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# Dr Youming Qiao

### Biography

Youming Qiao obtained his PhD in computer science from the Institute for Interdisciplinary Sciences, Tsinghua University in 2012. His advisor was Andrew Yao. Before that, he obtained his bachelor's degree in computer science from Tsinghua University. After working as a postdoc in the Centre for Quantum Technologies, National University of Singapore, he is currently a lecturer at the Centre for Quantum Computation & Intelligent Systems, University of Technology, Sydney.

### Professional

Discovery Early Career Researcher Award, Australian Research Council, to commence in 2015.

Project title: Algorithms and complexity for testing isomorphism of algebraic structures

Jiang NanXiang Scholarship (First Grade at Tsinghua University), Tsinghua University, 2011.

Senior Lecturer, School of Software
Core Member, Centre for Quantum Software and Information
BA Eng (Tsinghua Uni), PhD Eng (Tsinghua Uni)

Phone
+61 2 9514 4771
ORCID

### Research Interests

Complexity theory and algebraic computation, with an emphasis on the interaction with
group theory.
Can supervise: Yes
Algorithms, complexity, algebra, group theory.

## Conferences

Grochow, J.A., Mulmuley, K.D. & Qiao, Y. 2016, 'Boundaries of VP and VNP', Leibniz International Proceedings in Informatics, LIPIcs, International Colloquium on Automata, Languages, and Programming (ICALP), Schloss Dagstuhl--Leibniz-Zentrum fuer Informatik, Rome, Italy.
One fundamental question in the context of the geometric complexity theory approach to the VP vs. VNP conjecture is whether VP = !VP, where VP is the class of families of polynomials that can be computed by arithmetic circuits of polynomial degree and size, and VP is the class of families of polynomials that can be approximated infinitesimally closely by arithmetic circuits of polynomial degree and size. The goal of this article is to study the conjecture in (Mulmuley, FOCS 2012) that !VP is not contained in VP. Towards that end, we introduce three degenerations of VP (i.e., sets of points in VP), namely the stable degeneration Stable-VP, the Newton degeneration Newton-VP, and the p-definable one-parameter degeneration VP*. We also introduce analogous degenerations of VNP. We show that Stable-VP subseteq Newton-VP subseteq VP* subseteq VNP, and Stable-VNP = Newton-VNP = VNP* = VNP. The three notions of degenerations and the proof of this result shed light on the problem of separating VP from VP. Although we do not yet construct explicit candidates for the polynomial families in !VP\VP, we prove results which tell us where not to look for such families. Specifically, we demonstrate that the families in Newton-VP \VP based on semi-invariants of quivers would have to be nongeneric by showing that, for many finite quivers (including some wild ones), Newton degeneration of any generic semi-invariant can be computed by a circuit of polynomial size. We also show that the Newton degenerations of perfect matching Pfaffians, monotone arithmetic circuits over the reals, and Schur polynomials have polynomial-size circuits.
Kulkarni, R., Qiao, Y. & Sun, X. 2015, 'On the Power of Parity Queries in Boolean Decision Trees', Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), 12th Annual Conference, TAMC 2015, Springer, Singapore, pp. 99-109.
© Springer International Publishing Switzerland 2015. In an influential paper,Kushilevitz and Mansour (1993) introduced a natural extension of Boolean decision trees called parity decision tree (PDT) where one may query the sum modulo 2, i.e., the parity, of an arbitrary subset of variables. Although originally introduced in the context of learning, parity decision trees have recently regained interest in the context of communication complexity (cf. Shi and Zhang 2010) and property testing (cf. Bhrushundi, Chakraborty, and Kulkarni 2013). In this paper, we investigate the power of parity queries. In particular, we show that the parity queries can be replaced by ordinary ones at the cost of the total influence aka average sensitivity per query. Our simulation is tight as demonstrated by the parity function. At the heart of our result lies a qualitative extension of the result of O'Donnell, Saks, Schramme, and Servedio (2005) titled: Every decision tree has an influential variable. Recently Jain and Zhang (2011) obtained an alternate proof of the same. Our main contribution in this paper is a simple but surprising observation that the query elimination method of Jain and Zhang can indeed be adapted to eliminate, seemingly much more powerful, parity queries. Moreover, we extend our result to linear queries for Boolean valued functions over arbitrary finite fields.
Ivanyos, G., Karpinski, M., Qiao, Y. & Santha, M. 2014, 'Generalized Wong sequences and their applications to Edmonds' problems', Leibniz International Proceedings in Informatics, LIPIcs, pp. 397-408.
Grochow, J.A. & Qiao, Y. 2014, 'Algorithms for group isomorphism via group extensions and cohomology', Proceedings of the Annual IEEE Conference on Computational Complexity, pp. 110-119.
Decker, T., Ivanyos, G., Kulkarni, R., Qiao, Y. & Santha, M. 2014, 'An efficient quantum algorithm for finding hidden parabolic subgroups in the general linear group', Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), pp. 226-238.
Ivanyos, G., Kulkarni, R., Qiao, Y., Santha, M. & Sundaram, A. 2014, 'On the complexity of trial and error for constraint satisfaction problems', Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), pp. 663-675.
Qiao, Y., Sun, X. & Yu, N. 2013, 'Determinantal Complexities and Field Extensions', LNCS - Algorithms and Computation - Proceedings of the 24th International Symposium, ISAAC 2013, ISAAC, Springer, Hong Kong, pp. 119-129.
Let F be a field of characteristic ??2. The determinantal complexity of a polynomial P?F[x1,,xn] is defined as the smallest size of a matrix M whose entries are linear polynomials of x i s over F, such that P=det(M) as polynomials in F[x1,,xn]. To determine the determinantal complexity of the permanent polynomial is a long-standing open problem. Let K be an extension field of F; then P can be viewed as a polynomial over K. We are interested in the comparison between the determinantal complexity of P over K (denoted as dcK(P)), and that of P over F (denoted as dcF(P)). It is clear that dcK(P)=dcF(P), and the question is whether strict inequality can happen. In this note we consider polynomials defined over Q.
Kulkarni, R., Qiao, Y. & Sun, X. 2013, 'Any monotone property of 3-uniform hypergraphs is weakly evasive', LNCS - Theory and Applications of Models of Computation - Proceedings of the10th International Conference, TAMC 2013, Theory and Applications of Models of Computation, Springer, Hong Kong, pp. 224-235.
Our proof combines the combinatorial approach of Rivest and Vuillemin with the topological approach of Kahn, Saks, and Sturtevant. Interestingly, our proof makes use of Vinogradovs Theorem (weak Goldbach Conjecture), inspired by its recent use by Babai et. al. [1] in the context of the topological approach. Our work leaves the generalization to k-uniform hypergraphs as an intriguing open question.
Gupta, A., Kayal, N. & Qiao, Y. 2013, 'Random arithmetic formulas can be reconstructed efficiently', Proceedings of the Annual IEEE Conference on Computational Complexity, pp. 1-9.
Babai, L., Codenotti, P. & Qiao, Y. 2012, 'Polynomial-Time Isomorphism Test for Groups with No Abelian Normal Subgroups', LNCS - Automata, Languages, and Programming - Proceedings, Part I of the 39th International Colloquium on Automata, Languages and Programming, International Colloquium on Automata, Languages and Programming, Springer, Warwick, UK, pp. 51-62.
We consider the problem of testing isomorphism of groups of order n given by Cayley tables. The trivial n logn bound on the time complexity for the general case has not been improved upon over the past four decades. We demonstrate that the obstacle to efficient algorithms is the presence of abelian normal subgroups; we show this by giving a polynomial-time isomorphism test for groups without nontrivial abelian normal subgroups. This concludes a project started by the authors and J. A. Grochow (SODA 2011). Two key new ingredient are: (a) an algorithm to test permutational isomorphism of permutation groups in time, polynomial in the order and simply exponential in the degree; (b) the introduction of the twisted code equivalence problem, a generalization of the classical code equivalence problem by admitting a group action on the alphabet. Both of these problems are of independent interest
Babai, L. & Qiao, Y. 2012, 'Polynomial-time Isomorphism Test for Groups with Abelian Sylow Towers', 29th Symposium on Theoretical Aspects of Computer Science, Symposium on Theoretical Aspects of Computer Science, Dagstuhl Publishing, Paris, France, pp. 453-465.
We consider the problem of testing isomorphism of groups of order n given by Cayley tables. The trivial n^{log n} bound on the time complexity for the general case has not been improved over the past four decades. Recently, Babai et al. (following Babai et al. in SODA 2011) presented a polynomial-time algorithm for groups without abelian normal subgroups, which suggests solvable groups as the hard case for group isomorphism problem. Extending recent work by Le Gall (STACS 2009) and Qiao et al. (STACS 2011), in this paper we design a polynomial-time algorithm to test isomorphism for the largest class of solvable groups yet, namely groups with abelian Sylow towers, defined as follows. A group G is said to possess a Sylow tower, if there exists a normal series where each quotient is isomorphic to Sylow subgroup of G. A group has an abelian Sylow tower if it has a Sylow tower and all its Sylow subgroups are abelian. In fact, we are able to compute the coset of isomorphisms of groups formed as coprime extensions of an abelian group, by a group whose automorphism group is known. The mathematical tools required include representation theory, Wedderburn's theorem on semisimple algebras, and M.E. Harris's 1980 work on p'-automorphisms of abelian p-groups. We use tools from the theory of permutation group algorithms, and develop an algorithm for a parameterized versin of the graph-isomorphism-hard setwise stabilizer problem, which may be of independent interest
Babai, L., Codenotti, P., Grochow, J.A. & Qiao, Y. 2011, 'Code equivalence and group isomorphism', Proceedings of the twenty-second annual ACM-SIAM symposium on Discrete Algorithm, Annual ACM-SIAM symposium on Discrete Algorithm, ACM, San Francisco, USA, pp. 1395-1408.
The isomorphism problem for groups given by their multiplication tables has long been known to be solvable in time nlog n+O(1). The decades-old quest for a polynomial-time algorithm has focused on the very difficult case of class-2 nilpotent groups (groups whose quotient by their center is abelian), with little success. In this paper we consider the opposite end of the spectrum and initiate a more hopeful program to find a polynomial-time algorithm for semisimple groups, defined as groups without abelian normal subgroups. First we prove that the isomorphism problem for this class can be solved in time nO(log log n). We then identify certain bottlenecks to polynomial-time solvability and give a polynomial-time solution to a rich subclass, namely the semisimple groups where each minimal normal subgroup has a bounded number of simple factors. We relate the results to the filtration of groups introduced by Babai and Beals (1999). One of our tools is an algorithm for equivalence of (not necessarily linear) codes in simply-exponential time in the length of the code, obtained by modifying Luks's algorithm for hypergraph isomorphism in simply-exponential time in the number of vertices (FOCS 1999). We comment on the complexity of the closely related problem of permutational isomorphism of permutation groups.
Jansen, M., Qiao, Y. & Sarma, M.N.J. 2010, 'Deterministic Black-Box Identity Testing pi-Ordered Algebraic Branching Programs', IARCS International Conference on Foundations of Software Technology and Theoretical Computer Science, International Conference on Foundations of Software Technology and Theoretical Computer Science, Dagstuhl Publishing, Mumbai, India, pp. 296-307.
In this paper we study algebraic branching programs (ABPs) with restrictions on the order and the number of reads of variables in the program. An ABP is given by a layered directed acyclic graph with source $s$ and sink $t$, whose edges are labeled by variables taken from the set $\{x_1, x_2, \ldots, x_n\}$ or field constants. It computes the sum of weights of all paths from $s$ to $t$, where the weight of a path is defined as the product of edge-labels on the path. Given a permutation $\pi$ of the $n$ variables, for a $\pi$-ordered ABP ($\pi$-OABP), for any directed path $p$ from $s$ to $t$, a variable can appear at most once on $p$, and the order in which variables appear on $p$ must respect $\pi$. One can think of OABPs as being the arithmetic analogue of ordered binary decision diagrams (OBDDs). We say an ABP $A$ is of read $r$, if any variable appears at most $r$ times in $A$. Our main result pertains to the polynomial identity testing problem, i.e. the problem of deciding whether a given $n$-variate polynomial is identical to the zero polynomial or not. We prove that over any field $\F$, and in the black-box model, i.e. given only query access to the polynomial, read $r$ $\pi$-OABP computable polynomials can be tested in $\DTIME[2^{O(r\log r \cdot \log^2 n \log\log n)}]$. In case $\F$ is a finite field, the above time bound holds provided the identity testing algorithm is allowed to make queries to extension fields of $\F$.
Qiao, Y., Jayalal, S.M.N. & Tang, B. 2011, 'On isomorphism testing of groups with normal hall subgroups', Leibniz International Proceedings in Informatics, LIPIcs, pp. 567-578.
Bogdanov, A. & Qiao, Y. 2009, 'On the security of Goldreich's one-way function', Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), pp. 392-405.
Qiao, Y. & Tartary, C. 2008, 'Counting Method for Multi-party Computation over Non-abelian Groups', LNCS - Cryptology and Network Security - Proceedings of the 7th International Conference, Cryptology and Network Security, Springer, Hong Kong, pp. 162-177.
In the Crypto'07 paper [5], Desmedt et al. studied the problem of achieving secure n-party computation over non-Abelian groups. The function to be computed is f G (x 1,...,x n ) :?=?x 1 ...x n where each participant P i holds an input x i from the non-commutative group G. The settings of their study are the passive adversary model, information-theoretic security and black-box group operations over G. They presented three results. The first one is that honest majority is needed to ensure security when computing f G . Second, when the number of adversary t=?n2?-1, they reduced building such a secure protocol to a graph coloring problem and they showed that there exists a deterministic secure protocol computing f G using exponential communication complexity. Finally, Desmedt et al. turned to analyze random coloring of a graph to show the existence of a probabilistic protocol with polynomial complexity when t?<?n/µ, in which µ is a constant less than 2.948

