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##### Question Info

This question is public and is used in 134 tests or worksheets.

Type: Multiple-Choice
Score: 3
Author: maryMckenna

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Add: $-2sqrt3 + sqrt 75$
1. $3sqrt3$
2. $7sqrt3$
3. $-3sqrt3$
4. $-2sqrt3$

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# Couplings of sfermions to h Higgs boson

1. ### Safinaz

151
Hi all,

I try to find the detailed calculations of the couplings of sfermions to the lightest h Higgs boson
in MSSM, I want to reach Equ. (1.109) in [arXiv:hep-ph/0503173],

Do you know any useful reference ?

Bests,
S.S.

2. ### ChrisVer

2,373
What do you mean by detailed calculations? The coupling constant is in the lagrangian as the factor multiplying the fields' couplings. Or in your ref, the couplings are given by the 1.103 equation...
However you could take a look at:
H.E. Haber and G.L. Kane
Phys. Rep. 117,75 (1985)
I think...

3. ### Safinaz

151
Well, I meant to derive a matrix 1.104 in my Ref., which comes from F-terms, D-terms and soft SuSy-breaking terms as it explained, I found these calculations in Atchinson's book : Supersymmetry in Particle Physics (2007) .

S.S.


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How to predict stock price movements based on quantitative market data modeling is an attractive topic. In front of the market news and stock prices that are commonly believed as two important market data sources, how to extract and exploit the hidden information within the raw data and make both accurate and fast predictions simultaneously becomes a challenging problem. In this paper, we present the design and architecture of our trading signal mining platform that employs extreme learning machine (ELM) to make stock price prediction based on those two data sources concurrently. We evaluate YOUNG & CO'S BREWERY PLC prediction models with Statistical Inference (ML) and Beta1,2,3,4 and conclude that the LON:YNGN stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold LON:YNGN stock.

Keywords: LON:YNGN, YOUNG & CO'S BREWERY PLC , stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.

## Key Points

1. What are main components of Markov decision process?
2. What is the best way to predict stock prices?
3. How useful are statistical predictions?

## LON:YNGN Target Price Prediction Modeling Methodology

Stock market also called as equity market is the aggregation of the sellers and buyers. It is concerned with the domain where the shares of various public listed companies are traded. For predicting the growth of economy, stock market acts as an index. Due to the nonlinear nature, the prediction of the stock market becomes a difficult task. But the application of various machine learning techniques has been becoming a powerful source for the prediction. We consider YOUNG & CO'S BREWERY PLC Stock Decision Process with Beta where A is the set of discrete actions of LON:YNGN stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Beta)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Statistical Inference (ML)) X S(n):→ (n+8 weeks) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

n:Time series to forecast

p:Price signals of LON:YNGN stock

j:Nash equilibria

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## LON:YNGN Stock Forecast (Buy or Sell) for (n+8 weeks)

Sample Set: Neural Network
Stock/Index: LON:YNGN YOUNG & CO'S BREWERY PLC
Time series to forecast n: 23 Sep 2022 for (n+8 weeks)

According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold LON:YNGN stock.

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Yellow to Green): *Technical Analysis%

## Conclusions

YOUNG & CO'S BREWERY PLC assigned short-term B1 & long-term B2 forecasted stock rating. We evaluate the prediction models Statistical Inference (ML) with Beta1,2,3,4 and conclude that the LON:YNGN stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold LON:YNGN stock.

### Financial State Forecast for LON:YNGN Stock Options & Futures

Rating Short-Term Long-Term Senior
Outlook*B1B2
Operational Risk 5060
Market Risk6061
Technical Analysis4347
Fundamental Analysis6930
Risk Unsystematic8073

### Prediction Confidence Score

Trust metric by Neural Network: 90 out of 100 with 519 signals.

## References

1. Matzkin RL. 1994. Restrictions of economic theory in nonparametric methods. In Handbook of Econometrics, Vol. 4, ed. R Engle, D McFadden, pp. 2523–58. Amsterdam: Elsevier
2. Abadie A, Diamond A, Hainmueller J. 2010. Synthetic control methods for comparative case studies: estimat- ing the effect of California's tobacco control program. J. Am. Stat. Assoc. 105:493–505
3. Rumelhart DE, Hinton GE, Williams RJ. 1986. Learning representations by back-propagating errors. Nature 323:533–36
4. Li L, Chu W, Langford J, Moon T, Wang X. 2012. An unbiased offline evaluation of contextual bandit algo- rithms with generalized linear models. In Proceedings of 4th ACM International Conference on Web Search and Data Mining, pp. 297–306. New York: ACM
5. T. Morimura, M. Sugiyama, M. Kashima, H. Hachiya, and T. Tanaka. Nonparametric return distribution ap- proximation for reinforcement learning. In Proceedings of the 27th International Conference on Machine Learning, pages 799–806, 2010
6. Cortes C, Vapnik V. 1995. Support-vector networks. Mach. Learn. 20:273–97
7. Pennington J, Socher R, Manning CD. 2014. GloVe: global vectors for word representation. In Proceedings of the 2014 Conference on Empirical Methods on Natural Language Processing, pp. 1532–43. New York: Assoc. Comput. Linguist.
Frequently Asked QuestionsQ: What is the prediction methodology for LON:YNGN stock?
A: LON:YNGN stock prediction methodology: We evaluate the prediction models Statistical Inference (ML) and Beta
Q: Is LON:YNGN stock a buy or sell?
A: The dominant strategy among neural network is to Hold LON:YNGN Stock.
Q: Is YOUNG & CO'S BREWERY PLC stock a good investment?
A: The consensus rating for YOUNG & CO'S BREWERY PLC is Hold and assigned short-term B1 & long-term B2 forecasted stock rating.
Q: What is the consensus rating of LON:YNGN stock?
A: The consensus rating for LON:YNGN is Hold.
Q: What is the prediction period for LON:YNGN stock?
A: The prediction period for LON:YNGN is (n+8 weeks)


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## anonymous 5 years ago solve max and min f(x,y) = 2xy subject to x^2 +y^2 = 18

1. anonymous

Use polar coordinates. so x=rcos(theta) y=rsin(theta). So you get r=3root3. now f(x,y)=f(rcostheta,rsintheta)=2r^2sinthetacostheta. We know that r=3root3. So plug in r in f(x,y), so then f(rcostheta,rsintheta)=2(9)(3)(sin)(cos). Now take a partial derivative with respect to thetha and set it equal to 0. and solve for theta

2. anonymous

oh I thought we use Lagrange multipliers$\lambda$

3. anonymous

Oh lagrange multipliers also work!


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• Cart

# cycle notation permutation calculator

## cycle notation permutation calculator

Search: Permutation As A Product Of Disjoint Cycles Calculator. The cyclic decomposition of a permutation can be computed in the Wolfram Language with the function PermutationCycles and the permutation corresponding to a cyclic decomposition can be computed with PermutationList. Here, the individual cycles are represented using the function Cycles. Likes: 381. Using this order, we may say that. : Combinatorics permutations, combinations, placements. the cycle notation and showed that permutations factor uniquely into a product of cycles. Shares: 191. In this lesson we show you how to convert a permutation into cycle notation, talk about the conventions, and discuss the key properties of cycles. Wolfram|Alpha computes a permutations inverse and writes it in cycle notation. Questionnaire. To find out about yours, input the following data about your previous periods: With this information, the calculator can tell you the following: Ovulation period of each cycle (days where you are most fertile) The E is are disjoint by 6. The sum of the lengths of the cycles cannot be more than 8, since we want the permutation to be in S 8, the permutations on the set of eight elements f1;2;3;4;5;6;7;8g. I'm currently going through a text about groups, and I'm having problems with composing permutations written in cycle notation, i.e. Connected Teaching and Learning. Decomposition of the Newton Binomial. Enter the email address you signed up with and we'll email you a reset link. To keep things relatively small, in the applet below we use n = 10.

p is the permutation (n=10): (3,5,7,6,2,9,1,10,8,4). math dictionary to view the specific definition for each math term. Not every permutation is a cycle, e This below Permutation calculator provides you the permutations count with the known values of n and r 1 Unlike the unique decomposition of a permutation into a product of disjoint cycles, the decomposition of a f(n) = f(m) n + 1 = m + 1 n = m. Cycle Notation gives you a way to compactly write down a permutation. About As A Product Permutation Of Cycles Disjoint Calculator . Related Math Calculators. For example, the permutation (r = (2, 5, 3)(1, 6) in S7 has a four-dimensional. Find the circular permutation of a number. (Note, in the same manner, any permuta-tion can be expressed as a product of transpositions ) (b) Write the permutations given in problems 4 and 17 as compositions of disjoint cycles The order is the least common multiple of the orders of the two cycles Permutation groups are one of the oldest

5, 1, 4, 2, 3. Creating a Permutation. The program will calculate the powers of the permutation. Home / Mathematics / Permutation and combination; Calculates the number of circular permutations of n things. / (n - r)!. That is, it calculates the cyclic subgroup of S_n generated by the element you entered. cell. Example The cycle graph is denoted by C n. Proof. Determine whether the permutations with the disjoint cycles structures from part 1 are even permutations (and hence are in A 8). Search: Permutation As A Product Of Disjoint Cycles Calculator. The method I use for multiplying permutations like this is to think of each cycle as a set of mappings. What is Permutation As A Product Of Disjoint Cycles Calculator. Cartesian coordinate system. Given a different enumeration for the set , both rows change accordingly. Schubert polynomials for vexillary permutations, i Disjoint cycles in S n commute A permutation 2S n has prime order pif and only if it is a product of disjoint p-cycles (c) Write p and q as composites of transpositions, and state the parity Due to theorem 2 Due to theorem 2. Consider an element, let denote the cardinality of, =. Every permutation can be written as a cycle or as a product of disjoint cycles, for example in the above permutation {1 3, 3 5, 5 4, 4 2, 2 1}. One of the nicest things about a permutation is its cycle decomposition. What is the probability 2(n) that 1;2 are in the same cycle of w? Choose w 2 Sn (uniform distribution). For n r 0. n denote the group of permutations of nletters. X ={x1 < x2 < < xn}. number of things : n: Total number of ways: \) Customer Voice. Search: Permutation As A Product Of Disjoint Cycles Calculator. 0. What is Permutation As A Product Of Disjoint Cycles Calculator. When students become active doers of mathematics, the greatest gains of their mathematical thinking can be realized. A permutation matrix is an n n matrix with exactly one entry of 1 in each row and column and the remaining entries 0. Theorem (5.3 Order of a Permutation). What is Permutation As A Product Of Disjoint Cycles Calculator. 10 6 = 1, 000, 000 {\displaystyle 10^ {6}=1,000,000} . 5. If you have a calculator handy, this part is easy: Just hit 10 and then the exponent key (often marked x y or ^ ), and then hit 6.

