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# How to relate raw scores on five-point Likert to three conceptual categories?

My thesis is on employee productivity and I had a questionnaire which consists of a five-point Likert scale (5-strongly agree, 4-agree, 3-agree, 2-disagree, and 1-strongly disagree). On the other hand, I had a conceptual Framework of a 3 point scale (Highly Productive, Productive and Not Productive).

Is there any standard way to identify whether my respondents are highly productive, productive or not?

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I think you have an error in your description of the Likert scale, as you have 4 and 3 being identical. –  Peter Flom Oct 21 '12 at 10:21
I don't quite understand the question. Do you have data on the productivity of the respondents? If so, you can use it. If not, how could you possibly do this? –  Peter Flom Oct 21 '12 at 10:23

Many psychological tests convert numeric raw scores into categories. For example, Wikipedia mentions cut-offs for the Beck Depression Inventory:

• 0–9: indicates minimal depression
• 10–18: indicates mild depression
• 19–29: indicates moderate depression
• 30–63: indicates severe depression.

Or for example the BMI define various cut-offs (e.g., Cole et al, 2007).

In general, you lose information by collapsing categories or using cut-offs. Psychological reality tends to be more continuous. That said, categories do have heuristic value as decision aides.

A few options for converting scores to a collapsed set of categories

• Use the logical definition of the scale points: For example, you might use "strongly agree" as highly productive, "agree" as productive, and the other categories as "not productive". This is a simple approach that uses the scale anchor points to define the meaning of the categories.
• Use expert judgements: You can ask a set of experts to evaluate where they thinlk the cut-offs between categories should be. These can then be synthesised. This approach is often used to define acceptable standards for various tests.
• Use normative information: You could use information about the normative spread of the variable and an assumption about the prevalence of the phenomena to define cut-offs.
• Use prediction of external criterion: If the thing has objective existence, or if there are things related to it, you could use predictive models of this external criterion to define the categories.

### References

• Cole, T. J., Flegal, K. M., Nicholls, D., & Jackson, A. A. (2007). Body mass index cut offs to define thinness in children and adolescents: international survey. Bmj, 335(7612), 194. FULL TEXT
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text:
Applied Physics
Vol. 09  No. 05 ( 2019 ), Article ID: 30200 , 7 pages
10.12677/APP.2019.95027

Theoretical Study of the Probabilities for the Reaction H + DBr (v = 0, j = 0) on the 12A' Potential Energy Surface

Yuliang Wang*, Hui Li, Dezhi Su, Mingyu Zhou

School of Basic Sciences for Aviation, Naval Aviation University, Yantai Shandong

Received: Apr. 25th, 2019; accepted: May 7th, 2019; published: May 14th, 2019

ABSTRACT

Quasi-classical trajectory calculations were carried out for H + DBr (v = 0, j = 0) over the collision energy range 0 - 2.0 eV on the 12A' potential energy surface. The probabilities for both the exchange and abstraction channels of the reaction at total angular momentum J = 0, 10, 20, 30, 40 are predicted in this work. The attack angle dependent probabilities of H + DBr at J = 0 are calculated. It was found that the influence of attack angle on the distribution of probabilities for both two channels is significant because of the different effective potential barrier heights at different attack angles.

Keywords:H + DBr Reaction, Attack Angle, Reaction Probability, Reaction Dynamics

1. 引言

2. 计算理论

Figure 1. The definition figure of the attack angle

3. 结果与讨论

Figure 2. Minimum energy pathways of abstraction for the collinear configuration

Figure 3. Comparison of integral cross section for abstraction between present work and other theoretical results in the text

Figure 4. The reaction probabilities for exchange channel at different J values

Figure 5. The reaction probabilities for abstraction channel at different J values

Figure 6. The attack angle dependent probability of exchange channel

Figure 7. The attack angle dependent probability of abstraction channel

4. 结论

Theoretical Study of the Probabilities for the Reaction H + DBr (v = 0, j = 0) on the 12A' Potential Energy[J]. 应用物理, 2019, 09(05): 226-232. https://doi.org/10.12677/APP.2019.95027

1. 1. Aker, P.M., Germann, G.J. and Valentini, J.J. (1989) State-to-State Dynamics of H+HX Collisions. I. The H+HX→H2+X (X=Cl, Br, I) Abstraction Reactions at 1.6 eV Collision Energy. Journal of Chemical Physics, 90, 4795-4808. https://doi.org/10.1063/1.456574

2. 2. Pomerantz, A.E., Camden, J.P., Chiou, A.S., Ausfelder, F., Chawla, N., Hase, W.L. and Zare, R.N. (2005) Reaction Products with Internal Energy beyond the Kinematic Limit Result from Trajectories Far from the Minimum Energy Path: An Example from H+HBrH2 + Br. Journal of American Society, 127, 16368-16369. https://doi.org/10.1021/ja055440a

3. 3. Zhang, J.Y., Jankunas, J., Bartlett, N.C.M., Goldberg, N.T. and Zare, R.N. (2010) Search for Br Production in the D + DBr Reaction. Journal of Chemical Physics, 132, 084301. https://doi.org/10.1063/1.3319717

4. 4. Lynch, G.C., Truhlar, D.G., Brown, F.B. and Zhao, J.G. (1995) A New Potential Energy Surface for H2Br and Its Use to Calculate Branching Ratios and Kinetic Isotope Effects for the H + HBr Reaction. Journal of Chemical Physics, 99, 207-225. https://doi.org/10.1021/j100001a034

5. 5. Mielke, S.L., Tawa, G.J., Truhlar, D.G. and Schwenke, D.W. (1995) Quantum Photochemistry. Accurate Quantum Scattering Calculations for an Electronically Nonadiabatic Reaction. Chemical Physics Letters, 234, 57-63. https://doi.org/10.1016/0009-2614(94)01515-w

6. 6. Kurosaki, Y. and Takayanagi, T. (2003) Global ab initio Potential Energy Surfaces for the Lowest Three Doublet States (12A′, 22A′, and 12A′′) of the BrH2 System. Journal of Chemical Physics, 119, 7838-7856. https://doi.org/10.1063/1.1609398

7. 7. Kurosaki, Y. and Takayanagi, T. (2005) A Modified Version of the Ana-lytical Potential Function for the Global ab initio Ground-State Potential Energy Surface of the BrH2 System. Chemical Physics Letters, 406, 121-125. https://doi.org/10.1016/j.cplett.2005.02.106

8. 8. Jiang, B., Xie, C.J. and Xie, D.Q. (2011) New ab initio Potential Energy Surface for BrH2 and Rate Constants for the H + HBrH2 + Br Abstraction Reaction. Journal of Chemical Physics, 134, Article ID: 114301. https://doi.org/10.1063/1.3563750

9. 9. Quan, W.L., Song, Q. and Tang, B.Y. (2007) The Effects of the Rotational Excitation on the Br+H2 Reaction and Its Dependence on the Potential Energy Surfaces. Chemical Physics Letters, 442, 228-232. https://doi.org/10.1016/j.cplett.2007.05.105

10. 10. Zhang, W.Q., Cong, S.L., Zhang, C.H., Xu, X.S. and Chen, M.D. (2009) Theoretical Study of Dynamics for the Abstraction Reaction H′ + HBr (v=0, j=0)H′H + Br. The Journal of Physical Chemistry A, 113, 4192-4197. https://doi.org/10.1021/jp8105716

11. 11. Fu, B.N. and Zhang, D.H. (2007) A Time-Dependent Quantum Dynamical Study of the H + HBr Reaction. The Journal of Physical Chemistry A, 111, 9516-9521. https://doi.org/10.1021/jp073811z

12. 12. Xie, C.J., Jiang, B. and Xie, D.Q. (2011) State-to-State Quantum Dynamics of the H + HBr Reaction: Competition between the Abstraction and Exchange Reactions, Journal of Chemical Physics, 134, Article ID: 184303. https://doi.org/10.1063/1.3589407

13. 13. Zhang, A.J., Zhang, P.Y., Chu, T.S., Han, K.L. and He, G.Z. (2012) Quantum Dynamical Study of the Electronic Nonadiabaticity in the D + DBr→Br(Br*) + D2 Reaction on New Diabatic Potential Energy Surfaces. Journal of Chemical Physics, 137, Article ID: 194305. https://doi.org/10.1063/1.4766355

NOTES

*通讯作者。

format: plain


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M A T H 2 1 B
Mathematics Math21b Fall 2010
Linear Algebra and Differential Equations
Exhibit: damped harmonic oscillator
 The differential equation x'' = - x - a x' is a damped harmonic oscillator. The left hand side is the acceleration, the right hand side a sum of the spring force and a damping term. This system can be written as a system of first order differential equations: x' = y y' = - x - a y which is X' = A X for the matrix A = | 0 1 | | -1 -a | For a=0, the harmonic oscillator case, the eigenvalues of A are -i,i For positive a, the real part of the eigenvalues becomes negative and the eigenvalues move to the left in the complex plane. There is the moment, when the eigenvalues are both -1. This is the critically damped situation. For larger a, the eigenvalues are different. It is the overdamped situation.


