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Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > KEYWORD > DEPTH-$3$ CIRCUITS:
Reports tagged with depth-$3$ circuits:
TR04-005 | 19th January 2004
Stasys Jukna

#### On Graph Complexity

A boolean circuit $f(x_1,\ldots,x_n)$ \emph{represents} a graph $G$
on $n$ vertices if for every input vector $a\in\{0,1\}^n$ with
precisely two $1$'s in, say, positions $i$ and $j$, $f(a)=1$
precisely when $i$ and $j$ are adjacent in $G$; on inputs with more


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How do I figure out the coproduct on graded algebras?

I have to figure out the duals to a couple of graded algebras. This requires a comultiplication (also called a coproduct in Hatcher). Hatcher's book shows what form the comultiplication must take using a cohomology argument for the cohomology of H-spaces. I do not know that these algebras are the cohomology of some H-spaces. Is there a purely algebraic argument to figure out a comultipication for an arbitrary graded algebra or do I need some extra data.

If you were wondering, I need to figure the duals of $\Lambda (y)$ and $\Gamma [ \gamma ]$.

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There is no way to "figure out" a comultiplication for an arbitrary graded algebra. Some algebras are not bialgebras in any way, and those algebras that can be made into bialgebras can usually turned into bialgebras in many different ways.

You will probably have to be more specific about the algebras you have in mind (at the very least, explain the notation you are using!)

The following, earlier text answers another question, the one in the title...

Doesn't the following obvious construction work? If $A$ and $B$ are graded algebras you can present them as quotients of free algebras $T(V)$ and $T(W)$ modulo homogeneous ideals generated by sets of homogeneous elements $R_A\subset T(V)$ and $R_B\subset T(W)$, so that $A=T(V)/\langle R_A\rangle$ and $B=T(W)/\langle R_B\rangle$. Then $A\sqcup B$ is $T(V\oplus W)/\langle R_A\cup R_B\rangle$.

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So how does this give me a graded homomorphism $\Delta : A \to A \otimes A$? This nicely satisfies the categoric definition but I have no idea how this gives me a product structure on the dual. –  Sven Apr 13 '11 at 12:47
Hmmm. What you want it to define a comultiplicationon algebras. I answered the question you asked in your title: how does the coproduct of algebras work. You should rewrite your question so that it asks clearly what you want... –  Mariano Suárez-Alvarez Apr 13 '11 at 15:40
Thank you for the clarification. The two books I was referencing called the map a coproduct. –  Sven Apr 13 '11 at 16:11
@Sven: your first comment to this answer confuses me: are you aware that the two uses of the work coproduct in "categorical coproduct" and "coproduct of a bialgebra/Hopf algebra" are rather unrelated, right? –  Mariano Suárez-Alvarez Apr 13 '11 at 16:27

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Definition 15.10.1. A Zariski pair is a pair $(A, I)$ such that $I$ is contained in the Jacobson radical of $A$.

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Select Board & Class

Board Paper of Class 10 2015 Maths (SET 1) - Solutions

General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculated is not permitted.

• Question 1
If the quadratic equation ${\mathrm{px}}^{2}-2\sqrt{5}\mathrm{px}+15=0$ has two equal roots, then find the value of p. VIEW SOLUTION

• Question 2
In Figure 1, a tower AB is 20 m high and BC, its shadow on the ground, is $20\sqrt{3}$ m long. Find the Sun's altitude.
 Figure 1
VIEW SOLUTION

• Question 3
Two different dice are tossed together. Find the probability that the product of the two number on the top of the dice is 6. VIEW SOLUTION

• Question 4
In Figure 2, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠ PRQ.  Figure 2
VIEW SOLUTION

• Question 5
In Figure 3, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.  Figure 3
VIEW SOLUTION

• Question 6
In figure 4, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments RD and DC are respectively of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm2, then find the lengths of sides AB and AC.  Figure 4
VIEW SOLUTION

• Question 7
Solve the following quadratic equation for x :
4x2 + 4bx − (a2b2) = 0 VIEW SOLUTION

• Question 8
In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms. VIEW SOLUTION

• Question 9
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B find the value of p. VIEW SOLUTION

• Question 10
Find the relation between x and y if the points A(x, y), B(–5, 7) and C(–4, 5) are collinear. VIEW SOLUTION

• Question 11
The 14th term of an AP is twice its 8th term. If its 6th terms is –8, then find the sum of its first 20 terms. VIEW SOLUTION

• Question 12

Solve for x :

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$ VIEW SOLUTION

• Question 13
The angle of elevation of an aeroplane from a point A on the ground 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 1500$\sqrt{3}$ m, find the speed of the plane in km/hr. VIEW SOLUTION

• Question 14
If the coordinates of points A and B are (–2, –2) and (2, –4) respectively, find the coordinates of P such that $\mathrm{AP}=\frac{3}{7}\mathrm{AB}$, where P lies on the line segment AB. VIEW SOLUTION

• Question 15
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is $\frac{1}{4}.$ The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}.$ If the jar contains 10 orange balls, find the total number of balls in the jar. VIEW SOLUTION

• Question 16
Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment. VIEW SOLUTION

• Question 17
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount, the associations will have to pay. What values are shown by these associations? VIEW SOLUTION

• Question 18
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer. VIEW SOLUTION

• Question 19
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per 100 sq. cm. [Use π =3.14] VIEW SOLUTION

• Question 20
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use $\mathrm{\pi }=\frac{22}{7}$] VIEW SOLUTION

• Question 21
The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field. VIEW SOLUTION

• Question 22
Find the 60th term of the AP 8, 10, 12, ...., if it has a total of 60 terms and hence find the sum of its last 10 terms. VIEW SOLUTION

• Question 23
A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed ? VIEW SOLUTION

• Question 24
Prove that the lengths of tangents drawn from an external point to a circle are equal. VIEW SOLUTION

• Question 25
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. VIEW SOLUTION

• Question 26
Construct a Δ ABC in which AB = 6 cm, ∠ A = 30° and ∠ B = 60°. Construct another Δ AB'C' similar to Δ ABC with base AB' = 8 cm. VIEW SOLUTION

• Question 27
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A. VIEW SOLUTION

• Question 28
A card is drawn at random from a well-shuffled deck of playing cards.
Find the probability that the card drawn is
(i) a card of spade or an ace
(ii) a black king
(iii) neither a jack nor a king
(iv) either a king or a queen VIEW SOLUTION

• Question 29
Find the values of k so that the area of the triangle with vertices (1, –1), (–4, 2k) and (–k, –5) is 24 sq. units.
VIEW SOLUTION

• Question 30
In Figure 5. PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).  Figure 5
VIEW SOLUTION

• Question 31
From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. VIEW SOLUTION
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ISSN 0869-6632 (Print)
ISSN 2542-1905 (Online)

#### For citation:

Bakhanova Y. V., Kazakov A. O., Karatetskaia E. Y., Kozlov A. D., Safonov K. A. On homoclinic attractors of three-dimensional flows. Izvestiya VUZ. Applied Nonlinear Dynamics, 2020, vol. 28, iss. 3, pp. 231-258. DOI: 10.18500/0869-6632-2020-28-3-231-258

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Article type:
Article
UDC:
517.925 + 517.93

# On homoclinic attractors of three-dimensional flows

Autors:
Bakhanova Yulia Viktorovna, Lobachevsky State University of Nizhny Novgorod
Kazakov Aleksej Olegovich, National Research University "Higher School of Economics"
Karatetskaia E. Yu., National Research University "Higher School of Economics"
Kozlov Aleksandr Dmitrievich, Lobachevsky State University of Nizhny Novgorod
Safonov K. A., National Research University "Higher School of Economics"
Abstract:

The main goal is to construct a classification of such attractors and to distinguish among them the classes of pseudohyperbolic attractors which chaotic dynamics is preserved under perturbations of the system. The main research method is a qualitative method of saddle charts, which consists of constructing an extended bifurcation diagram on the plane of the system parameters in the form $\dot x=y+g_1(x,y,z), \dot y=z+g_2(x,y,z), \dot z=Ax+By+Cz+g_3(x,y,z), \;\; g_i(0,0,0) = (g_i)^\prime_x(0,0,0) = (g_i)^\prime_y(0,0,0) = (g_i)^\prime_z(0,0,0) = 0, \; i = 1, 2, 3$,  the linearization matrix of which is represented in the Frobenius form, and the eigenvalues that determine the type of equilibrium state are expressed only through the coefficients A, B, C. The pseudohyperbolicity of the attractors under consideration is verified by means of a numerical method which helps to check the continuity of the subspaces of strong contractions and volume expansion on the attractor. The homoclinic nature of attractors is established using the numerical method of calculating the distance from an attractor to a saddle equilibrium. Results. An extended bifurcation diagram is constructed on the parameter plane (A,B), on which the stability region of the equilibrium state is highlighted, as well as six regions corresponding to two different types of spiral figure-eight attractors, Shilnikov attractor, Lorenz-like attractor, Shilnikov saddle attractor, and Lyubimov–Zaks–Rovella attractor. The pseudohyperbolicity of the Lorenz-like attractor is confirmed numerically. For the attractors of Lyubimov– Zaks–Rovella, it is shown that despite the continuity of strong contracting and volume-expanding subspaces such attractors cannot be pseudohyperbolic. The paper discusses that in three-dimensional flows, in addition to Lorenz-like attractors, only Shilnikov saddle attractors containing a saddle equilibrium state with a two-dimensional unstable manifold can be pseudohyperbolic. However, we currently do not know examples of such attractors.

