text:
Intuitive explanation for formula of maximum length of a pipe moving around a corner?

For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):

My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.

The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?

I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.

If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?

Here's how he found the formula, if it's useful:

The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:

$$L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\$$

At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:

$$\cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\$$

The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$.

• This is phrased as a maximization problem but it's really a minimization problem. It's asking what's the length of the shortest line segment through that corner point having endpoints on the two opposite walls. That may be the easiest way to treat it. Obviously at the two opposite extremes the length approaches $\infty$, so it has a minimum somewhere between those. – Michael Hardy Nov 27 '13 at 20:37
• There is a very nice explanation of this problem in the free online course Calculus One by Bart Snapp of The Ohio State University. The video for Week 8 number 14 How large of an object can you carry around a corner? covers this problem. – Mufasa Nov 27 '13 at 20:37
• I'm not sure what you are asking. The formula for the longest pipe $L(\theta)$ that can fit across the bend at angle $\theta$ is straightforward to work out. A pipe can go around the bend iff the pipe length is less than $L(\theta)$ for all $\theta \in [0, \frac{\pi}{2}]$, which is where the minimization comes in. By parameterizing the problem differently, you can avoid the angles, but end up with the same result. – copper.hat Nov 27 '13 at 21:19
• This reminds me of : en.wikipedia.org/wiki/Moving_sofa_problem – K. Rmth Nov 29 '13 at 18:47

Here's a derivation without trigonometry. There is also a short discussion of the geometric significance of the exponents in the answer after the derivation.

Let $L$ be the length of a line touching the inner corner of the two corridors and extending to the outer wall of each corridor. Where this line meets the outer wall of each corridor, draw a line perpendicular to the corridor. Let $x$ be the distance from the inner corner to the perpendicular line across corridor $B$ and let $y$ be the distance from the inner corner to the perpendicular line across corridor $A$. The result is the figure shown above.

Applying the Pythagorean Theorem to the obvious right triangle with hypotenuse $L$, $$L^2 = (A + x)^2 + (B + y)^2.$$

The longest pipe that can fit around the corner is the smallest value of $L$ for any value $x > 0$, so let's minimize $L$ as a function of $x$. But $L$ is minimized when $L^2$ is minimized, so we would like to set $$\frac{d}{dx} L^2 = 0.$$

That is, $$\begin{eqnarray} 0 = \frac{d}{dx} L^2 &=& \frac{d}{dx}\left((A + x)^2 + (B + y)^2\right) \\ &=& 2(A + x) + 2(B + y)\frac{dy}{dx}. \end{eqnarray}$$

Now, by similar triangles, $\dfrac Bx = \dfrac yA .$ That is, $xy = AB.$ Differentiating both sides of this by $x$, $$x\frac{dy}{dx} + y = 0,$$ $$\frac{dy}{dx} = -\frac yx.$$

Substituting for $\frac{dy}{dx}$ in our earlier equation for $\frac{d}{dx}L^2,$ $$0 = 2(A + x) + 2(B + y)\left(-\frac yx\right),$$

from which it follows that $$(A + x)x = (B + y)y,$$

But (again by similar triangles) $$\frac Bx = \frac{B+y}{A+x},$$ $$B+y = \frac Bx (A+x),$$ and combining this with the fact that $y = \dfrac{AB}{x},$ $$(B + y)y = \left(\frac Bx (A+x)\right) \frac{AB}{x} = \frac{AB^2}{x^2}(A + x).$$

That is, $$(A + x)x = \frac{AB^2}{x^2}(A + x),$$ $$x^3 = AB^2,$$ $$x = A^{1/3}B^{2/3},$$ $$y = \frac{AB}{x} = A^{2/3}B^{1/3},$$

$$A + x = A + A^{1/3}B^{2/3} = \left(A^{2/3} + B^{2/3} \right)A^{1/3},$$ and $$B + y = B + A^{2/3}B^{1/3} = \left(A^{2/3} + B^{2/3} \right)B^{1/3}.$$

Therefore at the value of $x$ that minimizes $L$,

$$\begin{eqnarray} L^2 &=& (A + x)^2 + (B + y)^2 \\ &=& \left(A^{2/3} + B^{2/3} \right)^2 A^{2/3} + \left(A^{2/3} + B^{2/3} \right)^2 B^{2/3} \\ &=& \left(A^{2/3} + B^{2/3} \right)^2 \left(A^{2/3} + B^{2/3}\right) \\ &=& \left(A^{2/3} + B^{2/3} \right)^3 \end{eqnarray}$$

and $$L = \left(A^{2/3} + B^{2/3} \right)^{3/2}.$$

Regarding the symmetry of the solution, clearly the length of the pipe does not change if we swap the labels $A$ and $B$ on the two corridors. We should suspect from this that there might be some expression for $L$ in which $A$ and $B$ appear in completely symmetric roles.

Regarding the geometric interpretation of the exponents, the key facts that lead to the $\frac23$ exponents are the findings that $x^3 = AB^2$ and (by symmetry) $y^3 = BA^2.$ This is what leads to the conclusion that $L$ is the hypotenuse of a right triangle with legs $A + A^{1/3}B^{2/3}$ and $B + B^{1/3}A^{2/3}$; it is from the squares of those two lengths that we get terms in $A^{2/3}$ and $B^{2/3}$.

Finally, note that if we double both $A$ and $B$ simultaneously, we should expect the entire figure showing the "tightest" angle of the pipe around the corner to scale up by a factor of $2$, hence $L$ also doubles. But we see that $A^{2/3}$ and $B^{2/3}$ scale up only by a factor of $2^{2/3}$ when we double $A$ and $B$; it is the exponent of $\frac32$ outside the parentheses that gives us a scaling factor of $\left(2^{2/3}\right)^{3/2} = 2.$ So it is not a mere coincidence that the exponents $\frac23$ and $\frac32$ are multiplicative inverses.

Another way to approach this is to apply the fact that $B+y = \dfrac Bx (A+x)$ before differentiating $L^2$ rather than afterwards. That is, we simply substitute for $B+y$ in the formula for $L^2$, obtaining

$$\begin{eqnarray} L^2 &=& (A + x)^2 + \frac{B^2}{x^2}(A + x)^2 \\ &=& \left(1 + \frac{B^2}{x^2}\right) (A + x)^2. \end{eqnarray}$$

We can differentiate this with respect to $x$ to obtain an analytic solution, or apply numeric methods using $x$ as an independent variable.

• Is there anyway you could set up an equation of some sort? As in a function where you can see the minimum on the graph, and that minimum is your answer to the question? With this, it seems that this solves the problem analytically – user121615 May 21 '15 at 11:19
• @user121615 I added a formula for $L^2$ with just $x$ as a variable; that should give you something that can be graphed. – David K May 21 '15 at 13:11

As far as I know, this isn't a case where there's an intuitive explanation of this problem on its own. Here are two substitutes:

First, there are simpler calculations using more advanced techniques, and maybe these calculations will make the form of the answer seem less mysterious. For example, if we rephrase the question as

Find the length of the shortest line segment from $(x,0)$ to $(0,y)$ that passes through $(a,b)$.

then with a little coordinate geometry we can rephrase it again as

Minimize $(x^2+y^2)^{1/2}$ subject to the constraint $\frac ax + \frac by = 1$.

Using the method of Lagrange multipliers from multivariable calculus on this version of the problem yields quickly that the optimum has $(x^3,y^3)$ proportional to $(a,b)$.

Second, there is a bigger picture about this kind of problem — if we rephrase the question again by replacing $x,y$ with their reciprocals, we get the $p=-2$ case of the problem

Minimize $(x^p+y^p)^{-1/p}$ subject to the constraint $ax+by=1$.

Expressions like $(x^p+y^p)^{1/p}$ can be thought of as a more general notion of "distance". With $p=2$ it's the usual distance, as in Pythagoras' theorem; with $p=1$ you get taxicab distance; for $p\ge 1$ you get various weird notions of "distance" which still act in some essential ways like the familiar notion; for $0<p<1$ you get notions which are still somewhat distance-ish, but not very; for $p<0$ the feeling of "distance" has been left far behind, but we continue with stout hearts. (More.)

The relationship between the $p=-2$ notion of "distance" which occurs in the (rephrased) problem statement and the $p=\frac23$ notion of "distance" which occurs in the solution is a much studied one; these notions of distance are said to be "dual" to each other. For people who have studied this subject, the outcome here is quite natural — it's "because" $\frac23$ is the conjugate exponent of $-2$. (We need a version of Hölder's inequality for negative exponents, which is not always taught, but it exists.)

(The notion of "duality" between notions of distance has its roots in 19th century geometry, in efforts to understand geometric shapes not just as collections of points but also as envelopes of their tangents. From this point of view, we could say that your problem is dual to this one: if a ladder standing straight against a wall slides down to the ground (its top sliding down the wall and its base sliding along the ground), what curve is tangent to the ladder at all times? See Astroid, where you will see your $\frac23$.)

So, no, the form of the answer is not a coincidence, but as far as I know it is only "intuitive" in retrospect, which really means only that it has become familiar.

If this result could be explained intuitively it wouldn't have been necessary to go through the calculations. A priori you could expect $L$ to be homogeneous of degree $1$ in $A$ and $B$, as well as being symmetric in $A$ and $B$. As a matter of fact the result of the calculation could as well have been $$L=2\min\{A,B\},\quad{\rm or}\quad L=\sqrt{A^2+AB+B^2}\ ,$$ or an expression involving some other exponent in place of ${2\over3}$.


text:
# [Solution] White-Black Balanced Subtrees solution codeforces

White-Black Balanced Subtrees solution codeforces – You are given a rooted tree consisting of 𝑛n vertices numbered from 11 to 𝑛n. The root is vertex 11. There is also a string 𝑠s denoting the color of each vertex: if 𝑠𝑖=𝙱si=B, then vertex 𝑖i is black, and if 𝑠𝑖=𝚆si=W, then vertex 𝑖i is white.