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 another kind of
 Recently, there comes another kind of cross-platformed communication protocol–HBWSP (Hessian Binary Web Service Protocol). 当前,出现了另一种跨平台的通信协议HBWSP(Hessian Binary Web Service Protocol)。 短句来源 Z-Wave technology and its main features are introduced first, and its protocol stack is analyzed, and compared it with another kind of short distance wireless communication technology which named ZigBee, Then the structure and function of Z-Wave module are analyzed, and Z-Wave module of ZM2102 is introduced, Finally the prospect of Z-Wave technology is showed. 介绍了Z-Wave技术及其主要特征及协议栈,并与另一种短距无线通信技术ZigBee作出了比较,分析了Z-Wave模块的结构和功能,并对ZM2102Z-Wave模块进行了详细的介绍,最后对Z-Wave技术的应用前景进行了展望。 短句来源 Bone mesenchymal stem cells (BMSCs), which are different from hematopoietic stem cells (HSCs), are another kind of bone marrow stem cells. BMSCs can not only support the hemopoiesis by participating in the formation of hemopoietic micro-environment, but also differentiate to different marrow stromal cells which can also support hematopoiesis in the suitable conditions. 而骨髓间充质干细胞( bone marrow mesenchymal stem cells,BMSCs)是存在于骨髓中除造血干细胞(hemopoietic stem cells,HSCs)以外的另一种干细胞,不仅直接参与生成造血微环境支持造血,在适宜的条件下还可发育、分化为具有支持造血功能的多种骨髓基质细胞。 短句来源 Jacobi elliptic function expansion method is extended to construct the exact solutions to another kind of KdV equations, which have variable coefficients or forcing terms. Jacobi椭圆函数展开法被推广并用于求解另一种形式的KdV方程的新的精确解,所求解的这类KdV方程包括一种典型的变系数的KdV方程和具有强迫项(随机项)的KdV方程。 短句来源 There is a great deal of difference between Chinese national culture and British and American culture, Translation , that is to reflect another kind of language and culture sincerely and truly through one kind of language . 汉民族文化与英美文化之间存在着很大的差异 ,翻译透过一种语言忠实、准确地再现另一种语言所反映的思想和文化 短句来源 更多
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 Wang's theory for determining the approximate configurational partition function of the adsorbed layer is modified in two different ways. One is to assume that the configurational energy should be corrected: the other to advocate that the deficiency due to a wrong expression for the a priori probability of the. central site is more significant.The configurational partition function is evaluated is both methods and the adsorptipn isotherin and the beat of adsorption computed for the case of quadratic lattice... Wang's theory for determining the approximate configurational partition function of the adsorbed layer is modified in two different ways. One is to assume that the configurational energy should be corrected: the other to advocate that the deficiency due to a wrong expression for the a priori probability of the. central site is more significant.The configurational partition function is evaluated is both methods and the adsorptipn isotherin and the beat of adsorption computed for the case of quadratic lattice With dipole interaction. values for the last two quantities when a uniform continuous distribution of the distant adsorbed particles is assumed are further given for comparison. The second method, which surpasses the first, is compared with Kirkwood's method. in the case of hexagonal lattice with neighbour interaction. Numerical work is also carried out in this case. 本文就王竹溪先生之理论用两种不同方法加以改良。一种为改变能量公式,另一种为改变中心格点之或然率。用此二法皆可求得近似配分函数。所得关於方格在偶极作用情形下之粘附等温曲线及粘附热之数值用图比较之,并与旧日平均连续分布之理论比较。本文所用之第二法实较第一法为佳,故应用之以讨论一三角格在相邻作用情形下之粘附现象,且应用Kirkwood理论以为比较。数值结果亦有图表之。 The changes of pH, glucose consumption, nitrogen assimilation, aureomycin production, formation of organic acids and ammonia as well as mycelial respiration during aureomycin fermentation by Streptomyces aureofaciens were studied with two media. One medium(Ⅰ) contained as organic constituents glucose and casein hydrolysate, while to the other medium(Ⅱ) besides the mentioned ingredients meat extract and corn steep liquor were also added. The course of glucose consumption, nitrogen assimilation, ammonia formation... The changes of pH, glucose consumption, nitrogen assimilation, aureomycin production, formation of organic acids and ammonia as well as mycelial respiration during aureomycin fermentation by Streptomyces aureofaciens were studied with two media. One medium(Ⅰ) contained as organic constituents glucose and casein hydrolysate, while to the other medium(Ⅱ) besides the mentioned ingredients meat extract and corn steep liquor were also added. The course of glucose consumption, nitrogen assimilation, ammonia formation and aureomycin production did not vary considerably for both fermentation media. Aureomycin formation, however, was five times as high for medium Ⅱ as for medium Ⅰ. Acid formation and mycelial respiration differed between the two media. In experiments with medium Ⅱ two kinds of organisms were encountered. The one possessing a light coloured mycelium had an aerobic metabolism, while the other of dark mycelial colouration possessed a fermentative metabolism. The metabolic changes occuring during the fermentation period may be differentiated into three phases. The growth phase, which lasts to the second or third day, is followed by the stationary phase, which extends to the fourth day, and the phase of autolysis hereafter. Aureomycin in the medium reached maximal concentration during the second phase. 在兩種培養基中觀察了金黴菌培養過程中pH的變化、葡萄糖的消耗,氮的同化、金黴素的產生、氨和有機酸濃度的變化、菌絲的呼吸等代謝變化,這兩種培養基的區別,即在一種培養基中另加入肉湯粉和玉蜀黍漿。 兩種培養基發酵過程中,葡萄糖的消耗,氮的同化,氨和金黴素的產生等變化的一般趨勢,大致相似。加肉湯粉和玉蜀黍漿的培養基中,產生的金黴素量均為不加的5倍。 兩種培養基發酵過程中,有機酸的產生和菌絲的呼吸變化的趨勢有顯著不同。含肉湯粉和玉蜀黍漿的培養基中,培養出的菌絲有兩種類型:一種淺色的氧化代謝特強,另一種菌絲深褐色的呼吸低,代謝變化屬於發酵型,但兩種類型的金黴素的產生量是一樣的。在搖瓶內金黴菌的發酵過程,按代謝可以分為三個階段,第二天到第三天以前為“旺盛生長期”,接着到第四天止菌絲開始自溶為“開始自溶期”最後為“迅速自溶期”。培養基中,金黴素的濃度,在第二期最高。 A new method is proposed in this paper for the evaluation of the following three important kinds of three-center and four-center integrals needed in molecular quantum mechanics: (1) (2) (3) Obviously, integral (2) is a special case of integral (3), so we need only to evaluate the first and the third. The following equations have been obtained for the evaluation of the above-mentioned two kinds of integrals (4) (5) where (6) (7) (8) (9) R_0 in equation (4) is the distance of point a apart from the origin located... A new method is proposed in this paper for the evaluation of the following three important kinds of three-center and four-center integrals needed in molecular quantum mechanics: (1) (2) (3) Obviously, integral (2) is a special case of integral (3), so we need only to evaluate the first and the third. The following equations have been obtained for the evaluation of the above-mentioned two kinds of integrals (4) (5) where (6) (7) (8) (9) R_0 in equation (4) is the distance of point a apart from the origin located in bc line, while that in equation (5) is the distance between the two chosen origins separately located in ab and cd lines. θ_0 is the angle made byand, and θ_(10), θ_(20) and θ_(12) are respectively the angles made by and,and,and, The fourth term of equation (5) is, in general, negligible except in the case of R_0, which is less than two Bohr units. We propose two methods for the evaluation of and : the first one is to choose the origin at the end-point of the bond and evaluate the integral strictly inside and outside the sphere of radius, while the second one is to choose the origin at the mid-point of the bond and evaluate its value inside and outside the ellipsoid passing through the end-point of. The calculation involved in the second method is quite simple and, of course, a small error is introduced in changing the region of integration from the sphere to the ellipsoid, but it is quite negligible in comparison with the result of our first method. Equation (4) is exact in all cases, while equation (5) is exact in many cases but also involves certain errors in some other cases. From our actual calculations, we draw the conclusion that equation (5) is almost exact in the evaluation of integrals L_(aa, bc) and L_(ab, cd) and that certain error is involved in the evaluation of integral L_(ab, bc), but the error introduced does not exceed ten per cent. 在本篇文章中,我們建議了一種新方法來計算量子力學中的三中心和四中心積分;這方法此以往的好,因為計算簡單,應用廣闊,結果也比較可靠。我們用來計算三中心吸引能的公式[方程(5)]是在任何情况下都是正確的,而用來計算三中心和四中心的排斥能積分公式[方程(18)]在某些情况下是正確的,在另一些情况却能引進一些誤差。在計算非相隣鍵的積分時引進的誤差很小,可以忽略不計;在計算相隣鍵的積分時引進的誤差此較大,但不超過百分之十。我們建議兩種計算A_u和B_u的方法,一種方法是以鍵的一個端點爲原點,嚴格按照球內外的區域積分;另一種方法是以鍵的中點為原點,按照橢圓體的內外區域積分。前一種方法理論上嚴密,然而後一種方法計算簡單,收斂性快,引進的誤差也不大;尤其在計算相隣鍵的三中心排斥能的積分時,似乎後一方法得到的結果還比前一方法好。在本文中,為了容易說明起見,常常引用吸引能和排斥能這兩個名詞,實際我們的方法,是用來計算下列三類積分:它們不僅包括吸引能和排斥能積分,也把交换積分包括在內,甚至可以在更廣泛的意義上看待上列積分。若σ_1,σ_2也是Φ_1和Φ_2的函數時,仍可以用我們的一般展開理論處理,不過要此本文複雜。 << 更多相关文摘
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2008 CNKI－中国知网

2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社


text:
# Magnetic Flux Formula with Solved Problems

In this tutorial, we show you how to use the magnetic flux formula in physics to find flux through loops, solenoids, coils, and other geometries.

## Definition of magnetic flux:

The amount of magnetic field lines passing through a loop of area A which is tilted at angle $\theta$ from the field is defined as magnetic flux and is found by the below formula $\Phi_m=BA \cos \theta$ This definition is similar to that of electric field flux

The SI unit of magnetic flux is the weber which can be seen from the formula above as $1 \,weber={\rm 1\, Wb =1\, T\cdot m^2}$ $\theta$ in the magnetic flux formula is the angle between magnetic field vector $\vec{B}$ and the normal (perpendicular) $\hat{n}$ to the plane of the loop.

The above magnetic flux formula is similar to the vector scalar (dot) product of two vectors as $\vec{A}\cdot\vec{B}=AB\cos\theta$.

Thus, the equation for magnetic flux can be written as a dot product of two vectors magnetic field vector $\vec{B}$ and area vector $\vec{A}$ $\Phi_m=\vec{B}\cdot\vec{A}$ where $\vec{A}$, area vector, is defined as a vector perpendicular to the plane of the loop with magnitude equal to the area of the loop $A$.

