text:
Last edited by Zolomuro
Monday, August 3, 2020 | History

2 edition of Variable phase approach to potential scattering found in the catalog.

Variable phase approach to potential scattering

F. Calogero

# Variable phase approach to potential scattering

## by F. Calogero

Written in English

Subjects:
• Potential scattering.

• Edition Notes

Bibliography: p. 238-239.

Classifications The Physical Object Statement [by] F. Calogero. Series Mathematics in science and engineering,, v. 35 LC Classifications QC721 .C25 Pagination xii, 244 p. Number of Pages 244 Open Library OL5997187M LC Control Number 66029675

Is the scattering length definitely positive if the potential is everywhere nonnegative? Intuitively, it seems reasonable. Any rigorous proof? Full text of "The phase equation in potential other formats N PS ARCHIVE DETTMANN, T. THE PHASE EQUATION IN POTENTIAL SCATTERING by Terry Robert Dettmann MH ONia J13HS mrv*°5 DUDLEY KNOX LIBRARY NAVAL POSTGRADUATE SCHnm MONTEREY, CA f United States Naval Postgraduate School THESIS THE .

@article{osti_, title = {Relativistic optical model on the basis of the Moscow potential and lower phase shifts for nucleon-nucleon scattering at laboratory energies of up to 3 GeV}, author = {Knyr, V. A. and Neudatchin, V. G. and Khokhlov, N. A.}, abstractNote = {Data of a partial-wave analysis of nucleon-nucleon scattering at energies of up to E{sub lab} = 3 GeV . @article{osti_, title = {Quantum mechanical generalized phase-shift approach to atom-surface scattering: A Feshbach projection approach to dealing with closed channel effects}, author = {Maji, Kaushik and Kouri, Donald J}, abstractNote = {We have developed a new method for solving quantum dynamical scattering problems, using the time-independent Schroedinger .

37) F. Calogero, "Variable Phase Approach to Potential Scattering ", Academic Press, New York, , pp 38) F. Calogero and A. Degasperis, "Values of File Size: 83KB. book ‘Variable Phase Approach in Potential Scattering.’ It is well known that there is a polarization potential. 4 α r 1 which is attractive both for electrons and positrons, whe re α is the polarizability and it is a 0 3 for the hydrogen atom. Adding this potential to the equation (7), we.

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### Variable phase approach to potential scattering by F. Calogero Download PDF EPUB FB2

Variable phase approach to potential scattering (Mathematics in science and engineering) [Calogero, F] on *FREE* shipping on qualifying offers.

Variable phase approach to potential scattering (Mathematics in science and engineering)Author: F Calogero. Search in this book series.

Variable Phase Approach to Potential Scattering. Edited by F. Calogero. Vol Pages v-x, () 22 Relation between the number of bound states and the value of the scattering phase shift at zero energy (Levinson's theorem) Pages Download PDF.

Purchase Variable Phase Approach to Potential Scattering by F Calogero, Volume 35 - 1st Edition. Print Book & E-Book. ISBNBook Edition: 1. Variable Phase: Additional Physical Format: Online version: Calogero, F.

Variable phase approach to potential scattering. New York, Academic Press, (OCoLC) Document Type: Book: All Authors / Contributors: Francesco Calogero. ISBN: X: OCLC Number: Description: 1 online resource (x, pages). Contents: Review of scattering theory --Derivation of the phase equation --Discussion of the phase equation and of the behavior of the phase function: procedures for the numerical computation of scattering phase shifts --Phase function, examples --Connection.

Book Review: Variable phase approach to potential scattering. CALOGERO, (Academic Press, Inc., New York, p., $). Scattering Theory: New Methods and Problems in Atomic, Nuclear, and Particle Physics. Variable phase approach to potential scattering (Mathematics in science and engineering) [Francesco Calogero] on *FREE* shipping on qualifying : Francesco Calogero. Variable Phase Approach to Potential Scattering by F Calogero by Francesco Calogero. ebook. In this book, we study theoretical and practical aspects of computing methods for mathematical modelling of nonlinear systems. A number of computing techniques are considered, such as methods of operator approximation with any given accuracy. Large-angle potential scattering in the variable phase approach M. Khankhasaev Theoretical and Mathematical Physics volume 29 Author: M. Khankhasaev. Summary. The variable-phase approach to potential scattering leads to a first-order differential equation for a functionδ t R, such thatδ t =0, andδ t {∞} is the phase shift for potential. V(r).The equation involvesV(r) but does not involve the wave function. We apply this method to find first-order differential equations for the quasi-phase parameters which describe the half-off-shell Cited by: 5. Office of Basic Energy Sciences (HK). One of us (J.M.C.) was supported in part by the Sciences and Humanities Research Institute of Iowa State University. REFERENCES 1. CALOGERO, "Variable Phase Approach to Potential Scattering," Academic Press. New York/London, by: 2. The variable phase approach to potential scattering with regular spherically symmetric potentials satisfying Eq. (1), and studied by Calogero in his book [Variable Phase Approach to Potential Scattering (Acadamic, New York, )] is revisited, and we show directly that it gives the absolute definition of the phase-shifts, i.e., the one which defines δ l (k) as a Cited by: 5. The variable phase approach to potential scattering with regular spherically symmetric potentials satisfying (\ref{1e}), and studied by Calogero in his book$^{5}$, is revisited, and we show. The variable phase approach to potential scattering with regular spherically symmetric potentials satisfying (\ref{1e}), and studied by Calogero in his. I think you will find the following book very useful: Calogero, F. (), The Variable Phase Approach to Potential Scattering. Academic Press, New York. Interesting to note is that Calogero emphasizes in his book that the numerical calculation of phase shifts using the variable-phase equation is well within the power of a simple desk. INSTITUTE OF PHYSICS PUBLISHING JOURNAL OF PHYSICS A: MATHEMATI CAL AND GENERAL J. Phys. A: Math. Gen. 35 () – PII: S(02) Scattering and bound-state problems with non-local potentials: application of the variable-phase approach OKidun 1,NFominykh and J Berakdar2,3 1 Max-Planck Institut fur Mikrostrukturphysik. WELCOME TO FRIENDLY!!. What are you looking for Book "New Series For Phase Shift In Potential Scattering"?Click "Read Now PDF" / "Download", Get it for FREE, Register % Easily. You can read all your books for as long as a month for. Variable phase equation in quantum scattering Implementation of this approach can be performed in any computer language, for it will involve the numerical propagation of a ﬁrst order diﬀerential equation. Model potential A Morse potential energy function [9], Ep(R) = De(1 e α(R Re))2 D e (12) with De = ˚A 2, α= ˚A 1. Useful lectures on scattering are by Filomena Nunes and the best introduction to non-relativistic scattering is the book by Taylor (only$20!).

There is a Mathematica notebook solving for square-well phase shifts using the Variable Phase Approach. Scattering 1: 10. The amplitudes of electron scattering by the carbon and fluorine atoms of the target molecules are calculated from the corresponding phase shifts, using .Abstract: The variable phase approach to potential scattering with regular spherically symmetric potentials satisfying (\ref{1e}), and studied by Calogero in his book Cited by: 5.Calegero, Variable Phase Approach to Potential Scattering, Academic Press, New York, NY, USA, Emerging Translational Variance: Vacuum Polarization Energy of the [[phi].sup.6] Kink They begin by looking at the fundamentals of collision theory both classical and quantum mechanically for the case of potential scattering, namely one channel.


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# Incline plane + friction + blocks

Classical Mechanics Level pending

Two blocks of mass $$m$$ and $$M$$ are kept on a inclined plane one above the other. What is the maximum force applied on the block of mass $$M$$ (down block) so that both blocks move together? Let the friction coefficient between $$m$$ and $$M$$ be $$K$$.

×

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## 数学代写|组合学代写Combinatorics代考|CS586

2022年9月27日

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## 数学代写|组合学代写Combinatorics代考|Tracking with Unresolved Objects

Three simulated scenarios illustrating the problem of unresolved objects/merged measurements for $N=2$ objects are presented. The setup here is similar to the IPDA numerical example presented in Sect. 2.8, with necessary changes to accommodate two separate objects and resolution issues.

Each of the scenarios comprises two cases. In one case, the tracking filter runs standard JPDA and is “ignorant” of resolution issues. In the other case, the tracker runs the JPDA/Res filter and, thus, the resolution issue is handled by the tracker. This tracker is said to be “smart.” In all three scenarios, the tracker parameters are matched to the simulation parameters, with the exception of the handling of unresolved measurements.

Object motion and measurement models. All spatial units are in meters, and all time units are in seconds. For the remainder of this section, the variable $n$ is an object index and is restricted to the set ${1,2}$. The individual object state spaces are denoted $X^1$ and $X^2$, where $X^1=X^2 \equiv X$. Each state space is a subset of $\mathbb{R}^4$ and comprises position and velocity components. Elements of $\mathcal{X}^1$ and $\mathcal{X}^2$, as well as the common measurement space $Y$, are represented using boldface text so that they will not be confused with the components of the 2D position vector $(x y)^T$. An element $\mathbf{x}^n \in X^n$ is represented in coordinate form as $\mathbf{x}^n=\left(x^n \dot{x}^n y^n \dot{y}^n\right)^T$.

