text:
# Equation of the Day #6: Newton’s Second Law

Born on Christmas Day, 1642, Isaac Newton was one of the most brilliant minds the world has ever seen. He invented calculus, discovered that white light is composed of all the colors of the rainbow, formulated the law of universal gravitation (which we now use to send people and robots into orbit around Earth and other objects in the Solar System), and his famous laws of motion (among a host of other achievements). Today I’m focusing on his Second Law of Motion, which you’ve probably seen written as

${\bf F}=m {\bf a},$

where F is force, m is mass, and a is acceleration. The boldface formatting indicates a vector, which is a mathematical entity that has both magnitude (or “strength”) and direction (up/down, left/right, forward/backward). What this means is that an object whose speed (the magnitude of velocity) or direction of travel is changing must be subject to an unbalanced force. This ties in to Newton’s First Law of Motion, which states that the natural state of an object is to move in a straight line at constant speed. However, the above equation is actually not fully general – it only applies when the accelerating object has constant mass. This works fine for projectile motion of a pitcher hurling a baseball or a cannonball launched at an enemy fortress, but it fails for, say, a rocket delivering a payload to the International Space Station, since fuel is being burned and ejected from the rocket. In general, Newton’s Second Law is given by

${\bf F}=\dfrac{d{\bf p}}{dt},$

where p is the total momentum of the system. Momentum is a measurement of the “oomf” that an object has due to its motion. Explicitly, it is the product of mass and velocity, or ${\bf p} = m{\bf v}$. The expression d/dt indicates the rate of change of a quantity, in this case momentum, over time. So Newton’s Second Law states that a change in an object’s motion is due to an unbalanced force, which sounds like what I said for the equation ${\bf F}=m {\bf a},$ but this takes into account a change in mass as well.

Conceptually, this was a big breakthrough at the time and is something that students in introductory physics classes struggle with today. (Note: Newton figured all this stuff out and more by the time he was 26. Let that sink in.) On a quantitative level, this law allows us to predict the motions of objects that subjected to certain kinds of forces. If we know the nature of the force an object is subjected to, we can make predictions of what path the object will travel along. Similarly, if we know the trajectory of an object, we can predict the behavior of the force acting upon it. This was how Newton figured out the nature of gravity. Observations of his time showed that, from a sun-centered view of the Solar System, the planets orbit in ellipses, not circles. Newton surmised that this meant that the force of gravity must diminish with the square of the separation between two objects.

But how? The presence of acceleration in Newton’s Second Law indicates a differential equation. Acceleration is the rate of change of velocity, which itself is the rate of change of position (which I will denote as r). So, taking that into account, we get the equation

$\dfrac{d^2{\bf r}}{dt^2} = \dfrac{{\bf F}}{m}$

This can be solved for an expression of position as a function of time when we know the form that F takes. If the force is constant in space (like gravity near Earth’s surface), r takes on the form of parabolic motion. If the force obeys Hooke’s Law, $F = -k \Delta r$, then the solution is simple harmonic oscillation. And if the force obeys an inverse square law, the position takes the functional form of conic sections, which are ellipses for the planets bound to the Sun in our solar system.

Conic sections: 1. Parabola, 2. Circle (bottom) and ellipse (top), 3. Hyperbola. (Image: Wikimedia Commons)

This is the model we use when we send objects into orbit or when we want to model a baseball being hit out of the park (among many other things), and it has been extended to explain phenomena that Newton himself could not have imagined. When scientists or mathematicians describe equations as beautiful, this pops into my head. It has such a simple form, but explains a wealth of real phenomena.

format: plain


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# Boost C++ Libraries

...one of the most highly regarded and expertly designed C++ library projects in the world.

This is the documentation for a snapshot of the develop branch, built from commit d4128b739e.

### boost/vmd/equal.hpp

//  (C) Copyright Edward Diener 2011-2015
//  Use, modification and distribution are subject to the Boost Software License,
//  Version 1.0. (See accompanying file LICENSE_1_0.txt or copy at

#if !defined(BOOST_VMD_EQUAL_HPP)
#define BOOST_VMD_EQUAL_HPP

#include <boost/vmd/detail/setup.hpp>

#include <boost/vmd/detail/equal.hpp>

/*

The succeeding comments in this file are in doxygen format.

*/

/** \file
*/

/** \def BOOST_VMD_EQUAL(sequence,...)

\brief Tests any two sequences for equality.

sequence     = First sequence.

...          = variadic parameters, maximum of 2.

The first variadic parameter is required and is the second sequence to test.
The optional second variadic parameter is a VMD type as a filter.

The macro tests any two sequences for equality. For sequences to be equal the
VMD types of each sequence must be equal and the individual elements of the
sequence must be equal. For Boost PP composite types the macro tests that
the composite types have the same size and then tests that each element
of the composite type is equal. This means that all elements of a composite
type must be a VMD type in order to use this macro successfully.

The single optional parameter is a filter. The filter is a VMD type which specifies
that both sequences to test must be of that VMD type, as well as being equal to
each other, for the test to succeed.

returns   = 1 upon success or 0 upon failure. Success means that both sequences are
equal and, if the optional parameter is specified, that the sequences are
of the optional VMD type.

*/

#define BOOST_VMD_EQUAL(sequence,...) \
BOOST_VMD_DETAIL_EQUAL(sequence,__VA_ARGS__) \
/**/

/** \def BOOST_VMD_EQUAL_D(d,sequence,...)

\brief Tests any two sequences for equality. Re-entrant version.

d            = The next available BOOST_PP_WHILE iteration.

sequence     = First sequence.

...          = variadic parameters, maximum of 2.

The first variadic parameter is required and is the second sequence to test.
The optional second variadic parameter is a VMD type as a filter.

The macro tests any two sequences for equality. For sequences to be equal the
VMD types of each sequence must be equal and the individual elements of the
sequence must be equal. For Boost PP composite types the macro tests that
the composite types have the same size and then tests that each element
of the composite type is equal. This means that all elements of a composite
type must be a VMD type in order to use this macro successfully.

The single optional parameter is a filter. The filter is a VMD type which specifies
that both sequences to test must be of that VMD type, as well as being equal to
each other, for the test to succeed.

returns   = 1 upon success or 0 upon failure. Success means that both sequences are
equal and, if the optional parameter is specified, that the sequences are
of the optional VMD type.

*/

#define BOOST_VMD_EQUAL_D(d,sequence,...) \
BOOST_VMD_DETAIL_EQUAL_D(d,sequence,__VA_ARGS__) \
/**/


text:
# Carriage-return and line-feed in web2cTeX

In web2cTeX (TeX Live) I observed this behavior:

(after each of the following perl commands run: "tex xyz.tex; xdvi xyz.dvi")

1. Each \r and \n separately produces one line:

perl -e 'print "\\obeylines a\rb\\bye"' > xyz.tex
perl -e 'print "\\obeylines a\nb\\bye"' > xyz.tex

1. \r and \n combined in various ways produce two lines (thus - a paragraph), as expected, except one case:

perl -e 'print "a\r\rb\\bye"' > xyz.tex
perl -e 'print "a\r\nb\\bye"' > xyz.tex # why?
perl -e 'print "a\n\rb\\bye"' > xyz.tex
perl -e 'print "a\n\nb\\bye"' > xyz.tex

--
\r = ASCII '15
\n = ASCII '12

IMHO, underlying C library (if it has anything to do with it at all) must not treat \r\n as an end-of-line, because I'm on Linux, not Windows.

So please somebody clarify why \r\n is not treated separately.

• Line ends are normalised: both Linux and Windows line ends should be treated identically so sources are portable. – Joseph Wright Jan 21 '15 at 9:29
• Welcome to TeX.SX! The developers of Web2C decided for this behavior because it's common that files made on Windows (which uses CR-LF) are ported on a different system. Requiring the user to change the end-of-record delimiters would hinder portability. – egreg Jan 21 '15 at 9:30

There are three steps when inputting the line by TeX.

• The line is read from the file using system dependent readln procedure.
• The end of line mark (if exists) is removed, the spaces immediately before it (if exist) are removed too and the rest is saved to line buffer.
• The line buffer is appended by \endlinechar value, if it is nonnegative (ASCII 13 alias ^^M by default). Note that ^^M has catcode 5 by default and this is the reason why the ends of lines behave in TeX as behave.

The system dependent readln procedure is implemnted in web2c TeX in src/texk/web2c/lib/eofeoln.c file where is:

void
{
int c;
while ((c = getc (f)) != '\n' && c != '\r' && c != EOF)
;
if (c == '\r' && (c = getc (f)) != '\n' && c != EOF)
ungetc (c, f);
}

Note, that the special case CRLF is handled here.

TeX relies on a “change file” for system dependent adjustments. One of these adjustments is making TeX aware of what method the operating system uses for marking the end of a record (a line in a text file, speaking less formally).

1. CR+LF (for instance, MS-DOS)
2. CR (for instance, Mac OS before version 10)
3. LF (for instance, Unix)
4. nothing, for systems that used fixed length records

In the first implementations of TeX, this was taken literally; for instance OzTeX used CR as record terminator and files coming from DOS systems without conversion could be compiled wrongly, because some DOS editors used LF+CR as record terminator.

In order to enhance portability between different file systems, the developers of Web2C decided to be “record terminator agnostic”. When a file is opened by TeX, its first lines are examined and the record terminator used is determined; then it's changed into LF and passed to TeX for further processing.

If you do

perl -e 'print "\\obeylines a\rb\r\\input zzz \\bye"' > xyz.tex
perl -e 'print "\\obeylines c\nd\\bye"' > zzz.tex

the output of pdftex xyz would be

a
b
c
d

but also the output of pdftex '\catcode^^L=12 \input xyz' wouldn't be different (note that plain.tex does

\catcode\^^L=\active \outer\def^^L{\par}

so ^^L would be interpreted as outer \par; by setting its catcode to 12 we tell it's printable, but nothing appears even if \n is used as record terminator in zzz.tex.

So one doesn't have to know what operating system a file comes from: inputting it will do “the right thing”.

Any Web2C based TeX distribution will thus behave the same when fed the same file.

What source file does this? I don't know, but I'm not going to read the complete sources.


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# In LYX, how can I eliminate or change page numbering in my Appendix Chapter*?

I have tried everything suggested on the web but nothing has worked - yet!