The SI unit of the area vector is $m^2$.

One of the most important applications of magnetic flux is in Faraday's law of induction

That law states a changing magnetic flux with time creates a voltage.

Below, using the magnetic flux formula some magnetic flux problems are solved for more understanding.

## Magnetic flux formula Problems

Problem (1): A square loop of side 3 cm is positioned in a uniform magnetic field of magnitude 0.5 T so that the plane of the loop makes an angle of $60^\circ$ with the magnetic field. Find the flux passing through the square loop.

Solution: Putting the known values into the magnetic flux equation $\Phi_m=BA\cos \theta$, we get \begin{align*}\Phi_m&=BA\cos \theta\\&=(0.5)(0.03\times 0.03)\cos 30^\circ\\&=0.39\quad {\rm mWb}\end{align*} Note that the $\theta$ in the formula above is the angle between $\vec{B}$ and a unit vector perpendicular to the surface.

In this example, the given angle $60^\circ$ is with the surface of the loop not with a vector perpendicular to the surface that is $\hat{n}$.

Problem (2): A circular loop of area $200\,{\rm cm^2}$ sits in the $xz$-plane. Then, a uniform magnetic field of $\vec{B}=0.3\,\hat{i}+0.4\,\hat{j}\, {\rm T}$ applied on it. Determine

(a) Magnitude of the magnetic field
(b) Magnetic flux through the square loop?

Solution
(a) The magnitude of a vector such as $\vec{R}=R_x \hat{i}+R_y\hat{j}$ is found by formula $R=\sqrt{R_x^2+R_y^2}$ so strength (magnitude) of the magnetic field is determined as $B=\sqrt{(0.3)^2+(0.4)^2}=0.5\quad {\rm T}$

(b) This circular loop is positioned at the right angle with the $y$ axis so a unit vector perpendicular to it is written as $\hat{n}=\hat{j}$. Now we use the scalar definition of magnetic flux as $\Phi_m=\vec{B}\cdot\hat{n}A$ to find it as below \begin{align*}\Phi_m&=\vec{B}\cdot\hat{n}A\\&=(0.3\hat{i}+0.4\hat{j})\cdot (\hat{j})\left(200\times 10^{-4}\right)\\&=(0.3\,\hat{i}\cdot\hat{j}+0.4\,\hat{j}\cdot\hat{j})(0.2)\\&=0.4\times 0.2\\&=0.08\quad {\rm T}\end{align*} Where in above we used the definition of scalar product of Cartesian unit vectors. In addition, the SI unit of area, squared meter, is determined by the following conversion rule $\rm{1\, cm^2=10^{-4}\,m^2}$

Problem (3): A rectangular loop of dimensions 3 cm by 5 cm is placed perpendicular in a uniform magnetic field of magnitude 0.1 T. Find the magnetic flux through the loop.

Solution: magnetic flux is defined as the product of magnetic field, surface area, and the angle between B and a unit vector perpendicular to the surface with the formula $\Phi_m=BA\cos \theta$.

Here, the loop is positioned perpendicular to the magnetic field $B$ namely the angle between $B$ and a unit vector normal to the surface $\hat{n}$ is zero $\theta=0^\circ$. Putting the values in the magnetic flux formula, we get \begin{align*} \Phi_m&=BA\cos \theta \\&=(0.1)(0.03\times 0.05)\cos 0^\circ\\&=15\times 10^{-5}\quad{\rm Wb}\\ or &=0.15\quad{\rm mWb}\end{align*}

Problem (4): Find the magnetic flux through a 100-turn coil of a radius of 10 cm that carries a current of 3.00 A.

Solution: the magnetic flux through one loop of the coil is $\Phi_1=BA\cos \theta$ where B is the magnetic field produced by the loop.

Recall that the magnitude of the magnetic field of a loop is $B=\frac{\mu_0 i}{2 R}$ which is perpendicularly directed into or out of the plane of the loop. Thus, the angle between $\vec{B}$ and the normal to the plane of the loop is $0^\circ$.

Substituting these values into the magnetic flux equation above, we get the flux through one loop as \begin{align*} \Phi_1&=BA\cos \theta \\ \\&=\frac{\mu_0 i}{2 R}\left(\pi R^2\right)\cos 0^\circ\\ \\&=\frac{4\pi\times 10^{-7}\times 3}{(2)(0.10)}\left(\pi\times(0.1)^2\right)\\ \\&=6\pi^2 \times 10^{-8}\quad {\rm Wb}\end{align*} This is the flux passes through one loop of a coil with $N$ such loops, so the total flux through the coil is \begin{align*}\Phi_m&=N\Phi_1\\&=100\times \left(6\pi^2\times 10^{-8}\right)\\&=0.6 \quad {\rm \mu Wb}\end{align*}

Problem (5): Find the magnetic flux through a 1000-turn solenoid of radius 10 cm and a length equal to 30 cm so that a current of 3 A passes through it.

Solution: Recall that the magnetic field inside a tightly wrapped long solenoid (neglecting the end effects) is uniform with magnitude $B=\mu_0 \frac{N}{L}i$ where $N$ and $L$ are the number of turns and the length of the solenoid, respectively. \begin{align*}B&=\mu_0 \frac NL i\\&=(4\pi\times 10^{-7})\left(\frac{1000}{0.3}\right)(3)\\&=12.56\quad{\rm mT}\end{align*}

On the other hand, each loop of the long solenoid is perpendicular to its uniform magnetic field so $\theta =0^\circ$. The area of one loop is also $A=\pi r^2=\pi(0.1)^2=0.0314\quad {\rm m^2}$ Substituting the above given values into the equation for magnetic flux, we have \begin{align*} \Phi_1&=BA\cos \theta\\&=(12.56\times 10^{-3})(0.0314)\cos 0^\circ\\&=0.4\times 10^{-3}\quad {\rm Wb}\end{align*}This is the total magnetic flux through $N$ turns of a long solenoid.

Problem (6): A circular loop of a radius of 10 cm is placed perpendicular to a uniform magnetic field with an unknown strength. The magnitude of the flux through the loop is $5\times 10^{-3}\,{\rm T.m^2}$.

(a) What is the magnitude of the magnetic field?
(b) Now, rotate the magnetic field B such that it is parallel to the plane of the loop. How many magnetic field lines pass through the loop?

Solution: the product of magnetic field, surface area, and the angle between $\vec{B}$ and the direction perpendicular to the plane of the surface $\hat{n}$ gives the magnetic flux formula, $\Phi_m=BA\cos \theta$.

(a) $B$ is perpendicular to the surface, so the angle between them is zero, $\theta =0$. Substituting the values into the equation above and solving for magnetic field $B$, we get \begin{align*} \Phi_m&=BA\cos \theta\\ 5\times 10^{-3}&=B (\pi\times (0.1)^2)\cos 0\\ \Rightarrow B&=\frac{5\times 10^{-3}}{\pi \times 10^{-2}}\\ \\&= 0.16 \quad {\rm T}\end{align*}

(b) $B$ is parallel to the loop, namely, it makes an angle of $\theta=90^\circ$ with a vector perpendicular to the loop surface. In this case, the flux through the loop is \begin{align*} \Phi_m&=BA\cos \theta\\&=(0.16)(\pi\times(0.1)^2)\cos 90^\circ\\&=0\end{align*}

Problem (7): A square coil of side length $a=2\,{\rm m}$ is placed in the $xy$-plane in a uniform magnetic field of magnitude 0.1 T as shown in the figure. The direction of the magnetic field is toward the $y$ direction. Find the total magnetic flux through the coil.

Solution: magnetic flux is the scalar product of the magnetic field and the area vector with formula $\Phi_m=\vec{B}\cdot\vec{A}$. As you can see, all factors involved in this equation are vectors.

In this example, the magnetic field vector and area vectors are written as $\vec{B}=0.1\,\hat{j},{\rm T}$ and $\vec{A}=4\,\hat{i},{\rm m^2}$. Putting these vectors in the formula above and using the definition of scalar product, we get \begin{align*} \Phi_m&=\vec{B}\cdot\vec{A}\\&=(0.1\,\hat{j})\cdot (4\,\hat{i})\\&=0.4\,(\hat{j}\cdot\hat{i})\\&=0.4\times 0\\&=0\end{align*}

Problem (8): A uniform magnetic field of magnitude B is passing through the base of a hemisphere with radius R. How many field lines penetrate the spherical surface of the hemisphere?

Solution: magnetic flux is a measure of how many magnetic field lines pass through a surface which is computed by the formula $\Phi_m=BA\cos \theta$.

The magnetic flux through the base of the hemisphere is \begin{align*}\Phi_m&=B(A)\cos 180^\circ\\&=-B(\pi R^2)\end{align*} To find the flux through the spherical surface we can use directly the definition of magnetic flux in spherical coordinates (which is hard) or use Gauss's law for magnetism as $\oint\vec{B}\cdot \hat{n}\,dA=0$ According to this law, the magnetic flux through ANY closed surface is always ZERO

A hemisphere is composed of two parts, one is a flat base and the other is a spherical part. Total flux $\Phi_m$ through it, which is the sum of the fluxes through the base $\Phi_{m-b}$ and the curved surface $\Phi_{m-S}$, must be zero so the flux passing through the curved surface is obtained as below \begin{align*} \Phi_m&=\Phi_{m-b}+\Phi_{m-S}\\ 0&=-B\pi R^2 +\Phi_{m-S} \\\Rightarrow \Phi_{m-S}&=+B\pi R^2 \end{align*}

Author: Dr. Ali Nemati

Published: 2/7/2021


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# GATE Papers >> ECE >> 2011 >> Question No 14

Question No. 14

Consider the following statements regarding the complex Poynting vector $\stackrel{\to }{P}$ for the power radiated by a point source in an infinite homogeneous and lossless medium. Re$\left(\stackrel{\to }{P}\right)$ denotes the real part of $\stackrel{\to }{P}$, S denotes a spherical surface whose centre is at the point source, and $\stackrel{^}{n}$ denotes the unit surface normal on S. Which of the following statements is TRUE?

