text:
Home

>

English

>

Class 12

>

Physics

>

Chapter

>

Jee Mains

>

In conducting wire of radius 5...

# In conducting wire of radius 5mm, resistivity p-1.1xx10^(-8)Omega//m and current of 5A is flowing. Drift velocity of free electron is 1.1xx10^(-3)m//s find out mobility of free elctron.

Updated On: 27-06-2022

Text Solution

1.57m^(2) volt/sec.1.25m^(2) volt/sec.1.2m^(2) volt/sec.2 m^(2) volt/sec.

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

Hindi speaking questions Ada in a conducting wire of radius 5 mm christava 30 1.1 into 10 power minus 8 ohm and the current is 5 ampere is flowing in the Different cities given our free electron which is 1.1 into 10 power minus 3 we need to find out the mobility of electron we know the formula of mobility of electron is given velocity velocity upon electric field Suvidha is a drift velocity which is given in the question ladiss Vidhi one point 1.1 into 10 ki power minus 3 metre per second and and current is here that is equal to 5 a and radius

5 Mili metre which is 5 into 10 to the power minus 3 and the resistivity is given 1.1 into 10 power minus 8 ohm per metre to this is the reason given in the question and we need to find out the mobility the Drift velocity we have but electric field we don't have so we can explain the electric field in terms of conductivity and current density so that IT field is equal to resistivity in to current upon area so we will put this equation in equation this this equation that is equal to very different city upon row into a phone area

we have all the values of this we have the given City we have we have difficulty with average tivity map current and for the area it is Pi R Square and R is 5 into 10 ki power minus 3 from here is equal to buy into 25 into 10 power minus 6 putting all values to very 1.1 into 10 to power minus 3 upon resistivity P han 1.1 into 10 power minus 8 into current we have 5 a upon area which is bhai into 25 into 10 power minus 6 now equal to

this will be go disturb will go upside to 1.1 into 10 to the power minus 3 into 5 into 25 - 6 upon 1.1 into 10 power minus 8 9 calculating this we will get the mobility which is equal to 1.57 m square v per second normal AC which option in word sweety's option 1.57 M kal V per S hockey weather


text:
# In light-cone gauge, are open strings born with excitations of massless gauge boson?

+ 2 like - 0 dislike
52 views

For bosonic strings, on the wordsheet there are bosonic fields $X^\mu(\tau,\sigma),\ \mu=0,1,...,D-1$. In the light-cone coordinate, we take,
$$X^+=\frac{1}{\sqrt{2}}(X^0+X^1),\\ X^-=\frac{1}{\sqrt{2}}(X^0-X^1).$$
Within the light-cone gauge, we can fix for the string that $X^+\sim\tau$. Therefore $X^+$ plays the role of time. The non-trivial point for strings is that $X^-$ can be solved in terms of $X^+$ and $X^I$ with $I=2,...,D-1$ and turns out to be $\sigma-$independent. That means that there are no dynamical degrees of freedom for the field $X^-$. Therefore, there are only  transverse oscillations on the strings. For the mode-1 excitations of the string, we have $D-2$ independent creation operators corresponding to the transverse oscillations:
$$a^I|\text{ground state}\rangle,\ \text{for}\ I=2,...,D-1.$$
All these states carry the spacetime indices and there are $D-2$ states in total. Thus it must represent a massless gauge field. It turns out that, indeed, we can select the dimension of spacetime such that the above states are really massless.

My question is that, what is the insightful way to see that string theory was born with massless gauge states? I guess the fact that $X^-$ has no degrees of freedom comes from the reparametrization invariance. Can somebody provide an insightful explanation?

What if it turns out that, we can not set the mass for above states to be vanishing for any dimension $D$ of spacetime (this is possible, since the the dimension $D$ of spacetime must be integer which is a rather strong constraint). In string theory, we now know that, it is not the case. Is it simply a coincidence?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.

format: plain


text:
Other molecules like fullerenes?

When I first heard about buckminsterfullerene, I thought it was a very fascinating molecule, with its interesting complex ring structure and pseudo-symmetry. I wonder though, can you get similar structures with other atoms? Maybe neighbors of similar size, like boron, nitrogen or oxygen? Or maybe silicon, one row down? What would their chemistry be like?

• This looks kinda too broad. There are various similar molecules. – Mithoron Oct 22 '17 at 16:25
• Check out en.wikipedia.org/wiki/Borospherene – Mithoron Oct 22 '17 at 16:27
• Simple P4O10 has a hollow, if not spherical structure. en.wikibooks.org/wiki/Introduction_to_Inorganic_Chemistry/… – DrMoishe Pippik Oct 22 '17 at 21:27
• So does $\ce{P4O6}$. Hexamethylenetetramine is isoelectronic and isostructural with $\ce{P4O6}$ apart from the outer hydrogen atoms. – Oscar Lanzi Oct 22 '17 at 23:34
• There are plenty heteroatomic molecular clusters with even more complex structures. Or do you want a single element assembly only? – andselisk Oct 23 '17 at 2:45

The trick is that $$\ce{B-N}$$ has very similar properties to $$\ce{C=C}$$, with very little difference in molecular weight, and so many analogous compounds can be built using either monomer. However, boron nitride is different enough to have unique applications in certain areas. Cool, right?


text:
# MA3H2 Content

Content:

Let us briefly explain the mathematical setting using the example of Bernoulli percolation. Percolation is a simple probabilistic model which exhibits a phase transition. The simplest version of percolation takes place on $\mathbb{Z} ^2$ , which we view as a graph with edges between neighbouring vertices. All edges of $\mathbb{Z}^2$ are, independently of each other, chosen to be open with probability $p$ and closed with probability $1-p$ . A basic question in this model is What is the probability that there exists an open path from the origin to the exterior of the square $S_n=[-n,n]^2$?' A limit as $n\to\infty$of the question raised above is What is the probability that there exists an open path from $0$to infinity?' This probability is called the percolation probability and is denoted by $\theta(p)$ . Clearly $\theta(0)=0$ and $\theta(1)=1$, since there are no open edges at all when $p=0$ and all edges are open when $p=1$ . For some models there is a $0<p_{c}<1$ such that the global behaviour of the system is quite different for $p<p_{c}$ and for $p>p_{c}$. Such a sharp transition in global behaviour of a system at some parameter value is called a phase transition or a critical phenomenon, and the parameter value at which the transition takes place is called a critical value.

Books:

We will not follow a particular book. However, there are several sets of lecture notes used in the course, which can be downloaded from the Moodle page. The list below is a selection of books for a much deeper study of the subject.

H.O. Georgii: Stochastics: introduction to probability theory and statistics, de Gruyter (2008). [basic introduction to stochastics and Markov chains (discrete time)]

J. Norris: Markov chains, Cambridge University Press [standard reference treating the topic with mathematical rigor and clarity, and emphasizing numerous applications to a wide range of subjects]

G. Grimmett, D. Stirzaker: Probability and Random Processes, OUP Oxford (2001) [chapter 6 on Markov chains]

G. Grimmett: Probability on Graphs, Cambridge University Press (2010). [Available Online, contains a nice introduction to processes on graphs and percolation]

B. Bollabás, O. Riordan: Percolation, Cambridge University Press (2006). [a modern treatment of percolation. The introduction and the chapter on basic techniques are relevant for the lecture]

G. Grimmett: Percolation, 2nd ed., Springer (1999). [the standard reference on percolation. It contains much more than covered in the lecture. The first two chapters are relevant for the lecture]


text:
Change search
Refine search result
1 - 26 of 26
Cite
Citation style
• apa
• ieee
• modern-language-association-8th-edition
• vancouver
• Other style
More styles
Language
• de-DE
• en-GB
• en-US
• fi-FI
• nn-NO
• nn-NB
• sv-SE
• Other locale
More languages
Output format
• html
• text
• asciidoc
• rtf
Rows per page
• 5
• 10
• 20
• 50
• 100
• 250
Sort
• Standard (Relevance)
• Author A-Ö
• Author Ö-A
• Title A-Ö
• Title Ö-A
• Publication type A-Ö
• Publication type Ö-A
• Issued (Oldest first)
• Created (Oldest first)
• Last updated (Oldest first)
• Disputation date (earliest first)
• Disputation date (latest first)
• Standard (Relevance)
• Author A-Ö
• Author Ö-A
• Title A-Ö
• Title Ö-A
• Publication type A-Ö
• Publication type Ö-A
• Issued (Oldest first)
• Created (Oldest first)
• Last updated (Oldest first)
• Disputation date (earliest first)
• Disputation date (latest first)
Select
The maximal number of hits you can export is 250. When you want to export more records please use the Create feeds function.
• 1.
Lunds universitet.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. IMFM, Ljubljana, Slovenia.
Comment on "self-indexed sort"1996In: SIGPLAN notices, ISSN 0362-1340, E-ISSN 1558-1160, Vol. 31, no 8, p. 40-41Article in journal (Other academic)
• 2.
University of Waterloo.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Architecture of KRPAN1996In: Elektrotehniski Vestnik, ISSN 0013-5852, Vol. 63, no 2, p. 65-68Article in journal (Refereed)

In the paper we address the operation of an agency that provides information in a newspaper-like form on the information highway. We describe the software architecture and the physical layout of KRPAN, a kernel that provides the support necessary to operate such an agency. KRPAN is a distributed system which employs intelligent caching to improve space and network utilization. The implementation of KRPAN relies on standardized formats of data which permits usage of commonly available tools. At the end we touch legal and ethical questions and describe how KRPAN helps to solve them.

• 3.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. Luleå tekniska universitet. Luleå University of Technology, Department of Engineering Sciences and Mathematics, Mathematical Science.
Minimum curvature variation B-splines: validation of a path-planning model2004Report (Other academic)
• 4.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. University of Primorska, Slovenia. . Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. Luleå tekniska universitet. Luleå University of Technology, Department of Engineering Sciences and Mathematics, Mathematical Science.
Planning smooth and obstacle-avoiding b-spline paths for autonomous mining vehicles2010In: IEEE Transactions on Automation Science and Engineering, ISSN 1545-5955, E-ISSN 1558-3783, Vol. 7, no 1, p. 167-172Article in journal (Refereed)

We study the problem of automatic generation of smooth and obstacle-avoiding planar paths for efficient guidance of autonomous mining vehicles. Fast traversal of a path is of special interest. We consider four-wheel four-gear articulated vehicles and assume that we have an a priori knowledge of the mine wall environment in the form of polygonal chains. Computing quartic uniform B-spline curves, minimizing curvature variation, staying at least at a proposed safety margin distance from the mine walls, we plan high speed paths.

