4& M\\ 23.5& F\\ 27& F\\ 25.1& F\\ 28.4& M\\ 23.1& F\\ 23.8& F\\ 26& F\\ 25.4& M\\ 23.8& F\\ 24.8& M\\ 25.1& F\\ 24.8& F\\ 26& M\\ 25.4& F\\ 26& M\\ 27& M\\ 25.7& F\\ 27& M\\ 23.5& F\\ 29& F\\ \hline\end{array}
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A credit card contains 16 digits between 0 and 9. However, only 100 million numb
A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number?
{10}^{16}
strings of 16 digits between 0 and 9 (for each of the 16 digits we have 10 choices). Also, 100 million =
100\cdot {10}^{6}={10}^{8}
. So, the probability that the random entered number is valid is
number of valid numbers/all strings of 16 digits=
\frac{{10}^{8}}{{10}^{16}}=\frac{1}{{10}^{8}}
which is almost zero.
\frac{{z}^{2}-1}{{z}^{2}+z+1}
{i}^{13}
Write each complex number in standard form.
2\left(\mathrm{cos}{30}^{\circ }+i\mathrm{sin}{30}^{\circ }\right)
\frac{5-i}{3+3i}
Find the product of the complex numbers.Leave answers in polar form.
{z}_{1}=4\left({\mathrm{cos}15}^{\circ }+i{\mathrm{sin}15}^{\circ }\right)
{z}_{2}=7\left({\mathrm{cos}25}^{\circ }+i{\mathrm{sin}25}^{\circ }\right)
\frac{1}{2}\left(\mathrm{cos}\left(\frac{\pi }{3}\right)+i\mathrm{sin}\left(\frac{\pi }{3}\right)\right)÷3\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}g\left(\frac{\pi }{6}\right)\right)
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ListAttachments - MapleSim Help
Home : Support : Online Help : MapleSim : MapleSim Application Programming Interface : API Commands : ListAttachments
list the attachments of the linked MapleSim model
A := MapleSim:-LinkModel(options);
A:-ListAttachments(options)
list : list of the attachments in the linked model
showcategories = true or false
True means each element in the returned list is a sublist containing the name of the attachment and its category. False means each element is the name of the attachment. The default is false.
A:-ListAttachments returns the list of names of the existing attachments in the linked model A.
Link to a MapleSim model (located in .msim file):
A≔\mathrm{MapleSim}:-\mathrm{LinkModel}\left('\mathrm{filename}'=\mathrm{cat}\left(\mathrm{kernelopts}\left('\mathrm{toolboxdir}'=\mathrm{MapleSim}\right),"/data/examples/RLCcircuit.msim"\right)\right):
Set the attachment to the linked MapleSim model.
\mathrm{data}≔"this will be the ship"
\textcolor[rgb]{0,0,1}{\mathrm{data}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{"this will be the ship"}
A:-\mathrm{SetAttachment}\left("data.txt",\mathrm{data},'\mathrm{category}'="other"\right)
List the attachments of the linked MapleSim model.
A:-\mathrm{ListAttachments}\left(\right)
[\textcolor[rgb]{0,0,1}{"data.txt"}]
List the attachments with the categories.
A:-\mathrm{ListAttachments}\left(\mathrm{showcategories}\right)
[[\textcolor[rgb]{0,0,1}{"data.txt"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"other"}]]
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Euler simulation of stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, or SDEMRD models - MATLAB simByEuler - MathWorks Australia
simByEuler
Simulate Equity Markets Using simByEuler
Euler Simulation for a CIR Object
Quasi-Monte Carlo Simulation with simByEuler Using a CIR Model
Quasi-Monte Carlo Simulation with simByEuler Using a Heston Model
Quasi-Monte Carlo Simulation with simByEuler Using GBM Model
Euler simulation of stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, or SDEMRD models
[Paths,Times,Z] = simByEuler(MDL,NPeriods)
[Paths,Times,Z] = simByEuler(___,Name,Value)
[Paths,Times,Z] = simByEuler(MDL,NPeriods) simulates NTrials sample paths of NVars correlated state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods. simByEuler uses the Euler approach to approximate continuous-time stochastic processes.
[Paths,Times,Z] = simByEuler(___,Name,Value) specifies options using one or more name-value pair arguments in addition to the input arguments in the previous syntax.
Load the Data and Specify the SDE Model
Simulate a Single Path Over a Year
Simulate 10 trials and examine the SDE model
Plot the paths
d{X}_{t}=S\left(t\right)\left[L\left(t\right)-{X}_{t}\right]dt+D\left(t,{X}_{t}^{\frac{1}{2}}\right)V\left(t\right)dW
D
d{X}_{t}=0.2\left(0.1-{X}_{t}\right)dt+0.05{X}_{t}^{\frac{1}{2}}dW
CIR = cir(0.2, 0.1, 0.05) % (Speed, Level, Sigma)
Simulate a single path over a year using simByEuler.
[Paths,Times] = simByEuler(CIR,nPeriods,'Method','higham-mao','DeltaTime', dt)
Paths = 250×1
Times = 250×1
d{X}_{t}=S\left(t\right)\left[L\left(t\right)-{X}_{t}\right]dt+D\left(t,{X}_{t}^{\frac{1}{2}}\right)V\left(t\right)dW
D
d{X}_{t}=0.2\left(0.1-{X}_{t}\right)dt+0.05{X}_{t}^{\frac{1}{2}}dW
cir_obj = cir(0.2, 0.1, 0.05) % (Speed, Level, Sigma)
cir_obj =
Define the quasi-Monte Carlo simulation using the optional name-value arguments for 'MonteCarloMethod' and 'QuasiSequence'.
[paths,time,z] = simByEuler(cir_obj,10,'ntrials',4096,'method','basic','montecarlomethod','quasi','quasisequence','sobol');
d{X}_{1t}=B\left(t\right){X}_{1t}dt+\sqrt{{X}_{2t}}{X}_{1t}d{W}_{1t}
d{X}_{2t}=S\left(t\right)\left[L\left(t\right)-{X}_{2t}\right]dt+V\left(t\right)\sqrt{{X}_{2t}}d{W}_{2t}
Create a heston object.
heston_obj = heston (0.1, 0.2, 0.1, 0.05) % (Return, Speed, Level, Volatility)
heston_obj =
[paths,time,z] = simByEuler(heston_obj,10,'ntrials',4096,'montecarlomethod','quasi','quasisequence','sobol');
Create a univariate gbm object to represent the model:
d{X}_{t}=0.25{X}_{t}dt+0.3{X}_{t}d{W}_{t}
gbm_obj = gbm(0.25, 0.3) % (B = Return, Sigma)
gbm_obj =
gbm objects display the parameter B as the more familiar Return.
[paths,time,z] = simByEuler(gbm_obj,10,'ntrials',4096,'montecarlomethod','quasi','quasisequence','sobol');
Stochastic differential equation model, specified as an sde, bm, gbm, cev, cir, hwv, heston, sdeddo, sdeld, or sdemrd object.
Example: [Paths,Times,Z] = simByEuler(SDE,NPeriods,'DeltaTime',dt)
Method — Method to handle negative values
'basic' (default) | character vector with values'basic', 'absorption', 'reflection', 'partial-truncation', 'full-truncation', or 'higham-mao' | string with values "basic", "absorption", "reflection", "partial-truncation", "full-truncation", or "higham-mao"
Method to handle negative values, specified as the comma-separated pair consisting of 'Method' and a character vector or string with a supported value.
The Method argument is only supported when using a CIR object. For more information on creating a CIR object, see cir.
The simByEuler function partitions each time increment dt into NSteps subintervals of length dt/NSteps, and refines the simulation by evaluating the simulated state vector at NSteps − 1 intermediate points. Although simByEuler does not report the output state vector at these intermediate points, the refinement improves accuracy by allowing the simulation to more closely approximate the underlying continuous-time process.
Antithetic — Flag to indicate whether simByEuler uses antithetic sampling to generate the Gaussian random variates
Flag indicates whether simByEuler uses antithetic sampling to generate the Gaussian random variates that drive the Brownian motion vector (Wiener processes). This argument is specified as the comma-separated pair consisting of 'Antithetic' and a scalar logical flag with a value of True or False.
When you specify True, simByEuler performs sampling such that all primary and antithetic paths are simulated and stored in successive matching pairs:
Direct specification of the dependent random noise process used to generate the Brownian motion vector (Wiener process) that drives the simulation. This argument is specified as the comma-separated pair consisting of 'Z' and a function or as an (NPeriods ⨉ NSteps)-by-NBrowns-by-NTrials three-dimensional array of dependent random variates.
If StorePaths is True (the default value) or is unspecified, simByEuler returns Paths as a three-dimensional time series array.
If StorePaths is False (logical 0), simByEuler returns the Paths output array as an empty matrix.
If you specify an input noise process (see Z), simByEuler ignores the value of MonteCarloMethod.
simByEuler makes no adjustments and performs no processing (default) | function | cell array of functions
{X}_{t}=P\left(t,{X}_{t}\right)
The simByEuler function runs processing functions at each interpolation time. They must accept the current interpolation time t, and the current state vector Xt, and return a state vector that may be an adjustment to the input state.
If you specify more than one processing function, simByEuler invokes the functions in the order in which they appear in the cell array. You can use this argument to specify boundary conditions, prevent negative prices, accumulate statistics, plot graphs, and more.
For a given trial, each row of Paths is the transpose of the state vector Xt at time t. When the input flag StorePaths = False, simByEuler returns Paths as an empty matrix.
Dependent random variates used to generate the Brownian motion vector (Wiener processes) that drive the simulation, returned as a (NPeriods ⨉ NSteps)-by-NBrowns-by-NTrials three-dimensional time series array.
This function simulates any vector-valued SDE of the form
d{X}_{t}=F\left(t,{X}_{t}\right)dt+G\left(t,{X}_{t}\right)d{W}_{t}
simByEuler simulates NTrials sample paths of NVars correlated state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods, using the Euler approach to approximate continuous-time stochastic processes.
This simulation engine provides a discrete-time approximation of the underlying generalized continuous-time process. The simulation is derived directly from the stochastic differential equation of motion. Thus, the discrete-time process approaches the true continuous-time process only as DeltaTime approaches zero.
The input argument Z allows you to directly specify the noise-generation process. This process takes precedence over the Correlation parameter of the sde object and the value of the Antithetic input flag. If you do not specify a value for Z, simByEuler generates correlated Gaussian variates, with or without antithetic sampling as requested.
The end-of-period Processes argument allows you to terminate a given trial early. At the end of each time step, simByEuler tests the state vector Xt for an all-NaN condition. Thus, to signal an early termination of a given trial, all elements of the state vector Xt must be NaN. This test enables a user-defined Processes function to signal early termination of a trial, and offers significant performance benefits in some situations (for example, pricing down-and-out barrier options).
[1] Deelstra, G. and F. Delbaen. “Convergence of Discretized Stochastic (Interest Rate) Processes with Stochastic Drift Term.” Applied Stochastic Models and Data Analysis., 1998, vol. 14, no. 1, pp. 77–84.
[2] Higham, Desmond, and Xuerong Mao. “Convergence of Monte Carlo Simulations Involving the Mean-Reverting Square Root Process.” The Journal of Computational Finance, vol. 8, no. 3, 2005, pp. 35–61.
[3] Lord, Roger, et al. “A Comparison of Biased Simulation Schemes for Stochastic Volatility Models.” Quantitative Finance, vol. 10, no. 2, Feb. 2010, pp. 177–94
simByTransition | simBySolution | | simulate | sde | bm | gbm | sdeddo | sdeld | cev | cir | heston | hwv | sdemrd
Implementing Multidimensional Equity Market Models, Implementation 5: Using the simByEuler Method
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Transience and Recurrence of Markov Chains | Brilliant Math & Science Wiki
A Markov chain with one transient state and two recurrent states A stochastic process contains states that may be either transient or recurrent; transience and recurrence describe the likelihood of a process beginning in some state of returning to that particular state. There is some possibility (a nonzero probability) that a process beginning in a transient state will never return to that state. There is a guarantee that a process beginning in a recurrent state will return to that state.
Transience and recurrence issues are central to the study of Markov chains and help describe the Markov chain's overall structure. The presence of many transient states may suggest that the Markov chain is absorbing, and a strong form of recurrence is necessary in an ergodic Markov chain.
In a Markov chain, there is probability
1
of eventually (after some number of steps) returning to state
x
. Must the expected number of returns to state
x
be infinite?
1
x
. Must the expected number of steps to return to state
x
be finite?
Positive vs Null Recurrent
Intuitively, transience attempts to capture how "connected" a state is to the entirety of the Markov chain. If there is a possibility of leaving the state and never returning, then the state is not very connected at all, so it is known as transient. In addition to the traditional definition regarding probability of return, there are other equivalent definitions of transient states.
\{X_0, \, X_1, \, \dots\}
be a Markov chain with state space
S
. The following conditions are equivalent.
i \in S
is transient.
T_i = \text{min}\{n \ge 1 \mid X_n = i\}
\mathbb{P}(T_i < \infty \mid X_0 = i) < 1.
The following sum converges:
\displaystyle\sum_{n = 1}^\infty \mathbb{P}(X_n = i \mid X_0 = i) < \infty.
\mathbb{P}(X_n = i \text{ for infinitely many } n \mid X_0 = i) = 0.
Similarly, a classification exists for recurrent states as well.
\{X_0, \, X_1, \, \dots\}
S
i \in S
is recurrent.
T_i = \text{min}\{n \ge 1 \mid X_n = i\}
\mathbb{P}(T_i < \infty \mid X_0 = i) =1.
The following sum diverges:
\displaystyle\sum_{n = 1}^\infty \mathbb{P}(X_n = i \mid X_0 = i) = \infty.
\mathbb{P}(X_n = i \text{ for infinitely many } n \mid X_0 = i) = 1.
An aperiodic Markov chain with positive recurrent states
While a recurrent state has the property that the Markov chain is expected to return to the state an infinite number of times, the Markov chain is not necessarily expected to return even once within a finite number of steps. This is not good, as a lot of the intuition for recurrence comes from assuming that it will. In order to fix that, there is a further classification of recurrence states.
T_i = \text{min}\{n \ge 1 \mid X_n = i\}
be the time of the first return to
i
. Then, the expected value
\mathbb{E}(T_i \mid X_0 = i)
is the property discussed above.
is known as positive recurrent if
\mathbb{E}(T_i \mid X_0 = i) < \infty
is known as null recurrent if
\mathbb{E}(T_i \mid X_0 = i) = \infty
If a state is periodic, it is positive recurrent. However, as shown to the right, there exist plenty of aperiodic Markov chains with only positive recurrent states.
The following is a depiction of the Markov chain known as a random walk with reflection at zero.
p + q = 1
p < \tfrac{1}{2}
, all states in the Markov chain are positive recurrent. With
p = \tfrac{1}{2}
, all states in the Markov chain are null recurrent. With
p > \tfrac{1}{2}
, all states in the Markov chain are transient.
Cite as: Transience and Recurrence of Markov Chains. Brilliant.org. Retrieved from https://brilliant.org/wiki/transience-and-recurrence/
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Calculate relative atmospheric ratios - Simulink - MathWorks í•œêµ
Ïo
sqrt(θ)
Square root of theta
Calculate relative atmospheric ratios
The Relative Ratio block computes the relative atmospheric ratios, including the relative temperature ratio (θ),
\sqrt{\mathrm{θ}}
, relative pressure ratio (δ), and relative density ratio (σ).
θ represents the ratio of the air stream temperature at a chosen reference station relative to sea level standard atmospheric conditions:
\mathrm{θ}=\frac{T}{{T}_{0}}.
δ represents the ratio of the air stream pressure at a chosen reference station relative to sea level standard atmospheric conditions:
\mathrm{δ}=\frac{P}{{P}_{0}}.
σ represents the ratio of the air stream density at a chosen reference station relative to sea level standard atmospheric conditions:
\mathrm{Ï}=\frac{\mathrm{Ï}}{{\mathrm{Ï}}_{0}}.
The Relative Ratio block icon displays the input units selected from the Units parameter.
For cases in which total temperature, total pressure, or total density ratio is desired (Mach number is nonzero), the total temperature, total pressure, and total densities are calculated assuming perfect gas (with constant molecular weight, constant pressure specific heat, and constant specific heat ratio) and dry air.
Mach — Mach number
γ — Ratio
Ratio between the specific heat at constant pressure (Cp) and the specific heat at constant volume (Cv), specified as a scalar. For example, (γ = Cp/Cv).
To — Static temperature
Static temperature, specified as a scalar.
Po — Static pressure
Static pressure, specified as a scalar.
Ïo — Static density
Static density, specified as a scalar.
θ — Relative temperature ratio
Relative temperature ratio (θ), returned as a scalar.
To enable this port, select Theta.
sqrt(θ) — Square root of relative temperature ratio
Square root of the relative temperature ratio (
\sqrt{\mathrm{θ}}
), returned as a scalar.
To enable this port, select Square root of theta.
δ — Relative pressure ratio
Relative pressure ratio, (δ), returned as a scalar.
To enable this port, select Delta.
σ — Relative density ratio
Relative density ratio, (σ), returned as a scalar.
To enable this port, select Sigma.
Input units, specified as:
Pstatic
rho_static
Metric (MKS) Kelvin Pascal Kilograms per cubic meter
English Degrees Rankine Pound force per square inch Slug per cubic foot
Theta — Relative temperature ratio
When selected, the block calculates the relative temperature ratio (θ) and static temperature is a required input.
Block Parameter: theta
Square root of theta — Square root of relative temperature ratio
When selected, the block calculates the square root of relative temperature ratio (
\sqrt{\mathrm{θ}}
) and static temperature is a required input.
Selecting this check box enables the sqrt(θ) output port.
Block Parameter: sq_theta
Delta — Relative pressure ratio
When selected, the block calculates the relative pressure ratio (δ) and static pressure is a required input.
Block Parameter: delta
Sigma — Relative density ratio
When selected, the block the relative density ratio (σ) and static density is a required input.
Block Parameter: sigma
[1] Aeronautical Vestpocket Handbook, United Technologies Pratt & Whitney, August, 1986.
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Effective sample size - Wikipedia
Notion in statistics
It has been suggested that this article be merged into Design effect. (Discuss) Proposed since May 2021.
In statistics, effective sample size is a notion defined for a sample from a distribution when the observations in the sample are correlated or weighted. In 1965, Leslie Kish defined it as the original sample size divided by the design effect to reflect the variance from the current sampling design as compared to what would be if the sample was a simple random sample[1][2]: 162, 259
2 Weighted samples
Correlated observations[edit]
Suppose a sample of several independent identically distributed observations
{\displaystyle Y_{1},\dots ,Y_{n}}
is drawn from a distribution with mean
{\displaystyle \mu }
{\displaystyle \sigma }
. Then the mean of this distribution is estimated by the mean of the sample:
{\displaystyle {\hat {\mu }}={\frac {1}{n}}\sum _{i=1}^{n}Y_{i}.}
In that case, the variance of
{\displaystyle {\hat {\mu }}}
{\displaystyle \operatorname {Var} ({\hat {\mu }})={\frac {\sigma ^{2}}{n}}}
However, if the observations in the sample are correlated (in the intraclass correlation sense), then
{\displaystyle \operatorname {Var} ({\hat {\mu }})}
is somewhat higher. For instance, if all observations in the sample are completely correlated (
{\displaystyle \rho _{(i,j)}=1}
{\displaystyle \operatorname {Var} ({\hat {\mu }})=\sigma ^{2}}
regardless of
{\displaystyle n}
The effective sample size
{\displaystyle n_{\text{eff}}}
is the unique value (not necessarily an integer) such that
{\displaystyle \operatorname {Var} ({\hat {\mu }})={\frac {\sigma ^{2}}{n_{\text{eff}}}}.}
{\displaystyle n_{\text{eff}}}
is a function of the correlation between observations in the sample.
Suppose that all the (non-trivial) correlations are the same and greater than
{\displaystyle -1/(n-1)}
{\displaystyle i\neq j}
{\displaystyle \rho _{(i,j)}=\rho >-1/(n-1)}
{\displaystyle {\begin{aligned}\operatorname {Var} ({\hat {\mu }})&=\operatorname {Var} \left({\frac {1}{n}}Y_{1}+{\frac {1}{n}}Y_{2}+\cdots +{\frac {1}{n}}Y_{n}\right)\\[5pt]&=\sum _{i=1}^{n}{\frac {1}{n^{2}}}\operatorname {Var} (Y_{i})+\sum _{i=1}^{n}\sum _{j=1,j\neq i}^{n}{\frac {1}{n^{2}}}\operatorname {Cov} (Y_{i},Y_{j})\\[5pt]&=n{\frac {\sigma ^{2}}{n^{2}}}+n(n-1){\frac {\sigma ^{2}\rho }{n^{2}}}\\[5pt]&=\sigma ^{2}{\frac {1+(n-1)\rho }{n}}.\end{aligned}}}
{\displaystyle n_{\text{eff}}={\frac {n}{1+(n-1)\rho }}.}
{\displaystyle \rho =0}
{\displaystyle n_{\text{eff}}=n}
{\displaystyle \rho =1}
{\displaystyle n_{\text{eff}}=1}
. And if
{\displaystyle -1/(n-1)<\rho <0}
{\displaystyle n_{\text{eff}}>n}
The case where the correlations are not uniform is somewhat more complicated. Note that if the correlation is negative, the effective sample size may be larger than the actual sample size. If we allow the more general form
{\displaystyle {\hat {\mu }}=\sum _{i=1}^{n}a_{i}y_{i}}
{\displaystyle \sum _{i=1}^{n}a_{i}=1}
) then it is possible to construct correlation matrices that have an
{\displaystyle n_{\text{eff}}>n}
even when all correlations are positive. Intuitively, the maximal value of
{\displaystyle n_{\text{eff}}}
over all choices of the coefficients
{\displaystyle a_{i}}
may be thought of as the information content of the observed data.