## [Solution] White-Black Balanced Subtrees solution codeforces

A subtree of the tree is called balanced if the number of white vertices equals the number of black vertices. Count the number of balanced subtrees.

tree is a connected undirected graph without cycles. A rooted tree is a tree with a selected vertex, which is called the root. In this problem, all trees have root 11.

The tree is specified by an array of parents 𝑎2,,𝑎𝑛a2,…,an containing 𝑛1n−1 numbers: 𝑎𝑖ai is the parent of the vertex with the number 𝑖i for all 𝑖=2,,𝑛i=2,…,n. The parent of a vertex 𝑢u is a vertex that is the next vertex on a simple path from 𝑢u to the root.

The subtree of a vertex 𝑢u is the set of all vertices that pass through 𝑢u on a simple path to the root. For example, in the picture below, 77 is in the subtree of 33 because the simple path 75317→5→3→1 passes through 33. Note that a vertex is included in its subtree, and the subtree of the root is the entire tree.

The picture shows the tree for 𝑛=7n=7𝑎=[1,1,2,3,3,5]a=[1,1,2,3,3,5], and 𝑠=𝚆𝙱𝙱𝚆𝚆𝙱𝚆s=WBBWWBW. The subtree at the vertex 33 is balanced.

## [Solution] White-Black Balanced Subtrees solution codeforces

The first line of input contains an integer 𝑡t (1𝑡1041≤t≤104) — the number of test cases.

The first line of each test case contains an integer 𝑛n (2𝑛40002≤n≤4000) — the number of vertices in the tree.

The second line of each test case contains 𝑛1n−1 integers 𝑎2,,𝑎𝑛a2,…,an (1𝑎𝑖<𝑖1≤ai<i) — the parents of the vertices 2,,𝑛2,…,n.

The third line of each test case contains a string 𝑠s of length 𝑛n consisting of the characters 𝙱B and 𝚆W — the coloring of the tree.

It is guaranteed that the sum of the values 𝑛n over all test cases does not exceed 21052⋅105.

Output

For each test case, output a single integer — the number of balanced subtrees.

## [Solution] White-Black Balanced Subtrees solution codeforces

Example
input

Copy
3
7
1 1 2 3 3 5
WBBWWBW
2
1
BW
8
1 2 3 4 5 6 7
BWBWBWBW

output

Copy
2
1
4

## [Solution] White-Black Balanced Subtrees solution codeforces

The first test case is pictured in the statement. Only the subtrees at vertices 22 and 33 are balanced.

In the second test case, only the subtree at vertex 11 is balanced.

In the third test case, only the subtrees at vertices 113355, and 77 are balanced.

format: plain


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# Decision-LWE to Search-LWE

Regev requires $q$ to be prime on lemma 4.2 of his paper for LWE.

Why does he require that and how this effect the proof of lemma 4.2?

He needs to add something uniformly random to the second coordinate, to get a distribution that is uniformly random. (The idea is that if he guesses the correct key, his rerandomization turns real instances into real instances. If he guesses the wrong key, his rerandomization should turn real instances into random instances.)

If $p$ is a prime, then multiplication by any non-zero value is a bijection, so a uniformly distributed $l$ maps to a uniformly distributed product $l(k-s_1)$.

If $p$ is composite, then multiplication by a value that isn't relatively prime to $p$ would not be a bijection, so a uniformly distributed $l$ might not map to a uniformly distributed product $l(k-s_1)$.

• Thank you for your answer, can you recommend to me a paper with a mathematical proof that these results relied on? It would be very helpfull. – Mike Anast Jun 14 '18 at 18:24
• @MikeAnast I would not look for papers, but for an introduction to number theory and perhaps also an introduction to statistics. – K.G. Jun 22 '18 at 2:05
• I think i found the answer on lemma 4.2 in [eprint.iacr.org/2011/521.pdf]. I would appreciate your opinion to this. – Mike Anast Jun 23 '18 at 14:46


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# TITLES AND ABSTRACTS OF TALKS

Mahsa Allahbakhshi,   University of Victoria
PhD student, advised by Anthony Quas
Factor triples and relative entropy

Tim Austin,   UCLA
PhD student, advised by Terry Tao
Some recent approaches to multiple recurrence.
Furstenberg's 1977 proof of Szemeredi's Theorem using a conversion to an assertion of multiple recurrence' for probability-preserving systems gave rise to a whole subdiscipline of ergodic theory called Ergodic Ramsey Theory', which went on to provide proofs for several other extremal results in different combinatorial settings. In recent years, a number of alternatives to the original basic strategies of Ergodic Ramsey Theory have emerged within ergodic theory, and offered a clearer insight into the connections between this field and purely combinatorial approaches to the same results. In this talk I will describe Tao's infinitary version of the hypergraph removal lemma from extremal combinatorics, and then sketch how its conjunction with variants of the machinery of pleasant extensions gives a new approach to both the multidimensional multiple recurrence phenomenon and Furstenberg and Katznelson's density version of Hales-Jewett Theorem.

Francesco Cellarosi,   Princeton University
PhD student, advised by Yakov Sinai
On the limit curlicue process for theta sums
I shall discuss a random process achieved as the limit for the ensemble of curves generated by interpolating the values of theta sums. The existence and the properties of this process are established by means of purely dynamical tools and rely on generalizations of a result by Marklof and Jurkat and van Horne. (joint work with Jens Marklof).

Vaughn Climenhaga,   The Pennsylvania State University
PhD student, advised by Yakov Pesin
Multifractal analysis of Birkhoff averages
One important goal in understanding a dynamical system is to describe the asymptotic behaviour of time averages of an observable function---the Birkhoff averages. Given an ergodic invariant measure for the system, the Birkhoff ergodic theorem gives us information on a full measure set of "typical" points. However, there are reasons to want information about the asymptotics at every point, not just almost every point. We mention some of these reasons and describe the key tools of multifractal analysis, which uses dimension theory and thermodynamic formalism to give a more detailed picture.

Van Cyr,   The Pennsylvania State University
PhD student, advised by Omri Sarig
Conformal measures for transient Markov shifts

Andres del Junco ,   University of Toronto
A survey of almost continuous orbit equivalence
Suppose T and S are finite measure-preserving homeomorphisms of topological spaces. By Dye's theorem they are measurably orbit equivalent. If the map implementing the orbit equivalence is continuous when restricted to some set of full measure one could call it almost continuous. It turns out that this definition is too weak but appropriate strengthenings give rise to an interesting theory. We will see that there is an almost continuous version of Dye's theorem, as well as some results for non-singular homeomorphisms and analogues of other theorems in the measurable theory. In particular I will say something about the very interesting notion of almost continuous Kakutani equivalence. Dan Rudolph was quite involved in this field in his last few years.

Matt Foreman,   U.C. Irvine
The classification problem for ergodic measure preserving transformations
In 1932 von Neumann formulated the problem of classifying ergodic measure preserving transformations of [0,1]. In the nearly 80 years that have followed, there has been considerable positive progress on this problem; notably Ornstein's work using the Sinai-Kolmogorov invariant "entropy" to classify Bernoulli shifts and the theorem of von Neumann and Halmos classifying discrete spectrum transformations. After reviewing the history, this talk will focus on joint work with Dan Rudolph and Benjamin Weiss that established some sweeping anti-classification" results. These theorems show that there are fundamental reasons that any classification is impossible.

Hillel Furstenberg ,   Hebrew University
On Meiri Sequences
David Meiri in his Master's thesis inaugurated a study of what he called "generalized correlation sequences". These are sequences whose averages appear in so-called non-conventional ergodic theorems. They are also closely related to what Host and Kra call "Nil-Bohr sets", and they deserve to be analyzed on their own right.

Mike Hochman ,     Princeton University
Geometric rigidity of times-m invariant measures
In 1990 Rudolph showed that if a measure on [0,1] is invariant under times-2 and times-3 mod 1, ergodic, and has positive entropy under the individual maps, then it is Lebesgue measure, settling the positive-entropy case of Furstenberg's "times-2 times-3 conjecture". In particular, this says that if mu is a times-2 ergodic measure of positive entropy, and nu is times-3 ergodic and of positive entropy, and neither is Lebesgue, then mu and nu are not equal. From a result of Host, it follows that in fact mu,nu must be mutually singular. In this talk I will discuss a geometric generalization of this result, showing that it is robust under smooth perturbations: that is, under the same assumptions, if mu',nu' are obtained by applying (different) smooth maps to mu and nu, respectively, then mu',nu' are still singular, and one can identify which is which from the spectrum of a certain flow associated to the measures.

Chris Hoffman ,   University of Washington
A K but not Bernoulli Z^2 subshift of finite type
Domino tilings of the plane have been studied extensively for both their statistical properties and their dynamical properties. In particular Rick Kenyon has shown that the certain properties of a domino tiling are conformally invariant. We construct a subshift of finite type using a collection of colored dominoes. Using these results about domino tilings we will show that our subshift is K (every factor has positive entropy) but is not isomorphic to a Bernoulli shift.

Aimee Johnson ,     Swarthmore College
Recurrence on Infinite Measure Dynamical Systems
Here we consider Z^2 actions on infinite measure spaces and investigate the possible directions at which we can see recurrence. Extending the notion of recurrence to the unit circle, we consider the structure of the set of recurrent directions. This is joint work with Dan Rudolph and Ayse Sahin.

