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Finding Domain and Range | Brilliant Math & Science Wiki
Jason Dyer and Jimin Khim contributed
Finding the domain and range of a function is a process that can often be done with algebra or with the aid of graphical means.
Formally, a function is a relation between a set of inputs (called the domain) that generate a particular set of outputs (called the range). For example,
f(x) = x^2
has a domain of all real numbers (since any value of
x
is possible) and a range of 0 and positive numbers (since the output cannot be negative).
In practice, domain is most commonly restricted by either a) division by zero or b) invalid input of square roots and logarithms.
f(x) = \frac{x+1}{x}
cannot have an input of 0, since this causes the denominator to be 0. Hence the domain is all real numbers but 0.
If we assume real numbers only,
f(x) = \sqrt{x}
has a domain of 0 and positive numbers, since taking the square root of a negative number is not possible in real numbers (although it is possible in complex numbers).
x<1
x=1
x>1
x \ne 1
y = \frac 1{x-1}
, find the domain of
x
The variety of circumstances range is restricted is larger than with the domain. Here is an example:
f(x) = 5 ,
the horizontal line at
y = 5 ,
has only an output of
5,
so that is the range. If a line is not horizontal, the range is all real numbers. We can justify this algebraically by starting at the
y = mx + b
form and finding the inverse: swapping the
x
y
x = my + b ,
and isolating the
y
y = \frac{x-b}{m} ;
division by zero only happens when
m = 0
(that is, the original graph is a horizontal line!).
(-\infty, 0)
(0, \infty)
[0, \infty)
What is the range of the function
f(x)=e^x
defined on real numbers?
Cite as: Finding Domain and Range. Brilliant.org. Retrieved from https://brilliant.org/wiki/finding-domain-and-range/
|
Trying to find the Laplace transform of \(\displaystyle{\frac{{{\cos{{t}}}}}{{{t}}}}\) It
Mary Bates 2022-04-08 Answered
Trying to find the Laplace transform of
\frac{\mathrm{cos}t}{t}
It comes out as infinity, but that doesn't make any sense.
Does this mean that this function doesn't have a Laplace transform or is something wrong here?
Jax Burns
Well, we are trying to find (solving a more general problem):
{\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\cdot \mathrm{exp}\left(-\text{s}t\right)\text{d}x
Using the 'frequency-domain integration' property of the Laplace transform, we can write:
{\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}={\int }_{\text{s}}^{\mathrm{\infty }}{\mathcal{L}}_{t}{\left[\mathrm{cos}\left(\text{n}\cdot t\right)\right]}_{\left(\sigma \right)}\text{d}\sigma
Using the table of selected Laplace transforms, we find:
{\mathcal{L}}_{t}{\left[\mathrm{cos}\left(\text{n}\cdot t\right)\right]}_{\left(\sigma \right)}=\frac{\sigma }{{\sigma }^{2}+{\text{n}}^{2}}
{\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}={\int }_{\text{s}}^{\mathrm{\infty }}\frac{\sigma }{{\sigma }^{2}+{\text{n}}^{2}}\phantom{\rule{1em}{0ex}}\text{d}\sigma
Let's substitute
\text{u}\phantom{\rule{0.222em}{0ex}}={\sigma }^{2}+{\text{n}}^{2}
{\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}{\int }_{{\text{s}}^{2}+{\text{n}}^{2}}^{{ϵ}^{2}+{\text{n}}^{2}}\frac{1}{\text{u}}\phantom{\rule{1em}{0ex}}\text{d}\text{u}=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}{\left[\mathrm{ln}|\text{u}|\right]}_{{\text{s}}^{2}+{\text{n}}^{2}}^{{ϵ}^{2}+{\text{n}}^{2}}=
\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}\left(\mathrm{ln}|{ϵ}^{2}+{\text{n}}^{2}|-\mathrm{ln}|{\text{s}}^{2}+{\text{n}}^{2}|\right)=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}|\frac{{ϵ}^{2}+{\text{n}}^{2}}{{\text{s}}^{2}+{\text{n}}^{2}}|\to \mathrm{\infty }
I want to get a particular solution to the differential equation
y{}^{″}+2{y}^{\prime }+2y=2{e}^{x}\mathrm{cos}\left(x\right)
and therefore I would like to 'complexify' the right hand side. This means that I want to write the right hand side as
q\left(x\right){e}^{\alpha x}
with q(x) a polynomial. How is this possible?
Find the inverse Laplace transform of the followin
\frac{s+4}{{s}^{2}+4s+8}
Find the inverse Laplace transform of the following:
\left(\frac{7}{2}\right)\mathrm{log}\left(\frac{s-8}{s+8}\right)
L\left\{{t}^{4}+{t}^{3}-{t}^{2}-t+\mathrm{sin}\sqrt{2}t\right\}
L\left\{{\int }_{0}^{t}\frac{1-{e}^{u}}{u}du\right\}
Solve initial value problem using laplace transform tables
{y}^{\prime }=t+3,y\left(0\right)=2
Use Laplace transform to evaluate the integral
{\int }_{0}^{\mathrm{\infty }}t{e}^{-2t}\mathrm{sin}\left(2t\right)dt
\frac{1}{B}
\frac{4s}{\left({s}^{2}+4{\right)}^{2}}
\frac{4\left(s+2\right)}{\left({s}^{2}+4s+8{\right)}^{2}}
|
I am to create a six character password that consists of 2 lowercase letters and 4 numbers. The lett
kulisamilhh 2022-05-03 Answered
{26}^{2}×{10}^{4}
{}^{6}{P}_{2}
You're just a little wrong on the letters-numbers arrangement part; it's
\left(\genfrac{}{}{0}{}{6}{2}\right)=15
{}^{6}{P}_{2}=30
(since we do the arrangement before the character assignment).
Now multiply this with your (correct) value of
{26}^{2}×{10}^{4}
character assignments to get the final answer:
{26}^{2}×{10}^{4}×\left(\genfrac{}{}{0}{}{6}{2}\right)
A
B
S
S=A\cup B
A\cap B=\mathrm{\varnothing }
\mathcal{P}\left(X\right)
X
|Y|
Y
|\mathcal{P}\left(A\right)|+|\mathcal{P}\left(B\right)|=|\mathcal{P}\left(A\right)\cup \mathcal{P}\left(B\right)|
\mathcal{P}\left(A\right)
\mathcal{P}\left(B\right)
\mathcal{P}\left(X\right)
|Y|
\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}
f
A\to B
|A|=4,|B|=3
{3}^{4}-\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}+\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
|
the failure of certain electronic device is suspected
Hypothesis test is a formal produce for investigatiny our ideas about the word using statistic. It is most often useb by scientists to test specific productions, called hypothesis, that arrives from theories.
1) State research hypothesis as a null hypothesis (
{H}_{0}
) and alternate hypothesis (
{H}_{a}
{H}_{1}
2) Collect data in a way designed to test the hypothesis.
3) Perfom am appropriate statistical test.
4) Decide whether to reject or fail to reject null hypothesis.
5) Present the finding in results and discuss on section.
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
Find the principal unit normal vector to the curve at the specified value of the parameter
r\left(t\right)=ti+\frac{6}{t}j,t=3
The three components of velocity in a flow field are given by
u={x}^{2}+{y}^{2}+{z}^{2}
v=xy+yz+{z}^{2}
w=-3xz-\frac{{z}^{2}}{2}+4
(a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field?
A=\left(1,0,-3\right),B=\left(-2,5,1\right),\text{ }and\text{ }C=\left(3,1,1\right)
. Calculate the following, expressing your answers as ordered triples (three comma-separated numbers).
\left(2\overline{B}\right)\left(3\overline{C}\right)
\left(\overline{B}\right)\left(\overline{C}\right)
\stackrel{\to }{A}\left(\stackrel{\to }{B}×\stackrel{\to }{C}\right)
(F)If
{\overline{v}}_{1}\text{ and }{\overline{v}}_{2}
are perpendicular,
|{\overline{v}}_{1}×{\overline{v}}_{2}|
(G) If
{\overline{v}}_{1}\text{ and }{\overline{v}}_{2}
are parallel,
|{\overline{v}}_{1}×{\overline{v}}_{2}|
|
Difference between PCA and Time-ordered linear model. Vector
\stackrel{\to }{AB}
\stackrel{\to }{BC}
existing in a 2-D space, represent a cell departed form state A, bypassing state B, finally reached to state C. Left: Principal component analysis generates two features, PC1 corresponds to a line that passes through the mean and minimizes sum squared error; PC2 is perpendicular to PC1. On PC1 and PC2, the order of two vectors may be lost. Right: Time-ordered linear model generates one cell state line
\stackrel{\to }{{e}_{all}}
. On this line, the order of two vectors are preserved perfectly, meanwhile, the distance ratio is also preserved:
\stackrel{\to }{AB\text{'}}:\stackrel{\to }{B\text{'}C\text{'}}=\stackrel{\to }{AB}:\stackrel{\to }{BC}
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Triangles - Incenter Practice Problems Online | Brilliant
The incenter of a triangle is the intersection of which 3 lines in the triangle?
Perpendicular Bisector of sides of triangle Angle Bisectors of triangle Altitudes of triangle Medians of triangle
Which of the following statements is true about the incenter?
It is the midpoint of the orthocenter and the circumcenter It is equidistant from the sides of the triangle It is equidistant from the midpoints of the sides of the triangle It is equidistant from the vertices of the triangle
Which of the following triangles doesn't have the incenter?
How can we find the point equidistant from all sides in a triangle?
by finding the intersection of the angle bisectors of the triangle by finding the intersection of the medians of the triangle by finding the intersection of the altitudes of the triangle by finding the intersection of the perpendicular bisectors of the sides of the triangle
I
in the diagram above is the incenter of
\triangle{ABC}
\angle{IAB}=40^\circ
\angle{IBC}=30^\circ
\angle{IBA}
35^\circ
40^\circ
45^\circ
30^\circ
|
Interface between electrical and mechanical translational domains - MATLAB - MathWorks United Kingdom
Translational Electromechanical Converter
Interface between electrical and mechanical translational domains
The Translational Electromechanical Converter block provides an interface between the electrical and mechanical translational domains. It converts electrical energy into mechanical energy in the form of translational motion, and vice versa. The converter is described with the following equations:
F=K·I
V=K·U
V Voltage across the electrical ports of the converter
I Current through the electrical ports of the converter
K Constant of proportionality
The Translational Electromechanical Converter block represents a lossless electromechanical energy conversion, therefore the same constant of proportionality is used in both equations.
Connections + and – are conserving electrical ports corresponding to the positive and negative terminals of the converter, respectively. Connections C and R are conserving mechanical translational ports. If the current flowing from the positive to the negative terminal is positive, then the resulting force is positive acting from port C to port R. This direction can be altered by using a negative value for K.
Constant of proportionality K
Constant of proportionality for electromechanical conversions. The default value is 0.1 V/(m/s).
Electrical conserving port associated with the converter positive terminal.
Electrical conserving port associated with the converter negative terminal.
Mechanical translational conserving port.
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Configurational entropy (qualitative) Practice Problems Online | Brilliant
In the above circuit, the voltage across the resistor
R = 13 \ \Omega
V = 0 \text{ V}.
If the voltage
V
is suddenly increased to
V' = 13 \text{ V},
what has happened to the entropy of the system of electrons in the circuit?
Don't know Decreased Increased Not changed
If a warm object contacts a cold object, in which direction will the heat flow?
Heat will not flow Can't say From cold to warm From warm to cold
100
air particles are flying around a room. Suddenly, all the particles are moved to one side of the room. What has happened to the entropy of the system?
Don't know Increased Decreased Not changed
If we put a cup of water into a freezer, what will happen to its entropy?
If ice cream is melted in a warm room, what happens to the entropy of the system?
Not changed Increased Decreased Don't know
|
A large tank is partially filled with 100
purbahassan 2022-04-16
A large tank is partially filled with 100 gallons of fluid in which 10 pounds of
salt is dissolved. Brine containing ½ pound of salt per gallon is pumped into
the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out
at a slower rate of 4 gal/min. Find the number of pounds of salt in the tank
For what value of c, is the function continuous on (−∞, ∞)
f(x)={𝑐𝑥^2+2𝑥 𝑖𝑓 𝑥<2
𝑥^3−𝑐𝑥 𝑖𝑓 𝑥 ≥2
{X}_{1},{X}_{2},\dots ,{X}_{n}
p
p
\stackrel{^}{P}
p
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}
\stackrel{^}{P}
\stackrel{―}{X}
{f}_{x}\left(\overline{X}\right)=\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)*{p}^{\sum _{i=1}^{n}{X}_{i}}*\left(1-p{\right)}^{n-\sum _{i=1}^{n}{X}_{i}}
\mathrm{ln} {f}_{X}\left(X\right)=\mathrm{ln}\left(\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)\right)+\sum _{i=1}^{n}{X}_{i}\mathrm{ln}\left(p\right)+ \left(n-\sum _{i=1}^{n}{X}_{i}\right)\mathrm{ln}\left(1-p\right)
\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^{n}{X}_{i}}{p}-\frac{\left(n-{\sum }_{i=1}^{n}{X}_{i}\right)}{\left(1-p\right)}=\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}
\left(\frac{\partial ln {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}=\left(\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}{\right)}^{2}=\frac{{n}^{2}{p}^{2}-2{n}^{2}p\overline{X}+{n}^{2}{\overline{X}}^{2}}{{p}^{2}\left(1-p{\right)}^{2}}
E\left[{\stackrel{―}{X}}^{2}\right]={\mu }^{2}+\frac{{\sigma }^{2}}{n}
E\left[X\right]=\mu =p=E\left[\stackrel{―}{X}\right]
Var\left[X\right]={\sigma }^{2}=p\left(1-p\right)
E\left[\left(\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}\right]=\frac{{n}^{2}{p}^{2}-2{n}^{2}pE\left[\overline{X}\right]+{n}^{2}E\left[{\overline{X}}^{2}\right]}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{{n}^{2}{p}^{2}-2{n}^{2}{p}^{2}+{n}^{2}\left({p}^{2}+\frac{p\left(1-p\right)}{n}\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{np\left(1-p\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{n}{p\left(1-p\right)}
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}=\frac{1}{n\frac{n}{p\left(1-p\right)}}=\frac{p\left(1-p\right)}{{n}^{2}}
\frac{p\left(1-p\right)}{n}
n\mid \varphi \left({a}^{n}-1\right)
. Any natural solution that uses
Aut\left(G\right)
|
CybergeoNetworks
Full-text Semantic network
Semantic area
Geo-semantic Networks
Cybergeo | 1996-2016
Cybergeo turns 20: it’s time to look back for reflection and to anticipate future evolution! cybergeo.revues.org/
First entirely electronic journal for social sciences in the world, peer reviewed, European, open (free of charge for authors and readers), with a focus on geography and widely open to the diversity of research agendas and methodologies in all countries. Cybergeo is a success story with now more than one million papers downloaded every year.
An app to look back
This app builds on 20 years of publication in Cybergeo. You can play with data, drawing geographical networks of authoring, studying and citing through countries, analyzing semantic networks per key words and articles’ content, you can review twenty years of epistemological and thematic trends in a variety of fields of scientific interest. The networks tell who studies what, where and how. Data are regularly updated.
All data, materials and source codes are freely available on this repository: github.com/AnonymousAuthor3/cybergeo20
scientific articles from authors
authoring countries
citations from other articles
citations of other articles
Indicator to map
Authoring countries Countries Studied Countries Studies by Locals
Search and select a cybergeo paper in the table.
Citation network neighborhood
This tab allows to explore the disciplinary neighborhood of cybergeo. It combines citation data with semantic content of articles. The exploration can be at the article level or at an aggregated level with the full semantic network.
The citation network is constituted by articles citing cybergeo articles, cited by cybergeo articles, and citing the same articles than cybergeo (citing cited). Structure and number of papers is summed up by the following figure :
The data is collected from google scholar (allowing to get only citing articles, thus the particular network structure).
For around 200000 articles of the citation network, abstracts were collected via the Mendeley API. The extraction of significant keywords (n-grams), following [Chavalarias and Cointet, 2013], allows the construction of a semantic network by co-occurrence analysis, which nodes are keywords. Community detection (Louvain algorithm), optimized on filtering parameters (hub filtering and low edge filtering), yield disciplines which are given in the following figure :
The tab Network Exploration provides article-level exploration. The user can search an article in cybergeo with the datatable. Once one is selected, different informations are displayed if available [note that ~600 papers among 900 have citation and semantic information].
citation neighborhood of the selected article
semantic content of the selected article as a colored wordcloud (disciplines legend above), and semantic content for the neighborhood
geo espace territoire environnement société réseau interaction aménagement urbanisme carte modèle système SIG fractale durabilité représentation migration quantitatif qualitatif post-moderne
Climate and environment Development, territory, system of cities Economic geography Emotional geography History and epistemology Imagery and GIS Mobility and transportation Statistics and modeling Sustainability, risk, planning Urban dynamics
Nodes proportional to:
Nodes degree
Edges proportional to:
Edges size
accessibility agriculture air anthropic applied assessment behavior bibliometrics biodiversity biogeography border cartography centrality city climate coastal cognition communication community complex system concept conflict crisis culture data decision making decision support demographics density development didactics diffusion distance dynamics ecology economic activity economics emotion energy environment environmental change environmental degradation epistemology evolution festival finance flooding forest frontier game gender geography geohistory geology geomorphology geopolitics gis globalization graph theory harbor health hierarchy history housing ideology imagery imaginary inequality innovation institution interaction knowledge labor land use landscape learning lifestyle linguistics location logics logistics management manufacturing methodology metropolization migration mobility model morphology mountain movie multi-scale network participatory perception periphery physics planning policy politics population poverty property prospective public space publishing qualitative method region resource restoration ecology risk rural scale science scientist sea security segregation service simulation software space spatial analysis spatial statistics stakeholder state statistics suburb sustainability system of cities technical network technology territory time geography tourism trade transportation urban urban decline urban growth urban-rural urbanization virtual visualization voyage vulnerability water web web mapping
Font size (min)
Font size (max)
Exploratory Analysis of Cybergeo Keywords
Vertices and nodes attributes
The vertices are described by two variables: frequency and degree. The frequency is the number of articles citing the keyword. The degree is the total degree of the nodes in the network, that is the number of edges linking thiw keyword to the others (there is no distinction between in- and out- degree as the network is undirected). Both variables are distinct but correlated.
The edges are described by two variables: observed weight and relative residual. For two given keywords the observed weight is the number of articles citing both keywords. The relative residual is the ratio between the observed weight and the expected weigth of the edge. For a given edge the expected weight is the probability that this edge exists considering the degree of the nodes. It is computed as the union of two dependant probabilities.
The probability of drawing a vertex i equals
\frac{{w}_{i}}{w}
{w}_{i}
is the degree of vertex i (weighted degree) and w the half sum of weights.
Then the probability of drawing a vertex j distinct from i equals
\frac{{w}_{j}}{w-{w}_{i}}
The probability of existence of an edge between i and j is:
{P}_{i->j}=\frac{{w}_{i}}{w}×\frac{{w}_{j}}{w-{w}_{i}}
The probability of existence of an edge between j and i:
{P}_{j->i}=\frac{{w}_{j}}{w}×\frac{{w}_{i}}{w-{w}_{j}}
The probability of existence of an undirected edge is the union of both probabilities:
{P}_{i<->j}={P}_{i->j}+{P}_{j->i}
Eventually the expected weight is:
{w}^{e}=w\left(\frac{{P}_{i<->j}}{2}\right)
Algorithme de détection de communautés
The community detection is computed with the Louvain algorithm which finds an optimum of modularity. See Blondel et al. 2008.
Citations Keywords Semantic
Authoring Studied
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Second-order tunable notching and peaking IIR filter - MATLAB - MathWorks í•œêµ
Specify the filter’s 3 dB bandwidth as a finite positive numeric scalar in Hz. The value must be a scalar between 0 and half the sample rate.
Specify the filter’s center frequency (for both the notch and the peak) as a finite positive numeric scalar in Hz. The value must be a scalar between 0 and half the sample rate.
Specify the value that determines the filter’s 3 dB bandwidth as a finite numeric scalar in the range [-1 1]. The value -1 corresponds to the maximum 3 dB bandwidth (SampleRate/4), and 1 corresponds to the minimum 3 dB bandwidth (0 Hz, an allpass filter).
Specify the coefficient that determines the filter’s center frequency as a finite numeric scalar in the range [-1 1]. The value -1 corresponds to the minimum center frequency (0 Hz), and 1 corresponds to the maximum center frequency (SampleRate/2 Hz). The default is 0, which corresponds to SampleRate/4 Hz.
\frac{\mathrm{Fc}}{\mathit{Q}}
Bnotch = 1×3
Anotch = 1×3
Bpeak = 1×3
Apeak = 1×3
H\left(z\right)=\left(1âb\right)\frac{1â{z}^{â2}}{1â2b\mathrm{cos}{w}_{0}{z}^{â1}+\left(2bâ1\right){z}^{â2}}
H\left(z\right)=b\frac{1â2\mathrm{cos}{w}_{0}{z}^{â1}+{z}^{â2}}{1â2b\mathrm{cos}{w}_{0}{z}^{â1}+\left(2bâ1\right){z}^{â2}}
b=\frac{1}{1+\mathrm{tan}\left(\mathrm{Î}w/2\right)}
where ω0 = 2πf0/fs is the center frequency in radians/sample (f0 is the center frequency in Hz and fs is the sampling frequency in Hz). Δω = 2πΔf/fs is the 3 dB bandwidth in radians/sample (Δf is the 3 dB bandwidth in Hz). Note that the two filters are complementary:
\begin{array}{l}{H}_{\text{notch}}\left(z\right)+{H}_{\text{peak}}\left(z\right)=1\\ \text{they can be written as:}\\ {H}_{\text{peak}}\left(z\right)=\frac{1}{2}\left[1âA\left(z\right)\right]\\ {H}_{\text{notch}}\left(z\right)=\frac{1}{2}\left[1+A\left(z\right)\right]\\ \text{where }A\text{(z) is a }{2}^{\text{nd}}\text{ order allpass filter}\text{.}\\ A\left(z\right)=\frac{{a}_{2}+{a}_{1}{z}^{â1}+{z}^{â2}}{1+{a}_{1}{z}^{â1}+{a}_{2}{z}^{â2}}\\ \text{and}\\ {a}_{1}=â2b\mathrm{cos}{\mathrm{Ï}}_{0}\\ {a}_{2}=2bâ1\end{array}
\begin{array}{l}{G}_{\text{3dB}}={a}_{2}=2bâ1\\ {G}_{\text{cf}}=\frac{{a}_{1}â{a}_{1}{a}_{2}}{1â{a}_{2}{}^{2}}=â\mathrm{cos}{w}_{0}\end{array}
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Pictures of Julia and Mandelbrot Sets/Julia and Mandelbrot sets for transcendental functions - Wikibooks, open books for an open world
Pictures of Julia and Mandelbrot Sets/Julia and Mandelbrot sets for transcendental functions
{\displaystyle sin(z),cos(z),exp(z),...}
{\displaystyle sin(z)+c}
{\displaystyle sin(z)+c}
{\displaystyle sin(z)}
{\displaystyle cos(z)}
{\displaystyle Sin(z)}
{\displaystyle sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y)}
{\displaystyle sin(z)+c}
{\displaystyle cos(x)}
{\displaystyle sin(z)}
{\displaystyle sin(z)+c}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c}
{\displaystyle 1/cos(z1+c)+c}
{\displaystyle 1/cos(z2+c)+c}
{\displaystyle 2\pi }
{\displaystyle 1/cos(x)-1}
{\displaystyle cos(z)}
{\displaystyle 1-z^{2}/2!+z^{4}/4!-z^{6}/6!+z^{8}/8!-...}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))-1}
{\displaystyle (1-z^{2})/(z-z^{2}cos(z))-1}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Pictures_of_Julia_and_Mandelbrot_Sets/Julia_and_Mandelbrot_sets_for_transcendental_functions&oldid=3703414"
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Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three.
Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?
My approach:
N\left(A\cup B\cup C\right)=N\left(A\right)+N\left(B\right)+N\left(C\right)-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+N\left(A\cap B\cap C\right)
21=9+10+7-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+5
N\left(A\cap B\right)+N\left(A\cap C\right)+N\left(B\cap C\right)=10
Now the LHS has counted
N\left(A\cap B\cap C\right)
three times, so I will remove it two times as:-
Number of persons eating at least two dishes
=N\left(A\cap B+B\cap C+A\cap C\right)-2\ast N\left(A\cap B\cap C\right)=10-2\ast 5=0
Now it contradicts the questions that there are 5 eating all three dishes.
Is this anything wrong in my approach?
As far as I can see there is something wrong with your approach, but also the question does not have enough information to give a definite answer.
Your mistake is to assume that
N\left(A\cup B\cup C\right)=21
. However presumably the 21 people may include some who eat neither vegetables nor fish nor eggs: if there are n such people then you should have
N\left(A\cup B\cup C\right)=21-n\text{ }.
If you now follow your method you should get the number of people you want also to be
n
n
is not known, the problem cannot be answered. In fact as you point out, the answer must be at least 5, so we know that
n\ge 5
. You can find solutions by trial and error with
n=5,6,7,8,9,10
, so any of these is a possible answer.
f
A\to B
|A|=4,|B|=3
The number of surjective functions by applying the principle of inclusion exclusion is given by:
{3}^{4}-\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}+\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
The rationale is that we begin with the set of all possible functions, and subtract off the functions with one element in the codomain that is not in the range, and add back the functions with two elements in the codomain not in the range.
However, I don't understand why we are adding back
\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}
already include functions with two elements from the codomain not in the range, then wouldn't it be done already, as that's all the non-surjective functions?
Let n be a positive integer. Find the number of permutations of
\left(1,2,...,n\right)
such that no number remains in its original place.
Solution: To do this, first we are going to count the number of permutations where at least one number remains in its place, according to the inclusion-exclusion principle, we must first add the permutations where a given number is fixed, then subtract the permutations where 2 given numbers are fixed and so on.
To find a permutation that fixes
k
given elements, we only have to arrange the rest, which can be done in
\left(n-k\right)!
ways. However, if we do this for every choice of
k
elements, we are counting
\left(\genfrac{}{}{0}{}{n}{k}\right)\left(n-k\right)!=\frac{n!}{k!}
permutations. Since in total there are
n!
permutations we get as our result:
n!-\left(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+\left(-1{\right)}^{n+1}\frac{n!}{n!}\right)
I'm a bit confused about a point of the solution, when we fix
k
elements and rearrange the other
n-k
, some of the remaining elements will stay fixed in their position right? so why does this work?
A stained glass window consists of nine squares of glass in a 3x3 array. Of the nine squares, k are red, the rest blue. A set of windows is produced such that any possible window can be formed in just one way by rotating and/or turning over one of the windows in the set. Altogether there are more than 100 red squares in the set. Find k.
first, there are 8 Isometries of a square.
Identity, three rotations (90,-90,180) four reflections (vertical, horizontal, two diagonal axis). let G be the permutation group, then |G|=8, and I can find fix(g) for every g.
can someone give me a hint of how to proceed from there.
I'm missing something here. Let
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
by conjugation (inner automorphisms) and
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
A catering service offers 12 appetizers, 9 main courses, and 6 desserts. A customer is to select 5 appetizers, 3 main courses, and 4 desserts for a banquet. In how many ways can this be done?
Ministry of Education are inviting tender for four categories promoting the use of IT in education. Each category consists of 5, 4, 3, and 7 projects, respectively. Each project appears on exactly one category. How many possible projects are there to choose from? Explain your answer.
\left(5+4\right)+\left(4+4\right)+\left(3+3\right)+\left(7+4\right)=9+8+6+11=34
possible projects to choose from. I used the sum rule here.
I came across this question where I was asked to calculate the probability of drawing 2 identical socks from a drawer containing 24 socks (7 black, 8 blue, and 9 green). When solving this question, I tried tackling it using the counting principle directly as opposed to simply applying the combinatorics formula in hopes to get comfortable with that principle. Though, when I looked up the solution at the end of the book, the answer was completely different, and I am not sure why that is.
What I did was as follows: First, I noticed that when we draw the first sock, we have 24 different options to choose from and then only 1 in order to match it with the first one we chose. Then, I divided it by the total number of ways we can draw 2 socks out of 24 and got the expression:
\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}
The solution at the back of the book took care of each case separately (choosing 2 out of 7 blue socks, OR choosing 2 out of 8 blue socks OR choosing 2 out of 9 green socks), which makes sense, but what is wrong with my approach?
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Prove if F(\sqrt[n]{a}) is unramified or totally ramified in certain
F\left(\sqrt[n]{a}\right)
Tatairfzk
For (1)
{x}^{n}-a
is separable in the residue field
\frac{{O}_{F}}{\left({\pi }_{F}\right)}
\frac{F\left({a}^{\frac{1}{n}}\right)}{F}
is automatically unramified.
Note that Hensel lemma gives that
{\zeta }_{q-1}\in F\left({a}^{\frac{1}{n}}\right)
where q is the cardinality of the residue field, and
{x}^{n}-a
is separable in the residue field so Hensel lemma again gives that
{a}^{\frac{1}{n}}\in F\left({\zeta }_{q-1}\right)
F\left({a}^{\frac{1}{n}}\right)=F\left({\zeta }_{q-1}\right)
For (2) take
nl+mv\left(a\right)=1
b={a}^{m}{\pi }_{F}^{nl},\text{ }v\left(b\right)=1,\text{ }F\left({a}^{\frac{1}{n}}\right)=F\left({b}^{\frac{1}{n}}\right)
{x}^{n}-b
is Eisenstein over
{O}_{F}
\frac{F\left({b}^{\frac{1}{n}}\right)}{F}
has degree n and
v\left({b}^{\frac{1}{n}}\right)=\frac{1}{n}
, it is totally ramified.
\left\{171,172,173,..,286\right\}
⇒
Deriving the confidence interval
P\left(-{\mathrm{\Phi }}_{\alpha }^{-1}<X<{\mathrm{\Phi }}_{\alpha }^{-1}\right)=1-2\alpha
P\sim \mathcal{N}\left(0,1\right)
\begin{array}{rlrlr} & P\left(-{\mathrm{\Phi }}_{\alpha }^{-1}<X<{\mathrm{\Phi }}_{\alpha }^{-1}\right)& & & \left(1\right)\\ & P\left(X<{\mathrm{\Phi }}_{\alpha }^{-1}\right)-P\left(X<-{\mathrm{\Phi }}_{\alpha }^{-1}\right)& & P\left(A<X<B\right)=P\left(X<B\right)-P\left(X<A\right)& \left(2\right)\\ & \mathrm{\Phi }\left({\mathrm{\Phi }}_{\alpha }^{-1}\right)\right)-\mathrm{\Phi }\left(-{\mathrm{\Phi }}_{\alpha }^{-1}\right)\right)& & P\left(X<A\right)=\mathrm{\Phi }\left(A\right)& \left(3\right)\\ & \mathrm{\Phi }\left({\mathrm{\Phi }}_{\alpha }^{-1}\right)\right)-\left(1-\mathrm{\Phi }\left({\mathrm{\Phi }}_{\alpha }^{-1}\right)\right)\right)& & \mathrm{\Phi }\left(-x\right)=1-\mathrm{\Phi }\left(x\right)& \left(4\right)\\ & \alpha -1+\alpha & & f\left({f}^{-1}\right)=id& \left(5\right)\\ & \textcolor[rgb]{1,0,0}{-}\textcolor[rgb]{1,0,0}{1}×\left(1-2\alpha \right)& & & \left(6\right)\end{array}
Find the area of the described region.
region enclosed by one petal of
r = 8 cos(9𝜃)
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How do you find the critical point and determine whether
Addison Gross 2022-01-25 Answered
How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for
f\left(x,y\right)={x}^{2}+4x+{y}^{2}
trasahed
The first-order partial derivatives of
z=f\left(x,y\right)={x}^{2}+4x+{y}^{2}
\frac{\partial z}{\partial x}=2x+4
\frac{\partial z}{\partial y}=2y
Setting both of these equal to zero results in a system of equations whose unique solution is clearly
\left(x,y\right)=\left(-2,0\right)
, so this is the unique critical point of
The second-order partials are
\frac{{\partial }^{2}z}{\partial {x}^{2}}=2,\frac{{\partial }^{2}z}{\partial {y}^{2}}=2
\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{{\partial }^{2}z}{\partial y\partial x}=0
This makes the discriminant for the (multivariable) Second Derivative Test equal to
D=\frac{{\partial }^{2}z}{\partial {x}^{2}}\cdot \frac{{\partial }^{2}z}{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}z}{\partial x\partial y}\right)}^{2}=2\cdot 2-{0}^{2}=4>0
which means the critical point is either a local max or a local min (it's not a saddle point).
\frac{{\partial }^{2}z}{\partial {x}^{2}}=2>0
, the critical point is a local min.
lilwhitelieyc
f\left(x,y\right)={x}^{2}+4x+{y}^{2}
{f}_{x}\left(x,y\right)=2x+4
{f}_{y}\left(x,y\right)=2y
The critical point makes both partial derivatives 0 (simultaneously).
For this function there is one critical point:
\left(-2,0\right)
To determine whether f has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.)
{f}_{xx}\left(x,y\right)=2
{f}_{xy}\left(x,y\right)=0
(As usual, this is also
{f}_{yx}\left(x,y\right)
{f}_{yy}\left(x,y\right)=2
Evaluate the second partials at the critical point (In this case they are all constant, but in general we cannot skip this step.)
At the critical point
\left(-2,0\right)
A={f}_{xx}\left(-2,0\right)=2
B={f}_{xy}\left(-2,0\right)=0
C={f}_{yy}\left(-2,0\right)=2
D=AC-{B}^{2}
D=\left(2\right)\left(2\right)-{\left(0\right)}^{2}=4
Apply the second derivative test:
Since D is positive, we look at A and with
D>0
A>0
, we have a local minimum at the critical point.
\left(-2,0\right)=4-8=-4
f has a local minimum of−4 at
\left(-2,0\right)
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
In the given equation as follows, use Taylor’s Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error|:-
see the equation as attached here:-
\mathrm{cos}\left(0.3\right)\approx 1-\frac{{\left(0.3\right)}^{2}}{2!}+\frac{{\left(0.3\right)}^{4}}{4!}
How do you find critical points of multivariable function
f\left(x,y\right)={x}^{3}+xy-{y}^{3}
Find the indefinite integral by making a change of variables
\int x\sqrt{\left(3x-4\right)}dx
\mathbb{R}\to \mathbb{R}
g\left(t\right)=f\left(x+t\left(y-x\right)\right)
f:{\mathbb{R}}^{n}\to \mathbb{R}
x,y\in {\mathbb{R}}^{n}
g\prime \left(t\right)={\left(y-x\right)}^{T}\mathrm{\nabla }f\left(x+t\left(y-x\right)\right)
\begin{array}{|ccccccc|}\hline x& & & & & & \\ p\left(x\right)& & & & & & \\ \hline\end{array}
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Laplace transform of \(\displaystyle{f{{\left({t}^{{2}}\right)}}}\) Suppose we know the
Quinn Moses 2022-04-16 Answered
f\left({t}^{2}\right)
F\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt
muthe2ulj
You can change variables
{t}^{2}=x
H\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left({t}^{2}\right){e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)}{2\sqrt{x}}{e}^{-s\sqrt{x}}dx
Now we can express
\frac{{e}^{-s\sqrt{x}}}{\sqrt{x}}
as a Laplace transform, so:
={\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)}{2\sqrt{\pi t}}{e}^{-\frac{{s}^{2}}{4t}}{e}^{-xt}\text{ }dt\text{ }dx
={\int }_{0}^{\mathrm{\infty }}\frac{F\left(t\right)}{2\sqrt{\pi t}}{e}^{-\frac{{s}^{2}}{4t}}dt
L\left\{9+6t\right\}
L\left\{k\right\}=\frac{k}{s}
L\left\{t\right\}=\frac{1}{{s}^{2}}
You drink a Starbucks Coffee in SM Seaside with a 120 mg. of caffeine. Each hour, the caffeine in your body system reduces by about 12 percent. How long will you have a 10 mg. of caffeine?
Use Laplace transforms to solve the following initial value problem
y"-{y}^{\prime }-6y=0
y\left(0\right)=1
{y}^{\prime }\left(0\right)=-1
L\left\{{u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right\}
a\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{4}}-\frac{5}{{s}^{3}}+\frac{6}{{s}^{2}}\right)
b\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{3}}-\frac{5}{{s}^{2}}+\frac{6}{s}\right)
c\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{4}}
d\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{3}}
e\right)F\left(s\right)={e}^{-3s}\frac{2-11s+30{s}^{2}}{{s}^{3}}
y=\mathrm{tan}\sqrt{7t}
y=\frac{3-{v}^{2}}{3+{v}^{2}}
Solve the ff Differential equations. Indicate the type of differential equation
y\mathrm{ln}x\mathrm{ln}ydx+dy=0
u\left(t-2\right)
\frac{1}{s}+2
\frac{1}{s}-2
{e}^{2}\frac{s}{s}\left(d\right)\frac{{e}^{-2s}}{s}
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Revision as of 13:36, 30 July 2013 by NikosA (talk | contribs) (→Deriving Physical Quantities: conversion factors constants tables cleaned-up)
{\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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Arpan Sadhu
| x^2- 5 x | >| x^2 |- | 5 x |
Find domain of x
solve inequality using wavy curve method (x-1/x+1)-x
9th one pls fast!!!
solve the following system of inequations:
\left(2\right) 2\left(2x+3\right)-10<6\left(x-2\right), \frac{2x-3}{4}+6\ge 2+\frac{4x}{3}\phantom{\rule{0ex}{0ex}}\left(4\right) 11-5x>-4, 4x+13\le -11\phantom{\rule{0ex}{0ex}}\left(6\right) 2\left(x-6\right)<3x-7, 11-2x<6-x
Q. Solve the following system of inequations
(2) 2 (2x + 3) - 10 < 6 (x - 2),
\frac{2x-3}{4}+6\ge 2+\frac{4x}{3}
Q). Solve the following system of equations
\frac{5x}{4}+\frac{3x}{8}>\frac{39}{8}, \frac{2x-1}{12}-\frac{x-1}{3}<\frac{3x+1}{4}
Guntas Dhillon
Q 13 and 14 pls......pls ans fast I have test tomorrow
\frac{\left|x\right|-10}{\left|x\right|+20}>0
\frac{\left|x\right|-10}{\left|x\right|-20}\le 0
Yatharth Patel
\mathbf{8}\mathbf{.}\mathbf{ } Let a = \left(\frac{1}{9}\right){ }^{-2 {\mathrm{log}}_{3} 7} and b = {{{2}^{-\mathrm{log}}}_{\frac{1}{2}}}^{\left(7\right)} then a = {\left(b\right)}^{k} where k is equal to :\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{9}\mathbf{.} If {\mathrm{log}}_{3}5 = x and {\mathrm{log}}_{25}11 = y then the value of {\mathrm{log}}_{3} \left(\frac{11}{3}\right) in terms of x and y is
Log 3 to the base 20 lies between which intervals?
Is this sol. Correct???
Q9. Solve the inequation: 37 – (3x + 5)
\ge
9x – 8 (x – 3)
37 – 3x – 5
\ge
9x – 8x + 24
32-3\mathrm{x}\ge \mathrm{x}++24\phantom{\rule{0ex}{0ex}}32-3\mathrm{x}-32\ge \mathrm{x}+24-32\phantom{\rule{0ex}{0ex}}-3\mathrm{x}\ge \mathrm{x}-8\phantom{\rule{0ex}{0ex}}-3\mathrm{x} -\mathrm{x}\ge \mathrm{x}-8-\mathrm{x}\phantom{\rule{0ex}{0ex}}-4\mathrm{x}\ge -8\phantom{\rule{0ex}{0ex}}-4\mathrm{x}-4\ge -8/-4\phantom{\rule{0ex}{0ex}} \mathrm{x}\ge 2
If the eqn is formed by decreasing each root of ax^2+bx+c=0 by 1 is2x^2+8x+2=0 then find values of a/b and b/c
what is the sum of 1^2+2.2^2 + 3^2 + 2.4^2+5^2+2.62+....
Pls solve 15 question.....it's urgent...pls solve it in detail
Q15. If –
\sqrt{2x + 3} \ge x -1
\in
\left[\frac{-3}{2},\infty \right)
\left[2-\sqrt{6 }, \infty \right)
(c)
\left[\frac{-3}{2},0\right)
\left[-\frac{3}{2},2-\sqrt{6}\right]
Plz ans soon...dnt provide similar ink...
29 Solve the equation
{x}^{2 }+ \frac{x}{\sqrt{2}}+2 =0
{x}^{2}+\frac{x}{\sqrt{2}}+2=0
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Revision as of 09:41, 27 July 2013 by NikosA (talk | contribs) (various changes in structure, typos, added source to L7 handbook)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\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})}
|
Zitterbewegung - Wikipedia
In physics, the zitterbewegung ("jittery motion" in German) is the predicted rapid oscillatory motion of elementary particles that obey relativistic wave equations. The existence of such motion was first discussed by Gregory Breit in 1928[1][2] and later by Erwin Schrödinger in 1930[3][4] as a result of analysis of the wave packet solutions of the Dirac equation for relativistic electrons in free space, in which an interference between positive and negative energy states produces what appears to be a fluctuation (up to the speed of light) of the position of an electron around the median, with an angular frequency of 2mc2/ℏ, or approximately 1.6×1021 radians per second. For the hydrogen atom, zitterbewegung can be invoked as a heuristic way to derive the Darwin term, a small correction of the energy level of the s-orbitals.
1.1 Free fermion
1.2 Interpretation as an artifact
2 Experimental simulation
Free fermionEdit
The time-dependent Dirac equation is written as
{\displaystyle H\psi (\mathbf {x} ,t)=i\hbar {\frac {\partial \psi }{\partial t}}(\mathbf {x} ,t)}
{\displaystyle \hbar }
is the (reduced) Planck constant,
{\displaystyle \psi (\mathbf {x} ,t)}
is the wave function (bispinor) of a fermionic particle spin-½, and H is the Dirac Hamiltonian of a free particle:
{\displaystyle H=\beta mc^{2}+\sum _{j=1}^{3}\alpha _{j}p_{j}c}
{\textstyle m}
is the mass of the particle,
{\textstyle c}
{\textstyle p_{j}}
{\displaystyle \beta }
{\displaystyle \alpha _{j}}
are matrices related to the Gamma matrices
{\textstyle \gamma _{\mu }}
{\textstyle \beta =\gamma _{0}}
{\textstyle \alpha _{j}=\gamma _{0}\gamma _{j}}
In the Heisenberg picture, the time dependence of an arbitrary observable Q obeys the equation
{\displaystyle -i\hbar {\frac {\partial Q}{\partial t}}=\left[H,Q\right].}
In particular, the time-dependence of the position operator is given by
{\displaystyle {\frac {\partial x_{k}(t)}{\partial t}}={\frac {i}{\hbar }}\left[H,x_{k}\right]=c\alpha _{k}}
where xk(t) is the position operator at time t.
The above equation shows that the operator αk can be interpreted as the k-th component of a "velocity operator".
{\displaystyle \left\langle \left({\frac {\partial x_{k}(t)}{\partial t}}\right)^{2}\right\rangle =c^{2}}
as if the "root mean square speed" in every direction of space is the speed of light.
To add time-dependence to αk, one implements the Heisenberg picture, which says
{\displaystyle \alpha _{k}(t)=e^{\frac {iHt}{\hbar }}\alpha _{k}e^{-{\frac {iHt}{\hbar }}}}
The time-dependence of the velocity operator is given by
{\displaystyle \hbar {\frac {\partial \alpha _{k}(t)}{\partial t}}=i\left[H,\alpha _{k}\right]=2\left(i\gamma _{k}m-\sigma _{kl}p^{l}\right)=2i\left(p_{k}-\alpha _{k}H\right)}
{\displaystyle \sigma _{kl}\equiv {\frac {i}{2}}\left[\gamma _{k},\gamma _{l}\right].}
Now, because both pk and H are time-independent, the above equation can easily be integrated twice to find the explicit time-dependence of the position operator.
{\displaystyle \alpha _{k}(t)=\left(\alpha _{k}(0)-cp_{k}H^{-1}\right)e^{-{\frac {2iHt}{\hbar }}}+cp_{k}H^{-1}}
{\displaystyle x_{k}(t)=x_{k}(0)+c^{2}p_{k}H^{-1}t+{\tfrac {1}{2}}i\hbar cH^{-1}\left(\alpha _{k}(0)-cp_{k}H^{-1}\right)\left(e^{-{\frac {2iHt}{\hbar }}}-1\right)}
The resulting expression consists of an initial position, a motion proportional to time, and an oscillation term with an amplitude equal to the reduced Compton wavelength. That oscillation term is the so-called zitterbewegung.
Interpretation as an artifactEdit
In quantum mechanics, the zitterbewegung term vanishes on taking expectation values for wave-packets that are made up entirely of positive- (or entirely of negative-) energy waves. The standard relativistic velocity can be recovered by taking a Foldy–Wouthuysen transformation, when the positive and negative components are decoupled. Thus, we arrive at the interpretation of the zitterbewegung as being caused by interference between positive- and negative-energy wave components.
In quantum electrodynamics the negative-energy states are replaced by positron states, and the zitterbewegung is understood as the result of interaction of the electron with spontaneously forming and annihilating electron-positron pairs.[5]
More recently, it has been noted that in the case of free particles it could just be an artifact of the simplified theory. Zitterbewegung appear as due to the "small components" of the dirac 4-spinor, due to a little bit of antiparticle mixed up in the particle wavefunction for a nonrelativistic motion. It doesn't appear in the correct second quantized theory, or rather, it is resolved by using Feynman propagators and doing QED. Nevertheless, it is an interesting way to understand certain QED effects heuristically from the single particle picture. [6]
Experimental simulationEdit
Zitterbewegung of a free relativistic particle has never been observed directly, although some authors believe they have found evidence in favor of its existence.[7] It has also been simulated twice in model systems that provide condensed-matter analogues of the relativistic phenomenon. The first example, in 2010, placed a trapped ion in an environment such that the non-relativistic Schrödinger equation for the ion had the same mathematical form as the Dirac equation (although the physical situation is different).[8][9] Then, in 2013, it was simulated in a setup with Bose–Einstein condensates.[10]
Other proposals for condensed-matter analogues include semiconductor nanostructures, graphene and topological insulators.[11][12][13][14]
^ Breit, Gregory (1928). "An Interpretation of Dirac's Theory of the Electron". Proceedings of the National Academy of Sciences. 14 (7): 553–559. doi:10.1073/pnas.14.7.553. ISSN 0027-8424. PMC 1085609. PMID 16587362.
^ Greiner, Walter (1995). Relativistic Quantum Mechanics. doi:10.1007/978-3-642-88082-7. ISBN 978-3-540-99535-7.
^ Schrödinger, E. (1930). Über die kräftefreie Bewegung in der relativistischen Quantenmechanik [On the free movement in relativistic quantum mechanics] (in German). pp. 418–428. OCLC 881393652.
^ Schrödinger, E. (1931). Zur Quantendynamik des Elektrons [Quantum Dynamics of the Electron] (in German). pp. 63–72.
^ Zhi-Yong, W., & Cai-Dong, X. (2008). Zitterbewegung in quantum field theory. Chinese Physics B, 17(11), 4170.
^ "Dirac equation - is Zitterbewegung an artefact of single-particle theory?".
^ Catillon, P.; Cue, N.; Gaillard, M. J.; et al. (2008-07-01). "A Search for the de Broglie Particle Internal Clock by Means of Electron Channeling". Foundations of Physics. 38 (7): 659–664. doi:10.1007/s10701-008-9225-1. ISSN 1572-9516. S2CID 121875694.
^ Wunderlich, Christof (2010). "Quantum physics: Trapped ion set to quiver". Nature News and Views. 463 (7277): 37–39. doi:10.1038/463037a. PMID 20054385.
^ Gerritsma; Kirchmair; Zähringer; Solano; Blatt; Roos (2010). "Quantum simulation of the Dirac equation". Nature. 463 (7277): 68–71. arXiv:0909.0674. Bibcode:2010Natur.463...68G. doi:10.1038/nature08688. PMID 20054392. S2CID 4322378.
