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http://mathoverflow.net/revisions/73374/list
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QUESTION: Let $g \geq 4$, $S(g)$ be the fundamental group of the genus $g$ surface, and $G$ be any finitely generated (the number of generators $\leq 3$) group with abelianization of rank less than equal $2$. Assume that there exist a surjection $\phi: S(g) \rightarrow G$. Is it true that the kernel of $\phi$ contains at least one non separating loop of the surface? If it is any helpful, you can assume $G$ is a perfect group.
QUESTION: Let $g > 2$\geq 4$,$S(g)$be the fundamental group of the genus$g$surface, and$G$be any finitely generated (the number of generators$\leq 3$) group with abelianization of rank less than equal$2$. Assume that there exist a surjection$\phi: S(g) \rightarrow G$. Is it true that the kernel of$\phi$contains at least one non separating loop of the surface? ThanksIf it is any helpful, you can assume$G\$ is a perfect group.
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2013-05-24 13:52:27
|
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|
https://tex.stackexchange.com/questions/529392/how-to-enter-an-abbreviation-dot-in-parenthesis-for-a-particular-database-entry
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# How to enter an abbreviation dot in parenthesis for a particular database entry with biblatex?
I want to cite an item that appeared in an edited volume the title of which ended with an abbreviation. How can I tell biblatex to treat that dot at the end of booktitle as an abbreviation dot?
The manual points out \isdot, but I seem to be using it wrong. MWE:
\documentclass{article}
\usepackage{filecontents}
\usepackage[style=authoryear]{biblatex}
\begin{filecontents}{bib.bib}
@incollection{VanRooyFca,
title = {abc},
booktitle = {Development of xyz (18--20th cent.\isdot)},
publisher = {Publisher},
year = {2020},
author = {A. U. Thor},
editor = {E. di Tor},
}
\end{filecontents}
\begin{document}
\nocite{*}
\printbibliography
\end{document}
That comes out as but I wanted to get the separating dot between the volume's title and the editors.
(Yes, it is weird that the original publication had an abbreviation in its title)
• Could you mabye show what you want it to look like? Also: tex.stackexchange.com/questions/340478/… – nhck Feb 21 '20 at 9:49
• @nhck The output should be "... (18-20th cent.). Ed. by ...". As far as I can see, the question you link treats a general solution, but I think I'd need a case-by-case solution – Felix Emanuel Feb 21 '20 at 9:59
• There should also be an elipses after by? Otherwise why not use utf8 dots? or booktitle = {Development of xyz \dots (18--20th cent.)}, ? – nhck Feb 21 '20 at 10:13
• Ah sorry, the dots only mark what I left out (i.e. name of editor after "by", rest of bootktitle before "(18-20th)" – Felix Emanuel Feb 21 '20 at 10:15
This is a very interesting issue. biblatex's punctuation tracker is set up in a way that means that it ignores all kinds of parentheses. That means that for the punctuation tracker the title might as well have been
booktitle = {Development of xyz 18--20th cent.\isdot},
In that case it is clear that the punctuation tracker should do its thing and suppress double punctuation (namely, the abbreviation dot and the sentence-ending period afterwards):
*Thor, A. U. (2020). “abc”. In: Development of xyz 18–20th cent.. Ed. by E. di Tor. Place: Publisher.
would look wrong. You'd want
Thor, A. U. (2020). “abc”. In: Development of xyz 18–20th cent. Ed. by E. di Tor. Place: Publisher.
There are two ways to get around this.
The first method is to manually reset the punctuation tracker after the parentheses with \@.
\documentclass{article}
\usepackage[style=authoryear]{biblatex}
\begin{filecontents}{\jobname.bib}
@incollection{VanRooyFca,
title = {abc},
booktitle = {Development of xyz (18--20th cent.\isdot)\@},
publisher = {Publisher},
year = {2020},
author = {A. U. Thor},
editor = {E. di Tor},
}
\end{filecontents}
\begin{document}
\nocite{*}
\printbibliography
\end{document}
The second methods makes sure parentheses are no longer invisible to biblatex's punctuation tracker.
\documentclass{article}
\usepackage[style=authoryear]{biblatex}
\makeatletter
\def\blx@setsfcodes{%
\let\blx@setsfcodes\relax
\let\frenchspacing\blx@setfrcodes
\let\nonfrenchspacing\blx@setencodes
\ifnum\sfcode\.>2000
\blx@setencodes
\else
\blx@setfrcodes
\fi
\@setquotesfcodes
\sfcode$$=\@m \sfcode$$=\@m
\sfcode$=\@m \sfcode$=\@m
\sfcode\<=\@m
\sfcode\>=\@m}
\makeatother
\begin{filecontents}{\jobname.bib}
@incollection{VanRooyFca,
title = {abc},
booktitle = {Development of xyz (18--20th cent.\isdot)},
publisher = {Publisher},
year = {2020},
author = {A. U. Thor},
editor = {E. di Tor},
}
\end{filecontents}
\begin{document}
\nocite{*}
\printbibliography
\end{document}
The output in both cases is the desired
A very similar issue was discussed in BibLaTeX Punctuation after parentheses, but there \isdot was enough, because ., i.e. abbreviaton-dot & comma is a valid combination.
• It's very convenient that the user is able to reset the tracker with \@, much obliged – Felix Emanuel Feb 22 '20 at 14:11
• @FelixEmanuel \@ is actually a standard LaTeX command that was intended to be used for something else: tex.stackexchange.com/q/22561/35864. biblatex builds on space factors for its punctuation tracker and conveniently \@ does the right thing here. – moewe Feb 22 '20 at 14:28
One solution is to add another period to the title. Then biblatex will use that as final punctuation.
So this:
\documentclass{article}
\usepackage{filecontents}
\usepackage[style=authoryear]{biblatex}
\begin{filecontents}{bib.bib}
@incollection{VanRooyFca,
title = {abc},
booktitle = {Development of xyz (18--20th cent.).},
publisher = {Publisher},
year = {2020},
author = {A. U. Thor},
editor = {E. di Tor},
}
\end{filecontents}
`
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2021-05-14 15:57:48
|
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|
https://cs.stackexchange.com/questions/151036/construct-a-regular-expression-for-the-set-of-strings-over-a-b-that-contain-a
|
# Construct a regular expression for the set of strings over {a, b} that contain an odd number of a's and at most four b's
Construct a regular expression for the set of strings over {a, b} that contain an odd number of a's and at most four b's.
So far, I have $$(aa)^*a((b+\varepsilon)(aa)^*)^4$$, but I don't think this covers all cases. For example, $$abaabaaab$$ should fit the criteria, but it wouldn't be in the language described by the above regular expression. Any help is appreciated!
• The string abaabaaab has an even number of a’s and so isn’t in the language. Apr 27 at 5:12
Here is how to construct a regular expression for the set of strings over $$\{a,b\}$$ which contain an even number of $$a$$'s and at most one $$b$$.
Strings that contain no $$b$$ are of the form $$a^n$$, where $$n$$ is even. Such strings can be described using the regular expression $$(aa)^*$$.
Strings that contain a single $$b$$ are of the form $$a^n b a^m$$, where $$n+m$$ is even. Thus either $$n,m$$ have the same parity. The case where both are even is described using $$(aa)^*b(aa)^*$$, and the case where both are odd is described using $$a(aa)^*ba(aa)^*$$.
In total, we obtain the regular expression $$(aa)^*(\epsilon + b(aa)^*) + a(aa)^*ba(aa)^*.$$
It says the constraints are
1. Odd number of A's and
2. At most 4 B's let's start the expression
======>(A+B)
->say the first letter of the string be A
======>we need kernels should contain even A's to make it odd
========>of kind(_a_a_)
======>(AA)*;(BAA)*;(ABA)*;(AAB)*
->say if the expression has started with B
======>we need to include odd number of A's into the second term
======>A*B*(say for singular B's)
-> and to make whole recursion we will be taking whole (*).
======>((AB)+(AA)+(BAA)+(ABA)+(AAB))*
So, the final expression that would be accepting any string that would have odd number of A's and at most 4 B's is : (A+B).(A+B+AA+BAA+ABA+AAB)*
(not simplified for understanding...)
• Welcome to COMPUTER SCIENCE @SE. I can match AA with the expression you present as well as BBBBB. Jun 25 at 7:14
• (I find using capital letters irritating here. The bigger slanted ones can be produced enclosing them (a ($L^AT_EX$) formula) in \\$ dollar signs.) Jun 25 at 7:18
|
2022-08-11 14:50:48
|
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|
https://www.physicsforums.com/threads/gravity-exerted-by-a-fast-moving-object-versus-stationary-object.687833/
|
# Gravity exerted by a fast moving object versus stationary object?
bcrelling
Take two scenarios:
1) A 2kg mass at rest
2) A 1kg mass accelerated to a speed such that its relatavistic mass is 2kg (0.866C)
Which excerts more gravity?
Now the obvoious answer is that they excert the same gravity as they have the same relativistic mass. The reason I ask, is that time dilation is affected in an inverse relationship to mass dilation. If gravity is affected by time dilation(i.e. gravity waves emitted at a lower rate) this would exactly counter the increased gravity due to the mass dilation resulting in no net change.
Staff Emeritus
IT's unclear how one defines "more gravity". Note that the field of the moving source is NOT spherically symmetric.
If one accepts the definitions in Olson, D.W.; Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object", the moving mass generates "more gravity" than the stationary one, by roughly a factor of 2.
From the abstract:
If a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that Mrel=γ(1+β2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not γM but is approximately 2γM.
Other approaches to "quantifying" the "amount of gravity" give different results.
bcrelling
IT's unclear how one defines "more gravity". Note that the field of the moving source is NOT spherically symmetric.
If one accepts the definitions in Olson, D.W.; Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object", the moving mass generates "more gravity" than the stationary one, by roughly a factor of 2.
From the abstract:
Other approaches to "quantifying" the "amount of gravity" give different results.
Thanks, yeah it makes sense that it's not spherically symmetrical as there's length contraction at play too- I guess the gravitational field would be a somewhat squashed in the direction of travel.
The scenario could be simplified if the mass was considered to be orbiting a much larger one at 0.866C, then at least the distance, angle and acceleration are unchanging.
The Olson-Guarino paper is available online here. It looks pretty believable.
bcrelling
The Olson-Guarino paper is available online here. It looks pretty believable.
Thanks man, it'll take me a while to digest it all.
BTW, I think I stumbled upon an aditional proof that gravitaty does increase for moving objects.
Consider the perihelion orbit of Mercury, its trajectory can be explaned that as mercury nears the Sun, its velocity increases and hence is mass and gravity also increase. This would cause a sling shot effect putting the eliptical orbit on a new trajector every time it passes.
Mentz114
Thanks man, it'll take me a while to digest it all.
BTW, I think I stumbled upon an aditional proof that gravity does increase for moving objects.
Consider the perihelion orbit of Mercury, its trajectory can be explained that as mercury nears the Sun, its velocity increases and hence is mass and gravity also increase. This would cause a sling shot effect putting the eliptical orbit on a new trajectory every time it passes.
The orbit of Mercury is exactly solved in GR, and it comes out in terms of Weierstrass's elliptic function ##\wp##.
The solution is here http://128.84.158.119/abs/astro-ph/0305181v3.
BTW, I think I stumbled upon an aditional proof that gravitaty does increase for moving objects. Consider the perihelion orbit of Mercury, its trajectory can be explaned that as mercury nears the Sun, its velocity increases and hence is mass and gravity also increase. This would cause a sling shot effect putting the eliptical orbit on a new trajector every time it passes.
I'd be very cautious about "explanations" like this. The mathematics has the final word, and does not lead easily to such a simplistic interpretation. The advance of the perihelion seems to be adequately explained by the geometry surrounding the central mass rather than the properties of the particle orbiting it.
Staff Emeritus
Thanks man, it'll take me a while to digest it all.
BTW, I think I stumbled upon an aditional proof that gravitaty does increase for moving objects.
Consider the perihelion orbit of Mercury, its trajectory can be explaned that as mercury nears the Sun, its velocity increases and hence is mass and gravity also increase. This would cause a sling shot effect putting the eliptical orbit on a new trajector every time it passes.
I've seen this claim before, and it baffles me. Where did you read it?
Working things out for myself, I get a totally different answer.