Weighted samples[edit]
If the data has been weighted (the weights don't have to be normalized, i.e. have their sum equal to 1 or n, or some other constant), then several observations composing a sample have been pulled from the distribution with effectively 100% correlation with some previous sample. In this case, the effect is known as Kish's Effective Sample Size[3][2]: 162, 259
{\displaystyle n_{\text{eff}}={\frac {n}{D_{\text{eff}}}}={\frac {n}{\frac {\overline {w^{2}}}{{\overline {w}}^{2}}}}={\frac {n}{\frac {{\frac {1}{n}}\sum _{i=1}^{n}w_{i}^{2}}{\left({\frac {1}{n}}\sum _{i=1}^{n}w_{i}\right)^{2}}}}={\frac {n}{\frac {n\sum _{i=1}^{n}w_{i}^{2}}{(\sum _{i=1}^{n}w_{i})^{2}}}}={\frac {(\sum _{i=1}^{n}w_{i})^{2}}{\sum _{i=1}^{n}w_{i}^{2}}}}
^ Tom Leinster (December 18, 2014). "Effective Sample Size".
^ a b Kish, Leslie (1965). "Survey Sampling". New York: John Wiley & Sons, Inc. ISBN 0-471-10949-5. {{cite journal}}: Cite journal requires |journal= (help)
^ "Design Effects and Effective Sample Size".
M. B., Priestley (1981), Spectral Analysis and Time Series 1, Academic Press , §5.3.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Effective_sample_size&oldid=1081187717"
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Triangles - Circumcenter Practice Problems Online | Brilliant
The circumcenter is the intersection of which 3 lines in a triangle?
Medians of triangle Angle Bisectors of triangle Altitudes of triangle Perpendicular Bisector of sides of triangle
Which of the following triangles has its circumcenter outside of it?
Obtuse triangle Acute triangle Right triangle None of these
Which diagram shows a circumcenter?
Given the coordinates of the vertices of a triangle, how can we get the circumcenter of the triangle?
by finding a point whose coordinates are the averages of those of the three vertices by finding a point whose coordinates are the sums of those of the three vertices by finding a point that has the same distance from the three vertices by finding a point that has the same distance from the three sides
ABC
A = (0,0)
B = (0, 4 )
C = (14, 0)
O
, the circumcenter of triangle
ABC
(2, 7)
(14, 4 )
(4, 14 )
(7, 2 )
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Tell whether the function represents exponential growth or exponential decay. Then graph the function. f(x)=(1.5)^x
f\left(x\right)={\left(1.5\right)}^{x}
Answer is given on video below
Tell whether the function represents exponential growth or exponential decay. Identify the percent increase or decrease. Then graph the function.
f\left(x\right)={\left(\frac{1}{3}\right)}^{x}
y=a{\left(\frac{5}{4}\right)}^{t}
In the exponential growth or decay function
y={y}_{0}{e}^{k}t
{y}_{0}
represent? What does k represent?
Communicate Precisely How are exponential growth functions similar to exponential decay functions? How are they different?
Rewrite the function to determine whether it represents exponential growth or exponential decay.
y=2{\left(1.06\right)}^{9t}
Determine if it represents exponential growth or exponential decay.
f\left(x\right)=50{\left(1.1\right)}^{3}x
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How to Convert Metric Weight to Pounds: 7 Steps (with Pictures)
1 Converting Kilograms to Pounds
2 Converting Other Metric Weights into Pounds (grams, milligrams, etc.)
The metric system is the most common measurement system in the world, but it is not the only one. Several nations, most notably the United States, are still using pounds to measure weight instead of grams. However, since the metric system is so easy to comprehend, this makes conversions from metric weight to pounds simple and easy.
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Convert every kilogram to 2.2 pounds. A kilogram is a unit of weight equal to 2.2 pounds: 1kg = 2.2 lbs.[1] X Research source This means that 2 kg = 4.4 lbs, 3 kg = 6.6 lbs, and so on.
{\displaystyle 1kg*2.2{\frac {lb}{kg}}=2.2lbs}
Kilograms are abbreviated kg and pounds are abbreviated lbs.
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Multiply your weight in kilograms by 2.2 to get your weight in pounds.[2] X Research source Since each kilogram is 2.2 pounds, all you have to do is multiply the number of kilograms by this number to get your weight in pounds:
{\displaystyle 100kg=2.2{\frac {lb}{kg}}*100kg=220lbs}
{\displaystyle 38kg=2.2{\frac {lb}{kg}}*38kg=83.6lbs}
{\displaystyle 69.42kg=2.2{\frac {lb}{kg}}*69.42kg=152.724lbs}
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Use a more precise conversion. One kilogram is actually slightly heavier than 2.2 pound. This difference is so small that it doesn't matter in everyday life. If precision is very important, such as in a chemical reaction, use a more precise conversion:
{\displaystyle 1kilogram=2.20462pounds}
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Estimate the conversion in your head. Quickly, without a calculator — what is 78.4 kilograms converted into pounds? The following estimation methods will help you get a close answer if you just need to approximate something, though you should remember they aren't always accurate:[4] X Research source
Rounding up and down: Round to the closest number you can easily multiply for a rough estimate. In this example, round 78.4 kg to 80 kg, and 2.2 lb/kg to 2 lb/kg:
Estimated Answer: 80 kg * 2 lb/kg = 160 lbs.
Double Decimals: Remember that
{\displaystyle 78.4*2.2=(78.4*2)+(78.4*0.2)}
This is still pretty difficult, so round 78.4 to the nearest whole number to get (78 * 2) + (78 * 0.2). Multiply 78 by 2 to get 156. For the second part, you can use the same result (156), then just slide the decimal one place to the left: 156 becomes 15.6. Add the two results together and voila!
Estimated Answer: (78 * 2) + (78 * 0.2) = 156 + (156 * 0.1) = 156 + 15.6 = 171.6 lb
Actual Answer (for comparison): 78.4 kg * 2.2 lb/kg = 172.48 lbs.
Converting Other Metric Weights into Pounds (grams, milligrams, etc.)
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Convert between metric weights with powers of ten. The basic metric unit of weight is the gram. All other metric units of weight are equal to the gram multiplied by a power of ten, as described by the prefix. For example, "kilo-" means 1000, and a kilogram weighs exactly 1000 grams. "Milli" means "1/1000," so each milligram is one one-thousandth the weight of a gram (thus, 1000 milligrams = 1 gram).[5] X Research source
{\displaystyle 1kilogram=1,000grams}
{\displaystyle 1milligram={\frac {1}{1000}}gram}
= 0.001 grams
This article uses American notation, writing decimal marks with a period (.).
Convert your measurement into kilograms by moving the decimal the appropriate direction. Converting between metric units doesn't even require arithmetic. Instead, move the decimal point one place to the left to divide by 10, or one space to the right to multiply by 10.[6] X Research source Here's an example:
To convert 450 grams into kilograms, first work out the conversion for a single gram: 1000 grams = 1 kg, so therefore 1 gram =
{\displaystyle {\frac {1}{1000}}}
kg → 1 gram = 0.001 kg.
The conversion from 1 gram to 0.001 kg involves moving the decimal place to the left three times (1 → 0.1 → 0.01 → 0.001).
Do the same to convert your measurement from grams to kilograms. Start with 450 grams, move the decimal to the left three times, and you get 0.45. Therefore 450 grams = 0.45 kg.
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Multiply your converted kilograms by 2.2 to change them into pounds. Once you've got your measurement in kilograms, the process is the same as described in the section above. Simply multiply the kilograms by 2.2 to convert them into pounds. In the last example above, multiply 0.45 kg * 2.2 lb/kg = 0.99 lb. Here's another example to show you the whole process from start to finish:
Convert 3045 grams into pounds.
First, convert grams into kilograms:
Next, convert kilograms to pounds:
3.045 kg * 2.2 lb/kg = 6.699 lb
What is 83.9 kg. in pounds?
1 kg = 2.2 approximately, but the real number is 2.20462. So if you want to convert 83.9 kg to pounds, multiply 83.9 kg by 2.2 (or 2.20462) and you've got your answer.
How do I change pounds to kilograms?
Multiply the number of pounds by 0.4536.
How do you convert weight in pounds to metric equivalent?
One pound is the equivalent of 0.45359 kilogram.
How do I convert oz to lbs?
Divide ounces by 16.
80 ml equals what in ounces?
To convert milliliters to fluid ounces, multiply milliliters by 0.0338.
What is 43.7 kg in stones?
Multiply kilograms by 0.157473.
One square yard is how many square mm?
What is the conversion of inches to meters?
Multiply inches by 0.0254.
What is 3.2 kilograms in pounds and ounces?
(3.2 kg)(2.2 lb/kg) = 7.04 pounds = 7 pounds 0.64 ounce.
There are 28.35 grams in one ounce.
Since 1 July 1959, the international avoirdupois pound has been defined as exactly 0.45359237 kilogram.[7] X Research source
This implies that there are 1/0.45359237 ≈ 2,20462262 pounds in a kg.
Kilograms are denoted by the sign "kg".
Pounds are denoted by the sign "lb".
By using the rounded value (0.45 or 2.2) instead of the exact value above you will be making an error of less than 1%.
Convert Tenths of a Pound to Ounces
↑ https://www.bbc.com/bitesize/guides/zt2c82p/revision/3
↑ http://www.rapidtables.com/convert/weight/kg-to-pound.htm
↑ https://www.factmonster.com/math/measurement/metric-weights-and-measures
↑ https://www.purplemath.com/modules/metric.htm
↑ https://units.fandom.com/wiki/Pound_(international_avoirdupois)
Русский:конвертировать граммы и килограммы в фунты (и обратно)
中文:将公制重量单位转换为磅
Français:convertir des kilogrammes et des grammes en livres
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Solve the equations and inequalities. Write the solution sets to the inequalitie
Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. 9^{2m-3}=27^{m+1}
{9}^{2m-3}={27}^{m+1}
4{x}^{4}+3{x}^{2}-1=0
2x<30
solve each problem by setting up and solving an appropriate system of equations. In a class of 50 students, the number of females is two more than five times the number of males. How many females are there in the class?
h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. h(x) = (x - 2)^3 + 2
A bird species in danger of extinction has a population that is decreasing exponentially
A={A}_{0}{e}^{kt}
Five years ago the population was at 1400 and today only 1000 of the birds are alive. Once the population drops below 100, the situation will be irreversible. When will this happen?
d\frac{P}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{P}\right)\right)
\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\left(1-\frac{m}{P}\right)
Solve this differential equation using separation of variables. Only separate the equations and integration (not solving for P)
M= carrying capacity (maximum value)
m= threshold value (minimum value)
k= the proportionality constant
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Plot Markov chain redistributions - MATLAB distplot - MathWorks Switzerland
Animate Evolution of Markov Chain
Plot Markov chain redistributions
distplot(mc,X,Name,Value)
distplot(ax,___)
h = distplot(___)
distplot(mc,X) creates a heatmap from the data X showing the evolution of a distribution of states in the discrete-time Markov chain mc.
distplot(mc,X,Name,Value) uses additional options specified by one or more name-value arguments. For example, specify the type of plot or the frame rate for animated plots.
distplot(ax,___) plots on the axes specified by ax instead of the current axes (gca) using any of the input argument combinations in the previous syntaxes.
h = distplot(___) returns a handle to the distribution plot. Use h to modify properties of the plot after you create it.
P=\left[\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\ 0& 0& 1/3& 0& 2/3& 0& 0\\ 0& 0& 0& 0& 0& 1/3& 2/3\\ 0& 0& 0& 0& 0& 1/2& 1/2\\ 0& 0& 0& 0& 0& 3/4& 1/4\\ 1/2& 1/2& 0& 0& 0& 0& 0\\ 1/4& 3/4& 0& 0& 0& 0& 0\end{array}\right].
Compute the state redistributions at each step for 20 discrete time steps.
X = redistribute(mc,20);
Animate the redistributions in a histogram. Specify a half-second frame rate.
distplot(mc,X,'Type','histogram','FrameRate',0.5);
Evolution of state probabilities, specified as a (1 + numSteps)-by-NumStates nonnegative numeric matrix returned by redistribute. The first row is the initial state distribution. Subsequent rows are the redistributions at each step. distplot normalizes the rows by their respective sums before plotting.
By default, distplot plots to the current axes (gca).
Example: 'Type','graph','FrameRate',3 creates an animated plot of the redistributions using a frame rate of 3 seconds.
'evolution' (default) | 'histogram' | 'graph'
Evolution of the initial distribution. The plot is a (1 + NumSteps)-by-NumStates heatmap. Row i displays the redistribution at step i.
'histogram'
Animated histogram of the redistributions. The vertical axis displays probability mass, and the horizontal axis displays states. The 'FrameRate' name-value pair argument controls the animation progress.
Animated graph of the redistributions. distplot colors the nodes by their probability mass at each step. The 'FrameRate' name-value pair argument controls the animation progress.
h — Handle to distribution plot
Handle to the distribution plot, returned as a graphics object. h contains a unique plot identifier, which you can use to query or modify properties of the plot.
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d
{k}^{d}
k
k
⌊{n}^{\frac{1}{d}}⌋
; consequently, the number of points generated may be smaller than
d
d
n
\mathrm{points1},\mathrm{data1}≔\mathrm{Interpolation}:-\mathrm{Kriging}:-\mathrm{GenerateSpatialData}\left(\mathrm{Spherical}\left(1,10,1\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{points1}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{data1}}\textcolor[rgb]{0,0,1}{≔}\begin{array}{c}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{0.814723686393179}& \textcolor[rgb]{0,0,1}{0.706046088019609}\\ \textcolor[rgb]{0,0,1}{0.905791937075619}& \textcolor[rgb]{0,0,1}{0.0318328463774207}\\ \textcolor[rgb]{0,0,1}{0.126986816293506}& \textcolor[rgb]{0,0,1}{0.276922984960890}\\ \textcolor[rgb]{0,0,1}{0.913375856139019}& \textcolor[rgb]{0,0,1}{0.0461713906311539}\\ \textcolor[rgb]{0,0,1}{0.632359246225410}& \textcolor[rgb]{0,0,1}{0.0971317812358475}\\ \textcolor[rgb]{0,0,1}{0.0975404049994095}& \textcolor[rgb]{0,0,1}{0.823457828327293}\\ \textcolor[rgb]{0,0,1}{0.278498218867048}& \textcolor[rgb]{0,0,1}{0.694828622975817}\\ \textcolor[rgb]{0,0,1}{0.546881519204984}& \textcolor[rgb]{0,0,1}{0.317099480060861}\\ \textcolor[rgb]{0,0,1}{0.957506835434298}& \textcolor[rgb]{0,0,1}{0.950222048838355}\\ \textcolor[rgb]{0,0,1}{0.964888535199277}& \textcolor[rgb]{0,0,1}{0.0344460805029088}\\ \textcolor[rgb]{0,0,1}{⋮}& \textcolor[rgb]{0,0,1}{⋮}\end{array}]\\ \hfill \textcolor[rgb]{0,0,1}{\text{30 × 2 Matrix}}\end{array}\textcolor[rgb]{0,0,1}{,}\begin{array}{c}[\begin{array}{c}\textcolor[rgb]{0,0,1}{-1.31317888309841}\\ \textcolor[rgb]{0,0,1}{3.78399452938781}\\ \textcolor[rgb]{0,0,1}{-4.07906747556730}\\ \textcolor[rgb]{0,0,1}{2.81033657021080}\\ \textcolor[rgb]{0,0,1}{3.07159908082332}\\ \textcolor[rgb]{0,0,1}{0.128958765233144}\\ \textcolor[rgb]{0,0,1}{-3.21737272238246}\\ \textcolor[rgb]{0,0,1}{0.707245165710619}\\ \textcolor[rgb]{0,0,1}{0.0877877303791926}\\ \textcolor[rgb]{0,0,1}{0.937296621856498}\\ \textcolor[rgb]{0,0,1}{⋮}\end{array}]\\ \hfill \textcolor[rgb]{0,0,1}{\text{30 element Vector[column]}}\end{array}
\mathrm{k1}≔\mathrm{Interpolation}:-\mathrm{Kriging}\left(\mathrm{points1},\mathrm{data1}\right)
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\mathrm{ComputeGrid}\left(\mathrm{k1},[0..1,0..1],0.1,\mathrm{output}=\mathrm{plot}\right)
\mathrm{points2},\mathrm{data2}≔\mathrm{Interpolation}:-\mathrm{Kriging}:-\mathrm{GenerateSpatialData}\left(\mathrm{RationalQuadratic}\left(0.1,10,4\right),216,\mathrm{dimension}=3,\mathrm{grid}=\mathrm{true}\right)
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Under which of the following conditions is the relation
∆\mathrm{H}\quad =∆\mathrm{E}\quad +\quad \mathrm{P}∆\mathrm{V}
valid for a closed system ?
Constant temperature and pressure
Constant temperature, pressure and composition
∆\mathrm{p}\quad =0\phantom{\rule{0ex}{0ex}}∆\mathrm{H}=∆\mathrm{E}\quad +\mathrm{p}∆\mathrm{V}\quad \mathrm{is}\quad \mathrm{valid}.
The second law of thermodynamic says that in a cyclic process
work cannot be converted into heat
heat cannot be converted into work
work cannot be completely converted into heat
heat cannot be completely converted into work
Second law of thermodynamics states that in a cyclic process heat cannot be converted completely into work because attainment of 0 K temperature is impossible.
For the Paschen series the values of n1 and n2 in the expression
∆\mathrm{E}\quad =\mathrm{Rhc}\left(\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right)\quad \mathrm{are}
n1= 1, n2 = 2, 3, 4, ...
n1= 2, n2 = 3, 4, 5,...
n1= 3, n2 = 4,5,6,...
Paschen series is the third series of hydrogen spectrum and lies in the infrared region, therefore for this series, n1 = 3 and n2 > n1,ie, 4, 5, 6,....
The equilibrium constant (K) of a reaction may be written as
\mathrm{K}\quad =\quad {\mathrm{e}}^{-∆\mathrm{G}/\mathrm{RT}}
\mathrm{K}\quad ={\mathrm{e}}^{-∆{\mathrm{G}}^{°}/\mathrm{RT}}
\mathrm{K}\quad =\quad {\mathrm{e}}^{-∆\mathrm{H}/\mathrm{RT}}
\mathrm{K}\quad =\quad {\mathrm{e}}^{-∆{\mathrm{H}}^{°}/\mathrm{RT}}
\mathrm{K}\quad ={\mathrm{e}}^{-∆{\mathrm{G}}^{°}/\mathrm{RT}}
\mathrm{Gibbs}\quad \mathrm{free}\quad \mathrm{energy}\quad (∆{\mathrm{G}}^{°})\quad \mathrm{is}\quad \mathrm{related}\quad \mathrm{with}\quad \mathrm{equilibrium}\quad \mathrm{constant}\quad \left(\mathrm{K}\right)\quad \mathrm{as}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆{\mathrm{G}}^{°}\quad =\quad -\mathrm{RT}\quad \mathrm{ln}\quad \mathrm{K}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad -\quad \frac{∆{\mathrm{G}}^{°}}{\mathrm{RT}}\quad =\quad \mathrm{ln}\quad \mathrm{K}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{K}\quad =\quad {\mathrm{e}}^{\frac{-∆{\mathrm{G}}^{°}}{\mathrm{RT}}}
Which of the following will decrease the pH of a 50 mL solution of 0.01 M HCl?
Addition of 5 mL of 1 M HCl
Addition of 50 mL of 0.01 M HCl
Addition of 50 mL of 0.002 M HCl
Addition of Mg
Millimoles of 50 mL of 0.01 M HCl
= 50 x 0.01 M = 0.5
pH of the solution =-log (0.5) = 0.3010
(a) On adding 5 mL of 1 M HCl,
Concentration of the resultant solution
\frac{50\quad \times \quad 0.01+\quad 5\quad \times 1}{50+5}=0.1
:. pH of the solution = -log (0.1) = 1
(b) On adding 50 mL or 0.01 M HCl,
\frac{50\quad \mathrm{x}\quad 0.01\quad +\quad 50\quad \mathrm{x}\quad 0.01}{50\quad +\quad 50}=\quad 0.01
:. pH of the solution = -log (0.01) = 2
(c) On adding 50 mL of 0.002 M HCl,
\frac{50\quad \times 0.01+\quad 50\quad \mathrm{x}\quad 0.002}{50\quad +50}=\quad 6\times {10}^{-3}
:. pH of the solution = -log (6 x 10-3)= 2.22
(d) On adding Mg,
\mathrm{Mg}\quad +2\mathrm{HCl}\quad \to \quad {\mathrm{MgCl}}_{2}\quad +\quad 2{\mathrm{H}}^{+}\phantom{\rule{0ex}{0ex}}\quad \quad \mathrm{Since},\quad \mathrm{pH}\quad \infty \quad \frac{1}{\left[{\mathrm{H}}^{+}\right]}\phantom{\rule{0ex}{0ex}}:.\quad \mathrm{In}\quad \mathrm{this}\quad \mathrm{case}\quad \mathrm{pH}\quad \mathrm{decreases}.\phantom{\rule{0ex}{0ex}}
1 mole of photon, each of frequency 2500 s-1, would have approximately a total energy of
Frequency of one photon = 2500 s-1
:. Frequency of 1 mole of photon
= 6.023 x 1023 x 2500 s-1
\mathrm{Total}\quad \mathrm{energy}\quad \mathrm{of}\quad 1\quad \mathrm{mol}\quad \mathrm{of}\quad \mathrm{photon},\quad \mathrm{E}=\mathrm{h\nu }\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 6.626\quad \times \quad {10}^{-34}\mathrm{Js}\quad \times 6.023\quad \times {10}^{23}\times \quad 2500\quad {\mathrm{s}}^{-1}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 9.97\quad \times \quad {10}^{-7}\mathrm{J}\simeq \quad 10\quad \mathrm{erg}
An organic compound made of C, H and N contains 20% nitrogen. Its molecular weight is
Atomic mass of nitrogen= 14
Since the compound contains C, H and N, at
least one N atom must be there in the molecule.
20\%\quad \mathrm{N}\quad =14\phantom{\rule{0ex}{0ex}}100\quad \%\quad =\frac{14}{20\quad }\times \quad 100\quad =\quad 70
:. The minimum molecular weight of the organic compound = 70.
Equal volumes of molar hydrochloric acid 'and sulphuric acid are neutralized by dilute NaOH solution and x kcal and y kcal of heat are liberated respectively. Which of the following is true ?
x =y/2
Heat of neutralisation= Heat of formation of water (1 mol).