Uijin Jung , Korea Advanced Institute of Science and Technology, Daejeon
Extension theorems for codes between shifts of finite type

Byungsoo Kim ,   Colorado State University
Rolling molecules: stationray states, stability and statistical physics
When molecules of a gas or liquid move along a surface, it is commonly assumed that the molecular motion is purely translational along that surface. On the other hand, recent studies of molecular mono-layers and molecular devices indicate a strong possibility of rolling motion as well. In this talk, we consider the limiting case in which the rolling motion dominates the translational motion. The molecular monolayer is represented as an interacting system of rolling balls, with off-center center of mass and interacting through the charges and dipoles that are positioned in the center of mass. Two ranges of parameters for the system are studied: pure Lennard-Jones interaction that prevents particles from colliding, and dipole+Lennard-Jones, modeling water monolayers. We consider stationary states of the system and their stability. Next, we analyze (numerically) long-term motion of an ensemble of these molecules and show that classical notions like temperature can be applied to a system with even a small number of balls.

Patrick LaVictoire ,   U.C.Berkeley
PhD student, advised by Michael Christ
Singular Integral Theory and L^1 Ergodic Theorems
The harmonic analysis approach to proving nonstandard ergodic theorems has been remarkably successful. By transferring maximal and oscillation inequalities from dynamical systems to the integers, one finds many such problems transformed into questions of singular integral estimates. In this talk, I will sketch several instances of this approach, focusing on a technique of Christ and Fefferman via which several nonstandard ergodic theorems of interest have recently been extended from L^2 to L^1.

Douglas Lind ,   University of Washington and Yale University
Specification for Algebraic Actions
Specification is an important orbit tracing property of group actions. For expansive actions by automorphisms of compact abelian groups I'll describe the role homoclinic points play in establishing specification, and discuss recent joint work with Schmidt and Verbitskiy extending these ideas to certain nonexpansive higher rank actions.

Elon Lindenstrauss ,   Hebrew University and Princeton University
Classifying invariant measures on homogeneous spaces
A well known result of Rudolph states that Lebesgue measure is the unique x2, x3 invariant and ergodic measure on R/Z with positive entropy. Extending this theorem to diagonalizable flows on homogeneous spaces turned out to require rather different tools, and I will explain one such tool- the low entropy method - and how it relates to another important paper of Rudolph on the ergodic theoretic properties of the Patterson Sullivan measure on geometrically finite hyperbolic surfaces.

Alejandro Maass ,   University of Chile, Santiago
Nilsequences and a structure theorem for topological dynamical systems
In this talk, a characterization of inverse limits of nilsystems in topological dynamics via a structure theorem for topological dynamical systems is given. This is an analog of the structure theorem for measure preserving systems constructed by B. Host and B. Kra. Two applications of the structure are discussed. The firrst is to nilsequences: we give a characterization that detects if a given sequence is a nilsequence by only testing properties on finite intervals. The second application is the construction of the maximal nilfactor of any order in a distal minimal topological dynamical system. We show that this factor can be defined via a certain generalization of the regionally proximal relation that is used to produce the maximal equicontinuous factor and corresponds to the case of order 1. (This is a joint work with Bernard Host and Bryna Kra.)

Don Ornstein ,   Stanford University
(Retrospective on some of Dan Rudolph's work)
The speaker will discuss some of Dan Rudolph's early work including his work on the Bernoulli isomorphism theory and some of his ingenious counter exambles.

Ayse Sahin ,   DePaul University
The geometry and directional entropy of rank one Z^d actions
We study the dynamical properties of rank one Z^d actions as a function of the geometry of the shapes of the towers generating the action. The main results relate the geometry to directional entropy and the orbit equivalence classification of the action. This is joint work with E. A. Robinson.

Klaus Schmidt ,   University of Vienna and Erwin Schrodinger Institute
Principal Algebraic Actions
Let G be a countable discrete group. Every element f in the integral group ring ZG of G acts by right convolution rho_f on the compact group T^G of all maps from G to T=R/Z. The corresponding 'principal algebraic action' alpha_f of G is the (left) shift action of G on the subgroup X_f of T^G which is the kernel of rho_f. This talk is about elementary dynamical properties of this class of actions, and is largely based on joint work with Christopher Deninger and/or Doug Lind.

Bethany Springer ,   Colorado State University
PhD student, advised by Daniel Rudolph
New results in nearly continuous Kakutani equivalence
It was previously known that a conjugacy between induced systems on Polish probability spaces (X,T,\mu) and (Y,S, \nu) on nearly clopen subsets $A \subset X$ and $B \subset Y$ of same relative measure can be extended to a measurable orbit equivalence between the systems. This talk gives an algorithm for modifying the conjugacy in order to establish nearly continuous Kakutani equivalence. We also show that in the case $\mu(A) > \nu(B)$, we can establish conjugacy between $T$ and an induced system $S_{\bar B}$ where the systems are nearly uniquely ergodic.

Jean-Paul Thouvenot ,   Universite' de Paris VI, France
A criterion for a transformation to satisfy the weak Pinsker Property
The proofs of the deep structure theorems of Dan Rudolph concerning the finite (and isometric) extensions of Bernoulli shifts relied on the identification of \bar d separated distributions which had then to be extremal. We shall describe a criterion for a process which allows to produce separated distributions in a more general situation than the compact case, equivalent to the weak Pinsker property.

Raul Ures ,   University of the Republic, Montevideo, Uruguay
Entropy maximizing measures for some partially hyperbolic systems


text:
Courses

Test: Maxwell Law - 2

20 Questions MCQ Test Electromagnetic Theory | Test: Maxwell Law - 2

Description
This mock test of Test: Maxwell Law - 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Maxwell Law - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Maxwell Law - 2 quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Maxwell Law - 2 exercise for a better result in the exam. You can find other Test: Maxwell Law - 2 extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

The charge density of a electrostatic field is given by

Solution:

Explanation: From the Gauss law for electric field, the volume charge density is the divergence of the electric flux density of the field. Thus Div(D) = ρv.

QUESTION: 2

In the medium of free space, the divergence of the electric flux density will be

Solution:

Explanation: In free space or air, the charge density will be zero. In other words, the conduction is possible in mere air medium. By gauss law, since the charge density is same as the divergence of D, the Div(D) in air/free space will be zero.

QUESTION: 3

In a medium other than air, the electric flux density will be

Solution:

Explanation: In any medium other than the air, the conduction is possible, due to the charge carriers. Thus charge density is also non-zero. We can write from Gauss law that Div(D) is non-zero. When the divergence is said to be non-zero, the field is not solenoidal or called as divergent field.

QUESTION: 4

For a solenoidal field, the surface integral of D will be,

Solution:

Explanation: For a solenoidal field, the divergence will be zero. By divergence theorem, the surface integral of D and the volume integral of Div(D) is same. So as the Div(D) is zero for a solenoidal field, the surface integral of D is also zero.

QUESTION: 5

In a dipole, the Gauss theorem value will be

Solution:

Explanation: The Gauss theorem for an electric field is given by Div(D)= ρ. In a dipole only static charge exists and the divergence will be zero. Thus the Gauss theorem value for the dipole will be zero.

QUESTION: 6

Find the electric flux density of a material whose charge density is given by 12 units in a volume region of 0.5 units.

Solution:

Explanation: By Gauss law, Div(D) = ρv. To get D, integrate the charge density given. Thus D = ∫ρv dv, where ρv = 12 and ∫dv = 0.5. We get, D = 12 x 0.5 = 6 units.

QUESTION: 7

From the Gauss law for electric field, we can compute which of the following parameters?

Solution:

Explanation: From the Gauss law for electric field, we can find the electric flux density directly. On substituting, D= ε E, the electric field intensity can be calculated.

QUESTION: 8

The charge density of a system with the position vector as electric flux density is

Solution:

Explanation: The divergence of the electric flux density is the charge density. For a position vector xi + yj + zk, the divergence will be 1 + 1 + 1 = 3. Thus by Gauss law, the charge density is also 3.

QUESTION: 9

The sequence for finding E when charge density is given is

Solution:

Explanation: From the given charge density ρv, we can compute the electric flux density by Gauss law. Since, D = εE, the electric field intensity can also be computed. Thus the sequence is E-D-ρv.

QUESTION: 10

The Gauss law employs which theorem for the calculation of charge density?

Solution:

Explanation: The Gauss divergence theorem is given by ∫ D.ds = ∫Div(D).dv. From the theorem value, we can compute the charge density. Thus Gauss law employs the Gauss divergence theorem.

QUESTION: 11

Which quantity is solenoidal in the electromagnetic theory?

Solution:

Explanation: The divergence of the magnetic flux density is zero. This is the Maxwell fourth equation. As the divergence is zero, the quantity will be solenoidal or divergent less.

QUESTION: 12

Which equation will be true, if the medium is considered to be air?

Solution:

Explanation: From the Gauss law for magnetic field, the divergence of the magnetic flux density is zero. Also B = μH. Thus divergence of H is also zero, i.e, Div(H) = 0 is true.

QUESTION: 13

Find the sequence to find B when E is given.

Solution:

Explanation: From E, D can be computed as D = εE. Using the Ampere law, H can be computed from D. Finally, B can be calculated from H by B = μH.

QUESTION: 14

The Gauss law for magnetic field is valid in

Solution:

Explanation: The Gauss law for magnetic field states that the divergence of B is always zero. This is valid for all cases like free space, dielectric medium etc.

QUESTION: 15

The sequence for finding H from E is

Solution:

Explanation: From E, we can compute B using the Maxwell first law. Using B, the parameter H can be found since B = μH. Thus the sequence is E-B-H is true.

QUESTION: 16

The reason for non existence of magnetic monopoles is

Solution:

Explanation: Practically monopoles do not exist, due to the connection between north and south poles. But theoretically, they exist. The reason for their non- existence practically is that, the magnetic field confined to two poles cannot be split or confined to a single pole.

QUESTION: 17

The non existence of the magnetic monopole is due to which operation?

Solution:

Explanation: The Maxwell fourth law or the Gauss law for magnetic field states that the divergence of B is zero, implies the non existence of magnetic monopoles. Thus the operation involved is divergence.