^ Leblanc; Beeler; Jimenez-Garcia; Perry; Sugawa; Williams; Spielman (2013). "Direct observation of zitterbewegung in a Bose–Einstein condensate". New Journal of Physics. 15 (7): 073011. arXiv:1303.0914. doi:10.1088/1367-2630/15/7/073011. S2CID 119190847.
^ Schliemann, John (2005). "Zitterbewegung of Electronic Wave Packets in III-V Zinc-Blende Semiconductor Quantum Wells". Physical Review Letters. 94 (20): 206801. arXiv:cond-mat/0410321. doi:10.1103/PhysRevLett.94.206801. PMID 16090266. S2CID 118979437.
^ Katsnelson, M. I. (2006). "Zitterbewegung, chirality, and minimal conductivity in graphene". The European Physical Journal B. 51 (2): 157–160. arXiv:cond-mat/0512337. doi:10.1140/epjb/e2006-00203-1. S2CID 119353065.
^ Dóra, Balász; Cayssol, Jérôme; Simon, Ference; Moessner, Roderich (2012). "Optically engineering the topological properties of a spin Hall insulator". Physical Review Letters. 108 (5): 056602. arXiv:1105.5963. doi:10.1103/PhysRevLett.108.056602. PMID 22400947. S2CID 15507388.
^ Shi, Likun; Zhang, Shoucheng; Cheng, Kai (2013). "Anomalous Electron Trajectory in Topological Insulators". Physical Review B. 87 (16). arXiv:1109.4771. doi:10.1103/PhysRevB.87.161115. S2CID 118446413.
Messiah, A. (1962). "XX, Section 37" (pdf). Quantum Mechanics. Vol. II. pp. 950–952. ISBN 9780471597681.
Geometric Algebra in Quantum Mechanics
Retrieved from "https://en.wikipedia.org/w/index.php?title=Zitterbewegung&oldid=1057317634"
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BAL for Gas was terminated on Oct 27, 2021 after the launch of the Balancer Gnosis Protocol, which enabled gasless trades.
The BAL for Gas program has become an important mechanism for distributing Balancer protocol governance power to one of its widest user bases: traders.
To ensure lasting robustness of the governance process, both liquidity providers and traders should have a say in how the protocol evolves. While liquidity providers receive BAL as a function of the amount of liquidity provided in the system, traders earn BAL for swapping tokens on the Balancer Exchange dApp.
Every eligible trade made through the Balancer Exchange Proxy results in some BAL being allocated to the address (EOAs only) that sent the transaction. An eligible trade is a trade containing one or more eligible swaps, where an eligible swap is one between any two tokens on the allowlist. Claims are made available at the BAL claims interface on Wednesday (00:00 UTC) following the end (00:00 UTC Monday) of the weekly period in which the trade occurred.
The amount of BAL awarded to a trade is a function of the number of eligible swaps in the trade (N), which determines a number of gas units (G), the median gas price [1] of the block the transaction was included in (M), and the BAL/ETH price provided by CoinGecko closest to the block time (P).
The amount of BAL to be received by the user for a trade is computed as
\text{BAL received} = \frac{G \times M}{P}
Balancer V1 Trades
The BAL for Gas program also applies to trades which are executed on Balancer V1. The program is much the same as for V2 except for changes to the number of gas units reimbursed per swap.
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Quantitative Finance Problem: The Money Never Runs Out - Andy Hayes | Brilliant
Fred is an incredibly talented investor: so talented, in fact, that he doubles the amount of money in his bank account each year.
Fred doesn't worry about running out of money, so he spends increasing amounts every year, and never makes any deposits other than his original deposit. At the beginning of the first year (right after the original deposit), he spends $1. At the beginning of the second year, he spends $2. At the beginning of the third year, he spends $3, and so on.
What is the minimum amount of money (in dollars) Fred needs in his original deposit so that he never runs out of money?
Note: As an explicit example, this is what would happen if the initial deposit was $20:
\begin{array}{|c|c|c|} \hline & \text{Account at the} & & \text{Account after} & \text{Account after} \\ \text{Year} & \text{beginning of year} & \text{Amount spent} & \text{spending} & \text{doubling} \\ \hline 1 & \$20 & \$1 & \$19 & \$38 \\ 2 & \$38 & \$2 & \$36 & \$72 \\ 3 & \$72 & \$3 & \$69 & \$138 \\ 4 & \$138 & \$4 & \$134 & \$268 \\ \hline \end{array}
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高一米唐 人 2022-03-22
Galois theory tells us that
\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1
can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be
{\zeta }^{1},{\zeta }^{2},\dots ,{\zeta }^{10}
, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta }^{2}{x}_{3}+{\zeta }^{3}{x}_{4}+{\zeta }^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta }^{2}{x}_{2}+{\zeta }^{4}{x}_{3}+\zeta {x}_{4}+{\zeta }^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta }^{3}{x}_{2}+\zeta {x}_{3}+{\zeta }^{4}{x}_{4}+{\zeta }^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta }^{4}{x}_{2}+{\zeta }^{3}{x}_{3}+{\zeta }^{2}{x}_{4}+\zeta {x}_{5}\end{array}
Once one has
{A}_{0},\dots ,{A}_{4}
one easily gets
{x}_{1},\dots ,{x}_{5}
. It's easy to find
{A}_{0}
. The point is that
\tau
{A}_{j}
{\zeta }^{-j}{A}_{j}
and so takes
{A}_{j}^{5}
{A}_{j}^{5}
{A}_{j}^{5}
can be written down in terms of rationals (if that's your starting field) and powers of
\zeta
. Alas, here is where the algebra becomes difficult. The coefficients of powers of
\zeta
{A}_{1}^{5}
are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have
{A}_{1}
as a fifth root of a certain explicit complex number. Then one can express the other
{A}_{j}
{A}_{1}
. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.
{\beta }_{0}+{\beta }_{1}{x}_{0}
{x}_{0}
\stackrel{^}{{Y}_{0}}=\stackrel{^}{{\beta }_{0}}+\stackrel{^}{{\beta }_{1}}{x}_{0}
\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}}
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}}
E\left(\stackrel{^}{{Y}_{0}}\right)
var\left(\stackrel{^}{{Y}_{0}}\right)
{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)
Sally has caught covid but doesn’t know it yet. She is testing herself with rapid antigen kits which have an 80% probability of returning a positive result for an infected person. For the purpose of this question you can assume that the results of repeated tests are independent.
a) If sally uses 3 test kits what is the probability that at least one will return a positive result?
b) In 3 tests, what is the expected number of positive results?
c) Sally has gotten her hands on more effective tests, these ones have a 90% probability of returning a positive result for an infected person. If she tested herself
twice with the new tests, how many positive results would she expect to see?
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Tire with longitudinal behavior given by Magic Formula coefficients - MATLAB - MathWorks Nordic
Tire Kinematics and Response
Tire and Wheel Dynamics
Parameterize by
Peak longitudinal force at rated load
Slip at peak force at rated load (percent)
Magic Formula B coefficient
Magic Formula C coefficient
Magic Formula D coefficient
Magic Formula E coefficient
Tire nominal vertical load
Magic Formula C-coefficient parameter, p_Cx1
Magic Formula D-coefficient parameters, [p_Dx1 p_Dx2]
Magic Formula E-coefficient parameters, [p_Ex1 p_Ex2 p_Ex3 p_Ex4]
Magic Formula BCD-coefficient parameters, [p_Kx1 p_Kx2 p_Kx3]
Magic Formula H-coefficient parameters, [p_Hx1 p_Hx2]
Magic Formula V-coefficient parameters, [p_Vx1 p_Vx2]
Tire with longitudinal behavior given by Magic Formula coefficients
The Tire (Magic Formula) block models a tire with longitudinal behavior given by the Magic Formula [1], an empirical equation based on four fitting coefficients. The block can model tire dynamics under constant or variable pavement conditions.
The longitudinal direction of the tire is the same as its direction of motion as it rolls on pavement. This block is a structural component based on the Tire-Road Interaction (Magic Formula) block.
To increase the fidelity of the tire model, you can specify properties such as tire compliance, inertia, and rolling resistance. However, these properties increase the complexity of the tire model and can slow down simulation. Consider ignoring tire compliance and inertia if simulating the model in real time or if preparing the model for hardware-in-the-loop (HIL) simulation.
The Tire (Magic Formula) block models the tire as a rigid wheel-tire combination in contact with the road and subject to slip. When torque is applied to the wheel axle, the tire pushes on the ground (while subject to contact friction) and transfers the resulting reaction as a force back on the wheel. This action pushes the wheel forward or backward. If you include the optional tire compliance, the tire also flexibly deforms under load.
The figure shows the forces acting on the tire. The table defines the tire model variables.
Tire Model Variables
Description and Unit
u Tire longitudinal deformation
Ω′ Contact point angular velocity. If there is no tire longitudinal deformation, that is, if
u=0
{\Omega }^{\prime }=\Omega
{r}_{w}{\Omega }^{\prime }
Tire tread longitudinal velocity
{V}_{sx}={r}_{w}\Omega -{V}_{x}
Wheel slip velocity
{{V}^{\prime }}_{sx}={r}_{w}{\Omega }^{\prime }-{V}_{x}
Contact slip velocity. If there is no tire longitudinal deformation, that is, if
u=0
{{V}^{\prime }}_{sx}={V}_{sx}
k=\frac{{V}_{sx}}{|{V}_{x}|}
Wheel slip
{k}^{\prime }=\frac{{{V}^{\prime }}_{sx}}{|{V}_{x}|}
Contact slip. If there is no tire longitudinal deformation, that is, if
u=0
{k}^{\prime }=k
Vth Wheel hub threshold velocity
Fz Vertical load on tire
Fx Longitudinal force exerted on the tire at the contact point
{C}_{{F}_{x}}={\left(\frac{\partial {F}_{x}}{\partial u}\right)}_{0}
Tire longitudinal stiffness under deformation
{b}_{{F}_{x}}={\left(\frac{\partial {F}_{x}}{\partial \stackrel{˙}{u}}\right)}_{0}
Tire longitudinal damping under deformation
Iw Wheel-tire inertia, such that the effective mass is equal to
\frac{{I}_{w}}{{r}_{w}^{2}}
τdrive Torque applied by the axle to the wheel
Roll and Slip
A nonslipping tire would roll and translate as
{V}_{x}={r}_{w}\Omega
. However, as tires do slip, they respond by developing a longitudinal force, Fx.
The wheel slip velocity is
{V}_{sx}={r}_{w}\Omega -{V}_{x}
. The wheel slip is
k=\frac{{V}_{sx}}{|{V}_{x}|}
. For a locked, sliding wheel,
k=-1
. For perfect rolling,
k=0
Slip at Low Speed
For low speeds, as defined by
|{V}_{x}|\le |{V}_{th}|
, the wheel slip becomes:
k=\frac{2{V}_{sx}}{\left({V}_{th}+\frac{{V}_{x}^{2}}{{V}_{th}}\right)}
This modification allows for a nonsingular, nonzero slip at zero wheel velocity. For example, for perfect slipping, that is, in the case of a nontranslating spinning tire,
{V}_{x}=0
k=\frac{2{r}_{w}\Omega }{{V}_{th}}
If the tire is modeled with compliance, it is also flexible. In this case, because the tire deforms, the tire-road contact point turns at a slightly different angular velocity, Ω′, from the wheel, Ω, and requires, instead of the wheel slip, the contact point or contact patch slip κ'. The block models the deforming tire as a translational spring-damper of stiffness, CFx, and damping, bFx.
If you model a tire without compliance, that is, if
u=0
, then there is no tire longitudinal deformation at any time in the simulation and:
{k}^{\prime }=k
{{V}^{\prime }}_{sx}={V}_{sx}
{\Omega }^{\prime }=\Omega
The full tire model is equivalent to this Simscape™/Simscape Driveline™ component diagram. It simulates both transient and steady-state behavior and correctly represents starting from, and coming to, a stop. The Translational Spring and Translational Damper are equivalent to the tire stiffness CFx and damping bFx. The Tire-Road Interaction (Magic Formula) block models the longitudinal force Fx on the tire as a function of Fz and k′ using the Magic Formula, with k′ as the independent slip variable.
The Wheel and Axle radius is the wheel radius rw. The Mass value is the effective mass,
\frac{{I}_{w}}{{r}_{w}^{2}}
. The tire characteristic function f(k′, Fz) determines the longitudinal force Fx. Together with the driveshaft torque applied to the wheel axis, Fx determines the wheel angular motion and longitudinal motion.
Without tire compliance, the Translational Spring and Translational Damper are omitted, and contact variables revert to wheel variables. In this case, the tire effectively has infinite stiffness, and port P of Wheel and Axle connects directly to port T of Tire-Road Interaction (Magic Formula).
Without tire inertia, the Mass is omitted.
The Tire (Magic Formula) block assumes longitudinal motion only and includes no camber, turning, or lateral motion.
Tire compliance implies a time lag in the tire response to the forces on it. Time lag simulation increases model fidelity but reduces simulation performance. See Adjust Model Fidelity.
M — Magic formula coefficients
physical signal | vector | [B, C, D, E]
Physical signal input port associated with the Magic Formula coefficients. Provide the Magic Formula coefficients as a four-element vector, specified in the order [B, C, D, E].
Port M is exposed only if the Main > Parameterize by parameter is set to Physical signal Magic Formula coefficients. For more information, see Main Parameter Dependencies.
The table shows how the visibility of some Main parameters depends on the options that you choose for other parameters. To learn how to read the table, see Parameter Dependencies.
Parameterize by — Choose Peak longitudinal force and corresponding slip, Constant Magic Formula coefficients, Load-dependent Magic Formula coefficients, or Physical signal Magic Formula coefficients
Peak longitudinal force and corresponding slip Constant Magic Formula coefficients Load-dependent Magic Formula coefficients Physical signal Magic Formula coefficients — Exposes physical signal input port M for providing the Magic Formula coefficients to the block as an array of elements in this order [B, C, D, E].
Parameterize by — Parameterization method
Peak longitudinal force and corresponding slip (default) | Constant Magic Formula coefficients | Load-dependent Magic Formula coefficients | Physical signal Magic Formula coefficients
Magic Formula tire-road interaction model.
To model tire dynamics under constant pavement conditions, select one of these models:
Peak longitudinal force and corresponding slip — Parameterize the Magic Formula with physical characteristics of the tire.
Constant Magic Formula coefficients — Specify the parameters that define the constant B, C, D, and E coefficients as scalars, with these default values.
Load-dependent Magic Formula coefficients — Specify the parameters that define the load-dependent C, D, E, K, H, and V coefficients as vectors, one for each coefficient, with these default values.
C p_Cx1 1.685
D [ p_Dx1 p_Dx2 ] [ 1.21 –0.037 ]
E [ p_Ex1 p_Ex2 p_Ex3 p_Ex4 ] [ 0.344 0.095 –0.02 0 ]
K [ p_Kx1 p_Kx2 p_Kx3 ] [ 21.51 –0.163 0.245 ]
H [ p_Hx1 p_Hx2 ] [ –0.002 0.002 ]
V [ p_Vx1 p_Vx2 ] [ 0 0 ]
To model tire dynamics under variable pavement conditions, select Physical signal Magic Formula coefficients. Selecting this model exposes a physical signal port M. Use the port to input the Magic Formula coefficients as a four-element vector, specified in the order [B, C, D,E].
Each parameterization method option exposes related parameters and hides unrelated parameters. Selecting Physical signal Magic Formula coefficients exposes a physical signal input port. For more information, see Main Parameter Dependencies.
Rated vertical load — Rated load force
3000 N (default) | positive scalar
Rated vertical load force Fz0.
This parameter is exposed only when you select the Peak longitudinal force and corresponding slip parameterization method. For more information, see Main Parameter Dependencies.
Peak longitudinal force at rated load — Maximum longitudinal force at rated load
Maximum longitudinal force Fx0 that the tire exerts on the wheel when the vertical load equals its rated value Fz0.
Slip at peak force at rated load (percent) — Percent slip at maximum longitudinal force and rated load
Contact slip κ'0, expressed as a percentage (%), when the longitudinal force equals its maximum value Fx0 and the vertical load equals its rated value Fz0.
Magic Formula B coefficient — Constant Magic Formula B coefficient
Load-independent Magic Formula B coefficient.
This parameter is exposed only when you select the Constant Magic Formula coefficients parameterization method. For more information, see Main Parameter Dependencies.
Magic Formula C coefficient — Constant Magic Formula C coefficient
Load-independent Magic Formula C coefficient.
Magic Formula D coefficient — Constant Magic Formula D coefficient
Load-independent Magic Formula D coefficient.
Magic Formula E coefficient — Constant Magic Formula E coefficient
Load-independent Magic Formula E coefficient.
Tire nominal vertical load — Normal force
Nominal normal force Fz0 on tire.
This parameter is exposed only when you select the Load-dependent Magic Formula coefficients parameterization method. For more information, see Main Parameter Dependencies.
Magic Formula C-coefficient parameter, p_Cx1 — Variable Magic Formula C coefficient
Load-dependent Magic Formula C coefficient.
Magic Formula D-coefficient parameters, [p_Dx1 p_Dx2] — Variable Magic Formula D coefficient
[1.21, -.037] (default) | vector
Load-dependent Magic Formula D coefficient.
Magic Formula E-coefficient parameters, [p_Ex1 p_Ex2 p_Ex3 p_Ex4] — Variable Magic Formula E coefficient
[.344, .095, -.02, 0] (default) | vector
Load-dependent Magic Formula E coefficient.
Magic Formula BCD-coefficient parameters, [p_Kx1 p_Kx2 p_Kx3] — Variable Magic Formula K coefficient
[21.51, -.163, .245] (default) | vector
Load-dependent Magic Formula K coefficient.
Magic Formula H-coefficient parameters, [p_Hx1 p_Hx2] — Variable Magic Formula H coefficient
[-.002, .002] (default) | vector
Load-dependent Magic Formula H coefficient.
Magic Formula V-coefficient parameters, [p_Vx1 p_Vx2] — Variable Magic Formula V coefficient
[0, 0] (default) | vector
Load-dependent Magic Formula V coefficient.
Unloaded tire-wheel radius, rw.
The table shows how the visibility of some Dynamics parameters depends on the options that you choose for other parameters. To learn how to read the table, see Parameter Dependencies.
Dynamics Parameter Dependencies
Selecting the Specify stiffness and damping parameterization method, exposes stiffness and damping parameters. For more information, see Dynamics Parameter Dependencies.
Selecting Specify stiffness and damping for the Compliance parameter, exposes this parameter. For more information, see Dynamics Parameter Dependencies.
Selecting the Specify inertia and initial velocity parameterization method, exposes inertia and velocity parameters. For more information, see Dynamics Parameter Dependencies.
Selecting Specify inertia and initial velocity for the Inertia parameter, exposes this parameter. For more information, see Dynamics Parameter Dependencies.
Rolling Resistance Parameter Dependencies
Selecting On exposes rolling resistance parameters. For more information, see Rolling Resistance Parameter Dependencies.
Each Resistance model option exposes related parameters. For more information, see Rolling Resistance Parameter Dependencies.
Selecting On for the Rolling resistance parameter and Constant coefficient for the Resistance model parameter exposes this parameter. For more information, see Rolling Resistance Parameter Dependencies.
Selecting On for the Rolling resistance parameter and Pressure and velocity dependent for the Resistance model parameter exposes this parameter. For more information, see Rolling Resistance Parameter Dependencies.
Exponent of the tire pressure in the model equation.
Exponent of the normal force model equation.
Wheel hub velocity Vth below which the slip calculation is modified to avoid singular evolution at zero velocity. Must be positive.
[1] Pacejka, H. B. Tire and Vehicle Dynamics. Elsevier Science, 2005.
Tire (Friction Parameterized) | Tire (Simple) | Tire-Road Interaction (Magic Formula)
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Density | Brilliant Math & Science Wiki
(\rho)
is the amount of mass
(m)
per unit volume
(V)
of a substance.
\rho = \frac{m}{V}
Density is an intensive property, which means the density does not change as the amount of the substance present changes. This contrasts with mass (an extensive property), which appears quite often in physics, but becomes cumbersome to apply to fluids macroscopically.
The density of gold is 19,320
\frac{\text{kg}}{\text{m}^3}
. What is the density of a 12.5
\text{kg}
sample of gold?
Since density does not depend on the amount of gold in the sample, the mass is meaningless, and the density is always 19,320
\frac{\text{kg}}{\text{m}^3}.
The density of a substance is generally given near the surface of the earth. At this location, the pressure is 1
\text{atm}
1.01\times 10^5 \text{Pa}.
However, if the pressure changes, the density will change as well. Boyle's law says that the product of the pressure and the volume must remain constant if the amount of the substance present remains constant.
If the temperature and amount of an ideal gas remain constant, the product of the pressure and volume must also remain constant, or:
P_1V_1 = P_2V_2.
A rigid box is filled with air. If the pressure is doubled but the volume remains the same, does the density of the air increase or decrease?
By Boyle's law,
PV
must remain constant if the amount of the substance is unchanged. In this case,
PV
is doubled, which means the amount of air in the box must have changed. In order to increase the pressure, the amount of air must increase. Hence, the density of the air in the box increases.
Cite as: Density. Brilliant.org. Retrieved from https://brilliant.org/wiki/density/
|
In this paper, we study the control system associated with the incompressible 3D Euler system. We show that the velocity field and pressure of the fluid are exactly controllable in projections by the same finite-dimensional control. Moreover, the velocity is approximately controllable. We also prove that 3D Euler system is not exactly controllable by a finite-dimensional external force.
Classification : 35Q35, 93C20
Mots clés : controllability, 3D incompressible Euler equations, Agrachev-Sarychev method
author = {Nersisyan, Hayk},
title = {Controllability of {3D} incompressible {Euler} equations by a finite-dimensional external force},
AU - Nersisyan, Hayk
TI - Controllability of 3D incompressible Euler equations by a finite-dimensional external force
Nersisyan, Hayk. Controllability of 3D incompressible Euler equations by a finite-dimensional external force. ESAIM: Control, Optimisation and Calculus of Variations, Tome 16 (2010) no. 3, pp. 677-694. doi : 10.1051/cocv/2009017. http://www.numdam.org/articles/10.1051/cocv/2009017/
[1] A. Agrachev and A. Sarychev, Navier-Stokes equations controllability by means of low modes forcing. J. Math. Fluid Mech. 7 (2005) 108-152. | Zbl 1075.93014
[2] A. Agrachev and A. Sarychev, Controllability of 2D Euler and Navier-Stokes equations by degenerate forcing. Comm. Math. Phys. 265 (2006) 673-697. | Zbl 1105.93008
[3] J.T. Beale, T. Kato and A. Majda, Remarks on the breakdown of smooth solutions for the 3-D Euler equations. Comm. Math. Phys. 94 (1984) 61-66. | Zbl 0573.76029
[4] P. Constantin and C. Foias, Navier-Stokes Equations. University of Chicago Press, Chicago, USA (1988). | Zbl 0687.35071
[5] J.-M. Coron, On the controllability of 2-D incompressible perfect fluids. J. Math. Pures Appl. 75 (1996) 155-188. | Zbl 0848.76013
[6] D.E. Edmunds and H. Triebel, Function Spaces, Entropy Numbers, Differential Operators. Cambridge University Press, Cambridge, UK (1996). | Zbl 1143.46001
[7] E. Fernández-Cara, S. Guerrero, O.Yu. Imanuvilov and J.P. Puel, Local exact controllability of the Navier-Stokes system. J. Math. Pures Appl. 83 (2004) 1501-1542.
[8] A.V. Fursikov and O.Yu. Imanuvilov, Exact controllability of the Navier-Stokes and Boussinesq equations. Russian Math. Surveys 54 (1999) 93-146. | Zbl 0970.35116
[9] O. Glass, Exact boundary controllability of 3-D Euler equation. ESAIM: COCV 5 (2000) 1-44. | Numdam | Zbl 0940.93012
[10] G. Lorentz, Approximation of Functions. Chelsea Publishing Co., New York, USA (1986). | Zbl 0643.41001
[11] S.S. Rodrigues, Navier-Stokes equation on the rectangle: controllability by means of low mode forcing. J. Dyn. Control Syst. 12 (2006) 517-562. | Zbl 1105.35085
[12] A. Shirikyan, Approximate controllability of three-dimensional Navier-Stokes equations. Comm. Math. Phys. 266 (2006) 123-151. | Zbl 1105.93016
[13] A. Shirikyan, Exact controllability in projections for three-dimensional Navier-Stokes equations. Ann. Inst. H. Poincaré, Anal. Non Linéaire 24 (2007) 521-537. | Numdam | Zbl 1119.93021
[14] A. Shirikyan, Euler equations are not exactly controllable by a finite-dimensional external force. Physica D 237 (2008) 1317-1323. | Zbl 1143.76393
[15] M.E. Taylor, Partial Differential Equations, III. Springer-Verlag, New York (1996). | Zbl 1206.35004
[16] R. Temam, Local existence of
{C}^{\infty }
solution of the Euler equation of incompressible perfect fluids. Lect. Notes Math. 565 (1976) 184-194. | Zbl 0355.76017
|
Oxygen saturation - wikidoc
WikiDoc Resources for Oxygen saturation
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In medicine, oxygen saturation (SO2) measures the percentage of hemoglobin binding sites in the bloodstream occupied by oxygen. At low partial pressures of oxygen, most hemoglobin is deoxygenated. At around 90% (the value varies according to the clinical context) oxygen saturation increases according to an oxygen-haemoglobin dissociation curve and approaches 100% at partial oxygen pressures of >10 kPa. A pulse oximeter relies on the light absorption characteristics of saturated hemoglobin to give an indication of oxygen saturation.
An SaO2 (arterial oxygen saturation) value below 90% is termed hypoxemia. This may be due to various medical conditions.
Annual mean sea surface dissolved oxygen for the World Ocean. Data from the World Ocean Atlas 2001.
In aquatic environments, oxygen saturation is a relative measure of the amount of oxygen (O2) dissolved in the water. Dissolved oxygen (DO) is measured in standard solution units such as millimoles O2 per liter (mmol/L), milligrams O2
Solubility tables (based upon temperature) and corrections for different salinities and pressures can be found at the USGS web site.
Tables such as these of DO in milliliters per liter (ml/L) are based upon empirical equations that have been worked out and tested (e.g.,Weiss, 1970):
{\displaystyle ln(DO)=A1+A2*100/T+A3*ln(T/100)+A4*T/100+S*[B1+B2*T/100+B3*(T/100)^{2}]}
{\displaystyle A1=-173.4292}
{\displaystyle A2=249.6339}
{\displaystyle A3=143.3483}
{\displaystyle A4=-21.8492}
{\displaystyle B1=-0.033096}
{\displaystyle B2=0.014259}
{\displaystyle B3=-0.001700}
T = temperature in kelvins, and S = salinity in g/kg. DO = dissolved oxygen in ml/L. ln is the natural log. Multiply DO by 1.4276 to obtain mg/L.
Supersaturation can sometimes be harmful for organisms and cause gas bubble disease.
DO - at least 4 - 6 mg/L (ppm), but more is better
Weiss, R. (1970) Deep-Sea Res. 17, 721-735
Wastewater quality indicators discusses both Biochemical oxygen demand (BOD) and Chemical oxygen demand (COD) as indicators of wastewater quality.
da:Iltmætning de:Sauerstoffsättigung fi:Happisaturaatio
Retrieved from "https://www.wikidoc.org/index.php?title=Oxygen_saturation&oldid=720738"
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Consider two functions f: S Tandh:T> U for non-empty sets S, T, U.
Sam Johnston 2022-03-17
A 6.0 cm diameter horizontal pipe gradually narrows to 2.0 centimeters. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow ?
A penny of mass 3.1 g rests on a small 29.1 g block supported by a spinning disk of radius 8.3 cm. The coefficients of friction between block and disk are 0.742 (static) and 0.64 (kinetic) while those for the penny and block are 0.617 (static) and 0.45 (kinetic).
to 4x^2y"+17y=0, y(1)=-1, y'(1)= -1/2
{\beta }_{0}+{\beta }_{1}{x}_{0}
{x}_{0}
\stackrel{^}{{Y}_{0}}=\stackrel{^}{{\beta }_{0}}+\stackrel{^}{{\beta }_{1}}{x}_{0}
\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}}
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}}
E\left(\stackrel{^}{{Y}_{0}}\right)
var\left(\stackrel{^}{{Y}_{0}}\right)
{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)
Find the exponential model that fits the points shown in the table. (Round the exponent to four decimal places.)
Sharon draws a random card, from a regular deck of cards, and rolls a regular 6 sided die. What is the probability that Sharon draws a card that is a Heart and rolls a 3?
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JEE Sequence | Brilliant Math & Science Wiki
JEE Sequence
This page will teach you how to master JEE Sequences. 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 main concepts under sequences are arithmetic progression, geometric progression, harmonic progression, AM-GM-HM inequality and its applications.
General term:
t_n=a+(n-1)d
Sum of the terms:
S_n=\frac n2 \left( 2a+(n-1)d \right)
Arithmetic mean and its property:
(\text{AM of any } n \text{ numbers } a_1,a_2,\ldots,a_n)=\frac{a_1+a_2+\cdots+a_n}{n}
Insertion of arithmetic means between two numbers:
and
b
are two given numbers and
a,A_1,A_2,\ldots,A_n,b
A_1,A_2,\ldots,A_n
n
AM's between
and
b:
\begin{array}{c}&A_1=a+\frac{b-a}{n+1}, &A_2=a+\frac{2(b-a)}{n+1}, &\ldots, &A_n=a+\frac{n(b-a)}{n+1}.\end{array}
t_n=ar^{n-1}
S_n-\dfrac{a(r^n-1)}{r-1}
Geometric mean and its property:
(\text{GM of any } n \text{ numbers } a_1,a_2,\ldots,a_n)=\left( a_1 \cdot a_2 \cdot \cdot \cdot a_n \right)^{\frac 1n}
Insertion of geometric means between two numbers:
and
b
a, G_1, G_2, \ldots, G_n, b
are in GP, then
G_1,G_2,\ldots,G_n
n
GM's between
and
b:
\begin{array}{c}&G_1=a\left( \frac ba \right)^{\frac{1}{n+1}}, &G_2=a\left( \frac ba \right)^{\frac{2}{n+1}}, &\ldots, &G_n=a\left( \frac ba \right)^{\frac{n}{n+1}}. \end{array}
\dfrac{1}{a_n}=\dfrac{1}{a}+(n-1)d
Harmonic mean and its property:
If HM of any
numbers
a_1,a_2,\ldots,a_n
H
H=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}.
Insertion of harmonic means between two numbers:
and
b
a,H_1,H_2,\ldots,H_n,b
are in HP, then
H_1,H_2,\ldots,H_n
n
HM's between
and
b:
\begin{array}{c}&\frac{1}{H_1}=\frac{1}{a}+\frac{\left( \frac 1b - \frac 1a \right)}{n+1}, &\frac{1}{H_2}=\frac{1}{a}+\frac{ 2 \left( \frac 1b - \frac 1a \right)}{n+1}, &\ldots, &\frac{1}{H_n}=\frac{1}{a}+\frac{n \left( \frac 1b - \frac 1a \right)}{n+1}. \end{array}
AM-GM-HM inequality and its applications
a_1,a_2,\ldots,a_n
are all positive real numbers, then
\dfrac{a_1+a_2+...+a_n}{n} \geq \left( a_1 \cdot a_2 \cdot \cdot \cdot a_n \right)^{\frac 1n} \geq \dfrac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}.
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
\begin{array} { l l } A) \, & \quad \quad \quad \quad \quad & B) \, \\ C) \, & & D) \, \\ \end{array}
Once you are confident of JEE Sequences, move on to JEE Series.
Cite as: JEE Sequence. Brilliant.org. Retrieved from https://brilliant.org/wiki/jee-sequence/
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Total relation - Wikipedia
For relations R where xRy or yRx for all x and y, see connected relation.
It has been suggested that this article be merged into Serial relation. (Discuss) Proposed since May 2022.
In mathematics, a binary relation R ⊆ X×Y between two sets X and Y is total (or left total) if the source set X equals the domain {x : there is a y with xRy }. Conversely, R is called right total if Y equals the range {y : there is an x with xRy }.
When f: X → Y is a function, the domain of f is all of X, hence f is a total relation. On the other hand, if f is a partial function, then the domain may be a proper subset of X, in which case f is not a total relation.
"A binary relation is said to be total with respect to a universe of discourse just in case everything in that universe of discourse stands in that relation to something else."[1]
Algebraic characterizationEdit
Total relations can be characterized algebraically by equalities and inequalities involving compositions of relations. To this end, let
{\displaystyle X,Y}
be two sets, and let
{\displaystyle R\subseteq X\times Y.}
{\displaystyle A,B,}
{\displaystyle L_{A,B}=A\times B}
be the universal relation between
{\displaystyle A}
{\displaystyle B,}
{\displaystyle I_{A}=\{(a,a):a\in A\}}
be the identity relation on
{\displaystyle A.}
{\displaystyle R^{\top }}
for the converse relation of
{\displaystyle R.}
{\displaystyle R}
is total iff for any set
{\displaystyle W}
{\displaystyle S\subseteq W\times X,}
{\displaystyle S\neq \emptyset }
{\displaystyle SR\neq \emptyset .}
{\displaystyle R}
is total iff
{\displaystyle I_{X}\subseteq RR^{\top }.}
{\displaystyle R}
is total, then
{\displaystyle L_{X,Y}=RL_{Y,Y}.}
The converse is true if
{\displaystyle Y\neq \emptyset .}
{\displaystyle R}
{\displaystyle {\overline {RL_{Y,Y}}}=\emptyset .}
{\displaystyle Y\neq \emptyset .}
[note 2][2]: 63
{\displaystyle R}
{\displaystyle {\overline {R}}\subseteq R{\overline {I_{Y}}}.}
{\displaystyle Y\neq \emptyset .}
[2]: 54 [3]
{\displaystyle R}
is total, then for any set
{\displaystyle Z}
{\displaystyle S\subseteq Y\times Z,}
{\displaystyle {\overline {RS}}\subseteq R{\overline {S}}.}
{\displaystyle Y\neq \emptyset .}
{\displaystyle Y=\emptyset \neq X,}
{\displaystyle R}
will be not total.
^ Observe
{\displaystyle {\overline {RL_{Y,Y}}}=\emptyset \Leftrightarrow RL_{Y,Y}=L_{X,Y},}
and apply the previous bullet.
^ Take
{\displaystyle Z=Y,S=I_{Y}}
and appeal to the previous bullet.
^ Functions from Carnegie Mellon University
^ a b c d e Schmidt, Gunther; Ströhlein, Thomas (6 December 2012). Relations and Graphs: Discrete Mathematics for Computer Scientists. Springer Science & Business Media. ISBN 978-3-642-77968-8.
^ Gunther Schmidt (2011). Relational Mathematics. Cambridge University Press. doi:10.1017/CBO9780511778810. ISBN 9780511778810. Definition 5.8, page 57.
Gunther Schmidt & Michael Winter (2018) Relational Topology
C. Brink, W. Kahl, and G. Schmidt (1997) Relational Methods in Computer Science, Advances in Computer Science, page 5, ISBN 3-211-82971-7
Gunther Schmidt & Thomas Strohlein (2012)[1987] Relations and Graphs, p. 54, at Google Books
Gunther Schmidt (2011) Relational Mathematics, p. 57, at Google Books
Retrieved from "https://en.wikipedia.org/w/index.php?title=Total_relation&oldid=1088334911"
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Simulate multivariate stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, SDEMRD, Merton, or Bates models - MATLAB simulate - MathWorks Switzerland
Antithetic Sampling to a Path-Dependent Barrier Option
Simulate multivariate stochastic differential equations (SDEs) for SDE, BM, GBM, CEV, CIR, HWV, Heston, SDEDDO, SDELD, SDEMRD, Merton, or Bates models
[Paths,Times,Z] = simulate(MDL)
[Paths,Times,Z] = simulate(___,Optional)
[Paths,Times,Z] = simulate(MDL) simulates NTrials sample paths of NVars correlated state variables, driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods, approximating continuous-time stochastic processes.
simulate accepts any variable-length list of input arguments that the simulation method or function referenced by the SDE.Simulation parameter requires or accepts. It passes this input list directly to the appropriate SDE simulation method or user-defined simulation function.
[Paths,Times,Z] = simulate(___,Optional) adds optional input arguments.
Consider a European up-and-in call option on a single underlying stock. The evolution of this stock's price is governed by a Geometric Brownian Motion (GBM) model with constant parameters:
The stock volatility is 30% per annum.
The option strike price is 100.
The option barrier is 120.
The risk-free rate is constant at 5% per annum.
The goal is to simulate various paths of daily stock prices, and calculate the price of the barrier option as the risk-neutral sample average of the discounted terminal option payoff. Since this is a barrier option, you must also determine if and when the barrier is crossed.
This example performs antithetic sampling by explicitly setting the Antithetic flag to true, and then specifies an end-of-period processing function to record the maximum and terminal stock prices on a path-by-path basis.
Create a GBM model using gbm.
barrier = 120; % barrier
strike = 100; % exercise price
rate = 0.05; % annualized risk-free rate
sigma = 0.3; % annualized volatility
nPeriods = 63; % 63 trading days
dt = 1 / 252; % time increment = 252 days
T = nPeriods * dt; % expiration time = 0.25 years
obj = gbm(rate, sigma, 'StartState', 105);
Perform a small-scale simulation that explicitly returns two simulated paths.
rng('default') % make output reproducible
[X, T] = obj.simBySolution(nPeriods, 'DeltaTime', dt, ...
'nTrials', 2, 'Antithetic', true);
Perform antithetic sampling such that all primary and antithetic paths are simulated and stored in successive matching pairs. Odd paths (1,3,5,...) correspond to the primary Gaussian paths. Even paths (2,4,6,...) are the matching antithetic paths of each pair, derived by negating the Gaussian draws of the corresponding primary (odd) path. Verify this by examining the matching paths of the primary/antithetic pair.
plot(T, X(:,:,1), 'blue', T, X(:,:,2), 'red')
xlabel('Time (Years)'), ylabel('Stock Price'), ...
title('Antithetic Sampling')
legend({'Primary Path' 'Antithetic Path'}, ...
To price the European barrier option, specify an end-of-period processing function to record the maximum and terminal stock prices. This processing function is accessible by time and state, and is implemented as a nested function with access to shared information that allows the option price and corresponding standard error to be calculated. For more information on using an end-of-period processing function, see Pricing Equity Options.
Simulate 200 paths using the processing function method.
rng('default') % make output reproducible
barrier = 120; % barrier
strike = 100; % exercise price
rate = 0.05; % annualized risk-free rate
sigma = 0.3; % annualized volatility
nPeriods = 63; % 63 trading days
dt = 1 / 252; % time increment = 252 days
T = nPeriods * dt; % expiration time = 0.25 years
nPaths = 200; % # of paths = 100 sets of pairs
f = Example_BarrierOption(nPeriods, nPaths);
simulate(obj, nPeriods, 'DeltaTime' , dt, ...
'nTrials', nPaths, 'Antithetic', true, ...
'Processes', f.SaveMaxLast);
Approximate the option price with a 95% confidence interval.
optionPrice = f.OptionPrice(strike, rate, barrier);
standardError = f.StandardError(strike, rate, barrier,...
lowerBound = optionPrice - 1.96 * standardError;
upperBound = optionPrice + 1.96 * standardError;
displaySummary(optionPrice, standardError, lowerBound, upperBound);
Up-and-In Barrier Option Price: 6.6572
Standard Error of Price: 0.7292
Confidence Interval Lower Bound: 5.2280
Confidence Interval Upper Bound: 8.0864
function displaySummary(optionPrice, standardError, lowerBound, upperBound)
fprintf(' Up-and-In Barrier Option Price: %8.4f\n', ...
optionPrice);
fprintf(' Standard Error of Price: %8.4f\n', ...
standardError);
fprintf(' Confidence Interval Lower Bound: %8.4f\n', ...
lowerBound);
fprintf(' Confidence Interval Upper Bound: %8.4f\n', ...
Stochastic differential equation model, specified as an sde, batesbm, gbm, cev, cir, hwv, heston,merton sdeddo, sdeld, or sdemrd object.
Optional — (Optional) Any variable-length list of input arguments that the simulation method or function referenced by the SDE.Simulation parameter requires or accepts
Any variable-length list of input arguments that the simulation method or function referenced by the SDE.Simulation parameter requires or accepts, specified as a variable-length list of input arguments. This input list is passed directly to the appropriate SDE simulation method or user-defined simulation function.
Paths — Three-dimensional time series array, consisting of simulated paths of correlated state variables
Three-dimensional time series array, consisting of simulated paths of correlated state variables, returned as an (NPeriods + 1)-by-NVars-by-NTrials array.
For a given trial, each row of Paths is the transpose of the state vector Xt at time t.
Observation times associated with the simulated paths, returned as a (NPeriods + 1)-by-1 column vector.
Z — Three-dimensional time series array of dependent random variates used to generate the Brownian motion vector (Wiener processes)
array | matrix | table | timetable
Three-dimensional time series array of dependent random variates used to generate the Brownian motion vector (Wiener processes) that drove the simulated results found in Paths, returned as a NTimes-by-NBrowns-by-NTrials array.
NTimes is the number of time steps at which the simulate function samples the state vector. NTimes includes intermediate times designed to improve accuracy, which simulate does not necessarily report in the Paths output time series.
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 | simBySolution | simBySolution | sde | merton | bates | bm | gbm | sdeddo | sdeld | cev | cir | heston | hwv | sdemrd
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Turbulence kinetic energy - Wikipedia
Find sources: "Turbulence kinetic energy" – news · newspapers · books · scholar · JSTOR (March 2009) (Learn how and when to remove this template message)
In fluid dynamics, turbulence kinetic energy (TKE) is the mean kinetic energy per unit mass associated with eddies in turbulent flow. Physically, the turbulence kinetic energy is characterised by measured root-mean-square (RMS) velocity fluctuations. In the Reynolds-averaged Navier Stokes equations, the turbulence kinetic energy can be calculated based on the closure method, i.e. a turbulence model.
TKE, k
J/kg = m2⋅s−2
{\displaystyle k={\frac {1}{2}}\left(\,{\overline {(u')^{2}}}+{\overline {(v')^{2}}}+{\overline {(w')^{2}}}\,\right)}
Generally, the TKE is defined to be half the sum of the variances (square of standard deviations) of the velocity components:
{\displaystyle k={\frac {1}{2}}\left(\,{\overline {(u')^{2}}}+{\overline {(v')^{2}}}+{\overline {(w')^{2}}}\,\right),}
where the turbulent velocity component is the difference between the instantaneous and the average velocity
{\displaystyle u'=u-{\overline {u}}}
, whose mean and variance are
{\textstyle {\overline {u'}}={\frac {1}{T}}\int _{0}^{T}(u(t)-{\overline {u}})\,dt=0}
{\textstyle {\overline {(u')^{2}}}={\frac {1}{T}}\int _{0}^{T}(u(t)-{\overline {u}})^{2}\,dt\geq 0}
TKE can be produced by fluid shear, friction or buoyancy, or through external forcing at low-frequency eddy scales (integral scale). Turbulence kinetic energy is then transferred down the turbulence energy cascade, and is dissipated by viscous forces at the Kolmogorov scale. This process of production, transport and dissipation can be expressed as:
{\displaystyle {\frac {Dk}{Dt}}+\nabla \cdot T'=P-\varepsilon ,}
Dk/Dt is the mean-flow material derivative of TKE;
∇ · T′ is the turbulence transport of TKE;
P is the production of TKE, and
ε is the TKE dissipation.
Assuming that molecular viscosity is constant, and making the Boussinesq approximation, the TKE equation is:
{\displaystyle \underbrace {\frac {\partial k}{\partial t}} _{\begin{smallmatrix}{\text{Local}}\\{\text{derivative}}\end{smallmatrix}}+\underbrace {{\overline {u}}_{j}{\frac {\partial k}{\partial x_{j}}}} _{\begin{smallmatrix}{\text{Advection}}\end{smallmatrix}}=-\underbrace {{\frac {1}{\rho _{o}}}{\frac {\partial {\overline {u'_{i}p'}}}{\partial x_{i}}}} _{\begin{smallmatrix}{\text{Pressure}}\\{\text{diffusion}}\end{smallmatrix}}-\underbrace {{\frac {1}{2}}{\frac {\partial {\overline {u_{j}'u_{j}'u_{i}'}}}{\partial x_{i}}}} _{\begin{smallmatrix}{\text{Turbulent}}\\{\text{transport}}\\{\mathcal {T}}\end{smallmatrix}}+\underbrace {\nu {\frac {\partial ^{2}k}{\partial x_{j}^{2}}}} _{\begin{smallmatrix}{\text{Molecular}}\\{\text{viscous}}\\{\text{transport}}\end{smallmatrix}}\underbrace {-{\overline {u'_{i}u'_{j}}}{\frac {\partial {\overline {u_{i}}}}{\partial x_{j}}}} _{\begin{smallmatrix}{\text{Production}}\\{\mathcal {P}}\end{smallmatrix}}-\underbrace {\nu {\overline {{\frac {\partial u'_{i}}{\partial x_{j}}}{\frac {\partial u'_{i}}{\partial x_{j}}}}}} _{\begin{smallmatrix}{\text{Dissipation}}\\\varepsilon _{k}\end{smallmatrix}}-\underbrace {{\frac {g}{\rho _{o}}}{\overline {\rho 'u'_{i}}}\delta _{i3}} _{\begin{smallmatrix}{\text{Buoyancy flux}}\\b\end{smallmatrix}}}
By examining these phenomena, the turbulence kinetic energy budget for a particular flow can be found.[2]
1.1 Reynolds-averaged Navier–Stokes equations
Computational fluid dynamicsEdit
In computational fluid dynamics (CFD), it is impossible to numerically simulate turbulence without discretizing the flow-field as far as the Kolmogorov microscales, which is called direct numerical simulation (DNS). Because DNS simulations are exorbitantly expensive due to memory, computational and storage overheads, turbulence models are used to simulate the effects of turbulence. A variety of models are used, but generally TKE is a fundamental flow property which must be calculated in order for fluid turbulence to be modelled.
Reynolds-averaged Navier–Stokes equationsEdit
Reynolds-averaged Navier–Stokes (RANS) simulations use the Boussinesq eddy viscosity hypothesis [3] to calculate the Reynolds stress that results from the averaging procedure:
{\displaystyle {\overline {u'_{i}u'_{j}}}={\frac {2}{3}}k\delta _{ij}-\nu _{t}\left({\frac {\partial {\overline {u_{i}}}}{\partial x_{j}}}+{\frac {\partial {\overline {u_{j}}}}{\partial x_{i}}}\right),}
{\displaystyle \nu _{t}=c\cdot {\sqrt {k}}\cdot l_{m}.}
The exact method of resolving TKE depends upon the turbulence model used; k–ε (k–epsilon) models assume isotropy of turbulence whereby the normal stresses are equal:
{\displaystyle {\overline {(u')^{2}}}={\overline {(v')^{2}}}={\overline {(w')^{2}}}.}
This assumption makes modelling of turbulence quantities (k and ε) simpler, but will not be accurate in scenarios where anisotropic behaviour of turbulence stresses dominates, and the implications of this in the production of turbulence also leads to over-prediction since the production depends on the mean rate of strain, and not the difference between the normal stresses (as they are, by assumption, equal).[4]
Reynolds-stress models (RSM) use a different method to close the Reynolds stresses, whereby the normal stresses are not assumed isotropic, so the issue with TKE production is avoided.
Initial conditionsEdit
Accurate prescription of TKE as initial conditions in CFD simulations are important to accurately predict flows, especially in high Reynolds-number simulations. A smooth duct example is given below.
{\displaystyle k={\frac {3}{2}}(UI)^{2},}
where I is the initial turbulence intensity [%] given below, and U is the initial velocity magnitude;
{\displaystyle \varepsilon ={c_{\mu }}^{\frac {3}{4}}k^{\frac {3}{2}}l^{-1}.}
Here l is the turbulence or eddy length scale, given below, and cμ is a k–ε model parameter whose value is typically given as 0.09;
{\displaystyle I=0.16Re^{-{\frac {1}{8}}}.}
The turbulent length scale can be estimated as
{\displaystyle l=0.07L,}
with L a characteristic length. For internal flows this may take the value of the inlet duct (or pipe) width (or diameter) or the hydraulic diameter.[5]
^ Pope, S. B. (2000). Turbulent Flows. Cambridge: Cambridge University Press. pp. 122–134. ISBN 978-0521598866.