\mathrm{HCl}\quad +\quad \mathrm{NaCl}\quad \to \quad \mathrm{NaCl}\quad +\quad {\mathrm{H}}_{2}\mathrm{O};\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆\mathrm{H}\quad =-\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{or}\underset{1\mathrm{mol}\quad }{\quad {\mathrm{H}}^{+}}+\underset{1\mathrm{mol}}{\quad {\mathrm{OH}}^{-}\quad }\to \quad {\mathrm{H}}_{2}\mathrm{O};\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆\mathrm{H}\quad =\quad \mathrm{x}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\quad +\quad 2\mathrm{NaOH}\quad \to {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}\quad +\quad 2{\mathrm{H}}_{2}\mathrm{O}\quad ;\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆\mathrm{H}\quad =\quad -\mathrm{y}\phantom{\rule{0ex}{0ex}}\mathrm{or}\quad 2{\mathrm{H}}^{+}+\quad 2{\mathrm{OH}}^{-}\to 2{\mathrm{H}}_{2}\mathrm{O};\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆\mathrm{H}\quad =-\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}^{+}\quad +\quad {\mathrm{OH}}^{-}\quad \to \quad {\mathrm{H}}_{2}\mathrm{O};\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ∆\mathrm{H}\quad =-\frac{\mathrm{y}}{2}\phantom{\rule{0ex}{0ex}}\quad \quad ∆\mathrm{H}\quad =\quad ∆\mathrm{H}\phantom{\rule{0ex}{0ex}}\quad \quad \mathrm{x}\quad =\frac{\mathrm{y}}{2}
Hybridisation of central atom in NF3 is
Number of hybrid orbitals = number of
\mathrm{\sigma }
bonds + number of lone pairs
\Rightarrow 3\mathrm{\sigma }
bonds + 1 lone pair = 4
.:. Hybridisation of N is sp3 and geometry of the molecule is pyramidal.
\mathrm{For}\quad \mathrm{the}\quad \mathrm{reaction}\quad \quad {\mathrm{SO}}_{2}\quad +\quad \frac{1}{2}{\mathrm{O}}_{2}\quad \rightleftharpoons \quad {\mathrm{SO}}_{3}\quad ,\quad \mathrm{if}\quad \mathrm{we}\quad \mathrm{write}\quad {\mathrm{K}}_{\mathrm{p}\quad }=\quad {\mathrm{K}}_{\mathrm{c}}(\mathrm{RT}{)}^{\mathrm{x}},\quad \mathrm{then}\quad \mathrm{x}\quad \mathrm{become}
-\frac{1}{2}
\frac{1}{2}
-\frac{1}{2}
\mathrm{In}\quad \mathrm{the}\quad \mathrm{relation},\quad {\mathrm{K}}_{\mathrm{p}}=\quad {\mathrm{K}}_{\mathrm{c}}(\mathrm{RT}{)}^{\mathrm{x}},\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{x}=\quad {\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{r}}\phantom{\rule{0ex}{0ex}}\mathrm{For}\quad \mathrm{the}\quad \mathrm{reaction},\quad \phantom{\rule{0ex}{0ex}}\quad \quad \quad {\mathrm{SO}}_{2}(\mathrm{g})\quad +\quad \frac{1}{2}{\mathrm{O}}_{2}(\mathrm{g})\quad \rightleftharpoons {\mathrm{SO}}_{3}(\mathrm{g})\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{x}\quad =\quad 1\quad -\quad \left(1\quad +\frac{1}{2}\right)\quad =-\frac{1}{2}
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Allie collected 15 more cans of food than Peyton. Ling collected 8 more than All
Allie collected 15 more cans of food than Peyton. Ling collected 8 more than Allie. Ling collected 72 cans of food. How many cans of food did Allie, Peyton, and Ling collect in all?
Let x be the number of cans of food Peyton collected so that Allie collected x+15 and Ling collected
\left(x+15\right)+8=x+23
cans of food. Given that Ling collected 72 cans of food, we can write:
x+23=72
Solve for xx. Subtract 23 from both sides:
x=49
So, Peyton collected 4949 and Allie collected
49+15=6449+15=64
cans of food. Overall, they collected a total of:
49+64+72=185
cans of food
{x}^{2}+x-20
-4<\frac{2y-3}{3}<4
|5-3t|=\frac{1}{2}
Solve the equation system:
\left\{\begin{array}{ccccc}3x& -& 4y& =& -14\\ 5x& +& 7y& =& 45\end{array}
Write each number in standard form.
3×1+\left(3×\frac{1}{10}\right)+\left(9×\frac{1}{1000}\right)
The length of a rectangle is 3 ft longer than its width. If the perimeter of the rectangle is 46 ft, find its area.
Mt. Everest is 8707.37 feet higher than Mt. McKinley. What is the elevation of Mt. Everest?
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CLASSICAL MECHANICS - Encyclopedia Information
Classical mechanics Information
{\displaystyle {\textbf {F}}={\frac {d}{dt}}(m{\textbf {v}})}
Classical mechanics [note 1] is a physical theory describing the motion of macroscopic objects, from projectiles to parts of machinery, and astronomical objects, such as spacecraft, planets, stars, and galaxies. For objects governed by classical mechanics, if the present state is known, it is possible to predict how it will move in the future (determinism), and how it has moved in the past (reversibility).
angular position/ angle unitless (radian)
The position of a point particle is defined in relation to a coordinate system centered on an arbitrary fixed reference point in space called the origin O. A simple coordinate system might describe the position of a particle P with a vector notated by an arrow labeled r that points from the origin O to point P. In general, the point particle does not need to be stationary relative to O. In cases where P is moving relative to O, r is defined as a function of t, time. In pre-Einstein relativity (known as Galilean relativity), time is considered an absolute, i.e., the time interval that is observed to elapse between any given pair of events is the same for all observers. [3] In addition to relying on absolute time, classical mechanics assumes Euclidean geometry for the structure of space. [4]
{\displaystyle \mathbf {v} ={\mathrm {d} \mathbf {r} \over \mathrm {d} t}\,\!}
{\displaystyle \mathbf {u} '=\mathbf {u} -\mathbf {v} \,.}
{\displaystyle \mathbf {v'} =\mathbf {v} -\mathbf {u} \,.}
{\displaystyle \mathbf {u} '=(u-v)\mathbf {d} \,.}
{\displaystyle u'=u-v\,.}
{\displaystyle \mathbf {a} ={\mathrm {d} \mathbf {v} \over \mathrm {d} t}={\mathrm {d^{2}} \mathbf {r} \over \mathrm {d} t^{2}}.}
{\displaystyle x'=x-ut\,}
{\displaystyle y'=y\,}
{\displaystyle z'=z\,}
{\displaystyle t'=t\,.}
Forces and Newton's second law
Newton was the first to mathematically express the relationship between force and momentum. Some physicists interpret Newton's second law of motion as a definition of force and mass, while others consider it a fundamental postulate, a law of nature. [5] Either interpretation has the same mathematical consequences, historically known as "Newton's Second Law":
{\displaystyle \mathbf {F} ={\mathrm {d} \mathbf {p} \over \mathrm {d} t}={\mathrm {d} (m\mathbf {v} ) \over \mathrm {d} t}.}
The quantity mv is called the ( canonical) momentum. The net force on a particle is thus equal to the rate of change of the momentum of the particle with time. Since the definition of acceleration is a = dv/dt, the second law can be written in the simplified and more familiar form:
{\displaystyle \mathbf {F} =m\mathbf {a} \,.}
{\displaystyle \mathbf {F} _{\rm {R}}=-\lambda \mathbf {v} \,,}
{\displaystyle -\lambda \mathbf {v} =m\mathbf {a} =m{\mathrm {d} \mathbf {v} \over \mathrm {d} t}\,.}
{\displaystyle \mathbf {v} =\mathbf {v} _{0}e^{{-\lambda t}/{m}}}
Important forces include the gravitational force and the Lorentz force for electromagnetism. In addition, Newton's third law can sometimes be used to deduce the forces acting on a particle: if it is known that particle A exerts a force F on another particle B, it follows that B must exert an equal and opposite reaction force, −F, on A. The strong form of Newton's third law requires that F and −F act along the line connecting A and B, while the weak form does not. Illustrations of the weak form of Newton's third law are often found for magnetic forces.[ clarification needed]
If a constant force F is applied to a particle that makes a displacement Δr, [note 2] the work done by the force is defined as the scalar product of the force and displacement vectors:
{\displaystyle W=\mathbf {F} \cdot \Delta \mathbf {r} \,.}
{\displaystyle W=\int _{C}\mathbf {F} (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} \,.}
{\displaystyle E_{\mathrm {k} }={\tfrac {1}{2}}mv^{2}\,.}
{\displaystyle W=\Delta E_{\mathrm {k} }=E_{\mathrm {k_{2}} }-E_{\mathrm {k_{1}} }={\tfrac {1}{2}}m\left(v_{2}^{\,2}-v_{1}^{\,2}\right).}
{\displaystyle \mathbf {F} =-\mathbf {\nabla } E_{\mathrm {p} }\,.}
{\displaystyle \mathbf {F} \cdot \Delta \mathbf {r} =-\mathbf {\nabla } E_{\mathrm {p} }\cdot \Delta \mathbf {r} =-\Delta E_{\mathrm {p} }\,.}
{\displaystyle -\Delta E_{\mathrm {p} }=\Delta E_{\mathrm {k} }\Rightarrow \Delta (E_{\mathrm {k} }+E_{\mathrm {p} })=0\,.}
{\displaystyle \sum E=E_{\mathrm {k} }+E_{\mathrm {p} }\,,}
When both quantum mechanics and classical mechanics cannot apply, such as at the quantum level with many degrees of freedom, quantum field theory (QFT) is of use. QFT deals with small distances, and large speeds with many degrees of freedom as well as the possibility of any change in the number of particles throughout the interaction. When treating large degrees of freedom at the macroscopic level, statistical mechanics becomes useful. Statistical mechanics describes the behavior of large (but countable) numbers of particles and their interactions as a whole at the macroscopic level. Statistical mechanics is mainly used in thermodynamics for systems that lie outside the bounds of the assumptions of classical thermodynamics. In the case of high velocity objects approaching the speed of light, classical mechanics is enhanced by special relativity. In case that objects become extremely heavy (i.e., their Schwarzschild radius is not negligibly small for a given application), deviations from Newtonian mechanics become apparent and can be quantified by using the parameterized post-Newtonian formalism. In that case, general relativity (GR) becomes applicable. However, until now there is no theory of quantum gravity unifying GR and QFT in the sense that it could be used when objects become extremely small and heavy. [4] [5]
{\displaystyle \mathbf {p} ={\frac {m\mathbf {v} }{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\,,}
{\displaystyle \mathbf {p} \approx m\mathbf {v} \,.}
{\displaystyle f=f_{\mathrm {c} }{\frac {m_{0}}{m_{0}+{\frac {T}{c^{2}}}}}\,,}
where fc is the classical frequency of an electron (or other charged particle) with kinetic energy T and ( rest) mass m0 circling in a magnetic field. The (rest) mass of an electron is 511 keV. So the frequency correction is 1% for a magnetic vacuum tube with a 5.11 kV direct current accelerating voltage.
{\displaystyle \lambda ={\frac {h}{p}}}
The first published causal explanation of the motions of planets was Johannes Kepler's Astronomia nova, published in 1609. He concluded, based on Tycho Brahe's observations on the orbit of Mars, that the planet's orbits were ellipses. This break with ancient thought was happening around the same time that Galileo was proposing abstract mathematical laws for the motion of objects. He may (or may not) have performed the famous experiment of dropping two cannonballs of different weights from the tower of Pisa, showing that they both hit the ground at the same time. The reality of that particular experiment is disputed, but he did carry out quantitative experiments by rolling balls on an inclined plane. His theory of accelerated motion was derived from the results of such experiments and forms a cornerstone of classical mechanics. In 1673 Christiaan Huygens described in his Horologium Oscillatorium the first two laws of motion. [6] The work is also the first modern treatise in which a physical problem (the accelerated motion of a falling body) is idealized by a set of parameters then analyzed mathematically and constitutes one of the seminal works of applied mathematics. [7]
Newton had previously invented the calculus, of mathematics, and used it to perform the mathematical calculations. For acceptability, his book, the Principia, was formulated entirely in terms of the long-established geometric methods, which were soon eclipsed by his calculus. However, it was Leibniz who developed the notation of the derivative and integral preferred [8] today. Newton, and most of his contemporaries, with the notable exception of Huygens, worked on the assumption that classical mechanics would be able to explain all phenomena, including light, in the form of geometric optics. Even when discovering the so-called Newton's rings (a wave interference phenomenon) he maintained his own corpuscular theory of light.
Since the end of the 20th century, classical mechanics in physics has no longer been an independent theory. Instead, classical mechanics is now considered an approximate theory to the more general quantum mechanics. Emphasis has shifted to understanding the fundamental forces of nature as in the Standard model and its more modern extensions into a unified theory of everything. [9] Classical mechanics is a theory useful for the study of the motion of non-quantum mechanical, low-energy particles in weak gravitational fields. Also, it has been extended into the complex domain where complex classical mechanics exhibits behaviors very similar to quantum mechanics. [10]
^ The "classical" in "classical mechanics" does not refer classical antiquity, as it might in, say, classical architecture; indeed, the (European) development of classical mechanics involved substantial change in the methods and philosophy of physics. [1] The qualifier instead attempts to distinguish classical mechanics from physics developed after the revolutions of the early 20th century, which revealed classical mechanics' limits of validity. [2]
^ Jesseph, Douglas M. (1998). " Leibniz on the Foundations of the Calculus: The Question of the Reality of Infinitesimal Magnitudes". Perspectives on Science. 6.1&2: 6–40. Retrieved 31 December 2011.
Horbatsch, Marko, " Classical Mechanics Course Notes".
Rosu, Haret C., " Classical Mechanics". Physics Education. 1999. [arxiv.org : physics/9909035]
Retrieved from " https://en.wikipedia.org/?title=Classical_mechanics&oldid=1084929858"
Classical Mechanics Videos
Classical Mechanics Websites
Classical Mechanics Encyclopedia Articles
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Indeterminate system - Wikipedia
In mathematics, particularly in algebra, an indeterminate system is a system of simultaneous equations (e.g., linear equations) which has more than one solution (sometimes infinitely many solutions).[1] In the case of a linear system, the system may be said to be underspecified, in which case the presence of more than one solution would imply an infinite number of solutions (since the system would be describable in terms of at least one free variable[2]), but that property does not extend to nonlinear systems (e.g., the system with the equation
{\displaystyle x^{2}=1}
An indeterminate system by definition is consistent, in the sense of having at least one solution.[3] For a system of linear equations, the number of equations in an indeterminate system could be the same as the number of unknowns, less than the number of unknowns (an underdetermined system), or greater than the number of unknowns (an overdetermined system). Conversely, any of those three cases may or may not be indeterminate.
2 Conditions giving rise to indeterminacy
3 Finding the solution set of an indeterminate linear system
The following examples of indeterminate systems of equations have respectively, fewer equations than, as many equations as, and more equations than unknowns:
{\displaystyle {\text{System 1: }}x+y=2}
{\displaystyle {\text{System 2: }}x+y=2,\,\,\,\,\,2x+2y=4}
{\displaystyle {\text{System 3: }}x+y=2,\,\,\,\,\,2x+2y=4,\,\,\,\,\,3x+3y=6}
Conditions giving rise to indeterminacy[edit]
In linear systems, indeterminacy occurs if and only if the number of independent equations (the rank of the augmented matrix of the system) is less than the number of unknowns and is the same as the rank of the coefficient matrix. For if there are at least as many independent equations as unknowns, that will eliminate any stretches of overlap of the equations' surfaces in the geometric space of the unknowns (aside from possibly a single point), which in turn excludes the possibility of having more than one solution. On the other hand, if the rank of the augmented matrix exceeds (necessarily by one, if at all) the rank of the coefficient matrix, then the equations will jointly contradict each other, which excludes the possibility of having any solution.
Finding the solution set of an indeterminate linear system[edit]
Let the system of equations be written in matrix form as
{\displaystyle Ax=b}
{\displaystyle A}
{\displaystyle m\times n}
coefficient matrix,
{\displaystyle x}
{\displaystyle n\times 1}
vector of unknowns, and
{\displaystyle b}
{\displaystyle m\times 1}
vector of constants. In which case, if the system is indeterminate, then the infinite solution set is the set of all
{\displaystyle x}
vectors generated by[4]
{\displaystyle x=A^{+}b+[I_{n}-A^{+}A]w}
{\displaystyle A^{+}}
is the Moore-Penrose pseudoinverse of
{\displaystyle A}
{\displaystyle w}
{\displaystyle n\times 1}
^ "Indeterminate and Inconsistent Systems: Systems of Equations". TheProblemSite.com. Retrieved 2019-12-02.
^ Gustafson, Grant B. (2008). "Three Possibilities (of a Linear System)" (PDF). math.utah.edu. Retrieved 2019-12-02. {{cite web}}: CS1 maint: url-status (link)
^ "Consistent and Inconsistent Systems of Equations | Wyzant Resources". www.wyzant.com. Retrieved 2019-12-02.
^ James, M., "The generalised inverse", Mathematical Gazette 62, June 1978, 109–114.
Lay, David (2003). Linear Algebra and Its Applications. Addison-Wesley. ISBN 0-201-70970-8.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Indeterminate_system&oldid=1069731028"
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Prove {F}{left({s}right)}=frac{1}{{{s}-{2}}} text{ then } f{{left({t}right)}} is a)
Prove {F}{left({s}right)}=frac{1}{{{s}-{2}}} text{ then } f{{left({t}right)}} is a) {e}^{{{2}{t}}}{u}{left({t}right)} b) u(t+2) c) u(t-2) d) {e}^{{-{2}{t}}}{u}{left({t}right)}
F\left(s\right)=\frac{1}{s-2}\text{ }\text{ then }\text{ }f\left(t\right)
{e}^{2t}u\left(t\right)
u\left(t+2\right)
u\left(t-2\right)
{e}^{-2t}u\left(t\right)
F\left(s\right)=\frac{1}{s-2}
We have to find the f(t)
Use definition of inverse Laplace transform, which is given below
f\left(t\right)={L}^{-1}\left\{F\left(s\right)\right\}\dots \left(1\right)
Take inverse Laplace transform of
F\left(s\right)=\frac{1}{s-2}
{L}^{-1}\left\{F\right\}\left(s\right)={L}^{-1}\left\{\frac{1}{s-2}\right\}\dots \left(2\right)
f\left(t\right)={L}^{-1}\left\{\frac{1}{s-2}\right\}
={e}^{2t}u\left(t\right)
f\left(t\right)={e}^{2t}u\left(t\right)
Assume that the rate of evaporation for water is proportional to the initial amount of water in a bowl. If Jill places 1 cup of water outside and notices that there is
\frac{1}{2}
cup of water 6 hours later. Can she use a differential equation to find when there is no water left?
Find the solution of the Differentional equation by using Laplace Transformation
2{y}^{\prime }-3y={e}^{2t},y\left(0\right)=1
y"+y=t,y\left(0\right)=0\text{ }and\text{ }{y}^{\prime }\left(0\right)=2
xydx-{\left(x+2y\right)}^{2}dy=0
F{}^{″}\left(x\right)=\mathrm{cos}x,{F}^{\prime }\left(0\right)=3,F\left(\pi \right)=5
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform
L\left\{{t}^{4}+{t}^{3}-{t}^{2}-t+\mathrm{sin}\sqrt{2}t\right\}
Solve the following differential equations using the Laplace transform
y2{y}^{\prime }+y=t+3
y\left(0\right)=1,{y}^{\prime }\left(0\right)=0
use the Laplace transform to solve the given initial-value problem.
y"+{y}^{\prime }-2y=10{e}^{-t},y\left(0\right)=0,{y}^{\prime }\left(0\right)=1
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Physics Study Guide/Vectors and scalars - Wikibooks, open books for an open world
Physics Study Guide/Vectors and scalars
Vectors are quantities that are characterized by having both a numerical quantity (called the "magnitude" and denoted as |v|) and a direction. Velocity is an example of a vector; it describes the time rated change in position with a numerical quantity (meters per second) as well as indicating the direction of movement.
The definition of a vector is any quantity that adds according to the parallelogram law (there are some physical quantities that have magnitude and direction that are not vectors).
Scalars are quantities in physics that have no direction. Mass is a scalar; it can describe the quantity of matter with units (kilograms) but does not describe any direction.
1 Multiplying vectors and scalars
2 Frequently Asked Questions about Vectors
2.1 When are scalar and vector compositions essentially the same?
2.2 What is a "dot-product"? (work when force not parallel to displacement)
2.3 What is a "cross-product"? (Force on a charged particle in a magnetic field)
2.4 How to draw vectors that are in or out of the plane of the page (or board)
Multiplying vectors and scalarsEdit
A scalar times a scalar gives a scalar result.
A vector scalar-multiplied by a vector gives a scalar result (called the dot-product).
A vector cross-multiplied by a vector gives a vector result (called the cross-product).
A vector times a scalar gives a vector result.
Frequently Asked Questions about VectorsEdit
When are scalar and vector compositions essentially the same?Edit
Answer: when multiple vectors are in same direction then we can just add the magnitudes.so, the scalar and vector composition will be same as we do not add the directions.
What is a "dot-product"? (work when force not parallel to displacement)Edit
A Man walking up a hill
Answer: Let's take gravity as our force. If you jump out of an airplane and fall you will pick up speed. (for simplicity's sake, let's ignore air drag). To work out the kinetic energy at any point you simply multiply the value of the force caused by gravity by the distance moved in the direction of the force. For example, a 180 N boy falling a distance of 10 m will have 1800 J of extra kinetic energy. We say that the man has had 1800 J of work done on him by the force of gravity.
Notice that energy is not a vector. It has a value but no direction. Gravity and displacement are vectors. They have a value plus a direction. (In this case, their directions are down and down respectively) The reason we can get a scalar energy from vectors gravity and displacement is because, in this case, they happen to point in the same direction. Gravity acts downwards and displacement is also downwards.
When two vectors point in the same direction, you can get the scalar product by just multiplying the value of the two vectors together and ignoring the direction.
But what happens if they don't point in the same direction?