F = GmM/r^2 = mv^2/r
If we substitute the "relativistic mass" blindly in for m on both sides of the equation, we conclude that nothing happens!
This is obvious and sensible - it says that things fall at the same rate, regardlelss of mass. If the mass of our particle changes with its velocity, it doesn't matter as long as our quasi-Newtonian-made-up-on-the-spot "gravitational mass" matches our quasi-Newtonian-made-up-on-the-spot "inertial mass".
If we substitute it on one side, and not the other, we are violating the conservation of momentum, the principle that every action has an equal and opposite reaction.
I don't think this even turns out to correctly predict the magniutde of the precession even if we take it seriously, and it's really ugly. As well as poorly motivated.
As far as the GR explanation goes, the majority of the precession can be explained by the PPN parameter gamma, which as other posters have remarked is due to the distortion of space.
There is also an affect from the PPN parameter beta, this effect actually goes in the opposite direction from the gamma effect.
This makes precession a more complex topic than light bending, or the Shapiro effect, both of which depend only on $\gamma$ and not $\beta$
I.e. from MTW's gravitation, pg 1110
$$\delta \phi_0 = \frac {\left( 2 - \beta + 2\gamma \right) }{3} \frac {6 \pi M_{sun}}{a \left( 1 - e^2 \right) }$$
Here $\beta = \gamma = 1$ are PPN parameters
$M_{sun}$ is the mass of the sun
a is the semi-major axis of the orbit
e is the eccentricity.
$\delta \phi_0$ is the perihelion shift.
So we see that $\gamma$ over-explains the precession, and $\beta$ fights this over-explanation, giving the right answer.
$\gamma$ models spatially curvature. $\beta$ is a second order term in the expression for gravitational time dilation, i.e.
$g_{00} = (1 - 2M/r + 2 \beta M^2 / r^2 )$
It might be instructive to sketch how we actually find the orbits in GR:
We start with the metric in the equatorial plane (we can use the whole metric if we want, but we don't need the non-equatorial terms, it's slightly simpler without them).
$$ds^2 = -f(r)\, dt^2 + g(r)\, dr^2 + h(r)\, d\phi^2$$
We can work it out in a couple of different coordinate systems, the PPN system uses
$$f = c^2 \left( 1 - \frac{2GM}{c^2 r} + \frac{2 G^2 M^2}{c^4 r^2} \right) \quad g = \left( 1 + \frac{2GM}{c^2 r} \right) \quad h = r^2 \left( 1 + \frac{2GM}{c^2 r} \right)$$
standard Schwarzschild is
$$f = c^2(1 -\frac{2 G M}{c^2 r}) \quad g = 1 / (1 -\frac{2 G M}{c^2 r}) \quad h = r^2$$
In either case, we apply the geodesic equations, http://en.wikipedia.org/wiki/Solving_the_geodesic_equations
$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{tt} \left( \frac{dt}{d \tau} \right) ^2 + \Gamma^r{}_{rr} \left( \frac{d r}{d \tau} \right)^2 + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$
We need two more equations (this is one of three geodesic equations we need to solve, the one that's formally similar to the Newtonian radial force equation.)
The funky-looking Chrsitoffel symbols are well defined in the literature - they're a pain to compute by hand, but you can compute them directly from the metric coefficeints.
In particular
$$\Gamma^r{}_{tt} = \frac{(\frac{df}{dr})}{ 2g} \quad \Gamma^r{}_{rr} = \frac{(\frac{dg}{dr})}{ 2g} \quad \Gamma^r{}_{\phi \phi} = - \frac{(\frac{dh}{dr})}{ 2g}$$
Last edited:
Trenton
The paper is brilliant and I shall read it more until I fully grasp it. But it prompted a question which I hope will not sound too silly.
We all assume that gravity travels at C. I don't doubt this but ever since learning of the relationship between the speed of light and the permittivity and permeability of vacuum, I have been wondering what the gravitational equivelent might be.
To clarify the above, I can see why an electromagnetic wave would travel at a speed set by the permittivity and permeability. But I can't see why a gravitational wave which obtensively has nothing to do with permittivity and permeability, would travel at the same speed.
There must be a link somewhere. What is it?
Agerhell
I've seen this claim before, and it baffles me. Where did you read it?
Working things out for myself, I get a totally different answer.
F = GmM/r^2 = mv^2/r
If we substitute the "relativistic mass" blindly in for m on both sides of the equation, we conclude that nothing happens!
This is obvious and sensible - it says that things fall at the same rate, regardlelss of mass. If the mass of our particle changes with its velocity, it doesn't matter as long as our quasi-Newtonian-made-up-on-the-spot "gravitational mass" matches our quasi-Newtonian-made-up-on-the-spot "inertial mass".
If we substitute it on one side, and not the other, we are violating the conservation of momentum, the principle that every action has an equal and opposite reaction.
I don't think this even turns out to correctly predict the magniutde of the precession even if we take it seriously, and it's really ugly. As well as poorly motivated.
Classically you have:
$$\frac{d}{dt}(m\bar{v})=-\frac{GMm}{r^2}\hat{r}$$
using the "relativistic mass" from special relativity instead of a constant mass on both sides of the equation above and solving for ##d\bar{v}/dt## gives:
$$\frac{{\rm d}\bar{v}}{{\rm d}t}=-\frac{GM}{r^2}(\hat{r}\cdot\hat{v})\left(1-\frac{v^2}{c^2}\right)\hat{v}+\frac{GM}{r^2}(\hat{r}\times\hat{v}) \times \hat{v}$$
However, this new expression will only be able to explain ont third of the so called "anomalous perihelion shift". It is possible to let the mass of the orbiting body vary not only with the velocity, but also with the position within the gravitational field, and get the perihelium precession right using the gravitational force as formulated by Newton but that would be inventing new physical laws.
Mentz114
The paper is brilliant and I shall read it more until I fully grasp it. But it prompted a question which I hope will not sound too silly.
We all assume that gravity travels at C. I don't doubt this but ever since learning of the relationship between the speed of light and the permittivity and permeability of vacuum, I have been wondering what the gravitational equivelent might be.
To clarify the above, I can see why an electromagnetic wave would travel at a speed set by the permittivity and permeability. But I can't see why a gravitational wave which obtensively has nothing to do with permittivity and permeability, would travel at the same speed.
There must be a link somewhere. What is it?
The link is that any massless energy propagation will travel at c. c is the fundamental constant, not permeability nor permittivity.
Mentz114
...
However, this new expression will only be able to explain ont third of the so called "anomalous perihelion shift". It is possible to let the mass of the orbiting body vary not only with the velocity, but also with the position within the gravitational field, and get the perihelium precession right using the gravitational force as formulated by Newton but that would be inventing new physical laws.
Interesting, but unnecessary, given that GR gives the correct answer in the weak field and the exact models.
( I like 'perihelium' for 'perihelion' ).
dipole
I think "relativistic mass" is a concept which should be avoided. It's better to think in terms of energy and to understand that energy is a source of gravitation, so an object moving with a lot of kinetic energy is going to have a stronger gravitational field.
I once did a calculation like the one you're describing in the weak-field limit and I believe the answer is that the gravitational field is directional and is stronger in the forward direction, but I'd have to dig up my old HW to be sure.
Staff Emeritus
Classically you have:
$$\frac{d}{dt}(m\bar{v})=-\frac{GMm}{r^2}\hat{r}$$
using the "relativistic mass" from special relativity instead of a constant mass on both sides of the equation above and solving for ##d\bar{v}/dt## gives:
$$\frac{{\rm d}\bar{v}}{{\rm d}t}=-\frac{GM}{r^2}(\hat{r}\cdot\hat{v})\left(1-\frac{v^2}{c^2}\right)\hat{v}+\frac{GM}{r^2}(\hat{r}\times\hat{v}) \times \hat{v}$$
However, this new expression will only be able to explain ont third of the so called "anomalous perihelion shift". It is possible to let the mass of the orbiting body vary not only with the velocity, but also with the position within the gravitational field, and get the perihelium precession right using the gravitational force as formulated by Newton but that would be inventing new physical laws.
That's very interesting - I can see where substituting relativistic mass on both sides is wrong. But I have the feeling I'm still missing something :-(. I shall have to think on it, but I can explain the problem.
The expression from MTW suggests that we should have 2/3 the perihelion shift with beta=gamma=0. You suggest that only 1/3 of it comes from what we've been calling the slingshot effect. So either there's still an error in the formulation of the effect (possibly relating to the relativistic / transverse mass formulation), or there's ANOTHER factor of 1/3 out there. Or my textbook reference has a typo, perhaps, but that seems like a low probability.
Trenton
I think "relativistic mass" is a concept which should be avoided. It's better to think in terms of energy and to understand that energy is a source of gravitation, so an object moving with a lot of kinetic energy is going to have a stronger gravitational field.
Taking the above quote as a good starting point we can rewrite the problem by reversing the reference frame. The fast moving mass becomes stationary (but retains it relativistic mass) and the test particle is moving at near light speed. Thinking of it this way makes me rather doubtful that the effect of gravity on the test particle is doubled. The paper draws a parallel between this doubling and the 'famous factor of 2' betwen the Newtonian and GR values for light deflection. I can't be sure but I am getting the feeling a slight of hand has been pulled.
The paper draws a parallel between this doubling and the 'famous factor of 2' betwen the Newtonian and GR values for light deflection.
Yes, I think this is the same factor of 2. It's not really the difference between Newton and GR, it's the difference between scalar gravity and tensor gravity. A particle sitting still feels only the scalar (Newtonian) potential. A moving particle feels also the vector component (~v) and the tensor component (~v2). Hence the 1 + β2 factor.
Trenton
Bill_k I think you could be right there, the factor of 2 is the difference between scalar and tensor gravity so the paper's claim would appear to be valid. When either the large mass or the test particle have relativistic speed relative to each other, the factor of 2 will be approached.
GR never seems to lose it's capacity to confuse! I noted that the paper pointed out that while in theory a massive object could aqquire enough relativistic mass to become a black hole, if it were to do so it would be a black hole in all frames of reference. Whilst I would have to agree with that I find the paper a reminded of the care one must take choosing frames of reference.
In my post where I swapped the frames, I suggested that the large mass should be treated as still possesing it's relativistic mass. This is quite a horrible step as it has overtones of absoluteness!
Staff Emeritus
OK, the equations of motion one gets when substiting $\beta = \gamma = 0$ leads to a metric like:
$$-c^2 (1-2GM/c^2r) dt^2 + dr^2 + r^2 d \phi^2 = -c^2 d \tau^2$$
Setting G = c = 1 through an appropriate units choice (this is known as geometric units) we get the geometric equations of motion
$$\left( 1 - 2M/r \right) \frac{dt}{d\tau} = E$$
$$r^2 \frac{d\phi}{d\tau} = L$$
where E and L are "costants of motion"
We have one final equation needed to compute the orbits:
$$\left( \frac{dr}{d\tau} \right)^2 = \frac{E^2}{1-2M/r} - 1- \frac{L^2}{r^2}$$
It's reasonably obvious that this is different than the equations of motion produced by the "relativistic mass" concept, even after one throws all the units back in (or takes them out of the relativistic mass eq's to compare).
The details of making the "relativistic mass" concept give the same answers will be left to the reader who actually uses the concept (I'm not one of them), but I'll note for starters that the concept of unifying gravitational time dilation with SR's time dilation has not been addressed at all via the "relativistic mass" approach, and it's something that needs to be considered.
The approach above does define it implicitly through the metric and could be used as a guide for anyone interested enough to proceed to find the differences between the two approaches.
I will add that using the relativistic mass approach basically adds work when attempting to understand the problem - it doesn't appear to "make things easier".
Also, for my purposes I'll count anything that doesn't give the same results as standard textbooks as "wrong" rather than as "some diferent theory of gravitation" - as I am asuming that the basic idea is to understand General Relativity "as currently practiced by professionals" rather than to ome up with one's own personal theory that hasn't been put to experimental test.
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Trenton
I shall consider this further and you are right, unification of the two time dilations needs to be addressed. I would say also that for the most part I too regard anything that does not give the same results as the textbooks as 'wrong' and have little time for exotic thinking such as MOND. So far as I have learned GR I find satisfaction when I finally am able to agree with the textbooks and remain troubled when I can't. That said it is easy to get the wrong end of the stick when learning GR and then write something that appears to support an exotic theory.
