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## Alternating iteration of morphisms and the Kolakovski sequence.(English)Zbl 0766.68073

Lindenmayer systems. Impacts on theoretical computer science, computer graphics, and developmental biology, 93-106 (1992).
Summary: [For the entire collection see Zbl 0755.00005.]
Alternating iteration of morphisms is a natural extension of iteration of morphisms to define infinite words. We study properties of two infinite words thus obtained, one being the famous Kolakovski sequence.

### MSC:

 68Q45 Formal languages and automata

Zbl 0755.00005


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• A joint measurement is presented of the branching fractions $B^0_s\to\mu^+\mu^-$ and $B^0\to\mu^+\mu^-$ in proton-proton collisions at the LHC by the CMS and LHCb experiments. The data samples were collected in 2011 at a centre-of-mass energy of 7 TeV, and in 2012 at 8 TeV. The combined analysis produces the first observation of the $B^0_s\to\mu^+\mu^-$ decay, with a statistical significance exceeding six standard deviations, and the best measurement of its branching fraction so far. Furthermore, evidence for the $B^0\to\mu^+\mu^-$ decay is obtained with a statistical significance of three standard deviations. The branching fraction measurements are statistically compatible with SM predictions and impose stringent constraints on several theories beyond the SM.
• ### photon+jet process application for setting the absolute scale of jet energy and determining the gluon distribution at the LHC(hep-ex/0207034)

July 15, 2002 hep-ex
The possibility of jet energy scale setting at the CMS calorimeter by using photon+jet process is studied. The estimation of the number of photon+jet events suitable for determination of gluon distribution inside a proton in a new kinematic region of x, Q^2 variables beyond the one covered by HERA data is also presented.

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# Signed-digit representation

In mathematical notation for numbers, signed-digit representation is a positional system with signed digits.

Signed-digit representation can be used to accomplish fast addition of integers because it can eliminate chains of dependent carries.[1] In the binary numeral system, a special case signed-digit representation is the non-adjacent form, which can offer speed benefits with minimal space overhead.

## Notation and use

Denoting the positional notation number base as ${\displaystyle b}$, every integer ${\displaystyle a}$ can be written uniquely as

${\displaystyle a=\sum _{i=0}^{n}d_{i}(b)^{i}}$

where each digit ${\displaystyle d_{k}}$ is an integer such that ${\displaystyle -\alpha \leq d_{k}\leq \beta }$. given that ${\displaystyle -\alpha \leq 0}$ is the minimum digit value and ${\displaystyle \beta \geq 0}$ is the maximum digit value, where the leading digit ${\displaystyle d_{n}>0}$ unless ${\displaystyle n=0}$, and where the base ${\displaystyle b=\alpha +\beta +1}$. Negative-integer digits are typically represented with a bar over the digit, i.e. ${\displaystyle -\alpha ={\bar {\alpha }}}$.

### Balanced form

Balanced form representations are representations where ${\displaystyle \alpha =\beta }$, or equivalently, where there are an equal number of negative and positive digits in the set of digits for the representation. Only odd bases can have balanced form representations, as when ${\displaystyle \alpha =\beta }$, ${\displaystyle b=\alpha +\beta +1}$ is always an odd number.

For example, consider the ternary numeral system with only three digits. In standard ternary, the digits are ${\displaystyle \lbrace 0,1,2\rbrace }$; however, the balanced form of ternary, balanced ternary, uses digits ${\displaystyle \lbrace {\bar {1}},0,1\rbrace }$. This convention is adopted in finite fields of odd prime order ${\displaystyle q}$:[2]

${\displaystyle GF(q)=\lbrace 0,1,{\bar {1}}=-1,...\beta ={\frac {q-1}{2}},\ {\bar {\beta }}={\frac {1-q}{2}}\ |\ q=0\rbrace .}$

In balanced form, truncation and rounding become the same operation.

### Signed-digit decimal

Signed-digits are commonly associated with the decimal number system. The following table displays the signed-digit representations of the integers ${\displaystyle -10\leq n\leq 10}$ for base ${\displaystyle b=10=\alpha +\beta +1}$.

Decimal for minimum digit value ${\displaystyle 0\leq \alpha
${\displaystyle \alpha =0}$ (Unsigned decimal)${\displaystyle \alpha =1}$${\displaystyle \alpha =2}$${\displaystyle \alpha =3}$${\displaystyle \alpha =4}$${\displaystyle \alpha =5}$${\displaystyle \alpha =6}$${\displaystyle \alpha =7}$${\displaystyle \alpha =8}$${\displaystyle \alpha =9}$
${\displaystyle -10}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$${\displaystyle {\bar {1}}0}$
${\displaystyle -9}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {1}}1}$${\displaystyle {\bar {9}}}$
${\displaystyle -8}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {1}}2}$${\displaystyle {\bar {8}}}$${\displaystyle {\bar {8}}}$
${\displaystyle -7}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {1}}3}$${\displaystyle {\bar {7}}}$${\displaystyle {\bar {7}}}$${\displaystyle {\bar {7}}}$
${\displaystyle -6}$${\displaystyle {\bar {1}}4}$${\displaystyle {\bar {1}}4}$${\displaystyle {\bar {1}}4}$${\displaystyle {\bar {1}}4}$${\displaystyle {\bar {1}}4}$${\displaystyle {\bar {6}}}$${\displaystyle {\bar {6}}}$${\displaystyle {\bar {6}}}$${\displaystyle {\bar {6}}}$
${\displaystyle -5}$${\displaystyle {\bar {1}}5}$${\displaystyle {\bar {1}}5}$${\displaystyle {\bar {1}}5}$${\displaystyle {\bar {1}}5}$${\displaystyle {\bar {5}}}$${\displaystyle {\bar {5}}}$${\displaystyle {\bar {5}}}$${\displaystyle {\bar {5}}}$${\displaystyle {\bar {5}}}$
${\displaystyle -4}$${\displaystyle {\bar {1}}6}$${\displaystyle {\bar {1}}6}$${\displaystyle {\bar {1}}6}$${\displaystyle {\bar {4}}}$${\displaystyle {\bar {4}}}$${\displaystyle {\bar {4}}}$${\displaystyle {\bar {4}}}$${\displaystyle {\bar {4}}}$${\displaystyle {\bar {4}}}$
${\displaystyle -3}$${\displaystyle {\bar {1}}7}$${\displaystyle {\bar {1}}7}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$${\displaystyle {\bar {3}}}$
${\displaystyle -2}$${\displaystyle {\bar {1}}8}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$${\displaystyle {\bar {2}}}$
${\displaystyle -1}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$${\displaystyle {\bar {1}}}$
${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$${\displaystyle 0}$
${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle 1}$${\displaystyle -{\bar {1}}}$
${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 2}$${\displaystyle 1{\bar {8}}}$${\displaystyle -{\bar {2}}}$
${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 3}$${\displaystyle 1{\bar {7}}}$${\displaystyle 1{\bar {7}}}$${\displaystyle -{\bar {3}}}$
${\displaystyle 4}$${\displaystyle 4}$${\displaystyle 4}$${\displaystyle 4}$${\displaystyle 4}$${\displaystyle 4}$${\displaystyle 1{\bar {6}}}$${\displaystyle 1{\bar {6}}}$${\displaystyle 1{\bar {6}}}$${\displaystyle -{\bar {4}}}$
${\displaystyle 5}$${\displaystyle 5}$${\displaystyle 5}$${\displaystyle 5}$${\displaystyle 5}$${\displaystyle 1{\bar {5}}}$${\displaystyle 1{\bar {5}}}$${\displaystyle 1{\bar {5}}}$${\displaystyle 1{\bar {5}}}$${\displaystyle -{\bar {5}}}$
${\displaystyle 6}$${\displaystyle 6}$${\displaystyle 6}$${\displaystyle 6}$${\displaystyle 1{\bar {4}}}$${\displaystyle 1{\bar {4}}}$${\displaystyle 1{\bar {4}}}$${\displaystyle 1{\bar {4}}}$${\displaystyle 1{\bar {4}}}$${\displaystyle -{\bar {6}}}$
${\displaystyle 7}$${\displaystyle 7}$${\displaystyle 7}$${\displaystyle 1{\bar {3}}}$${\displaystyle 1{\bar {3}}}$${\displaystyle 1{\bar {3}}}$${\displaystyle 1{\bar {3}}}$${\displaystyle 1{\bar {3}}}$${\displaystyle 1{\bar {3}}}$${\displaystyle -{\bar {7}}}$
${\displaystyle 8}$${\displaystyle 8}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle 1{\bar {2}}}$${\displaystyle -{\bar {8}}}$
${\displaystyle 9}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle 1{\bar {1}}}$${\displaystyle -{\bar {9}}}$
${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle 10}$${\displaystyle -{\bar {1}}0}$

### Table

The following table displays the set of digits for each signed-digit representations for ${\displaystyle \alpha \leq 5}$ and ${\displaystyle \beta \leq 5}$.