• 5.
Luleå tekniska universitet.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Formal verification of a trie-based data structure2005Conference paper (Refereed)
• 6.
School of Computer Science, Carleton University, Ottawa.
University of Karlskona/Ronneby. Massachusetts Institute of Technology, Laboratory for Computer Science, Cambridge. Department of Computer Science, Hong Kong University of Science and Technology. Department of Computer Science, University of Waterloo, Ontario. School of Computer Science, Carleton University, Ottawa. Department of Computer Science, University of Waterloo, Ontario.
Online routing in convex subdivisions2002In: International journal of computational geometry and applications, ISSN 0218-1959, Vol. 12, no 4, p. 283-295Article in journal (Refereed)

We consider online routing algorithms for finding paths between the vertices of plane graphs. We show (1) there exists a routing algorithm for arbitrary triangulations that has no memory and uses no randomization, (2) no equivalent result is possible for convex subdivisions, (3) there is no competitive online routing algorithm under the Euclidean distance metric in arbitrary triangulations, and (4) there is no competitive online routing algorithm under the link distance metric even when the input graph is restricted to be a Delaunay, greedy, or minimum-weight triangulation

• 7. Bose, Prosenjit
Luleå tekniska universitet.
Online routing in convex subdivisions2000In: Algorithms and Computation: 11th International Conference, ISAAC 2000 Taipei, Taiwan, December 18-20, 2000 Proceedings / [ed] D.T. Lee; Shang-Hua Teng, Berlin: Encyclopedia of Global Archaeology/Springer Verlag, 2000, p. 47-59Conference paper (Refereed)

We consider online routing algorithms for finding paths between the vertices of plane graphs. We show (1) there exists a routing algorithm for arbitrary triangulations that has no memory and uses no randomization, (2) no equivalent result is possible for convex subdivisions, (3) there is no competitive online routing algorithm under the Euclidean distance metric in arbitrary triangulations, and (4) there is no competitive online routing algorithm under the link distance metric even when the input graph is restricted to be a Delaunay, greedy, or minimum-weight triangulation.

• 8. Brodnik, Andrej
Luleå tekniska universitet.
Sub-linear decoding of Huffman codes almost in-place1998In: Preprint, IMFM , 1998Conference paper (Refereed)

We present a succinct data structure storing the Huffman encoding that permits sublinear decoding in the number of transmitted bits. The size of the extra storage except for the storage of the symbols in the alphabet for the new data structure is O(l log N) bits, where l is the longest Huffman code and N is the number of symbols in the alphabet. We present a solution that typically decodes texts of sizes ranging from a few hundreds up to 68 000 with only one third to one fifth of the number of memory accesses of that of regular Huffman implementations. In our solution, the overhead structure where we do all but one memory access to, is never more than 342 bytes. This will with a very high probability reside in cache, which means that the actual decoding time compares even better

• 9. Brodnik, Andrej
Luleå tekniska universitet. Department of Computer Science, University of Waterloo, Ontario. Department of Computer Science, University of Waterloo, Ontario. Department of Computer Science, Princeton University.
Resizable arrays in optimal time and space1999In: Algorithms and Data Structures: 6th International Workshop, WADS'99 Vancouver, Canada, August 11-14, 1999 Proceedings / [ed] Frank Dehne, Berlin: Encyclopedia of Global Archaeology/Springer Verlag, 1999, p. 37-48Conference paper (Refereed)

We present simple, practical and efficient data structures for the fundamental problem of maintaining a resizable one-dimensional array, A[l..l + n - 1], of fixed-size elements, as elements are added to or removed from one or both ends. Our structures also support access to the element in position i. All operations are performed in constant time. The extra space (i.e., the space used past storing the n current elements) is O(√n) at any point in time. This is shown to be within a constant factor of optimal, even if there are no constraints on the time. If desired, each memory block can be made to have size 2k - c for a specified constant c, and hence the scheme works effectively with the buddy system. The data structures can be used to solve a variety of problems with optimal bounds on time and extra storage. These include stacks, queues, randomized queues, priority queues, and deques.

• 10. Brodnik, Andrej
Blekinge Institute of Technology, Karlskrona. Department of Computer Science, Rutgers University, New Brunswick, NJ. Luleå tekniska universitet. School of Computer Science, University of Waterloo, Waterloo, Ontario.
Worst case constant time priority queue2005In: Journal of Systems and Software, ISSN 0164-1212, E-ISSN 1873-1228, Vol. 78, no 3, p. 249-156Article in journal (Refereed)

We present a new data structure of size 3M bits, where M is the size of the universe at hand, for realizing a discrete priority queue. When this data structure is used in combination with a new memory topology it executes all discrete priority queue operations in O(1) worst case time. In doing so we demonstrate how an unconventional, but practically implementable, memory architecture can be employed to sidestep known lower bounds and achieve constant time performance.

• 11. Brodnik, Andrej
Luleå tekniska universitet. Luleå tekniska universitet. Department of Computer Science, University of Waterloo, Ontario.
Worst case constant time priority queue2001In: Proceedings of the twelfth annual ACM-SIAM symposium on Discrete algorithms, IEEE Communications Society, 2001, p. 523-528Conference paper (Refereed)

We present a new data structure of size 3M +o(M) bits for solving the "discrete priority queue " problem. When this data structure is used in combination with a new memory topology it provides an O(1) worst case time solution.

• 12. Brodnik, Andrej
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Interoperability and semantic filtering2006In: Je-LKS: Journal of e-Learning and Knowledge Society, ISSN 1826-6223, E-ISSN 1971-8829, Vol. 2, no 2, p. 165-175Article in journal (Refereed)

Interoperability is often seen only as a technological problem related to reusability. We would like to give interoperability a new role in the structure of the learning environment and in collaborative activities. For us interoperability is also connected with didactical, semiotic and cultural aspects. It improves the interaction between the materials produced with different tools, the construction of kaleidoscopic artifacts and the reﬂ ection on the learning process.

• 13.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Bitwise Operations under RAMBO2006Report (Other academic)

In this paper we study the problem of computing w-bit bitwise operations using only O(1) memory probes. We show that under the RAM model there exists a Omega(2^w) space lower bound while under the RAMBO model this space bound goes down to O(w) bits. We present algorithms that use four different RAMBO memory topologies to perform bitwise boolean operations and shift operations.

• 14. Brodnik, Andrej
Luleå tekniska universitet.
Multiprocess time queue2001In: Algorithms and Computation: 12th International Symposium, ISAAC 2001, Christchurch, New Zealand, December 19-21, 2001. / [ed] Peter Eades; Tadao Takaoka, Berlin: Encyclopedia of Global Archaeology/Springer Verlag, 2001, p. 599-609Conference paper (Refereed)

We show how to implement a bounded time queue for two different processes. The time queue is a variant of a priority queue with elements from a discrete universe. The bounded time queue has elements from a discrete bounded universe. One process has time constraints and may only spend constant worst case time on each operation while the other process may spend more time. The time constrained process only has to be able to perform some of the time queue operations while the other process has to be able to perform all operations. We show how to do a deamortization of the deleteMin cost and to provide mutual exclusion for the parts of the data structure that both processes maintain.

• 15.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. Faculty of Education, University of Primorska. Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
An O(1) solution to the prefix sum problem on a specialized memory architecture2006In: PFourth IFIP International Conference on Theoretical Computer Science - TCS 2006: IFIP 19th World Computer Congress, TC-1, Foundations of Computer Science, August 23-24, 2006, Santiago, Chile / [ed] Gonzalo Navarro; Leopoldo Bertossi; Yoshiharu Kohayakawa, New York: Encyclopedia of Global Archaeology/Springer Verlag, 2006, p. 103-114Conference paper (Refereed)

In this paper we study the Prefix Sum problem introduced by Fredman. We show that it is possible to perform both update and retrieval in O(1) time simultaneously under a memory model in which individual bits may be shared by several words. We also show that two variants (generalizations) of the problem can be solved optimally in Θ(lgN) time under the comparison based model of computation.

• 16. Brodnik, Andrej
BRICS, Centre of the Danish National Research Foundation, University of Aarhus. Department of Computer Science, University of Waterloo, Ontario.
Trans-dichotomous algorithms without multiplication: some upper and lower bounds1997In: Algorithms and Data Structures: 5th International Workshop, WADS'97 Halifax, Nova Scotia, Canada August 6-8, 1997 Proceedings / [ed] Frank Dehne; Andrew Rau-Chaplin; Jörg-Rüdiger Sack; Roberto Tamassia, Berlin: Encyclopedia of Global Archaeology/Springer Verlag, 1997, p. 426-439Conference paper (Refereed)

We show that on a RAM with addition, subtraction, bitwise Boolean operations and shifts, but no multiplication, there is a transdichotomous solution to the static dictionary problem using linear space and with query time log n(log log n)1+0(1). On the way, we show that two w-bit words can be multiplied in time (log w)1+0(1) and that time (log w) is necessary, and that (log log w) time is necessary and sufficient for identifying the least significant set bit of a word.

• 17. Brodnik, Andrej
Membership in constant time and almost-minimum space1999In: SIAM journal on computing (Print), ISSN 0097-5397, E-ISSN 1095-7111, Vol. 28, no 5, p. 1627-1640Article in journal (Refereed)

This paper deals with the problem of storing a subset of elements from the bounded universe $\mathcal{M} = \{0, \ldots, M-1\}$ so that membership queries can be performed efficiently. In particular, we introduce a data structure to represent a subset of $N$ elements of $\mathcal{M}$ in a number of bits close to the information-theoretic minimum, $B = \left\lceil \lg {M\choose N} \right\rceil$, and use the structure to answer membership queries in constant time.

• 18.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
University of Waterloo, Ontario.
Neighbours on a grid1996In: Algorithm Theory - SWAT'96: Proceedings of the 5th Scandinavian Workshop on Algorithm Theory / [ed] R. Karlsson; Andrzej Lingas, Berlin: Encyclopedia of Global Archaeology/Springer Verlag, 1996, p. 309-320Conference paper (Refereed)
• 19. Brodnik, Andrej
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
A static data structure for discrete advance bandwidth reservations on the Internet2003In: Proceedings of Swedish National Computer Networking Workshop: SNCNW 2003, Stockholm, 2003Conference paper (Refereed)
• 20. Brodnik, Andrej
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
An efficient data structure for advance bandwidth reservations on the Internet2002In: Proceedings of CSEE 2002: Second Annual Conference on Computer Science and Electrical Engineering. The third Annual Symposium on Computer Science and Electrical Engineering, 2002Conference paper (Refereed)

In this contribution we present a problem of resource reservation during some time. We show that the problem has a lower bound of Ω (log n) per operation on average and also give a matching upper bound algorithm.