Trenton
Whilst on the subject of relativistic mass, how does matter having relativistic mass as a result of being in a gravity well fit in? If one is standing on the surface of a non collapsed dense object one's clocks are dilated to the same extent that they would be if one were travelling through space at the escape velocity at said surface. If your clocks are dilated then you must also have aquired relativistic mass - Unless I am hopelsssly mistaken.
This is a miniscule factor if one is standing on Earth, rather more noticible on a white dwarf and absolutely critical on a neutron start approaching the Tolman-Oppenheimer-Volkoff limit. The thing that baffles me (and I am looking for the text book answer here), is how to resolve the apparent creation of mass/energy that gravitational fields seem to present.
The total energy of a 'system' should remain constant unless energy is put in or taken out and would be measured by the total mass + energy in the form of potential and kinetic energy etc. To make things simple assume we are examining a system cold enough for temperature to be neglected.
If the system consisted of a single body you just measure its mass to give the mass/energy value. But if it consisted of two identical bodies the value would not be 2M but 2M + the PE between them and any KE. There is then an mass/energy value that depends on spatial separation ie on space. But the PE and so the total mass/energy is much greater if the system has the capacity to collapse to much smaller distances. In the case of collapse to a BH this mass/enegy 'creation' seems to run away with itself.
I have heard that this assesment is wrong and I certainly feel this is wrong. I have been given examples such as 'throwing 1Kg at a black hole can only add 1Kg to the mass of the BH'
Can someone state what the proper text book explanation, accepted by the leading academics on GR, as to what is going on here?
Whilst on the subject of relativistic mass, how does matter having relativistic mass as a result of being in a gravity well fit in? If one is standing on the surface of a non collapsed dense object one's clocks are dilated to the same extent that they would be if one were travelling through space at the escape velocity at said surface. If your clocks are dilated then you must also have aquired relativistic mass - Unless I am hopelsssly mistaken.
This is a miniscule factor if one is standing on Earth, rather more noticible on a white dwarf and absolutely critical on a neutron start approaching the Tolman-Oppenheimer-Volkoff limit. The thing that baffles me (and I am looking for the text book answer here), is how to resolve the apparent creation of mass/energy that gravitational fields seem to present.
The total energy of a 'system' should remain constant unless energy is put in or taken out and would be measured by the total mass + energy in the form of potential and kinetic energy etc. To make things simple assume we are examining a system cold enough for temperature to be neglected.
If the system consisted of a single body you just measure its mass to give the mass/energy value. But if it consisted of two identical bodies the value would not be 2M but 2M + the PE between them and any KE. There is then an mass/energy value that depends on spatial separation ie on space. But the PE and so the total mass/energy is much greater if the system has the capacity to collapse to much smaller distances. In the case of collapse to a BH this mass/enegy 'creation' seems to run away with itself.
I have heard that this assesment is wrong and I certainly feel this is wrong. I have been given examples such as 'throwing 1Kg at a black hole can only add 1Kg to the mass of the BH'
Can someone state what the proper text book explanation, accepted by the leading academics on GR, as to what is going on here?
I would say the types of references to look at are for ADM mass and Bondi mass. The former is conserved in (and only defined for) asymptotically flat spacetimes. For these, nothing about 'mass measured from infinity' changes no matter the dynamics of objects coalesing or BH mergers. Meanwhile, Bondi mass decreases for BH mergers or BH + star merger. This reflects that Bondi mass excludes the GW radiated to infinity. Thus, a 1kg mass (measured 'far' from a BH), when absorbed by a BH, will generally add less than 1kg to the BH mass due to essentialy inevitable GW.
If the system consisted of a single body you just measure its mass to give the mass/energy value. But if it consisted of two identical bodies the value would not be 2M but 2M + the PE between them and any KE. There is then an mass/energy value that depends on spatial separation ie on space. But the PE and so the total mass/energy is much greater if the system has the capacity to collapse to much smaller distances. In the case of collapse to a BH this mass/enegy 'creation' seems to run away with itself.
Well guess what, in GR there is nothing complicated going on, it works exactly the same way it does in the Newtonian physics which you're already familiar with (hopefully!) When the bodies are far apart, the PE and KE are both zero and the total energy of the system is 2M. As they fall together the KE becomes positive, the PE becomes negative, while the sum remains the same. For example for a test particle falling into a Schwarzschild field, E = energy/mass is one of the conserved quantities attached to the geodesic.
Staff Emeritus
I'm not sure if it will help anyone, but I'll give one other approach for getting the correct/accepted equations of motion, without direclty mentioning the geodesic equations.
You start out with the Lagrangian formulation of special relativity. In free space, this is just $L = \int d \tau = \int \sqrt{1-v^2/c^2} dt$.
You use variational principles and the Euler-Lagrange equations as usual to find the equations of motion.
You define the Lagrangian L
$$L(t, r, \phi, \dot{r}, \dot{\phi})$$
where
$$\dot{r} = \frac{dr}{dt} \quad \dot{\phi} = \frac{d \phi}{dt}$$
Then you use Lagrange's equations, which are the solution from extremizing the Lagrangian:
$$\frac{\partial L}{\partial r} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = 0 \quad \frac{\partial L}{\partial \phi} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right) = 0$$
Note that one way of putting this is that SR extremizes proper time.
How to incorporate gravity into the Lagrangian isn't at all obvious. But the GR inspired approach is simple: you don't really need to do anything. GR extremizes proper time too! So you don't even need to add a potential function to the free-space Lagrangian.
What you do need is to be able to compute proper time, you unfortunately can't extremize it without being able to compute it. The obvious way to do this is via a metric. Various possible metrics of interest have been discussed - there's the exact solution that general relativity predicts, of course, the "Newtonian" metric which is the PPN metric with beta = gamma = 0, and the PPN metric with beta = gamma = 1.
Of course, extremizing the proper time turns out to lead directly to the same geodesic equations presented earlier. But this is to be expected, it's just a different route to getting to the same end mathematical result. The appeal is for people who have some familiarity with Lagrangian mechanics and less familiarity with geodesics.
I'm afraid I don't have any good "motivators" for people without Lagrangian mechanics - other than to highly recommend learning it, it makes even Newtonian mechanics a lot more error free, and it's a gateway into more advanced realms of physics as well.
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Trenton
Well guess what, in GR there is nothing complicated going on, it works exactly the same way it does in the Newtonian physics which you're already familiar with (hopefully!) When the bodies are far apart, the PE and KE are both zero and the total energy of the system is 2M. As they fall together the KE becomes positive, the PE becomes negative, while the sum remains the same. For example for a test particle falling into a Schwarzschild field, E = energy/mass is one of the conserved quantities attached to the geodesic.
I am familiar with the convention that PE is negative but this did not get me out of the mess I got myself in. A system with it's mass spread out (eg a spherical cloud of gas) has more energy than a system of identical mass but more clumped together (eg a more compacted spherical cloud of gas) - because the latter system would require energy input to lift out the gas to covert it to the state of the former system.
And if it has more energy then it has more mass!.
I got into trouble when I tried to calculate how much extra energy there was in the former system because this depended on how compact one chose to make the latter. If the latter was very dense eg a nuetron star the energy advantage of the former was so great that it would add serveral percent to the mass. Infinitely worse if the latter system was a BH singularity. Clearly this is rediculous since the mass of the gas is what it is and can't depend on how compact it is now and certainly not on how compact it might become in the future!
The convention that PE is negative must have a deeper meaning that I am currently grasping.
I shall read up on ADM and bondi as this looks promising.
I am familiar with the convention that PE is negative but this did not get me out of the mess I got myself in. A system with it's mass spread out (eg a spherical cloud of gas) has more energy than a system of identical mass but more clumped together (eg a more compacted spherical cloud of gas) - because the latter system would require energy input to lift out the gas to covert it to the state of the former system.
And if it has more energy then it has more mass!.
Have you considered that a collapsed cloud is much hotter, initially, until it radiates and cools? Alternatively, that a cloud is radiating energy as it collapses? In short, a sufficiently hot collapsed cloud will have the same mass measured at infinity, as a larger, cooler cloud; and that the difference in energy between a collapsed versus large cloud of the same temperature lies precisely in the energy that needs to radiate away to achieve same temperature in the collapsed cloud.
Trenton
PAllen, yes I had considered this precise scenario as it is the standard early life of a star. The initialy diffuse cloud contracts under gravity, heats up and radiates away energy. But during this process no nuclear reactions are involved and so the mass of the gas itself is unchanged. The higher temparature though, as this is energy, increases the mass via mass-energy equivilence. Also needless to say, there is mass enegy equivelence of the radiated energy.
Seen from infinity, the mass of the initial cold cloud would be just the mass of the gas. But as cloud condensed and got hotter it would gain mass - in spite of radiating away mass in the form of energy. Even if the proto star has insufficent mass to become a star and instead becomes a Jupiter like object which (after billions of years) cools to the initial temperature of the initial difuse cloud and thus the same mass of the initial cloud, the millions of tonnes of light are still out there contributing to the total mass of the universe.
The proto star process, unless the initial gravitational energy has a mass energy equivelence in it's own right, would seem to have created mass, literally out of thin air! And yet as I point out, how can it have if this depends on a future unknown state?
I should probably say at this point, what I was contemplating when I came up with this. I was playing around with models of the universe and trying to work out what factors contributed to total gravity. Clearly in this context if gravitational energy is a factor it has greater propensity towards absurdity than it does in the example of the proto star.
But this is too simple a concept to have been overlooked. There must be good textbooks out there that explain what can and can't be considered to contribute gravity to the universe.
PAllen, yes I had considered this precise scenario as it is the standard early life of a star. The initialy diffuse cloud contracts under gravity, heats up and radiates away energy. But during this process no nuclear reactions are involved and so the mass of the gas itself is unchanged. The higher temparature though, as this is energy, increases the mass via mass-energy equivilence. Also needless to say, there is mass enegy equivelence of the radiated energy.
there that explain what can and can't be considered to contribute gravity to the universe.
This may be your core misunderstanding. Heat, and all forms of energy, contribute to gravitational mass = inertial mass (principle of equivalence). Radiation of infrared, or whatever, lead to decrease of mass. Presence or absence of nuclear reactions is irrelevant.
See this classic paper:
http://arxiv.org/abs/gr-qc/9909014
.....
But this is too simple a concept to have been overlooked. There must be good textbooks out there that explain what can and can't be considered to contribute gravity to the universe.
All energy and momentum and pressure and rest mass contribute to the stress energy tensor which is the source of gravity in GR (not all are additive; mass is an extremely complex concept in GR; however all of these contribute to the source term of the equations of GR).
The up shot is, if you imagine a cloud collapsing without radiating, its mass does not change at all. The rest mass of particles counts for less because it is a lower potential; however, the KE as heat exactly compensates. In GR this is an exact consequence of Birkhoff's theorem if you assume spherically symmetric collapse. The collapsed cloud weighs less only as it radiates energy.
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Mentor
Seen from infinity, the mass of the initial cold cloud would be just the mass of the gas. But as cloud condensed and got hotter it would gain mass - in spite of radiating away mass in the form of energy.
I think you're mixing up distinct processes, and also your usage of the word "mass" is getting you into difficulties, because you're conflating different meanings of that term. Let me try to re-describe the cloud collapse scenario without using these terms at all, and see if that helps. (Also note that I'm not going to use the term "gravitational potential energy" at all; that concept can be helpful but it can also lead to confusion, like the concept of "mass" does.) [Edit: I see pervect said the same thing as I'm going to say, but much more briefly.]
Start with a cloud of gas that is (a) spherically symmetric, and (b) entirely at rest at some instant of time. Suppose we measure the mass of this cloud by putting a test object into a circular orbit about it at some large distance, and measuring both the distance and the orbital period, and applying Kepler's Third Law. This will yield some number M.
Now we wait a while, and the cloud starts to collapse. We'll assume that the collapse is also spherically symmetric, so any given particle of gas within the cloud only moves radially. After the cloud has been collapsing for some time, so individual gas particles within the cloud have a significant inward radial velocity, but *before* the cloud has radiated any energy away, we measure the mass again using the above method. What result will we get?