Set of digits for ${\displaystyle \alpha \leq 5}$ and ${\displaystyle \beta \leq 5}$
${\displaystyle \alpha ,\beta }$012345
0 ${\displaystyle \lbrace 0\rbrace }$${\displaystyle \lbrace 0,1\rbrace }$${\displaystyle \lbrace 0,1,2\rbrace }$${\displaystyle \lbrace 0,1,2,3\rbrace }$${\displaystyle \lbrace 0,1,2,3,4\rbrace }$${\displaystyle \lbrace 0,1,2,3,4,5\rbrace }$
1 ${\displaystyle \lbrace {\bar {1}},0\rbrace }$${\displaystyle \lbrace {\bar {1}},0,1\rbrace }$${\displaystyle \lbrace {\bar {1}},0,1,2\rbrace }$${\displaystyle \lbrace {\bar {1}},0,1,2,3\rbrace }$${\displaystyle \lbrace {\bar {1}},0,1,2,3,4\rbrace }$${\displaystyle \lbrace {\bar {1}},0,1,2,3,4,5\rbrace }$
2 ${\displaystyle \lbrace {\bar {2}},{\bar {1}},0\rbrace }$${\displaystyle \lbrace {\bar {2}},{\bar {1}},0,1\rbrace }$${\displaystyle \lbrace {\bar {2}},{\bar {1}},0,1,2\rbrace }$${\displaystyle \lbrace {\bar {2}},{\bar {1}},0,1,2,3\rbrace }$${\displaystyle \lbrace {\bar {2}},{\bar {1}},0,1,2,3,4\rbrace }$${\displaystyle \lbrace {\bar {2}},{\bar {1}},0,1,2,3,4,5\rbrace }$
3 ${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0\rbrace }$${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0,1\rbrace }$${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0,1,2\rbrace }$${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3\rbrace }$${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4\rbrace }$${\displaystyle \lbrace {\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4,5\rbrace }$
4 ${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0\rbrace }$${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1\rbrace }$${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2\rbrace }$${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3\rbrace }$${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4\rbrace }$${\displaystyle \lbrace {\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4,5\rbrace }$
5 ${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0\rbrace }$${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1\rbrace }$${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2\rbrace }$${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3\rbrace }$${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4\rbrace }$${\displaystyle \lbrace {\bar {5}},{\bar {4}},{\bar {3}},{\bar {2}},{\bar {1}},0,1,2,3,4,5\rbrace }$

Considered as a matrix, the table itself is a skew-symmetric matrix. The representations down the diagonal are the balanced form signed-digit representations, while the representations on either side of the diagonal are isomorphic to its counterpart on the other side of the diagonal by negation, where positive numbers on one side correspond to negative numbers on the other, and vice versa. When ${\displaystyle \alpha =0}$, the representation can only represent non-negative integers without the use of a sign, and when ${\displaystyle \beta =0}$, the representation can only represent non-positive integers without the use of a sign. When both ${\displaystyle \alpha =0}$ and ${\displaystyle \beta =0}$, only zero can be represented, yielding the trivial ring.

### Non-uniqueness of rationals

Just as with unsigned positional number systems, when the representations are extended to rational numbers, uniqueness is lost. For example, consider signed-digit decimal with ${\displaystyle \alpha =5}$

${\displaystyle {(0.4444\ldots )}_{10}={\frac {4}{9}}={(1.{\bar {5}}{\bar {5}}{\bar {5}}{\bar {5}}\ldots )}_{10}}$

Such examples can be shown to exist by considering the greatest and smallest possible representations with integral parts 0 and 1 respectively, and then noting that they are equal.

## History

Challenges in calculation stimulated early authors Colson (1726) and Cauchy (1840) to use signed-digit representation. The further step of replacing negated digits with new ones was suggested by Selling (1887) and Cajori (1928).

In 1928, Florian Cajori noted the recurring theme of signed digits, starting with Colson (1726) and Cauchy (1840).[3] In his book History of Mathematical Notations, Cajori titled the section "Negative numerals".[4] For completeness, Colson[5] uses examples and describes addition (pp 163,4), multiplication (pp 165,6) and division (pp 170,1) using a table of multiples of the divisor. He explains the convenience of approximation by truncation in multiplication. Colson also devised an instrument (Counting Table) that calculated using signed digits.

Eduard Selling[6] advocated inverting the digits 1, 2, 3, 4, and 5 to indicate the negative sign. He also suggested snie, jes, jerd, reff, and niff as names to use vocally. Most of the other early sources used a bar over a digit to indicate a negative sign for a it. Another German usage of signed-digits was described in 1902 in Klein's encyclopedia.[7]

## Other systems

There exist other signed-digit bases such that the base ${\displaystyle b\neq \alpha +\beta +1}$. A notable examples of this is Booth encoding, which has ${\displaystyle \alpha =1}$ and ${\displaystyle \beta =1}$, with digits of value ${\displaystyle \lbrace {\bar {1}},0,1\rbrace }$, but which uses a base ${\displaystyle b=2<3=\alpha +\beta +1}$. The standard binary numeral system would only use digits of value ${\displaystyle \lbrace 0,1\rbrace }$.

Note that non-standard signed-digit representations are not unique. For instance:

${\displaystyle 0111_{2}=4+2+1=7}$
${\displaystyle 10{\bar {1}}1_{2}=8-2+1=7}$
${\displaystyle 1{\bar {1}}11_{2}=8-4+2+1=7}$
${\displaystyle 100{\bar {1}}_{2}=8-1=7}$

The non-adjacent form (NAF) of Booth encoding does guarantee a unique representation for every integer value. However, this only applies for integer values. For example, consider the following repeating binary numbers in NAF,

${\displaystyle (0.101010\ldots )_{2}={\frac {2}{3}}=(1.0{\bar {1}}0{\bar {1}}0{\bar {1}}\ldots )_{2}}$

## In written and spoken language

The oral and written forms of numbers in the Punjabi language use a form of a negative numeral one written as una or un.[8] This negative one is used to form 19, 29, …, 89 from the root for 20, 30, …, 90. Explicitly, here are the numbers:

• 19 unni, 20 vih, 21 ikki
• 29 unatti, 30 tih, 31 ikatti
• 39 untali, 40 chali, 41 iktali
• 49 unanja, 50 panjah, 51 ikvanja
• 59 unahat, 60 sath, 61 ikahat
• 69 unattar, 70 sattar, 71 ikhattar
• 79 unasi, 80 assi, 81 ikiasi
• 89 unanve, 90 nabbe, 91 ikinnaven.

Similarly, the Sesotho language utilizes negative numerals to form 8's and 9's.

• 8 robeli (/Ro-bay-dee/) meaning "break two" i.e. two fingers down
• 9 robong (/Ro-bong/) meaning "break one" i.e. one finger down

In the English language it is common to refer to times as, for example, 'seven til three', 'til' performing the negation.


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# Homophone riddle 8

It feels like such a long time that I didn't post one...

 It might be something 10 feet away from you
Or it could be a verb
In movies, this would be an escape route
I chose these because of me

• Wait, homophone or homonym??? – Kevin L Oct 22 '18 at 3:19
• Homophone, if you saw my other 7 – Duck Oct 22 '18 at 3:34

Maybe a

Duct

It might be something 10 feet away from you

they're common in most buildings?

Or it could be a verb

"I Ducked" or maybe it's a ventilation duct and I "vented"

In movies, this would be an escape route

Escape through a noisy thin metal tunnel because the aliens/zombies/bad-guys are hard of hearing in Hollywood

I chose these because of me

Duck-t

• Homophone of... rot13[qhpxrq] :P – jafe Oct 22 '18 at 7:30
• Yes! I knew the "I chose these because of me" was a dead giveaway – Duck Oct 22 '18 at 14:37
• I actually got it from the movie escape route more than anything – Brent Hackers Oct 25 '18 at 7:05

Is it:

CEILING / SEALING?

It might be something 10 feet away from you

The ceiling is about 10 feet from the ground.

Or it could be a verb

"Sealing", that is to seal something is a verb, and a homophone because they are pronounced the same.

In movies, this would be an escape route

I'm sure I've seen movies where people have escaped through the ducts in the ceiling.

I chose these because of me

Well, your name is Duck. This could either because you have to "duck" for a low ceiling (or if you crawled through a ceiling space); or perhaps because of the "ducts" that run through ceilings?

• At the end: Or for the tape? – Chronocidal Oct 22 '18 at 13:56

Even though the answer is supposed to be a homophone, I found a homonym that fits all the descriptions

Water

It might be something 10 feet away from you

Well, there's a sewer right outside of my house and it has water

Or it could be a verb

Water as a verb like to "water" plants

In movies, this would be an escape route

Well, many actors jump into the water when they have to escape from a ship in a movie

I chose these because of me

Ducks, just like your beautiful name, swim in water

• nice answer, but that isn't a homophone though – Shahriar Mahmud Sajid Oct 22 '18 at 3:24
• @ShahriarMahmudSajid Yea, I know it's not. But it fits everything right? XD – Kevin L Oct 22 '18 at 3:25
• Sorry, nope! Think about homophones and movies – Duck Oct 22 '18 at 3:34
• @Duck Well, back to the beginning then :D – Kevin L Oct 22 '18 at 4:40

Could it be a

pole/poll?

It might be something 10 feet away from you

One end of it, at least, since a 10-ft. pole is the measuring device for things you want to keep away, as in the phrase "I wouldn't touch that soup with a ten foot pole."

Or it could be a verb

To take a survey is to poll a group.

In movies, this would be an escape route

Because real firefighters don't use them to get down from high places, but fictional people are often portrayed as doing so.

I chose these because of me

You are not really a duck. You are a decoy, and thus could use one of these to support yourself. (I see no evidence that you are from Poland, or I would have gone with that.)

May it be...

stairs / stares

It might be something 10 feet away from you

In a house or in most of the buildings

Or it could be a verb

It is, and it may also be rude :)

In movies, this would be an escape route

Often, when bad guys (or policemen according to the kind of movies) reach the front door

I chose these because of me

You are not a real duck, like WAF said... so I think you have a fix gaze... a stare!


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# Complex Spectral Phase Evolution (CSPE) Performance depending on signal windowing?

I am look into CSPE. "Signal Analysis Using the Complex Spectral Phase Evolution (CSPE) Method"

The method is simple. It compares the original signal's FFT and shifted signal FFT in phase domain so that it can get an estimate of frequency. The original purpose of this paper is to improve the accuracy. However, I am wondering if it can be used to detect if there is a tone around certain FFT bin.

One way to do this is to get the $$\delta$$ value for each FFT bin. If $$|\delta| < .5$$ indicates a potential tone around the frequency. One simulation is to do it on pure noise. However, I found that the results depend largely on the window you added to your signal. Here is my code:

#!/usr/bin/env python3

import numpy as np
import matplotlib.pyplot as plt
from scipy import signal

nfft = 512
nsamples = 513

noise = np.random.randn(nsamples) + 1j * np.random.randn(nsamples)
noise = np.sqrt(.5) * noise

SNR = 10
noise = noise * 10 ** (-SNR/20)

recv = noise # pure noise

s0 = recv[:nsamples-1]
s1 = recv[1:]

S0 = np.fft.fft(s0 * signal.chebwin(len(s0), at = 80), nfft)
S1 = np.fft.fft(s1 * signal.chebwin(len(s1), at = 80), nfft)

#S0 = np.fft.fft(s0, nfft) # square window
#S1 = np.fft.fft(s1, nfft) # square window

SS = np.conj(S0) * S1
aSS = np.angle(SS)
idx = np.where(aSS < 0)
aSS[idx] = aSS[idx] + 2 * np.pi
cSS = aSS * (nfft/2/np.pi)
bSS = cSS - np.arange(nfft)
print(np.sum(np.abs(bSS) < .5)) # estimation of potential # of tones

Some results:

1. If chebwin is used, I usually got 300 potential tones, which is bad.
2. If I used squared window, I usually got 20+, which is not bad
3. If I reduce chebwin's attenuation, the number also reduces.