QUESTION: 18

Will dielectric breakdown lead to formation of magnetic monopole?

Solution:

Explanation: When dielectric breakdown occurs, the material loses its dielectric property and becomes a conductor. When it is subjected to a magnetic field, north and south flux lines coexists, giving magnetic force. Thus there exists magnetic dipole. Suppose if the conductor is broken into very small pieces, still there exist a magnetic dipole in every broken part. In other words, when a piece is broken into half, there cannot exist a north pole in one half and a south pole in the other. Thus monopoles never exist.

QUESTION: 19

Which equation will hold good for a magnetic material?

Solution:


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Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. 1 1. There are exactly six simple connected graphs with only four vertices. 2 2 2 2 <- step 5, subtract 1 from the left 3 degrees. Now we have a cycle, which is a simple graph, so we can stop and say 3 3 3 3 2 is a simple graph. If the degree of each vertex in the graph is two, then it is called a Cycle Graph. There are 4 non-isomorphic graphs possible with 3 vertices. a) deg (b). It is tough to find out if a given edge is incoming or outgoing edge. Sufficient Condition . we have a graph with two vertices (so one edge) degree=(n-1). We can create this graph as follows. so every connected graph should have more than C(n-1,2) edges. 7) A connected planar graph having 6 vertices, 7 edges contains _____ regions. (b) Draw all non-isomorphic simple graphs with four vertices. Show transcribed image text. Since n(n −1) must be divisible by 4, n must be congruent to 0 or 1 mod 4; for instance, a 6-vertex graph … (d) None Of The Other Options Are True. Example graph. Sum of degree of all vertices = 2 x Number of edges . eg. Question: Suppose A Simple Connected Graph Has Vertices Whose Degrees Are Given In The Following Table: Vertex Degree 0 5 1 4 2 3 3 1 4 1 5 1 6 1 7 1 8 1 9 1 What Can Be Said About The Graph? Calculation: Two graphs are G and G’ (with vertices V ( G ) and V (G ′) respectively and edges E ( G ) and E (G ′) respectively) are isomorphic if there exists one-to-one correspondence such that [u, v] is an edge in G ⇔ [g (u), g (v)] is an edge of G ′.We are interested in all nonisomorphic simple graphs with 3 vertices. The graph can be either directed or undirected. How many simple non-isomorphic graphs are possible with 3 vertices? A simple graph with 6 vertices, whose degrees are 2, 2, 2, 3, 4, 4. Given information: simple graphs with three vertices. Let X - Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.a)4b)8c)16d)32Correct answer is option 'C'. An n-vertex self-complementary graph has exactly half number of edges of the complete graph, i.e., n(n − 1)/4 edges, and (if there is more than one vertex) it must have diameter either 2 or 3. 1 Connected simple graphs on four vertices Here we brie°y answer Exercise 3.3 of the previous notes. actually it does not exit.because according to handshaking theorem twice the edges is the degree.but five vertices of degree 3 which is equal to 3+3+3+3+3=15.it should be an even number and 15 is not an even number and also the number of odd degree vertices in an undirected graph must be an even count. Find the in-degree and out-degree of each vertex for the given directed multigraph. Figure 1: An exhaustive and irredundant list. 0 0 <- everything is a 0 after going through the full Havel-Hakimi algo, so yes, 3 3 3 3 2 is a simple graph. Each of these provides methods for adding and removing vertices and edges, for retrieving edges, and for accessing collections of its vertices and edges. Fig 1. In Graph 7 vertices P, R and S, Q have multiple edges. Then G contains at least one vertex of degree 5 or less. Answer to Draw the following: a. K3 b. a 2-regular simple graph c. simple graph with = 5 & = 3 d. simple disconnected graph with 6 vertices e. graph that is Active 2 years ago. a) a graph with five vertices each with a degree of 3 b) a graph with four vertices having degrees 1,2,2,3 c) a graph with a three vertices having degrees 2,5,5 d) a SIMPLE graph with five vertices having degrees 1,2,3,3,5 e. A 4-regualr graph with four vertices Suppose a simple graph has 15 edges, 3 vertices of degree 4, and all others of degree 3. Let us start by plotting an example graph as shown in Figure 1.. # Create a directed graph g = Graph(directed=True) # Add 5 vertices g.add_vertices(5). Simple Graphs :A graph which has no loops or multiple edges is called a simple graph. Ask Question Asked 2 years ago. a) 15 b) 3 c) 1 d) 11 Answer: b Explanation: By euler’s formula the relation between vertices(n), edges(q) and regions(r) is given by n-q+r=2. Or keep going: 2 2 2. The search for necessary or sufficient conditions is a major area of study in graph theory today. (b) This Graph Cannot Exist. 22. Which of the following statements for a simple graph is correct? If you are considering non directed graph then maximum number of edges is $\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}$. Graphs; Discrete Math: In a simple graph, every pair of vertices can belong to at most one edge and from this, we can estimate the maximum number of edges for a simple graph with {eq}n {/eq} vertices. There is a closed-form numerical solution you can use. Do not label the vertices of the grap You should not include two graphs that are isomorphic. We know that the sum of the degree in a simple graph always even ie, $\sum d(v)=2E$ Question 96490: Draw the graph described or else explain why there is no such graph. Therefore the degree of each vertex will be one less than the total number of vertices (at most). Since K 3,3 has 6 vertices and 9 edges and no triangles, it follows from Corollary 2 that 9 ≤ (2×6) - 4 = 8. a) Every path is a trail b) Every trail is a path c) Every trail is a path as well as every path is a trail ... 14. Viewed 993 times 0 $\begingroup$ I'm taking a class in Discrete Mathematics, and one of the problems in my homework asks for a Simple Graph with 5 vertices of degrees 2, 3, 3, 3, and 5. In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. (a) Draw all non-isomorphic simple graphs with three vertices. 3 vertices - Graphs are ordered by increasing number of edges in the left column. E.1) Vertex Set and Counting / 4 points What is the cardinality of the vertex set V of the graph? 3 = 21, which is not even. Thus, Total number of vertices in the graph = 18. How can I have more than 4 edges? deg (b) b) deg (d) _deg (d) c) Verify the handshaking theorem of the directed graph. Your task is to calculate the number of simple paths of length at least $$1$$$in the given graph. The list contains all 4 graphs with 3 vertices. We have that is a simple graph, no parallel or loop exist. All graphs in simple graphs are weighted and (of course) simple. 2n = 42 – 6. Substituting the values, we get-3 x 4 + (n-3) x 2 = 2 x 21. 8 vertices (3 graphs) 9 vertices (3 graphs) 10 vertices (13 graphs) 11 vertices (21 graphs) 12 vertices (110 graphs) 13 vertices (474 graphs) 14 vertices (2545 graphs) 15 vertices (18696 graphs) Edge-4-critical graphs. ie, degree=n-1. Note that paths that differ only by their direction are considered the same (i. e. you have to calculate the number of undirected paths). Corollary 3 Let G be a connected planar simple graph. A simple graph has no parallel edges nor any Theorem 1.1. We’ll start with directed graphs, and then move to show some special cases that are related to undirected graphs. For example, paths $$[1, 2, 3]$$$ and \$[3… Assume that there exists such simple graph. Notation − C n. Example. (b) A simple graph with five vertices with degrees 2, 3, 3, 3, and 5. It has two types of graph data structures representing undirected and directed graphs. Let G be a connected planar simple graph with 20 vertices and degree of each vertex is 3. Denote by y and z the remaining two vertices… 8)What is the maximum number of edges in a bipartite graph having 10 vertices? This is a directed graph that contains 5 vertices. (c) 4 4 3 2 1. This contradiction shows that K 3,3 is non-planar. A graph is a set of points, called nodes or vertices, which are interconnected by a set of lines called edges.The study of graphs, or graph theory is an important part of a number of disciplines in the fields of mathematics, engineering and computer science.. Graph Theory. Use contradiction to prove. Problem Statement. (n-1)=(2-1)=1. It is impossible to draw this graph. 4 3 2 1 Graph G has n nodes n=(n-1)+1 A graph to be disconnected there should be at least one isolated vertex.A graph with one isolated vertex has maximum of C(n-1,2) edges. Let GV, E be a simple graph where the vertex set V consists of all the 2-element subsets of {1,2,3,4,5). 2n = 36 ∴ n = 18 . Definition − A graph (denoted as G = (V, E)) consists of a non-empty set of vertices or nodes V and a set of edges E. Please come to o–ce hours if you have any questions about this proof. Dirac's Theorem Let G be a simple graph with n vertices where n ≥ 3 If deg(v) ≥ 1/2 n for each vertex v, then G is Hamiltonian. This question hasn't been answered yet Ask an expert. Simple Graph with 5 vertices of degrees 2, 3, 3, 3, 5. Directed Graphs : In all the above graphs there are edges and vertices. Let x be any vertex of such 3-regular graph and a, b, c be its three neighbors. 23. Graph 1, Graph 2, Graph 3, Graph 4 and Graph 5 are simple graphs. Since through the Handshaking Theorem we have the theorem that An undirected graph G =(V,E) has an even number of vertices of odd degree. Jan 08,2021 - Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. A simple graph with 'n' vertices (n >= 3) and 'n' edges is called a cycle graph if all its edges form a cycle of length 'n'. Find the number of regions in G. Solution- Given-Number of vertices (v) = 20; Degree of each vertex (d) = 3 . Given two integers N and M, the task is to count the number of simple undirected graphs that can be drawn with N vertices and M edges.A simple graph is a graph that does not contain multiple edges and self loops. The Number Of Non-isomorphic Simple Graphs With 3 Vertices Is Select One: O A.3 O B.6 O 0.4 O D.5; Question: The Number Of Non-isomorphic Simple Graphs With 3 Vertices Is Select One: O A.3 O B.6 O 0.4 O D.5. O(C) Depth First Search Would Produce No Back Edges. How many vertices does the graph have? WUCT121 Graphs: Tutorial Exercise Solutions 3 Question2 Either draw a graph with the following specified properties, or explain why no such graph exists: (a) A graph with four vertices having the degrees of its vertices 1, 2, 3 and 4. Solution. The vertices will be labelled from 0 to 4 and the 7 weighted edges (0,2), (0,1), (0,3), (1,2), (1,3), (2,4) and (3,4). There is an edge between two vertices if the corresponding 2-element subsets are disjoint. 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# Bent's rule for Sulfur trioxide and Sulfur dioxide

How is Bent's rule applied on $\ce{SO3}$ and $\ce{SO2}$?