## Journal articles

Mukhopadhyay, P. & Qiao, Y. 2017, 'Sparse multivariate polynomial interpolation on the basis of Schubert polynomials', Computational Complexity, pp. 1-29.
© 2016 Springer International PublishingSchubert polynomials were discovered by A. Lascoux and M. Schützenberger in the study of cohomology rings of flag manifolds in 1980s. These polynomials generalize Schur polynomials and form a linear basis of multivariate polynomials. In 2003, Lenart and Sottile introduced skew Schubert polynomials, which generalize skew Schur polynomials and expand in the Schubert basis with the generalized Littlewood–Richardson coefficients. In this paper, we initiate the study of these two families of polynomials from the perspective of computational complexity theory. We first observe that skew Schubert polynomials, and therefore Schubert polynomials, are in #P (when evaluating on nonnegative integral inputs) and VNP. Our main result is a deterministic algorithm that computes the expansion of a polynomial f of degree d in (Formula presented.) on the basis of Schubert polynomials, assuming an oracle computing Schubert polynomials. This algorithm runs in time polynomial in n, d, and the bit size of the expansion. This generalizes, and derandomizes, the sparse interpolation algorithm of symmetric polynomials in the Schur basis by Barvinok and Fomin (Adv Appl Math 18(3):271–285, 1997). In fact, our interpolation algorithm is general enough to accommodate any linear basis satisfying certain natural properties. Applications of the above results include a new algorithm that computes the generalized Littlewood–Richardson coefficients.
Ivanyos, G., Qiao, Y. & Subrahmanyam, K.V. 2016, 'Non-commutative Edmonds' problem and matrix semi-invariants', Computational Complexity, pp. 1-47.
Ivanyos, G., Karpinski, M., Qiao, Y. & Santha, M. 2015, 'Generalized Wong sequences and their applications to Edmonds' problems', Journal of Computer and System Sciences, vol. 81, no. 7, pp. 1373-1386.
Kulkarni, R., Qiao, Y. & Sun, X. 2015, 'Any monotone property of 3-uniform hypergraphs is weakly evasive', Theoretical Computer Science, vol. 588, pp. 16-23.
In this paper, we provide a characterization of multipartite entanglement in terms of equivalence classes of states under local unitary transformation (LU) by demonstrating a simple method to construct all homogenous polynomials that are invariant under local unitary group(LUIPs) for any fixed degree. We give an upper bound on the degree of the LUIP such that the ring of LUIPs can be generated by LUIPs with degree lower than the bound. Our characterization can directly generate an algorithm whose output is a generating set of LUIPs. By employing the concept of LUIPs, we prove that multipartite entanglement is additive in the sense that two multipartite states are LU equivalent if and only if $n$-copies of these two states are LU equivalent for some $n$. The method for studying LU equivalence is also used to classify the different types of multipartite mixed entanglement according to equivalence classes of states under stochastic local operations and classical communication (SLOCC), where the pure states case was previously studied by Gour and Wallach using another approach.


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# A Ladder on the Platform of a Fire Brigade Van Can Be Elevated at an Angle of 70° to the Maximum. the Length of the Ladder Can Be Extended Upto 20 M. If the Platform is 2m Above the Ground - Geometry

ConceptHeights and Distances

#### Question

A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

#### Solution

Suppose the maximum height from the ground upto which the ladder can reach be h m.

Here, AB = 2 m, AD = 20 m, ∠DAE = 70º and AE ⊥ CD.
DE = CD − CE = (h − 2) m             (CE = AB)
In right ∆AED,
$\sin70^\circ = \frac{DE}{AD}$
$\Rightarrow 0 . 94 = \frac{h - 2}{20}$
$\Rightarrow h - 2 = 20 \times 0 . 94 = 18 . 8$
$\Rightarrow h = 18 . 8 + 2 = 20 . 8 m$

Thus, the maximum height from the ground upto which the ladder can reach is 20.8 m.

Is there an error in this question or solution?

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Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Problem set 6 | Q: 9 | Page no. 139

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Solution A Ladder on the Platform of a Fire Brigade Van Can Be Elevated at an Angle of 70° to the Maximum. the Length of the Ladder Can Be Extended Upto 20 M. If the Platform is 2m Above the Ground Concept: Heights and Distances.
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# CBSE Physics 2014 Exam Questions

#### Short Answer Type

1.

Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?