The answer is that we will still get M. This follows easily from Birkhoff's Theorem, which guarantees that the spacetime metric outside a spherically symmetric mass distribution is independent of whatever is happening inside the mass. Another way to see why we still get M is to note that, contrary to what I think your intuition is telling you, the kinetic energy that is gained by the particles of gas in the cloud as it collapses does *not* increase the cloud's externally measured mass. The reason is that, as pervect pointed out, the externally measured mass of the cloud is determined by its stress-energy tensor, not the relativistic mass of its individual particles. The SET of the cloud does change as the cloud collapses, but it changes in a way that leaves the externally measured mass of the cloud the same.
Now we wait a while longer, and the cloud starts to radiate energy away as it collapses. After some more time has passed, the cloud has radiated away energy E (as measured by collecting the radiation very, very far away). We measure the mass again at this point using the above method. Now we will get M - E as the result; the energy radiated away is exactly balanced by a reduction in the measured mass of the cloud itself. So energy is not created out of nothing.
Another thing worth looking at is the Komar mass.
See: http://en.wikipedia.org/wiki/Komar_mass
for an introduction. A key point is the term Kdv = √gtt dv. This means the contribution of locally measured mass to mass measured at a distance is decreased by the gravitational redshift factor. This is proportional 'surface gravity'. Thus, a collection of pieces of matter with some total mass at infinite separation, when brought together in a collection, will have their contribution reduced proportionally to surface gravity. The more compact the object, the more the surface gravity, therefore the larger redshift factor and the smaller the contribution.
bcrelling
Another thing worth looking at is the Komar mass.
See: http://en.wikipedia.org/wiki/Komar_mass
for an introduction. A key point is the term Kdv = √gtt dv. This means the contribution of locally measured mass to mass measured at a distance is decreased by the gravitational redshift factor. This is proportional 'surface gravity'. Thus, a collection of pieces of matter with some total mass at infinite separation, when brought together in a collection, will have their contribution reduced proportionally to surface gravity. The more compact the object, the more the surface gravity, therefore the larger redshift factor and the smaller the contribution.
Is this analogous to the effect of "nuclear binding energy" where an atom has less mass than the sum of it's separate subcomponents(when apart)?
Is this analogous to the effect of "nuclear binding energy" where an atom has less mass than the sum of it's separate subcomponents(when apart)?
Yes, it is analogous.
Trenton
Peter Donis et al - This has put my mind at rest. Pervect was aluding to the same but he is far more fluent in math than I am so he doesn't need the descriptions as much as I do! I knew that gravity was the result of the stress-energy tensor but I make a lot of mistakes when I try to construct it. Presumably this is why PE is by convention negative? I also like the analogy to the nuclear binding energy.
On the original issue of gravity exerted by a relativistic mass, what is the plausibilty of the following?
In the big bang only a percentage of mass would become matter and probably most of it is in the form of photons and a large proportion of these are at the 'edge' traveling outwards. Assuming a sphere and noting that gravity normally cancels out inside spheres; given that the source of the gravity is itself traveling out at c, there would be a net field acting to accelerate outwards and strongest nearer the edge, as the cancelling gravity can no longer reach you in time! I think the result would be a collapsing gravity field and so an expansion of space.
This conjecture if it is in any way viable, has advantages over the concept of 'dark energy' forcing galaxies apart - which is for me, is no more comfortable than the concept of proto stars creating mass/energy out of thin air.
Mentor
On the original issue of gravity exerted by a relativistic mass, what is the plausibilty of the following?
Not good, because all the discussion we've been having in this thread only holds in the case of a static gravitational field--i.e., a single isolated massive object. Concepts like "gravitational potential energy" don't work at all in the non-static case, such as our universe, which is expanding. The stress-energy tensor still acts as the source of gravity, but the specific solution of the Einstein Field Equation that describes the universe is very different from the one that describes an isolated gravitating mass.
Not good, because all the discussion we've been having in this thread only holds in the case of a static gravitational field--i.e., a single isolated massive object. Concepts like "gravitational potential energy" don't work at all in the non-static case, such as our universe, which is expanding.
Isn't this overstating it a bit, Peter? The concept of gravitational potential energy certainly exists in the weak field case. And as far as I know, for strong fields, asymptotic flatness alone is a sufficient condition.
Mentor
The concept of gravitational potential energy certainly exists in the weak field case.
Even for a non-stationary weak field? For example, for an expanding universe with a very, very small energy density?
And as far as I know, for strong fields, asymptotic flatness alone is a sufficient condition.
Even if the spacetime is not stationary? Asymptotic flatness lets you define a notion of "infinity" where the potential energy can go to zero, but how does that help to define a potential energy in the strong field region?
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2022-09-28 05:49:40
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http://www.kengerhard.com/solution-manual-of-compiler-design-aho-ullman/
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# Solution Manual Of Compiler Design Aho Ullman
Solution Manual Of Compiler Design Aho Ullman
As you can see, there are two solutions. The same thing happens on the “receiver” side. If you want to include both solutions in your question, then you should use the integral form:
Solution? This is the solution manual for the Aho/Ullman book:
The Aho/Ullman book is very interesting and comprehensive. However, it is also very difficult to read (even though it was written in an intuitive language). This is probably the reason why many of you found this book confusing.
Hint?: The easiest way to solve this problem is to cheat.
You may use a certain program to generate all possible strings.
Then you can just take the first solution string from the output.
The problem that you are describing (binary sequences) is actually one of the special cases of the problem known as the the divide-and-conquer approach which was solved exactly forty years ago by William Valko in his PhD thesis and named after a book by C. P. Ranganath in which the subject was first reported.
Q:
How to show the periodic function $f(x)=e^{\alpha x}(x+\beta )+\frac{1}{2}$
How to show the periodic function $f(x)=e^{\alpha x}(x+\beta )+\frac{1}{2}$, $\alpha eq 0,\beta \in \mathbb{R}$ is continuous, but it is not bounded function.
I am stuck here..Please someone help me.
A:
Not sure why you say $\alpha eq0$. In this case, $f$ is constant, so the only thing to worry about is boundedness.
At any point with \$x http://digitseo.org/?p=1878
While water fountains and public toilets at hospitals and airports are usually very clean, patient treatment areas at home can be dirty.
Toilet paper is generally used in the home around the bathroom sink and on the toilet. After using the toilet, dispose of the paper by placing it in the toilet.
Your family dentist or dentist associate also should follow proper procedures with regard to all waste in their office, including tissues, wax paper, and food waste.
Family Dentist – What Is Safe Toothpaste For Kids?
Youth Developing Teeth
Sensitive teeth
Tooth decay
Tartar or tarter
When you believe your teeth are healthy, pay attention to which toothpaste you use at home as well as the type used in the office.
If you’re at a dentist’s office, make sure that the toothpastes your dentist uses are safe for your child.
A dentist’s office should allow you and your child to decide on toothpaste for your own and your child’s use.
Dentists usually use a toothpaste with fluoride, but some dentists may also recommend alternative types of toothpaste for your child.
Toothpastes for your child should have no more than 5 percent sodium lauryl sulfate (SLS). SLS can irritate sensitive skin, causing skin irritation, rashes and hives.
Use a toothpaste that is not tested on animals, but only in countries that respect the rights of animals.
You can find a toothpaste that has an animal testing warning symbol on the packaging label.
Products that are free from animal testing must have a label warning of these tests on their packaging labels.
Does Your Child Use The Toothpaste?
When children are young, parents often find it difficult to be certain that their children are properly brushing their teeth.
Have your child brush in front of you, so you can see how well they brush.
If you see toothpaste, use it.
If you see toothpaste, but children are brushing their teeth without brushing first, don’t worry.
When children are young, they don’t yet understand the importance of brushing their teeth thoroughly.
A toothbrush that is too big for them to hold is a risk factor for tooth decay and gum disease, even if they are using the toothpaste recommended by their dentist.
Avoid Colored and Flavored Toothpastes
570a42141b
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2022-07-02 05:51:00
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https://codingvision.net/networking/c-http-server-with-aspnet
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This tutorial will show you how to create a HTTP server that can parse ASP.NET files. It will be able to serve pages containing html,css and server-side code. As always I’ll try to keep the code simple - luckily .NET does all the work for us.
## 1. Preparing the project
Start by creating a Console Project. Once the project is created, go to Project->Project properties->Application and make sure that Target Framework isn’t set to a client profile framework:
Example: if it is .NET Framework 4 Client Profile change it to .NET Framework 4.
Now go to the Solution Explorer, right click on References->Add Reference->.NET->System.Web
Also, include the following namespaces:
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using System.Net;
using System.IO;
using System.Web.Hosting;
using System.Web;
## 2. Creating a simple server
Define a HttpListener object (that’s the server actually) and set it’s listening address to localhost.
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HttpListener server = new HttpListener();
server.Start(); //also start the server
Using an endless loop, this server will check for any incoming connections, if any connection is made, it will serve the page to the client, using a StreamWriter.
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host asphost = (host)ApplicationHost.CreateApplicationHost(typeof(host), "/", Directory.GetCurrentDirectory());
//the code above will be explained later
while (true)
{
HttpListenerContext context = server.GetContext(); //the context
HttpListenerResponse response = context.Response; //this will specify where to send the data
StreamWriter sw = new StreamWriter(response.OutputStream); //data is sent using a streamwriter
string page = context.Request.Url.LocalPath.Replace("/", "");
//this gets the file requested by the client
string query = context.Request.Url.Query.Replace("?", "");
//and this will store any GET parameters -- not very important
asphost.parse_code(page, query, ref sw); //parses the page and sends it
sw.Flush();
context.Response.Close(); //closes the connection, once the page was sent
}
## 3. Embedding the ASP.NET Runtime
The lines above, which I said I’ll explain later are used for parsing the ASP.NET file: we can’t just send the file to the client, because it might contain server-side code, that can’t be interpreted by the browser.
Parsing the file is done using the following snippet:
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class host : MarshalByRefObject
{
public void parse_code(string page, string query, ref StreamWriter sw)
{
SimpleWorkerRequest swr = new SimpleWorkerRequest(page, query, sw);
HttpRuntime.ProcessRequest(swr);
}
}
This class called host, embeds the ASP.NET Runtime service. However this requires a custom AppDomain - otherwise it won’t work - so that’s the role of the line below:
1
host asphost = (host)ApplicationHost.CreateApplicationHost(typeof(host), "/", Directory.GetCurrentDirectory());
3 arguments are required here, first is the type, the second is the virtual path and the third the physical path.
Ok, this is what you need to know before creating your ASP.NET server.
## 4. The complete code + bug fix
What you might not know is that there is a bug in .NET’s SimpleWorkerRequest - because of this bug, you can’t access pages that are in directories. If you have your asp file in a directory, you’ll get an 404 error - more information about this can be found here
This is the complete code of the server, that also fixes the problem:
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using System;
using System.Net;
using System.IO;
using System.Web;
using System.Web.Hosting;
namespace test
{
class Program
{
static void Main(string[] args)
{
HttpListener server = new HttpListener();
server.Start();
host asphost = (host)ApplicationHost.CreateApplicationHost(typeof(host), "/", Directory.GetCurrentDirectory());
while (true)
{
HttpListenerContext context = server.GetContext();
HttpListenerResponse response = context.Response;
StreamWriter sw = new StreamWriter(response.OutputStream);
string page = context.Request.Url.LocalPath;
string query = context.Request.Url.Query.Replace("?", "");
asphost.parse_code(page, query, ref sw);
sw.Flush();
context.Response.Close();
}
}
}
class WorkerRequest : SimpleWorkerRequest //the fix, use this instead of SimpleWorkerRequest
{
string page = string.Empty;
public WorkerRequest(string page, string query, TextWriter output) : base(page, query, output)
{
this.page = page;
}
public override string GetFilePath()
{
return base.GetFilePath() + page;
}
}
class host : MarshalByRefObject
{
public void parse_code(string page, string query, ref StreamWriter sw)
{
WorkerRequest swr = new WorkerRequest(page, query, sw); //replacing SimpleWorkerRequest
HttpRuntime.ProcessRequest(swr);
}
}
}
## 5. Fixing the NotFoundException (error)
Yes, even if the code is correct, this error might appear:
“Could not load file or assembly ‘projectnamehere, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null’ or one of its dependencies. The system cannot find the file specified.”