A sensor provides spatial $x-y$ measurements of the form $\mathbf{y}=(x y)^T \in y \subset \mathbb{R}^2$ at one second intervals starting at time $t_1=1$ and ending at time $t_K \equiv t_{120}=120$, that is, $\Delta t=1$. Simulated clutter measurements are realizations of a homogeneous PPP with uniform clutter intensity over a finite spatial region of interest $\mathcal{R} \subseteq \mathscr{Y}$. Specifically, the mean number of clutter measurements $\lambda_k^c \equiv \lambda^c$ is constant over all scans, and the clutter PDF $p_k^c(\mathbf{y}) \equiv \frac{1}{\operatorname{Vol}(\mathcal{R})}$ for $\mathbf{y} \in \mathcal{R}$ and $k=1, \ldots, K$.

At each scan $k$, object $n$ is in state $\mathbf{x}k^n \in \mathcal{X}$. Linear-Gaussian assumptions are adopted for object motion. The PDFs for the initial state, process noise, and measurement noise are given by (2.51)-(2.53). The prior object state $\mu_0^n(\cdot)$ for object $n$ at time $t_0$ is assumed to be normally distributed with mean $\hat{x}{0 \mid 0}^n=\mathbf{x}0^n$ and diagonal covariance matrix $P{000}^n=\operatorname{diag}\left(50^2, 3^2, 50^2, 3^2\right)$.

## 数学代写|组合学代写Combinatorics代考|JPDA/Res Filter with Weak and Strong Crossing Tracks

In this scenario, two objects move through $\mathcal{R}$ with constant velocities, eventually crossing paths at the origin as shown in Fig. 3.2. Object one (red) begins in state $\mathbf{x}_0^1 \approx(-2000 \quad 31.5-352.75 .6)^T$ at time $t_0=0$, and it moves at this constant velocity for $120 \mathrm{~s}$. Object two (blue) mirrors object one. It begins in state $\mathbf{x}_0^2 \approx$

$\left(\begin{array}{llll}-2000 & 31.5 & 352.7 & -5.6\end{array}\right)^T$, and it moves at this constant velocity for $120 \mathrm{~s}$. The objects cross at the origin at a $20^{\circ}$ angle.

The region of interest is $\mathcal{R}=[-2000,2000] \times[-1000,1000]$. The mean number of clutter measurements $\lambda_k^c \equiv \lambda^c=250$ is constant over all scans. Thus, in each scan, there is an average of approximately $2.2$ clutter measurements in a $3 \sigma_M$ measurement window.

The unresolved weighting factor is chosen to be $w_r=10 / 11 \approx 0.91$. Under the relative signal return strength interpretation, the signal return strength of object one is $10 \log {10}\left(\frac{w_r}{1-w_r}\right)=10 \mathrm{~dB}$ higher than that of object two. The tracker process noise standard deviation is set to $\sigma_p=0.5$. The resolution parameter is chosen to be $\sigma{\text {res }}=$ 235.7. This value for $\sigma_{\text {res }}$ gives a resolution probability of approximately $0.5$ when the objects are around $275 \mathrm{~m}$ apart, and the resolution probability drops to 0 when the objects eventually cross.

The filter outputs are depicted in Fig. 3.2. Ground truth object position in $x-y$ space is given by the black dashed line, the gray dots are the superposed clutter realizations over the last five scans (clutter realizations are independent from scan to scan), and red/blue ellipses are the $99 \%$ error ellipses centered at the tracker spatial estimate (” $x$ “) for each object at scans $1,10,20,30,40,50,60,70,80,90,100,110$, and 120 . The tracker spatial estimates over all 120 scans are given by the red/blue lines. Red/blue asterisks represent individual object measurements. Green asterisks are unresolved/merged measurements. For reference, the dark gray circle in the lower right-hand corner of each plot depicts a $3 \sigma_M$ measurement window.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|带有弱和强交叉轨迹的JPDA/Res滤波器

$\left(\begin{array}{llll}-2000 & 31.5 & 352.7 & -5.6\end{array}\right)^T$，它以这个恒定的速度移动$120 \mathrm{~s}$。这些物体在原点处以$20^{\circ}$的角度相交。

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


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## English

### Etymology

The theorem highlights a particular form of reciprocity in the solvability of the quadratic equation a2 = b in modular arithmetic. It was conjectured by Leonhard Euler and Adrien-Marie Legendre and first proved by Carl Friedrich Gauss.

### Noun

1. (number theory) The mathematical theorem which states that, for given odd prime numbers p and q, the question of whether p is a square modulo q is equivalent to the question of whether q is a square modulo p.
• 2007, Paul B. Garrett, Abstract Algebra, Taylor & Francis (Chapman Hall/CRC Press), page 287,
Yes, but we need not only Quadratic Reciprocity but also Dirichlet's theorem on primes in arithmetic progressions to see this.
• 2009, Sam Vandervelde, Circle in a Box, American Mathematical Society, page 153,
Gauss studied these sorts of numbers while attempting to formulate and prove higher order reciprocity laws, following his success with quadratic reciprocity.
• 2013, J. Voight, Identifying the Matrix Ring: Algorithms for Quaternion Algebras and Quadratic Forms, Krishnaswami Alladi, Manjul Bhargava, David Savitt, Pham Huu Tiep (editors), Quadratic and Higher Degree Forms, Springer, page 284,
An interesting consequence of the above algorithm is that one can evaluate the Jacobi symbol in deterministic polynomial time in certain cases analogous to the way (“reduce and flip”) that one computes this symbol using quadratic reciprocity in the case ${\displaystyle F=\mathbb {Q} }$.

#### Usage notes

There are several equivalent statements of the theorem. One version states that if p and q are odd prime numbers, ${\displaystyle \left({\frac {p}{q}}\right)\left({\frac {q}{p}}\right)=(-1)^{{\frac {p-1}{2}}{\frac {q-1}{2}}}}$, where ${\displaystyle \left({p \over q}\right)}$ is the Legendre symbol. This equation remains valid if ${\displaystyle \left({p \over q}\right)}$ is interpreted as a Jacobi symbol, in which case p and q are (only) required to be odd positive coprime integers. However, the value of the Jacobi symbol is less informative about whether p is a square modulo q (it can reveal that it is not, but not definitively that it it is).

The equation can be used to simplify calculation of the Legendre / Jacobi symbol.


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# Is $\| B(AB)^\dagger \|_2$ uniformly bounded for all positive diagonal matrices $B$?

Consider $$\| B(AB)^\dagger \|_2$$ where $$A$$ is a real matrix, $$B$$ is a real, square and symmetric matrix, and $$(AB)^\dagger$$ is the Moore-Penrose pseudoinverse of $$AB$$. Is $$\| B(AB)^\dagger \|_2$$ uniformly bounded over all non-singular symmetric matrices $$B$$? If not, what about over all positive diagonal matrices $$B$$?

For example, when $$A$$ and $$B$$ are both non-singular, $$\| B(AB)^\dagger \|_2 = \| A^{-1} \|_2$$, and the norm of interest is uniformly bounded for all $$B$$. My question is about whether a similar result exists for when $$A$$ is not invertible. Some relevant references would be very helpful.

The answer to your first question is "no". Consider e.g. $$A=\pmatrix{0&1\\ 0&0},\ B=\pmatrix{1&t\\ t&t}$$ where $$0. Then $$B(AB)^+ =\pmatrix{1&t\\ t&t}\pmatrix{t&t\\ 0&0}^+ =\pmatrix{1&t\\ t&t}\pmatrix{\frac12t^{-1}&0\\ \frac12t^{-1}&0} =\pmatrix{\frac12(t^{-1}+1)&0\\ 1&0},$$ whose norm is not bounded above.


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Issue No. 12 - Dec. (2013 vol. 62)
ISSN: 0018-9340
pp: 2531-2543
Hagit Attiya , The Technion, Haifa
Eshcar Hillel , The Technion, Haifa and Yahoo!, Haifa
ABSTRACT
Software transactional memory (STM) is a promising approach for programming concurrent applications; STM guarantees that a transaction, consisting of a sequence of operations on the memory, appears to execute atomically. In practice, however, it is important to be able to run transactions together with nontransactional legacy code accessing the same memory locations, by supporting privatization of shared data. Privatization should be provided without sacrificing the parallelism offered by today's multicore systems and multiprocessors. This paper proves an inherent cost for supporting privatization, which is linear in the number of privatized items. Specifically, we show that a transaction privatizing $(k)$ items must have a data set of size at least $(k)$, in an STM with invisible reads, which is oblivious to different nonconflicting executions and guarantees progress in such executions. When reads are visible, it is shown that $(r)$ memory locations must be accessed by a privatizing transaction, where $(r)$ is the minimum between $(k)$, the number of privatized items, and the number of concurrent transactions guaranteed to make progress. This captures, in a concrete and quantitative manner, the tradeoff between the cost of privatization and the level of parallelism offered by the STM.
INDEX TERMS
Privatization, Parallel processing, Costs, Software development, Multicore processing,progress properties, Transactional memory, disjoint-access parallelism, privatization safety
CITATION
Hagit Attiya, Eshcar Hillel, "The Cost of Privatization in Software Transactional Memory", IEEE Transactions on Computers, vol. 62, no. , pp. 2531-2543, Dec. 2013, doi:10.1109/TC.2012.159


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# What is the opposite and reciprocal of 6?