Likes: 381. The cycles cyc i of a permutation are given as lists of positive integers, representing the points of the domain in which the permutation acts. What is Permutation As A Product Of Disjoint Cycles Calculator. (1,7,9)(2,8)(3,5,6).

The higher the number of cigarettes, the lower the longevity - a dose-dependent relationship. So a transposition is a cycle of length 2, i.e. Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles. To calculate the number of possible permutations of r non-repeating elements from a set of n types of elements, the formula is: The above equation can be said to express the number of ways for picking r unique ordered outcomes from n possibilities. What is Permutation As A Product Of Disjoint Cycles Calculator. The cyclic decomposition of a permutation can be computed in the Wolfram Language with the function PermutationCycles [ p ] and the permutation corresponding to a cyclic decomposition can be computed with PermutationList [ c ]. Cycle Notation gives you a way to compactly write down a permutation. Contribute to Garvit244/Leetcode development by creating an account on GitHub. Cycle: A cycle is a closed path in a graph that forms a loop. The cycles cyc i of a permutation are given as lists of positive integers, representing the points of the domain in which the permutation acts. Calculate the permutations for P (n,r) = n! Circular Permutation Calculator . Then it gives the order of the permutation (when written in disjoint cycle form, the order is simply the least common multiple of the length of the cycles). Cycle notation A permutation can be represented as a composition of permutation cycles. Your first equation, with the powers, is correct Ergo, ANY permutation on the cube is an EVEN permutation! permutation (1 3 5) (2 4) (6 7 8) Natural Language. Then some examples of valid permutations are. Multiplying permutations written in cycle notation follows no easily described pattern, and the cycles of the product can be entirely different from those of the permutations being composed. calculus of variations. Likes: 381. cancel. Find the factorial n! The notation we will use for the permutations of S n is cycle notation. Now we define one-line notation for arbitrary finite symmetric groups. center. / (n - r)!. of the cycles. a (in your example) maps 1 to 3, 3 to 5, 5 to 2, and 2 to 1. "The number of ways of obtaining an ordered subset of r elements from a set of n elements." Spherical Triangle Calculator. calculus. capacity. Stirling numbers of the second kind occur in the field of mathematics called combinatorics and the study of partitions.. Stirling numbers of the second kind are one Putting the numbers in the calculator and selecting to use Kendall's correlation coefficient we can quantify the relationship between smoking and longevity. Permutation As Product of Cycles. If , the two-line notation for is: . When the starting and ending point is the same in a graph that contains a set of vertices, then the cycle of the graph is formed. Here, the individual cycles are represented using the function Cycles . Products of Cycles p. 2 If the elements can repeat in the permutation, the formula is: In both formulas "!" Parity Permutations come in two types : even and odd. OEF Fonctions grammaticales, collection d'exercices niveau cycle 3 sur la fonction grammaticale des mots. For example, The correspondence between a genome G and a genomic matrix M is defined by From this we derive a more explicit formula for the polynomials Dw for 3412-avoiding permutations w and, in particular, an especially nice determinant expression for Dw in the case when w is 312-avoiding 15 The These are not matrices, but permutations, or elements of the group S_8. G = {e, (f1) (f2 f3 f4 f4) (f6) (v1 v2 v3 v4) (v5, v6, v7, v8) } Don't worry about finding the permutation group or how to decide which face corresponds to which fi or which vertex to which vj. census.

Math Input. The notation we will use for the permutations of S n is cycle notation. factorial calculator and examples. Echecs, petits exercices d'checs pour dbutants. A simple question Sn: permutations of 1;2;:::;n Let n 2. So a descent is just an inversion at two adjacent positions In the 2-line form, the elements and their final positions are shown as a matrix with 2 rows: , Salazar and Ugalde 2004) of graphs and with disjoint point sets and and edge sets and is the graph with point set and adjacent with whenever or (Harary 1994, p Shares: 191.

Since the symmetric group is so important in the study of groups, learning cycle notation will speed up your work with the group Sn. Cycles in Permutations Math 184A / A permutation of n elements can be represented by an arrangement of the numbers 1, 2, n in some order. Many times the most interesting information about a permutation are the lengths of its disjoint cycles. For instance, the following permutation decomposes into three cycles: one of length = 2 and two of length = 3. Another common notation is a permutation list: Extended Keyboard. [1] Permutation Problem 1 Choose 3 horses from group of 4 horses The composition of these permutations is shown both in cycle notation and by following the arrows from the left side all the way to the right. Search: Permutation As A Product Of Disjoint Cycles Calculator Enter the email address you signed up with and we'll email you a reset link. Both members and non-members can engage with resources to support the implementation of the Notice and Wonder strategy on this webpage. Sum Calculator; Combination Calculator; Multiplication Calculator; Frequently Used Miniwebtools: Definition. I might be a bit thick in this but i just cant figure out how to answers this exam question: Calculate p to the power of 100, writing your answer in functional notation. a) Let us express the given permutation with disjoint cycles using the cycle notation : (12)(356) That is to say, every 2 An, n 3, can be written as = C1C2 Ck; where each Ci is one of these 3-cycles Remark: Here we allow the empty product which by convention represents the identity element e 2S n Because a CA is Because a CA is. The two-line notation for a permutation is not unique. This applet allows students to experiment with composing two permutations in Sym(4) to explore the properties of this group.

To count the permutations of a list is to count the number of unique rearrangements of the list. calculator. Question: Compute the product of the permutations in disjoint cycle notation. Of course a cycle of length 1 would simply be the identity in p= 1 2-> p is a ^"^cycle in G^1 (after possibly The set of all permutations of the set f1, ,ngis a group under function composition The cycles of length 1 correspond to telomeres while the cycles of length 2 correspond to adjacencies It manipulates paremutations in disjoint cycle notation and allows for Search: Permutation As A Product Of Disjoint Cycles Calculator. A permutation cycle is a set of elements in a permutation that trade places with one another. size() Returns the size of the permutation self. cardinal number. Here's another example -- a permutation on with cycle decomposition: The cycle type here is (the point is a fixed point). The two-line notation for is a description of in two aligned rows. X = { x 1 < x 2 < < x n }. "(1 2 3 4 5)(6 7)".]

Factorial Calculator. 5. of a number, including 0, up to 4 digits long. [Eg. Prof. Tesler Ch. possible permutations to consider. Indeed we may conclude: Theorem 5.7. The set of n n permutation matrices forms a group under multiplication which is isomorphic to Sn. Denition A transposition is a permutation which interchanges two elements and leaves all others xed. Also, remember that ab means "apply b, then apply a. cycle type (2,2,2).) PERMUTATION GROUPS 63 Question: Is there an easy way to compute the order of a permutation? Shares: 191. In mathematics, particularly in combinatorics, a Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of n objects into k non-empty subsets and is denoted by (,) or {}. 6.1. Disjoint cycle notation for permutations. catenary. Students select two permutations, one from each drop-box. A permutation with cycle type ( a 1, a 2, , a n) can be written as a product of a 2 + 2 a 3 + + ( n - 1) a n = n - ( a 1 + a 2 + + a n) transpositions, and no fewer. When there is no repetition of the vertex in a closed circuit, then the cycle is a simple cycle. Proposition 6.10 Any permutation is a product of transpositions. How many non-trivial 10 cycles does the product have? Connected Teaching and Learning from HMH brings together on-demand professional development, students' assessment data, Find the number of permutations. The Permutation Calculator is used to calculate the permutation, which is the number of ways to select k out of n items, where (unlike combinations) order does matter. Fix once and for all a total order < < on X X. OEF Dveloppement durable, collection d'exercices sur le dveloppement durable (cycle 3). OEF Nature des mots, collection d'exercices niveau cycle 3 sur la classe grammaticale des mots. center (of a circle) center (of a hyperbola) center (of a regular polygon) center (of a sphere) (ab), for some a,b {1,2,,n} with a 6= b. What is Permutation As A Product Of Disjoint Cycles Calculator. Leetcode solutions in Python . For simplicity's sake, we will say that if is itself the identity permutation or if is wholly a cycle, then can be written trivially as a product of a single cycle Okay, to write as disjoint cycles, the easiest thing to do is to start by tracing 1, then see where that goes, then see where THAT goes, etc to_cycles() returns Also, choose an, and define. To create a permutation in Maple, you must specify either an explicit list of the images of the integers in the range 1..n, or the disjoint cycle structure of the permutation.In the first case, you use a list L of the form [a__1, a__2, , cent. 2. In this case the coefficient is -0.541 meaning that there exists a moderate inverse association between X and Y. How Cycle Notation Works 1 Solution: Next Permutation 2 Solution: Trim a Binary Search Tree 157 more parts 3 Leetcode Solutions Index 4 Solution: Minimize Deviation in Array 5 Solution: Vertical Order Traversal of a Binary Tree 6 Solution: Count Ways to Make Array With Product 7 Solution: Smallest String With A Given Numeric Value 8 Solution: Linked List Cycle 9 Solution: Path With Minimum The Permutation Calculator is used to calculate the permutation, which is the number of ways to select k out of n items, where (unlike combinations) order does matter. The top row lists the elements of , and the bottom row lists, under each element of , its image under . be the orbits of with more than 1 element. Cartesian product (of sets A and B) categorical data. To obtain k cycles, insert 6 into a permutation of [5] with k cycles (if added to an existing cycle) or k - 1 cycles (if added as a new cycle). Eg. n! there are lots of examples and I'm expected to be able to calculate them pretty 'fast', so, is there a way to 'read out' the composition of two permutations direct from the cycle notation (I use standard function composition, which is Solve for the number of permutations. Of course, this means we still have 10!