Bond length of $\ce{SO3}$ < $\ce{SO2}$. Is this a direct consequence of applying the Bent's rule?

• I really don't see why you would need to apply Bent's rule. – M.A.R. Mar 19 '15 at 20:40

Here are structures of dominant resonance structure (only one of 3 and 2 differencing only by rotation) for sulfur tri- and dioxide. Despite common misconception, d-orbitals apaprantely play very insignificant role in chemistry of p-block elements, so the structures include formal charges.

As can be seen, the average order of sulfur-oxigen bond lowers with move from di- to trioxide from $1.5$ to $1.(3)$. However, average formal charges grows from $+1$ to $+2$ on sulfur and $-0.5$ to $-0.(6)$, on oxygen, thus greatly increasing ionic bonding (roughly from $1*0.5 = 0.5$ to $2*0.(6)=1.(3)$) Generally, ionic bonding greatly outperforms covalent bonding, and here charges grown considerably so bond shortening may be attributed to that. The point, however, is, that in this particular structure more than one effect is in play, and their effects outperform whatever small effects are considered when Bent's rule is applied.

• But how would you apply Bent's rule here? – Shaurya Gupta Sep 20 '14 at 6:40
• @shauryagupta I wouldn't. There are more significant factors in effect here. – permeakra Sep 20 '14 at 12:16

There are several factors which play a role in determining the bond length in a molecule.

Here, in $\ce{SO3}$, we have no lone pairs on the sulfur atom whereas, in $\ce{SO2}$, the central atom has a pair of lone electrons.

According to Bent's rule, the lone pairs tend to occupy the orbital with more s-character which in turn decreases the s-character in the $\ce{S-O}$ bonds and increases the p-character. Also, since there's an increment in p-character, there's an increment in bond length and thus S-O bond length in $\ce{SO2}$ > $\ce{SO3}$.

I hope this might get you on clear grounds of how Bent's rule can be applied in this case.


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# Subsets of $\omega$-regular lanuages accepted by automata with special acceptance condition

Let $\mathcal A = (X, Q, \delta, q_0, F)$ be a deterministic finite automata with the following acceptance condition on infinite words:

The automata accepts $\xi \in X^{\omega}$ with respect to $F$ iff $$\forall i : \delta(q_0, \xi[0...i]) \in F.$$ Meaning that every prefix $\xi[0...i]$ of $\xi$ goes to an acceptance state. With $L'(\mathcal A)$ I denote the set of all accepted infinite words.

Now such automata have a special form, a word is accepted iff it just passes through certain allowed states in $F$, otherwise it is rejected if it ever enters a state from $Q\setminus F$. This by the way means that if $\mathcal A$ is reduced and complete, it has exactly one non-accepting trapping state $s$, because every word which enters $s$ needs to ultimately stay there, because the word gets never accepted, regardless of what comes there after $s$ was entered.

Now I define $F_n(\xi) := \{ w \in X^* : w \in \mbox{infix}(\xi) \cap X^n \}$, the set of all factors (or infixes) of $\xi$ of length $n$. Then $$K_n(\xi) := \xi[0...n] \cdot X^{\omega} \cap \{ \eta \in X^{\omega} : F_n(\eta) = F_n(\xi) \}$$ is the set of all infinite words which share with $\xi$ a common prefix of length $n$ and which have the same set of factors of length $n$.

Now I conjecture: If $\xi$ is accepted by $\mathcal A$ according to the above mentioned acceptance condition, then there exists a $n > 0$ such that $$K_n(\xi) \subseteq L'(\mathcal A)$$ meaning that if $\xi$ is accepted there exists a $n > 0$ such that every word which share with $\xi$ a common prefix of length $n$ and all factors of length $n$ (and up to $n$) is also accepted.

Intuitively I guess this is right, because if $\xi$ is accepted, and because $\mathcal A$ is a finite automata, then as $\xi$ goes through the states it eventually ends in some cycle, which I guesss bound the possible factors, and the prefix condition could be used to ensure that another word would end in the same cycle.

If $\xi = uv^{\omega}$ for finite $u, v$, i.e. if $\xi$ is ultimately periodic I guess a construction would be to determine the smallest $k$ such that $F_k(\xi) = F_{k+1}(\xi)$, so the number of factors of length $k' > k$ is the same as $|F_k(\xi)|$, and set $n := k + 1$. Then I conjecture $$K_{n}(\xi) \subseteq L'(\mathcal A).$$ (I have no proof) Maybe this works also for non-ultimately periodic words, but I could not proof it. So any ideas? Or maybe an idea how to proof my conjecture?

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something is wrong with the definition of $K_n(\xi)$. –  Victor Oct 28 '13 at 13:05
ok, i think i see what is $K_n(\xi)$, but you should delete the formula describing it -- it's confusing –  Victor Oct 28 '13 at 13:17
yes, there was an error in the definition, now I fixed it, thank you. Still confusing? (then I would delete the formulae for $K_n(\xi)$) –  Stefan Oct 28 '13 at 13:20
+1. This is a very nice question. –  Joel David Hamkins Oct 28 '13 at 15:10

For a counterexample, consider the language consisting of all infinite binary strings $\xi$, such that every infix maximal finite block of $0$s has even length. This language can be accepted by a machine according to your criterion, since we can design an automata that when reading a $0$ toggles between two (accepting) states to check the parity, and if you have a $1$ from the odd state, you go to a rejecting state, but a $1$ from the even state returns to the even state. Any infinite string with an odd block will reach that rejecting state, but strings with only even blocks will always be in the acceptable states.
Now, let $\xi$ be a string with arbitrarily long even-length blocks of $0$s, separated by individual $1$s. For example, $\xi=11001000010000001\cdots$. I claim that there is no $n$ for which your desired property holds. Fix any $n$, and let $\eta$ be obtained from $\xi$ by inserting an extra $0$ into one of the very long blocks of $0$s in $\xi$, well beyond $n$. Since the affected block was very long, it follows that $\eta$ and $\xi$ have exactly the same infix strings of length $n$, and they agree on their first $n$ characters. But $\xi$ was accepted and $\eta$ will be rejected.
In the case of eventually periodic input, however, I claim that the conjecture is correct. Suppose the input $\xi=uv^\omega$ is eventually periodic. Let $n$ be extremely large in comparison with the lengths of $u$ and $v$. Now, I claim that any string $\eta$ that starts with $u$ and has the same length $n$ infix strings as $\xi$, must in fact be equal to $\xi$. The reason is that $\eta$ must begin with something like $uv^k$, and if it subsequently deviates from $\xi$, then this deviation will be at the end of a length $n$ string that does not arise in $\xi$. (Please correct me if this is mistaken.) By using this $n$, we attain your desired criterion trivially, since the only string $\eta$ involved is $\xi$ itself, which is in the language.


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1. ## probability

hey all can anyone help me with these problems?

1)a die is tossed 150 times, the rolling of a 6 is counted as a success:

a) find the mean and standard deviation of this binominal experiment and

b) what is the probability of throwing between 30 and 35 sixes?

2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

a)between 90 and 100kg?
b)between 70 and 100kg?
c)is over 101kg?
d)weighs less than 65kgs?
e)how many in the group would weigh less than 65kgs?

any help will be appreciated.....thanks

2. You should consider asking only one question in a single post
it makes it easier to tell what has been answered, as it is
not always possible to answer multiple questions in one go

Originally Posted by amber79
hey all can anyone help me with these problems?

1)a die is tossed 150 times, the rolling of a 6 is counted as a success:

a) find the mean and standard deviation of this binominal experiment and

b) what is the probability of throwing between 30 and 35 sixes?
Part a)
Assume a fair dice, the the probability $p$ of success is $1/6$.

In $N$ trials the mean number of successes is:

$m=p.N=150/6=25$.

In a binomial experiment with probability of success $p$ and $N$
trials the standard deviation of the number of success is:

$\sigma=\sqrt{N.p.(1-p)}=\sqrt{150.\frac{1}{6} .\frac{5}{6}}\approx 4.56$

Part b)
If we use the normal approximation we want to find the probability of a
normally distributed random variable $X$ with mean $25$ and
standard deviation $\approx 4.56$ taking values between $29.5$ and $35.5$
(we extend the range from $(30, 35)$ to $(29.5,35.5)$ as a continuity
correction) or:

$prob(29.5 $P((35.5-25)/4.56)-P((29.5-25)/4.56)$,

where $P$ denotes the cumulative distribution function for the
standard normal distribution.

$prob(29.5 $P(2.30)-P(0.99)\approx 0.9893-0.8389=0.1504 \approx 15 \%$.

RonL

3. If I make a mistake in how to slove the problem do not blame me I am not familiar with Probabilty and Statistics but I think I might be able to help with some of them.

Part 2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

a)between 90 and 100kg?
b)between 70 and 100kg?
c)is over 101kg?
d)weighs less than 65kgs?
e)how many in the group would weigh less than 65kgs?