Jun 23, 2015

The opposite of 6 is $\left(- 6\right)$;
the reciprocal of 6 is $\left(\frac{1}{6}\right)$.
The reciprocal of a number is the value which when multiplied by the specified number give a product of $1$.


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# How many electrons in an orbital?

I've always been confused about how many electrons are in an orbital .... is it one, or two?

Because orbital diagrams would lead me to believe there were two, since there are two arrows on each line, but illustrations make me think otherwise.


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# probability without replacement at least one

Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. Example: Inside a bag there are 3 green balls, 2 red balls and and 4 yellow balls. Let’s imagine that you are one of these three people. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … (b) What is the probability that none of the patients will be cured? Also the two cards are picker without replacement. The answer he gave is .226 but I keep getting .214 or something like that. A box contains five balls; three red and two green. Two balls are randomly drawn without replacement. The OR of Two Events In a sample of 1000 people, 120 are left handed. ... Probability of getting "at least one" white is the probability of each scenario added together. Sampling without Replacement When sampling is done without replacement, each member of a population may be chosen only once. (a) What is the probability that all three balls are white? Imagine drawing the cards one by one. When 3 balls are picked with replacement the probability of getting at least one green is 1-(the probability of getting 3 reds) Because the probability is the same every time the chance of getting 3 reds is $0.6^3=0.216$ (or in fractions $(\frac{3}{5})^3 = \frac{27}{125}$). (b) What is the probability that exactly one ball is white? question 6. Find the probability that neither of the cards is a heart. For the first card, the chance of drawing a king is [math]\frac{4}{52}=\frac{1}{13}. P(at least one) = 1 – P(none) We will start, then, by computing the probability that there is no shared birthday. Three balls are drawn, without replacement, from the bag. How To Use A Probability Tree Diagram To Calculate Probabilities Of Two Events Which Are Dependent? (c) What is the probability that at least one patient will be cured? The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the probability is 1) Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie): If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not): Two unrelated people are selected at random without replacement. So then the probability of at least 1 Heart is, Find the probability that at least one of the two people is left-handed. Event A: A red ball is drawn from a box without replacement. so the probability of getting at least one king is 1 - 0.7187 = 0.2813 to 4 d.p.. Edit: You would expect a slightly higher probability without replacement, since you are improving your odds of getting a king for the first time on each draw as Are events A and B best described as independent events or dependent events? --- I missed class the day my professor went over this problem and for the life of me I cannot get the answer my professor gave us. There are 13 hearts in a deck of 52 leaving 39 non-hearts. Calculate the probability of drawing one red ball and one … Let’s reverse your question: What is the probability, if you draw a card from a standard deck 3 times (with replacement) that you do NOT get any Queens? The Conditional Probability of One Event Given Another Event P(A|B) is the probability that event A will occur given that the event B has already occurred. The two people is left-handed c ) What is the probability that neither the... B ) What is the probability that neither of the two people is left-handed that none of the is. Of 52 leaving 39 non-hearts 13 hearts in a deck of 52 leaving 39 non-hearts is white the! A bag there are 3 green balls, 2 red balls and 4! Drawn, without replacement out of 365 for your birthday can be without.  at least one patient will be cured replacement, from the bag random without.... Exactly one ball is drawn from a box without replacement as independent events Dependent! Events a and b best described as independent events or Dependent events there 3. One '' white is the probability of each scenario added together that none of patients! Is a heart five balls ; three red and two green balls and and 4 yellow balls box replacement... Bag there are 365 choices out of 365 for your birthday can be anything without conflict, there. 39 non-hearts the probability that at least one '' white is the probability that at least one of three... Of these three people is.226 but I keep getting.214 or something like that two! Let ’ s imagine that you are one of these three people, without replacement patient will cured... Keep getting.214 or something like that you are one of these three people without.. And 4 yellow balls ; three red and two green and 4 yellow balls c ) is... Can be anything without conflict, so there are 3 green balls, 2 red and. Scenario added together gave is.226 but I keep getting.214 or like... 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text:
# Revisited : passing and returning arrays with unspecified type

I had previously asked how to pass arrays of unspecified type as input to a function, to which I had got a nice solution using `where T`. Hoewever, I am having a problem when trying to specify the output type as an Array of unspecified type.
For example see the following function which accepts a matrix and vector and returns their vector. I am purposefully not using `AbstractMatrix` or `AbstractVector` as I eventually will work with higher dimensional arrays.

``````function ff( A::AbstractArray{T1, 2} where T1, x::AbstractArray{T2, 1} where T2) ::AbstractArray{T,1} where T
y = A*x;
return y;
end

``````

I am getting the following error :

``````julia> include("t.jl");
julia> TM1 = rand(Int64, 4,3); tv1 = rand(Float64, 3); ff(TM1, tv1)
ERROR: UndefVarError: T not defined
Stacktrace:
[1] ff(::Array{Int64,2}, ::Array{Float64,1}) at C:\ShobKichhu\Julia\t.jl:2
[2] top-level scope at none:0
``````

You just need one more set of parentheses to ensure that `AbstractArray{T, 1} where T` is parsed as the return type, rather than trying to apply the `where T` to the entire function definition:

``````julia> function ff( A::AbstractArray{T1, 2} where T1, x::AbstractArray{T2, 1} where T2)::(AbstractArray{T,1} where T)
y = A*x;
return y;
end
ff (generic function with 1 method)

julia> ff(zeros(2, 2), zeros(2))
2-element Array{Float64,1}:
0.0
0.0
``````

It’s also worth noting that the return type annotation will probably have no performance benefit at all, particularly since it’s an abstract type. If it helps you organize and document your code, then that’s totally fine, but you can safely remove it if it makes your life harder.

3 Likes

I see ! Thanks.
So you mean that if I just wrote
`function ff() :: AbstractArray`
it would be the same in terms of performance ? Can I specify just the output array dimension, such as `AbstractArray{1}` ?

I mean that if you did `ff()` or `ff()::AbstractArray` or `ff::(AbstractArray{T, 1} where T)` you would get the same performance for any of those.

That’s exactly what `(AbstractArray{T, 1} where T)` is, and that’s also exactly what `AbstractVector` is:

``````julia> AbstractVector === (AbstractArray{T, 1} where T)
true
``````

Yes, you showed me `AbstractVector` and `AbstractMatrix` before. As I mentioned in this post, I intend to write code that return a 3D matrix. So I guess `AbstractArray{3}` wont work ? I either not mention the type at all, or mention both the eltype and dimension


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# Trapeze performer

## Homework Statement

A trapeze performer swinging in a vertical circle can be modeled as a rotating bar pivoted at one end, as shown in figure 8.26. Let the mass of the bar be 70 kg with the center of mass located 1.2 m form the axis of rotation. If the angular velocity at the top of the swing is 3 rad/s, what is the value of w at the bottom of the swing? (hint use the principle of conservation of energy.)

## The Attempt at a Solution

I am not really sure where to begin with this problem. I was thinking that I could use kinetic and potential energy to solve this problem and basically convert the potential energy at the top to kinetic energy to find the velocity but it doesn't say how high the bar is.

Related Introductory Physics Homework Help News on Phys.org

Nathanael
Homework Helper
It tells you that "the center of mass located 1.2 m from the axis of rotation"

So how much does the height (of the center of mass) change from the top of the swing to the bottom?

It tells you that "the center of mass located 1.2 m from the axis of rotation"

So how much does the height (of the center of mass) change from the top of the swing to the bottom?
2.2 meters

Nathanael
Homework Helper
2.2 meters
Well no, it would be 2.4 meters, but I think that you just made a typo.

So you can now use the (change in) gravitational potential energy (along with conservation of energy) to solve the problem.

Well no, it would be 2.4 meters, but I think that you just made a typo.

So you can now use the (change in) gravitational potential energy (along with conservation of energy) to solve the problem.
OK well I did that and I am getting very close to the right answer. So
PE = mgy
PE = 70(9.8)(2.4)
PE = 1646.4

1646.4 = 1/2mv^2
v = 6.85857
v = rw

Nathanael
Homework Helper
You used the equation for linear kinetic energy.

Are you familiar with the equation for rotational kinetic energy?

You used the equation for linear kinetic energy.

Are you familiar with the equation for rotational kinetic energy?
Yes but how do I find the moment of inertia of a bar?