• 21. Brodnik, Andrej
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
Data structure for a time-based bandwidth reservations problem2003Report (Other academic)

We discuss a problem of handling resource reservations. The resource can be reserved for some time, it can be freed or it can be queried what is the largest amount of reserved resource during a time interval. We show that the problem has a lower bound of $\Omega(\log n)$ per operation on average and we give a matching upper bound algorithm. Our solution also solves a dynamic version of the related problems of a prefix sum and a partial sum.

• 22. Brodnik, Andrej
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab.
Static data structure for discrete advance bandwidth reservations on the internet2003Report (Other academic)

In this paper we present a discrete data structure for reservations of limited resources. A reservation is defined as a tuple consisting of the time interval of when the resource should be reserved, $I_R$, and the amount of the resource that is reserved, $B_R$, formally $R=\{I_R,B_R\}$. The data structure is similar to a segment tree. The maximum spanning interval of the data structure is fixed and defined in advance. The granularity and thereby the size of the intervals of the leaves is also defined in advance. The data structure is built only once. Neither nodes nor leaves are ever inserted, deleted or moved. Hence, the running time of the operations does not depend on the number of reservations previously made. The running time does not depend on the size of the interval of the reservation either. Let $n$ be the number of leaves in the data structure. In the worst case, the number of touched (i.e. traversed) nodes is in any operation $O(\log n)$, hence the running time of any operation is also $O(\log n)$.

• 23.
Luleå tekniska universitet.
Luleå tekniska universitet. Luleå University of Technology.
Small forwarding tables for fast routing lookups1997In: Computer communication review, ISSN 0146-4833, E-ISSN 1943-5819, Vol. 27, no 4, p. 3-14Article in journal (Refereed)

For some time, the networking community has assumed that it is impossible to do IP routing lookups in software fast enough to support gigabit speeds. IP routing lookups must find the routing entry with the longest matching prefix, a task that has been thought to require hardware support at lookup frequencies of millions per second.We present a forwarding table data structure designed for quick routing lookups. Forwarding tables are small enough to fit in the cache of a conventional general purpose processor. With the table in cache, a 200 MHz Pentium Pro or a 333 MHz Alpha 21164 can perform a few million lookups per second. This means that it is feasible to do a full routing lookup for each IP packet at gigabit speeds without special hardware.The forwarding tables are very small, a large routing table with 40,000 routing entries can be compacted to a forwarding table of 150-160 Kbytes. A lookup typically requires less than 100 instructions on an Alpha, using eight memory references accessing a total of 14 bytes.

• 24.
Luleå tekniska universitet.
Luleå tekniska universitet. Luleå University of Technology.
Small forwarding tables for fast routing lookups1997Report (Other academic)

For some time, the Internet community has believed that it is impossible to do IP routing lookups in software fast enough to support gigabit speeds. IP routing lookups must find the routing entry with the longest matching prefix, a task that has been thought to require hardware support at lookup frequencies of millions per second. We present a forwarding table data structure designed for quick routing lookups. Forwarding tables are small enough to fit in the cache of a conventional general purpose processor. With the table in cache, a 200 MHz Pentium Pro or a 333 MHz Alpha 21164 can perform a few million lookups per second. This means that it is feasible to do a full routing lookup for each IP packet at gigabit speeds without special hardware. The forwarding tables are very small, a large routing table with 40,000 routing entries can be compacted to a forwarding table of 150--160 Kbytes. A lookup typically requires less than 100 instructions on an Alpha, using eight memory references accessing a total of 14 bytes.

• 25.
Luleå tekniska universitet.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Embedded Internet Systems Lab. Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
Extended expedited forwarding: the in-time PHB group2003In: Computers and Communication, 2003. (ISCC 2003): Proceedings of the Eighth IEEE International Symposium on Computers and Communications, IEEE Computer Society Press , 2003, Vol. 1, p. 291-298Conference paper (Refereed)

This paper presents a new set of forwarding behaviorsthat fits rate-adaptive and delay-sensitive applications withlimited loss tolerance. We consider an application to havelimited loss tolerance if it needs loss-free forwarding of specificpackets up to a certain rate. The new set of forwardingbehaviors are attractive for developing real-time applicationsfor the Internet. In particular, such applications canbe designed to use reserved forwarding capacity efficientlyand compete for more bandwidth while being fair to best-efforttraffic. To provide the new set of forwarding behaviors,we define a scheduling mechanism that can be implementedefficiently. Through simulations, we show that thismechanism supports the defined forwarding behaviors.

• 26.
Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science.
University of Ljubljana, Faculty of Computer and Information Science. Luleå University of Technology, Department of Computer Science, Electrical and Space Engineering, Computer Science. University of Ljubljana, Faculty of Computer and Information Science.
Reality considerations when designing a TDMA-FDMA based link-layer for real-time WSN2012In: Multiple Access Communications: 5th International Workshop, MACOM 2012, Maynooth, Ireland, November 19-20, 2012. Proceedings / [ed] Boris Bellalta, Encyclopedia of Global Archaeology/Springer Verlag, 2012, p. 93-96Conference paper (Refereed)

In this article we elaborate on reality considerations when deploying wireless sensor networks with strict end-to-end delay requirement. Our improvements are particularly important for a design and implementation of strict real-time systems while at the same time we decrease overall power consumption. Firstly, we design and implement application tailored TDMA-FDMA medium access protocol with guaranteed end-to-end delay. Secondly, we integrate in the implementation of the protocol the bootstrapping mechanism and thirdly, the time synchronization mechanism. Next, we show that by combining medium access protocol, bootstrapping, and time synchronization mechanisms within the link-layer, we can limit on average clock drift in the network to 0.5 micro seconds, as well as achieve 81 % energy efficiency while keeping collision probability at its minimum of 1 %. Finally, we conclude with challenges and lessons learned in real-world deployment of TDMA/FDMA based link-layer with guaranteed end-to-end delay in wireless sensor networks.

1 - 26 of 26
Cite
Citation style
• apa
• ieee
• modern-language-association-8th-edition
• vancouver
• Other style
More styles
Language
• de-DE
• en-GB
• en-US
• fi-FI
• nn-NO
• nn-NB
• sv-SE
• Other locale
More languages
Output format
• html
• text
• asciidoc
• rtf


text:
# Calculating and balancing challenge ratings

Are there standard rules for calculating CR of groups? For example, in my Pathfinder bestiary, a single Stirge is listed as CR 1/2, and a group of 2d4 (average 5) of them is listed as an average CR 4.

Also, how do you scale CRs? I'm aware of the rule of thumb that says:

A party of 5 players at level 9 should be able to face 5 challenges of rating 9 in a day.

But how do you scale that to different levels? How many CR 4 challenges is suitable for a level 6 party?

-

The Challenge Rating of a monster is a very useful guide for judging the difficulty of an encounter, but it is not an exact science. Experience with your own players will tell you how tough an encounter they can handle, and what kinds of encounters they are best at.

When the entire party can gang up on a single monster (even one with multiple attacks) tactics on the battlefield can be less important than when they are outnumbered. However, the more monsters you have, the less certain the estimate of Encounter Level (EL) / "CR Equivalency" becomes. You should also be careful when advancing monsters (adding extra hit dice to increase their CR), as this too can lead to some nasty surprises.

I find this online calculator to be very useful for estimating EL, especially when I'm using monsters with a range of different CR's in a single encounter: http://www.d20srd.org/extras/d20encountercalculator/

As far as multiple encounters per day are concerned, this too depends on a number of things. A fighter or a warlock can likely go for longer than a wizard, but even they likely have limited use magic items (particularly healing potions) that they will run out of eventually. When they are near a settlement where they can go shopping and are likely to get an uninterrupted 8 hours of rest, it isn't a problem. However, when they're deep in the Underdark, surrounded by enemies, things can get a bit tough.

Rather than having multiple EL 4 encounters in a row, you should aim for variety. Table 3-2 Encounter Difficulty, in the 3.5 edition DMG, recommends the following:

10% easy (EL lower than party level)
20% "easy if handled properly" (whatever that means...)
50% challenging (EL equals Average Party Level)
15% very difficult (EL 1-4 higher)
5% overwhelming (EL 5+ higher)

In Pathfinder, the range of encounters is a little tighter:

Easy (APL –1)
Average (=APL)
Challenging (APL +1)
Hard (APL +2)
Epic (APL +3)
(APL = Average Party Level)

However, I think that the advice still holds: don't aim to make every enounter "average". Now and then, throw something "epic" at them to keep 'em from getting too cocky. You should drop some pretty heavy hints, though, that they need to run away to avoid a TPK (especially the first time you do it). In a level-based game like D&D, it's always good to have something to strive for.

-
That calculator is very helpful. I know raw numbers aren't the complete answer, but having them to hand makes it much easier. – Marcus Downing Nov 14 '10 at 0:14

The way the D&D power curve is that encounters too far under your APL do not deplete your abilities at all. The "5 encounter rule" is based on an average of 20% depletion of spells/hit points/whatnot. So really an APL 6 party can fight a near infinite number of CR3 encounters (at least until boredom sets in).

You can use the normal "CR math" in terms of total amount to toss in, like 2 CR4s = a CR6 and use that to gauge off, but once you get more than APL+2 the answer is "one" and once you go below APL-2 the answer is "very many."

Oh, and for the standard CR math, basically double the number of creatures means CR+2. A creature with a sidekick of CR-1 means the encounter is CR+1.

-

Be sure to remember that CR and EL are not the same thing. CR is a property of a creature. EL is a property of a group of creatures and the environment you face off against them in. Creatures on their home turf getting bonuses for that make for a higher EL, and creatures who are out of their element reduce the EL (usually just by +/- 1). For a single creature, CR = EL. For groups of creatures, EL can be calculated using the following rules of thumb:

• Two things of the same CR have an EL of the CR+2. This can be repeated again, and sometimes a third time, before the players are too far ahead to be threatened by whatever is showing up in a huge group.
• Fractional CR creatures are intended to be added together until you get a full CR 1, and then treat that "squad" as a single unit at that point. 3 dogs = 2 orcs = 4 kobolds = 1 gnoll = 2 goblins and 1 dog.

So yeah, two Stirges is EL 1, four of them is EL 3, and five of them is pushing slightly above that (but it'd take more like eight to be a 'proper' EL4).