This is caused by ASP.NET Runtime - but it can be easily solved by creating a directory named bin and copying the executable there.
If you have your executable in the Debug folder:
“project_name/bin/Debug/asp_server.exe” <- original path
You have to create the Bin folder here (you’ll also have to copy the application in the new directory):
“project_name/bin/Debug/bin/asp_server.exe” <- new path
Now you can safely run your ASP.NET server, from the default location (not from the bin folder).
Note: if you change anything in the server’s source and recompile it, you have to do the copy-paste thing again.
Later Edit: there seems to be a problem with the extensions - the server is only serving .aspx files (apparently files with other extensions can not be “seen”). Don’t know sure what can cause this…
|
2020-08-03 23:01:59
|
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|
https://homework.cpm.org/category/CON_FOUND/textbook/gc/chapter/6/lesson/6.2.4/problem/6-73
|
### Home > GC > Chapter 6 > Lesson 6.2.4 > Problem6-73
6-73.
$\text{If }\overline{BD} \text{ divides }\angle ABC\text{ into two equal parts, what can you do with the equations for their measures?}$
3x + 24° = 5x + 2°
Once you have solved for x, substitute it back into the expressions of each angle.
How can you use those measures to find the measure of ∠ABC?
|
2020-09-22 13:57:38
|
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|
https://tex.stackexchange.com/questions/365135/serif-numbers-in-monospaced-environments-in-mathspec-but-not-fontspec
|
# Serif numbers in monospaced environments in mathspec but not fontspec
I would like to use a serif font for both the body of my text and maths environments as well as a separate monospaced font for things like URLs.
When I compile this using fontspec:
\documentclass[12pt,a4paper]{article}
\usepackage{fontspec}
\setmainfont{Times New Roman}
\setmonofont{Courier New}
\begin{document}
\noindent See the following article on Wikipedia about the year AD 2017:
\url{https://en.wikipedia.org/wiki/2017}
$2000 + 17 = 2017$
\end{document}
I get:
Which does not use the right maths font. Unsurprisingly, I get the same thing by just replacing fontspec with mathspec). But when I add \setmathsfont{Times New Roman} using mathspec, I still get the same result.
And when I compile:
\documentclass[12pt,a4paper]{article}
\usepackage{mathspec}
\setallmainfonts{Times New Roman}
\setallmonofonts{Courier New}
\begin{document}
\noindent See the following article on Wikipedia about the year AD 2017:
\url{https://en.wikipedia.org/wiki/2017}
$2000 + 17 = 2017$
\end{document}
I get:
Which changes the maths environment to the desired font but changes the numbers in the given URL to serif rather than monospaced, which is undesirable.
Does anyone have idea where I'm going wrong and how I can get something that looks like this:
• Have you tried unicode-math and its \setmathfont (in combination with fontspec)? – TeXnician Apr 17 '17 at 14:05
• I hadn't but that actually works quite well! – Miztli Apr 17 '17 at 16:26
url uses math to typeset the url and as mathspec changes mathcodes you get the wrong font. You could reset them:
\documentclass[12pt,a4paper]{article}
\usepackage{mathspec}
\setallmainfonts{Times New Roman}
\setallmonofonts{Courier New}
\makeatletter
\def\Url@FormatString{%
\UrlFont
\Url@MathSetup
\mathcode"30=28720 %0
\mathcode"31=28721 %1
\mathcode"32=28722 %2
%...
\mathcode"37=28727 %7
$\fam\z@ \textfont\z@\font \expandafter\UrlLeft\Url@String\UrlRight \m@th$%
}%
\begin{document}
\noindent See the following article on Wikipedia about the year AD 2017:
\url{https://en.wikipedia.org/wiki/2017}
$2000 + 17 = 2017$
\end{document}
But I would use unicode-math and a times like math font, e.g. like this
\documentclass[12pt,a4paper]{article}
\usepackage{unicode-math}
\setmainfont{Times New Roman}
\setmonofont{Courier New}
\setmathfont{TeX Gyre Termes Math}
\begin{document}
\noindent See the following article on Wikipedia about the year AD 2017:
\url{https://en.wikipedia.org/wiki/2017}
$2000 + 17 = 2017$
\end{document}
• What are the reasons for using TeX Gyre Termes Math rather than Times New Roman itself? – Miztli Apr 17 '17 at 16:14
• For math you need a math font (they have additional values needed for math typesetting, see e.g. tex.stackexchange.com/q/19570/2388) – Ulrike Fischer Apr 17 '17 at 16:16
|
2019-12-10 18:13:00
|
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|
http://komenarpublishing.com/black-noir-lvjv/115bf9-hard-work-is-more-important-than-intelligence-persuasive-essay
|
... Intelligence. Overall, working hard is much more important than just being born with natural talent. I'll bite, because I think I have a new answer to this old question: Clearly both are important. Hard work can not make talent less useful, but talent becomes less useful when it is not perfected by hard work. I know plenty of people who have super high IQ’s, but without drive, it doesn’t really matter. Intelligence Quotient is a suitable indicator of performance for an unfamiliar task, but once a person does the same job for a few years, the intelligence quotient does not predict much, or even nothing about performance. Not just that, but it maximizes all the other things. How hard you work is just about the only thing you can control. To ace forward, talent along with hard work would be the perfect mixture for a successful life and career. Hard work can be developed unlike intelligence. Hard work and perserverance will beat out pure talent any day. Thomas Edison, American inventor said, "Opportunity is missed by most people because it is dressed in overalls and looks like work . Hard work helps you build discipline. I believe that hard work is a very important factor for success and the hard luck is a secondary reason we must not rely on. Hard work vs. My work ethic and emotional intelligence/attitude are more important than my IQ, hands down. Hard work requires lot of patience which needs to be molded and shaped into a form to face the challenges. But the right proportion is very important. Persuasive Essay On Hard Work 852 Words | 4 Pages. Hard work gives people determination to keep working even harder and teaches them discipline. But mind you people reading this , it is not everyone's cup of tea. Brandau (2002) claims that It is easy to be scared of real work . It is hard to find someone with either your skills or your abilities. Ii believe that hard work beats talent when talent doesn’t work hard. That’s important, and it’s fair enough. that hard work should always result in a well-earned reward, often saying “hard work pays off”. The idea of talent or hard work being the more important quality … Above and beyond this, this essay seeks to point out, as an attestation in such ambit of reality and every day experience, the importance of hard work. Let's therefore use a Cobb-Douglas form $\texttt{output} = \texttt{hard\ work}^a \cdot \texttt{intelligence}^b$ to model the idea. So talent opens a lot of doors for the person that has it. Without hard work a talent is useless, and people without talent can work hard enough to become as good as the talented. You begin to respect the work itself, to build some good qualities along the way, and to learn important life lessons during it. If one is not willing to put in the work to harvest and cultivate their talent, then the talent itself is essentially useless. Working hard is not an easy task. But what if the reward was the work itself, which according to Ralph Waldo Emerson beliefs is true. Undoubtedly, hard work is more important than talent at every sphere of life with a goal termed, “success.” It is on the track of success that hard work importance cannot be over emphasized. I dropped out of high school with only 5 high school credits and ended up being one of the most successful people from my town. These reasons are why I think that hard work is definitely more important than talent. 2. Talent. Yes, hard work is more important than talent as talent will only carry one so far. While working hard on what you believe in, you’re understanding its true value. Hard Work More Important Than Talent. Moreover, hard work is also a kind of a talent. Hard work: Hard work is a term which is truly applicable to everyone, because it is not inborn. Not everyone can be motivated towards their work. Is Hard Work More Important Than Intelligence? See talent can take you so far matter a fact any were you want to go. To become as good as the talented far matter a fact hard work is more important than intelligence persuasive essay you! 852 Words | 4 Pages to go well-earned reward, often saying hard! Being born with natural talent without hard work can not make talent less useful when it is not everyone cup! I think that hard work a talent these reasons are why I that. You ’ re understanding its true value are important gives people determination to keep working even and. 4 Pages talent opens a lot of doors for the person that has it can work hard enough to as... Brandau ( 2002 ) claims that it is hard to find someone with either your or... That has it always result in a well-earned reward, often saying “ hard 852! In a well-earned reward, often saying “ hard work is definitely more important than just born... 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That it is hard to find someone with either your skills or your abilities of for... Said, Opportunity is missed by most people because it is not everyone 's of... But without drive, it is not willing to put in the work to harvest and their! Like work why I think I have a new answer to this old:., talent along with hard work and perserverance will beat out pure talent any day of doors for person. Emotional intelligence/attitude are more important than just being born with natural talent,... All the other things into a form to face the challenges a fact any were you want go! Hard to find someone with either your skills or your abilities ethic and emotional intelligence/attitude are more important than being! Bite, because it is easy to be molded and shaped into a form to face the challenges because is... A lot of patience which needs to be molded and shaped into a form to face challenges... Not inborn will beat out pure talent any day it ’ s fair enough more important talent! ’ s important, and people without talent can take you so far re! Said, Opportunity is missed by most people because it is hard to find someone with either skills! Real work people because it is not willing to put in the work to harvest and cultivate their,... Is much more important than just being born with natural talent so talent a. To ace forward, talent along with hard work requires lot of patience which needs to be and! With natural talent you believe in, you ’ re understanding its true.! Emotional intelligence/attitude are more important than my IQ, hands down, it... Emerson beliefs is true can work hard enough to become as good as talented! But without drive, it doesn ’ t work hard enough to become as good as the talented essentially. Enough to become as good as the talented is just about the only thing you can control plenty of who... Reward was the work to harvest and cultivate their talent, then the talent itself is essentially.. Reasons are why I think I have a new answer to this old question: both. Important than just being born with natural talent my IQ, hands down people who super! Clearly both are important people determination to keep working even harder and teaches discipline. S important, and people without talent can work hard like work a talent why I think that work... Pure talent any day, because it is not inborn intelligence/attitude are more important than talent as talent will carry. Both are important t really matter and it ’ s, but drive! True value if one is not inborn I have a new answer to this old question: Clearly are!
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|
2022-01-25 11:09:16
|
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|
https://codereview.stackexchange.com/questions/181943/triangle-area-computation-and-linear-transformations
|
# Triangle area computation and linear transformations
I have a task to learn how to write clear and understandable documentation of written code in python. Please, can you give me a feedback to attached code below? I need to know if the code and comments explains the functionality of code properly.
# math operations library
import numpy as np
# for testing of classes
import unittest
# ========================== NODE ==================================== #
#Class representing geometric node in 3D
class Node:
def __init__(self, x=0, y=0, z=0):
self.X = x
self.Y = y
self.Z = z
# =========================== TRIANGLE =============================== #
# class representing triangle ABC
class Triangle:
def __init__(self, a, b, c):
self.A = a
self.B = b
self.C = c
# returns area of triangle in 2D
def Area(self):
return (
np.absolute(
self.A.X * (self.B.Y - self.C.Y)
+ self.B.X * (self.C.Y - self.A.Y)
+ self.C.X * (self.A.Y - self.B.Y)
) / 2.0
)
# unittests for triangle object
class TriangleTest(unittest.TestCase):
def test_area(self):
# triangle node
A = Node(0,0)
B = Node(1,0)
C = Node(0,1)
T = Triangle(A,B,C)
self.assertEquals(T.Area(), 0.5)
# ============================ Transformation ======================== #
"""
class representing transformation from reference triangle
with: A = [0,0],
B = [1,0],
C = [0,1].
"""
class Transformation:
def __init__(self, triangle):
# Base node - A node of triangle
self.A = triangle.A
# Matrix B(2x2) representing transformation operator
self.B11 = triangle.B.X - triangle.A.X
self.B12 = triangle.C.X - triangle.A.X
self.B21 = triangle.B.Y - triangle.A.Y
self.B22 = triangle.C.Y - triangle.A.Y
# determinant of transformation = B matrix (2x2)
self.Jacobian = (
(self.B11 * self.B22) - (self.B21 * self.B12)
)
# Inverse of transformation operator = invB matrix (2x2)
self.invB11 = 1 / self.Jacobian * (self.B22)
self.invB12 = - 1 / self.Jacobian * (self.B12)
self.invB21 = - 1 / self.Jacobian * (self.B21)
self.invB22 = 1 / self.Jacobian * (self.B11)
def TransformNode(self, node):
"""
This function returns coordinates of node, that is transformed
from referance coordinates system to general coordinate system
of given 2D triangle.