The glass bob would reach the ground earlier. Glass bob which is non-conducting in nature will only experience Earth’s gravitational pull unlike the metallic bob which is conducting.

Since the metallic bob is conducting in nature, eddy current is induced as it falls through the magnetic field of the Earth. As per Lenz’s law, current is induced in a direction opposite to the motion of the metallic bob. Hence, there is a delay.

2.

State Kirchhoff's rules. Explain briefly how these rules are justified.

Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”

This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule.

Here, I1I2 I3, and I4 are the currents flowing through the respective wires.

Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative.

I3 + (− I1) + (− I2) + (− I4) = 0

Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and the currents flowing through them.

OR

“The algebraic sum of all the potential drops and emfs along any closed path in a network is zero.”

For the closed loop BACB:

E1 − E2 = I1R1 + I2R2 − I3R3

For the closed loop CADC:

E2 = I3R3 + I4R4 + I5R5

This law is based on the law of conservation of energy.

3.

Given a uniform electric field E =5 ×10 N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?

Given, Electric flux,

We have, Electric flux, ϕ = E.A =

= 50 Weber

When the plane makes a 30° angle with the x-axis, the area vector makes 60° with the x-axis

ϕ = E. A

⇒              ϕ=EA cos θ

⇒              ϕ= (5×103)(10−2) cos 60°

⇒              ϕ=

⇒             ϕ = 25 Weber

4.

Write the expression, in a vector form, for the Lorentz magnetic force  due to a charge moving with velocity   in a magnetic field B. What is the direction of the magnetic force?

Lorentz magnetic force is given by:

; q is the magnitude of the moving charge.
The direction of the magnetic force is perpendicular to the plane containing the velocity vector and the magnetic field vector.

5.

A capacitor 'C', a variable resistor 'R' and a bulb 'B' are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance?

(i) When a dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C  ) thereby decreasing the potential drop across the bulb because, both the bulb and capacitor is connected in series. So, brightness of the bulb will increase.

(ii) When resistance (R) is increased keeping the capacitance same, the potential drop across the resistor will increases. Therefore, the potential drop across the bulb will decrease because both are connected in series. So, brightness of the bulb will decrease.

6.

The carrier wave is given by C (t) = 2sin (8πt) Volt.
The modulating signal is a square wave as shown. Find modulation index.

The generalized equation of a carrier wave is given by,

The generalized equation of a modulating signal is given by,

On comparing the given equation of carrier wave with the generalized equation, we get,

Amplitude of modulating signal, Am = 1 V

Amplitude of carrier wave, Ac = 2 V

Modulation index is the ratio of the amplitude of modulating signal to the amplitude of carrier wave .

It is denoted by,

So,

7.

Define the term 'Mobility' of charge carriers in a conductor. Write its S.I. unit

Drift velocity per unit applied electric field is known as the mobility of charge carriers in a conductor.

Mathematically,

Mobility,

S.I. unit of mobility:  or

8.

Show variation of resistivity of copper as a function of temperature in a graph.

The graph below shows variation of resistivity of copper with temperature. The graph is parabolic in nature.
'

9.

"For any charge configuration, equipotential surface through a point is normal to the electric field." Justify.

Work done (W) in moving a test charge along an equipotential surface is zero.

Work done is given by,

F is the electric force and s is the magnitude of displacement.

For non-zero displacement, this is possible only when cos   is equal to 0.

= 90°

Thus, the force acting on the point charge is perpendicular to the equipotential surface.

Electric field lines give us the direction of electric force on a charge.

Thus, for any charge configuration, equipotential surface through a point is normal to the electric field.

10.

Out of the two magnetic materials, 'A' has relative permeability slightly greater than unity while 'B' has less than unity. Identify the nature of the materials 'A' and 'B'. Will their susceptibilities be positive or negative?

A is a paramagnetic material because its permeability is greater than unity and its susceptibility is positive. The relative permeability of a paramagnetic material is between   and susceptibility is

For a diamagnetic material, the relative permeability lies between 0 ≤ μr < 1 and its susceptibility lies between −1<  χ<  0.

Therefore, 'B' is a diamagnetic material and its susceptibility is negative.

This is because its relative permeability is less than unity.

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# Kolmogorov's example of a measurable function not (generally) differentiable

In [1, page 7], the author says.

Kolmogorov showed that if the function $$f(x) = \sum_{n=1}^{\infty} \frac{\cos 3^n x}{3^n}$$ has a finite or infinite generalized derivative on a set of positive measure, then the function is nonmeasurable.

Where can I find a proof/explanation of this result (and/or other similar results) in English? The reference doesn't have to be to Kolmogorov's orignal paper; for example, a modern exposition would suffice (and might very well be better).

[1] A. N. Shiryaev, "Andrei Nikolaevich Kolmogorov", Theory of Probability and Applications, vol 34, no. 1, 1988.

-
Note that I also posted this question on math.SE: math.stackexchange.com/questions/103000/… –  Quinn Culver Jan 27 '12 at 18:41
add comment

## 2 Answers

A translation in english is in "Selected Works of A.N. Kolmogorov I" : "On the possibility of a general definition of derivative, integral and summation of divergent series" (page 33 and 34).

-
add comment

Kolmogorov's original paper:

'Sur la possibilite de la definition generale de la derivee, de l'integrale et de la sommation des series divergentes', C.R. Acad. Sci. Paris 180(1925), 362-364.

-
Thanks. I suppose that reference is in French, which I cannot read. Do you know if there's an English translation? –  Quinn Culver Jan 27 '12 at 15:38
Yes, see: V.M. Tikhomirov (Ed.): Selected works of A.N. Kolmogorov, Vol.1, Kluwer, pp.51-52 –  Ljubomir Cukic Jan 29 '12 at 8:22
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# Calculating Derivative – Third root – Exercise 1106

Exercise

Find the derivative of the following function:

$$f(x)=\sqrt[3]{x+\sqrt{x}}$$

$$f'(x)=\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{(x+\sqrt{x})}^2}$$

Solution

First, we simplify the function:

$$f(x)={(x+\sqrt{x})}^{\frac{1}{3}}$$

Using Derivative formulas, we get the derivative:

$$f'(x)=\frac{1}{3}{(x+\sqrt{x})}^{-\frac{2}{3}}\cdot (1+ \frac{1}{2\sqrt{x}})=$$

$$=\frac{1}{3\sqrt[3]{{(x+\sqrt{x})}^2}}\cdot \frac{2\sqrt{x}+1}{2\sqrt{x}}=$$

$$=\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{(x+\sqrt{x})}^2}$$

Have a question? Found a mistake? – Write a comment below!
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# 5.2 Electromotive force: terminal voltage

 Page 1 / 12
• Compare and contrast the voltage and the electromagnetic force of an electric power source.
• Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source increases (due to aging of batteries, for example).
• Explain why it is beneficial to use more than one voltage source connected in parallel.

When you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they simply blink off when the battery’s energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted.

Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery’s output voltage is reduced by the overload.

The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance    . Let us examine both.

## Electromotive force

You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences.

A few voltage sources are shown in [link] . All such devices create a potential difference    and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the name electromotive force , abbreviated emf.

Emf is not a force at all; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a source when no current is flowing. Units of emf are volts.

Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device’s output voltage can be measured without drawing current, then output voltage will equal emf (even for a very depleted battery).

## Internal resistance

As noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcycle battery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance $r$ . Internal resistance is the inherent resistance to the flow of current within the source itself.

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Preparation and Applications of Nanomaterial for Drug Delivery
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LITNING
scanning tunneling microscope
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# Prove that if $f(x)$ divides $g(x)$ and $f(x)$ divides $h(x)$, then $f(x)$ divides $s(x)g(x) +t(x)h(x)$.

Let $$F$$ be a field. Prove that for all polynomials $$f(x), g(x), h(x) \in {F}[x]$$, if $$f(x)$$ divides $$g(x)$$ and $$f(x)$$ divides $$h(x)$$, then for all polynomials $$s(x), t(x)\in {F}[x]$$, $$f(x)$$ divides $$s(x)g(x) +t(x)h(x)$$.

How do I prove this question? I know that $$f(x)=g(x)q(x)+r(x)$$ but I'm not sure if I use that at all in this question.

• $f(x)$ is a factor of both $g(x)$ and $h(x)$. Consider the case where $f(x)=2$, $g(x)=4$, and $h(x)=20$. Does $f(x)$ divide $4\cdot s + 20 \cdot t$? – Andrew Tawfeek Apr 2 at 1:49

Hint: How would you prove an equivalent statment for the integers?

• Would I use DIC for polynomials? – Sania Apr 2 at 1:40
• What is DIC?... – Maria Mazur Apr 2 at 1:41
• If a divides b and a divides c then for integers x,y a divides bx+cy – Sania Apr 2 at 1:41
• Actualy this is an equivalent statement, you need to remember how do you prove it. – Maria Mazur Apr 2 at 1:43

If $$f(x)$$ divides $$g(x)$$, then $$g(x) = f(x)q(x)$$ and likewise $$h(x) = f(x)r(x)$$.