-
I think you mean EL and there's no such thing in Pathfinder. – mxyzplk Nov 12 '10 at 0:42
Ah, yes, EL not ECL, my mistake. Write it wrong once and keep making the same mistake the whole way through, silly me. As to the idea that "EL doesn't exist in Pathfinder", well that's just absurd. Pathfinder is based on 3.5 with a million little edits. There aren't enough changes to the fundamental math to affect how the CR system works (that would break backwards compatibility), and so the same rules apply to all of 3.0/3.5/PF. – Lokathor Nov 12 '10 at 6:06

rule of thumb is: when doubling the monsters add 2 EL/CR so one 8th level monster is a 8 EL/CR and two 8th level monsters is 10 EL/CR four 8th level would be 12 EL/CR

-


text:
## Typography

Some material relating to Typography and Typesetting in general.

Between 1947 and 1949 the famous German typographer Jan Tschichold worked at Penguin Books, standardizing the cover design of Penguin’s books and developing a set of rules to ensure consistency in typography across Penguin’s output. The result of the latter was a four page leaflet, Penguin Composition Rules.
Files: penguinrules.tex, penguinrules.pdf.

Glossary from ‘Introduction to Typography’ by Oliver Simon (1963 edition):
Files: Glossary63.tex, Glossary63.pdf.

The \entry macro used in Glossary63.tex is from p. 443 of ‘A Plain TeX Primer’ by Malcolm Clark. The colon in \def\entry#1:#2 serves as a delimiter between the italic of the head word and the roman of the definition.

I made another version of this glossary to show how \mark can be used with \headline to put reference marks in the header. I put the \marks close to the middle of their pages, because when I originally put them at the ends of their page beside the \entry that they got their name from, the \mark in the header usually ended up on the wrong page. This is probably because until TeX outputs the page, it is fuzzy about where the page boundaries will be.
Files: Glossary63m.tex, Glossary63m.pdf.

Some more modern glossaries:

A collection of links from TypeCulture:

One Typeface, many fonts by W.F. Adams: from the TeX Showcase.

Glossary of printing terms:

Another glossary from Harper Collins:

Some specimen books and machine manuals:

The ATF (American Type Founders Company) Specimen Book:
1923 Edition
1912 Edition

Lanston Monotype Specimen Book of Typefaces, 1922:

Stephenson Blake Specimen, 1908:

Barnhart & Spindler Specimen, 1907:

MacKellar, Smiths & Jordan Specimen Books: 1885 1892

Specimen Book of the Bauersche Giesserei, 1900:

V. & J. Figgins, 1845. Some Porson Greek types, supplied with upright or slanting capitals. Also Saxon, Gaelic, Hebrew, Syriac, Persian Arabic, and Bengali fonts (page 111 and ff.):

Fry Specimen, 1816:

Binny & Ronaldson Specimen, 1812 (the first type specimen printed in North America):

Caslon Specimen Books:
1785 1798 1841 1915
William Caslon I’s 1734 Specimen:
In the 1798 Specimen the long s (ſ) has disappeared, and some of the types come with lining figures; see Figures on p. 127.

The Egenolff-Berner specimen (1592), showing Claude Garamond’s roman types and Robert Granjon’s italics:

The Manual of Linotype Typography (1923):

Linotype machine manual (1892):

The Mechanism of the Linotype (1902):

Linotype Keyboard Practice (1947):

The Monotype system: a book for owners & operators of Monotypes (1912)

A list of metal Monotype typefaces in alphabetical order →, and numerical order , from the website of Alembic Press.

More Monotype stuff.

Printers’ Rulers

Old style pica pole rulers (used in the hot metal days): , .
A transparent plastic ruler with picas, points, rules and a leading gauge: (cheaper here ).
Or print out one of these: , , .
See also ‘The TeX ruler’ (search your TeX Live or MiKTeX installation for ‘ruler’, you will find the source file of the PDF), and ‘pointruler’, a printer’s ruler done in LaTeX: .

Videos

Hand setting (1959):

A vocational film for printers from 1946:

A short film about the Linotype machine from TypeCulture.
A longer film (35 mins) from c. 1960 that describes the working of the Linotype machine in detail.

A short film about the Monotype system from TypeCulture.
Videos from the National Print Museum of Ireland showing the operation of the Keyboard and Caster. This channel also has videos about hand setting, Linotype, Ludlow (another kind of hot metal typesetting system), Wharfdale, Heidelberg and Albion presses and other things.

The Monotype Vault:

A film from 1947 showing the complete process of turning an author’s manuscript into a book. The material is first typeset on a Linotype machine and locked up in a small forme. Then, because hot metal type is softer than foundry type and tends to wear out over long print runs, a copper plate is formed from a wax mould of this forme by the electrotype process. The copper plates are cut up into individual pages and imposed on the bed of the printing press. After the printing is done, the large sheets are folded into signatures, collated and stitched, then the book block is trimmed and finally the cover is put on .

Offset lithography, how most books are printed nowadays: , .

Film from Edwards Brothers Malloy showing the modern process of book printing and binding: . This is the 21st century equivalent of the film from 1947 linked to above.

Donald Knuth explaining why he decided to create his own typesetting program: and following videos. Lots of other interesting videos on this channel.


text:
# 22 students is what percent of 55?

Nov 16, 2016

40%

#### Explanation:

The equation for percents is

$\text{Percent" xx "Whole" = "Part}$

So all you have to do is identify the parts of the equation. We don't know the percent, but we know everything else, so we can plug what we know into the equation

$P \times 55 = 22$

$\frac{55}{22} = P$

P = 0.4=40%


text:
# Raw moments of 3-parameter Weibull distribution

#### mehdi_m990

##### New Member
Hi,
I know that the formula for computing the raw moments of 2-parameter Weibull distribution is:
Mu'n=b^n*Gamma(1+n/c), where b and c are scale and shape parameters, respectively.

However, I couldn't find any exact formula for a 3-parameter Weibull distribution. Is there any simple formula for it???

#### BGM

##### TS Contributor
First I need to make sure that the 3rd parameter you mentioned is the location parameter.

https://en.wikipedia.org/wiki/Weibull_distribution#Related_distributions

If that's the case, then we have the following:

Let $$X \sim \text{Weibull}(k, \lambda, \theta)$$ and $$Y \sim \text{Weibull}(k, \lambda, 0)$$ be two 3-parameters Weibull random variable.

Then
$$Y$$ is a 2-parameter Weibull random variable; and

$$X \stackrel {d} {=} Y + \theta$$

As you know raw moments

$$m_n \triangleq E[Y^n] = \lambda^n \Gamma\left(1 + \frac {n} {k}\right)$$

Therefore,

$$E[X^n] = E[(Y + \theta)^n]$$

$$= E\left[\sum_{r=0}^n \binom {n} {r} Y^r \theta^{n-r}\right]$$

$$= \sum_{r=0}^n \binom {n} {r} m_r \theta^{n-r}$$

$$= \sum_{r=0}^n \binom {n} {r} \lambda^r \Gamma\left(1 + \frac {r} {k}\right) \theta^{n-r}$$

This is not simple or very nice, but at least this is exact and can be computed.


text:
This website requires JavaScript.

# Molecular gas in super spiral galaxies

Mar 2023

At the highest stellar masses (log(\mstar) $\gtrsim$ 11.5 \msun), only asmall fraction of galaxies are disk-like and actively star-forming objects.These so-called `super spirals' are ideal objects to better understand howgalaxy evolution proceeds and to extend our knowledge about the relationbetween stars and gas to a higher stellar mass regime. We present new CO(1-0)data for a sample of 46 super spirals and for 18 slightly lower-mass(log(\mstar) $>$ 11.0 \msun ) galaxies with broad HI lines -- HI fast-rotators(HI-FRs). We analyze their molecular gas mass, derived from CO, in relation totheir star formation rate (SFR) and stellar mass, and compare the results tovalues and scaling relations derived from lower-mass galaxies. We confirm thatsuper spirals follow the same star-forming main sequence (SFMS) as lower-massgalaxies. We find that they possess abundant molecular gas, which lies abovethe extrapolation of the scaling relation with stellar mass derived fromlower-mass galaxies, but within the relation between \mmol/\mstar and thedistance to the SFMS. The molecular gas depletion time, \taudep = \mmol/SFR, ishigher than for lower-mass galaxies on the SFMS (\taudep = 9.30 $\pm$ 0.03,compared to \taudep = 9.00 $\pm$ 0.02 for the comparison sample) and seems tocontinue an increasing trend with stellar mass. HI-FR galaxies have anatomic-to-molecular gas mass ratio that is in agreement with that of lower-massgalaxies, indicating that the conversion from the atomic to molecular gasproceeds in a similar way. We conclude that the availability of molecular gasis a crucial factor to enable star formation to continue and that, if gas ispresent, quenching is not a necessary destiny for high-mass galaxies. Thedifference in gas depletion time suggests that the properties of the moleculargas at high stellar masses are less favorable for star formation.

Q1论文试图解决什么问题？
Q2这是否是一个新的问题？
Q3这篇文章要验证一个什么科学假设？
0


text:
# Tag Info

## Hot answers tagged momentum

12

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + ...

6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...

5

That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$v = \frac{pc^2}{E}$$ When $m = 0$, $E ... 5 The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ... 4 In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ... 4 Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object. A massive object can have any momentum (at least as long as its velocity ... 4 Momentum: The resistance of an object to a change in its state of motion. That sounds like a fishy definition of momentum to me. A slightly better definition, at least at your level, is that momentum represents the "amount of motion" an object has. Granted, "amount of motion" is a very vague term, but it stands to reason that if "amount of motion" were ... 4 A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons. Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by: $$E^2 = p^2c^2 + m^2c^4$$ where$p$is the ... 4 Yes,$F=ma$, but also$v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at$9.8 m/s$or$35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach$98 m/sec$or$350 km/h$. For a free-falling human, the air resistance actually limits you to about$200 km/h$. When you hit the ... 3 You are right. If$q$is a generalized coordinate then$\dot{q}$is the generalized velocity and hence the generalized momentum is $$p = \frac{\partial L}{\partial\dot{q}}$$ Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the$p_{\theta_i}$are indeed ... 3 Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ... 2 The thrust coming from a rocket engine is exerted on the engine bell, and it is directed along its axis of symmetry. It's not completely clear how you're modelling your ship but it is probably more realistic to apply the force to the "thruster fire" block, whatever that is. It's important to note, though, that if applying the force to the engine bell and ... 2 If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ... 2 Your analysis will need to be a combination of theory and experiment. You are not maximizing impulse alone: you need to think about drag as well. So you need to think about the shape of your rocket as well as the volume of water and initial pressure. The drag on the rocket is probably well modelled by the ram pressure equation ... 2 Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ... 2 Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ... 1 This seems like a list question which aren't very good for SE. Here is a proof of your statement:$m_1v_1 = m_2v_2\dfrac{m_1}{m_2} = \dfrac {v_2}{v_1}$Let us assume that$m_1$<$m_2\dfrac{(m_1)^2}{(m_2)^2} = \dfrac {(v_2)^2}{(v_1)^2} < 1v_1$must be greater than$v_2$The only case the energy of a higher velocity is beaten is shown by ... 1 If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear. For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the ... 1 I will bring an example from classical electrodynamics. In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the ... 1 You didn't cancel out the mass$m$properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision,$a=b\$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...