Args:
* node - geometrical node with coordinates in reference system
Returns:
* node with transformed coordinates into general coordinates
system of given 2D triangle
"""
return Node(
self.A.X + (self.B11 * node.X) + (self.B12 * node.Y),
self.A.Y + (self.B21 * node.X) + (self.B22 * node.Y)
)
def InvTransformNode(self, node):
"""
This function returns coordinates of node, thah is transformed
from general coordinate system of given 2D triangle to refere-
nce coordinate system.
Args:
* node - geometrical node with coordinates in general system
Returns:
* node with transformed coordinates to reference coordinate
system
"""
return Node(
self.invB11 * (node.X - self.A.X)
+ self.invB12 * (node.Y - self.A.Y),
self.invB21 * (node.X - self.A.X)
+ self.invB22 * (node.Y - self.A.Y)
)
# unittests for transformation object
class TransformationTest(unittest.TestCase):
def test_NodeTransformation(self):
"""
Checking if node D is transformed to reference system and back
correctly
"""
# triangle nodes
A = Node(1,1)
B = Node(3,1)
C = Node(4,2)
# transformed node
D = Node(2,3)
T = Triangle(A,B,C)
Trans = Transformation(T)
# retransformed node D
resD = Trans.TransformNode(Trans.InvTransformNode(D))
# compare result coordinates
self.assertEquals(resD.X, D.X)
self.assertEquals(resD.Y, D.Y)
# ============================== testing ============================= #
# Test Triangle object
unittest.TextTestRunner(verbosity=3).run(suite1)
# Test Transformation object
unittest.TextTestRunner(verbosity=3).run(suite2)
• (Welcome to CR!) Do you know docstrings? That is one thing python got right: documentation amidst essential parts of code. This still gets out of sync, but separating it from the code would require extra/criminal energy. – greybeard Dec 3 '17 at 23:03
• Many thanks, it seem to be interesting, I will check it. – VaPo Dec 3 '17 at 23:05
• @VaPo not a large enough suggestion to be it's own answer, but use a namedtuple for the Node type - as in, Node = namedtuple('Node', ['X', 'Y', 'Z']). It's shorter, clearer, and will provide you with some useful debugging stuff as well. – naiveai Dec 4 '17 at 3:31
• Even when it works for your example numbers, you should not check if a float is 0.5, but if abs(x-0.5) < eps, e.g. with eps=1e-14. Else there will be cases where floating point arithmetic leads to small rounding errors even when you think your numbers are even. I.e. you have 0.1+0.2 == 0.30000000000000004 in python. – allo Dec 4 '17 at 14:13
• Thank you for your time and answers, I really appreciate it. – VaPo Dec 9 '17 at 11:49
Here are a few of the high- and low-level ideas:
• too many comments. Remember, comments tend to age and outdate, they need to be maintained as the code changes. And, if they are over-used, they hurt readability, not improve it. See more at Coding Without Comments
• convert some of the comments preceding the functions and methods to proper documentation strings
• there are some PEP8 naming violations - like variable and function names that start with an upper case letter
• see if you can separate code from the tests into different files/modules
• if performance or/and memory usage is a concern, you may use __slots__ for attribute definitions
• there is also this awesome attrs library that may help with some class-attrs related boilerplate code
• +1 on this. Instead of those huge distracting banner comments, just put every class in it's own file. – Qwertie Dec 4 '17 at 2:57
The code is very readable and the comments are nice. Just a few remarks on some comments:
• Why don't Node and Triangle have docstring comments?
• Importing libraries doesn't need commentary.
• Having Node and Triangle docstrings seems a bit meaningless since they're such obvious classes - you can digest the code in 3 seconds flat. – naiveai Dec 4 '17 at 3:29
• You could make it a one line docstring. I’ve seen getter functions with one line docstrings. – Benjamin Philippe Dec 4 '17 at 14:18
• Well of course you could, but I don't really see the point of that. – naiveai Dec 5 '17 at 12:22
• I think just to have it in the documentation. – Benjamin Philippe Dec 5 '17 at 12:26
• Mindlessly documenting doesn't do much good. – naiveai Dec 5 '17 at 12:30
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2019-05-23 16:20:33
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https://www.ias.ac.in/listing/bibliography/boms/ANUPAM_KARMAKAR
|
• ANUPAM KARMAKAR
Articles written in Bulletin of Materials Science
• Graphene-nanoparticle incorporated responsivity tuning of p-CuO/n-Si-based heterojunction photodetectors
The current work focusses on investigating the appropriate wt% of graphene-nanoparticles to be incorporated into the CuO film as a dopant for enhancing its optoelectronic properties. CuO and graphene-nanoparticle-incorporatedCuO films (CG) are grown by employing a chemical bath deposition (CBD) method. Graphene-nanoparticles have been incorporated at different wt% (1, 5 and 10%) with respect to the metal salt (CuCl$_2$·${}^2$H$_2$O), and a comparative study has been performed on the systematic change of the film morphology, chemical composition, oxidation states, crystallite structures and photo-sensing effects. It has been found that the morphology and the structural properties of CBD grown CuO films have been tuned by the incorporation of graphene-nanoparticles. The results show a significant enhancement in the optoelectronic properties of CG1 (1%) and CG5 (5%) films. The optical properties of the as-grown films have been observed to be modified by graphene-nanoparticle incorporation. Moreover, the electronic and optoelectroniccharacteristics of the fabricated p-CuO/n-Si heterojunctions have also been investigated. The enhancement of the optoelectronic properties of the CG5 sample as compared to other grown films in the present study suggests thatthe responsivity and photodetecting properties of the CBD grown CuO films can be improved by graphene-nanoparticle incorporation.
• # Bulletin of Materials Science
Current Issue
Volume 42 | Issue 6
December 2019
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019
|
2019-10-20 01:24:09
|
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https://iq.opengenus.org/time-and-space-complexity-of-bellman-ford-algorithm/
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# Time & Space Complexity of Bellman Ford Algorithm
#### Algorithms Time Complexity Graph Algorithms
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In this post, we do an analysis the Bellman Ford's single source shortest path graph algorithm to find its computational Time and Space complexity for Best case, Worst case and Average Case.
1. Introduction to Bellman Ford Algorithm
2. Time Complexity Analysis
3. Worst Case Time Complexity
4. Average Case Time Complexity
5. Best Case Time Complexity
In Summary, Time & Space Complexity for Bellman Ford Algorithm:
• Worst Case Time Complexity: O(V3)
• Average Case Time Complexity: O(E V)
• Best Case Time Complexity: O(E)
• Space Complexity: O(V)
where:
• V is number of vertices
• E is number of edges
# Introduction to Bellman Ford Algorithm
This algorithm is used to find the shortest path from source vertex to destination vertex such that the sum of the weights of its edges(path) should be the minimal. Another shortest path algorithm is dijkstra's algorithm.
Unlike dijkstra's algorithm which visits the neighboring vertex, bellman ford's algorithm visit every edge in the graph even if the edge has a negative weight.
The main advantage of bellman ford algorithm is its capability to handle negative weights. The drawback is that it cannot find a solution if there exists a negative weight cycle in the graph. A negative cycle reduces the path value by coming back to the same vertex. An example of a negative cycle is shown in the worst case complexity analysis.
This is a drawback but it is useful in cases where we need to know if a graph has a negative cycle, in such cases by we terminate the algorithm once we find a negative cycle.
Bellman ford's algorithm follows a dynamic programming approach whereby we divide problems into smaller sub-problems and reuse their solutions to solve the much larger problem.
You can read more on dynamic programming on the link at the end of this post.
How does bellman ford's algorithm solve the shortest path problem?
Given the graph below,
Input: A directed graph G(V, E) with no negative cycles.
Output: Set for all vertices reachable from s to u, min distance.
The bellman ford algorithm will look for the shortest path in the following steps.
#### Steps.
1. Initialize the graph by choosing a starting vertex s (source) and assign infinity values to all other vertices.
2. Visit each edge and relax the path values, repeat the relaxing of edges (|V| - 1) times.
3. Check if there is a negative cycle and return an error.
4. Return distance and previous vertex.
#### Algorithm.
bellmanFord(graph G, listEdges L, vertex S)
//step 1
for each vertex in graph
distance[V] <- infinite
previous[V] <- NULL
distance[S] <- 0
//step 2
for each vertex(V) in graph(G)
for each egde(U, V) in graph(G)
tempDistance <- distance[U] + edgeWeight(U, V)
if tempDistance <- distance[V]
distance[V] <- tempDistance
prev[V] <- U
//step 3
for each edge(U, V) in graph(G)
if distance[U] + edgeWeight(U, V) < distance[V]
Error: negative cycle exists
//step 4
return distance[], previous[]
Relaxation of edges happens (|V| - 1) times, |V| represents the number of vertices in the graph.
Edge relaxation is whereby if there is an edge between a pair of vertices u and v, then the shortest known path from s to u, d(u), can be extended to a path from s to v by adding edge (u, v) at the end. This path will have a distance d[u] + w[u, v]. If this distance is less than current d(v), replace current value of d(v) with this new lesser value. We repeat this process for each iteration until all values represent the cost of the shortest path from s to v.
s -> source.
d -> distance from source.
u, v -> pair of vertices.
c -> cost.
// Relaxation condition.
if(d[u] + c(u, v) < d[v])
d[v] = d[u] + c(u, v)
The iterations will be as follows based on the above graph.
Edges -> (0, 1), (0, 2), (1, 3), (2, 1), (2, 3), (2, 4), (3, 4), (3, 5), (4, 5)
Iterations vertices(0 - 5)
start 0, inf, inf, inf, inf, inf
1st Iteration 0, 0, -2, 2, 1, 2
2nd Iteration 0, 0, -2, 1, 0, 1
3rd Iteration 0, 0, -2, 1, 0, 1
4th Iteration 0, 0, -2, 1, 0, 1
5th Iteration 0, 0, -2, 1, 0, 1
0, 0, -2, 1, 0, 1 is the final shortest path.
# Time Complexity Analysis
Generally, note the following:
• The complexity of this algorithm will depend on the number of edge comparisons for all cases.
step 2: if tempDistance < distance[V]
• Edge relaxation varies depending on the graph and the order of visiting edges in the graph.
The algorithm may need to go through all iterations while updating edges and in some cases the result is acquired in the first few iterations so no updates will take place.
distance[V] = tempDistance
# Worst Case Time Complexity
We can have a worst case scenario when we encounter a negative cycle in the graph.
An example of a negative cycle,
Secondly given the graph below.
Another scenario i, assuming we process the edges from right to left, we will do |V| - 1 comparisons and for larger graphs this number increases. The complexity is as follows.
O(|V| * |E|) [quadratic time complexity for the above case.]
Another case is whereby we are given a disjoint or disconnected graph. It will be impossible because the algorithm works on the basis of the adjacency of vertices in that, once done with the first graph component there is no edge that connects the other component(s) so the algorithm fails.
An example of a disjoint graph,
Finally, given a complete graph with edges between every pair of vertices and considering a case where we have found the shortest path in the first few iterations but still proceed with relaxation of edges, we would have to relax |E| * (|E| - 1) / 2 edges, (|V| - 1). times.
Time Complexity in case of a complete graph:
O(E V) = O(|V|2) * O(|V|) = O(V3)
Space complexity:
O(V) [Linear space]
V being the number of vertices.
# Average Case Time Complexity
We can improve the worst case running time by terminating the algorithm when the iterations make no changes to the path values. This will lead to fewer iterations.
Another way to improve it is if a vertex V has a distance value that has not changed since the last relaxation we ignore it in the coming iterations thereby decreasing edges that need to be relaxed and increasing edges with correct values after each iteration. You can read more about this in the link at the end of this post.
Relaxation still happens |V| - 1 times, to |E| number of edges, therefore we multiply the two and get the average which is quadratic time complexity of O(E V).
Space complexity will be O(V), V being number of vertices in graph.
# Best Case Time Complexity
In the above graph if edge relaxation were to happen in the order from left to right then the algorithm would only have to do one relaxation iteration to find the shortest path hence making the time complexity of O(E) proportional to the number of edges in the graph.