Key words:
Reference:

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# qEI with QMCPy

Quasi-Monte Carlo methods (QMC) are a valuable tool for sampling random variables in a structured fashion; this allows for computing key statistics of random variables more efficiently than with i.i.d. sampling.  Such quantities can play fundamental roles in larger algorithms, making their efficient computation fundamental to practical implementations of numerous applications.  This was the motivation for creating the QMCPy library.  In this post, we demonstrate the use of QMC methods in computing a key quantity in Bayesian optimization.

### Bayesian optimization

Bayesian optimization (also called many other names, including sequential model based optimization or Gaussian process optimization) is a broad class of algorithms which involves alternately building statistical models of scattered data and optimally sampling for further data to try to optimize a given function.  To conduct this optimal sampling process, a strategy must be applied to take the statistical model and appropriately balance a desire to explore the optimization domain against a desire to exploit the high performing values found thus far and search in their proximity.  We refer to this strategy as the acquisition function.

Probably the most popular acquisition function is Expected Improvement (EI), which is widely used both because of its closed form (for a Gaussian process surrogate model) and because of its effectiveness.  When trying to push beyond the sequential model of Bayesian optimization to allow for parallelism, the EI acquisition function can be redefined for optimally choosing the next q points at which to sample.  This qEI acquisition function (first defined here, with more recent discussion here), unfortunately, does not have a simple closed form, and is generally computed through Monte Carlo estimation.

### qEI estimation

We consider a simple problem to demonstrate the value of QMC in estimating this qEI quantity.  The left panel of the figure below depicts 5 noisy observations of a 1D function.  We fit a Gaussian process (GP) model to this data, and provide examples of posterior draws in the center panel.

The right panel of the figure above depicts the qEI quantity for $q=2$ future points to be sampled; qEI is defined (for a maximization problem) with a $q$ dimensional integral as

$\qEI(x_1,\ldots,x_q) = \int_{\RR^q} \max_{1\leq i\leq q}(y_i – y^*)_+ p_{Y_{x_1,\ldots, x_q}}(y_1, \ldots, y_q)\; \text{d}y_1\cdots\text{d}y_q,$

where $Y_{x_1,\ldots, x_q}$ is the joint posterior GP distribution at $x_1,\ldots, x_q$ and $y^*$ is the best value observed thus far.  Note the symmetric nature of the qEI — since both of the $q=2$ points will be evaluated simultaneously, the designation of “first” and “second” is arbitrary.  The highest qEI value occurs when sampling near $x_1=0$ and $x_2=.6$, points which are near the highest valued draws of the posterior.

This integral is generally estimated using Monte Carlo (discussed here and here, among other places) with i.i.d. draws.  Using Quasi-Monte Carlo, we can reach the same level of accuracy with fewer GP posterior draws.  The figure below shows the results of using different quasi-random sequences, available in QMCPy, to estimate qEI for $q=5$ next points, located at $0.158, 0.416, 0.465, 0.718, 0.935$.

The integrals present in BO are another example of a computation which benefits from QMC.  More efficient acquisition function computation gives us the ability to conduct more complicated strategies in a feasible amount of time.  Check out our docs, or contact us, to learn more about the QMCPy project and how QMC can help your work.

##### Michael McCourt
Website | + posts

I work at SigOpt, a SF startup, developing and deploying Bayesian optimization tools for customers in finance, AI, consumer goods and general research and development.


text:
#### Chapter 5 Geometry

Section 5.2 Angles and Angle Measurement

# 5.2.2 Angles

Two rays (half-lines) $g$ and $h$ in the plane starting from the same initial point $S$ enclose an angle $\measuredangle \left(g,h\right)$.

Angle enclosed by the rays $g$ and $h$.

For the notation of the angle $\measuredangle \left(g,h\right)$, the order of $g$ and $h$ is relevant. $\measuredangle \left(g,h\right)$ denotes the angle shown in the figure above. It is defined by rotating the half-line $g$ counter-clockwise to the half-line $h$.
In contrast, $\measuredangle \left(h,g\right)$ denotes the angle from $h$ to $g$ as illustrated by the figure below.

Angle enclosed by the rays $h$ and $g$

The point $S$ is called a vertex of the angle, and the two half-lines enclosing the angle are called the arms of the angle. If $A$ is a point on the line $g$ and $B$ is a point on the line $h$, then the angle $\measuredangle \left(g,h\right)$ can also be denoted by $\measuredangle \left(ASB\right)$. In this way, angles between line segments $\stackrel{‾}{SA}$ and $\stackrel{‾}{SB}$ are described.
Angles are often denoted by lower-case Greek letters to distinguish them from variables, which are generally denoted by lower-case Latin letters (see Table 1.1.8 in module 1). Further angles can be found by considering angles formed by intersecting lines.
##### Vertical Angles and Supplementary Angles 5.2.1

Let $g$ and $h$ be two lines intersecting in a point $S$.

• The angles $\phi$ and $\phi \text{'}$ are called vertical angles.
• The angles $\phi$ and $\psi$ are called supplementary angles with respect to $g$.

The figure above contains further vertical and supplementary angles.
##### Exercise 5.2.2
Find all vertical and supplementary angles occurring in the figure above.

Some special angles have their own dedicated name. For example, the angle bisector $w$ is the half-line whose points have the same distance from the two given half-lines $g$ and $h$. Then, it can be said that $w$ bisects the angle between $g$ and $h$.
##### Names of Special Angles 5.2.3
Let $g$ and $h$ be half-lines with the intersection point $S$.
• The angle covering the entire plane is called the complete angle.
• If the rays $g$ and $h$ form a line, the angle between $g$ and $h$ is called a straight angle.
• The angle between two half-lines bisecting a straight angle is called the right angle. One also says that $g$ and $h$ are perpendicular (or orthogonal) to each other.

Next, three lines are considered. Two of the three lines are parallel, while the third line is not parallel to the others. It is called a transversal. These lines form eight cutting angles. Four of the eight angles are equal.
##### Angles at Parallel Lines 5.2.4
Let two parallel lines $g$ and $h$ be given cut by another transversal line $j$.

• Then the angle $\alpha \text{'}$ is called a corresponding angle of $\alpha$ and
• the angle $\beta \text{'}$ is called an alternate angle of $\beta$.
Since the lines $g$ and $h$ are parallel, the angles $\alpha$ and $\alpha \text{'}$ are equal. Likewise, the angles $\beta$ and $\beta \text{'}$ are equal.

##### Exercise 5.2.5
The figure shows two parallel lines $g$ and $h$ cut by another line $j$. Explain which angles are equal and which angles are corresponding angles or alternate angles to each other, respectively.


text:
# Using the Cosmo Hammer with Monte Python¶

Monte Python can now use the software Cosmo Hammer written by J. Akeret and S. Seehars, which is based on the emcee sampler, itself based on the Affine Invariant Markov Chain Monte Carlo

Note

By using the Cosmo Hammer, you agree to abide by the GNU General Public License v3.0 or higher (see their website)

\$] export OMP_NUM_THREADS=4


text:
# Journal of Operator Theory

Volume 73, Issue 2, Spring 2015  pp. 491-514.

Groupoids associated to Ore semigroup actions

Authors:  Jean Renault (1) S. Sundar (2)
Author institution:(1) Universite d'Orleans et CNRS (UMR 7349 et FR 2964), Departement de Mathematiques, 45067, Orleans Cedex 2, France
Summary: In this paper, we consider actions of locally compact Ore semigroups on compact topological spaces. Under mild assumptions on the semigroup and the action, we construct a semi-direct product groupoid with a Haar system. We also show that it is Morita-equivalent to a transformation groupoid. We apply this construction to the Wiener--Hopf $C^{*}$-algebras.
Keywords: Wiener-Hopf $C^*$-algebras, semigroups, groupoids


text:
1. ## Trig Brain Cramp

The problem says:

An airplane is flying on a horizontal path at a height of 3800 ft. At what rate is the distance s between the airplane and a fixed point P (on the ground) changing with respect to theta when theta equals 30 degrees? Express your answer in units of ft/degree.

I understand the problem, and got -13,163.2 feet/??? Here is my dilemma:

The answer key says that this answer is in ft/rad and that I need to convert it to ft/deg, which is approximately -230 ft/deg. My question: WHY? If I plug 30 DEGREES into my calculator and get -13,163.2 feet, isn't that in DEGREES? I know HOW to do the conversion. I am not clear on why. Help?