1

Hint: The center of mass of pieces A and B moves in the same path as the intact shell would. (This arises from the conservation of momentum.) Edit: The gravitational force is only acting along the vertical direction. So there is net acceleration only in the vertical direction. Looking at the horizontal one, if we neglect things like air resistance (which ...

1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...

1

As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# arccos(x) - arccos(y) = arccos(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

We will learn how to prove the property of the inverse trigonometric function arccos(x) - arccos(y) = arccos(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

Proof:

Let, cos$$^{-1}$$ x = α and cos$$^{-1}$$ y = β

From cos$$^{-1}$$ x = α we get,

x = cos α

and from cos$$^{-1}$$ y = β we get,

y = cos β

Now, cos (α - β) = cos α cos β + sin α sin β

⇒ cos (α - β) = cos α cos β + $$\sqrt{1 - cos^{2} α}$$ $$\sqrt{1 - cos^{2} β}$$

⇒ cos (α - β) = (xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

⇒ α - β = cos$$^{-1}$$(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

or, cos$$^{-1}$$ x - cos$$^{-1}$$ y = cos$$^{-1}$$(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

Therefore, arccos(x) - arccos(y) = arccos(xy) + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)       Proved.

Note: If x > 0, y > 0 and x$$^{2}$$ + y$$^{2}$$ > 1, then the cos$$^{-1}$$ x + sin$$^{-1}$$ y may be an angle more than π/2 while cos$$^{-1}$$(xy - $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$), is an angle between – π/2 and π/2.

Therefore, cos$$^{-1}$$ x - cos$$^{-1}$$ y = π - cos$$^{-1}$$(xy + $$\sqrt{1 - x^{2}}$$$$\sqrt{1 - y^{2}}$$)

`

Inverse Trigonometric Functions

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.


text:
# A trignometric limit going to infinity

## Homework Statement:

Problem -: lim cos(pi(n^2+n)^(1/2)) where n tends to infinity.

## Relevant Equations:

Costheta=cottheta/cosectheta
L'Hospital rule
I wrote cos(pi(n^2+n)^(1/2)) as cot(pi(n^2+n)^(1/2))/cosec(pi(n^2+n)^(1/2)) and as we know cot(npi)=infinity and cosec(npi)=infinity , so i applied L'Hospital.After i differentiated i again got the same form but this time cosec/cot which is again infinity/infinity.But if i differentiate it i will get the same form again.So what to do now.

Related Calculus and Beyond Homework Help News on Phys.org
i dont think you have to use L'hospital here have you tried other methods

i dont think you have to use L'hospital here have you tried other methods
Can you suggest any other method i tried to use expansion also but it didn't worked

did you try expanding
##
\sqrt{n^2 +n}
##
for large n

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You cannot apply l’Hopital. Your function is not a function of a continuous parameter.

did you try expanding
##
\sqrt{n^2 +n}
##
for large n
Tried it but it didn't helped.

You cannot apply l’Hopital. Your function is not a function of a continuous parameter.
So sir what should i do i am struggling over it.

Tried it but it didn't helped.
it does help how did you expand it

it does help how did you expand it
What is the use of treating it as capital N.After it what will i do.

epenguin
Homework Helper
Gold Member
Seems to me obvious what the answer is considering ##lim\text{ } cos(f(n)) = cos (lim(f(n)) ## as ##n→∞ ##

timetraveller123
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Seems to me obvious what the answer is considering ##lim\text{ } cos(f(n)) = cos (lim(f(n)) ## as ##n→∞ ##
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.

Tried it but it didn't helped.
Show us what you did. If you do not we cannot help you.

timetraveller123
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.

Show us what you did. If you do not we cannot help you.
I took out n^2 outside underroot and then opened the expansion of (1+1/n)^(1/2).Further i was left with cos(npi+pi/2) which will be equal to cos(pi/2) or -cos(pi/2) which gives me the answer as 0.
But i want to know is there any other method of solving it without expansion as we are not taught about this expansion in our school.

timetraveller123
i am not sure of this would work some one please correct me
and since op already knows answer just want another method i posting another "solution"

##
\sqrt{n^2+n} -n = x\\
0<x<1\\
0 = x^2 + 2nx -n\\
0=\frac{x^2}{n} + 2x - 1
##
first term goes arbitrarily close to zero
so x tends to ##
\frac{1}{2}
##

Physics lover
i am not sure of this would work some one please correct me
and since op already knows answer just want another method i posting another "solution"

##
\sqrt{n^2+n} -n = x\\
0<x<1\\
0 = x^2 + 2nx -n\\
0=\frac{x^2}{n} + 2x - 1
##
first term goes arbitrarily close to zero
so x tends to ##
\frac{1}{2}
##
Thanks for this one.It is far better than my previous attempt.

epenguin
Homework Helper
Gold Member
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.
Ah, true. Ulseful maybe as conjecture when stuck even if wrong, to get somewhere like #13

wait i am not understanding why does the function not have well defined limit doesn't it do this
##
\sqrt{n^2 + n} = n+\frac{1}{2}
##

Orodruin
Staff Emeritus
Homework Helper
Gold Member
wait i am not understanding why does the function not have well defined limit doesn't it do this
##
\sqrt{n^2 + n} = n+\frac{1}{2}
##
The limit of that is infinity. You are basically suggesting to compute ##f(\infty)## where ##f## is an oscillating function.

timetraveller123
but n is integer here which makes f(infinity) = 0
i am not understanding something here

wait i said the limit is
##
n + 0.5
##
and not infinity
ok this is probably wrong

Orodruin
Staff Emeritus
Homework Helper
Gold Member
but n is integer here which makes f(infinity) = 0
i am not understanding something here

wait i said the limit is
##
n + 0.5
##
and not infinity
ok this is probably wrong
That is the asymptote, not the limit. The limit is infinity. The appropriate way of doing this is noting that
$$\sqrt{n^2+n} = n \sqrt{1+1/n} = n+1/2+ \mathcal O(1/n).$$
Therefore
\begin{align*}
\cos(\pi\sqrt{n^2+n}) &= \cos(\pi[n+1/2+ \mathcal O(1/n)]) \\
&= \cos(\pi n + \pi/2 + \mathcal O(1/n)) \\
&= \cos(\pi n + \pi/2) \cos(\mathcal O(1/n)) - \sin(\pi n + \pi/2) \sin(\mathcal O(1/n)).
\end{align*}
The first term is zero and the second term is ##\mathcal O(1/n)## since the sine function is bounded. Thus, the limit of the expression is 0.

sysprog, Mark44 and timetraveller123
ah i see thanks


text:
# zeta(2k+1) is a rational multiple of pi^{2k} zeta'(-2 k) ?

Probably this is well know and elementary and will delete it, but couldn't find it on the web.

Got a sketch of proof and numerical evidence that $\zeta(2k+1)$ is a rational multiple of $\pi^{2k} \zeta'(-2k)$

An identity from Derivatives of the Hurwitz Zeta Function for Rational Arguments p.7

$$\zeta'(-n,x) + (-1)^n \zeta'(-n,1-x) = \pi i \frac{B_{n+1}(x)}{n+1} + \frac{n!}{(2 \pi) ^ n} e^{-\pi i n / 2} \operatorname{Li}_{n+1}(e^{2\pi i x}). \qquad (21)$$

For even $n$ and $x=\frac12$ (21) is:

$$2 \zeta'(-n,\frac12) = \pi i \frac{B_{n+1}(\frac12)}{n+1} \pm \frac{n!}{(2 \pi) ^ n} \operatorname{Li}_{n+1}(-1).$$

The choise of $\pm$ depends on $e^{-\pi i n / 2}$.

According to Wolfram Alpha $Li_{n+1}(-1)$ is an integer multiple of $\zeta(n+1)$ and $B_{n+1}(\frac12)$ vanishes.

$\zeta(s,\frac12) = (2^s-1) \zeta(s)$. Taking derivative and having in mind $\zeta(-2k)=0$ we have $\zeta'(s,\frac12)$ is a rational multiple of $\zeta'(s)$.

For even $n$ and $x=\frac12$ (21) simplifies to:

$$\mathbb{Q} \zeta'(-n) = \pm \mathbb{Z} \frac{n!}{(2 \pi) ^ n} \zeta(n+1). \qquad (1)$$

In particular,

$$\zeta(3) = -4 \zeta'(-2) \pi^2$$ $$\zeta(5) = 4/3 \zeta'(-4) \pi^4$$

The last two hold with precision of 1000 digits.

Is this true?

Is this known?

-
Did you try the usual functional equation (relating ζ(s) and ζ(1−s))? –  Carl May 7 '13 at 10:59
Don't delete your question as I gave an answer. –  GH from MO May 7 '13 at 11:54

$$\zeta(2k+1)=\frac{(-1)^k2^{2k+1}}{(2k)!}\pi^{2k}\zeta'(-2k).$$
Following Carl Dettmann's suggestion, let us start from the functional equation $$\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s).$$ For $s=-2k+\epsilon$ this gives $$\zeta(2k+1-\epsilon)=\pi^{2k+\frac{1}{2}-\epsilon}\frac{\Gamma(-k+\epsilon/2)}{\Gamma(k+1/2-\epsilon/2)}\zeta(-2k+\epsilon).$$ On the right hand side, for $\epsilon\to 0$, $$\Gamma(-k+\epsilon/2)\sim \frac{2(-1)^k}{k!\epsilon}$$ and $$\zeta(-2k+\epsilon)\sim\epsilon\zeta'(-2k),$$ hence we have at $\epsilon=0$ i.e. at $s=-2k$ $$\zeta(2k+1)=\pi^{2k+\frac{1}{2}}\frac{2(-1)^k}{k!\Gamma(k+1/2)}\zeta'(-2k).$$ On the right hand side $$\Gamma\left(k+\frac{1}{2}\right)=\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\cdots\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{(2k-1)(2k-3)\cdots 1}{2^k}\pi^{\frac{1}{2}},$$ whence the stated identity follows.
Thank you GH ${}{}$ –  joro May 7 '13 at 12:07
GH, can you express $\zeta''(-2)$ in simpler terms? Might have solution with zeta'(3),zeta(3),log,pi,gamma –  joro May 7 '13 at 13:52
@joro: It is possible to generate such expressions by looking at the Taylor series of $Z(s)=\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)$ at $s=-2$ and at $s=3$. The Taylor coefficients will be the same up to $\pm$ signs due to the symmetry $Z(s)=Z(1-s)$. Each Taylor coefficient in these identities can be expressed from the Laurent series coefficients at $s=-2$ and at $s=3$ of the factors involved in $Z(s)$. –  GH from MO May 7 '13 at 17:05


text:
# Determine the bond order of the "C"-"O" bond in the following molecules, and then arrange the "C"-"O" bond lengths in increasing order? "CO", "CO"^(+), "CO"^(2+), "CO"_2, "CO"_3^(2-)

Apr 30, 2017

Bond length increases from left to right on your list, i.e.