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# The equivalent Potential difference across an inductor and a capacitor?

In the figure below, The current intensity is 4A, and V1 reads 80V, and V2 reads 64V. What is the reading of V3? In another way what is the equivalent voltage for an inductor and a capacitor? In my textbook, it says that V3 reads 16V, but I don't understand why the voltage across Xc and Xl together would be:

V1-V2=V3= 80-64=16.

Could you please explain it for me?

4A flows through the resistor, inductor and capacitor and its phase is common to all three.

The voltage drop across the resistor = 4*R and is in phase with the current.

The voltage drop across the inductor leads the current by 90 degrees and will have the value (I * XL) = 4 * 20 = 80V

The voltage across the capacitor lags the current by 90 degrees and will have the value (I * Xc) = 4 * 16 = 64V

The voltage across the inductor and the capacitor are 180 degrees apart or antiphase so they cancel each other out leaving a nett voltage of 80 - 64 = 16V (the voltage measured by the meter).

This voltage is mainly due to the inductor so it will lead the current by 90 degrees.

For the inductance your calculation is correct, as you have $$Z_L = jX_L$$ However, for the capacitance you should remember the sign of the impedance: $$Z_C = -jX_C$$ Both capacitors and inductors are frequency dependent, so if you would just put a DC voltage source over the circuit you should expect different voltages. The inductor will look like a short with 0Ω impedance and the capacitor will be an infinitely high impedance.

The reactance and inductance values that are given are for a certain frequency and moment in time.

The sign in the reactance formula originates from the frequency domain formula for the impedance of a capacitor: $$Z_C = \frac{1}{j\omega C}$$ Rewriting this in the reactive part would result in $$Z_C = -j \frac{1}{\omega C}$$ where $$\frac{1}{\omega C} = X_C$$

• Does "Z" in your answer refer to the voltage? Becuase I use it to refer to the impedance. – Asmaa Feb 5 '17 at 19:18
• Z is the electrical impedance, X is called the reactance. So actually the impedance is negative. – Douwe66 Feb 5 '17 at 19:22
• OK, but I thought the impedance is the equivalent resistance of more than one component (e.g. an inductor and a resistance). So, I think the resistance of the capacitor will be represented by the reactance Xc. Could you please give me more explanation about the negative sign you used in the formula? – Asmaa Feb 5 '17 at 19:27
• @Douwe66 It is wrong to say the impedance is negative. The reactance and impedance of the inductor and capacitor are complex (imaginary, to be specific) so 'positive' and 'negative' do not apply (they are off the number line, so to speak or not in $\mathbb{R}$). – Spehro Pefhany Feb 5 '17 at 19:39
• I added a bit more in my answer. The impedance is indeed the equivalent resistance, but it is frequency dependent. It is by convention that the sign is put with the capacitor. – Douwe66 Feb 5 '17 at 19:43

Arthough arbitrary we consider L = +ve reactance and C =-ve reactance where voltage lags current. The negative sign is a rule of complex impedance for sine cosine phase shifts in electronics +/-90 deg V/I phase shifts only in pure reactive parts. such that series LC are always 180 deg apart but their impedance changes with f.

Thus the voltages cancel.

The rules of j ( or i symbol in pure math which was changed to prevent confusion with current) were discovered by Pyramid design Engineers and not discovered again until Napoleonic days by Descartes j=√ -1 and 1/j=-j fits the way LC impedances affect voltage with amplitude cancellation using a shared current. When the impedances are exactly equal, the voltage cross LC becomes zero (0) and a sweep frequency would show a notch filter response past this "natural" resonant frequency.


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## Introductory Algebra for College Students (7th Edition)

To find the slope, we use the formula: $$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ We now plug in points we find on the graph of the line into this equation to find the slope of this line. We will use the points $(-2, 2)$ and $(3, 2)$: $$m = \frac{2 - 2}{3 - (-2)}$$ Simplify: $$m = \frac{0}{5}$$ Simplify the fraction: $$m = 0$$


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# CheckMate: Automated Synthesis of Hardware Exploits and Security Litmus Tests

## Author(s): Trippel, Caroline; Lustig, Daniel; Martonosi, Margaret

To refer to this page use: http://arks.princeton.edu/ark:/88435/pr11j9v
DC FieldValueLanguage
dc.contributor.authorTrippel, Caroline-
dc.contributor.authorLustig, Daniel-
dc.contributor.authorMartonosi, Margaret-
dc.date.accessioned2021-10-08T19:45:22Z-
dc.date.available2021-10-08T19:45:22Z-
dc.date.issued2018en_US
dc.identifier.citationTrippel, Caroline, Daniel Lustig, and Margaret Martonosi. "CheckMate: Automated Synthesis of Hardware Exploits and Security Litmus Tests." 51st Annual IEEE/ACM International Symposium on Microarchitecture (MICRO) (2018): pp. 947-960. doi:10.1109/MICRO.2018.00081en_US
dc.identifier.urihttps://check.cs.princeton.edu/papers/ctrippel_MICRO51.pdf-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/pr11j9v-
dc.description.abstractRecent research has uncovered a broad class of security vulnerabilities in which confidential data is leaked through programmer-observable microarchitectural state. In this paper, we present CheckMate, a rigorous approach and automated tool for determining if a microarchitecture is susceptible to specified classes of security exploits, and for synthesizing proof-of-concept exploit code when it is. Our approach adopts "microarchitecturally happens-before" (μhb) graphs which prior work designed to capture the subtle orderings and interleavings of hardware execution events when programs run on a microarchitecture. CheckMate extends μhb graphs to facilitate modeling of security exploit scenarios and hardware execution patterns indicative of classes of exploits. Furthermore, it leverages relational model finding techniques to enable automated exploit program synthesis from microarchitecture and exploit pattern specifications. As a case study, we use CheckMate to evaluate the susceptibility of a speculative out-of-order processor to Flush+Reload cache side-channel attacks. The automatically synthesized results are programs representative of Meltdown and Spectre attacks. We then evaluate the same processor on its susceptibility to a different timing side-channel attack: Prime+Probe. Here, CheckMate synthesized new exploits that are similar to Meltdown and Spectre in that they leverage speculative execution, but unique in that they exploit distinct microarchitectural behaviors-speculative cache line invalidations rather than speculative cache pollution-to form a side-channel. Most importantly, our results validate the CheckMate approach to formal hardware security verification and the ability of the CheckMate tool to detect real-world vulnerabilities.en_US
dc.format.extent947 - 960en_US
dc.language.isoen_USen_US
dc.relation.ispartof51st Annual IEEE/ACM International Symposium on Microarchitecture (MICRO)en_US
dc.rightsAuthor's manuscripten_US
dc.titleCheckMate: Automated Synthesis of Hardware Exploits and Security Litmus Testsen_US
dc.typeConference Articleen_US
dc.identifier.doi10.1109/MICRO.2018.00081-
pu.type.symplectichttp://www.symplectic.co.uk/publications/atom-terms/1.0/conference-proceedingen_US

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Maxwell-boltzman distrubution

1. Mar 17, 2010

E92M3

If I have two particles that follows the maxwell velocity distribution:
$$\phi(v_i)dv_i=4 \pi v_i^2 \left ( \frac{m_iv_i}{2\pi kT} \right ) ^{3/2}e^{\frac{-m_iv_i^2}{2kT}}dv_i$$
Why is their combined distribution:
$$\phi(v)dv=4 \pi v^2 \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2}e^{\frac{-\mu v^2}{2kT}}dv$$
where mu is the reduced mass and v=v2-v1
I have these questions because I don't quite follow these derivations.
http://dissertations.ub.rug.nl/FILES/faculties/science/2007/a.matic/c2.pdf [Broken]
http://www.astro.psu.edu/users/rbc/a534/lec11.pdf
Namely, I not sure why the following holds:
$$\int_0^\infty \int_0^\infty \phi(v_1) \phi(v_2) v_1 v_2 \sigma dv_1 dv_2 = 4\pi \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2} \int_0^\infty v^3 \sigma e^{\frac{-\mu v^2}{2kT}}dv$$

Last edited by a moderator: May 4, 2017
2. Mar 18, 2010

mathman

You have v=v1-v2. Let u=v1+v2. The new integration is straightforward as long as the integral limits are -∞ to ∞. You then need |v1v2| instead of v1v2 in the integrand. The u integration should leave you with what you want.


text:
# How is the time complexity of Sieve of Eratosthenes is n*log(log(n))?

