text:
# CAT Quantitative Aptitude Questions | CAT Ratios, Mixtures, Alligations and Averages Questions

###### CAT Questions | Ratio and Proportion | Alligation

The question is from CAT Ratio, Mixtures and Averages. Three mixtures with different capcity are mixed together. What would be the ratio in the resultant solution? CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2020.

Question 37: What would be the ratio of milk and water in a final mixture formed by mixing milk and water that are present in three vessels of capacity 1l, 2l, and 3l respectively and in the ratios 5:1, 3:2 and 4:3 respectively?

1. 747:443
2. 787:1260
3. 787:473
4. 747:473

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##### Method of solving this CAT Question from Ratio, Mixtures and Averages: When we mix two mixtures in a particular ratio, we get a third mixture. Given the third mixture how does one find the ratio in which they were mixed.

Solve this type of questions by taking 2 at a time. Take the first 2 vessels,
In 1st, fraction of milk = $$frac{5}{6}\\$ In 2nd, fraction of milk = $\frac{3}{5}\\$ Therefore, => $\frac{3/5 - x}{x - 5/6}\\$ = $\frac{1}{2}\\$ => $\frac{6}{5}\\$ - 2x = x - $\frac{5}{6}\\$ => $\frac{6}{5}\\$ + $\frac{5}{6}\\$ = 3x => x = $\frac{61}{90}\\$$And the volume of mixture after mixing = 1 + 2 = 3 l)
Therefore,

=> $$frac{4/7 - x}{x - 61/90}\\$ = $\frac{3}{3}\\$ => $\frac{4}{7}\\$ - x = x + $\frac{61}{90}\\$ => $\frac{4}{7}\\$ + $\frac{61}{90}\\$ = 2x => 2x = $\frac{787}{630}\\$ => x = $\frac{787}{1260}\\$ Therefore, Milk : Water = $\frac{787}{$1260 - 787$}$ = $\frac{787}{473}\\$ The question is "What would be the ratio of milk and water in a final mixture formed by mixing milk and water that are present in three vessels of capacity 1l, 2l, and 3l respectively and in the ratios 5:1, 3:2 and 4:3 respectively?" ##### Hence, the answer is "787:473". Choice C is the correct answer. ###### 🎉 Price Slashed by Rs. 3000 on VARC, DILR, and Quant Courses! ###### Already have an Account? ###### CAT Coaching in ChennaiCAT 2021Enroll at 49,000/- 44,000/- Online Classroom Batches Starting Now! ###### Best CAT Coaching in Chennai Introductory offer of 5000/- Attend a Demo Class ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Phone:$91) 44 4505 8484
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text:
# Write short notes on gangue and anode mud.

Verified
147k+ views
Hint: We can explain the given terms if we go through the metallurgical process involving the process of leaching and electrolysis for purification of the ores of different metals. Gangue is a component obtained during the process of an electrolysis involving the anode mud.

It is earthly material and does not have any electrical or the magnetic properties. It was calculated at $500{}^\circ C$ in extraction of the mineral.

format: plain


text:
ID 683750
Date 9/13/2021
Public

## 6.7. ALTFP_SQRT Parameters

Table 38.  ALTFP_SQRT Parameters
Parameter Name Type Required Description
WIDTH_EXP Integer Yes Specifies the precision of the exponent. If this parameter is not specified, the default is 8. The bias of the exponent is always set to 2 (WIDTH_EXP -1) -1, that is, 127 for the single-precision format and 1023 for the double-precision format. The value of the WIDTH_EXP parameter must be 8 for the single-precision format, 11 for the double-precision format, and a minimum of 11 for the single-extended precision format. The value of the WIDTH_EXP parameter must be less than the value of the WIDTH_MAN parameter, and the sum of the WIDTH_EXP and WIDTH_MAN parameters must be less than 64.
WIDTH_MAN Integer Yes Specifies the value of the mantissa. If this parameter is not specified, the default is 23. When the WIDTH_EXP parameter is 8 and the floating-point format is single-precision, the WIDTH_MAN parameter value must be 23. Otherwise, the value of the WIDTH_MAN parameter must be a minimum of 31. The value of the WIDTH_MAN parameter must be greater than the value of the WIDTH_EXP parameter. The sum of the WIDTH_EXP and WIDTH_MAN parameters must be less than 64.
ROUNDING String Yes Specifies the rounding mode. The default value is TO_NEAREST. Other rounding modes are not supported.
PIPELINE Integer Yes Specifies the number of clock cycles for the square root results of the result[] port. Values are WIDTH_MAN + 5 and ((WIDTH_MAN + 5/2)+2) as specified by truncating the radix point.


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# filePath

0th

Percentile

##### Usage
## S3 method for class 'default}(..., fsep=.Platform\$file.sep, removeUps=TRUE, expandLinks=c("none", "any", "local", "relative", "network"), mustExist=FALSE, verbose=FALSE)':
filePathundefined

Construct the path to a file from components and expands Windows Shortcuts along the pathname from root to leaf.  This function is backward compatible with
file.path() when argument removeUps=FALSE and

This function exists on all platforms, not only Windows systems.

...{Arguments to be pasted together to a file path and then be
parsed from the root to the leaf where Windows shortcut files are
recognized and expanded according to argument which in each
step.}
fsep{the path separator to use.}
removeUps{If TRUE, relative paths, for instance "foo/bar/../"
are shortend into "foo/", but also "./" are removed from the final
pathname, if possible.}
expandLinks{A character string. If "none", Windows
Shortcut files are ignored.  If "local", the absolute target
on the local file system is used. If "relative", the relative
target is used. If "network", the network target is used. If
"any", the first the local, then the relative and finally the
network target is searched for.}
mustExist{If TRUE and if the target does not exist, the original
pathname, that is, argument pathname is returned. In all other
cases the target is returned.}
verbose{If TRUE, extra information is written while reading.}

Returns a character string.

If expandLinks==TRUE, each component, call it parent, in
the absolute path is processed from the left to the right as follows:
1. If a "real" directory of name parent exists, it is followed.
2. Otherwise, if Microsoft Windows Shortcut file with name
parent.lnk exists, it is read. If its local target exists, that
is followed, otherwise its network target is followed.
3. If no valid existing directory was found in (1) or (2), the expanded
this far followed by the rest of the pathname is returned quietly.
4. If all of the absolute path was expanded successfully the expanded
absolute path is returned.

On speed{
Internal file.exists() is call while expanding the pathname.
This is used to check if the exists a Windows shortcut file named
'foo.lnk' in 'path/foo/bar'. If it does, 'foo.lnk' has to be followed,
and in other cases 'foo' is ordinary directory.
The file.exists() is unfortunately a bit slow, which is why
this function appears slow if called many times.
}

# Default
print(file.path("foo", "bar", "..", "name")) # "foo/bar/../name"

# Shorten pathname, if possible
print(filePath("foo", "bar", "..", "name"))  # "foo/name"
print(filePath("foo/bar/../name"))           # "foo/name"

# Recognize Windows Shortcut files along the path, cf. Unix soft links
filename <- system.file("data-ex/HISTORY.LNK", package="R.utils")
print(filename)
print(filename)

stopifnot(file.exists(filename))

[object Object]

IO


text:
# problem at drawing 3D square for right angle with tikz, pgfplots

I'm drawing a 3D square for right angle following https://tex.stackexchange.com/a/163384/14423. That works fine, but I try filling that square and the results isn't the spected:

\documentclass{memoir}
\usepackage{tikz,pgfplots}
\usetikzlibrary{calc}

\pgfplotsset{ /pgfplots/area style/.style={%
area cycle list,
area legend,
axis on top,
}}

\newcommand{\RightAngle}[4][5pt]{%
\draw[fill] ($#3!#1!#2$)
--($#3!2!($($#3!#1!#2$)!.5!($#3!#1!#4$)$)$)
--($#3!#1!#4$) ;
}

\begin{document}
\begin{tikzpicture}
\begin{axis}[scale=0.65,view={130}{30},axis equal,axis on top,axis lines=center,xlabel=$x(t)$,ylabel=$y(t)$,zlabel=$z(t)$,xtick=\empty,ytick=\empty,ztick=\empty]
\addlegendentryexpanded{$T(t)$}
\addlegendentryexpanded{$T'(t)$}
\addlegendentryexpanded{$N(t)$}
%\addlegendentryexpanded{$\Ce$}
\coordinate (P) at (0.261,1.047,1);
\coordinate (T) at (0.511,2.047,1);
\coordinate (N) at (0.261,1.047,-0.125);
\RightAngle{(T)}{(P)}{(N)};
\end{axis}
\end{tikzpicture}

\end{document}

As you can see there is a dotted node and the square for right angle is not filled completely. What am I doing wrong? and how get square filled with dashed lines?

• Where is the dotted node? – cfr Dec 25 '15 at 2:09
• I don't get the output you posted from the code you posted. My output looks rather different. – cfr Dec 25 '15 at 2:11
• How different is? I obtain that picture (maybe I miss some command). Can you show me your output please? – juanuni Dec 25 '15 at 2:26
• The dotted node is below z(t) – juanuni Dec 25 '15 at 2:27
• Oh, OK. It is a dot and not a dotted node. I was looking for dotted somewhere. You need to specify a suitable compatibility option to get the output you posted. If you run the code you posted, you will probably not get the output you posted unless you've configured defaults somewhere where are not the standard ones. – cfr Dec 25 '15 at 2:51

Because you are using something optimised for an older version of pgfplots which gave wrong output, you need to specify a version to get the correct output. At least, that is how I roughly understood the console warnings. Adding the compatibility option newest gave me the output posted in the question. So, with advice from Alenanno, this could be set to e.g. 1.12 which seems to be the version I get if I ask for newest.

When you specify the right angle, you specify from left to right of the bottom and then up to the top right and nothing else. TikZ completes the path using the shortest possible route i.e. from top right to bottom left of the square and fills the result i.e. half of the square. You need to specify that the path goes through P before returning to the start point in order to fill the square.