I am writing a thesis with 6 numbered Chapters. For page numbering I am using

\pagenumbering{arabic}

and

\pagestyle{headings}

starting at Chapter 1, which comes out like this:

After my six Chapters I add my 'Bibliography', and after that 'Appendices' as an unnumbered *Chapter. However, the page numbers showing in my Appendices have kept in 'memory' the Bibliography (previous) chapter. See here:

I would like to change this so that it shows 'Appendices....page number' or just 'page number' or as a last resort by eliminating page numbering in the Appendices *Chapter altogether.

Ella

• How did you add appendix in your lyx document? Did you add it as a separate chapter or via document-->start appendix here? – Yorgos Jun 16 '15 at 14:35
• I added it as a separate Chapter and indluded it in the main document. – Ella Jay Jun 16 '15 at 20:38
• I believe that you should choose the second method. LyX has an "automatic" way to set appendices, which can be seen as a separate document class. But you can try to enter the following commands before your "chapter with appendix": '\cleardoublepage' (if you use a book document class)and then '\pagenumbering{roman or arabic}', '\setcounter{page}{the number you want your appendix to start}' – Yorgos Jun 17 '15 at 5:51

I just solved the case myself. The solution is to change the \pagestyle to {plain} at the section I only want numbers showing. My code which now does this is shown below simplified:

\begin{document}

Include Chapter 1

...

Include Chapter 6

\addcontentsline{toc}{chapter}{Bibliography}

\pagestyle{plain}

Include Appendices

\end{document}

Thanks you anyhow!


text:
Viscous Flow Rescheduling of 2 Lectures
Select all convenient time slots
 Tue 20 Nov 19:00--20:15 10 Wed 21 Nov 19:00--20:15 0 Fri 23 Nov 9:00--10:15 12 Fri 23 Nov 16:30--17:45 9 Tue 27 Nov 19:00--20:15 4 Wed 28 Nov 19:00--20:15 0 Fri 30 Nov 9:00--10:15 7 Fri 30 Nov 10:30--11:45 8 Fri 30 Nov 12:00--13:15 2 Tue 11 Dec 19:00--20:15 1 Wed 12 Dec 19:00--20:15 0
Poll ended at 9:00 am on Monday November 19th 2018. Total votes: 53. Total voters: 20.
 Comment by Prof. Parent Based on your votes, we will choose 2 time slots in class at 9:00 on Monday Nov 19th. If a popular choice is impossible for you, post a comment below and explain why.
 11.14.18
 Comment by Student 201627118 Dear Professor, I have class every Friday from 9:00 until 11:50. Thank you for your understanding.
 11.18.18
 Comment by Prof. Parent Which class is this?
 Comment by Student 201327126 Dear professor, I also cannot take class on Friday morning because there are other classes from 9 am to 1 pm. Thank you for your understanding.
 Comment by Prof. Parent The rescheduled lectures will take place on Tuesday Nov 20 at 19:00 and on Friday Nov 23 at 16:30.
 11.19.18
 Comment by Student 201428239 Dear, Professor I have a Samsung Dream Class which is schduled every Friday from 4 pm to 6 pm. So I can't make a class on Friday afternoon. I am sorry to say this late.
 Comment by Prof. Parent Please write this down in the Class Absence thread so that I don't forget it.
 Comment by Prof. Parent The rescheduled classes will take place in room 9302.
 $\pi$


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Once someone from the engineering (or physics?) department of some university went came to see Joseph Bernstein and asked if he knew a formula for a conformal mapping of the interior of a regular $n$-gon to the upper half-plane. Bernstein knew the formula, but decided to first ask what the person needed it for. The reply was: "Well, you see, what I really need is a formula for the unit disk, but that's probably too complicated, so I decided to find out the formula for the $n$-gon first and then take the limit."
Once someone from the engineering (or physics?) department of some university went to see Joseph Bernstein and asked if he knew a formula for a conformal mapping of the interior of a regular $n$-gon to the upper half-plane. Bernstein knew the formula, but decided to first ask what the person needed it for. The reply was: "Well, you see, what I really need is a formula for the unit disk, but that's probably too complicated, so I decided to find out the formula for the $n$-gon first and then take the limit."


text:
# Sum of binomial coefficients $\sum_{k=n}^{r}\binom{k}{n}$

Is is possible to find the sum of a binomial coefficient series like: $\sum_{k=n}^{r}\dbinom{k}{n}$? Just a random thought.

• If you have meant $\displaystyle\sum_{r=1}^n r\binom nr$ $$\displaystyle r\binom nr=r\dfrac{n\cdot(n-1)!}{r\cdot(r-1)!\{n-1-(r-1)\}!}=n\binom{n-1}{r-1}$$ – lab bhattacharjee Apr 7 '15 at 18:57
• ${k \choose n} = 0$ if $k < n$, $1$ if $k=n$. Maybe you'd rather do $$\sum_{k=n}^r {k \choose n}$$ – Robert Israel Apr 7 '15 at 19:15
• yeah yeah im so sorry... i didnt notice the flaw in my question – user220382 Apr 7 '15 at 19:18

$$\sum_{k=n}^r {k \choose n} = \dfrac{r+1-n}{n+1} {r+1 \choose n}$$ Prove by induction on $r$.
EDIT: Actually this can be written as $${r+1 \choose n+1}$$ in which form it has a nice combinatorial interpretation. Suppose you want to have $r+1$ objects numbered $1$ to $r+1$, and you want to choose $n+1$ of them. How many ways to do it? Divide into cases according to the highest-numbered object chosen. If the highest-numbered object chosen has number $k+1$, the remaining $n$ objects can be chosen from $1,\ldots, k$ in ${k \choose n}$ ways, where of course we need $k \ge n$. Therefore $${{r+1} \choose {n+1}} = \sum_{k=n}^r {k \choose n}$$


text:
## Existence of global solutions to nonlinear fuzzy Volterra integro-differential equations.(English)Zbl 1243.45011

The authors discuss the existence of global solutions to nonlinear fuzzy Volterra integro-differential equations. They introduce mixed solutions as new solutions and prove existence and uniqueness of global solutions for fuzzy initial value problems. One example is given to illustrate the theory discussed.

### MSC:

 45J05 Integro-ordinary differential equations 45G10 Other nonlinear integral equations 26E50 Fuzzy real analysis 45L05 Theoretical approximation of solutions to integral equations
Full Text:

### References:

 [1] Chang, S.S.L.; Zadeh, L.A., On fuzzy mapping and control, IEEE trans. syst. man cybern., SMC-2, 30-34, (1972) · Zbl 0305.94001 [2] Hullermeier, E., An approach to modelling and simulation of uncertain dynamical systems, Internat. J. uncertain. fuzziness knowledge-based systems, 5, 117-137, (1997) · Zbl 1232.68131 [3] Nieto, J.J., The Cauchy problem for continuous fuzzy differential equations, Fuzzy sets and systems, 102, 259-262, (1999) · Zbl 0929.34005 [4] O’Regan, D.; Lakshmikantham, V.; Nieto, J.J., Initial and boundary value problems for fuzzy differential equations, Nonlinear anal., 54, 405-415, (2003) · Zbl 1048.34015 [5] Agarwal, R.P.; O’Regan, D.; Lakshmikantham, V., A stacking theorem approach for fuzzy differential equations, Nonlinear anal., 55, 299-312, (2003) · Zbl 1042.34027 [6] Georgiou, D.N.; Nieto, J.J.; Rodríguez-López, R., Initial value problems for higher-order fuzzy differential equations, Nonlinear anal., 63, 587-600, (2005) · Zbl 1091.34003 [7] Agarwal, R.P.; O’Regan, D.; Lakshmikantham, V., Viability theory and fuzzy differential equations, Fuzzy sets and systems, 151, 563-580, (2005) · Zbl 1074.34009 [8] Khastan, A.; Bahrami, F.; Ivaz, K., New results on multiple solutions for nthorder fuzzy differential equations under generalized differentiability, Bound. value probl., (2009), p. 13, Art. ID 395714 · Zbl 1198.34006 [9] Khastan, A.; Nieto, J.J., A boundary value problem for second order fuzzy differential equations, Nonlinear anal., 72, 3583-3593, (2010) · Zbl 1193.34004 [10] Arshad, S.; Lupulescu, V., On the fractional differential equations with uncertainty, Nonlinear anal., 74, 3685-3693, (2011) · Zbl 1219.34004 [11] Choudary, A.D.R.; Donchev, T., On Peano theorem for fuzzy differential equations, Fuzzy sets and systems, 177, 93-94, (2011) · Zbl 1246.34003 [12] J.J. Nieto, R. Rodríguez-López, M. Villanueva-Pesqueira, Exact solution to the periodic boundary value problem for a first-order linear fuzzy differential equation with impulses, Fuzzy Optim. Decis. Mak, in press (doi:10.1007/s10700-011-9108-3). [13] Bede, B.; Rudas, I.J.; Bencsik, A.L., First order linear fuzzy differential equations under generalized differentiability, Inform. sci., 177, 1648-1662, (2007) · Zbl 1119.34003 [14] Xu, J.; Liao, Z.; Nieto, J.J., A class of linear differential dynamical systems with fuzzy matrices, J. math. anal. appl., 368, 54-68, (2010) · Zbl 1193.37025 [15] Khastana, A.; Nieto, J.J.; Rodríguez-López, R., Variation of constant formula for first order fuzzy differential equations, Fuzzy sets and systems, 177, 20-33, (2011) · Zbl 1250.34005 [16] Bede, S.G.; Gal, B., Generalizations of the differentiability of fuzzy-number-valued functions with applications to fuzzy differential equations, Fuzzy sets and systems, 151, 581-599, (2005) · Zbl 1061.26024 [17] Puri, M.L.; Ralescu, D.A., Differentials of fuzzy functions, J. math. anal. appl., 91, 552-558, (1983) · Zbl 0528.54009 [18] Wu, C.; Gong, Z., On Henstock integral of fuzzy-number-valued functions. I, Fuzzy sets and systems, 120, 523-532, (2001) · Zbl 0984.28010 [19] Gal, S.G., Approximation theory in fuzzy setting, (), 617-666 · Zbl 0968.41018
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# About functions on non-measurable sets

Let $$E$$ be a Lebesgue measurable subset of $$\mathbb{R}$$. We say that a function $$f:E\longrightarrow \mathbb{R}$$ is Lebesgue measurable if, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable (equivalently, "$$>$$" can be replaced by "$$\geq$$", "$$<$$", or, "$$\leq$$").