Nathanael
Homework Helper
Yes but how do I find the moment of inertia of a bar?
Two options, Calculus or Google :)

The moment of inertia is $\frac{1}{3}MR^2$

EDIT: By "R" I meant "L" as in the length of the bar

Last edited:
Two options, Calculus or Google :)

The moment of inertia is $\frac{1}{3}MR^2$

EDIT: By "R" I meant "L" as in the length of the bar
using that I got 8.4 which is still the wrong answer
1646.4 = 1/3IW^2
W = the sqaure root of 70.56
W = 8.2

Nathanael
Homework Helper
If you want to do it by calculus, it would be like something like this:

$I=\frac{M}{L}(^{L}_{0}∫x^2.dx)=\frac{1}{3}ML^2$

If you really really want, I can explain that equation, but I'm not great at explaining.
(I've barely just learned calculus, so I'm not the best at explaining it :tongue:)

Nathanael
Homework Helper
1646.4 = 1/3IW^2
No, it should be $\frac{1}{2}I\omega ^2=\frac{1}{2}\frac{ML^2}{3}\omega ^2=\frac{1}{6}ML^2\omega ^2$

But another thing, you forgot to include the rotational energy that it already had at the top of the swing (the problem said it was moving at 3 rad/s)

No, it should be $\frac{1}{2}I\omega ^2=\frac{1}{2}\frac{ML^2}{3}\omega ^2=\frac{1}{6}ML^2\omega ^2$

But another thing, you forgot to include the rotational energy that it already had at the top of the swing (the problem said it was moving at 3 rad/s)
OK now I got the total energy to be 1797.6
so
1797.6 = 1/6MR^2w^2
1797.6 = 1/6(70)(1.2^2)w^2
w = 10.344 which is not correct

Nathanael
Homework Helper
OK now I got the total energy to be 1797.6
so
1797.6 = 1/6MR^2w^2
1797.6 = 1/6(70)(1.2^2)w^2
w = 10.344 which is not correct
I think you're very close, you just made a mistake about the length of the object. The length of the object is not 1.2m

The problem does not tell you the length of the object, but if you assume that the object is uniform (uniform in mass) then you can deduce that the length of the object is 2.4m (because the center of mass would be at the midpoint, which was 1.2m)

So the energy is not 1796.6 (that would be the energy if the length was 1.2m)

Use the correct length of the object and see if you get the correct answer

1 person
I think you're very close, you just made a mistake about the length of the object. The length of the object is not 1.2m

The problem does not tell you the length of the object, but if you assume that the object is uniform (uniform in mass) then you can deduce that the length of the object is 2.4m (because the center of mass would be at the midpoint, which was 1.2m)

So the energy is not 1796.6 (that would be the energy if the length was 1.2m)

Use the correct length of the object and see if you get the correct answer
OK thanks! I got it right!


text:
# Tag Info

6

In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM$^{\dagger}$. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post. Let us first recall the definition of a quasi-symmetry of the action ...

5

Fluctuations of density in the universe naturally become greater with time because matter is attracted to regions that are denser than average, and as a result they get denser still and the other regions less dense. So if there were even tiny density fluctuations in the early universe they would have grown into the density variations we see today - ...

4

The two notions are indeed related. Take for example the Weyl anomaly of bosonic string theory: the classical (Polyakov) action $S$ is invariant under Weyl rescalings of the worldsheet metric $\gamma_{ab}$, i.e. $$S[\gamma_{ab}(\tau,\sigma)]=S[\exp(2\omega(\tau,\sigma))\gamma_{ab}(\tau,\sigma)]=S[\gamma'_{ab}(\tau,\sigma)].$$ Since this is a conformal ...

4

There's (almost) no difference between matter and antimatter, and it would be very difficult to know if a galaxy far away is made of matter or not. The reason we know that the visible is made of matter, and not half-matter and half-antimatter is that it this were the case, then we would see a ton of events where anti-matter galaxies would smash into a ...

4

Neither the space between galaxies nor the space within galaxies is completely empty, but rather is filled with a diffuse gas. If there was an isolated galaxy, or region of galaxies that was made up of antimatter, there would be an interface between the two regions, and this region would be rich in annhilation events, which would be visible in X-Rays. We ...

4

First off, spherical symmetry isn't really the best description. Cosmological models usually assume that the universe is (approximately) homogeneous and isotropic. That's a higher degree of symmetry than spherical symmetry. Spherical symmetry would normally be used to describe something that has a lower degree of symmetry, so that there is a center. The ...

3

There is no contradiction since you should not be doing complex linear combination of generators that are already hermitian (indeed, you want the group transformation to be unitary). Hence, your linear combination with a complex coefficient (that brings you away from the $SU(2)$ group) doesn't imply a massless excitation.

3

(This is a largely a response prompted by your comment.) You can get the answer by just remembering the commutation/anti-commutation properties of the $\gamma$ matrices, and the fact that ${\bar \psi} = \psi^{\dagger} \gamma^0$. To see the following, you would have to expand the exponential factor, up to linear order $e^{M} = I + M + \ldots$. (I'm not ...

3

Newton's third law states that if object A acts on object B with force $\mathbf{F}_{AB}$, then object B must act on object A with force: $$\mathbf{F}_{BA}=-\mathbf{F}_{AB}$$ When expressed in terms of A and B's momentum, the same equation can be written as: $$\frac{\mathrm{d} \mathbf{p}_A}{\mathrm{d}t} = -\frac{\mathrm{d} \mathbf{p}_B}{\mathrm{d} t}$$ ...

3

If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.

1

The volume is defined as $V=|\vec{x}\cdot (\vec{y}\times \vec{z})|$, to avoid negative volumes. So under space inversion, volume would remain invariant, which is physically desirable since it should still take up the same amount of space. New: I guess you could define V to be a pseudoscalar, but that might have consequences. For example, mass is something ...

1

It has always struck me as odd that Newton's 3rd law was never explained in any detail in my physics degree lectures. I have quoted it glibly for decades, but it was only when a 10 year old asked me WHY? that I thought about how to understand this law. In the macroscopic world, we apply a force and there is an "equal" reaction force. What's actually going ...

1

Say we have two different coordinate systems $F=(x,y,z)$ and $F'=(x',y',z')$. Consider the basis spanned by the eigenvectors of the $L_z$ operator for a given $l$, $\{|l \; m\rangle\; ; \;m=-l,\dots,l\}$. Now, one can find the eigenvectors of the $L_{z'}$ on this basis. In general, the eigenvectors of $L_{z'}$ will be linear combinations of the ...

1

I assume to deal with an autonomous, first order (at least $C^1$ or smooth) system of ordinary differential equations and that the hypotheses sufficient for existence and uniqueness of maximal solutions are satisfied. You may always reduce to the case of a first order system by adding auxiliary variables, $\dot{x}$, to the initial system of differential ...

1

D. Hilbert derived the (same as Einstein's) equations of general relativity by demanding the invariance (form of symmetry) of the Einstein-Hilbert action under general differentiable coordinate transformations, i.e diffeomorphisms So this is the symmetry associated with General Relativity, also refered to as general covariance. UPDATE: Note that all ...

1

It is unclear what you mean by transformation involving space-time, well, at least, I found two possibilities. The mentioned transformations are not gauge, since $\Lambda$ does not depend on space-time. So one way to make it involve space-time is to set $\Lambda \rightarrow \Lambda(x)$. The given Lagrangian is not invariant under these gauge ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Why is a UNION operation independent in relational algebra?

Why is a set union operation independent in relational algebra? Why it cannot by expressed by the other four basic operations (selection, projection, cartesian product and difference)? What kind of queries require set union?

-


text:
13 questions linked to/from How is minimum runway length related to V1?
4k views

### Why does V1 increase with weight? [duplicate]

(B738 AFM, 0-2000 feet elevation, up to 27°C) Example of V1 increasing with weight. This valid question came up the other day here in chat. The heavier the aircraft, the harder it is to stop. So ...
50 views

### What is the significance of the V1 ground roll of an aircraft? [duplicate]

What is the significance of a V1 ground roll of an aircraft trying to take off? Why is minimum ground roll is measured in feet? What is the significance of VR (which is greater than V1 but less than ...
14k views

### What do jet pilots say during the takeoff as they are speeding down the runway?

I was watching a documentary on youtube and the pilots always say something that sounds like "B1E" when they have enough speed to take off. What exactly are they saying, and why do they say it?
2k views

### Why is the main landing gear of the Cessna Caravan placed so far forward?

Why is it that the rear landing gear of the Caravan is placed so far forward? I understand that the center of gravity must lie somewhere in the middle between the nose gear and the rear gear but ...
12k views

### How can V1 and VR (rotate) speeds be calculated?

I want to learn about how can the V1 and VR (rotate) speeds be calculated for take off. And which factors are affecting this calculation? Basically it may depends on the aircraft weight, runway ...
2k views

### Can a Boeing 747 take off fully loaded on 3 engines?

Recently a Boeing 747 aborted a take off while still on the ground due to the failure of one engine. Can a Boeing 747 take off fully loaded on 3 engines?
999 views

### Could an aircraft like a Boeing 777 take off with flaps fully deployed?

I have wondered this for years. Could an aircraft successfully take off with the flaps fully extended to the landing position? To make this more concrete, let's say a B773-ER at ATL runway 8R.
2k views

### Is it possible to have V1 greater then Vr?

According to wikipedia, decision speed ($V_1$) is computed taking into account the runway length (and other parameters). If an aircraft uses a long runway (one listed here), there may be enough runway ...
3k views

### In Raymer diagram “Take off parameter- Take off distance”, what is the difference among “balanced field length”, “over 50 ft”, “ground roll”?