I can't figure how this can be related to the window function.

Thanks

Yan


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• This Question is Open
1. Michele_Laino

hint: you have to shift the decimal point of eight positions to left

2. anonymous

Thank you for the hint.

3. Michele_Laino

for example, if we have this number: 753,100,000 in scientific notation it becomes: $\Large 7.531 \times {10^8}$ I wrote 10^8 since I have shifted the decimal point eight positions to left

4. anonymous

$1.082 x 10 ^8$


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# Finding the minimum value.

I'm struck on this question, I tried hard but couldn't solve it.

Question: if a quadratic equation in $$x$$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $$5a + b$$.

So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.

Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $$x.$$

Any help will be appreciated.

$$D\leq0$$ $$\Longrightarrow b^2\leq20a$$ $$\Longrightarrow 5a\geq\frac{b^2}{4}$$ $$\Longrightarrow 5a+b\geq \frac{b^2}{4}+b\geq -1$$ where equality holds if $$b = -2\ and\ a = \frac{1}{5}$$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $$x$$, hence, it must be concave upwards.
• Good answer! Just to make things clearer, that last $\frac{b^2}{4}+b\geq -1$ can be easily checked by completing the square. – s0ulr3aper07 Dec 8 '18 at 12:59


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# Quantum entanglement faster than speed of light?

recently i was watching a video on quantum computing where the narrators describes that quantum entanglement information travels faster than light!

Is it really possible for anything to move faster than light? Or are the narrators just wrong?

Regards,

-
I must unfortunately state that, at the present day, anything you read or hear in the popular media about quantum computing should be treated with deep suspicion. (I say this as someone who works in the field!) The problem is the media is absolutely full of total garbage about the subject, in part because of the existing culture surrounding popular presentations of QM (which is also largely garbage, with a few notable exceptions: e.g. Penrose, Hawking, and other such luminaries). If something said about QC sounds fantastic, then you should expect that it is close to being totally false! – Niel de Beaudrap Oct 1 '11 at 18:33
(I would like to add: models of quantum computing do have intriguing properties which surpass anything we know how to do with classical computers, and it's realistic to hope that we build them some day. However, they are not magical, nor paradoxical. Their properties are just bold extensions of the properties of classical computers, when you add one or two extra ingredients. Entanglement, for instance, is an exotic sort of correlation; but that's all that it is --- correlation of random results --- albeit one of a peculiar sort, which one could not even describe in "classical" probability.) – Niel de Beaudrap Oct 1 '11 at 18:39
@Niel: The problem with describing entanglement as probability correlation (although it is the direct quantum analog) is that correlation can be always interpreted as ignorance of hidden variables, while quantum entanglement has no local ignorance interpretation. – Ron Maimon Dec 13 '11 at 14:06
@Ron: I am not describing it as being a merely classical correlation, though. If we define "correlated" as just being "not independent", the fact that entanglement is a form of correlation immediately follows. The fact that there is no intuitive ignorance interpretation doesn't really affect this. – Niel de Beaudrap Dec 13 '11 at 20:08

Collapsing an entangled pair occurs instantaneously but can never be used to transmit information faster than light. If you have an entangled pair of particles, A and B, making a measurement on some entangled property of A will give you a random result and B will have the complementary result. The key point is that you have no control over the state of A, and once you make a measurement you lose entanglement. You can infer the state of B anywhere in the universe by noting that it must be complementary to A.

The no-cloning theorem stops you from employing any sneaky tricks like making a bunch of copies of B and checking if they all have the same state or a mix of states, which would otherwise allow you to send information faster than light by choosing to collapse the entangled state or not.

On a personal note, it irks me when works of sci-fi invoke quantum entanglement for superluminal communication (incorrectly) and then ignore the potential consequences of implied causality violation...

-
so, it means whatever you do with quantum entanglement, the information flow is always bound to $c$. Change of state of spin(for example) does travel at speed of light to affect at the other end. – Vineet Menon Oct 2 '11 at 5:18
You cannot change the spin on one particle and get a corresponding change on the second particle. Niel de Beaudrap sums it up very well as a 'correlation of random results'. Once you make a measurement (i.e. interact meaningfully with one of the entangled particles) the entanglement is collapsed. – Richard Terrett Oct 2 '11 at 5:42
so, is it that one cannot use entanglement for any use? since a single measurement destroys the entanglement itself! – Vineet Menon Oct 2 '11 at 6:50
Sure, you can use entanglement. You can use it for the same things that you can use correlated random results for: for example, you can use it to turn insecure public communication between distant parties, into secure private communication. And a number of other intriguing theoretical applications. Just not for instantaneous communication, or anything similar to it. – Niel de Beaudrap Oct 2 '11 at 9:24
@KonradRudolph - Both appear to be correct: the no-cloning theorem forbids superluminal communication through cloning of states. However this is 'sufficient but not necessary', as the article states, as the theorem does not say anything about other possible techniques that don't employ cloning of states. Perhaps I should have said 'any sneaky tricks (that employ cloning of states)' to be more clear. – Richard Terrett Feb 27 '12 at 3:05

Even though I'm not a physicist, I don't agree with "deepthought". A change in the shadow is a change in the propagation of light, and does carry information, it is not possible for the change of light propagation to travel faster than light, it travels at light speed.

Back to the OP, from my understating of quantum entanglement, it is not transfer of energy, you have no control whatsoever on the collapse of the wave function so u can't transfer information either. it is akin to the following analogy: Suppose you and your friend have two boxes, each has an apple, one green and one red and you both know that for sure. Each takes a box and travels to a distant location, the moments one of you opens his box knows what is in the other box instantaneously. Although there is a slight difference between the two scenarios since the interpretation of quantum mechanics says that the wave function collapses only at the experiment, while here the two apples had a specified state prior to the experiment. What happens at a deeper level and how two distant related events happen instantaneously I know nothing about.

-
The shadow can move faster than light speed; but the shadow is in no way controlled by the objects that it falls upon, but by the fly which is far away. The shadow carries information about the fly, which is propagating at light speed from the fly, not about the objects it falls upon. So the shadow can sweep across distances faster than light, but the shadow is not an information-bearing signal across those distances. – Niel de Beaudrap Oct 1 '11 at 18:52
-1: shadows can move faster than light, it is obvious. The shadow is an object which is inferred from the way light hits a surface, it is not a physical information carrier. – Ron Maimon Dec 13 '11 at 14:05
Right. A shadow is nothing more than "a point". I can state "point X is here!" and then "point X is now here!" and it can "over" faster than light. For example, the "point" on a pair of scissors (where the blades cross) actually "moves" incredibly quickly - but nothing is moving, you're talking about "nothing" .. it's just "a point" we're describing. – Joe Blow Mar 15 '15 at 4:11

I am not sure I agree with the statement that this is not possible. Here is my scenario. We have two space explorers who both set off from Earth at the same time. Their destination is in opposite directions. The travel at the same speed. When they set-off from earth, at that time, a beam of entangled electrons are generated. One particle is transmitted to destination 1, its entangled particle is transmitted to destination 2. This transmission is completed at sub-light speed.

When our two explorers reach their destination and find a beam of electrons constantly arriving. They then try and communicate at superluminal speed. Remember, they have left earth and travelled at the same speed. The atomic clocks they carry will need to be in line. The rules for communication are: -

Traveller A always starts communication. Communication starts at a fixed time. Communication ends at a fixed time. After communication has ended Traveller B starts at a fixed time.

Clearly time needs to be a constant between the two travellers. Difficult, but not impossible to maintain. Any fluctuation in gravity would need to adjust the travellers clock. Anyway, the communication could be done a slow enough speed to enable some time fluctuations.

Traveller A examines an electron. He detects its Spin. This collapses the entanglement and the Traveller B will "see" the spin of the entangled electron (This article indicates that we are able to examine the spin of an individual electron: http://www.sciencedaily.com/releases/2004/08/040811080351.htm). What the spin is, is irrelevant. It is the gaps in the collapse of the entanglement that will provide the communication.

A simple Morse code effect could then be generated (based on the fixed clocks of the travellers) to enable faster than light communication.

My only grey area is can Traveller B have some apparatus that only reacts when the entanglement is collapsed - e.g. some form of magnetic device that when traveller A collapses the entanglement the other particle then moves to a particular part of the magnetic detector??

I AM NOT A PHYSICIST!!

-
This will not work because "some apparatus that only reacts when the entanglement is collapsed" fundamentally cannot exist. There is no way to tell whether a particle is entangled with some other particle or not. – Nathaniel Aug 11 '15 at 8:22
We already know they are entangled. That was set when they left earth. What I want to do is to detect when the entanglement has been collapsed. Could this not create a magnetic response? – Matt Luckham Aug 11 '15 at 8:31
When the entanglement collapses, the particles are no longer entangled, and measuring whether the collapse has happened would be the same as measuring whether they're still entangled. There is no measurable change of any kind when the other particle is measured, magnetic or otherwise. Entanglement, or the collapse thereof, cannot be detected or measured in any way, except by comparing the results of the two measurements - which you can't do without sending a signal between the two points. – Nathaniel Aug 11 '15 at 8:35

There are many things that happen faster than speed of light. For example when big bang happened at the beginning of universe, the expansion of Universe is faster than speed of light. If you have studied Bell's theorem, it states and proved by experimentation that nature itself is fundamentally nonlocal. Nonlocality is in the form of instantaneous collapse of wave function. Another example is, if a bug flies across the beam of a movie projector , the speed of its shadow is proportional to the distance to the screen: in principle that distance can be as large as you like and hence the shadow can move arbitrarily at a high velocity. Note: The shadow of the bug moves across the screen at a velocity greater than c, provided the screen is far far enough away. Its true. However the shadow does not carry any energy or transmit any message. Another example is ethereal influences in EPR Experiment. Likewise there are many examples but the important point nothing carries energy or a message from point A to point B.