Also note that $$f(x)$$ divides $$s(x)g(x)$$, since $$s(x)g(x) = s(x) f(x)q(x)$$.


text:
# Demo - Lid driven cavity¶

Authors

Mikael Mortensen (mikaem at math.uio.no)

Date

Nov 14, 2019

Summary. The lid driven cavity is a classical benchmark for Navier Stokes solvers. This is a demonstration of how the Python module shenfun can be used to solve the lid driven cavity problem with full spectral accuracy using a mixed (coupled) basis in a 2D tensor product domain. The demo also shows how to use mixed tensor product spaces for vector valued equations. Note that the regular lid driven cavity, where the top wall has constant velocity and the remaining three walls are stationary, has a singularity at the two upper corners, where the velocity is discontinuous. Due to their global nature, spectral methods are usually not very good at handling problems with discontinuities, and for this reason we will also look at a regularized lid driven cavity, where the top lid moves according to $$(1-x)^2(1+x)^2$$, thus removing the corner discontinuities.

Velocity vectors for $$Re=100$$

## Bases and tensor product spaces¶

With the Galerkin method we need basis functions for both velocity and pressure, as well as for the nonlinear right hand side. A Dirichlet basis will be used for velocity, whereas there is no boundary restriction on the pressure basis. For both two-dimensional bases we will use one basis function for the $$x$$-direction, $$\mathcal{X}_k(x)$$, and one for the $$y$$-direction, $$\mathcal{Y}_l(y)$$. And then we create two-dimensional basis functions like

(1)$v_{kl}(x, y) = \mathcal{X}_k(x) \mathcal{Y}_l(y),$

and solutions (trial functions) as

(2)$u(x, y) = \sum_{k}\sum_{l} \hat{u}_{kl} v_{kl}(x, y).$

For the homogeneous Dirichlet boundary condition the basis functions $$\mathcal{X}_k(x)$$ and $$\mathcal{Y}_l(y)$$ are chosen as composite Legendre polynomials (we could also use Chebyshev):

(3)$\mathcal{X}_k(x) = L_k(x) - L_{k+2}(x), \quad \forall \, k \in \mathbf{k}^{N_0-2},$
(4)$\mathcal{Y}_l(y) = L_l(y) - L_{l+2}(y), \quad \forall \, l \in \mathbf{l}^{N_1-2},$

where $$\mathbf{k}^{N_0-2} = (0, 1, \ldots, N_0-3)$$, $$\mathbf{l}^{N_1-2} = (0, 1, \ldots, N_1-3)$$ and $$N = (N_0, N_1)$$ is the number of quadrature points in each direction. Note that $$N_0$$ and $$N_1$$ do not need to be the same. The basis funciton (3) satisfies the homogeneous Dirichlet boundary conditions at $$x=\pm 1$$ and (4) the same at $$y=\pm 1$$. As such, the basis function $$v_{kl}(x, y)$$ satisfies the homogeneous Dirichlet boundary condition for the entire domain.

With shenfun we create these homogeneous spaces, $$D_0^{N_0}(x)=\text{span}\{L_k-L_{k+2}\}_{k=0}^{N_0-2}$$ and $$D_0^{N_1}(y)=\text{span}\{L_l-L_{l+2}\}_{l=0}^{N_1-2}$$ as

N = (51, 51)
family = 'Legendre' # or use 'Chebyshev'
quad = 'LG'         # for Chebyshev use 'GC' or 'GL'

The spaces are here the same, but we will use D0X in the $$x$$-direction and D0Y in the $$y$$-direction. But before we use these bases in tensor product spaces, they remain identical as long as $$N_0 = N_1$$.

Special attention is required by the moving lid. To get a solution with nonzero boundary condition at $$y=1$$ we need to add one more basis function that satisfies that solution. In general, a nonzero boundary condition can be added on both sides of the domain using the following basis

(5)$\mathcal{Y}_l(y) = L_l(y) - L_{l+2}(y), \quad \forall \, l \in \mathbf{l}^{N_1-2}.$
(6)$\mathcal{Y}_{N_1-2}(y) = (L_0+L_1)/2 \quad \left(=(1+y)/2\right),$
(7)$\mathcal{Y}_{N_1-1}(y) = (L_0-L_1)/2 \quad \left(=(1-y)/2\right).$

And then the unknown component $$N_1-2$$ decides the value at $$y=1$$, whereas the unknown at $$N_1-1$$ decides the value at $$y=-1$$. Here we only need to add the $$N_1-2$$ component, but for generality this is implemented in shenfun using both additional basis functions. We create the space $$D_1^{N_1}(y)=\text{span}\{\mathcal{Y}_l(y)\}_{l=0}^{N_1-1}$$ as

D1Y = Basis(N[1], family, quad=quad, bc=(1, 0))

where bc=(1, 0) fixes the values for $$y=1$$ and $$y=-1$$, respectively. For a regularized lid driven cavity the velocity of the top lid is $$(1-x)^2(1+x)^2$$ and not unity. To implement this boundary condition instead, we can make use of sympy and quite straight forward do

import sympy
x = sympy.symbols('x')

Uncomment the last line to run the regularized boundary conditions. Otherwise, there is no difference at all between the regular and the regularized lid driven cavity implementations.

The pressure basis that comes with no restrictions for the boundary is a little trickier. The reason for this has to do with inf-sup stability. The obvious choice of basis functions are the regular Legendre polynomials $$L_k(x)$$ in $$x$$ and $$L_l(y)$$ in the $$y$$-directions. The problem is that for the natural choice of $$(k, l) \in \mathbf{k}^{N_0} \times \mathbf{l}^{N_1}$$ there are nullspaces and the problem is not well-defined. It turns out that the proper choice for the pressure basis is simply the regular Legendre basis functions, but for $$(k, l) \in \mathbf{k}^{N_0-2} \times \mathbf{l}^{N_1-2}$$. The bases $$P^{N_0}(x)=\text{span}\{L_k(x)\}_{k=0}^{N_0-3}$$ and $$P^{N_1}(y)=\text{span}\{L_l(y)\}_{l=0}^{N_1-3}$$ are created as

PX = Basis(N[0], family, quad=quad)
PX.slice = lambda: slice(0, N[0]-2)
PY.slice = lambda: slice(0, N[1]-2)

Note that we still use these spaces with the same $$N_0 \cdot N_1$$ quadrature points in real space, but the two highest frequencies have been set to zero.

We have now created all relevant function spaces for the problem at hand. It remains to combine these spaces into tensor product spaces, and to combine tensor product spaces into mixed (coupled) tensor product spaces. From the Dirichlet bases we create two different tensor product spaces, whereas one is enough for the pressure

(8)$V_{1}^{\mathbf{N}}(\mathbf{x}) = D_0^{N_0}(x) \otimes D_1^{N_1}(y),$
(9)$V_{0}^{\mathbf{N}}(\mathbf{x}) = D_0^{N_0}(x) \otimes D_0^{N_1}(y),$
(10)$P^{\mathbf{N}}(\mathbf{x}) = P^{N_0}(x) \otimes P^{N_1}(y).$

With shenfun the tensor product spaces are created as

V1 = TensorProductSpace(comm, (D0X, D1Y))
V0 = TensorProductSpace(comm, (D0X, D0Y))
P = TensorProductSpace(comm, (PX, PY))

These tensor product spaces are all scalar valued. The velocity is a vector, and a vector requires a mixed basis like $$W_1^{\mathbf{N}} = V_1^{\mathbf{N}} \times V_0^{\mathbf{N}}$$. The mixed basis is created in shenfun as

W1 = MixedTensorProductSpace([V1, V0])
W0 = MixedTensorProductSpace([V0, V0])

Note that the second mixed basis, $$W_0^{\mathbf{N}} = V_0^{\mathbf{N}} \times V_0^{\mathbf{N}}$$, uses homogeneous boundary conditions throughout.

## Mixed variational form¶

We now formulate a variational problem using the Galerkin method: Find $$\mathbf{u} \in W_1^{\mathbf{N}}$$ and $$p \in P^{\mathbf{N}}$$ such that

(11)$\int_{\Omega} (\nu \nabla^2 \mathbf{u} - \nabla p ) \cdot \mathbf{v} \, dxdy = \int_{\Omega} (\nabla \cdot \mathbf{u}\mathbf{u}) \cdot \mathbf{v}\, dxdy \quad\forall \mathbf{v} \, \in \, W_0^{\mathbf{N}},$
(12)$\int_{\Omega} \nabla \cdot \mathbf{u} \, q \, dxdy = 0 \quad\forall q \, \in \, P^{\mathbf{N}}.$

Note that we are using test functions $$\mathbf{v}$$ with homogeneous boundary conditions.