\newcommand{\RightAngle}[4][5pt]{%
\draw[fill] ($#3!#1!#2$) -- ($#3!2!($($#3!#1!#2$)!.5!($#3!#1!#4$)$)$) -- ($#3!#1!#4$) -- #3  -- cycle;
}

I also moved the legend a bit so I could see what I was doing.

To fill the square with lines, try the patterns library and use pattern rather than fill for the path. For example:

\newcommand{\RightAngle}[4][5pt]{%
\draw[pattern=north east lines] ($#3!#1!#2$) -- ($#3!2!($($#3!#1!#2$)!.5!($#3!#1!#4$)$)$) -- ($#3!#1!#4$) -- #3  -- cycle;
}

Complete code:

\documentclass[tikz,border=10pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.12}
\usetikzlibrary{calc,patterns}

\pgfplotsset{ /pgfplots/area style/.style={%
area cycle list,
area legend,
axis on top,
}}

\newcommand{\RightAngle}[4][5pt]{%
\draw[pattern=north east lines] ($#3!#1!#2$) -- ($#3!2!($($#3!#1!#2$)!.5!($#3!#1!#4$)$)$) -- ($#3!#1!#4$) -- #3  -- cycle;
}
\begin{document}
\begin{tikzpicture}
\begin{axis}
[
scale=0.65,
view={130}{30},
axis equal,
axis on top,
axis lines=center,
xlabel=$x(t)$,
ylabel=$y(t)$,
zlabel=$z(t)$,
xtick=\empty,
ytick=\empty,
ztick=\empty,
legend style={anchor=east}
]
\addlegendentryexpanded{$T(t)$}
\addlegendentryexpanded{$T'(t)$}
\addlegendentryexpanded{$N(t)$}
%\addlegendentryexpanded{$\Ce$}

• I have to admit I didn't understand the question very well, but using newest isn't advisable, because compiling the same code with a newer version that is released later might break the output. I think even the author advised against it somewhere. Better choose a version that gives the desired output and stick with it. – Alenanno Dec 25 '15 at 3:08
• @Alenanno I just used it because I am not, frankly, familiar with the package and didn't know what version to use. Should I put 1.12? Is that the current version? – cfr Dec 25 '15 at 3:14
• I think so, but it's not mandatory to use the latest in every case. If 1.10 worked better, then I'd go for it. :D you can try and test. newest is not even a problem but it might not be reliable as time goes on. – Alenanno Dec 25 '15 at 3:17
• @Alenanno I've put 1.12.1 which is what I seem to have. I was really just trying to reproduce the problem in the question. Since this reproduced the problem, for my purposes, no other version could work better ;). – cfr Dec 25 '15 at 3:19


text:
# Essential norm of generalized composition operators from weighted Dirichlet or Bloch type spaces to Q_K type spaces

Document Type : Research Paper

Authors

1 Islamic Azad University, Iran

2 Tarbiat Moallem University, Iran

Abstract

In this paper we obtain lower and upper estimates for the
essential norms of generalized composition operators from weighted
Dirichlet spaces or Bloch type spaces to $Q_K$ type
spaces.

Keywords

Main Subjects

### History

• Receive Date: 14 April 2011
• Revise Date: 13 September 2011
• Accept Date: 13 September 2011
• First Publish Date: 01 March 2013


text:
# Math Help - Number in Q

1. ## Number in Q

prouve :

2. Originally Posted by dhiab
prouve :

Hi

If you consider the equation $x^3+5x-6=0$ you can find that x=1 is the only real solution since $x^3+5x-6=(x-1)(x^2+x+6)$

Now if you try to solve it using Cardan method Méthode de Cardan - Wikipédia you will find that one real solution is

$\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}$

Since 1 is the only real solution you get $\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}} = 1$


text:
# Infinite set of positive integers such that the greatest common divisor of any two distinct numbers in $B$ is $p$

Let $A$ be an infinite set of positive integers such that every $n \in A$ is the product of at most $k$ prime numbers where $k$ is a positive integer. Prove that there are an infinite set $B \subset A$ and a number $p$ such that the greatest common divisor of any two distinct numbers in $B$ is $p$.

I thought about proving this by induction on $k$. If $k = 1$, then $A$ must be an infinite set of distinct primes and so we let $B$ be an infinite set of primes. Thus, we see that $p = 1$. How do we do the inductive step?

Before proving it, let's resolve an ambiguity in the statement of the problem. The statement "$n$ is the product of at most $k$ prime numbers" must be taken to mean "$n$ can be written as the product of at most $k$ prime numbers, not necessarily distinct." (If we took it to mean "$n$ has at most $k$ prime factors", then $A=$ the set of all powers of $2$ would be a counterexample.)

It was pointed out in a comment (apparently now deleted) that the problem asks for a proper subset of $A.$ In fact, it's sufficient to show that for any set $A$ with the given properties, we can find an infinite $B \subseteq A$ such that any two members of $B$ have the same gcd. After all, if there is such a $B,$ then $B$ with its first element removed is a proper subset of $A$ with the required property.

We'll prove the statement by induction on $k,$ as the OP suggests.

For $k=1,$ take $B=A.$

Now assume the statement is true for $k,$ and let $A$ be an infinite set of positive integers such that each $n \in A$ is the product of at most $k+1$ primes.

Case 1: There exists some number $m > 1$ such that infinitely many multiples of $m$ belong to $A.$ Set $A' = \lbrace n \;\vert\; mn \in A \rbrace,$ and observe that $A'$ is infinite and each member of $A'$ is the product of at most $k$ primes. Let $B$ be an infinite set contained in $A'$ such that the gcd of any two distinct members of $B$ is some fixed number $g.$ Then $\lbrace mb \;\vert\; b \in B \rbrace$ is an infinite subset of $A$ such that any two distinct members of $B$ have gcd $mg.$

Case 2: For every $m,$ only finitely many multiples of $m$ belong to $A.$ Define $b_0, b_1, \dots \in A$ by induction, as follows: let $b_n$ be the least member of $A$ that is greater than 1 and that is not a multiple of any prime that divides some $b_k$ for $k<n.$ Then $B = \lbrace b_0, b_1, \dots \rbrace$ is an infinite subset of $A$ such that any two distinct members of $B$ have gcd $1.$

• How do you know an infinite set $B$ contained in $A'$ exists such that the greatest common divisor of any two distinct members of $B$ is some fixed number $g$? – user19405892 Jul 25 '16 at 23:15
• By the induction hypothesis. Every member of $A'$ is the product of at most $k$ primes, and we've assumed the statement is true for $k.$ – Mitchell Spector Jul 25 '16 at 23:17

HINT: If there is a prime $p$ that is a factor of infinitely many members of $A$, let $B=\{n\in A:p\mid n\}$, and apply the induction hypothesis to $\left\{\frac{n}p:n\in B \right\}$.

If there is no such $p$, show that there is an infinite $B\subseteq A$ such that the members of $B$ are pairwise relatively prime.

For each prime $q$, let $A_q$ be the set of members of $A$ divisible by $q$. Each member of $A$ is in at most $k$ of the sets $A_q$.

Case 1: if all $A_q$ are finite, then it's not hard to show that there are infinitely many nonempty and mutually disjoint $A_q$; taking one member of each of them, you have your $B$ with $p=1$.

Case 2: If some $A_q$ is infinite, replace $A$ by $C_q = \{a/q: a \in A_q\}$. This satisfies the same conditions as $A$, but with $k$ replaced by $k-1$. Using induction, there exist an infinite set $B_q \subseteq C_q$ and $p_q$ which is the gcd of any two members of $B_q$. Let $B = \{bq: b \in B_q$ and $p = q p_q$.

If infinitely many elements of $A$ share a prime factor this is easy.

If they don't, you could try to find elements of $A$ such that they share no prime factors with any other elements of $A$. There are infinitely many primes, so it's easy to think of 'nice' $A$ where this could be possible. This should be true for general $A$ because each $n\in A$ is the product of at most $k$ prime numbers and $A$ is infinite. Thus you must have elements that share no prime factors with other elements - otherwise you would either get a finite set, or an infinite subset that shares a prime factors (which isn't true for this case).


text:
How to compare two approximation algorithms

We say that "An algorithm is a $\beta$-approximation to problem $X$, if for any instance of problem $X$, the solution returned by this algorithm is within a factor $\beta$ of the optimal solution".

Now assume that we have a $\log(N)$-approximation algorithm (called it Alg 1) for problem $X$ where $N$ is the size of problem $X$. Further assume there is one other algorithm (called it Alg 2) for problem $X$, but for that algorithm we could find an instance of problem $X$ with size $N$ such that the solution returned by Alg 2 is $N/2$ times of the optimal solution.

What is a correct way of saying this statement that clearly shows that Alg 1 is much more better than Alg 2? I was thinking about the following one:

"Alg 2 provides a $\beta$-approximation to problem $X$ where $\beta \geq N/2$."

Is this correct? Any better suggestion?

I would write something like this: whereas Alg 1 is a $\log(N)$-approximation algorithms, there are instances on which the approximation ratio of Alg 2 is only $N/2$. So Alg 1 is much better (in the worst case) than Alg 2.


text:
# But what is a continuous function?

I have a very basic problem. I am confused about "continuous function" term.

What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?

Let's say:

$$\ln|x|$$ - the graph clearly says it's continuous for all real numbers except for $$0$$ which is not part of the domain. So is this function continuous or not? I could say same about $$\tan{x}$$ or $$\frac{x+1}{x}$$

$$\ln{x}$$ - the graph clearly says it's continuous for all of its domain: $$(0; \infty)$$ - so is this $$f$$ continuous or not?

Thanks for clarification.