-
I don't think this really clarifies anything. I appreciate that you are saying that nothing "physical" is transmitted at light-speed or faster (which is true). But you undermine yourself by talking about "ethereal influences" in the EPR experiment. Which is it --- is there an actual 'influence, or not? – Niel de Beaudrap Oct 1 '11 at 18:48
There are ethereal influences. That's how the conservation of angular momentum is done. Yes there is an actual influence. – deepthought Oct 1 '11 at 19:36
The usual interpretation of Bell's inequality violations is that the universe is non-real (i.e. no hidden variables), not non-local. – user2963 Oct 1 '11 at 20:49

## protected by Community♦Oct 3 '12 at 7:09

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text:
Chapter 5 - Section 5.2 - Adding and Subtracting Decimals - Exercise Set - Page 348: 38

$35.8$

Work Step by Step

Subtract 9.2 from 45 can be changed into the equation: $45-9.2=x$, solving for $x$, we get $x=35.8$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Graphing can fool you 2

Algebra Level 5

$\large \left \lceil x^2 \right \rceil = \lfloor 2|x| \rfloor$

There are $$m$$ integral values and $$n$$ ranges of values of $$x$$ that satisfy the equation above. What is $$m+n$$?

Bonus: Find all these $$m$$ integers and all the intervals of each of the $$n$$ ranges of $$x$$.

×


text:
Linear algebra basis question

1. Jan 8, 2014

jimmycricket

1. The problem statement, all variables and given/known data
Let $v_1,...,v_k$ be vectors in a vector space $V$. If $v_1,...,v_k$ span $V$ and after removing any of the vectors the remaining $k-1$ vectors do not span $V$ then $v_1,...,v_k$ is a basis of $V$?

2. Relevant equations

3. The attempt at a solution
If $v_1,...,v_k$ span $V$ but $v_1,...,v_{k-1}$ do not then $v_1,...,v_k$ are linearly independent.
If $v_1,...,v_k$ span $V$ and are linearly independent the $v_1,...,v_k$ is a basis of $V$
Is this reasoning correct?

2. Jan 8, 2014

HallsofIvy

Staff Emeritus
Yes, your reasoning is correct. If any subset of this set of vectors does not span the vector space, then the original set is independent.

3. Jan 8, 2014

1MileCrash

If I were writing a proof I would want to emphasize that $v_{k}$ is any arbitrary vector of the set and not a named one.

Personally I would say:
{${v_{1}, v_{2}, ... v_{k}}$} \ {${v_{i}}$} is linearly dependent for all i in {1,2,..,k}.

But I'm just being nitpicky.

4. Jan 8, 2014

Office_Shredder

Staff Emeritus
This isn't a mathematical point but given the level of the exercise I would guess you are expected to prove this part (but obviously you are the only one who can know what level of detail is required in your homework)


text:
# Variational quantum amplitude estimation

Kirill Plekhanov, Matthias Rosenkranz, Mattia Fiorentini, Michael Lubasch

## Main idea

• This work proposes a novel variational quantum amplitude estimation (VQAE) algorithm with the potential to deliver a quantum advantage on near-term quantum computers.
• The algorithm combines the ideas of the original quantum amplitude estimation (QAE) algorithm and the variational quantum algorithmic method.

## Background

• Monte Carlo (MC) is a technique commonly used to calculate expectation values of real-valued functions with some probability distributions. QAE provides a quadratic quantum speedup over MC method by using Grover’s quantum amplification technique.
• Variational quantum algorithms use parametrized quantum circuits trained via some classical optimizer. This approach allows to overcome constrains of currently available NISQ devices, such as a restricted qubit number and a limited maximum circuit depth.
• VQAE is based on the maximum maximum likelihood version of the QAE algorithm considered in the work (Y. Suzuki et al., Quantum Inf. Process. 19, 75 (2020)). Furthermore, VQAE simplifies a step which relies on repeatedly applying a Grover operator.
• VQAE uses the Grover operator only a fixed number of times and performs a variational optimization once the circuit depth treshold is reached.

## Results

• Shallow variational quantum circuits can accurately approximate many amplitude amplification steps of the QAE algorithm.
• Variational approximation cost represents the main source of the quantum advantage loss.
• By using an adaptive rescaling of the amplitude, variational approximation cost can be significantly reduced, making it possible for the VQAE to outperform classical MC sampling.

tags: variational-quantum-algorithms


text:
## Simeon Reich, Alexander J. Zaslavski, A porosity result regarding fixed points for a class of nonexpansive mappings

Full Text: PDF
DOI: 10.23952/asvao.4.2022.3.10
Volume 4, Issue 3, 1 December 2022, Pages 375-380

Abstract. In one of our recent papers, we considered a complete metric space of nonexpansive mappings taking a bounded and closed subset of a complete hyperbolic space into the space so that the boundary of this subset is mapped back into the subset itself. Using the Baire category approach, we proved that most of these mappings possess a unique fixed point which attracts all their iterates. In the present paper, we improve upon this result by showing that the complement of the set of mappings which have a fixed point is not only of the first Baire category, but also is $\sigma$-porous.


text:
# Shell sort

Shell sort is a generalisation of insertion sort. It sorts all $h$th elements in a list as if they were $h$ interleaved lists. Shell sort begins with a large value of $h$ which is then reduced in a sequence.

## Gap sequences

These gap sequences are difficult to pick and to analyse. Each sequence that contains 1, does a full insertion sort and is guaranteed a correct sort. One such sequence is Marcin Ciura's $1, 4, 10, 23, 57, 132, 301, 701$

Ciura, Marcin (2001). "Best Increments for the Average Case of Shellsort". In Freiwalds, Rusins. Proceedings of the 13th International Symposium on Fundamentals of Computation Theory. London: Springer-Verlag. pp. 106–117. ISBN 3-540-42487-3.

For this example we use the sequence $1, 3, 5$


text:
Wednesday, July 22, 2015

## Introduction

In the previous two articles, we've talked about linear and logistic regression, covering by it most linear models. And that would be it, if only these models could process data with numerous features, that is x1, x2... xn-1, xn. Let's us see how our classification model deals with Olivetti faces classification dataset:

from sklearn import datasets, metrics, linear_model, cross_validation
import matplotlib.pyplot as plt
import time

faces = datasets.fetch_olivetti_faces()

start = time.time()

X = faces.data
Y = faces.target

x_train, x_test, y_train, y_test = cross_validation.train_test_split(X, Y)

# Create a classifier: a support vector classifier
model = linear_model.LogisticRegression(C=10000)

# We learn the faces of train set
model.fit(x_train, y_train)

# Now predict the person on test set
expected = y_test
predicted = model.predict(x_test)

end = time.time()
print("%.2f seconds" % (end - start)) # 8.85 seconds

# Reshape data into images
x_train_images = x_train.reshape(len(x_train), 64, 64)
x_test_images = x_test.reshape(len(x_test), 64, 64)

# Plot train images
for index, (image, label) in enumerate(
zip(x_train_images, y_train)[:4]):
plt.subplot(2, 4, index + 1)
plt.axis('off')
plt.imshow(image, cmap=plt.cm.gray)
plt.title('Train: %i' % label)

# Plot test images
for index, (image, prediction, expectation) in enumerate(
zip(x_test_images, predicted, expected)[:4]):
plt.subplot(2, 4, index + 5)
plt.axis('off')
plt.imshow(image, cmap=plt.cm.gray)
plt.title('Predict: %i/%i' % (prediction, expectation))

The dataset contains 64x64 different images of 40 distinct subjects, resulting each item of the dataset containing 4096 features. Despite the good performance of our classifier, we won't dive into measures now, it took almost 9 seconds to process this toy dataset, which is not good enough if we want to learn really big datasets.

The reason for this delay, is the way the classifier solves the optimization problem. Our models analytically solved the values of the model's parameters, which minimize the cost function with the following equation:

X is the matrix of the values of the explanatory variables for each training example. The dot product inside the brackets results a square matrix with dimensions nxn, where n is the number of features. The computational complexity of inverting this square matrix is nearly cubic in the number of explanatory variables. Thus the more features we have, the more complex the computation will be, consequently taking more time.

Given a function defined by a set of parameters, gradient descent iteratively moves toward a set of parameter values, which minimize the function. This iterative minimization is achieved by taking steps in the negative direction of the function gradient. As though a blindfolded man, who is trying to find his way from somewhere on a mountainside to the lowest point of the valley. He takes a step in the direction with the steepest decline and the sizes of his steps are proportional to the steepness of the terrain. If he were to take large steps, he may have accidentally stepped over the valley's lowest point. In opposite case, while he would reach the desired point, however since the steps are too small, it would take him a considerable amount of time. The proportion is specified by the learning rate provided to the model. When used to minimize the function, a standard (or "batch") gradient descent method would perform the following iterations until it converges:

Standard gradient descent evaluates the sum-gradient, which may require expensive evaluations of the gradients from all sums and functions. When the training set is enormous and no simple formulas exist, evaluating the sums of gradients becomes very expensive, because evaluating the gradient requires evaluating all the summand functions' gradients. In stochastic (or "on-line") gradient descent, the true gradient of Q(w) is approximated by a gradient at a single example:

We compute an estimate or approximation to this direction. The most simple way is to just look at one training example (or subset of training examples) and compute the direction to move only on this approximation. It is called as Stochastic because the approximate direction that is computed at every step can be thought of a random variable of a stochastic process. Stochastic method converges much faster compared to standard, but the error function is not as well minimized as in the case of the latter. Often in most cases, the close approximation that you get in stochastic method for the parameter values are enough because they reach the optimal values and keep oscillating there.

A compromise between the two forms called "mini-batches" computes the gradient against more than one training examples at each step. This can perform significantly better than true stochastic gradient descent, because the code can make use of vectorization libraries rather than computing each step separately. It may also result in smoother convergence, as the gradient computed at each step uses more training examples.