Jeannine

2. The answer key says that this answer is in ft/rad and that I need to convert it to ft/deg, which is approximately -230 ft/deg. My question: WHY? If I plug 30 DEGREES into my calculator and get -13,163.2 feet, isn't that in DEGREES? I know HOW to do the conversion. I am not clear on why. Help?
because the derivatives of the trig functions you have learned are only valid when $\theta$ is in radians ... substituting in 30 degrees at the end of the problem using degree mode in your calculator only gave a specific value of the trig ratio in your expression that determined the desired rate of change, you would have gotten the same result substituting in $\frac{\pi}{6}$ in radian mode.


text:
anonymous 3 years ago Based on the following results what is the experimental probability of rolling a 1? Outcome Frequency 1 6 2 2 3 2 4 3 5 2 6 0

1. anonymous

@garrett_payne

2. anonymous

Add up all the frequencies. Divide the frequency of the target outcome by the sum of all frequencies.

3. anonymous

let me try..

4. anonymous

5. anonymous

No.. divide 6 by 6+2+2+3+...

6. anonymous

I know. 1-6 means all the numbers between 0 and 6.

7. anonymous

1+2+3+4+5+6/6

8. anonymous

?

9. anonymous

divide x by y means x / y not y / x

10. anonymous

I got '16'.

11. anonymous

It makes no sense for probability to be over 1.

12. anonymous

wait,sorry.

13. anonymous

can you give me the equation?

14. anonymous

o wait, you mean 6/1+2+3+4+5+6 ?

15. anonymous

probability of an event A happening in N number of experiments: $P(A)=\frac{\text{number of times A event happens}}{N}$

16. anonymous

so - what would the numerator be?

17. anonymous

event = the dice rolls a "1" how many times does that happen?

18. anonymous

6.

19. anonymous

bingo

20. anonymous

oooh.

21. anonymous

:)

22. anonymous

tyty!

23. anonymous

and the denominator is

24. anonymous

6. right?


text:
Option greeks: sensitivity to 1% move

In a Black&Scholes framework how can I compute the following sensitivities:

• to 1% move in the underlying price
• to 1% move in implied volatility

I would like the greeks to tell me how many dollars I lose/gain if the underlying/implied volatility moves by 1%. In particular, I would like to calculate the delta and gamma (to 1% move in underlying price) and vega and volga (to 1% move in implied volatility).

For the vanna I would like to consider a 1% move in both underlying and implied volatility.

Can you please suggest how to modify Black&Sholes greeks and also how to compute the sensitivities numerically?

A reference would also be very welcome. Thank you.

2 Answers

Try Finite Differences to calculate your Greeks, it will give all the greeks for that specific underlying moviment. In order to back out the dollar change in your pnl just multiply each greek by the amount held in that position.

In the BS model, everything is explicit.

If your spot increases by $h\%$, the price will increase by $\Delta_{rel,h}\%$ where $$\Delta_{rel,h} = \frac{C_{BS}(S(1+h),T,K,\sigma)}{C_{BS}(S,T,K,\sigma)} - 1$$ That's high school math.

• Thank you. Can you suggest a closed-form formula also? – mickG Jan 24 '15 at 19:37
• It's already a closed form formula! Just replace C_BS by the Black Scholes formula – AFK Jan 25 '15 at 16:25
• It is not in the sense that you need to calculate the price twice while analytic delta just need one calculation. – mickG Jan 25 '15 at 16:28


text:
Tags » Maximum Principle

First Bourbaki seminar of 2015 (II): Carron's talk

For the second installment of this series of posts (which started here) on the first Bourbaki seminar of 2015, we will discuss Gilles Carron… 2,384 more words

Mathematics

A strong maximum principle due to Tolksdorf and application to the prescribed scalar curvature equation

In this note, we consider a strong maximum principle due to Tolksdorf and application to the prescribed scalar curvature equation. In the context of closed and compact Riemannian manifolds of dimension , it can be stated as follows… 397 more words

PDEs

The generalized maximum principle

Let us start our notes with a very fundamental maximum principle for any strongly second order elliptic operator. We have

Theorem (Maximum principle). Let satisfy the differential inequality…

571 more words
PDEs

Monge-Ampere equation and bounday behavior of domain

Suppose is a bounded domain and is . If there exists a convex function satisfies

Then is uniformly convex. In other words, the principle curvature  of every point on , namely , are positive. 151 more words

PDE

Improvement of Aleksandrov maximum principle

For we have

where .

There is an easy proof when the RHS is . We will adjust the original proof.

By replacing with , it suffices to assume on . 92 more words

PDE

Strong barrier function and generalized linear elliptic equation

Suppose is a domain in . satisfies the exterior sphere condition. That is there exists a ball such that . Then the function

is a barrier function for at . 111 more words

PDE

One simple maximum principle

If is elliptic with in a bounded domain , and satisfies in , then

Let

Then and on . By the maximum principle, we conclude, … 12 more words

PDE


text:
# 5 Seconds to Find Pie

Pi times e (or Pie if you like ambiguous notation) to 100 decimal places is:

8.5397342226735670654635508695465744950348885357651149618796011301792286111573308075725638697104739439...

Your task is to write a program that takes in a positive integer N, and outputs Pi*e truncated to N decimal places. e.g. if N = 2, then the output should be 8.53.

This is an optimization problem, so the submission that can give the correct output for the highest value of N will win.

To ensure all submissions are judged using the same computing power, your code must be run on ideone, using any language they support. According to the ideone faq, there is a 5 second run time limit for not logged in users. This 5 second limit is the one you must use, not the 15 second limit for logged in users. (See faq for other limits like memory, code size, etc.)

Specifically, anyone not logged into ideone should be able to run your program on ideone for all values of N from 1 to some maximum Nmax, and see correct output nearly all of the time. without any Time limit exceeded or Memory limit exceeded, etc. errors. The submission with the largest Nmax wins.

(Whether the actual time taken is a smidge over 5 seconds doesn't matter as long as ideone doesn't give errors. "Nearly all of the time" is defined as more than 95% of the time for any particular N.)

# Details

• You may use any mathematical method you like to calculate Pi*e, but you may not hardcode the output beyond the first dozen digits of Pi, e or Pi*e.
• Your program should be able to work for any N, given unlimited resources.
• You may use built in Pi or e constants if your language happens to have them.
• You may not access websites or resources external to your code (if ideone even allows this).
• Beyond hardcoding and accessing external resources, anything that ideone allows is almost certainly fine.
• Your input and output must (obviously) work with whatever ideone provides for i/o (stdin/stdout only it seems). It's fine if you need quotes around the input N or the output is something like ans = ..., etc.
• If there happens to be a tie (only likely if multiple submissions hit the 64kB output character limit) the highest votes answer wins.
• Mmm... ambiguous pie. – Dennis Sep 23 '14 at 4:16
• This can very easily be a code-golf and would rather be more fun if it is. – Optimizer Sep 23 '14 at 6:53
• @Optimizer It could be code-golf but then it would be pretty much like every other digit generation code-golf. I wanted to try a time based contest that could be verified online. (Though a more computationally complex problem might have been better.) – Calvin's Hobbies Sep 23 '14 at 7:07
• if this is code golf APL would probably win (minus the arbitrary precision part) – TwiNight Sep 23 '14 at 7:24
• I suspect that these programs will be entirely IO bound trying to write out the digits to stdout. Five seconds is a long time to something like y-cruncher. – Will Sep 23 '14 at 13:43

## Python - 65535

http://ideone.com/knKRhn

from math import exp, log

def divnr(p, q):
"""
Integer division p/q using Newton-Raphson Division.
Assumes p > q > 0.
"""

sp = p.bit_length()-1
sq = q.bit_length()-1
sr = sp - sq + 1

s = []
t = sr
while t > 15:
s = [t] + s
t = (t>>1) + 1
# Base-case division
r = (1 << (t<<1)) / (q >> sq-t)

for u in s:
r = (r << u-t+1) - (r*r * (q >> sq-u) >> (t<<1))
t = u
return (r * (p >> sq)) >> sr

def pibs(a, b):
if a == b:
if a == 0:
return (1, 1, 1123)
p = a*(a*(32*a-48)+22)-3
q = a*a*a*24893568
t = 21460*a+1123
return (p, -q, p*t)
m = (a+b) >> 1
p1, q1, t1 = pibs(a, m)
p2, q2, t2 = pibs(m+1, b)
return (p1*p2, q1*q2, q2*t1 + p1*t2)

def ebs(a, b):
if a == b:
if a == 0:
return (1, 1)
return (1, a)
m = (a+b) >> 1
p1, q1 = ebs(a, m)
p2, q2 = ebs(m+1, b)
return (p1*q2+p2, q1*q2)

if __name__ == '__main__':
n = input()

pi_terms = int(n*0.16975227728583067)

# 10^n == e^p
p = n*2.3025850929940457

# Lambert W_0(p/e) a la Newton
k = log(p) - 1
w = k - (k-1)/(k+1)
while k > w:
k = w
w -= (k - p*exp(-k-1))/(k+1)

# InverseGamma(e^p) approximation
e_terms = int(p / w)

pp, pq, pt = pibs(0, pi_terms)
ep, eq = ebs(0, e_terms)

z = 10**n
p = 3528*z*ep*abs(pq)
q = eq*abs(pt)

pie = divnr(p, q)
print pie,

Ideone doesn't seem to have gmpy2 installed, which is unfortunate for at least two reasons. One, because it would make the calcuation a lot faster, and two, because it makes any formula requiring an arbitrary precision square root impractical.