• If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question? – Nap D. Lover Jan 29 at 21:16
• Not really. Question is: Is function $f(x) = \ln{|x|}$ continuous? I assume it is? – weno Jan 29 at 21:35
• continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $\log |x|$ is certainly continuous where it is defined. Is that clear? – Nap D. Lover Jan 29 at 21:38
• functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 \le x < 1$ and $f(x) = \sqrt x$ for $x \ge 1$. Is continuous at $x = -7$ and $x = \pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain. – fleablood Jan 29 at 22:49
• That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = \frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong. – fleablood Jan 29 at 22:51

Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $$f(x) = 1/x$$ and $$f(x)=\ln x$$ are continuous functions.

• Yes, but preferrably the domain should be specified if this is important. The functions $f : \mathbb{R}\setminus0\to\mathbb{R}$, $x\mapsto \tfrac1x$ and $g : ]0,\infty[\to\mathbb{R}$, $x\mapsto \ln x$ are continuous functions, and it's clear what's meant. – leftaroundabout Jan 30 at 11:07

"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.

It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.

• This is just not true. You can define continuity in terms of continuity at a given point, but in many contexts, that is not a very natural definition. I also can't recall any important mathematical concept for which continuity at a single point (as opposed to not being continuous anywhere) changes much. – tomasz Jan 30 at 12:53
• Once students get to a topology course, we do indeed teach them continuity as a property of a function, namely: A function is continuous if and only if the inverse image of each open subset of the range is an open subset of the domain. – Lee Mosher Jan 31 at 16:39

The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $$(a,b)$$ if for every number $$c$$ in $$(a,b)$$, $$f(x)=\lim_{x\to c} f(x)$$.

In your example $$f(x)=\ln{x}$$ is continuous on the interval $$(0,\infty)$$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $$f$$ to include the complex numbers or not.

In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).

• No. Not for most uses, a function is regarded as being continuous on an interval. – user370967 Jan 29 at 20:53
• There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply. – R. Burton Jan 29 at 21:01
• I have always seen continuity defined at a point and on a set of points. Every once in a while I saw a calculus textbook author assume continuity on an open interval to simplify the proof of a basic result. But in the context given I think it is exceedingly likely that "is the function continuous" implicitly refers either to "on $\mathbb R$" or "on its domain". I don't think the implicit meaning of "on an interval" is likely. – rschwieb Jan 30 at 16:04

It is, but what you are looking for might be the notion of continous extension.

Here both $$\frac{1}{⋅}$$ and $$\log$$ are continuous in the sense that they are (pointwise) continuous on their respective domains, as is $$\left| \begin{array}{lll} f : &\mathbb{R}^* ⟶ \mathbb{R}\\ & x \longmapsto x² \end{array} \right.$$ The difference is that there exists $$g: \mathbb{R}→\mathbb{R}$$ continuous such that $$g(x)=f(x)$$ for all $$x∈\mathbb{R}^*$$ — so $$f$$ has a continous extension on $$\mathbb{R}$$ — whereas $$\frac{1}{⋅}$$ has no such thing.


text:
# Chapter2-Equilibrium and Kinetics¶

## Ex1-pg13¶

In [1]:
## Calculate the entropy increase
import math
del_h = 6.02 ## Heat added in kJ/mol
t_m = 273.15 ## mean temperature in kelvin
print("\n Example 2.1")
del_s = del_h*1e3/t_m
print"%s %.2f %s"%("\n Increase in entropy is ",del_s," J mol^-1 K^-1")

 Example 2.1

Increase in entropy is  22.04  J mol^-1 K^-1

## Ex2-pg16¶

In [5]:
## Calculation of fraction of atoms with energy equal to or greater than 1eV at temperatures
import math
E = 1. ## energy in electron volt
e = 1.6e-19 ## charge on electron
k = 1.38e-23 ## constant
t1 = 300. ## temperature in K
t2 = 1500. ## temperature in K
print("\n Example 2.2")
print("\n\n Part A:")
n_N = math.exp(-(e/(k*t1)))
print"%s %.2f %s %.2e %s "%("\n Fraction of atoms with energy equal to or greater than 1eV at temperature ",t1," K is" ,n_N, " ") ## numerical value of answer in book is 2e-17
print("\n\n Part B:")
n_N = math.exp(-(e/(k*t2)))
print"%s %.2f %s %.2e %s "%("\n Fraction of atoms with energy equal to or greater than 1eV at temperature",t2," K is ", n_N," ") ## numerical value of answer in book is 4.3e4

#ans are converted into units

 Example 2.2

Part A:

Fraction of atoms with energy equal to or greater than 1eV at temperature  300.00  K is 1.64e-17

Part B:

Fraction of atoms with energy equal to or greater than 1eV at temperature 1500.00  K is  4.40e-04


text:
# Why is it that when the score variance gets LARGER we get MORE confident about the MLE estimate?

I'm following Kashin's review of MLE

Suppose we have a simple Bernoulli distribution and we know its true parameter $p_0$. The Fisher expected information $I(p_0)$ is equal to the variance of the expected score, when evaluated at $p_0$.

It's also known that the MLE estimate $p_{mle}$ converges to $p_0$ like $N(p_0, I(p_0)^{-1})$. So as LARGER the information gets, the variance of the score gets LARGER, but our confidence in the MLE gets better! This doesn't make sense to me intuitively:

In the below graphs the X-axis is values of $p$, and both the log-likelihood $L$ and the score $S$ are plotted for 2 samples. On the left graph they have a larger observed fisher information than on the right, so their curvatures are steeper and indeed the score when evaluated at $p_0$ has a larger variance (in green).

So why is it that we are more confident that the MLEs on the left graph will converge towards $p_0$? We can see that their score variance in $p_0$ is larger, which means that they are farther away from the expected score $E(S(p_0)) = 0$.

Kashin has a formal proof, can anyone lay an intuitive explanation on this counter-intuitive (at least for me) phenomenon?

NOTE: I realize that the curvature of the likelhoods on the left is steeper than those on the right, and I understand the intuition behind this interpretation of the information - It is the other interpretation of the information - the score variance when evaluated at the true parameter - which I want to understand.

Thanks!

I'm not sure what is really puzzling you. The score indicates how sensitively a likelihood function depends on its parameter. So it grows (and so does the variance) with the amount of information that the likelihood carries about the parameter. On the other, the maximum likelihood estimate get better as the amount of information grows. Seems intuitive enough for me...

I made some maths to clarify things. You can also have a look at my notebook where I checked that the maths are working by generating some Bernoulli samples.

## Math

Let $X_1 ,...,X_T$ be an iid sample with $X_i \sim Bernoulli(\theta_0)$ with $\theta_0$ being the true parameter.

The joint density/likelihood function is given by: $$f( x ; \theta )= L ( \theta | x )= \prod_{i=1}^{T}{\theta^{x_i}(1-\theta)^{1-x_i}}=\theta^{\sum x_i}(1-\theta)^{n-\sum x_i}$$

The score function is given by: $$S(\theta | x)=\frac{\partial \ln L(\theta|x)}{\partial \theta}= \frac{1}{\theta}\sum x_i - \frac{1}{1-\theta}(T - \sum x_i)$$

The MLE of $\theta_0$ is: $$\hat{\theta}=\frac{1}{T}\sum_{i=1}^{T}x_i$$

And the sample Fisher Information is: $$I(\theta | x) = -\mathbf{E}[\frac{\partial^2 \ln L(\theta|x)}{\partial \theta^2}]$$

Evaluated at $\theta_0$: $$I(\theta_0 | x) = T(\frac{1}{\theta_0} + \frac{1}{1-\theta_0})$$

Then: $$\frac{1}{\sqrt{T}}S(\theta_0|x) \xrightarrow{D} \mathcal{N}(0, \frac{I(\theta_0)}{T})$$

and: $$\sqrt{T}(\hat{\theta}-\theta_0) \xrightarrow{D} \mathcal{N}(0, \{\frac{I(\theta_0)}{T}\}^{-1})$$

## Conclusion

• Situation 1.

$I(\theta_0 | x) = -\mathbf{E}[\frac{\partial^2 \ln L(\theta|x)}{\partial \theta^2}\big|_{\theta_0}]$ is large (hence variance of the mle will be small) then it means that the gradient of $\frac{\partial \ln L(\theta|x)}{\partial \theta}$ is steep. Hence even for small deviations from $\theta_0$ ,$\frac{\partial \ln L(\theta|x)}{\partial \theta}$ is likely to be far from zero. This means the mle $\hat{\theta}$ is likely to be in a close neighbourhood of $\theta_0$.

• Situation 2.

$I(\theta_0 | x) = -\mathbf{E}[\frac{\partial^2 \ln L(\theta|x)}{\partial \theta^2}\big|_{\theta_0}]$ is small (hence variance of the mle will be large). In this case the gradient of the likelihood $\frac{\partial \ln L(\theta|x)}{\partial \theta}$ is flatter and hence $\frac{\partial \ln L(\theta|x)}{\partial \theta}\approx 0$ for a large neighbourhood about the true parameter $\theta_0$. Therefore the mle $\hat{\theta}$ can lie in a large neighbourhood of $\theta_0$.

## Sources

http://faculty.washington.edu/ezivot/econ583/mleLectures.pdf

https://www.stat.tamu.edu/~suhasini/teaching613/STAT613.pdf

• Wow, @Adrien - you went to a lot of trouble for me and I really appreciate it! However, I'm still not sure what am I seeing there - I was interested in the variance of the score - and it looks like its the same = 11 in both situations. Also the histograms spread is about the same... Also - I think that the core of what you're saying is in the end - can you explain this to me "Hence even for small deviations from θ0 ,the score is likely to be far from zero. This means that the mle is likely to be in a close neighbourhood of θ0." - thanks again!!! Sep 12 '16 at 19:57

My intuition would be as follows: As $n\to\infty$, $p_{mle}\to_p p_0$ by consistency of the MLE, so the log-likelihood will be centered around the true value.