Consider a man walking up a hill. Obviously it takes energy to do this because you are going against the force of gravity. The steeper the hill, the more energy it takes every step to climb it. This is something we all know unless we live on a salt lake.
In a situation like this we can still work out the work done. In the diagram, the green lines represent the displacement. To find out how much work against gravity the man does, we work out the projection of the displacement along the line of action of the force of gravity. In this case it's just the y component of the man's displacement. This is where the cos θ comes in. θ is merely the angle between the velocity vector and the force vector.
If the two forces do not point in the same direction, you can still get the scalar product by multiplying the projection of one force in the direction of the other force. Thus:
{\displaystyle {\vec {a}}\cdot {\vec {b}}\equiv \|{\vec {a}}\|\ \|{\vec {b}}\|\ \cos \theta }
There is another method of defining the dot product which relies on components.
{\displaystyle {\vec {a}}\cdot {\vec {b}}\equiv a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}}
What is a "cross-product"? (Force on a charged particle in a magnetic field)Edit
Answer: Suppose there is a charged particle moving in a constant magnetic field. According to the laws of electromagnetism, the particle is acted upon by a force called the Lorentz force. If this particle is moving from left to right at 30 m/s and the field is 30 Tesla pointing straight down perpendicular to the particle, the particle will actually curve in a circle spiraling out of the plane of the two with an acceleration of its charge in coulombs times 900 newtons per coulomb! This is because the calculation of the Lorentz force involves a cross-product.when cross product can replace the sin0 can take place during multiplication. A cross product can be calculated simply using the angle between the two vectors and your right hand. If the forces point parallel or 180° from each other, it's simple: the cross-product does not exist. If they are exactly perpendicular, the cross-product has a magnitude of the product of the two magnitudes. For all others in between however, the following formula is used:
{\displaystyle \left\|{\vec {a}}\times {\vec {b}}\right\|=\left\|{\vec {a}}\right\|\ \left\|{\vec {b}}\right\|\sin \theta }
The right-hand rule: point your index finger along the first vector and your middle finger across the second; your thumb will point in the direction of the resulting vector
But if the result is a vector, then what is the direction? That too is fairly simple, utilizing a method called the "right-hand rule".
The right-hand rule works as follows: Place your right-hand flat along the first of the two vectors with the palm facing the second vector and your thumb sticking out perpendicular to your hand. Then proceed to curl your hand towards the second vector. The direction that your thumb points is the direction that cross-product vector points! Though this definition is easy to explain visually it is slightly more complicated to calculate than the dot product.
{\displaystyle (a_{x},\ a_{y},\ a_{z})\times (b_{x},\ b_{y},\ b_{z})=(a_{y}b_{z}-a_{z}b_{y}\ ,a_{x}b_{z}-a_{z}b_{x}\ ,a_{x}b_{y}-a_{y}b_{x})}
How to draw vectors that are in or out of the plane of the page (or board)Edit
How to draw vectors in the plane of the paper
Standard symbols of a vector going into or out of a page
Answer: Vectors in the plane of the page are drawn as arrows on the page. A vector that goes into the plane of the screen is typically drawn as circles with an inscribed X. A vector that comes out of the plane of the screen is typically drawn as circles with dots at their centers. The X is meant to represent the fletching on the back of an arrow or dart while the dot is meant to represent the tip of the arrow.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Physics_Study_Guide/Vectors_and_scalars&oldid=3989110"
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Trigonometric Equations - R method | Brilliant Math & Science Wiki
Christopher Boo, Mei Li, and Jimin Khim contributed
R
Method is a technique used to solve trigonometric equations of the form
a\sin x \pm b\cos x = c
by applying the addition formulae.
3\sin x + 4 \cos x
R\sin (x+\alpha)
Our goal is to solve for
R
\alpha
\begin{aligned} 3\sin x + 4 \cos x &\equiv R\sin (x+\alpha) \\ & = R\sin x \cos \alpha + R \cos x \sin \alpha\\\\ \Rightarrow R\cos \alpha &= 3,\quad R\sin \alpha = 4. \end{aligned}
Taking their ratio leads to
\displaystyle \tan \alpha = \frac{4}{3}\implies \alpha = \arctan \frac{4}{3} = 53^\circ 8'.
\alpha
into the previous equation gives
R\cos53^\circ 8' = 3\implies R = 5.
3\sin x + 4 \cos x \equiv 5\sin (x+53^\circ 8').\ _\square
Taking the most suitable form of trigonometric addition formula will save you a lot of work. The equations below state the best
R
method in different cases.
a\sin x + b \cos x \equiv R\sin (x+\alpha)
a\sin x - b \cos x \equiv R\sin (x-\alpha)
a\cos x + b \sin x \equiv R\cos (x-\alpha)
a\cos x - b \sin x \equiv R\cos (x+\alpha)
R
method can be very effective when solving trigonometric equations and inequalities.
3\sin x + 4 \cos x = 1
0\leq x \leq 180^\circ.
Applying the
R
\begin{aligned} 3 \sin x + 4 \cos x &= 5\sin (x+53^\circ 8') \\ 5\sin (x+53^\circ 8') &= 1 \\ \sin (x+53^\circ 8') &= \frac{1}{5} \\ x+53^\circ 8' &= 168^\circ 28' \\ x &= 115^\circ 20'.\ _\square \end{aligned}
3\sin x + 4 \cos x?
Since the maximum value of the sine function is
1
\begin{aligned} 3 \sin x + 4 \cos x &= 5\sin (x+53^\circ 8') \\ &\leq 5.\ _\square \end{aligned}
Cite as: Trigonometric Equations - R method. Brilliant.org. Retrieved from https://brilliant.org/wiki/trigonometric-equations-r-method/
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Lens (geometry) - Wikipedia
Lens (geometry)
For other uses, see Lens (optics).
A lens contained between two circular arcs of radius R, and centers at O1 and O2
In 2-dimensional geometry, a lens is a convex region bounded by two circular arcs joined to each other at their endpoints. In order for this shape to be convex, both arcs must bow outwards (convex-convex). This shape can be formed as the intersection of two circular disks. It can also be formed as the union of two circular segments (regions between the chord of a circle and the circle itself), joined along a common chord.
Example of two asymmetric lenses (left and right) and one symmetric lens (in the middle)
The Vesica piscis is the intersection of two disks with the same radius, R, and with the distance between centers also equal to R.
If the two arcs of a lens have equal radius, it is called a symmetric lens, otherwise is an asymmetric lens.
The vesica piscis is one form of a symmetric lens, formed by arcs of two circles whose centers each lie on the opposite arc. The arcs meet at angles of 120° at their endpoints.
The area of a symmetric lens can be expressed in terms of the radius R and arc lengths θ in radians:
{\displaystyle A=R^{2}\left(\theta -\sin \theta \right).}
The area of an asymmetric lens formed from circles of radii R and r with distance d between their centers is[1]
{\displaystyle A=r^{2}\cos ^{-1}\left({\frac {d^{2}+r^{2}-R^{2}}{2dr}}\right)+R^{2}\cos ^{-1}\left({\frac {d^{2}+R^{2}-r^{2}}{2dR}}\right)-2\Delta }
{\displaystyle \Delta ={\frac {1}{4}}{\sqrt {(-d+r+R)(d-r+R)(d+r-R)(d+r+R)}}}
is the area of a triangle with sides d, r, and R.
The two circles overlap if
{\displaystyle d<r+R}
. For sufficiently large
{\displaystyle d}
, the coordinate
{\displaystyle x}
of the lens centre lies between the coordinates of the two circle centers:
{\displaystyle d}
{\displaystyle x}
of the lens centre lies outside the line that connects the circle centres:
By eliminating y from the circle equations
{\displaystyle x^{2}+y^{2}=r^{2}}
{\displaystyle (x-d)^{2}+y^{2}=R^{2}}
the abscissa of the intersecting rims is
{\displaystyle x=(d^{2}+r^{2}-R^{2})/(2d)}
The sign of x, i.e.,
{\displaystyle d^{2}}
being larger or smaller than
{\displaystyle R^{2}-r^{2}}
, distinguishes the two cases shown in the images.
The ordinate of the intersection is
{\displaystyle y={\sqrt {r^{2}-x^{2}}}={\frac {\sqrt {[(R-d)^{2}-r^{2}][r^{2}-(R+d)^{2}]}}{2d}}}
Negative values under the square root indicate that the rims of the two circles do not touch because the circles are too far apart or one circle lies entirely within the other.
The value under the square root is a biquadratic polynomial of d. The four roots of this polynomial are associated with y=0 and with the four values of d where the two circles have only one point in common.
The angles in the blue triangle of sides d, r and R are
{\displaystyle \sin a_{r}=y/r;\quad \sin a_{R}=y/R}
where y is the ordinate of the intersection. The branch of the arcsin with
{\displaystyle a_{r}>\pi /2}
is to be taken if
{\displaystyle d^{2}<R^{2}-r^{2}}
{\displaystyle \Delta ={\frac {1}{2}}yd}
The area of the asymmetric lens is
{\displaystyle A=a_{r}r^{2}+a_{R}R^{2}-yd}
, where the two angles are measured in radians. [This is an application of the Inclusion-exclusion principle: the two circular sectors centered at (0,0) and (d,0) with central angles
{\displaystyle 2a_{r}}
{\displaystyle 2a_{R}}
have areas
{\displaystyle 2a_{r}r^{2}}
{\displaystyle 2a_{R}R^{2}}
. Their union covers the triangle, the flipped triangle with corner at (x,-y), and twice the lens area.]
A lens with a different shape forms part of the answer to Mrs. Miniver's problem, which asks how to bisect the area of a disk by an arc of another circle with given radius. One of the two areas into which the disk is bisected is a lens.
Lenses are used to define beta skeletons, geometric graphs defined on a set of points by connecting pairs of points by an edge whenever a lens determined by the two points is empty.
Lune, a related non-convex shape formed by two circular arcs, one bowing outwards and the other inwards
Lemon, created by a lens rotated around an axis through its tips.[2]
^ Weisstein, Eric W. "Lens". MathWorld.
^ Weisstein, Eric W. "Lemon". Wolfram MathWorld. Retrieved 2019-11-04.
Pedoe, D. (1995). "Circles: A Mathematical View, rev. ed". Washington, DC: Math. Assoc. Amer. MR 1349339.
Plummer, H. (1960). An Introductory Treatise of Dynamical Astronomy. York: Dover. Bibcode:1960aitd.book.....P.
Watson, G. N. (1966). A Treatise on the Theory of Bessel Functions, 2nd ed. Cambridge, England: Cambridge University Press. MR 1349110.
Fewell, M. P. (2006). "Area of common overlap of three circles". Defence Science and Technology Organisation.
Librion, Federico; Levorato, Marco; Zorzi, Michele (2012). "An algorithmic solution for computing circle intersection areas and its application to wireless communications". Wirel. Commun. Mobile Comput. 14 (18): 1672–1690. doi:10.1002/wcm.2305.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Lens_(geometry)&oldid=1075014536"
Piecewise-circular curves
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Angles A and B are supplementary. If mangle A= 78° what is mangle B?
Angles A and B are supplementary.If mangle A= 78° what is mangle B?
Angles A and B are supplementary.
m\mathrm{\angle }A=78°
m\mathrm{\angle }B
The measures of supplementary angles add up to
{180}^{\circ }
m\mathrm{\angle }A={78}^{\circ }
, then the measure of angle B must be
m\mathrm{\angle }B={180}^{\circ }-m\mathrm{\angle }A={180}^{\circ }-{78}^{\circ }={102}^{\circ }
m\mathrm{\angle }A=104\circ
m\mathrm{\angle }B?
m\mathrm{\angle }B=\circ degrees
T:{C}^{1}\left[-1,1\right]\to {R}^{1}
T\left(f\right)={f}^{\prime }\left(0\right)
On each of the given triangles, perform a rotation of
{180}^{\circ }
about point X. Shade the quadrilateral formed and give the most specific name for the quadrilateral in the spaces below.
{k}^{2}\text{ }+\text{ }{m}^{2}
\left[a,\text{ }b\right]\text{ }>\text{ }\left[c,\text{ }d\right]\text{ }\text{if and only if}\text{ }ab{d}^{2}\text{ }-\text{ }cd{b}^{2}\text{ }\in {D}^{+}
\text{Acoording to the definition of "greater than,"}\text{ }>\text{ }\text{is defined in Q by}\text{ }\left[a,\text{ }b\right]\text{ }>\text{ }\left[c,\text{ }d\right]\text{ }\text{if and only if}\text{ }\left[a,\text{ }b\right]\text{ }-\text{ }\left[c,\text{ }d\right]\in {Q}^{+}
:{P}_{2}\to {P}_{3}
L:\left(1\right)=1,L\left(t\right)={t}^{2},L\left({t}^{2}\right)={t}^{3}=t.
L\left(2{t}^{2}-5t=3\right).
L\left(a{t}^{2}-bt+c\right).
\text{Let}\text{ }over\to \left\{{e}_{1}\right\},over\to \left\{{e}_{2}\right\},over\to \left\{{e}_{3}\right\}\text{ }\text{be standard unit vectors along the coordinate axes in}\text{ }{R}^{3}.\text{Let S and T be the linear transformations defind in}\text{ }{R}^{3}.\text{Show that if}
S\left(over\to {\left\{e\right\}}_{1}\right)=T\left(over\to {\left\{e\right\}}_{1}\right),S\left(over\to {\left\{e\right\}}_{2}\right)=T\left(over\to {\left\{e\right\}}_{2}\right),S\left(over\to {\left\{e\right\}}_{3}\right)=T\left(over\to {\left\{e\right\}}_{3}\right)
S\left(over\to \left\{x\right\}\right)=T\left(over\to \left\{x\right\}\right)
over\to \left\{x\right\}\in {R}^{3}
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Fourier Series of the Periodic Bernoulli and Euler Functions
Cheon Seoung Ryoo, Hyuck In Kwon, Jihee Yoon, Yu Seon Jang, "Fourier Series of the Periodic Bernoulli and Euler Functions", Abstract and Applied Analysis, vol. 2014, Article ID 856491, 4 pages, 2014. https://doi.org/10.1155/2014/856491
Cheon Seoung Ryoo,1 Hyuck In Kwon,2 Jihee Yoon,2 and Yu Seon Jang 3
3Department of Applied Mathematics, Kangnam University, Yongin 446-702, Republic of Korea
We give some properties of the periodic Bernoulli functions and study the Fourier series of the periodic Euler functions which are derived periodic functions from the Euler polynomials. And we derive the relations between the periodic Bernoulli functions and those from Euler polynomials by using the Fourier series.
The numbers and polynomials of Bernoulli and Euler are very useful in classical analysis and numerical mathematics. Recently, several authors have studied the identities of Bernoulli and Euler numbers and polynomials [1–12]. The Bernoulli and Euler polynomials, , , , are defined, respectively, by the following exponential generating functions: When , these values and , , are called the Bernoulli numbers and Euler numbers, respectively [13].
Euler polynomials are related to the Bernoulli polynomials by where [3, 14].
Bernoulli polynomials and the related Bernoulli functions are of basic importance in theoretical numerical analysis. The periodic Bernoulli functions are Bernoulli polynomials evaluated at the fractional part of the argument as follows: where and is the greatest integer less than or equal to [13]. Periodic Bernoulli functions play an important role in several mathematical results such as the general Euler-McLaurin summation formula [1, 10, 15]. And also it was shown by Golomb et al. that the periodic Bernoulli functions serve to construct periodic polynomials splines on uniform meshes. For uniform meshes Delvos showed that Locher’s method of interpolation by translation is applicable to periodic -splines. This yields an easy and stable algorithm for computing periodic polynomial interpolating splines of arbitrary degree on uniform meshes via Fourier transform [15].
A Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions. Since these functions form a complete orthogonal system over , the Fourier series of a function is given by where The notion of a Fourier series can also be extended to complex coefficients [16, 17].
The complex form of the Fourier series can be written by the Euler formula, , as follows: where For a function periodic in , these become where
In this paper, we give some properties of the periodic Bernoulli functions and study the Fourier series of the periodic Euler functions which are derived periodic functions from the Euler polynomials. And we derive the relations between the periodic Bernoulli functions and those from Euler polynomials by using the Fourier series. We indebted this idea to Kim [6–9, 18–20].
2. Periodic Bernoulli and Euler Functions
The periodic Bernoulli functions can be represented as follows: satisfying From the definition of we know that for These can be rewritten as follows: by using the symbolic convention exhibited by [7].
Observe that for Since , , are periodic with period on , we have
The Apostol-Bernoulli and Apostol-Euler polynomials have been investigated by many researchers [1, 2, 10, 11]. In [1], Bayad found the Fourier expansion for Apostol-Bernoulli polynomials which are complex version of the classical Bernoulli polynomials. As a result of ordinary Bernoulli polynomials, we have the following lemma.
Lemma 1 (Bayad [1]; see also Luo [10]). The Fourier series of on is
From Lemma 1 we have the following theorem.
Theorem 2. For and one has
Proof. Since and under we have This implies the desired result.
The periodic Euler polynomials can be usually defined by and for [13]. Unlike the periodic Bernoulli, which have period , the have periodic and exhibit an even (versus odd) symmetry about zero [21].
As the above Bernoulli case, we consider the periodic Euler functions as the following: such that Then the functions , , are also periodic. From definition of Euler polynomials, we know that where is the th Euler number. These can be rewritten as follows: by using the symbolic convention exhibited by . When , these relations are given by where is Kronecker symbol and is interpreted as [9].
Remark 3. Observe that for As in the Bernoulli case, we have the following equation: This means that if is odd (even) number, then is odd (even) function.
Proof. Let be the Fourier series for , , on . Then From definition of , we have Since , if , then For , so we have the following recurrence relation: This implies that where is falling factorial. Since we have This is completion of the proof.
From Lemma 1 and Theorem 4 we have the following corollary.
Corollary 5. For one has where and is falling factorial.
Corollary 6. For one has where .
Proof. From Lemma 1, we have This becomes the desired result.
The authors would like to thank T. Kim for all the motivation and insightful conversations on this subject. The authors would also like to thank the referee(s) of this paper for the valuable comments and suggestions.
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Copyright © 2014 Cheon Seoung Ryoo 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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JEE Application of Derivatives | Brilliant Math & Science Wiki
JEE Application of Derivatives
Sandeep Bhardwaj, Prince Loomba, Arron Kau, and
This page will teach you how to master JEE Applications of Derivatives . We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.
As per JEE syllabus, the mains concepts are Derivative as Rate Measurement, Approximation, Rolle's Theorem and Lagrange's Mean Value Theorem.
Derivative as Rate Measurement
y=f(x)
\frac{dy}{dt}=f'(x) \frac{dx}{dt}
Differential of a function :
df=f'(x)dx
Linearisation of a function :
f(x)=f(x_0)+f'(x_0)(x-x_0)
Geometrical significance of approximation
Rolle's theorem and its applications
Lagrange's theorem and its applications
A ladder
\SI[per-mode=symbol]{10}{\meter}
long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate of
\SI[per-mode=symbol]{1}{\meter\per\second}
. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is
\SI[per-mode=symbol]{5}{\meter\per\second}
away from the wall ?
\begin{array}{lll} A) \, \SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second} & \quad \quad \quad \quad \quad \quad \quad & B) \, -\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second} \\ C) \, \SI[per-mode=symbol]{\sqrt{3}}{\meter\per\second} & & D) \, -\SI[per-mode=symbol]{\sqrt{3}}{\meter\per\second} \end{array}
Concepts tested: Rate measurement
\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}
Solution: We first draw the diagram for the given situation which will look like:
Now, relationship between
x
y
can be established by Pythagoras Theorem:
x^2+y^2=100
. Differentiating both sides with respect to
t
(time), we get :
2x \frac{dx}{dt}+2y\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}
x=5,
the Pythagoras Theorem gives
y=5\sqrt{3}
. Now putting the values of
x,y
\frac{dx}{dt}
\frac{dy}{dt}=-\frac{5}{5\sqrt{3}}(1)=\SI[per-mode=symbol]{-\frac{1}{\sqrt{3}}}{\meter\per\second}.
Point to be noted here is that
\frac{dy}{dt}
is negative means top of the ladder is sliding down the wall at the rate of
\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}
. So, the correct answer is
\SI[per-mode=symbol]{\frac{1}{\sqrt{3}}}{\meter\per\second}
If you considered the negative sign, then it will represent the rate with which top of the ladder will be sliding upwards the wall, while the question asks sliding rate down the wall.
If you tried to insert the values before differentiating, it will lead to zero rate, which is off course, wrong.
P(x)
and
b
(a<b)
, be two consecutive zeroes of
P(x)
a<c<b
P'(c)+10P(c)
\begin{array}{lll} A) \, 0 & \quad \quad \quad \quad \quad \quad \quad & B) \, -1 \\ C) \, 10 & & D) \, -10 \end{array}
Concepts tested: Applications of Rolle's theorem
0
Solution: Let us assume
f(x)=e^{10x}P(x)
f(a)=f(b)=0
P(a)=P(b)=0
) Also as
P(x)
is a polynomial,
f(x)
[a,b]
and differentiable in
(a,b)
. Hence Rolle's theorem can be applied to
f(x)
. Hence there exists
c \in (a,b)
f'(c)=0
f'(x)=e^{10x}[P'(x)+10P(x)]
\Rightarrow e^{10c}[P'(c)+10P(c)]=0
\Rightarrow P'(c)+10P(c)=0 \text{ for some } c \in (a,b)
\frac{1}{3}
\frac{1}{\sqrt{5}}
9
\frac{1}{9}
The ends
A
B
of a rod of length of
\sqrt{5}
are sliding along the curve
y=2x^2
x_A
x_B
x
-coordinates of the ends. At the moment when
A
(0,0)
B
(1,2)
. Find the value of the derivative
\dfrac{dx_B}{dx_A}
\lambda=\left( \frac{f(5102)-f(2015)}{f'(c)} \right) \left( \frac{f^2(2015)+f^2(5102)+f(2015)f(5102)}{f^2(c)} \right)
f:[2015,5102] \rightarrow [0,\infty)
be any continuous and differentiable function. Find the value of
\lambda
, such that there exists some
c\in [2015,5102]
f'(c)=g'(c)
2f'(c)=3g'(c)
f'(c)=2g'(c)
2f'(c)=g'(c)
If
g
are differentiable functions in
(0,1)
f(0)=2=g(1)
g(0)=0
f(1)=6
, then for some
c \in (0,1)
A man runs along a straight path at the speed of
\SI[per-mode=symbol]{4}{\meter\per\second}
. A searchlight is located on the ground
\SI[per-mode=symbol]{20}{\meter}
from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is
\SI[per-mode=symbol]{15}{\meter}
from the point on the path closest to the searchlight ?