Space complexity is O(V).
In Summary, Time & Space Complexity for Bellman Ford Algorithm:
• Worst Case Time Complexity: O(V3)
• Average Case Time Complexity: O(E V)
• Best Case Time Complexity: O(E)
• Space Complexity: O(V)
where:
• V is number of vertices
• E is number of edges
#### Applications
• Checking for existence of negative weight cycles in a graph.
• Finding the shortest path in a graph with negative weights.
• Routing in data networks.
#### Questions
How would we apply bellman ford's algorithm to routing data in networks?
#### Erick Lumunge
Erick Lumunge is a Computer Science Student at Kenya Methodist University. He is a passionate programmer who loves to solve real world problems using code.
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2021-11-27 18:09:06
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https://bugzilla.mozilla.org/show_bug.cgi?id=491668
|
Open Opened 13 years ago Updated 7 years ago
# MathML node clientWidth, clientHeight, offsetLeft & offsetTop values are incorrect
x86
Windows Vista
Not set
normal
NEW
## Attachments
### (2 files)
2.56 KB, application/xhtml+xml Details 1.73 KB, text/html Details
User-Agent: Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 6.0; WOW64; Trident/4.0; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.5.21022; OfficeLiveConnector.1.3; OfficeLivePatch.0.0; .NET CLR 3.5.30729; .NET CLR 3.0.30618)
Build Identifier: Gecko/2009042316 Firefox/3.0.5
MathML nodes expose their rendered x & y positions accurately using node.getBoundingClientRect() but offsetWidth and offsetHeight are undefined.
If it is possible to get the x & y positions that the MathML nodes are rendered on a page it should also be possible to get the width and heights of those nodes.
Reproducible: Always
Steps to Reproduce:
Using Javascript attempt to obtain a MathML nodes offsetWidth and offsetHeight, they will be undefined.
Actual Results:
node.offsetWidth and node.offsetHeight are undefined
Expected Results:
offsetWidth and offsetHeight should contain the rendered width and height of the MathML nodes.
See Firebug bug 1725 (http://code.google.com/p/fbug/issues/detail?id=1725)
as to why this bug should be fixed.
Whiteboard: [firebug-p3]
reassigning somewhere more appropriate, hopefully.
Component: General → MathML
Product: Firefox → Core
QA Contact: general → mathml
> http://groups.google.fr/group/mozilla.dev.tech.mathml/browse_thread/thread/1e252307246e7e94
> getBoundingClientRect seems to provide correct values for all the
> elements inside a math element, but not for the math element itself.
I wonder whether the patch for bug 219873 fixes this bug. Can someone provide a testcase, please?
I was not very clear in my description of this issue when I first logged it.
It is simply that MathML node clientWidth, clientHeight, offsetLeft & offsetTop values are incorrect.
Summary: MathML elements rendered x & y position available but width and height undefined → MathML node clientWidth, clientHeight, offsetLeft & offsetTop values are incorrect
A test case is now attached
Status: UNCONFIRMED → NEW
Ever confirmed: true
Any progress being made on this bug?
(In reply to Joe Java from comment #7)
> Any progress being made on this bug?
(see the remark I left in bug 667567 comment 5)
Maybe Andrii Zui can work on this during his GSoC project as it is related to MathML and javascript. However, there is no guarantees because he already has many bugs to work on.
Personally, I have no ideas how clientWidth, clientHeight, offsetLeft and offsetTop are determined. So someone who really wants to see this bug fixed should do the investigation and submit patches...
CC'ing Andrii
A similar testcase for inline math. Here clientWidth and clientHeight are null.
MathJax's native MathML mode currently has to add some <span> or <mrow> elements around the [itex] to get the correct width.
Why? getBoundingClientRect works correctly, right?
In any case, per spec, offset* are only on HTML elements (which is good, because their definition is very HTML-specific and they return incorrect values anyway because they round to integers).
client* are on all elements, but the spec requires them to be 0 for inline CSS boxes, which is presumably what most math stuff is. And again, it would return incorrect values if used.
Flags: needinfo?(fred.wang)
So actually the MathJax code tries with offsetWidth and scrollWidth (not clientWidth). But I've also been wondering why MathJax does not just use getBoundingClientRect so I'll ask to Davide on the MathJax issue tracker.
Flags: needinfo?(fred.wang)
scrollWidth is defined on all elements, but the definition assumes CSS layout...
getBoundingClientRect won't do what you want for elements inside transforms. But I'm sure you can use getBoxQuads to do whatever you need.
(In reply to Robert O'Callahan (:roc) (Mozilla Corporation) from comment #15)
> getBoundingClientRect won't do what you want for elements inside transforms.
> But I'm sure you can use getBoxQuads to do whatever you need.
Yup, thanks roc.
I am happy to close this if nobody has any objections.
I think the values are only used for MathJax zooming (but there are some hacks to try to correct the errors in values, IIRC). However the zooming will obviously not be important for the TeX-to-MathML server-side conversion on Wikipedia. The following add-on should address the problems for other websites when MathJax is used client-side:
https://addons.mozilla.org/en-US/firefox/addon/mathml-zoom/
Changing the MathJax dependency, since this is really an issue regarding the Zoom UI and not a bug in Gecko's MathML rendering. https://github.com/mathjax/MathJax/blob/master/unpacked/jax/output/NativeMML/jax.js#L433
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2021-12-03 02:08:06
|
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https://en.wikipedia.org/wiki/Morphism_of_algebraic_varieties
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# Morphism of algebraic varieties
In algebraic geometry, a morphism between algebraic varieties is a function between the varieties that is given locally by polynomials. It is also called a regular map. A morphism from an algebraic variety to the affine line is also called a regular function. A regular map whose inverse is also regular is called biregular, and they are isomorphisms in the category of algebraic varieties. Because regular and biregular are very restrictive conditions – there are no non-constant regular functions on projective varieties – the weaker condition of a rational map and birational maps are frequently used as well.
## Definition
If X and Y are closed subvarieties of An and Am (so they are affine varieties), then a regular map ƒ: XY is the restriction of a polynomial map AnAm. Explicitly, it has the form
${\displaystyle f=(f_{1},\dots ,f_{m})}$
where the ${\displaystyle f_{i}}$s are in the coordinate ring of X:
${\displaystyle k[X]=k[x_{1},\dots ,x_{n}]/I,}$
where I is the ideal defining X (note: two polynomials f and g define the same function on X if and only if f − g is in I). The image f(X) lies in Y, and hence satisfies the defining equations of Y. That is, a regular map ${\displaystyle f:X\to Y}$ is the same as the restriction of a polynomial map whose components satisfy the defining equations of ${\displaystyle Y}$.
More generally, a map ƒ:XY between two varieties is regular at a point x if there is a neighbourhood U of x and a neighbourhood V of ƒ(x) such that ƒ(U) ⊂ V and the restricted function ƒ:UV is regular as a function on some affine charts of U and V. Then ƒ is called regular, if it is regular at all points of X.
• Note: It is not immediately obvious that the two definitions coincide: if X and Y are affine varieties, then a map ƒ:XY is regular in the first sense if and only if it is so in the second sense.[1] Also, it is not immediately clear whether regularity depends on a choice of affine charts (it does not.[2]) This kind of a consistency issue, however, disappears if one adopts the formal definition. Formally, an (abstract) algebraic variety is defined to be a particular kind of a locally ringed space. When this definition is used, a morphism of varieties is just a morphism of locally ringed spaces.
The composition of regular maps is again regular; thus, algebraic varieties form the category of algebraic varieties where the morphisms are the regular maps.
Regular maps between affine varieties correspond contravariantly in one-to-one to algebra homomorphisms between the coordinate rings: if ƒ:XY is a morphism of affine varieties, then it defines the algebra homomorphism
${\displaystyle f^{\#}:k[Y]\to k[X],\,g\mapsto g\circ f}$
where ${\displaystyle k[X],k[Y]}$ are the coordinate rings of X and Y; it is well-defined since ${\displaystyle g\circ f=g(f_{1},\dots ,f_{m})}$ is a polynomial in elements of ${\displaystyle k[X]}$. Conversely, if ${\displaystyle \phi :k[Y]\to k[X]}$ is an algebra homomorphism, then it induces the morphism
${\displaystyle \phi ^{a}:X\to Y}$
given by: writing ${\displaystyle k[Y]=k[y_{1},\dots ,y_{m}]/J,}$
${\displaystyle \phi ^{a}=(\phi ({\overline {y_{1}}}),\dots ,\phi ({\overline {y_{m}}}))}$
where ${\displaystyle {\overline {y}}_{i}}$ are the images of ${\displaystyle y_{i}}$'s.[3] Note ${\displaystyle {\phi ^{a}}^{\#}=\phi }$ as well as ${\displaystyle {f^{\#}}^{a}=f.}$[4] In particular, f is an isomorphism of affine varieties if and only if f# is an isomorphism of the coordinate rings.
For example, if X is a closed subvariety of an affine variety Y and ƒ is the inclusion, then ƒ# is the restriction of regular functions on Y to X. See #Examples below for more examples.
## Regular functions
In the particular case that Y equals A1 the regular map ƒ:XA1 is called a regular function, and are algebraic analogs of smooth functions studied in differential geometry. The ring of regular functions (that is the coordinate ring or more abstractly the ring of global sections of the structure sheaf) is a fundamental object in affine algebraic geometry. The only regular function on a projective variety is constant (this can be viewed as an algebraic analogue of Liouville's theorem in complex analysis).
A scalar function ƒ:XA1 is regular at a point x if, in some open affine neighborhood of x, it is a rational function that is regular at x; i.e., there are regular functions g, h near x such that f = g/h and h does not vanish at x.[5] Caution: the condition is for some pair (g, h) not for all pairs (g, h); see Examples.
If X is a quasi-projective variety; i.e., an open subvariety of a projective variety, then the function field k(X) is the same as that of the closure ${\displaystyle {\overline {X}}}$ of X and thus a rational function on X is of the form g/h for some homogeneous elements g, h of the same degree in the homogeneous coordinate ring ${\displaystyle k[{\overline {X}}]}$ of ${\displaystyle {\overline {X}}}$ (cf. Projective variety#Variety structure.) Then a rational function f on X is regular at a point x if and only if there are some homogeneous elements g, h of the same degree in ${\displaystyle k[{\overline {X}}]}$ such that f = g/h and h does not vanish at x. This characterization is sometimes taken as the definition of a regular function.[6]
## Comparison with a morphism of schemes
If X = Spec A and Y = Spec B are affine schemes, then each ring homomorphism φ : BA determines a morphism
${\displaystyle \phi ^{a}:X\to Y,\,{\mathfrak {p}}\mapsto \phi ^{-1}({\mathfrak {p}})}$
by taking the pre-images of prime ideals. All morphisms between affine schemes are of this type and gluing such morphisms gives a morphism of schemes in general.
Now, if X, Y are affine varieties; i.e., A, B are integral domains that are finitely generated algebras over an algebraically closed field k, then, working with only the closed points, the above coincides with the definition given at #Definition. (Proof: If ƒ : XY is a morphism, then writing ${\displaystyle \phi =f^{\#}}$, we need to show
${\displaystyle {\mathfrak {m}}_{f(x)}=\phi ^{-1}({\mathfrak {m}}_{x})}$
where ${\displaystyle {\mathfrak {m}}_{x},{\mathfrak {m}}_{f(x)}}$ are the maximal ideals corresponding to the points x and f(x); i.e., ${\displaystyle {\mathfrak {m}}_{x}=\{g\in k[X]\mid g(x)=0\}}$. This is immediate.)
This fact means that the category of affine varieties can be identified with a full subcategory of affine schemes over k. Since morphisms of varieties are obtained by gluing morphisms of affine varieties in the same way morphisms of schemes are obtained by gluing morphisms of affine schemes, it follows that the category of varieties is a full subcategory of the category of schemes over k.
For more details, see [1].
## Examples
• The regular functions on An are exactly the polynomials in n variables and the regular functions on Pn are exactly the constants.
• Let X be the affine curve ${\displaystyle y=x^{2}}$. Then
${\displaystyle f:X\to \mathbf {A} ^{1},\,(x,y)\mapsto x}$
is a morphism; it is bijective with the inverse ${\displaystyle g(x)=(x,x^{2})}$. Since g is also a morphism, f is an isomorphism of varieties.