The formula I use for π was listed by Ramanujan as Formula (39):

which converges at the rate of ~5.89 digits per term. To my knowledge, this is the fastest converging series of its kind that doesn't require the evaluation of an arbitrary precision square root. Formula (44) in the same paper (convergence rate ~7.98 digits per term) is most often referred to as the Ramanujan Formula.

The formula I use for e is the sum of inverse factorials. The number of terms required is calculated as Γ-1(10n), using an approximation I found on mathoverflow. The Lambert W0 component is found using Newton's Method.

The calculation of each of these summations is done via the Fast E-function Evalution (more generally known as binary-splitting), originally devised by Karatsuba. The method reduces a summation to n terms to a single rational value p/q. These two values are then multiplied to produce the final result.

Update:
Profiling revealed that over half of the time needed for the calculation was spent in the final division. Only the upper most log2(10n) bits of q are needed to obtain full precision, so I trim a few off beforehand. The code now fills the Ideone output buffer in 3.33s.

Update 2:
Since this is an challenge, I decided to write my own division routine to combat CPython's slowness. The implementation of divnr above uses Newton-Raphson Division. The general idea is to calculate d = 1/q · 2n using Newton's Method, where n is the number of bits the result requires, and compute the result as p · d >> n. Runtime is now 2.87s - and this is without even chopping off bits before the calculation; it's unnecessary for this method.

## PARI/GP : 33000

This is basically the program given at OEIS, modified to take input and format output correctly. It should serve as a baseline to beat, if nothing else.

I assume this is accurate. I checked it at 100 and 20k against OEIS, and it matched for both. It's quite hard to find further digits online to check against.

For 33,000 it takes about 4.5s, so it could probably be bumped a small bit. I just got tired of fiddling with the input and ideone's slow-ass submit/compile/run loop.

{
m=eval(input());
default(realprecision, m+80);
x=Pi*exp(1);
s="8.";
d=floor(x);
x=(x-d)*10;
for (n=1, m, d=floor(x);
x=(x-d)*10;
s=Str(s,d));
print(s);
}

• Your digits match mine, so I'm going to go out on a limb and say they're probably correct. – primo Sep 25 '14 at 15:03
• This program spends essentially all of its time in the loop, generating digits one by one. If you just take Str(floor(frac(x)*10^m) it goes hundreds/thousands of times faster. – Charles Apr 22 '16 at 13:46

# Python 3

Since the built in pi and e don't have enough digits, I decided to calculate my own.

import decimal
import math
decimal.getcontext().prec=1000000
decimal=decimal.Decimal;b=2500
print(str(sum([decimal(1/math.factorial(x)) for x in range(b)])*sum([decimal(1/16**i*(4/(8*i+1)-2/(8*i+4)-1/(8*i+5)-1/(8*i+6))) for i in range(b)]))[0:int(input())+2])

### Output for STDIN = 1000:

8.5397342226735669504281233688422467204743749305568824722710929852470173635361001388261308713809518841081669216573834376992066672804294594807026609638293539437286935503772101801433821053915371716284188665787967232464763808892618434263301810056154560438283877633957941572944822034479453916753507796910068912594560500836608215235605783723340714760960119319145912948480279651779184356994356172418603464628747082162475871780202868607325544781551065680583616058471475977367814338295574582450942453416002008665325253385672668994300796223139976640645190237481531851902147391807396201201799703915343423499008135819239684881566321559967077443367982975103648727755579256820566722752546407521965713336095320920822985129589997143740696972018563360331663471959214120971348584257396673542429063767170337770469161945592685537660073097456725716654388703941509676413429681372333615691533682226329180996924321063261666235129175134250645330301407536588271020457172050227357541822742441070313522061438812060477519238440079

• Nmax is the largest input value you can give your program before ideone will no longer run it. – Calvin's Hobbies Sep 23 '14 at 6:58
• @Calvin'sHobbies I think that nmax is arbitrarily large though... – Beta Decay Sep 23 '14 at 7:35
• ideone doesn't give you infinite computing power. What is the largest input value your program can run on ideone? (Though in fact your program does not follow the should be able to work for any N, given unlimited resources rule. Most of the output is zeros at around N = 10000.) – Calvin's Hobbies Sep 23 '14 at 7:47
• That's not python3: NameError: name 'xrange' not defined. – Bakuriu Sep 23 '14 at 16:57

# Scala - 1790

IDEOne at http://ideone.com/A2CIto.

We use Wetherfield's formula for π (and Machin-formula code crudely ported from here). We calculate e using the ordinary power series.

object Main extends App {
import java.math.{BigDecimal => JDecimal}
import java.math.RoundingMode._
import scala.concurrent.Future
import scala.concurrent.Await
import scala.concurrent.ExecutionContext.Implicits._
import scala.concurrent.duration._
val precision = 1800

def acotPrecision(numDigits: Int)(x: BigDecimal) = {
val x1 = x.underlying
val two = JDecimal.valueOf(2)
val xSquared = x1 pow 2
val unity = JDecimal.ONE.setScale(numDigits, HALF_EVEN)
var sum = unity.divide(x1, HALF_EVEN)
var xpower = new JDecimal(sum.toString)
var term = unity

var n = JDecimal.valueOf(3).setScale(numDigits)
while (term.setScale(numDigits, HALF_EVEN).compareTo(JDecimal.ZERO) != 0) {
xpower = xpower.divide(xSquared, HALF_EVEN)
term = xpower.divide(n, HALF_EVEN)
}
sum
}

def ePrecision(numDigits: Int) = {
val zero = JDecimal.ZERO
var sum = zero
var term = JDecimal.ONE.setScale(numDigits, HALF_EVEN)
var n = JDecimal.ONE.setScale(numDigits, HALF_EVEN)
while(term.setScale(numDigits, HALF_EVEN).compareTo(zero) != 0) {
term = term.divide(n, HALF_EVEN)
}
sum
}

val acot = acotPrecision(precision) _

def d(x: Int) = JDecimal.valueOf(x)

def piFuture = Future(d(4) multiply (
(d(83) multiply acot(107)) add (d(17) multiply acot(1710)) subtract (d(22) multiply acot(103697))
subtract (d(24) multiply acot(2513489)) subtract (d(44) multiply acot(18280007883L))


text:
Calculate the definite integral of $\int_0^5 a x^3 dx$ for a given value of a and compare it with the analytic answer: $y=\frac{a}{4} 5^4$

In [0]:

xvec=arange(0,5,0.01) #vector from 0 to 5 in 0.01 step sizes

In [0]:

xvec[:10] #print the first 10 values

Now make a vector of intervals $dx$ using the diff function

In [0]:

dx=diff(xvec) #note that dx is one item shorter than xvec

integration is just the sum over rectangles of width $dx$ and height $ax^3$

In [0]:

a=5
integral=sum(a*xvec[0:-1]**3.*dx)
analytic=a/4.*5.**4.
print "analytic: %8.4f, approximate: %8.4f" % (analytic,integral)

## Now apply this to the Planck function (Stull Chapter 2, Equation 2.13, p. 36):

$E_\lambda^* = \frac{c1}{\lambda^5 \left ( exp(c_2/(\lambda T)) - 1 \right )}$

$E_\lambda* =\pi \frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}}-1}$

where

$h$=Planck's constant ($6.63 \times 10^{-34}$ Joule seconds})

$c$= Speed of light in a vacuum ($3.00 \times 10^{8}\ \mathrm{meters/second}$)

$k_b$ =Boltzman's constant ($1.38 \times 10^{-23}\ \mathrm{Joules/Kelvin}$)

(note that $E_\lambda*$ is the blackbody flux, or irradiance, with units of $W/m^2 /\mu m$) Wikipedia (and Wallace and Hobbs) give the blackbody radiance which differs by by a factor of $\pi$ and has units of $W/m^2/\mu m /sr$. We will talk a lot about this in the next few weeks, don't worry about it for now.