^ Baldocchi, D. (2005), Lecture 16, Wind and Turbulence, Part 1, Surface Boundary Layer: Theory and Principles , Ecosystem Science Division, Department of Environmental Science, Policy and Management, University of California, Berkeley, CA: USA.
^ Boussinesq, J. V. (1877). "Théorie de l'Écoulement Tourbillant". Mem. Présentés Par Divers Savants Acad. Sci. Inst. Fr. 23: 46–50.
^ Laurence, D. (2002). "Applications of Reynolds Averaged Navier Stokes Equations to Industrial Flows". In van Beeck, J. P. A. J.; Benocci, C. (eds.). Introduction to Turbulence Modelling, Held March 18–22, 2002 at Von Karman Institute for Fluid Dynamics. Sint-Genesius-Rode: Von Karman Institute for Fluid Dynamics.
^ Flórez Orrego; et al. (2012). "Experimental and CFD study of a single phase cone-shaped helical coiled heat exchanger: an empirical correlation". Proceedings of ECOS 2012 – The 25th International Conference on Efficiency, Cost, Optimization, Simulation and Environmental Impact of Energy Systems, June 26–29, 2012, Perugia, Italy. ISBN 978-88-6655-322-9.
Turbulence kinetic energy at CFD Online.
Absi, R. (2008). "Analytical solutions for the modeled k-equation". Journal of Applied Mechanics. 75 (44501): 044501. Bibcode:2008JAM....75d4501A. doi:10.1115/1.2912722.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Turbulence_kinetic_energy&oldid=1060608673"
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DEBUG - Maple Help
Home : Support : Online Help : Programming : Debugging : DEBUG
The Maple debugger breakpoint function
DEBUG(arg1, arg2,...)
The DEBUG function has the effect of a breakpoint when encountered in a Maple procedure. When the DEBUG function is executed, execution stops before the next statement executed, and the debugger (implemented by the Maple function debugger) is invoked. If the debugger returns anything other than NULL, the returned value is evaluated, and the debugger function is reinvoked.
The stopat function (and the stopat debugger command) set breakpoints by inserting a call to the DEBUG function before the statement at which the breakpoint is set. Such breakpoints can be removed using the unstopat function (or the unstopat debugger command).
A breakpoint can be inserted explicitly into the source code of a Maple procedure by inserting a call to the DEBUG function. Such breakpoints cannot be removed using the unstopat function (or the unstopat debugger command).
Arguments passed to the DEBUG function are interpreted in one of three ways:
A single argument that is of type boolean makes the breakpoint conditional. The argument is evaluated in the context in which execution is about to stop, and if it does not return true, execution continues instead. To avoid the condition being evaluated too soon, it should be enclosed in unevaluation (single) quotes.
A single argument that is a range, N..M, with integer literal lower and upper bounds (for example, 3..5) will cause execution to stop only at the N-th through M-th times that the breakpoint is encountered since a command was last invoked from the Maple (not the debugger) prompt.
Any other type of argument, or any sequence of two or more arguments, is displayed by the debugger. If no arguments were passed, the debugger displays the result of the previous computation.
The DEBUG function returns NULL so as not to affect %, %%, and %%%. Therefore, inserting it as the last statement of a procedure will hide the return value of the procedure, so this is not recommended. It is not possible to use stopat to set a breakpoint after the last statement in a procedure.
The DEBUG command is thread-safe as of Maple 15.
f := proc(x,y) local a;
a := x^2; DEBUG(); a := y^2; DEBUG(`Hello`); a := (x+y)^2
f\left(2,3\right)
3 a := y^2;
5 a := (x+y)^2;
\textcolor[rgb]{0,0,1}{25}
% ditto operator
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Agnishom Chattopadhyay, Aditya Raut, and Jimin Khim contributed
Chennai Mathematical Institute (CMI), a premier center for research and education in mathematical sciences, is organizing a nationwide (India) mathematics, physics, and computer science contest, the S.T.E.M.S. (Scholastic Test of Excellence in Mathematical Sciences), as a part of the college fest Tessellate.
Visit this wiki for S.T.E.M.S. 2019.
By participating in S.T.E.M.S., you get to show off your mathematics/physics/computer science skills at a national level and have the chance to attend a science camp at CMI, where you'd meet some of the finest mathematicians and physicists in the country, not to mention the gifts.
The camp will feature some of the best mathematicians, physicists, and computer scientists from some of the best research institutes in India such as CMI, ISI Bangalore, IMSc, IISc, the IIT's, etc. (The complete list of speakers shall be uploaded soon.) CMI shall provide travel fare, food, and accommodations for the selected participants. Students from all age groups will have fair representation. It promises to be a great opportunity for the students to interact with CMI-ites and get an insight into college mathematics and college life (probably in that order, though). The camp shall also feature student talks from students of the aforementioned institutes. They shall also receive certificates signed by some of the best mathematicians of the country, books, and other prizes. The top 100 participants shall receive certificates of participation.
Tessellate S.T.E.M.S 2018 Question Papers
Sample Problems - Mathematics
Sample Problems - Physics
Sample Problems - Computer Science
The entire contest will occur online in two parts for school and college students. The dates for the contest are the
13^\text{th}
14^\text{th}
of January, 2018.
Both parts will take place according to the following schedule: On the
13^\text{th},
the Computer Science exam shall start at 12:00 PM and submissions will be accepted till 3:00 PM. After an hour's break, the Physics exam shall begin at 4:00 PM and submissions must be handed in before 7:00 PM. The Mathematics exam is to be held on the
14^\text{th}.
We shall be uploading the question papers at 12:00 PM and the submission of answers must be made online by 6:00 PM.
For students who are writing MTRP: For students of class 9 and 11 who are appearing for MTRP, conducted by ISI, we understand that you might want to appear for both the examinations, and to avoid a clash with STEMS (school mathematics), we have decided that we can give those people a different time slot for writing STEMS. Please contact us in the given email IDs if you are interested in writing both exams, and we will give you a revised time slot for STEMS Mathematics. Note that this will only be done for students who are writing MTRP and STEMS Math together, upon showing appropriate evidence.
The dates and times are subject to change. We will inform the participants in case such a change takes place.
The event is divided into two sections: Section A is for students from
8^\text{th}
12^\text{th}
grade. Section B is for undergraduate and postgraduate students. The question paper, needless to say, shall be different for the two sections.
The question paper shall consist of 20 multiple choice questions and 6 subjective questions. The question papers shall be mailed to the registered participants exactly at the starting time.
Students are allowed to use books and electronic resources to solve the problems. However, your solutions must be strictly original. There might be an interview after the selection is made. In case of any discrepancies, the answer sheet of the student in question will be invalidated.
The solutions have to be submitted in the form of scanned copies or clear photographs of the answer sheets. Any electronic formats will also be accepted. They have to be mailed to tessellate.cmi@gmail.com from your registered email ID strictly before the ending time. Any submissions made after the deadline will not be accepted.
In the registration portal, please enter your name, gender, email ID, phone number, and permanent address in the mentioned format. Fill in the name of your school/college with its city in the 'College/City' tab.
In case of any issues regarding registration, send us a mail at tessellate.cmi@gmail.com.
Registration is hassle-free. Just click on our website http://tessellate.cmi.ac.in/#stems and register online. The fee is Rs. 100 per subject. For any queries, contact us again at tessellate.cmi@gmail.com or the following:
Ankita Sarkar: 8428532216 (ankita_s@cmi.ac.in)
Soham Chakraborty: 9884232190 (sochak@cmi.ac.in)
Srijan Ghosh: 9433777622 (srijang@cmi.ac.in)
Sarvesh Bandhaokar: 9405956066 (bandhaokar@cmi.ac.in)
Tessellate S.T.E.M.S 2018 - Mathematics - College
Tessellate S.T.E.M.S 2018 - Mathematics - School
Tessellate S.T.E.M.S 2018 - Physics - College
Tessellate S.T.E.M.S 2018 - Physics - School
Tessellate S.T.E.M.S 2018 - Computer Science - College
Tessellate S.T.E.M.S 2018 - Computer Science - School
Tessellate S.T.E.M.S - Mathematics - School - Set 1
Tessellate S.T.E.M.S - Mathematics - College - Set 1
Tessellate S.T.E.M.S. - Physics - School - Set 1
Tessellate S.T.E.M.S. - Physics - College - Set 1
Tessellate S.T.E.M.S. - Computer Science - School - Set 1
Tessellate S.T.E.M.S - Computer Science - School - Set 2
Tessellate S.T.E.M.S. - Computer Science - College - Set 1
Basic Counting (Rule of Sum, Rule of Product, Combinations, Permutations, etc.)
Elementary Recurrence Relations
Complex Numbers (Basics)
Arithmetic Progressions, Geometric Progression, etc.
Coordinate Geometry (Distance Formula, Equations of Straight Lines, Equation of Circles, etc.)
Conic Sections (Basics)
Modular Congruences (Euler's Theorem, Fermat's Little Theorem, etc. may be helpful.)
Trigonometry (Basics)
\big(
\mathbb{R}^n \to \mathbb{R}^m
\big)
(
)
Conduction in 1-dimension, Elementary concepts of Convection and Radiation
\frac{1}{2}
The objective of the exam is to test the student on their computational, algorithmic, and logical thinking abilities. Specific details about hardware architecture, operating systems, software systems, web technologies, programming languages, etc. will not be asked. To find the answer to a problem, one would not require programming.
Understanding correctness of algorithms
We do not expect any pre-requisite formal training of the candidates in any of these areas. Any relevant definitions and/or hints that are necessary for the understanding of the problem shall be provided. High school mathematical knowledge (and an inquisitive and computational mind!) should be enough to get started on any of the problems in this category.
Besides the sample problems and papers, these resources on Brilliant might be helpful:
Introduction to Algorithms - Concept Quiz
Algorithms - Wiki
Runtime of Algorithms
Computer Science Fundamentals - Course (Premium Content)
Computer Science Algorithms - Course (Premium Content)
Discrete Mathematics Warmup - Concept and Challenge Quizzes
Graph Theory - Concept and Challenge Quizzes
Graph Theory - Wiki
Discrete Probability - Warmup Quiz
Probability - Course (Premium Content)
Probability - Wiki
Combinatorial Games - Concept Quiz
Combinatorial Games - Wiki
Logic Puzzles - Quiz
Logic - Course (Premium Content)
The objective of the exam is to test the student on the theoretical aspects of computation. Specific details about hardware architecture, operating systems, web technologies, etc. will not be asked. To find the answer to a problem, one would not require programming.
Comprehensive understanding of algorithms and algorithmic paradigms such as greedy algorithms, dynamic programming, divide and conquer, and introductory graph algorithms. A preliminary knowledge of analysis of these algorithms is essential.
Understanding of data structures and various discrete structures such as graphs, trees, heaps, stacks, and queues
An understanding of finite state machines, pushdown systems, and turing machines, along with their properties and representations including grammars and computation models
An understanding of computation in terms of complexity and decidability
Greedy Algorithms - Wiki
Divide and Conquer - Wiki
Dynamic Programming - Wiki
Graphs - Wiki
Big O Notation - Wiki
You may refer to the material recommended for the school section for topics in discrete mathematics.
Cite as: Tessellate S.T.E.M.S. 2018. Brilliant.org. Retrieved from https://brilliant.org/wiki/tessellate-stems/
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An Improved Method to Analyze the Stress Relaxation of Ligaments Following a Finite Ramp Time Based on the Quasi-Linear Viscoelastic Theory | J. Biomech Eng. | ASME Digital Collection
Musculoskeletal Research Center, Department of Orthopaedic Surgery, Department of Bioengineering, University of Pittsburgh, E1641 Biomedical Science Tower, 210 Lothrop Street, P.O. BOX 71199, Pittsburgh, PA 15213 Phone: 412-648-2000 FAX: 412-648-2001
Contributed by the Bioengineering Division for publication in the JOURNAL OF BIOMECHANICAL ENGINEERING. Manuscript received by the Bioengineering Division April 2, 2003; revision received October 2, 2003. Associate Editor: L. Soslowsky.
Abramowitch , S. D., and Woo, S. L. (March 9, 2004). "An Improved Method to Analyze the Stress Relaxation of Ligaments Following a Finite Ramp Time Based on the Quasi-Linear Viscoelastic Theory ." ASME. J Biomech Eng. February 2004; 126(1): 92–97. https://doi.org/10.1115/1.1645528
The quasi-linear viscoelastic (QLV) theory proposed by Fung (1972) has been frequently used to model the nonlinear time- and history-dependent viscoelastic behavior of many soft tissues. It is common to use five constants to describe the instantaneous elastic response (constants A and B) and reduced relaxation function (constants C,
τ1,
τ2)
on experiments with finite ramp times followed by stress relaxation to equilibrium. However, a limitation is that the theory is based on a step change in strain which is not possible to perform experimentally. Accounting for this limitation may result in regression algorithms that converge poorly and yield nonunique solutions with highly variable constants, especially for long ramp times (Kwan et al. 1993). The goal of the present study was to introduce an improved approach to obtain the constants for QLV theory that converges to a unique solution with minimal variability. Six goat femur-medial collateral ligament-tibia complexes were subjected to a uniaxial tension test (ramp time of 18.4 s) followed by one hour of stress relaxation. The convoluted QLV constitutive equation was simultaneously curve-fit to the ramping and relaxation portions of the data
r2>0.99.
Confidence intervals of the constants were generated from a bootstrapping analysis and revealed that constants were distributed within 1% of their median values. For validation, the determined constants were used to predict peak stresses from a separate cyclic stress relaxation test with averaged errors across all specimens measuring less than 6.3±6.0% of the experimental values. For comparison, an analysis that assumed an instantaneous ramp time was also performed and the constants obtained for the two approaches were compared. Significant differences were observed for constants B, C,
τ1,
τ2,
τ1
differing by an order of magnitude. By taking into account the ramping phase of the experiment, the approach allows for viscoelastic properties to be determined independent of the strain rate applied. Thus, the results obtained from different laboratories and from different tissues may be compared.
biological tissues, physiological models, stress relaxation, viscoelasticity, elasticity, proteins, biomechanics
Relaxation (Physics), Stress, Biological tissues, Errors
Fung, Y. C., 1972, “Stress Strain History Relations of Soft Tissues in Simple Elongation,” in Biomechanics: Its Foundations and Objectives, eds., Y. C. Fung, N. Perrone, and M. Anliker, PrenticeHall, Englewood Cliffs, NJ, pp. 181–207.
Quasi-Linear Viscoelastic Theory Applied to Internal Shearing of Porcine Aortic Valve Leaflets
The Constitutive Behavior of Passive Heart Muscle Tissue: A Quasi-Linear Viscoelastic Formulation
Comparison of Viscoelastic Properties of the Pharyngeal Tissue: Human and Canine
The Viscoelastic Responses of the Human Cervical Spine in Torsion: Experimental Limitations of Quasi-Linear Theory, and a Method for Reducing These Effects
Elastic and Viscoelastic Material Behavior of Fresh and Glutaraldehyde-Treated Porcine Aortic Valve Tissue
Lin, H. C., Kwan, M. K., and Woo, S. L.-Y., 1987, “On the Stress Relaxation Properties of Anterior Cruciate Ligament (ACL),” Proceedings, ASME Adv Bioeng., pp. 5–6.
The Healing Medial Collateral Ligament Following a Combined Anterior Cruciate and Medial Collateral Ligament Injury—A Biomechanical Study in a Goat Model
Structure and Function of the Healing Medial Collateral Ligament in a Goat Model
The use of a Laser Micrometer System to Determine the Cross-Sectional Shape and Area of Ligaments: A Comparative Study With Two Existing Methods
Mechanical Properties of Tendons and Ligaments. I. Quasi-Static and Nonlinear Viscoelastic Properties
Press, W. H., Teukolsky, S. A., Vetterling, W. T., and Flannery, B. P., 1992, “Chapter 15: Modeling of Data,” in Numerical Recipies in C: The Art of Scientific Computing, Cambridge University Press, New York, NY, pp. 681–688.
An Approach to Quantification of Biaxial Tissue Stress-Strain Data
Lyon, R. M., Lin, H. C., Kwan, M. K., Hollis, J. M., Akeson, W. H., and Woo, S. L.-Y., 1988, “Stress Relaxation of the Anterior Cruciate Ligament (ACL) and the Patellar Tendon,” Proceedings, 34th Annual Meeting, Orthopaedic Research Society, Atlanta, GA, p. 81.
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Revision as of 14:05, 26 July 2013 by NikosA (talk | contribs) (→Pre-Processing Overview: added equation for at-sensor Radiance, needs "LaTeX" attention!)
In order to derive Reflectance values, likewise as with remotely sensed data acquired by other sensors, IKONOS raw image digital numbers (DNs) need to be converted to at-sensor spectral Radiance values. At-sensor spectral Radiance values are an important input for the equation to derive Reflectance values.
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
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(1-4)-a-D-glucan 1-a-D-glucosylmutase Wikipedia
(1-4)-a-D-glucan 1-a-D-glucosylmutase
(1->4)-alpha-D-glucan 1-alpha-D-glucosylmutase
(1->4)-alpha-D-glucan 1-alpha-D-glucosylmutase (EC 5.4.99.15, malto-oligosyltrehalose synthase, maltodextrin alpha-D-glucosyltransferase) is an enzyme with systematic name (1->4)-alpha-D-glucan 1-alpha-D-glucosylmutase.[1][2][3] This enzyme catalyses the following chemical reaction
4-[(1->4)-alpha-D-glucosyl]n-1-D-glucose
{\displaystyle \rightleftharpoons }
1-alpha-D-[(1->4)-alpha-D-glucosyl]n-1-alpha-D-glucopyranoside
The enzyme from Arthrobacter sp., Sulfolobus acidocaldarius acts on (1->4)-alpha-D-glucans containing three or more (1->4)-alpha-linked D-glucose units.
^ Maruta K, Nakada T, Kubota M, Chaen H, Sugimoto T, Kurimoto M, Tsujisaka Y (October 1995). "Formation of trehalose from maltooligosaccharides by a novel enzymatic system". Bioscience, Biotechnology, and Biochemistry. 59 (10): 1829–34. doi:10.1271/bbb.59.1829. PMID 8534970.
^ Nakada T, Maruta K, Tsusaki K, Kubota M, Chaen H, Sugimoto T, Kurimoto M, Tsujisaka Y (December 1995). "Purification and properties of a novel enzyme, maltooligosyl trehalose synthase, from Arthrobacter sp. Q36". Bioscience, Biotechnology, and Biochemistry. 59 (12): 2210–4. doi:10.1271/bbb.59.2210. PMID 8611744.
^ Nakada T, Ikegami S, Chaen H, Kubota M, Fukuda S, Sugimoto T, Kurimoto M, Tsujisaka Y (February 1996). "Purification and characterization of thermostable maltooligosyl trehalose synthase from the thermoacidophilic archaebacterium Sulfolobus acidocaldarius". Bioscience, Biotechnology, and Biochemistry. 60 (2): 263–6. doi:10.1271/bbb.60.263. PMID 9063973.
>4)-alpha-D-glucan%201-alpha-D-glucosylmutase (1->4)-alpha-D-glucan+1-alpha-D-glucosylmutase at the US National Library of Medicine Medical Subject Headings (MeSH)
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Home : Support : Online Help : Programming : ImageTools Package : Histogram
compute a histogram for each layer of an image
Histogram( img, buckets, opts )
(optional) posint; number of buckets/bins in the histogram (per layer)
(optional) equation(s) of the form option = value; specify options for the Histogram command
centers = truefalse
If true, return two elements: the Vector (or Matrix) corresponding to the histogram, and a Vector corresponding to the center value of each bucket in the histogram. The default is false.
cumulative = truefalse
Specifies that the count in each bucket is to include the counts in all the lower numbered buckets. This is especially useful in conjunction with the normalized option causing each bucket to end up containing a value indicating which fraction of the pixels in the image were less than or equal to the upper bound of the bucket. By selecting an appropriate number of buckets (4, 5, 100), the values correspond to concepts like quartiles, quintiles, and percentiles. The default is false.
If true, the value in each bucket of each layer is scaled so that the sum of the values in the buckets in a layer is 1.0. The default is false.
The Histogram command computes a histogram of the intensity of each layer of an image, and returns a Vector (for one-layer images) or Matrix (for multi-layer images) containing the histogram information.
The optional buckets parameter specifies the number of buckets per color channel desired. For most images, which were originally read from 8-bit per channel image files, the default of 256 is a suitable value.
For a single layer (grayscale) image, the Histogram command returns a column Vector with buckets elements. For multi-layer images, a buckets x N Matrix is returned, where N is the number of layers.
\mathrm{with}\left(\mathrm{ImageTools}\right):
Create a grayscale image with intensity varying from 1/4 to 3/4.
\mathrm{img}≔\mathrm{Create}\left(100,200,\left(r,c\right)↦\mathrm{evalf}\left(\frac{1}{2}+\frac{\mathrm{sin}\left(\frac{\mathrm{\pi }\cdot r}{50}\right)}{8}+\frac{\mathrm{cos}\left(\frac{\mathrm{\pi }\cdot c}{100}\right)}{8}\right)\right):
\mathrm{Histogram}\left(\mathrm{img},10\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{1339.}\\ \textcolor[rgb]{0,0,1}{3278.}\\ \textcolor[rgb]{0,0,1}{5324.}\\ \textcolor[rgb]{0,0,1}{5442.}\\ \textcolor[rgb]{0,0,1}{3278.}\\ \textcolor[rgb]{0,0,1}{1339.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\end{array}]
\mathrm{Histogram}\left(\mathrm{img},10,\mathrm{autorange},\mathrm{centers}\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{1339.}\\ \textcolor[rgb]{0,0,1}{1510.}\\ \textcolor[rgb]{0,0,1}{1768.}\\ \textcolor[rgb]{0,0,1}{2146.}\\ \textcolor[rgb]{0,0,1}{3178.}\\ \textcolor[rgb]{0,0,1}{3296.}\\ \textcolor[rgb]{0,0,1}{2146.}\\ \textcolor[rgb]{0,0,1}{1768.}\\ \textcolor[rgb]{0,0,1}{1510.}\\ \textcolor[rgb]{0,0,1}{1339.}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.275000000000000}\\ \textcolor[rgb]{0,0,1}{0.325000000000000}\\ \textcolor[rgb]{0,0,1}{0.375000000000000}\\ \textcolor[rgb]{0,0,1}{0.425000000000000}\\ \textcolor[rgb]{0,0,1}{0.475000000000000}\\ \textcolor[rgb]{0,0,1}{0.525000000000000}\\ \textcolor[rgb]{0,0,1}{0.575000000000000}\\ \textcolor[rgb]{0,0,1}{0.625000000000000}\\ \textcolor[rgb]{0,0,1}{0.675000000000000}\\ \textcolor[rgb]{0,0,1}{0.725000000000000}\end{array}]
\mathrm{Histogram}\left(\mathrm{img},10,\mathrm{range}=0..0.5,\mathrm{centers}\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{1339.}\\ \textcolor[rgb]{0,0,1}{1510.}\\ \textcolor[rgb]{0,0,1}{1768.}\\ \textcolor[rgb]{0,0,1}{2146.}\\ \textcolor[rgb]{0,0,1}{13237.}\end{array}]\textcolor[rgb]{0,0,1}{,}[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.0250000000000000}\\ \textcolor[rgb]{0,0,1}{0.0750000000000000}\\ \textcolor[rgb]{0,0,1}{0.125000000000000}\\ \textcolor[rgb]{0,0,1}{0.175000000000000}\\ \textcolor[rgb]{0,0,1}{0.225000000000000}\\ \textcolor[rgb]{0,0,1}{0.275000000000000}\\ \textcolor[rgb]{0,0,1}{0.325000000000000}\\ \textcolor[rgb]{0,0,1}{0.375000000000000}\\ \textcolor[rgb]{0,0,1}{0.425000000000000}\\ \textcolor[rgb]{0,0,1}{0.475000000000000}\end{array}]
\mathrm{Histogram}\left(\mathrm{img},10,\mathrm{range}=0..0.5,\mathrm{normalized}\right)
[\begin{array}{c}\textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.0669500000000000}\\ \textcolor[rgb]{0,0,1}{0.0755000000000000}\\ \textcolor[rgb]{0,0,1}{0.0884000000000000}\\ \textcolor[rgb]{0,0,1}{0.107300000000000}\\ \textcolor[rgb]{0,0,1}{0.661850000000000}\end{array}]
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The points P and P_0 are entered as
[x=a+b{t}^{r},y=\left(\mathrm{Laurent series}\mathrm{in}t\right)]
, where a and b are constants, and r is an integer. If r < 0, that is, if entering one of the points for
x=\mathrm{\infty }
, then a = 0.
\mathrm{with}\left(\mathrm{algcurves},\mathrm{AbelMap},\mathrm{genus},\mathrm{puiseux}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{AbelMap}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{genus}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{puiseux}}]
f≔{y}^{2}-\left({x}^{2}-1\right)\left({x}^{2}-4\right)\left({x}^{2}-9\right)\left({x}^{2}-16\right)
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{9}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{16}\right)
\mathrm{genus}\left(f,x,y\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{puiseux}\left(f,x=1,y,0,t\right)
{[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{720}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{720}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}]}
\mathrm{puiseux}\left(f,x=4,y,0,t\right)
{[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{10080}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{10080}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}]}
\mathrm{P_0},P≔\mathrm{op}\left(\right),\mathrm{op}\left(\right)
\textcolor[rgb]{0,0,1}{\mathrm{P_0}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{720}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{720}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{10080}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{10080}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}]
A≔\mathrm{AbelMap}\left(f,x,y,P,\mathrm{P_0},t,7\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{-0.5086732390}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1.395818333}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.5158465233}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.3733240360}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.00716997551}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.3585270829}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}]
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"Avoiding logic pitfalls in personal finance"
Avoiding logic pitfalls in personal finance
I've never particularly liked the way many people describe the tradeoffs between different investment accounts. You'll hear phrases like
"Pay the tax now, enjoy tax-free distributions later." (Roth contributions)
"Don't pay today and let a bigger principal compound, but face a huge tax bill." (deductible traditional contributions)
"Pay tax on contributions; earnings grow tax-deferred." (nondeductible traditional contributions)
The issue I take with these is that they're not very usable. It's not clear how to compare one type of tax structure to another. Indeed, it's all too easy to draw incorrect conclusions from loaded, prose approximations of what are actually mathematical statements. Take as examples the following two claims and their alleged proofs:
You should minimize the number of investment accounts you have. Why? To start, consider that the value of investments rises exponentially, and also that the rate of growth of an exponential curve increases as you move along it. Therefore, it's more powerful to have fewer accounts with bigger balances (so they'll accelerate upward faster) than to have more accounts, all growing slowly.
Roth contributions are preferable to traditional ones, since they incur a smaller tax bill. Why? Suppose you have $10,000 to invest, and you know that over the lifetime of your investment you'll see a return of 10x. Given the option, would you rather pay 25% tax on your principal today—that'd be $2,500—and enjoy the distributions tax free, or would you rather invest all $10,000 today and pay $25,000 in tax when you take distributions? I'd take a $2,500 tax liability over a $25,000 one any day.
One or both of these may seem valid arguments to you; they did at one time either to me or to people I know. They sound plausible! We'll see, however, that both are invalid.
These claims both attempt to suggest answers to the question "In what kind of account(s) should I put my savings?". In our conversations on that question my father is fond of saying "There are a lot of variables", and he's not wrong. To think clearly about such kinds of questions, it's incredibly helpful to be comfortable with letting variables be, well, variables! In the rest of this post, we'll explore using simple math to describe investment growth, with important quantities left as variables (e.g., time:
t
years). You'll find that in so doing, the value of a given type of investment will be clear and readily comparable to other types of investments.
Suppose an investor has $1,000 of cash in her pocket that she'd like to invest. She puts it in a bucket (also known as an account), within which she then trades it for securities: tradable financial assets like shares of a company's stock, or government bonds. Assuming that the securities are priced such that all she can trade every cent of her cash for some securities, she can and does make that trade:
Her investments, in the aggregate, appreciate 7% over the next year. The contents of her bucket are now valued at
The expressions on the right-hand side are all equivalent ways of writing the same thing. Take a moment to reacquaint yourself with some of the notation if you need to!
The investor lets it grow another year. Because she starts the year with $1,070, the math works out like
You may see a pattern forming! If we let
P
stand for her principal ($1,000),
r
stand for the annual rate of return (0.07), and
t
stand for the number of years elapsed since she invested the principal, then we have
where the multiplicative factors of
(1+r)
, of which one more appears with each additional year, are collapsed together via the exponent ("1.07 to the power of
t
"). This is where the saying "in investing, time is your best friend" comes from, since the account value increases exponentially with time.
You may see this equation called the "compound interest with reinvestment" formula. Though we've framed it as an investment scenario here, consider how compound interest might be modeled mathematically. Though the processes are not the same in a literal sense, the math describing the two is the same!
We now have all the tools we need to precisely describe basic investing scenarios. With them, let's disprove the claims given at the outset.
Claim #1: "It's more valuable to have fewer accounts with bigger balances. " Suppose
P_1
dollars are invested, and the assets they were traded for grow for one year. Then,
P_2
more dollars are invested. Another year passes. The result:
Does it make a difference whether we invested
P_2
in the same or a different account as
P_1
? Notice that we actually didn't make any assumption about exactly which account
P_2
was put in! You may have thought we were doing one or the other, but the math of either choice is the same. In general, an investment of
P
dollars is valued at
P(1+r)^t
t
years. Time invested matters, colocation with other assets does not. This can be understood by regarding each of your securities (a single share of stock, a bond, etc.) as being on its own personal journey of appreciation, simultaneous to one another. There's no reason that cohabitation in the same bucket should affect those separate journeys.
Claim #2: "Roth contributions are preferable, since they incur a smaller tax bill." Suppose we have
P
pretax dollars. In the traditional-vs.-Roth question, we're choosing when to subject our money to taxation: contribution-time, or distribution-time. Call the tax rates at those times
T_c
T_d
, respectively. What's the post-tax value of either choice, after
t
years of growth at annual rate of return
r
What do you notice about these results? Recall that in math, multiplication can happen in any order and not change the result. Bearing that in mind, you might see that the results are equal when
T_c = T_d
! In other words, if the tax rates are the same, taxing before growth or after leaves you with the same final amount in your pocket in either situation. This can be seen in the argument for Claim #2, for example, where you'll see that both contributions yield the same final post-tax value of $75,000 (the claim used a sleight of hand, suggesting that the minimum tax bill was the goal rather than maximum post-tax value). Finally, notice that if
T_c > T_d
—i.e., you're in a higher tax bracket earlier than you will be later—then the traditional contribution comes out ahead; vice versa, the opposite is true.
This result won't be surprising if you've already heard and accepted the common rule of thumb regarding traditional vs. Roth contributions:
Assuming you're investing the same portfolios for the same amount of time, you care most about the relation (>, <, =) between
T_c
T_d
We've just derived that rule.
The degree to which investments are colocated is inconsequential.
Under common assumptions, the post-tax values of a Roth and traditional contributions depend only on your marginal tax rates at contribution- and distribution-time, respectively.
You have a renewed appreciation for the tool that allowed us to convincingly prove these claims: math!
With this foundation, we can further compare actual account types (IRAs, 401(k)s, HSAs, etc.), while also incorporating some of the real-world considerations like contribution limits. All that and more in a future post! Stay tuned.
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Laplace transform of a exponential. we know that Laplace
Laplace transform of a exponential.
we know that Laplace transform of
{x}^{n}
\mathcal{L}\left[{x}^{n}\right]
provided n is a positive integer
but what is Laplace transform of
{a}^{x}
where a is some constant number
please help i searched on google so many times and try to solve this by its original defination that is
\mathcal{L}\left[{x}^{n}\right]={\int }_{0}^{\mathrm{\infty }}{a}^{x}{e}^{-sx}dx
but how to solve further ( i am using x insted of t here)
Latkaxu8j
{\int }_{0}^{\mathrm{\infty }}{a}^{x}{e}^{-sx}dx={\int }_{0}^{\mathrm{\infty }}{e}^{x\mathrm{ln}a}{e}^{-sx}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-\mathrm{ln}a\right)x}dx=\frac{1}{s-\mathrm{ln}a}
Drantumcem0
{a}^{x}{e}^{-sx}={\left(a{e}^{-s}\right)}^{x}
\mathcal{L}{\left(a{e}^{-s}\right)}^{x}={\int }_{0}^{\mathrm{\infty }}{\left(a{e}^{-s}\right)}^{x}=\frac{1}{\mathrm{ln}\left(a{e}^{-s}\right)}{\left(a{e}^{-s}\right)}^{x}{\mid }_{0}^{\mathrm{\infty }}=\frac{1}{\mathrm{ln}a-s}\left(0-1\right)=\frac{1}{s-\mathrm{ln}a}
Find the Laplace transform Y(s), of the solution of the IVP
y"+3{y}^{\prime }+2y=\mathrm{cos}\left(2t\right)
y\left(0\right)=0
{y}^{\prime }\left(0\right)=1
Do not solve the IVP
Derive the Laplace transform of the following unction: cos bt
y{y}^{\prime }=0
y\left(0\right)=2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=3
.Solve it using Laplace transforms
{y}^{\prime }-{\int }_{0}^{t}y\left(\tau \right)d\tau =4t{e}^{-t},y\left(0\right)=3
A part icle moves along the curve
y=2\mathrm{sin}\left(\pi \frac{x}{2}\right)
. As the particle passes through the point
\left(\frac{1}{3},1\right)
its x-coordinate increases at a rate of
\sqrt{10}
cm/s. How fast is the distance from the particle to the origin changing at this instant?
{y}^{\prime }=|x|,y\left(-1\right)=2
tu\left(t-2\right)={e}^{-2s}\left[\frac{s+2{s}^{2}}{{s}^{3}}\right]
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Penrose sought to reconcile general relativity and quantum theory using his own ideas about the possible structure of spacetime.[17][18] He suggested that at the Planck scale curved spacetime is not continuous, but discrete. He further postulated that each separated quantum superposition has its own piece of spacetime curvature, a blister in spacetime. Penrose suggests that gravity exerts a force on these spacetime blisters, which become unstable above the Planck scale of
{\displaystyle 10^{-35}{\text{m}}}
and collapse to just one of the possible states. The rough threshold for OR is given by Penrose's indeterminacy principle:
{\displaystyle \tau \approx \hbar /E_{G}}
{\displaystyle \tau }
is the time until OR occurs,
{\displaystyle E_{G}}
is the gravitational self-energy or the degree of spacetime separation given by the superpositioned mass, and
{\displaystyle \hbar }
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Find kk such that the following matrix M
Find k
such that the following matrix M
Evaluate the following limit ( IE do not use L'Hopital's rule).
lim𝑥→4 sin(𝑥2−4𝑥) sin(𝑥2−10𝑥+24) .
\underset{x\to 4}{lim}\frac{\text{ }\mathrm{sin}\left({x}^{2}-4x\right)\text{ }}{\text{ }\mathrm{sin}\left({x}^{2}-10x+24\right)\text{ }}.
U
\left(0,1\right). {X}_{1},...,{X}_{n}
U
f:\left(0,1{\right)}^{n}->R
E\left[\left(U-f\left({X}_{1},...,{X}_{n}\right){\right)}^{2}\right)\right]
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Control system in which an oscillator relies on a filter to calibrate its input
"PLL" redirects here. For other uses, see PLL (disambiguation).
3.1 Clock recovery
3.2 Deskewing
3.7 Jitter and noise reduction
5.4 Feedback path and optional divider
6.1 Time domain model of APLL
6.2 Phase domain model of APLL
6.3 Linearized phase domain model
6.4 Implementing a digital phase-locked loop in software
7 Practical analogies
7.1 Automobile race analogy
7.2 Clock analogy
Main article: Phase-locked loop ranges
Deskewing[edit]
Clock generation[edit]
Spread spectrum[edit]
Clock distribution[edit]
AM detection[edit]
Jitter and noise reduction[edit]
Frequency synthesis[edit]
Feedback path and optional divider[edit]
{\displaystyle N}
{\displaystyle M}
{\displaystyle N/M}
Time domain model of APLL[edit]
{\displaystyle f_{1}(\theta _{1}(t))}
{\displaystyle f_{2}(\theta _{2}(t))}
{\displaystyle \theta _{1}(t)}
{\displaystyle \theta _{2}(t)}
{\displaystyle f_{1}(\theta )}
{\displaystyle f_{2}(\theta )}
{\displaystyle \varphi (t)}
{\displaystyle \varphi (t)=f_{1}(\theta _{1}(t))f_{2}(\theta _{2}(t))}
{\displaystyle g(t)}
{\displaystyle {\dot {\theta }}_{2}(t)=\omega _{2}(t)=\omega _{\text{free}}+g_{v}g(t)\,}
{\displaystyle g_{v}}
{\displaystyle \omega _{\text{free}}}
{\displaystyle {\begin{array}{rcl}{\dot {x}}&=&Ax+b\varphi (t),\\g(t)&=&c^{*}x,\end{array}}\quad x(0)=x_{0},}
{\displaystyle \varphi (t)}
{\displaystyle g(t)}
{\displaystyle A}
{\displaystyle n}
{\displaystyle n}
{\displaystyle x\in \mathbb {C} ^{n},\quad b\in \mathbb {R} ^{n},\quad c\in \mathbb {C} ^{n},\quad }
{\displaystyle x_{0}\in \mathbb {C} ^{n}}
{\displaystyle {\begin{array}{rcl}{\dot {x}}&=&Ax+bf_{1}(\theta _{1}(t))f_{2}(\theta _{2}(t)),\\{\dot {\theta }}_{2}&=&\omega _{\text{free}}+g_{v}(c^{*}x)\\\end{array}}\quad x(0)=x_{0},\quad \theta _{2}(0)=\theta _{0}.}
{\displaystyle \theta _{0}}
Phase domain model of APLL[edit]
{\displaystyle f_{1}(\theta _{1}(t))}
{\displaystyle f_{2}(\theta _{2}(t))}
{\displaystyle 2\pi }
{\displaystyle f_{1}(\theta )}
{\displaystyle f_{2}(\theta )}
{\displaystyle \varphi (\theta )}
{\displaystyle G(t)}
{\displaystyle {\begin{array}{rcl}{\dot {x}}&=&Ax+b\varphi (\theta _{1}(t)-\theta _{2}(t)),\\G(t)&=&c^{*}x,\end{array}}\quad x(0)=x_{0},}
{\displaystyle G(t)-g(t)}
is small with respect to the frequencies) to the output of the Filter in time domain model. [17] [18] Here function
{\displaystyle \varphi (\theta )}
{\displaystyle \theta _{\Delta }(t)}
{\displaystyle \theta _{\Delta }=\theta _{1}(t)-\theta _{2}(t).}
{\displaystyle {\begin{array}{rcl}{\dot {x}}&=&Ax+b\varphi (\theta _{\Delta }),\\{\dot {\theta }}_{\Delta }&=&\omega _{\Delta }-g_{v}(c^{*}x).\\\end{array}}\quad x(0)=x_{0},\quad \theta _{\Delta }(0)=\theta _{1}(0)-\theta _{2}(0).}
{\displaystyle \omega _{\Delta }=\omega _{1}-\omega _{\text{free}}}
{\displaystyle \omega _{1}}
{\displaystyle \omega _{\text{free}}}
{\displaystyle f_{1}(\theta _{1}(t))=A_{1}\sin(\theta _{1}(t)),\quad f_{2}(\theta _{2}(t))=A_{2}\cos(\theta _{2}(t))}
{\displaystyle {\begin{aligned}{\dot {x}}&=-{\frac {1}{RC}}x+{\frac {1}{RC}}A_{1}A_{2}\sin(\theta _{1}(t))\cos(\theta _{2}(t)),\\[6pt]{\dot {\theta }}_{2}&=\omega _{\text{free}}+g_{v}(c^{*}x)\end{aligned}}}
{\displaystyle \varphi (\theta _{1}-\theta _{2})={\frac {A_{1}A_{2}}{2}}\sin(\theta _{1}-\theta _{2})}
{\displaystyle {\begin{aligned}{\dot {x}}&=-{\frac {1}{RC}}x+{\frac {1}{RC}}{\frac {A_{1}A_{2}}{2}}\sin(\theta _{\Delta }),\\[6pt]{\dot {\theta }}_{\Delta }&=\omega _{\Delta }-g_{v}(c^{*}x).\end{aligned}}}
{\displaystyle {\begin{aligned}x&={\frac {{\dot {\theta }}_{2}-\omega _{2}}{g_{v}c^{*}}}={\frac {\omega _{1}-{\dot {\theta }}_{\Delta }-\omega _{2}}{g_{v}c^{*}}},\\[6pt]{\dot {x}}&={\frac {{\ddot {\theta }}_{2}}{g_{v}c^{*}}},\\[6pt]\theta _{1}&=\omega _{1}t+\Psi ,\\[6pt]\theta _{\Delta }&=\theta _{1}-\theta _{2},\\[6pt]{\dot {\theta }}_{\Delta }&={\dot {\theta }}_{1}-{\dot {\theta }}_{2}=\omega _{1}-{\dot {\theta }}_{2},\\[6pt]&{\frac {1}{g_{v}c^{*}}}{\ddot {\theta }}_{\Delta }-{\frac {1}{g_{v}c^{*}RC}}{\dot {\theta }}_{\Delta }-{\frac {A_{1}A_{2}}{2RC}}\sin \theta _{\Delta }={\frac {\omega _{2}-\omega _{1}}{g_{v}c^{*}RC}}.\end{aligned}}}
Linearized phase domain model[edit]
{\displaystyle {\frac {\theta _{o}}{\theta _{i}}}={\frac {K_{p}K_{v}F(s)}{s+K_{p}K_{v}F(s)}}}
{\displaystyle \theta _{o}}
{\displaystyle \theta _{i}}
{\displaystyle K_{p}}
{\displaystyle K_{v}}
{\displaystyle F(s)}
{\displaystyle F(s)={\frac {1}{1+sRC}}}
{\displaystyle {\frac {\theta _{o}}{\theta _{i}}}={\frac {\frac {K_{p}K_{v}}{RC}}{s^{2}+{\frac {s}{RC}}+{\frac {K_{p}K_{v}}{RC}}}}}
{\displaystyle s^{2}+2s\zeta \omega _{n}+\omega _{n}^{2}}
{\displaystyle \zeta }
{\displaystyle \omega _{n}}
{\displaystyle \omega _{n}={\sqrt {\frac {K_{p}K_{v}}{RC}}}}
{\displaystyle \zeta ={\frac {1}{2{\sqrt {K_{p}K_{v}RC}}}}}
{\displaystyle RC={\frac {1}{2K_{p}K_{v}}}}
{\displaystyle \omega _{c}=K_{p}K_{v}{\sqrt {2}}}
{\displaystyle F(s)={\frac {1+sCR_{2}}{1+sC(R_{1}+R_{2})}}}
{\displaystyle \tau _{1}=C(R_{1}+R_{2})}
{\displaystyle \tau _{2}=CR_{2}}
{\displaystyle \omega _{n}={\sqrt {\frac {K_{p}K_{v}}{\tau _{1}}}}}
{\displaystyle \zeta ={\frac {1}{2\omega _{n}\tau _{1}}}+{\frac {\omega _{n}\tau _{2}}{2}}}
{\displaystyle \tau _{1}={\frac {K_{p}K_{v}}{\omega _{n}^{2}}}}
{\displaystyle \tau _{2}={\frac {2\zeta }{\omega _{n}}}-{\frac {1}{K_{p}K_{v}}}}
Implementing a digital phase-locked loop in software[edit]
Practical analogies[edit]
Automobile race analogy[edit]
Clock analogy[edit]
Retrieved from "https://en.wikipedia.org/w/index.php?title=Phase-locked_loop&oldid=1085889022"
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Method of Undetermined Coefficients | Brilliant Math & Science Wiki
Arulx Z, Rohit Udaiwal, and Jimin Khim contributed
Method of undetermined coefficients is used for finding a general formula for a specific summation problem. The method can only be used if the summation can be expressed as a polynomial function.
The method involves comparing the summation to a general polynomial function followed by simplification.
Let's start with an easy and well-known summation.
Find a general formula for
1 + 2 + 3 + \dots + n.
Let's start by assuming that the above summation can be expressed as a polynomial function
f\left( n \right)
f\left( n \right) = { A }_{ 0 }+{ A }_{ 1 }n+{ A }_{ 2 }{ n }^{ 2 }+{ A }_{ 3 }{ n }^{ 3 }+\cdots.
Let's equate this function to our original summation:
1 + 2 + 3 + \dots + n = { A }_{ 0 }+{ A }_{ 1 }n+{ A }_{ 2 }{ n }^{ 2 }+{ A }_{ 3 }{ n }^{ 3 }+\cdots. \qquad (1)
Now we want to simplify this equation. There's no easy way to simplify this because the left-hand side is not a proper expression. We first need to express the left-hand side as a proper expression.
The easiest way to do this is to compare the above expression to the summation
1 + 2 + 3 + \dots + n + \left( n+1 \right)
\begin{aligned} 1+2+3+\cdots +n+\left( n+1 \right) & = f\left( n+1 \right) \\ 1+2+3+\cdots +n+\left( n+1 \right) & = { A }_{ 0 }+{ A }_{ 1 }\left( n+1 \right) +{ A }_{ 2 }{ \left( n+1 \right) }^{ 2 }+{ A }_{ 3 }{ \left( n+1 \right) }^{ 3 }+\cdots. \qquad (2) \end{aligned}
(2)-(1)
n + 1 = { A }_{ 1 }+\left( 2n+1 \right) { A }_{ 2 }+\left( 3{ n }^{ 2 }+3n+1 \right) { A }_{ 3 }+\cdots.
Since this expression works for all positive integers
n
, we can compare the degrees of the left-hand side and right-hand side. But before we do that, it's worth noting that the highest degree on the left-hand side is 1. For this equation to satisfy all values of
n
, the degree of the right-hand side must also be 1. Hence we can eliminate all variables on the right except for
{ A }_{ 1 }+\left( 2n+1 \right) { A }_{ 2 }
because all other values are 0. Then the new equation is
n+1={ A }_{ 1 }+\left( 2n+1 \right) { A }_{ 2 },
\begin{aligned} n+1 & = { A }_{ 1 }+\left( 2n+1 \right) { A }_{ 2 } \\ & = 2{ A }_{ 2 }n+\left( { A }_{ 1 }+{ A }_{ 2 } \right)\\ \\ 2A_2 & = 1, ~A_1+A_2=1\\ \Rightarrow A_1&=A_2=\frac{1}{2}. \end{aligned}
A_1=A_2=\frac{1}{2}
(1)
\begin{aligned} 1+2+3+\dots +n & = { A }_{ 0 }+{ A }_{ 1 }n+{ A }_{ 2 }{ n }^{ 2 }+{ A }_{ 3 }{ n }^{ 3 }+\cdots \\ & = { A }_{ 0 }+\frac { 1 }{ 2 } n+\frac { 1 }{ 2 } { n }^{ 2 }+0\cdot { n }^{ 3 }+\cdots \\ &={ A }_{ 0 }+\frac { 1 }{ 2 } n+\frac { 1 }{ 2 } { n }^{ 2 }. \end{aligned}
Now we just need to find the value of
{ A }_{ 0 }
, which is easy. Since the above expression is true for all values of
n
, we can just try substituting a specific value of
n
into the equation and then find the value of
{ A }_{ 0 }
. For simplicity, let's try
n = 2
(but you can also try any other positive integer
n
\begin{aligned} 1+2 & = { A }_{ 0 }+\frac { 1 }{ 2 } \times 2+\frac { 1 }{ 2 } \times { 2 }^{ 2 } \\ 3 & = { A }_{ 0 }+1+2 \\ { A }_{ 0 } & = 0. \end{aligned}
Therefore, we have now completely simplified the equation. Expressing it in proper form, we have
\begin{aligned} 1+2+3+\cdots +n & = \frac { 1 }{ 2 } n+\frac { 1 }{ 2 } { n }^{ 2 } \\ & = \frac { { n }^{ 2 }+n }{ 2 } \\ & = \frac { n\left( n+1 \right) }{ 2 }. \ _\square \end{aligned}
Let's try a slightly harder version of this problem. Before seeing the solution, try figuring out the answer on your own!