${r}_{C O} < {r}_{C {O}^{+}} < \left[{r}_{C {O}^{2 +}} = {r}_{C {O}_{2}}\right] < {r}_{C {O}_{3}^{2 -}}$

• $\text{CO}$: $3$
• ${\text{CO}}^{+}$: $2.5$
• ${\text{CO}}^{2 +}$: $2$
• ${\text{CO}}_{2}$: $2$
• ${\text{CO}}_{3}^{2 -}$: $1. \overline{33}$

In order to determine this, we should reference an MO diagram.

We can see that the highest occupied molecular orbital (HOMO) is fully occupied, but the next-highest MOs are the ${\pi}^{\text{*}}$ antibonding lowest unoccupied molecular orbitals (LUMOs). Keep this in mind.

CO MOLECULE

$\text{CO}$ by default has a triple bond, so by chemical intuition, its bond order is $3$. This can be verified by the usual equation:

"BO" = 1/2("Bonding - Antibonding")

$= \frac{1}{2} \left(\left[\stackrel{2 \sigma}{\overbrace{2}} + \stackrel{1 \pi}{\overbrace{2 \left(2\right)}} + \stackrel{3 \sigma}{\overbrace{2}}\right] - \left[\stackrel{2 {\sigma}^{\text{*}}}{\overbrace{2}}\right]\right) = 3$

CO+ CATION

${\text{CO}}^{+}$ has one electron removed from a bonding MO, so its bond order decreases by $\frac{1}{2}$ to become $2.5$. Use the example above to verify via the equation.

CO""^(bb(2+)) DICATION

${\text{CO}}^{2 +}$ has an additional bonding electron removed relative to ${\text{CO}}^{+}$, so it has a bond order of $2$.

CO2 MOLECULE

${\text{CO}}_{2}$ looks like this:

:stackrel(..)("O")="C"=stackrel(..)"O":

So, its double bonds suggest a bond order of $2$.

CARBONATE ANION

By considering its resonance delocalization, the bond order on this molecular ion can be determined.

We should notice that for the $\pi$ bond order, there will be $2$ $\pi$ electrons delocalized onto three bonds, making each bond on average having $\frac{2}{3}$ of a $\pi$ electron, thereby giving a $\pi$ bond order of $\frac{1}{3}$.

Since perfect single bonds have a bond order of $1$, the total bond order on ${\text{CO}}_{3}^{2 -}$ is $\boldsymbol{1. \overline{33}}$.

OVERALL

Overall, we have the bond orders:

• $\text{CO}$: $3$
• ${\text{CO}}^{+}$: $2.5$
• ${\text{CO}}^{2 +}$: $2$
• ${\text{CO}}_{2}$: $2$
• ${\text{CO}}_{3}^{2 -}$: $1. \overline{33}$

You can choose how to order ${\text{CO}}^{2 +}$ and ${\text{CO}}_{2}$ however you like, but the rest is already in order. The shortest bond corresponds to the strongest bond.

Hence, bond length increases down this list.


text:
Kurtosis & its Application in Risk Evaluation

When analysing the performance of financial securities, we give a lot of importance to the mean and the standard deviation as measures of the average return and risk, respectively.

However, the risk-adjusted performance of financial securities also depends upon the risks which arise due to the shape of the distribution of their returns. These are the higher-moment risks such as the skewness and kurtosis risks, which need to be taken into consideration for proper evaluation.

In this blog, we discuss the concept of kurtosis and its application in understanding the risk profiles of financial securities. In addition, we also glance over some common misconceptions regarding the calculation and interpretation of kurtosis.

Contents

What is kurtosis?

Kurtosis is a statistical measure which quantifies the degree to which a distribution of a random variable is likely to produce extreme values or outliers relative to a normal distribution.

From extreme values and outliers, we mean observations that cluster at the tails of the probability distribution of a random variable. In other words, kurtosis measures the 'tailedness' of distribution relative to a normal distribution.

Along with variance and skewness, which measure the dispersion and symmetry, respectively, kurtosis helps us to describe the 'shape' of the distribution.

Mathematically, the kurtosis of a distribution of a random variable X, with a mean μ and standard deviation σ is defined as the ratio of the fourth moment to the square of the variance $σ^2$

ie. $kurtosis = \frac{\text{The fourth moment}}{σ^{4}} = \frac{\frac{\sum \limits _{j=1} ^{n} (X_{j}-μ)^{4}}{n}}{σ^{4}}$

From the above expression, we can already see that outliers will significantly impact this value.

This is because for any outlier $x_p$ the value $(x_p- μ)$ will tend to be large anyway, but then the outliers contribution to the sum is raised to the power of four

i.e. $(x_p-μ)^4$

In other words, the key factor that would determine the value of kurtosis is the number and size of the outliers/extreme values, which are reflected in heavier/fatter tails of the distribution.

Contrary to popular perception, kurtosis does not measure the peakedness of the distribution, and the only unambiguous interpretation of kurtosis is with regard to the heaviness or lightness of tails of the distribution, relative to a normal distribution (we will see an example of this in a subsequent section).

Excess kurtosis

There exists one more method of calculating the kurtosis called 'excess kurtosis'. As kurtosis is calculated relative to the normal distribution, which has a kurtosis value of 3, it is often easier to analyse in terms of excess kurtosis.

As the name suggests, it is the kurtosis value in excess of the kurtosis value of the normal distribution. This means that for a normal distribution with any mean and variance, the excess kurtosis is always 0.

$$Excess\ kurtosis = \frac{\text{The fourth moment}}{σ^{4}} - 3 = \frac{\frac{\sum \limits _{j=1} ^{n} (X_{j}-μ)^{4}}{n}}{σ^{4}} - 3$$

Many resources refer to 'excess kurtosis' as 'kurtosis' and hence to avoid any confusion one must clarify this point beforehand.

In the next section, we will learn about the three categories of distributions based on the kurtosis.

Classification of distributions based on kurtosis/excess kurtosis

Any statistical distribution can be categorised into one of the three categories based on its kurtosis/excess kurtosis:

Mesokurtic

When the kurtosis of distribution is 3, i.e. the excess kurtosis is equal to 0, the distribution is called mesokurtic. This means the distribution is a normal distribution. ('meso' stands for medium or middle). As the kurtosis of a mesokurtic distribution is neither positive nor negative, it serves as a baseline for the two other categories.

Platykurtic

The second category is that of platykurtic distributions which have negative excess kurtosis values. Negative values of excess kurtosis indicate that distribution has short/thin tails. A good example of such a distribution is the continuous uniform distribution.

Leptokurtic

The last category is that of leptokurtic distributions which have positive excess kurtosis. A large positive kurtosis indicates a distribution where more of the values are located away from the mean i.e. in the tails of the distribution. Thus, leptokurtic distributions are characterized by fatter tails. A good example is the Laplace distribution.

In the following Python code, we draw random samples from three example distributions with roughly equal mean, equal variance and which are almost symmetrically distributed around their means. (New to Python? Get started with our Python for trading course for better understanding.)

These distributions are:

• Standard Normal (mesokurtic)
• Uniform (platykurtic)
• Laplace (leptokurtic)

Now let us summarize the information about the means, standard deviations, skewness and kurtosis of the three distributions:

Results

It can be seen above that all the three standardized distributions have roughly the same mean (0) and standard deviation (1) and the skewness of almost 0* i.e. they are symmetrically distributed around their means.

However, when we look at the plot, we see that they are very different distributions and the mean, the standard deviation and the skewness failed to capture this difference in the shape of the distributions. This is where kurtosis comes into the picture.

We can observe in the above figure that the Laplace distribution (green plot) has a leptokurtic distribution and has fatter tails compared to the standard normal distribution (blue plot), which is also reflected in a large and positive excess kurtosis.

On the other hand, the uniform distribution has much shorter/thinner tails compared to a standard normal distribution, which is summarized by a negative excess kurtosis.

Kurtosis-risk/ tail-risk in financial securities

The normality of the distribution of asset returns is a common assumption in many quantitative finance models. This is a convenient assumption, as the normal distribution can be completely summarized by its mean and standard deviation/variance (and has a skewness and excess kurtosis of 0).

Under the assumption of normality, the standard deviation is often used as the only measure of risk. However, the actual distributions of asset returns are not normal and often display positive excess kurtosis.

Thus, assuming normality can severely underestimate the risk of extreme events as distributions with the same mean, same standard deviations and the same skewness can have very different tail profiles (we saw an example of this in the previous section). This risk is also referred to as 'kurtosis risk' or 'fat-tail risk’.

To capture and estimate the tail risk in a better way, risk managers often use a leptokurtic distribution such as the student's t distribution, which is symmetric like the normal distribution but has fatter tails.

In the following figure we generate and plot sample returns from both the standard normal and t distributions and compare them with the actual standardized returns of Indian Tobacco Company (ITC) for the past five years:

We can see in the above figure that the t distribution is better able to model the returns of the stock compared to normal distribution and at the same time captures the kurtosis/tail risk.

Thus, it is advantageous to use a leptokurtic distribution like the student's t distribution while calculating risk metrics such as the Value-at-risk (VaR) or the expected shortfall.

A common error in interpretation of kurtosis

One of the common misconceptions associated with kurtosis is that it is a measure of 'peakedness', as it is easy for us to think that fatter tails automatically implies higher peaks.

However, this is not always true. Take for example two t distributions, one with 3 degrees of freedom and the other with 10 degrees of freedom, that we plot using the following commands:

We can see from above that although the t_3 distribution has a smaller peak, it has much higher excess kurtosis compared to the t_10 distribution.