The first obvious issue with Eq (11) is the nonlinearity. In other words we will need to linearize and iterate to be able to solve these equations with the Galerkin method. To this end we will introduce the solution on iteration $$k \in [0, 1, \ldots]$$ as $$\mathbf{u}^k$$ and compute the nonlinearity using only known solutions $$\int_{\Omega} (\nabla \cdot \mathbf{u}^k\mathbf{u}^k) \cdot \mathbf{v}\, dxdy$$. Using further integration by parts we end up with the equations to solve for iteration number $$k+1$$ (using $$\mathbf{u} = \mathbf{u}^{k+1}$$ and $$p=p^{k+1}$$ for simplicity)

(13)$-\int_{\Omega} \nu \nabla \mathbf{u} \, \colon \nabla \mathbf{v} \, dxdy + \int_{\Omega} p \nabla \cdot \mathbf{v} \, dxdy = \int_{\Omega} (\nabla \cdot \mathbf{u}^k\mathbf{u}^k) \cdot \mathbf{v}\, dxdy \quad\forall \mathbf{v} \, \in \, W_0^{\mathbf{N}},$
(14)$\int_{\Omega} \nabla \cdot \mathbf{u} \, q \, dxdy = 0 \quad\forall q \, \in \, P^{\mathbf{N}}.$

Note that the nonlinear term may also be integrated by parts and evaluated as $$\int_{\Omega}-\mathbf{u}^k\mathbf{u}^k \, \colon \nabla \mathbf{v} \, dxdy$$. All boundary integrals disappear since we are using test functions with homogeneous boundary conditions.

Since we are to solve for $$\mathbf{u}$$ and $$p$$ at the same time, we formulate a mixed (coupled) problem: find $$(\mathbf{u}, p) \in W_1^{\mathbf{N}} \times P^{\mathbf{N}}$$ such that

(15)$a((\mathbf{u}, p), (\mathbf{v}, q)) = L((\mathbf{v}, q)) \quad \forall (\mathbf{v}, q) \in W_0^{\mathbf{N}} \times P^{\mathbf{N}},$

where bilinear ($$a$$) and linear ($$L$$) forms are given as

(16)$a((\mathbf{u}, p), (\mathbf{v}, q)) = -\int_{\Omega} \nu \nabla \mathbf{u} \, \colon \nabla \mathbf{v} \, dxdy + \int_{\Omega} p \nabla \cdot \mathbf{v} \, dxdy + \int_{\Omega} \nabla \cdot \mathbf{u} \, q \, dxdy,$
(17)$L((\mathbf{v}, q); \mathbf{u}^{k}) = \int_{\Omega} (\nabla \cdot \mathbf{u}^{k}\mathbf{u}^{k}) \cdot \mathbf{v}\, dxdy.$

Note that the bilinear form will assemble to a block matrix, whereas the right hand side linear form will assemble to a block vector. The bilinear form does not change with the solution and as such it does not need to be reassembled inside an iteration loop.

The algorithm used to solve the equations are:

• Set $$k = 0$$

• Guess $$\mathbf{u}^0 = (0, 0)$$

• while not converged:

• assemble $$L((\mathbf{v}, q); \mathbf{u}^{k})$$

• solve $$a((\mathbf{u}, p), (\mathbf{v}, q)) = L((\mathbf{v}, q); \mathbf{u}^{k})$$ for $$\mathbf{u}^{k+1}, p^{k+1}$$

• compute error = $$\int_{\Omega} (\mathbf{u}^{k+1}-\mathbf{u}^{k})^2 \, dxdy$$

• if error $$<$$ some tolerance then converged = True

• $$k$$ += $$1$$

## Implementation of solver¶

We will now implement the coupled variational problem described in previous sections. First of all, since we want to solve for the velocity and pressure in a coupled solver, we have to create a mixed tensor product space $$VQ = W_1^{\mathbf{N}} \times P^{\mathbf{N}}$$ that couples velocity and pressure

VQ = MixedTensorProductSpace([W1, P])    # Coupling velocity and pressure

We can now create test- and trialfunctions for the coupled space $$VQ$$, and then split them up into components afterwards:

up = TrialFunction(VQ)
vq = TestFunction(VQ)
u, p = up
v, q = vq

Note

The test function v is using homogeneous Dirichlet boundary conditions even though it is derived from VQ, which contains W1. It is currently not (and will probably never be) possible to use test functions with inhomogeneous boundary conditions.

With the basisfunctions in place we may assemble the different blocks of the final coefficient matrix. For this we also need to specify the kinematic viscosity, which is given here in terms of the Reynolds number:

Re = 100.
nu = 2./Re
G = inner(div(v), p)
D = inner(q, div(u))

Note

The inner products may also be assembled with one single line, as

AA = inner(grad(v), -nu*grad(u)) + inner(div(v), p) + inner(q, div(u))

But note that this requires addition, not subtraction, of inner products, and it is not possible to move the negation to -inner(grad(v), nu*grad(u)). This is because the inner() function returns a list of tensor product matrices of type TPMatrix, and you cannot negate a list.

The assembled subsystems A, G and D are lists containg the different blocks of the complete, coupled, coefficient matrix. A actually contains 4 tensor product matrices of type TPMatrix. The first two matrices are for vector component zero of the test function v[0] and trial function u[0], the matrices 2 and 3 are for components 1. The first two matrices are as such for

A[0:2] = inner(grad(v[0]), -nu*grad(u[0]))

Breaking it down the inner product is mathematically

(18)$\int_{\Omega}-\nu \left(\frac{\partial \mathbf{v}[0]}{\partial x}, \frac{\partial \mathbf{v}[0]}{\partial y}\right) \cdot \left(\frac{\partial \mathbf{u}[0]}{\partial x}, \frac{\partial \mathbf{u}[0]}{\partial y}\right) dx dy .$

We can now insert for test function $$\mathbf{v}[0]$$

(19)$\mathbf{v}[0]_{kl} = \mathcal{X}_k \mathcal{Y}_l, \quad (k, l) \in \mathbf{k}^{N_0-2} \times \mathbf{l}^{N_1-2}$

and trialfunction

(20)$\mathbf{u}[0]_{mn} = \sum_{m=0}^{N_0-3} \sum_{n=0}^{N_1-1} \hat{\mathbf{u}}[0]_{mn} \mathcal{X}_m \mathcal{Y}_n,$

where $$\hat{\mathbf{u}}$$ are the unknown degrees of freedom for the velocity vector. Notice that the sum over the second index runs all the way to $$N_1-1$$, whereas the other indices runs to either $$N_0-3$$ or $$N_1-3$$. This is because of the additional basis functions required for the inhomogeneous boundary condition.

Inserting for these basis functions into (17), we obtain after a few trivial manipulations

(21)$-\sum_{m=0}^{N_0-3} \sum_{n=0}^{N_1-1} \nu \Big( \underbrace{\int_{-1}^{1} \frac{\partial \mathcal{X}_k(x)}{\partial x} \frac{\partial \mathcal{X}_m}{\partial x} dx \int_{-1}^{1} \mathcal{Y}_l \mathcal{Y}_n dy}_{A[0]} + \underbrace{\int_{-1}^{1} \mathcal{X}_k(x) X_m(x) dx \int_{-1}^{1} \frac{\partial \mathcal{Y}_l}{\partial y} \frac{\partial \mathcal{Y}_n}{\partial y} dy}_{A[1]} \Big) \hat{\mathbf{u}}[0]_{mn}.$

We see that each tensor product matrix (both A[0] and A[1]) is composed as outer products of two smaller matrices, one for each dimension. The first tensor product matrix, A[0], is

(22)$\underbrace{\int_{-1}^{1} \frac{\partial \mathcal{X}_k(x)}{\partial x} \frac{\partial \mathcal{X}_m}{\partial x} dx}_{c_{km}} \underbrace{\int_{-1}^{1} \mathcal{Y}_l \mathcal{Y}_n dy}_{f_{ln}}$

where $$C\in \mathbb{R}^{N_0-2 \times N_1-2}$$ and $$F \in \mathbb{R}^{N_0-2 \times N_1}$$. Note that due to the inhomogeneous boundary conditions this last matrix $$F$$ is actually not square. However, remember that all contributions from the two highest degrees of freedom ($$\hat{\mathbf{u}}[0]_{m,N_1-2}$$ and $$\hat{\mathbf{u}}[0]_{m,N_1-1}$$) are already known and they can, as such, be moved directly over to the right hand side of the linear algebra system that is to be solved. More precisely, we can split the tensor product matrix into two contributions and obtain

$\sum_{m=0}^{N_0-3}\sum_{n=0}^{N_1-1} c_{km}f_{ln} \hat{\mathbf{u}}[0]_{m, n} = \sum_{m=0}^{N_0-3}\sum_{n=0}^{N_1-3}c_{km}f_{ln}\hat{\mathbf{u}}[0]_{m, n} + \sum_{m=0}^{N_0-3}\sum_{n=N_1-2}^{N_1-1}c_{km}f_{ln}\hat{\mathbf{u}}[0]_{m, n}, \quad \forall (k, l) \in \mathbf{k}^{N_0-2} \times \mathbf{l}^{N_1-2},$

where the first term on the right hand side is square and the second term is known and can be moved to the right hand side of the linear algebra equation system.