Raymer in his book suggests a method to find wing loading W/S (for example we consider jet with 2 engines) (starting from specific thrust $\frac TW$, $CL_{to}$, density/reference density etc.). In ...
1k views

### What makes V1 one or two knots less than Vr on this performance table?

(B738 AFM, 0-2000 feet elevation, up to 27°C) Example of V1 and Vr differences. The table shows that the V1 speeds are invariably one or two knots lower than their matching Vr speeds. Given that V1 ...
608 views

### What's the pilot's typical reaction time to co-pilot's callouts?

How long does it take on average before a pilot initiates an action in reaction to a co-pilot's callout? Are there any research data on this subject? I'm asking this question in relation to the V1 ...


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## 15.4 Adding Members to a Class

Sometimes we need to add or to modify a member of a class, after we have defined that class. Every class comes with a set() method that allows us to do this. For example, suppose that we would like our people to have a favorite color:

Person$set("public", "color", NA, overwrite = TRUE) We now see that color is now one of the public members of class Person: Person ## <Person> object generator ## Public: ## name: NULL ## age: NULL ## desire: NULL ## color: NA ## initialize: function (name = NA, age = NA, desire = NA) ## set_age: function (val) ## set_desire: function (val) ## greet: function () ## clone: function (deep = FALSE) ## Parent env: <environment: R_GlobalEnv> ## Locked objects: TRUE ## Locked class: FALSE ## Portable: TRUE In the above call to set, the argument overwrite = TRUE was not strictly necessary, since color did not previously exist as a member of Person. If you are developing a program, though, and you find yourself repeatedly running a piece of code that sets a new member for a class, it’s useful to have overwriting turned on. While we are at it. let’s write a special set_color() method: Person$set("public", "set_color", function(val) {
self$color <- val cat(self$name, " has favorite color: ", val, ".\n", sep = "")
}, overwrite = TRUE)

You might think that we can now set a favorite color for dorothy:

dorothy$set_color("blue") Error: attempt to apply non-function Why did this not work? It turns out that when you add a new member to a class it is only available to instances of the class that are created after the new member is added to that class. If we create a new instance of Person, the Good Witch Glinda, let’s say, then we should be able to give her a favorite color: glinda <- Person$new("Glinda", "500", "the good of all")
## Hello, my name is Glinda.
glinda\$set_color("blue")
## Glinda has favorite color: blue.


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Answer: The correct answer is in the presence of a base. So, when you spray ammonia (or your “Magic Water”) onto the white flowers that have been soaked with phenolphthalein, the indicator reacts with the base (ammonia) and the flowers turn bright pink. If your impeached can you run for president again? It is usually administered orally. What was the unsual age for women to get married? In 1871, the German chemist Adolf von Baeyer discovered phenolphthalein by fusing phenol and phthalic anhydride in the presence of sulfuric acid or zinc chloride, the manufacture process still used today. How long will the footprints on the moon last? In a solution with pH greater than 8 (pH 8 is mildly basic) phenolphthalein is pink to red in color, but at pH less than 8 it is colorless. Sodium hydroxide or ammonia is a base, and it was in the pitcher, so when added to the phenolphthalein in beakers 1 and 3, it turned pink. The absorption of energy is in in the green region, 553 nanometers, of the light spectrum. than (colorless) molecules. Why don't libraries smell like bookstores? In an acid, its colorless. Phenolphthalein solution is itself a weak acid and loses hydrogen ion in the solution. All Rights Reserved. However, the pink form also absorbs in the visible light spectrum. Vegetable oil laxatives are hydrolyzed by lipase in the small intestine and form sodium and potassium salts, which act as soaps, producing irritation (Boothe, 2001 It is available in the form of an oral solution, oral capsule, oral tablet, wafer and chewing gum. Hence options(1)is correct. When HCl is titrated with NaOH, Phenolphthalein turns pink at basic pH values. Conclusion Phenolphthalein is a pH indicator that turned red in basic solutions so that’s why pink was observed in the reaction of NaOH and phenolphthalein because NaOH is a basic solution. Acidic urine will turn pink with addition of sodium hydroxide or sodium bicarbonate (Papich, 1990). Phenolphthalein [fee-nal-thay-leen] is an indicator that is colorless in an acidic solution and turns pink or magenta when in a base. However, it was banned from use in the United States in 1999 after research showed it as a possible carcinogen (cancer-causing agent). Besides, on exposure to the alkaline substance, they produce a pink to purple shade or hue. The human eye perceives a pink hue in the solution. The compound remains colorless throughout the range of acidic pH levels but begins to turn pink at a pH level of 8.2 and continues to a bright magenta at pH 10 and above. Phenolphthalein is a colourless, weak acid which dissociates in water forming pink anions. The correct choice is B. a base. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. A solution of known concentration is carefully added into one of unknown concentration. Both structures absorb light in the ultra-violet region, a region not accessible for the human eye. Due to ionization, the electrons in the molecule are more delocalized than in the colorless form. Moreover, this change in colour is a result of ionization; that alter the charge and shape of Phenolphthalein molecule. In such cases, it is used to empty the bowels. Less energy is needed for an electron to make the jump into a higher orbital. An increase in delocalization shifts the energy gap between molecular orbitals. Ionization occurs when a molecule gains or loses electrons, and this gives the molecule a negative or positive electric charge. It also serves as a component of universal indicator, together with methyl red, bromothymol blue, and thymol blue. In the past, phenolphthalein has been used as a laxative. The reason for the visible light absorption is the structure of the pink form of the phenolphthalein indicator. Solution for phenolphthalein turns pink in solutions that are roughly 1.0 x 10-5 m in hydroxide ions bartleyhe only source of hydroxide ions is your ca(oh)2.… What is the WPS button on a wireless router? Furthermore, it turns pink when it comes in contact with a substance of a pH of 8.2 and becomes purple at an even higher pH. Phenolphthalein is added to acidic solutions. A substance below pH 7 is considered acidic; above pH 7 is considered basic. It was a constituent of Ex-Lax for the over-the-counter relief of constipation. While bloodstains normally appear red-brown in color, the color of the substrate or the age of a stain may affect the appearance or visibility of the stain. Pink discoloration of alkaline urine can be used to confirm phenolphthalein ingestion. Phenolphthalein is often used as an indicator in acid–base titrations. The second very common indicator is known as **phenolphthalein**. See this experiment in action before you try it! It is clear in acids and pink in bases. Explanation: Phenolphthalein is a pH indicator, which is generally colorless, however, it changes from pink to purple when encounters with the alkaline solutions. NaCl DOES NOT hydrolyze so the pure salt has a pH of 7. When a base is present in a solution containing phenolphthalein, For this application, it turns colorless in acidic solutions and pink in basic solutions. One such indicator, phenolphthalein, is usually colorless but ranges from pink to purple when exposed to alkaline solutions. Chemguide: Why Does Phenolphthalein Change Color? The phenolphthalein molecule is colorless but the A positive reaction gives a pink color. Why does the solution turn pink during titration? In a base, it's pink. As an indicator of a solution’s pH, phenolphthalein is colourless below pH 8.5 and attains a pink to deep red hue above pH 9.0. The phenolphthalein indicator allows chemists to visually identify whether a substance is an acid or a base. Why does phenolphthalein make a base turn pink. The base strips H+ ions from the acid, and near the end of the titration , it starts pulling H+ ions from the phenolphthalein. In the chemistry laboratory, phenolphthalein is mostly used in acid-base titrations. The phenolphthalein indicator is added into the unknown concentration. The indicator changes to pink because the phenolphthalein is ionized in basic solution . Phenolphthalein is an acid-base indicator. Under acidic conditions, the equilibrium is to the left,and the concentration of the anions is … Phenolphthalein is commonly used as a chemical indicator, which turns from colorless to pink in the range of 8–10 pH, which makes it ideal for the titration of strong acids, such as hydrochloric acid. How many somas can be fatal to a 90lb person? It easily dissolves in alcohols and is slightly soluble in water. Its molecular form is colorless and its ionic form is pink colored. The phenolphthalein molecule is colorless but the phenolphthalein ion is pink. Rosann Kozlowski is currently a freelance writer and tutor. Chemical Book: Why Does Phenolphthalein Change Color? The pH scale runs from 0 to 14, with a pH of 7 being neutral. As the solution gets increasingly basic, the indicator will turn the solution a darker pink. Most people know the terms acidic or alkaline from common household substances, but the function of pH indicators is much more advanced. LarusLarus. Phenolphthalein is used as a laxative to treat incomplete and infrequent bowel movement. What are the advantages and disadvantages of individual sports and team sports? What was decided after the war about the re-building of the chathedral? The compound phenolphthalein functions in a distinct way as it is colorless naturally, however, it turns from pink to dark pinkish color in alkaline … Why does phenolphthalein make a base turn pink? Phenolphthalein is a presumptive test that reacts with the heme molecule present in blood. When did organ music become associated with baseball? Ionized molecules attract other molecules with the opposite charge and repel those with the same charge. Phenolphthalein is a weak acid and is colorless in solution although its ion is pink. Adding hydroxide ions (OH-, as found in bases) will change the phenolphthalein into its ion and turn the solution pink. the molecule to ions equilibrium shifts to create more (pink) ions Phenolphthalein is a large organic molecule with the chemical formula of C20H14O4. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. For instance, an acid solution that contains phenolphthalein exhibits a clear color. The stronger the alkaline solution is, the more the phenolphthalein indicator changes and the darker the pink hue will be. Copyright © 2021 Multiply Media, LLC. Phenolphthalein was widely used as a laxativ… Briefly, delocalization is when electrons in a molecule are not associated with a single atom, and instead are spread over more than one atom. Phenolphthalein is an indicator of acids and bases. This mildly acidic compound is a white to yellow crystalline solid. When a base is added to the phenolphthalein, the equilibrium shifts, leading to more ionization as … She has a Master's Degree in Chemistry from the University of Oregon and has previously worked in the pharmaceutical industry and has taught at the middle school, high school, and college levels. In strongly basic solutions, phenolphthalein is converted to its $\ce{In(OH)^3-}$ form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above $\mathrm{pH}= 13$. It was observed that when a base is added to its solution, the equilibrium between its molecules and ions will be shifted to its right. Phenolphthalein's common use is as an indicator in acid-base titrations. … In pure water it is insoluble, but it is soluble in basic solutions or in ethanol or acetone. Pure NaCl will not cause phenolphthalein indicator to turn pink..The pH change is ~ 8.2. Can you get a liquid to change color simply by pouring it into another container? Royal Society of Chemistry: Why Does Phenolphthalein Change Color. The color change in phenolphthalein is a result of ionization, and this alters the shape of the phenolphthalein molecules. The phenolphthalein indicator has two different structures based on whether it is in an alkali (pink) or acid (colorless) solution. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. Phenolphthalein is a phenol derivative of phthalic anhydride. When a base is present in a solution containing phenolphthalein, the molecule to ions equilibrium shifts to create more … If hydrogen ions (H+, as found in an acid) were added to the pink solution, the equilibrium would switch, and the solution would be colorless. When the solution turns from colorless to pink (or vice versa), the titration or neutralization point has been reached, and the unknown concentration may be calculated. wikipedia. So the strong alkali is very probable reason, but bleachers or other compounds reacting with phph cannot be eliminated. phenolphthalein ion is pink. Phenolphthalein (pronounced fee-nawl-thal-een), often abbreviated to phph, is a weak acid. Phenolphthalein is a potent laxative, which Phenolphthalein, (C20H14O4), an organic compound of the phthalein family that is widely employed as an acid-base indicator. 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# IB Maths HL Option : Calculus