Analysing the tail risk in S&P 500 and Bitcoin

A general misconception is that highly volatile securities have a higher tail risk and hence a higher excess kurtosis value. However, this is not always true.

Take, for example, the return series for SPDR S&P 500 ETF and the bitcoin for the past five years, which we fetch in Python using the following commands:

We can clearly see that bitcoin is almost four times more volatile compared to the S&P 500.

Now let us check out the distributions of the two series using the following commands:

In the above figure, we can see that the two distributions are fairly symmetric around zero with the bitcoin (BTC-USD) returns having a much wider spread (heavier tails), with the extreme values well in excess of 20 percent.

Such magnitudes of extreme values lead to an impression that the bitcoin presents a much higher tail risk compared to the S&P 500. This would naturally imply that the kurtosis value for BTC should well exceed that of S&P 500.

Also, this is in line with the general perception that bitcoin being a cryptocurrency is prone to wild moves giving rise to a higher frequency of observing a tail event compared to equities.

Thus, it is natural that one might think that the distribution of bitcoin will have a higher kurtosis value compared to the S&P 500 equity index. However, when we calculate the excess kurtosis values, we might be in for a surprise:

We see that the kurtosis values above, the SPDR S&P 500 returns have an excess kurtosis which is almost double that of the bitcoin. The reason for this is that the value of kurtosis depends on the departures in the shape of a distribution from the normal distribution.

In order to get more clarity, we should compare the kurtosis of distributions with equal variance, so that we can observe this fact visually. In order to do this, we will first standardise both the distributions such that each one of them has a mean of zero and a variance of one.

Also, we add a third distribution, which is the random sample from a standard normal distribution:

Now, we plot the standardized returns of BTC and S&P 500 along with the random sample from a standard normal distribution:

We can see in the above that the standardized S&P 500 returns (blue) show much more extreme deviations from the standard normal (green). This becomes more evident when we capture these in the density plots using the following commands:

The above figure makes it clear why the standardised SPDR S&P 500 (SPY) returns distribution has a higher kurtosis value compared to the distribution of bitcoin, contrary to the perception.

The reason is the SPY's wider spread (heavier tails) relative to the standard normal distribution i.e. the departure of standardized S&P 500 returns distribution from the standard normal distribution is much more than that of standardized bitcoin (BTC) returns.

Thus, it is important to remember that kurtosis is always calculated relative to the normal distribution and it should be interpreted as such.

Conclusion

In this blog, we have seen how kurtosis/excess kurtosis captures the 'shape' aspect of distribution, which can be easily missed by the mean, variance and skewness. Furthermore, we discussed some common errors and misconceptions in the interpretation of kurtosis.

Kurtosis is a very useful metric to quantify the tail-risk in finance. Ignoring tail-risk can potentially lead to the overestimation of alphas, and hence tail-risk/kurtosis-risk evaluation should be a part of the overall performance evaluation for financial securities.

Similarly, there are other metrics like Kurtosis that if combined with Algorithmic trading can add value to your trading practises. However, if you wish to learn the what, how, why of algorithmic trading, our course on Getting Started with Algorithmic Trading is perfect for you! It not only builds a foundation in Algorithmic Trading but also empowers you with the knowledge of different algorithmic trading strategies and regulations for setting up an algorithmic trading business. Be sure to check it out!

Disclaimer: All investments and trading in the stock market involve risk. Any decisions to place trades in the financial markets, including trading in stock or options or other financial instruments is a personal decision that should only be made after thorough research, including a personal risk and financial assessment and the engagement of professional assistance to the extent you believe necessary. The trading strategies or related information mentioned in this article is for informational purposes only.

File in the zip archive: Complete blog with codes in a Python notebook format


text:
# Escape Velocity

by SoccaCrazy24
Tags: escape, velocity
 P: 183 Consider that, in circular motion: $$a=\frac{v^2}{r}=\omega^2r$$ If a > g, then what would happen? Does this help you? Regards, Sam


text:
# Introduction

The $$\n\$$-ellipse is a generalization of the ellipse with possibly more than two foci. Specifically, given $$\n\$$ points on the plane, called foci, the $$\n\$$-ellipse is the set of points of the plane whose sum of distances to the $$\n\$$ foci is equal to a constant $$\t\$$.

This challenge is about plotting the $$\n\$$-ellipse together with its interior; that is, the set of points whose sum of distances to the $$\n\$$ foci is less than or equal to equal to $$\t\$$. Note that the resulting set is always convex.

To simplify, only points with integer coordinates need to be considered.

# The challenge

Input:

• Number of points $$\n\$$: positive integer;
• List of $$\n\$$ foci (possibly repeated) with integer coordinates, $$\(x_1, y_1)\$$, ..., $$\(x_n, y_n)\$$;
• Threshold $$\t\$$: positive integer.

Output:

An image representing all points $$\P\$$ with integer coordinates such the sum of Euclidean distances from $$\P\$$ to the $$\n\$$ foci is less than or equal to $$\t\$$.

Each pixel in the image should correspond to a point with integer coordinates; that is, pixel size is $$\1 \times 1\$$. Alternatively, you can use a finer resolution (smaller pixels), or vector graphics.

The image should consistently use two different colours for points satisfying the condition ("active" pixels) or not satisfying it ("inactive").

• Graphical output is required, in any format. (ASCII art is not allowed because of size limitations and aspect ratio).

• Both axes should have the same scale. The $$\x\$$ axis should be horizontal, and can consistently increase left to right or right to left. Similarly, the $$\y\$$ axis should be vertical and can consistently increase upwards or downwards.

• Axis labels, auxiliary grid lines and similar elements are allowed.

• The output image can have an arbitrarily large "frame" of inactive pixels around the set of active pixels.

• The set of active points is guaranteed to be non-empty.

• The code should work in theory for inputs containing arbitrarily large numbers. In practice, it is acceptable if the program is limited by time, memory or data-type restrictions.

• Shortest code in bytes wins.

# Test cases

Input is shown as a list of $$\n\$$ foci defined as coordinate pairs, followed by $$\t\$$, and an optional comment.

In the output images, $$\x\$$ axis increases left to right, $$\y\$$ axis increases upwards.

(5,8), 100  % A circle

(0,0), (70, 150), 220  % An ellipse

(-100, 0), (100, 0), (100, 0), 480  % Repeated point

(-90, -90), (-90, 90), (90, -90), (90, 90), (90, -90), 670  % Repeated; arbitrary order

(200, 600), (320, -60), (-350, 220), (-410, 130), (40, -140), 2100

(-250, -250), (-250, 250), (250, -250), 1000

(-250, -250), (-250, 250), (250, -250), 1200

(-390, 0), (-130, 120), (130, -120), (390, 0), 1180  % Finally, a pair of lips?

• Sorry for a week title Mar 4 '20 at 11:11
• I read through the link on acceptable input/output methods. Can we output a gray-scale image in the format of a matrix of floats between 0 and 1?
– RGS
Mar 4 '20 at 14:31
• Now I know where guitar picks come from! Mar 4 '20 at 16:07
• @Kaddath Those were my thoughts too (guitar player here) Mar 4 '20 at 16:30
• @LuisMendo What about this? By the way, I like the title. Mar 7 '20 at 0:16

# Python 3 + numpy + pylab, 168163 156 bytes

import numpy,pylab
f,t=eval(input())
s=max(map(abs,sum(f,())))+t
k=numpy.r_[-s:s+1]
pylab.imsave('a.png',sum(((k-x)**2+(k[:,None]-y)**2)**.5for x,y in f)>t)

Outputs to a file called a.png, the y-axis increases downwards.

Sample outputs (click for full-res versions)

[(5, 8)], 100:

[(0, 0), (70, 150)], 220:

[(-90, -90), (-90, 90), (90, -90), (90, 90), (90, -90)], 670:

[(-390, 0), (-130, 120), (130, -120), (390, 0)], 1180:

# Python 2, 152 bytes

f,t=input()
s=max(map(abs,sum(f,())))+t
w=s*2+1
k=range(-s,s+1)
print'P1',w,w
for y in k:
for x in k:print+(sum(((a-x)**2+(b-y)**2)**.5for a,b in f)>t)

Try it online!

Outputs a PBM file to STDOUT. Sample output:

• Any reason not to output to a file called a to save 4 bytes? Mar 4 '20 at 15:48
• @Noodle9 the imsave function needs the file extension to determine the desired file format.
– ovs
Mar 5 '20 at 12:34