In [0]:

c=2.99792458e+08  #m/s -- speed of light in vacumn
h=6.62606876e-34  #J s  -- Planck's constant
kb=1.3806503e-23  # J/K  -- Boltzman's constant

def planckwavelen(wavel,Temp):
"""input wavelength in microns and Temp in K, output
bbr in W/m^2/micron
"""
wavel=wavel*1.e-6  #convert to meters
c1=2.*h*c**2.
c2=h*c/kb
Elambda=1.e-6*pi*c1/(wavel**5.*(np.exp(c2/(wavel*Temp)) -1))
return Elambda

In [0]:

the_wavelengths=linspace(0.1,50,5000) #microns
the_temp=300 #K
flux=planckwavelen(the_wavelengths,the_temp)
plot(the_wavelengths,flux)

Assignment: Use sum and diff along with the function planckwavelen to integrate the Planck function over wavelength and verify the Stefan-Boltzman equation on Stull chapter 2 p. 37:

$E^* = \int_0^\infty E_\lambda^* d\lambda = \sigma T^4$

where $\sigma=5.67 \times 10^-8$ $W/m^2/K^{4}$

# Write a set of python instructions that integrates the planck function and show that your integral agrees with the Stefan-Boltzman equation

In [0]:


text:
# Sequential separability on $C_p(X)$

Definition. Let $$E$$ be a topological space. Suppose that $$E$$ contains a sequence $$\{x_n\}$$ such that for every $$x\in E$$, there exists a subsequence $$\{x_{n_k}\}$$ of $$\{x_n\}$$ with $$x=\lim x_{n_k}$$. Then we say $$E$$ enjoys sequential separability.

For a given topological space $$X$$, let $$C_p(X)$$ be the space of continuous functions on $$X$$ endowed with the point-wise topology. It is proved that $$C_p(X)$$ is separable iff $$X$$ is separably submetrizable (see this paper).

Question.
$$C_p(X) ~\textrm{is sequentially separable} \Leftrightarrow X=?$$


text:
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Bibliometrics: Citation Count: 0 Uncertain data has been very common in many applications. In this paper, we consider the one-center problem for uncertain data on tree networks. In this problem, we are given a tree T and n (weighted) uncertain points each of which has m possible locations on T associated with probabilities. The ... Keywords: One-center, Algorithms and data structures, Uncertain data, Facility location, Tree network 2 May 2017 Algorithmica: Volume 78 Issue 1, May 2017 Publisher: Springer-Verlag New York, Inc. Bibliometrics: Citation Count: 0 We consider an interval coverage problem. Given n intervals of the same length on a line L and a line segment B on L, we want to move the intervals along L such that every point of B is covered by at least one interval and the sum of the ... Keywords: Algorithms and data structures, Mobile sensors, Barrier coverage, Computational geometry, Interval coverage, min-sum 3 April 2017 Discrete & Computational Geometry: Volume 57 Issue 3, April 2017 Publisher: Springer-Verlag New York, Inc. Bibliometrics: Citation Count: 0 For a polygonal domain with h holes and a total of n vertices, we present algorithms that compute the $$L_1$$L1 geodesic diameter in $$O(n^2+h^4)$$O(n2+h4) time and the $$L_1$$L1 geodesic center in $$O((n^4+n^2 h^4)\alpha (n))$$O((n4+n2h4)ź(n)) time, respectively, where $$\alpha (\cdot )$$ź(·) denotes the inverse Ackermann function. No algorithms were known for ... Keywords: Geodesic diameter, 68W05, Polygonal domains, Shortest paths, $$L_1$$L1 metric, Geodesic center, 68W40 4 January 2017 Algorithmica: Volume 77 Issue 1, January 2017 Publisher: Springer-Verlag New York, Inc. Bibliometrics: Citation Count: 0 Given a set $${\mathcal {P}}$$P of h pairwise-disjoint polygonal obstacles with a total of n vertices in the plane, we study the problem of computing the (weak) visibility polygon from a polygonal obstacle $$P^*$$Pź (an island) in $${\mathcal {P}}$$P. The problem was previously solved in $$O(n^4)$$O(n4) time, which has been ... Keywords: Algorithms, Computational geometry, Polygon with holes, Visibility polygons, Polygonal domains 5 May 2016 Algorithmica: Volume 75 Issue 1, May 2016 Publisher: Springer-Verlag New York, Inc. Bibliometrics: Citation Count: 0 In the classic k-center problem, we are given a metric graph, and the objective is to select k nodes as centers such that the maximum distance from any vertex to its closest center is minimized. In this paper, we consider two important generalizations of k-center, the matroid center problem and ... Keywords: k-median, Fault-tolerant, k-center 6 May 2016 Operations Research Letters: Volume 44 Issue 3, May 2016 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 0 We consider the one-dimensional one-center problem on uncertain data. We are given a set P of n (weighted) uncertain points on a real line L and each uncertain point is specified by a probability density function that is a piecewise-uniform function (i.e., a histogram). The goal is to find a ... Keywords: Lowest point, One-center, Upper envelope, Histogram, Algorithms, Uncertain data 7 April 2016 IEEE Transactions on Information Forensics and Security: Volume 11 Issue 4, April 2016 Publisher: IEEE Press Bibliometrics: Citation Count: 0 Geometric range search is a fundamental primitive for spatial data analysis in SQL and NoSQL databases. It has extensive applications in location-based services, computer-aided design, and computational geometry. Due to the dramatic increase in data size, it is necessary for companies and organizations to outsource their spatial data sets to ... 8 January 2016 Theoretical Computer Science: Volume 609 Issue P1, January 2016 Publisher: Elsevier Science Publishers Ltd. Bibliometrics: Citation Count: 0 Given a set P of n uncertain points on the real line, each represented by its one-dimensional probability density function, we consider the problem of building data structures on P to answer range queries of the following three types for any query interval I: (1) top-1 query: find a point ... Keywords: Data structures, Threshold queries, Top-k queries, Range queries, Algorithms, Uncertain data 9 October 2015 Theoretical Computer Science: Volume 602 Issue C, October 2015 Publisher: Elsevier Science Publishers Ltd. Bibliometrics: Citation Count: 2 Problems on uncertain data have attracted significant attention due to the imprecise nature of many measurement data. In this paper, we consider the k-center problem on one-dimensional uncertain data. The input is a set P of (weighted) uncertain points on a real line, and each uncertain point is specified by ... Keywords: Algorithms and data structures, Uncertain data, k-center, Computational geometry, Facility locations 10 August 2015 Theoretical Computer Science: Volume 593 Issue C, August 2015 Publisher: Elsevier Science Publishers Ltd. Bibliometrics: Citation Count: 0 We study the computational complexity of the puzzle Quell. The goal is to collect pearls by sliding a droplet of water over them in a map that is a two-dimensional grid. Each cell of the grid is either a free space (possibly with a pearl in it) or an obstacle. ... Keywords: Approximation algorithms, Hardness of approximation, Traveling salesman problem, Fixed-parameter tractability, Combinatorial game theory 11 August 2015 Theoretical Computer Science: Volume 592 Issue C, August 2015 Publisher: Elsevier Science Publishers Ltd. Bibliometrics: Citation Count: 2 We consider the problem of finding k centers for n weighted points on a real line. This (weighted) k-center problem was solved in O ( n log � n ) time previously by using Cole's parametric search and other complicated approaches. In this paper, we present an easier O ( ... Keywords: 2-D sublist LP queries, One-dimension, Algorithms, k-center, Computational geometry, Data structures, Facility locations 12 August 2015 Computational Geometry: Theory and Applications: Volume 48 Issue 6, August 2015 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 2 In this paper, we show that the L 1 geodesic diameter and center of a simple polygon can be computed in linear time. For the purpose, we focus on revealing basic geometric properties of the L 1 geodesic balls, that is, the metric balls with respect to the L 1 ... Keywords: Geodesic center, Simple polygon, Geodesic diameter, L1 metric 13 August 2015 Computational Geometry: Theory and Applications: Volume 48 Issue 6, August 2015 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 1 Given a simple polygon P in the plane, we present new data structures for computing the weak visibility polygon from any query line segment in P. We build a data structure in O ( n ) time and space that can compute the visibility polygon for any query line segment ... Keywords: Visibility queries, Visibility polygons, Weak visibility from line segments, Zone theorem 14 July 2015 Operations Research Letters: Volume 43 Issue 4, July 2015 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 0 We study a problem on splitting intervals. Let I be a set of n intervals on a line L , with non-negative weights. Given any integer k � 1 , we want to find k points on L to partition L into k + 1 segments, such that the maximum ... Keywords: Multi-version databases, Temporal databases, Interval splitting, Algorithms, Combinatorial optimizations 15 June 2015 Algorithmica: Volume 72 Issue 2, June 2015 Publisher: Springer-Verlag New York, Inc. Bibliometrics: Citation Count: 0 Given n points in a circular region C in the plane, we study the problems of moving the n points to the boundary of G to form a regular n-gon such that the maximum (min-max) or the sum (min-sum) of the Euclidean distances traveled by the points is minimized. These ... Keywords: Barrier coverage, Computational geometry, Circular convex bipartite graph, Dynamic maximum matching, Algorithms and data structures, Circular region coverage, Mobile sensors 16 May 2015 Discrete Applied Mathematics: Volume 187 Issue C, May 2015 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 0 We consider computing a maximum non-crossing matching in convex bipartite graphs. For a convex bipartite graph of n vertices and m edges, we present an O ( n log n ) time algorithm for finding a maximum non-crossing matching in the graph. The previous best algorithm takes O ( m ... Keywords: Convex bipartite graphs, Algorithms, Maximum non-crossing matching, Data structures, Maximum matching 17 April 2015 ACM Transactions on Algorithms (TALG): Volume 11 Issue 4, June 2015 Publisher: ACM Bibliometrics: Citation Count: 0 Downloads (6 Weeks): 5,   Downloads (12 Months): 38,   Downloads (Overall): 282 Full text available: PDF A fundamental problem in computational geometry is to compute an obstacle-avoiding Euclidean shortest path between two points in the plane. The case of this problem on polygonal obstacles is well studied. In this article, we consider the problem version on curved obstacles, which are commonly modeled as splinegons . A ... Keywords: tangent graphs, curved obstacles, Shortest paths, Voronoi diagrams 18 October 2014 Computational Geometry: Theory and Applications: Volume 48 Issue 2, February, 2015 Publisher: Elsevier Science Publishers B. V. Bibliometrics: Citation Count: 1 Given a polygonal domain (or polygon with holes) in the plane, we study the problem of computing the visibility polygon of any query point. As a special case of visibility problems, we also study the ray-shooting problem of finding the first point on the polygon boundaries that is hit by ... Keywords: Visibility queries, Visibility polygons, Polygonal domains, Ray-shootings, Algorithms 19 June 2014 SOCG'14: Proceedings of the thirtieth annual symposium on Computational geometry Publisher: ACM Bibliometrics: Citation Count: 2 Downloads (6 Weeks): 2,   Downloads (12 Months): 23,   Downloads (Overall): 153 Full text available: PDF Let P be a set of h pairwise-disjoint polygonal obstacles with a total of n vertices in the plane. In this paper, we consider the problem of building a data structure that can quickly compute an L1 shortest obstacle-avoiding path between any two query points s and t. We build ... Keywords: polygons with holes, L1 metric, Shortest paths, polygonal domains, two-point queries, weighted rectilinear obstacles 20 June 2014 ASIA CCS '14: Proceedings of the 9th ACM symposium on Information, computer and communications security Publisher: ACM Bibliometrics: Citation Count: 4 Downloads (6 Weeks): 2,   Downloads (12 Months): 40,   Downloads (Overall): 419 Full text available: PDF Cloud computing promises users massive scale outsourced data storage services with much lower costs than traditional methods. However, privacy concerns compel sensitive data to be stored on the cloud server in an encrypted form. This posts a great challenge for effectively utilizing cloud data, such as executing common SQL queries. ... Keywords: multiple dimension, tree structures, range search, encrypted cloud data