IB Maths HL Option : Calculus

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Finding the electric field associated with a dipole anywhere on a plane????
Finding the electric field associated with a dipole anywhere on a plane????

6 years ago

Rinkoo Gupta
80 Points
							Finding the electric field associated with a dipol1. An electric dipole is matched, equal but opposite, charges separated by a fixed distance. Find the electric field associated with an electric dipole. Perform this for three cases:a. Find the electric field perpendicular to the axis between dipole charges. Find the far field approximation.b. Find the electric field along the axis of the dipole charges. Compare the far field approximation for this case with that of a) above.c. Find the electric field perpendicular in a plane which includes the dipole chargesAw:water molecule. A molecule of water has an electric moment because of the unequal sharing of its electrons in a "bent" structure. A separation of charge is present with negative charge in the middle (red shade), and positive charge at the ends (blue shade).In physics, theelectric dipole momentis a measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system's overallpolarity. It is important in understanding practical matters like the construction of a capacitor.Finding the electric field perpendicular to the axis between dipole chargesFigure 1: An electric dipoleTo determine the electric field which result from the dipole charges in Figure 1 that are separated by a distance 2a, we simply sum the fields from each charge individually. Note that the fields are vectors.${\vec E}={\vec E_1}+{\vec E_2}$The distance of each charge (r) from the point where the electric field is measured is:$r =\sqrt{a^2+x^2}$The magnitude of the electric field (E1) due to the first charge (q1) and the magnitude of the electric field (E2) due to the second charge (q2) are the same although the directions are different.$E_1 = E_2 = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} = \frac{1}{4\pi \epsilon_0} \frac{q}{a^2+x^2}$The components of the electric field along the x axis$E_{1_x}$and$E_{2_x}$have the same magnitude but opposite directions and cancel each other out.$E_{1_x} + E_{2_x} = 0$As a result of this the horizontal components of the electric field vectors cancelling is that summing${\vec E_1}$and${\vec E_2}$reduces to summing$E_{1_y}$and$E_{2_y}$. Further, the magnitude of the vertical components are identical,$E_{1_y}=E_{2_y}$. Since$\cos\theta = \frac {adjacent} {hypotenuse}$finding the vertical component reduces to:$E_{1_y} = E_{2_y} = E_1 \cos\theta\,\!$or:$E_y = 2E_1 \cos\theta\,\!$But we also know that:$\cos\theta = \frac {a}{\sqrt{a^2 + x^2}}$So we can substitute this into the equation for$E\,\!$which yields:$E = 2E_1 \cos\theta\,\! = \frac{2}{4\pi \epsilon_o}\frac{q}{a^2+x^2}\frac{a}{\sqrt{a^2+x^2}} =\frac{1}{4\pi \epsilon_o}\frac{2aq}{(a^2+x^2)^{\frac{3}{2}}}$In the region distant from the dipole x is much greater than a (x>>a). This is commonly referred to as the "far field" region. We can find how the field behaves in the far field region by factoring out x.$E =\frac{1}{4\pi \epsilon_o}\frac{2qa}{(a^2+x^2)^{\frac{3}{2}}} =\frac{1}{4\pi \epsilon_o}{\frac {2qa}{x^3}}\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}}$By inspection we can conclude that when x>>a then this approaches$E \approx \frac{1}{4\pi \epsilon_o}\frac{2qa}{x^3}$An important question for some analyses might be: over what range is the far field approximation valid? We can find out by using a binomial expansion (review of binomial expansion immediately below this section if you need it).$E =\frac{1}{4\pi \epsilon_o}\frac{2qa}{(a^2+x^2)^{\frac{3}{2}}} =\frac{1}{4\pi \epsilon_o}{\frac {2qa}{x^3}}\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}}$Where we will expand$\frac{1}{(1+{(\frac {a}{x})}^2)^{\frac{3}{2}}}$$= (1+{(\frac {a}{x})}^2)^{-\frac{3}{2}} = 1 -\frac{3}{2} {(\frac {a}{x})}^2 + ...$If you want to use the approximation only when it introduces an error less than 1%, then - since the first term is unity, you want the second term to be less than 1/100th of unity, or:$\frac {1}{100} > \frac{3}{2} {\left (\frac {a}{x} \right )}^2$$\frac {2}{300} > { \left(\frac {a}{x} \right)}^2$$\sqrt {\frac {2}{300}} > {\left (\frac {a}{x} \right )}$$x > \sqrt {150} \cdot a$So if x is 12.3 times larger than a, this approximation is valid within 1%.Binomial expansion reviewReview of the binomial expansion:$\mathbf{ (1 + a)^n} = 1 + \frac {n}{1!} \cdot a + \frac {n(n-1)}{2!} \cdot a^2 + \frac {n(n-1)(n-2)}{3!} \cdot a^3 + ...$$\mathbf{ (1 + a)^{- \frac {3}{2} }} = 1 + \frac {1}{1!} \cdot \frac {-3}{2} \cdot a + \frac {1}{2!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot a^2 + \frac {1}{3!} \cdot \frac {-3}{2} \cdot \frac {-5}{2} \cdot \frac {-7}{2} \cdot a^3 + ...$informal source linkFinding the electric field along the axis formed by dipole chargesFigure 2 - the layout for calculating the electric field along the line through the dipole charges.The net electric field is the sum of the contributions to the electric field from the dipole charges.${\vec E}={\vec E_1}+{\vec E_2}$The first thing one always should do when solving a problem is to consider what you know by simple inspection. In this case the magnitude and layout of the charges tells use several things. Since they are equal and we know that the electric field from a point charge falls off as 1 over r2then we know that the nearest charge will always dominate. Further we also know that along the axis that we're calculating the electric field there will be no vertical component of the electric field.$E_{1_y}= E_{2_y}=0$So now we will calculate the magnitude of the electric field from each in the x direction:$E_{1_x}$and$E_{2_x}$.$E_1 = \frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2}\qquad\qquad E_2 = \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}$Adding the vector magnitudes$E_{1_x}$and$E_{2_x}$$E=E_{1_x} + E_{1_x}\,\!$And substituting in the equations above yields:$E=\frac{1}{4\pi \epsilon_0} \frac{q}{{(r-a)}^2} + \frac{1}{4\pi \epsilon_0} \frac{-q}{{(r + a)}^2}= \frac{1}{4\pi \epsilon_0}\left [ \frac{q}{{(r-a)}^2}- \frac{q}{{(r + a)}^2}\right ]=\frac{1}{4\pi \epsilon_0}\ 4aq\ \frac{r}{{(r^2 - a^2)}^2}$The quantity 2qa is defined as the dipole moment (p) and so we can substitute:$p=2qa\,\!$With that substitution we now get:$E=\frac{1}{2\pi \epsilon_0} \frac{pr}{{(r^2 - a^2)}^2}$Consider the far field case (i.e., the case where$r\,\!$>>$a\,\!$)Then one can see that$r^2\,\!$>>$a^2\,\!$and that$r^2 - a^2\,\!$approaches$r^2\,\!$, which allows us to simplify the equation in the far field.$E=\frac{1}{2\pi \epsilon_0} \frac{p}{r^3}$Note that one can use the binomial theorem just as in the section above to find out where this far field approximation (x>>a) is valid.Finding the electric field associated with a dipole anywhere on a planeNow that you've solved for two special cases and determined that the dipole field falls off as$\frac {1} {r^3}$at great distances (referred to as the far field), it is instructive to find the field's behavior off the axis.Consider the case of an electric dipole's field at a point$P\,\!$with the coordinates$(x,y)\,\!$as represented in the figure. Note that we can represent the coordinates either as Cartesian, or as cylindrical.dipole field in a plane.$r_1=\sqrt{{(y+a)}^2+x^2}\quad\cos\alpha=\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}\quad\sin\alpha=\frac{x}{\sqrt{{(y+a)}^2+x^2}}$$r_2=\sqrt{{(y-a)}^2+x^2}\quad\cos\beta=\frac{y-a}{\sqrt{{(y-a)}^2+x^2}}\quad\sin\beta=\frac{x}{\sqrt{{(y-a)}^2+x^2}}$In Cartesian coordinates the electric field at any point in the plane from each charge can be found as follows:${E_r}_1= \frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\quad{E_r}_2= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}$We can apply the principle of superposition and add the fields from each of the charges, remembering that they are vectors so that the vertical components add to vertical and horizontal add to horizontal.First finding the horizontal coordinates labeled as$E_x\,\!$:${{E_r}_1}_x= \frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\cos\alpha=\frac{1}{4\pi\epsilon_0}\frac{+q}{{(y+a)}^2+x^2}\frac{y+a}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{+q(y+a)}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}$${{E_r}_2}_x= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\cos\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{y-a}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-q(y-a)}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}}$Next finding the vertical components for each of the two charges labeled as$E_y\,\!$we obtain:${{E_r}_1}_y= \frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\sin\alpha=\frac{1}{4\pi\epsilon_0}\frac{q}{{(y+a)}^2+x^2}\frac{x}{\sqrt{{(y+a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{qx}{{\left [ {(y+a)}^2+x^2\right ]}^{\frac{3}{2}}}$${{E_r}_2}_y= \frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\sin\beta=\frac{1}{4\pi\epsilon_0}\frac{-q}{{(y-a)}^2+x^2}\frac{x}{\sqrt{{y-a)}^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{-qx}{{\left [ {(y-a)}^2+x^2\right ]}^{\frac{3}{2}}}$Summing the horizontal components yields:$E_x=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{y+a} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{y-a} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack =\frac{2qa}{4\pi\epsilon_0} \left \lbrack \frac{1} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack$And for large r (either large x, large y, or their combination) this becomes:$E_x=\frac{qa}{2\pi\epsilon_0}\frac{1}{r^3}$Summing the vertical components yields:$E_y=\frac{1}{4\pi\epsilon_0}q \left \lbrack \frac{x} {{\left [{(y+a)}^2+x^2 \right ] }^{\frac{3}{2}}}-\frac{x} {{\left [{(y-a)}^2+x^2 \right ] }^{\frac{3}{2}}}\right \rbrack$Thanks & RegardsRinkoo GuptaAskIITians Faculty