 (A)  $\mathbf{Re}\mathbf{\left(}\stackrel{\mathbf{\to }}{\mathbf{P}}\mathbf{\right)}$ remains constant at any radial distance from the source (B)  $\mathbf{Re}\mathbf{\left(}\stackrel{\mathbf{\to }}{\mathbf{P}}\mathbf{\right)}$ increases with increasing radial distance from the source (C) (D)


text:
# How to loop in a nested Association?

Considering we have an Association:

<|a -> tr, b -> <|c -> <|d -> 12, e -> del, f -> <|h -> true, i -> false|>|>|>|>

How can I iterate through it, and delete the key->val where val === del and update the values that are true or false to be True and False,so that having as a result:

  <|a -> tr, b -> <|c -> <|d -> 12, f -> <|h -> True, i -> False|>|>|>|>

Thanks!

You can use a combination of DeleteCases and ReplaceAll to accomplish this for your particular example:

data =  <|a -> tr, b -> <|c -> <|d -> 12, e -> del, f -> <|h -> true, i -> false|>|>|>|>;
DeleteCases[
ReplaceAll[data, {true -> True, false -> False}],
del,
Infinity]

<|a -> tr, b -> <|c -> <|d -> 12, f -> <|h -> True, i -> False|>|>|>|>

The Infinity, the last argument of DeleteCases, specifies that the deletion should occur at every level (and not just on the outer association). When you use DeleteCases on an association, it matches on the values of the association, so simply matching on del is sufficient in this case.

Note that if you explicitly want to check for a value being the same (=== or SameQ) as del, then you would want to replace the argument del to DeleteCases above with something like val_ /; val === del.

• And what if I would also like to check the Keys? e.g. if the key is e I don't want to do anything with that pair? – SuTron May 11 '15 at 9:15
• In that case, you'd probably want something like Association@DeleteCases[Normal[a], (Except[e] -> del)]; the Normal call just turns the association into a list of rules so you can match on the entire rule; the (Except[e] -> del) is just a pattern that matches anything but e as the lhs of the rule and del as the rhs. – nben May 11 '15 at 15:41


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# Data-security

## *Can* Users Consent to Data Mining, Ethically (Even if Legally)?

Having taken part this week in a panel discussion at the 2017 DLF Forum conference, I raise a question I’ve been struggling with: can users consent to data mining, if they can’t realistically be expected to substantively understand the risks?

## Slides and speaking notes from a Digital Library Federation Forum Panel on Data Governance and Ethics

Earlier this week, I presented as part of a panel discussion at the 2017 DLF Forum conference. I am releasing my slides and speaking notes for that presentation.

## A Self-Hosted Sync Solution for Zotero

Because of my preference not to use the Zotero server to sync my academic library, I have written a small bash (command-line) function to sync Zotero database files over any server of one’s choosing (including, e.g., OwnCloud or Dropbox).

## KeePass2 KPScript Installation on Linux

KeePass2 comes with an extension that gives it a Command-Line Interface. I found it confusing to install in Linux, and so am posting instructions here.

## A Converter for Full KeePass2 XML Export to CSV

If you use KeePass2, you may have noticed that CSV exports do not include user-defined columns. I wrote a script to convert KeePass2 XML exports to CSV, allowing CSV exporting from KeePass2.

## A Note on Hard Drives and Protected Data

If you need to store protected information on an unencrypted hard drive, even temporarily, there is a quick step that you can take to protect your data from being written possibly irrevocably to the hard drive.

## A Converter for Full KeePass2 XML Export to CSV

If you use KeePass2, you may have noticed that CSV exports only have

## Paperback is Fascinating, Exciting, and Funnily Funny

Paperback is an open-source program that was apparently created as a joke for paper-based backups, but works, and well.


text:
• Subject:

...

• Topic:

...

During an adiabatic process. the pressure of a gas is found to be proportional to the cube; of its
temperature. The ratio of CP/CV for the gas is

(a) 4/3

(b) 2

(c) 5/3

(d) 3/2

NEET


text:
Question-and-Answer Resource for the Building Energy Modeling Community
Get started with the Help page

Setting room surface reflection with OpenStudio/Radiance

How can i setting the Reflection on Radiance Simulation model? where the setting? in my model i must no using reflection of the daylight, is there any solution?

i'am using OpenStudio Radiance 1.11 to simulate the illuminance map of my object. My research is focus on the Daylight potential on the building. its a good thing, if i can costumize the Reflection Factor of my model. Is it possible?

*image explain :

----- > is full window (in sketchup, i just break the wall. without install the 3D warehouse window)

_______> is full wall, there no ventilation

For example, i upload the result of the First Zone based on image (.osm) file. https://www.dropbox.com/sh/c4tu1u5gey...

edit retag close merge delete

Sort by » oldest newest most voted

(updated 2016.07.26) It is absolutely possible to change the reflectance of the room surfaces in your OpenStudio model. This is explained in the second 'graph, below.

Again, if using Radiance directly, the simplest method is to use a plastic material, and equal RGB values to define a grey material of n reflectance, definition e.g.:

void plastic my_60%_reflective_material

0 0 5

0.6 0.6 0.6 0 0

As you state you are using OpenStudio, then you'd want to make an opaque material whose absorptance is the inverse of the reflectance you desire, and place that on the innermost layer of your construction. (This will be translated to a similar plastic material definition in the corresponding Radiance model, when you run the Radiance measure in OpenStudio.)

The safest thing to do would be to locate the material(s) you are already using for your interior room surfaces and modify their absorptances to get you the reflectance you want. The screenshot below shows the material properties for 1/2" Gypsum, from your model (thanks for posting that); it's got a Visible Absorptance of 0.40 (see lower right), meaning it's a 60% reflective material, in the visible spectrum. If you change this value to the inverse of whatever you want for visible reflectance, you're all set. You can duplicate that one as many times as you need to get all the different values you desire. Then you'd apply this material as the innermost layer in your constructions, and that's it.

Now, as for the shape of the floor plan: it's non-convex. This is problematic both for EnergyPlus (shading calculations), and unfortunately for Radiance as well, at least in the OpenStudio / Radiance implementation. The illuminance maps, which you will use for your daylight evaluation, must be rectangular; OpenStudio does not (easily) support arbitrarily shaped lighting calculation grids. You will not be able to cover the entire area of your floor plan with uniformly arrayed calculation points. You can either place illuminance maps to cover what you can and evaluate relative changes in your designs, but obviously, the daylight metrics will not be spatially representative.

Hope this helps!

more


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# Shorten curved arrow proportionally to length

I am drawing curved arrows in tikz-cd, using the method devised by @AndréC in this answer. There are situations where I want to shorten these arrows, proportional to their arc length. However, the naïve solution (using shorten) is not suitable, as it creates distorted curves (as discussed, for instance, in this question). As an example (Example 1), the following two curves have the same input path, but the red curve has been shortened. What should happen is the red curve should overlay the black curve.

Note that it is not just the position of the shortened curve that is incorrect: the shape is incorrect too: in the example (Example 2) below, I've positioned the curves so they have the same apex, but the shortened curve does not overlay the original (the amount of distortion depends on the curve width and height).

The other issue with shorten is that it requires an absolute length to shorten by, whereas I want to specify it proportionally with respect to the length of the curve itself.

I attempted to manually draw an arrow head with decorations.markings, but this does not properly calculate the right size for the arrowhead (or take into account the existing style of the head or tail). In the example below (Example 3), the arrow head further up the curve should be the same size as the one at the tip.

My idea was to shorten this arrow using a custom dash pattern, but this seems like a hack that is likely in practice to fail in edge cases. Ideally, the curve path itself should be changed, which would handle the existing style/arrowheads/tails/etc. The option pgfpathcurvebetweentime seemed like a possible solution, but I could not figure out how to integrate it with the custom curve style in the two examples.

How might one add an option to the custom curve style to allow the curve to be shortened (e.g. curve={height=-40pt, shorten=0.2})? In practice, shortening symmetrically from both ends is usually sufficient, but having options to shorten the start and end differently would be even better if it doesn't add much extra complexity.

For a demonstration of what I intend by "shortening a curved arrow", see @Thruston's example.

Example 1

\documentclass{article}
\usepackage{tikz-cd}
\usetikzlibrary{calc}

\begin{document}

{\tikzset{curve/.style={settings={#1},to path={(\tikztostart)
.. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
.. (\tikztotarget)\tikztonodes}},
settings/.code={\tikzset{quiver/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}},
quiver/.cd,pos/.initial=0.35,height/.initial=0}

$\begin{tikzcd} {\bullet} & {\bullet} \arrow[from=1-1, to=1-2, curve={height=-40pt}, shorten <= 30pt, shorten >= 30pt, color=red] \arrow[from=1-1, to=1-2, curve={height=-40pt}] \end{tikzcd}$}

\end{document}

Example 2

\documentclass{article}
\usepackage{tikz-cd}
\usetikzlibrary{calc}

\begin{document}

{\tikzset{curve/.style={settings={#1},to path={(\tikztostart)
.. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
.. (\tikztotarget)\tikztonodes}},
settings/.code={\tikzset{quiver/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}},
quiver/.cd,pos/.initial=0.35,height/.initial=0}

$\begin{tikzcd} {\bullet} &&& {\bullet} \arrow[from=1-1, to=1-4, curve={height=-60pt}, shorten <= 30pt, shorten >= 30pt, color=red, yshift=-11pt] \arrow[from=1-1, to=1-4, curve={height=-60pt}] \end{tikzcd}$}

\end{document}

Example 3

\documentclass{article}
\usepackage{tikz-cd}
\usetikzlibrary{calc}
\usetikzlibrary{decorations.markings}

\begin{document}

{\tikzset{curve/.style={settings={#1},to path={(\tikztostart)
.. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
.. (\tikztotarget)\tikztonodes},
decoration={
markings,
mark=at position 0.8 with {\arrow{>}}}
, postaction={decorate}},
%
settings/.code={\tikzset{quiver/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}},
quiver/.cd,pos/.initial=0.35,height/.initial=0}

$\begin{tikzcd} {\bullet} & {\bullet} \arrow[Rightarrow, from=1-1, to=1-2, curve={height=-40pt}] \end{tikzcd}$}

\end{document}

• Isn't it enough to move this arrow with e.g. yshift=5pt? – AndréC Oct 28 '20 at 12:02
• @AndréC: unfortunately not: the actual curve shape is incorrect. I've added an example in the question. In some cases, it's close enough not to be noticeable, but that's not always the case. I also don't know of a way to deterministically calculate how much to shift by to align the apices, even if this was sufficient. – varkor Oct 28 '20 at 12:25
• What are the cases that prevent you from using such a move? – AndréC Oct 28 '20 at 12:44
• I'm outputting TikZ code from a diagram editor, so the TikZ diagram needs to match the original diagram as closely as possible, for arbitrary diagrams. This means I can't hard-code offsets, and need to be confident that the output will be reasonable regardless of the shape of the curve. – varkor Oct 28 '20 at 13:22
• I can add a parameter that shifts the curve perpendicular to the path, would that be okay with you? – AndréC Oct 28 '20 at 14:32

Is this what you mean?