## Implementation

Scikit-learn provides us with both linear and logistic regression implementation using gradient descent through SGDRegressor and SGDClassifier classes respectively. By default the models use stochastic method with batches of 5 examples. Since all scikit-learn models follow the same API, the only needed change to the previous example, is to change the model instantiation.

...
model = linear_model.SGDClassifier(alpha=0.00001)
...

After we change the classifier, the code finished within less then a second!

## Conclusion

The important point to remember is that gradient descent looks for local minima, which sometimes may differ from the global one. The outcome of the process is significantly dependent upon where it starts, as illustrated below. To gain more confidence in the results, one must run it several times.

Hope you've enjoyed the article and feel free to comment below. In the next article we'll look into techniques to choose the parameters provided to the models, to achieve better results.


text:
Lemma 42.3.2. Let $R$ be a Noetherian local ring. Let $x \in R$. If $M$ is a finite Cohen-Macaulay module over $R$ with $\dim (\text{Supp}(M)) = 1$ and $\dim (\text{Supp}(M/xM)) = 0$, then

$\text{length}_ R(M/xM) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}(M_{\mathfrak q_ i}).$

where $\mathfrak q_1, \ldots , \mathfrak q_ t$ are the minimal primes of the support of $M$. If $I \subset R$ is an ideal such that $x$ is a nonzerodivisor on $R/I$ and $\dim (R/I) = 1$, then

$\text{length}_ R(R/(x, I)) = \sum \nolimits _ i \text{length}_ R(R/(x, \mathfrak q_ i)) \text{length}_{R_{\mathfrak q_ i}}((R/I)_{\mathfrak q_ i})$

where $\mathfrak q_1, \ldots , \mathfrak q_ n$ are the minimal primes over $I$.

Proof. These are special cases of Lemma 42.3.1. $\square$

## Comments (1)

Comment #4980 by Kazuki Masugi on

Does this lemma hold in general $M$? (Is "Cohen-Macaulay module"-ness nessesary?)

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text:
# What’s the Forecast for Total Stock ahead of Q1 Earnings?

By

Updated

## Implied volatility in Total

In this part, we’ll look at Total’s (TOT) stock price forecast range based on its implied volatility for the ten days until its earnings. This price forecast will be based on the latest level of implied volatility in the stock. Total is expected to post its Q1 2019 earnings on April 26, 2019.

## Forecasted price range for Total stock for the ten days ending April 26

Total’s stock price range is estimated considering its implied volatility of 18.5%, assuming a normal distribution of prices (bell curve model) and standard deviation of one (which means a probability of 68.2%). Total stock price could close between \$57.4 and \$54.0 per share in the next ten calendar days ending on April 26.

## Implied volatility in peers

Implied volatility in Petrobras (PBR) also has fallen by 6.4 percentage points over January 2 to 40.4%. Also, implied volatilities in PetroChina (PTR) and YPF (YPF) have decreased by 10.5 percentage points and 11.4 percentage points, respectively, in the same period. PetroChina and YPF’s implied volatilities currently stand at 23.8% and 36.7%, respectively.

If we consider their stock prices, then Petrobras has risen by 9.7% since January 2. Also, PetroChina and YPF have surged by 5.9% and 8.7%, respectively, in the stated period.


text:
# Measure Theory: Prove Set is Measurable Question

## Homework Statement

The question is from Stein, "Analysis 2", Chapter 1, Problem 5:

Suppose E is measurable with m(E) < ∞, and E = E1 ∪ E2, E1 ∩ E2 = ∅.

Prove:

a) If m(E) = m∗(E1) + m∗(E2), then E1 and E2 are measurable.

b) In particular, if E ⊂ Q, where Q is a finite cube, then E is measurable if and only if m(Q) = m∗(E) + m∗(Q − E).

## Homework Equations

The definition of a 'measurable set' given in the book is that for any ε > 0 there exists an open set O with E ⊂ O and m∗(O − E) ≤ ε, so I'm looking for a set of implications that lead me back to this definition.

## The Attempt at a Solution

The problem seems suspiciously similar to the definition of a measurable set as one that satisfies the 'caratheodory criterion'. My attempt at a solution has been to try to show that what we are given in the problem must imply that the caratheodory criterion holds and from there show that if the caratheodory criterion holds then the set is measurable in the above sense. I'm having trouble knowing where to start filling in the details.

I also wonder though, if there is a simpler and neater way to solve the problem?

One way that I have looked to prove (a) is to use the lemma that:

If E ⊂ Rd, then m∗(E) = inf m∗(O), where the infimum is taken over all open sets O containing E.

To construct an open set O1 that encompasses E1 and such that m∗(E1) = inf m∗(O1). And likewise and open set that encompasses E2 such that m∗(E2) = inf m∗(O2).

Then if I can show that m*(O1-E1)=0 and m*(O2-E2)=0, then E1 and E2 differ from an open set by measure 0 and so by a theorem they are also measurable.

However, I'm unsure about how to do this and get to m*(O1-E1)=0 and m*(O2-E2)=0?

...Anyone out there?

Hey, still stuck on this one. Let me know if you need me to clarify anything.

Thanks!


text:
## Tight Bound for the Density of Sequences of Integers the Sum of No Two of which is a Perfect Square

### Authors: Ayman Khalfalah, Sachin Lodha and Endre Szemeredi

ABSTRACT

P. Erd\H{o}s and D. Silverman~\cite{eg-80} proposed the problem of determining the maximal density attainable by a set $S$ of positive integers having the property that no two distinct elements of $S$ sum up to a perfect square. J. P. Massias exhibited such a set consisting of all $x \equiv 1$ (mod $4$) with $x \equiv 14, 26, 30$ (mod $32$) in~\cite{m}. In~\cite{los-82}, J. C. Lagarias, A. M. Odylzko and J. B. Shearer showed that for any positive integer $n$, one cannot find more than $\frac{11}{32} n$ residue classes (mod $n$) such that the sum of any two is never congruent to a square (mod $n$), thus essentially proving that the Massias' set has the best possible density. They~\cite{los-83} also proved that the density of such a set $S$ is never more than $0.475$ when we allow general sequences.

We improve on the lower bound for general sequences, essentially proving that it is not $0.475$, but arbitrarily close to $\frac{11}{32}$, the same as that for sequences made up of only arithmetic progressions.

Paper Available at: ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/2000/2000-39.ps.gz


text:
## pH and Activity Corrections

pH Value

pH (Potential Hydrogen) is defined as the negative decimal logarithm of the hydrogen ion activity:

 (1) pH = – log {H+}

Curly brackets {..} symbolize the activity, whereas rectangular brackets [..] are reserved for concentrations. Inserting the relationship, {H+} = γ ∙ [H+], one gets

 (2) pH = – log γ – log [H+]

Formulas for the activation coefficient (log γ) are presented here. In diluted solutions we have γ = 1, and Eq.(2) simplifies to (because log 1 = 0):

 (3) pH = – log [H+] (only for diluted solutions)

Example

What is the pH of a 0.1 molar HCl solution without and with activity corrections?

HCl is a strong acid. The hydrogen ion concentration is then given by [H+] = 0.1 M. Inserting it into Eq.(3) yields an approximate pH value of

 (4) pH = – log 0.1 = 1.0 (without activity correction)

Now with activity correction. For this purpose we adopt the extended Debye-Hückel equation

 (5) $log \, \gamma_{i} = -\dfrac{Az^{2}_{i}\sqrt{I}}{1+Ba\sqrt{I}}$

with A = 0.5085 M-1/2, B = 3.281 M-1/2 nm-1 and the effective ion-size parameter a = 0.9 nm. With z = 1 and the ionic strength I = 0.1 M we obtain log γ = – 0.083. Inserting this result into Eq.(2) gives us the exact pH value:

 (6) pH   = – log γ – log [H+]   =   0.083 + 1.0   =   1.083

The pH of the 0.1 molar HCl solution is 1.083 (rather than 1.0 as often assumed). This exact pH value is also predicted by aqion – see the initial pH in the titration example.


text:
Math Help - Units of integers mod m

1. Units of integers mod m

Show that U(I_m) = {[k] in I_m : (k,m) = 1}.

I_m is integers mod m.

I know that if m is prime then the result is obvious. But do i need to break it into cases.???

2. Re: Units of integers mod m

How is the case (m prime) obvious?

Why not use set inclusion, and that (m, k) = 1 iff am + bk = 1 for some a, b...

3. Re: Units of integers mod m

Should I assume (m,k)= 1 to try to show that [k] is in I_m?

4. Re: Units of integers mod m

I think I got it. Say (k,m)= 1 then there exist a, b that ak+bm =1 I can put bm on the other side to get

ak-1=bm. Now we know that m| ak -1 and so ak is congruent to 1mod m. That proves [k] is in I_m. further more ak is congruent to 1 mod m therefore there exist a c such that kc =1 and that is the definition of a unit.

Did I miss something?

5. Re: Units of integers mod m

a couple of things:

[k] is, by definition, in I_m. what you are trying to prove is that [k] is in U(I_m). so you don't want a c such that kc = 1 (unless k =1, no such integer exists), you want a [c] such that [c][k] = [1].

now you have shown that [ak] = [1], so you're almost there. the [c] that you're looking for is clearly [a]. what you haven't done, is show if 0 < c < m such that [c] = [a], that (c,m) = 1.

6. Re: Units of integers mod m

I have this piace of a proof from Rotman advanced algebra book. I dont understand how or what it means by solve of [x].