{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+\cdots +{ n }^{ 2 }.
As in the previous problem, we have
\begin{aligned} { 1 }^{ 2 }+{ 2 }^{ 2 }+\cdots +{ n }^{ 2 }&={ A }_{ 0 }+{ A }_{ 1 }n+{ A }_{ 2 }{ n }^{ 2 }+{ A }_{ 3 }{ n }^{ 3 }+{ A }_{ 4 }{ n }^{ 4 }+\cdots \\ { 1 }^{ 2 }+{ 2 }^{ 2 }+\dots +{ n }^{ 2 }+{ \left( n+1 \right) }^{ 2 }&={ A }_{ 0 }+{ A }_{ 1 }\left( n+1 \right) +{ A }_{ 2 }{ \left( n+1 \right) }^{ 2 }+{ A }_{ 3 }{ \left( n+1 \right) }^{ 3 }+{ A }_{ 4 }{ \left( n+1 \right) }^{ 4 }+\cdots. \end{aligned}
On subtracting the equations,
{ \left( n+1 \right) }^{ 2 }={ A }_{ 1 }+{ A }_{ 2 }\left( 2n+1 \right) +{ A }_{ 3 }\left( 3{ n }^{ 2 }+3n+1 \right) +{ A }_{ 4 }\left( 4{ n }^{ 3 }+6{ n }^{ 2 }+4n+1 \right) +\cdots.
The variables on right-hand side with a degree greater than 2 can be discarded:
\begin{aligned} { \left( n+1 \right) }^{ 2 }&={ A }_{ 1 }+{ A }_{ 2 }\left( 2n+1 \right) +{ A }_{ 3 }\left( 3{ n }^{ 2 }+3n+1 \right) \\ { n }^{ 2 }+2n+1&={ A }_{ 1 }+{ A }_{ 2 }\left( 2n+1 \right) +{ A }_{ 3 }\left( 3{ n }^{ 2 }+3n+1 \right). \end{aligned}
On expressing in proper form,
{ n }^{ 2 }+2n+1=3{ A }_{ 3 }{ n }^{ 2 }+\left( 2{ A }_{ 2 }+3{ A }_{ 3 } \right) n+\left( { A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 } \right) .
On comparing the degrees,
\begin{aligned} { n }^{ 2 }=3{ A }_{ 3 }{ n }^{ 2 }&\Rightarrow { A }_{ 3 }=\frac { 1 }{ 3 } \\ 2n=\left( 2{ A }_{ 2 }+3{ A }_{ 3 } \right) n\Rightarrow 2=2{ A }_{ 2 }+3{ A }_{ 3 }&\Rightarrow A_2=\frac { 1 }{ 2 } \\ 1={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }&\Rightarrow { A }_{ 1 }=\frac { 1 }{ 6 }. \end{aligned}
On substituting this into the original equation,
\begin{aligned} { 1 }^{ 2 }+{ 2 }^{ 2 }+\cdots +{ n }^{ 2 } &={ A }_{ 0 }+{ A }_{ 1 }n+{ A }_{ 2 }{ n }^{ 2 }+{ A }_{ 3 }{ n }^{ 3 }+{ A }_{ 4 }{ n }^{ 4 }+\cdots \\ &={ A }_{ 0 }+\frac { 1 }{ 6 } n+\frac { 1 }{ 2 } { n }^{ 2 }+\frac { 1 }{ 3 } { n }^{ 3 }+{ A }_{ 4 }{ n }^{ 4 }+\cdots \\ &={ A }_{ 0 }+\frac { 1 }{ 6 } n+\frac { 1 }{ 2 } { n }^{ 2 }+\frac { 1 }{ 3 } { n }^{ 3 }, \end{aligned}
since variables with degree greater than 3 can be removed.
Now, since all values of
n
work, we can substitute a value of
n
{ A }_{ 0 }
n = 1
{ 1 }^{ 2 }={ A }_{ 0 }+\frac { 1 }{ 6 } \times 1+\frac { 1 }{ 2 } \times { 1 }^{ 2 }+\frac { 1 }{ 3 } \times { 1 }^{ 3 }={ A }_{ 0 }+\frac { 1 }{ 6 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \Rightarrow { A }_{ 0 }=0.
\begin{aligned} { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+\cdots +{ n }^{ 2 } & = \frac { 1 }{ 6 } n+\frac { 1 }{ 2 } { n }^{ 2 }+\frac { 1 }{ 3 } { n }^{ 3 } \\ & = \frac { 2{ n }^{ 3 }+3{ n }^{ 2 }+n }{ 6 } \\ & = \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }.\ _\square \end{aligned}
Can you now find a general formula for the sum of cubes or fourth powers?
Find a general formual for
1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1).
\begin{aligned} 1\cdot2+2\cdot3+\cdots+n(n+1)=& A_{0}+A_{1}n+A_{2}n^2+\cdots \\ 1\cdot2+2\cdot3+\cdots+n(n+1)+(n+1)(n+2)=& A_{0}+A_{1}(n+1)+A_{2} (n+1)^2+\cdots. \end{aligned}
By subtracting, we have
(n+1)(n+2)=A_{1}+A_{2}(2n+1)+A_{3}\left(3n^2+3n+1\right)+A_{4}\left(4n^3+6n^2+4n+1\right)+\cdots.
By comparing corresponding coefficients, we get
3A_{3}=1,\quad 3A_{3}+2A_{2}=3,\quad A_{1}+A_{2}+A_{3}=2,
A_{3}=\dfrac{1}{3},\quad A_{2}=1, \quad A_{1}=\dfrac{2}{3}.
Our sum then becomes
1\cdot2+2\cdot3+\cdots+n(n+1) = A_0 + \dfrac{2}{3}n+n^2+\dfrac{1}{3}n^3.
A_{0}
n=1
, then the series reduces to its first term:
2=A_{0}+2 \Rightarrow A_{0}=0.
On simplifying and factorizing,
1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1)=\dfrac{1}{3}n(n+1)(n+2).\ _\square
{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdots +{ 99 }^{ 2 } = \, ?
1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\dots +98\cdot 99\cdot 100 = \, ?
\sum _{ n=1 }^{ 50 }{ \left( 2n-1 \right) { \left( n+2 \right) }^{ 2 } } =1\cdot { 3 }^{ 2 }+3\cdot { 4 }^{ 2 }+5\cdot { 5 }^{ 2 }+\cdots +99\cdot { 52 }^{ 2 } = \, ?
Cite as: Method of Undetermined Coefficients. Brilliant.org. Retrieved from https://brilliant.org/wiki/method-of-undetermined-coefficients/
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PH - New World Encyclopedia
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The Hydrangea macrophylla blossoms in pink or blue, depending on soil pH. In acidic soils, the flowers are blue; in alkaline soils, the flowers are pink.
pH is a measure of the acidity and the basicity/alkalinity of a solution in terms of activity of hydrogen (H+) (strictly speaking, there is no such thing as the H+ ion; it is H3O+, called the hydroxonium ion). For dilute solutions, however, it is convenient to substitute the activity of the hydrogen ions with the concentration or molarity (mol/L) of the hydrogen ions (however, this is not necessarily accurate at higher concentrations[1][2]).
In aqueous systems, the hydrogen ion activity is dictated by the dissociation constant of water (Kw = 1.011 × 10−14 M2 at 25°C) and interactions with other ions in solution. Due to this dissociation constant, a neutral solution (hydrogen ion activity equals hydroxide ion activity) has a pH of approximately 7. Aqueous solutions with pH values lower than 7 are considered acidic, while pH values higher than 7 are considered basic.
The concept of pH was introduced by S.P.L. Sørensen in 1909, and some sources trace it to the Latin term pondus hydrogenii.[3] Other sources, however, attribute the name to the French term pouvoir hydrogène[4][5][6] or puissance d'hydrogène.
Though pH is generally expressed without units, it is not an arbitrary scale; the number arises from a definition based on the activity of hydrogen ions in the solution. The pH scale is a reverse logarithmic representation of relative hydrogen ion (H+) concentration. On this scale, an upward shift by one integral number represents a ten-fold decrease in value. For example, a shift in pH from 2 to 3 represents a 10-fold decrease in H+ concentration, and a shift from 2 to 4 represents a 100-fold (10 × 10-fold) decrease in H+ concentration.
The precise formula for calculating pH is as follows:
{\displaystyle {\mbox{pH}}=-\log _{10}{(a_{\mathrm {H^{+}} })}}
aH+ denotes the activity of H+ ions, and is dimensionless. In solutions that contain other ions, activity and concentration are not the same. The activity is an effective concentration of hydrogen ions, rather than the true concentration; it accounts for the fact that other ions surrounding the hydrogen ions will shield them and affect their ability to participate in chemical reactions. These other ions effectively change the hydrogen ion concentration in any process that involves H+.
In dilute solutions (such as tap water), the activity is approximately equal to the numeric value of the concentration of the H+ ion, denoted as [H+] (or more accurately written, [H3O+]), measured in moles per liter (also known as molarity). Therefore, it is often convenient to define pH as:
{\displaystyle {\mbox{pH}}\approx -\log _{10}{\frac {[\mathrm {H^{+}} ]}{1~\mathrm {mol/L} }}=-\log _{10}{\left|[\mathrm {H^{+}} ]\right|}}
For both definitions, log10 denotes the base-10 logarithm; therefore, pH defines a logarithmic scale of acidity. The straight bars, indicating absolute value, make pH a dimensionless quantity. For example, if one makes a lemonade with a H+ concentration of 0.0050 moles per liter, its pH would be:
{\displaystyle {\mbox{pH}}_{\mathrm {lemonade} }\approx -\log _{10}{\left|0.0050~\mathrm {mol/L} \right|}=-\log _{10}{(0.0050)}}
{\displaystyle {\mbox{pH}}_{\mathrm {lemonade} }\approx 2.3}
A solution of pH = 8.2 will have an [H+] concentration of 10−8.2 mol/L, or about 6.31 × 10−9 mol/L. Thus, its hydrogen activity aH+ is around 6.31 × 10−9. A solution with an [H+] concentration of 4.5 × 10−4 mol/L will have a pH value of 3.35.
In solution at 25°C, a pH of 7 indicates neutrality (i.e., the pH of pure water) because water naturally dissociates into H+ and OH− ions with equal concentrations of 1×10−7 mol/L. A lower pH value (for example pH 3) indicates increasing strength of acidity, and a higher pH value (for example pH 11) indicates increasing strength of basicity. Note, however, that pure water, when exposed to the atmosphere, will take in carbon dioxide, some of which reacts with water to form carbonic acid and H+, thereby lowering the pH to about 5.7.
Neutral pH at 25°C is not exactly 7. pH is an experimental value, so it has an associated error. Since the dissociation constant of water is (1.011 ± 0.005) × 10−14, the pH of water at 25°C would be 6.998 ± 0.001. The value is consistent, however, with neutral pH being 7.00 to two significant figures, which is near enough for most people to assume that it is exactly 7. The pH of water gets smaller with higher temperatures. For example, at 50°C, the pH of water is 6.55 ± 0.01. This means that a diluted solution is neutral at 50°C when its pH is around 6.55 and that a pH of 7.00 is basic.
Most substances have a pH in the range 0 to 14, although extremely acidic or extremely basic substances may have pH less than 0 or greater than 14. An example is acid mine runoff, with a pH = –3.6. Note that this does not translate to a molar concentration of 3981 M.
Arbitrarily, the pH is defined as
{\displaystyle -\log _{10}{([{\mbox{H}}^{+}])}}
{\displaystyle {\mbox{pH}}=-\log _{10}{[{{\mbox{H}}^{+}}]}}
{\displaystyle {\mbox{pH}}={\frac {\epsilon }{0.059}}}
The "pH" of any other substance may also be found (e.g., the potential of silver ions, or pAg+) by deriving a similar equation using the same process. These other equations for potentials will not be the same, however, as the number of moles of electrons transferred (n) will differ for the different reactions.
Under the Brønsted-Lowry theory, stronger or weaker acids are a relative concept. Here we define a strong acid as a species that is a much stronger acid than the hydronium (H3O+) ion. In that case, the dissociation reaction (strictly HX+H2O↔H3O++X− but simplified as HX↔H++X−) goes to completion, i.e., no unreacted acid remains in solution. Dissolving the strong acid HCl in water can therefore be expressed:
When calculating the pH of a weak acid, it is usually assumed that the water does not provide any hydrogen ions. This simplifies the calculation, and the concentration provided by water, ×10−7 mol/L, is usually insignificant.
{\displaystyle 1.6\times 10^{-4}={\frac {x^{2}}{0.1-x}}}
Hydrochloric Acid, 1M 0.1
By addition of a pH indicator into the solution under study. The indicator color varies depending on the pH of the solution. Using indicators, qualitative determinations can be made with universal indicators that have broad color variability over a wide pH range and quantitative determinations can be made using indicators that have strong color variability over a small pH range. Extremely precise measurements can be made over a wide pH range using indicators that have multiple equilibriums in conjunction with spectrophotometric methods to determine the relative abundance of each pH-dependent component that together make up the color of a solution, or
As the pH scale is logarithmic, it doesn't start at zero. Thus the most acidic of liquids encountered can have a pH of as low as −5. The most alkaline typically has a pH of 14.
There is also pOH, in a sense the opposite of pH, which measures the concentration of OH− ions, or the basicity. Since water self-ionizes, and notating [OH−] as the concentration of hydroxide ions, we have
{\displaystyle K_{w}=a_{{\rm {H}}^{\ }}a_{{\rm {OH}}^{-}}=10^{-14}}
{\displaystyle \log _{10}K_{w}=\log _{10}a_{{\rm {H}}^{+}}+\log _{10}a_{{\rm {OH}}^{-}}}
{\displaystyle -14={\rm {log}}_{\rm {10}}\,a_{{\rm {H}}^{\rm {+}}}+\log _{10}\,a_{{\rm {OH}}^{-}}}
{\displaystyle {\rm {pOH}}=-\log _{10}\,a_{{\rm {OH}}^{-}}=14+\log _{10}\,a_{{\rm {H}}^{+}}=14-{\rm {pH}}}
This formula is valid exactly for temperature = 298.15 K (25°C) only, but is acceptable for most lab calculations.
An indicator is used to measure the pH of a substance. Common indicators are litmus paper, phenolphthalein, methyl orange, phenol red, bromothymol blue, and bromocresol purple. To demonstrate the principle with common household materials, red cabbage, which contains the dye anthocyanin, is used.[7]
↑ http://www.jp.horiba.com/story_e/ph/ph01_03.htm. Retrieved December 14, 2007.
↑ http://chem.lapeer.org/Chem2Docs/pHFacts.html. Retrieved December 14, 2007.
↑ http://www.madsci.org/posts/archives/sep2001/1000136604.Sh.r.html. Retrieved December 14, 2007.
↑ Davis, R. E.; Metcalfe, H. C.; Williams, J. E.; Castka, J. F. et al. 2002. Aqueous Solutions and the Concept of pH. In Modern Chemistry, p. 485. Austin: Holt, Rinehart and Winston.
↑ http://encarta.msn.com/encyclopedia_761552883/pH.html. Retrieved December 14, 2007.
↑ http://bcn.boulder.co.us/basin/data/COBWQ/info/pH.html. Retrieved December 14, 2007.
↑ chemistry.about.com. Retrieved December 14, 2007.
Brown, Theodore E., H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy. 2005. Chemistry: The Central Science (10th Edition). Upper Saddle River, NJ: Prentice Hall. ISBN 0131096869
Corwin, C. H. 2001. Introductory Chemistry Concepts and Connections (3rd ed.). Upper Saddle River, NJ: Prentice Hall. ISBN 0130874701
McMurry, J., and R. C. Fay. 2004. Chemistry (4th ed.). Upper Saddle River, NJ: Prentice Hall. ISBN 0131402080
Moore, J. W., C. L. Stanitski, and P. C. Jurs. 2002. Chemistry: The Molecular Science. New York: Harcourt College. ISBN 0030320119
Oxlade, Chris (2002). Acids and Bases (Chemicals in Action). Heinemann Library. ISBN 1588101940 and ISBN 978-1588101945
CurTiPot pH calculation + virtual titration + acid-base titration curve data analysis + distribution diagram generation + pKa database.
History of "PH"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=PH&oldid=1017068
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Ministry of Education are inviting tender for four categories promoting the use of IT in education.
\left(5+4\right)+\left(4+4\right)+\left(3+3\right)+\left(7+4\right)=9+8+6+11=34
Since each project appears in exactly one category, the four lists of projects can be combined into one list of projects containing
5+4+3+7=19
projects. Hence, there are 19 projects from which to choose.
The Addition Principle (or Sum Rule) states that if there are
{n}_{1}
ways of performing one task and
{n}_{2}
ways of performing another task that cannot be performed at the same time, there are
{n}_{1}+{n}_{2}
ways of performing both tasks.
In this case, we can select a project from category one in five ways, a project from category two in four ways, a project from category three in three ways, or a project in category four in seven ways. Since it is not possible to choose the same project from more than one list, the Addition Principle applies. Hence, there is a total of
5+4+3+7=19
projects from which to choose.
{26}^{2}×{10}^{4}
{}^{6}{P}_{2}
X=\left\{\left(123\right),\left(132\right),\left(124\right),\left(142\right),\left(134\right),\left(143\right),\left(234\right),\left(243\right)\right\}
{A}_{4}
X
x=\left(123\right)
4=|\mathcal{O}\left(x\right)|=|G|/|{G}_{x}|=12/|{G}_{x}|
{G}_{x}=\left\{1\right\}
\left(1,2,...,n\right)
k
\left(n-k\right)!
k
\left(\genfrac{}{}{0}{}{n}{k}\right)\left(n-k\right)!=\frac{n!}{k!}
n!
n!-\left(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+\left(-1{\right)}^{n+1}\frac{n!}{n!}\right)
k
n-k
\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}
f
A\to B
|A|=4,|B|=3
{3}^{4}-\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}+\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{2}\right){1}^{4}
\left(\genfrac{}{}{0}{}{3}{1}\right){2}^{4}
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Cauchy distribution - Simple English Wikipedia, the free encyclopedia
In mathematics, the Cauchy-Lorentz distribution (after Augustin-Louis Cauchy and Hendrik Lorentz) is a continuous probability distribution with two parameters: a location parameter and a scale parameter.[1][2] As a probability distribution, it is usually called a Cauchy distribution. Physicists know it as a Lorentz distribution.
When the location parameter is 0 and the scale parameter is 1, the probability density function of the Cauchy distribution reduces to
{\displaystyle f(x)=1/[\pi (x^{2}+1)]}
. This is called the standard Cauchy distribution.[2]
The Cauchy distribution is used in spectroscopy to describe the spectral lines found there, and to describe resonance.[3] It is also often used in statistics as the canonical example of a "pathological" distribution, since both its mean and its variance are undefined. The look of a Cauchy distribution is similar to that of a normal distribution, though with longer "tails".[4]
Estimating the mean (middle column) and standard deviation (right column) through samples from a Cauchy distribution (bottom) and normal distribution (top). The top graphs converge, wheras the bottom graphs jump up and down
Due to this, estimating the mean value may not converge to any single value with more data (law of large numbers) unlike a normal distribution; due to a higher chance of getting extreme values (the tails of a frequency plot).
↑ 2.0 2.1 "1.3.6.6.3. Cauchy Distribution". www.itl.nist.gov. Retrieved 2020-10-13.
↑ "The Lorentz Oscillator Model". Archived from the original on 2014-04-22. Retrieved 2013-06-14.
↑ "Cauchy distribution | mathematics". Encyclopedia Britannica. Retrieved 2020-10-13.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Cauchy_distribution&oldid=7771360"
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Find the value of k where 31k2 is divisible by 6 - Maths - Introduction to Graphs - 11027009 | Meritnation.com
Applying the divisibility of 6 i.e a number should pass the test of 2 and 3
i.e it must be an even number and also the sum of digit is divisible by 3
31k2 \phantom{\rule{0ex}{0ex}}\mathrm{sin}ce the unit digit is even therefore it satisfies the divisibility rue of 2\phantom{\rule{0ex}{0ex}}also, 3+1+k+2=6+k\phantom{\rule{0ex}{0ex}}when k=0, 6+0=6\phantom{\rule{0ex}{0ex}} it satisfies the divisibility rue of 3\phantom{\rule{0ex}{0ex}}⇒k=0
Saarika answered this
Is the answer 0 ?
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urca3m403 2021-11-20 Answered
Letf\left(x\right)={x}^{5}+5{x}^{4}+6x+3.
Contract analysis. A contractor's financial outlay X and labor force Y are random variables with bivariate pdf given by:
{f}_{X,Y}\left(x,y\right)=kxy
10,000<x<100,000
10<y<20
=0
, elsewhere. (a) Evaluate constant k.
(b) Detemine the marginal pdf of X and Y.
-\frac{\pi }{4}\le \theta \le \frac{\pi }{4},-1\le r\le 1
Liesehf 2021-11-19 Answered
\theta =\frac{\pi }{3},-1\le r\le 3
oppvarmet16 2021-11-19 Answered
From 0-2
\mathrm{sin}\left({x}^{2}\right)dx
using trapezoidal rule with n=4 as well as midpoint rule with n=4
Idilwsiw2 2021-11-19 Answered
\begin{array}{cccccc}Source& DF& SS& MS& F& P\\ Regression& 1& 3445.9& 3445.9& 9.50& 0.005\\ Residual\text{ }Error& & & & & \\ Total& 29& 13598.3& & & \end{array}
{s}_{e}
{r}^{2}
hroncits8y 2021-11-18 Answered
Why is it very easy to conclude that
{\int }_{1}^{\propto }{x}^{2}dx
diverges without making any integration calculations?
smellmovinglz 2021-11-18 Answered
Quick decision analysis Please pick the best outcome after calculation,
a) Council would accept
\left({P}_{2}=0.6\right)
and the mayor would not veto
\left({P}_{3}=0.8\right)
, The probability of the compound event is
b) Council would accept
\left({P}_{2}=0.6\right)
and the mayor would veto
\left(1-{P}_{3}=0.2\right)
. The probability of the compound event is
c) Council would reject
\left(1-{P}_{2}=0.4\right)
and the council would retum the same budget
\left({P}_{4}=0.6\right)
. The probability of this compound event is
\mathrm{cos}\left(0.3\right)\approx 1-\frac{{\left(0.3\right)}^{2}}{2!}+\frac{{\left(0.3\right)}^{4}}{4!}
TokNeekCepTdh 2021-11-18 Answered
What is the advantage of saying that to multiply a number by 100, we shift the digits two places to the left, instead of saying move the decimal places to the right?
deliriwmfg 2021-11-17 Answered
T-Shirts The Physics Club sells
E=m{c}^{2}
T-shirts at the local flea market. Unfortunately, the club’s previous
administration has been losing money for years, so you decide to do an
analysis of the sales. A quadratic
regression based on old sales data reveals the following demand equation for the T-shirts:
q=-2p+33p\left(9\le p\le 15\right)2
Here, p is the price the club charges per T-shirt and q is the
number it can sell each day at the flea market. a. Obtain a formula for
the price elasticity of demand for
E=m{c}^{2}
T-shirts. b. Compute the
elasticity of demand if the price is set at $10 per shirt. Interpret the
result. c. How much should the Physics Club charge for the T-shirts to
obtain the maximum daily revenue? What will this revenue be?"
pavitorj6 2021-11-17 Answered
Loring Brace and colleagues at the University of Michigan Museum of Anthropology indicate that the size of human teeth continues to grow. Decrease, and that the evolutionary process is not yet stop. In northern Europeans, for example, tooth size the reduction is now 1% in 1000 years. What size will the teeth of our descendants be in 20,000 years? Now (as a percentage of our current tooth size)?
Javier earned $8.50 per hour for 32 hours and $12.75 per hour for 8 hour. What was his average hourly wage for those 40 hours?
Carolyn Moore 2021-11-17 Answered
A researcher calculates the variance of her sample to be
{s}^{2}=-6.
What would you tell the researcher about her analysis?
katelineliseua 2021-11-17 Answered
Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is
y=10t+\frac{1}{2}{t}^{2}-{t}^{3}
0\le t\le 8
a) After how many hours will the hourly number of units be maximized?
b) What is the maximum hourly output?
Discuss the continuity of function
f\left(x\right)=\frac{1}{{x}^{3}+1}
\mathbb{R}
{\int }_{0}^{3}3{x}^{2}dx
. Use Riemann sums with right endpoints to compute the exact value of the integral.
Bob wanted to reach the peak of a mountain.He started at te the base and started driving a car towards the peak of the mountain.
The car started at 0 mph and the speed increased over time to 40 mph.Halfway towards the peak of the mountain,the terrain became too rugged and Bob instantly stopped his car to analyze the terrain.After spending time observing the terrain, Bob decided to instantly climb the rocky terrain at a slov pase.OnceBob passed the rocky terrain he realized his shins were cut up so he immediately travelled back to the car at thesame slov pace as before. After spending time putting on bandages, he realized there was a side path to thepeak of the mountain. He immediately started running to the peak of the mountain and eventually reached the peak.
Graph the relationship between Bob's (a) distance and time and (b) position and time.
For the graph,make it clear what part of your function corresponds with his different parts of his travels,
(Label each segment of your graph)
Isaiah Alsup 2021-11-17 Answered
At HD Sport and Fitness gym, analysis shows that, as the demand of the gym, the number of members is 86 when annual membership fee is $25 per member and the number of members is 84 when annual membership fee is $33 per member. As the operator of the gym, you found out that number of members (q) and membership fee (p) have a linear relationship.
A) At what membership price, p , is the revenue maximized?
B) What is the maximum annual revenue?
chanyingsauu7 2021-11-17 Answered
(2) The blood of k people are mixed and the mixture is analysed. If the result is negative, then this single analysis is sufficient for k persons. But if it is positive, then the blood of each one must be subsequently investigated separately, and in toto for k people, k+1 analysis are needed. It is assumed that the probability of a positive result (p) is the same for all people and that the results of the analysis are independent in the probabilistic sense.
For what k is the minimum expected number of necessary analysis attained?
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I am stuck with this equation. If you can help me y''(t) + 12 y'(t) + 32 y(t) = 32 u(t) wi
y{}^{″}\left(t\right)+12{y}^{\prime }\left(t\right)+32y\left(t\right)=32u\left(t\right)
y\left(0\right)={y}^{\prime }\left(0\right)=0
Y\left(p\right)=x=\frac{32}{\left(p\left({p}^{2}+12p+32\right)\right)}
\frac{32}{p\left({p}^{2}+12p+32\right)}=\frac{32}{p\left({\left(p+6\right)}^{2}-{2}^{2}\right)}=\frac{32}{p\left(p+4\right)\left(p+8\right)}
Now use partial fractions method to get
\frac{a}{p}+\frac{b}{p+4}+\frac{c}{p+8}
Then the Laplace inverse would be,
a+{b}^{-4t}+{c}^{-8t}
Find the solution of the following Differential Equations by Exact
\left(2y+xy\right)dx+2xdy=0,y\left(3\right)=\sqrt{2}
Find the orthogonal trajectory of the family of curves
y=c{x}^{2}
How does a harmonic oscillator with nonlinear damping behave?
\stackrel{¨}{x}+c\stackrel{˙}{x}+x=0
with positive c, the amplitude of the oscillations decays exponentially when
c<2
. If it is higher than 2, the system fails to oscillate at all and is said to be overdamped.
\frac{1}{2a}\left(\mathrm{sin}at+at\mathrm{cos}at\right)
Evaluate the laplace transform for this
L\left\{4{x}^{2}-3\mathrm{cos}\left(2x\right)+5{e}^{x}\right\}
L\left\{4+\frac{1}{2}\mathrm{sin}\left(x\right)-{e}^{4x}\right\}
\left(x+2y-1\right)dx+\left(2x+4y-3\right)dy=0
I was reading some of my notes and I was not sure how the following works:
\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}=0
Solve the above with the condition
y\left(0\right)=0
⇒y\left(x\right)=A\left(1-{e}^{-x}\right)
with A an arbitrary constant.
I was just wondering how do you solve the above to get
y\left(x\right)=A\left(1-{e}^{-x}\right)
? Because when I tried it, I got:
y=A+B{e}^{-x}
, and using the condition
y\left(0\right)=0
, I got
A+B=0
|
Weighted Pools - Balancer
Weighted Pools are highly versatile and configurable pools. Weighted Pools use Weighted Math, which makes them great for general cases, including tokens that don't necessarily have any price correlation (ex. DAI/WETH). Unlike pools in other DeFi protocols that only provide 50/50 weightings, Balancer Weighted Pools enable users to build pools with different token counts and weightings, such as pools with 80/20 or 60/20/20 weightings.
Weighted Pools allow users to choose their levels of exposure to certain assets while still maintaining the ability to provide liquidity. The higher a token's weight in a pool, the less impermanent loss it will experience in the event of a price surge.
For example if a user wants to provide liquidity for WBTC and WETH, they can choose the weight that most aligns with their strategy. A pool more heavily favoring WBTC implies they expect bigger gains for WBTC, while a pool more heavily favoring WETH implies bigger gains for WETH. An evenly balanced pool is a good choice for assets that are expected to remain proportional in value in the long run.
Variable pool weight allows for fine-tuned exposure to assets
Impermanent Loss is the difference in value between holding a set of assets and providing liquidity for those same assets.
Some people find the word "Impermanent" misleading and prefer to call it "Divergence Loss" or "Rebalancing Loss" because one token may perpetually out-value another token, and the loss may become... permanent.
For pools that heavily weight one token over another, there is far less impermanent loss, but this doesn't come for free; very asymmetric pools do have higher slippage when making trades due to the fact that one side has much less liquidity. 80/20 pools have emerged as a happy medium when balancing liquidity an Impermanent Loss mitigation.
Since each token in a pool can be traded with any other token in a pool, the number of trading pairs grows significantly with each additional token. By providing more trading pairs, pools are able to facilitate more swaps, giving them more opportunities to collect fees.
The number of trading pairs in a pool follows the combinations equation
_nC_r = \frac{n!}{r!(n-r)!}
n
is 2 and
r
is the number of tokens in the pool.
Aave is a decentralized non-custodial liquidity protocol where users can participate as depositors or borrowers. Aave uses pool tokens from the 80/20 AAVE/WETH Weighted Pool to lock funds in their Safety Module (SM) while still providing liquidity for the AAVE token.
The Safety Module solves the issues with traditional staking systems and market liquidity: Tokens with locking/reward schemes tend to suffer from low market liquidity and extreme volatility when high percentages of the total supply are being locked. With the ability of contributing to the SM not only by locking AAVE, but also by contributing with liquidity [on Balancer], stakers create a trustless and decentralized market with deep liquidity for trading AAVE against ETH.
Source: Aave Docs [Safety Module in Detail]
mStable Staking
mStable is a decentralized stablecoin protocol whose flagship product, mUSD, represents a basket of underlying stablecoins: DAI, sUSD, USDC and USDT.
Similar to how Aave's Safety Module works, staked pool tokens from the 80/20 MTA/WETH Weighted Pool can be used in the mStable protocol as a backstop in case of a re-collateralization event to protect users from peg failures. By allowing users to stake pool tokens, mStable gains a more robust backstop and decreases the volatility of the MTA token.
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Rigid Assembly of Model Components - MATLAB & Simulink - MathWorks Italia
interface uses 'dual assembly' formulation to connect the components. In the concept of dual assembly, the global set of degrees of freedom (DoFs)
\mathit{q}
is retained and the physical coupling is expressed as consistency and equilibrium constraints at the interface. For rigid connections, these constraints are of the form:
\mathit{g}
is the vector of internal forces at the interface, and the matrix
\mathit{B}
is permutable to
\left[\mathit{I}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\mathit{I}\right]
. For a pair of matching DOFs with indices
{\mathit{i}}_{\mathit{i}}
{\mathit{i}}_{2}
{\mathit{i}}_{\mathit{i}}
selects a DOF in the first component while
{\mathit{i}}_{2}
selects the matching DOF in the second component, enforces consistency of displacements:
while enforces equilibrium of the internal forces g at the interface:
Combining these constrains with the uncoupled equations leads to the following dual assembly model for the coupled system:
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Pass Function to Another Function - MATLAB & Simulink - MathWorks Benelux
You can use function handles as input arguments to other functions, which are called function functions . These functions evaluate mathematical expressions over a range of values. Typical function functions include integral, quad2d, fzero, and fminbnd.
For example, to find the integral of the natural log from 0 through 5, pass a handle to the log function to integral.
q1 = integral(@log,a,b)
Similarly, to find the integral of the sin function and the exp function, pass handles to those functions to integral.
q2 = integral(@sin,a,b)
q3 = integral(@exp,a,b)
q3 = 147.4132
Also, you can pass a handle to an anonymous function to function functions. An anonymous function is a one-line expression-based MATLAB® function that does not require a program file. For example, evaluate the integral of
x/\left({e}^{x}-1\right)
on the range [0,Inf]:
fun = @(x)x./(exp(x)-1);
q4 = integral(fun,0,Inf)
Functions that take a function as an input (called function functions) expect that the function associated with the function handle has a certain number of input variables. For example, if you call integral or fzero, the function associated with the function handle must have exactly one input variable. If you call integral3, the function associated with the function handle must have three input variables. For information on calling function functions with more variables, see Parameterizing Functions.
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Solve for w in the equation y-1=4wy-6.
catcher1307ieh
Rewrite the equation as 4wy-6=y-1
4wy-6=y-1
4wy=y+5
Divide each term in 4wy=y+5 by 4y and simplify.
w=\frac{1}{4}+\frac{5}{4y}
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
P\left(x\right)=-12{x}^{2}+2136x-41000
{\int }_{0}^{3}3{x}^{2}dx
w=2\mathrm{sin}xy
A salesperson earns $20 a day plus 10% commission on sales over $200. If her daily earnings are $25.40, how much money in merchandside did she sell?
f\left(x,y\right)=xy+2x+y-36
f\left(2,-3\right)
b) Find all x-values such that
f\left(x,x\right)=0
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Revision as of 10:12, 27 July 2013 by NikosA (talk | contribs) (→Pre-Processing Overview: lower case \theta and link to {{wikipedia|Solar_zenith_angle}} -- link does not work?)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\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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Revision as of 15:22, 17 January 2020 by Munich (talk | contribs) (→Distribution of turbulent kinetic energy and its budgets terms: production, diffusive transport, dissipation and convection)
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle \langle u\rangle }
{\displaystyle \langle w\rangle }
{\displaystyle \langle u'_{i}u'_{j}\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle ||{\vec {U}}||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {PIV} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle ||{\vec {U}}_{\mathrm {LES} }||={\sqrt {\langle u^{2}\rangle +\langle w^{2}\rangle }}/u_{\mathrm {b} }}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle x/D}
{\displaystyle z/D}
{\displaystyle -0.788}
{\displaystyle 0.03}
{\displaystyle -0.843}
{\displaystyle 0.037}
{\displaystyle -0.918}
{\displaystyle 0}
{\displaystyle -1.1}
{\displaystyle 0}
{\displaystyle -0.533}
{\displaystyle 0}
{\displaystyle -0.534}
{\displaystyle 0}
{\displaystyle -0.507}
{\displaystyle 0.036}
{\displaystyle -0.50}
{\displaystyle 0.04}
{\displaystyle -0.697}
{\displaystyle 0.051}
{\displaystyle -0.735}
{\displaystyle 0.06}
{\displaystyle -0.513}
{\displaystyle 0.017}
{\displaystyle -0.513}
{\displaystyle 0.02}
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }={\frac {x-x_{\mathrm {Cyl} }}{x_{\mathrm {Cyl} }-x_{\mathrm {V1} }}}}
{\displaystyle x_{\mathrm {Cyl} }=-0.5D}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {V1} }}
{\displaystyle \langle u(z)\rangle /u_{\mathrm {b} }}
{\displaystyle u(z)}
{\displaystyle x_{\mathrm {adj} }=-0.25}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle {\frac {\partial \langle u\rangle }{\partial z}}}
{\displaystyle \langle u_{i}'u_{j}'(z)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(z)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle x_{\mathrm {adj} }=-1.5}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle x_{\mathrm {adj} }=-0.5}
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle u'u'\rangle }
{\displaystyle \langle w'w'\rangle }
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u'w'\rangle }
{\displaystyle \langle w(x)\rangle /u_{\mathrm {b} }}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle w(x)\rangle }
{\displaystyle x-}
{\displaystyle x_{\mathrm {adj} }\approx -0.1}
{\displaystyle x_{\mathrm {adj} }=-0.65}
{\displaystyle \langle u_{i}'u_{j}'(x)\rangle /u_{\mathrm {b} }^{2}}
{\displaystyle \langle k(x)\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle z_{\mathrm {V1} }/D}
{\displaystyle \langle u_{i}'u_{j}'\rangle }
{\displaystyle \langle k\rangle }
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k\rangle =0.5(\langle u'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {PIV,inplane} }\rangle =0.074u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,inplane} }\rangle =0.079u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.5(\langle u'^{2}\rangle +\langle v'^{2}\rangle +\langle w'^{2}\rangle )/u_{\mathrm {b} }^{2}}
{\displaystyle \langle k_{\mathrm {LES,total} }\rangle =0.09u_{\mathrm {b} }^{2}}
{\displaystyle 0=P+\nabla T-\epsilon +C}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle C}
{\displaystyle v}
{\displaystyle P=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}}
{\displaystyle T=\underbrace {-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle } _{\text{turbulent fluctuations}}\underbrace {-{\frac {1}{\rho }}\langle u_{i}'p'\rangle } _{\text{pressure transport}}\underbrace {+2\nu \langle u_{j}'s_{ij}\rangle } _{\text{viscous diffusion}}}
{\displaystyle \epsilon =2\nu \langle s_{ij}s_{ij}\rangle }
{\displaystyle s_{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}'}{\partial x_{j}}}+{\frac {\partial u_{j}'}{\partial x_{i}}}\right)}
{\displaystyle \epsilon _{\mathrm {total} }=\epsilon _{\mathrm {res} }+\epsilon _{\mathrm {SGS} }=2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle }
{\displaystyle C=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}}
{\displaystyle D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {PIV} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P_{\mathrm {LES} }=-\langle u_{i}'u_{j}'\rangle {\frac {\partial \langle u_{i}\rangle }{\partial x_{j}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle 0.3u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {LES} }\approx 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {PIV} }\approx 0.2u_{\mathrm {b} }^{3}/D}
{\displaystyle x=-0.7D}
{\displaystyle P}
{\displaystyle \nabla T_{\mathrm {turb,PIV} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb,LES} }=-{\frac {1}{2}}{\frac {\partial \langle u_{i}'u_{j}'u_{j}'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle x=-0.75D}
{\displaystyle 0.4u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb,LES} }\approx 0.35u_{\mathrm {b} }^{3}/D}
{\displaystyle \nabla T_{\mathrm {press,LES} }=-{\frac {1}{\rho }}{\frac {\partial \langle u_{i}'p'\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {visc,LES} }=2\nu {\frac {\partial \langle u_{j}'s_{ij}\rangle }{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \nabla T_{\mathrm {turb} }}
{\displaystyle \nabla T_{\mathrm {press} }}
{\displaystyle \langle w\rangle <0}
{\displaystyle w-}
{\displaystyle w'}
{\displaystyle p'<0}
{\displaystyle \nabla T_{\mathrm {visc} }}
{\displaystyle |0.05|u_{\mathrm {b} }^{3}/D}
{\displaystyle P}
{\displaystyle \nabla T}
{\displaystyle \epsilon }
{\displaystyle \epsilon _{\mathrm {PIV} }=2\nu \langle s_{ij}s_{ij}\rangle \cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle \epsilon _{\mathrm {LES,total} }=(2\nu \langle s_{ij}s_{ij}\rangle +2\langle \nu _{\mathrm {t} }s_{ij}s_{ij}\rangle )\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle P}
{\displaystyle \epsilon _{\mathrm {LES} }=0.066u_{\mathrm {b} }^{3}/D}
{\displaystyle P_{\mathrm {max} }}
{\displaystyle \epsilon _{\mathrm {max} }}
{\displaystyle C=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}}
{\displaystyle C_{\mathrm {PIV} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle C_{\mathrm {LES} }=-\langle u_{i}\rangle {\frac {\partial k}{\partial x_{i}}}\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle x\approx -0.63D}
{\displaystyle C}
{\displaystyle R_{\mathrm {PIV} }=P+\nabla T_{\mathrm {turb} }-\epsilon +C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle -\nabla T_{\mathrm {press,LES} }}
{\displaystyle R_{\mathrm {LES} }=P+\nabla T-\epsilon _{\mathrm {total} }+C\cdot D/u_{\mathrm {b} }^{3}}
{\displaystyle <|0.01|u_{\mathrm {b} }^{3}/D}
{\displaystyle T_{\mathrm {turb} }=-{\frac {1}{2}}\langle u_{i}'u_{j}'u_{j}'\rangle }
{\displaystyle c_{\mathrm {p} }(x)}
{\displaystyle c_{\mathrm {f} }(x)}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }={\frac {\langle p\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }={\frac {\langle \tau _{\mathrm {w} }\rangle }{{\frac {\rho }{2}}u_{\mathrm {b} }^{2}}}}
{\displaystyle z_{1}\approx 0.0036D\approx 10\mathrm {px} }
{\displaystyle z_{1}\approx 0.0005D}
{\displaystyle z-}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {p} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle x_{\mathrm {adj} }=-1.0}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle |c_{\mathrm {f} }|=0.01}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle c_{\mathrm {f} }}
{\displaystyle 50\times 171(n\times m)}
{\displaystyle 143\times 131(n\times m)}
{\displaystyle n\cdot m}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle -}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle -}
{\displaystyle \epsilon {\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle -}
{\displaystyle x_{\mathrm {adj} }}
{\displaystyle {\frac {x}{D}}}
{\displaystyle {\frac {z}{D}}}
{\displaystyle {\frac {\langle u\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle v\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle w\rangle }{u_{\mathrm {b} }}}}
{\displaystyle {\frac {\langle u'u'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle w'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'v'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle u'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle {\frac {\langle v'w'\rangle }{u_{\mathrm {b} }^{2}}}}
{\displaystyle P{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle C{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {turb} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {press} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \nabla T_{\mathrm {visc} }{\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle \epsilon {\frac {D}{u_{\mathrm {b} }^{3}}}}
{\displaystyle c_{\mathrm {p} }}
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Eichowa4 2021-11-26 Answered
{x}^{2}+{y}^{2}+4x-6y-3=0
in a rectangular coordinate system. If two functions are indicated, graph both in the same system.Then use your graphs to identify the relation’s domain and range.
Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results
y=3{x}^{\frac{2}{3}}-2x
Halkadvalseln 2021-11-23 Answered
f\left(x\right)=2{x}^{3}-33{x}^{2}+168x+9
has one local minimum and one local maximum.
Use a graph of the function to estimate these local extrema.
This function has a local minimum at
x=?
with output value:
x=?
vousetmoiec 2021-11-23 Answered
Graph the sets of points whose polar coordinates satisfy the given polar equation.
The given polar equation is written as follows:
\frac{\pi }{4}\le \theta \le \frac{3\pi }{4},0\le r\le 1
tbbfiladelfia6l 2021-11-23 Answered
The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients, oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats and rice it should include in each box of cereal to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes 8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes 6 milligrams of A and 2 milligrams of B. An ounce of oats costs $0.05, and an ounce of rice costs $0.03.
b. Solve this model by using graphical analysis.
c. What would be the effect on the optimal solution if the cost of rice increased from $0.03 per ounce to $0.06 per ounce?
kdgg0909gn 2021-11-22 Answered
A large number N of people are subjected to a blood investigation. This investigation can be organized in two ways.
(1) The blood of each person is investigated separately. In this case N analyses are needed.
(2) The blood of k people are mixed and the mixture is analysed. If the result is negative, then this single analysis is sufficient for k persons. But if it is positive, then the blood of each one must be subsequently investigated separately, and in toto for k people, k + 1 analysis are needed. It is assumed that the probability of a positive result (p) is the same for all people and that the results of the analysis are independent in the probabilistic sense.
What is the probability that the analysis of the mixed blood of k people is positive?
What is the expectation of the number of analysis necessary in the second method of testing?
For what k is the minimum expected number of necessary analysis attained?"
Virginia Caron 2021-11-22 Answered
\theta =\frac{2\pi }{3},r\le -2
Sheelmgal1p 2021-11-22 Answered
A commodity has a demand function modeled by
p=126-0.5x
and a total cost function modeled by
C=50x+39.75
, where x is the number of units.
(a)Use the first marginal analysis criterion presented in class to find the production level that yields a maximum profit. Then use the second criterion to ensure that level yields a maximum and not a minimum. Finally, what unit price (in dollars) yields a maximum profit?
(b)When the profit is maximized, what is the average total cost (in dollars) per unit? (Round your answer to two decimal places.)
ambiguit 2021-11-22 Answered
r\ge 1
deliredejoker7m 2021-11-22 Answered
-\frac{\pi }{2}\le \theta \le \frac{\pi }{2},1\le r\le 2
Find the solution set for the system
{x}^{2}+{y}^{2}=25
25{x}^{2}+{y}^{2}=25
by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
Graph the sets of points whose polar coordinates satisfy the given polar equation. The given polar equation is written as follows:
\theta =\frac{\pi }{2},r\le 0
What is the difference between a relative maximum and an absolute maximum on an interval |?
cleritere39 2021-11-21 Answered
Forest Fires and Acres Burned Numbers (in thousands) of forest fires over the year and the number (in hundred thousands) of acres burned for 7 recent years are shown.
\begin{array}{cccccccc}Number\text{ }of\text{ }fires\text{ }x& 62& 57& 72& 69& 58& 47& 84\\ Number\text{ }of\text{ }acres\text{ }burned\text{ }y& 61& 56& 71& 68& 57& 46& 83\end{array}
The correlation coefficient for the data is r=1 and a=0.05, Should regression analysis be done?
{y}^{{}^{\prime }}=a+bx
{y}^{{}^{\prime }}
when x = 60.
villane0 2021-11-21 Answered
Explain about Correlation test for Scale/ Ordinal Variables in terms of
a. Analysis background - a brief introduction on the objective of the analysis, hypothesis statements, and the variables.
b. Analysis procedures - the flow chart of the analysis procedures or the step-by - step procedure in performing the analysis.
It has been reported that many university students are increasingly stressed out as a result of the pandemic and current online learning environment. To respond to the situation, a mental health center at a university is offering online meditation sessions for students. They are planning to investigate if attendance in the sessions can help students reduce their stress levels.
a) What type of experimentation and statistical analysis among the following: ANOVA (one-way, two-way), hypothesis testing (t-test, chi-squared test, F-test, ...), regression analysis, is needed to conduct such investigation?
b) If your answer to part (a) is hypothesis testing, specify the type of the test (e.g. t-test, chi squared test, F-test, ...) that is needed and then write the null and alternative hypothesis for your test.
If your answer to part (a) is one-way or two-way ANOVA, define the factor(s) and their level(s).
If your answer to part (a) is regression analysis, define your random variables.
osi4a2nxk 2021-11-20 Answered
(a) In a regression analysis, the sum of squares for the predicted scores is 100 and the sum of squares error is 200, what is
{R}^{2}
(b) In a different regression analysis, 40% of the variance was explained. The sum of squares total is 1000. What is the sum of squares of the predicted values?
Question: Do laying hens have a comparable average weight depending on the food they eat ?
Find the type of analysis and the studied population used to answer the question above
information provided to help:
type of analysis: choose the right type of analysis to answer the questions
studied population: describe which population is concerned in each question (Ex: cows of all regions, or cows in X region, etc.)
elchatosarapage 2021-11-20 Answered
In a regression analysis, the variable that is being predicted is the
wurmiana6d 2021-11-20 Answered
The analysis of shafts for a compressor is summarized by conformance to specifications. Suppose that a shaft is selected at random. What is the probability that the selected shaft conforms to surface finish requirements or to roundness requirements?
\begin{array}{|ccc|}\hline & & Roundness\\ & & Conforms& Does\text{ }not\\ & & & Conforms\\ Surface& Conform& 345& 5\\ Finish& Does\text{ }not& 1& 8\\ & Conform\\ \hline\end{array}
|
< Splashing
This article details how to splash. Splashing is the act of training magic by casting combat spells against a NPC while having too low magic accuracy to be able to actually hit. Although no damage is dealt, each Magic combat spell will still give the player a constant base amount of experience.