All the parts of the matrices that are to be moved to the right hand side can be extracted from A, G and D as follows

# Extract the boundary matrices
bc_mats = extract_bc_matrices([A, G, D])

These matrices are applied to the solution below (see BlockMatrix BM). Furthermore, this leaves us with square submatrices (A, G, D), which make up a symmetric block matrix

(23)$\begin{split} M = \begin{bmatrix} A[0]+A[1] & 0 & G[0] \\ 0 & A[2]+A[3] & G[1] \\ D[0] & D[1] & 0 \end{bmatrix}\end{split}$

This matrix, and the matrix responsible for the boundary degrees of freedom, can be assembled from the pieces we already have as

M = BlockMatrix(A+G+D)
BM = BlockMatrix(bc_mats)

We now have all the matrices we need in order to solve the Navier Stokes equations. However, we also need some work arrays for iterations and we need to assemble the constant boundary contribution to the right hand side

# Create Function to hold solution
uh_hat = Function(VQ)
ui_hat = uh_hat[0]
D1Y.bc.apply_after(ui_hat[0], True) # Fixes the values of the boundary dofs

# New solution (iterative)
uh_new = Function(VQ)
ui_new = uh_new[0]
D1Y.bc.apply_after(ui_new[0], True)

# Compute the constant contribution to rhs due to nonhomogeneous boundary conditions
bh_hat0 = Function(VQ)
bh_hat0 = BM.matvec(-uh_hat, bh_hat0) # Negative because moved to right hand side
bi_hat0 = bh_hat0[0]

Note that bh_hat0 now contains the part of the right hand side that is due to the non-symmetric part of assembled matrices. The line with D1Y.bc.apply_after(ui_hat[0], True) ensures the known boundary values of the solution are fixed for ui_hat.

The nonlinear right hand side also requires some additional attention. Nonlinear terms are usually computed in physical space before transforming to spectral. For this we need to evaluate the velocity vector on the quadrature mesh. We also need a rank 2 Array to hold the outer product $$\mathbf{u}\mathbf{u}$$. The required arrays and spaces are created as

bh_hat = Function(VQ)

# Create arrays to hold velocity vector solution
ui = Array(W1)

# Create work arrays for nonlinear part
QT = MixedTensorProductSpace([W1, W0])  # for uiuj
uiuj = Array(QT)
uiuj_hat = Function(QT)

The right hand side $$L((\mathbf{v}, q);\mathbf{u}^{k});$$ is computed in its own function compute_rhs as

def compute_rhs(ui_hat, bh_hat):
global ui, uiuj, uiuj_hat, V1, bh_hat0
bh_hat.fill(0)
ui = W1.backward(ui_hat, ui)
uiuj = outer(ui, ui, uiuj)
uiuj_hat = uiuj.forward(uiuj_hat)
bi_hat = bh_hat[0]
#bi_hat = inner(v, div(uiuj_hat), output_array=bi_hat)
bh_hat += bh_hat0
return bh_hat

Here outer() is a shenfun function that computes the outer product of two vectors and returns the product in a rank two array (here uiuj). With uiuj forward transformed to uiuj_hat we can assemble the linear form either as inner(v, div(uiuj_hat) or inner(grad(v), -uiuj_hat). Also notice that the constant contribution from the inhomogeneous boundary condition, bh_hat0, is added to the right hand side vector.

Now all that remains is to guess an initial solution and solve iteratively until convergence. For initial solution we simply set the velocity and pressure to zero and solve the Stokes equations:

from scipy.sparse.linalg import splu
uh_hat, Ai = M.solve(bh_hat0, u=uh_hat, constraints=((2, 0, 0),), return_system=True) # Constraint for component 2 of mixed space
Alu = splu(Ai)
uh_new[:] = uh_hat

Note that the BlockMatrix given by M has a solve method that sets up a sparse coefficient matrix Ai of size $$\mathbb{R}^{3(N_0-2)(N_1-2) \times 3(N_0-2)(N_1-2)}$$, and then solves using scipy.sparse.linalg.spsolve. The matrix Ai is then pre-factored for reuse with splu. Also note that the constraints=((2, 0, 0),) keyword argument ensures that the pressure integrates to zero, i.e., $$\int_{\Omega} pdxdy=0$$. Here the number 2 tells us that block component 2 in the mixed space (the pressure) should be integrated, dof 0 should be fixed, and it should be fixed to 0.

With an initial solution from the Stokes equations we are ready to start iterating. However, for convergence it is necessary to add some underrelaxation $$\alpha$$, and update the solution each time step as

\begin{split}\begin{align*} \hat{\mathbf{u}}^{k+1} &= \alpha \hat{\mathbf{u}}^* + (1-\alpha)\hat{\mathbf{u}}^{k},\\ \hat{p}^{k+1} &= \alpha \hat{p}^* + (1-\alpha)\hat{p}^{k}, \end{align*}\end{split}

where $$\hat{\mathbf{u}}^*$$ and $$\hat{p}^*$$ are the newly computed velocity and pressure returned from M.solve. Without underrelaxation the solution will quickly blow up. The iteration loop goes as follows

converged = False
count = 0
alfa = 0.5
while not converged:
count += 1
bh_hat = compute_rhs(ui_hat, bh_hat)
uh_new = M.solve(bh_hat, u=uh_new, constraints=((2, 0, 0),), Alu=Alu) # Constraint for component 2 of mixed space
error = np.linalg.norm(ui_hat-ui_new)
uh_hat[:] = alfa*uh_new + (1-alfa)*uh_hat
converged = abs(error) < 1e-10 or count >= 10000
print('Iteration %d Error %2.4e' %(count, error))

up = uh_hat.backward()
u, p = up

X = V0.local_mesh(True)
plt.figure()
plt.quiver(X[0], X[1], u[0], u[1])

The last three lines plots the velocity vectors that are shown in Figure Velocity vectors for Re=100. The solution is apparently nice and smooth, but hidden underneath are Gibbs oscillations from the corner discontinuities. This is painfully obvious when switching from Legendre to Chebyshev polynomials. With Chebyshev the same plot looks like Figure Velocity vectors for Re=100 using Chebyshev. However, choosing instead the regularized lid, the solutions will be nice and smooth, both for Legendre and Chebyshev polynomials.

Velocity vectors for Re=100 using Chebyshev

## Complete solver¶

A complete solver can be found in demo NavierStokesDrivenCavity.py.


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# Insert Templated Code Blocks with Yasnippet

Published on Sep 28, 2021, by Junji Zhi

When editing org file with code blocks, sometimes we want to print out the stdout, pipe the output to a file. As a result, we may type :results option near the code block a lot.

This post is about using the power of Yasnippet to streamline the workflow.

Below is my snippet for inserting a scripting code block and set the result to output:

# key: <so
# name: src-script
# --
#+begin_${1:$$(yas-choose-value '("src"))} {2:$$(yas-choose-value '("bash" "ruby" "elisp"))} :results output$0
#+end_\$1

Now if I type <so and press TAB, it expands the block, prompts me to select language, or to enter the name of output file.

Example of going through the <so flow and selecting bash:

#+begin_src bash :results output
echo "test"
#+end_src

Note: There was a small quirks with #begin_src block when inserting as a plaintext, so I create the extra field to disable eval error. It results in an extra enter keystroke and a TAB, but it is good enough for my use.

Share:


text:
# Function satisfying $f^{-1} =f'$

How many functions are there which are differentiable on $(0,\infty)$ and they satisfy the relation $f^{-1}=f'$.

-
What is the motivation for this question? –  JBL Jul 31 '10 at 20:07
Unless I am missing something, this is an elementary differential equations question, and so might be more appropriately asked at math.stackexchange.com . –  Emerton Jul 31 '10 at 20:34
If $f^{-1}$ means $1/f$, then yes, it is an easy differential equations question. On the other hand if $f^{-1}$ is the functional inverse of $f$, then it looks pretty hard. –  Robin Chapman Jul 31 '10 at 21:18
$f^{-1}$ means inverse of $f$ –  S.C. Jul 31 '10 at 21:30
Dear Chandru1, My apologies; I misunderstood the notation (in exactly the way that Robin Chapman suggested). –  Emerton Aug 1 '10 at 1:20

Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum_{k \ge 1} c_k t^k$ we obtain from $(*)$ a recursion formula for the $c_k$, and one can show that $0< c_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.

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I know little about analysis, so my apologies if this is a silly question. Are you solving a version of the problem where the function f is not necessarily defined on (0,∞) but is defined on some interval (0,b) (which is also an interesting problem)? –  Tsuyoshi Ito Aug 1 '10 at 14:16
The function $f$ constructed here is a priori only defined in a neighborhood of the point $a$. –  Christian Blatter Aug 1 '10 at 14:57
Thanks for the clarification. –  Tsuyoshi Ito Aug 1 '10 at 14:59

Wow. I remember that I thought exactly the same problem out of curiosity as a high school student but did not reach an answer. In fact, I was thinking about posting this problem on MathOverflow!

At least it is easy to construct one solution: f(x)=xφφ−1, where φ=(1+√5)/2 is the golden ratio.

Edit: Corrected the calculation. Thanks to Aaron Meyerowitz for spotting the error!