If (a,m) = 1, then [a][x] = [1] can be solved for
[x] in Im
. Now(x,m) = 1, for rx + sm = 1 for some integer s, and so Proposition 1.13
on page 5 gives (x,m) = 1; therefore, [x] ∈ U(Im),

7. Re: Units of integers mod m

we are given that (k,m) = 1 by the definition of U(I_m).

so (by the defintion of gcd) we have a,b in Z such that ak + bm = 1.

if two integers are equal, then they remain equal mod m, so

[ak + bm] = [1]
[a][k] + [b][m] = [1]
[a][k] + [b][0] = [1]
[a][k] = [1].

now we just take the integer c, where 0 < c < m, so that [c] = [a] (in other words, reduce the integer a used in the gcd formula mod m).

to give a concrete example, suppose m = 6 and k = 5. (5,6) = 1, so [5] is in U(6). note that 5(-7) + 6(6) = 1.

so we can take a = -7. this means that [5][-7] = [1] (mod 6). but -1 isn't in the right range, so we add multiples of 6 until it is:

-7 + 2(6) = 5. so [-1] = [5], so [5] is its own inverse in U(6). and indeed, [5][5] = [25] = [1], since 25 = 1 mod 6.

well, here, we obviously have (5,6) = 1. ok, back to the general case:

we know that [a] ( = [c]) comes from ak + bm = 1, so it's trivial that (a,m) = 1 (just switch the role of a and k).

what's not so trivial, and what your proof from Rotman doesn't state explicitly, is that , in fact, (c,m) = 1.

but this isn't that much extra difficulty: [c] = [a], so c = a + tm, for some integer t.

therefore ak + bm = 1 --> (c - tm)k + bm = 1, and re-arranging slightly:

kc + (b - tk)m = 1, so since there exists integers k, b - tk such that kc + (b - tk)m = 1, (c,m) = 1.


text:
Squaring the impulse operator

$$\hat{p}^2=\hat{p}_r\hat{p}_r=(-i\hbar)^2\frac{1}{r}\frac{\partial}{\partial r}r\frac{1}{r}\frac{\partial}{\partial r}r=-\hbar^2\frac{1}{r}\frac{\partial^2}{\partial r^2}r$$

I'm trying to understand what is happening here. I understand the part $$(-i \hbar)^2=-\hbar^2$$, but not the rest. I would expect the result to be something like $$-\hbar^2 * \frac1 r^2 * \frac{d^2} {dr^2} * r^2$$.

The way it seems to me now, is that multiplying operators isn't the same as multiplying numbers? But instead it means applying the first operator to the result of the later operator? That would mean (here I leave out the constants) applying $$\frac1 r \frac{d} {dr} r$$ to $$\frac1 r \frac{d} {dr} r$$ and $$r * \frac1 r$$ is $$1$$ and then we get the result.

But then, why is $$\nabla * \nabla$$ a dot product, considering that multiplication doesn't work the simply anymore? For example, it seems like when multiplying operators, you can't switch around order anymore. So why does the dot product rule still apply? After all, $$\Delta$$ yields a number, but $$\nabla$$ is a vector and still yields a vecor right? Simply applying $$\nabla$$ twice instead of $$\nabla * \nabla = \Delta$$ would still yield a vector, right?

• For your last paragraph: you are applying $\Delta$ to a scalar $f$ and getting a scalar, right? Well, let's see what's happening here. $\nabla f$ is a vector (the gradient of a scalar field $f$). Then we apply another $\nabla$ from the left, we are getting the divergence of the vector field $\nabla f$, which is again, a scalar. – Yuriy S Sep 29 '18 at 21:50
• Will the identity $\nabla f(r)=\frac{\vec r}{r} \partial_r f(r)$ help you? – Cosmas Zachos Sep 30 '18 at 0:02

Nothing special has happened. To get from the second to last, to the last equation all you have to do is realize $$r\frac{1}{r}=1$$ and it follows immediately. In general, operator algebra is associative which means you may apply operators products of operators in any order you like as long as you preserve their sequence i.e. $$A(BC)=(AB)C$$ I see no reason to expect the result to be $$-\hbar^2 \frac{1}{r^2} \frac{\partial^2} {\partial r^2} r^2$$ as you suggest. Also remember that your result is still an operator, so you must apply $$\left(\frac{\partial^2} {\partial r^2} r^2\right)$$ to whatever is acted upon by $$\hat{p}^2$$.
As for your confusion with the Laplacian or $$\nabla^2$$, you have to recognize that this is all just shorthand. The Laplacian is defined as $$\nabla\cdot\nabla$$ so that's just what it is.
If you want $$\nabla \left(\nabla f\right)$$ to be defined then you must take some care. $$\nabla f$$ represents the gradient of a scalar function which is itself a vector. So if you try to apply $$\nabla$$ to it again, how should you apply it? The gradient is defined acting on a scalar field, but now you are trying to apply it to a vector. Normally when you apply $$\nabla$$ to a vector, you use the divergence or $$\nabla\cdot\vec{f}$$ so then we'd have $$\nabla\cdot \left(\nabla f\right)$$ or $$\left(\nabla\cdot \nabla\right) f$$ which is exactly what we've defined the Laplacian as.
• I just understood why you were thinking the result would be $$-\hbar^2 \frac{1}{r^2} \frac{\partial^2} {\partial r^2} r^2$$ You just moved the $r$'s through the derivatives. But you know this can't work from basic calculus. $$r \frac{\partial}{\partial r} r = \frac{\partial}{\partial r} r^2 = 2r?$$ Or even more extreme $$r^2\frac{\partial}{\partial r} 1 = \frac{\partial}{\partial r} r^2 = 2r?$$ You can only pull something through a derivative when it's a constant w.r.t. the derivative. – bRost03 Sep 30 '18 at 0:59


text:
## GCOEセミナー

### 2012年02月22日(水)

16:15-17:15   数理科学研究科棟(駒場) 270号室
Oleg Emanouilov 氏 (Colorado State University)
Determination of first order coefficient in semilinear elliptic equation by partial Cauchy data. (ENGLISH)
[ 講演概要 ]
In a bounded domain in $R^2$, we consider a semilinear elliptic equation $¥Delta u +qu +f(u)=0$.
Under some conditions on $f$, we show that the coefficient $q$ can be uniquely determined by the following partial data
$${¥mathcal C}_q=¥{(u,¥frac{¥partial u}{¥partial¥nu})¥vert_{\\\\tilde Gamma}¥vert - ¥Delta u +qu +f(u)=0, ¥,¥,¥, u¥vert_{¥Gamma_0}=0,¥,¥, u¥in H^1(¥Omega)¥}$$
where $¥tilde ¥Gamma$ is an arbitrary fixed open set of
$¥partial¥Omega$ and $¥Gamma_0=¥partial¥Omega¥setminus¥tilde¥Gamma$.


text:
## anonymous one year ago Which property is shown? 5+{(-3x)+4}={5+(-3x)}+4

1. anonymous

Associative Property.

2. anonymous

3. anonymous

symbols stay in the same order parentheses move that is the associative law as for addition vs multiplication, what operation you see there, $$+$$ or $$\times$$?

4. anonymous

I see both? which I'm not very good at math. So I don't really know which one it is?

5. anonymous

there are a bunch of plus signs there right?

6. anonymous

Yes!

7. anonymous

I'm extra stupid tonight lol THOSE ARE X's

8. anonymous

lol

9. anonymous

Its been a long day..

10. anonymous

Its addition right? just to be 100% sure lol


text:
# The Count of Monte Cristo

## Chapter 38

The Compact.

The first words that Albert uttered to his friend, on the following morning, contained a request that Franz would accompany him on a visit to the count; true, the young man had warmly and energetically thanked the count on the previous evening; but services such as he had rendered could never be too often acknowledged. Franz, who seemed attracted by some invisible influence towards the count, in which terror was strangely mingled, felt an extreme reluctance to permit his friend to be exposed alone to the singular fascination that this mysterious personage seemed to exercise over him, and therefore made no objection to Albert's request, but at once accompanied him to the desired spot, and, after a short delay, the count joined them in the salon. "My dear count," said Albert, advancing to meet him, "permit me to repeat the poor thanks I offered last night, and to assure you that the remembrance of all I owe to you will never be effaced from my memory; believe me, as long as I live, I shall never cease to dwell with grateful recollection on the prompt and important service you rendered me; and also to remember that to you I am indebted even for my life."

"My very good friend and excellent neighbor," replied the count, with a smile, "you really exaggerate my trifling exertions. You owe me nothing but some trifle of 20,000 francs, which you have been saved out of your travelling expenses, so that there is not much of a score between us; -- but you must really permit me to congratulate you on the ease and unconcern with which you resigned yourself to your fate, and the perfect indifference you manifested as to the turn events might take."

"Upon my word," said Albert, "I deserve no credit for what I could not help, namely, a determination to take everything as I found it, and to let those bandits see, that although men get into troublesome scrapes all over the world, there is no nation but the French that can smile even in the face of grim Death himself. All that, however, has nothing to do with my obligations to you, and I now come to ask you whether, in my own person, my family, or connections, I can in any way serve you? My father, the Comte de Morcerf, although of Spanish origin, possesses considerable influence, both at the court of France and Madrid, and I unhesitatingly place the best services of myself, and all to whom my life is dear, at your disposal."

"Monsieur de Morcerf," replied the count, "your offer, far from surprising me, is precisely what I expected from you, and I accept it in the same spirit of hearty sincerity with which it is made; -- nay, I will go still further, and say that I had previously made up my mind to ask a great favor at your hands."

"Oh, pray name it."

"I am wholly a stranger to Paris -- it is a city I have never yet seen."

"Is it possible," exclaimed Albert, "that you have reached your present age without visiting the finest capital in the world? I can scarcely credit it."

"Nevertheless, it is quite true; still, I agree with you in thinking that my present ignorance of the first city in Europe is a reproach to me in every way, and calls for immediate correction; but, in all probability, I should have performed so important, so necessary a duty, as that of making myself acquainted with the wonders and beauties of your justly celebrated capital, had I known any person who would have introduced me into the fashionable world, but unfortunately I possessed no acquaintance there, and, of necessity, was compelled to abandon the idea."

"So distinguished an individual as yourself," cried Albert, "could scarcely have required an introduction."

"You are most kind; but as regards myself, I can find no merit I possess, save that, as a millionaire, I might have become a partner in the speculations of M. Aguado and M. Rothschild; but as my motive in travelling to your capital would not have been for the pleasure of dabbling in stocks, I stayed away till some favorable chance should present itself of carrying my wish into execution. Your offer, however, smooths all difficulties, and I have only to ask you, my dear M. de Morcerf" (these words were accompanied by a most peculiar smile), "whether you undertake, upon my arrival in France, to open to me the doors of that fashionable world of which I know no more than a Huron or a native of Cochin-China?"