Despite the NPC remaining alive indefinitely, after 20 minutes the player will stop auto-retaliating. As a consequence, timely interaction with the game is needed to not be logged out due to inactivity.
3 Splashing rates by spell
3.1 Standard Spellbook
3.3 Arceuus spellbook
4 30 to 99 strategies
It is necessary to wear items adding up to at most -64 Magic attack.
A typical example setup (40+ Range):
Full bronze/iron/steel/black/mithril/adamant/rune armour set (Platebody, Full Helm, Platelegs/Plateskirt and Kiteshield)
Bronze/iron/steel/black/mithril/adamant/rune/dragon boots, Fremennik sea boots, Snakeskin boots, or Fancy/Fighting boots
Any d'hide vambraces or Snakeskin vambraces if you are a member and use some Granite equipment
Any elemental staff to reduce runes needed and to enable autocast
Players without 40+ Range for d'hide vambraces can achieve the desired Magic attack by swapping the elemental staff for a Cursed goblin staff (obtained from Diango in Draynor). This precludes the rune savings of an elemental staff, but achieves -68 Magic attack.
The exceptions are the Granite body, as it gives -22 instead of -30 with a regular Platebody, and the Granite boots as they only give a -1 to magic attack.
Any monster that cannot deal damage sufficient to cumulate to a player's health going to zero can be used to splash (given the defensive bonuses of the splash equipment the player is wearing). Commonly used monsters are rats, seagulls, and spiders, though rats in member's worlds can be killed by unwanted cats.
Popular locations include seagulls in Port Sarim and in Port Piscarilius.
Players may also elect to check the Bestiary/Levels 1 to 10 and find a suitable creature/location themselves.
The only exception is the area in Lumbridge as shown in the map, including the castle basement. Here, spells yield no experience for casting if a player's magic accuracy is too low to deal damage. This was done because the large amount of players splashing on rats looked odd to new players. [1]
The lvl 1 chickens just east of East Ardougne east bank right next to Captain Barnaby are a good choice, as they are close to a bank and there will be few (if any) other players.
Should reaching -64 Magic attack prove difficult, attempting to splash on monks at the Edgeville Monastery is a viable alternative. This is due to their high health, relatively high defence for their level, and regular healing. As a result, low level strike spells are unlikely to kill them even when they connect. This will also provide slightly higher Magic XP per cast and some Hitpoints XP when damage is dealt.
Splashing rates by spell[edit | edit source]
Base exp/cast
Exp/hr*1
(Harmonised nightmare staff)*2
Raw cost/cast*3
Cost/cast
(basic elemental staff*4)
Gp/exp
(combination elemental staff*5)
Wind Strike 1 1 5.5 6600 8250 8 3 3,600 0.55 3 3,600 0.55
Water Strike 1 1 1 7.5 9000 11250 13 8 9,600 1.07 3 3,600 0.4
Earth Strike 1 2 1 9.5 11400 14250 18 8 9,600 0.84 3 3,600 0.32
Fire Strike 2 3 1 11.5 13800 17250 28 13 15,600 1.13 3 3,600 0.26
Wind Bolt 2 1 13.5 16200 20250 79 69 82,800 5.11 69 82,800 5.11
Water Bolt 2 2 1 16.5 19800 24750 89 79 94,800 4.79 69 82,800 4.18
Earth Bolt 2 3 1 19.5 23400 29250 94 79 94,800 4.05 69 82,800 3.54
Fire Bolt 3 4 1 22.5 27000 33750 104 84 100,800 3.73 69 82,800 3.07
Wind Blast 3 1 25.5 30600 38250 207 192 230,400 7.53 192 230,400 7.53
Water Blast 3 3 1 28.5 34200 42750 222 207 248,400 7.26 192 230,400 6.74
Earth Blast 3 4 1 31.5 37800 47250 227 207 248,400 6.57 192 230,400 6.1
Fire Blast 4 5 1 34.5 41400 51750 237 212 254,400 6.14 192 230,400 5.57
Wind Wave 5 1 36 43200 54000 401 376 451,200 10.44 376 451,200 10.44
Water Wave 5 7 1 37.5 45000 56250 436 401 481,200 10.69 376 451,200 10.03
Earth Wave 5 7 1 40 48000 60000 436 401 481,200 10.03 376 451,200 9.4
Fire Wave 5 7 1 42.5 51000 63750 436 401 481,200 9.44 376 451,200 8.85
Wind Surge 7 1 44.5 53400 66750 339 304 364,800 7.15 304 364,800 7.15
Water Surge 7 10 1 46.5 55800 69750 389 339 406,800 7.29 304 364,800 6.54
Earth Surge 7 10 1 48.5 58200 72750 389 339 406,800 6.99 304 364,800 6.27
Fire Surge 7 10 1 50.5 60600 75750 389 339 406,800 6.71 304 364,800 6.02
*1 Each game tick is 0.6s, a magic combat spell takes 5 ticks, so in 1 hour a player can cast ("splash") 1200 combat spells.
*2 Using a Harmonised nightmare staff (584,629,443), which autocasts with a speed of 4 ticks.
*3 The "raw cost" assumes that all runes needed are bought.
*4 The "cost" assumes that the Basic elemental staff is used for the core elemental runes and air runes are also used for non-Wind spells. (No requirements to wield)
*5 The "cost" assumes that Mist Battlestaff (39,475), Dust Battlestaff (17,542), Smoke Battlestaff (1,847,277) or their Mystic variants are used for Water, Earth, Fire spells respectively to completely cover the core elemental runes. (This requires at least level 30 Magic , which is quickly obtained).
Cost/cast*6
Cost/hr*6
Gp/exp*6
Smoke Rush 1 1 2 2 30 542 542 36000 650,400 18.07
Shadow Rush 1 2 2 1 31 717 717 37200 860,400 23.13
Blood Rush 1 2 2 33 898 898 39600 1,077,600 27.21
Ice Rush 2 2 2 34 532 522 40800 626,400 15.35
Smoke Burst 2 2 4 2 36 695 695 43200 834,000 19.31
Shadow Burst 1 4 2 2 37 1,035 1,035 44400 1,242,000 27.97
Blood Burst 2 4 2 39 1,412 1,412 46800 1,694,400 36.21
Ice Burst 4 4 2 40 680 660 48000 792,000 16.5
Smoke Blitz 2 2 2 2 42 1,171 1,171 50400 1,405,200 27.88
Shadow Blitz 2 2 2 2 43 1,521 1,521 51600 1,825,200 35.37
Blood Blitz 4 2 45 1,888 1,888 54000 2,265,600 41.96
Ice Blitz 3 2 2 46 1,151 1,136 55200 1,363,200 24.7
Smoke Barrage 4 4 2 4 48 1,585 1,585 57600 1,902,000 33.02
Shadow Barrage 4 2 4 3 49 2,105 2,105 58800 2,526,000 42.96
Blood Barrage 4 4 1 51 2,452 2,452 61200 2,942,400 48.08
Ice Barrage 6 2 4 52 1,550 1,520 62400 1,824,000 29.23
*6 The "cost" assumes that a Kodai Wand (112,525,024) is used, which supplies infinite water runes. Elemental staves cannot be used as they do not provide auto-cast for ancient spells.
The demonbane spells of the Arceuus spellbook can only be cast on demonic monsters. Magic attack bonus needs to be -64 or lower to always splash, and equipping a tome of fire or skull sceptre will make that more difficult to reach while saving runes. Weaker demonic monsters such as icefiends and pyrefiends are ideal given their low stats, however armour will need to maximise Magical melee while keeping Magic accuracy low enough to splash.
(Tome of Fire*10)
Raw Cost/hr*9
gp/xp*9
Inferior Demonbane 4 1 27 32400 89 69 106,800 82,800 3.3 2.56
Superior Demonbane*11 8 1 36 43200 220 180 264,000 216,000 6.11 5
Dark Demonbane*11 12 2 43.5 52200 420 360 504,000 432,000 9.66 8.28
*9 Includes the cost of Fire Runes.
*10 Tome of fire does not consume pages when autocasting Arceuus Spells. Therefore the cost of burnt pages is ignored.
*11 Requires completion of the quest A Kingdom Divided.
30 to 99 strategies[edit | edit source]
Total experience required: 13021068.
# of splashes
Cost - Combination Staff
Cost - Elemental Staff (F2P)
2367467 1973.0 7,102,400.73 7,102,400.73
1736143 1446.9 5,208,427.2 13,889,139.2
1370639 1142.3 4,111,916.21 10,965,109.89
1132267 943.6 3,396,800.35 14,719,468.17
{\displaystyle \rightarrow }
579314 482.8 39,891,685.4 48,575,260.3
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
381648 318.1 71,839,516.68 79,411,218.1
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
317135 264.3 110,644,042.59 Members Only Spells
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
{\displaystyle \rightarrow }
301246 251.1 98,591,805.73 Members Only Spells
^ Jagex. "Last Man Standing Beta and Splashing Restrictions". 11 July 2019. (Archived from the original.) Old School RuneScape News.
Retrieved from ‘https://oldschool.runescape.wiki/w/Splashing/Guide?oldid=14280301’
12m ago - YoshiFan12
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I am trying to take the laplace transform of \cos(t)u(t-\pi).
\mathrm{cos}\left(t\right)u\left(t-\pi \right)
\left(\left(\mathrm{cos}\left(t\right)+\pi \right)-\pi \right)u\left(t-\pi \right)
\mathrm{cos}\left(t\right)-\pi
utloverej
\mathcal{L}\left(\mathrm{cos}\left(t\right)\right)=\frac{s}{{s}^{2}+1}
\mathcal{L}\mathrm{cos}\left(t\right)u\left(t-\pi \right)={e}^{-\pi s}\mathcal{L}\left(\mathrm{cos}\left(t-\pi \right)\right)={e}^{-\pi s}\mathcal{L}\left(-\mathrm{cos}\left(t\right)\right)=-{e}^{-\pi s}\frac{s}{{s}^{2}+1}
Recall from the sum formula:
\mathrm{cos}\left(t-\pi \right)=\mathrm{cos}t\mathrm{cos}\pi +\mathrm{sin}t\mathrm{sin}\pi =-\mathrm{cos}t
\mathrm{cos}\left(t\right)=\mathrm{cos}\left(\left(t-\pi \right)+\pi \right)=-\mathrm{cos}\left(t-\pi \right)
\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta \text{ }\text{ where }\text{ }\alpha =t-\pi \text{ }\text{ and }\text{ }\beta =\pi
\mathcal{L}\left\{\mathrm{cos}\left(t\right)u\left(t-\pi \right)\right\}=-\mathcal{L}\left\{\mathrm{cos}\left(t-\pi \right)u\left(t-\pi \right)\right\}=\dots
\left(2x-5y+1\right)dx-\left(5y-2x\right)dy=0
y={\left(\frac{x}{x+1}\right)}^{5}
\frac{dy}{dx}=
y"+4y=4t
y\left(0\right)=1
{y}^{\prime }\left(0\right)=5
L\left[{e}^{-3t}{t}^{4}\right]
{s}^{-\frac{3}{2}}
\frac{2}{\sqrt{\pi }}\frac{\sqrt{\pi }}{2{s}^{\frac{3}{2}}}=2\sqrt{\frac{t}{\pi }}
I know the first part
\frac{2}{\sqrt{\pi }}
is obtained using the gamma function
\mathrm{\Gamma }\left(\frac{3}{2}\right)
, but not quite sure how the rest is obtained.
Evaluate this laplace transform of
\frac{1-{e}^{-t}}{t}
and the integral exists according to wolfram
What is the differential equation of the orthogonal trajectories of the family of curves
{x}^{2}-3xy-2{y}^{2}=C
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h is related to one of the six parent functions. (a) Identify the parent function f. (b) Describe the sequence of transformations from f to h. (c) Sketch the graph of h by hand. (d) Use function notation to write h in terms of the parent function f. h(x)=√x−1+4
For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.
\text{Misplaced \hline}
f:R\to R
Prove that f is onto.
g\left(x\right)=-{\left(x+3\right)}^{3}-10
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ f\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -3& -2& -1& 0& 1\\ g\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ h\left(x\right)& -2& -4& 3& 1& 0\\ \hline\end{array}
Contain linear equations with constants in denominators. Solve each equation. 3x/5-x=x/10-5/2
f\left(x\right)=\left\{\begin{array}{lll}1-x& if& x<0\\ 1& if& x\ge 0\end{array}
g\left(x\right)=\surd \frac{1}{4}x
=\frac{2{x}^{2}-3x-20}{x-4},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\ne 4
=kx-15, if x=4
Evaluate the constant k that makes the function continuous.
Given an indicated variable:
S=180n-360
Solve it for n
h\left(x\right)={\left(x-2\right)}^{3}+5
h\left(x\right)=\mid 2x+8\mid -1
g is related to one of the parent functions. Describe the sequence of transformations from f to g. g(x) = -x^3 - 1
g\left(x\right)=\frac{1}{2}\mid x-2\mid -3
f:R->R
f\left(x\right)=8x-2
f\left(a\right)=f\left(b\right)
a,b\in R
Prove that f is one-to-one.
Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f(x).
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ f\left(x\right)& -2& -1& -3& 1& 2\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -1& 0& 1& 2& 3\\ g\left(x\right)& -2& -1& -3& 1& 2\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ h\left(x\right)& -1& 0& -2& 2& 3\\ \hline\end{array}
f\left(x\right)=\sqrt{2x-{x}^{2}}
The function g is related to one of the parent functions described in an earlier section.
g\left(x\right)=\frac{1}{6}\sqrt{x}
b)Describe the sequence of transformations from f to g.
c)Use function notation to write g in terms of f.
g\left(x\right)=\left(?\right)f\left(x\right)
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Engineering Tables/Laplace Transform Properties - Wikibooks, open books for an open world
Engineering Tables/Laplace Transform Properties
{\displaystyle {\mathcal {L}}\left\{af(t)+bg(t)\right\}=aF(s)+bG(s)}
{\displaystyle {\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0^{-})}
{\displaystyle {\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0^{-})-f'(0^{-})}
{\displaystyle {\mathcal {L}}\left\{f^{(n)}\right\}=s^{n}{\mathcal {L}}\{f\}-s^{n-1}f(0^{-})-\cdots -f^{(n-1)}(0^{-})}
Frequency Division
{\displaystyle {\mathcal {L}}\{tf(t)\}=-F'(s)}
{\displaystyle {\mathcal {L}}\{t^{n}f(t)\}=(-1)^{n}F^{(n)}(s)}
Frequency Integration
{\displaystyle {\mathcal {L}}\left\{{\frac {f(t)}{t}}\right\}=\int _{s}^{\infty }F(\sigma )\,d\sigma }
Time Integration
{\displaystyle {\mathcal {L}}\left\{\int _{0}^{t}f(\tau )\,d\tau \right\}={\mathcal {L}}\left\{u(t)*f(t)\right\}={1 \over s}F(s)}
{\displaystyle {\mathcal {L}}\left\{f(at)\right\}={1 \over a}F\left({s \over a}\right)}
{\displaystyle f(0^{+})=\lim _{s\to \infty }{sF(s)}}
{\displaystyle f(\infty )=\lim _{s\to 0}{sF(s)}}
Frequency Shifts
{\displaystyle {\mathcal {L}}\left\{e^{at}f(t)\right\}=F(s-a)}
{\displaystyle {\mathcal {L}}^{-1}\left\{F(s-a)\right\}=e^{at}f(t)}
Time Shifts
{\displaystyle {\mathcal {L}}\left\{f(t-a)u(t-a)\right\}=e^{-as}F(s)}
{\displaystyle {\mathcal {L}}^{-1}\left\{e^{-as}F(s)\right\}=f(t-a)u(t-a)}
{\displaystyle {\mathcal {L}}\{f(t)*g(t)\}=F(s)G(s)}
{\displaystyle f(t)={\mathcal {L}}^{-1}\{F(s)\}}
{\displaystyle g(t)={\mathcal {L}}^{-1}\{G(s)\}}
{\displaystyle s=\sigma +j\omega }
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Tables/Laplace_Transform_Properties&oldid=3257556"
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Discrete Probability Distribution - Uniform Distribution Practice Problems Online | Brilliant
Jill has a set of
33
cards labelled with integers from 1 through
33.
She faces all the cards down, shuffles the deck repeatedly and then picks the card on the top. What is the probability that the card she picks shows a number larger than
19?
Note: No two cards have the same number.
\frac{14}{33}
\frac{13}{33}
\frac{196}{1089}
\frac{169}{1089}
Which of the following represents the probability density function of a random variable
X
that takes on integers following the discrete uniform distribution
X\sim U\{9,48\}?
f(x;40,\frac{1}{40})=\binom{40}{x}\left(\frac{1}{40}\right)^x\left(\frac{39}{40}\right)^{1-x}
f(x)=\frac{1}{40}
f(x)=\frac{1}{40 x}
f(x)=\frac{1}{40}x
Harry spins a roulette which consists of integers from 1 through
14
(not necessarily in order). What is the variance of the outcome he gets?
Note: The areas that correspond to each number are equal.
\frac{197}{12}
\frac{99}{10}
\frac{97}{10}
\frac{65}{4}
Find the variance of a random variable
X
which follows the discrete uniform distribution
X \sim U\{5,18\}?
\frac{65}{4}
\frac{195}{16}
\frac{197}{16}
\frac{197}{12}
What is the expected value of the outcome when throwing a fair die?
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In the solution to Q 15 in tbe step no 1 they have arranged x before 8 and after 5 - Maths - Statistics - 12677111 | Meritnation.com
In the solution to Q 15 in tbe step no 1 they have arranged x before 8 and after 5. But how did they come to know that 5
Q15. If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.
Sol. Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,
Here n = 7 which is odd.
\therefore \mathrm{Median} = {\frac{n +1}{2}}^{\mathrm{th}} \mathrm{term} =\frac{7+1}{2}= {4}^{\mathrm{th}} \mathrm{term} = x\phantom{\rule{0ex}{0ex}} \therefore \mathrm{But} \mathrm{median} = 6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}} \therefore x = 6
Here we can see that we have odd number of data , We know middle one is the median .
And given median = 6 and that is missing in given data
So, we assume that missing number ' x ' is median of given data .
In Mathematics, It is very important to understand the concepts and memorise the formulae. In order to achieve that we need to practise mathematical problems a lot . If you have any doubts or queries , this platform will try its best to provide you the best solution .
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chebsort - Maple Help
Home : Support : Online Help : Mathematics : Numerical Computations : Approximations : numapprox Package : chebsort
sort the terms in a Chebyshev series
chebsort(e)
expression assumed to be a Chebyshev series
The input expression e is assumed to be a polynomial expressed in terms of a Chebyshev basis
T\left(0,x\right),...
First the expression e is collected in 'T'. Then the terms in the collected polynomial expression are sorted in ``Chebyshev order''; i.e. the
T\left(k,x\right)
basis polynomials are ordered in ascending order with respect to the first argument.
If some basis polynomials
T\left(k,x\right)
have non-numeric first argument then ordering will be attempted using the ``is'' predicate. If that is not successful then ordering is performed only with respect to numeric first arguments (other terms are left as trailing terms).
Note that chebsort is a destructive operation because it invokes the Maple sort function (see sort); i.e. the input expression is sorted ``in-place''.
The command with(numapprox,chebsort) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{numapprox}\right):
\mathrm{Digits}≔3:
a≔\mathrm{chebyshev}\left(\mathrm{sin}\left(x\right),x\right):
b≔\mathrm{chebyshev}\left(\mathrm{cos}\left(x\right),x\right):
c≔a+b
\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{0.880}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.0391}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.000500}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.765}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.230}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.00495}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{chebsort}\left(c\right)
\textcolor[rgb]{0,0,1}{0.765}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.880}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.230}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.0391}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.00495}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.000500}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{assume}\left(5<j,j<k\right)
d≔1.2y+\mathrm{cj}T\left(j,x\right)+a+\mathrm{ck}T\left(k,x\right)
\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{1.2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{cj}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{j~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.880}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.0391}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.000500}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ck}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{k~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{chebsort}\left(d\right)
\textcolor[rgb]{0,0,1}{0.880}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{0.0391}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.000500}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{cj}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{j~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ck}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{k~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1.2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}
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Simulating Electricity Prices with Mean-Reversion and Jump-Diffusion - MATLAB & Simulink - MathWorks Deutschland
Calibration of the Market Price of Risk
Simulation of Risk-Neutral Prices
Pricing a Bermudan Option
This example shows how to simulate electricity prices using a mean-reverting model with seasonality and a jump component. The model is calibrated under the real-world probability using historical electricity prices. The market price of risk is obtained from futures prices. A risk-neutral Monte Carlo simulation is conducted using the calibrated model and the market price of risk. The simulation results are used to price a Bermudan option with electricity prices as the underlying.
Electricity prices exhibit jumps in prices at periods of high demand when additional, less efficient electricity generation methods, are brought on-line to provide a sufficient supply of electricity. In addition, they have a prominent seasonal component, along with reversion to mean levels. Therefore, these characteristics should be incorporated into a model of electricity prices [2].
In this example, electricity price is modeled as:
log\left({P}_{t}\right)=f\left(t\right)+{X}_{t}
{P}_{t}
is the spot price of electricity. The logarithm of electricity price is modeled with two components:
f\left(t\right)
{X}_{t}
. The component
f\left(t\right)
is the deterministic seasonal part of the model, and
{X}_{t}
is the stochastic part of the model. Trigonometric functions are used to model
f\left(t\right)
f\left(t\right)={s}_{1}\mathrm{sin}\left(2\pi t\right)+{s}_{2}\mathrm{cos}\left(2\pi t\right)+{s}_{3}\mathrm{sin}\left(4\pi t\right)+{s}_{4}\mathrm{cos}\left(4\pi t\right)+{s}_{5}
{s}_{i},i=1,...,5
are constant parameters, and
t
is the annualized time factors. The stochastic component
{X}_{t}
is modeled as an Ornstein-Uhlenbeck process (mean-reverting) with jumps:
d{X}_{t}=\left(\alpha -\kappa {X}_{t}\right)dt+\sigma d{W}_{t}+J\left({\mu }_{J},{\sigma }_{J}\right)d\Pi \left(\lambda \right)
\alpha
\kappa
are the mean-reversion parameters. Parameter
\sigma
is the volatility, and
{W}_{t}
is a standard Brownian motion. The jump size is
J\left({\mu }_{J},{\sigma }_{J}\right)
, with a normally distributed mean
{\mu }_{J}
, and a standard deviation
{\sigma }_{J}
. The Poisson process
\Pi \left(\lambda \right)
has a jump intensity of
\lambda
Sample electricity prices from January 1, 2010 to November 11, 2013 are loaded and plotted below. Prices contain the electricity prices, and PriceDates contain the dates associated with the prices. The logarithm of the prices and annual time factors are calculated.
% Load the electricity prices and futures prices.
load('electricity_prices.mat');
PriceDates = datetime(PriceDates,'ConvertFrom','datenum');
FutExpiry = datetime(FutExpiry,'ConvertFrom','datenum');
FutValuationDate = datetime(FutValuationDate,'ConvertFrom','datenum');
% Plot the electricity prices.
plot(PriceDates, Prices);
title('Electricity Prices');
% Obtain the log of prices.
logPrices = log(Prices);
% Obtain the annual time factors from dates.
PriceTimes = yearfrac(PriceDates(1), PriceDates);
First, the deterministic seasonality part is calibrated using the least squares method. Since the seasonality function is linear with respect to the parameters
{s}_{i}
, the backslash operator (mldivide) is used. After the calibration, the seasonality is removed from the logarithm of price. The logarithm of price and seasonality trends are plotted below. Also, the de-seasonalized logarithm of price is plotted.
% Calibrate parameters for the seasonality model.
seasonMatrix = @(t) [sin(2.*pi.*t) cos(2.*pi.*t) sin(4.*pi.*t) ...
cos(4.*pi.*t) t ones(size(t, 1), 1)];
C = seasonMatrix(PriceTimes);
seasonParam = C\logPrices;
% Plot the log price and seasonality line.
plot(PriceDates, logPrices);
title('log(price) and Seasonality');
ylabel('log(Prices)');
plot(PriceDates, C*seasonParam, 'r');
legend('log(Price)', 'seasonality');
% Plot de-seasonalized log price
X = logPrices-C*seasonParam;
plot(PriceDates, X);
title('log(price) with Seasonality Removed');
The second stage is to calibrate the stochastic part. The model for
{X}_{t}
needs to be discretized to conduct the calibration. To discretize, assume that there is a Bernoulli process for the jump events. That is, there is at most one jump per day since this example is calibrating against daily electricity prices. The discretized equation is:
{X}_{t}=\alpha \Delta t+\varphi {X}_{t-1}+\sigma \xi
\left(1-\lambda \Delta t\right)
{X}_{t}=\alpha \Delta t+\varphi {X}_{t-1}+\sigma \xi +{\mu }_{J}+{\sigma }_{J}{\xi }_{J}
\lambda \Delta t
\xi
{\xi }_{J}
are independent standard normal random variables, and
\varphi =1-\kappa \Delta t
. The density function of
{X}_{t}
{X}_{t-1}
is [1,4]:
f\left({X}_{t}|{X}_{t-1}\right)=\left(\lambda \Delta t\right){N}_{1}\left({X}_{t}|{X}_{t-1}\right)+\left(1-\lambda \Delta t\right){N}_{2}\left({X}_{t}|{X}_{t-1}\right)
{N}_{1}\left({X}_{t}|{X}_{t-1}\right)=\left(2\pi \left({\sigma }^{2}+{\sigma }_{J}^{2}\right){\right)}^{-\frac{1}{2}}\mathrm{exp}\left(\frac{-\left({X}_{t}-\alpha \Delta t-\varphi {X}_{t-1}-{\mu }_{J}{\right)}^{2}}{2\left({\sigma }^{2}+{\sigma }_{J}^{2}\right)}\right)
{N}_{2}\left({X}_{t}|{X}_{t-1}\right)=\left(2\pi {\sigma }^{2}{\right)}^{-\frac{1}{2}}\mathrm{exp}\left(\frac{-\left({X}_{t}-\alpha \Delta t-\varphi {X}_{t-1}{\right)}^{2}}{2{\sigma }^{2}}\right)
\theta =\left\{\alpha ,\varphi ,{\mu }_{J},{\sigma }^{2},{\sigma }_{J}^{2},\lambda \right\}
can be calibrated by minimizing the negative log likelihood function:
mi{n}_{\theta }-\sum _{t=1}^{T}\mathrm{log}\left(f\left({X}_{t}|{X}_{t-1}\right)\right)
subject\phantom{\rule{0.5em}{0ex}}to\phantom{\rule{0.5em}{0ex}}\varphi <1,{\sigma }^{2}>0,{\sigma }_{J}^{2}>0,0\le \lambda \Delta t\le 1
The first inequality constraint,
\varphi <1
, is equivalent to
\kappa >0
. The volatilities
\sigma
{\sigma }_{J}
must be positive. In the last inequality,
\lambda \Delta t
is between 0 and 1, because it represents the probability of a jump occurring in
\Delta t
time. In this example, assume that
\Delta t
is one day. Therefore, there is at most 365 jumps in one year. The mle function from the Statistics and Machine Learning Toolbox™ is well suited to solve the above maximum likelihood problem.
% Prices at t, X(t).
Pt = X(2:end);
% Prices at t-1, X(t-1).
Pt_1 = X(1:end-1);
% Discretization for the daily prices.
% PDF for the discretized model.
mrjpdf = @(Pt, a, phi, mu_J, sigmaSq, sigmaSq_J, lambda) ...
lambda.*exp((-(Pt-a-phi.*Pt_1-mu_J).^2)./ ...
(2.*(sigmaSq+sigmaSq_J))).* (1/sqrt(2.*pi.*(sigmaSq+sigmaSq_J))) + ...
(1-lambda).*exp((-(Pt-a-phi.*Pt_1).^2)/(2.*sigmaSq)).* ...
(1/sqrt(2.*pi.*sigmaSq));
% Constraints:
% phi < 1 (k > 0)
% sigmaSq > 0
% sigmaSq_J > 0
% 0 <= lambda <= 1
lb = [-Inf -Inf -Inf 0 0 0];
ub = [Inf 1 Inf Inf Inf 1];
% Initial values.
x0 = [0 0 0 var(X) var(X) 0.5];
% Solve the maximum likelihood.
params = mle(Pt,'pdf',mrjpdf,'start',x0,'lowerbound',lb,'upperbound',ub,...
'optimfun','fmincon');
% Obtain the calibrated parameters.
alpha = params(1)/dt
alpha = -20.1060
kappa = (1-params(2))/dt
kappa = 188.2535
mu_J = params(3)
mu_J = 0.2044
sigma = sqrt(params(4)/dt);
sigma_J = sqrt(params(5))
sigma_J = 0.2659
lambda = params(6)/dt
lambda = 98.3357
The calibrated parameters and the discretized model allow us to simulate electricity prices under the real-world probability. The simulation is conducted for approximately 2 years with 10,000 trials. It exceeds 2 years to include all the dates in the last month of simulation. This is because the expected simulation prices for the futures contract expiry date is required in the next section to calculate the market price of risk. The seasonality is added back on the simulated paths. A plot for a single simulation path is plotted below.
% Simulate for about 2 years.
nPeriods = 365*2+20;
n1 = randn(nPeriods,nTrials);
n2 = randn(nPeriods, nTrials);
j = binornd(1, lambda*dt, nPeriods, nTrials);
SimPrices = zeros(nPeriods, nTrials);
SimPrices(1,:) = X(end);
for i=2:nPeriods
SimPrices(i,:) = alpha*dt + (1-kappa*dt)*SimPrices(i-1,:) + ...
sigma*sqrt(dt)*n1(i,:) + j(i,:).*(mu_J+sigma_J*n2(i,:));
% Add back seasonality.
SimPriceDates = PriceDates(end) + days(0:(nPeriods-1))';
SimPriceTimes = yearfrac(PriceDates(1), SimPriceDates);
CSim = seasonMatrix(SimPriceTimes);
logSimPrices = SimPrices + repmat(CSim*seasonParam,1,nTrials);
% Plot the logarithm of Prices and simulated logarithm of Prices.
plot(SimPriceDates(2:end), logSimPrices(2:end,1), 'red');
seasonLine = seasonMatrix([PriceTimes; SimPriceTimes(2:end)])*seasonParam;
plot([PriceDates; SimPriceDates(2:end)], seasonLine, 'green');
title('Actual log(price) and Simulated log(price)');
ylabel('log(price)');
legend('market', 'simulation');
% Plot the prices and simulated prices.
PricesSim = exp(logSimPrices);
plot(SimPriceDates, PricesSim(:,1), 'red');
title('Actual Prices and Simulated Prices');
Up to this point, the parameters were calibrated under the real-world probability. However, to price options, you need the simulation under the risk-neutral probability. To obtain this, calculate the market price of risk from futures prices to derive the risk-neutral parameters. Suppose that there are monthly futures contracts available on the market, which are settled daily during the contract month. For example, such futures for the PJM electricity market are listed on the Chicago Mercantile Exchange [5].
The futures are settled daily during the contract month. Therefore, you can obtain daily futures values by assuming the futures value is constant for the contract month. The expected futures prices from the real-world measure are also needed to calculate the market price of risk. This can be obtained from the simulation conducted in the previous section.
% Obtain the daily futures prices.
FutPricesDaily = zeros(size(SimPriceDates));
idx = find(year(SimPriceDates(i)) == year(FutExpiry) & ...
month(SimPriceDates(i)) == month(FutExpiry));
FutPricesDaily(i) = FutPrices(idx);
% Calculate the expected futures price under real-world measure.
SimPricesExp = mean(PricesSim, 2);
To calibrate the market price of risk against market futures values, use the following equation:
\mathrm{log}\left(\frac{{F}_{t}}{{E}_{t}}\right)=-\sigma {e}^{-kt}{\int }_{0}^{t}{e}^{ks}{m}_{s}ds
{F}_{t}
is the observed futures value at time
t
{E}_{t}
is the expected value under the real-world measure at time
t
. The equation was obtained using the same methodology as described in [3]. This example assumes that the market price of risk is fully driven by the Brownian motion. The market price of risk,
{m}_{t}
, can be solved by discretizing the above equation and solving a system of linear equations.
% Setup system of equations.
t0 = yearfrac(PriceDates(1), FutValuationDate);
tz = SimPriceTimes-t0;
b = -log(FutPricesDaily(2:end) ./ SimPricesExp(2:end)) ./ ...
(sigma.*exp(-kappa.*tz(2:end)));
A = (1/kappa).*(exp(kappa.*tz(2:end)) - exp(kappa.*tz(1:end-1)));
A = tril(repmat(A', size(A,1), 1));
% Precondition to stabilize numerical inversion.
P = diag(1./diag(A));
b = P*b;
% Solve for the market price of risk.
riskPremium = A\b;
{m}_{t}
is obtained, risk-neutral simulation can be conducted using the following dynamics:
{X}_{t}=\alpha \Delta t+\varphi {X}_{t-1}-\sigma {m}_{t-1}\Delta t+\sigma \xi
\left(1-\lambda \Delta t\right)
{X}_{t}=\alpha \Delta t+\varphi {X}_{t-1}-\sigma {m}_{t-1}\Delta t+\sigma \xi +{\mu }_{J}+{\sigma }_{J}{\xi }_{J}
\lambda \Delta t
sigma*sqrt(dt)*n1(i,:) - sigma*dt*riskPremium(i-1) + ...
j(i,:).*(mu_J+sigma_J*n2(i,:));
% Convert the log(Price) to Price.
The expected values from the risk-neutral simulation are plotted against the market futures values. This confirms that the risk-neutral simulation closely reproduces the market futures values.
% Obtain expected values from the risk-neutral simulation.
SimPricesExp = mean(PricesSim,2);
fexp = zeros(size(FutExpiry));
for i = 1:size(FutExpiry,1)
idx = SimPriceDates == FutExpiry(i);
if sum(idx)==1
fexp(i) = SimPricesExp(idx);
% Plot expected values from the simulation against market futures prices.
plot(FutExpiry, FutPrices(1:size(FutExpiry,1)),'-*');
plot(FutExpiry, fexp, '*r');
title('Market Futures Prices and Simulated Futures Prices');
legend('market', 'simulation', 'Location', 'NorthWest');
plot(SimPriceDates(2:end), riskPremium);
title('Market Price of Risk');
ylabel('Market Price of Risk');
The risk-neutral simulated values are used as input into the function optpricebysim in the Financial Instruments Toolbox™ to price a European, Bermudan, or American option on electricity prices. Below, the price is calculated for a two-year Bermudan call option with two exercise opportunities. The first exercise is after one year, and the second is at the maturity of the option.
% Settle, maturity of option.
Settle = FutValuationDate;
Maturity = FutValuationDate + calyears(2);
% Create the interest-rate term structure.
'EndDates', Maturity, 'Rate', riskFreeRate, 'Compounding', ...
% Cutoff the simulation at maturity.
endIdx = find(SimPriceDates == Maturity);
SimPrices = PricesSim(1:endIdx,:);
Times = SimPriceTimes(1:endIdx) - SimPriceTimes(1);
% Bermudan call option with strike 60, two exercise opportunities, after
% one year and at maturity.
ExerciseTimes = [Times(366) Times(end)];
Price = optpricebysim(RateSpec, SimPrices, Times, OptSpec, Strike, ...
ExerciseTimes)
[1] Escribano, Alvaro, Pena, Juan Ignacio, Villaplana, Pablo. "Modeling Electricity Prices: International Evidence." Universidad Carloes III de Madrid, Working Paper 02-27, 2002.
[2] Lucia, Julio J., Schwartz, Eduaro. "Electricity Prices and Power Derivatives: Evidence from the Nordic Power Exchange." Review of Derivatives Research. Vol. 5, Issue 1, pp 5-50, 2002.
[3] Seifert, Jan, Uhrig-Homburg, Marliese. "Modelling Jumps in Electricity Prices: Theory and Empirical Evidence." Review of Derivatives Research. Vol. 10, pp 59-85, 2007.
[4] Villaplana, Pablo. "Pricing Power Derivatives: A Two-Factor Jump-Diffusion Approach." Universidad Carloes III de Madrid, Working Paper 03-18, 2003.
[5] https://www.cmegroup.com
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Applying Kepler's Laws | Brilliant Math & Science Wiki
Adam Strandberg, Matt DeCross, and Jimin Khim contributed
Kepler's laws describe the orbits of planets around the sun or stars around a galaxy in classical mechanics. They have been used to predict the orbits of many objects such as asteroids and comets , and were pivotal in the discovery of dark matter in the Milky Way. Violations of Kepler's laws have been used to explore more sophisticated models of gravity, such as general relativity. While Newton's laws generalize Kepler's laws, most problems related to the periods of orbits are still best solved using Kepler's laws, since they are simpler.
Recall the statements of Kepler's laws:
Planets move in elliptical orbits with the sun at one focus.
The line joining planets to either focus sweeps out equal areas in equal times.
The square of the period is proportional to the cube of the semi-major axis (half the longer side of the ellipse):
T^2 \propto a^3.
These laws can be applied to model natural objects like planets, stars, or comets, as well as man-made devices like rockets and satellites in orbit.
Although Kepler originally developed his laws in the context of planetary orbits, the results hold for any system with a radial force obeying the inverse square law. Coulomb's law holds that the electric force between two charged particles in an inverse square law like gravity (assuming that the particles have opposite charge). In spite of the fact that quantum mechanics is needed to fully model how electrons orbit nuclei, electrons with very high energy behave as though they had Keplerian orbits, and atoms containing such electrons are known as Rydberg atoms.
Most Efficient Route to Mars
Straight Lines are Degenerate Ellipses
Deriving Velocity of Circular motion
Violating Kepler's Laws
Picture of Halley's Comet taken from the European spacecraft Giotto in 1986 [1]
Halley's Comet is the first comet that astronomers realized had a periodic orbit. It passes within sight of Earth once every 75 years. Kepler's third law determines the length of the semimajor axis of this orbit:
The highly elliptical orbit of Halley's Comet, compared to the relatively circular orbits of Jupiter, Saturn, Uranus and Neptune
\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ \Rightarrow a&= \sqrt[3]{\frac{G M T^2}{4 \pi^2}}\\ &= 2.7 \times 10^{12} \text{ (m)}. \end{aligned}
Above, the more precise form of Kepler's third law
T^2 = \frac{4 \pi^2}{G M} a^3
has been used, where the proportionality constant between
T^2
a^3
has been solved for. Obtaining this constant requires an extensive derivation.
At the point in its orbit where it is closest to the sun, Halley's comet is only
8.8 \times 10^{8} \text{ m}
from the sun, coming between the orbits of Mercury and Venus. It is furthest away from the sun at a distance of approximately
2a = 5.4 \times 10^{12} \text{ m}
, past the orbit of Neptune.
The Hohmann Transfer Orbit [2]
The most efficient route from Earth to Mars is called the Hohmann transfer orbit [2]. By Kepler's third law, it must be the ellipse that touches both orbits with the shortest possible semimajor axis. This is intuitive because without any acceleration, a rocket on Earth would stay in the orbit of Earth. The most efficient route is just the (elliptical) orbit which starts at the Earth's orbit and ends at the Mars' orbit.
The orbits of Earth and Mars are approximately circular, with radii
r_{E} = 1 \text{ AU}
r_{M} = 1.542 \text{ AU}
. An ellipse touching both of these circles will have semimajor axis
a = \frac{1}{2} (r_{E} + r_{M}) = 1.262 \text{ AU}.
a
is in units of
\text{AU}
T
is in units of years, Kepler's third law simplifies to the expression
T^{2} = a^{3} \implies T = \sqrt{2.0992} = 1.41 \text{ (yr)}.
An astronaut stranded out in space wants to figure out how far she is from Earth. The only thing that she has with her is her handy dandy indestructible stopwatch that can survive the heat of re-entry and the shock of impact with Earth. She starts the stopwatch and lets it go with approximately zero velocity in the reference frame of Earth. After the watch crashes to Earth, mission control retrieves it and finds that the stopwatch was falling for 7 days. How far away from Earth is the astronaut?
This problem can be solved by integrating the Newtonian gravitational force over a straight line going towards Earth to find the position as a function of time and vice versa, but using the trick of applying Kepler's third law is easier. Suppose the astronaut's distance from Earth is some number
R
, and consider an ellipse with major axis of length slightly larger than
R
, with a minor axis
b \ll R
. This is an ellipse with eccentricity very near one. Taking the limit of this ellipse as the eccentricity goes to one yields a straight line of length R. This suggests that the properties of ellipses may be used to derive quantities related to straight lines in general, since any straight line may be approximated well by an ellipse of eccentricity near one.
t
it takes to fall down this line is half of the period
T
it would take to go around the corresponding near-unit-eccentricity ellipse. Since the semimajor axis of the ellipse is
\frac{R}{2}
, by Kepler's third law
\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ (2t)^2 &= \frac{4 \pi^2}{G M} \left(\frac{R}{2}\right)^3\\ \Rightarrow R &= 2 \sqrt[3]{\frac{G M t^{2}}{\pi^2}}. \end{aligned}
G = 6.67 \times 10^{-11} \frac{\text{m}^3}{\text{ kg} \ \text{s}^2}
M = 5.97 \times 10^{24} \text{ kg}
t = 6.048 \times 10^{5} \text{ s}
R = 3.4 \times 10^{10} \text{ m}
for this astronaut's distance from Earth. This derivation was much more straightforward than the burdensome calculus required to solve Newton's second law for the gravitational force.
Astronomers define a galaxy rotation curve to be the plot of the tangential velocity of stars as a function of their distance from the center of any given galaxy. Kepler's laws can be used to derive the shape of this curve assuming the stars have circular (or low eccentricity) orbits.
By Kepler's second law, the area swept out by the line from the galaxy center to a star in a given time must be constant. Since the radius
r
to such a star is constant in circular motion, the velocity
v
of these stars in orbit must also be constant. In that case, the period of rotation
T
\frac{2 \pi r}{v}
. Plugging this into Kepler's third law gives
\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ \left(\frac{2 \pi r}{v}\right)^{2} &= \frac{\pi^2}{G M} r^{3}\\ v^{2} &= \frac{G M}{r}\\ \Rightarrow v(r) &= \sqrt{\frac{G M}{r}}. \end{aligned}
M
is also a function of the radius
r
, since even if the mass density in a galaxy is homogeneous this means that the mass contained in some radius scales as the radius cubed. That is, a cluster of stars with roughly homogeneous mass density will have mass that scales like
M(r) = M_{0} r^{3}
. Plugging this into the above expression for
v
v(r) = r \sqrt{G M_{0}}
A galaxy with a cluster of stars in the middle and sparsely spaced stars further out will therefore have a galaxy rotation curve that starts out linear and then falls off like
\frac{1}{\sqrt{r}}
, like curve A in the picture.
Astronomers have found that the galaxy rotation curve of the Milky Way does not follow this curve if we take into account all of the known mass in the Milky Way. It looks more like curve B, where the velocity stays approximately constant even far away from the center of the galaxy. This implies that either Kepler's laws are wrong or there is mass we haven't accounted for. While there are cases in which it is known that Kepler's laws are incomplete or not a good description of nature (see below), physicists are very confident that Kepler's laws should apply to these stars because their dynamics satisfy the relevant assumptions for Newtonian gravitation. The data predicts the existence of dark matter: matter that we can't see, but must be there due to its influence on gravitation.
Perihelion shift of mercury [3]
One of the greatest mysteries in early twentieth-century astronomy was the precession of the perihelion of orbit of the planet Mercury. The perihelion is the point in a planet's orbit at which it is closest to the sun. Kepler's first law predicts that the perihelion is constant in time for every orbit. However, the perihelion of Mercury moves every year by a small angle. Even accounting for perturbations due to the gravitational effects of other planets, there was still an unaccounted-for rotation of 43 arc-seconds per century when the orbit was solved in Newtonian gravity. The size of this effect is equivalent to a full rotation of the perihelion in 3 million years [4]. This slight precession of the orbit is an effect of general relativity, in which not all orbits are closed ellipses. Einstein's correct computation of the precession of the perihelion of Mercury in general relativity was hailed as one of the great early theoretical successes for the theory.
[1] Image retrieved from http://apod.nasa.gov/apod/ap100104.html on 26 Feb 2016.
[2] Stern, David. Kepler's Three Laws of Planetary Motion. NASA. 23 Mar 2005. Retrieved on 9 Mar 2016 from http://www-spof.gsfc.nasa.gov/stargaze/Kep3laws.htm
[3] Image retrieved from https://commons.wikimedia.org/wiki/File:Mercure_orbite_precession_sk.JPG on 26 Feb 2016.
[4] Wudka, J. Precession of the perihelion of mercury. 24 Sep 1998. http://physics.ucr.edu/~wudka/Physics7/Notes_www/node98.html Retrieved on 18 Feb 2016.
Cite as: Applying Kepler's Laws. Brilliant.org. Retrieved from https://brilliant.org/wiki/applying-keplers-laws/
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2.1 Find Poles of f(z)
2.2 Choose Path of Contour Integral
2.3 Compute Residues of f at z0= exp{i\pi /n}
2.4 Bound Arc Portion (B) of Integral
2.5 Parametrize (C) in terms of (A)
2.6 Apply Cauchy Integral Formula
4.1 Lemma: Two fixed points imply identity
4.2 Shift Points to Create Fixed Points
4.3 Use Riemann Mapping Theorem
4.4 Define Composition Function
6.1 Choose any compact set K in D
6.2 Apply Maximum Modulus Principle to find |f(z0)|
6.3 Apply Cauchy's Integral Formula to f^2(z0)
6.4 Integrate with respect to r
6.5 Bound |f(z0)| by using hypothesis
6.6 Apply Montel's Theorem
{\displaystyle \int _{0}^{\infty }{\frac {1}{x^{3}+1}}dx\!\,}
We will compute the general case:
{\displaystyle \int _{0}^{\infty }{\frac {1}{x^{n}+1}}dx\!\,}
Find Poles of f(z)Edit
{\displaystyle f(z)={\frac {1}{z^{n}+1}}\!\,}
are just the zeros of
{\displaystyle z^{n}+1\!\,}
, so we can compute them in the following manner:
{\displaystyle z=re^{i\theta }\!\,}
{\displaystyle z^{n}+1=0\!\,}
{\displaystyle z^{n}=r^{n}e^{i\theta n}=-1\!\,}
{\displaystyle \Rightarrow r=1\!\,}
{\displaystyle in\theta =i(\pi +2\pi k)\!\,}
{\displaystyle \Rightarrow \theta ={\frac {\pi +2\pi k}{n}}\!\,}
, k=0,1,2,...,n-1.