I do not know a or b I could approximate what the answer is but I do not know if that is what the problem is asking for.

d)There is a rule in statistics the first deviation Gives 68% two Deviations Give 95% and three give 99.7%. Since 74 is the mean one standard deviation over is 74-9 and 74+9 which is 65 and 83. Thus, 68% of the people are between 65 and 83. Thus are are 32% left. Which means there are 32% of people below 65 or more the than 83. But the Standard Deviation curve is symetrically, thus 16% are below 65 and 16% are above 83. But the problem is asking for below 65 thus the answer is 16%.
e) Since 16% of the people are below 65 and there are 200 we have 16% of 200 = 32 people.
c)Same reasoning as in part c this is why I did it out of order. Between 65 and 83 you have 68% of all the people. Move another 9 units (because that is the standard deviation) to get 65-9 and 83+9 thus 56 and 92 which means between 56 and 92 there are 95% of the people. Move the standard deviation for the third time thus 56-9 and 92+9 which is 47 and 101 which means between 47 amd 101 there are 99.7% of all the people. Thus .3% remains which means .3% of all the people are below 47 or above 101. But the standard deviation curve is symetrically thus .15% are above 101 and .15% are below 101. The problem is asking for above 101 thus the answer is .15%.

4. Originally Posted by amber79
2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

a)between 90 and 100kg?
b)between 70 and 100kg?
c)is over 101kg?
d)weighs less than 65kgs?
e)how many in the group would weigh less than 65kgs?
Here you are required to assume that weight is normally distributed, and
as 200 is a reasonably large number in this context we may use 74kg as the
mean and 9kg as the standard deviation of the weight distribution.

Then if $P$ is the cumulative standard normal distribution:

$prob(weight

$prob(weight>x)=1-P\left ( \frac{x-mean}{standard\ deviation}\right)$

Also:

$prob(y $-\ P\left ( \frac{y-mean}{standard\ deviation}\right)$

$P$ is what is tabulated in a normal probability distribution table.

With the above you should be able to answer a-e by plugging in the relevant
values given in the problem.

RonL


text:
# Homework Help: Linear differential equation hwk check

1. Sep 20, 2016

### Ashley1nOnly

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

2. Relevant equations

Posted
3. The attempt at a solution
Posted

I just need the work to be checked.

2. Sep 20, 2016

### Ashley1nOnly

I integrated to get the last part.

3. Sep 20, 2016

### LCKurtz

2. Your answer should be an equation with $y$ and $t$.


text:
breja 2 years ago P = 2l + 2w

1. jazy

Yes, the perimeter of a rectangle is (2 times the length) multiplied by (2 times the width). What is it you need?

2. breja

w=P-I OR W=P/2-I OR W=P/2+I

3. jazy

Okay: p = 2l + 2w Subtract 2l from both sides. p - 2l = 2w Divide both sides by 2. 1/2p - l = w

4. jazy

$w = \frac{ 1 }{ 2 }p - l$


text:
# zbMATH — the first resource for mathematics

Prediction of the maximum size in Wicksell’s corpuscle problem. (English) Zbl 0986.62075
Summary: In the Wicksell corpuscle problem [S. D. Wicksell, Biometrika 17, 84–99 (1925; JFM 51.0392.02)] the maximum size of random spheres in a volume part is to be predicted from the sectional circular distribution of spheres cut by a plane. The size of the spheres is assumed to follow the generalized gamma distribution. Some prediction methods according to measurement methods on the sectional plane are proposed, and their performances are evaluated by simulation. The prediction method based on the $$r$$ largest sizes and the total number of the sectional circles is recommended, because of its satisfactory performance.

##### MSC:
 62M30 Inference from spatial processes 60D05 Geometric probability and stochastic geometry 62M99 Inference from stochastic processes
Full Text:


text:
login  home  contents  what's new  discussion  bug reports help  links  subscribe  changes  refresh  edit

 Submitted by : (unknown) at: 2007-11-17T23:05:32-08:00 (15 years ago) Name : Axiom Version : default friCAS-20090114 Axiom-20050901 OpenAxiom-20091012 OpenAxiom-20110220 OpenAxiom-Release-141 Category : Axiom Aldor Interface Axiom Compiler Axiom Library Axiom Interpreter Axiom Documentation Axiom User Interface building Axiom from source lisp system MathAction Doyen CD Reduce Axiom on Windows Axiom on Linux Severity : critical serious normal minor wishlist Status : open closed rejected not reproducible fix proposed fixed somewhere duplicate need more info Optional subject :   Optional comment :

The axiom command:

axiom
x^1 abc[y]
 (1)
Type: Fraction(Polynomial(Integer))

does not produce any syntax error while

axiom
x abc[y]
There are no library operations named x
Use HyperDoc Browse or issue
)what op x
to learn if there is any operation containing " x " in its name.
Cannot find a definition or applicable library operation named x
with argument type(s)
Symbol
Perhaps you should use "@" to indicate the required return type,
or "$" to specify which version of the function you need. does produce an error message. It seems that 1 abc[y] is parsed as something unusual axiom 1 abc[y]  (2) Type: UnivariatePolynomial(*01abc(y),Integer) What is Type: UnivariatePolynomial(*01abc y,Integer)? To explain this, you have to understand several syntax conventions in Axiom. When two expressions are separated by a space, it means "apply" as in f x meaning f(x). Exactly what "apply" does depends on f and x. When f is a symbol and x is a list, this application produces a subscripted symbol. Also "apply" has higher precedence than any other operation. Finally, a subscripted symbol cd when displayed on the Type info line is written as *01c d. So abc[y] is the same as abc [y] (because clearly the left '[' terminates the symbol abc) and so this is a subscripted symbol abcy. axiom abc [y]  (3) Type: Symbol axiom x [1,2,3]  (4) Type: Symbol Now if b is a symbol, then 1 b is 1 applied to b. Axiom tries to find what operation this application is, and finds one in UP(b,INT) where the application means evaluate the element 1::UP(b,INT) as a function at the value b (substitution), which gives 1. axiom 1 (b::Symbol)  (5) Type: UnivariatePolynomial(b,Integer) This is the same as axiom elt(1::UP(b,INT),b::UP(b,INT))$UP(b,INT)
 (6)
Type: UnivariatePolynomial(b,Integer)

The results reported above are therefore all obtained by the these rules. The input 'x abc[y]' produces an error because this apply can only take place in UP (the fact that UP has only one variable makes the substitution unambiguous) and although it actually makes sense under suitable domains, the interpreter is not smart enough to find it.

axiom
(x::UP(x, POLY INT)) (b::UP(x, POLY INT))
 (7)
Type: UnivariatePolynomial(x,Polynomial(Integer))

and actually, the construction UP(x, POLY INT) is not kosher, but somehow the extra parentheses helped.

axiom
x::UP(x, POLY INT) b::UP(x, POLY INT)
There are no library operations named Type
Use HyperDoc Browse or issue
)what op Type
to learn if there is any operation containing " Type " in its
name.
Cannot find application of object of type Type to argument(s) of
type(s)
Variable(b)

Not a syntax error
Thu, 17 Feb 2005 06:09:33 -0600 reply
Status: open => closed

First, I find it rather annoying that the Status was changed by an anonymous user to closed without further discussion. Maybe I would agree with a status of pending but I think it would at least be courteous to identify oneself by clicking preferences before changing a status. That way at least I know who to blame. ;) Maybe we should make it so that a user id must be specified in order to click the Change button?

Second, it is clear from the original description that it is "not a syntax error". The bug report said: "should be a syntax error". So maybe you mean "should not be a syntax error"?

Third most of your "explanation" is quite easy to find in the Axiom book (see page 818)

"Univariate polynomials can also be used as if they were functions."

but the fact that abc[y]? when appearing in a Type is displayed as *01abc y is not described anywhere. This notation is especially odd since it is not a valid input form. So perhaps you might have proposed to change the category of this issue to Axiom Documentation and leave the status Open?

But I am not at all convinced that Axiom should behave in this strange manner. The idea that:

  1 x[k]

is the unit of the domain of univariate polynomials over the variable x[k]? is extraordinarily obscure and does not seem to have any obvious application to me. Further the fact that:

  1 x

is not defined (because the type of x is of type *Variable x and not Symbol and for some reason the interpreter does not consider the obvious coercion of x to Symbol while something more complicated like 'x[k]?' is of type Symbol by defautl) is at best very awkward and bordering on the bizarre.

I am not sure whether to blame the interpreter (which is apparently only following a fairly well defined search strategy) or perhaps better to blame the implementation of UnivariatePolynomial for allowing this form of abbreviated composition of polynomials (which I think should be seen as at best idiosyncratic) as something so ubiquitous in Axiom as function application. I think there are very good reasons to distinguish polynomials and expressions in general from functions (i.e. the lambda abstraction).

[#99 x^1 abc[y]? should be a syntax error] annoying --William Sit, Thu, 17 Feb 2005 19:19:21 -0600 reply
Bill Page wrote:

First, I find it rather annoying that the Status was changed by an anonymous user to closed without further discussion. Maybe I would agree with a status of pending but I think it would at least be courteous to identify oneself by clicking preferences before changing a status. That way at least I know who to blame. ;) Maybe we should make it so that a user id must be specified in order to click the Change button?

Sorry, Bill. I was too hasty in closing it. But I don't see why anyone should be "annoyed". If you don't think it is right, just change it back to "open". Isn't that the reason for Wiki?

By the way, I did look all over to find where to login
I advocated that --but putting it under "preference" does not ring a bell (neither would "user options"). Login and setting preferences are two distinct functions that should not be put under one button. Preferences should be set once when one registers, but login should done each time.
Second, it was clear from the original description that it is not a syntax error. The bug report said: should be a syntax error. So maybe you mean should not be a syntax error?

Third most of your "explanation" is quite easy to find in the Axiom book (see page 818)

"Univariate polynomials can also be used as if they were functions."