Now, if we have more curvature, that means that the log-likelihood bends away from the true value more quickly, meaning that other candidate values $p$ for the true value $p_0$ are assigned much less likelihood than the MLE, meaning that we are much more confident to prefer the MLE over other values of $p$.

Conversely, if the log-likelihood is relatively flat like on the right, we still have that, by construction, the MLE is at the peak of the log-likelihood, but other values receive about the same likelihood, so that we would not be very confident to prefer the MLE over the others.

• Thanks! but your intuition relies solely on the curvature of the log-likelihood which is one interpretation for the fisher information, you ignored the score variance at the true parameter, which is another interpretation of the fisher information - and that's what my question is about... Sep 9 '16 at 13:30
• Fair point - here, my intuition to relate these two ideas would be that, if the curvature is higher, a small change in the data will imply a large chance in the score, because even if we move just a little along the axis, the slope will change a lot. Sep 9 '16 at 13:33


text:
# Private Tutors

The following persons have informed the Department of Mathematical Sciences that they are interested in tutoring mathematics at the college level. To help students who are looking for a tutor, the department provides this information as it was received. Please note that the department has not checked the qualifications of the listed tutors, nor does it recommend the listed tutors over others who did not request listing.

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text:
## anonymous one year ago I need help this is sooo hard 8th grade math!

1. anonymous

it is hard, isn't it?

2. anonymous

aalright haha im sure if you tae it in steps everything willl be simplar

3. anonymous

What is the value of the expression $\frac{ 1 }{ 4^{-3}}$

4. anonymous

alright easy lets start by looking at the problem

5. anonymous

a.12 b.64 c. 1/64 d. 1/12

6. anonymous

we have a 4^-3 in the denominator. what does that tell you?

7. anonymous

Its either a or b

8. anonymous

umm idk

9. anonymous

10. anonymous

alright i will let you know a nice rule. when you bring a number from the denominator to the numerator you slip the exponent

11. anonymous

so something like 1/5^-2 = 5^2

12. anonymous

do you understand?

13. anonymous

no

14. anonymous

ok thats fine. its just a rule. i just want to make sure you know 5^2 is 5 squared

15. anonymous

do you understand that?

16. anonymous

yes

17. anonymous

alright cool

18. anonymous

so with your problem we have 1/4^-3

19. anonymous

we bring the 4^-3 to the numerator

20. anonymous

and we get?....what?

21. anonymous

umm

22. anonymous

...

23. anonymous

ok with our rule we change the sign of the exponent

24. anonymous

so what is it?

25. anonymous

4* -3 ???????

26. anonymous

oh i see what you are confused about. when i am talking about sign i mean positive or negative

27. anonymous

so all i mean by change the sign of the exponent is change it from -3 to 3

28. anonymous

so you have just 4^3

29. anonymous

now what is 4 to the power of 3?

30. anonymous

Its 64!

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text:
# Secure Science issues preview of their upcoming block cipher

Fri Mar 25 12:23:24 EST 2005

Jerrold Leichter wrote:

> I can come up with a cipher provably just as secure as AES-128 very quickly....
>
> (Actually, based on the paper a while back on many alternative ways to
> formulate AES - it had a catchy title something like "How Many Ways Can You
> Spell AES?", except that I can't find one like that now - one could even
> come up with a formulation that is (a) probably as secure as AES-128; (b)
> actually faster in hardware or simpler to implement or whatever...)

You're probably looking for [1] by Barkan and Biham. What they do is
replacing the irreducible polynomial and all the constants involved in
Rijndael to get what they call "dual ciphers"; basically those ciphers
are isomorphic to Rijndael. All in all they get 240 dual ciphers which
are listed in [2]. What I found more interesting back then was that they
also give square dual and log dual ciphers of Rijndael. I.e. let E be
the Rijndael encryption and E' be the encryption function of the
square/log dual Rijndael construction. Furthermore let f be a function
that either performs bytewise squaring in GF(2^8) or replaces each byte
with a logarithmic representation (relative to a generator g. you also
need to fix log_g(0) = -\infty for this to make sense). Then

E'(f(plaintext), f(key)) = f(E(plaintext, key))

holds. The squaring construction then also naturally extends to what
they call "higher-order self dual ciphers": meaning you can apply the
squaring multiple times.

In 2004 Wu, Lu and Laih then demonstrated that using Barkan's and
Biham's method can indeed lead to more efficient implementations of
AES/Rijndael in hardware.

Cheers,
Ralf

[1] Elad Barkan and Eli Biham:
In How Many Ways Can You Write Rijndael?
ASIACRYPT 2002, Springer
note: also on ePrint as http://eprint.iacr.org/2002/157
if you don't have Springer Link access

[2] Elad Barkan and Eli Biham:
The Book of Rijndaels
http://eprint.iacr.org/2002/158

[3] Shee-Yau Wu and Shih-Chuan Lu and Chi Sung Laih:
Design of AES Based on Dual Cipher and Composite Field
Topics in Cryptology, CT-RSA 2004, Springer

--


text:
Dipole

Last updated

In electromagnetism, there are two kinds of dipoles:

Electromagnetism is a branch of physics involving the study of the electromagnetic force, a type of physical interaction that occurs between electrically charged particles. The electromagnetic force usually exhibits electromagnetic fields such as electric fields, magnetic fields, and light, and is one of the four fundamental interactions in nature. The other three fundamental interactions are the strong interaction, the weak interaction, and gravitation. At high energy the weak force and electromagnetic force are unified as a single electroweak force.

Contents

• An electric dipole is a separation of positive and negative charges. The simplest example of this is a pair of electric charges of equal magnitude but opposite sign, separated by some (usually small) distance. A permanent electric dipole is called an electret.
• A magnetic dipole is a closed circulation of electric current. A simple example of this is a single loop of wire with some constant current through it. [1] [2]

The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity. The SI units for electric dipole moment are coulomb-meter (C.m); however, the most common unit is the debye (D).

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charges; positive and negative. Like charges repel and unlike attract. An object with an absence of net charge is referred to as neutral. Early knowledge of how charged substances interact is now called classical electrodynamics, and is still accurate for problems that do not require consideration of quantum effects.

An electret is a dielectric material that has a quasi-permanent electric charge or dipole polarisation. An electret generates internal and external electric fields, and is the electrostatic equivalent of a permanent magnet. Although Oliver Heaviside coined this term in 1885, materials with electret properties were already known to science and had been studied since the early 1700s. One particular example is the electrophorus, a device consisting of a slab with electret properties and a separate metal plate. The electrophorus was originally invented by Johan Carl Wilcke in Sweden and again by Alessandro Volta in Italy.

Dipoles can be characterized by their dipole moment, a vector quantity. For the simple electric dipole given above, the electric dipole moment points from the negative charge towards the positive charge, and has a magnitude equal to the strength of each charge times the separation between the charges. (To be precise: for the definition of the dipole moment, one should always consider the "dipole limit", where, for example, the distance of the generating charges should converge to 0 while simultaneously, the charge strength should diverge to infinity in such a way that the product remains a positive constant.)

For the current loop, the magnetic dipole moment points through the loop (according to the right hand grip rule), with a magnitude equal to the current in the loop times the area of the loop.

In addition to current loops, the electron, among other fundamental particles, has a magnetic dipole moment. That is because it generates a magnetic field that is identical to that generated by a very small current loop. However, the electron's magnetic moment is not due to a current loop, but is instead an intrinsic property of the electron. [3] It is also possible that the electron has an electric dipole moment although it has not yet been observed (see electron electric dipole moment for more information).

The electron is a subatomic particle, symbol
e
or
β
, whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure. The electron has a mass that is approximately 1/1836 that of the proton. Quantum mechanical properties of the electron include an intrinsic angular momentum (spin) of a half-integer value, expressed in units of the reduced Planck constant, ħ. As it is a fermion, no two electrons can occupy the same quantum state, in accordance with the Pauli exclusion principle. Like all elementary particles, electrons exhibit properties of both particles and waves: they can collide with other particles and can be diffracted like light. The wave properties of electrons are easier to observe with experiments than those of other particles like neutrons and protons because electrons have a lower mass and hence a longer de Broglie wavelength for a given energy.

A magnetic field is a vector field that describes the magnetic influence of electrical currents and magnetized materials. In everyday life, the effects of magnetic fields are often seen in permanent magnets, which pull on magnetic materials and attract or repel other magnets. Magnetic fields surround and are created by magnetized material and by moving electric charges such as those used in electromagnets. Magnetic fields exert forces on nearby moving electrical charges and torques on nearby magnets. In addition, a magnetic field that varies with location exerts a force on magnetic materials. Both the strength and direction of a magnetic field varies with location. As such, it is an example of a vector field.

In science and engineering, an intrinsic property is a property of a specified subject that exists itself or within the subject. An extrinsic property is not essential or inherent to the subject that is being characterized. For example, density is an intrinsic property of any physical object, whereas belonging to something is an extrinsic property that depends on another object.

A permanent magnet, such as a bar magnet, owes its magnetism to the intrinsic magnetic dipole moment of the electron. The two ends of a bar magnet are referred to as poles (not to be confused with monopoles), and may be labeled "north" and "south". In terms of the Earth's magnetic field, they are respectively "north-seeking" and "south-seeking" poles: if the magnet were freely suspended in the Earth's magnetic field, the north-seeking pole would point towards the north and the south-seeking pole would point towards the south. The dipole moment of the bar magnet points from its magnetic south to its magnetic north pole. The north pole of a bar magnet in a compass points north. However, that means that Earth's geomagnetic north pole is the south pole (south-seeking pole) of its dipole moment and vice versa.

In particle physics, a magnetic monopole is a hypothetical elementary particle that is an isolated magnet with only one magnetic pole. A magnetic monopole would have a net "magnetic charge". Modern interest in the concept stems from particle theories, notably the grand unified and superstring theories, which predict their existence.