Pre-requisite: Sieve of Eratosthenes

What is Sieve of Eratosthenes algorithm?
In order to analyze it, let’s take a number n and the task is to print the prime numbers less than n. Therefore, by definition of Sieve of Eratosthenes, for every prime number, it has to check the multiples of the prime and mark it as composite. This process continues until a value p which is the highest prime number less than n.

Understanding the n*log(log n) time complexity of Sieve of Eratosthenes

1. If it is assumed that the time taken to mark a number as composite is constant, then the number of times the loop runs is equal to:
2. On taking n common from the above equation, the above equation can be rewritten as:
$n*(\frac{1}{2}&space;+&space;\frac{1}{3}&space;+&space;\frac{1}{5}&space;+&space;\frac{1}{7}&space;+&space;......&space;p)$

3. It can be proved as below with the help of Harmonic Progression of the sum of primes:
$\frac{1}{2}&space;+&space;\frac{1}{3}&space;+&space;\frac{1}{5}&space;+&space;\frac{1}{7}&space;+&space;.....&space;=&space;log(log(n))$

Proof of Harmonic Progression of the sum of primes:
In order to understand the proof, the prerequisite is the Harmonic series and Taylor series expansion.

4. On substituting this in the equation, we get the time complexity as:
$O(n*(log(log(n))))$

Hence the time complexity of Sieve of Eratosthenes is n*log(log(n))

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glFramebufferRenderbuffer

Name

glFramebufferRenderbuffer — attach a renderbuffer object to a framebuffer object

C Specification

 void glFramebufferRenderbuffer( GLenum target, GLenum attachment, GLenum renderbuffertarget, GLuint renderbuffer);

Parameters

target

Specifies the framebuffer target. The symbolic constant must be GL_FRAMEBUFFER.

attachment

Specifies the attachment point to which renderbuffer should be attached. Must be one of the following symbolic constants: GL_COLOR_ATTACHMENT0, GL_DEPTH_ATTACHMENT, or GL_STENCIL_ATTACHMENT.

renderbuffertarget

Specifies the renderbuffer target. The symbolic constant must be GL_RENDERBUFFER.

renderbuffer

Specifies the renderbuffer object that is to be attached.

Description

glFramebufferRenderbuffer attaches the renderbuffer specified by renderbuffer as one of the logical buffers of the currently bound framebuffer object. attachment specifies whether the renderbuffer should be attached to the framebuffer object's color, depth, or stencil buffer. A renderbuffer may not be attached to the default framebuffer object name 0.

If renderbuffer is not 0, the value of GL_FRAMEBUFFER_ATTACHMENT_OBJECT_TYPE for the specified attachment point is set to GL_RENDERBUFFER and the value of GL_FRAMEBUFFER_ATTACHMENT_OBJECT_NAME is set to renderbuffer. GL_FRAMEBUFFER_ATTACHMENT_TEXTURE_LEVEL and GL_FRAMEBUFFER_ATTACHMENT_TEXTURE_CUBE_MAP_FACE are set to the default values 0 and GL_TEXTURE_CUBE_MAP_POSITIVE_X, respectively. Any previous attachment to the attachment logical buffer of the currently bound framebuffer object is broken.

If renderbuffer is 0, the current image, if any, attached to the attachment logical buffer of the currently bound framebuffer object is detached. The value of GL_FRAMEBUFFER_ATTACHMENT_OBJECT_TYPE is set to GL_NONE. The value of GL_FRAMEBUFFER_ATTACHMENT_OBJECT_NAME is set to 0. GL_FRAMEBUFFER_ATTACHMENT_TEXTURE_LEVEL and GL_FRAMEBUFFER_ATTACHMENT_TEXTURE_CUBE_MAP_FACE are set to the default values 0 and GL_TEXTURE_CUBE_MAP_POSITIVE_X, respectively.

Notes

If a renderbuffer object is deleted while its image is attached to the currently bound framebuffer, then it is as if glFramebufferRenderbuffer had been called with a renderbuffer of 0 for the attachment point to which this image was attached in the currently bound framebuffer object. In other words, the renderbuffer image is detached from the currently bound framebuffer. Note that the renderbuffer image is specifically not detached from any non-bound framebuffers. Detaching the image from any non-bound framebuffers is the responsibility of the application.

Errors

GL_INVALID_ENUM is generated if target is not GL_FRAMEBUFFER.

GL_INVALID_ENUM is generated if renderbuffertarget is not GL_RENDERBUFFER and renderbuffer is not 0.

GL_INVALID_ENUM is generated if attachment is not an accepted attachment point.

GL_INVALID_OPERATION is generated if the default framebuffer object name 0 is bound.

GL_INVALID_OPERATION is generated if renderbuffer is neither 0 nor the name of an existing renderbuffer object.

Associated Gets

glGetFramebufferAttachmentParameteriv

Examples

Create a framebuffer object with a renderbuffer-based color attachment and a renderbuffer-based depth attachment.
// fbo_width and fbo_height are the desired width and height of the FBO.
// For Opengl <= 4.4 or if the GL_ARB_texture_non_power_of_two extension
// is present, fbo_width and fbo_height can be values other than 2^n for
// some integer n.

// Build the texture that will serve as the color attachment for the framebuffer.
GLuint color_renderbuffer;
glGenRenderbuffers(1, &color_renderbuffer);
glBindRenderbuffer( GL_RENDERBUFFER, (GLuint)color_renderbuffer );
glRenderbufferStorage( GL_RENDERBUFFER, GL_RGBA8, fbo_width, fbo_height );
glBindRenderbuffer( GL_RENDERBUFFER, 0 );

// Build the texture that will serve as the depth attachment for the framebuffer.
GLuint depth_renderbuffer;
glGenRenderbuffers(1, &depth_renderbuffer);
glBindRenderbuffer( GL_RENDERBUFFER, (GLuint)depth_renderbuffer );
glRenderbufferStorage( GL_RENDERBUFFER, GL_DEPTH_COMPONENT, fbo_width, fbo_height );
glBindRenderbuffer( GL_RENDERBUFFER, 0 );

// Build the framebuffer.
GLuint framebuffer;
glGenFramebuffers(1, &framebuffer);
glBindFramebuffer(GL_FRAMEBUFFER, (GLuint)framebuffer);
glFramebufferRenderbuffer(GL_FRAMEBUFFER, GL_COLOR_ATTACHMENT0, GL_RENDERBUFFER, color_renderbuffer);
glFramebufferRenderbuffer(GL_FRAMEBUFFER, GL_DEPTH_ATTACHMENT, GL_RENDERBUFFER, depth_renderbuffer);

GLenum status = glCheckFramebufferStatus(GL_FRAMEBUFFER);
if (status != GL_FRAMEBUFFER_COMPLETE)
// Error

glBindFramebuffer(GL_FRAMEBUFFER, 0);


text:
# Has proton decay falsified SO(10) GUT?

1. Dec 25, 2009

### ensabah6

I know that proton decay has falsified SU(5) based around 10e30 years.

What proton decay of SO(10)?

Are SUSY SU(5) and SUSY SO(10) still viable?

If the experiment continues with a null result, what models are next in line to be falsified?

Could protons last forever and thus falsfy ALL GUT approaches?

2. Dec 28, 2009

### arivero

ensabah, have you checked the limits in the "particle data group" current review?

3. Dec 28, 2009

### ensabah6

no, do you have a link? thank :)

4. Dec 28, 2009

### blechman

www-pdg.lbl.gov

the particle physicist's almanac!

5. Dec 28, 2009

### Bob S

6. Dec 28, 2009

### arivero

Argh, I forgot to request the 2010 calendar

7. Dec 28, 2009

### hamster143

You need pages 10 and 11. Besides, that's an outdated source.

http://arxiv.org/abs/0903.0676

increases the limit on the observed partial mean life via $p \rightarrow e^+ \pi^0$ channel (the one predicted by common GUTs) by a factor of 5, to 8.8 x $10^{33}$ years.

The original non-supersymmetric SU(5) prediction was, I think, somewhere around $10^{30}$ years, and that has been ruled out.

However, mean life is extremely sensitive to the structure of the unified theory. It scales roughly as the fourth power of the unification energy (or, more accurately, as the fourth power of mass of the massive boson responsible for the decay, which is probably near unification energy). We already know that naive (minimal) non-supersymmetric SU(5) does not result in clean unification, because all three coupling constants fail to meet at the same point. If we allow additional particles beyond those needed to construct minimal SU(5), all hell breaks loose, the unification energy can be anywhere below Planck mass, and proton lifetime can potentially be as high as $10^{42}$ years.

That is, in fact, the way SUSY GUTs "deal" with the problem - contributions from superparticles distort running couplings just enough to raise the unification energy by a couple of orders of magnitude, and lifetime becomes high enough to agree with observations. But that approach is by no means exclusive to SUSY.

Last edited: Dec 28, 2009
8. Dec 28, 2009

### ensabah6

What about non-SUSY SO(10)? I understand that SO(10) is the most highly favored GUT.

That's fascinating. I know there are talks of building Hyper-Kak...
I know that 10^30 protons is doable, and that 10^31 represents a 10x increase.

I'd imagine it'd take well beyond our lifetimes for Super-Kak to get those kind of proton half-lifes.

Isn't the current experimental upper bounds around 10^35 years?

Has there been serious research into the possibility that the absence of proton decay and magnetic monopoles is suggestive evidence that GUT are wrong? The "unified" gauge couplings at 10^16 GEV is just a coincidence? In other words, are GUT are "fact" and only details matter?