"Oh, that I do, and with infinite pleasure," answered Albert; "and so much the more readily as a letter received this morning from my father summons me to Paris, in consequence of a treaty of marriage (my dear Franz, do not smile, I beg of you) with a family of high standing, and connected with the very cream of Parisian society."

"Connected by marriage, you mean," said Franz, laughingly.

"Well, never mind how it is," answered Albert, "it comes to the same thing in the end. Perhaps by the time you return to Paris, I shall be quite a sober, staid father of a family! A most edifying representative I shall make of all the domestic virtues -- don't you think so? But as regards your wish to visit our fine city, my dear count, I can only say that you may command me and mine to any extent you please."

"Then it is settled," said the count, "and I give you my solemn assurance that I only waited an opportunity like the present to realize plans that I have long meditated." Franz did not doubt that these plans were the same concerning which the count had dropped a few words in the grotto of Monte Cristo, and while the Count was speaking the young man watched him closely, hoping to read something of his purpose in his face, but his countenance was inscrutable especially when, as in the present case, it was veiled in a sphinx-like smile. "But tell me now, count," exclaimed Albert, delighted at the idea of having to chaperon so distinguished a person as Monte Cristo; "tell me truly whether you are in earnest, or if this project of visiting Paris is merely one of the chimerical and uncertain air castles of which we make so many in the course of our lives, but which, like a house built on the sand, is liable to be blown over by the first puff of wind?"

"I pledge you my honor," returned the count, "that I mean to do as I have said; both inclination and positive necessity compel me to visit Paris."

"When do you propose going thither?"

"Have you made up your mind when you shall be there yourself?"

"Certainly I have; in a fortnight or three weeks' time, that is to say, as fast as I can get there!"

"Nay," said the Count; "I will give you three months ere I join you; you see I make an ample allowance for all delays and difficulties.

"And in three months' time," said Albert, "you will be at my house?"

"Shall we make a positive appointment for a particular day and hour?" inquired the count; "only let me warn you that I am proverbial for my punctilious exactitude in keeping my engagements."

"Day for day, hour for hour," said Albert; "that will suit me to a dot."

"So be it, then," replied the count, and extending his hand towards a calendar, suspended near the chimney-piece, he said, "to-day is the 21st of February;" and drawing out his watch, added, "it is exactly half-past ten o'clock. Now promise me to remember this, and expect me the 21st of May at the same hour in the forenoon."

"Capital," exclaimed Albert; "your breakfast shall be waiting."

"Where do you live?"

"No. 27, Rue du Helder."

"Have you bachelor's apartments there? I hope my coming will not put you to any inconvenience."

"I reside in my father's house, but occupy a pavilion at the farther side of the court-yard, entirely separated from the main building."

"Quite sufficient," replied the count, as, taking out his tablets, he wrote down "No. 27, Rue du Helder, 21st May, half-past ten in the morning."

"Now then," said the count, returning his tablets to his pocket, "make yourself perfectly easy; the hand of your time-piece will not be more accurate in marking the time than myself."

"Shall I see you again ere my departure?" asked Albert.

"That depends; when do you leave?"

"To-morrow evening, at five o'clock."

"In that case I must say adieu to you, as I am compelled to go to Naples, and shall not return hither before Saturday evening or Sunday morning. And you, baron," pursued the count, addressing Franz, "do you also depart to-morrow?"

"Yes."

"For France?"

"No, for Venice; I shall remain in Italy for another year or two."

"Then we shall not meet in Paris?"

"I fear I shall not have that honor."

"Well, since we must part," said the count, holding out a hand to each of the young men, "allow me to wish you both a safe and pleasant journey." It was the first time the hand of Franz had come in contact with that of the mysterious individual before him, and unconsciously he shuddered at its touch, for it felt cold and icy as that of a corpse. "Let us understand each other," said Albert; "it is agreed -- is it not? -- that you are to be at No. 27, in the Rue du Helder, on the 21st of May, at half-past ten in the morning, and your word of honor passed for your punctuality?"

"The 21st of May, at half-past ten in the morning, Rue du Helder, No. 27," replied the Count. The young men then rose, and bowing to the count, quitted the room. "What is the matter?" asked Albert of Franz, when they had returned to their own apartments; "you seem more than commonly thoughtful."

"I will confess to you, Albert," replied Franz, "the count is a very singular person, and the appointment you have made to meet him in Paris fills me with a thousand apprehensions."

"My dear fellow," exclaimed Albert, "what can there possibly be in that to excite uneasiness? Why, you must have lost your senses."

"Whether I am in my senses or not," answered Franz, "that is the way I feel."

"Listen to me, Franz," said Albert; "I am glad that the occasion has presented itself for saying this to you, for I have noticed how cold you are in your bearing towards the count, while he, on the other hand, has always been courtesy itself to us. Have you anything particular against him?"

"Possibly."

"Did you ever meet him previously to coming hither?"

"I have."

"And where?"

"Will you promise me not to repeat a single word of what I am about to tell you?"

"I promise."

"Upon my honor."

"Then listen to me." Franz then related to his friend the history of his excursion to the Island of Monte Cristo and of his finding a party of smugglers there, and the two Corsican bandits with them. He dwelt with considerable force and energy on the almost magical hospitality he had received from the count, and the magnificence of his entertainment in the grotto of the "Thousand and One Nights." He recounted, with circumstantial exactitude, all the particulars of the supper, the hashish, the statues, the dream, and how, at his awakening, there remained no proof or trace of all these events, save the small yacht, seen in the distant horizon driving under full sail toward Porto-Vecchio. Then he detailed the conversation overheard by him at the Colosseum, between the count and Vampa, in which the count had promised to obtain the release of the bandit Peppino, -- an engagement which, as our readers are aware, he most faithfully fulfilled. At last he arrived at the adventure of the preceding night, and the embarrassment in which he found himself placed by not having sufficient cash by six or seven hundred piastres to make up the sum required, and finally of his application to the count and the picturesque and satisfactory result that followed. Albert listened with the most profound attention. "Well," said he, when Franz had concluded, "what do you find to object to in all you have related? The count is fond of travelling, and, being rich, possesses a vessel of his own. Go but to Portsmouth or Southampton, and you will find the harbors crowded with the yachts belonging to such of the English as can afford the expense, and have the same liking for this amusement. Now, by way of having a resting-place during his excursions, avoiding the wretched cookery -- which has been trying its best to poison me during the last four months, while you have manfully resisted its effects for as many years, -- and obtaining a bed on which it is possible to slumber, Monte Cristo has furnished for himself a temporary abode where you first found him; but, to prevent the possibility of the Tuscan government taking a fancy to his enchanted palace, and thereby depriving him of the advantages naturally expected from so large an outlay of capital, he has wisely enough purchased the island, and taken its name. Just ask yourself, my good fellow, whether there are not many persons of our acquaintance who assume the names of lands and properties they never in their lives were masters of?"

"But," said Franz, "the Corsican bandits that were among the crew of his vessel?"

"Why, really the thing seems to me simple enough. Nobody knows better than yourself that the bandits of Corsica are not rogues or thieves, but purely and simply fugitives, driven by some sinister motive from their native town or village, and that their fellowship involves no disgrace or stigma; for my own part, I protest that, should I ever go to Corsica, my first visit, ere even I presented myself to the mayor or prefect, should be to the bandits of Colomba, if I could only manage to find them; for, on my conscience, they are a race of men I admire greatly."

"Still," persisted Franz, "I suppose you will allow that such men as Vampa and his band are regular villains, who have no other motive than plunder when they seize your person. How do you explain the influence the count evidently possessed over those ruffians?"

"My good friend, as in all probability I own my present safety to that influence, it would ill become me to search too closely into its source; therefore, instead of condemning him for his intimacy with outlaws, you must give me leave to excuse any little irregularity there may be in such a connection; not altogether for preserving my life, for my own idea was that it never was in much danger, but certainly for saving me 4,000 piastres, which, being translated, means neither more nor less than 24,000 livres of our money -- a sum at which, most assuredly, I should never have been estimated in France, proving most indisputably," added Albert with a laugh, "that no prophet is honored in his own country."

"Talking of countries," replied Franz, "of what country is the count, what is his native tongue, whence does he derive his immense fortune, and what were those events of his early life -- a life as marvellous as unknown -- that have tinctured his succeeding years with so dark and gloomy a misanthropy? Certainly these are questions that, in your place, I should like to have answered."

"My dear Franz," replied Albert, "when, upon receipt of my letter, you found the necessity of asking the count's assistance, you promptly went to him, saying, My friend Albert de Morcerf is in danger; help me to deliver him.' Was not that nearly what you said?"

"It was."

"Well, then, did he ask you, Who is M. Albert de Morcerf? how does he come by his name -- his fortune? what are his means of existence? what is his birthplace! of what country is he a native?' Tell me, did he put all these questions to you?"

"I confess he asked me none."

"No; he merely came and freed me from the hands of Signor Vampa, where, I can assure you, in spite of all my outward appearance of ease and unconcern, I did not very particularly care to remain. Now, then, Franz, when, for services so promptly and unhesitatingly rendered, he but asks me in return to do for him what is done daily for any Russian prince or Italian nobleman who may pass through Paris -- merely to introduce him into society -- would you have me refuse? My good fellow, you must have lost your senses to think it possible I could act with such cold-blooded policy." And this time it must be confessed that, contrary to the usual state of affairs in discussions between the young men, the effective arguments were all on Albert's side.

"Well," said Franz with a sigh, "do as you please my dear viscount, for your arguments are beyond my powers of refutation. Still, in spite of all, you must admit that this Count of Monte Cristo is a most singular personage."

"He is a philanthropist," answered the other; "and no doubt his motive in visiting Paris is to compete for the Monthyon prize, given, as you are aware, to whoever shall be proved to have most materially advanced the interests of virtue and humanity. If my vote and interest can obtain it for him, I will readily give him the one and promise the other. And now, my dear Franz, let us talk of something else. Come, shall we take our luncheon, and then pay a last visit to St. Peter's?" Franz silently assented; and the following afternoon, at half-past five o'clock, the young men parted. Albert de Morcerf to return to Paris, and Franz d'Epinay to pass a fortnight at Venice. But, ere he entered his travelling carriage, Albert, fearing that his expected guest might forget the engagement he had entered into, placed in the care of a waiter at the hotel a card to be delivered to the Count of Monte Cristo, on which, beneath the name of Vicomte Albert de Morcerf, he had written in pencil -- "27, Rue du Helder, on the 21st May, half-past ten A.M."


text:
# Axes Limits – Scream Louder, I Can’t Hear You!6

Posted by Loren Shure,

Today's guest post comes from Sean de Wolski, one of Loren's fellow Application Engineers. You might recognize him from MATLAB answers and the pick of the week blog!