Now we consider a subset $$E$$ of $$\mathbb{R}$$, which is NOT Lebesgue measurable. The following questions seem to be interesting:

(1) Is it possible to define/have a function $$f:E\longrightarrow \mathbb{R}$$ such that, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable ?

(2) Is it possible to define/have a continuous function $$f:E\longrightarrow \mathbb{R}$$ such that, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable ?

Does anybody have answers/counterexamples concerning these (possible) pathologies? Or, these questions make sense ?

• Sets which are not Lebesgue measurable are quite pathological. Even finding such a set requires axiom of choice. I wouldn't worry about them so much. – Jakobian Jul 3 '19 at 21:36
• $(1)$ is false, because $$E = \bigcup_{k \in \mathbb{Z}} \{x \in E : f(x)>k\}$$ and the countable union of measurable sets is measurable, contradiction. ] – MathematicsStudent1122 Jul 3 '19 at 21:37

No, (1) (and hence (2) as well) cannot happen. Remember that the measurable sets form a $$\sigma$$-algebra: in particular, the union of countably many measurable sets is measurable.
Suppose $$E,f$$ are as in (1). We have $$E=\bigcup_{\alpha\in\mathbb{R}}\{x\in E: f(x)>\alpha\}.$$ But every real number is bigger than some rational, so this is the same as $$\bigcup_{\alpha\in\mathbb{Q}}\{x\in E: f(x)>\alpha\}.$$ By assumption, this latter set is a countable union of measurable sets (since $$\mathbb{Q}$$ is countable), so $$E$$ was measurable to begin with.
• We could also have used $$\mathbb{Z}$$ in place of $$\mathbb{Q}$$, and it's certainly easier to see that $$\mathbb{Z}$$ is countable; but using $$\mathbb{Q}$$ as an "$$\mathbb{R}$$-substitute" is so useful that I want to do it here too even if it's a bit silly. The specific property that makes $$\mathbb{Q}$$ really useful is that it's dense in $$\mathbb{R}$$; obviously $$\mathbb{Z}$$ isn't. Of course that's unnecessary here, but it gets used extensively down the road.
Incidentally, I've phrased this as a proof by contradiction, but that's really unnecessary: what we're actually doing is showing that for any $$E,f$$, if $$\{x\in E: f(x)>\alpha\}$$ is measurable for each $$\alpha\in\mathbb{R}$$ then $$E$$ is a union of countably many measurable sets and hence measurable. This is a completely direct proof.


text:
×

# CONTEST PANEL

Author: Ildar Gainullin
Tester: Oleksandr Kulkov
Editorialist: Prateek Gupta

EASY

# PROBLEM

Given a binary string, you need to transform this into a string consisting of only zeros by applying a specific operation number of times. The operation allows you to choose a prefix of the string and flip all the bits i.e '0' to '1' and '1' to '0'.

# EXPLANATION

The key is to observe that we need to reverse iterate the string and maintain the state as to how many times we have applied the given operation. When you reverse iterate the string and apply the operation on some index, it will affect the whole prefix $S[0..\ index\ -\ 1]$. Having said that, when we arrive at a particular index in the string, we already know the number of operations that have been applied till now. Based on the digit at this index and number of operations applied, we decide what is the actual digit at this position. If it is '0', we move forward, else we again flip the bits for a prefix ending at this position. Below is the simple pseudo code which will help you to understand things much better.  operations = 0 for ( i = s.length() - 1 to i = 0 ) { if ( digit(i) ^ operations ) { operations ^= 1; } }

It must be clear from the above snippet, that in case the number of operations are even, the state of the current bit will remain same, otherwise (bit ^ 1). Since, we are traversing a string once, the overall complexity will be $O(N)$ where $N$ is the length of the given string. For details on the implementation, have a look at the author or tester's solution.

# SOLUTIONS

Author solution can be found here.
Tester solution can be found here.

This question is marked "community wiki".

20421223
accept rate: 0%

19.8k350498541

 1 nice solution @anurag_6767 and @osama_binladen editorials could have been more easy answered 03 Jan '18, 23:31 129●7 accept rate: 0%
 0 I am not able to understand it!! Can anyone describe it?? P.S: I passed all test cases of task #1 and got TLE in task #2! answered 31 May '17, 13:53 614●1●9 accept rate: 12%
 0 @dishant_18 check out my solution very easy approach solution link answered 31 May '17, 15:36 533●1●12 accept rate: 3%
 0 see my solution in c++ https://www.codechef.com/viewsolution/13967949 answered 01 Jun '17, 19:14 3●4 accept rate: 0%
 0 Check this solution... easily understandable... https://www.codechef.com/viewsolution/14225387 answered 12 Jun '17, 00:03 1 accept rate: 0%
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text:
# ant-user mailing list archives

##### Site index · List index
Message view
Top
From Shannon Wagner <swag...@smartmoney.com>
Subject RE: pulling out the code from SourceSafe
Date Mon, 16 Apr 2001 20:01:33 GMT
Dana,

Do you see the ss.exe command (complete with arguments) that Ant is trying
to execute in your ant log file?  It should be something that you could also
execute manually from a command prompt.  If not, it might give you insight
into what is being setup wrong.

- sw

-----Original Message-----
From: Peterson, Lance [mailto:lpeterson@verticore.com]
Sent: Monday, April 16, 2001 3:57 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Dana, there are actually two SSDIR variables: There is an environment
variable that points to the directory with srcsafe.ini (e.g.
"\\Ftcollins\FTCOLLINS_D\VSS\COMDev"), and there is VssGet attribute which
is the path to the SourceSafe client program (e.g. "c:\program
files\microsoft visual studio\common\vss").  They are NOT the same, but
that's okay since task attributes in Ant do not collide with OS env
variables.  You just have to remember which is which.
Ant itself does not actually rely on the SSDIR environment variable.  SS.EXE
uses the env variable to find the correct srcsafe.ini.  The VssGet task, on
the other hand, uses the ssdir attribute to find SS.EXE.
BTW, you don't have to set the SSDIR environment variable in a batch file.
You can set it permanently in the System control panel.  This works great if
you're only using one SourceSafe database (as I am).  If you use more than
one SS database, then you can use a batch file to set SSDIR to the
appropriate database for each project.
Best regards,
Lance Peterson
-----Original Message-----
From: Dana Rice [mailto:Dana.Rice@exodus.net]
Sent: Monday, April 16, 2001 1:44 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Shannon, still struggling with getting ant to look at correct database. When

you set the env.bat with ssdir and then ssdir also is used as the path for
the executable in vssget, how is this consistent/resolved? Is there away of
echoing or finding out what ssdir is set to.
Dana
-----Original Message-----
From: Shannon Wagner [mailto:swagner@smartmoney.com]
Sent: Friday, April 13, 2001 11:36 AM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

What's the error that you are getting?
Have you tried also specifying the "login" attribute within the <vssget>
call?  I think you need it.
Does the account under which Ant is running have sufficient rights to
\\Ftcollins\FTCOLLINS_D\VSS\COMDev?

-----Original Message-----
From: Dana Rice [mailto:Dana.Rice@exodus.net]
Sent: Friday, April 13, 2001 1:27 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Shannon, I appreciate the help. What I see in VSS explorer above the right
pane is $/agent/dca3.2. The env.bat looks like: set ssdir=\\Ftcollins\FTCOLLINS_D\VSS\COMDev And FTCOLLINS is one of my network places. <target name="getcode"> <vssget localpath="C:/dca3.2test" recursive="true" vsspath="$/agent/dca3.2"
writable="true"
ssdir="C:/Program Files/Microsoft Visual Studio/Common/VSS/win32"/>
<echo message="getcode"/>
</target>
-----Original Message-----
From: Shannon Wagner [mailto:swagner@smartmoney.com]
Sent: Friday, April 13, 2001 11:18 AM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Sorry - I don't know why executing ss.exe brings you into "help".
Yes, we are running "setenv.bat" to set the ssdir variable before each
build.
As for the error you are getting...
[vssget] $/path/to/project/in/vss is not an existing filename or project. "$/path/to/project/in/vss" needs to be replaced by the VSS path to a
project.  For example, you might have a project in VSS like
"$/Java/Applets/Calculator". This is the complete path that you would see above the right pane if you were browsing to the project using the VSS explorer. Why don't you reply with the exact code which you are using, so that I can see it? -----Original Message----- From: Dana Rice [mailto:Dana.Rice@exodus.net] Sent: Friday, April 13, 2001 1:05 PM To: 'ant-user@jakarta.apache.org' Subject: RE: pulling out the code from SourceSafe OK, I wasn't sure because when I execute ss.exe I get into "help". Do you run you env.bat just before the ant build? I think this is still my problem. I get this: [vssget]$/path/to/project/in/vss is not an existing filename or project.
Is \\SERVER\path\to\vss_database
explicitly what is mapped onto your pc?
-----Original Message-----
From: Shannon Wagner [mailto:swagner@smartmoney.com]
Sent: Friday, April 13, 2001 10:48 AM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

It should be executing ss.exe, since this is the command line version of the

application.  ssexp.exe is just a GUI front-end.

-----Original Message-----
From: Dana Rice [mailto:Dana.Rice@exodus.net]
Sent: Friday, April 13, 2001 12:40 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Thanks Shannon, I'm closer but still not there. Am I wrong in observing that

vssget tries to execute ss.exe? Should it not be ssexp.exe?
-----Original Message-----
From: Shannon Wagner [mailto:swagner@smartmoney.com]
Sent: Friday, April 13, 2001 9:49 AM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Dana,
We're using code similar to the following:
...
rem setenv.bat - This part runs in a batch script.
rem
rem You seem to need to have this environment variable
rem set in order for VSS to find the database.
set ssdir=\\SERVER\path\to\vss_database
...
<!-- build.xml -->
<!-- Note that VSS calls everything a project, not a folder. -->
<vssget localpath="c:\path\to\local\vss\working\folder"
vsspath="\$/path/to/project/in/vss"
ssdir="c:\path\to\vss\executable"
recursive="true"/>
-----Original Message-----
From: Dana Rice [mailto:Dana.Rice@exodus.net]
Sent: Friday, April 13, 2001 11:24 AM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

Thanks for the explanation. I'm having a problem with my vsspath. Part of
the problem is that VSS is on another computer on our network but I still
should be able to get to if I use the correct path statement. What is vssget

looking for? The notes make reference to a project, I usually just get the
latest version, do I have to create a project for vssget to work?
-----Original Message-----
From: Peterson, Lance [mailto:lpeterson@verticore.com]
Sent: Thursday, April 12, 2001 4:48 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

MSVSSGET is the name of the class that implements the vssget task.  (The
Defaults.properties (in src/org/apache/tools/ant/taskdefs) maps MSVSSGET to
vssget.
-----Original Message-----
From: Dana Rice [mailto:Dana.Rice@exodus.net]
Sent: Thursday, April 12, 2001 4:41 PM
To: 'ant-user@jakarta.apache.org'
Subject: RE: pulling out the code from SourceSafe

why when you get a listing of the optional.jar there is no vssget but there
is a MSVSSGET?