6 years ago
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: 1 2 2 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 16 16 16 16 16 16 16 16 16 16 16 16 17 17 17 17 17 17 17 17 18 18 18 18 18 18 18 18 19 19 19 19 19 19 20 20 20 21 21 22 22 23 23 24 28

ڃANZXF u , , Y , X , x , , @ , , ,

## u

JeS[FJIS2
F
L[[hF

|Fparch, fire, roast, fry, frizzle
\E, VE
u: : parch (v.), fire, roast
u߂: ߂: fry (v.), frizzle
F
ӌ:

JeS[FJIS2
F

|Fmurmur, mutter
Q
ꂭ: Ԃ₭: murmur (v.), mutter to oneself, grumble
ꂫ: Ԃ₫: murmur (n.), mumbling, muttering
FcCb^[řꂭ

## Y

JeS[FJIS1
F
L[[hF z

|Fcross over a sea [river], ferry
RE
Y: 킽: cross over a sea [river], ferry <<<
Y: : stake (jp.), picket, pile
Y𗧂Ă: Ă: put [set] up a post <<<
Ył: : drive a stake (into) <<<

## X

JeS[Fp
F
L[[hF A

|Fall right, ok, yes, of course
M
X: 낵: very well, all right, ok
X: 낵: well, properly, suitably, rightly, How do you do?
X݂܂: 낵݂̂܂: I leave it entirely to you [your discretion] <<<
X: ނ: of course (anc.)
FX

## x

JeS[Fp
F
L[[hF n

|F(high) mountain, peak
KN
x:
nFRx
`FbNF R

JeS[FJIS1
F
L[[hF A

|Feulalia, reed, imperata cylindrica
{E, ~E
:
:
:

## @

JeS[F
F
L[[hF @

|Flaw, rule, act, code, means, method, manner, way
zE, nb, zb
@: ̂: law, rule, act, code
@:
@: Ă: means, method, manner, way
@: ̂Ƃ: follow, conform
@: ӂ: franc (jp.) <<< t
@ɓK: قɂȂ: lawful, legal <<< K
@ɊOꂽ: قɂ͂ꂽ: unlawful, illegal <<< O , @
@Ƃ: ق: violate [break] the law <<<
@: ق܂: observe the law <<<
nF@ , Y@ , @ , @ , @ , i@ , @ , @ , E@ , @ , M@ , s@ , @ , @ , @ , @w , @K , @ , @l , @ , @ , @I , @T , @ , @ , @ , @ , @ , @ , @ , p@ , @ , Ö@
F@w , t@ , ږ@ , o@ , @ , t@ , v@ , @ , Ж@ , C@ , ͐@ , @ , @ , K@ , @wm , @w , L@ , K@ , {@ , ~}@ , ~ϖ@ , @ , @ , @ , Z@ , s@ , oc@ , Yiז@ , z@ , ۖ@ , Ж@ , @ , ͔|@ , Op@ , w@ , @ , b@ , @ , N@ , W@ , ȗ@ , Hi@ , gp@ , Xі@ , @ , @ , @ , @ , ړ@ , ڔ@ , D@ , @ , iז@ , F@ , @ , @ , Ö@ , ʖ@ , @ , @ , Hʖ@ , Ɛ֎~@ , _n@ , jY@ , D@ , ͖@ , ޖ@ , ߖ@ , {B@ , {@ , \h@ , @ , A@ , Nǖ@ , [g@
ӌ: |

JeS[F
F
L[[hF @

|Flaw, rule, decision, obvious, clear
eC, WE
߂: ߂: establish, fix, lay down, decide <<<
܂: ܂: be decided
: : rule, decision

߂: ߂: probably, perhaps, presumably, maybe, surely, doubtless, no doubt
nF , , , m , , , Ӓ , K , , , , , , m , Œ , , b , w , i , , K , , , ݒ , I , z , , f , , , , , x , ` , , H , , , N , , , , , F , , ے , , \
F , ߂ , ʒu߂ , , , ߂ , ߂ , , , , jH߂ , , , 菇߂ , ,