text:
A finite classical group is best thought of as a group of linear operators on some vector space defined over a finite field. Which means, of course, that I can take a basis for this vector space and then represent the elements of my group as matrices.

However, I have almost never seen anyone do this in the literature. “Taking a basis” is thought of as a rather crude thing to do when doing linear algebra – one generally tries to write general arguments that do not refer to any particular basis. However, speaking for myself, when I’m trying to work out what the hell is going on inside a finite matrix group, I often end up trying to write down individual elements as matrices… And then hiding all this when it comes to writing up the paper!

This means that I have never publicly written down any of these calculations, despite using many of them over and over again. So this page is designed to be a little repository for me to note down interesting observations about such calculations… And perhaps they’ll be useful for someone else some time…

## Working with $\Omega_4^+(q)$

This family of groups has some weird behaviour, especially when $q$ is even. For instance, let us write $\mathcal{U}$ for the set of maximal totally singular subspaces in a formed space of type $O+$ and dimension $2m$. If $q$ is even, then, provided $(m,q)\neq (2,2)$, we can define $\Omega_{2m}^+(q)$ to be the group inducing odd permutations on $\mathcal{U}$…. If $(m,q)=(2,2)$, however, the group so defined is not $\Omega_4^+(2)$, and one needs to define it in a completely different way (see Kleidman and Liebeck, p.31).

# Finding some elements in $\Omega_4^+(q)$.

Let $(e_1,f_1)$ and $(e_2, f_2)$ be hyperbolic pairs, and consider the basis $\mathcal{B}={e_1, e_2, f_2, f_1}$ maintaining order. Then, one can calculate directly that $O_4^+(q)$ contains elements which are written with respect to $\mathcal{B}$ in the form

these elements form a subgroup $S$ of size $q^2$; indeed, for $q$ odd, they form a Sylow $p$-subgroup of $O_4^+(q)$.

When $q$ is odd, one can look at orders to see directly that $S$ is actually a subgroup of $\Omega_4^+(q)$. Indeed, the same is true when $q$ is even, but to see this it is easiest to observe that $S$ is normalized by the element

This element clearly takes one element $U=\langle e_1, e_2\rangle \in \mathcal{U}_2$ to another element $W=\langle e_1, f_2\rangle\in \mathcal{U}_2$. What is more $U\cap W$ has codimension $1$ in both $U$ and $W$. This allows us to conclude that $g\in SO_4^+(q)\setminus \Omega_4^+(q)$ (see Kleidman and Liebeck, p. 30). Now order considerations imply that $S$ must be a subgroup of $\Omega_4^+(q)$.

# Finding all elements in $\Omega_4^+(q)$.

Let $R$ be the set of elements whose transpose is in $S$. One sees immediately that both $R$ and $S$ lie in $\Omega_4^+(q)$ and hence so does $\langle R,S\rangle$.

If one sets $a$ to equal $0$, while $b$ ranges across $\mathbb{F}_q$, in both $R$ and $S$, then one immediately observes a copy of $SL_2(q)$ in $\Omega_4^+(q)$. The same is true setting $b$ to equal $0$. Since these two copies effectively “avoid interaction” one immediately obtains a copy of $SL_2(q)\circ SL_2(q)$ inside $\Omega_4^+(q)$. (The central product is due to the fact that both copies of $SL_2(q)$ share $-I$ as an element.)

Now one can observe that $\Omega_4^+(q)$ also contains the element

Now order considerations allow us to conclude that $\Omega_4^+(q)\cong (SL_2(q)\circ SL_2(q)):2$, and we have written down all elements of the group.

## $Sp_{n-2}(2^a) < \Omega_n^+(2^a)$

When $q$ is even, $\Omega_n^+(q)$ has a maximal subgroup – the stabilizer of a non-degenerate $1$-space – isomorphic to $Sp_{n-2}(q)$. I will write down the elements of this subgroup for $n=4$; the general case follows similarly.

First, we adjust our basis from before to be $\mathcal{C}={z_1, x_1, x_2, y_2}$ where $z_1=x_1+y_1$.

With respect to this basis, our quadratic form becomes

and we see that $z_1$ is non-singular. Now simply observe that the following elements stabilize $z_1$:

Doing the “transpose trick” like before, one immediately obtains a copy of $SL_2(q)$ in this stabilizer and (using the perfectness of $SL_2(q)$ for $q>3$ if necessary), one obtains that the stabilizer of $z_1$ in $\Omega_4^+(q)$ is isomorphic to $SL_2(q)\cong Sp_{2}(q)$.

One can do this more generally for larger $n$ – one simply has to exhibit the root groups of $Sp_{n-2}(q)$ inside the stabilizer of $z_1$ in $\Omega_4^+(q)$. There are two sorts of root group here: the first already ocur in $\Omega_{n-2}^+(q)$ and so are easy to write down in $\Omega_n^+(q)$; the second all take the form given above.


text:
170 Answered Questions for the topic Ratios And Proportions

Ratios And Proportions Scale Drawings

05/25/17

#### If 1 3/4 inch = 17 1/12 feet, then how much of 1 inch = 1 foot

Trying to figure out the dimensions of a hallway in a poorly provided blue print. The only dimensions provided are the living room and kitchen size.
Ratios And Proportions

05/17/17

#### a shark can swim 100 meters in 4 seconds and marlin can swim 70 meters in 3.5 seconds

The shark is faster but by how much seconds
Ratios And Proportions

05/01/17

#### 100 possible points each week, week 6 avg. is 78%, what is lowest score student must get for next three weeks of 9 week period in order to raise score to 84%

Mrs Cooley gives 100 points each week during a nine week grading period. At the end of the sixth week, Carlos has a 78% average. What is the lowest average score he can receive each week during the... more
Ratios And Proportions

04/26/17

A children’s library has 5 storybooks for every 3 science books it has. The library also has the same number of picture books as science books. The library added 50 more picture books so now there... more
Ratios And Proportions

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#### If 12 cubic yards of wood are burnt in 9 fireplaces in 16 days, then in how many days of 9 cubic yards of wood burnt in 12 fireplaces?