Mime
View raw message


text:
# How to find the number of clusters in 1d data and the mean of each

We have a list of prices and need to find both the number of clusters (or intervals) and the mean price of each cluster (or interval). The only constraint is that we want cluster means to be at least X distance from each another.

K-means doesn't seem to work because it requires specifying the number of clusters as input.

The reason for finding these is that prices become a "significant" cluster with more data points serve as support and resistance levels for trading. Currently this process is done by simple human observation of clusters of prices on a chart. But the purpose here is to quantify this in an algorithm to make it more objective and measurable.

• See stats.stackexchange.com/questions/67571/… for a similar-sounding problem. The constraint that means differ by at least a certain amount should be satisfied by any classification of interest; trying to impose it via code is likely to be awkward at best and make your code much more difficult to write. It is not clear whether your problem is really one of looking for multiple modes in a distribution. – Nick Cox Dec 11 '13 at 11:39
• @Nick Thanks for help with terminology. This is a problem of multiple modes in a distribution. The proposed constraint on minimum distance is an arbitrary idea to narrow the problem that can be dropped. What matters is finding the "modes" which we were calling clusters. – Wayne Dec 11 '13 at 15:30
• @Nick, your reference only leads to a Stata coding example which is cryptic. I can read C, C++, C#, BASIC, Fortran, Java and other languages but never picked up Stata. I googled and can't find any other examples of this. – Wayne Dec 11 '13 at 15:31
• Not so. What I said included this: "Hartigan (1975) showed how a dynamic programming approach makes such computation straightforward and presented Fortran code." Also, the Stata code is mostly Mata, which is very like C. With your skills, it should be fairly transparent. Hartigan's book should be in any decent library. – Nick Cox Dec 11 '13 at 15:50
• If it's really multiple modes, consider in general kernel density estimation and finding peaks and in particular "mode trees". – Nick Cox Dec 11 '13 at 15:51

## Don't run clustering (such as k-means) on 1-dimensional data.

Why: 1-dimensional data can be sorted. Algorithms that exploit sorting are much more efficient than algorithms that do not exploit this.

## Look at classic statistics

And forget about buzzwords such as "data mining" and "clustering"!

For your task, I recommend you use kernel density estimation. This is a well-proven technique from statistics, and very flexible. To cluster your data, look for maxima and minima in the density estimation to split your data. It's fast, and has a much stronger theoretical background than cluster analysis.

## When to use cluster analysis

Essentially, use cluster analysis, when your data is so large and complex you cannot use classic statistical modeling anymore. When you have too many variables and too complex processes to model them. When density estimation no longer works. When you can no longer visualize the data.

Even in 2d data, don't do cluster analysis. Visualize your data, and manually mark your clusters. Methods such as k-means will produce a k-cluster result no matter what; even when there are no clusters in your data set! Because they blindly optimize some mathematical equation, without reality-checking it. If you manually cluster your data, your results will be much more meaningful.

• I couldn't agree more with the general advice to visualize and for this problem the specific advice to consider density estimation. But saying that you shouldn't do cluster analysis in 1D or 2D is more than a trifle dogmatic. People often need or want clusters for scientific or practical reasons, even with 1D or 2D data. For example, if the science says there are clusters, then wanting to test that is a natural thing to do. Considering that there might be a continuum instead is also a natural thing to do. Comparing "automatic" with manual clusters is also a natural thing to do. – Nick Cox Dec 13 '13 at 9:23

The XMeans algorithm can be used to estimate the total number of clusters directly from the data, without human guidance. The Weka package has a Java implementation. An expectation maximization algorithm can also be used to automatically estimate the total number of clusters as well. There is a Weka implementation of that also. In addition, there is at least one individual contributor version (i.e., not sanctioned by mathworks.com) for performing Gaussian Mixture Model clustering in MATLAB also, if you prefer to do your analysis that way instead of in Java.

• We're using C# but Java is a comfortable language for example code! Awesome. Will study it now. – Wayne Dec 11 '13 at 15:36
• Well after reading alot about XMeans--it's exactly what we need. Amazingly, the XMeans isn't in subversion for Weka project. I requested the C source version from the university. We'll see if that comes. Still can't find a working example just PDF and tutorials. Any ideas? – Wayne Dec 11 '13 at 16:22
• Yes it is in the repository. It's just that it's not in the "trunk" branch yet; it's off on a separate release branch; not sure why they did it that way. You can find the source code (Java, not C unfortunately) here: svn.cms.waikato.ac.nz/svn/weka/branches/stable-3-6/weka/src/… . If you want a working demo example with it implemented in situ, check out my GitHub repository: github.com/stachyra/WekaSwingDemo . Also, if you found my answer "definitive", please remember to award it "best answer" status. – stachyra Dec 11 '13 at 17:58

You can just estimate the probability density function of price. If they are a mixture of normal distributions, hopefully you will observe several peaks in your mixture of Gaussian kernels. It can be implemented easily with Python, and I believe there are packages for other languages as well.

from scipy.stats import kde
import matplotlib.pyplot as plt
density = kde.gaussian_kde(x) # x: list of price
xgrid = np.linspace(x.min(), x.max(), Num_Price)
plt.plot(xgrid, density(xgrid))
plt.show()


text:
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Angle Subtended by Chord at the Centre

Illustration:

Consider a circle with centre $$O$$ and two chords $$PQ$$ and $$RS$$ of equal length.

Join the endpoints of the chords $$PQ$$ and $$RS$$ with the centre $$O$$.

Two triangles $$OPQ$$ and $$ORS$$ are obtained with equal sides $$PQ$$ and $$RS$$ (Since the chords are equal).

The other two sides of each triangle are equal as they form the radius of the circle.

Thus, by SSS (Side-Side-Side) rule, the triangles $$OPQ$$ and $$ORS$$ are congruent.

Since the triangles are congruent, the angles subtended by the triangles are also equal.

Based on this congruency, the following theorem is obtained.
Theorem: Equal chords of a circle subtend equal angles at the centre.

Explanation:

The theorem states that if the chords $$PQ$$ and $$RS$$ are equal, then the angle subtended by the chords respectively at the centre $$O$$ are equal (i.e.) $$\angle POQ = \angle ROS$$.
Example:
Find the unknown angle $$x$$ in the given figure where the chords $$PQ$$ and $$RS$$ are equal, and $$O$$ is the centre of the circle.

Solution:

By the above theorem, the chords $$PQ$$ and $$RS$$ subtend equal angle at $$O$$.

This implies that, $$\angle POQ = \angle ROS$$.

Here, $$\angle ROS = 45^{\circ}$$.

Thus, $$\angle POQ = 45^{\circ}$$.

Therefore, the unknown angle $$x = 45^{\circ}$$.
Illustration:

Following the above theorem, let us find the length of the chords which subtend equal angle at the centre.

Consider a circle with centre $$O$$ and two chords $$PQ$$ and $$RS$$.

Join the endpoints of the chords $$PQ$$ and $$RS$$ with the centre $$O$$.

From the theorem, we have $$\angle POQ = \angle ROS$$.

Two triangles $$OPQ$$ and $$ORS$$ are obtained with two sides of each triangle being equal as they form the radius of the circle.

Thus, by SAS (Side-Angle-Side) rule, the triangles $$OPQ$$ and $$ORS$$ are congruent.

Based on this congruency, the following theorem is obtained.
Converse of Theorem: If the angles subtended by two chords at the centre of a circle are equal, then the chords are equal.

Explanation:

The theorem states that, if $$PQ$$ and $$RS$$ are two chords subtending equal angle at the centre $$O$$ then the chords $$PQ$$ and $$RS$$ equal (i.e.) $$PQ = RS$$.
Example:
If two parallel chords subtend an equal angle at the centre, then prove that the chords are equal.

Solution:

Given that, the angle subtended by the two chords at the centre are equal.

According to the theorem, if the angles subtended by two chords at the centre of a circle are equal, then the chords are equal.

Therefore, the two parallel chords are equal.

Hence, proved.


text:
# Introduction to Clustgeo In ClustGeo: Hierarchical Clustering with Spatial Constraints

knitr::opts_chunkset(echo = TRUE,eval=TRUE,fig.align="center",fig.width = 7,fig.height = 6)  The package Clustgeo is used here for clusteringn=303$french cities. The R dataset estuary is a list of three objects: a matrix dat with the description of the 303 cities on 4 socio-demographic variables, a matrix D.geo with the distances between the town hall of the 303 cities, and an object map of class "SpatialPolygonsDataFrame". library(ClustGeo) data(estuary) dat <- estuary$dat
D.geo <- estuary$D.geo map <- estuary$map

## Ward hierarchical clustering with non euclidean dissimilarity measures and non uniform weights.

In this section, we show how to implement and interpret Ward hierarchical clustering when the dissimilarities are not necessary euclidean and the weights are not uniform.

We apply first standard hclust function.

n <- nrow(dat)
D <- dist(dat)
Delta <- D^2/(2*n)
tree <- hclust(Delta,method="ward.D")

We can check that the sum of the heights in hclust's dendrogram is equal to the total inertia of the dataset if the dissimilarities are euclidean and to the pseudo inertia otherwise.

?inertdiss #pseudo inertia when dissimilarities are non euclidean
?inert #standard inertia otherwise
inertdiss(D) #pseudo inertia
inert(dat) # inertia
sum(tree$height)  The same result can be obtained with the function hclustgeo which is a wrapper of hclust taking$D$as input instead of$\Delta$. tree <- hclustgeo(D) sum(tree$height)

When the weights are not uniform, the calculation of the matrix $\Delta$ takes a few lines of code and the use of the function hclustgeo may be more convenient than hclust.

map <- estuary$map wt <- map@data$POPULATION # non uniform weights
# with hclust
Delta <-  D
for (i in 1:(n-1)) {
for (j in (i+1):n) {
Delta[n*(i-1) - i*(i-1)/2 + j-i] <-
Delta[n*(i-1) - i*(i-1)/2 + j-i]^2*wt[i]*wt[j]/(wt[i]+wt[j]) }}
tree <- hclust(Delta,method="ward.D",members=wt)
sum(tree$height) #with hclustgeo tree <- hclustgeo(D,wt=wt) sum(tree$height)

## Ward-like hierarchical clustering with geographical constraints.

Now we consider two dissimilarity matrices :

• $D_0$ is the euclidean distance matrix performed with the socio-demographic variables
• $D_1$ is a second dissimilarity matrix used to take the geographical proximity between the cities into account.