I was at first very puzzled by how the outputs in the original post are obtained. Perhaps if whoever posted it knew how the outputs came about, that should be included in the post so others do not have to waste time figuring that out. I interpreted the heading (ok wrongly) as some of the expressions have syntax error. I figured that they do not. Rather, it is the inability of the interpreter to coerce some parameters to the "correct" type and find a reasonable signature. But this is nothing new and short of analyzing the interpreter code to make it even smarter, what can we do in the meantime? That is why I thought "case closed".

but the fact that abd[y]? when appearing in a Type is displayed as *01abc y is not described anywhere. This notation is especially odd since it is not a valid input form. So perhaps you might have proposed to change the category of this issue to Axiom Documentation and leave the status Open?

I disagree with your objection to *01abc y in the output message. Sure, it would be better to be able to display that as a subscripted symbol, but this is merely a device to make outputting a subscript on a line-oriented device easier. Output in most cases cannot be fed back as input (that is on the wish list). This is not part of the Axiom language, just as how TeXForm? is coded is not part of Axiom.

On the otherhand, one can use easily

axiom
t:=1 x[y[z]]
 (8)
Type: UnivariatePolynomial(*01x(*01y(z)),Integer)
axiom
dom:=typeOf t
 UnivariatePolynomial}\ } ( 01 x ( 01 y (z)) , \hbox{\axiomType{Integer}\ })" class="equation" src="images/3421699712761778963-16.0px.png" align="bottom" Style="vertical-align:text-bottom" width="352" height="18"/> (9)
Type: Type
axiom
h:dom:=monomial(2,3)
 (10)
Type: UnivariatePolynomial(*01x(*01y(z)),Integer)

In fact, precisely because Axiom wants to preserve this structural information on the variable that it outputs it this way! Had it used a two dimensional method, it would not be able recapture the type information.

But I am not at all convinced that Axiom should behave in this strange manner. The idea that:

  1 x[k]

is the unit of the domain of univariate polynomials over the variable x[k]? is extraordinarily obscure and does not seem to have any obvious application to me. Further the fact that:

  1 x

is not defined (because the type of x is of type *Variable x and not Symbol and for some reason the interpreter does not consider the obvious coercion of x to Symbol while something more complicated like 'x[k]?' is of type Symbol by defautl) is at best very awkward and bordering on the bizarre.

I am not sure whether to blame the interpreter (which is apparently only following a fairly well defined search strategy) or perhaps better to blame the implementation of UnivariatePolynomial for allowing this form of abbreviated composition of polynomials (which I think should be seen as at best idiosyncratic) as something so ubiquitous in Axiom as function application. I think there are very good reasons to distinguish polynomials and expressions in general from functions (i.e. the lambda abstraction).

You must have a pretty bad day? (three "blame" words in one message? :-)

I agree 100% with you that one should distinguish polynomials or expressions from functions. But that is on the mathematics or theory level. In computing, we need to be able to make substitutions (compose if you like under certain cases) into the polynomials or expressions -- even if you use eval, you are still considering the polynomials as functions without any change in domain or representation. It would be another nightmare of coercion if a polynomial has to be coerced into a function before any evaluation. Moreover, if a polynomial can be thought of as a function, then ground elements are constant functions. Axiom is consistent in this interpretation and any one writing 1 x intentionally would understand that to be the constant function 1 evaluated at x and the interpreter is to find the domain in which this makes sense. One way is to treat x as in a univariate polynomial ring (a multivariate polynomial ring would be ok too except that the function that does the evaluation must operate on all cases, and unless you are in a univariate situation, 1 x would only be a special case, and (a+b) x would be ambiguous because x may replace a or replace b).

While it may be unintuitive why anyone would perform an operation like 1 x, we should always bear in mind that this computation may have come about as an intermediate computation (piping), when most of the time it would be a b where a, b are both polynnomials in POLY INT, and in some situation, a becomes 1. You cannot treat 1 x differently from a b, and certainly should not be treated as a syntax error (when suitably placed, as in code intended for compilation) because it is not.

The interpreter succeeds in seemingly more complicated 1 x[y] and not 1 x may perhaps be understood the following way. More complicated expressions provide more information to the interpreter and this guides it: to apply x to [y], the interpreter figures out the correct map with target domain Symbol, which helps it to locate UP(xy,INT). Converting a Variable(x) element to Symbol is "counter-intuitive" to the interpreter, because the domain Variable(x:Symbol) starts with a symbol. Variable(x) is an unexposed domain, and the conversion (either coerce or variable) are unexposed (possibly to avoid the possibility to change it back from Symbol to Variable(x) again). So unless you explicitly make the conversion, the interpreter does not do it.

The Wiki pages are supposed to be for experts as well as for novices. When simple input like this causes some apparent problem that is "unexpected", we should explain it. For some this may sound trivial, but for others, it may not.

forwarded from 99X1AbcYShouldBeASyntaxError#msg20050217150552-0600@page.axiom-developer.org

William

bad day? not really :) --Bill Page, Thu, 17 Feb 2005 22:14:54 -0600 reply
axiom
)clear all
All user variables and function definitions have been cleared.

Today I felt like taking what I think might be the point of view of a new user of Axiom. I agree completely that the Wiki pages are supposed to be for experts as well as for novices and when simple input like this causes some apparent problem that is "unexpected", that we should explain it. I think you did a good job of that. But in my opinion the explanation is rather bizarre (especially from the point of view of a new user). So this is were the experts come in: How can we make Axiom's behaviour less bizarre to new users?

I originally saw this behaviour in an simple test in the SandBox where a new user wrote something like this

axiom
integrate(z^2 ln[z],z)
 (11)
Type: Union(Expression(Integer),...)

I was quite sure that what they intended to compute was

axiom
integrate(z^2*log(z),z)
 (12)
Type: Union(Expression(Integer),...)

but I had a hard time understanding the original answer and I wondered why Axiom did not at least produce some kind of syntax error.

About the status: Closed looks like you have decided that there should be no more discussion. Yes this is a wiki and so I could simply Open it again, but I think it is better to try to develop some "rules of netiquette" so that the open collaboration allowed by the wiki can work smoothly. We are all still fairly new at this.

Perhaps I was too direct in using the word "annoyed" since I am well aware that it is very hard to correctly interpret other peoples' motives and emotional state when reading what they wrote alone. I was annoyed :) but if you had seen me in person you would have noticed that I also smiled when I said that. Of course I used the word "blame" in the same way. I did not mean it to sound like I was "finding fault" with anyone, only that I wanted to identify what might be changed to make Axiom's behaviour more intuitive.

About 'preferences': It seems that "wiki people" generally would like to avoid any perceived "login barrier". Setting preferences is just a matter of saying how you would like to be identified (and a few other details) - it is not supposed to be like "logging in". And you are right that in principle this only needs to be done once. Setting preferences sends some cookies to your browser which, depending on whether you have set your browser security to permit it or not, will allow the system to remember your preferences the next time you connect to the wiki. But maybe you are right that there might be a better label than preferences. How about something like Who Am I?

About the use of *01abc y as a symbol in a type. Maybe this notation starts to make a little more sense if you see an expression like

axiom
1 t[x,y,z]
 (13)
Type: UnivariatePolynomial(*03t(x,y,z),Integer)

but I still don't see why, if there is a need for some linear syntax, there would be any objection to using the same syntax as the input. Surely:

             Type: UnivariatePolynomial(t[x,y,z], Integer)

would be self explanatory, no?

About functions and expressions. There is no problem making substitutions into polynomials like this:

axiom
subst((1+b^2)::UP(b,INT),b=(t^2::UP(t,INT)))::UP(t,INT)
 (14)
Type: UnivariatePolynomial(t,Integer)

or even this

axiom
subst((1+c*b^2)::MPOLY([b,c],INT),b=(t^2::UP(t,INT)))::MPOLY([t,c],INT)
 (15)
Type: MultivariatePolynomial?([t,c],Integer)

We can also explicitly use eval. So I don't understand the need to define application for univariate polynomials and to treat them like functions.

Finally, could you explain further your reasoning about why the interpreter should not coerce Variable x to Symbol? Axiom has no trouble coercing x to a Symbol in the case of an expresion like UP(x,INT). Surely it should be able to solve the problem of coercing x to Symbol in order that Symbol can be coerced to UP(x,INT) in cases like 1 x and integrate(x)?

Or is it the case that the interpreter never tries harder than just one coercion?

BTW, I tried specifying

axiom
)expose Variable
Variable is now explicitly exposed in frame initial

but I did not see any change in Axiom's behaviour.

axiom
1 x
Cannot find a definition or applicable library operation named 1
with argument type(s)
Variable(x)
Perhaps you should use "@" to indicate the required return type,
or "$" to specify which version of the function you need. integrate(x) There are 4 exposed and 2 unexposed library operations named integrate having 1 argument(s) but none was determined to be applicable. Use HyperDoc Browse, or issue )display op integrate to learn more about the available operations. Perhaps package-calling the operation or using coercions on the arguments will allow you to apply the operation. Cannot find a definition or applicable library operation named integrate with argument type(s) Variable(x) Perhaps you should use "@" to indicate the required return type, or "$" to specify which version of the function you need.

--------- Bill Page

[#99 x^1 abc[y]? should be a syntax error] badday? not really :) --William Sit, Mon, 21 Feb 2005 01:30:46 -0600 reply
Bill Page wrote:
Today I felt like taking what I think might be the point of view of a new user of Axiom. I agree completely that the Wiki pages are supposed to be for experts as well as for novices and when simple input like this causes some apparent problem that is "unexpected", that we should explain it. I think you did a good job of that. But in my opinion the explanation is rather bizarre (especially from the point of view of a new user). So this is were the experts come in: How can we make Axiom's behaviour less bizarre to new users?

This is a tall order even though I would agree that such bizarre results may drive many away.