Please solve and show how you arrived at the solution
Ratios And Proportions Percents

04/11/17

#### mack burns 95 calories when running for 10 minutes how many calories would he burn if he ran 18 minutes

mack burns 95 calories when running 10 minutes on a treadmill. how many calories would he burn if he ran for 18 minutes on the treadmill?
Ratios And Proportions

04/03/17

#### Anna can type 123 words in 3 minutes. Caroline can type 220 words in 5 minutes . Who can type the fastest

How would I solve this problem ?
Ratios And Proportions

03/22/17

#### What is the ratio of 3 1/2 and 5 1/4

This ajd sofjos djfasodjfos djsoa osajfods jofsjaoid osjf osi
Ratios And Proportions

03/22/17

#### it took 8 hours to clean 12 houses how many houses could be claen in 12 hours

how do i do it
Ratios And Proportions

03/13/17

Three insurance companies, A, B, and C, agree to jointly insure a cargo ship for $28,600,000. They allocate the premiums received in the ratio 7 : 3 : 12 respectively. How will the annual premium... more Ratios And Proportions 03/13/17 #### calculate the missing term 3:12 = 6:x x:1.6 =3.2:6.4 1: 5 1/5 = x: 3 3/4 Ratios And Proportions Ratio 03/07/17 #### The ratio of actors to audience in a theatre is 2:17. There are 391 people in he audience. How many actors are there? I do not know how to solve the word problem. Ratios And Proportions 03/03/17 #### Sebastian swam laps. He swam 45 minutes each day, 5 days a week, for 12 weeks. In that time he swam 1,800 laps. What was his average rate in laps per hour? You need to show the steps in how you did it and the answer. Ratios And Proportions 02/12/17 #### Ratio and proportions mrs and mr K decided to share$120 equally between their children for summer vacation. When the eldest got holiday job, he divided his share equally between all his brothers and sisters. If each... more
Ratios And Proportions

02/01/17

#### You mix 1/4 cup of juice concentrate for every 2 cups of water to make 18 cups of juice. How much juice concentrate and water do you use?

How much juice concentrate and water do you use?
Ratios And Proportions

01/22/17

#### Distance/Rate Time question

Aubrey and Isabella are both riding their bikes in the same direction on the same road. At 11 am Aubrey is 7 miles ahead of Isabella. Aubrey covers 1 mile in 6 minutes and Isabella covers 3 miles... more
Ratios And Proportions Geometry Word Problems

01/19/17

#### The ratio of a bunkbed's weight to the weight it can bear is 3:13. How many 70-lb children can a 60-lb bed hold?

I am trying to figure this problem out but I can't seem to do it.
Ratios And Proportions

01/11/17

#### if 3/4 cup of packed brown sugar is needed for one batch of chocolate chip cookies, how much packed sugar is needed for five batches?

I dont know if i got the right answer but can you show it step by step please?
Ratios And Proportions

12/06/16

#### Do the ratios in each pair form a proportion?

1) 1/3:3, 1/4:4 2) 2/3:2/9, 1/4:1/12 3) 4/5:1/2, 2/3:1/4
Ratios And Proportions

12/06/16

#### Do the ratios in each pair form a proportion?

1) 1/3:3, 1/4:4   2) 2/3:2/9, 1/4:1/12   3) 4/5:1/2, 2/3:1/4
Ratios And Proportions

12/05/16

#### Francis drove to her mothers house which is 375 miles away if it took her five hours what was her average speed

this is a ratios and proportions question i don't know how to solve i need help. thanks

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text:
## Elementary Algebra

Debra's mother is $2x-3$ years old.
In order to get from Debra's age, $x$, to her mother's age, we must first multiply by $2$, because the word "twice" is used. Then, we subtract $3$, because the words "$3$ less than" are used. This leads us to her mother's age, which is $2x-3$ years old.


text:
# Symplectic Koopmanism

Let $(M, \omega)$ be a $2n$-dimensional symplectic manifold and let $L_2(M,|\omega^n|)$ be the Hilbert space of complex-valued functions on $M$ that are square integrable with respect to the Liouville measure.

A classical (but still wonderful) remark due to Koopman is that given a measure preserving map $\phi: M \rightarrow M$, the operator $$U_\phi : L_2(M,|\omega^n|) \longrightarrow L_2(M,|\omega^n|)$$ defined by $f \mapsto f \circ \phi$ is unitary. The operator $U_\phi$ is sometimes called the Koopman operator of the map $\phi$. Since symplectomorphisms are measure preserving, it seems natural to ask the following

Vague question. Is there anything at all particular about Koopman operators of symplectomorphisms?

• Assuming that by "uniform closure", you mean "norm closure", then operators that you describe form a closed discrete set (any tow of them have distance 2). – André Henriques Jun 27 '14 at 12:39
• Of course !! It was a dumb way to try to make precise a vague question. I'm editing the question to leave it vague. ;-) – alvarezpaiva Jun 27 '14 at 12:48
• What is the motivation for your question? – André Henriques Jun 27 '14 at 15:32
• @AndréHenriques: I'm trying to see if there is any symplectic version of ergodic theory, one that uses that the maps are not just measure preserving. – alvarezpaiva Jun 27 '14 at 17:13
• Which results/concepts in ergodic theory do you expect to be affected by the fact that the transformation is a symplectomorphism? – André Henriques Jun 27 '14 at 19:05

The Koopman operator is bracket preserving with respect to the canonical Poisson bracket on $C^{\infty}(M)$. Formally, this implies the dual operator (the Frobenius-Perron operator) is a Poisson automorphism with respect to the Lie-Poisson structure on $\bigwedge^{2n}(M)$.


text:
# Coherent states

vanhees71
Gold Member
2019 Award
Of course, the ground state is an eigenstate of ##\hat{b}## with eigenvalue 0. Just calculate the derivative to see it.

Cthugha
Of course, the ground state is an eigenstate of ##\hat{b}## with eigenvalue 0. Just calculate the derivative to see it.
Now I am somewhat puzzled. Are you sure about that? Maybe I just missed something, but isn't ##\hat{b}## the equivalent of the creation operator? I always thought that the annihilation operator had the plus sign.

vanhees71
Gold Member
2019 Award
You are right. I've misread the sign.

The annihilation operator is
$$\hat{a}=\sqrt{\frac{m \omega}{2}} \hat{x}+\mathrm{i} \frac{1}{\sqrt{2m \omega}} \hat{p}.$$
In position representation that maps to
$$\hat{a}=\sqrt{\frac{m \omega}{2}} x +\frac{1}{\sqrt{2 m \omega}} \partial_x = \frac{1}{\sqrt{2}} (\xi + \partial_{\xi}).$$
The creation operator is its adjoint,
$$\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\xi-\partial_{\xi})=\hat{b},$$
which of course has no eigenstate.

The annihilation operator has many eigenstates, the coherent states. The special one with eigenvalue 0 is the ground state. This gives the eigenvector equation
$$\hat{a} u_0(\xi)=0 \; \Rightarrow \; \partial_{\xi} u_0=-\xi u_0 \xi \; \Rightarrow \; u_0(\xi)=N \exp(-\xi^2/2)$$
with ##N## a normalization factor.


text:
# derivative trigonometry

• September 11th 2008, 08:51 PM
jvignacio
derivative trigonometry
$

f(x) = tanxsinx

$

$

f '(x) = (sec^2x)(sinx) + (tanx)(cosx)

$

this correcT?
• September 11th 2008, 08:58 PM
11rdc11
Yep

$
f '(x) = (sec^2x)(sinx) + (tanx)(cosx)
$

You could simplify it a little more

$tan(x)(secx+cosx)$
• September 11th 2008, 09:00 PM
jvignacio
Quote:

Originally Posted by 11rdc11
Yep

can we simplify this ?
if so how would u do it?
• September 11th 2008, 09:02 PM
Chris L T521
Quote:

Originally Posted by jvignacio
$

f(x) = tanxsinx

$

$

f '(x) = (sec^2x)(sinx) + (tanx)(cosx)

$

this correcT?

Yes, but it can be simplified further:

Note that $\sec^2x\sin x=\frac{1}{\cos^2x}\sin x=\tan x \sec x$ and $\tan x\cos x=\frac{\sin x}{\cos x}\cos x=\sin x$

So we now see that $f '(x) = (\sec^2x)(\sin x) + (\tan x)(\cos x)=\sec x\tan x+\sin x$

--Chris
• September 11th 2008, 09:07 PM
jvignacio
Quote:

Originally Posted by Chris L T521
Yes, but it can be simplified further:

Note that $\sec^2x\sin x=\frac{1}{\cos^2x}\sin x=\tan x \sec x$ and $\tan x\cos x=\frac{\sin x}{\cos x}\cos x=\sin x$

So we now see that $f '(x) = (\sec^2x)(\sin x) + (\tan x)(\cos x)=\sec x\tan x+\sin x$

--Chris

what if the possible answers were:

$
sinx -
sinxsec^2x -
sinx(sec^2x+1) -
secx -
none of these
$

what would you go for?
• September 11th 2008, 09:16 PM
11rdc11
I did this kind of fast but I'm pretty sure it the 3rd option

$
f '(x) =\sec x\tan x+\sin x
$

$\frac{sinx}{cos^2x} + sinx$

$\frac{sinx+sinxcos^2x}{cos^2x}$

$\frac{sinx(1+cos^2x)}{cos^2x}$

$(sinx)(sec^2x + 1)$
• September 11th 2008, 09:27 PM
jvignacio
Quote:

Originally Posted by 11rdc11
I did this kind of fast but I'm pretty sure it the 3rd option

yeah thats what i put but i wasnt sure why we got rid of the tanx and cosx


text:
# Multi-Digit Division

## Objective

Solve division word problems with remainders.