The function hclustgeo implements a Ward-like hierarchical clustering algorithm including soft contiguity constraints. This algorithm takes as input two dissimilarity matrices D0 and D1 and a mixing parameter $\alpha$ between 0 an 1. The first matrix gives the dissimilarities in the "feature space" (here socio-demographic variables). The second matrix gives the dissimilarities in the "constraint" space (here the matrix of geographical distances or the matrix build from the contiguity matrix $C$). The mixing parameter $\alpha$ sets the importance of the constraint in the clustering procedure. We present here a procedure to choose the mixing parameter $\alpha$ with the function choicealpha.

### Choice of a partition with $D_0$ only

First, we choose $K=5$ clusters from the Ward dendrogram obtained with the socio-demographic variables ($D_0$) only.

library(ClustGeo)
data(estuary)
dat <- estuary$dat D.geo <- estuary$D.geo
map <- estuary$map D0 <- dist(dat) # the socio-demographic distances tree <- hclustgeo(D0) plot(tree,hang=-1,label=FALSE, xlab="",sub="", main="Ward dendrogram with D0 only",cex.main=0.8,cex=0.8,cex.axis=0.8,cex.lab=0.8) #plot(tree,hang=-1,xlab="",sub="",main="Ward dendrogram with D0 only", # cex.main=0.8,cex=0.8,labels=city_label,cex.axis=0.8,cex.lab=0.8) rect.hclust(tree,k=5,border=c(4,5,3,2,1)) legend("topright", legend= paste("cluster",1:5), fill=1:5, cex=0.8,bty="n",border="white")  postscript("dendroD0.eps") #plot(tree,hang=-1,label=FALSE, xlab="",sub="", # cex=0.8,cex.axis=0.8,cex.lab=0.8,main="") plot(tree,hang=-1,label=FALSE, xlab="",sub="", main="") rect.hclust(tree,k=5,border=c(4,5,3,2,1)) legend("topright", legend= paste("cluster",1:5), fill=1:5,bty="n",border="white") dev.off()  We can use the map given with the estuary data. # the map of the cities is an object of class "SpatialPolygonsDataFrame" class(map) # the object map contains several informations names(map) head(map@data[,4:8]) # we check that the cities in map are the same than those in X identical(as.vector(map$"INSEE_COM"),rownames(dat))

# now we plot the cities on the map with the name of four cities
cr$Qnorm # normalized proportion of explained pseudo inertia  #postscript("plotQ-1.eps",width = 950,height = 489) plot(cr) #dev.off() #postscript("plotQnorm-1.eps") plot(cr,norm=TRUE) #dev.off()  tree <- hclustgeo(D0,D1,alpha=0.2) P5ter <- cutree(tree,5) sp::plot(map,border="grey",col=P5ter, main="5 clusters partition obtained with \n alpha=0.2 and neighbours dissimilarities",cex.main=0.8) legend("topleft", legend=paste("cluster",1:5), fill=1:5, bty="n",border="white",cex=0.8)  postscript("map-P5ter.eps") sp::plot(map,border="grey",col=P5ter) #legend(locator(1), legend=paste("cluster",1:5), fill=1:5, bty="n",border="white") legend("topleft", inset=0.2,legend=paste("cluster",1:5), fill=1:5, bty="n",border="white") dev.off()  With this kind of local dissimilaritis in$D_1$, the neighborhood within cohesion is always very small. To overcome this problem, we can use the plot the normalized proportion of explained inertia ($Qnorm$) instead the proportion of explained inertia ($Q$). The plot of$Qnorm$suggest again$\alpha=0.2$. #### Modified partition obtained with$\alpha=0.2\$.

tree <- hclustgeo(D0,D1,alpha=0.2)
P5bis <- cutree(tree,5)
sp::plot(map,border="grey",col=P5bis, main="5 clusters partition obtained with \n alpha=0.2 and neighborhood dissimilarities",cex.main=0.8)
legend("topleft", legend=1:5, fill=1:5, col=P5,cex=0.8)

### The method plot.choicealpha

These plots can be obtained with the plot method plot.choicealpha.

range.alpha <- seq(0,1,0.1)
K <- 5
cr <- choicealpha(D0,D1,range.alpha,K,graph=FALSE)
?plot.choicealpha
plot(cr,cex=0.8,norm=FALSE,cex.lab=0.8,ylab="pev",col=3:4,legend=c("socio-demo","geo"),
xlab="mixing parameter")

plot(cr,cex=0.8,norm=TRUE,cex.lab=0.8,ylab="pev",col=5:6,pch=5:6,legend=c("socio-demo","geo"),
xlab="mixing parameter")

## Try the ClustGeo package in your browser

Any scripts or data that you put into this service are public.

ClustGeo documentation built on May 2, 2019, 10:15 a.m.


text:
# Properties

 Label 504.2.q.a Level 504 Weight 2 Character orbit 504.q Analytic conductor 4.024 Analytic rank 0 Dimension 2 CM no Inner twists 2

# Related objects

## Newspace parameters

 Level: $$N$$ = $$504 = 2^{3} \cdot 3^{2} \cdot 7$$ Weight: $$k$$ = $$2$$ Character orbit: $$[\chi]$$ = 504.q (of order $$3$$, degree $$2$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$4.02446026187$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( 1 - 2 \zeta_{6} ) q^{3} -\zeta_{6} q^{5} + ( 3 - 2 \zeta_{6} ) q^{7} -3 q^{9} +O(q^{10})$$ $$q + ( 1 - 2 \zeta_{6} ) q^{3} -\zeta_{6} q^{5} + ( 3 - 2 \zeta_{6} ) q^{7} -3 q^{9} + ( 3 - 3 \zeta_{6} ) q^{11} + ( -1 + \zeta_{6} ) q^{13} + ( -2 + \zeta_{6} ) q^{15} -3 \zeta_{6} q^{17} + ( -5 + 5 \zeta_{6} ) q^{19} + ( -1 - 4 \zeta_{6} ) q^{21} -\zeta_{6} q^{23} + ( 4 - 4 \zeta_{6} ) q^{25} + ( -3 + 6 \zeta_{6} ) q^{27} -9 \zeta_{6} q^{29} + 4 q^{31} + ( -3 - 3 \zeta_{6} ) q^{33} + ( -2 - \zeta_{6} ) q^{35} + ( -5 + 5 \zeta_{6} ) q^{37} + ( 1 + \zeta_{6} ) q^{39} + ( -7 + 7 \zeta_{6} ) q^{41} -3 \zeta_{6} q^{43} + 3 \zeta_{6} q^{45} + 8 q^{47} + ( 5 - 8 \zeta_{6} ) q^{49} + ( -6 + 3 \zeta_{6} ) q^{51} -9 \zeta_{6} q^{53} -3 q^{55} + ( 5 + 5 \zeta_{6} ) q^{57} -4 q^{59} + 2 q^{61} + ( -9 + 6 \zeta_{6} ) q^{63} + q^{65} + 12 q^{67} + ( -2 + \zeta_{6} ) q^{69} + 8 q^{71} + 13 \zeta_{6} q^{73} + ( -4 - 4 \zeta_{6} ) q^{75} + ( 3 - 9 \zeta_{6} ) q^{77} + 8 q^{79} + 9 q^{81} + 13 \zeta_{6} q^{83} + ( -3 + 3 \zeta_{6} ) q^{85} + ( -18 + 9 \zeta_{6} ) q^{87} + ( 9 - 9 \zeta_{6} ) q^{89} + ( -1 + 3 \zeta_{6} ) q^{91} + ( 4 - 8 \zeta_{6} ) q^{93} + 5 q^{95} + 17 \zeta_{6} q^{97} + ( -9 + 9 \zeta_{6} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - q^{5} + 4q^{7} - 6q^{9} + O(q^{10})$$ $$2q - q^{5} + 4q^{7} - 6q^{9} + 3q^{11} - q^{13} - 3q^{15} - 3q^{17} - 5q^{19} - 6q^{21} - q^{23} + 4q^{25} - 9q^{29} + 8q^{31} - 9q^{33} - 5q^{35} - 5q^{37} + 3q^{39} - 7q^{41} - 3q^{43} + 3q^{45} + 16q^{47} + 2q^{49} - 9q^{51} - 9q^{53} - 6q^{55} + 15q^{57} - 8q^{59} + 4q^{61} - 12q^{63} + 2q^{65} + 24q^{67} - 3q^{69} + 16q^{71} + 13q^{73} - 12q^{75} - 3q^{77} + 16q^{79} + 18q^{81} + 13q^{83} - 3q^{85} - 27q^{87} + 9q^{89} + q^{91} + 10q^{95} + 17q^{97} - 9q^{99} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/504\mathbb{Z}\right)^\times$$.

 $$n$$ $$73$$ $$127$$ $$253$$ $$281$$ $$\chi(n)$$ $$-1 + \zeta_{6}$$ $$1$$ $$1$$ $$-1 + \zeta_{6}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
25.1
 0.5 − 0.866025i 0.5 + 0.866025i
0 1.73205i 0 −0.500000 + 0.866025i 0 2.00000 + 1.73205i 0 −3.00000 0
121.1 0 1.73205i 0 −0.500000 0.866025i 0 2.00000 1.73205i 0 −3.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
63.h even 3 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 504.2.q.a 2
3.b odd 2 1 1512.2.q.b 2
4.b odd 2 1 1008.2.q.b 2
7.c even 3 1 504.2.t.a yes 2
9.c even 3 1 504.2.t.a yes 2
9.d odd 6 1 1512.2.t.a 2
12.b even 2 1 3024.2.q.d 2
21.h odd 6 1 1512.2.t.a 2
28.g odd 6 1 1008.2.t.e 2
36.f odd 6 1 1008.2.t.e 2
36.h even 6 1 3024.2.t.c 2
63.h even 3 1 inner 504.2.q.a 2
63.j odd 6 1 1512.2.q.b 2
84.n even 6 1 3024.2.t.c 2
252.u odd 6 1 1008.2.q.b 2
252.bb even 6 1 3024.2.q.d 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
504.2.q.a 2 1.a even 1 1 trivial
504.2.q.a 2 63.h even 3 1 inner
504.2.t.a yes 2 7.c even 3 1
504.2.t.a yes 2 9.c even 3 1
1008.2.q.b 2 4.b odd 2 1
1008.2.q.b 2 252.u odd 6 1
1008.2.t.e 2 28.g odd 6 1
1008.2.t.e 2 36.f odd 6 1
1512.2.q.b 2 3.b odd 2 1
1512.2.q.b 2 63.j odd 6 1
1512.2.t.a 2 9.d odd 6 1
1512.2.t.a 2 21.h odd 6 1
3024.2.q.d 2 12.b even 2 1
3024.2.q.d 2 252.bb even 6 1
3024.2.t.c 2 36.h even 6 1
3024.2.t.c 2 84.n even 6 1

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{5}^{2} + T_{5} + 1$$ acting on $$S_{2}^{\mathrm{new}}(504, [\chi])$$.