This is done in Metapost, purely for comparison, and in the hope that some one else here can show us how to do the equivalent in TikZ.

\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
input colorbrewer-rgb
beginfig(1);
z0 = -z1 = 34 left;
path a; a = z0 .. controls (-13, 55) and (13, 55) .. z1;
ahangle := 30;
ahlength := 2;
for s=1 upto 8:
drawarrow a
cutbefore fullcircle scaled 10s shifted z0
cutafter  fullcircle scaled 10s shifted z1
withcolor Blues[9][s];
endfor
drawdot z0 withpen pencircle scaled dotlabeldiam;
drawdot z1 withpen pencircle scaled dotlabeldiam;
endfig;
\end{mplibcode}
\end{document}

(Compile with lualatex).

What I have done each is draw the same curve each time, but with more of it "cut off" at each end.

• + 1 very nice ! – AndréC Oct 29 '20 at 12:23
• Yes, thank you, this is precisely what I mean. I will add a link to your answer in the original question for clarity. – varkor Oct 29 '20 at 13:12
• Great answer...Thruston and don't forget that I'm still waiting for your answer on the feymmp package arrows. Otherwise I won't vote for you any more ahahah. – Sebastiano Oct 29 '20 at 22:13

This is a topic I also struggle with for some time. I found something that could be useful on this topic: Tikz Graphics: Curved arrow drawn parallel to curved line

I adapted it to the current question and am almost happy with it:

\documentclass{article}
\usepackage{tikz}
\begin{document}
\thispagestyle{empty}

\begin{tikzpicture}

% Nodes for start and end
\node at (0,0) (A) {};
\node at (5,0) (B) {};

% Path from start to end, using Bezier definition (no to[bend] allowed)
% The path is cut into 20 parts, using 'pos'
\path  (A) .. controls +(2,4) and +(-2,4).. (B)
{\foreach \t [count=\i] in {0.05,0.10,0.15,...,0.85,0.90,0.95,1}
{
coordinate[pos=\t] (p\i)
}
};

% naming p0 and p20 to allow the loop to draw the complete first curve
\coordinate (p0) at (A);
\coordinate (p20) at (B);

% Drawing the 9 other curves
\foreach \k in {0,...,9}
{
\pgfmathtruncatemacro\l{20-\k}      % p\k is the start point, p\l is the end point
\pgfmathtruncatemacro\clr{10*\k}    % fading the colour of the curves

% Now draw (segments) between all p\i points from p\k to p\l
\draw[cap=round,line width=1.5pt,-stealth,blue!\clr] (p\k)
{\foreach \i in {\k,...,\l} {-- (p\i) }};
}

\end{tikzpicture}
\end{document}

Now, the main issue of this solution is that it doesn't draw curves but broken lines. This could easily be fixed by computing more points on the original path. You may want to play with it, while finding a better solution.

• Thanks for sharing this method! This achieves the effect I was going for, but I'm concerned that it may not be compatible with existing TikZ/tikz-cd arrow styles, like double arrows and squiggly arrows, because it relies upon manually drawing the edges. Perhaps I'm wrong, though? – varkor Jan 2 at 1:18
• I don't understand why it wouldn't be compatible with any TikZ decorations. You may use the decorations.markings TikZ library with postaction to place any arrow tip at the end of the partial path. I'm working on another better solution, maybe you could be more precise about what you exactly want to draw. – SebGlav Jan 2 at 11:28
• I'm just not familiar enough with tikzpicture to know how to apply the Rightarrow and squiggly styles, for instance, to arrows drawn with this method. – varkor Jan 2 at 12:18

Here is my proposal:

On the previous style, I added the shift parameter which allows me to shift the curve perpendicular to the path. To do this, I applied the idea I developed in my answer to the question Annotating a table with arrows.

The new style is called curve 1, I put as default height=-40 pt as in your examples.

\documentclass{article}
\usepackage{tikz-cd}
\usetikzlibrary{calc}

\begin{document}

\tikzset{curve 1/.style={settings={#1},
to path={($(\tikztostart)!\pv{shift}!90:(\tikztotarget)$) .. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
.. ($(\tikztotarget)!\pv{shift}!270:(\tikztostart)$)\tikztonodes}},
settings/.code={\tikzset{quiver/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}},
quiver/.cd,pos/.initial=0.35,height/.initial=-40pt,shift/.initial=0pt}

$\begin{tikzcd} {\bullet} & {\bullet} \arrow[from=1-1, to=1-2, curve 1={shift=2mm},color=blue,thick] \arrow[from=1-1, to=1-2, curve 1={shift=4mm},color=cyan,thick] \arrow[from=1-1, to=1-2, curve 1={shift=6mm},color=green,thick] \arrow[from=1-1, to=1-2, curve 1={shift=8mm},color=yellow,thick] \arrow[from=1-1, to=1-2, curve 1={shift=10mm},color=orange,thick] \arrow[from=1-1, to=1-2, curve 1={shift=12mm},color=red,thick] \end{tikzcd}$

\end{document}

• Thanks for your answer. Unfortunately, this doesn't solve the problem mentioned above, that using shorten, combined with shift will have distorted shape compared to the original curve. Also, even if the shape were correct, there's no well to tell how much to shift by to align the two curves perfectly. – varkor Oct 28 '20 at 18:17

Since I don't really understand what still refrains you to use this method, I made some adjustments, in order to make it easier to use. In this new version, you can decide the position of both starting and ending points of your arrow, and the number of points to be computed, in order to avoid the uggly broken line effect on strongly curved lines.

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\thispagestyle{empty}

\begin{tikzpicture}

% First, the starting and ending nodes
\draw (0,0) node[circle, inner sep=0.8pt, fill=black, label={below:{$A$}}] (A) {};
\draw (5,1) node[circle, inner sep=0.8pt, fill=black, label={above:{$B$}}] (B) {};

% Define the path between the two nodes (Bezier curve)
\def\route{(A) .. controls +(2,5) and +(-2,-3).. (B)}

% Drawing the complete path
\draw[orange] \route;

%%%%%%%%%%%%%%
% Blue arrow %
%%%%%%%%%%%%%%

\def\start{0.5}     % Starting point position
\def\len{0.45}      % Arrow length (percentage of total line length)
\def\numpoints{30}  % Number of points to draw segments between them

% Computing the 30 points from the arrow start and the arrow end
\path  \route
{\foreach \i in {1,...,\numpoints} { coordinate[pos=\start+\len*\i/\numpoints] (p\i) } };

% Drawing a broken line linking all 30 points, with the arrow decoration at the end
\draw[blue, cap=round, opacity=0.5, line width=1.5pt,->>] (p1)
{ \foreach \j in {2,...,\numpoints} {-- (p\j) } };

%%%%%%%%%%%%%
% Red arrow %
%%%%%%%%%%%%%
\def\start{0.4} \def\len{-0.3} \def\numpoints{30}
\path  \route
{\foreach \i in {1,...,\numpoints} { coordinate[pos=\start+\len*\i/\numpoints] (p\i) } };
\draw[red, cap=round, opacity=0.5, line width=1.5pt,->>] (p1)
{ \foreach \j in {2,...,\numpoints} {-- (p\j) } };

\end{tikzpicture}

\end{document}

Hope this will give you a better insight on this.

edit: I realize now that you would want to do this, using @AndréC 's method of creating curves between points. If you don't find how to implement this, let us know, sure someone will find a way.


text:
# Project 7 (TLS)

## Objectives

• Learn how websites you frequently use establish TLS connections.
• Learn about the openssl TLS debug utility

The following openssl command can be used to debug TLS connections. It will output information regarding the TLS handshake. Port 443 is the default port for HTTPS connections. Run this command and notice the key exchange method that is used for this particular BYU website. Use control-c to break out of the TLS connection established by this command.

openssl s_client -connect booklist.byu.edu:443

Consider also using ssllabs test too, and other command line options to openssl s_client. for instance, you can examine server certificates, see how they are signed, by what 3rd parties etc.

## Requirements

• Use the openssl command to connect to five secure websites that you regularly use (email, social networking, banking, etc.)
• Write a brief report that lists the key cryptographic properties used by these websites to establish TLS connections. It might be convenient to put this in a table.
• key exchange method
• key size
• cipher suite chosen
• is session resumption supported?
• key sizes
• perfect forward secrecy support
• tls extensions supported
• etc.
• Summarize any interesting differences or common features that you observe


text:
# Drywall Finishing Cost (Room)

Not Reviewed
Equation / Last modified by KurtHeckman on 2018/12/26 20:10
F =
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KurtHeckman.Drywall Finishing Cost (Room)
UUID
d4204d0f-0909-11e9-8682-bc764e2038f2

The Drywall Finishing Cost for a Room calculator estimates the cost to finish drywall in a room based on the dimensions of the room and the cost per unit (4x8) for finishing. Dimensions of a Room

INSTRUCTIONS: Choose units and enter the following:

• (L)  This is the length of the room.
• (W) This is the width of the room.
• (H)  This is the height of the room.
• (PPS)  This is the finishing price per drywall (4x8) sheet.