Thus, the poles of
{\displaystyle f(z)\!\,}
{\displaystyle z=e^{\frac {\pi +2\pi k}{n}}\!\,}
{\displaystyle k=0,1,...,n-1\!\,}
Choose Path of Contour IntegralEdit
In order to get obtain the integral of
{\displaystyle f(x)\!\,}
{\displaystyle \infty \!\,}
, let us consider the path
{\displaystyle \gamma \!\,}
consisting in a line
{\displaystyle A\!\,}
going from 0 to
{\displaystyle R\!\,}
, then the arc
{\displaystyle B\!\,}
{\displaystyle R\!\,}
from the angle 0 to
{\displaystyle {\frac {2\pi }{n}}\!\,}
and then the line
{\displaystyle C\!\,}
joining the end point of
{\displaystyle B\!\,}
and the initial point of
{\displaystyle A\!\,}
{\displaystyle R\!\,}
is a fixed positive number such that
the pole
{\displaystyle z_{0}=e^{i{\frac {\pi }{n}}}\!\,}
is inside the curve
{\displaystyle \gamma \!\,}
. Then , we need to estimate the integral
{\displaystyle \int _{\gamma }f(z)=\underbrace {\int _{0}^{R}f(z)} _{A}+\underbrace {\int _{S(R)}f(z)} _{B}+\underbrace {\int _{Re^{i{\frac {2\pi }{n}}}}^{0}f(z)} _{C}\!\,}
Compute Residues of f at z0= exp{i\pi /n}Edit
{\displaystyle {\begin{aligned}Res(f,z_{0})&=\left.{\frac {1}{(z^{n}+1)'}}\right|_{z=z_{0}}\\&={\frac {1}{nz_{0}^{n-1}}}\\&={\frac {1}{ne^{{\frac {i\pi }{n}}(n-1)}}}\\&={\frac {e^{\frac {i\pi }{n}}}{ne^{i\pi }}}\\&={\frac {-1}{n}}e^{\frac {i\pi }{n}}\end{aligned}}\!\,}
Bound Arc Portion (B) of IntegralEdit
{\displaystyle {\begin{aligned}|B|&=\left|\int _{S(R)}{\frac {dz}{z^{n}+1}}\right|\\&\leq \int _{S(R)}{\frac {dz}{|z^{n}+1|}}\\&=\int _{S(R)}{\frac {dz}{|R^{n}+1|}}\\&\leq {\frac {1}{R^{n}}}\int _{S(R)}dz\\&={\frac {1}{R^{n}}}{\frac {2\pi }{n}}R\\&={\frac {2\pi }{nR^{n-1}}}\end{aligned}}}
Hence as
{\displaystyle R\rightarrow \infty \!\,}
{\displaystyle |B|\rightarrow 0\!\,}
Parametrize (C) in terms of (A)Edit
{\displaystyle z=re^{i{\frac {2\pi }{n}}}\!\,}
{\displaystyle r\!\,}
is real number. Then
{\displaystyle dz=e^{i{\frac {2\pi }{n}}}dr\!\,}
{\displaystyle {\begin{aligned}C&=\int _{Re^{i{\frac {2\pi }{n}}}}^{0}{\frac {dz}{1+z^{n}}}\\&=-\int _{0}^{Re^{i{\frac {2\pi }{n}}}}{\frac {dz}{1+z^{n}}}\\&=-\int _{0}^{R}{\frac {e^{i{\frac {2\pi }{n}}}}{1+(re^{i{\frac {2\pi }{n}}})^{n}}}\\&=-\int _{0}^{R}{\frac {e^{i{\frac {2\pi }{n}}}}{1+r^{n}}}\\&=-e^{i{\frac {2\pi }{n}}}\underbrace {\int _{0}^{R}{\frac {dr}{1+r^{n}}}} _{A}\\\end{aligned}}}
Apply Cauchy Integral FormulaEdit
From Cauchy Integral Formula, we have,
{\displaystyle A+B+C=2\pi i{\frac {-e^{i{\frac {\pi }{n}}}}{n}}\!\,}
{\displaystyle R\rightarrow \infty \!\,}
{\displaystyle B\rightarrow 0\!\,}
{\displaystyle C\!\,}
{\displaystyle A\!\,}
{\displaystyle {\begin{aligned}A+B+C&=A+C\\&=(1-e^{i{\frac {2\pi }{n}}})A\end{aligned}}}
{\displaystyle {\begin{aligned}A&={\frac {2\pi i}{n}}{\frac {e^{i{\frac {\pi }{n}}}}{e^{i{\frac {2\pi }{n}}}-1}}\\&={\frac {\pi }{n}}{\frac {2ie^{i{\frac {\pi }{n}}}}{e^{i{\frac {\pi }{n}}}(e^{i{\frac {\pi }{n}}}-e^{-i{\frac {\pi }{n}}})}}\\&={\frac {\pi }{n}}{\frac {1}{\sin({\frac {\pi }{n}})}}\end{aligned}}}
{\displaystyle S=\{z:-{\frac {\pi }{2}}<\Im (z)<{\frac {\pi }{2}}\}\!\,}
and there is an entire function
{\displaystyle g\!\,}
{\displaystyle g(S)\subset S\!\,}
{\displaystyle g(-1)=0\!\,}
{\displaystyle g(0)=1\!\,}
{\displaystyle g(z)=z+1\!\,}
Lemma: Two fixed points imply identityEdit
{\displaystyle f\!\,}
be analytic on the unit
{\displaystyle D\!\,}
{\displaystyle |f(z)|<1\!\,}
on the disc. Prove that if there exist two distinct points
{\displaystyle a\!\,}nd
{\displaystyle b\!\,}
in the disc which are fixed points, that is,
{\displaystyle f(a)=a\!\,}
{\displaystyle f(b)=b\!\,}
{\displaystyle f(z)=z\!\,}
{\displaystyle h:D\rightarrow D\!\,}
be the automorphism defined as
{\displaystyle h(z)={\frac {a-z}{1-{\overline {a}}z}}\!\,}
Consider now
{\displaystyle F(z)=h\circ f\circ h^{-1}(z)\!\,}
. Then, F has two fixed points, namely
{\displaystyle F(0)=h\circ f\circ h^{-1}(0)=h\circ f(a)=h(a)=0\!\,}
{\displaystyle F\left({\frac {a-b}{1-{\overline {a}}b}}\right)=h\circ f\circ h^{-1}\left({\frac {a-b}{1-{\overline {a}}b}}\right)=h\circ f(b)=h(b)={\frac {a-b}{1-{\overline {a}}b}}\!\,}
{\displaystyle F(0)=0\!\,}
{\displaystyle {\frac {a-b}{1-{\overline {a}}b}}\neq 0\!\,}
{\displaystyle a\!\,}
is different to
{\displaystyle b\!\,}
{\displaystyle \left|F\left({\frac {a-b}{1-{\overline {a}}b}}\right)\right|=\left|{\frac {a-b}{1-{\overline {a}}b}}\right|\!\,}
by Schwarz Lemma,
{\displaystyle F(z)=\alpha z\!\,}
But, replacing
{\displaystyle {\frac {a-b}{1-{\overline {a}}b}}\!\,}
into the last formula, we get
{\displaystyle \alpha =1\!\,}
{\displaystyle h\circ f\circ h^{-1}(z)=z\!\,}
{\displaystyle f(z)=z\!\,}
Shift Points to Create Fixed PointsEdit
{\displaystyle f(z)=g(z)-1\!\,}
{\displaystyle f(0)=0\!\,}
{\displaystyle f(-1)=-1\!\,}
{\displaystyle S\!\,}
is an infinite horizontal strip centered around the real axis with height
{\displaystyle \pi \!\,}
{\displaystyle f(z)\!\,}
is a unit horizontal shift left,
{\displaystyle f(S)\subset S\!\,}
Use Riemann Mapping TheoremEdit
From the Riemann mapping theorem, there exists a biholomorphic (bijective and holomorphic) mapping
{\displaystyle h\!\,}
, from the open unit disk
{\displaystyle D\!\,}
{\displaystyle S\!\,}
Define Composition FunctionEdit
{\displaystyle F=h^{-1}\circ f\circ h\!\,}
{\displaystyle F\!\,}
{\displaystyle D\!\,}
{\displaystyle D\!\,}
From the lemma, since
{\displaystyle F(z)\!\,}
has two fixed points,
{\displaystyle F(z)=z\!\,}
{\displaystyle f(z)=z\!\,}
{\displaystyle g=z+1\!\,}
{\displaystyle {\mathcal {F}}\!\,}
be the family of functions
{\displaystyle f\!\,}
{\displaystyle \{|z|<1\}\!\,}
{\displaystyle \int \int _{|z|<1}|f(x+iy)|^{2}dxdy\leq 1\!\,}
{\displaystyle {\mathcal {F}}\!\,}
is a normal family on
{\displaystyle \{|z|<1\}\!\,}
Choose any compact set K in DEdit
Choose any compact set
{\displaystyle K\!\,}
in the open unit disk
{\displaystyle D\!\,}
{\displaystyle K\!\,}
is compact, it is also closed and bounded.
We want to show that for all
{\displaystyle f\in {\mathcal {F}}\!\,}
{\displaystyle z\in K\!\,}
{\displaystyle |f(z)|\!\,}
is bounded i.e.
{\displaystyle |f(z)|<B_{K}\!\,}
{\displaystyle B_{K}\!\,}
is some constant dependent on the choice of
{\displaystyle K\!\,}
Apply Maximum Modulus Principle to find |f(z0)|Edit
{\displaystyle z_{0}\!\,}
that is the shortest distance from the boundary of the unit disk
{\displaystyle D\!\,}
. From the maximum modulus principle,
{\displaystyle |f(z_{0})|=\max _{z\in K}|f(z|\!\,}
{\displaystyle z_{0}\!\,}
{\displaystyle f\in {\mathcal {F}}\!\,}
Apply Cauchy's Integral Formula to f^2(z0)Edit
We will apply Cauchy's Integral formula to
{\displaystyle f^{2}(z_{0})\!\,}
{\displaystyle f(z_{0})\!\,}
) to take advantage of the hypothesis.
Choose sufficiently small
{\displaystyle r_{0}>0\!\,}
{\displaystyle D(z_{0},r_{0})\!\,\in D}
{\displaystyle {\begin{aligned}f^{2}(z_{0})&={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r_{0}}{\frac {f^{2}(z)}{z-z_{0}}}dz\\&={\frac {1}{2\pi i}}\int _{0}^{2\pi }{\frac {f^{2}(z_{0}+r_{0}e^{i\theta })ir_{0}e^{i\theta }}{(z_{0}+r_{0}e^{i\theta })-z_{0}}}d\theta \\&={\frac {1}{2\pi }}\int _{0}^{2\pi }f^{2}(z_{0}+r_{0}e^{i\theta })d\theta \end{aligned}}}
Integrate with respect to rEdit
{\displaystyle {\begin{aligned}\int _{0}^{r_{0}}rf^{2}(z_{0})dr&=\int _{0}^{r_{0}}\int _{0}^{2\pi }f^{2}(z_{0}+re^{i\theta })rdrd\theta \\&={\frac {1}{2\pi }}\int \int _{D(z_{0},r)}f^{2}(x+iy)dxdy\end{aligned}}}
Integrating the left hand side, we have
{\displaystyle \int _{0}^{r_{0}}rf^{2}(z_{0})dr={\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\!\,}
{\displaystyle {\frac {r_{0}^{2}}{2}}f^{2}(z_{0})={\frac {1}{2\pi }}\int \int _{D(z_{0},r)}f^{2}(x+iy)dxdy\!\,}
Bound |f(z0)| by using hypothesisEdit
{\displaystyle {\begin{aligned}\left|{\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\right|&=\left|{\frac {1}{2\pi }}\int \int _{D(z_{0},r_{0})}f^{2}(x+iy)dxdy\right|\\&\leq {\frac {1}{2\pi }}\int \int _{D(z_{0},r_{0})}|f^{2}(x+iy)|dxdy\\&\leq {\frac {1}{2\pi }}\int \int _{D}|f^{2}(x+iy)|dxdy\\&\leq {\frac {1}{2\pi }}\\\\{\mbox{Then }}\\\left|{\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\right|&\leq {\frac {1}{2\pi }}\\\\{\mbox{This implies}}\\\\|f(z_{0})|&\leq {\frac {1}{r_{0}{\sqrt {\pi }}}}\end{aligned}}}
Apply Montel's TheoremEdit
Then, since any
{\displaystyle f\in {\mathcal {F}}\!\,}
is uniformly bounded in every compact set, by Montel's Theorem, it follows that
{\displaystyle {\mathcal {F}}\!\,}
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University of Texas, Mathematics, Arlington, Texas, USA.
Dragan, I. (2018) Egalitarian Allocations and the Inverse Problem for the Shapley Value. American Journal of Operations Research, 8, 448-456. doi: 10.4236/ajor.2018.86025.
\left(N,v\right)
v:P\left(N\right)\to R
P\left(N\right)
v\left(\varnothing \right)=0
S\subset N
v\left(S\right)
N-S
v\left(N\right)
S{H}_{i}\left(N,v\right)=\underset{S:i\in S\subseteq N}{\sum }\frac{\left(s-1\right)!\left(n-s\right)!}{n!}\left[v\left(S\right)-v\left(S-\left\{i\right\}\right)\right],\forall i\in N,
s=|S|
n=|N|
E{A}_{i}\left(N,v\right)=\frac{v\left(N\right)}{n},\forall i\in N.
v\left(N\right)
\underset{i\in S}{\sum }{x}_{i}\ge v\left(S\right),\forall S\subseteq N,\underset{i\in N}{\sum }{x}_{i}=v\left(N\right).
v\left(1\right)=v\left(2\right)=v\left(3\right)=0,v\left(1,2\right)=v\left(1,3\right)=v\left(2,3\right)=v\left(1,2,3\right)=1,
SH\left(N,v\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right),EA\left(N,v\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right),
v\left(1\right)=v\left(2\right)=v\left(3\right)=0,v\left(1,2\right)=22,v\left(1,3\right)=v\left(2,3\right)=18,v\left(1,2,3\right)=25,
SH\left(N,v\right)=\left(9,9,7\right),EA\left(N,v\right)=\left(\frac{25}{3},\frac{25}{3},\frac{25}{3}\right),
{w}_{T},\forall T\subseteq N,T\ne \varnothing
{w}_{T}\left(T\right)=\frac{1}{{p}_{t}^{t}},{w}_{{}_{T}}\left(S\right)=\underset{l=0}{\overset{l=s-t}{\sum }}\frac{{\left(-1\right)}^{l}\left(\begin{array}{c}s-t\\ l\end{array}\right)}{{p}_{t+l}^{t+l}},\forall S\supset T.
{p}_{s}^{t-1}={p}_{s}^{t}+{p}_{s+1}^{t},s=1,2,\cdots ,t-1,
v=\underset{i\in N}{\sum }{c}_{\left\{i\right\}}{w}_{\left\{i\right\}}+\underset{S:|S|=2}{\sum }{c}_{S}{w}_{S}+{c}_{N}{w}_{N},
w=\underset{i\in N}{\sum }{c}_{\left\{i\right\}}{w}_{\left\{i\right\}}+{c}_{N}\left({w}_{N}+\underset{i\in N}{\sum }{w}_{N-\left\{i\right\}}\right)-\underset{i\in N}{\sum }{L}_{i}{w}_{N-\left\{i\right\}},
w={c}_{N}\left({w}_{N}+\underset{i\in N}{\sum }{w}_{N-\left\{i\right\}}\right)-\underset{i\in N}{\sum }{L}_{i}{w}_{N-\left\{i\right\}},
\begin{array}{l}w\left(1\right)=w\left(2\right)=w\left(3\right)=0,w\left(1,2,3\right)={L}_{1}+{L}_{2}+{L}_{3},\\ w\left(1,2\right)=2\left({c}_{123}-{L}_{3}\right),w\left(1,3\right)=2\left({c}_{123}-{L}_{2}\right),w\left(2,3\right)=2\left({c}_{123}-{L}_{1}\right).\end{array}
{c}_{123}
w\left(N\right)=v\left(N\right)
\frac{w\left(N\right)}{n}\ge \frac{w\left(N-\left\{i\right\}\right)}{n-1},\forall i\in N.
\underset{j\in N-\left\{i\right\}}{\sum }E{A}_{j}\left(N,w\right)=\left(n-1\right)\frac{w\left(N\right)}{n}\ge w\left(N-\left\{i\right\}\right),\forall i\in N,
{c}_{123}\le \frac{1}{2}\left\{{L}_{1}+{L}_{2}+{L}_{3}+Mi{n}_{i}{L}_{i}\right\}=\alpha ,
L=SH\left(N,v\right)
M=EA\left(N,v\right)
{c}_{123}\le \frac{1}{3}\left({L}_{1}+{L}_{2}+{L}_{3}\right)+Mi{n}_{i}{L}_{i}=\beta .
{c}_{123}\in \left[0,\alpha \right]
{c}_{123}\in \left[0,\beta \right]
\alpha
\beta
\alpha -\beta =\frac{1}{6}\left({L}_{1}+{L}_{2}+{L}_{3}\right)-\frac{1}{2}Mi{n}_{i}{L}_{i},
\alpha \ge \beta
{c}_{123}\in \left[0,\beta \right]
\alpha =\beta =\frac{2}{3}
w\left(1\right)=w\left(2\right)=w\left(3\right)=0,w\left(1,2\right)=w\left(1,3\right)=w\left(2,3\right)=\frac{2}{3},w\left(1,2,3\right)=1.
\alpha =16
\beta =\frac{46}{3}
\left[0,\beta \right]
w\left(1\right)=w\left(2\right)=w\left(3\right)=0,w\left(1,2\right)=\frac{50}{3},w\left(1,3\right)=w\left(2,3\right)=\frac{38}{3},w\left(1,2,3\right)=25.
S\subset n,|S|\le n-2
\begin{array}{l}w\left(S\right)=0,\forall S\subset N,S\ne \varnothing ,|S|\le n-2,\\ w\left(N-\left\{i\right\}\right)=\left(n-1\right)\left({c}_{N}-{L}_{i}\right),\forall i\in N,w\left(N\right)=\underset{i\in N}{\sum }{L}_{i},\end{array}
L=SH\left(N,v\right)
n=3
{c}_{N}
{c}_{N}\le \frac{1}{n-1}\underset{i\in N}{\sum }{L}_{i}+\frac{n-2}{n-1}Mi{n}_{i}{L}_{i}=\alpha ,
L-SH\left(N,v\right)
n=3
w\left(N\right)=v\left(N\right)
\left(n-1\right)\frac{w\left(N\right)}{n}\ge w\left(N-\left\{i\right\}\right)=\left(n-1\right)\left({c}_{N}-{L}_{i}\right),\forall i\in N,
{c}_{N}\le \frac{1}{n}\underset{i\in N}{\sum }{L}_{i}+Mi{n}_{i}{L}_{i}=\beta .
n=3
\alpha -\beta =\frac{1}{n-1}\left(\frac{1}{n}\underset{i\in N}{\sum }{L}_{i}-Mi{n}_{i}{L}_{i}\right)\ge 0,
\left(N,v\right)
SH\left(N,v\right)
EA\left(N,v\right)
{c}_{N}
{c}_{N}\in \left[0,\alpha \right]
{c}_{N}
{c}_{N}\in \left[0,\beta \right]
\alpha \ge \beta
\left[0,\beta \right]
v\left(i\right)=0,\forall i\in N,v\left(i,j\right)=\frac{1}{2},\forall i,j\in N,v\left(i,j,k\right)=1,\forall i,j,k\in N,v\left(N\right)=1,
i\ne j\ne k
SH\left(N,v\right)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right),EA\left(N,v\right)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right),
\beta =\frac{1}{2}
\alpha =\frac{1}{2}
\left[0,\frac{1}{2}\right]
w\left(i\right)=0,\forall i\in N,w\left(i,j\right)=0,\forall i,j\in N,w\left(i,j,k\right)=\frac{3}{4},\forall i,j,k\in N,w\left(N\right)=1.
v\left(1,2,3\right)=v\left(1,2,4\right)=33,v\left(1,3,4\right)=v\left(2,3,4\right)=27,v\left(1,2,3,4\right)=32.
SH\left(N,v\right)=\left(9,9,7,7\right),EA\left(N,v\right)=\left(8,8,8,8\right),
\alpha =\frac{46}{3}
\beta =15
{c}_{N}\in \left[0,15\right]
{c}_{N}=15
w\left(1,2,3\right)=w\left(1,2,4\right)=24,w\left(1,3,4\right)=w\left(2,3,4\right)=18,w\left(1,2,3,4\right)=32.
\begin{array}{l}ENS{C}_{i}\left(N,v\right)\\ =v\left(N\right)-v\left(N-\left\{i\right\}\right)+\frac{1}{n}\left\{v\left(N\right)-\underset{j\in N}{\sum }\left[v\left(N\right)-v\left(N-\left\{j\right\}\right)\right]\right\},\forall i\in N,\end{array}
v\left(N\right)-v\left(N-\left\{1\right\}\right)=v\left(N\right)-v\left(N-\left\{2\right\}\right)=7,v\left(N\right)-v\left(N-\left\{3\right\}\right)=3,
ENSC\left(N,v\right)=\left(\frac{29}{3},\frac{29}{3},\frac{17}{3}\right),
w\left(1,2\right)=2{c}_{123}-14,w\left(1,3\right)=w\left(2,3\right)=2{c}_{123}-18,w\left(1,2,3\right)=25.
ENSC\left(N,v\right)=E
{E}_{1}+{E}_{2}=\frac{58}{3}\ge w\left(1,2\right)=2{c}_{N}-14,{E}_{1}+{E}_{3}={E}_{2}+{E}_{3}=\frac{46}{3}\ge 2{c}_{N}-18,
{c}_{N}\le \frac{50}{3}=\gamma
\alpha =16,\beta =\frac{46}{3}
\gamma \ge \alpha \ge \beta
\left[0,\beta \right]
{c}_{123}=\frac{46}{3}
[2] Hart, S. and Mas-Colell, A. (1989) Potential, Value and Consistency. Econometrica, 57, 589-614.
[3] Dragan, I. (2014) On the Coalitional Rationality of the Shapley Value and Other Efficient Values of Cooperative TU Games. American Journal of Operations Research, 4, 228-234.
[4] Dragan, I. (2005) On the Inverse Problem for Semivalues of Cooperative TU Games. IJPAM, 22, 545-561.
[5] Dragan, I. (2017) On the Coalitional Rationality and the Inverse Problem for Shapley Value and the Semivalues. Applied Mathematics, 8, 2152-2164.
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m (→Automatising Conversions: some "wrong" comment removed from script)
WorldView is a commercial earth observation satellite. Details about the sensor are provided at Digital Globe's [http://www.digitalglobe.com/sites/default/files/ ].
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\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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Substitution and elimination, and matrix methods such as the Gauss-Jordan method and Cramers
Describe how the given functions can be obtained from their basic (or parent) function f using transformations.
f\left(x\right)=\sqrt{x},g\left(x\right)=\sqrt{2x-10}
For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain. y=300(1−t)^5
y=|x|-2
y=1-2x+{x}^{2}
y={x}^{2}
\frac{{\left(x-2\right)}^{2}}{3}+4
y=\mathrm{cos}x
y=-\mathrm{cos}\left(3x-\frac{\pi }{6}\right)+4
Armorikam 2021-08-17 Answered
y={x}^{2}
y=x\left(6+x\right)\right)
y=\mathrm{sin}x
3\cdot \mathrm{sin}\left(\frac{x}{2}\right)
f\left(x\right)=\sqrt{c}
m\left(x\right)=\sqrt{7x-3.5}-10
j\left(x\right)=-2\sqrt{12x}+4
f\left(x\right)={\mathrm{log}}_{2}x
f\left(x\right)=1+\sqrt{x}
f\left(x\right)={\mathrm{log}}_{2}x
a.f\left(x\right)=-2\mathrm{cos}\left(2x\right)+3\phantom{\rule{0ex}{0ex}}b.g\left(x\right)=3\mathrm{sin}\left(x-\pi \right)-1
f\left(x\right)={\mathrm{log}}_{2}x.
Then use transformations of this graph to graph the given function. What is the graph's x-intercept? What is the vertical asymptote? Use the graphs to determine each functions domain and range.
h\left(x\right)=-1+{\mathrm{log}}_{2}x
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Collision Theory - Course Hero
General Chemistry/Chemical Kinetics/Collision Theory
Collision theory describes how chemical reactions occur through molecular collisions and why reaction rates vary between reactions. It can be used to explain why certain factors affect reaction rate. According to collision theory, collisions lead to chemical change. However, not all collisions result in a chemical change. The collisions must occur with a specific orientation and must occur with sufficient energy. The majority of collisions do not result in a chemical change. The following describes the reaction between nitric oxide (NO) and ozone (O3).
\rm{NO}(\mathit g)+{\rm O}_3(\mathit g)\rightarrow{\rm{NO}}_2(\mathit g)+{\rm O}_2(\mathit g)
According to collision theory, this chemical reaction occurs through collisions between nitric oxide (NO) and ozone (O3) molecules. For a reaction to occur, the nitrogen atom of a nitric oxide (NO) molecule must strike one of the oxygen atoms in the ozone (O3) molecule. Furthermore, the angle of the two molecules also plays a role. A reaction will occur when the two molecules strike each other at a specific range of angles. A strike on the central oxygen on the ozone or a strike between two oxygen atoms does not result in a chemical reaction. A strike between correct atoms but at an unfavorable angle will also not result in a chemical reaction.
The collisions of molecules must occur at an effective orientation to lead to chemical change.
Atoms in a molecule have bonds. Bonds release energy as they form: Atoms in a molecule have less energy than those same atoms if the molecular bond is broken. A molecule is a state of low potential energy of the atoms.
Recall that kinetic energy, the energy of motion, is related to mass and speed. If two molecules collide at sufficient speeds, the kinetic energy of the collision can break some existing bonds. This can lead to formation of new bonds. Thus, according to collision theory, another factor that affects whether or not a reaction will occur during a collision is the kinetic energy, or speed, of the colliding molecules. If the molecules do not have enough energy, they will bounce off each other without a chemical reaction occurring.
During a collision, the minimum energy needed for a chemical reaction to initiate is called the activation energy. Activation energy is an increase in potential energy relative to the reactants and represents a conversion from kinetic energy to potential energy. Activation energy depends on the nature of the chemical reaction. Some reactions have higher activation energy compared to others.
This increase in energy can be demonstrated using the chemical reaction between nitric oxide (NO) and ozone (O3) as an example.
\rm{NO}(\mathit g)+{\rm O}_3(\mathit g)\rightarrow{\rm{NO}}_2(\mathit g)+{\rm O}_2(\mathit g)
The total potential energy of the reactants is at one level when they are outside a collision. During a collision, the kinetic energy is converted to potential energy. The collision between reactants, such as NO and O3, must have more energy than the activation energy of the reaction in order for a reaction to occur. If there is sufficient energy in the collision, then the two reactants form an activated complex. An activated complex, also called a transition state, is an intermediate configuration of atoms during a chemical reaction with high potential energy. An activated complex is an inherently unstable structure. It will break down to the products of the reaction, with formation of new bonds. For this reaction, the total potential energy of the products is at a lower level than that of the reactants, and energy is released. In some reactions the reverse is true. Energy is absorbed, and products have more energy than the reactants.
If there is sufficient energy in a collision, the reactants form an unstable, intermediate configuration of atoms called an activated complex. The activated complex will break down into the products.
<Reaction Rates>Rate Laws and Reaction Order
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1 Nishi-Harima Astronomical Observatory, Center for Astronomy, University of Hyogo, Sayo, Hyogo, Japan.
2 Faculty of Education, Saitama University, Saitama, Japan.
Abstract: Star-forming regions are often associated with nebulosity. In this study, we investigated infrared diffuse emission in Spitzer IRAC images. The infrared nebula L1527 traces outflows emanating from a low-mass protostar. The nebular color is consistent with the color of a stellar photosphere with large extinction. Nebulae around the HII region W5-East are bright in the infrared. These colors are consistent with the model color of dust containing polycyclic aromatic hydrocarbon (PAH). The strength of ultraviolet irradiation of the nebulae and the small dust fraction were deduced from the infrared colors of the nebulae. We found that the edges of the nebulae are irradiated by strong ultraviolet radiation and have abundant small dust. Dust at the surface of the molecular cloud is thought to be destroyed by ultraviolet radiation from an early-type star.
Keywords: Interstellar Medium, Dust, Extinction, HII Regions
0.47%<q<4.58%
{U}_{\text{ISRF}}
\mathrm{log}\left(U\right)=5.15
\mathrm{log}\left(U\right)=5.30
q=1.5%
q=3.0%
q=1.5%
Cite this paper: Itoh, Y. and Oasa, Y. (2019) Spitzer IRAC Colors of Nebulae Associated with Star-Forming Regions. International Journal of Astronomy and Astrophysics, 9, 39-50. doi: 10.4236/ijaa.2019.91004.
[1] Mathis, J.S., Rumpl, W. and Nordsieck, K.H. (1977) The Size Distribution of Interstellar Grains. Astrophysical Journal, 217, 425-433.
[2] Nakajima, Y., et al. (2003) Deep Imaging Observations of the Lupus 3 Cloud: Dark Cloud Revealed as Infrared Reflection Nebula. Astrophysical Journal, 125, 1407-1417.
[3] Tanii, R., et al. (2012) High-Resolution Near-Infrared Polarimetry of a Circumstellar Disk around UX Tau A. Publ. of Astronomical Society of Japan, 64, Article ID: 124.
[4] Boersma, C., Bregman, J.D. and Allamandola, L.J. (2013) Properties of Polycyclic Aromatic Hydrocarbons in the Northwest Photon Dominated Region of NGC 7023. I. PAH Size, Charge, Composition, and Structure Distribution. Astrophysical Journal, 769, Article ID: 117.
[5] Draine, B.T. and Li, A. (2007) Infrared Emission from Interstellar Dust. IV. The Silicate-Graphite-PAH Model in the Post-Spitzer Era. Astrophysical Journal, 657, 810-837.
[6] Draine, B.T., et al. (2007) Dust Masses, PAH Abundances, and Starlight Intensities in the SINGS Galaxy Sample. Astrophysical Journal, 663, 866-894.
[7] Megeath, S.T. et al. (2004) Initial Results from the Spitzer Young Stellar Cluster Survey. Astrophysical Journal, 154, 367-373.
[8] Chapman, N.L. and Mundy, L.G. (2009) Deep JHKs and Spitzer Imaging of Four Isolated Molecular Cloud Cores. Astrophysical Journal, 699, 1866-1882.
[9] Hartmann, L., et al. (2005) IRAC Observations of Taurus Pre-Main-Sequence Stars. Astrophysical Journal, 629, 881-896.
[10] Tamura, M., et al. (1996) Interferometric Observations of Outflows from Low-Mass Protostars in Taurus. Astrophysical Journal, 112. 2076-2085.
[11] Koenig, X.P., et al. (2008) Clustered and Triggered Star Formation in W5: Observations with Spitzer. Astrophysical Journal, 688, 1142-1158.
[12] Sugitani, K., et al. (1989) Star formation in Bright-Rimmed Globules—Evidence for Radiation-Driven Implosion. Astrophysical Journal, 342, L87-L90.
[13] Miao, J., et al. (2009) An Investigation on the Morphological Evolution of Bright-Rimmed Clouds. Astrophysical Journal, 692, 382-401.
[14] Matsuyanagi, I., et al. (2006) Sequential Formation of Low-Mass Stars in the BRC 14 Region. Publications of the Astronomical Society of Japan, 58, L29-L34.
[15] Weingartner, J.C. and Draine, B.T. (2001) Dust Grain-Size Distributions and Extinction in the Milky Way, Large Magellanic Cloud, and Small Magellanic Cloud. Astrophysical Journal, 548, 296-309.
[16] Mathis, J.S., Mezger, P.G. and Panagia, N. (1983) Interstellar Radiation Field and Dust Temperatures in the Diffuse Interstellar Matter and in Giant Molecular Clouds. Astronomy & Astrophysics, 128, 212-229.
[17] Salpeter, E.E. (1977) Formation and Destruction of Dust Grains. Annual Review of Astronomy & Astrophysics, 15, 267-293.
[18] Draine, B.T. and Salpeter, E.E. (1979) Destruction Mechanisms for Interstellar Dust. The Astrophysical Journal, 231, 438-455.
[19] Verstraete, L., et al. (1996) SWS Spectroscopy of Small Grain Features across the M17-Southwest Photo Dissociation Front. Astronomy & Astrophysics, 315, L337-L340.
[20] Pavlyuchenkov, Y.N., Kirsanova, M.S. and Wiebe, D.S. (2013) Infrared Emission and the Destruction of Dust in HII Regions. Astronomy Reports, 57, 573-585.
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Angular speed required for a body to complete one orbit
In orbital mechanics, mean motion (represented by n) is the angular speed required for a body to complete one orbit, assuming constant speed in a circular orbit which completes in the same time as the variable speed, elliptical orbit of the actual body.[1] The concept applies equally well to a small body revolving about a large, massive primary body or to two relatively same-sized bodies revolving about a common center of mass. While nominally a mean, and theoretically so in the case of two-body motion, in practice the mean motion is not typically an average over time for the orbits of real bodies, which only approximate the two-body assumption. It is rather the instantaneous value which satisfies the above conditions as calculated from the current gravitational and geometric circumstances of the body's constantly-changing, perturbed orbit.
Mean motion is used as an approximation of the actual orbital speed in making an initial calculation of the body's position in its orbit, for instance, from a set of orbital elements. This mean position is refined by Kepler's equation to produce the true position.
1.1 Mean motion and Kepler's laws
1.2 Mean motion and the constants of the motion
1.3 Mean motion and the gravitational constants
1.4 Mean motion and mean anomaly
Define the orbital period (the time period for the body to complete one orbit) as P, with dimension of time. The mean motion is simply one revolution divided by this time, or,
{\displaystyle n={\frac {2\pi }{P}},\qquad n={\frac {360^{\circ }}{P}},\quad {\mbox{or}}\quad n={\frac {1}{P}},}
with dimensions of radians per unit time, degrees per unit time or revolutions per unit time.[2][3]
The value of mean motion depends on the circumstances of the particular gravitating system. In systems with more mass, bodies will orbit faster, in accordance with Newton's law of universal gravitation. Likewise, bodies closer together will also orbit faster.
Mean motion and Kepler's laws[edit]
Kepler's 3rd law of planetary motion states, the square of the periodic time is proportional to the cube of the mean distance,[4] or
{\displaystyle {a^{3}}\propto {P^{2}},}
where a is the semi-major axis or mean distance, and P is the orbital period as above. The constant of proportionality is given by
{\displaystyle {\frac {a^{3}}{P^{2}}}={\frac {\mu }{4\pi ^{2}}}}
where μ is the standard gravitational parameter, a constant for any particular gravitational system.
If the mean motion is given in units of radians per unit of time, we can combine it into the above definition of the Kepler's 3rd law,
{\displaystyle {\frac {\mu }{4\pi ^{2}}}={\frac {a^{3}}{\left({\frac {2\pi }{n}}\right)^{2}}},}
and reducing,
{\displaystyle \mu =a^{3}n^{2},}
which is another definition of Kepler's 3rd law.[3][5] μ, the constant of proportionality,[6][note 1] is a gravitational parameter defined by the masses of the bodies in question and by the Newtonian gravitational constant, G (see below). Therefore, n is also defined[7]
{\displaystyle n^{2}={\frac {\mu }{a^{3}}},\quad {\text{or}}\quad n={\sqrt {\frac {\mu }{a^{3}}}}.}
Expanding mean motion by expanding μ,
{\displaystyle n={\sqrt {\frac {G(M+m)}{a^{3}}}},}
where M is typically the mass of the primary body of the system and m is the mass of a smaller body.
This is the complete gravitational definition of mean motion in a two-body system. Often in celestial mechanics, the primary body is much larger than any of the secondary bodies of the system, that is, M ≫ m. It is under these circumstances that m becomes unimportant and Kepler's 3rd law is approximately constant for all of the smaller bodies.
Kepler's 2nd law of planetary motion states, a line joining a planet and the Sun sweeps out equal areas in equal times,[6] or
{\displaystyle {\frac {\mathrm {d} A}{\mathrm {d} t}}={\text{constant}}}
for a two-body orbit, where dA/dt is the time rate of change of the area swept.
See also: Leibniz's notation
Letting 't = P, the orbital period, the area swept is the entire area of the ellipse, dA = πab, where a is the semi-major axis and b is the semi-minor axis of the ellipse.[8] Hence,
{\displaystyle {\frac {\mathrm {d} A}{\mathrm {d} t}}={\frac {\pi ab}{P}}.}
Multiplying this equation by 2,
{\displaystyle 2\left({\frac {\mathrm {d} A}{\mathrm {d} t}}\right)=2\left({\frac {\pi ab}{P}}\right).}
From the above definition, mean motion n = 2π/P. Substituting,
{\displaystyle 2{\frac {\mathrm {d} A}{\mathrm {d} t}}=nab,}
and mean motion is also
{\displaystyle n={\frac {2}{ab}}{\frac {\mathrm {d} A}{\mathrm {d} t}},}
which is itself constant as a, b, and dA/dt are all constant in two-body motion.
Mean motion and the constants of the motion[edit]
Because of the nature of two-body motion in a conservative gravitational field, two aspects of the motion do not change: the angular momentum and the mechanical energy.
The first constant, called specific angular momentum, can be defined as[8][9]
{\displaystyle h=2{\frac {\mathrm {d} A}{\mathrm {d} t}},}
and substituting in the above equation, mean motion is also
{\displaystyle n={\frac {h}{ab}}.}
The second constant, called specific mechanical energy, can be defined,[10][11]
{\displaystyle \xi =-{\frac {\mu }{2a}}.}
Rearranging and multiplying by 1/a2,
{\displaystyle {\frac {-2\xi }{a^{2}}}={\frac {\mu }{a^{3}}}.}
From above, the square of mean motion n2 = μ/a3. Substituting and rearranging, mean motion can also be expressed,
{\displaystyle n={\frac {1}{a}}{\sqrt {-2\xi }},}
where the −2 shows that ξ must be defined as a negative number, as is customary in celestial mechanics and astrodynamics.
Mean motion and the gravitational constants[edit]
Two gravitational constants are commonly used in Solar System celestial mechanics: G, the Newtonian gravitational constant and k, the Gaussian gravitational constant. From the above definitions, mean motion is
{\displaystyle n={\sqrt {\frac {G(M+m)}{a^{3}}}}\,\!.}
By normalizing parts of this equation and making some assumptions, it can be simplified, revealing the relation between the mean motion and the constants.
Setting the mass of the Sun to unity, M = 1. The masses of the planets are all much smaller, m ≪ M. Therefore, for any particular planet,
{\displaystyle n\approx {\sqrt {\frac {G}{a^{3}}}},}
and also taking the semi-major axis as one astronomical unit,
{\displaystyle n_{1\;{\text{AU}}}\approx {\sqrt {G}}.}
The Gaussian gravitational constant k = √G,[12][13][note 2] therefore, under the same conditions as above, for any particular planet
{\displaystyle n\approx {\frac {k}{\sqrt {a^{3}}}},}
and again taking the semi-major axis as one astronomical unit,
{\displaystyle n_{1{\text{ AU}}}\approx k.}
Mean motion and mean anomaly[edit]
Mean motion also represents the rate of change of mean anomaly, and hence can also be calculated,[14]
{\displaystyle n={\frac {M_{1}-M_{0}}{t}}}
where M1 and M0 are the mean anomalies at particular points in time, and t is the time elapsed between the two. M0 is referred to as the mean anomaly at epoch, and t is the time since epoch.
For Earth satellite orbital parameters, the mean motion is typically measured in revolutions per day. In that case,
{\displaystyle n={\frac {d}{2\pi }}{\sqrt {\frac {G(M+m)}{a^{3}}}}=d{\sqrt {\frac {G(M+m)}{4\pi ^{2}a^{3}}}}\,\!}
d is the quantity of time in a day,
M and m are the masses of the orbiting bodies,
To convert from radians per unit time to revolutions per day, consider the following:
{\displaystyle {\rm {{\frac {radians}{time\ unit}}\times {\frac {1\ revolution}{2\pi \ radians}}\times }}{\frac {d\ {\rm {time\ units}}}{1{\rm {\ day}}}}={\frac {d}{2\pi }}{\rm {\ revolutions\ per\ day}}}
From above, mean motion in radians per unit time is:
{\displaystyle n={\frac {2\pi }{P}},}
therefore the mean motion in revolutions per day is
{\displaystyle n={\frac {d}{2\pi }}{\frac {2\pi }{P}}={\frac {d}{P}},}
where P is the orbital period, as above.
^ Do not confuse μ, the gravitational parameter with μ, the reduced mass.
^ The Gaussian gravitational constant, k, usually has units of radians per day and the Newtonian gravitational constant, G, is usually given in the SI system. Be careful when converting.
^ Roy, A.E. (1988). Orbital Motion (third ed.). Institute of Physics Publishing. p. 83. ISBN 0-85274-229-0.
^ a b Brouwer, Dirk; Clemence, Gerald M. (1961). Methods of Celestial Mechanics. Academic Press. pp. 20–21.
^ Vallado, David A. (2001). Fundamentals of Astrodynamics and Applications (second ed.). El Segundo, CA: Microcosm Press. p. 29. ISBN 1-881883-12-4.
^ Battin, Richard H. (1999). An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition. American Institute of Aeronautics and Astronautics, Inc. p. 119. ISBN 1-56347-342-9.
^ a b Vallado, David A. (2001). p. 31.
^ Vallado, David A. (2001). p. 53.
^ Bate, Roger R.; Mueller, Donald D.; White, Jerry E. (1971). Fundamentals of Astrodynamics. Dover Publications, Inc., New York. p. 32. ISBN 0-486-60061-0.
^ Bate, Roger R.; Mueller, Donald D.; White, Jerry E. (1971). p. 28.
^ U.S. Naval Observatory, Nautical Almanac Office; H.M. Nautical Almanac Office (1961). Explanatory Supplement to the Astronomical Ephemeris and the American Ephemeris and Nautical Almanac. H.M. Stationery Office, London. p. 493.
^ Smart, W. M. (1953). Celestial Mechanics. Longmans, Green and Co., London. p. 4.
Glossary entry mean motion at the US Naval Observatory's Astronomical Almanac Online
Retrieved from "https://en.wikipedia.org/w/index.php?title=Mean_motion&oldid=1076598709"
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1Institut des Matériaux Jean Rouxel (IMN), Université de Nantes, Nantes, France.
2IFREMER, Laboratoire Ecosystèmes Microbiens et Molécules Marines pour les Biotechnologies, Nantes, France.
3IPREM Equipe de Physique et Chimie des Polymères, Université de Pau et des Pays de l’Adour, Pau, France.
4UNIV Bretagne-Loire, Université d’Angers, Université de Brest, Groupe d’Etude des Interactions Hote-Pathogène, Angers, France.
5Laboratoire de Parasitologie-Mycologie, Centre Hospitalier Universitaire, Institut de Biologie en Santé, Angers, France.
Understanding the molecular mechanism of the protein assembly still remains a challenge in the case of many biological systems. In this frame, the mechanism which drives RodA hydrophobins to self-assemble onto the surface of the conidia of the human fungal pathogen Aspergillus fumigatus into highly ordered nanorods known as rodlets, is still unresolved. Here, AFM investigations were combined with Monte Carlo simulations to elucidate how these small amphiphilic proteins self-assemble into tightly packed rodlets and how they are further organized in nanodomains. It becomes that the assembly of RodA hydrophobins into rodlets and their parallel alignment within nanodomains result from their anisotropic properties. Monte Carlo simulations allowed us to confirm that anisotropic interactions between macromolecules are sufficient to drive them to assembly into rodlets prior to nanodomains formation. Better knowledge of the mechanism of hydrophobins assembly into rodlets offers new prospects for the development of novel strategies leading to inhibition of rodlet formation, which should allow more rapid detection of the conidia by the immune system.
Protein Assembly, Aspergillus fumigatu, Anisotropic Interactions, Monte Carlo Simulations
Cuenot, S. , Zykwinska, A. , Radji, S. and Bouchara, J. (2017) Understanding the Assembly Mechanism of Proteins from Monte Carlo Simulations. Applied Mathematics, 8, 280-292. doi: 10.4236/am.2017.83023.
H=-J\underset{<k,l>}{\sum }|{\stackrel{\to }{S}}_{i,j}{\stackrel{\to }{r}}_{ij-kl}||{\stackrel{\to }{S}}_{k,l}{\stackrel{\to }{r}}_{ij-kl}|
where J, is the interaction strength between two nearest-neighbour particles (kept constant in the simulations); , represents a sum over the nearest-neighbours sites identified by the row k and column l on the two-dimensi- onal square lattice;
{\stackrel{\to }{S}}_{i,j}
, is the occupation and orientation vector of site identified by the row i and column j (similar definitions are used for the occupation and orientation vector of nearest-neighbours sites identified by the row k and column l);
{\stackrel{\to }{r}}_{ij-kl}
, corresponds to the inter-particles vector connecting the sites (i, j) and (k, l).
P=\mathrm{min}\left[1,\mathrm{exp}\left(-\Delta H/{k}_{B}T\right)\right]
\Delta H={H}_{\text{final}}-{H}_{\text{initial}}
, is the difference between the Hamiltonians of the final and initial states. If
\Delta H<0
, the energy of the system is minimized, the orientation change is accepted and the process is repeated by choosing another site. If
\Delta H>0
, the Boltzmann weight is compared to a random number comprised between 0 and 1. If this random number is lower than the Boltzmann weight, the orientation change is accepted; otherwise this change is rejected. At each time step, a new particle is added on the lattice and its orientation is either chosen at random if there is no particle in its neighbouring or chosen to minimize the total energy by considering the interactions with the nearest particles. It is important to note that the configurational entropy is implicitly taken into account in such Monte Carlo simulations with a jointly optimization of the total energy and configurational entropy of the system [31] . The simulations were performed at constant temperature T, but the aggregation kinetics could also be simulated using Monte Carlo dynamics for different values of kBT (where kB is the Boltzmann constant) [7] . Monte Carlo simulations using a deposition-eva- poration algorithm of particles could also be an alternative to our model but the final result should not be significantly different [34] .
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Liquidity Provider Rewards | LooksRare Docs
As announced here, the liquidity rewards program will be discontinued starting from 2022/02/15, 10:15 AM (UTC) and the tokens will be reallocated to be used for long-lasting Uniswap V3 liquidity managed by the team.
We will be discontinuing the Liquidity Rewards Program at Ethereum Block 14211012 on Day 37 (Approximately 2022/02/15, 10:15 AM (UTC)), leaving 2,660,000 LOOKS to reallocate to V3.
We’ll reallocate the leftover tokens to pair with ETH to add as V3 liquidity ourselves. The ETH for the LP will come from directly from our Treasury.
LP token stakers can continue to earn rewards for another week. We are giving you a one-week heads up before implementing the change to give you enough time to reassess your positions going forward.
If you haven’t unstaked your LP tokens by the time rewards end, don’t sweat. The pool will be open and available on the Rewards page for everyone to withdraw from for the foreseeable future.
We understand that this may be an inconvenience for those who provided liquidity for the sake of earning LP rewards on a longer timeframe, but we are confident that this change is for the best of both the project and LOOKS holders/traders long-term.
LooksRare is incentivizing liquidity for the LOOKS token by giving rewards for users that stake LOOKS-ETH UniV2 LP tokens.
The initial liquidity program will continue for 500,000 blocks (approximately 77 days) with the rate of rewards set at 10 LOOKS per block for LP stakers. We will assess the performance and requirement of a further liquidity program after the 500,000 blocks have passed.
How are liquidity provider rewards calculated?
User A’s LP rewards per block are calculated by:
\frac{\text{User A's LP tokens staked}}{\text{Total amount of LP tokens staked}}\times{\text{LOOKS Rewarded per block}}
User A stakes 100 LOOKS/ETH LP tokens on Day 5
The total amount of total staked LOOKS/ETH LP tokens is now 10,000
Based on the above, User A would receive:
\frac{1,000}{10,000}\times{10}=1
LOOKS per block.
How do I set up liquidity on UniV2?
To learn how to add liquidity on Uniswap V2, read our guide on adding liquidity
How are liquidity provider rewards calculated?
How do I set up liquidity on UniV2?
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PUSCH resource element indices - MATLAB ltePUSCHIndices - MathWorks Nordic
ltePUSCHIndices
Generate PUSCH RE Indices
PUSCH resource element indices
[ind,info] = ltePUSCHIndices(ue,chs)
[ind,info] = ltePUSCHIndices(ue,chs,opts)
[ind,info] = ltePUSCHIndices(ue,chs) returns a column vector of resource element indices given the UE-specific settings structure, ue, and channel transmission configuration, chs. It returns a column vector of Physical Uplink Shared Channel (PUSCH) resource element (RE) indices and a structure, info, containing information related to the PUSCH indices. By default, the indices are returned in 1-based linear indexing form that can directly index elements of a resource matrix. These indices are ordered as the PUSCH modulation symbols should be mapped. Alternative indexing formats can also be generated.
Support of PUSCH frequency hopping is provided by the function lteDCIResourceAllocation, which creates PRBSet from a DCI Format 0 message.
[ind,info] = ltePUSCHIndices(ue,chs,opts) formats the returned indices using options specified by opts.