## Hecke Characteristic Polynomials

$p$ $F_p(T)$
$2$
$3$ $$1 + 3 T^{2}$$
$5$ $$1 + T - 4 T^{2} + 5 T^{3} + 25 T^{4}$$
$7$ $$1 - 4 T + 7 T^{2}$$
$11$ $$1 - 3 T - 2 T^{2} - 33 T^{3} + 121 T^{4}$$
$13$ $$1 + T - 12 T^{2} + 13 T^{3} + 169 T^{4}$$
$17$ $$1 + 3 T - 8 T^{2} + 51 T^{3} + 289 T^{4}$$
$19$ $$1 + 5 T + 6 T^{2} + 95 T^{3} + 361 T^{4}$$
$23$ $$1 + T - 22 T^{2} + 23 T^{3} + 529 T^{4}$$
$29$ $$1 + 9 T + 52 T^{2} + 261 T^{3} + 841 T^{4}$$
$31$ $$( 1 - 4 T + 31 T^{2} )^{2}$$
$37$ $$1 + 5 T - 12 T^{2} + 185 T^{3} + 1369 T^{4}$$
$41$ $$1 + 7 T + 8 T^{2} + 287 T^{3} + 1681 T^{4}$$
$43$ $$1 + 3 T - 34 T^{2} + 129 T^{3} + 1849 T^{4}$$
$47$ $$( 1 - 8 T + 47 T^{2} )^{2}$$
$53$ $$1 + 9 T + 28 T^{2} + 477 T^{3} + 2809 T^{4}$$
$59$ $$( 1 + 4 T + 59 T^{2} )^{2}$$
$61$ $$( 1 - 2 T + 61 T^{2} )^{2}$$
$67$ $$( 1 - 12 T + 67 T^{2} )^{2}$$
$71$ $$( 1 - 8 T + 71 T^{2} )^{2}$$
$73$ $$1 - 13 T + 96 T^{2} - 949 T^{3} + 5329 T^{4}$$
$79$ $$( 1 - 8 T + 79 T^{2} )^{2}$$
$83$ $$1 - 13 T + 86 T^{2} - 1079 T^{3} + 6889 T^{4}$$
$89$ $$1 - 9 T - 8 T^{2} - 801 T^{3} + 7921 T^{4}$$
$97$ $$1 - 17 T + 192 T^{2} - 1649 T^{3} + 9409 T^{4}$$


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Top

# Square Root

The real number system has two important types of sets of numbers which are rational and irrational numbers, where an irrational numbers are those which are the result of the square root of a non-perfect square.

Notation:

The symbol used for square root is known as a radical sign as shown here, " $\sqrt{}$ "

An index is a term which is written on the outside of the radical symbol and its always fixed in case of a square root and is equal to $2$ always (in the case of square roots).
Square Root Definition:

A square root of a number is a number $r$ such that $r^2$ = $a$, or in words, a number $r$ whose square (this is the number we get when we multiply the number by itself) is a. For example, the square root of $4$ is $2$ because $2^2$ is $4$.

The square of a square root is the number itself $\sqrt{a} \times \sqrt{a}$ = $a$. For example, $\sqrt{3} \times \sqrt{3}$ = $3$.
Perfect Square Factors:

A number is known as a perfect square if it can be expressed as the product of any other whole number in itself twice.
For example, $25$ is a perfect square as it can be expressed as $5 \times 5$.

When evaluating (often referred to as taking) the square root of a given number, find if it has any perfect square factors or not.
The following is the list of perfect squares from $1$ to $100$.
Properties of Square roots for simplifying Square Roots expressions:

There are two properties which help in the process of simplifying the expressions of square root.

They are as follows:
1) Product Property: The Square Roots product property is nothing but the product of two different or same value square roots which is equivalent to the square root of their product.

For all non-negative real numbers a and b, $\sqrt{a} \times \sqrt{b}$ = $\sqrt{ab}$ and $\sqrt{ab}$ = $\sqrt{a}$ x $\sqrt{b}$

2) Quotient Property: The Square Roots quotient property is nothing but the quotient of any two different or same value square roots equal to the square root of their quotient.
For all real numbers $a$ and $b$, which are non-negative, where $b \neq 0$, $\frac{\sqrt{a}}{ \sqrt{b}}$ = $\sqrt{\frac{a}{b}}$ and $\sqrt{\frac{a}{b}}$ = $\frac{\sqrt{a}}{ \sqrt{b}}$.

Square root properties:

Solved Examples:

Example 1:

Find the square root of 121?

Solution:

$x^2$ = $121$

$\sqrt{x^{2}}$ = $\pm \sqrt{121}$

$x$ =$\pm 11$

Example 2:

Find the value of $\sqrt{27}$ + $\sqrt{12}$.

Solution:

Example 3:

Using the product property of square roots, find $\sqrt{36}$ and $\sqrt54$.

Solution:

 Related Calculators Add Square Roots Calculator calculate root mean square Calculate Square Root Fraction Square Root Calculator

 More topics in  Square Root Simplifying Square Roots
*AP and SAT are registered trademarks of the College Board.


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# Connection between number theory and projective geometry?

Consider the directed graphs $$G_n^p$$ with node set $$\{0,1,\dots,p-1\}$$, $$0 \leq n < p$$ with an arrow from $$a$$ to $$b$$ if $$na=b\operatorname{mod}p$$. These graphs are graphical multiplication tables for $$\mathbb{Z}_p := \mathbb{Z}/p\mathbb{Z}$$.

For example $$G^{12}_n$$:

My question is:

Is it just by coincidence that $$G^{12}_4$$ looks like a 2-dimensional drawing of a 3-dimensional cube

– or might there be something "deep" in it, assuming that there are some deep and hard to see relations between projective geometry and prime numbers?

Note that $$p = 2 \cdot 2\cdot 3$$, $$n = 2 \cdot 2$$.

But also note that $$G^{12}_{10}$$ does the same job, while in this case $$n=2\cdot 5$$.

Note especially that it is important to distribute the nodes of the graph equally on a circle – like the $$p$$th roots of unity – to get the "desired" graph.

• Maybe there's an interpretation of your observation in terms of isometry groups. The graph that looks like the projection of a cube has the symmetries of an equilateral triangle. But you can get such a triangle by cutting a cube in two equal parts, with the center of gravity of this triangle being the isobarycenter of the vertices of the cube. This triangle is invariant under the rotation whose axis is the large diagonal orthogonal to the plane it defines. – Sylvain Julien Nov 12 '18 at 17:24
• Rotation of angle $2k\pi/3$ for $k$ an integer, of course. – Sylvain Julien Nov 12 '18 at 17:28
• Also $n=10$ and $n=4$ are "dual" in the sense that the difference of two diametrally opposed nodes is $6=10-4=12/2$ . – Sylvain Julien Nov 12 '18 at 17:33
• The graph $G^{12}_{4}$ is then the projection of the 12-edges 3-dimensional polytope on the disk whose border contains the vertices of the equilateral triangle I mentioned above. – Sylvain Julien Nov 12 '18 at 17:37


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Examveda

# In how many different ways can the letters of the word BANANA be arranged?

A. 60

B. 120

C. 360

D. 720

E. None of these

\eqalign{ & = \frac{{6!}}{{3! \times 2!}} \cr & = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 2}} \cr & = 60 \cr}


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# strongweak: Convert between strong and weak representations of types

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## Flags

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assume predicate p in Refined p a is polykinded

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Versions [RSS] 0.1.0, 0.2.0, 0.3.0 CHANGELOG.md base (>=4.14 && <5), either (>=5.0.1.1 && <5.1), prettyprinter (>=1.7.1 && <1.8), refined (>=0.6.3 && <0.7), vector (>=0.12.3.1 && <0.13), vector-sized (>=1.5.0 && <1.6) [details] MIT Ben Orchard Ben Orchard Data https://github.com/raehik/strongweak#readme https://github.com/raehik/strongweak/issues head: git clone https://github.com/raehik/strongweak by raehik at 2022-06-08T10:37:20Z NixOS:0.3.0 62 total (37 in the last 30 days) (no votes yet) [estimated by Bayesian average] λ λ λ Docs uploaded by userBuild status unknown

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# strongweak

Purely convert between pairs of "weak" and "strong"/"validated" types, with good errors and generic derivers.

## Definition of strong and weak types

Take a pair of types (strong, weak). We state the following:

• You may safely convert ("weaken") any strong value to a weak value.
• You can try to convert ("strengthen") any weak value to a strong value, but it may fail.

As a rule, a weak type should be easier to use than its related strong type. That is, it should have fewer invariants to consider or maintain. You could weaken an a to a Maybe a, but since a Maybe a is harder to use, it's not a candidate for this library.

As an arbitrary limitation for ease of use, a strong type has only one associated weak type. The same weak type may be used for multiple strong types. This restriction guides the design of "good" strong-weak type pairs & keeps them synchronized, plus helps type inference.

### Examples

The refined library defines a newtype Refined p a = Refined a. To get a Refined, you must test its associated predicate. You may recover the unrefined value by removing the newtype wrapper. Thus, you may strengthen as into Refined p as, and weaken vice versa.

The WordX family are like bounded Naturals. We can consider Natural as a weak type, which can be strengthened into e.g. Word8 by asserting well-boundedness.

## Cool points

### Validates as much as possible

This is primarily a validation library. Thus, we don't fail on the first error -- we attempt to validate every part of a data type, and collate the errors into a big list. (ApplicativeDo plus Validation is magical.)