# Jelly, 36 35 bytes

«/_⁹r»/+¥,þ/_Ɱ⁸²§½S>⁹µ;L;ZL$;1ṚK”P;  Try it online! A full program that takes a list of lists of integers as the first argument, representing the foci, and the threshold as an integer as its second argument. Prints a PBM file to STDOUT. ### Example • First argument: (0,0), (7, 15) • Second argument: 220 • Output (converted to PNG): # Haskell + Gloss, 165 bytes import Graphics.Gloss i=fromIntegral h t=[i(-t)*3..i t*3] d f t=display(InWindow""(t*3,t*3)(0,0))red$Line[(x,y)|x<-h t,y<-h t,i t>sum[sqrt$(x-u)^2+(y-v)^2|(u,v)<-f]]  Draws the shapes to the screen. It sometimes draws the images REALLY BIG, so I have cropped accordingly. main = d [(5,8)] 100 main = d [(0,0), (70, 150)] 220 main = d [(-100, 0), (100, 0), (100, 0)] 480 main = d [(-250, -250), (-250, 250), (250, -250)] 1000 ## JavaScript (ES7), 212 196 bytes with(Math)f=(c,l,...p)=>{m=l-max(...p) c.height=c.width=s=min(...p)+l+m for(i=0;i<s;i++)for(j=0;j<s;j++){for(d=k=0;1/p[k];)d+=hypot(p[k++]+m-i,p[k++]+m-j) d>l||c.getContext2d.fillRect(i,j,1,1)}} eval(f(c,${e.value}))
input{position:fixed}
<input id=e oninput=eval(f(c,${e.value})) value="100,-25,-25,-25,25,25,-25"><canvas id=c> Outputs via first parameter which is a canvas element to draw on, second parameter is the threshold, then the remaining parameters are the loci. Edit: Saved 16 bytes thanks to @Arnauld, however the code is now dead slow and will trigger slow script warnings for large thresholds. • @LuisMendo Running the snippet itself doesn't actually trigger the input event that calls the function, so the suggested input doesn't run unless you edit the value e.g. by appending a space. – Neil Mar 4 '20 at 20:53 • 196 bytes, I think (not really tested). Mar 4 '20 at 23:53 • @LuisMendo I've added an extra line to calculate the suggestion automatically. – Neil Mar 5 '20 at 0:39 • @Arnauld Boy that's slow :-( – Neil Mar 5 '20 at 0:39 # APL (Dyalog Unicode), 67 bytes (SBCS) ⎕IO←0 +{(⎕NEW'Bitmap'(⊂'Cbits'(⍵≥+⌿⍺∘.(.5*⍨-+.×-)(⊢,-)⍳2×⍵ ⍵))).MakePNG}⊢  ## NB • This image is very low contrast (made from a boolean matrix) - the output can be white interior shape on black background by multiplying the boolean matrix: +{(⎕NEW'Bitmap'(⊂'Cbits'(16777215×~⍵≥+⌿⍺∘.(.5*⍨-+.×-)(⊢,-)⍳2×⍵ ⍵))).MakePNG}⊢ • Images are transposed compared to many other solutions, but you can flip the indices of each focus • Only works on Dyalog for Windows ## Explanation This function is a 3-train (fork), the arguments to the inner function are the results of the outer functions: +  + (plus): Translate the image by adding distance to the foci (please comment if this is a misinterpretation of the challenge) ⊢  ⊢ (right): given two arguments (left and right), return the right argument APL is parsed (glossing over some binding rules) as Brackets, long Right scope, short Left scope. We break it down as follows: ⍳2×⍵ ⍵  Indices from 0 to 2×⍵ in 2 dimensions (our integer pixels) (⊢,-)⍳2×⍵  (⊢,-) (Same catenate negate) to cover positive and negative coordinates (.5*⍨-+.×-)  Euclidean distance from focus to pixel... ⍺∘.(.5*⍨-+.×-)  ... applied between each focus and every pixel +⌿⍺∘.(.5*⍨-+.×-)  Sum the distances between all foci and each pixel (n-ellipse) ⍵≥+⌿⍺∘.(.5*⍨-+.×-)  Boolean matrix where 1 indicates interior (active) pixels (i.e. sum of distances between foci and pixel is less than t) (⎕NEW'Bitmap'(⊂'Cbits'(*boolean matrix*))).MakePNG  Output signed integer PNG. This can be written to file using ∇ WritePNG png;tn tn←'img.png'⎕NCREATE 0 png ⎕NAPPEND tn 83 ⎕NUNTIE tn ∇  ## Output Images First is to demonstrate, the written PNG has very low contrast f ← +{(⎕NEW'Bitmap'(⊂'Cbits'(⍵≥+⌿⍺∘.(.5*⍨-+.×-)(⊢,-)⍳2×⍵ ⍵))).MakePNG}⊢ WritePNG 5 8 f 100  Below I'm using the black and white version. WritePNG (0 0)(70 150) f 220  WritePNG (¯100 0)(100 0)(100 0) f 480  Very resource hungry algorithm so I did a low res version of this WritePNG (20 60)(32 ¯6)(¯35 22)(¯41 13)(4 ¯14) f 210  • Can you perhaps include a couple of screenshots of your results? – RGS Mar 4 '20 at 23:20 # Red, 300 bytes Red[]f: func[p t][a: :absolute m: last sort collect[foreach n p[keep a n/x keep a n/y]]s: t + m * 2 u: as-pair s s i: make image! reduce[u]k: 0 repeat y s[repeat x s[d: 0 foreach a p[d: d +(sqrt x - t - m - a/x ** 2 +(y - t - m - a/y ** 2))]k: k + 1 if d <= t[i/:k: 0.0.255]]]save/as %1.jpg i 'jpeg]  f [5x8] 100 f [0x0 70x150] 220 f [-100x0 100x0 100x0] 480 f [200x600 320x-60 -350x220 -410x130 40x-140] 2100 f [-390x0 -130x120 130x-120 390x0] 1180 Note: I scaled all the images to 256x256 pixels and croped the last two images so that they don't take too much space - they had a large "frame" # Applescript + grapher, 601 This will almost certainly be the longest answer, but I thought this would be a fun challenge in Applescript. on f(l, t) set y to 0 tell application "Grapher" to activate tell application "System Events" keystroke "n" using command down keystroke return keystroke "a" using command down repeat with d in l keystroke "sqrt((x-"&d's item 1&")^2)+(y-"&d's item 2&")^2))+" set y to y+d's item 2 end repeat click application process "Grapher"'s menu bar 1's menu bar item 8's menu 1's menu item 3 set c to l's items count keystroke " "&(y-t)/c as text&" "&(y+t)/c as text&" " keystroke "c" using command down keystroke " " keystroke "v" using command down keystroke return&"0<"&t as text&return end tell end f  This is an applescript function. It may be called as follows for the testcases: my f({{5, 8}}, 100) my f({{0, 0}, {70, 150}}, 220) my f({{-100, 0}, {100, 0}, {100, 0}}, 480) my f({{-90, -90}, {-90, 90}, {90, -90}, {90, 90}, {90, -90}}, 670) my f({{200, 600}, {320, -60}, {-350, 220}, {-410, 130}, {40, -140}}, 2100) my f({{-250, -250}, {-250, 250}, {250, -250}}, 1000) my f({{-250, -250}, {-250, 250}, {250, -250}}, 1200) my f({{-390, 0}, {-130, 120}, {130, -120}, {390, 0}}, 1180)  Output looks like this, e.g. for the 5th testcase: • This looks a lot like natural language:-) Mar 5 '20 at 19:28 • @LuisMendo Apparently that's the idea en.wikipedia.org/wiki/AppleScript#Natural_language_metaphor Mar 5 '20 at 19:32 • Very much Apple-like Mar 5 '20 at 20:10 # MATLAB + DIPimage toolbox, 133121117 111 bytes (shaved off 12 16 bytes thanks to flawr and LuisMendo) function s(c,r),c=c-min(c)+r;a=zeros(max(c)+r);l='corner';for x=c',a=a+hypot(xx(a,l)-x(1),yy(a,l)-x(2));end,a>r  Use: c = [-390, 0; -130, 120; 130, -120; 390, 0]; r = 1180; s(c,r)  In a more readable fashion: function s(c,r) c=c-min(c)+r; % offset coordinates so all points within shape have non-negative coordinates a=zeros(max(c)+r); % compute image size and create empty output image l='corner'; % long argument used twice below for x=c' a=a+hypot(xx(a,l)-x(1),yy(a,l)-x(2)); % add distance to each point end a>r % threshold and display (leaving semi-colon off causes graphical display of image)  # Bash + Standard utilities (but no graphics library), 194190 186 bytes t=$1
shift
L=H=$1 for n do L=$[n<L?n:L]
H=$[n>H?n:H] done echo P2$[L-=t,H+=t+1,H-L] $[H-L] 1 for((x=y=L;y<H;++x>=H?x=L,y++:0)){ bc -l<<<printf "sqrt(($x- %d)^2+($y- %d)^2)+"$@0\>\$t;}

Try it online! (If you do, though, try a tiny example: even the circle test times out on TIO.)

Input is passed in the arguments: first $$\t\$$, then $$\x_1\$$, $$\y_1\$$, $$\x_2\$$, $$\y_2\$$, ..., $$\x_n\$$, $$\y_n\$$.

Output is a PGM image on stdout. Save it in a file named something.pgm and view it in your favorite graphics program.

The x-axis goes from left to right, and the y-axis goes from top to bottom. White is the background, and black is the n-ellipse being drawn.

For example, to view the first example (the circle), you could execute:

lips 100 5 8 > lips-circle.pgm

(where the program is named lips). Then open the imagelips-circle.pgm. Here's the result:

Here's the second test, the output of lips 220 0 0 70 150 > lips-ellipse.pgm:

Next is lips 480 -100 0 100 0 100 0 > lips-repeatedpoint.pgm:

Here is lips 670 -90 -90 -90 90 90 -90 90 90 90 -90 > lips-repeated-arbitaryOrder.pgm:

The next test is lips 2100 200 600 320 -60 -350 220 -410 130 40 -140 > lips-shape1.pgm:

This script is very slow (but runnable). I will post more sample images as my computer gets around to finishing them :) ....

Well, after waiting overnight again, the other tests are finally done!

Here's the guitar pick: lips 1000 -250 -250 -250 250 250 -250 > lips-guitarpick.pgm

The next test is lips 1200 -250 -250 -250 250 250 -250 > lips-shape2.pgm:

And finally the pair of lips! lips 1180 -390 0 -130 120 130 -120 390 0 > lips-pairOfLips.pgm:

# Shadertoy (GLSL, non-competing) 125 bytes + input data

vec2 a[]=vec2[](vec2(-390, 0),vec2(-130,120),vec2(130,-120),vec2(390,0));
float d=1180.;

void mainImage(out vec4 o,vec2 c){c-=iResolution.xy/2.;o-=o;for(int i=0;i<a.length();o.a+=length(c-a[i++]));o.b=step(o.a,d);}

As the input array's size is known, this could be further reduced by replacing the a.length() by n.

• Why is this non-competing? Mar 29 '20 at 21:26
• Also, why don't you do the further reduction you suggest in your answer? Mar 29 '20 at 21:27
• it doesn't receive input, I could change it to a uniform value but that's still not like in other languages. Mar 29 '20 at 21:57

# Desmos, 50 49 bytes

\sum_{n=1}^c \sqrt{(x-a[n].x)^2+(y-a[n].y)^2}<=t

-1 byte from finding a new quirk in Desmos' syntax while working on another problem.

View it online

It just kinda... works. Input n in the variable c, t in the variable t, and the points as an array a of coordinates.

Lips example:

Guitar pick example:

Note: If you try to copy the equation from the editor, you'll find a bunch of \lefts and \rights and such. These have been golfed out of this answer, as you can paste this text in and it works. The editor automatically re-adds those when parsing input from the clipboard, but it accepts the code without them.

Note two: I'm not 100% sure if this counts due to the infill being transparent, but hopefully it's ok. If not, there's a couple options:

1. Stack a bunch of whole bunch of these equations on top of each other. I tried this and it turns out that a quirk of their rendering engine is that the fill color can never drop below #010101, while the outline is at #000000. Additionally, the area is a little larger than would be ideal because the outline looks like part of the fill.

2. Rewrite the n-ellipse equation in parametric form. I don't know if this is possible or not, but if it is, Desmos's enhanced visual options for parametric equations allows us to remove the outline.

• Right tool for the job :-) Fill being half-transparent is ok Aug 23 '20 at 1:01