\SI[per-mode=symbol]{0.128}{\radian\per\second}
Step 1: Drawing figure and variable selection
We draw the figure and let
x
be the distance from the man to the point on the path closest to the searchlight and
\theta
be the distance from the man to the point on the path closest to the searchlight and perpendicular to the path. The figure will look like: Step 2: Establishing relation between
x
\theta
and finding the required rate
The equation that relates
x
\theta
can be written from the figure:
\frac{x}{20}=\tan\theta \Rightarrow x=20 \tan\theta
. Now differentiating both sides with respect to
t
\frac{dx}{dt}=\SI[per-mode=symbol]{20}{\second\squared}\theta \frac{d\theta}{dt} \Rightarrow \frac{d\theta}{dt}=\frac{\cos^2\theta}{20} \frac{dx}{dt}
Step 3: Putting the values
We're given with
\frac{dx}{dt}=\SI[per-mode=symbol]{4}{\meter\per\second}
x=\SI[per-mode=symbol]{15}{\meter}
, the length of the beam becomes
\SI[per-mode=symbol]{25}{\meter}
using Pythagoras theorem. So,
\cos\theta=\frac{4}{5}
\Rightarrow \frac{d\theta}{dt}=\frac{1}{20} (\frac{4}{5})^2 (4)=\SI[per-mode=symbol]{0.128}{\radian\per\second}
If you didn't the value of
\frac{dx}{dt}
, it will give you linear rate, not the rate at which searchlight is rotating.
f(x)
be a non-constant thrice differentiable function defined on real numbers such that
f(x)=f(6-x)
f'(0)=0=f'(2)=f'(5)
. Find the minimum number of values of
p \in [0,6]
which satisfy the equation
(f''(p))^2+f'(p)f'''(p)=0
f'(p)=\left( \frac{df(x)}{dx} \right)_{x=p}
f''(p)=\left( \frac{d^2f(x)}{dx^2} \right)_{x=p}
f'''(p)=\left( \frac{d^3f(x)}{dx^3} \right)_{x=p}
Once a couple was caught in an unending argument due to some unknown reasons. They got irritated from each other and decided to move away. They started moving with velocities
\SI[per-mode=symbol]{\sqrt{2+\sqrt{2}}}{\meter\per\second}
at the same time from the junction of two roads inclined to
45^\circ
to each other. If they travel by different roads, find the rate at which they are being separated?
Cite as: JEE Application of Derivatives. Brilliant.org. Retrieved from https://brilliant.org/wiki/application-of-derivatives/
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Write a proof for the following \overline{AB}\cong \overline{DE} and C
Write a proof for the following \overline{AB}\cong \overline{DE} and C is the midpoint of both \overline{AB}\ and\ \overline{BE}
Write a proof for the following:
\stackrel{―}{AB}\stackrel{\sim }{=}\stackrel{―}{DE}
and C is the midpoint of both
\stackrel{―}{AB}\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\stackrel{―}{BE}
\mathrm{△}ABC\stackrel{\sim }{=}\mathrm{△}DEC
curwyrm
AB\stackrel{\sim }{=}DE
and C is the mid point of both AD and BE.
AC=DC...(1) (
\because
C is mid point of AD)
CE=BC...(2) (
\because
C is the mid point of BE)
\mathrm{\angle }ACB=\mathrm{\angle }BCE
...(3) (
\because
vertically opposite
\mathrm{\angle }s
AB\stackrel{\sim }{=}DE
...(4) (Given)
Therefore from (1), (2) and (4)
\mathrm{△}ABC\stackrel{\sim }{=}\mathrm{△}DEC
(by SSS similarity)
Also from (1), (2) and (4)
\mathrm{△}ABC\stackrel{\sim }{=}\mathrm{△}DEC
(by SAS similarity)
y=2x+3
DE=\sqrt{AD\cdot EB}
To determine: Whether the triangle ABC and GHI are similar to each other.
Triangle ABC that is 75% of its corresponding side in triangle DEF.
Triangle GHI that is 32% of its corresponding side in triangle DEF.
Determine whatever each pair if figures is similar. If so, write the similarity statement and scale factor. If not, explain your reasoning.
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Merge_algorithm Knowpia
Merge algorithms are a family of algorithms that take multiple sorted lists as input and produce a single list as output, containing all the elements of the inputs lists in sorted order. These algorithms are used as subroutines in various sorting algorithms, most famously merge sort.
An example for merge sort
The merge algorithm plays a critical role in the merge sort algorithm, a comparison-based sorting algorithm. Conceptually, the merge sort algorithm consists of two steps:
Recursively divide the list into sublists of (roughly) equal length, until each sublist contains only one element, or in the case of iterative (bottom up) merge sort, consider a list of n elements as n sub-lists of size 1. A list containing a single element is, by definition, sorted.
Repeatedly merge sublists to create a new sorted sublist until the single list contains all elements. The single list is the sorted list.
The merge algorithm is used repeatedly in the merge sort algorithm.
An example merge sort is given in the illustration. It starts with an unsorted array of 7 integers. The array is divided into 7 partitions; each partition contains 1 element and is sorted. The sorted partitions are then merged to produce larger, sorted, partitions, until 1 partition, the sorted array, is left.
Merging two listsEdit
Merging two sorted lists into one can be done in linear time and linear or constant space (depending on the data access model). The following pseudocode demonstrates an algorithm that merges input lists (either linked lists or arrays) A and B into a new list C.[1][2]: 104 The function head yields the first element of a list; "dropping" an element means removing it from its list, typically by incrementing a pointer or index.
algorithm merge(A, B) is
inputs A, B : list
C := new empty list
while A is not empty and B is not empty do
if head(A) ≤ head(B) then
append head(A) to C
drop the head of A
append head(B) to C
drop the head of B
// By now, either A or B is empty. It remains to empty the other input list.
while A is not empty do
while B is not empty do
When the inputs are linked lists, this algorithm can be implemented to use only a constant amount of working space; the pointers in the lists' nodes can be reused for bookkeeping and for constructing the final merged list.
In the merge sort algorithm, this subroutine is typically used to merge two sub-arrays A[lo..mid], A[mid+1..hi] of a single array A. This can be done by copying the sub-arrays into a temporary array, then applying the merge algorithm above.[1] The allocation of a temporary array can be avoided, but at the expense of speed and programming ease. Various in-place merge algorithms have been devised,[3] sometimes sacrificing the linear-time bound to produce an O(n log n) algorithm;[4] see Merge sort § Variants for discussion.
K-way mergingEdit
k-way merging generalizes binary merging to an arbitrary number k of sorted input lists. Applications of k-way merging arise in various sorting algorithms, including patience sorting[5] and an external sorting algorithm that divides its input into k = 1/M − 1 blocks that fit in memory, sorts these one by one, then merges these blocks.[2]: 119–120
Several solutions to this problem exist. A naive solution is to do a loop over the k lists to pick off the minimum element each time, and repeat this loop until all lists are empty:
Input: a list of k lists.
Loop over the lists to find the one with the minimum first element.
Output the minimum element and remove it from its list.
In the worst case, this algorithm performs (k−1)(n−k/2) element comparisons to perform its work if there are a total of n elements in the lists.[6] It can be improved by storing the lists in a priority queue (min-heap) keyed by their first element:
Re-heapify h.
Searching for the next smallest element to be output (find-min) and restoring heap order can now be done in O(log k) time (more specifically, 2⌊log k⌋ comparisons[6]), and the full problem can be solved in O(n log k) time (approximately 2n⌊log k⌋ comparisons).[6][2]: 119–120
A third algorithm for the problem is a divide and conquer solution that builds on the binary merge algorithm:
If k = 1, output the single input list.
If k = 2, perform a binary merge.
Else, recursively merge the first ⌊k/2⌋ lists and the final ⌈k/2⌉ lists, then binary merge these.
When the input lists to this algorithm are ordered by length, shortest first, it requires fewer than n⌈log k⌉ comparisons, i.e., less than half the number used by the heap-based algorithm; in practice, it may be about as fast or slow as the heap-based algorithm.[6]
Parallel mergeEdit
A parallel version of the binary merge algorithm can serve as a building block of a parallel merge sort. The following pseudocode demonstrates this algorithm in a parallel divide-and-conquer style (adapted from Cormen et al.[7]: 800 ). It operates on two sorted arrays A and B and writes the sorted output to array C. The notation A[i...j] denotes the part of A from index i through j, exclusive.
algorithm merge(A[i...j], B[k...ℓ], C[p...q]) is
inputs A, B, C : array
i, j, k, ℓ, p, q : indices
let m = j - i,
n = ℓ - k
swap A and B // ensure that A is the larger array: i, j still belong to A; k, ℓ to B
swap m and n
if m ≤ 0 then
return // base case, nothing to merge
let r = ⌊(i + j)/2⌋
let s = binary-search(A[r], B[k...ℓ])
let t = p + (r - i) + (s - k)
C[t] = A[r]
merge(A[i...r], B[k...s], C[p...t])
merge(A[r+1...j], B[s...ℓ], C[t+1...q])
The algorithm operates by splitting either A or B, whichever is larger, into (nearly) equal halves. It then splits the other array into a part with values smaller than the midpoint of the first, and a part with larger or equal values. (The binary search subroutine returns the index in B where A[r] would be, if it were in B; that this always a number between k and ℓ.) Finally, each pair of halves is merged recursively, and since the recursive calls are independent of each other, they can be done in parallel. Hybrid approach, where serial algorithm is used for recursion base case has been shown to perform well in practice [8]
The work performed by the algorithm for two arrays holding a total of n elements, i.e., the running time of a serial version of it, is O(n). This is optimal since n elements need to be copied into C. To calculate the span of the algorithm, it is necessary to derive a Recurrence relation. Since the two recursive calls of merge are in parallel, only the costlier of the two calls needs to be considered. In the worst case, the maximum number of elements in one of the recursive calls is at most
{\textstyle {\frac {3}{4}}n}
since the array with more elements is perfectly split in half. Adding the
{\displaystyle \Theta \left(\log(n)\right)}
cost of the Binary Search, we obtain this recurrence as an upper bound:
{\displaystyle T_{\infty }^{\text{merge}}(n)=T_{\infty }^{\text{merge}}\left({\frac {3}{4}}n\right)+\Theta \left(\log(n)\right)}
{\displaystyle T_{\infty }^{\text{merge}}(n)=\Theta \left(\log(n)^{2}\right)}
, meaning that it takes that much time on an ideal machine with an unbounded number of processors.[7]: 801–802
Note: The routine is not stable: if equal items are separated by splitting A and B, they will become interleaved in C; also swapping A and B will destroy the order, if equal items are spread among both input arrays. As a result, when used for sorting, this algorithm produces a sort that is not stable.
Some computer languages provide built-in or library support for merging sorted collections.
The C++'s Standard Template Library has the function std::merge, which merges two sorted ranges of iterators, and std::inplace_merge, which merges two consecutive sorted ranges in-place. In addition, the std::list (linked list) class has its own merge method which merges another list into itself. The type of the elements merged must support the less-than (<) operator, or it must be provided with a custom comparator.
C++17 allows for differing execution policies, namely sequential, parallel, and parallel-unsequenced.[9]
Python's standard library (since 2.6) also has a merge function in the heapq module, that takes multiple sorted iterables, and merges them into a single iterator.[10]
Join (Unix)
^ a b Skiena, Steven (2010). The Algorithm Design Manual (2nd ed.). Springer Science+Business Media. p. 123. ISBN 978-1-849-96720-4.
^ a b c Kurt Mehlhorn; Peter Sanders (2008). Algorithms and Data Structures: The Basic Toolbox. Springer. ISBN 978-3-540-77978-0.
^ Katajainen, Jyrki; Pasanen, Tomi; Teuhola, Jukka (1996). "Practical in-place mergesort". Nordic J. Computing. 3 (1): 27–40. CiteSeerX 10.1.1.22.8523.
^ Kim, Pok-Son; Kutzner, Arne (2004). Stable Minimum Storage Merging by Symmetric Comparisons. European Symp. Algorithms. Lecture Notes in Computer Science. Vol. 3221. pp. 714–723. CiteSeerX 10.1.1.102.4612. doi:10.1007/978-3-540-30140-0_63. ISBN 978-3-540-23025-0.
^ Chandramouli, Badrish; Goldstein, Jonathan (2014). Patience is a Virtue: Revisiting Merge and Sort on Modern Processors. SIGMOD/PODS.
^ a b c d Greene, William A. (1993). k-way Merging and k-ary Sorts (PDF). Proc. 31-st Annual ACM Southeast Conf. pp. 127–135.
^ a b Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2009) [1990]. Introduction to Algorithms (3rd ed.). MIT Press and McGraw-Hill. ISBN 0-262-03384-4.
^ Victor J. Duvanenko (2011), "Parallel Merge", Dr. Dobb's Journal
^ "std:merge". cppreference.com. 2018-01-08. Retrieved 2018-04-28.
^ "heapq — Heap queue algorithm — Python 3.10.1 documentation".
Donald Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89685-0. Pages 158–160 of section 5.2.4: Sorting by Merging. Section 5.3.2: Minimum-Comparison Merging, pp. 197–207.
High Performance Implementation of Parallel and Serial Merge in C# with source in GitHub and in C++ GitHub
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Eliminate variables from rational equations - MATLAB eliminate
Eliminate Variables from Rational Equations
Eliminate Variables from Polynomial Equations
expr = eliminate(eqns,vars)
expr = eliminate(eqns,vars) eliminates the variables vars from the rational equations eqns. The result is a vector of symbolic expressions that is equal to zero.
Create two rational equations that contain the variables x and y.
eqns = [x*y/(x-2) + y == 5/(y - x), y-x == 1/(x-1)]
\left(\begin{array}{cc}y+\frac{x y}{x-2}=-\frac{5}{x-y}& y-x=\frac{1}{x-1}\end{array}\right)
Eliminate the variable x. The result is a symbolic expression that is equal to zero.
expr = eliminate(eqns,x)
\left[6 {y}^{2}-5 y-75\right]
Create two polynomial equations that contain the variables x and y.
eqns = [2*x+y == 5; y-x == 1]
\left(\begin{array}{c}2 x+y=5\\ y-x=1\end{array}\right)
Eliminate the variable x from the equations. The result is a symbolic expression that is equal to zero.
\left[3 y-7\right]
Now, create three polynomial equations that contain the variables x, y, and z. Eliminate the variable x. The result is a vector of symbolic expressions that is equal to zero.
eqns = [x^2 + y-z^2 == 2;
x - z == y;
x^2 + y^2-z == 4];
\left[5 {z}^{3}-5 {z}^{2}-8 z+4 y-8,5 {z}^{4}-11 {z}^{2}-18 z-8\right]
To eliminate both x and y, use the eliminate function and specify the two variables as the vector [x y].
expr = eliminate(eqns,[x y])
\left[5 {z}^{4}-11 {z}^{2}-18 z-8\right]
eqns — Rational equations
Rational equations, specified as a vector of symbolic equations or symbolic expressions. A rational equation is an equation that contains at least one fraction in which the numerator and the denominator are polynomials.
The relation operator == defines symbolic equations. If a symbolic expression eqn in eqns has no right side, then a symbolic equation with a right side equal to 0 is assumed.
vars — Variables to eliminate
Variables to eliminate, specified as a vector of symbolic variables.
gbasis | solve
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What is the frequency of a X-ray photon whose momentum is 1.1 x 10-23 kg-ms-2 ?
\mathrm{\nu }
\frac{{\mathrm{mc}}^{2}}{\mathrm{h}}
\frac{1.1\quad \times \quad {10}^{-23}\quad \times \quad 3\quad \times \quad {10}^{8}}{6.6\quad \times \quad {10}^{-34}}
= 5 x 1018 Hz
The bond energy is the energy required to :
dissociate one mole of the substance
dissociate bond in 1 kg of the substance
break one mole of similar bonds
break bonds in one mole of the substance
Bond energy : It may be defined as the energy released when two atoms get bonded or it is equal to the energy required to break the bond to form the neutral atoms .
e.g , H-H (g)
\to
H + H ,
∆
H = + 103 Kcal/mol
The magnitude of orbital angular momentum of an electron of azimuthal quantum number 2 is :
\frac{2\mathrm{h}}{2\mathrm{\pi }}
\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}
\frac{\sqrt{2}\mathrm{h}}{2\mathrm{\pi }}
\frac{6\mathrm{h}}{2\mathrm{\pi }}
\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}
Orbital angular momentum =
\sqrt{\mathrm{l}\quad (\mathrm{l}\quad +\quad 1\quad )}
\frac{\mathrm{h}}{2\mathrm{\pi }}
where l = azimuthal quantum number
\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}
Heat of neutralisation of an acid with a base is 13. 7 kcal when :
both acid and base are weak
acid is weak and base is strong
both acid and base are strong
acid is strong and base is weak
Heat of neutralisation for strong acid and strong base is always close to 13.7 kcal .
\underset{\mathrm{strong}\quad \mathrm{acid}}{\mathrm{HA}}
\underset{\mathrm{strong}\quad \mathrm{base}}{\mathrm{BOH}}
⇌
H2O + AB
∆
H = - 13.7 Kcal
The heat of neutralisation of strong acid and strong base is actually the heat of formation of 1 gmole of water from 1 g of H+ and 1 g of OH- .
Hess's law is based on :
Hess's law is based upon conservation of energy .
Heisenberg's uncertainty principle can be explained as :
∆
⩾
\frac{∆\mathrm{p}\quad \times \quad \mathrm{h}}{4\mathrm{\pi }}
\mathrm{∆}
∆
⩾
\frac{\mathrm{h}}{4\mathrm{\pi }}
∆
∆
⩾
\frac{\mathrm{h}}{\mathrm{\pi }}
∆
⩾
\frac{\mathrm{\pi h}}{∆\mathrm{x}}
\mathrm{∆}
∆
⩾
\frac{\mathrm{h}}{4\mathrm{\pi }}
Heisenberg's uncertainty principle : It is not possible to determine precisely both the position and the momentum (or velocity) of a small moving particle (e.g , electron , proton etc)
∆
∆
⩾
\frac{\mathrm{h}}{4\mathrm{\pi }}
∆
∆
p are the uncertainties with regard to position , momentum respectively .
The solubility product of Mg(OH)2 at 25°C is 1.4 x 10-11 .What is the solubility of Mg(OH)2 in g/L ?
⇌
Ksp = [Mg2+][OH-]2
1.4 X 10-11 = (s)(2s)2
1.4 X 10-11 = 4s3
s = 1.5 x 10-4 mol/L
\because
Mol. wt. of Mg(OH)2 = 52)
= 58 X 1.5 x 10-4
= 0.0087 g/L
The pH of a 0.02 M solution of HCl is :
0.02 M HCl = 2 x 10-2 M HCl
= - log 2 + 2 log 10
= - 0.3010 + 2
The heat of formarion of CO2 is - 393 kJ mol-1 .The amount of heat evolved in the formation of 0.156 kg of CO2 .
+ 1165.5 kJ
- 1165.5 kJ
\because
Heat evolved during the formation of 44 g of CO2 = - 93kJ
\therefore
Heat evolved in the formation of 0.156 kg of CO2
\frac{-\quad 393\quad \times \quad 156}{44}
= - 1393 kJ
The geometry of methane molecule is :
Methane (CH3) shows the sp3 hybridisation .So , the geometry of methane molecule is tetrahedral .
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Math of Voting Warmup Practice Problems Online | Brilliant
Based on the results above, from a survey of basketball fans about the best player of all time, who did more fans vote for? (Note: Assume the number of voters under 30 was the same as the number of voters 30 or over.)
Lebron MJ Not enough info
Three states are being allocated 62 seats in a legislative body. State A has 17% of the population, State B has 37%, and State C has 46%.
Applying these percentages to the 62 seats and rounding to the nearest integer means State A gets 11 seats, State B gets 23 seats, and State C gets 29 seats.
The algorithm will generally be biased towards larger populations. The algorithm will generally be biased towards smaller populations. The number of seats doesn't add up to 62.
In the map above, the Blue party wins none of the five voting regions. If voting regions must contain 5 connected squares, what is the most voting regions the Blue party could win?
For example, if the voting regions were drawn like this, they would win 1:
A political score
P
is a number between 0 and 100 where
P=0
represents the extreme left-wing and
P=100
represents the extreme right-wing. A candidate's positions are also assigned a score, and each voter will vote for whichever candidate has positions with a score closest to their own.
In a two-candidate race where each candidate wants to get as many votes as possible, what position(s) will they take?
Note: In the case where two candidates are equidistant from a voter, assume the voter decides randomly.
They will take positions corresponding to political scores of 0 and 100 They will take positions corresponding to political scores of 25 and 75 They will both take positions corresponding to the median voter political score They will both take positions corresponding to the average voter political score
There are four candidates in the beginning of an election: 1, 2, 3, and 4. Given the preferences of the four voters below, the only head-to-head race that will not end in a tie is between candidates numbered
A
B.
A \times B?
\begin{aligned} \text{Alexa:} \ 1 &> 2 > 3 > 4 \\ \text{Benny:} \ 2 &> 1> 4> 3 \\ \text{Celina:} \ 3 &> 4 > 2 > 1 \\ \text{Dakota:} \ 4&>1>3>2 \\ \end{aligned}
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For an electron , if the uncertainty in velocity is
∆
v , the uncertainty in its position (
∆
x) is given by :
\frac{\mathrm{hm}}{4\mathrm{\pi }∆\mathrm{\nu }}
\frac{4\mathrm{\pi }}{\mathrm{hm}∆\mathrm{\nu }}
\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}
\frac{4\mathrm{\pi m}}{\mathrm{h}∆\mathrm{\nu }}
\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}
∆
∆
⩾
\frac{\mathrm{h}}{4\mathrm{\pi }}
∆
∆\mathrm{\nu }
∆
x . m
∆
⩾
\frac{\mathrm{h}}{4\mathrm{\pi }}
∆
\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}
The enthalpy change (
∆
H) for the neutralisation of M HCl by caustic potash in dilute solution at 298 K is :
\underset{(\mathrm{strong}\quad \mathrm{acid})}{\mathrm{HCl}}
\underset{(\mathrm{strong}\quad \mathrm{base})}{\mathrm{KOH}}
\stackrel{\mathrm{neutralisation}}{⇌}
KCl + H2O
In this reaction , HCl is the strong acid and KOH is the base and it has been found that the heat of neutralisation of a strong acid with a strong base is always constant i.e. , 57.3 kJ .