A compass is an instrument used for navigation and orientation that shows direction relative to the geographic cardinal directions. Usually, a diagram called a compass rose shows the directions north, south, east, and west on the compass face as abbreviated initials. When the compass is used, the rose can be aligned with the corresponding geographic directions; for example, the "N" mark on the rose points northward. Compasses often display markings for angles in degrees in addition to the rose. North corresponds to 0°, and the angles increase clockwise, so east is 90° degrees, south is 180°, and west is 270°. These numbers allow the compass to show magnetic North azimuths or true North azimuths or bearings, which are commonly stated in this notation. If magnetic declination between the magnetic North and true North at latitude angle and longitude angle is known, then direction of magnetic North also gives direction of true North.

The only known mechanisms for the creation of magnetic dipoles are by current loops or quantum-mechanical spin since the existence of magnetic monopoles has never been experimentally demonstrated.

The term comes from the Greek δίς (dis), "twice" [4] and πόλος (polos), "axis". [5] [6]

Classification

A physical dipole consists of two equal and opposite point charges: in the literal sense, two poles. Its field at large distances (i.e., distances large in comparison to the separation of the poles) depends almost entirely on the dipole moment as defined above. A point (electric) dipole is the limit obtained by letting the separation tend to 0 while keeping the dipole moment fixed. The field of a point dipole has a particularly simple form, and the order-1 term in the multipole expansion is precisely the point dipole field.

Although there are no known magnetic monopoles in nature, there are magnetic dipoles in the form of the quantum-mechanical spin associated with particles such as electrons (although the accurate description of such effects falls outside of classical electromagnetism). A theoretical magnetic point dipole has a magnetic field of exactly the same form as the electric field of an electric point dipole. A very small current-carrying loop is approximately a magnetic point dipole; the magnetic dipole moment of such a loop is the product of the current flowing in the loop and the (vector) area of the loop.

Any configuration of charges or currents has a 'dipole moment', which describes the dipole whose field is the best approximation, at large distances, to that of the given configuration. This is simply one term in the multipole expansion when the total charge ("monopole moment") is 0—as it always is for the magnetic case, since there are no magnetic monopoles. The dipole term is the dominant one at large distances: Its field falls off in proportion to 1/r3, as compared to 1/r4 for the next (quadrupole) term and higher powers of 1/r for higher terms, or 1/r2 for the monopole term.

Molecular dipoles

Many molecules have such dipole moments due to non-uniform distributions of positive and negative charges on the various atoms. Such is the case with polar compounds like hydrogen fluoride (HF), where electron density is shared unequally between atoms. Therefore, a molecule's dipole is an electric dipole with an inherent electric field that should not be confused with a magnetic dipole which generates a magnetic field.

The physical chemist Peter J. W. Debye was the first scientist to study molecular dipoles extensively, and, as a consequence, dipole moments are measured in units named debye in his honor.

For molecules there are three types of dipoles:

Permanent dipoles
These occur when two atoms in a molecule have substantially different electronegativity: One atom attracts electrons more than another, becoming more negative, while the other atom becomes more positive. A molecule with a permanent dipole moment is called a polar molecule. See dipole–dipole attractions.
Instantaneous dipoles
These occur due to chance when electrons happen to be more concentrated in one place than another in a molecule, creating a temporary dipole. These dipoles are smaller in magnitude than permanent dipoles, but still play a large role in chemistry and biochemistry due to their prevalence. See instantaneous dipole.
Induced dipoles
These can occur when one molecule with a permanent dipole repels another molecule's electrons, inducing a dipole moment in that molecule. A molecule is polarized when it carries an induced dipole. See induced-dipole attraction.

More generally, an induced dipole of any polarizable charge distribution ρ (remember that a molecule has a charge distribution) is caused by an electric field external to ρ. This field may, for instance, originate from an ion or polar molecule in the vicinity of ρ or may be macroscopic (e.g., a molecule between the plates of a charged capacitor). The size of the induced dipole moment is equal to the product of the strength of the external field and the dipole polarizability of ρ.

Dipole moment values can be obtained from measurement of the dielectric constant. Some typical gas phase values in debye units are: [7]

Potassium bromide (KBr) has one of the highest dipole moments because it is an ionic compound that exists as a molecule in the gas phase.

The overall dipole moment of a molecule may be approximated as a vector sum of bond dipole moments. As a vector sum it depends on the relative orientation of the bonds, so that from the dipole moment information can be deduced about the molecular geometry.

For example, the zero dipole of CO2 implies that the two C=O bond dipole moments cancel so that the molecule must be linear. For H2O the O−H bond moments do not cancel because the molecule is bent. For ozone (O3) which is also a bent molecule, the bond dipole moments are not zero even though the O−O bonds are between similar atoms. This agrees with the Lewis structures for the resonance forms of ozone which show a positive charge on the central oxygen atom.

Cis isomer, dipole moment 1.90 D
Trans isomer, dipole moment zero

An example in organic chemistry of the role of geometry in determining dipole moment is the cis and trans isomers of 1,2-dichloroethene. In the cis isomer the two polar C−Cl bonds are on the same side of the C=C double bond and the molecular dipole moment is 1.90 D. In the trans isomer, the dipole moment is zero because the two C−Cl bonds are on opposite sides of the C=C and cancel (and the two bond moments for the much less polar C−H bonds also cancel).

Another example of the role of molecular geometry is boron trifluoride, which has three polar bonds with a difference in electronegativity greater than the traditionally cited threshold of 1.7 for ionic bonding. However, due to the equilateral triangular distribution of the fluoride ions about the boron cation center, the molecule as a whole does not exhibit any identifiable pole: one cannot construct a plane that divides the molecule into a net negative part and a net positive part.

Quantum mechanical dipole operator

Consider a collection of N particles with charges qi and position vectors ri. For instance, this collection may be a molecule consisting of electrons, all with chargee, and nuclei with charge eZi, where Zi is the atomic number of the ith nucleus. The dipole observable (physical quantity) has the quantum mechanical dipole operator:[ citation needed ]

${\displaystyle {\mathfrak {p}}=\sum _{i=1}^{N}\,q_{i}\,\mathbf {r} _{i}\,.}$

Notice that this definition is valid only for non-charged dipoles, i.e. total charge equal to zero. To a charged dipole we have the next equation:

${\displaystyle {\mathfrak {p}}=\sum _{i=1}^{N}\,q_{i}\,(\mathbf {r} _{i}-\mathbf {r} _{c})\,.}$

where ${\displaystyle \mathbf {r} _{c}}$ is the center of mass of the molecule/group of particles. [8]

Atomic dipoles

A non-degenerate (S-state) atom can have only a zero permanent dipole. This fact follows quantum mechanically from the inversion symmetry of atoms. All 3 components of the dipole operator are antisymmetric under inversion with respect to the nucleus,

${\displaystyle {\mathfrak {I}}\;{\mathfrak {p}}\;{\mathfrak {I}}^{-1}=-{\mathfrak {p}},}$

where ${\displaystyle {\mathfrak {p}}}$ is the dipole operator and ${\displaystyle {\mathfrak {I}}}$ is the inversion operator.

The permanent dipole moment of an atom in a non-degenerate state (see degenerate energy level) is given as the expectation (average) value of the dipole operator,

${\displaystyle \left\langle {\mathfrak {p}}\right\rangle =\left\langle \,S\,|{\mathfrak {p}}|\,S\,\right\rangle ,}$

where ${\displaystyle |\,S\,\rangle }$ is an S-state, non-degenerate, wavefunction, which is symmetric or antisymmetric under inversion: ${\displaystyle {\mathfrak {I}}\,|\,S\,\rangle =\pm |\,S\,\rangle }$. Since the product of the wavefunction (in the ket) and its complex conjugate (in the bra) is always symmetric under inversion and its inverse,

${\displaystyle \left\langle {\mathfrak {p}}\right\rangle =\left\langle \,{\mathfrak {I}}^{-1}\,S\,|{\mathfrak {p}}|\,{\mathfrak {I}}^{-1}\,S\,\right\rangle =\left\langle \,S\,|{\mathfrak {I}}\,{\mathfrak {p}}\,{\mathfrak {I}}^{-1}|\,S\,\right\rangle =-\left\langle {\mathfrak {p}}\right\rangle }$

it follows that the expectation value changes sign under inversion. We used here the fact that ${\displaystyle {\mathfrak {I}}}$, being a symmetry operator, is unitary: ${\displaystyle {\mathfrak {I}}^{-1}={\mathfrak {I}}^{*}\,}$ and by definition the Hermitian adjoint ${\displaystyle {\mathfrak {I}}^{*}\,}$ may be moved from bra to ket and then becomes ${\displaystyle {\mathfrak {I}}^{**}={\mathfrak {I}}\,}$. Since the only quantity that is equal to minus itself is the zero, the expectation value vanishes,

${\displaystyle \left\langle {\mathfrak {p}}\right\rangle =0.}$

In the case of open-shell atoms with degenerate energy levels, one could define a dipole moment by the aid of the first-order Stark effect. This gives a non-vanishing dipole (by definition proportional to a non-vanishing first-order Stark shift) only if some of the wavefunctions belonging to the degenerate energies have opposite parity; i.e., have different behavior under inversion. This is a rare occurrence, but happens for the excited H-atom, where 2s and 2p states are "accidentally" degenerate (see article Laplace–Runge–Lenz vector for the origin of this degeneracy) and have opposite parity (2s is even and 2p is odd).

Field of a static magnetic dipole

Magnitude

The far-field strength, B, of a dipole magnetic field is given by

${\displaystyle B(m,r,\lambda )={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}{\sqrt {1+3\sin ^{2}(\lambda )}}\,,}$

where

B is the strength of the field, measured in teslas
r is the distance from the center, measured in metres
λ is the magnetic latitude (equal to 90°  θ) where θ is the magnetic colatitude, measured in radians or degrees from the dipole axis [note 1]
m is the dipole moment, measured in ampere-square metres or joules per tesla
μ0 is the permeability of free space, measured in henries per metre.