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Which does not answer the question, as far as I can tell. –  Did Nov 18 '12 at 12:53
@Didier Piau: It clearly does not. In case anyone is wondering, the asker posted the answer shortly after Christian Blatter posted his related analysis, and deleted it after I asked him if he had posted the question knowing the answer. –  Tsuyoshi Ito Nov 20 '12 at 1:04


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# 5/5/4

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5/5/4 [#permalink]  14 May 2018, 14:10
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Question Stats:

87% (00:10) correct 12% (00:00) wrong based on 8 sessions
$$\frac{5}{\frac{5}{4}}$$

A. $$\frac{1}{5}$$

B. $$\frac{1}{4}$$

C. $$4$$

D. $$5$$

E. $$\frac{25}{4}$$
[Reveal] Spoiler: OA

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Kudos [?]: 18 [1] , given: 2

Re: 5/5/4 [#permalink]  15 Jun 2018, 07:59
1
KUDOS
I got the answer C. Although it looks funky, the equation is dividing 5 by 5/4. Dividing by a fraction can also be written as multiplying by its multiplicative inverse. So: 5 / 5/4 is the same as 5 * 4/5, which is simplified to 20/5, or 4.
Re: 5/5/4   [#permalink] 15 Jun 2018, 07:59
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text:
5
Q:

# The compound interest accrued on an amount of Rs.25,500 at the end of three years is Rs.8,440.5. What would be the simple interest accrued on the same amount at the same rate in the same period?

 A) Rs.4,650 B) Rs.5,650 C) Rs.6,650 D) Rs.7,650

Explanation:
Subject: Verbal Reasoning - Mental Ability
Exam Prep: Bank Exams
Job Role: Bank PO
Q:

What would be the next number in the series

9, 8, 18, 6, 36, 4, 72, 2, 144, ?

 A) 288 B) 1 C) 0 D) 576

Explanation:

Here there are combination of two series.

1) 9, 18, 36, 72, 144

2) 8, 6, 4, 2, 0

$\fn_jvn&space;\small&space;\therefore$ The next number in the given series is 0.

Filed Under: Verbal Reasoning - Mental Ability
Exam Prep: Bank Exams
Job Role: Bank Clerk

7 120
Q:

The following groups of alphabets form a certain pattern with regard to their position in the English alphabetical series. Based upon the pattern, which of the following five alternatives shall replace the question mark?

 A) OOP B) PPP C) SNO D) QLM

Explanation:

Filed Under: Verbal Reasoning - Mental Ability
Exam Prep: Bank Exams
Job Role: Bank PO

4 480
Q:

813  967  429  295  648

If the positions of the second and third digits in each of the numbers are interchanged , which of the following will be the sum of the sum of the three digits of the highest number?

 A) 12 B) 15 C) 16 D) 22

Explanation:

Filed Under: Verbal Reasoning - Mental Ability
Exam Prep: Bank Exams
Job Role: Bank PO

3 414
Q:

How many such digits are there in the number 84315269 each of which is as far away from the beginning of the number as when the digits are rearranged in ascending order?

 A) None B) one C) two D) three

Explanation:

Filed Under: Verbal Reasoning - Mental Ability
Exam Prep: Bank Exams
Job Role: Bank PO

6 600
Q:

Among P,Q, R, S, and T,each having a different weight, R is heavier than only P. S is lighter than Q and heavier than T. Who among them is the heaviest?

 A) Q B) P C) S D) Data inadequate


text:
NEW
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Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### WB JEE 2010

In a $$\Delta$$ABC, $$2ac\sin \left( {{{A - B + C} \over 2}} \right)$$ is equal to

A
a2 + b2 $$-$$ c2
B
c2 + a2 $$-$$ b2
C
b2 $$-$$ a2 $$-$$ c2
D
c2 $$-$$ a2 $$-$$ b2

## Explanation

$$2ac\sin \left( {{{A - B + C} \over 2}} \right) = 2ac\,.\,\sin \left( {{{\pi - B - B} \over 2}} \right)$$

$$= 2ac\cos B = 2ac\,.\,{{{c^2} + {a^2} - {b^2}} \over {2ca}} = {c^2} + {a^2} - {b^2}$$

2

### WB JEE 2010

If $${{\cos A} \over 3} = {{\cos B} \over 4} = {1 \over 5}, - {\pi \over 2} < A < 0, - {\pi \over 2} < B < 0$$, then value of $$2\sin A + 4\sin B$$ is

A
4
B
$$-$$ 2
C
$$-$$ 4
D
0

## Explanation

$$\because$$ $$- {\pi \over 2} < A,B < 0$$

$$\therefore$$ $$\sin A < 0,\sin B < 0$$

$$\because$$ $${{\cos A} \over 3} = {{\cos B} \over 4} = {1 \over 5}$$

$$\therefore$$ $$\cos A = {3 \over 5},\cos B = {4 \over 5}$$

$$\Rightarrow \sin A = - \sqrt {1 - {{\cos }^2}A} = - \sqrt {1 - {9 \over {25}}} = - {4 \over 5}$$

$$\Rightarrow \sin B = - \sqrt {1 - {{\cos }^2}B} = - \sqrt {1 - {{16} \over {25}}} = - {3 \over 5}$$

$$\therefore$$ $$2\sin A + 4\sin B = - {8 \over 5} - {{12} \over 5} = - 4$$

3

### WB JEE 2010

In a right-angled triangle, the sides are a, b and c, with c as hypotenuse, and c $$-$$ b $$\ne$$ 1, c + b $$\ne$$ 1. Then the value of $$({\log _{c + b}}a + {\log _{c - b}}a)/(2{\log _{c + b}}a \times {\log _{c - b}}a)$$ will be

A
2
B
$$-$$ 1
C
$${1 \over 2}$$
D
1

## Explanation

Given expression $$= {1 \over 2}\left[ {{1 \over {{{\log }_{c - b}}a}} + {1 \over {{{\log }_{c + b}}a}}} \right]$$

$$= {1 \over 2}[{\log _a}(c - b) + {\log _a}(c + b)]$$

$$= {1 \over 2}{\log _a}({c^2} - {b^2}) = {1 \over 2}{\log _a}{a^2} = 1$$ [$$\because$$ $${c^2} = {a^2} + {b^2}$$]

4

### WB JEE 2010

In a triangle PQR, $$\angle$$R = $$\pi$$/2. If $$\tan \left( {{P \over 2}} \right)$$ and $$\tan \left( {{Q \over 2}} \right)$$ are roots of ax2 + bx + c = 0, where a $$\ne$$ 0, then which one is true?

A
c = a + b
B
a = b + c
C
b = a + c
D
b = c

## Explanation

$$\angle R = {\pi \over 2} \Rightarrow P + Q = {\pi \over 2}$$

$$\Rightarrow \tan \left( {{P \over 2} + {Q \over 2}} \right) = \tan {\pi \over 4} \Rightarrow {{\tan {P \over 2} + \tan {Q \over 2}} \over {1 - \tan {P \over 2} + \tan {Q \over 2}}} = 1$$

$$\Rightarrow - {b \over a} = 1 - {c \over a}$$ $$\therefore$$ $$c = a + b$$

[$$\because$$ $$\tan {P \over 2},\,\tan {Q \over 2}$$ are roots of $$a{x^2} + bx + c = 0$$

$$\therefore$$ $$\tan {P \over 2} + \tan {Q \over 2} = - {b \over a}$$ and $$\tan {P \over 2}.\,\tan {Q \over 2} = {c \over a}$$]

### Joint Entrance Examination

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text:
# downloading home version for mac

This is I think a trivial question but I did check against those already asked.

I am downloading Mathematica (home use). Very slow connection. So my Mac displayed the progress bar but this morning the bar is replaced in the downloads window with "mounting failed." When I double-click this icon it informs me the program can't be accessed because it is still downloading.

When I right-clicked "get info" on the dmg file it went from 2.07G to 4 KB over the course of a few minutes.

a. Has the download stopped? Been reversed? b. Is it possible to get the progress bar back to see what's happening?

Thanks for any advice.

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I guess this is more of a question to be asked to the fellows of Wolfram- support. (or your ISP if this is just a question of connection quality) –  Sascha Dec 1 '12 at 14:32
It's a weekend, so I went out on a limb. It's a support issue but if someone has an answer it might help others similarly situated. It's not about the connection. –  daniel Dec 1 '12 at 14:44
Daniel, welcome to Mathematica.SE. As Sascha says this is sort of off-topic but I'll leave it open for a while since you're anxious for an answer. Once you have a bit higher "reputation" you can make use of the chat room for this kind of question. –  Mr.Wizard Dec 1 '12 at 14:52
Okay thanks and Sascha's point taken. –  daniel Dec 1 '12 at 14:53
Mr. Wizard already indicated this question was slated for eventual closure. No need to spend precious moments downvoting. –  daniel Dec 1 '12 at 15:16

## 1 Answer

Here is a possible solution which might work for you too. When you use the Home use edition, then I assume you have another Mathematica licence at work or in university. What you can do is, that you download the dmg over a VPN connection with the fast speed of your office/university.

After that, you can use scp or an ftp or something similar to transfer it to your home at any speed you like without being afraid, that the browser or user portal does something weird.

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No in fact I do not have another Mathematica license. Not sure why you assume that. –  daniel Dec 1 '12 at 15:05
He's talking about the Home Edition, not a home-use license. –  Brett Champion Dec 1 '12 at 15:10
@daniel As Brett commented: I assumed that because you wrote home use which is very different from Home edition. Sorry, but my answer still can help you, if you have access to VPN wich more bandwidth. –  halirutan Dec 1 '12 at 16:01
Your answer is helpful and I had something like that in mind as a possibility. Thanks. –  daniel Dec 1 '12 at 19:44
I was able to download using your suggestion. Thanks. –  daniel Dec 4 '12 at 0:59


text:
Please go over all steps needed to complete a, b, and c. Thank you so much.