Generate 0-based PUSCH resource element (RE) indices in linear form.
puschIndices = ltePUSCHIndices(frc,frc.PUSCH,{'0based','ind'});
puschIndices(1:4)
UE-specific settings, specified as a structure having the following fields.
{N}_{\text{RB}}^{\text{UL}}
Channel transmission configuration, specified as a structure. It contains the following fields.
PRB indices, specified as a column vector or a 2-column matrix, containing the Physical Resource Block indices (PRBs) corresponding to the resource allocations for this PUSCH.
Modulation Optional 'QPSK', '16QAM', '64QAM', or '256QAM'
Modulation format, specified as a character vector or string scalar for one codeword, or a cell array of character vectors or string array for two codewords.
ind — PUSCH resource element indices
PUSCH resource element (RE) indices, returned as column vector of integers.
info — Information related to PUSCH indices
Information related to the PUSCH indices, returned as a structure having the following fields.
1- or 2-element vector of integers
A one- or two-element vector, specifying the number of coded and rate matched UL-SCH data bits for each codeword
Number of coded and rate matched UL-SCH data symbols, equal to the number of rows in the PUSCH indices
ltePUSCH | ltePUSCHDecode | ltePUSCHPrecode | ltePUSCHDeprecode | ltePUSCHDRS | ltePUSCHDRSIndices | lteDCIResourceAllocation
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KVPY Exam Preparation | Brilliant Math & Science Wiki
Andrew Ellinor, Satyabrata Dash, Nanayaranaraknas Vahdam, and
Priyansh Sangule
KVPY (Kishore Vaigyanik Protsahan Yojana) is an exam-based fellowship in India. This page outlines the exam details and math topics covered, providing relevant wikis and quizzes for training and practice. Other sciences are covered on the exam, but are not currently covered on this page.
The first round of the selection procedure is an aptitude test. Generally, the cut off is seen to be around 45%-60% of the total marks but it may vary according to the toughness of the paper. Then, based on the performance in the aptitude test, short-listed students are called for an interview which is the final stage of the selection procedure. For receiving a fellowship, both aptitude test and interview marks are considered.
These are the basic topics covered by KVPY. This is a great place to start learning if you’re new to KVPY!
\large \textbf{Algebra}
\large \textbf{Geometry}
\large \textbf{Combinatorics (Counting and Probability)}
\large \textbf{Calculus}
These are the intermediate topics covered by KVPY. This is a great place to hone your skills if you are already comfortable with solving the first few problems on KVPY.
\large \textbf{Algebra}
\large \textbf{Geometry}
\large \textbf{Combinatorics (Counting and Probability)}
\large \textbf{Calculus}
These are the advanced topics covered by KVPY, which usually appear in the later problems. This is a great place to learn to solve the hardest problems on the KVPY if you're shooting for a perfect score!
\large \textbf{Algebra}
\large \textbf{Geometry}
\large \textbf{Combinatorics (Counting and Probability)}
\large \textbf{Calculus}
Cite as: KVPY Exam Preparation. Brilliant.org. Retrieved from https://brilliant.org/wiki/kvpy-exam-preparation/
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Jackscrew (5541 views)
A jackscrew is a type of jack that is operated by turning a leadscrew. In the form of a screw jack it is commonly used to lift moderately heavy weights, such as vehicles. More commonly it is used as an adjustable support for heavy loads, such as the foundations of houses, or large vehicles. These can support a heavy load, but not lift it.
3D CAD Models - Jackscrew
Licensed under Creative Commons Attribution-Share Alike 3.0 (Johnalden at the English language Wikipedia).
A jackscrew is a type of jack that is operated by turning a leadscrew. In the form of a screw jack it is commonly used to lift moderately heavy weights, such as vehicles. More commonly it is used as an adjustable support for heavy loads, such as the foundations of houses, or large vehicles. These can support a heavy load, but not lift it.[citation needed]
4.1 Industrial and technical applications
4.2 In electronic connectors
A screw jack consists of a heavy-duty vertical screw with a load table mounted on its top, which screws into a threaded hole in a stationary support frame with a wide base resting on the ground. A rotating collar on the head of the screw has holes into which the handle, a metal bar, fits. When the handle is turned clockwise, the screw moves further out of the base, lifting the load resting on the load table. In order to support large load forces, the screw usually has either square threads or buttress threads.
An advantage of jackscrews over some other types of jack is that they are self-locking, which means when the rotational force on the screw is removed, it will remain motionless where it was left and will not rotate backwards, regardless of how much load it is supporting. This makes them inherently safer than hydraulic jacks, for example, which will move backwards under load if the force on the hydraulic actuator is accidentally released.
The ideal mechanical advantage of a screw jack, the ratio of the force the jack exerts on the load to the input force on the lever ignoring friction is
{\displaystyle {\frac {F_{\text{load}}}{F_{\text{in}}}}={\frac {2\pi r}{l}}\,}
{\displaystyle F_{\text{load}}\,}
is the force the jack exerts on the load
{\displaystyle F_{\text{in}}\,}
is the rotational force exerted on the handle of the jack
{\displaystyle r\,}
is the length of the jack handle, from the screw axis to where the force is applied
{\displaystyle l\,}
is the lead of the screw.
The screw jack consists of two simple machines in series; the long operating handle serves as a lever whose output force turns the screw. So the mechanical advantage is increased by a longer handle as well as a finer screw thread. However, most screw jacks have large amounts of friction which increase the input force necessary, so the actual mechanical advantage is often only 30% to 50% of this figure.
Screw jacks are limited in their lifting capacity. Increasing load increases friction within the screw threads. A fine pitch thread, which would increase the advantage of the screw, also reduces the size and strength of the threads. Longer operating levers soon reach a point where the lever will simply bend at their inner end.
Screw jacks have now largely been replaced by hydraulic jacks. This was encouraged in 1858 when jacks by the Tangye company to Bramah's hydraulic press concept were applied to the successful launching of Brunel's SS Great Britain, after two failed attempts by other means. The maximum mechanical advantage possible for a hydraulic jack is not limited by the limitations on screw jacks and can be far greater. After World War II, improvements to the grinding of hydraulic rams and the use of O ring seals reduced the price of low-cost hydraulic jacks and they became widespread for use with domestic cars. Screw jacks still remain for minimal cost applications, such as the little-used tyre-changing jacks supplied with cars.
The large area of sliding contact between the screw threads means jackscrews have high friction and low efficiency as power transmission linkages, around 30%–50%. So they are not often used for continuous transmission of high power, but more often in intermittent positioning applications.
In technical application such as actuators, an Acme thread is used, although it has higher friction, because it is easy to manufacture, wear can be compensated for, it is stronger than a comparably sized square thread and it makes for smoother engagement.
The ball screw is a more advanced type of leadscrew that uses a recirculating-ball nut to minimize friction and prolong the life of the screw threads. The thread profile of such screws is approximately semicircular (commonly a "gothic arch" profile) to properly mate with the bearing balls. The disadvantage to this type of screw is that it is not self-locking. Ball screws are prevalent in powered leadscrew actuators.
Jackscrews are also used extensively in aviation, and are used to raise and lower horizontal stabilizers. The failure of a jackscrew on a Yakovlev Yak-42 airliner due to design flaws resulted in the crash of Aeroflot Flight 8641 in 1982. A similar failure on a McDonnell Douglas MD-80, resulting from a lack of grease, brought down Alaska Airlines Flight 261 in 2000. In 2013, National Airlines Flight 102 crashed as a result of a MRAP armoured vehicle breaking loose and breaking through the rear bulkhead, severing hydraulics lines and most importantly, destroying the jackscrew, rendering the aircraft uncontrollable. Investigators found that had the jackscrew not been destroyed, the 747-400F may have been able to perform a successful landing at Bagram Airfield.
In electronic connectors
The term jackscrew is also used for the captive screws that draw the two parts of D-subminiature electrical connectors together and hold them mated. When unscrewed, they allow the connector halves to be taken apart. These small jackscrews may have ordinary screw heads or extended heads (also making them thumbscrews) that allow the user's fingers to turn the jackscrew. Furthermore, the head sometimes has an internal female thread, with the male externally threaded screw shaft extending from that. The threaded-head type can be used to panel-mount one connector and provide a means to attach the mating connector to the first connector.
The jackscrew figured prominently in the classic novel Robinson Crusoe. It was also featured in a recent History Channel program as the saving tool of the Pilgrims' voyage – the main crossbeam, a key structural component of their small ship, cracked during a severe storm. A farmer's jackscrew secured the damage until landfall.
ScrewDrillTorxScrew threadSelf-tapping screwList of screw drivesScrew conveyor
This article uses material from the Wikipedia article "Jackscrew", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Modulate - Maple Help
Home : Support : Online Help : Programming : Audio Processing : Modulate
modulate one audio recording using another
Modulate(audArray, maskArray)
Array, Vector, or Matrix containing the audio to be modulated
Array, Vector, or Matrix specifying the modulation mask
The Modulate command uses one audio recording to modulate another.
The audArray parameter specifies the audio to be modulated, and must be a dense, rectangular, one or two dimensional Array, Vector, or Matrix with datatype=float[8].
The maskArray parameter specifies the modulation mask, and must be a dense rectangular Array, Vector, or Matrix with datatype=float[8], and the same dimensions as audArray.
The Modulate operation consists of multiplying each sample in the audArray by the corresponding sample in the maskArray, and writing the result to the output. Notice that this operation is commutative; the data and mask can be interchanged and will still give the same result.
Samples with a value of 1.0 in the maskArray will cause the corresponding audArray sample to be copied verbatim into the output. Samples of value 0.0 in the mask will result in 0.0 in the output, regardless of the value of the corresponding audio sample. Samples of value -1.0 in the mask will invert the value of the corresponding audio sample.
For example, consider a mask M that gradually transitions from 1.0 on the left to 0.0 on the right. Using the Modulate command with this mask and audio data A produces audio data that starts out sounding like A and fades to silence. Modulating audio B with the inverse of M, namely
1-M
, will produce audio data that starts out silent and intensifies to full volume.
Combining these two audio objects by simple addition,
A+B
, yields audio data that fades from A to B.
\mathrm{audiofile}≔\mathrm{cat}\left(\mathrm{kernelopts}\left(\mathrm{datadir}\right),"/audio/stereo.wav"\right):
\mathrm{with}\left(\mathrm{AudioTools}\right):
\mathrm{aud}≔\mathrm{Read}\left(\mathrm{audiofile}\right)
\textcolor[rgb]{0,0,1}{\mathrm{aud}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{dims}≔[\mathrm{rtable_dims}\left(\mathrm{aud}\right)]
\textcolor[rgb]{0,0,1}{\mathrm{dims}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{19962}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{2}]
\mathrm{left}≔\mathrm{aud}[\mathrm{dims}[1],1]
\textcolor[rgb]{0,0,1}{\mathrm{left}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{right}≔\mathrm{aud}[\mathrm{dims}[1],2]
\textcolor[rgb]{0,0,1}{\mathrm{right}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{weird}≔\mathrm{Modulate}\left(\mathrm{left},\mathrm{right}\right)
\textcolor[rgb]{0,0,1}{\mathrm{weird}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
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Free answers to College College algebra questions
Recent questions in College algebra
f\left(x\right)=\mathrm{log}2x
y=3+{\mathrm{log}}_{2}x
y={\mathrm{log}}_{2}x
y={\mathrm{log}}_{3}x
The two linear equations shown below are said to be dependent and consistent:
2x-5y=3
6x-15y=9
Explain in algebraic and graphical terms what happens when two linear equations are dependent and consistent.
Write an equation in terms of x and y for the function that is described by the given characteristics. A sine curve with a period of
\pi
, an amplitude of 3, a right phase shift of
\frac{\pi }{2}
, and a vertical translation up 2 units.
2x-5y=3
6x-15y=9
What happens if you use a graphical method?
Harlen Pritchard 2021-07-03 Answered
The sum of three times a first number and twice a second number is 43. If the second number is subtracted from twice the first number, the result is -4. Find the numbers.
The graph below expresses a radical function that can be written in the form
f\left(x\right)=a\left(x+k{\right)}^{\frac{1}{n}}+c
. What does the graph tell you about the value of k in this function?
5{e}^{0.2}x=7
Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f (x).
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ f\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -3& -2& -1& 0& 1\\ g\left(x\right)& -1& -3& 4& 2& 1\\ \hline\end{array}
\begin{array}{|cccccc|}\hline x& -2& -1& 0& 1& 2\\ h\left(x\right)& -2& -4& 3& 1& 0\\ \hline\end{array}
\left(1+\mathrm{cos}\theta \right)\left(1-\mathrm{cos}\theta \right)=\left({\mathrm{sin}}^{2}\right)\theta
A back-to-back stemplot is particularly useful for (A) identifying outliers. (B) comparing two data distributions. (C) merging two sets of data. (D) graphing home runs. (E) distinguishing stems from leaves.
\mathrm{tan}\left(\frac{14\pi }{3}\right)
using the unit circle
{x}^{2}-3x-4÷\left(x+2\right)
C\mathrm{\infty }\left[a,b\right]
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
\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)
\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)
\left(DI\right)\left(f\right)=\left(ID\right)\left(f\right)\left(DI\right)\left(f\right)=\left(ID\right)\left(f\right)
(a) Identify the parent function f.
(b) Describe the sequence of transformations from f to h.
(c) Sketch the graph of h by hand.
(d) Use function notation to write h in terms of the parent function f.
h\left(x\right)={\left(x-2\right)}^{3}+5
{x}^{2}+8x+15
The two linear equations shown below are said to be dependent and inconsistent: 3x−5y=3
−9x+15y=8
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Solve multiobjective goal attainment problems - MATLAB fgoalattain - MathWorks France
\underset{x,\gamma }{\text{minimize}}\text{ }\gamma \text{ such that }\left\{\begin{array}{c}F\left(x\right)-\text{weight}\cdot \gamma \le \text{goal}\\ c\left(x\right)\le 0\\ ceq\left(x\right)=0\\ A\cdot x\le b\\ Aeq\cdot x=beq\\ lb\le x\le ub.\end{array}
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+\left(x-3{\right)}^{2}}{5+{x}^{2}/4}\right].
{F}_{1}\left(x\right)
x=3
{F}_{2}\left(x\right)
x=0
F\left(x\right)
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
{x}_{1}+{x}_{2}\le 4
F\left(x\right)
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
0\le {x}_{1}\le 3
2\le {x}_{2}\le 5
F\left(x\right)
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
‖x{‖}^{2}\le 4
F\left(x\right)
{F}_{1}\left(x\right)
{F}_{2}\left(x\right)
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
{x}_{1}+{x}_{2}\le 4
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
{x}_{1}+{x}_{2}\le 4
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
{x}_{1}+{x}_{2}\le 4
F\left(x\right)=\left[\genfrac{}{}{0}{}{2+‖x-{p}_{1}{‖}^{2}}{5+‖x-{p}_{2}{‖}^{2}/4}\right].
{x}_{1}+{x}_{2}\le 4
{F}_{i}\left(x\right)-{\text{goal}}_{i}\le {\text{weight}}_{i}\text{\hspace{0.17em}}\gamma .
{F}_{i}\left(x\right)-{\text{goal}}_{i}\le {\text{weight}}_{i}\text{\hspace{0.17em}}\gamma .
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R. Murray, J. Doyle Issued: 2 May 2016 (Mon)
(updated 4 May)
CDS 240, Spring 2016 Due: 11 May 2016 (Thu)
MLS, Problem 4.1
Derive the equations of motion for a pendulum on a wire: an idealized planar pendulum whose pivot is free to slide along a horizontal wire. Assume that the top of the pendulum can move freely on the wire.
Calculate the dynamics of a spherical pendulum using the Lagrange-d'Alambert equations. The system consists of a mass
{\displaystyle m}
suspended from a rigid wire that is free to pivot in any direction at its point of attachment (a spherical joint). Choose as your primary coordinates the xy position of the bottom of the pendulum.
(updated 4 May) Consider a constrained Lagrangian system with
{\displaystyle L(q,{\dot {q}})={\frac {1}{2}}{\dot {q}}^{T}{\dot {q}},\qquad \omega (q){\dot {q}}={\dot {q}}_{3}-q_{2}{\dot {q}}_{1}=0.}
Show that substituting the constraints into the Lagrangian and then applying Lagrange's equations (without constraints) gives the incorrect equations of motion.
A planar pendulum (in the x-z plane) of mass
{\displaystyle m}
{\displaystyle \ell }
hangs from a support point that moves according to
{\displaystyle x=a\cos(\omega t)}
. Find the Lagrangian, the Hamiltonian, and write the first-order equations of motion for the pendulum.
Perko, Section 2.14, problem 1(a)
Show that the system
{\displaystyle {\begin{aligned}{\dot {x}}&=a_{11}x+a_{12}y+Ax^{2}-2Bxy+Cy^{2}\\{\dot {y}}&=a_{21}x-a_{11}y+Dx^{2}-2Axy+By^{2}\end{aligned}}}
is a Hamiltonian system with one degree of freedom; i.e., find the Hamiltonian function
{\displaystyle H(x,y)}
for this system.
Show that the differential constraint in
{\displaystyle \mathbb {R} ^{5}}
{\displaystyle \left[{\begin{matrix}0&1&\rho \sin q_{5}&\rho \cos q_{3}&\cos q_{5}\end{matrix}}\right]{\dot {q}}=0}
is nonholonomic.
A firetruck can be modeled as a car with one trailer, with the difference that the trailer is steerable, as shown in the figure below.
The constraints on the system are similar to that of the car in Section 7.3 of MLS, with the difference that back wheels are steerable. Derive the nonlinear control system for a firetruck corresponding to the control inputs for driving the cab and steering both the cab and the trailer, and show that it represents a controllable system.
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Residue of x in polynomial ring \mathbb{Z}[x]/(f) When we say that
Sullivan Pearson 2022-04-30 Answered
\mathbb{Z}\frac{x}{f}
\mathbb{Z}\frac{x}{f}
f={x}^{4}+{x}^{3}+{x}^{2}+x
\alpha
x=0\text{ }f+x
The elements of a quotient ring R/I (where I is a two-sided ideal of the ring R) are the cosets of I, i.e. the sets of the form
a+I\phantom{\rule{0.222em}{0ex}}=\left\{a+i:i\in I\right\}
. In particular, the elements of R/I are subsets of R!
When we talk about the residue of an element
x\in R
in the quotient R/I, we mean the coset
x+I
So, the residue of x in
\mathbb{Z}\frac{x}{f}
is (by definition) the coset
x+\left(f\right)=\left\{x+i:i\in \left(f\right)\right\}=\left\{x+gf:g\in \mathbb{Z}\left[x\right]\right\}.
Be warned: sometimes it gets annoying to keep writing "
+I
everywhere, so people will sometimes just use "x" to refer to the residue of x in some quotient. E.g. someone might say "
{x}^{2}=0
in R/I" to mean"
{x}^{2}+I=0+I
. It'll be up to you to keep track of the context and interpret elements as cosets when necessary!
Another approach, at least for an irreducible polynomial f, is to consider a hypothetical element
\alpha
f\left(\alpha \right)=0
, and extend the original ring R by
\alpha
(and whatever formal elements that makes a ring), analogously as one extends R by an element i which satisfies
{i}^{2}+1=0
to obtain the field of complex numbers C.
Actually the quotient ring construction (as the set of cosets of the ideal I generated by f) shows that the above hypothetical construction is feasible, namely we can set
\alpha :=x+I
\mathbb{Z}+\left(3x\right)
\mathbb{Z}\left[x\right]
\mathbb{Z}\left[x\right]\to \mathbb{Z}+\left(3x\right)
\frac{a}{1}\in {S}^{-1}A
a\in A
{\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)
M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}
\mathbb{R}\left[x\right]
\alpha n=1
\left(\frac{3}{5}+\frac{4}{5i}\right)
\alpha
\alpha
Jacobson radical of the ring of lower triangular
n×n
\mathbb{Z}
{S}_{p}
{S}_{p}
\left(p-1\right)!+1
\left(p-1\right)!\equiv -1
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Rendering equation - Wikipedia
In computer graphics, the rendering equation is an integral equation in which the equilibrium radiance leaving a point is given as the sum of emitted plus reflected radiance under a geometric optics approximation. It was simultaneously introduced into computer graphics by David Immel et al.[1] and James Kajiya[2] in 1986. The various realistic rendering techniques in computer graphics attempt to solve this equation.
1 Equation form
Equation form[edit]
The rendering equation may be written in the form
{\displaystyle L_{\text{o}}(\mathbf {x} ,\omega _{\text{o}},\lambda ,t)=L_{\text{e}}(\mathbf {x} ,\omega _{\text{o}},\lambda ,t)\ +\int _{\Omega }f_{\text{r}}(\mathbf {x} ,\omega _{\text{i}},\omega _{\text{o}},\lambda ,t)L_{\text{i}}(\mathbf {x} ,\omega _{\text{i}},\lambda ,t)(\omega _{\text{i}}\cdot \mathbf {n} )\operatorname {d} \omega _{\text{i}}}
{\displaystyle L_{\text{o}}(\mathbf {x} ,\omega _{\text{o}},\lambda ,t)}
is the total spectral radiance of wavelength
{\displaystyle \lambda }
directed outward along direction
{\displaystyle \omega _{\text{o}}}
{\displaystyle t}
, from a particular position
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {x} }
is the location in space
{\displaystyle \omega _{\text{o}}}
is the direction of the outgoing light
{\displaystyle \lambda }
is a particular wavelength of ligh{\displaystyle t}
{\displaystyle L_{\text{e}}(\mathbf {x} ,\omega _{\text{o}},\lambda ,t)}
is emitted spectral radiance
{\displaystyle \int _{\Omega }\dots \operatorname {d} \omega _{\text{i}}}
is an integral over
{\displaystyle \Omega }
{\displaystyle \Omega }
is the unit hemisphere centered around
{\displaystyle \mathbf {n} }
containing all possible values for
{\displaystyle \omega _{\text{i}}}
{\displaystyle f_{\text{r}}(\mathbf {x} ,\omega _{\text{i}},\omega _{\text{o}},\lambda ,t)}
is the bidirectional reflectance distribution function, the proportion of light reflected from
{\displaystyle \omega _{\text{i}}}
{\displaystyle \omega _{\text{o}}}
{\displaystyle \mathbf {x} }
{\displaystyle t}
, and at wavelength
{\displaystyle \lambda }
{\displaystyle \omega _{\text{i}}}
is the negative direction of the incoming light
{\displaystyle L_{\text{i}}(\mathbf {x} ,\omega _{\text{i}},\lambda ,t)}
is spectral radiance of wavelength
{\displaystyle \lambda }
coming inward toward
{\displaystyle \mathbf {x} }
from direction
{\displaystyle \omega _{\text{i}}}
{\displaystyle t}
{\displaystyle \mathbf {n} }
is the surface normal at
{\displaystyle \mathbf {x} }
{\displaystyle \omega _{\text{i}}\cdot \mathbf {n} }
is the weakening factor of outward irradiance due to incident angle, as the light flux is smeared across a surface whose area is larger than the projected area perpendicular to the ray. This is often written as
{\displaystyle \cos \theta _{i}}
Two noteworthy features are: its linearity—it is composed only of multiplications and additions, and its spatial homogeneity—it is the same in all positions and orientations. These mean a wide range of factorings and rearrangements of the equation are possible. It is a Fredholm integral equation of the second kind, similar to those that arise in quantum field theory.[3]
Note this equation's spectral and time dependence —
{\displaystyle L_{\text{o}}}
may be sampled at or integrated over sections of the visible spectrum to obtain, for example, a trichromatic color sample. A pixel value for a single frame in an animation may be obtained by fixing
{\displaystyle t;}
motion blur can be produced by averaging
{\displaystyle L_{\text{o}}}
over some given time interval (by integrating over the time interval and dividing by the length of the interval).[4]
Note that a solution to the rendering equation is the function
{\displaystyle L_{\text{o}}}
{\displaystyle L_{\text{i}}}
{\displaystyle L_{\text{o}}}
via a ray-tracing operation: The incoming radiance from some direction at one point is the outgoing radiance at some other point in the opposite direction.
Solving the rendering equation for any given scene is the primary challenge in realistic rendering. One approach to solving the equation is based on finite element methods, leading to the radiosity algorithm. Another approach using Monte Carlo methods has led to many different algorithms including path tracing, photon mapping, and Metropolis light transport, among others.
Although the equation is very general, it does not capture every aspect of light reflection. Some missing aspects include the following:
Transmission, which occurs when light is transmitted through the surface, such as when it hits a glass object or a water surface,
Subsurface scattering, where the spatial locations for incoming and departing light are different. Surfaces rendered without accounting for subsurface scattering may appear unnaturally opaque — however, it is not necessary to account for this if transmission is included in the equation, since that will effectively include also light scattered under the surface,
Polarization, where different light polarizations will sometimes have different reflection distributions, for example when light bounces at a water surface,
Phosphorescence, which occurs when light or other electromagnetic radiation is absorbed at one moment and emitted at a later moment, usually with a longer wavelength (unless the absorbed electromagnetic radiation is very intense),
Interference, where the wave properties of light are exhibited,
Fluorescence, where the absorbed and emitted light have different wavelengths,
Non-linear effects, where very intense light can increase the energy level of an electron with more energy than that of a single photon (this can occur if the electron is hit by two photons at the same time), and emission of light with higher frequency than the frequency of the light that hit the surface suddenly becomes possible, and
Relativistic Doppler effect, where light that bounces on an object that is moving in a very high speed will get its wavelength changed; if the light bounces at an object that is moving towards it, the impact will compress the photons, so the wavelength will become shorter and the light will be blueshifted and the photons will be packed more closely so the photon flux will be increased; if it bounces at an object that is moving away from it, it will be redshifted and the photons will be packed more sparsely so the photon flux will be decreased.
For scenes that are either not composed of simple surfaces in a vacuum or for which the travel time for light is an important factor, researchers have generalized the rendering equation to produce a volume rendering equation[5] suitable for volume rendering and a transient rendering equation[6] for use with data from a time-of-flight camera.
^ Immel, David S.; Cohen, Michael F.; Greenberg, Donald P. (1986), "A radiosity method for non-diffuse environments" (PDF), SIGGRAPH 1986: 133, doi:10.1145/15922.15901, ISBN 978-0-89791-196-2
^ Kajiya, James T. (1986), "The rendering equation" (PDF), SIGGRAPH 1986: 143–150, doi:10.1145/15922.15902, ISBN 978-0-89791-196-2
^ Watt, Alan; Watt, Mark (1992). "12.2.1 The path tracing solution to the rendering equation". Advanced Animation and Rendering Techniques: Theory and Practice. Addison-Wesley Professional. p. 293. ISBN 978-0-201-54412-1.
^ Owen, Scott (September 5, 1999). "Reflection: Theory and Mathematical Formulation". Retrieved 2008-06-22.
^ Kajiya, James T.; Von Herzen, Brian P. (1984), "Ray tracing volume densities", SIGGRAPH 1984, 18 (3): 165, CiteSeerX 10.1.1.128.3394, doi:10.1145/964965.808594
^ Smith, Adam M.; Skorupski, James; Davis, James (2008). Transient Rendering (PDF) (Technical report). UC Santa Cruz. UCSC-SOE-08-26.
Lecture notes from Stanford University course CS 348B, Computer Graphics: Image Synthesis Techniques
Retrieved from "https://en.wikipedia.org/w/index.php?title=Rendering_equation&oldid=1088384242"
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How do I factorize x^{6}-1 over GF(3)? I know that
Kaiya Hardin 2022-04-25 Answered
{x}^{6}-1
{\left(x+1\right)}^{3}{\left(x+2\right)}^{3}
n=±
is a multiple of p then over
GF\left(p\right)
{x}^{n}-1={\left({x}^{m}-1\right)}^{p}
so the problem reduces swiftly to the case where n is co' to p.
If p is not a factor of n then over an algebraic closure of GF(p)
{x}^{n}-1=\prod _{k=0}^{n-1}\left(x-{\zeta }^{j}\right)
\zeta
is a primitive n-th root of unity. One makes this into a factorization over GF(p) by combining conjugate factors together. For each k, the polynomial
\left(x-{\zeta }^{k}\right)\left(x-{\zeta }^{pk}\right)\left(x-{\zeta }^{{p}^{2}k}\right)\cdots \left(x-{\zeta }^{{p}^{r-1}k}\right)
has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with
{p}^{r}k\equiv k
(mod n)
Using this, it's easy to work out the degrees of the irreducible factors of
{x}^{n}-1
, but to find the factors themselves needs a bit more work, using for instance Berlekamp's algorithm.
Factoring the polynomials
{x}^{n}-1
is very different from factoring general polynomials, since you already know what the roots are; they are, in a suitably large finite extension, precisely the elements of order dividing n. Since you know that the multiplicative group of a finite field is cyclic, the conclusion follows from here.
Module with matrix multiplication
Define the rational
3×3
A:=\left(\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 0& 1& 2\end{array}\right)
We then define a
\mathbb{Q}\left[X\right]
-module on
V={\mathbb{Q}}^{3}
\mathbb{Q}\left[X\right]×V\to V,\left(P,v\right)\to P\left(A\right)\cdot v
P\left(A\right)\in {\mathbb{Q}}^{3×3}
is achieved by plugging in matrix A into polynomial P and
P\left(A\right)\cdot v
is therefore the matrix-vector-multiplication. Call this module
{V}_{A}
Argue if
{V}_{A}=\mathbb{Q}\left[X\right]\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)\oplus \mathbb{Q}\left[X\right]\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)
Notation of homeomorphism from B(H) to B(K), corresponding to unitary transformation of Hilbert spaces
Let U be a unitary transformation from Hilbert space H to Hilbert space K.
How do you call a *-homomorphism f from B(H) to B(K), defined by
f\left(a\right)=Ua{U}^{-1}
I'm interested both in a symbol, which can be used in formulas, and a name for it, which can be used in texts or in speech.
Prove that if "a" is the only elemnt of order 2 in a group, then "a" lies in the center of the group.
{\mathbb{Q}}_{2}\left(\sqrt{3},\text{ }\sqrt{2}\right)
If N is normal in G, show
{Z}_{i}\left(G\right)\frac{N}{N}\le {Z}_{i}\left(\frac{G}{N}\right)
{Z}_{i}\left(G\right)
is the i-th term in the upper central series for G.
K⇝K\left(X\right)
preserve the degree of field extensions?
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Proof of Nagumo's Theorem of invariance
Given a continuous system
{x}^{\prime }\left(t\right)=p\left(x\left(t\right)\right),t\ge 0
{\mathbb{R}}^{n}
x\left(0\right)\in {\mathbb{R}}^{n}
and assuming that solutions exist and are unique in
{\mathbb{R}}^{n}
S\subset {\mathbb{R}}^{n}
be a closed set. Then, S is positively invariant under the flow of the system
x\left(t\right)\in S
t\ge 0
p\left(x\right)\in {K}_{x}\left(S\right)
x\in \partial S
(the boundary of S) where
{K}_{x}\left(S\right)
is the set of all sub-tangential vectors to S at x, i.e.,
{K}_{x}\left(S\right)\phantom{\rule{0.222em}{0ex}}=\left\{z\in {\mathbb{R}}^{n}:\underset{h-\to 0}{lim}\frac{d\left(x+hp\left(x\right);S\right)=0}{h}\right\},
where d(w;S) is the distance from the vector w to the subset S.
luminarc24lry 2022-04-28 Answered
\frac{dy}{dt}=k\left(a-y\right)\left(b-y\right)
y\left(0\right)=0
a=b
0<a<b
b⇒a
\to \mathrm{\infty }
of a solution to an ODE
Consider the Cauchy problem:
\left\{\begin{array}{l}{y}^{\prime }\left(t\right)=\frac{{t}^{2}}{2+\mathrm{sin}\left({t}^{2}\right)}cos\left(y\left(t{\right)}^{2}\right)\\ y\left(0\right)=0\end{array}
Solve this nonhomegenous ode
y{}^{″}+4y=\mathrm{cos}\left(2x\right)
Solve the heat equation using a transform method
k\frac{{\partial }^{2}U}{\partial {x}^{2}}=\frac{\partial U}{\partial t}
U\left(0,t\right)=1,t>0
U\left(x,0\right)={e}^{-x},x>0
Solve the following differential equation by the form of homogeneous equation. Letting
y=vx
{x}^{2}\frac{dy}{dx}+xy+1=0
I am unable to solve the differential equation of first order given below,
\left(\mathrm{sin}y\mathrm{cos}y+x{\mathrm{cos}}^{2}y\right)dx+xdy=0
How to solve this equation by integration
\frac{{x}^{″}}{x}-\frac{{x}^{\prime }}{{x}^{2}}-\frac{{x}^{\prime 2}}{{x}^{2}}=0,
coraletsmmh 2022-04-28 Answered
How to solve this diffrential equation using parts formula?
\frac{dy}{dx}+\frac{y}{20}=50\left(1+\mathrm{cos}x\right)
Molecca89g 2022-04-27 Answered
f\in {C}_{o}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)
. Propose formulas of the form
u\left(x\right)={\int }_{ }^{ }\frac{\stackrel{^}{f}\left(\delta \right){e}^{ix\delta }}{p\left(\delta \right)}d\delta
\sum _{j=1}^{n}\frac{{\partial }^{4}u\left(x\right)}{\partial {x}_{j}^{4}}+u\left(x\right)=f\left(x\right)
\sum _{k,l}^{ }\frac{{\partial }^{4}u\left(x\right)}{\partial {x}_{k}^{2}\partial {x}_{l}^{2}}-2·\sum _{k=1}^{n}\frac{{\partial }^{2}u\left(x\right)}{\partial {x}_{k}^{2}}+u\left(x\right)=f\left(x\right)
\left\{\begin{array}{rl}& {x}^{2}{x}^{\prime }={\mathrm{sin}}^{2}\left({x}^{3}-3t\right)\\ & x\left(0\right)=1\end{array}.
Zack Wise 2022-04-27 Answered
I have two problems for which I know the answers (and working) but am still confused about the method used to solve them. The equations are
y{y}^{\left(2\right)}={\left({y}^{\left(1\right)}\right)}^{2}\text{ }\text{and}\text{ }{x}^{\left(2\right)}+{\left({x}^{\left(1\right)}\right)}^{2}=0
In my notes it instructs that in the case of the independent variable missing from the equation (x and t, respectively), the substitution to reduce the order is made as follows:
p={y}^{\left(1\right)}
{y}^{\left(2\right)}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy}
Then, for the first equation:
y{y}^{\left(2\right)}={\left({y}^{\left(1\right)}\right)}^{2}⇒yp\frac{dp}{dy}={p}^{2}
But for the second equation, the substitution is made as:
{x}^{\left(2\right)}+{\left({x}^{\left(1\right)}\right)}^{2}=0⇒\frac{dp}{dt}=-{p}^{2}
I've tested this with online calculators and even they compute these two differential equations in the two different ways.
Any explanation of where I'm going wrong would be greatly appreciated.
Solve the series:
\underset{n\to \mathrm{\infty }}{lim}\frac{k+1}{{3}^{k}}
x\left(2y+1\right)dx=y\left({x}^{2}-3x+2\right)dy
Poemslore8ye 2022-04-27 Answered
I am trying to find the common ratio of
\sum _{n=0}^{\mathrm{\infty }}{2}^{-n}{z}^{{n}^{2}}
\frac{d}{dx}\left(2y{y}^{\prime }\right)={\left({y}^{\prime }\right)}^{2}
y{}^{″}-qy
q:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}^{\cdot }
{x}^{3}y\text{'}\text{'}\text{'}+xy\text{'}-y=x\mathrm{ln}\left(x\right)
using shift
x={e}^{z}
and differential operator
Dz=\frac{d}{ dz }
Dz=\frac{d}{ dz }
\left({e}^{z}{\right)}^{3}y\text{'}\text{'}\text{'}+{e}^{z}y\text{'}-y={e}^{z}\mathrm{ln}\left({e}^{z}\right)
\left({e}^{3z}\right)y\text{'}\text{'}\text{'}+{e}^{z}y\text{'}-y={e}^{z}z
y={z}^{r}
{e}^{3}r\left({r}^{2}-r-2\right){z}^{r-3}+{e}^{z}r{z}^{r-1}-{z}^{r}=0
y\text{'}=\frac{x-{e}^{x}}{x+{e}^{y}}
misangela4gi 2022-04-27 Answered
{y}^{\prime }=2y\left(x\sqrt{y}-1\right)
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Gaussian Fitting with an Exponential Background - MATLAB & Simulink - MathWorks Australia
Fit the data using this equation
y\left(x\right)=a{e}^{-bx}+{a}_{1}{e}^{-{\left(\frac{x-{b}_{1}}{{c}_{1}}\right)}^{2}}+{a}_{2}{e}^{-{\left(\frac{x-{b}_{2}}{{c}_{2}}\right)}^{2}}
where ai are the peak amplitudes, bi are the peak centroids, and ci are related to the peak widths. Because unknown coefficients are part of the exponential function arguments, the equation is nonlinear.
load gauss3
xpeak is a vector of predictor values.
ypeak is a vector of response values.
In the Curve Fitter app, on the Curve Fitter tab, in the Data section, click Select Data. In the Select Fitting Data dialog box, select xpeak as the X Data value and ypeak as the Y Data value. Enter Gauss2exp1 as the Fit name value.
On the Curve Fitter tab, in the Fit Type section, click the arrow to open the gallery. In the fit gallery, click Custom Equation in the Custom group.
In the Fit Options pane, replace the example text in the equation edit box with these terms:
a*exp(-b*x) + a1*exp(-((x-b1)/c1)^2) + a2*exp(-((x-b2)/c2)^2)
The fit is poor (or incomplete) at this point because the starting points are randomly selected and no coefficients have bounds.
Specify reasonable coefficient starting points and constraints. Deducing the starting points is particularly easy for the current model because the Gaussian coefficients have a straightforward interpretation and the exponential background is well defined. Additionally, as the peak amplitudes and widths cannot be negative, constrain a1, a2, c1, and c2 to be greater than 0.
In the Fit Options pane, click Advanced Options.
In the Coefficient Constraints table, change the Lower bound for a1, a2, c1, and c2 to 0, as the peak amplitudes and widths cannot be negative.
Enter the StartPoint values as shown for the specified coefficients.
As you change the fit options, the Curve Fitter app updates the fit.
Observe the fit and residuals plots. To create a residuals plot, click Residuals Plot in the Visualization section of the Curve Fitter tab.
|
Finding Extrema Practice Problems Online | Brilliant
Finding the largest and smallest possible values of a function, or extremizing, is one of the most important practical applications of calculus.
Finding the optimal values of a function of several variables is even more useful since many successful real-life models depend on two or more inputs.
This quiz will build off our knowledge of extremizing single-variable functions to give us a glimpse of how we'll optimize many-variable functions in the Derivatives and Optimization chapters.
Let's start with a classic two-variable problem.
Imagine a box with base length
x
y.
The box doesn't have a lid. The cost of manufacturing is directly proportional to the amount of material (or surface area) of the box.
What is the surface area of the outside of the box as a function of base width
x
y?
Note: Count the area of each box side only once.
5 x y
2 x^2 + 4 xy
x^2 y
x^2+ 4 xy
The surface area of the box
A
depends on the base width
x
y.
We express this relationship explicitly as
A(x,y) = x^2 + 4 x y.
Writing the surface area as
A (x,y)
tells us that it is a function of both base width and height.
A company wants to produce such a box with base width at least
4
units and a height of at least
1
unit. If box material is
4
dollars per unit area, what's the cost of the cheapest box that can be produced?
Hint: This problem actually doesn't require any calculus, just some intuition.
Now let's put our calculus to good use.
Suppose the company wants to produce the cheapest possible box with a fixed volume of
x^2 y = 4
Besides being positive, there's no restriction on the base width
x
or the height
y
this time.
If the cost is
C(x,y) = 4 x^2 + 16 xy,
what is the minimum output in this case?
A constraint on
x
y
can reduce a two-variable function down to a single-variable function.
In the last problem, fixing the volume tied the dimensions of the box together and effectively made the cost a function of just
x,
which we optimized using single-variable calculus.
However, there are many examples of optimization problems where we have two legitimately independent variables
x
y ;
i.e. there's no constraint tying them together.
To close out this quiz, we'll look at one such example unrelated to our box problem. In the process we'll discover the need for a new kind of tool: the partial derivative.
We'll work our way up to our new example by looking at an analogous problem first.
In an upcoming chapter, we'll chart the bottom of an unexplored lakebed.
A series of boats move across the lake's surface, lowering bobs to measure the depth of the bed. Here's a table displaying the depths recorded by each boat in the survey.
Depth records from survey of part of Square Lake
We want the largest overall depth, but we're going to find it in an unconventional way that'll help us understand partial derivatives.
Find this largest depth by finding the largest depth in each row, then taking the maximum of these values.
To find the largest depth in the table, we maximized row by row, and then took the largest of these values to find the global maximum depth.
The depth table data was actually built using the function
f(x,y) = 2 x y e^{1-x^2-y^2} \ \ \text{for} \ \ -2 \leq x \leq 2, \ 0 \leq y \leq 2
(not to scale) so we expect it has a maximum. Let's combine our table strategy with what we know of single-variable calculus to find it.
By analogy, let's fix one variable, say
, though we could also fix
and get the same answer. To keep things simple, let's take
y = 1.
This reduces the number of variables down to one, which we can handle!
f(x,1) = 2 x e^{-x^2},\ \ -2 \leq x \leq 2,
which is like maximizing over a single "row."
Note: If you'd like to refresh your memory, here's a wiki on optimization in the single-variable case.
2
\sqrt{2} e^{-\frac{1}{2}}
e^{-\frac{1}{2}}
\sqrt{2} e^{\frac{1}{2}}
Generalizing last problem's result, for
0 \leq y \leq 2,
the max value of
f(x,y) = 2 x y e^{1-x^2-y^2} \ \ \text{for} \ \ -2 \leq x \leq 2
M(y) = \sqrt{2} y e^{\frac{1}{2}-y^2}.
By analogy, this is like compiling the list
Boat 1
2
3
4
5
6
Largest Depth Recorded 130
214
199
133
157
130
for the depth table
M(y) = \sqrt{2} y e^{\frac{1}{2}-y^2}= (\sqrt{2} e^{\frac{1}{2}} ) y e^{-y^2}
maximized on
0 \leq y \leq 2\, ?
(This is like maximizing over the max value of each row in the depth table.)
Hint: Up to a constant, this function of
y
is the same as the function of
x
we maximized in the last problem.
y=\frac{1}{2}
y=\frac{1}{3}
y=\frac{1}{\sqrt{2}}
y=\frac{1}{\sqrt{3}}
Now for the bad news: our approach only works because
f
and its domain have very special properties. We need a versatile method for finding extrema of general multivariable functions.
Here's an equivalent way of looking at how we optimized
f(x,y) = 2 x y e^{1-x^2-y^2} \ \ \text{for} \ \ -2 \leq x \leq 2, \ 0\leq y \leq 2:
Let's introduce a new kind of derivative
\frac{\partial f}{\partial x},
meaning we take the derivative of
as if it were just a function of
x;
y
is fixed. Similarly,
\frac{\partial f}{\partial y}
means we differentiate with respect to
y
x
Then to optimize
f ,
we really solved
\frac{\partial f}{\partial x} = 0, \ \frac{\partial f}{\partial y} = 0,
and then checked these critical values against the values of
on the edges of the rectangle
-2 \leq x \leq 2, \ 0 \leq y \leq 2
forming the boundary of
's domain.
If this is the right way to optimize, what is the largest value of
f(x,y) = 6 - x^2-y^2
(x,y)
inside the circle of radius
2
centered at the origin; i.e. all points with
\sqrt{x^2+y^2} \leq 2 \ ?
By introducing partial derivatives
\frac{\partial f}{\partial x }
\frac{\partial f}{\partial y},
we took our first steps into the larger world of multivariable optimization.
We'll have a whole chapter on these derivatives, but we first need to know more about functions and their graphs.
Graphs are visual tools that can help us in our quest for extrema. Take a look at the animations below. The surface represents the graph of
f(x,y) = 2 x y e^{1-x^2-y^2}
on the larger square
-2 \leq x, y \leq 2.
The mountain tops correspond to maxima, and the pits represent minima: with a graph we can identify extrema at a glance.
Surface graphs can also help us begin to understand one of the other major pillars of multivariable calculus: integration.
To graph surfaces, we need a coordinate system in space, which is the topic of our next quiz.
With this detour out of the way, we'll come back to many-variable integration to finish off our intro chapter.
|
Cubic Discriminant | Brilliant Math & Science Wiki
Alan Enrique Ontiveros Salazar, Pi Han Goh, shashwat kasliwal, and
We can compute the discriminant of any power of a polynomial. For example, the quadratic discriminant is given by
\Delta_2 = b^2 - 4ac
. But it gets more complicated for higher-degree polynomials.
The discriminant of a cubic polynomial
ax^3 + bx^2 + cx + d
\Delta_3 = b^2 c^2 - 4ac^3 - 4b^3 d - 27a^2 d^2 + 18abcd.
\Delta_3 > 0
, then the equation has three distinct real roots.
\Delta_3 = 0
, then the equation has a repeated root and all its roots are real.
\Delta_3 < 0
, then the equation has one real root and two non-real complex conjugate roots.
Alternatively, we can compute the value of the cubic determinant if we know the roots to the polynomial. Let
\alpha,\beta,
\gamma
denote the roots of a certain cubic polynomial, then its discriminant is equal to
\Delta_3 = a^4(\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 \; .
Here's a good example that requires us to use cubic discriminant:
f(x)=x^{3}-x^{2}-2x+1
Let the zeros of the function above be
\alpha,\beta,
\gamma.
f'(\alpha)\times f'(\beta)\times f'(\gamma).
f'(x)
f(x).
Definition of Discriminant
Computing the Discriminant for a Cubic Polynomial
Relation with Cardano's Formula
P(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0
having roots
x_1,x_2,\ldots,x_n
(counting multiplicity), its discriminant is
\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2.
When there's a repeated root, the discriminant vanishes.
If the coefficients of
P(x)
are all real, the discriminant is always a real number.
If all roots are real and distinct, the discriminant is always positive.
For the case when the polynomial is cubic, i.e.
P(x)=ax^3+bx^2+cx+d,
\Delta=a^4(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2
Since the discriminant is symmetric in the roots of the polynomial, we can express it as elementary symmetric polynomials, i.e. the coefficients of
P(x)
. Although there is a general method to derive the discriminant of any polynomial, this is an elementary and algebraic approach.