Conversion to cylindrical coordinates is achieved using r2 = z2 + ρ2 and

${\displaystyle \lambda =\arcsin \left({\frac {z}{\sqrt {z^{2}+\rho ^{2}}}}\right)}$

where ρ is the perpendicular distance from the z-axis. Then,

${\displaystyle B(\rho ,z)={\frac {\mu _{0}m}{4\pi \left(z^{2}+\rho ^{2}\right)^{\frac {3}{2}}}}{\sqrt {1+{\frac {3z^{2}}{z^{2}+\rho ^{2}}}}}}$

Vector form

The field itself is a vector quantity:

${\displaystyle \mathbf {B} (\mathbf {m} ,\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\left({\frac {3(\mathbf {m} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {m} }{r^{3}}}\right)+{\frac {2\mu _{0}}{3}}\mathbf {m} \delta ^{3}(\mathbf {r} )}$

where

B is the field
r is the vector from the position of the dipole to the position where the field is being measured
r is the absolute value of r: the distance from the dipole
= r/r is the unit vector parallel to r;
m is the (vector) dipole moment
μ0 is the permeability of free space
δ3 is the three-dimensional delta function. [note 2]

This is exactly the field of a point dipole, exactly the dipole term in the multipole expansion of an arbitrary field, and approximately the field of any dipole-like configuration at large distances.

Magnetic vector potential

The vector potential A of a magnetic dipole is

${\displaystyle \mathbf {A} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}{\frac {\mathbf {m} \times {\hat {\mathbf {r} }}}{r^{2}}}}$

with the same definitions as above.

Field from an electric dipole

The electrostatic potential at position r due to an electric dipole at the origin is given by:

${\displaystyle \Phi (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\,{\frac {\mathbf {p} \cdot {\hat {\mathbf {r} }}}{r^{2}}}}$

where

${\displaystyle {\hat {\mathbf {r} }}}$ is a unit vector in the direction of r, p is the (vector) dipole moment, and ε0 is the permittivity of free space.

This term appears as the second term in the multipole expansion of an arbitrary electrostatic potential Φ(r). If the source of Φ(r) is a dipole, as it is assumed here, this term is the only non-vanishing term in the multipole expansion of Φ(r). The electric field from a dipole can be found from the gradient of this potential:

${\displaystyle \mathbf {E} =-\nabla \Phi ={\frac {1}{4\pi \epsilon _{0}}}\left({\frac {3(\mathbf {p} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {p} }{r^{3}}}\right)-{\frac {1}{3\epsilon _{0}}}\mathbf {p} \delta ^{3}(\mathbf {r} )}$

where E is the electric field and δ3 is the 3-dimensional delta function. [note 2] This is formally identical to the magnetic H field of a point magnetic dipole with only a few names changed.

Torque on a dipole

Since the direction of an electric field is defined as the direction of the force on a positive charge, electric field lines point away from a positive charge and toward a negative charge.

When placed in an electric or magnetic field, equal but opposite forces arise on each side of the dipole creating a torque τ}:

${\displaystyle {\boldsymbol {\tau }}=\mathbf {p} \times \mathbf {E} }$

for an electric dipole moment p (in coulomb-meters), or

${\displaystyle {\boldsymbol {\tau }}=\mathbf {m} \times \mathbf {B} }$

for a magnetic dipole moment m (in ampere-square meters).

The resulting torque will tend to align the dipole with the applied field, which in the case of an electric dipole, yields a potential energy of

${\displaystyle U=-\mathbf {p} \cdot \mathbf {E} }$.

The energy of a magnetic dipole is similarly

${\displaystyle U=-\mathbf {m} \cdot \mathbf {B} }$.

In addition to dipoles in electrostatics, it is also common to consider an electric or magnetic dipole that is oscillating in time. It is an extension, or a more physical next-step, to spherical wave radiation.

In particular, consider a harmonically oscillating electric dipole, with angular frequency ω and a dipole moment p0 along the direction of the form

${\displaystyle \mathbf {p} (\mathbf {r} ,t)=\mathbf {p} (\mathbf {r} )e^{-i\omega t}=p_{0}{\hat {\mathbf {z} }}e^{-i\omega t}.}$

In vacuum, the exact field produced by this oscillating dipole can be derived using the retarded potential formulation as:

{\displaystyle {\begin{aligned}\mathbf {E} &={\frac {1}{4\pi \varepsilon _{0}}}\left\{{\frac {\omega ^{2}}{c^{2}r}}\left({\hat {\mathbf {r} }}\times \mathbf {p} \right)\times {\hat {\mathbf {r} }}+\left({\frac {1}{r^{3}}}-{\frac {i\omega }{cr^{2}}}\right)\left(3{\hat {\mathbf {r} }}\left[{\hat {\mathbf {r} }}\cdot \mathbf {p} \right]-\mathbf {p} \right)\right\}e^{\frac {i\omega r}{c}}e^{-i\omega t}\\\mathbf {B} &={\frac {\omega ^{2}}{4\pi \varepsilon _{0}c^{3}}}({\hat {\mathbf {r} }}\times \mathbf {p} )\left(1-{\frac {c}{i\omega r}}\right){\frac {e^{i\omega r/c}}{r}}e^{-i\omega t}.\end{aligned}}}

For /c  1, the far-field takes the simpler form of a radiating "spherical" wave, but with angular dependence embedded in the cross-product: [9]

{\displaystyle {\begin{aligned}\mathbf {B} &={\frac {\omega ^{2}}{4\pi \varepsilon _{0}c^{3}}}({\hat {\mathbf {r} }}\times \mathbf {p} ){\frac {e^{i\omega (r/c-t)}}{r}}={\frac {\omega ^{2}\mu _{0}p_{0}}{4\pi c}}({\hat {\mathbf {r} }}\times {\hat {\mathbf {z} }}){\frac {e^{i\omega (r/c-t)}}{r}}=-{\frac {\omega ^{2}\mu _{0}p_{0}}{4\pi c}}\sin(\theta ){\frac {e^{i\omega (r/c-t)}}{r}}\mathbf {\hat {\phi }} \\\mathbf {E} &=c\mathbf {B} \times {\hat {\mathbf {r} }}=-{\frac {\omega ^{2}\mu _{0}p_{0}}{4\pi }}\sin(\theta )\left({\hat {\phi }}\times \mathbf {\hat {r}} \right){\frac {e^{i\omega (r/c-t)}}{r}}=-{\frac {\omega ^{2}\mu _{0}p_{0}}{4\pi }}\sin(\theta ){\frac {e^{i\omega (r/c-t)}}{r}}{\hat {\theta }}.\end{aligned}}}

The time-averaged Poynting vector

${\displaystyle \langle \mathbf {S} \rangle =\left({\frac {\mu _{0}p_{0}^{2}\omega ^{4}}{32\pi ^{2}c}}\right){\frac {\sin ^{2}(\theta )}{r^{2}}}\mathbf {\hat {r}} }$

is not distributed isotropically, but concentrated around the directions lying perpendicular to the dipole moment, as a result of the non-spherical electric and magnetic waves. In fact, the spherical harmonic function (sin θ) responsible for such toroidal angular distribution is precisely the l = 1 "p" wave.

The total time-average power radiated by the field can then be derived from the Poynting vector as

${\displaystyle P={\frac {\mu _{0}\omega ^{4}p_{0}^{2}}{12\pi c}}.}$

Notice that the dependence of the power on the fourth power of the frequency of the radiation is in accordance with the Rayleigh scattering, and the underlying effects why the sky consists of mainly blue colour.

A circular polarized dipole is described as a superposition of two linear dipoles.

Notes

1. Magnetic colatitude is 0 along the dipole's axis and 90° in the plane perpendicular to its axis.
2. δ3(r) = 0 except at r = (0, 0, 0), so this term is ignored in multipole expansion.

Related Research Articles

In quantum mechanics, a Hamiltonian is an operator corresponding to the sum of the kinetic energies plus the potential energies for all the particles in the system. It is usually denoted by H, also Ȟ or Ĥ. Its spectrum is the set of possible outcomes when one measures the total energy of a system. Because of its close relation to the time-evolution of a system, it is of fundamental importance in most formulations of quantum theory.

A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the dimensions of the source are reduced to zero while keeping the magnetic moment constant. It is a magnetic analogue of the electric dipole, but the analogy is not complete. In particular, a magnetic monopole, the magnetic analogue of an electric charge, has never been observed. Moreover, one form of magnetic dipole moment is associated with a fundamental quantum property—the spin of elementary particles.

Synchrotron radiation is the electromagnetic radiation emitted when charged particles are accelerated radially, i.e., when they are subject to an acceleration perpendicular to their velocity. It is produced, for example, in synchrotrons using bending magnets, undulators and/or wigglers. If the particle is non-relativistic, then the emission is called cyclotron emission. If, on the other hand, the particles are relativistic, sometimes referred to as ultrarelativistic, the emission is called synchrotron emission. Synchrotron radiation may be achieved artificially in synchrotrons or storage rings, or naturally by fast electrons moving through magnetic fields. The radiation produced in this way has a characteristic polarization and the frequencies generated can range over the entire electromagnetic spectrum which is also called continuum radiation.

In atomic physics, hyperfine structure refers to small shifts and splittings in the energy levels of atoms, molecules, and ions, due to interaction between the state of the nucleus and the state of the electron clouds.

The magnetic moment is a quantity that represents the magnetic strength and orientation of a magnet or other object that produces a magnetic field. Examples of objects that have magnetic moments include: loops of electric current, permanent magnets, elementary particles, various molecules, and many astronomical objects.

In physics, the Rabi cycle is the cyclic behaviour of a two-level quantum system in the presence of an oscillatory driving field. A great variety of physical processes belonging to the areas of quantum computing, condensed matter, atomic and molecular physics, and nuclear and particle physics can be conveniently studied in terms of two-level quantum mechanical systems, and exhibit Rabi flopping when coupled to an oscillatory driving field. The effect is important in quantum optics, magnetic resonance and quantum computing, and is named after Isidor Isaac Rabi.