Image text transcribed for accessibility: For the schematic shown below, derive an expression for lout. Ignore base current and base width modulation. Assume that the transistors are the same size and operate in forward active. Derive an expression for the maximum value of lout.(Hint: Solve for ln by setting ). Plot lout as a function of lin from 1 muA. Assume that R = 10kOhm.


text:
## minimum bounding ellipsoid example does not work in 12.1.1.0

I am surprised that WRI never checks previous examples before shipping new versions. The following example is mentioned and highlighted on the webpage. It does not work in 12.1.1.0 and as far as I remember it used to work in older version 12.0 perhaps. Does someone else has trouble running this example with a fresh kernel. I have filed a bug report nevertheless Case number 4696657

https://www.wolfram.com/language/12/convex-optimization/minimal-bounding-ellipsoid.html?product=mathematica

I get this output below:

## Example of Dependent Types?

Say you have 3 objects, a global `MemoryStore`, which has an array of `MemorySlabCache` objects, and each `MemorySlabCache` has an array of `MemorySlab` objects. Sort of like this:

``class MemoryStore {   caches: Array<MemorySlabCache> = [] }  class MemorySlabCache {   size: Integer   slabs: Array<MemorySlab> = [] }  class MemorySlab {    } ``

But the thing is, this doesn’t capture everything. It also needs to capture the fact that each `MemorySlabCache` has a size, which is used to tell what size the `MemorySlab` objects are it contains. So it’s more like this:

``class MemoryStore {   caches: Array<MemorySlabCache> = [] }  class MemorySlabCache {   size: Integer   slabs: Array<MemorySlab<size>> = [] }  class MemorySlab<size: Integer> {    } ``

Then we create our caches:

``let 4bytes = new MemorySlabCache(size: 4) let 8bytes = new MemorySlabCache(size: 8) ... let 32bytes = new MemorySlabCache(size: 32) ... store.caches.push(4bytes, 8bytes, ..., 32bytes, ...) ``

Does this count as a "dependent type", "a type whose definition depends on a value"? Since the type of the `Array<MemorySlab<size>>` is dependent on the value assigned to the `size` field on `MemorySlabCache`. If not, what is this? What would make it into an example of dependent types?

## What is the earliest example of a varint monster in a published adventure?

In an answer to another question I made the point that using non-standard variants of published monsters has been common practice since the early days of RPGs. This was based on my own experience, but I am certain I have seen the practice in published aventures. What is the earliest instance of a variant monster in a published adventure?

How I am defining the term "variant monster":

• A variant monster must be based on a published, official monster but differs from the official monster in a significant way. By significant way, I mean any change in physical statisics (including hit points outside the range normally possible) or a change in the monster’s physical description that might cause players to misidentify it or not notice it (example: red slime with green slime stats).
• Includes any monster with abilities not accounted for in its official description, such as a spell-casting medusa or a psionic basilisk.
• Includes any monster that behaves in a way that would normally be impossible for that monster (example: a sentient iron golem that acts on its own free will). But this does not include a creature that has been turned into a monster and still behaves as its natural self (example: a gnome trapped inside an iron golem’s body).
• Does not include new monsters that are not based on existing oficial monsters.
• Does not include new types of old monsters that receive a full description in the module and possibly later were published as monsters in supplements. Example: the drow (mentioned briefly, not statted in the 1e Monster Manual) was fully described and statted in the appendix of the first adventure in which they appeared (G3 Hall of the Fire Giant King) and was later published in the 1e Fiend Folio, so drow is not a variant monster.
• Does not include monsters that are physically and statistically the same as their official type but behave in an unusual way (e.g., a good-aligned red dragon or a cunning, educated ogre).
• Does not include monsters equipped in an unusual way.

## How do I see what services are running on a server (for example google dot com) [closed]

I used nmap and I get that these services are running on there: ssh, http etc, but what if I want to check for a specific service like icmp or http over tls? How do I do it?

## Using an example to comprehend why “safely” erasing a drive yields better results than filling it up with meaningless data

A hypothetical 1GB USB stick is full of sensitive documents/images/etc. and it is not encrypted.

The owner wishes to discard it and is aware of having to safely erase it first.

There are several tools and utilities to do this. Some can be configured to do it “faster yet less safely”, others do it “slower but more safely”.

As opposed to have it erased using all the different ways known to do this, the owner chooses to simply drag all the current items to the recycle bin and then paste one 1GB (~2-hour) black screen movie file to the USB stick.

Again, no fancy erase utilities are used. The USB stick is then discarded.

If it falls into the wrong hands, can any of the sensitive files (that filled the stick before the movie file was pasted) be retrieved?

(1) If no, why do complex hard drive erase utilities exist? Some of them feature “safe” erase procedures that take houuurs, when simply filling a soon to be discarded HD with meaningless files, can do the job?

(2) If yes, how can 2GB (movie file + sensitive files) co-exist in a 1GB stick? Seems to me like the only logical explanation is (a) the movie file was in fact less than 1GB, (b) the USB stick was secretly larger than 1GB as stated, or (c) the movie file was copy-pasted only partially and the owner did not notice.

## Potential Example of an algorithm failure

I was looking an algorithm to solve a problem of finding whether and array contains a quadruple with sum = k,(k is input) mentioned at GeeksforGeeks. In one solution the approach mentioned is below,which i think i not correct:

1. Store all possible two elements pairs of array in some another auxiliary array and also store i,j indices of the elements.There will be $$C(n,2)$$ pairs.
2. Sort the auxiliary array.
3. Take two pointers,left=0 and right=n-1 to auxiliary array.

3.1. If auxiliary[left]+auxiliary[right]==k and there is no common index in elements represented by left and right, we return true.

3.2 else If auxiliary[left]+auxiliary[right]<k, we do left++

3.3 else we do right– 3.4 And this goes in continues loop until left<right or we get element at 3.1

Now my doubt is that in case sum of left and right elements=k and there is some common element between left and right,then this algo will do right–. I want to understand that why we do right– here and no left++ and the other way.I think there must be some example where this scenario will make algorithm fail but i am not able to produce that.Can some tell that whether it will indeed fail or prove that it is always correct?

## example of bad case for perptron algorithm

I am trying to think of an example using 3 data points where the margin produced by a perceptron algorithm can be arbitrarily bad. But I got stuck on how to approach this example.

## What is a make sense (meaningful) example of language that an unrestricted grammar could generate?

I have learned that:

1. Unrestricted grammar is used to define (or describe) a formal language.
2. Unrestricted grammar is used to define recursively enumerable set [https://en.wikipedia.org/wiki/Recursively_enumerable_set][1].

I’d like to find a meaningful example for #1 case which is similar to below context sensitives grammar example to learn the purpose of unrestricted grammar. I could find a meaningful example for context sensitives grammar but I could not find a one for the unrestricted grammar yet. Could you help me?

Language for “Network racing game record” with below record instances:

Mr. Bean Male Player 1

Ms. Emma Female Player 2

Mr. Hải n/a Computer 3

Ms. Tú n/a Computer 4

Production rule:

S ⟶ Title Name TAB Sex UserType TAB Rank

Title WomanName ⟶ "Ms. " WomanName

Title ManName ⟶ "Mr. " ManName

WomanName TAB Sex "Player" ⟶ WomanName TAB "Female" "Player"

ManName TAB Sex "Player" ⟶ ManName TAB "Male" "Player"

Name TAB Sex "Computer" ⟶ Name TAB "n/a" "Computer"

Name ⟶ WomanName

Name ⟶ ManName

Sex ⟶ "Male"

Sex ⟶ "Female"

UserType ⟶ "Player"

UserType ⟶ "Computer"

Rank ⟶ "1"

Rank ⟶ "2"

Rank ⟶ "3"

Rank ⟶ "4"

WomanName ⟶ "Emma"

WomanName ⟶ "Tú"

ManName ⟶ "Bean"

ManName ⟶ "Hải"

TAB ⟶ "\t"

## Can you give an example of a problem in \$EXP^{RE}\$ but not In \$P^{RE}\$

Can you give an example of a problem in $$EXP^{RE}$$ but not In $$P^{RE}$$?

## Are My Answers to This Hamming Code Example Correct?

I have attempted to answer this question on Hamming code but i am very new to it and would appreciate feedback on my answers, Thank you!


text:
# $q$-bic Hypersurfaces

### Raymond Cheng (Columbia)

 Fri Jul 16, 19:00-20:00 (5 weeks from now) Login required for livestream access

Abstract: Let’s count: 1, $q$, $q+1$; here, $q$ is a power of a prime $p$. In this talk, I will sketch an analogy between the geometry of a class of hypersurfaces over a field of positive characteristic $p$, which I call $q$-bic hypersurfaces, and the geometry of low degree hypersurfaces, such as quadrics and cubics, over the complex numbers. For instance, a smooth $q$-bic threefold has a smooth Fano surface of lines, and the intermediate Jacobian of the threefold is isogenous to the Albanese of the Fano surface.

algebraic geometry

Audience: researchers in the topic