## Common Core Standards

### Core Standards

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• 4.OA.A.3 — Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.

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• 3.OA.C.7

• 3.OA.D.8

## Criteria for Success

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1. Understand that a remainder is the number left over when one number is divided by another.
2. Solve division word problems within 100 that involve a remainder, using an array, an area model, or a tape diagram to represent the problem (MP.4).
3. Check the solution to division problems using inverse operations, multiplying the quotient with the dividend and adding the remainder.
4. Understand that the units of the quotient and remainder are different, as the quotient represents either the number of groups or the size of the groups, but the remainder represents what remains of the whole.

## Tips for Teachers

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• Students have likely come across division situations in real life that do not result in a simple whole number quotient (especially considering that problems with 6 as a divisor, for example, will result in a whole number only one time out of six). Because of this, students may choose to interpret the remainder in the problems they encounter in this lesson. You don’t necessarily want to discourage students from doing so, but note that interpretation will happen in Lesson 2.
• “To compute and interpret remainders in word problems (4.OA.A.3), students must reason abstractly and quantitatively (MP.2), make sense of problems (MP.1), and look for and…search for the structure (MP.7) in problems with similar interpretations of remainders” (PARCC Model Content Frameworks, Mathematics, Grade 3—11).
• “The result of division within the system of whole numbers is frequently written as:

$84 \times 10 = 8\space \mathrm{R} \space 4$ and $44 \div 5 = 8 \space \mathrm{R}\space 4$

Because the two expressions on the right are the same, students should conclude that $84\times10$ is equal to $44\div5$, but this is not the case. (Because the equal sign is not used appropriately, this usage is a non-example of Standard for Mathematical Practice 6.) Moreover, the notation $8\space \mathrm{R} \space 4$ does not indicate a number. Rather than writing the result of division in terms of a whole-number quotient and remainder, the relationship of whole-number quotient and remainder can be written as:

$84 = 8 \times 10 + 4$ and $44 = 8\times 5 + 4$

(Progressions for the Common Core State Standards in Mathematics, Number and Operations in Base Ten, K-5, p. 16). Thus, you should avoid using the “R” notation immediately following the equal sign.

#### Fishtank Plus

• Problem Set
• Student Handout Editor
• Vocabulary Package

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### Problem 1

There are 13 students to be split among 4 teams. How many students will be on each team?

#### References

EngageNY Mathematics Grade 4 Mathematics > Module 3 > Topic E > Lesson 14Concept Development Problem 1

Grade 4 Mathematics > Module 3 > Topic E > Lesson 14 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by The Match Foundation, Inc.

### Problem 2

Kristy has 22 roses to sell. She arranges 6 roses in each vase.

1. How many vases will she be able to fill completely? Will there be any roses left over?
2. What unit should Kristy use for the quotient and remainder?

#### References

EngageNY Mathematics Grade 4 Mathematics > Module 3 > Topic E > Lesson 14Concept Development Problem 3

Grade 4 Mathematics > Module 3 > Topic E > Lesson 14 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by The Match Foundation, Inc.

## Problem Set & Homework

#### Discussion of Problem Set

• Look at #4 and #5. These problems seem very similar, but the units used for their quotients are different. Why is that?
• Look at #2. How many total tables would be needed to seat everyone? Why?
• Look at #3. How many total pies can Caleb make? Why?

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Fifty-three students are going on a field trip. The students are divided into groups of 6 students. How many groups of 6 students will there be? How many remaining children will there be?

#### References

EngageNY Mathematics Grade 4 Mathematics > Module 3 > Topic E > Lesson 14Exit Ticket

Grade 4 Mathematics > Module 3 > Topic E > Lesson 14 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by The Match Foundation, Inc.

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text:
My watch list
my.chemeurope.com

# Poynting vector

In physics, the Poynting vector can be thought of as representing the energy flux (W/m2) of an electromagnetic field. It is named after its inventor John Henry Poynting. Oliver Heaviside independently co-discovered the Poynting vector. Usually, it is defined as

$\mathbf{S} = \mathbf{E} \times \mu_r \mathbf{H} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B},$

where E is the electric field, H the auxiliary magnetic field, B the magnetic field, μ0 the permeability of vacuum, and μr the dimensionless relative permeability of the surrounding medium. (All bold letters represent vectors.)

For example, the Poynting vector within the dielectric insulator of a coaxial cable is nearly parallel to the wire axis (assuming no fields outside the cable) - so electric energy is flowing through the dielectric between the conductors. If the core conductor was replaced by a wire having significant resistance, then the Poynting vector would become tilted toward that wire, indicating that energy flows from the e/m field into the wire, producing resistive Joule heating in the wire.

## Interpretation

The Poynting vector appears in the energy-conservation law[1], or Poynting's theorem,

$\frac{\partial u}{\partial t} = - \mathbf{\nabla}\cdot\mathbf{S} -\mathbf{J} \cdot \mathbf{E},$

where J is the current density and u is the electromagnetic energy density,

$u = \frac{1}{2}\left(\epsilon_0 \mathbf{E}^2 + \frac{1}{\mu_0} \mathbf{B}^2\right),$

where ε0 is the permittivity of free space and εr the relative dielectric constant of the medium. The first term in the right-hand side represents the net electromagnetic energy flow into a small volume, while the second term represents negative of the work done by electrical currents that are not necessarily converted into electromagnetic energy.

## The Poynting vector in electromagnetic waves

In a propagating sinusoidal electromagnetic plane wave of a fixed frequency, the Poynting vector oscillates, always pointing in the direction of propagation. The time-averaged magnitude of the Poynting vector is

$\langle S \rangle = \frac{1}{2 \mu_0 c} E_0^2 = \frac{\epsilon_0 c}{2} E_0^2=\frac{1}{2}\sqrt{\frac{\epsilon_0}{\mu_0}} E_0^2,$

where E0 is the maximum amplitude of the electric field and c is the speed of light. This time-averaged value is also called the irradiance or intensity I.

### Derivation

In an electromagnetic plane wave, $\mathbf{E}$ and $\mathbf{B}$ are always perpendicular to each other and the direction of propagation. Moreover, their amplitudes are related according to

$B_0 = \frac{E_0}{c},$

and their time and position dependences are

$E\left(t,{\mathbf r}\right) = E_0\,\cos\left(\omega\,t- {\mathbf k} \cdot {\mathbf r} \right),$
$B\left(t,{\mathbf r}\right) = B_0\,\cos\left(\omega\,t- {\mathbf k} \cdot {\mathbf r} \right),$

where ω is the frequency of the wave and $\mathbf{k}$ is wave vector. The time-dependent and position magnitude of the Poynting vector is then

$S(t) = \frac{1}{\mu_0} E_0\,B_0\,\cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r}\right) = \frac{1}{\mu_0 c} E_0^2 \cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r} \right) = \epsilon_0 c E_0^2 \cos^2\left(\omega t-{\mathbf k} \cdot {\mathbf r} \right).$

In the last step, we used the equality $\epsilon_0\,\mu_0 = c^{-2}$. Since the time- or space-average of $\cos^2\left(\omega\,t-{\mathbf k} \cdot {\mathbf r}\right)$ is $\frac{1}{2}$, it follows that

$\left\langle S \right\rangle = \frac{\epsilon_0 c}{2} E_0^2.$

S divided by the square of the speed of light in a vacuum is the density of the linear momentum of the electromagnetic field. The time-averaged intensity $\langle\mathbf{S}\rangle$ divided by the speed of light in a vacuum is the radiation pressure exerted by an electromagnetic wave on the surface of a target:

$P_{rad}=\frac{\langle S\rangle}{c}.$

## References

• "Poynting Vector" from ScienceWorld (A Wolfram Web Resource) by Eric W. Weisstein
• Richard Becker, Electromagnetic Fields and Interactions, Dover Publications, 1964
1. ^ J.D. Jackson (1975). Classical Electrodynamics, 2nd Ed..


text:
# Sech

## Description

The sech(x) function returns hyperbolic secant of x. An absolute value larger than about 710 is computationally out of bounds and returns missing value.

## Syntax

double sech(double x)

## Parameter

x

can be any real number. An absolute value larger than about 710 is computationally out of bounds and returns missing value.

## Return

Returns hyperbolic secant of x.

## Example

aa = sech(10);
aa = ; //should return 9.0799859337817E-5