I arrived at work early one day and immediately got a call from Brett, who doesn't face the same commute I do, wondering why his listeners were not working on axes' 'XLim', 'YLim' and 'ZLim'.

This seemed a little surprising, surely a listener pointed at the wrong property? Here's a minimal working example

% Plot something
ax = gca;
plot(humps);

% Add listener to axes' 'XLim' 'PostSet' event (so it fires after axes
% change).  Update the title so it's a visible change.

% Change Plot
plot(sin(1:0.1:100))

Huh? No Title! What happened?

What's interesting is that it allowed me to set the listener on the 'XLim'. Usually, if a property is not 'SetObservable', it would error.

try
% Add listener to non-SetObservable property.
catch ME
fprintf(2,'\n%s\n',ME.message);
end

While adding a PostSet listener, property 'TightInset' in class 'matlab.graphics.axis.Axes' is not defined to be SetObservable.

So what's happening?

When MATLAB versions R2014b and newer are automatically calculating the limits, the limits frequently change several times as MATLAB inspects each member of the axes and updates the limits. Allowing listeners to be triggered during the automatic calculation (even if no listener was attached) would have caused performance degradation for all plot updates even if there are no listeners. Thus the ability to trigger listeners with automatically calculated property value changes was disabled for performance reasons.

So how do we listen for limit changes?

The answer I provided to Brett was to write my own events class that has events for each of the limits changing.

classdef LimitsChangedNotifier < handle
% This class has events of limits changes to be used for updating when axes
% limits change.

events
% Events for each axes limit property
XLimChanged;
YLimChanged;
ZLimChanged;
end
end

Now every time I update the plot, or call x/y/zlim('auto'), I would notify afterward.

% Build the events class
limnotifier = LimitsChangedNotifier;

% Attach a listener to it

% Draw a new plot and let any listeners know you did
plot(sin(linspace(-2*pi,2*pi,100)));
notify(limnotifier,'XLimChanged')

Note: By doing this, I retain control of letting the listeners know the limits changed. This wouldn't work if the user did this on their own, say within plottools.

For a more detailed and advanced explanation of events and listeners, please see this documentation page.

Get the MATLAB code

Published with MATLAB® R2015b

### Note

Adam replied on : 1 of 6
Very interesting. Is there a list anywhere of which properties are observable (or rather can have listeners attached) in builtin graphics classes like axes? Since I was interested I had a quick mess around with the metaclass and this does give a list of properties that are 'SetObservable' though it is a little messy. This list contained 165 properties though (I also filtered out ones without public set access so there were slightly more), some of which I have never heard of. But it does also contain 'XLim', 'YLim' etc. So as your code and comments above suggest, these properties are actually SetObservable but cannot have listeners attached. I've only ever really used listeners with custom classes rather than graphics objects, but they would be useful assuming I wouldn't keep running into examples where the properties appear to allow listeners, but actually don't!
Yaroslav replied on : 2 of 6
Hi, The offered solution seems to be a bit overkill. Defining a whole class—just to manually notify a listener—may result in performance issues, code obfuscation, design complexity, etc. A simpler way (since we are doing it manually anyways) would be to call xlim: % Change Plot plot(sin(1:0.1:100)) xlim(ax.XLim); % <-- calling xlim invokes the XLim listener This will invoke the listener without changing the new x-axis limits. Kind regards,
Brett replied on : 3 of 6
Sean, Thanks for the interesting post. I have to admit, I'm still a bit puzzled though. If I have to _manually_ trigger a callback with a |notify| command, that seems no more useful than a |disp(ax.XLim)| call. In fact, in your example above, the limnotifier has no association with _ax_, and has no notion of the current value of the XLim of ax. (And as Yaroslav points out, calling ax.XLim explicitly will successfully trigger the desired action anyway.) So two things: * How do I trigger a callback without an explicit call to notify or ax.XLim (by plotting a different range of data, for instance)? * How do I get that callback to reflect the current value of the axes XLim, so I can do something like this: plot(humps); ax = gca; addlistener(ax,... 'XLim','PostSet',... @(~,x) title(sprintf('XLIM: [%d, %d]',x.AffectedObject.XLim))); and trigger the title to update with a plot(ax,1:10) command the same way that it would with a call to ax.XLim = [1,10]; ? Brett
Sean de Wolski replied on : 4 of 6
@Adam: As far as I know, there is no comprehensive list. I'll put in an enhancement request for that. @Yaroslav: You're right, setting
ax.XLim = ax.XLim
does fire the listener because 'AbortSet' is not true for the 'XLim' property. Thus, your solution is probably better (!). It also takes care of Brett's second question below. @Brett: 1) To my knowledge, this is not currently possible. 2) You can always create your own event data for any event you make. Here's an example class that I would use with my above example: Custom Event Data Class
classdef (ConstructOnLoad) CustomEventData < event.EventData
% CustomEventData class
% https://www.mathworks.com/help/releases/R2015b/matlab/matlab_oop/events-and-listeners--syntax-and-techniques.html#brb6i_k
properties
AffectedObject
Message
end
methods
function obj = CustomEventData(AffectedObject,Message)
obj.AffectedObject = AffectedObject;
obj.Message = Message;
end
end
end

And using it:
% Build the events class
limnotifier = LimitsChangedNotifier;

% Attach a listener to it

% Draw a new plot and let any listeners know you did
plot(sin(linspace(-2*pi,2*pi,100)));
notify(limnotifier,'XLimChanged',CustomEventData(gca,'Updated With plot()'))

Results:
  CustomEventData with properties:

AffectedObject: [1x1 Axes]
Message: 'Updated With plot()'
Source: [1x1 LimitsChangedNotifier]
EventName: 'XLimChanged

Brett replied on : 5 of 6
Sean, In my (procedural) brain, if I have to “notify” the listener that something is happening, it’s not really “listening” in the sense that I want it to. And I don’t necessarily want to “make” events. I want to call events that are pre-made, and have those calls trigger an action that I can define, and that is aware of the effects of the event. Moreover, the plot thickens: In your blog discussion, you say that XLim- and YLim- listeners were disable for performance reasons; hence this does not trigger the callback: plot(1:10);plot(1:100); But consider this (two puzzles in one!): clear;close all; ax = gca; addlistener(ax,'XLim','PostSet',@(~,~)title('Listener Fired')); imshow('peppers.png') pause(1) title('') pause(1) imshow('peppers.png') Why does the listener fire in the first place, and why does it re-fire in the second? Curiouser and curiouser!
bshoelso replied on : 6 of 6
@Adam: I'm also unaware of any pre-built lists of listenable properties. However, this snippet should allow you to query listenability: function [observable,nonObservable] = getListenable(hndl) % Return an alphabetical list of Observable properties in handle HNDL % % Brett Shoelson, PhD % brett.shoelson@mathworks.com % 11/11/2015 % %%% EXAMPLE Property Listener: % f = figure; % d = dir; % a = addlistener(f,... % 'ApplicationData','PostSet',... % @(~,~) disp('appdatachanged')); %%% This triggers the callback: % f.ApplicationData.test = d; %%% This does not. WHY? % setappdata(f,'test',d) % Copyright 2015 The MathWorks, Inc. tmp = metaclass(hndl); properties = tmp.PropertyList; names = {properties(:).Name}'; observability = {properties(:).SetObservable}'; observability = [observability{:}]'; observable = sort(names(observability)); if nargout > 1 nonObservable = sort(names(~observability)); end Hope that helps! Brett


text:
Section 1: Dimensional relationships—the Power model $y=\text{scale}\cdot{x}^{\text{power}}$

Processes in which the relationship of input to output depends on volume, area, or distance often can be modeled by formulas in which output is computed by raising the input variable to a particular exponent power, then multiplying the result by a scaling factor. This differs from the earlier exponential formula because here the exponent is a parameter rather than being the input variable x, while x is used as the base rather than the exponent.

In addition to obvious relationships such as those between size, area, and volume for geometric shapes, power-function models are useful for many situations where the geometry involved is indirect. The rate at which animals use food, for example, depends on size based on interactions between weight (which grows rapidly with size) and breathing rate (which slows down), resulting in power-function relationships with fractional powers. The concentration of a fixed amount of a chemical dissolved in water has the reciprocal power-function relationship (where the power equals –1) to the amount of water. If the power equals exactly 1, the power function is a straight line through the origin; if the power is exactly 2, the power function is a parabola whose vertex is at the origin.

The power parameter may have any value, including fractional and negative numbers. However, for fractional powers (such as the x0.5 for square roots) power functions are well-defined only for values of x that are greater than or equal to zero. The scale parameter can be any value, although it is usually positive (a zero value will make all output values zero and a negative value will flip all values around the x axis); the effect of the scale parameter is to uniformly stretch or shrink the graph vertically.

 Power models can take many shapes, depending on the value of the power.

Examples of power-formula models:

y = 122.3 x0.667 predicts the surface area of a steel ball in mm2 based on its mass in grams.

y = 0.018 x3 predicts the weight of a cantaloupe in pounds based on its diameter in inches.

y = 2×108 x-1 predicts daily visitors to web sites based on the order of popularity of the sites.

y = 1.80 x0.667 predicts a planet’s orbit radius in miles based on the length of its year in days.

The process of fitting a power model to a dataset is the same as for the other models you have studied—put the data into a worksheet in Models.xls, then use Solver to find the best-fit parameters. But in this case you will not have a preset worksheet template in which C3 already contains the right kind of formula. Instead, you will need to make a modeling worksheet yourself, or modify a copy of one of the ones you used earlier in Models.xls. The description below assumes that you will use the same row and column numbers as in Model.xls for the same kind of purposes, although you may vary them as long as you do so consistently.