By Vieta's formula we have
\begin{aligned} x_1+x_2+x_3 &=-\dfrac{b}{a}\\\\ x_1x_2+x_1x_3+x_2x_3 &=\dfrac{c}{a}\\\\ x_1x_2x_3&=-\dfrac{d}{a}. \end{aligned}
Let's expand the discriminant:
\Delta=a^4\Big(\big(x_1x_2^2+x_2x_3^2+x_3x_1^2\big)-\big(x_1^2x_2+x_2^2x_3+x_3^2x_1\big)\Big)^2.
m=x_1x_2^2+x_2x_3^2+x_3x_1^2
n=x_1^2x_2+x_2^2x_3+x_3^2x_1
\Delta=a^4(m-n)^2
. We can't directly compute
m
n
, but since they are cyclic polynomials, it turns out that the elementary symmetric polynomials of
m
n
are symmetric in
x_1,x_2,
x_3
and hence expressable in terms of the coefficients of
P(x)
. So, let's find them:
\begin{aligned} m+n&=x_1x_2(x_1+x_2)+x_2x_3(x_2+x_3)+x_3x_1(x_3+x_1) \\\\ &=(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-3x_1x_2x_3 \\\\ &=\dfrac{-bc+3ad}{a^2}. \end{aligned}
mn
requires more tedious expansion:
\begin{aligned} mn=&\color{#3D99F6}{(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3}+3(x_1x_2x_3)^2+x_1x_2x_3\big(\color{#D61F06}{x_1^3+x_2^3+x_3^3}\color{#3D99F6}\big)\\\\ =&\color{#3D99F6}{(x_1x_2+x_1x_3+x_2x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)(x_1x_2x_3)+3(x_1x_2x_3)^2}+3(x_1x_2x_3)^2\\ &{\color{#3D99F6}{+x_1x_2x_3\big(}}\color{#D61F06}{(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)+3x_1x_2x_3}\color{#3D99F6}\big)\\\\ =&\dfrac{c^3}{a^3}-\dfrac{3bcd}{a^3}+\dfrac{6d^2}{a^2}+\dfrac{b^3d}{a^4}-\dfrac{3bcd}{a^3}+\dfrac{3d^2}{a^2}\\\\ =&\dfrac{ac^3-6abcd+9a^2d^2+b^3d}{a^4}. \end{aligned}
Finally, use the identity
(m-n)^2=(m+n)^2-4mn:
\begin{aligned} \Delta &=a^4(m-n)^2=a^4\left(\dfrac{-bc+3ad}{a^2}\right)^2-4a^4\left(\dfrac{ac^3-6abcd+9a^2d^2+b^3d}{a^4}\right)\\\\ &=b^2c^2-6abcd+9a^2d^2-4ac^3+24abcd-36a^2d^2-4b^3d\\\\ &=\boxed{b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd}. \end{aligned}
Cite as: Cubic Discriminant. Brilliant.org. Retrieved from https://brilliant.org/wiki/cubic-discriminant/
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Free answers to multivariable calculus questions
Get help with multivariable calculus equations
Recent questions in Multivariable calculus
Prove multivariable function is surjective
f:{\mathbb{N}}^{2}\to \mathbb{N}
f\left(a,b\right)={a}^{b}
multivariable functions works when the partial derivatives are continous but if the function is just differentiable does the chain rule work ? I mean if
z=f\left(x,y\right)
is differentiable function and
x=g\left(t\right),y=h\left(t\right)
are also differentiable functions can we write:
\frac{dz}{dt}=\frac{\partial f}{\partial x}×\frac{dx}{dt}+\frac{\partial f}{\partial y}×\frac{dy}{dt}
I have the multivariable function
\mathrm{log}\left({y}^{2}+4{x}^{2}-4\right)
and I have found the maximal domain to be
{x}^{2}+\frac{{y}^{2}}{4}>1
let x be a differentiable function
x:\left[a,b\right]\to R
which satisfies:
\frac{dx\left(t\right)}{dt}=f\left(t,x\left(t\right)\right)
x\left(a\right)={x}_{a}
In particular I am trying to understand what
f\left(t,x\left(t\right)\right)
means, I understand this represents a multivariable function with parameters
t
x\left(t\right)
, but I can't think of what this would mean in the given context?
Walter Clyburn 2022-01-05 Answered
Differentiation of multivariable function proof
\frac{d}{dx}{\int }_{v\left(x\right)}^{u\left(x\right)}f\left(t,x\right)dt={u}^{\prime }\left(x\right)f\left(u\left(x\right),x\right)-{v}^{\prime }\left(x\right)f\left(v\left(x\right),x\right)+{\int }_{v\left(x\right)}^{u\left(x\right)}\frac{\partial }{\partial x}f\left(t,x\right)dt
Is it possible to graph out a multivariable function (3 variables at max) on a graphing calculator? If yes, how?
garnentas3m 2022-01-04 Answered
Sketch the level curves for
f\left(x,y\right)={x}^{2}-y
together with the gradient
\mathrm{\nabla }f
at a few typical points. Write an equation for the tangent line of the level curve
f\left(x,y\right)=1
\left(\sqrt{2},1\right)
Stefan Hendricks 2022-01-04 Answered
Derivative of a multivariable function
evaluate the derivative of a function
\mathbb{R}\to \mathbb{R}
g\left(t\right)=f\left(x+t\left(y-x\right)\right)
f:{\mathbb{R}}^{n}\to \mathbb{R}
is a multivariable function and
x,y\in {\mathbb{R}}^{n}
g\prime \left(t\right)={\left(y-x\right)}^{T}\mathrm{\nabla }f\left(x+t\left(y-x\right)\right)
Donald Johnson 2022-01-04 Answered
First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be
{R}^{2}
and its image is going to be all real numbers, the graph of that function is defined as a subset of
{R}^{3}
in which the x and y axis are going to receive the inputs, and the output is going to be in
z\left(x,y,f\left(x,y\right)\right)
? Is that correct? Will its graph, in this example, be some kind of surface?
How do you define iterations of multivariable functions?
To be clear(example):
f:{\mathbb{R}}^{2}\to \mathbb{R}
f\circ f,\text{or}f\circ \dots \circ f?
{f}_{xx}
{f}_{yy}
f\left(x,y\right)=\mathrm{ln}\left({x}^{2}y\right)+{y}^{3}{x}^{2}
A camera shop stocks eight different types of batteries, one of which is type A7b.Assume there are at least 30 batteries.
of each type.a. How many ways can a total inventory of 30 batteries be distributed among the eight different types? b.
How many way can a total inventory of 30 batteries be distributed among the eight different types of the inventory must
include at least four A76 batteries?c. How many ways can a total inventory of 30 batteries be distributed among the eight
different types of the inventory includes at most three A7b batteries
Sho 2021-12-13 Answered
Consider the following statement of masses,
If both masses start from rest,
A) find a1? B) find a2 ? C) How fast is m1 moving after sliding the 2.00 m shown?
*Assume both masses are simultaneously released from the rest*
Answer the following questions about an ANOVA analysis involving three samples.
a. In this ANOVA analysis, what are we trying to determine about the three populations they’re taken from?
b. State the null and alternate hypotheses for a three-sample ANOVA analysis.
c. What sample statistics must be known to conduct an ANOVA analysis?
d. In an ANOVA test, what does an F test statistic lower than its critical value tell us about the three populations we’re examining?
Madeline Lott 2021-12-10 Answered
INVESTMENT ANALYSIS Paul Hunt is considering two business ventures. The anticipated returns (in thousands of dollars) of each venture are described by the following probability distributions:
\overline{)\begin{array}{cc}Venture\text{ }A& \\ Earnings& Probability\\ -20& 3\\ 50& 4\\ 50& 3\end{array}}
\overline{)\begin{array}{cc}Venture\text{ }B& \\ Earnings& Probability\\ -15& 2\\ 30& 5\\ 40& 3\end{array}}
a. Compute the mean and variance for each venture.
b. Which investment would provide Paul with the higher expected return (the greater mean)?
c. In which investment would the element of risk be less (that is, which probability distribution has the smaller variance)?
lunnatican4 2021-12-09 Answered
For each of the following test, outline a research scenario and use any hypothetical data to show how the test is conducted
ii. Chi-square for goodness-of-fit test
iii. Chi-square test for independence
On the final examination in Engineering Data Analysis, the mean was 72 and the standard deviation was 15.
a. Determine the standard scores (z-values) of students receiving the grade of 65.
b. Determine the standard scores (z-values) of students receiving the grade of 85
c. Determine the probability that a student will score greater than 75.
d. Determine the probability that a student will score lower than 65.
e. Find the grade corresponding to a standard score of 1.
A geologist has collected 15 specimens of basaltic rock and 15 specimens of granite. The geologist instructs a laboratory assistant to randomly select 25 of specimens for analysis.
a) What is the pmf of the number of granite specimens selected for analysis? (Round your probabilities to four decimal places.)
\begin{array}{|ccccccc|}\hline x& & & & & & \\ p\left(x\right)& & & & & & \\ \hline\end{array}
b) What is the probability that all specimens of the two types of rock are selected for analysis? (Round your answer to four decimal places.)
c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? (Round your answer to four decimal places.)
rastafarral6 2021-12-02 Answered
{\int }_{1}^{0}\mathrm{tan}\left({x}^{2}\right)dx+2{\int }_{1}^{0}{x}^{2}{\mathrm{sec}}^{2}\left({x}^{2}\right)dx=\frac{\pi }{4}
philosphy111of 2021-11-26 Answered
{x}^{2}-2x+1>0
and graph the solution set on a real number line. Express the solution set in interval notation.
As you start exploring calculus and analysis, you will encounter multivariable calculus equations that are self-explanatory as well because all of them will contain at least two questions related to each variable being involved. See our multivariable calculus examples to receive more help and information regarding how these are used. The answers to solving these multivariable calculus questions should be based on finding the deterministic behavior. These are used in engineering and those fields where the parametric equations solver will provide you an optimal control of time dynamic systems. As an interesting subject, applying at least one equation in practice will keep you inspired!
|
Write in words how to read each of
Write in words how to read each of the following:
{x∈I ├|0<x<1┤}
{x∈R├|x≤0 or x≥1┤}
Given set
\left\{x\in {R}^{+}|0<x<1\right\}
can be written in words as:
"The set of all positive real numbers
x
x
is greater than zero but less than one'"
\left\{x\in R|x\le 0 \mathrm{or} x\ge 1\right\}
"The set of all real numbers
x
x
is greater than or equal to one (or) less than or equal to zero"
Let G be a finite group with
card\left(G\right)={p}^{2}q
p<q
two ' numbers. We denote
{s}_{q}
the number of q-Sylow subgroups of G and similarly for p. I have just shown that
{s}_{q}\in \left\{1, {p}^{2}\right\}
. Now I want to show that
\underset{S\in Sy{l}_{q}\left(G\right)}{\cup }S\setminus \left\{1\right\}={\stackrel{˙}{\cup }}_{S\in Sy{l}_{q}\left(G\right)}S\setminus \left\{1\right\}
i.e. that for
S,\text{ }T\in Sy{l}_{q}\left(G\right)
S\ne T
we that
\mathrm{S}\setminus \left\{1\right\}\cap \mathrm{T}\setminus \left\{1\right\}=\varnothing
Given you have an independent random sample
{X}_{1},{X}_{2},\dots ,{X}_{n}
of a Bernoulli random variable with parameter
p
, estimate the variance of the maximum likelihood estimator of
p
using the Cramer-Rao lower bound for the variance
So, with large enough sample size, I know the population mean of the estimator
\stackrel{^}{P}
p
, and the variance will be:
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}
Now I'm having some trouble calculating the variance of
\stackrel{^}{P}
, this is what I have so far:
since the probability function of
\stackrel{―}{X}
is binomial, we have:
{f}_{x}\left(\overline{X}\right)=\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)*{p}^{\sum _{i=1}^{n}{X}_{i}}*\left(1-p{\right)}^{n-\sum _{i=1}^{n}{X}_{i}}
\mathrm{ln} {f}_{X}\left(X\right)=\mathrm{ln}\left(\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)\right)+\sum _{i=1}^{n}{X}_{i}\mathrm{ln}\left(p\right)+ \left(n-\sum _{i=1}^{n}{X}_{i}\right)\mathrm{ln}\left(1-p\right)
\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^{n}{X}_{i}}{p}-\frac{\left(n-{\sum }_{i=1}^{n}{X}_{i}\right)}{\left(1-p\right)}=\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}
\left(\frac{\partial ln {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}=\left(\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}{\right)}^{2}=\frac{{n}^{2}{p}^{2}-2{n}^{2}p\overline{X}+{n}^{2}{\overline{X}}^{2}}{{p}^{2}\left(1-p{\right)}^{2}}
E\left[{\stackrel{―}{X}}^{2}\right]={\mu }^{2}+\frac{{\sigma }^{2}}{n}
, and for a Bernoulli random variable
E\left[X\right]=\mu =p=E\left[\stackrel{―}{X}\right]
Var\left[X\right]={\sigma }^{2}=p\left(1-p\right)
E\left[\left(\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}\right]=\frac{{n}^{2}{p}^{2}-2{n}^{2}pE\left[\overline{X}\right]+{n}^{2}E\left[{\overline{X}}^{2}\right]}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{{n}^{2}{p}^{2}-2{n}^{2}{p}^{2}+{n}^{2}\left({p}^{2}+\frac{p\left(1-p\right)}{n}\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{np\left(1-p\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{n}{p\left(1-p\right)}
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}=\frac{1}{n\frac{n}{p\left(1-p\right)}}=\frac{p\left(1-p\right)}{{n}^{2}}
However, I believe the true value I should have come up with is
\frac{p\left(1-p\right)}{n}
Find the area of the largest that can be inscribed in a semicircle of radius r.
the manager of an electrical supply store measured th diameters of the rolls of wire in the inventory. The diameter of the rolls are listed below 0.56, 0.622, 0.154, 0.412, 0.287, 0.118
find an equation in rectangular coordinates p+3=e and r-2 = 3cosΘ
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Applying the Distributed Hydrological Model for Tropical Monsoon Basins by Using Earth Observation Data (Case Studies: Kone and Ba River Basins) ()
Department of Surface Water Resources, Institute of Geography, Vietnam Academy of Science and Technology, Hanoi, Vietnam.
Thi Thanh Hang, P. (2019) Applying the Distributed Hydrological Model for Tropical Monsoon Basins by Using Earth Observation Data (Case Studies: Kone and Ba River Basins). Journal of Geoscience and Environment Protection, 7, 23-37. doi: 10.4236/gep.2019.71003.
NSE=1-\frac{{{\displaystyle \sum }}_{t=1}^{T}{\left({Q}_{s}^{t}-{Q}_{o}^{t}\right)}^{2}}{{{\displaystyle \sum }}_{t=1}^{T}{\left({Q}_{o}^{t}-\stackrel{¯}{{Q}_{o}}\right)}^{2}}
{Q}_{o}^{t}
\stackrel{¯}{{Q}_{o}}
[1] CCI Program (2017). Land Cover CCI. Version 2.0.
[2] Deng, Z. M., Zhang, X., Li, D., & Pan, G. Y. (2015). Simulation of Land Use/Land Cover Change and Its Effects on the Hydrological Characteristics of the Upper Reaches of the Hanjiang Basin. Environmental Earth Sciences, 73, 1119-1132.
[3] FAO (2006). Global Soil Maps.
[4] Fu, G. B., & Chen, S. L. (2005). Geostatistical Analysis of Observed Streamflow and It Response to Precipitation and Temperature Changes in the Yellow River. Regional Hydrological Impacts of Climate Change—Hydroclimatic Variability. IAHS Publ., 296, 238-245.
[5] Geospatial Information Authority of Japan (2012). NDVI-Data.
[6] Ishidaira, H., Ishikawa, Y., Funada, S., & Takeuchi, K. (2008). Estimating the Evolution of Vegetation Cover and Its Hydrological Impact in the Mekong River Basin in the 21st Century. Hydrological Processes, 22, 1395-1405.
[7] Khoi, D. N., & Thom, V. (2015). Impacts of Climate Variability and Land-Use Change on Hydrology in the Period 1981-2009 in the Central Highlands of Vietnam. Global NEST Journal, 17, 870-881.
[8] Nash, J. E., & Sutcliffe, J. V. (1970). River Flow Forecasting through Conceptual Models Part I-A Discussion of Principles. Journal of Hydrology, 10, 282-290.
[9] Quyen, N. T. N., Lie, N. D., & Loi, N. K. (2015). Effect of Land Use Change on Water Discharge in Srepok Watershed, Central Highland, Vietnam. International Soil and Water Conservation Research, 2, 74-86.
[10] Sun, W. C., Wang, J., Li, Z. J., Yao, X. L., & Yu, J. S. (2014). Influences of Climate Change on Water Resources Availability in Jinjiang Basin, China. The Scientific World Journal, 2014, Article ID 908349.
[11] Takeuchi, K., Hapuarachchi, P., Zhou, M., Ishidaira, H., & Magome, J. (2008). A BTOP Model to Extend TOPMODEL for Distributed Hydrological Simulation of Large Basins. Hydrological Processes, 22, 3236-3251.
[12] Takeuchi, K., Tianqi, A., & Ishidaira, H. (1999). Introduction of Block-Wise of TOPMODEL and Muskingum-Cunge Method for Hydro-Environmental Simulation of Large Ungauged Basins. Hydrological Science Journal, 44, 633-646.
[13] Tan, M. L., Ibrahim, A. L., Yusop, Z., Duan, Z., & Ling, L. (2015). Impacts of Land-Use and Climate Variability on Hydrological Components in the Johor River Basin, Malaysia. Hydrological Sciences Journal, 60, 873-889.
[14] University of East Anglia (2002). CRU CL v. 2.0.
[15] Vietnam Meteorological and Hydrological Administration. Annual Observed Data.
[16] Wang, J., Wang, H., Ning, S. W., & Hiroshi, I. (2018). Predicting Future Land Cover Change and Its Impact on Streamflow and Sediment Load in a Trans-Boundary River Basin. Proceedings of the International Association of Hydro-logical Sciences, 379, 217-222.
[17] Zhou, M. C., Ishidaira, H., Hapuarachchi, H. P., Magome, J., Kiem, A. S., & Takeuchi, K. (2006). Estimating Potential Evapotranspiration Using Shuttleworth-Wallace Model and NOAA-AVHRR NDVI Data to Feed a Distributed Hydrological Model over the Mekong River Basin. Journal of Hydrology, 327, 151-173.
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Is k[[x]] ever a finitely generated k[x](x) module? For k a field, the localization k[x]_{(x)
et3atissb 2022-04-23 Answered
Is k[[x]] ever a finitely generated k[x](x) module?
For k a field, the localization
k{\left[x\right]}_{\left(x\right)}
naturally includes into k[[x]]. I can prove that if
k=\mathbb{C},
, then this inclusion is not surjective, and k[[x]] is not even finitely generated over
k{\left[x\right]}_{\left(x\right)}
\mathbb{C}{\left[x\right]}_{\left(x\right)}
corresponds to rational functions holomorphic at 0, and
\mathbb{C}\left[\left[x\right]\right]
corresponds to germs of all functions holomorphic at 0, and so with some complex analysis you can see the finite generation is impossible. But over an arbitrary field, is this claim still true?
haarplukxjf
In fact, k[[x]] is never even countably generated as a
k{\left[x\right]}_{\left(x\right)}
To prove this, suppose k[[x]] were generated by countably many elements
{\left\{{c}_{i}\right\}}_{i\in \mathbb{N}}
k{\left[x\right]}_{\left(x\right)}
-module, and let
{k}_{0}\subseteq k
be the subfield generated by all the coefficients of the
{c}_{i}
f\in k\left[\left[x\right]\right]
, then, there exist
a,{b}_{1},\cdots ,{b}_{n}\in k\left[x\right]
a\ne 0
af=\sum _{i=1}^{n}{b}_{i}{c}_{i}
(here a is the common denominator of the coefficients when you write f as a
k{\left[x\right]}_{\left(x\right)}
-linear combination of the
{c}_{i}
f\in {k}_{0}\left[\left[x\right]\right]
, we can consider this equation as a system of linear equations with coefficients in
{k}_{0}
(coming from the coefficients of f and the
{c}_{i}
) in the (finitely many) coefficients of a and the
{b}_{i}
. A basis for the solutions of such a system can be computed by Gaussian elimination, and this process is unchanged by extending the base field. In particular, since there exists a solution to this system over k where one of the coefficients of a is nonzero, such a solution also exists over k0. This just means that f is also a
{k}_{0}{\left[x\right]}_{\left(x\right)}
{c}_{i}
{k}_{0}\left[\left[x\right]\right]
is also countably generated as a
{k}_{0}{\left[x\right]}_{\left(x\right)}
-module (by the same countably many elements
{c}_{i}
{k}_{0}
is countable, so
{k}_{0}{\left[x\right]}_{\left(x\right)}
is countable and thus so is any countably generated
{k}_{0}{\left[x\right]}_{\left(x\right)}
-module. Since
{k}_{0}\left[\left[x\right]\right]
is uncountable, this is a contradiction.
When does the produt of two polynomials
={x}^{k}
Suppose f and g are are two polynomials with complex coefficents (i.e
f,\text{ }g\in \mathbb{C}\left[x\right]
). Let m be the order of f and let n be the order of g.
Are there some general conditions where
fg=\alpha {x}^{n+m}
for some non-zero
\alpha \in \mathbb{C}
\frac{\mathbb{Z}}{2\mathbb{Z}}?
In group theory (abstract algebra), is there a special name given either to the group, or the elements themselves, if
{x}^{2}=e
for all x?
We were learning about centralizers in my abstract algebra class, and I had a thought about centralizers.
Is every subgroup of a group G the centralizer of some element? And if not, how many of the subgroups are?
Find the isomorphic ring with
{\mathbb{Z}}_{\mathbb{6}}\frac{x}{⟨{x}^{3}-x⟩}
(Chinese remainder thm)
Group structure of given
2×2
matrix group over
{\mathbb{F}}_{3}
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y\text{'}=|\frac{1}{1+{x}^{2}}+\mathrm{sin}|{x}^{2}+\mathrm{arctan}{y}^{2}||, y\left({x}_{0}\right)={y}_{0}
\left({x}_{0},{y}_{0}\right)\in \mathbb{R}×\mathbb{R}
\mathbb{R}
\sum \left[{\left(j+t\right)}^{-1}-{j}^{-1}\right]=\sum _{k\ge 1}\zeta \left(k+1\right){\left(-t\right)}^{k}
f:\mathbb{R}\to \mathbb{R}
k
{x}_{1},{x}_{2},...,{x}_{k}
\left[a,b\right]
z
\left[a,b\right]
f\left(z\right)=\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k
a<{x}_{1}<b,a<{x}_{2}<b,...,a<{x}_{k}<b
f\left(a\right)<f\left({x}_{1}\right)<f\left(b\right),...,f\left(a\right)<f\left({x}_{k}\right)<f\left(b\right)
k.f\left(a\right)<f\left({x}_{1}\right)+...+f\left({x}_{k}\right)<k.f\left(b\right)
f\left(a\right)<\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k<f\left(b\right)
f\left(a\right)<f\left({x}_{k}\right)<f\left(b\right)
I=\left[a,b\right]
a<b
u:I\to \mathbb{R}
Va{r}_{I}u=sup\left\{\sum _{i=1}^{n}|u\left({x}_{i}\right)-u\left({x}_{i-1}\right)|\right\}<\mathrm{\infty }
P=\left\{a={x}_{0}<{x}_{1}<...<{x}_{n-1}<b={x}_{n}\right\}
u
u
u
{f}_{1},{f}_{2}
{f}_{1},{f}_{2}
{f}_{1},{f}_{2}
\frac{dy}{dt}\frac{{d}^{2}x}{{dt}^{2}}=\frac{dx}{dt}\frac{{d}^{2}y}{{dt}^{2}}
Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients)
Find the general solution of the following Differential equation
y{}^{″}-2{y}^{\prime }+10y={e}^{x}\mathrm{cos}\left(3x\right)
. We know that the general solution for 2nd order Nonhomogeneous differential equations is the sum of
{y}_{p}+{y}_{c}
{y}_{c}
is the general solution of the homogeneous equation and
{y}_{p}
the solution of the nonhomogeneous. Therefore
{y}_{c}={e}^{x}\left({c}_{1}\mathrm{cos}\left(3x\right)+{c}_{2}\mathrm{cos}\left(3x\right)\right)
. Now we have to find
{y}_{p}
. I know in fact that
{y}_{c}={e}^{x}\left({c}_{1}\mathrm{cos}\left(3x\right)+{c}_{2}\mathrm{cos}\left(3x\right)\right)
. Now we have to find yp. I know in fact that
{y}_{p}={e}^{×}\frac{\mathrm{sin}\left(3x\right)}{6}
but i do not know how to get there.
\frac{dv}{dt}=10-0.1{v}^{2}
t=\frac{1}{2}\mathrm{ln}|\frac{10+v}{10-v}|
t=\frac{1}{2}\mathrm{ln}|\frac{10+v}{10-v}|
2t=\mathrm{ln}|\frac{10+v}{10-v}|
{e}^{2t}=\frac{10+v}{10-v}
±{e}^{2t}=\frac{10+v}{10-v}
Question on a differential equation
x:\mathbb{R}⇒\mathbb{R}
a solution of the differenzial equation:
5x{ }^{″}\left(t\right)+10{x}^{\prime }\left(t\right)+6x\left(t\right)=0
Proof that the function:
f:\mathrm{ℝ}⇒\mathrm{ℝ}
{\left(\stackrel{˙}{r}\right)}^{2}=\frac{2\mu }{r}+2h
Where mu and h are constants.
I have no idea how to solve it, maybe there is a trick I didn't know.
The only thing that came in mind is to integrate
\int \frac{dr}{\sqrt{\frac{2\mu }{r}+2h}}=\int dt
but I don't think this is really a solution, since I don't know too how to evaluate the integral in terms of elementary functions.
Let's say for example these two:
\frac{3n}{2n+3}{\left(-1\right)}^{n+1}-\frac{{\left(-1\right)}^{n}}{2n}{\left({\left(-1\right)}^{n}+\frac{1}{n}\right)}^{2}
Let u be a solution of the differential equation
{y}^{\prime }+xy=0
\varphi =u\psi
be a solution of the differential equation
{y}^{″}+2x{y}^{\prime }+\left({x}^{2}+2\right)y=0
\varphi \left(0\right)=1,{\varphi }^{\prime }\left(0\right)=0
\varphi \left(x\right)
\left({\mathrm{cos}}^{2}\left(x\right)\right){e}^{\frac{-{x}^{2}}{2}}
\left(\mathrm{cos}\left(x\right)\right){e}^{\frac{-{x}^{2}}{2}}
\left(1+{x}^{2}\right){e}^{\frac{-{x}^{2}}{2}}
\left(\mathrm{cos}\left(x\right)\right){e}^{-{x}^{2}}
f:\mathbb{R}⇒\mathbb{R}
be a differentiable function, and suppose
f={f}^{\prime }
f\left(0\right)=1
. Then prove
f\left(x\right)\ne 0
x\in \mathbb{R}
wuntsongo0cy 2022-04-29 Answered
Let N be a positive integer. Find all real numbers a such that the differential equation
\frac{{d}^{2}y}{d{x}^{2}}-4a\frac{dy}{dx}+3y=0
has a nontrivial solution satisfying the conditions
y\left(0\right)=0
y\left(2N\pi \right)=0
I have to solve this recurrence using substitutions:
\left(n+1\right)\left(n-2\right){a}_{n}=n\left({n}^{2}-n-1\right){a}_{n-1}-{\left(n-1\right)}^{3}{a}_{n-2}
{a}_{2}={a}_{3}=1
I have the differential equation
\frac{d}{dx}\left(p\left(x\right)\frac{df}{dx}\right)+\cdots =0
and I want to perform a generic change of variable from x to
y=y\left(x\right)
I have encountered a question that goes like
\frac{dy}{dx}={y}^{3}+x\mathrm{cos}\left(x\right),\text{ }y\left(0\right)=0.
Find the approximate solution which is
o\left({x}^{5}\right)
c\in \mathbb{R}
f:{\mathbb{R}}^{n}×\left[0,t\right)\to \mathbb{R}
\left\{\begin{array}{ll}{u}_{t}+b\cdot \left({D}_{x}u\right)+f\left(x,t\right)u=h\left(x,t\right)& \text{in }\phantom{\rule{0.167em}{0ex}}{\mathbb{R}}^{n}×\left(0,\mathrm{\infty }\right)\\ u\left(x,0\right)=g\left(x\right)& \text{on }\phantom{\rule{0.167em}{0ex}}{\mathbb{R}}^{n}×\left\{t=0\right\}\end{array}
How to you find the general solution of
\frac{dy}{dx}=x{\mathrm{cos}x}^{2}
How to solve this ordinary differential equation
tx{}^{‴}+3x{}^{″}-t{x}^{\prime }-x=0
For equation
tx{}^{‴}+3x{}^{″}-t{x}^{\prime }-x=0
, we know a special solution
{x}_{1}=\frac{1}{t}
, how to general solution?
I firstly attempted
d\left(tx{}^{″}+2{x}^{\prime }-tx\right)=0
tx2{x}^{\prime }-tx=C
, C is a constant.But in next step , I found that my solution is wrong.Since
{x}_{1}=\frac{1}{t}
is a special solution of
tx{}^{‴}+3x{}^{″}-t{x}^{\prime }-x=0
, we found
{x}_{2}=-\frac{1}{t}
is a solution of equation.Then
x={x}_{1}-{x}_{2}=\frac{2}{t}
tx{}^{″}+2{x}^{\prime }-tx=0
. As you can see ,the step is wrong.
Then I attemped other way to solve this equation ,but all failed.Could help me solve this equation?
Maximillian Patterson 2022-04-29 Answered
y{}^{″}+16y=-\frac{2}{\mathrm{sin}\left(4x\right)}
{r}^{2}+16=0
{u}_{1}\left(x\right)=\mathrm{sin}\left(4x\right),{u}_{2}\left(x\right)=\mathrm{cos}\left(4x\right)
y\left(x\right)={c}_{1}\left(x\right)\mathrm{sin}\left(4x\right)+{c}_{2}\left(x\right)\mathrm{cos}\left(4x\right)
{c}_{1}^{\prime }\left(x\right)\mathrm{sin}\left(4x\right)+{c}_{2}^{\prime }\mathrm{cos}\left(4x\right)=0
{c}_{1}^{\prime }\left(x\right)\mathrm{cos}\left(4x\right)-{c}_{2}^{\prime }\mathrm{sin}\left(4x\right)=-\frac{2}{\mathrm{sin}\left(4x\right)}
\mathrm{cos}\left(4x\right)
{c}_{1}^{\prime }\left(\mathrm{sin}\left(4x\right)\mathrm{cos}\left(4x\right)-\mathrm{cos}\left(4x\right)\mathrm{sin}\left(4x\right)\right)+{c}_{2}^{\prime }\left({\mathrm{cos}}^{2}\left(4x\right)+{\mathrm{sin}}^{2}\left(4x\right)\right)=2
{c}_{2}^{\prime }=2⇒{c}_{2}=2x
{c}_{1}^{\prime }=-\frac{2\mathrm{cos}\left(4x\right)}{\mathrm{sin}\left(4x\right)}=-2\mathrm{cot}\left(4x\right)⇒{c}_{1}=-\frac{1}{2}\mathrm{ln}|\mathrm{sin}\left(4x\right)|.
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Find an answer to any Сalculus 2 practice problems
Calculus 2 questions and answers
Recent questions in Calculus 2
Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
f\left(x\right)={\mathrm{tan}}^{-1}4x,a=0
\sum _{n=1}^{\mathrm{\infty }}\frac{{4}^{n}\left(x-1{\right)}^{n}}{n}
(a) write the repeating decimal as a geometric series, and (b) write its sum as the ratio of two integers.
0.\stackrel{―}{81}
Use Theorem Alternating Series remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001.
\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}}{{n}^{5}}
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
f\left(x\right)=\text{cosh}\left(2x-2\right),a=1
Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{{n}^{3}+2n}}
Determine whether the series converges or diverges and justify your answer. If the series converges, find its sum
\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n-2}}{{4}^{n+1}}
Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series).
\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}
Determine the convergence or divergence of the series.
\sum _{n=1}^{\mathrm{\infty }}\mathrm{ln}\left(\frac{n+1}{n}\right)
Determine the radius of convergence of this series.
\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{2n+1}}{2n+1}
Find the Maclaurin series for using the definition of a Maclaurin series. [Assume that has a power series expansion.Do not show that Rn(x) tends to 0.] Also find the associated radius of convergence.
f\left(x\right)=\left(1-x{\right)}^{-2}
Use the Limit Comparison Test to determine the convergence or divergence of the series.
\sum _{n=1}^{\mathrm{\infty }}\frac{5}{{4}^{n}+1}
f=xy-4ze+35
Find the sum of the convergent series.
\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{5}{\right)}^{n}
Find the Taylor series centered at zero for the function
f\left(x\right)=\mathrm{ln}\left(2+{x}^{2}\right)
. Determine the radius of convergence of this series.
Does the series
\sum _{n=1}^{\mathrm{\infty }}\frac{n+5}{n\sqrt{n+3}}
converges of diverges
\sum _{n=1}^{\mathrm{\infty }}\left(1+\frac{1}{n}{\right)}^{n}
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion.
{R}_{n}\left(x\right)\to 0
f\left(x\right)=2\mathrm{cos}\left(x\right),a=3\pi
Find the series
{y}^{\mathrm{sin}x}={x}^{{\mathrm{cos}}^{2}x},f\in d:\frac{dx}{dy}
When you are dealing with any Calculus 2 homework, it is vital to have a look at the various questions and answers that will help you see whether you are correct in your approach to finding solutions. Even if you are dealing with analytical aspects of Calculus 2, it will be helpful as you are looking at provided equations and learn how the answers relate to original questions and problems specified.
Do not be afraid to take a look at the basic integration and related application if Calculus 2 does not sound clear or start with the Calculus 1 first.
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Let \(\displaystyle{x}^{{\ast}}{\left({s}\right)}={\left({\frac{{{1}}}{{{\left({s}+\mu_{{1}}+\mu_{{2}}\right)}{\left({s}+\hat{{\lambda}}_{{2}}\right)}{\left({s}+\lambda_{{1}}+\lambda_{{2}}\right)}}}}\right)}\) be the laplace transform in question, where
Leia Sullivan 2022-04-13 Answered
{x}^{\ast }\left(s\right)=\left(\frac{1}{\left(s+{\mu }_{1}+{\mu }_{2}\right)\left(s+{\stackrel{^}{\lambda }}_{2}\right)\left(s+{\lambda }_{1}+{\lambda }_{2}\right)}\right)
be the laplace transform in question, where
{\mu }_{1},{\mu }_{2},{\lambda }_{1},{\lambda }_{2},{\stackrel{^}{\lambda }}_{2}
are just real parameters.
How can i get its inverse by means of convolution,
x\left(t\right)={\mathcal{L}}^{-1}\left({x}^{\ast }\left(s\right)\right)
abangan85s0
You right, first thing here is a convolution
x\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{\left(s+{\mu }_{1}+{\mu }_{2}\right)\left(s+{\stackrel{^}{\lambda }}_{2}\right)\left(s+{\lambda }_{1}+{\lambda }_{2}\right)}\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_{1}+{\mu }_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\stackrel{^}{\lambda }}_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}}\right)=f\left(t\right)*g\left(t\right)*w\left(t\right)
f\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_{1}+{\mu }_{2}}\right)
g\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\stackrel{^}{\lambda }}_{2}}\right)
w\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}}\right)
Just to note, since convolution is associative and commutative, you can evaluate it in any order (which is why no brackets were needed above),
x\left(t\right)=\left(f\left(t\right)\cdot g\left(t\right)\right)\cdot w\left(t\right)=f\left(t\right)\cdot \left(g\left(t\right)\cdot w\left(t\right)\right)=\left(g\left(t\right)\cdot f\left(t\right)\right)\cdot w\left(t\right)=f\left(t\right)\cdot \left(w\left(t\right)\cdot g\left(t\right)\right)=\left(f\left(t\right)\cdot w\left(t\right)\right)\cdot g\left(t\right)\right)
Now use the formula
\left(f\cdot g\right)\left(t\right)={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau
and finish the transform.
cab65699m
By the residue theorem:
\frac{1}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}=\frac{1}{\left(a-b\right)\left(a-c\right)\left(x-a\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)\left(x-b\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)\left(x-c\right)}
{\mathcal{L}}^{-1}\left(\frac{1}{x-c}\right)={e}^{cs}
y{}^{″}\left(t\right)+12{y}^{\prime }\left(t\right)+32y\left(t\right)=32u\left(t\right)
y\left(0\right)={y}^{\prime }\left(0\right)=0
Y\left(p\right)=x=\frac{32}{\left(p\left({p}^{2}+12p+32\right)\right)}
\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
f\left(t\right)=\frac{{e}^{-5t}}{\sqrt{t}}
Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification.
2x{\mathrm{sin}}^{2}ydx-\left({x}^{2}-9\right)\mathrm{cos}ydy=0
With the aid of Laplace Transform, solve the Initial Value Problem
y\left(t\right)-{y}^{\prime }\left(t\right)={e}^{t}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=0
L\left\{3{e}^{-4t}-{t}^{2}+6t-9\right\}
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Effect of Boundary Conditions on Nonlinear Vibration and Flutter of Laminated Cylindrical Shells | Journal of Vibration and Acoustics | ASME Digital Collection
, Kluyverweg 1, 2629 HS Delft, The Netherlands
e-mail: E.L.Jansen@TUDelft.nl
Jansen, E. L. (November 12, 2007). "Effect of Boundary Conditions on Nonlinear Vibration and Flutter of Laminated Cylindrical Shells." ASME. J. Vib. Acoust. February 2008; 130(1): 011003. https://doi.org/10.1115/1.2775512
A nonlinear vibration analysis of laminated cylindrical shells is presented in which the effect of the specified boundary conditions at the shell edges, including nonlinear fundamental state deformations, can be accurately taken into account. The method is based on a perturbation expansion for both the frequency parameter and the dependent variables. The present theory includes the effects of finite vibration amplitudes, initial geometric imperfections, and a nonlinear static deformation. Nonlinear Donnell-type equations formulated in terms of the radial displacement
W
and an Airy stress function
F
are used, and classical lamination theory is employed. Furthermore, an extension of the theory is presented to analyze linearized flutter in supersonic flow, based on piston theory. The effect of different types of boundary conditions on the nonlinear vibration and linearized flutter behavior of cylindrical shells is illustrated for several characteristic cases.
buckling, laminations, mechanical stability, nonlinear equations, perturbation techniques, shells (structures), supersonic flow, vibrations
Boundary-value problems, Flutter (Aerodynamics), Nonlinear vibration, Pipes, Shells, Stress, Vibration, Buckling, Supersonic flow, Lamination
Influence of Geometric Imperfections and In-Plane Constraints on Nonlinear Vibrations of Simply Supported Cylindrical Panels
Non-Linear Free Vibration and Postbuckling of Symmetrically Laminated Orthotropic Imperfect Shallow Cylindrical Panels With Two Adjacent Edges Simply Supported and the Other Edges Clamped
Small Vibrations of an Imperfect Panel in the Vicinity of a Non-Linear Static State
Vibration of Geometrically Imperfect Panels Subjected to Thermal and Mechanical Loads
Theory and Experiments for Large-Amplitude Vibrations of Circular Cylindrical Panels With Geometric Imperfections
Theory and Experiments for Large-Amplitude Vibrations of Empty and Fluid-Filled Circular Cylindrical Shell With Imperfections
Supersonic Flutter of Circular Cylindrical Shells Subjected to Internal Pressure and Axial Compression
Recent Contributions to Experiments on Cylindrical Shell Panel Flutter
, Leyden, The Netherlands.
Nonlinear Supersonic Flutter of Circular Cylindrical Shells
Aeroelastic Stability Characteristics of Cylindrical Shells Considering Imperfections and Edge Constraint
Koiter’s Stability Theory in a Computer-Aided Engineering (CAE) Environment
Hierarchical High-Fidelity Analysis Methodology for Buckling Critical Structures
A Comparison of Analytical-Numerical Models for Nonlinear Vibrations of Cylindrical Shells
Computational Fluid and Solid Mechanics, Proc. of 2nd MIT Conference on Computational Fluid and Solid Mechanics
Free Vibration of Rotationally Symmetric Shells
Computer Analysis of Asymmetric Buckling of Ring-Stiffened Orthotropic Shells of Revolution
Nonlinear Vibrations of Imperfect Thin-Walled Cylindrical Shells
,” Ph.D. thesis, Faculty of Aerospace Engineering, Delft University of Technology, The Netherlands.
Nonlinear Vibrations of Cylindrical Shells
Nonlinear Free Vibrations of Elastic Structures
Vibrations of Geometrically Imperfect Beam and Shell Structures
, 2001, Nonlinear Vibrations of Anisotropic Cylindrical Shells, Ph.D. thesis, Faculty of Aerospace Engineering, Delft University of Technology, The Netherlands.
The Effect of Geometric Imperfections on the Vibrations of Anisotropic Cylindrical Shells
Effect of a Nonlinear Prebuckling State on the Post-Buckling Behavior and Imperfection Sensitivity of Elastic Structures
REDUCE User's Manual
, Version 3.5. RAND Publication CP78,
, Santa Monica, CA.
Buckling of Imperfect Anisotropic Cylinders under Combined Loading
,” UTIAS Report 203, Institute for Aerospace Studies, University of Toronto.
, 1972, Nonlinear Vibrations of Cylindrical Shells, Ph.D. thesis, California Institute of Technology, Pasadena.
Nonlinear Flutter of Circular Cylindrical Shell in Supersonic Flow
Field-Consistent Element Applied to Flutter Analysis of Circular Cylindrical Shells
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2.2 Brief Review of UFR Studies and Choice of Test Case
Flows around bluff bodies such as circular cylinders are among the basic flow configurations which are not yet fully understood. The field of application of this configuration is broad, e.g. turbomachinery, aeronautical engineering, or scour of bridge piers embedded in river beds.
This flow situation is highly three-dimensional. The presented data, however, are restricted to the vertical symmetry plane upstream of a circular cylinder mounted on a flat plate. The oncoming flow is fully-developed turbulent open-channel flow, and due to the deceleration by the obstacle an adverse pressure gradient occurs in the main flow direction. The approaching boundary layer causes a vertical pressure gradient at the cylinder front that leads to a downward deflection of the flow towards the cylinder-wall junction transporting high-momentum fluid. While the down-flow approaches the bottom wall, a boundary layer develops along the cylinder resulting in a pressure gradient between the stagnation point S3 (see Fig. 6) and the cylinder-wall junction. The down-flow impinges at the bottom plate at stagnation point S3 and is deflected in all directions: (i) towards the cylinder rolling up to a corner vortex V3; (ii) in the spanwise direction around the cylinder; and (iii) in the upstream direction forming a wall-parallel jet (Dargahi 1989). This jet accelerates from the point of deflection onwards and exerts a large shear stress on the bottom wall. Some parts of the down-flow that are deflected in the upstream direction are not contributing to this jet but form a main vortex: the well-known horseshoe vortex (V1) (Baker 1980). The saddle point S1 devides the fluid entrained into the main vortex from the one contributing to the near-wall jet.
Brief Review of UFR Studies and Choice of Test Case
This jet-vortex interaction is the characterizing feature of the cylinder-wall junction flow, and was therefore in the focus of numerous studies in the past. Since the configuration of a body-wall junction is of interest in various fields, the backgrounds and motivations of these studies differ likewise. Many studies focus on scour research related to the daunting task of predicting the depth of a scour hole, such as Laursen & Toch (1956), Melville & Raudkivi (1977), Roulund et al. (2005), Ettema et. al (2006), and Link et al. (2008) just to mention a few. Since the bed erosion is a highly complex process, the corresponding models vary widely in predicting the final scour depth at the cylinder front (Roulund et al. 2005; Pfleger 2011). Furthermore, the effect of time plays a major role when extrapolating empirical models for estimating the equilibrium scour depth. Baghbadorani et al. 2017 observed that common scour models underpredicted the equilibrium scour depth by at least
{\displaystyle 29\%}
and therefore suggest to take a time factor into account to reduce this error.
Dargahi (1989), Escauriaza & Sotiropoulos (2011), Apsilidis et al. (2015), and Schanderl et al. (2017), for example, studied the flow around a circular cylinder while focussing on the turbulence structure of the horseshoe vortex system. This system consists of a set of vortices that interact with each other and the number of individual vortices that appear depends on the cylinder Reynolds number (
{\displaystyle Re_{D}}
) (Escauriaza & Sotiropoulos 2011).
According to Simpson (2001), the geometry of the body plays a minor role and the flow field does not significantly change for different body shapes. Therefore, the work of Martinuzzi & Tropea (1993), Devenport & Simpson (1990) and Paik et al. (2007) are mentioned here as well, who studied the flow around a prismatic or a wing body and observed mechanisms similar to those of a cylinder, for example. Furthermore, the observations of the dynamics of the wall-parallel jet showing a bi-modal probability density function of the streamwise velocity component goes back to Devenport & Simpson (1990). They found the upstream-pointing jet to be either in the back-flow or the zero-flow mode. The first describes a strong wall-parallel flow in the upstream direction in which the wall-parallel velocity component dominates the flow. In the zero-flow mode, the jet breaks and erupts the fluid away from the wall. Corresponding to the dynamics of the jet, the vortex oscillates horizontally generating large levels of turbulent kinetic energy (TKE).
The TKE distribution in the vertical symmetry plane upstream of a cylinder has a characteristic c-shaped distribution (Paik et al., 2007; Escauriaza & Sotiropoulos, 2011; Kirkil & Constantinescu, 2015; Apsilidis et al., 2015; Schanderl & Manhart, 2016). The horizontal oscillations of the HV cause mainly wall-normal fluctuations in the region of the vortex itself. On the other hand, the streamwise velocity fluctuations are concentrated at the lower branch of the c-shaped TKE and are caused by the dynamics of the jet.
The current contribution is based on the work reported in the publications listed at the beginning of the Test Case Studies section.
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Resonant circuits | Brilliant Math & Science Wiki
July Thomas, Derik K, a b, and
Resonance occurs in a circuit when the reactances within a circuit cancel one another out. As a result, the impedance is at a minimum and the current is at a maximum. Mathematically, the condition for resonance is
X_L = X_C.
Resonance allows for the maximum power output of an RLC circuit.
The current in a circuit peaks at the resonant frequency.
LC Circuit Resonance
RLC Circuit Resonance
Power at Resonance
Find the resonant frequency of an LC series circuit.
Resonance occurs when the reactance of the inductor is equal to that of the capacitor:
\begin{aligned} X_L &= X_C\\ \omega L &= \frac{1}{\omega C}\\ \Rightarrow \omega &= \frac{1}{\sqrt{LC} }. \end{aligned}
This is also the resonant frequency for an RLC series circuit.
_\square
20 \text{ s}
\frac{100}{pi^2} \text{ s}
\frac{25}{4\pi^2} \text{ s}
\frac{\pi^2}{5} \text{ s}
What is the period of an LC series circuit with inductance
L = \frac{25}{\pi^2} \text{ H}
C = 4 \text{ F}?
Resonance of an RLC circuit refers to the condition when the voltage across the inductor is the same as the voltage across the capacitor, or
{ V }_{ L } = { V }_{ C}
. As a result, the EMF of the battery is entirely consumed by the resistor and the current achieves its maximum value.
{ V }_{ L } = { V }_{ C}
\begin{aligned} I{ X }_{ L }&=I{ X }_{ C }\\ L{ \omega }_{ r }&=\frac { 1 }{ C{ \omega }_{ r } } \\ { \omega }_{ r }&=\frac { 1 }{ \sqrt { LC } }. \end{aligned}
{ \omega }_{ r }
{ X }_{ L }={ X }_{ C }
, from the formula for impedance of the circuit, we can easily derive the relation that
Z=R,
or in other words, the impedance of a circuit in case of resonance is minimum, or conversely, the current in the circuit is maximum.
_\square
This property of resonant circuits is used amazingly in television and radio sets. Quite basically, such a device can be viewed to consist of an LCR circuit in it. When it receives an electromagnetic signal of some frequency, this signal is converted into an electrical signal which tends to be the AC source for the circuit. Now, for every channel, there's a particular configuration of inductor and capacitor used. So, if the received frequency matches with resonant frequency for that particular channel, then the current in that circuit goes to maximum and the signal is said to accepted.
\frac{2\sqrt{13}}{13}
\frac{\sqrt{13}}{4}
\frac{9}{4}
\frac{3}{2}
\frac{1}{2}
An RLC circuit with components related by
R=\sqrt{\frac{L}{C}}
is tuned to half of its resonant frequency. What is the ratio of the current to the maximum current?
ELI the ICE man is a mnemonic to determine if an RLC circuit is mainly inductive or capacitative. If voltage
E
leads current
I
, the circuit is inductive; if
I
E
the circuit is capacitative.
ELI the ICE man is a mnemonic used to remember the relationship between the inductor and capacitor in an RLC circuit. [1]
We know that the average power of any LCR circuit can be given by
\bar { P } ={ V }_\text{rms}{ I }_\text{rms}\cos { \phi }.
But, for a circuit in which the inductive and capacitive reactance are equal, it can be easily inferred that
\phi=0
\cos { \phi }=1
\bar { { P }_{ r } } ={ V }_\text{rms}{ I }_\text{rms}.
Z
attains its minimum, the current in the circuit
{ I }_\text{rms}
attains its maximum. Hence, real power in case of a resonating circuit is maximized.
10\text{ A}
5\, \Omega,
2\text{ mH},
20\text{ pF}.
\omega
{V}_\text{rms}.
\frac {\omega} {{10}^{7}} + {V}_\text{rms}.
Wikieditor4321, . Lagging-leading. Retrieved June 29, 2016, from https://commons.wikimedia.org/wiki/File:Lagging-Leading.jpg
Cite as: Resonant circuits. Brilliant.org. Retrieved from https://brilliant.org/wiki/resonant-circuits/
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