I originally saw this behaviour in an simple test in the SandBox where a new user wrote something like this
axiom
integrate(z^2 ln[z],z)
 (16)
Type: Union(Expression(Integer),...)

A new user has to learn the basic operations and syntax. The above user may be used to Mathematica and Maple or just simply use intuition. I have the same experience myself: I am used to Mathematica and I can't stand the syntax and especially the worksheet interface of Maple (my prejudice). I made so many mistakes and got frustrated every time. But this is not to say Maple is not great for a lot of other people.

I was quite sure that what they intended to compute was
axiom
integrate(z^2*log(z),z)
 (17)
Type: Union(Expression(Integer),...)

but I had a hard time understanding the original answer and I wondered why Axiom did not at least produce some kind of syntax error.

I did not find it easy either. Just because it can be explained does not mean the problem is not there. On the other hand, Axiom was designed for mathematicians and perhaps there is some assumption built in that the users would follow the rules and would realize what their mistakes are. This "apply" notion permeates all levels of Axiom and I believe it would be impossible to modify it without causing a lot of other failures, creating more problems.

About the status: Closed looks like you have decided that there should be no more discussion. Yes this is a wiki and so I could simply Open it again, but I think it is better to try to develop some "rules of netiquette" so that the open collaboration allowed by the wiki can work smoothly. We are all still fairly new at this.

Agreed. I will not close any discussion from now on. One cannot foresee the implications sometimes when the examples illustrating the bugs are already traced to such a low level. I decided to close it because the example x^1 abc[y]? was clearly artificial; had I seen this one on integration, I probably wouldn't have.

Perhaps in addition to a pin-point example, there should also be the other examples on a more complex level to illustrate the implications. For example, I would classify the problem with '1 x[y]?' as not a severe problem, but 'integrate(z^2 ln[z]?, z)' as severe, even though the cause is the same. And the two should be combined into one page, may be with a heading: Beware of Functional Applications.

Perhaps I was too direct in using the word "annoyed" since I am well aware that it is very hard to correctly interpret other peoples' motives and emotional state when reading what they wrote alone. I was annoyed :) but if you had seen me in person you would have noticed that I also smiled when I said that. Of course I used the word "blame" in the same way. I did not mean it to sound like I was "finding fault" with anyone, only that I wanted to identify what might be changed to make Axiom's behaviour more intuitive.

While this is a worthwhile goal, it may be too late in the game because these issues should have been resolved from the moment the system was designed. The interpreter is already quite different from the compiler in its handling of user input. Increasing its user friendliness will only make it even harder to learn to use the compiler, especially if the grammar is changed. Right now, I think all the interpreter does is performing coercions, lifting domains, and matching signatures. I don't believe you would get any of these problems with a compiled package because one is forced by the compiler to know exactly what one is computing. So if the "behavior" of Axiom is to be changed, one has to rewrite the interpreter, making Axiom even more different than the compiler version.

It is probably easier to change user behavior :-). I remember how new students using Mathematica forget to start the trig functions (or any built-in function) with a capital letter and enter xy to mean x times y. I would say that to a new user, when (s)he has not yet got familiar with the nuiances, many answers or error messages would seem bizarre too. But in the end they almost always learn these. I also provided a calculator type pallette for them to just click on these functions to minimize the accidental mistakes. Perhpas that's the area we need to improve Axiom on: user input interface and better tutorials. I find Mathematica's interface the best (I have used Maple, and Matlab as well). It is easy to scroll back to a previous command and edit it, wrap it inside a function for more processing, or append afterthought functions. I think these pages, as we accumulate more and reorganize them systematically, can become a good introductory set of notes (or teasers) for users to hone their Axiom skills.

On the other hand, I think no software system can stand a deliberate scrutiny on all possible input errors (not to mention those that are syntactically correct but with wishful semantics). It is similar to anticipating all student errors in mathematics or their wishful laws on algebraic operations. We can't and instead we teach students the correct ways.

I recommend strongly that when users find "strange behavior" in the interpreter version that they try to analyse it by using )set mess bot on. If that does not help, then use a simple test package and compile it. They will then be forced to be explicit (and may find that the compiler uses a different set of assumptions) and find out what went wrong and thus continue with their computations in the correct way. When it is a bug in computation with the compiler, it becomes a genuine concern.

About 'preferences': It seems that "wiki people" generally would like to avoid any perceived "login barrier". Setting preferences is just a matter of saying how you would like to be identified (and a few other details) - it is not supposed to be like "logging in". And you are right that in principle this only needs to be done once. Setting preferences sends some cookies to your browser which, depending on whether you have set your browser security to permit it or not, will allow the system to remember your preferences the next time you connect to the wiki. But maybe you are right that there might be a better label than preferences. How about something like Who Am I?

That sounds like a better label. May be "Who Are You"? "whoami" is a Unix command for a different purpose. But of course, I know now where to do it. For new comers, that would certainly help.

About the use of *01abc y as a symbol in a type. Maybe this notation starts to make a little more sense if you see an expression like
axiom
1 t[x,y,z]
 (18)
Type: UnivariatePolynomial(*03t(x,y,z),Integer)

but I still don't see why, if there is a need for some linear syntax, there would be any objection to using the same syntax as the input. Surely:

             Type: UnivariatePolynomial(t[x,y,z], Integer)

would be self explanatory, no?

I suppose, I was only guessing at some logic behind this "design". And actually Axiom does use your suggested notation too (indeed, mixed), and so it is again not that consistent.

axiom
integrate(z^2*x[z],z)
 (19)
Type: Polynomial(Fraction(Integer))
axiom
1 x[%]
 (20)
Type: UnivariatePolynomial(*01x((1/3)**01x(z)*z^3),Integer)

Here in the Type message, x(z) is used instead of *01x z. You will see Axiom tries very hard to work out the type of the last one if you had )set mess bot on.

About functions and expressions. There is no problem making substitutions into polynomials like this:

axiom
subst((1+b^2)::UP(b,INT),b=(t^2::UP(t,INT)))::UP(t,INT)
 (21)
Type: UnivariatePolynomial(t,Integer)

or even this

axiom
subst((1+c*b^2)::MPOLY([b,c],INT),b=(t^2::UP(t,INT)))::MPOLY([t,c],INT)
 (22)
Type: MultivariatePolynomial?([t,c],Integer)

We can also explicitly use eval. So I don't understand the need to define application for univariate polynomials and to treat them like functions.

Univariate polynomials are just like finite sequences (streams would be infinite sequences) and all are just functions in the sense the domain is the set of indices (subscripts, exponents or more complicated scripting at times). Axiom is just treating them uniformly. In using subst, you explicitly spell out how the substitution is done. Axiom does not prevent you to use subst. But even in subst, Axiom's behavior is vastly different than, say Mathematica. For example, Axiom would not substitute every occurrence of b.

axiom
)clear all
All user variables and function definitions have been cleared.
subst(1+c[b^2],b=t^2)
 (23)
Type: Expression(Integer)
axiom
subst(1+c(b^2), b=t^2)
There are no library operations named c
Use HyperDoc Browse or issue
)what op c
to learn if there is any operation containing " c " in its name.
Cannot find a definition or applicable library operation named c
with argument type(s)
Polynomial(Integer)
Perhaps you should use "@" to indicate the required return type,
or "$" to specify which version of the function you need. whereas in Mathematica, 1+c[b^2]/.b->t^2 will give 1+c[t^4] and 1+Subscript[c,b^2] /.b->t^2 will give 1+Subscript[c,t^4] even when c is not defined. There is a lot of differences between languages and one cannot expect Axiom to behave "like" Mathematica, which has no concept of where (that is, in what domain) the operation is taking place. Finally, could you explain further your reasoning about why the interpreter should not coerce Variable x to Symbol? Axiom has no trouble coercing x to a Symbol in the case of an expresion like UP(x,INT). Surely it should be able to solve the problem of coercing x to Symbol in order that Symbol can be coerced to UP(x,INT) in cases like 1 x and integrate(x)? Nope, because I was wrong! I thought it might create an infinite mutual coercion, but that is apparently not the case. I don't have any documentation on the interpreter. Like what physicists practice, the theory must follow the experiments and will change when new evidence shows up. So I now find out a situation where the interpreter actually does the coercion from Variable(x) to Symbol. (So my previous statement that claim the interpreter does not do this conversion in my last message is wrong -- I was misguided by the hyperdoc documentation that the convert and variable functions are unexposed, even if Variable() is). ## See IssueTracker#110. Or is it the case that the interpreter never tries harder than just one coercion? No, the interpreter tries very hard sometimes. See above symbol with a complicated subscript. BTW, I tried specifying axiom )expose Variable Variable is already explicitly exposed in frame initial but I did not see any change in Axiom's behaviour. axiom 1 x Cannot find a definition or applicable library operation named 1 with argument type(s) Variable(x) Perhaps you should use "@" to indicate the required return type, or "$" to specify which version of the function you need.
integrate(x)
There are 4 exposed and 2 unexposed library operations named
integrate having 1 argument(s) but none was determined to be
applicable. Use HyperDoc Browse, or issue
)display op integrate
package-calling the operation or using coercions on the arguments
will allow you to apply the operation.
Cannot find a definition or applicable library operation named
integrate with argument type(s)
Variable(x)
Perhaps you should use "@" to indicate the required return type,
or "\$" to specify which version of the function you need.

I don't know. According to hyperdoc, even if Variable() is exposed, the two operations convert and variable are still unexposed. But I now do not think that is the problem. Apparently, the interpreter is able to use them when it wants to.

--------- Bill Page -- forwarded from "99X1AbcYShouldBeASyntaxError#msg20050217221454-0600@page.axiom-developer.org:99X1AbcYShouldBeASyntaxError#msg20050217221454-0600@page.axiom-developer.org

William

property change
Tue, 08 Mar 2005 04:52:29 -0600 reply

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