The term magnetic potential can be used for either of two quantities in classical electromagnetism: the magnetic vector potential, or simply vector potential, A; and the magnetic scalar potentialψ. Both quantities can be used in certain circumstances to calculate the magnetic field B.

In classical electromagnetism, magnetization or magnetic polarization is the vector field that expresses the density of permanent or induced magnetic dipole moments in a magnetic material. The origin of the magnetic moments responsible for magnetization can be either microscopic electric currents resulting from the motion of electrons in atoms, or the spin of the electrons or the nuclei. Net magnetization results from the response of a material to an external magnetic field, together with any unbalanced magnetic dipole moments that may be inherent in the material itself; for example, in ferromagnets. Magnetization is not always uniform within a body, but rather varies between different points. Magnetization also describes how a material responds to an applied magnetic field as well as the way the material changes the magnetic field, and can be used to calculate the forces that result from those interactions. It can be compared to electric polarization, which is the measure of the corresponding response of a material to an electric field in electrostatics. Physicists and engineers usually define magnetization as the quantity of magnetic moment per unit volume. It is represented by a pseudovector M.

The Larmor formula is used to calculate the total power radiated by a non relativistic point charge as it accelerates or decelerates. This is used in the branch of physics known as electrodynamics and is not to be confused with the Larmor precession from classical nuclear magnetic resonance. It was first derived by J. J. Larmor in 1897, in the context of the wave theory of light.

The electromagnetic wave equation is a second-order partial differential equation that describes the propagation of electromagnetic waves through a medium or in a vacuum. It is a three-dimensional form of the wave equation. The homogeneous form of the equation, written in terms of either the electric field E or the magnetic field B, takes the form:

In many-body theory, the term Green's function is sometimes used interchangeably with correlation function, but refers specifically to correlators of field operators or creation and annihilation operators.

Magnetic dipole–dipole interaction, also called dipolar coupling, refers to the direct interaction between two magnetic dipoles.

The transition dipole moment or transition moment, usually denoted for a transition between an initial state, , and a final state, , is the electric dipole moment associated with the transition between the two states. In general the transition dipole moment is a complex vector quantity that includes the phase factors associated with the two states. Its direction gives the polarization of the transition, which determines how the system will interact with an electromagnetic wave of a given polarization, while the square of the magnitude gives the strength of the interaction due to the distribution of charge within the system. The SI unit of the transition dipole moment is the Coulomb-meter (Cm); a more conveniently sized unit is the Debye (D).

Resonance fluorescence is the process in which a two-level atom system interacts with the quantum electromagnetic field if the field is driven at a frequency near to the natural frequency of the atom.

An LC circuit can be quantized using the same methods as for the quantum harmonic oscillator. An LC circuit is a variety of resonant circuit, and consists of an inductor, represented by the letter L, and a capacitor, represented by the letter C. When connected together, an electric current can alternate between them at the circuit's resonant frequency:

Electric dipole transition is the dominant effect of an interaction of an electron in an atom with the electromagnetic field.

Magnetic resonance is a phenomenon that affects a Magnetic dipole when placed in a uniform static magnetic field. Its energy is split into a finite number of energy levels, depending on the value of quantum number of angular momentum. This is similar to energy quantization for atoms, say
e
in H atom; in this case the atom, in interaction to an external electric field, transitions between different energy levels by absorbing or emitting photons. Similarly if a magnetic dipole is perturbed with electromagnetic field of proper frequency( ), it can transit between its energy eigenstates, but as the separation between energy eigenvalues is small, the frequency of the photon will be the microwave or radio frequency range. If the dipole is tickled with a field of another frequency, it is unlikely to transition. This phenomenon is similar to what occurs when a system is acted on by a periodic force of frequency equal to its natural frequency.

Magnetic current is, nominally, a current composed of fictitious moving magnetic monopoles. It has the dimensions of volts. The usual symbol for magnetic current is which is analogous to for electric current. Magnetic currents produce an electric field analogously to the production of a magnetic field by electric currents. Magnetic current density, which has the units of V/m², is usually represented by the symbols and . The superscripts indicate total and impressed magnetic current density. The impressed currents are the energy sources. In many useful cases, a distribution of electric charge can be mathematically replaced by an equivalent distribution of magnetic current. This artifice can be used to simplify some electromagnetic field problems. It is possible to use both electric current densities and magnetic current densities in the same analysis.

References

1. Brau, Charles A. (2004). Modern Problems in Classical Electrodynamics. Oxford University Press. ISBN   0-19-514665-4.
2. Griffiths, David J. (1999). Introduction to Electrodynamics (3rd ed.). Prentice Hall. ISBN   0-13-805326-X.
3. Griffiths, David J. (1994). Introduction to Quantum Mechanics. Prentice Hall. ISBN   978-0-13-124405-4.
4. δίς, Henry George Liddell, Robert Scott, A Greek-English Lexicon, on Perseus
5. πόλος, Henry George Liddell, Robert Scott, A Greek-English Lexicon, on Perseus
6. "dipole, n.". Oxford English Dictionary (second ed.). Oxford University Press. 1989.
7. Weast, Robert C. (1984). CRC Handbook of Chemistry and Physics (65th ed.). CRC Press. ISBN   0-8493-0465-2.
8. David J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1999, page 447


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Fabrication of Microsensor for Detection of Low-Concentration Formaldehyde Gas in Formalin-Treated Fish
S. Das, R. Kumar, J. Singh,
Published in Institute of Electrical and Electronics Engineers Inc.
2020
Volume: 67

Issue: 12
Pages: 5710 - 5716
Abstract
Here, an ultrafast direct laser patterning technique to fabricate a low-cost microsensor and its application for formaldehyde detection are reported. The patterns of microheater and interdigitated electrodes (IDEs) were realized using laser micromachining techniques by ablation of gold thin film on alumina substrate. The thin film of gold microheater showed good stability up to 300 °C with a fast response time of 80 s and temperature coefficient of resistance (TCR) was calculated as $1.37\times 10^{-{3}}/^{\circ }\text{C}$. Moreover, gold microheater exhibited long-term reliability under self-heating mode with a negligible resistance drift < 0.5% over a period of 330 h at 250 °C through consuming low power with a heating efficiency of 0.23 °C/mW. Thermal imaging camera revealed the uniform temperature distribution with negligible heat gradient profile over the whole microsensor platform. To state-of-the-art gas sensing application of this coplanar sensing platform, a nanostructured SnO2 was deposited on IDE, which exhibited high sensitivity (13.96% ppm $^{-{1}}$ ) to formaldehyde even to detect sub-ppm concentrations with fast response (32 s) and recovery kinetics (72 s). Moreover, the microsensor was also used on-site rapid screening for the detection and quantification of formaldehyde concentration in formalin-treated fish sample. © 1963-2012 IEEE.


text:
# Definition of Continuity

The definition of continuity I am using is the following:

Let $f$ be a real function, $a\in D(f)$. If for any sequence $\{x_n\}$ in $D(f)$ converging to $a$: $$\lim f(x_n)=f(a)$$ then $f$ is continuous at the point $a$.

Now there's one detail that bothers me. The text I am using (lecture notes) then briefly mentions:

Notice that we can only consider monotonic sequences

My question is simple: why only monotonic?

Thanks.

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I suppose the text is implicitly asserting that the definition holds if and only if the restricted definition for all monotone sequences holds. Using this restricted definition probably simplifies some argument later on. So all you have to do is prove that the two definitions are equivalent. –  user108903 Jan 15 '13 at 8:35

Assuming the domain of the function is a subset of $\mathbb R$ (otherwise it is not clear what monotonous should mean) I think the remark should be "Notice that it suffices to consider only monotone sequences". Clearly, if the condition holds for any sequence that it holds for a monotone one and so the the more general condition implies the one for monotone sequences. On the other hand, any sequence in $\mathbb R$ contains a monotone subsequence and you can use this to show that the more restrictive condition implies the more general one. Thus they are equivalent.
Because you can always take a monotone sequence from an arbitrary sequence. So if for some sequence ${x_n}$ in D(f) converging to a : $limf(x_n )$ not converges to f(a), then there is a subsequence $y_n$ in $x_n$, such that $|f(y_n)-f(a)|$>r>0, then take a monotone subsequence of ${y_n}$ you get that there is a monotone sequence such that the condition is not satisfied, this shows that you just need to consider the monotone sequence.


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# Compute Time Complexity of Neural network, SVM and other classification algorithms

I would like to know what is the asymptotic time complexity analysis for general models of Back-propagation Neural Network, SVM and Maximum Entropy. Does it just depend on number of features included and training time complexity is the only stuff that really matters.

And does it real matter when applying on large chunk of text classification like twitter data or blog data

• SVMs contain an underlying optimization step that is solved heuristically, so for any actual algorithm that purports to solve SVMs, the answer is undefined. A number like $O(n^3)$ is generally bandied around for implementations like libsvm, which means something like time/iteration * #iterations (where #iterations is assumed to be constant) – Suresh Venkat Aug 25 '13 at 17:50
• @SureshVenkat: That should be an answer, not a comment. – Jeffε Aug 31 '13 at 18:14
• @JɛﬀE it is done. – Suresh Venkat Sep 1 '13 at 5:17

SVMs contain an underlying optimization step that is solved heuristically, so for any actual algorithm that purports to solve SVMs, the answer is undefined. A number like $O(n^3)$ is generally bandied around for implementations like libsvm, which means something like time/iteration * #iterations (where #iterations is assumed to be constant)
Relevance Vector Machines need a $O(n^3)$ time too, but the main overhead there is in the learning phase, it needs to repeatedly compute the inverse of a Hessian Matrix (that requires $O(n^3)$ computations.
However, a preprocessing step is often useful and it also costs you $O(nd)$, specially if you want faster classification speeds. This is a good paper to read if you want to know more.