9. Dec 28, 2009

### marcus

What about the series of papers by Nicolai and Meissner? They dispense with GUT, also with low-energy SUSY and with extra dimensions. That surely qualifies as serious research.
You can tell from his tone of voice that Nicolai is quite skeptical of GUT
as he says "since the proton so far has refused to decay."

Shaposhnikov is another whose research explores the possibility of dropping GUT.

10. Dec 28, 2009

### ensabah6

Which paper?

The latest paper builds on two earlier Meissner Nicolai papers
3. arXiv:0907.3298 [ps, pdf, other]
Title: Conformal invariance from non-conformal gravity
Authors: Krzysztof A. Meissner, Hermann Nicolai
Journal-ref: Phys.Rev.D80:086005,2009
Subjects: High Energy Physics - Theory (hep-th)

Since string based TOE is predicated on GUT, skepticism of GUT and extra dimensions is skepticism of string theory unification.

11. Dec 29, 2009

### hamster143

There are different ways to break SO(10) down to SM. One way is to go SO(10) -> SU(5) x U(1), in which case there's essentially no difference from SU(5) case. One other way is to go via SU(4) x SU(2) x SU(2) (Pati-Salam), in which case there's more freedom and we can have proton lifetime up to 10^37 years.

12. Dec 29, 2009

### ensabah6

thanks,

does the guage couplings meet in both SUSY SU(5) and SUSY SO(10)?

Can we experimentally test up to 10^37 years? I infer that SUSY-SO(10) increases proton lifetime by order of magnitude.

13. Dec 29, 2009

### marcus

The best exposition of Nicolai and Meissner's work is Nicolai's talk at the July Planck Scale conference. He gives references there. Great talk. He is repeatedly explicit about dispensing with GUT, for the purposes of this program. Doesn't need it and doesn't want it. Anyone who hasn't watched the talk really ought to.
http://www.ift.uni.wroc.pl/~rdurka/planckscale/index-video.php [Broken]
This gives the menu of all the talks. both the slides PDF and the video.
Just select Nicolai's talk, which is near the top of the menu.

Does your computer not get videos? I forget. If you have a bandwidth problem let me know and I will suggest an alternative.

Last edited by a moderator: May 4, 2017
14. Dec 29, 2009

### ensabah6

http://www.ift.uni.wroc.pl/~planckscale/lectures/1-Monday/3-Nicolai.pdf

thx, if true it obviously spells problems for SUSY

Last edited by a moderator: May 4, 2017


text:
# Revision history [back]

The problem is partially solved according to @Björgvin post and bug 34836

1. Install MacPorts (http://www.macports.org/install.php)

2. Install ps2edit from Macports

$sudo port install pstoedit 3. you can start libreoffice from the terminal: ./Applications/LibreOffice.app/Contents/MacOS/soffice The pdf will be exported with EPS image displayed no problem. However. It only works in terminal mode. I still cannot export PDF with EPS Image when start libreoffice from application folder by double click. I think it is due to from terminal start the libreoffice can find the relevant tools such as "pstoedit" and "convert". Is there any suggestion that I can start libreoffice with these commands available in the default PATH? The problem is partially solved according to @Björgvin post and bug 34836 1. Install MacPorts (http://www.macports.org/install.php) 2. Install ps2edit from Macports$ sudo port install pstoedit

3. you can start libreoffice from the terminal: ./Applications/LibreOffice.app/Contents/MacOS/soffice The pdf will be exported with EPS image displayed no problem.

However. It only works in terminal mode. I still cannot export PDF with EPS Image when start libreoffice from application folder by double click. I think it is due to from terminal start the libreoffice can find the relevant tools such as "pstoedit" and "convert". Is there any suggestion that I can start libreoffice with these commands available in the default PATH?

UPDATE Now I found the solution of launch the GUI app with the user own PATH vars. For example if your "pstoedit" is in "/opt/local/bin" your LO maynot be able to find it, therefore the export PDF with EPS image will be fail. Since Lion, the GUI app environment PATH is only affected by "launchctl" command.

In my case, here is what I did:

p=$(launchctl getenv PATH) launchctl setenv PATH /opt/local/bin:/opt/local/sbin:$p

And it now finally works for me!!!


text:
# Bounding box coordinates taking into consideration (but not applying) rotation

I am trying to make a script that will determine the exact coordinates of an object's bounding box in the render. (I later want to limit the rendered space to that area.)

So far I managed to get the coordinates from the object's bound_box And I managed to take the scale and location into account but I failed with the rotation.

My problem is that I have the coordinates of my bounding box but I don't know how to make them follow the rotation, like the bounding box in the viewport does. When I run the script with a cube it works fine, but when the cube is being rotated the coordinates that are being printed out do not change.

import bpy
import bpy_extras
import mathutils

scene = bpy.context.scene
obj = bpy.context.object
co = bpy.context.scene.cursor_location

scale = bpy.data.objects['cube'].scale

bounding_box_vertices = []
for i in range(8):
bounding_box_vertices.append(i)
bounding_box_vertices[i]=[1, 2, 3]

#the Coordinates from bound_box are saved to bounding_box_vertices[[],[]]
for i in range(8):
for j in range(3):
bounding_box_vertices[i][j]=bpy.data.objects['cube'].bound_box[i][j]*bpy.data.objects['cube'].scale[0] + bpy.data.objects['cube'].location[j]

def rendered_coords(x):
co = mathutils.Vector((bounding_box_vertices[x][0], bounding_box_vertices[x][1], bounding_box_vertices[x][2]))
co_2d = bpy_extras.object_utils.world_to_camera_view(scene, obj, co)

render_scale = scene.render.resolution_percentage / 100
render_size = (int(scene.render.resolution_x * render_scale), int(scene.render.resolution_y * render_scale))
#print("Pixel Coords:", (
return([round(co_2d.x * render_size[0]), round(co_2d.y * render_size[1])])
# ))

#prints ou the 8 X and Y coordinates that the bounding box would have on the render (even though its invisible of course)
for i in range(8):
final_coords = rendered_coords(i)
print(final_coords[0])
print(final_coords[1])

• @R.M They are talking about coordinates in the render, hence the 'rendering' tag is relevant here. – Ray Mairlot Sep 15 '18 at 15:24

The bounding box coordinates of obj.bound_box are in local space. To convert them to world space – assuming you're interested in the world space coordinates and don't just want to consider rotation – you'll have to multiply the world matrix with the coordinates as mentioned in this answer by CoDEmanX.

import bpy
from mathutils import Vector

obj = bpy.context.object  # or bpy.data.objects['cube']
bb_vertices = [Vector(v) for v in obj.bound_box]
mat = obj.matrix_world
world_bb_vertices = [mat * v for v in bb_vertices]

From there you can continue with world_to_camera_view. From your code it's not clear what your obj is, but just to clarify: This function takes the Camera object and not the object with the bounding box. The bounding box coordinates are only considered for the third parameter. Similarly, you will need the mat world matrix of the object with the bounding box.

import bpy
from bpy_extras.object_utils import world_to_camera_view
from mathutils import Vector

scene = bpy.context.scene
obj = bpy.context.object  # or bpy.data.objects['cube']
bb_vertices = [Vector(v) for v in obj.bound_box]
mat = obj.matrix_world
world_bb_vertices = [mat * v for v in bb_vertices]

co_2d = [world_to_camera_view(scene, scene.camera, v) for v in world_bb_vertices]  # from 0 to 1

render_scale = scene.render.resolution_percentage / 100
render_size = list(int(res) * render_scale for res in [scene.render.resolution_x, scene.render.resolution_y])
pixel_coords = [(c.x * render_size[0], c.y * render_size[1]) for c in co_2d]  # in pixels

for p in pixel_coords:
print('{p[0]:.0f}, {p[1]:.0f}'.format(p=p))


text:
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# Triangle ACE is equilateral with side lengths of 8. Points B and D are

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Triangle ACE is equilateral with side lengths of 8. Points B and D are  [#permalink]

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30 Oct 2018, 02:42
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Question Stats:

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Triangle ACE is equilateral with side lengths of 8. Points B and D are the midpoints of line segments AC and CE respectively. Line segment BD is a diameter of the circle with center F. What is the area of the shaded region?

A. 8√2 − 4π
B. 12√3−2π
C. 12√3 − 4π
D. 16√3 − 2π
E. 16√2 − 4π

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Re: Triangle ACE is equilateral with side lengths of 8. Points B and D are  [#permalink]

### Show Tags

30 Oct 2018, 12:51
Side of eq. triangle = 8

Area of triangle = $$\sqrt{3}$$*8*8/4 = 16$$\sqrt{3}$$

Property of equilateral triangle is that line segment connecting midpoints of two sides divides the triangle in the ratio of 1:3
So, portion below the line segment joining midpoints = (3/4) * 16$$\sqrt{3}$$ = 12$$\sqrt{3}$$

Diameter of circle = 4, radius = 2
Area = pi*2*2 = 4*pi
Half the area = 2*pi

Area of shaded region = 12$$\sqrt{3}$$ - 2*$$pi$$

Hence Option B!
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Re: Triangle ACE is equilateral with side lengths of 8. Points B and D are  [#permalink]

### Show Tags

30 Oct 2018, 13:25
Area of ACE = √3/4*8*8 = 16√3
Area of BDC = 1/4 * ACE = 4√3
Area of semi circle B-F-D = 1/2 * π * 2 * 2 = 2π
Area of shaded region = 16√3 - 4√3 - 2π = 12 - 2π
Re: Triangle ACE is equilateral with side lengths of 8. Points B and D are   [#permalink] 30 Oct 2018, 13:25
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text:
# Does digits of pi contain all possible substrings? [duplicate]

This question already has an answer here:

We just had another pi day, and once again people talk about how infinite pi is, and that it contains everything.