A metal surface is exposed to solar radiations :
the emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations
the emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiations
the emitted electrons have zero energy
the emitted electrons have energy equal to energy of photons of incident light
A metal surface is exposed to solar radiations ,the emitted electrons have energy less than maximum value of energy depending upon freauency of incident radiations .
The Ksp of Mg(OH)2 is 1 x 10-12 , 0.01 M Mg(OH)2 will precipitate at the limited pH :
⇌
Mg2+ + 2OH-
the solubility product Ksp of
Mg(OH)2 = [Mg2+][OH-]2
1 x 10-12 = 0.01 [OH-]2
[OH-]2 = 1 x 10-10
[H+][OH-] = 10-14
[H+][10-5] = 10-14
[H] = 10-14/10-5 = 10-9
pH = - log[H+] = - log 10-9 = 9
Which of the following transitions have minimum wavelengths ?
\to
\to
\to
\to
\to
\frac{\mathrm{hc}}{\mathrm{\lambda }}
or E
\propto
\frac{1}{\mathrm{\lambda }}
Thus a decrease in wave length represents an increase in energy For n4
\to
n1 transition .These are greater the. energy difference and lesser will be wavelength .Thus , for n4
\to
n1 transition has minimum wavelength .
A radioactive sample is emitting 64 times radiations than non-hazardous limit .If its half-life is 2 hours , after what time it becomes non- hazardous ?
Nt = No x
{\left(\frac{1}{2}\right)}^{\mathrm{n}}
where Nt = amount left after expiry of 'n' half lives
No = initial amount
n = number of half lives elapsed
\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_{\mathrm{o}}}
{\left(\frac{1}{2}\right)}^{\mathrm{n}}
\frac{1}{64}
{\left(\frac{1}{2}\right)}^{\mathrm{n}}
{\left(\frac{1}{2}\right)}^{6}
{\left(\frac{1}{2}\right)}^{\mathrm{n}}
Time taken (T) = t1/2 x n = 2 x 6 = 12 h
Orbital is :
circular path around the nucleus in which the electrons revolves
space around the nucleus where the probability of finding the electron is maximum
amplitude of electron wave
An atomic orbital is a three dimensional region of definite shape around the nucleus where the probability of finding the electron is maximum .Therefore , an atom has a both characteristic energy and a characteristic shape .
Which of the following sequence is correct as per Aufbau principle ?
3s < 3d < 4s < 4p
ls < 2p < 4s < 3d
2s < 5s < 4p < 5d
2s < 5s < 4p < 5f
According to aufbau principle , "sub-shells are filled with electrons in the increasing order of their energies , "i.e. , sub-shell of lower energy will be filled first with electrons .
Thus correct order is -
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d ... etc
Ionic compounds are formed most easily with :
low electron affinity , high ionisation energy
high electron affinity , low ionisation energy
low electron affinity , low ionisation energy
high electron affinity , high ionisation energy
An ionic bond is formed between an electropositive metal and a electronegative non metal .The electropositive metal converts into a cation , while the electronegative non metal converts into a anion .
Thus the formation ofionic bond is favoured by
(I) Low ionisation potential of metal
(II) Greater valtie of electron affinity of non metal
(III) Higher value of lattice energy of the resulting ionic compound .
Equation of Boyle's law is :
\frac{\mathrm{dP}}{\mathrm{P}}
\frac{\mathrm{dV}}{\mathrm{V}}
\frac{\mathrm{dP}}{\mathrm{P}}
= +
\frac{\mathrm{dV}}{\mathrm{V}}
\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}
\frac{\mathrm{dV}}{\mathrm{dT}}
\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}
\frac{{\mathrm{d}}^{2}\mathrm{V}}{\mathrm{dT}}
\frac{\mathrm{dP}}{\mathrm{P}}
\frac{\mathrm{dV}}{\mathrm{V}}
According to Boyle's law , "for a given mass of a gas , at constant temperature the volume of a gas is inversely proportional to its pressure" .
\propto
\frac{1}{\mathrm{P}}
or PV = constant
on differentiating the equation
d(PV) = d (constant)
= PdV + VdP =0
= VdP = - PdV
\frac{\mathrm{dP}}{\mathrm{P}}
\frac{\mathrm{dV}}{\mathrm{V}}
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Heisenberg Uncertainty Principle | Brilliant Math & Science Wiki
Andrew Ellinor, Matt DeCross, Satyabrata Dash, and
The Heisenberg Uncertainty Principle is a relationship between certain types of physical variables like position and momentum, which roughly states that you can never simultaneously know both variables exactly. Informally, this means that both the position and momentum of a particle in quantum mechanics can never be exactly known.
Mathematically, the Heisenberg uncertainty principle is a lower bound on the product of uncertainties of a pair of conjugate variables. The most well-known expression takes the position and momentum to be the conjugate variables:
\sigma_x \sigma_p \geq \frac{\hbar}{2}.
Suppose the position of a particle is known to very high precision, so that
\sigma_x
is very small. Then the uncertainty principle shows that
\sigma_p
must be large, i.e. the momentum is not known precisely. Furthermore, neither uncertainty can vanish, so neither position nor momentum can ever be exactly measured.
Top: states with a high position uncertainty (red) have a small momentum uncertainty (blue) and vice versa (bottom).
States that saturate the inequality in the Heisenberg uncertainty principle are called squeezed states, of which coherent states like the photon states in a laser are a subclass.
Derivation of the Uncertainty Principle
Generalization: Energy-Time Uncertainty
The uncertainty principle relates the standard deviations of two conjugate variables, which are any two variables related to each other by the Fourier transform. For example, the wavefunction of a free particle is:
\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}dp,
\phi (p)
are states of definite momentum, i.e. eigenstates of the momentum operator.
\Phi(p,t)={\phi(p)}{e}^{iEt/ \hbar}
, the integral becomes
\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Phi(p,t)}{e}^{ipx/ \hbar}dp.
Therefore, the position-space wavefunction
\Psi(x,t)
is indeed the Fourier transform of another wavefunction that is dependent on momentum instead. This wavefunction,
\Phi(p,t)
, is called the momentum-space wavefunction.
The momentum-space wavefunction can be recovered given the position-space wavefunction by taking the inverse Fourier transform:
\Phi(p,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Psi(x,t)}{e}^{-ipx/ \hbar}dx.
This Fourier analysis achieves the interesting result of displaying the direct relationship between two variables
x
p
that describe the same wave-packet in different spaces. In this example, a free particle is measured in the position space by
\Psi(x,t)
, but measured in the momentum space by
\Phi(p,t)
. Regardless of which space one chooses to measure the particle, the physics doesn't change. Therefore, this relationship of conjugate variables allows one to measure particles in physical experiments in two ways. In fact, much of crystallography and solid-state physics relies on measurements made in momentum space in lieu of position space.
\sigma_x
of some variable
x
\sigma_{x}^{2} = \left\langle{(x-\langle x \rangle)}^{2}\right\rangle = \left\langle x^2 \right\rangle - \langle x \rangle^2.
\sigma_x^2
gives the average squared difference from the mean
\langle x \rangle
The uncertainty principle holds for any quantum mechanical state
|\Psi\rangle
. To derive the uncertainty principle thus requires consideration of the expectation value of
(x - \langle x \rangle)^2
in some arbitrary state
|\Psi\rangle
. This expectation value is given by:
{\sigma}_{x}^{2}=\left\langle\Psi\left|\left(\hat{x}-\langle\hat{x}\rangle\right)^2\right|\Psi\right\rangle.
x
is Hermitian, setting
f=(\hat{x}-\langle\hat{x}\rangle)\Psi
{\sigma}_{x}^{2}=\langle f|f\rangle
Similarly, the momentum uncertainty is written:
{\sigma}_{p}^{2}=\left\langle\Psi \left|\left(\hat{p}-\langle\hat{p}\rangle\right)^2 \right|\Psi\right\rangle =\langle g|g\rangle,
g=(\hat{p}-\langle\hat{p}\rangle)\Psi
A clever application of the Cauchy-Schwarz inequality gives the rough form of the Heisenberg uncertainty principle:
{\sigma}_{x}^{2}{\sigma}_{p}^{2} = \langle f|f\rangle\langle g|g\rangle \ge {|\langle f|g\rangle |}^{2}.
To obtain the final expression, compute the right-hand side
\langle f|g\rangle
\begin{aligned} \langle f|g\rangle &=\langle(\hat{x}-\langle\hat{x}\rangle)\Psi|(\hat{p}-\langle \hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|(\hat{x}-\langle\hat{x}\rangle)(\hat{p}-\langle\hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|(\hat{x}\hat{p}-\hat{x}\langle\hat{p}\rangle)-\hat{p}\langle\hat{x}\rangle+\langle\hat{x}\rangle\langle\hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|\hat{x}\hat{p}\Psi\rangle-\langle p\rangle\langle\Psi|\hat{x}\Psi\rangle-\langle x\rangle\langle\Psi|\hat{p}\Psi \rangle+\langle x\rangle\langle p\rangle\langle\Psi|\Psi\rangle\\ &=\langle\hat{x}\hat{p}\rangle-\langle\hat{p}\rangle\langle\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle+\langle\hat{x}\rangle\langle\hat{p}\rangle\\ &=\langle\hat{x}\hat{p}\rangle-\langle\hat{p}\rangle\langle\hat{x}\rangle. \end{aligned}
Taking the complex conjugate,
\langle g|f\rangle=\langle\hat{p}\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle.
Using the following property of complex numbers:
{z}^{*}z \ge {\left[\frac{1}{2i}(z-{z}^{*})\right]}^{2},
z
\langle f|g\rangle
{z}^{*}
\langle g|f\rangle
{\sigma}_{x}^{2}{\sigma}_{p}^{2}\ge {\left[\frac{1}{2i}(\langle f|g\rangle-\langle g|f\rangle)\right]}^{2}.
Substituting back in for all of the inner products previously computed, this translates to
{\sigma}_{x}^{2}{\sigma}_{p}^{2} \ge {\left|\frac{1}{2i}\left\langle [\hat{x},\hat{p}]\right\rangle \right|}^{2}.
This is the generalized Heisenberg uncertainty principle. Note that no part of the above derivation uses the fact that
x
p
represent position and momentum, only the fact that they are conjugate variables. Therefore, the above inequality holds for any two conjugate variables.
The canonical commutation relation gives
[\hat{x},\hat{p}]=i\hbar
. Substituting in above and taking the square root yields the most well-known form of the Heisenberg uncertainty principle:
{\sigma}_{x}{\sigma}_{p} \ge \frac{\hbar}{2}.
\frac{\hbar}{2}
\hbar
\hbar \sqrt{\text{ abs}(\langle \hat{p}^2 \rangle )}
\hbar \text{ abs}(\langle \hat{p} \rangle )
Which of the following gives the correct lower bound
B
in the uncertainty relation for the operators
\hat{x}
\hat{p}^2
The relevant uncertainty relation is
\sigma_x \sigma_{p^2} \geq B.
Another uncertainty relation which is often referenced in discussion of quantum mechanics is the energy-time uncertainty principle,
\sigma_E \sigma_t \geq \frac{\hbar}{2}.
It is tempting to interpret this equation as the statement that a system may fluctuate in energy by an arbitrarily large amount over a sufficiently short time scale. This explanation is often given as a description for particle-antiparticle production and annihilation, where a particle and its antiparticle appear spontaneously from the vacuum briefly via "borrowed" energy before colliding and returning to vacuum. However, this explanation is not very precise and the given inequality is not so well-defined in quantum mechanics despite the nice physical interpretation. The reason it is not well-defined is because there is no operator in quantum mechanics corresponding to the measurement of time, although the Hamiltonian is the operator corresponding to energy. Nevertheless, as explained in this section, there is some way to make sense of an energy-time uncertainty principle by considering how the measurement of an arbitrary operator changes in time.
Since time is not an operator, it is unclear how time enters quantum mechanics at all. The answer is that time is incorporated into the Schrödinger equation, where it describes the time rate of change of a wavefunction. Physically, the passage of time is recorded by noting that certain physical observables are changing over time: for instance, perhaps the position of a particle is changing, which one interprets as motion over time, or the momentum of a particle is changing, which one interprets as accelerating or decelerating over time.
To quantify this statement, consider the Ehrenfest theorem governing the dynamics of the expectation value of an operator
\hat{A}
in terms of the commutator with the Hamiltonian:
\frac{d}{dt}\left\langle\hat{A}\right\rangle =\frac{i}{\hbar}\left\langle\left[\hat{H},\hat{A}\right]\right\rangle +\left\langle\frac{\partial \hat{A}}{\partial t}\right\rangle.
\hat{A}
does not depend explicitly on time (which is typically the case), the second term vanishes and the expectation value of the
\left\langle \left[\hat{H} ,\hat{A}\right]\right\rangle
is generally nonzero. Therefore,
\hat{H}
\hat{A}
satisfy the generalized Heiseinberg uncertainty relation:
\sigma_H^2 \sigma_A^2 \geq {\left|\frac{1}{2i}\left\langle \left[\hat{H},\hat{A}\right]\right\rangle \right|}^{2} = {\left|\frac{1}{2i} \frac{\hbar}{i} \frac{d\left\langle \hat{A} \right\rangle}{dt} \right|}^{2} = \frac{\hbar^2}{4} \left| \frac{d\left\langle \hat{A} \right\rangle}{dt} \right|^2
Notably, since the Hamiltonian is the energy operator,
\sigma_H
corresponds to the uncertainty in energy.
Taking square roots now gives the relation:
\sigma_H \sigma_A \geq \frac{\hbar}{2} \left|\frac{d\left\langle \hat{A} \right\rangle}{dt} \right|.
\sigma_t
\sigma_A = \left|\frac{d\left\langle \hat{A} \right\rangle}{dt} \right| \sigma_t.
\sigma_t
corresponds to the time it takes the measured value of
\left\langle \hat{A} \right\rangle
to shift by
\sigma_A
, for any operator
\hat{A}
Given this definition, substituting in yields the relation:
\sigma_H \sigma_t \geq \frac{\hbar}{2},
which is a more rigorous expression of the energy-time uncertainty principle. Formally, using the definition of
\sigma_t
one can see that if the energy changes slowly, the rate of change of all expectation values of any observable operator must also be slow. This makes sense because if the energy (i.e., the Hamiltonian) changes slowly, the wavefunction itself that is evolved by the Hamiltonian changes slowly, so the expectation value of observables corresponding to measurements of that wavefunction must change slowly as well.
3.3 \times 10^{-3}
3.3 \times 10^{-26}
3.3 \times 10^{12}
3.3 \times 10^{-12}
Suppose an exotic particle with a very large mass of
20 \pm 0.01 \text{ TeV}
is measured at the Large Hadron Collider. What is the correct lower bound for the lifetime of this exotic particle given no additional information, in seconds? (What does this say about the detectability of very heavy particles?)
\frac{5\hbar}{\mu B}
\frac{\hbar}{2\mu B}
\frac{2\hbar}{\mu B}
\frac{\hbar}{10 \mu B}
The energy of an electron in a magnetic field
B
is, in some unit system:
E = \pm \frac{1}{2} \mu B,
where choice of positive or negative sign corresponds to spin-down or spin-up respectively, and
\mu
is the spin magnetic moment of the electron.
In an adiabatic transition, the parameters of a quantum system are gradually changed to bring a system smoothly from one state to another state. Suppose an electron starts in the spin-up ground state in a magnetic field of strength
B
. The magnetic field is then reduced slowly to strength
\frac{B}{10}
and then increased slowly again back to strength
B
. Find the minimum time for the process of tuning the magnetic field to occur for which the electron is expected to remain in the spin-up ground state after the process ends. Hint: consider the energy-time uncertainty principle.
Note: this is a very simple demonstration of the fact that adiabatically tuning electron spins requires relatively long time scales.
Cite as: Heisenberg Uncertainty Principle. Brilliant.org. Retrieved from https://brilliant.org/wiki/heisenberg-uncertainty-principle/
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Spark-ignition engine model using lookup tables - Simulink - MathWorks 日本
Æ’(Tcmd,N) Æ’(Tcmd,N,TempEng)
Æ’(Tbrake,N) Æ’(Tbrake,N,TempEng)
\begin{array}{l}{M}_{Nom}=\frac{{P}_{std}{V}_{d}}{{N}_{cyl}{R}_{air}{T}_{std}}\\ L=\frac{\left(\frac{60s}{min}\right)Cpsâ
{\stackrel{Ë}{m}}_{air}}{\left(\frac{1000g}{Kg}\right){N}_{cyl}â
Nâ
{M}_{Nom}}\end{array}
{M}_{Nom}
Cps
{P}_{std}
{T}_{std}
{R}_{air}
{V}_{d}
{N}_{cyl}
{\stackrel{Ë}{m}}_{intk}
\frac{d{T}_{brake}}{dt}=\frac{1}{{\mathrm{Ï}}_{eng}}\left({T}_{stdy}â{T}_{brake}\right)
{\mathrm{Ï}}_{bst}=\left\{\begin{array}{c}{\mathrm{Ï}}_{bst,rising}\text{ when }{T}_{stdy}>{T}_{brake}\\ {\mathrm{Ï}}_{bst,falling}\text{ when }{T}_{stdy}â¤{T}_{brake}\end{array}
{\mathrm{Ï}}_{eng}=\left\{\begin{array}{c}{\mathrm{Ï}}_{thr}\text{ when }{T}_{brake}<{f}_{bst}\left(N\right)\\ {\mathrm{Ï}}_{bst}\text{ when }{T}_{brake}â¥{f}_{bst}\left(N\right)\end{array}
Ï„thr Time constant during throttle control
Æ’bst(N) Boost torque speed line
{Q}_{fuel}=\frac{{\stackrel{Ë}{m}}_{fuel}}{\left(\frac{1000kg}{{m}^{3}}\right)S{g}_{fuel}}
{\stackrel{Ë}{m}}_{fuel}
â{\mathrm{Ï}}_{eng}\mathrm{Ï}
{\stackrel{Ë}{m}}_{fuel}LHV
{\mathrm{Ï}}_{eng}\mathrm{Ï}â{\stackrel{Ë}{m}}_{fuel}LHV
{\stackrel{Ë}{m}}_{fuel}
Torque, Tcmd, in N·m.
\underset{0}{\overset{\left(360\right)Cps}{â«}}EngSpd\frac{180}{30}d\mathrm{θ}
Cps
{T}_{brake}
\frac{d{T}_{brake}}{dt}=\frac{1}{{\mathrm{Ï}}_{eng}}\left({T}_{stdy}â{T}_{brake}\right)
{\mathrm{Ï}}_{bst}=\left\{\begin{array}{c}{\mathrm{Ï}}_{bst,rising}\text{ when }{T}_{stdy}>{T}_{brake}\\ {\mathrm{Ï}}_{bst,falling}\text{ when }{T}_{stdy}â¤{T}_{brake}\end{array}
{\mathrm{Ï}}_{eng}=\left\{\begin{array}{c}{\mathrm{Ï}}_{thr}\text{ when }{T}_{brake}<{f}_{bst}\left(N\right)\\ {\mathrm{Ï}}_{bst}\text{ when }{T}_{brake}â¥{f}_{bst}\left(N\right)\end{array}
Time constant below boost line, Ï„thr, in s.
Rising torque boost time constant, Ï„bst,rising, in s.
Falling torque boost time constant, Ï„bst,falling, in s.
The engine torque lookup table is a function of commanded engine torque and engine speed, T = Æ’(Tcmd, N), where:
T is engine torque, in N·m.
Tcmd is commanded engine torque, in N·m.
The engine torque lookup table is a function of commanded engine torque, engine speed, and engine temperature, T = Æ’(Tcmd, N, TempEng), where:
{\stackrel{Ë}{m}}_{intk}
= Æ’(Tcmd, N), where:
{\stackrel{Ë}{m}}_{intk}
{\stackrel{Ë}{m}}_{intk}
= Æ’(Tcmd, N, TempEng), where:
{\stackrel{Ë}{m}}_{intk}
The engine fuel mass flow lookup table is a function of commanded engine torque and engine speed, MassFlow = Æ’(Tcmd, N), where:
The engine fuel mass flow lookup table is a function of commanded engine torque, engine speed, and engine temperature, MassFlow = Æ’(Tcmd, N, TempEng), where:
The engine exhaust temperature lookup table is a function of commanded engine torque and engine speed, Texh = Æ’(Tcmd, N), where:
The engine exhaust temperature lookup table is a function of commanded engine torque, engine speed, and engine temperature, Texh = Æ’(Tcmd, N, TempEng), where:
The brake-specific fuel consumption (BSFC) efficiency is a function of commanded engine torque and engine speed, BSFC = Æ’(Tcmd, N), where:
The brake-specific fuel consumption (BSFC) efficiency is a function of commanded engine torque, engine speed, and engine temperature, BSFC = Æ’(Tcmd, N, TempEng), where:
The engine-out hydrocarbon emissions are a function of commanded engine torque and engine speed, EO HC = Æ’(Tcmd, N), where:
The engine-out hydrocarbon emissions are a function of commanded engine torque, engine speed, and engine temperature, EO HC = Æ’(Tcmd, N, TempEng), where:
The engine-out carbon monoxide emissions are a function of commanded engine torque and engine speed, EO CO = Æ’(Tcmd, N), where:
The engine-out nitric oxide and nitrogen dioxide emissions are a function of commanded engine torque and engine speed, EO NOx = Æ’(Tcmd, N), where:
The engine-out nitric oxide and nitrogen dioxide emissions are a function of commanded engine torque, engine speed, and engine temperature, EO NOx = Æ’(Tcmd, N, TempEng), where:
The engine-out carbon dioxide emissions are a function of commanded engine torque and engine speed, EO CO2 = Æ’(Tcmd, N), where:
The engine-out carbon dioxide emissions are a function of commanded engine torque, engine speed, and engine temperature, EO CO2 = Æ’(Tcmd, N, TempEng), where:
Mapped Motor | Mapped CI Engine
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Presample Data for Conditional Mean Model Estimation - MATLAB & Simulink - MathWorks Benelux
In a conditional mean model, the distribution of εt is conditional on historical information. Historical information includes past responses,
{y}_{1},{y}_{2},\dots ,{y}_{t-1}
, past innovations,
{\epsilon }_{1},{\epsilon }_{2},\dots ,{\epsilon }_{t-1}
, and, if you include them in the model, past and present exogenous covariates,
{x}_{1},{x}_{2},\dots ,{x}_{t-1},{x}_{t}
The number of past responses and innovations that a current innovation depends on is determined by the degree of the AR or MA operators, and any differencing. For example, in an AR(2) model, each innovation depends on the two previous responses,
{\epsilon }_{t}={y}_{t}-c-{\varphi }_{1}{y}_{t-1}-{\varphi }_{2}{y}_{t-2}.