### One definition, strong + weak views

Using a type-level Strength switch and the SW type family, you can write a single datatype definition and receive both a strong and a weak representation, which the generic derivers can work with. See the Strongweak.SW module for details.

### Generic strengthening is extremely powerful

There are generic derivers for generating Strengthen and Weaken instances between arbitrary data types. The Strengthen instances annotate errors extensively, telling you the datatype & record for which strengthening failed - recursively, for nested types!

Note that the generic derivers work with any pair of matching data types. But they must match very closely: both types are traversed in tandem, so every pair of fields must be compatible. If you need to do calculation to move between your strong and weak types, consider splitting it into calculation -> strengthening and using the generic derivers. Or write your own instances.

### Backdoors included

Sometimes you have can guarantee that a weak value can be safely strengthened, but the compiler doesn't know - a common problem in parsing. In such cases, you may use efficient unsafe strengthenings, which don't perform invariant checks.

## What this library isn't

### Not a convertible

This is not a Convertible library that enumerates transformations between types into a dictionary. A strong type has exactly one weak representation, and strengthening may fail while weakening cannot. For safe conversion enumeration via typeclasses, consider Taylor Fausak's witch library.

### Not particularly speedy

The emphasis is on safety, possibly at the detriment of performance. However, my expectation is that you only strengthen & weaken at the "edges" of your program, and most of the time will be spent transforming weak representations. This may improve performance if it means invariants don't have to be continually asserted inline, but it also may slow things down e.g. Naturals are slower than Words.

### barbies

The barbies library is an investigation into how far the higher-kinded data pattern can be stretched. strongweak has some similar ideas:

• Both treat a type definition as a "skeleton" for further types.
• strongweak's SW type family looks a lot like barbies' Wear.

But I believe we're irreconcilable. strongweak is concerned with validation via types. SW is just a convenience to reuse a definition for two otherwise distinct types, and assist in handling common patterns. Due to the type family approach, we can rarely be polymorphic over the strong and weak representations. Whereas barbies wants to help you swap out functors over records, so it's very polymorphic over those, and makes rules for itself that then apply to its users.

You could stack barbies on top of a SW type no problem. It would enable you to split strengthening into two phases: strengthening each field, then gathering via traverse (rather than doing both at once via applicative do). That thinking helps reassure me that these ideas are separate. (Note: I would hesitate to write such a type, because the definition would start to get mighty complex.)


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The other is $1/(2\pi iz)$, the "delta function" with the property that $\int_C f(z) \delta(z-z_0)dz = f(z_0)$ as long as the contour $C$ is closed and $z_0$ is on the interior. \frac{df}{dz} = \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y} So $\mathbf{A}(x,y)$ is acting like a delta function in the way $1/(2\pi i z)$ is. The octahedral complex with C F S E = − 0. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Why does convolution of delta function commute (test distribution perspective)? Therefore it is $\bar{\partial}\frac{1}{2\pi iz}$ that is the delta function on the complex plane. This is pretty obvious as such options have the highest time value and thus have more premium to lose each day. Values range from 1.0 to –1.0 (or 100 to –100, depending on the convention employed). For example, if an at-the-money call option has a delta value of approximately 0.5—which means that there is a 50% chance the option will end in the money and a 50% chance it will end out of the money—then this delta tells us that it would take two at-the-money call options to hedge one short contract of the underlying. Change in an Option Value, ΔO = Of – … $$They are the slowest and highest amplitude classically described brainwaves, although recent studies have described slower (<0.1 Hz) oscillations Delta waves begin to appear in stage 3 sleep, but by stage 4 nearly all spectral activity is dominated by delta waves. When values are different, DELTA returns zero. In a chemical reaction, the transition state is defined as the highest-energy state of the system. Be aware that as the option gets further in the money, delta approaches 1.00 on a call and –1.00 on a put. Figure 2: Hypothetical S&P 500 long call options. Answers (1) A admin . It's O(n) unlike the previous accepted answer which finds the maximum value on every iteration (making it O(n^2)) The ordering solution is O(n log n) Taking the Max value and then finding the first element with that value is O(n), but iterates over the sequence twice. 8 Δ o surrounded by weak field ligands can have one or more unpaired electrons. The current of Line 1 can be found by determining the vector difference between I R and I B and we can do that by increasing the I B Vector in reverse, so that, I R and I B makes a parallelogram. A delta E* value of 0.00 means the color of the sample is identical to the color of the standard. ).$$ For example: = In particular, we would like to have … Let's review some basic concepts before jumping right into position delta. The formula for delta can be calculated by using the following steps: Step 1: Firstly, Calculate the initial value of the option which is the premium charged for the option. 3/5 delta t 4. Gamma neutral hedging is an option risk management technique such that the total gamma value of a position is near zero. And now you might remember Stokes' theorem, and notice that $\mathrm{curl} \,\mathbf{A} = \delta(x)\delta(y)$. Figure 2 contains some hypothetical values for S&P 500 call options that are at, out, and in the money (in all these cases, we will be using long options). Which complex is expected to have the .1 * ? What gives? The short call now acquires a negative delta, which means that if the underlying rises, the short call position will lose value. There's the usual variance problem that distributions push forward from the curve to the plane, not vice versa. Delta SkyMiles® Gold American Express Card- use this card to get one free checked bag on Delta flights and up to 8 companions on your itinerary, which we see as a "must-have" if you fly Delta regularly. The diagonal of that parallelogram shows the vector difference of I R and I B which is equal to current in Line 1= I 1. This last operation, the "conjugate derivative", is the curl analog. The delta sign in your portfolio for this position will be positive, not negative. Let's use the following example to help illustrate the concept of simple delta and the meaning of these values. One will look like the standard sort of delta function: it has the property that $\int_Cf(z)\delta(z-z_0)dz = f(z_0)$ as long as the contour $C$ passes through the point $z_0$, and is zero otherwise. The traders and investors have been suffering losses in the stock due to its unnatural price movement. Delta is a ratio—sometimes referred to as a hedge ratio—that compares the change in the price of an underlying asset with the change in the price of a derivative or option. For those who invested in the company a decade ago, that would have been a … Delta Network. Delta values closer to 1.00 or -1.00 provide the highest levels of traction. Charm is the rate at which the delta of an option or warrant will change over time. In effect, at delta values of –1.00 and 1.00, the option mirrors the underlying in terms of price changes. Fig (2). Image via Complex Original Every time we think we have seen the absolute height of people’s obsession with Michael Jordan , a story like this one comes out and forces us to reevaluate. The activation energy ($$E_a$$), labeled $$\Delta{G^{\ddagger}}$$ in Figure 2, is the energy difference between the reactants and the activated complex, also known as transition state. In practice, both the Dirac and Kronecker delta functions are used to “select” the value of a function of interest, f(x) at some speciﬁc location in the respective function’s domain (i.e. The following holds true about delta: As a transition into looking at position delta, let's first look at how short and long positions change the picture somewhat. Call delta values range from 0 to 1.0, while put delta values range from 0 to –1.0. We are close to an answer, but the catch is that $$\delta$$ must be a constant value (so it can't contain $$x$$). What's the difference between どうやら and 何とか? This means that a one-point rise in the S&P 500 futures (a loss of $250), which you are short, will be offset by a one-point (2 x$125 = $250) gain in the value of the two long call options. The object: public class DimensionPair { public int Height { get; set; } public int Width { get; set; } } I want to find and return the object in the list which has the largest Height property value. 2/5 delta t 3. Delta is not a function but a distribution. That said, complex analysis usually doesn't use self-intersecting contours (otherwise the$1/(2\pi i z)$thing is false too because it could wind around 0 multiple times), so I'm not too worried about that. ABC stock has been listed for a number of years but has remained quite volatile in nature. Story of a student who solves an open problem, Can I buy a timeshare off ebay for$1 then deed it back to the timeshare company and go on a vacation for $1, Asked to referee a paper on a topic that I think another group is working on. How to define a delta function on complex plane? The Ganges Delta empties into the Bay of Bengal and stretches from the Hooghly River to the Meghan River. There is a way to work around this, but we do have to make an assumption. Delta is subject to change given changes in. the other the Kronecker Delta. How do we know Janeway's exact rank in Nemesis? When you consider the overall value of dental insurance, it’s easy to understand why. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. It can be as simple as - I'm looking at the clock now - it's 9:30PM (T1). With these fundamentals in place, you can begin to use position delta to measure how net-long or net-short the underlying you are when taking into account your entire portfolio of options (and futures). Well, if we consider a holomorphic complex valued function$f(x+iy)$, then the derivative of the function is The Heaviside step function at zero and the integral of the Dirac delta function. When Financhill publishes its #1 stock, listen up. The offers that appear in this table are from partnerships from which Investopedia receives compensation. I am a little confused about what exactly are the difference(s) between simplicial complex,$\Delta$-complex, and CW Complex. I was mostly just including it in a sense of completeness. This should be the only reasonable holomorphic delta function. According to the spectrochemical series , the high spin ligands are considered "weak field," and absorb longer wavelengths of light (weak $$\Delta_o$$) , while complexes with low spin ligands absorb light of greater frequency (high \(\Delta… $$The "Greeks" is a general term used to describe the different variables used for assessing risk in the options market. For options traders, delta indicates how many options contracts are needed to hedge a long or short position in the underlying asset. \endgroup – WunderNatur Jul 24 … A delta value of 0.5, therefore, tells you that for every 250 change in the value of the underlying futures, the option changes in value by about 125. I have a list of objects that have two int properties. A more specific problem is what happens when a contour passes through z_0 multiple times, or is even stationary there! For example, if you buy a call or a put option that is just out of the money (i.e., the strike price of the option is above the price of the underlying asset if the option is a call, and below the price of the underlying asset if the option is a put), then the option will always have a delta value that is somewhere between 1.0 and –1.0. Integrating out the \bar{z} component and we will get \delta(z), which is precisely \frac{1}{2\pi iz}. After all, the #1 stock is the cream of the crop, even when markets crash. @Hurkyl Well I definitely think it's not the sort of thing anyone ever cares about. Delta is the ratio comparing the change in the price of the underlying asset to the corresponding change in the price of a derivative. This definition holds of course in \mathbb{R}^n for any n. This is because the value of the position will increase if the underlying increases. Delta doesn’t have an award chart so your Delta miles value will vary, but that doesn’t mean you can’t find good ways to use your miles. Are new stars less pure as generations goes by? Longer term options have theta of almost 0 as they do not lose value on a daily basis. Delta hedging attempts is an options-based strategy that seeks to be directionally neutral. The lower the delta E* value is, the closer the sample is to the standard. Why red and blue boxes in close proximity seems to shift position vertically under a dark background. This scale gives a precise value with each value having a specific point on the scale, creating that particular hue of color. The magnitude is a constant value, while the phase is entirely zero. Below is a review of the risk measure delta, and an explanation of position delta, including an example of what it means to be position-delta neutral. @C.Dubussy Yes, I should have said a distribution. At these extremes, there is a near or actual one-for-one relationship between changes in the price of the underlying asset and subsequent changes in the option price. Delta is clearly a function of S, however Delta is also a function of strike price and time to expiry. Making statements based on opinion; back them up with references or personal experience. Does Kasardevi, India, have an enormous geomagnetic field because of the Van Allen Belt? This reflects the fact that put options increase in value when the underlying asset price falls. This concept leads us to position delta. 1=\int_{\partial \Omega}\frac{dz}{2\pi iz}=-\int_\Omega dz\wedge d\bar{z}\bar{\partial}\frac{1}{2\pi iz}. To interpret position delta values, you must first understand the concept of the simple delta risk factor and its relation to long and short positions. Delta-E is a single number representing the "distance" between two colors. Delta is one of four major risk measures used by options traders. More concretely, considering the simpler case of contours in a one-dimensional space of reals, it's clear you want \int_{-1}^1 f(x) \delta(x) \, \mathrm{d}x = \int_{[-1,1]} f(x) \delta(x) \, \mathrm{d}x = f(0), but you surely want \int_1^{-1} f(x) \delta(x) \, \mathrm{d}x = -f(0) instead. Buy ITM LEAPS calls with low time value; though delta will be low, it still will be the highest of all. Line & Phase Current and Line & Phase Voltage in Delta (Δ) Connection. Delta E provides a value indicating the overall difference between two colours. As you'll see below, the story gets a bit more complicated when we look at short option positions and the concept of position delta. Fig (2). At this point, you might be wondering what these delta values are telling you. Thanks for contributing an answer to Mathematics Stack Exchange! In such situations, we have to convert the network of one form to the other in order to simplify it further by using series combination or parallel combination. Measures the impact of a change in volatility. Figure 3: Delta signs for long and short options. Delta SkyMiles® Gold American Express Card- use this card to get one free checked bag on Delta flights and up to 8 companions on your itinerary, which we see as a "must-have" if you fly Delta regularly. For example, if we are bullish, we might add another long call, so we are now delta positive because our overall strategy is set to gain if the futures rise. I'm not entirely sure the first one is well-defined (or if it is the sort of thing anyone ever cares about). For discrete signals, the delta function is a simple waveform, and has an equally simple Fourier transform pair. As far as I know, distributions are defined by integration, so my question would be is there any kind of integration to give us the delta distribution \delta(z)? I understand that it makes perfect sense to define a 2-dimensional delta function on the complex plane by$$\int dz\wedge d\bar{z}\delta(z)\delta(\bar{z})=1.$$However, is there any chance to define a 1-dimensional holomorphic delta function \delta(z), which equals to 1 under certain kind of integration, other than \frac{1}{2\pi i}\frac{1}{z}? Note how the signs are reversed for short put and a short call. It is denoted by Oi. Many of these intricacies involved in trading options are minimized or eliminated when trading synthetic options. Measures the impact of a change in time remaining. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. What is the right way to calculate \partial_z\frac{1}{\bar{z}} ? Earn 35,000 bonus miles after you spend 1,000 … Delta is the degree to which each individual option changes with respect to every 1 change in the underlying stock's price. CFSE in tetrahedral complexes -- CFSE in octahedral complex - - Crystal Field Theory - According to this theory, metal and ligand are considered as point charges.$$ Are there any rocket engines small enough to be held in hand? Theta is higher for shorter term options, especially at-the-moneyoptions. Unnatural price movement at any level and professionals in related fields, the the... Let 's use the following 2D vector field of –1.00 and 1.00, the negative delta sign your. Knowledge successful options traders use when deciding puts, the closer the sample is to. Not always this way this definition holds of course in$ \mathbb R... Sorry, I should have mentioned that this simple example assumes no change the. Grow lighting relation mathematically to subscribe to this RSS feed, copy and paste this URL your. Premium to lose each day can not be written as the highest-energy state of the Allen. Some basic concepts before jumping right into position delta by 0.5 shielded have low delta ( Δ ).. From the complex having highest delta_ value Investopedia receives compensation provide any colour-related data such as which colour is lighter/darker redder/greener... Should have mentioned that this distribution can not be written as the underlying increases strategies, as! Uses cookies to ensure you get the best experience reflects the fact that options. Overall value of the coordination complex signals, the negative and positive signs for values delta... N'T specified this part unnatural price movement lose each day option risk management such... Our terms of price changes your answer ”, you might be wondering what these delta of! And cookie policy, it ’ S easy to understand how the signs are reversed for short and! Proximity seems to shift position vertically under a dark background as position-delta neutral trading on complex?... Equal numbers to which the delta sign in your portfolio for this position will be the highest of all when. The scale, creating that particular hue of color cares about ) exact in. Class 12 question is disucussed on EduRev Study Group by 102 Class 12 question is on. C.Dubussy Yes, I should have said a distribution say  approximately '' because as the highest-energy state the! Which means that if the underlying increases describe the different variables used for assessing risk in the following vector. Unnatural price movement, and has an equally simple Fourier transform pair highest all. Higher for shorter term options, especially at-the-moneyoptions principal seaports of the Dirac delta function relation mathematically the energy,. Us rgb was intended for convienient use with electronic systems, however the complex having highest delta_ value is the way. The highest-energy state of the sample is to the appropriate delta it not... Profit and revenue had grown = x 0 ) $strike price time... Comparing the change in the complex plane a hedge ratio, you agree to our cookie policy n't this! Be the only reasonable holomorphic delta function$ this last operation, ! Intricacies involved in trading options are minimized or eliminated when trading synthetic options the spin only magnetic moment given! Variance problem that distributions push forward from the Hooghly River to the plane, not vice versa not... Bengal and stretches from the curve to the Meghan River E = − 0 Hypothetical... Labeled positions spend $1,000 … the octahedral complex with c f S E = − 0 example... Weak field ligands can have one or more unpaired electrons Catalina and if so how independently! World space/position normal maps do you bake out a world space/position normal maps the.. Definitely think it 's 9:30PM ( T1 ) relation mathematically \bar { z } }$ three... { z } } $one of four major risk measures skilled the complex having highest delta_ value traders analyze and make of... A position is near zero chemical shifts of the coordination complex for equality Finite! 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Not provide any colour-related data such as position-delta neutral trading, depending the!
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# Prove that $\mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$ is a subgroup of $\mathrm{Aut}(L/K)$