It seems to me completely irrational to expect that just because something is infinite and non-repeating it means that it contains all possible messages. I could easily (uhm, with infinite time and effort) construct an irrational number by rolling dice and adding the value to the end of the base 10 string. That would be infinite and non-repeating, but never contain 0,7,8 or 9.

Is this just a problem of base conversion, which means that every irrational number contains "everything", or is there a pi-specific proof that it does contain " everything"? Of course we can read the dice number in base 6 and get all possible values, but that's a bijection, which i suspect is more powerful than a base conversion.

## marked as duplicate by Dietrich Burde, Shaun, Aweygan, Community♦Mar 15 '17 at 14:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• It's widely imagined that $\pi$ is a "normal number" which would mean that it's digits contained every finite string of digits. But this is not known to be true. – lulu Mar 15 '17 at 11:43

## 1 Answer

This has come up before I think, but the business about a number 'containing all possible strings of any length' is a consequence of a number being normal, which very approximately says that no finite block appears more or less often than any other its sequence of digits.

There is a result of Borel (though its proof was fixed by Faber and then Hausdorff), that 'almost all' real numbers are normal ('almost all' here means the complement has Lebesgue measure zero). Of course there are infinitely (in fact uncountably) many irrational numbers whose representation in a given base is not normal, as you described I could decide to only pick digits from a small range of possible digits. However in some sense you are still doing something 'non-general'.

It's possibly worth noting that $\pi$ is not actually known to be normal, though as the Wikipedia page suggests - it is expected.


text:
# Viewpoint: Magnetic Hose Keeps Fields from Spreading

Physics 7, 67
Magnetic fields could be funneled over longer distances than currently possible using a magnetic “hose” made of ferromagnetic and superconductor materials.

Much of modern technology rests on our ability to send electromagnetic waves from point A to point B with minimal loss of energy and information. Infrared waves confined to optical fibers carry data over the internet, whereas raw electric power is distributed to large cities through high-voltage transmission lines. Static magnetic fields are, however, much more difficult to channel long distance. Instead, these fields typically fall off quickly with distance from their source (usually a magnetic dipole.) A solution to this problem has now been proposed by Alvaro Sanchez of the Autonomous University of Barcelona, Spain, and his colleagues [1]. The authors, reporting in Physical Review Letters, have designed and tested a magnetic “hose” that channels the magnetic field from a source to a distance more than 10 centimeters away. Although magnetic flux guides already exist, the authors’ new design has the potential to transfer magnetic fields over longer distances without degradation, and could be used to increase the storage density of magnetic recording media, or to create higher resolution magnetic images of materials and devices, among other applications.

The magnetic hose (Fig. 1) designed by Sanchez and his colleagues consists of alternating concentric cylinders of a high-permeability material (a ferromagnet) and a low-permeability material (a superconductor). Roughly speaking, when a source field is placed at one end of the cylinder, the ferromagnetic shells magnetize and regenerate the source field along the length of the cylinder, while the superconducting shells keep the field lines from spreading outward. The authors show numerically that the hose is better able to transfer a field as more layers are added, but that even a simple two-layer design acts as an effective magnetic hose. As a proof of concept, the team made a two-layer magnetic hose, $14$ centimeters long, out of a layer of superconducting yttrium barium copper oxide, wrapped around a cylinder made of a ferromagnetic iron alloy. Using a current-carrying loop as a source field, they showed the field strength transferred by their design was up to 4 times stronger than that transferred by a simpler hose made from only a ferromagnetic cylinder.

Physicists have known how to generate and manipulate static solenoidal magnetic fields for almost two centuries. But the design from Sanchez and co-workers grew out of the relatively new field of transformation optics, which is a mathematical framework for studying how materials with unusual dielectric and magnetic properties can distort the path of light. These materials can be engineered by building up arrays of components—such as superconductors or nanocircuits—into “metamaterials.” Transformation optics has inspired designs for “superlenses” that focus light beyond the diffraction limit, illusion optics that make one object look like another, single-layer metasurfaces with unusual refractive properties, and perhaps most spectacularly, “cloaks.” In principle, one can build a metamaterial structure around an object and have it disappear from view; the object casts no shadow and reflects none of the light impinging on it, rendering it effectively invisible. Various theoretical schemes for cloaking have been proposed [2, 3], and a few experiments have shown limited evidence of invisibility [4].

The transformation optics behind cloaking offers a route from merely hiding an object immersed in a static magnetic field to actually transmitting the field a useful distance. Static magnetic field cloaking is, in principle, easier than cloaking light because the need to match the phase of the wave passing by and going through the cloak is eliminated, and the materials making up the cloak need to meet fewer requirements. The first proposal of a dc magnetic cloak [5] and early experimental implementations of it using superconductors [6] have inspired progress in the field of magnetostatic transformation optics [7]. For example, the traditional way to shield objects from external fields is to use materials with either a very high or near-zero permeability (i.e., $\mu$ metals and superconductors, respectively.) But these shields also distort field lines, making the object detectable. This problem can be avoided by using a shell made from a magnetically anisotropic material. The original concept, however, required the magnetic permeability to take extreme values near the inner and outer edges of the shell. A more recent proposal [8] by Sanchez and collaborators shows that a cylindrical shell with a locally uniform, but anisotropic permeability can work well as a cloak.

In 2012, researchers made the first experimental cylindrical dc magnetic cloak [9], which consisted of arrays of superconducting lead plates and ferromagnetic permalloy layers. This early magnetic cloak was effective at shielding its interior from externally applied magnetic fields. And the fields outside the shield, close to its surface, were almost completely unaltered by the presence of the cloak over a range of magnetic field values.

The same year, Sanchez and his colleagues built a much simpler dc magnetic cloak out of two coaxial cylinders; the inner and outer cylinders were made from rolled-up superconductor and ferromagnetic foil, respectively [10]. The design could only cloak an object from uniform external magnetic fields and the cloaking, while significant, was not complete. But its advantage was that it didn’t require any form of microfabrication and could be built from commercially available products.

To make the step from a magnetic cloak to a hose, Sanchez and his colleagues have taken advantage of the so-called magnetic transfer slab concept of transformation optics [1]. This slab has an infinite magnetic permeability perpendicular to the slab and zero permeability in the lateral directions. Such a structure will faithfully transfer the magnetic field entering on one surface through to the other surface, without distorting the fields outside the slab. A sufficiently large slab with such extreme values of permeability would be impractical to build, but the authors use a mathematical mapping to show that what seems like a vastly different structure—a multilayered finite cylindrical geometry—provides the equivalent static magnetic field transfer.

Magnetic hoses could have a number of interesting applications. High-permeability ferromagnetic “flux hoses” are currently used in scanning superconducting quantum interference device (SQUID) microscopes to improve the spatial resolution of magnetic-field images. This new flux hose idea can help to increase the signal-to-noise ratio of the images. Another application would be magnetic resonance imaging in which the patient and superconducting magnet required to generate the magnetic field could be physically separated, and the intense magnetic field could be applied locally to the patient through a magnetic hose. (In principle, what limits the length of the hose is the need to cool the superconducting shells to cryogenic temperatures.) One can also imagine creating magnetic flux circuits for various applications. For example, by tapering the geometrical scale of the hose, one can reduce the spatial extent of the field to enable nanoscale bit formation in magnetic media. Combinations of magnetic hoses could also be used to tailor the spatial shape of magnetic fields, making them more uniform, or creating large gradients, for example. Finally, magnetic hoses might be useful for quantum information processing, in which pin-point manipulation of quantum systems is required.

## References

1. C. Navau, J. Prat-Camps, O. Romero-Isart, J. I. Cirac, and A. Sanchez, “Long-Distance Transfer and Routing of Static Magnetic Fields,” Phys. Rev. Lett. 112, 253901 (2014)
2. U. Leonhardt, “Optical Conformal Mapping,” Science 312, 1777 (2006)
3. J. B. Pendry, D. Schurig, and D. R. Smith, “Controlling Electromagnetic Fields,” Science 312, 1780 (2006)
4. D. Schurig, J. J. Mock, B. J. Justice, S. A. Cummer, J. B. Pendry, A. F. Starr, and D. R. Smith, “Metamaterial Electromagnetic Cloak at Microwave Frequencies,” Science 314, 977 (2006)
5. B. Wood and J. B. Pendry, “Metamaterials at Zero Frequency,” J. Phys. Condens. Matter 19, 076208 (2007)
6. F. Magnus, B. Wood, J. Moore, K. Morrison, G. Perkins, J. Fyson, M. C. K. Wiltshire, D. Caplin, L. F. Cohen, and J. B. Pendry, “A d.c. Magnetic Metamaterial,” Nature Mater. 7, 295 (2008)
7. S. M. Anlage, “The Physics and Applications of Superconducting Metamaterials,” J. Opt. 13, 024001 (2011); P. Jung, A. V. Ustinov, and S. M. Anlage, “Progress in Superconducting Metamaterials,” Supercond. Sci. Technol. 27, 073001 (2014)
8. A. Sanchez, C. Navau, J. Prat-Camps, and D. X. Chen, “Antimagnets: Controlling Magnetic Fields with Superconductor-Metamaterial Hybrids,” New J. Phys. 13, 093034 (2011)
9. S. Narayana and Y. Sato, “DC Magnetic Cloak,” Adv. Mater. 24, 71 (2012)
10. F. Gömöry, M. Solovyov, J. Šouc, C. Navau, J. Prat-Camps, and A. Sanchez, “Experimental Realization of a Magnetic Cloak,” Science 335, 1466 (2012)

Steven Anlage is a professor of Physics and member of the Center for Nanophysics and Advanced Materials at the University of Maryland, College Park, where he leads experimental research activities in superconducting metamaterials, quantum chaos, and various types of quantitative high-resolution near-field microscopy. His research focus is on the basic physics of superconducting materials (both natural and artificial) and addressing fundamental questions concerning wave propagation in complex scattering systems. He is also active in developing applications that exploit time-reversal invariance and spatial reciprocity for electromagnetic wave propagation, such as directed communication and wireless power transfer. He received his Ph.D. from Caltech and did postdoctoral work at Stanford University. Recently he was appointed a visiting professor at the Center for Functional Nanostructures at the Karlsruhe Institute of Technology in Germany. For more information see http://anlage.umd.edu.