Cost to Finish Drywall (F):  The calculator returns the estimated cost in U.S. dollars.  However, this can be automatically converted to compatible units via the pull-down menu.

#### The Math / Science

1. computes the surface area of the room including the walls and ceiling,
2. divides the area by 32 square feet for a 4x8
3. multiplies the number of sheets calculated above by the price per unit for finishing.

Note: in 2019 the U.S. national average for drywall finishing ranged between $45 to$70 per 4x8 sheet.

#### Drywall Pricing

Pricing for drywall is an estimate.  The pricing per sheet used in this estimator is as follows:


text:
# 2020-24 Divisions of Fibonacci numbers and their remainders

For each $$i \in \mathbb{N}$$, let $$F_i$$ be the $$i$$-th Fibonacci number where $$F_0=0, F_1=1$$ and $$F_{i+1}=F_{i}+F_{i-1}$$ for each $$i\geq 1$$.
For $$n>m$$, we divide $$F_n$$ by $$F_m$$ to obtain the remainder $$R$$. Prove that either $$R$$ or $$F_m-R$$ is a Fibonacci number.

GD Star Rating

# 2020-23 The area of a random polygon

Suppose we choose a point on the unit circle in the plane at random with the uniform probability measure on the circle. When we choose n points in that way, what is the probability of the n-gon obtained as the convex hull of the chosen points has the area bigger than $$\pi/2$$ in terms of n?

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# 2020-22 Regular simplex

Let $$S$$ be the unit sphere in $$\mathbb{R}^n$$, centered at the origin, and $$P_1 P_2 \dots P_{n+1}$$ a regular simplex inscribed in $$S$$. Prove that for a point $$P$$ inside $$S$$,
$\sum_{i=1}^{n+1} (PP_i)^4$
depends only on the distance $$OP$$ (and $$n$$).

GD Star Rating

# 2020-21 이적이 부릅니다. 거짓말 거짓말 거짓말(A lie, a lie, a lie)

Alice and Bob play the following game with $$S=\{1,\dots, 777\}$$.

Alice picks a number $$x \in S$$ without telling anyone and Bob will guess what the number is at the end of the game. Alice is malicious so that she can always change her number $$x$$ at any time until the end of the game.

In each round, Bob picks a subset $$T\subseteq S$$ and asks a following question to Alice: “is your $$x$$ belong to $$T$$?” Alice must say either Yes or No. At the end of the game, Bob guesses her $$x$$ first and then Alice reveals her number $$x$$ (Alice can still change her number after she listen to Bob’s guess and before revealing her number). According to her final number $$x$$, each of her previous answers are determined to be either a truth or a lie.

Bob wins if Alice end up lying more than three times or his answer is correct. Alice wins if Bob’s answer is wrong and at most three of her answers are lies. Prove that if a game consists of twenty rounds, then no matter what Bob does Alice can always win.

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# 2020-20 Efficient triangulation of surfaces

Let $$S_g$$ denote the closed orientable connected surface of genus $$g$$. Suppose we glue triangles along the edges so that the resulting space is $$S_g$$ and the intersection of any two triangles are either empty or a single edge. Let $$n(g)$$ be the minimum number of triangles one needs to make $$S_g$$ while satisfying the above rule. What are $$n(1), n(2), n(3)$$? Does the limit $$\lim_{g \to \infty} n(g)/g$$ exist?

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# 2020-19 Continuous functions

Let $$n$$ be a positive integer. Determine all continuous functions $$f: [0, 1] \to \mathbb{R}$$ such that
$f(x_1) + \dots + f(x_n) =1$
for all $$x_1, \dots, x_n \in [0, 1]$$ satisfying $$x_1 + \dots + x_n = 1$$.

GD Star Rating

# 2020-18 A way of shuffling cards

Consider the cards with labels $$1,\dots, n$$ in some order. If the top card has label $$m$$, we reverse the order of the top $$m$$ cards. The process stops only when the card with label $$1$$ is on the top. Prove that the process must stop in at most $$(1.7)^n$$ steps.

GD Star Rating

# 2020-17 Endomorphisms of abelian groups

Prove or disprove that a surjective homomorphism from a finitely generated abelian group to itself is an isomorphism.

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# 2020-16 A convex function of matrices

Let $$A$$ be an $$n \times n$$ Hermitian matrix and $$\lambda_1 (A) \geq \lambda_2 (A) \geq \dots \geq \lambda_n (A)$$ the eigenvalues of $$A$$. Prove that for any $$1 \leq k \leq n$$
$A \mapsto \lambda_1 (A) + \lambda_2 (A) + \dots + \lambda_k (A)$
is a convex function.

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Let $$m_0=n$$. For each $$i\geq 0$$, choose a number $$x_i$$ in $$\{1,\dots, m_i\}$$ uniformly at random and let $$m_{i+1}= m_i – x_i$$. This gives a random vector $$\mathbf{x}=(x_1,x_2, \dots)$$. For each $$1\leq k\leq n$$, let $$X_k$$ be the number of occurrences of $$k$$ in the vector $$\mathbf{x}$$.
For each $$1\leq k\leq n$$, let $$Y_k$$ be the number of cycles of length $$k$$ in a permutation of $$\{1,\dots, n\}$$ chosen uniformly at random. Prove that $$X_k$$ and $$Y_k$$ have the same distribution.


text:
Back to PCP
Back to notes

# Historical Development of PCP

40%

## History of results

The way we present the historical development of PCP theorems is based on the survey of Sudan (Sudan, 2009).

### Phase 0: Folklore

• $\NP=\PCP[0,\poly(n)]$
• $\NP=\PCP[O(\log n),\poly(n)]$

## References

1. Sudan, M. (2009). Probabilistically checkable proofs. Communications of the ACM, 52(3), 76–84.
2. Babai, L., Fortnow, L., & Lund, C. (1990). Nondeterministic exponential time has two-prover interactive protocols. In Foundations of Computer Science, 1990. Proceedings., 31st Annual Symposium on (pp. 16–25). IEEE.
3. Babai, L., Fortnow, L., Levin, L. A., & Szegedy, M. (1991). Checking computations in polylogarithmic time. In Proceedings of the twenty-third annual ACM symposium on Theory of computing (pp. 21–32). ACM.
4. Arora, S., & Safra, S. (1998). Probabilistic checking of proofs: A new characterization of NP. Journal of the ACM (JACM), 45(1), 70–122.
5. Arora, S., Lund, C., Motwani, R., Sudan, M., & Szegedy, M. (1998). Proof verification and the hardness of approximation problems. Journal of the ACM (JACM), 45(3), 501–555.
6. Håstad, J. (2001). Some optimal inapproximability results. Journal of the ACM (JACM), 48(4), 798–859.
7. Guruswami, V., Lewin, D., Sudan, M., & Trevisan, L. (1998). A tight characterization of NP with 3 query PCPs. In Foundations of Computer Science, 1998. Proceedings. 39th Annual Symposium on (pp. 8–17). IEEE.
8. Polishchuk, A., & Spielman, D. A. (1994). Nearly-linear size holographic proofs. In Proceedings of the twenty-sixth annual ACM symposium on Theory of computing (pp. 194–203). ACM.
9. Karloff, H., & Zwick, U. (1997). A 7/8-approximation algorithm for MAX 3SAT? In Foundations of Computer Science, 1997. Proceedings., 38th Annual Symposium on (pp. 406–415). IEEE.


text:
# Homework Help: Find the minimal polynomial

1. Nov 12, 2008

### Office_Shredder

Staff Emeritus
1. The problem statement, all variables and given/known data

Find the minimal polynomial of $$\frac{\sqrt{3}}{1+2^{1/3}}$$ over Q

we'll call this x

2. Relevant equations
I wish I knew some :(

3. The attempt at a solution

By taking powers of x, I was able to show that the extension Q(x) has degree six (since 21/3 and sqrt(3) are both independently inside Q(x)) and hence the minimal polynomial has degree six. So then I took a general degree six polynomial, plugged in x, and got six equations in six unknowns. This is less than elegant, and I ended with a system of equations (luckily three of the six unknown coefficients were zero):

255b + 261d + 595f = -171
150b + 222d + 460f = -144
105b + 159d + 375f = -108

solving for b,d,f. An attempt at a numerical solution doesn't convince me this has a rational solution, but I wouldn't be surprised if there was a computational error preceding this. Is there a better way to do this?

Last edited: Nov 12, 2008
2. Nov 12, 2008

### Dick

How about starting from the equation x=sqrt(31)+2^(1/3)? So (x-sqrt(31))^3=2. Expand the left side and move all of the terms involving sqrt(31) to one side and everything else to the other. Now square.

3. Nov 12, 2008

### Office_Shredder

Staff Emeritus
Sorry, you posted before I fixed the latex. Different value for x

EDIT: Same principle worked though. Thanks a ton

4. Nov 12, 2008

### Dick

Fast, aren't I? Same idea. x=sqrt(3)/(1+2^(1/3)) -> x2^(1/3)=sqrt(3)-x. Cube, rearrange and then square.


text:
Remark 97.20.2 (Strong effectiveness). Let $S$ be a locally Noetherian scheme. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume we have

1. an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$,

2. an inverse system $(R_ n)$ of $\Lambda$-algebras with surjective transition maps whose kernels are locally nilpotent,

3. a system $(\xi _ n)$ of objects of $\mathcal{X}$ lying over the system $(\mathop{\mathrm{Spec}}(R_ n))$.

In this situation, set $R = \mathop{\mathrm{lim}}\nolimits R_ n$. We say that $(\xi _ n)$ is effective if there exists an object $\xi$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(R)$ whose restriction to $\mathop{\mathrm{Spec}}(R_ n)$ gives the system $(\xi _ n)$.