In ARIMAX models, the current innovation also depends on the current value of the exogenous covariate (unlike distributed lag models). For example, in an ARX(2) model with one exogenous covariate, each innovation depends on the previous two responses and the current value of the covariate,
{\epsilon }_{t}={y}_{t}-c-{\varphi }_{1}{y}_{t-1}-{\varphi }_{2}{y}_{t-2}+{x}_{t}.
In general, the likelihood contribution of the first few innovations is conditional on historical information that might not be observable. How do you estimate the parameters without all the data? In the ARX(2) example,
{\epsilon }_{2}
explicitly depends on
{y}_{1},
{y}_{0},
{x}_{2},
{\epsilon }_{1}
{y}_{0},
{y}_{-1},
{x}_{1}
. Implicitly,
{\epsilon }_{2}
{x}_{1}
{x}_{0},
{\epsilon }_{1}
{x}_{0}
{x}_{-1}.
However, you cannot observe
{y}_{0},
{y}_{-1},
{x}_{0},
{x}_{-1}.
The amount of presample data that you need to initialize a model depends on the degree of the model. The property P of an arima model specifies the number of presample responses and exogenous data that you need to initialize the AR portion of a conditional mean model. For example, P = 2 in an ARX(2) model. Therefore, you need two responses and two data points from each exogenous covariate series to initialize the model.
One option is to use the first P data from the response and exogenous covariate series as your presample, and then fit your model to the remaining data. This results in some loss of sample size. If you plan to compare multiple potential models, be aware that you can only use likelihood-based measures of fit (including the likelihood ratio test and information criteria) to compare models fit to the same data (of the same sample size). If you specify your own presample data, then you must use the largest required number of presample responses across all models that you want to compare.
The property Q of an arima model specifies the number of presample innovations needed to initialize the MA portion of a conditional mean model. You can get presample innovations by dividing your data into two parts. Fit a model to the first part, and infer the innovations. Then, use the inferred innovations as presample innovations for estimating the second part of the data.
For a model with both an autoregressive and moving average component, you can specify both presample responses and innovations, one or the other, or neither.
By default, estimate generates automatic presample response and innovation data. The software:
Generates presample responses by backward forecasting.
Sets presample innovations to zero.
Does not generate presample exogenous data. One option is to backward forecast each exogenous series to generate a presample during data preprocessing.
|
Emmanuel Bengio, Moksh Jain, Maksym Korablyov, Doina Precup, Yoshua Bengio
arXiv preprint, code
also see the GFlowNet Foundations paper
and a more recent (and thorough) tutorial on the framework.
What follows is a high-level overview of this work, for more details refer to our paper. Given a reward
R(x)
and a deterministic episodic environment where episodes end with a ``generate
x
'' action, how do we generate diverse and high-reward
x
s?
We propose to use Flow Networks to model discrete
p(x) \propto R(x)
from which we can sample sequentially (like episodic RL, rather than iteratively as MCMC methods would). We show that our method, GFlowNet, is very useful on a combinatorial domain, drug molecule synthesis, because unlike RL methods it generates diverse
x
s by design.
A flow network is a directed graph with sources and sinks, and edges carrying some amount of flow between them through intermediate nodes -- think of pipes of water. For our purposes, we define a flow network with a single source, the root or
s_0
; the sinks of the network correspond to the terminal states. We'll assign to each sink
x
an ``out-flow''
R(x)
Given the graph structure and the out-flow of the sinks, we wish to calculate a valid flow between nodes, e.g. how much water each pipe is carrying. Generally there can be infinite solutions, but this is not a problem here -- any valid solution will do. For example above, there is almost no flow between
s_7
s_{13}
that goes through
s_{11}
, it all goes through
s_{10}
, but the reverse solution would also be a valid flow.
Why is this useful? Such a construction corresponds to a generative model. If we follow the flow, we'll end up in a terminal state, a sink, with probability
p(x) \propto R(x)
. On top of that, we'll have the property that the in-flow of
s_0
--the flow of the unique source--is
\sum_x R(x)=Z
, the partition function. If we assign to each intermediate node a state and to each edge an action, we recover a useful MDP.
F(s,a)=f(s,s')
be the flow between
s
s'
T(s,a)=s'
s'
is the (deterministic) state transitioned to from state
s
and action
a
\begin{aligned}\pi(a|s) = \frac{F(s,a)}{\sum_{a'}F(s,a')}\end{aligned}
then following policy
\pi
s_0
, leads to terminal state
x
R(x)
(see the paper for proofs and more rigorous explanations).
Approximating Flow Networks
As you may suspect, there are only few scenarios in which we can build the above graph explicitly. For drug-like molecules, it would have around
10^{16}
nodes!
Instead, we resort to function approximation, just like deep RL resorts to it when computing the (action-)value functions of MDPs.
Our goal here is to approximate the flow
F(s,a)
. Earlier we called a valid flow one that correctly routed all the flow from the source to the sinks through the intermediary nodes. Let's be more precise. For some node
s'
, let the in-flow
F(s')
be the sum of incoming flows:
\begin{aligned}F(s') = \sum_{s,a:T(s,a)=s'} F(s,a)\end{aligned}
Here the set
\{s,a:T(s,a)=s'\}
is the set of state-action pairs that lead to
s'
. Now, let the out-flow be the sum of outgoing flows--or the reward if
s'
is terminal:
\begin{aligned}F(s') = R(s') + \sum_{a'\in\mathcal{A}(s')} F(s',a').\end{aligned}
Note that we reused
F(s')
. This is because for a valid flow, the in-flow is equal to the out-flow, i.e. the flow through
s'
F(s')
\mathcal{A}(s)
is the set of valid actions in state
s
, which is the empty set when
s
is a sink.
R(s)
is 0 unless
is a sink, in which case
R(s)>0
We can thus call the set of these equalities for all states
s'\neq s_0
the flow consistency equations:
\begin{aligned}\sum_{s,a:T(s,a)=s'} F(s,a) = R(s') + \sum_{a'\in\mathcal{A}(s')} F(s',a').\end{aligned}
Here the set of parents
\{s,a:T(s,a)=s_3\}
\{(s_0, a_1), (s_1, a_7), (s_2, a_3)\}
\mathcal{A}(s_3)=\{a_2,a_4,a_8\}
By now our RL senses should be tingling. We've defined a value function recursively, with two quantities that need to match.
A TD-Like Objective
Just like one can cast the Bellman equations into TD objectives, so do we cast the flow consistency equations into an objective. We want
F_\theta
that minimizes the square difference between the two sides of the equations, but we add a few bells and whistles:
\begin{aligned}\mathcal{L}_{\theta,\epsilon}(\tau) = \sum_{\mathclap{s'\in\tau\neq s_0}}\,\left(\log\! \left[\epsilon+{\sum_{\mathclap{s,a:T(s,a)=s'}}} \exp F^{\log}_\theta(s,a)\right]- \log\! \left[\epsilon + R(s') + \sum_{\mathclap{a'\in{\cal A}(s')}} \exp F^{\log}_\theta(s',a')\right]\right)^2.\end{aligned}
First, we match the
\log
of each side, which is important since as intermediate nodes get closer to the root, their flow will become exponentially bigger (remember that
F(s_0) = Z = \sum_x R(x)
), but we care equally about all nodes. Second, we predict
F^{\log}_\theta\approx\log F
for the same reasons. Finally, we add an
\epsilon
value inside the
\log
; this doesn't change the minima of the objective, but gives more gradient weight to large values and less to small values.
We show in the paper that a minimizer of this objective achieves our desiderata, which is to have
p(x)\propto R(x)
when sampling from
\pi(a|s)
GFlowNet as Amortized Sampling with an OOD Potential
It is interesting to compare GFlowNet with Monte-Carlo Markov Chain (MCMC) methods. MCMC methods can be used to sample from a distribution for which there is no analytical sampling formula but an energy function or unnormalized probability function is available. In our context, this unnormalized probability function is our reward function
R(x)=e^{-energy(x)}
Like MCMC methods, GFlowNet can turn a given energy function into samples but it does it in an amortized way, converting the cost a lot of very expensive MCMC trajectories (to obtain each sample) into the cost training a generative model (in our case a generative policy which sequentially builds up
x
). Sampling from the generative model is then very cheap (e.g. adding one component at a time to a molecule) compared to an MCMC. But the most important gain may not be just computational, but in terms of the ability to discover new modes of the reward function.
MCMC methods are iterative, making many small noisy steps, which can converge in the neighborhood of a mode, and with some probability jump from one mode to a nearby one. However, if two modes are far from each other, MCMC can require exponential time to mix between the two. If in addition the modes occupy a tiny volume of the state space, the chances of initializing a chain near one of the unknown modes is also tiny, and the MCMC approach becomes unsatisfactory. Whereas such a situation seems hopeless with MCMC, GFlowNet has the potential to discover modes and jump there directly, if there is structure that relates the modes that it already knows, and if its inductive biases and training procedure make it possible to generalize there.
GFlowNet does not need to perfectly know where the modes are: it is sufficient to make guesses which occasionally work well. Like for MCMC methods, once a point in the region of new mode is discovered, further training of GFlowNet will sculpt that mode and zoom in on its peak.
Note that we can put
R(x)
to some power
\beta
, a coefficient which acts like a temperature, and
R(x)^\beta = e^{-\beta\; energy(x)}
, making it possible to focus more or less on the highest modes (versus spreading probability mass more uniformly).
Generating molecule graphs
The motivation for this work is to be able to generate diverse molecules from a proxy reward
R
that is imprecise because it comes from biochemical simulations that have a high uncertainty. As such, we do not care about the maximizer as RL methods would, but rather about a set of ``good enough'' candidates to send to a true biochemical assay.
Another motivation is to have diversity: by fitting the distribution of rewards rather than trying to maximize the expected reward, we're likely to find more modes than if we were being greedy after having found a good enough mode, which again and again we've found RL methods such as PPO to do.
Here we generate molecule graphs via a sequence of additive edits, i.e. we progressively build the graph by adding new leaf nodes to it. We also create molecules block-by-block rather than atom-by-atom.
We find experimentally that we get both good molecules, and diverse ones. We compare ourselves to PPO and MARS (an MCMC-based method).
Figure 3 shows that we're fitting a distribution that makes sense. If we change the reward by exponentiating it as
R^\beta
\beta>1
, this shifts the reward distribution to the right.
Figure 4 shows the top-
k
found as a function of the number of episodes.
Finally, Figure 5 shows that using a biochemical measure of diversity to estimate the number of distinct modes found, GFlowNet finds much more varied candidates.
Active Learning experiments
The above experiments assume access to a reward
R
that is cheap to evaluate. In fact it uses a neural network proxy trained from a large dataset of molecules. This setup isn't quite what we would get when interacting with biochemical assays, where we'd have access to much fewer data. To emulate such a setting, we consider our oracle to be a docking simulation (which is relatively expensive to run, ~30 cpu seconds).
In this setting, there is a limited budget for calls to the true oracle
O
. We use a proxy
M
initialized by training on a limited dataset of
(x, R(x))
D_0
R(x)
is the true reward from the oracle. The generative model (
\pi_{\theta}
) is then trained to fit
R
but as predicted by the proxy
M
. We then sample a batch
B=\{x_1, x_2, \dots x_k\}
x_i\sim \pi_{\theta}
, which is evaluated with the oracle
O
. The proxy
M
is updated with this newly acquired and labeled batch, and the process is repeated for
N
By doing this on the molecule setting we again find that we can generate better molecules. This showcases the importance of having these diverse candidates.
For more figures, experiments and explanations, check out the paper, or reach out to us!
|
Solving for Equilibrium | Brilliant Math & Science Wiki
Beakal Tiliksew, Sravanth C., Mahindra Jain, and
Onkar Gulati
When a force is applied on a body, it may change the state of motion of the body. Motions created by the applied force can be translational or rotational, but not all forces cause translational or rotational motion. You may like understanding the concept of equilibrium.
Understanding balanced and unbalanced forces
Gravitational force or weight
Two or more forces are said to be balanced if the net force of the all the forces is equal to zero. Balanced forces do not cause any motion in the object they are acting on.
_\square
Unbalanced Force:
Two or more forces are said to be unbalanced if the net force of the all the forces is equal to not equal to zero. Unbalanced forces always cause motion in the object they are acting on.
_\square
A rigid body is one that does not deform under the application of an external force. For a rigid body to be in a state of equilibrium, the net force on the object must be zero. That is, if an object remains motionless, then Newton's second law,
F=ma
, tells us that the total external force acting on the object is zero. The total external force on an object is also zero, if the body moves with a constant nonzero velocity.
The whole goal when solving for equilibrium is to find out what the various forces have to do so that there is zero net force on each body
\left(\sum { F=0 }\right).
Since force is a vector, we have to take into consideration the direction in which the force acts.
Consider an object that is on top of a frictionless table. From the figure below what single force should be acted so that the object is in equilibrium? What will be the direction of the force?
For convention let the right direction be positive, and left direction negative. Also, let the force required be
X.
Now balance all the forces:
\begin{aligned} \sum {F} &=0 \\ X+4\text{ N}+3.5\text{ N}-3\text{ N}-1.5\text{ N}&=0\\ \Rightarrow X&=4.5\text{ N}-7.5\text{ N}\\ &=-3\text{ N}. \end{aligned}
Since the negative sign means left direction, a force of
3N
must act to the left so that the object is in equilibrium.
_\square
A common force when solving for equilibrium is weight. Weight is the direct application of Newton's law of gravitation and for an object near the surface of the earth it is given by
W=mg,
m
is the mass of the body, and
g
is the gravitational acceleration and is equal to approximately
9.8\text{ m/s}^2
. Note that weight always acts downwards toward the center of the earth.
Another common force when solving for equilibrium is tension. Tension is the general name for the pulling force of a rope, stick, cable, etc. Tension has a unit of force and can be measured in newtons.
If a block of mass
3\text{ kg}
is hanging from a ceiling by a massless rope, what will be the tension on the rope?
g
9.8\text{ m/s}^{2}
fkjashf
We know that the block is acted upon by a gravitational force equal to
mg
, and since the object is motionless there must be another force to balance it. That force is tension, which we can get as follows:
\begin{aligned} \sum F&=0 \\ T-mg&=0\\ \Rightarrow T&=(3\text{ kg})\left(9.8\text{ m/s}^{2}\right)\\ &=29.4\text{ N}. \ _\square \end{aligned}
A stick-man who has a mere mass of
3\text{ kg}
is sitting on a swing and is at rest, as shown in the figure below. If the platform he is sitting upon has a mass of
5\text{ kg},
what will be the tension in each individual rope that holds the swing?
Assume that the weight is equally divided by the two ropes.
{: .center}
Consider the system of the stick-man and the platform. Since the mass is equally distributed, let the tension in each rope be
T.
dslkf
Then since the system is in equilibrium, we have
\begin{aligned} \sum F&=0 \\ T+T-(5\text{ kg})(g)-(3\text{ kg})(g)&=0\\ 2T&=8g\cdot\text{ kg}\\ \Rightarrow T&=4g\cdot \text{ kg}\\ &\approx4\times 9.8 \text{ m/s}^2 \cdot \text{ kg}\\ &= 39.2\text{ N}. \ _\square \end{aligned}
Cite as: Solving for Equilibrium. Brilliant.org. Retrieved from https://brilliant.org/wiki/solving-for-equilibrium/
|
Predicting aqueous solubility with neural networks | Research @ WDSS
Created 2021-10-19| Updated 2021-10-19 |Aleksander Dukaczewski, Annie Stevenson, Escher Luton|Natural SciencesChemistry
Aqueous solubility is a key physical property of interest in the medicinal and agrochemical industry. Low aqueous solubility of compounds can be a major problem in drug development, as more than 40% of newly developed chemicals are practically insoluble in water. For a drug to be absorbed it needs to be contained in a solution at the site of the absorption and solubility is the main parameter that influences the bioavailability of a molecule. Many drug manufacturers are inclined more to produce oral drug products due to their ease of administration, lack of sterility constraints and high patient compliance, which causes them to prioritise bioavailability and therefore aqueous solubility.
As designing and approving a new drug is an expensive, nearly decade-long process, new methods for the prediction of a compound’s aqueous solubility prior to its synthesis could greatly facilitate the process of drug development. Aqueous solubility is also a major factor in the development of insecticides, fungicides and herbicides as highly soluble substances are more likely to move through the soil in water than less-soluble substances, therefore they are able to reach the plants and take effect more easily. This suggests that the agrochemical industry can also greatly benefit from new methods of estimating aqueous solubility of compounds without the presence of a physical sample.
Machine learning is a branch of computer science that focuses on the use of algorithms and data to imitate the way that humans learn. It allows us to create models that can predict a compound’s aqueous solubility straight from its molecular structure without the presence of a physical sample. Not having to run sophisticated and computationally expensive physical simulations can greatly reduce the amount of time and funds spent in the process of drug discovery.
Machine learning algorithms can be grouped into regressive algorithms used to predict a continious value (age, salary, etc.) and classification algorithms used to predict discrete values (male, female, etc.). Predicting a compound’s aqueous solubility is a regression problem rather than a classification one and the model can be taught to predict a value based on a set of inputs in a process called supervised learning.
The entire project was completed using Python. We used the dataset ESOL, containing the solubilities of 1144 chemical compounds and their structures in the SMILES format. SMILES is a commonly used notation used for describing the structure of chemical compounds with short ASCII strings. The compounds contained in the dataset are mostly pesticide products, low molecular weight organic compounds, and heavy compounds.
The library rdkit was used to extract features from the dataset. It provides methods to calculate:
logP
- partition coefficient; measure of differential solubility in a hydrophobic and hydrophilic solvent
Molecular weight (
MWT
Number of rotatable bonds (
RB
Number of H-bond donors (
HBD
Number of H-bond acceptors (
HBA
Polar surface area - the amount of molecular surface arising from polar atoms (
TPSA
Aromatic proportion - number of atoms in an aromatic ring divided by the number of heavy atoms in the molecule (
AP
Non-carbon proportion - number of non-carbon atoms divided by the number of heavy atoms in the molecule (
NCP
In order to observe which features have the most predictive power it was decided to train a linear regression model from the library sklearn to predict log(solubility). Multiple linear regression is a statistical technique that uses several explanatory variables to predict the outcome of a response variable. The goal of multiple linear regression is to model the linear relationship between the explanatory (independent) variables and response (dependent) variables. Data was split into training and testing datasets with 80% of the data used for training the model. Metrics we used to quantify how well a model fits the dataset were RMSE and R-Squared.
RMSE = \sqrt{\frac{\Sigma_{i=1}^{N}{\left(y_i - \hat{y}_i\right)^2}}{N}}
. RMSE is the square root of the average of squared residuals / errors. When used on the holdout dataset, RMSE serves to aggregate the magnitudes of the errors in predictions for various data points into a single measure of predictive power. It tells us how far apart predicted values are from those in a dataset (on average). The lower the value, the better the data fits the model.
R^2 = 1 - \frac{\text{RSS}}{\text{TSS}}
, where RSS = sum of squares of residuals, TSS = total sum of squares of the values in the dataset (proportional to the variance of the data). It tells us about the proportion of the variation in the dependent variable that is predictable from the independent variables.
R^2
takes the values between 0 and 1. An
R^2
of 1 indicates that the regression predictions perfectly fit the data.
After training, this linear regression model resulted in training
RMSE
R^2
values of approximately
0.964
78.8
%, and testing
RMSE
R^2
0.993
77.9
%. The predictive power of this simple linear model inspired us to create a more sophisticated neural network model for the same purpose.
The final equation given by the trained model is
LogS = 0.12 -0.78logP -0.0065 MWT + 0.0158 RB -0.05 HBD + 0.20 HBA -0.02 TPSA -0.25 AP + 1.16 NCP
The coefficients of the linear model can help us identify which features might have the most predictive power. From the above equation we deduce that these are non-carbon proportion and aromatic proportion.
These two-dimmensional plots represent how the equation fits the data points from the training and testing datasets respectively. The line is a graph of the equation and the dots represent entries from the datasets.
The linear regression model is only able to capture linear relationships between the independent variables and the target variable. If the relationship between the variables is non-linear, features should be additionally transformed, for example by taking the logarithm of the predictor variables or introducing higher order polynomial terms.
Artificial neural networks are machine learning models designed to loosely model neural networks present in animal brains. They are collections of nodes called artificial neurons, which are models that imitate neurons in real brains. In contrast to linear regression, neural networks can model more complex, non-linear relationships between predictive variables and the response variable without assuming that a fixed equation can fit the model.