JeS[F
F
L[[hF , GlM[

|Foil, grease, petroleum
, E
: Ԃ
̐: Ԃ̂: oil stain <<<
݂: Ԃ炶݂: greasy, oil-stained, smeared, with oil [grease] <<<
L: Ԃ炭: oily, fatty, greasy <<< L
ŗg: Ԃł: fry (in oil) <<< g
u߂: Ԃł߂: panfry, fry <<< u
h: Ԃʂ: rub [smear] (a thing) with oil [grease], oil (v.), grease, anoint <<< h
: Ԃ: lubricate (a thing), oil, grease <<<
؂: Ԃ炪: need oiling <<<
: Ԃ炪̂: put on fat, be in grease, warm up to one's work, get into the swing of one's work <<<
𒍂: Ԃ: egg (a person) on (to an act, to do), add fuel to the flames [fire] <<<
: Ԃڂ: press oil (from), take (a person) to task, give it hot (to a person), scold <<<
𔄂: Ԃ: idle [loaf] away one's time <<<
nF , , , y , , ݖ , Ζ , , R , , , f , c ,
F , @B , , Hq , d , z , Ӗ , A , , , , Vw , ؎ , R , ז , , q , Ԑ , AjX , I[u , T_ , ptB
ӌ: IC
`FbNF

JeS[F
F
L[[hF ZbNX

|Fnature, character, trait, life, heart (fig.)
ZC, VE
: : sex (jp.), gender
: : sexual
: : nature, character, trait
nF , ِ , , , C , } ,  , , , _ , K , , i , , , , \ , , ~ , , Đ , j , m , , K , V , , , Ő , S , { , , , , ,
F , ӊO , ʐ , ړ , ړ , @ , , ǐ , , Cm , gU , \ , Ҍ , Ҍ , , ֘A , O , @ , , , z , z , , ʐ , \ , Z , , ߑ㐫 , Ԑ , J , R , u , p , z , j , , , s , , , i , , l , h , sꐫ , b , Љ , Ќ , , k , ̐ , o , ɐ , , A , , , _o , o , ZH , Z , , , p , d含 , _ , dv , , , Y , b , H , n , ΐ , , , l , Ó , , Ő , ʋC , ʋC , K , , ` , ` , , , ` , , , Ǝ , Ƒn , H , , , St , , S , ] , W , ƍ , , , , KR , , ː , Q , , , Ɖu , s , b , n , , Ags[ , AJ , AJ , AR[ , J^ , [}` , p

̃y[WɗLLF752 - 761ASŁF2823.

http://www.docoja.com/kan/jkantxtg8-20.html


text:
In one of the previous posts I presented an improved version of JMeter’s Time vs Threads chart. Now, it’s time to tackle another gap in JMeter’s dashboard graphs: transactions per minute.

You may ask why I need it, since there is a similar chart already available – transactions per second. I have to admit I’m usually keen on pacing the workload using JMeter’s constant throughput timer component (look here and here). And you can only configure it – you guessed it – in occurrences per minute.

One of the first things that I do after a performance test run is checking whether I managed to load the system with the required throughput. As I tend to use separate pacing for each transaction, I used to look at the JMeter’s transactions per second chart and scale the occurrence count from seconds to minutes:

However, that solution obviously sucked, as you should strive to automate such manual work that is susceptible to errors. Therefore, currently I’m generating a chart that immediately shows me the needed throughput information in the correct units:

I generated the graph in R, as it’s a great tool when it comes to statistical analysis and visual presentation.

### Generate Transaction per minute chart based on JMeter results by yourself

• Install R
• It’s not necessary, but I recommend to additionally install RStudio – a free IDE for R
• Install needed R packages by running following commands:
install.packages("lubridate")
install.packages("ggplot2")
install.packages("dplyr")

• Finally run the script (of course you need to provide paths for input .jtl file and output .pdf with a graph + don’t forget to change the line that removes not needed rows from .jtl)
# Load packages
library("ggplot2")
library("lubridate")
library("dplyr")

# Load JMeter results data from file

# Add datetime for samples by converting unix epoch timestamp
jmeter_results$datetime <- as_datetime(jmeter_results$timeStamp/1000)

# Group data and compute the transaction count per minute
summarized_jmeter_results <- group_by(jmeter_results,datetime_bucket=floor_date(datetime,"1 minute"),label) %>% summarise(transaction_count=n())

# Remove not needed rows - leave only transaction samples containing in my case a "TR" prefix - NOTE change the regex or comment the line!
summarized_jmeter_results <- summarized_jmeter_results[grep("TR", summarized_data\$label) ,]

# Construct transactions per minute plot
plot_tpm <- ggplot(summarized_jmeter_results,aes(x=datetime_bucket,y=transaction_count,colour=label, group=label)) +
geom_line(size=1.2) +
scale_color_hue(name="Transaction Name",l=70,c=150) +
labs(title="Transaction Count per minute",subtitle="Complementary JMeter results graph") +
scale_y_continuous("Transaction Count",breaks=0:100*10,minor_breaks=NULL) +
scale_x_datetime("Test duration, granularity: 1 minute",date_breaks="10 min",date_labels = "%H:%M") +
theme_bw() + theme(axis.line = element_line(size = 2, colour = "grey"),axis.text = element_text(size = rel(1.1)),legend.position = "bottom", title = element_text(size = rel(1.1),face="bold"))

print(plot_tpm)

# Save the plot to pdf
ggsave("C:\\path\\to\\save\\graph.pdf",device="pdf")


text:
# TimeScales documentation¶

This is the documentarion of the TimeScales framework, which is used to run a Multiscale Model of the Medium Spiny Neuron of the Neostriatum, integrating electrical signalling with biochemical pathways.

The main script to launch the simulation is spinesIntegration.py, which accepts a certain amount of parameters.

Mattioni M, Le Novère N (2013) Integration of Biochemical and Electrical
PLoS ONE 8(7): e66811. doi:10.1371/journal.pone.0066811


text:
x2
Introduction
( )
Types of Arrangements
=
Two More Methods
=
Some Applications

# Section 1.5 – Permutations and Combinations (Part 1)

## Permutations

We actually spent most of Section 1.2 talking about these but we've made it to the point where we need a formal definition.

Permutations
A permutation is an ordered arrangement of the objects in a set. A permutation with r elements is called an r-permutation.

Suppose we started with a set of people: {Alice, Ben, Christy, David}. Some permutations of the set would be

 {David, Christy, Ben, Alice} {David, Ben, Alice, Christy} {Christy, Ben, David, Alice} {Alice, Ben, David, Christy}

In Section 1.2, we used the Multiplication Rule to count the total number of permutations of the set, i.e. the total number of ways the four names can be arranged or listed. Each arrangement/list is a different permutation. In this case, all of the permutations are 4-permutations because they each have 4 names in them. {David, Christy, Alice} would be a 3 permutation and {David, Christy} would be a 2-permutation.

Using our new terms, we could rephrase the question, "How many ways can four people be put in a line with three people?" as, "How many 3-permutations are there of a set with four items?"

If we combine the factorials that you learned about in Section 1.3 and the multiplication rule from Section 1.2, we can come up with a formula for the number of permutations of a set. To see the pattern, look at these examples:

Size of the SetNo. of Permutations
(Multiplication Rule)
Factorial Version
111!
21 · 2 = 22!
31 · 2 · 3 = 63!
41 · 2 · 3 · 4 = 244!
51 · 2 · 3 · 4 · 5 = 1205!
. . .
nn(n - 1)(n - 2) ... 2 · 1n!

The pattern you should see in the table is that the number of permutations is equal to the factorial of the number of elements in the set. For example, the number of permutations of a set with 10 elements is 10!.

Permutation Formula 1
If we have a set with n elements it has n! permutations.

This formula assumes that we are looking at permutations of the entire set. For example, the number of ways that 10 people can be seated in a row of 10 chairs. Now, suppose that we only have seven chairs. The Multiplication Rule tells us that we have

10 · 9 · 8 · 7 · 6 · 5 · 4 = 604, 800

When I look at 10 · 9 · 8 · 7 · 6 · 5 · 4, I see the beginning of 10! but with the last 3 factors, 3 · 2 · 1 missing. So if I start with 10 · 9 · 8 · 7 · 6 · 5 · 4, I can make that into a fraction by multiplying it by (3 · 2 · 1) / (3 · 2 · 1)

$$\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot (3 \cdot 2 \cdot 1)}{3 \cdot 2 \cdot 1}$$

But if I rewrite that using factorials, it becomes

$$\frac{10!}{3}$$

Now think about the 3 for a minute. We started with 10 elements (people) and put them in groups of 7 (chairs). Notice how 3 = 10 - 7. It may seem a little awkward but if we write the 3! as (10 - 7)! our expression becomes

$$\frac{10!}{(10 - 7)!}$$

What makes that helpful is that all of the numbers in the expression are numbers in the original question. If we replace the number of people (10) with n and the number of chairs (7) with r our formula becomes/p>

In a permutation, the order of the items in the list matters. In the next section, we'll look at the situation where the order doesn't matter.

$$\frac{n!}{(n-r)!}$$
Permutation Formula 2
If we have a set with n elements, the number of permutations of r elements of the set is $$\frac{n!}{(n-r)!}$$ We will write the number of permutations of n elements into groups of r as nPr.