There are also:

• 2 comment(s) on Section 97.20: Strong formal effectiveness

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Questions about the Schwarzschild metric

1. Nov 5, 2012

### nick41

Hello everybody! I have some questions concerning the structure of the Schwarzschild metric, which is given by
$$ds^2=-(1- \frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+ r^2(d\theta^2+ \sin^2(\theta) d\phi^2)$$
where we set $c=1$. I would like to know the following: \\
\\
1. Why is it reasonable to consider $M$ as the mass of the black hole? What is the motivation behind this? \\
\\
2. What are the geodesics in the Schwarzschild solution? Is there any good way to visualize them all at once?
\\
\\

2. Nov 5, 2012

### Staff: Mentor

Put simply, M is the mass you will obtain if you put a small test object in orbit about the hole, measure its orbital parameters, and apply Kepler's Third Law.

Probably not; there are too many different kinds. A couple of types that are commonly considered are:

(1) Purely radial timelike geodesics, particularly infalling ones; these represent the worldlines of objects that are free-falling into the hole.

(2) Tangential timelike geodesics, which represent the worldlines of objects that are orbiting the hole.

3. Nov 5, 2012

### Staff: Mentor

There doesn't have to be a black hole at all - the Schwarzchild solution works just fine in the empty space around any stationary spherically symmetric mass. You only get a black hole if the object is so dense that r=2M lies above its surface.

4. Nov 5, 2012

### pervect

Staff Emeritus
As others have pointed out, if you look at the orbits of bodies far away from the black hole, where gravity is nearly Newtonian, you can find that the Newtonian limit gives M as the mass of the black hole.

The geodesics of the Schwarzschild solution are just the various orbits of unaccelerated test particles. These include orbits that go around the black hole, go into it, or fly by it, as long as the test particle in question a) isn't accelerating like a rocket and b) isn't so big that its own gravity perturbs the metric and c) you don't have significant emission of gravitational radiation, which would also cause the test particle to accelerate.

5. Nov 6, 2012

6. Nov 6, 2012

### pervect

Staff Emeritus
Note though that while in the exterior region the Komar formula gives you the Schwarzschild mass parameter M, it doesn't give you the same distribution of mass in the interior region as the fairly well known integral, valid only in Schwarzschild coordinates

$m(r) = \int \rho(r) 4 \pi r^2 dr$

does. The difference won't matter with a Schwarzschild black hole, but it will matter with a a Schwarzschild non-black hole solution in the interior region (not the exterior region though).

The easiest way of convincing oneself of this is direct calculation, the particular case that convinced me was the "box of light" case. I'm not sure I've seen much comment on this difference, though the fact that Townsend mentions specifically showing it's equal in the exterior region is suggestive of the issues that arise in the interior region.

In the box of light case, the Schwarzschild formula will suggest a mass that increases monotonically as r increases. In the idealized "box of light" case, where the box is a sphere of exotic matter with no mass but the tension required to hold the box together, you get a picture where the contribution of the exotic matter shell to the total mass is actually negative, obviously different from the monotonic increase in mass predicted by the Schwarzschild formula.

The total exterior mass remains the same in both cases, but the details of how it's distributed (and in particular, the contribution of the shell to the mass) varies with the two aproaches.

This may sound funky, but the Komar approach correlates better to the gravitational fields one would measure by measuring the 4-acceleration of a static observer. This is because you can cast the Komar approach into a rather Gauss-law like integral (see Wald, for example, around pg 288).


text:
# Formulas from Geometry

$V=\text{Volume},\text{ and }$

If then

where

# Formulas from Trigonometry

## Trigonometric Functions of Important Angles

 0° 0 0 1 0 30° 45° 1 60° 90° 1 0 —


text:
# Central Element in Sklyanin Algebras?

I'm interested in Sklyanin Algebras or Artin-Shelter regular algebras of type A. These are generated in degree 1 by three variables x,y,z, and have three defining relations in degree 2, which you can get by multiplying the matrix $$M=\begin{pmatrix} cx & bz & ay \\ az & cy & bx \\ by & ax & cz\end{pmatrix}$$ either on the left or right by the row/column vector of variables. (For every triple of numbers $a$, $b$, $c$, excluding a finite set, you get one of these algebras.)

Such algebras $A$ contain a central regular element of degree $3$, let me call it $g$. It has the property that $A/gA$ is the twisted homogeneous coordinate ring of the scheme of point modules of $A$, which is the cubic curve defined by $\det M$. $$g=c(c^3-b^3)y^3+b(c^3-a^3)yxz+a(b^3-c^3)xyz+c(a^3-c^3)x^3$$

This central element $g$ seems somewhat mysterious. In the original paper by Artin-Shelter it was apparently found by a computer search, in the foundational paper by Artin-Tate-Van den Bergh, only its existence (and its properties) are proved.

My question is: is there a conceptual way to get this element $g$? Sorry, it's a little vague. Thanks for any insight into this matter.

-
If translation by the point $\tau$ of the elliptic curve $E$ has infinite order, then the center of $A$ is $k[g]$. That's the only intrinsic description of $g$ I know. There might be more known about the two central elements in the Sklyanin algebra of GKdim 4. Paul Smith has some notes that might be useful there: math.washington.edu/~smith/Research/Skly-survey.pdf – Graham Leuschke Apr 25 '13 at 20:22


text:
regtools

Novel tools tools for linear, nonlinear and nonparametric regression.

These tools are associated with my forthcoming book, From Linear Models to Machine Learning: Modern Statistical Regression and Classification, N. Matloff, CRC, 2017.

The tools are useful in general, independently of the book.

FEATURES:

• Nonparametric regression for general dimensions in predictor and response variables, using k-Nearest Neighbors. Local-linear option. Allows for user-specified smoothing method. Allows for accelerated exploration of multiple values of k at once. Tool to aid in choosing k.

• Innovative tools for assessing fit in linear and nonlinear parametric models, via nonparametric methods. Model evaluation, examination of quadratic effects, investigation of nonhomogeneity of variance.

• Tools for multiclass classification, parametric and nonparametric. One vs. All and All vs. All. Novel adjustment for artificially balanced data.

• Linear regression, PCA and log-linear model estimation in missing-data setting, via the Available Cases method.

• Nicer implementation of ridge regression, with more meaningful scaling and better plotting.

• Extension to nonlinear parametric regression with of Eickert-White technique to handle heteroscedasticity.

• Misc. tools, e.g. Method of Moments estimation (including for nonregression settings).

EXAMPLE: MODEL FIT ASSESSMENT

Let’s take a look at the data set prgeng, some Census data for California engineers and programmers in the year 2000. The response variable in this example is wage income, and the predictors are age, number of weeks worked, and dummy variables for MS and PhD degrees. (Some data wrangling was performed first; type ?knnest for the details.)

The fit assessment techniques in regtools gauge the fit of parametric models by comparing to nonparametric ones. Since the latter are free of model bias, they are very useful in assessing the parametric models.

The function nonparvsxplot() plots the nonparametric fits against each predictor variable, for instance to explore nonlinear effects. Here is the plot for wage versus (scaled) age:

Of course, the effects of the other predictors don’t show up here, but there does seem to be a quadratic effect. The same was true for the predictor measuring the number of weeks worked (slightly concave up, not shown here). In our linear parametric model, then, we will include squared terms for these two predictors.

So, after fitting the linear model, run parvsnonparplot(), which plots the fit of the parametric model against the nonparametric one. Here is the result:

There is quite a bit suggested in this picture:

• There seems to be some overfitting near the low end, and underfitting at the high end.

• The outliers, meaning points far from the fitted linear model, are almost all below the linear fit.

• There are intriguing “sreaks” or “tails” of points, suggesting the possible existence of important subpopulations.

• There appear to be a number of people with 0 wage income. Depending on the goals of our analysis, we might consider removing them.

Finally, let’s check the classical assumption of homoscedasticity, meaning that the conditional variance of Y given X is constant. The function nonparvarplot() plots the estimated conditional variance against the estimated conditional mean, both computed nonparametrically::

Wow, a hockey stick! Though there is a mild rise in coefficient of determination, i.e. standard deviation relative to the mean, up to about $80K, the slope increases sharply after that. What to do? As long as our linear regression model assumption holds, violation of the homoscedasticity assumption won’t invalidate our estimates; they still will be statistically consistent. But the standard errors we compute, and thus the statistical inference we perform, will be affected. This is correctible using the Eickert-White procedure, which for linear models is available in the car package, included in regtools. Our package also extends this to nonlinear parametric models, in our function nlshc() (the validity of this extension is shown in the book). Of course, the “hockey stick” form is another indication that we should further investigate the model itself. It may well be useful to fit two separate linear models, one for incomes below$80K and the other for the higher incomes. For a more formal approach to this, we might consider changepoint methods, such as in the CRAN package chngpt.

What is different:

Note carefully that the above graph is unaffected by the validity of the parametric model; it is based purely on nonparametric analysis. This is in contrast to classic regression fit methods, most of which are based on examination of residuals of the fitted model.

EXAMPLE: OVA VS. AVA IN MULTICLASS PROBLEMS

A very popular prediction method in 2-class problems is to use logistic (logit) regression. In analyzing click-through patterns of Web users, for instance, we have 2 classes, Click and Nonclick. We might fit a logistic model for Click, given user Web history, demographics and so on. Note that logit actually models probabilities, e.g. the probability of Click given the predictor variables.

But the situation is much less simple in multiclass settings. Suppose our application is recognition of hand-written digits (a famous machine learning example). The predictor variables are pixel patterns in images. There are two schools of thought on this:

• One vs. All (OVA): We would run 26 logistic regression models, one for predicting ‘0’ vs. non-‘0’, one for ‘1’ vs. non-‘1’, and so on. For a particular image, we would thus obtain 26 estimated probabilities. Let imax be the image that yields the largest probability; we would then guess the digit for the image to be ‘i’.

• All vs. All (AVA): Here we would run C(10,2) = 45 logit analyses, one for each pair of digits. There would be one for ‘0’ vs. ‘1’, one for ‘0’ vs. ‘2’, etc., all the way up through ‘8’ vs. ‘9’. Many in the machine learning literature recommend AVA over OVA, on the grounds that might be linearly separable (in the statistical sense) in pairs but not otherwise. My book counters by positing that such a situation could be remedied under OVA by adding quadratic terms to the logit models.