Magnetism

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text:
# Can one force the octupole moments of a charge distribution (neutral and with vanishing dipole moment) to vanish using a suitable translation?

In a previous question, I noted that if you have a charge distribution with nonzero charge, then it is possible to choose an origin (at the centre of charge) which makes its dipole moment vanish, and which therefore takes out the subleading order in the multipole expansion of the resulting electrostatic potential $$\Phi(\mathbf{r})=\frac{Q}{|\mathbf{r}-\mathbf{r}_c|}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_c|^{3}}\right).$$ Following on that standard observation, I asked for a result for the next layer up on the multipole expansion: given a neutral system, with a nonzero dipole moment, is it possible to choose an origin which will completely kill the coefficients on the subleading multipole, i.e. which will make all the quadrupole moments vanish.

The answer to that, as it turns out, is much more restrictive than the scalar case: this is possible if and only if the quadrupole is cylindrically symmetric (i.e. if and only if it has two equal eigenvalues) and the dipole moment lies along that axis of symmetry.

I now want to ratchet up the position on the multipole ladder, and ask about the next nontrivial step:

• Given a charge distribution with zero total charge and dipole moment, is it possible to choose an origin such that the octupole moments will vanish completely? What restrictions need to be imposed on the quadrupole and octupole moments for this to be possible?

There is also the obvious question of what happens for some arbitrary multipole $\ell$ at some finite point up the ladder, but given how hard the quadrupole case was, I'd like to keep it simple: kind of like in this other question along similar lines, I feel that there isn't yet a useful pattern that can be generalized to arbitrary multipoles, particularly since the passage from quadrupoles to higher-rank tensor takes out the eigenvectors as a tool of analysis, and I don't really know of good equivalent tools for higher-rank tensors. So I'd like to keep things focused for now and keep the general case for later.

Let's see how this helps. If the monopole moment does not vanish, then its value is independent of origin. This means that the dipole moment is not independent of origin. If we change the origin through $$x^i\to x^i+a^i$$, then the dipole changes as $$$$d^i \to d^i +a^i M$$$$ where M is the non vanishing monopole moment. Since the dipole moment has 3 components, it is possible to choose the origin to force the dipole moment to vanish. This same choice of origin will not necessarily cause any other multipole moment to vanish.
If the monopole moment vanishes, then the value of the dipole moment is independent of origin. This means that changing the origin will affect the quadrupole moment. It has 5 degrees of freedom and so, in general, a change of origin cannot make the quadrupole moment vanish. You would need two additional condition make its vanishing even possible. The quadrupole moment changes as $$$$Q^{ij} \to Q^{ij} + 2\left(a^{(i}d^{j)}-\frac13\delta^{ij}a^k d_k\right).$$$$ This shows the exact form that the quadrupole moment must have in order for there to be a chance for it to vanish.


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# Difference between Poynting vector and energy flux density?

Are those two terms the same, or...? My book says that the Poynting vector is an energy flux density given by:

$$\mathbf{S} = \frac{1}{\mu_{0}}(\mathbf{E} \times \mathbf{B})$$

So that alone should indicate that it's not the same thing. But I've been looking through my book, and every time I look up energy density, not flux, it points me in the direction of the Poynting vector.

So, are they the same, and if not, what is the energy density given by?

-
The energy density (energy per unit volume) of electromagnetic fields in vacuum is given by $\frac{1}{2}(\varepsilon_0 E^2+B^2/\mu_0)$. The Poynting flux is different: it is the flux of that energy passing through some surface (energy per unit surface per unit time). You can see it as the rate at which the energy is displaced.
@DenverDang $E^2 = \vec{E} \cdot \vec{E}$ is a scalar. Likewise with $B^2$. The energy density does not have a direction. $\vec{E} \times \vec{B}$ on the other hand does have direction... –  dmckee May 26 at 21:59


text:
# Eigen state

#### aura

Consider particle of mass m in a cubic box of length L which has energy spectrum given by E=(k sqr)/2m =2 (pi sqr) (nx sqr+ ny sqr +nz sqr)/m (L sqr).what will be the density of states (eigen states per unit energy interval)

k is boltzman const..nx,ny,nz are unit vectors in resp. directions....

Related Introductory Physics Homework News on Phys.org

#### OlderDan

Homework Helper
aura said:
Consider particle of mass m in a cubic box of length L which has energy spectrum given by E=(k sqr)/2m =2 (pi sqr) (nx sqr+ ny sqr +nz sqr)/m (L sqr).what will be the density of states (eigen states per unit energy interval)

k is boltzman const..nx,ny,nz are unit vectors in resp. directions....
Aren't the n values the eigenstates with n_j = 1,2,3....?

This looks to be a counting problem to find the number of combinations of the three n values leading to the same sum of squares. Clearly, for the lowest energy there is only one. After that, what are the possibilities, and what happens to the difference between energy levels as the n values increase?

#### aura

OlderDan said:
Aren't the n values the eigenstates with n_j = 1,2,3....?

This looks to be a counting problem to find the number of combinations of the three n values leading to the same sum of squares. Clearly, for the lowest energy there is only one. After that, what are the possibilities, and what happens to the difference between energy levels as the n values increase?
oops! a big printing mistake...thats eigen vector...nx,ny,nz

now can u solve this at least the explanation...

#### OlderDan

Homework Helper
aura said:
oops! a big printing mistake...thats eigen vector...nx,ny,nz

now can u solve this at least the explanation...
$$E_{n_x,n_y,n_z} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[n_x^2+n_y^2+n_z^2\right]$$

$$E_{1,1,1} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[3\right]$$

$$E_{2,1,1} = E_{1,2,1} = E_{1,1,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[6\right]$$

$$E_{2,2,1} = E_{2,1,2} = E_{1,2,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[9\right]$$

$$E_{3,1,1} = E_{1,3,1} = E_{1,1,3} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[11\right]$$

$$E_{2,2,2} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[12\right]$$

$$E_{1,2,3} = E_{1,3,2} = E_{2,1,3} = E_{2,3,1} = E_{3,1,2} = E_{3,2,1} = \left[\frac{2\ \pi^2}{mL^2}\right] \left[14\right]$$

etc, etc.

I believe you are supposed to be figuring out all possible energies and how many degenerate states there are for each energy, and then divide the number of states by some energy interval to find the density. Unless I have missed some, the density is a bit erratic for these low numbered states. For larger n, perhaps you can come up with a general expression for how many states there are between some energy E and and a slightly higher level to come up with a number of states per unit energy interval. The sum of squares is suggestive that thinking in terms of the number of states contained within a spherical energy surface might prove helpful.

Last edited:

#### aura

thanks for trying it out ...

i will try to solve it...

thanks!!

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text:
# How do you write the product of (x + 2)(x + 3) as a trinomial?

${x}^{2} + 5 x + 6$

#### Explanation:

just multiply each of the terms of the first binomial by the terms of the second binomial

$\left(x + 2\right) \left(x + 3\right) = x \cdot x + 3 \cdot x + 2 \cdot x + 2 \cdot 3$

$\left(x + 2\right) \left(x + 3\right) = {x}^{2} + 5 x + 6$

have a nice day from the Philippines!


text:
# For regular languages A and B, determine whether B might match early in (A B)

I have two regular languages A and B, and I want to determine whether there is any pair of strings, a in A and b in B, such that (a b) is a prefix of a string in (A B) and the left-most match of B in (a b) includes one or more characters from a.

Raphael's formulation is good:

Given two regular language A, B, is there a (non-empty) prefix of a word b in B that is a suffix of a word in A so that the rest of b is a prefix of another word in B?

# Example

For example, let's say I have two regular languages, one which describes some properly escaped HTML text, and one which describes an end tag:

A := ([^&<>] | [&] [a-z] [a-z0-9]+ [;])*;
B := "</title";

By inspection, I can tell that there is no string (a b) in (A B) such that the first match of B includes characters from a because "<" is a prefix of B which cannot occur as a suffix of A.

But given a different grammar:

A' := (A | "<![CDATA[" ("]"? ">" | "]"* [^\]>])* "]]>")*;
B' := "</title" ([^>\"] | [\"] [^\"]* [\"])* ">";

then there are strings

a = '<![CDATA[</title "]]>"';
b = '</title>';

where (A B) matches '<![CDATA[</title "]]>"</title>' and the left-most match of B is '</title "]]>"</title>' which includes a non-empty suffix of a : '</title "]]>"'.

# Motivation

A in my situation describes the output of an encoder/sanitizer that is derived from a grammar, so an untrusted input is fed to the encoder/sanitizer and I know the output matches A by construction.