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# Math Help - proof involving lipschitz and uniformly continuity. is it correct?

1. ## proof involving lipschitz and uniformly continuity. is it correct?

Let f: D-R satisfy the Lipschitz condition: there exists an L such that for all x,y $\in$D |f(x)-f(y)|<=L|x-y|

Prove that f is uniformly continuous on D

Here is what I have, is this correct?

Let $\epsilon$>0. Choose $\delta$= $\epsilon$/L. Then by the definition of uniform continuity, if |x-y|< $\delta$= $\epsilon$/L, then |f(x)-f(y)|<L $\epsilon$/L= $\epsilon$.

2. Your logic is correct but your writing is misleading. When you say "by the definition of uniform continuity" you are signalling that you're about to use uniform continuity as a hypothesis. But instead you use Lipschitz continuity---which is what you should do!

A proper write-up would look something like this...

Let $f\to R" alt="f\to R" /> satisfy a Lipschitz condition with constant $L$, and let $\epsilon>0$. Then there is $\delta:=\epsilon/L>0$ such that for any $x,y\in D$, if $|x-y|<\delta$ then $|f(x)-f(y)|. Hence $f$ is uniformly continuous.


text:
Translate

Translate to:

Chaos Theory…

Predictability: Does the Flap of a Butterfly’s Wings in Brazil set off a Tornado in Texas?
The title of a paper given by Edward Lorenz in 1972

How many of have misquoted that title I wonder?

The Drake Equation

The Drake equation states that:

$N = R^* \times f_p \times n_e \times f_l \times f_i \times f_c \times L$

Do you know what it is?

The Drake Equation is  a means by which to establish how many intelligent, communicating civilizations might be likely in our galaxy.

Try calculating The Drake Equation for yourself

About The Drake Equation


text:
# Construction Steps

## Step1-Background Color

Change background color as blue.

## Step2-Background Image

Paste it to game. Edit the size and position of image by pulling from points A and B.

## Step3- Table of Information

• Click on and write
Weather Condition Snowy Rainy Sunny
• Add second text and write
Percentage of Water Vapor in the Air 60%-100% 35%-60% 0%-35%

## Step4

Click on. And write; Days, Water Vapor, Weather Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday

## Step5-Input Variables

• Construct table
Click on the and design a table which contains texts that we constructed in previous step.
• Input different variables for water vapor column
By using input tool at the left-bottom corner, create two different variables for each day. Write code ı= RandomBetween(0,5) for numerator of the ratios and i= RandomBetween(5,10) for denominator of the ratios.

## Step6

• Create texts for Water Vapor column
Click right on the and go settings. On settings part, click on text button. Write codes \frac{ı}{i} to the space for every text. To write ı and i, click on the button of geogebra symbol and choose different variables.
• Create a button to change numerator and denumerator of the ratios.
Click on the button and name as "Next Week".
• Click right on the button, go settings.
Write its's Scripting that ı= RandomBetween(0,5) and i=RandomBetween(5,10) for each variable. azö=0 azöö=0

## Step7

• Create polygons in weather columns.
Click on and create 4-vertice polygons fitting in the each box of weather columns. Click right on the polygons and go settings. Click on basic button and click on show object.
Paste pictures to the game and edit their sizes.

## Step8

• Download pictures of sunny cloud, rainy cloud, snowy cloud and lightning on the top of the screen.
• Click on input bar and write ha= IsInRegion(R_2,d1), hb=IsInRegion(P_2,d1), hc=IsInRegion(N_2,d1) for each polygon.

## Step9

• Write input bar f=round((ı)/(i)) for each variable.
• Write ht ≠ 0 ∧ 0.6 ≤ ggc ≤ 1 under "Condition To Show Object".
• Write input bar azö as variable. Add 3  name as "Next Week", "Try Again", "Calculate Your Score".
• Click right on "Try Again" and write its set condition to Show part 0<azö<5.
• Click right on "Next Week" and write its set condition to Show part azö>5.
• Click right on "Calculate Your Score" button, go scripting and
• write If(ha ≠ 0 ∧ 0.35 < f < 0.6 ∨ hd ≠ 0 ∧ 0.35 < gg < 0.6 ∨ hn ≠ 0 ∧ 0.35 < ggb < 0.6 ∨ hr ≠ 0 ∧ 0.35 < ggc < 0.6 ∨ hz ≠ 0 ∧ 0.35 < gge < 0.6 ∨ hu ≠ 0 ∧ 0.35 < ggd < 0.6 ∨ hk ≠ 0 ∧ 0.35 < gga < 0.6,SetValue(azö,azö+1))
Click right and go settings/text. Write azö.


text:
# $P(X)$ is locally compact if $X$ is?

Let us assume, as in here Measurable structure on the space of probability measures that $X$ is a locally compact Polish space. Then can the same thing be said of $P(X)$, its probability measures with weak * convergence? I am only missing local compactness and completeness relative to one metrization now, as I realized that appeal to the 1 point compactification of X helps. (This process gives a bigger space that is actually a metric space.)

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@EnjoysMath I'm studying from Aldous' notes on exchangeability in preparation to work on one of his research problems. My advisor and him have different ways of doing things (See the question I linked for how it differs but how I hope they are equivalent.) – Jeff Sep 27 '13 at 5:39

That $P(X)$ is Polish is true and can be found in many books; see e.g. Chapter 17 of Kechris' Classical descriptive set theory. One way to prove it is to observe that, denoting by $\widehat X=X\cup\{\infty\}$ the one-point compactification of $X$, the space $P(X)$ can be identified with $G=\{ \mu\in P(\widehat X);\; \mu(\{\infty\})=0\}$, which is $G_\delta$ in the compact metrizable space $P(\widehat X)$, and hence Polish.

On the other hand, $P(X)$ is not locally compact unless $X$ is compact.

For simplicity, consider $X=[0,\infty)$. Take any neighbourhood $\mathcal U$ of $\delta_0$ in $P(X)$. Then you will have no problem to show that one can find $\varepsilon >0$ such that $(1-\varepsilon)\delta_0+ \varepsilon \delta_a\in\mathcal U$ for every $a\in\mathbb R$.

Now take a sequence $(a_n)$ tending to $\infty$ and put $\mu_n:=(1-\varepsilon)\delta_0+\varepsilon\delta_{a_n}$. Then $(\mu_n)$ is a sequence in $\mathcal U$ which has no subsequence converging in $P(X)$. (The sequence $(\mu_n)$ is of course convergent in $P([0,\infty])$, but the limit $(1-\varepsilon)\delta_0+\varepsilon\delta_\infty$ is not in $P(X)$). So $\mathcal U$ is certainly not compact.

Exactly the same argument gives the result for a general locally compact and non-compact $X$.

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Local compactness is required before one can talk about iterating this procedure, because the Riesz representation theorem is used. I suppose then you are saying that I misunderstood my advisor because it is simply not possible to iterate the procedure unless $X$ is compact. Thanks. – Jeff Sep 28 '13 at 7:41
By "iterate the procedure" you mean consider $P(X)$, $P(P(X))$ and so on? This you can still do with $X$ Polish. – Etienne Sep 28 '13 at 10:11
how can one consider $PP(X)$ when $X$ is only Polish? There is no Riesz correspondence of measures with functionals. I suppose the map of measures into functionals is still injective, but just not onto. So then you inherit the topology this way I guess. – Jeff Sep 28 '13 at 15:33
Wel, $Z=P(X)$ is Polish. $P(Z)$ is perfectly well-defined (Borel probability measures on $Z$), and the topology as well (it is the topology on $P(Z)$ generated by the evaluation maps $\mu\mapsto \int fd\mu$, where $f$ ranges over the bounded continuous functions on $Z$; usually, this topology is called the Prokhorov topology). Then $P(Z)$ is Polish whenever $Z$ is, so you can indeed iterate. – Etienne Sep 28 '13 at 17:36


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# Continuous strictly increasing function with derivative infinity at a measure 0 set

Let $E\subset [0,1]$ with $\mu(E)=0$. Does there exist a continuous, strictly increasing function $f$ on $[0,1]$ so that $f'(x)=\infty$ for all $x\in E$ (in Lebesgue sense)?

I think there exist such a function, but I don't know how to construct.

• Is E given and you want to construct f or do you just need an example of such an E and f? Do you expect something from f? Differentiable everywhere away feom E ...? – YannickSSE May 10 '16 at 15:10
• @user3808066 E is given and I need to construct an f for it. f is differentiable at every points since it is a strictly increasing function. – Peter Liu May 10 '16 at 15:12
• Strictly increasing only implies allmost everywhere differentiability. You can have edges. – YannickSSE May 10 '16 at 15:14
• And what does the in "Lebesque sense" mean? It is allready a condition on a zero set. – YannickSSE May 10 '16 at 15:16
• See my 18 April 2007 sci.math post (follow-up comment here) and the stackexchange question Absolute continuous function with given set of discontinuities of derivatives.. – Dave L. Renfro May 12 '16 at 14:39

I couldn't figure how the general case works but if $E$ is countable consider $E=\{q_n\;:\;n\in\mathbb{N}\}$ and take $f$ a function which is $0$ on $(-\infty,-1/2]$, $1$ on $[1/2,\infty)$, monotonic and has a derivative $f':\mathbb{R} \to [0,\infty]$ with $f'(0)=\infty$, additionally take a sequence $(a_n)_{n\in\mathbb{N}}\in \ell^1$, $a_n>0$. Then the function $$g(x) = \sum_{n\in\mathbb{N}} a_nf(x-q_n)$$ is continuous and monotone. Further since for all $n$ $a_nf'(x-q_n)$ is a non negative function $$g(x) = \int_{-1}^x \sum_n a_n f'(s-q_n)\mathrm{d}s$$ and hence at every $q_m$ $$g'(x) = f'(0) + \sum_{n\not= m} f'(q_m-q_n) =\infty.$$