At any rate, the regtools package gives you a choice, OVA or AVA, for both parametric and nonparametric methods. For example, avalogtrn() and avalogpred() do training and prediction operations for logit with AVA.

Another feature concerns adjustment of class probabilities. In many multiclass data sets, the numbers of points in each class is the same, or least not reflective of the population class probabilities. In regtools, the user can specify estimates of the latter, for logit and nonparametric methods.

So, let’s look at an example, using the UCI Letter Recognition data set, another image recognition example. Again, the code below was preceded by some data wrangling, which changed the letter data from character to numeric, and which divided the data set into training and test sets. Here is the OVA run:

> ologout <- ovalogtrn(26,lrtrn[,c(2:17,1)])
> ypred <- ovalogpred(ologout,lrtest[,-1])
> mean(ypred == lrtest[,1])
[1] 0.7193333

So, we get about a 72% rate of correct classification. Now let’s try AVA:

> alogout <- avalogtrn(26,lrtrn[,c(2:17,1)])
> ypred <- avalogpred(26,alogout,lrtest[,-1])
> mean(ypred == lrtest[,1])
[1] 0.8355

AVA did considerably better, 84%. So, apparently AVA fixed a poor model. But of course, it’s better to make a good model in the first place. Based on our previous observation that the boundaries may be better approximated by curves than lines, let’s try a quadratic model.

There were 16 predictors, thus 16 possible quadratic terms, and C(16,2) = 120 possible interaction terms. Inclusion of all such variables would probably produce too rich a model for the 14000 points in our training set. We’ll settle for adding just the squared terms (not shown):

> ologout <- ovalogtrn(26,lrtrn[,c(2:33,1)])
> ypred <- ovalogpred(ologout,lrtest[,-1])
> mean(ypred == lrtest[,1])
[1] 0.8086667

Ah, much better, though still not quite as good as AVA.


text:
# COVID-19/Mathematical Modelling/Logistical Growth

Example of an Epidemic: Ebola cases (red) and mortality (black) during the Ebola Epidemics in Westafrika until the end July 2014 (approximately a logistic growth)

## Comparison Exponential Growth - logistical growth

In school you might be exposed to exponential growth. In terms of epidemiology the exponential growth model assumes that there are not limits of the growth. In fact for epidemiology the maximum of infected people is given by the total population. If we consider the other biological processes of growth (e.g. cell division) we have also limitations of growth (e.g. limits of resources, limits of space, ...). The logistical growth incorporates a capacity in the modelling. A logistic function or logistic curve is a common "S" shape (sigmoid curve), with equation:[1]

${\displaystyle f(x)={\frac {M}{1+e^{-k(x-x_{0})}}}}$

where

• ${\displaystyle e}$ = the base of exponential function, which is also known as Euler's number,
• ${\displaystyle x_{0}}$ = the ${\displaystyle x}$-value of the sigmoid's midpoint, which is in epidemiology the point in time with maximum growth rate (maximum value of the derivation).
• ${\displaystyle M}$ = the curve's capacity of the growth, and
• ${\displaystyle k}$ = the logistic growth rate or steepness of the curve.[2]

## References

1. Wikipedia contributors (February 5, 2020). Logistic function.
2. Verhulst, Pierre-François (1838). "Notice sur la loi que la population poursuit dans son accroissement" (PDF). Correspondance Mathématique et Physique 10: 113–121. Retrieved 3 December 2014.


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# Search for $s$ channel single top quark production in pp collisions at $\sqrt{s}$=7 and 8 TeV

CMS Collaboration; Canelli, F; Chiochia, V; Kilminster, B; Robmann, P; et al (2016). Search for $s$ channel single top quark production in pp collisions at $\sqrt{s}$=7 and 8 TeV. Journal of High Energy Physics, 2016(9):27.

## Abstract

A search is presented for single top quark production in the s channel in proton-proton collisions with the CMS detector at the CERN LHC in decay modes of the top quark containing a muon or an electron in the final state. The signal is extracted through a maximum-likelihood fit to the distribution of a multivariate discriminant defined using boosted decision trees to separate the expected signal contribution from background processes. The analysis uses data collected at centre-of-mass energies of 7 and 8 TeV and corresponding to integrated luminosities of 5.1 and 19.7 fb$^{−1}$, respectively. The measured cross sections of 7.1 ${\pm}$ 8.1 pb (at 7 TeV) and 13.4 ${\pm}$ 7.3 pb (at 8 TeV) result in a best fit value of 2.0 ${\pm}$ 0.9 for the combined ratio of the measured and expected values. The signal significance is 2.5 standard deviations, and the upper limit on the rate relative to the standard model expectation is 4.7 at 95% confidence level.

## Abstract

A search is presented for single top quark production in the s channel in proton-proton collisions with the CMS detector at the CERN LHC in decay modes of the top quark containing a muon or an electron in the final state. The signal is extracted through a maximum-likelihood fit to the distribution of a multivariate discriminant defined using boosted decision trees to separate the expected signal contribution from background processes. The analysis uses data collected at centre-of-mass energies of 7 and 8 TeV and corresponding to integrated luminosities of 5.1 and 19.7 fb$^{−1}$, respectively. The measured cross sections of 7.1 ${\pm}$ 8.1 pb (at 7 TeV) and 13.4 ${\pm}$ 7.3 pb (at 8 TeV) result in a best fit value of 2.0 ${\pm}$ 0.9 for the combined ratio of the measured and expected values. The signal significance is 2.5 standard deviations, and the upper limit on the rate relative to the standard model expectation is 4.7 at 95% confidence level.

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# Tag Info

8

What seems to help at least with NSum is increasing the number of terms that are computed explicitly via the option NSumTerms. This way one can get agreement for all first $100$ digits: NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400, WorkingPrecision -> 100] 2....

6

First determine for what value of $r$ the argument is less than $10^{-5}$: η = 0.05211184484645051; sol = Solve[{η/(r + 1) + ArcTan[η/(r + 1)] == 1*^-5, r > 0}, r] (* Out: {{r -> 10421.4}} *) Then use that value as the upper limit in your summation: Sum[η/(r + 1) + ArcTan[η/(r + 1)], {r, 0, Ceiling[r /. First@sol]}] (* Out: 1.02436 *)

4

Sum[\[Eta]/(1 + r) + ArcTan[\[Eta]/(1 + r)], {r, 0, n}] is given by \[Eta]*EulerGamma - (1/2)*I*LogGamma[1 - I*\[Eta]] + (1/2)*I*LogGamma[1 + I*\[Eta]] + (1/2)*I*LogGamma[2 - I*\[Eta] + n] - (1/2)*I*LogGamma[2 + I*\[Eta] + n] + \[Eta]*PolyGamma[0, 2 + n] The sum and the analytic expression diverge for n->Infinity. For n > 10421 each additional ...

4

Clear["Global*"] To find the number of terms required, find the minimum value of n for which the last term is less than 10^-100 min = MinValue[{n, 2^(-n + Sin[n Pi/5]) < 10^-100}, n, Integers] (* 334 *) sum1 = NSum[2^(-n + Sin[n Pi/5]), {n, 0, Infinity}, NSumTerms -> 400, WorkingPrecision -> 100] (* 2....

4

Accumulate /@ matrix1 produces {{a, a + b, a + b + c, a + b + c + d}, {e, e + f, e + f + g, e + f + g + h}, {i, i + j, i + j + k, i + j + k + l}, {m, m + n, m + n + o, m + n + o + p}}

3

Note, a double sum can be written with a single Sum. With this we make a replacement: Sum[a[k], {k, 1, n}]*Sum[b[k], {k, 1, n}] /. Sum[a1_, {a2_, 1, n}] Sum[b1_, {b2_, 1, n}] :> Sum[(a1 /. a2 -> k1) (b1 /. b2 -> k2), {k1, 1, n}, {k2, 1, n}]

3

You can look at the Sin[] term in the exponent as some periodic noise. It would be good to smooth that out before trying numeric methods by summing over a period at a time: sum = NSum[ Sum[2^(-(10 n + k) + Sin[(10 n + k) Pi/5]), {k, 0, 9}], {n, 0, Infinity}, WorkingPrecision -> 100] (* 2....

1

Instead of evaluating a single sum, you could break up the sum into two or more sums. L = 24; sind = Range[-Pi, Pi, 2*Pi/L]; a[x_, y_] := x y g[x_, y_] := x + y f = Cos[x]*Sin[y]*Sin[x + x1]*Cos[y + y1]*a[x, y] + g[x, y]*Cos[x2 - y2]*Sin[x + x1 + x2]; AbsoluteTiming[sx = Sum[f, {x, sind}, {x1, sind}, {x2, sind}] // Simplify; sxy = Sum[sx, {y, sind}, {y1, ...

1

Posting as an answer at the request of the OP There is probably no built-in way to achieve what you want because the result you are after is in general not correct: in order to get it you need to rewrite and reorder the summations, which is only possible under special convergence conditions. That being said, if you are sure beforehand that all your ...

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## a. 1000 \Omegab. 0.003 s

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for part A were asked to find the value of an R in an R L circuit in an R C circuit if they have the exact same time. Constant. Okay, well, Tao then in both cases is going to be equal to each other. So it Tao in an RC circuit is equal to our C and an L C circuit is equal to hell over our therefore solving for are we want to know the resistance Mrs Part A. So it's indicated. A such resistance is equal to the square root of induct Inso over capacity. It's which we know both plugging those values. In this expression, we find that this is equal to one times 10 to the third owes or one killer oh, become box and it is their solution For part A for part B were asked to go ahead and find this common time constant so I can simply be done by using either our time see or l over are. So we'll just go ahead and use our time. See, carrying out this expression, we find that this time constant is equal to three times 10 to the minus three seconds. You can box it in as their solution for party. That's the conclusion of our question

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