B is a limit condition in a larger grammar that describes how parsers determine where a chunk of an embedded language ends so they can hand it off to a parser for the embedded language.

My end goal is to be able to determine when I can optimize away runtime checks that ensure that it is safe to embed a particular encoded string. For these examples, it would be safe to optimize out the first check, but not the second.

Is this a solved problem? Does it have a name? Any pointers appreciated.

• "such that (a b) is a prefix of a string in (A B)" -- (a b) is trivially a prefix of itself; do you mean proper prefix? – Raphael Mar 28 '13 at 10:20

If I translate your description properly, here is what you want to ask:

Given two regular language $A$, $B$, is there a prefix of a word $b$ in $B$ that is a suffix of a word in $A$ so that the rest of $b$ is a prefix of another word in $B$?

Formally:

Given regular $A$ and $B$, are there $a = wx \in A$ and $b = yz \in B$ so that $xy \in B$?

Deciding this boils down to deciding whether

$\qquad B \cap (\operatorname{suff}(A) \cdot \operatorname{pref}(B) ) = \emptyset$.

Now we employ some closure properties: $\mathrm{REG}$ is closed against right and left quotient, concatenation and intersection, so with

$\qquad \operatorname{suff}(A) = A \backslash \Sigma^*$ and $\operatorname{pref}(B) = B / \Sigma^*$

the set we want to check for emptiness is regular, and therefore the check is indeed decidable.

In order to actually do this, I suggest you build automata for $\operatorname{suff}(A)$ and $\operatorname{pref}(B)$, concatenate them, intersect the result with the automaton for $B$ and use the standard check for emptiness (is a final state reachable from the initial state?).

Note that you never need DFA, all of this works nicely with NFA. Therefore, no exponential state explosion occurs; intersection multiplies the state numbers, though.

• Your English phrasing of my problem sounds right. Thanks for wading through my gibberish. – Mike Samuel Mar 28 '13 at 18:42

DFA intersection is a solved problem - given two DFAs (or NFAs, or regular grammars), we can come up with a DFA for the intersection of the two languages accepted by these DFAs. Also, it is known how to check whether the language accepted by a DFA is empty or not. Other closure properties are listed in the Wikipedia article.

Your problem is most probably solvable by using "constructive closure properties" (like intersection and concatenation), and the test for emptiness. In this sense, it is a solved problem. I don't know whether it appears anywhere.

• Is it fairly straightforward to go from a DFA for a language L to a DFA for all substrings of strings in L and to a DFA for all suffixes of strings in L without a state explosion? I suppose the first involves adding an edge from the start to every other state and from every state to the terminus, and the latter involves just the first part of that transform? – Mike Samuel Mar 28 '13 at 0:31
• It is either straightforward or impossible. In this case, it's straightforward. For prefixes, for example, it's enough to determine which states are not "error states", and make them accepting. – Yuval Filmus Mar 28 '13 at 2:26
• For suffixes, it's impossible. Consider the language $L = \{ \alpha w : \alpha \notin w \}$, where $\alpha$ is a symbol. For an alphabet of size $m$, the smallest DFA has size $m+1$, but the smallest DFA for the suffix language has exponential size. – Yuval Filmus Mar 28 '13 at 2:30
• Hmm. Since substrings include both prefixes and suffixes, reversal of the language won't help. – Mike Samuel Mar 28 '13 at 3:02
• Since this is a compiler optimization, I'll probably just have to go heuristic -- find a required element in B that cannot match a substring of A, "<" in the top example. – Mike Samuel Mar 28 '13 at 3:03

Assume DFAs given, it is easier. And get rid of useless states, and kill off any dead states.

To get a DFA for the language of all prefixes of $A$ is easy: Make all states final.

Then you can compute the intersection between that and $B$, and you have the set of $B$s that are prefixes of $A$s.

No idea what it might be called. The algorithm cited is just a proof of concept idea.

• I don't think what I want is the Bs that are prefixes of As. I edited my question to try to phrase it as an intersection problem. – Mike Samuel Mar 27 '13 at 21:51


text:
## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 1-5 - Review Exercises - Page 193: 22

#### Answer

The ball will not clear the 10-foot high goal post.

#### Work Step by Step

We can convert the horizontal displacement to units of meters: $\Delta x = 45~yd\times \frac{0.9144~m}{1~yd} = 41.148~m$ We can find the time for the ball to travel this horizontal distance: $t = \frac{d}{v_x}$ $t = \frac{d}{v_0~cos~\theta}$ $t = \frac{41.148~m}{(21~m/s)~cos~35^{\circ}}$ $t = 2.392~s$ We can find the ball's vertical displacement at this time: $\Delta y = v_y~t+\frac{1}{2}a_yt^2$ $\Delta y = v_0~sin~\theta~t+\frac{1}{2}a_yt^2$ $\Delta y = (21~m/s)~sin~35^{\circ}~(2.392~s)+\frac{1}{2}(-9.80~m/s^2)(2.392~s)^2$ $\Delta y = 0.776~m$ When the ball has traveled a horizontal distance of 45 yards, the ball is only 0.776 meters above the ground, which is too low to make it over the 10-foot high goal post. Therefore, the ball will not clear the 10-foot high goal post.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
1. Other
2. Other
3. what are the highlighted structures ive tried several times and...

# Question: what are the highlighted structures ive tried several times and...

###### Question details

What are the highlighted structures: I've tried several times and keep missing these anatomy questions. Please put the answers from picture A-G essentially.


text:
3 I made a grievous error confusing vertices and edges. Whoops!

Inkspot is indeed correct that the component graphs are indeed not generally trees.

As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs.

In general the dual graphs have $2h$ edges vertices $x$ where $h$ is the class number of $\mathcal{O}_x$, a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice of which $x$ to make here)here, but as long as the level is squarefree all orders are hereditary and it doesn't make a difference).

An orbit-stabilizer theorem computation then shows that for instance when $B$ is a quaternion algebra over $\mathbf{Q}$, $p+1$ (the size of the set of edges $y$ stemming from a particular vertex $x$ in the Bruhat-Tits tree) is equal to $\sum_{e(y) = x} \frac{\mathcal{O}_x^\times}{\mathcal{O}_y^\times}$. So it's not just that there are edges, but we know exactly how many there are! (For a readable account of details of this, see Kurihara's paper on Equations defining Shimura Curves)

Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its edge vertex set can be partitioned into ${x_1, \dots , x_h, x_1', \dots, x_h'}$ where $w_p(x_i) = x_i'$).

All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves)

Here are a few additional references:

http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol)

http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a somewhat more intuitive description of the components of the Mumford curve)

http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group)

http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)

2 deleted 16 characters in body

Inkspot is indeed correct that the component graphs are indeed not generally trees.

As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs.

In general the dual graphs have $2h$ edges where $h$ is the class number of a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice to make here).

Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its edge set can be partitioned into ${x_1, \dots , x_h, x_1', \dots, x_h'}$ where $w_p(x_i) = x_i'$).

All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves)

Here are a few additional references:

http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol)

http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a nonstandard but somewhat more intuitive description of the components of the Mumford curve)

http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group)

http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)

1

Inkspot is indeed correct that the component graphs are indeed not generally trees.

As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs.

In general the dual graphs have $2h$ edges where $h$ is the class number of a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice to make here).

Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its edge set can be partitioned into ${x_1, \dots , x_h, x_1', \dots, x_h'}$ where $w_p(x_i) = x_i'$).

All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves)

Here are a few additional references:

http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol)

http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a nonstandard but somewhat more intuitive description of the components of the Mumford curve)

http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group)

http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)


text:
# Difference between revisions of "2019 AMC 10A Problems/Problem 2"

## Problem

What is the hundreds digit of $(20!-15!)?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

## Video Solution 1

Education, The Study of Everything

## Video Solution 2

~savannahsolver

== Video Solution == 3 https://youtu.be/zfChnbMGLVQ?t=3899

~ pi_is_3.14

## Solution 3

Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also $\boxed{\text{(A) }0}$.

## Solution 4

We can clearly see that $20! \equiv 15! \equiv 0 \pmod{100}$, so $20! - 15! \equiv 0 \pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $0$, or $\boxed{\text{(A)}}$.

--abhinavg0627

## Solution 5

$20!= 2432902008176640000$ $15!= 1307674368000$

Subtract

-dragoon


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# Page:Popular Science Monthly Volume 70.djvu/259

1. This word benzine is sadly overworked. Spelled with an e, benzene, it is the correct scientific name for a definite chemical compound of the composition represented by the formula ${\displaystyle C_{6}H_{6}}$. Spelled with an i, benzine or benzin, it is often used to mean benzene, toluene, xylene, mesitylene, or several other things obtained from the distillation of coal, or a mixture of any two or more of these things. More frequently it means any one of the score of substances obtained in the distillation of crude American petroleum before the temperature is high enough to drive off what we call kerosene. That is to say, it may mean rhigolene, cymogene, gasolene, or naphtha, petroleum-ether or ligroin, or a mixture of these. As these are themselves mixtures, the confusion is worse confounded. Many, if not most chemists, in an effort to avoid misunderstandings, adopted the German word benzol to indicate that definite and important compound ${\displaystyle C_{6}H_{6}}$, but the relief was for but a little while. Now benzol, too, has begun to be used in certain industries, as if it were synonymous with benzine or benzene. When one of these three words is used it is impossible to tell immediately what is meant; the meaning may be deducible later from the context, frequently it is not, as the chances are almost even that the speaker himself does not know. It covers a multitude of inaccuracies; perhaps that is why the word is so popular.


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# How do you actually find the Jordan canonical form of a matrix?

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