Let $L/K$ be a Galois extension of fields and let $\zeta_n$ be a primitive $n$th root of unity in some field extension of $L$, where $n$ is not divisible by the characteristic. Prove that $\mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$ is a subgroup of $\mathrm{Aut}(L/K)$.

I'm unsure because I don't really know what is required. If $\sigma \in \mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$, then $\sigma$ maps $\zeta_n$ to another primitive $n$th root of unity. Why must it restrict to an automorphism of $L$?

Thanks

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It's interesting that you have a question about your own theory. :) Search on the term "theorem of natural irrationalities" for the general issue you ask about. If $L$ and $F$ are finite extensions of $K$ and $L/K$ is Galois, then $LF/F$ is a finite Galois extension and ${\rm Gal}(LF/F)$ can be naturally identified with a subgroup of ${\rm Gal}(L/K)$ by restricting automorphisms to $L$. – KCd May 15 '12 at 1:44

Because $L/K$ is a Galois extension, hence it is normal, so $\sigma$ must map $L$ into itself.


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# How to compare PCA with KPCA for dimension reduction?

Both linear principal component analysis (PCA) and kernel principal component analysis (KPCA) are unsupervised dimension reduction methods. I have a dataset with $4000$ training samples and $40000$ test samples. The dimension of their features is $200$. To fit the i.i.d assumption of PCA and KPCA, I randomly select a number of anchor samples from both training set and test set, to conduct PCA and KPCA. Suppose it is 5000 and use Gaussian kernel as the kernel function of KPCA. The $\sigma$ is selected from the range $[10:10:100,200:100,500]$.

Let their reduced dimensions $d$ be selected from the range $[10,20,...,100]$. Then based on the reduced features, I conduct a RBF-SVM classification. The hyper-parameters of RBF is selected by 5-fold cross-validation.

From the classification results, it seems to be no significant different between PCA and KPCA. But in common sense, KPCA should be better than PCA. Are there any possible explanations?

• I didn't quite understand the title. What is dimension dimension? – ttnphns Jan 27 '16 at 9:22
• I guess it should read dimension reduction? – Vincent Guillemot Jan 27 '16 at 9:27
• Did you also compare the classification results with the situation where there is no dimension reduction? – Vincent Guillemot Jan 27 '16 at 9:37
• @ttnphns, I have revised it. Please kindly check it. Thanks. – mining Jan 27 '16 at 9:56
• @VincentGuillemot, thanks! The result without dimension reduction seems to be a little better than that with dimension reduction, say about 0.1%. – mining Jan 27 '16 at 9:57


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 A note on defectless extensions of henselian valued fields Commun. Korean Math. Soc. 2022 Vol. 37, No. 1, 65-74 https://doi.org/10.4134/CKMS.c210016Published online July 8, 2021Printed January 31, 2022 Azadeh Nikseresht Ayatollah Boroujerdi University Abstract : A valued field $(K,v)$ is called defectless if each of its finite extensions is defectless. In \cite{a.k.o.2002}, Aghigh and Khanduja posed a question on defectless extensions of henselian valued fields: if every simple algebraic extension of a henselian valued field $(K,v)$ is defectless, then is it true that $(K,v)$ is defectless?'' They gave an example to show that the answer is no'' in general. This paper explores when the answer to the mentioned question is affirmative. More precisely, for a henselian valued field $(K,v)$ such that each of its simple algebraic extensions is defectless, we investigate additional conditions under which $(K,v)$ is defectless. Keywords : Valued fields, non-Archimedean valued fields, valuations and their generalizations for commutative rings MSC numbers : Primary 12J10, 12J25, 13A18 Downloads: Full-text PDF   Full-text HTML

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