A neural network model was created using keras and tensorflow. To build a neural network we first need to investigate the following features: vectors, layers and linear regression. Our data source has been stored in vectors and with the help of python can be converted into multidimensional arrays. These arrays are then used to calculate the dot product, which tells us how similar the data is in terms of direction. Weights and bias vectors are the main type of vectors present inside a neural network; these are used to predict outputs by comparing input data to previously seen inputs. Layers have been added to the neural network to extract representations found in the previous data, hidden layers are also present to take a set of weighted inputs and produce an output through an activation function. Linear regression can be thought of as a neural network consisting of just one layer, as every input is connected to every output. This layer can be seen as fully connected and can be used within the neural network.
ann.add(Dense(60, input_dim=8, activation='tanh'))
ann.add(Dense(40, input_dim=60, activation='tanh'))
ann.add(Dense(7, input_dim=20, activation='tanh'))
ann.add(Dense(1, input_dim=7, activation='linear'))
tf.keras.optimizers.Adam(learning_rate=0.001, beta_1=0.9, beta_2=0.99)
ann.compile(loss='mean_squared_error', optimizer='adam', metrics='mse')
ann.summary
#Fitting/ tuning the model
ann.fit(X_train, Y_train, epochs=100, batch_size=32, validation_split=0.10, verbose=True)
#Applying the model to the training data set
ann_Y_train_pred = ann.predict(X_train)
The artificial neural network was evaluated like the linear regression model and achieved better metric scores, with training
RMSE
R^2
0.721
88.1
RMSE
R^2
0.722
88.1
The neural network model resulted in lower values of root-mean-square error (
RMSE
) than the linear regression model, meaning that the values predicted by the linear regression model were farther apart. The neural network model also got higher values of the coefficient of determination (
R^2
), which also means that the model fits the dataset better. The models can be compared using this graph, showing the distribution of measured molecule solubility and predicted molecule solubility of both models.
As our problem is classed as regression, we needed to replace the categorical variables within the neural network to numeric dependent variables. Otherwise, the model would try categorise outputs. There are several limitations to using neural networks on regression problems which are mainly related to the data source properties and their optimalisation, these include:
Dataset size – how many training examples are present
Data structure - structured/ unstructured
To stabilise the model, a large dataset is required. Our dataset is well structure however, it is relatively small with only a 1000 plus records. Combining our dataset with other solubility data would increase the number of inputs and therefore improve the models ability to predict. As a further develop we could also investigate different validation methods such as cross validation, as our method reduced the size of the training dataset as holdout validation was used. Our models results proved more useful than those of the linear regression model; yet, improvements could still be made for better performance and other machine learning models should be investigated. This includes researching gradient boosted decision trees, which are said to work well in similar regression problems.
Molecular machine learning methods for the prediction of chemical properties can have far-reaching benefits for industries relating to chemistry. They could greatly reduce the cost and time used in the process of drug development. Methods similar to ours can also be used to predict other chemical properties like lipophilicity or solvation free energy.
While the accuracy of these methods may not always be perfect, they can still prove to be useful by giving researchers suggestions about whether the compounds they’re researching fit their criteria. Until sophisticated and reliable molecular learning methods are widely recognised we can’t greatly depend on them in research, however they can still serve as advisory tools. Conducting synthesis and running sophisticated tests isn’t always the most viable option, therefore these methods can prove to be useful where time and budget is limited.
Authors: Aleksander Dukaczewski, Annie Stevenson, Escher Luton
Permalink: https://research.wdss.io/chemical-properties/
machine learningmolecular machine learningchemistrysolubilityesolpython
|
Pastes a series of NDFs upon each other
This application copies a series of NDFs , in the order supplied and taking account of origin information, on to a ‘base’ NDF to produce an output NDF. The output NDF is therefore a copy of the base NDF obscured wholly or partially by the other input NDFs. This operation is analogous to pasting in publishing. It is intended for image editing and the creation of insets.
The dimensions of the NDFs may be different, and indeed so may their dimensionalities. The output NDF can be constrained to have the dimensions of the base NDF, so the pasted NDFs are clipped. Normally, the output NDF will have dimensions such that all the input NDFs are accommodated in full.
Bad values in the pasted NDFs are by default transparent, so the underlying data are not replaced during the copying.
Input NDFs can be shifted in pixel space before pasting them into the output NDF (see Parameter SHIFT).
paste in p1 [p2] ... [p25] out=?
CONFINE = _LOGICAL (Read)
This parameter controls the dimensions of the output NDF. If CONFINE is FALSE the output NDF just accommodates all the input NDFs. If CONFINE is TRUE, the output NDF’s dimensions matches those of the base NDF. [FALSE]
This parameter is either:
a) the base NDF on to which the other input NDFs supplied via Parameters P1 to P25 will be pasted; or
b) a group of input NDFs (of any dimensionality) comprising all the input NDFs, of which the first is deemed to be the base NDF, and the remainder are to be pasted in the order supplied.
The group should be given as a comma-separated list, in which each list element can be:
\ast
", "?", "[a-z]" etc.).
\text{^}
". Each line in the text file should contain a comma-separated list of elements, each of which can in turn be an NDF name (with optional wild-cards, etc.), or another file specification (preceded by an up-arrow). Comments can be included in the file by commencing lines with a hash character "#".
If the value supplied for this parameter ends with a hyphen "-", then you are re-prompted for further input until a value is given which does not end with a hyphen. All the NDFs given in this way are concatenated into a single group.
The group can contain no more than 1000 names.
The NDF resulting from pasting of the input NDFs on to the base NDF. Its dimensions may be different from the base NDF. See Parameter CONFINE.
P1-P25 = NDF (Read)
The NDFs to be pasted on to the base NDF. The NDFs are pasted in the order P1, P2, ... P25. There can be no missing NDFs, e.g. in order for P3 to be processed there must be a P2 given as well. A null value (!) indicates that there is no NDF. NDFs P2 to P25 are defaulted to !. At least one NDF must be pasted, therefore P1 may not be null.
P1 to P25 are ignored if the group specified through Parameter IN comprises more than one NDF.
SHIFT( * ) = _INTEGER (Read)
An incremental shift to apply to the pixel origin of each input NDF before pasting it into the output NDF. If supplied, this parameter allows a set of NDFs with the same pixel bounds to be placed ‘side-by-side’ in the output NDF. For instance, this allows a set of images to be pasted into a cube. The first input NDF is not shifted. The pixel origin of the second NDF is shifted by the number of pixels given in SHIFT. The pixel origin of the third NDF is shifted by twice the number of pixels given in SHIFT. Each subsequent input NDF is shifted by a further multiple of SHIFT. If null (!) is supplied, no shifts are applied. [!]
TRANSP = _LOGICAL (Read)
If TRANSP is TRUE, bad values within the pasted NDFs are not copied to the output NDF as if the bad values were transparent. If TRANSP is FALSE, all values are copied during the paste and a bad value will obscure an underlying value. [TRUE]
paste aa inset out=bb
This pastes the NDF called inset on to the arrays in the NDF called aa to produce the NDF bb. Bad values are transparent. The bounds and dimensionality of bb may be larger than those of aa.
paste aa inset out=bb notransp
As above except that bad values are copied from the NDF inset to NDF bb.
paste aa inset out=bb confine
As the first example except that the bounds of NDF bb match those of NDF aa.
paste in="aa,inset" out=bb
The same as the first example.
paste in="aa,inset,inset2,inset3" out=bb
Similar to first example, but now two further NDFs inset2 and inset3 are also pasted.
paste ccd fudge inset out=ccdc
This pastes the NDF called fudge, followed by NDF inset on to the arrays in the NDF called ccd to produce the NDF ccdc. Bad values are transparent. The bounds and dimensionality of ccd may be larger than those of ccdc.
paste in="canvas,
\text{^}
shapes.lis" out=collage confine
This pastes the NDFs listed in the text file shapes.lis in the order given on the NDF called canvas. Bad values are transparent. The bounds of NDF collage match those of NDF canvas.
paste in=
\text{^}
planes out=cube shift=[0,0,1]
Assuming the text file planes contains a list of two-dimensional NDFs, this arranges them into a cube, one behind the other.
This routine correctly processes the AXIS, DATA, QUALITY, VARIANCE, LABEL, TITLE, UNITS, WCS, and HISTORY, components of an NDF data structure and propagates all extensions. Propagation is from the base NDF.
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Angles A and B are complementary. If mangle A=27∘, what is the mangle B? mangle
Angles A and B are complementary.If mangle A=27∘, what is the mangle B?mangle B=□ degrees
Angles A and B are complementary. If
m\mathrm{\angle }A=27\circ
, what is the
m\mathrm{\angle }B?
m\mathrm{\angle }B=\circ degrees
m\mathrm{\angle }A+m\mathrm{\angle }B=90\circ
27\circ +m\mathrm{\angle }B=90\circ
m\mathrm{\angle }B=63\circ
Define Transformation.
Write down the properties of Linear transformation and rotational transformation.
Angles A and B are vertical angles.
m\mathrm{\angle }A=104\circ
m\mathrm{\angle }B?
m\mathrm{\angle }B=\circ degrees
1. On a sheet of graph paper, draw scalene acute
\mathrm{△}ABC
- Draw side
\stackrel{―}{BC}\text{ }\text{of}\text{ }\mathrm{△}ABC
along a grid line.
- Be sure all vertices are placed at the intersection of grid lines.
2. Draw an altitude,
\stackrel{―}{AD}\text{ }\text{of}\text{ }\mathrm{△}ABC
from point A to
\stackrel{―}{BC}
3. Label the figure you have drawn by indicating cingruent sides, angles, and measures.
4. Reflect
\mathrm{△}ABC
across the line containing
\stackrel{―}{BC}
5. On a separate sheet of graph paper, repeat steps 1-4 with a scalene obtuse
\mathrm{△}ABC
6. Discuss with your group the properties of the kites ACAB
y=-2\left(\frac{3}{2}-{e}^{3-x}\right)
a) Performing the necessary algebra so that the function is in the proper form (i.e., the transformations are in the proper order).
b) Listing the transformations in the order that they are to be applied.
c) Marking the key point and horizontal asymptote.
Give an example of an undefined term and a defined term in geometry. Explain the difference between an undefined term and a defined term.
The given statement, " If k is any odd integer and m is any even integer, then
{k}^{2}\text{ }+\text{ }{m}^{2}
is odd".
\mathrm{△}\text{ }CBF\text{ }\stackrel{\sim }{=}\mathrm{△}EDF
\mathrm{△}\text{ }CBF\text{ }\stackrel{\sim }{=}\text{ }\mathrm{△}\text{ }EDF
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Engineering Acoustics/Bessel Functions and the Kettledrum - Wikibooks, open books for an open world
Engineering Acoustics/Bessel Functions and the Kettledrum
2 What is a kettledrum
3 The math behind the kettledrum:the brief version
4 The math behind the kettledrum:the derivation
5 The math behind the kettledrum:the entire drum
In class, we have begun to discuss the solutions of multidimensional wave equations. A particularly interesting aspect of these multidimensional solutions are those of Bessel functions for circular boundary conditions. The practical application of these solutions is the kettledrum. This page will explore in qualitative and quantitative terms how the of the kettledrum works. More specifically, the kettledrum will be introduced as a circular membrane and its solution will be discussed in visual (e.g. visualization of Bessel functions, video of kettledrums and audio forms (wav files of kettledrums playing. In addition, links to more information about this material, including references, will be included.
What is a kettledrumEdit
A kettledrum is a percussion instrument with a circular drumhead mounted on a "kettle-like" enclosure. When one strikes the drumhead with a mallet, it vibrates which produces its sound. The pitch of this sound is determined by the tension of the drumhead, which is precisely tuned before playing. The sound of the kettldrum (called the Timpani in classical music) is present in many forms of music from many difference places of the world.
The math behind the kettledrum:the brief versionEdit
When one looks at how a kettledrum produces sound, one should look no farther than the drumhead. The vibration of this circular membrane (and the air in the drum enclosure) is what produces the sound in this instrument. The mathematics behind this vibrating drum are relatively simple. If one looks at a small element of the drum head, it looks exactly like the situation for the vibrating string (see:). The only difference is that there are two dimensions where there are forces on the element, the two dimensions that are planar to the drum. As this is the same situation, we have the same equation, except with another spatial term in the other planar dimension. This allows us to model the drumhead using a helmholtz equation. The next step (solved in detail below) is to assume that the displacement of the drumhead (in polar coordinates) is a product of two separate functions for theta and r. This allows us to turn the PDE into two ODES which are readily solved and applied to the situation of the kettledrum head. For more info, see below.
The math behind the kettledrum:the derivationEdit
So starting with the trusty general Helmholtz equation:
{\displaystyle \nabla ^{2}\Psi +k^{2}\Psi =0}
Where k is the wave number, the frequency of the forced oscillations divided by the speed of sound in the membrane.
Since we are dealing with a circular object, it make sense to work in polar coordinates (in terms of radius and angle) instead of rectangular coordinates. For polar coordinates the Laplacian term of the Helmholtz relation (
{\displaystyle \nabla ^{2}}
{\displaystyle \partial ^{2}\Psi /\partial r^{2}+1/r\partial ^{2}\Psi /\partial r+1/r^{2}\partial ^{2}\Psi /\partial \theta ^{2}}
Now lets assume that:
{\displaystyle \Psi (r,\theta )=R(r)\Theta (\theta )}
This assumption follows the method of separation of variables. (see Reference 3 for more info) Substituting this result back into our trusty Helmholtz equation gives the following:
{\displaystyle r^{2}/R(d^{2}R/dr^{2}+1/rdR/dr)+k^{2}r^{2}=-1/\Theta d^{2}\Theta /d\theta ^{2}}
Since we separated the variables of the solution into two one-dimensional functions, the partial derivatives become ordinary derivatives. Both sides of this result must equal the same constant. For simplicity, I will use
{\displaystyle \lambda }
as this constant. This results in the following two equations:
{\displaystyle d^{2}\Theta /d\theta ^{2}=-\lambda ^{2}\Theta }
{\displaystyle d^{2}R/dr^{2}+1/rdR/dr+(k^{2}-\lambda ^{2}/r^{2})R=0}
The first of these equations readily seen as the standard second order ordinary differential equation which has a harmonic solution of sines and cosines with the frequency based on
{\displaystyle \lambda }
. The second equation is what is known as Bessel's Equation. The solution to this equation is cryptically called Bessel functions of order
{\displaystyle \lambda }
of the first and second kind. These functions, while sounding very intimidating, are simply oscillatory functions of the radius times the wave number that are unbounded at when kr (for the function of the second kind) approaches zero and diminish as kr get larger. (For more information on what these functions look like see References 1,2, and 3)
Now that we have the general solution to this equation, we can now model a infinite radius kettledrum head. However, since i have yet to see an infinite kettle drum, we need to constrain this solution of a vibrating membrane to a finite radius. We can do this by applying what we know about our circular membrane: along the edges of the kettledrum, the drum head is attached to the drum. This means that there can be no displacement of the membrane at the termination at the radius of the kettle drum. This boundary condition can be mathematically described as the following:
{\displaystyle R(a)=0}
Where a is the arbitrary radius of the kettledrum. In addition to this boundary condition, the displacement of the drum head at the center must be finite. This second boundary condition removes the Bessel function of the second kind from the solution. This reduces the R part of our solution to:
{\displaystyle R(r)=AJ_{\lambda }(kr)}
{\displaystyle J_{\lambda }}
is a Bessel function of the first kind of order
{\displaystyle \lambda }
. Apply our other boundary condition at the radius of the drum requires that the wave number k must have discrete values, (
{\displaystyle j_{mn}/a}
) which can be looked up. Combining all of these gives us our solution to how a drumhead behaves (which is the real part of the following):
{\displaystyle y_{\lambda n}(r,\theta ,t)=A_{\lambda n}J_{\lambda n}(k_{\lambda n}r)e^{j\lambda \theta +jw_{\lambda n}t}}
The math behind the kettledrum:the entire drumEdit
The above derivation is just for the drum head. An actual kettledrum has one side of this circular membrane surrounded by an enclosed cavity. This means that air is compressed in the cavity when the membrane is vibrating, adding more complications to the solution. In mathematical terms, this makes the partial differencial equation non-homogeneous or in simpler terms, the right side of the Helmholtz equation does not equal zero. This result requires significantly more derivation, and will not be done here. If the reader cares to know more, these results are discussed in the two books under references 6 and 7.
As one can see from the derivation above, the kettledrum is very interesting mathematically. However, it also has a rich historical music tradition in various places of the world. As this page's emphasis is on math, there are few links provided below that reference this rich history.
A discussion of Persian kettledrums: Kettle drums of Iran and other countries
A discussion of kettledrums in classical music: Kettle drum Lit.
A massive resource for kettledrum history, construction and technique" Vienna Symphonic Library
Wikibooks sister cite, references under Timpani: Wikipedia reference
1.Eric W. Weisstein. "Bessel Function of the First Kind." From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
2.Eric W. Weisstein. "Bessel Function of the Second Kind." From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/BesselFunctionoftheSecondKind.html
3.Eric W. Weisstein. "Bessel Function." From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/BesselFunction.html
4.Eric W. Weisstein et al. "Separation of Variables." From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/SeparationofVariables.html
5.Eric W. Weisstein. "Bessel Differential Equation." From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/BesselDifferentialEquation.html
6. Kinsler and Frey, "Fundamentals of Acoustics", fourth edition, Wiley & Sons
7. Haberman, "Applied Partial Differential Equations", fourth edition, Prentice Hall Press
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Bessel_Functions_and_the_Kettledrum&oldid=3327032"
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Winter Orb | PoE Wiki
Winter OrbCold, Spell, Channelling, AoE, Duration, Projectile, Orb
Increases and Reductions to Cast Speed also apply to Projectile FrequencyPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. Acquisition
Metadata ID: Metadata/Items/Gems/SkillGemWinterOrb
Winter Orb is a channelling skill (a spell). Channeling the skill charges up an orb, which fires icy projectiles to deal cold damage enemies, which target the ground to deal area of effect damage.
1.1 Projectile Rate
While Winter Orb has the Projectile keyword, testing shows it appears to gain no benefit from the Fork SupportFork SupportSupport, Projectile
Projectiles from Supported Skills ForkThis is a Support Gem. It does not grant a bonus to your character, but to skills in sockets connected to it. Place into an item socket connected to a socket containing the Active Skill Gem you wish to augment. Right click to remove from a socket. , Pierce SupportPierce SupportSupport, Projectile
Supported Skills deal (0-19)% more Projectile DamageThis is a Support Gem. It does not grant a bonus to your character, but to skills in sockets connected to it. Place into an item socket connected to a socket containing the Active Skill Gem you wish to augment. Right click to remove from a socket. , or Chain SupportChain SupportSupport, Chaining, Projectile
Supported Skills deal (30-11)% less Damage with HitsThis is a Support Gem. It does not grant a bonus to your character, but to skills in sockets connected to it. Place into an item socket connected to a socket containing the Active Skill Gem you wish to augment. Right click to remove from a socket. gems, even though the Active Skills window shows them as compatible. Firing projectiles in a nova with Sire of ShardsSire of Shards
20% increased Light RadiusThat which was broken may yet break. causes multiple projects to launch spread out as expected but land not exactly the same distance away from the caster.
Base projectile rate at stage 1 without any increased cast speed is .625 per second.
The first stage does not gain projectile frequency. Therefore 10 stages gives (9 X 15) = 135% increased projectile frequency. The plus 2 maximum stages helmet enchantment adds 30% projectile frequency.
All cast speed modifiers are additive with the increased projectile frequency. Therefore a 10 stage winter orb with 100% cast speed has (135 + 100) = 235% total increase projectile frequency.
150% more projectile frequency will multiply your projectile fire rate by 2.5 while channeling.
Example of projectile fire rate while channeling at 10 stages and 100% increased cast speed:
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}0.625\times 3.35\times 2.5\approx 5.23{\text{ per second}}
1 28 67 2 30 to 37 0 0
2 31 73 2 35 to 44 199345 199345
5 40 92 3 59 to 74 554379 1440883
7 44 100 3 73 to 92 583786 2502106
8 46 104 3 82 to 102 710359 3212465
9 48 109 3 91 to 113 1355511 4567976
10 50 113 3 101 to 126 1138877 5706853
13 56 125 3 136 to 171 1956648 10670072
17 64 142 4 202 to 253 15138193 40240771
19 68 151 4 245 to 307 62620247 128944843
20 70 155 4 270 to 337 211708088 340652931
21 72 159 4 296 to 371 N/A N/A
22 74 N/A 4 326 to 407 N/A N/A
34 94 N/A 5 810 to 1013 N/A N/A
39 99 N/A 5 1011 to 1264 N/A N/A
40 100 N/A 5 1057 to 1321 N/A N/A
Winter Orb can drop anywhere.
Winter Orb can be created from the following recipes:
The following threshold jewels augment Winter Orb:
The following helmet enchantments affect Winter Orb.
Enchantment Frost Fury Additional Max Number Of Stages 1 75 Eternal Labyrinth Winter Orb has +2 Maximum Stages helmet 100
Enchantment Frost Fury Area Of Effect Per Stage 1 66 Merciless Labyrinth Winter Orb has 2% increased Area of Effect per Stage helmet 100
Enchantment Frost Fury Area Of Effect Per Stage 2 75 Eternal Labyrinth Winter Orb has 3% increased Area of Effect per Stage helmet 100
Enchantment Winter Orb Damage 1 66 Merciless Labyrinth Winter Orb deals 25% increased Damage helmet 100
Enchantment Winter Orb Damage 2 75 Eternal Labyrinth Winter Orb deals 40% increased Damage helmet 100
Winter Orb has the following alternate skill effects:
Celestial Winter Orb Effect Celestial Winter Orb EffectSkill Gem EffectYour Winter Orb becomes a Celestial Effect. Your Winter Orb becomes a Celestial Effect.
Winter Orb now deals 30 to 37 Cold Damage at gem level 1 (unchanged), up to 270 to 337 at gem level 20 (previously 232 to 291).
Now has 150% more Projectile Frequency while Channelling (previously 100%).
Now has 60% Effectiveness of Added Damage at all gem levels (previously 50%).
Got Orb tag.
Deals 30 to 37 cold damage at gem level 1 (from 25 to 31), up to 232 to 291 at gem level 20 (from 232 to 289).
Base duration is now 1.2 seconds (from 0 base duration, building solely on stages).
Now states that increases and reductions to cast speed now also apply to projectile frequency (actual behaviour is unchanged).
Now has a 20% reduced target acquisition range.
Now fires projectiles every 1.6 seconds (from every 0.8 seconds)
Now gains 25% increased duration per stage (from +0.4 seconds to base duration per stage).
Now has 100% more projectile frequency while channelling.
Winter Orb has been added to the game.
ru:Зимняя сфера
Retrieved from ‘https://www.poewiki.net/w/index.php?title=Winter_Orb&oldid=1156181’
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