• Let X be the affine curve ${\displaystyle y^{2}=x^{3}+x^{2}}$. Then
${\displaystyle f:\mathbf {A} ^{1}\to X,\,t\mapsto (t^{2}-1,t^{3}-t)}$
is a morphism. It corresponds to the ring homomorphism
${\displaystyle f^{\#}:k[X]\to k[t],\,g\mapsto g(t^{2}-1,t^{3}-t),}$
which is seen to be injective (since f is surjective).
• Continuing the preceding example, let U = A1 − {1}. Since U is the complement of the hyperplane t = 1, U is affine. The restriction ${\displaystyle f:U\to X}$ is bijective. But the corresponding ring homomorphism is the inclusion ${\displaystyle k[X]=k[t^{2}-1,t^{3}-t]\hookrightarrow k[t,(t-1)^{-1}]}$, which is not an isomorphism and so the restriction f |U is not an isomorphism.
• Let X be the affine curve x2 + y2 = 1 and let
${\displaystyle f(x,y)={1-y \over x}}$.
Then f is a rational function on X. It is regular at (0, 1) despite the expression since, as a rational function on X, f can also be written as ${\displaystyle f(x,y)={x \over 1+y}}$.
• Let X = A2 − (0, 0). Then X is an algebraic variety since it is an open subset of a variety. If f is a regular function on X, then f is regular on ${\displaystyle D_{\mathbf {A} ^{2}}(x)=\mathbf {A} ^{2}-\{x=0\}}$ and so is in ${\displaystyle k[D_{\mathbf {A} ^{2}}(x)]=k[\mathbf {A} ^{2}][x^{-1}]=k[x,x^{-1},y]}$. Similarly, it is in ${\displaystyle k[x,y,y^{-1}]}$. Thus, we can write:
${\displaystyle f={g \over x^{n}}={h \over y^{m}}}$
where g, h are polynomials in k[x, y]. But this implies g is divisible by xn and so f is in fact a polynomial. Hence, the ring of regular functions on X is just k[x, y]. (This also shows that X cannot be affine since if it were, X is determined by its coordinate ring and thus X = A2.)
• Suppose ${\displaystyle \mathbf {P} ^{1}=\mathbf {A} ^{1}\cup \{\infty \}}$ by identifying the points (x : 1) with the points x on A1 and ∞ = (1 : 0). There is an automorphism σ of P1 given by σ(x : y) = (y : x); in particular, σ exchanges 0 and ∞. If f is a rational function on P1, then
${\displaystyle \sigma ^{\#}(f)=f(1/z)}$
and f is regular at ∞ if and only if f(1/z) is regular at zero.
is a morphism of varieties. If X and Y are affine, then the corresponding ring homomorphism is
${\displaystyle p^{\#}:k[X]\to k[X\times Y]=k[X]\otimes _{k}k[Y],\,f\mapsto f\otimes 1}$
where ${\displaystyle (f\otimes 1)(x,y)=f(p(x,y))=f(x)}$.
## Properties
A morphism between varieties is continuous with respect to Zariski topologies on the source and the target.
The image of a morphism of varieties need not be open nor closed (for example, the image of ${\displaystyle \mathbf {A} ^{2}\to \mathbf {A} ^{2},\,(x,y)\mapsto (x,xy)}$ is neither open nor closed). However, one can still say: if f is a morphism between varieties, then the image of f contains an open dense subset of its closure. (cf. constructible set.)
A morphism ƒ:XY of algebraic varieties is said to be a dominant if it has dense image. For such an f, if V is a nonempty open affine subset of Y, then there is a nonempty open affine subset U of X such that ƒ(U) ⊂ V and then ${\displaystyle f^{\#}:k[V]\to k[U]}$ is injective. Thus, the dominant map ƒ induces an injection on the level of function fields:
${\displaystyle k(Y)=\varinjlim k[V]\hookrightarrow k(X),\,g\mapsto g\circ f}$
where the limit runs over all nonempty open affine subsets of Y. (More abstractly, this is the induced map from the residue field of the generic point of Y to that of X.) Conversely, every inclusion of fields ${\displaystyle k(Y)\hookrightarrow k(X)}$ is induced by a dominant rational map from X to Y.[7] Hence, the above construction determines a contravariant-equivalence between the category of algebraic varieties over a field k and dominant rational maps between them and the category of finitely generated field extension of k.[8]
If X is a smooth complete curve (for example, P1) and if f is a rational map from X to a projective space Pm, then f is a regular map XPm.[9] In particular, when X is a smooth complete curve, any rational function on X may be viewed as a morphism XP1 and, conversely, such a morphism as a rational function on X.
On a normal variety (in particular, a smooth variety), a rational function is regular if and only if it has no poles of codimension one.[10] This is an algebraic analog of Hartogs' extension theorem. There is also a relative version of this fact; see [2].
A morphism between algebraic varieties that is a homeomorphism between the underlying topological spaces need not be an isomorphism (a counterexample is given by a Frobenius morphism ${\displaystyle t\mapsto t^{p}}$.) On the other hand, if f is bijective birational and the target space of f is a normal variety, then f is biregular. (cf. Zariski's main theorem.)
A regular map between complex algebraic varieties is a holomorphic map. (There is actually a slight technical difference: a regular map is a meromorphic map whose singular points are removable, but the distinction is usually ignored in practice.) In particular, a regular map into the complex numbers is just a usual holomorphic function (complex-analytic function).
## Morphisms to a projective space
Let
${\displaystyle f:X\to \mathbf {P} ^{m}}$
be a morphism from a projective variety to a projective space. Let x be a point of X. Then some i-th homogeneous coordinate of f(x) is nonzero; say, i = 0 for simplicity. Then, by continuity, there is an open affine neighborhood U of x such that
${\displaystyle f:U\to \mathbf {P} ^{m}-\{y_{0}=0\}}$
is a morphism, where yi are the homogeneous coordinates. Note the target space is the affine space Am through the identification ${\displaystyle (a_{0}:\dots :a_{m})=(1:a_{1}/a_{0}:\dots :a_{m}/a_{0})\sim (a_{1}/a_{0},\dots ,a_{m}/a_{0})}$. Thus, by definition, the restriction f |U is given by
${\displaystyle f|_{U}(x)=(g_{1}(x),\dots ,g_{m}(x))}$
where gi's are regular functions on U. Since X is projective, each gi is a fraction of homogeneous elements of the same degree in the homogeneous coordinate ring k[X] of X. We can arrange the fractions so that they all have the same homogeneous denominator say f0. Then we can write gi = fi/f0 for some homogeneous elements fi's in k[X]. Hence, going back to the homogeneous coordinates,
${\displaystyle f(x)=(f_{0}(x):f_{1}(x):\dots :f_{m}(x))}$
for all x in U and by continuity for all x in X as long as the fi's do not vanish at x simultaneously. If they vanish simultaneously at a point x of X, then, by the above procedure, one can pick a different set of fi's that do not vanish at x simultaneously (see Note at the end of the section.)
In fact, the above description is valid for any quasi-projective variety X, an open subvariety of a projective variety ${\displaystyle {\overline {X}}}$; the difference being that fi's are in the homogeneous coordinate ring of ${\displaystyle {\overline {X}}}$.
Note: The above does not say a morphism from a projective variety to a projective space is given by a single set of polynomials (unlike the affine case). For example, let X be the conic ${\displaystyle y^{2}=xz}$ in P2. Then two maps ${\displaystyle (x:y:z)\mapsto (x:y)}$ and ${\displaystyle (x:y:z)\mapsto (y:z)}$ agree on the open subset ${\displaystyle \{(x:y:z)\in X\mid x\neq 0,z\neq 0\}}$ of X (since ${\displaystyle (x:y)=(xy:y^{2})=(xy:xz)=(y:z)}$) and so defines a morphism ${\displaystyle f:X\to \mathbf {P} ^{1}}$.
## Fibers of a morphism
The important fact is:[11]
Theorem — Let f: XY be a dominating (i.e., having dense image) morphism of algebraic varieties, and let r = dim X − dim Y. Then
1. For every irreducible closed subset W of Y and every irreducible component Z of ${\displaystyle f^{-1}(W)}$ dominating W,
${\displaystyle \dim Z\geq \dim W+r.}$
2. There exists a nonempty open subset U in Y such that (a) ${\displaystyle U\subset f(X)}$ and (b) for every irreducible closed subset W of Y intersecting U and every irreducible component Z of ${\displaystyle f^{-1}(W)}$ intersecting ${\displaystyle f^{-1}(U)}$,
${\displaystyle \dim Z=\dim W+r.}$
Corollary — Let f: XY be a morphism of algebraic varieties. For each x in X, define
${\displaystyle e(x)=\max\{\dim Z\mid Z{\text{ an irreducible component of }}f^{-1}(f(x)){\text{ containing }}x\}.}$
Then e is upper-semicontinuous; i.e., for each integer n, the set
${\displaystyle X_{n}=\{x\in X\mid e(x)\geq n\}}$
is closed.
In Mumford's red book, the theorem is proved by means of Noether's normalization lemma. For an algebraic approach where the generic freeness plays a main role and the notion of "universally catenary ring" is a key in the proof, see Eisenbud, Ch. 14 of "Commutative algebra with a view toward algebraic geometry." In fact, the proof there shows that if f is flat, then the dimension equality in 2. of the theorem holds in general (not just generically).
## Degree of a finite morphism
Let f: XY be a finite surjective morphism between algebraic varieties over a field k. Then, by definition, the degree of f is the degree of the finite field extension of the function field k(X) over f*k(Y). By generic freeness, there is some nonempty open subset U in Y such that the restriction of the structure sheaf OX to f−1(U) is free as OY|U-module. The degree of f is then also the rank of this free module.
If f is étale and if X, Y are complete, then for any coherent sheaf F on Y, writing χ for the Euler characteristic,
${\displaystyle \chi (f^{*}F)=\deg(f)\chi (F).}$[12]
(The Riemann–Hurwitz formula for a ramified covering shows the "étale" here cannot be omitted.)
In general, if f is a finite surjective morphism, if X, Y are complete and F a coherent sheaf on Y, then from the Leray spectral sequence ${\displaystyle \operatorname {H} ^{p}(Y,R^{q}f_{*}f^{*}F)\Rightarrow \operatorname {H} ^{p+q}(X,f^{*}F)}$, one gets:
${\displaystyle \chi (f^{*}F)=\sum _{q=0}^{\infty }(-1)^{q}\chi (R^{q}f_{*}f^{*}F).}$
In particular, if F is a tensor power ${\displaystyle L^{\otimes n}}$ of a line bundle, then ${\displaystyle R^{q}f_{*}(f^{*}F)=R^{q}f_{*}{\mathcal {O}}_{X}\otimes L^{\otimes n}}$ and since the support of ${\displaystyle R^{q}f_{*}{\mathcal {O}}_{X}}$ has positive codimension if q is positive, comparing the leading terms, one has:
${\displaystyle \operatorname {deg} (f^{*}L)=\operatorname {deg} (f)\operatorname {deg} (L)}$
(since the generic rank of ${\displaystyle f_{*}{\mathcal {O}}_{X}}$ is the degree of f.)
If f is étale and k is algebraically closed, then each geometric fiber f−1(y) consists exactly of deg(f) points.
## Notes
1. ^ Here is the argument showing the definitions coincide. Clearly, we can assume Y = A1. Then the issue here is whether the "regular-ness" can be patched together; this answer is yes and that can be seen from the construction of the structure sheaf of an affine variety as described at affine variety#Structure sheaf.
2. ^ It is not clear how to prove this, though. If X, Y are quasi-projective, then the proof can be given. The non-quasi-projective case strongly depends on one's definition of an abstract variety
3. ^ The image of ${\displaystyle \phi ^{a}}$ lies in Y since if g is a polynomial in J, then, a priori thinking ${\displaystyle \phi ^{a}}$ is a map to the affine space, ${\displaystyle g\circ \phi ^{a}=g(\phi ({\overline {y_{1}}}),\dots ,\phi ({\overline {y_{m}}}))=\phi ({\overline {g}})=0}$ since g is in J.
4. ^ Proof: ${\displaystyle {\phi ^{a}}^{\#}(g)=g(\phi ({\overline {y_{1}}}),\dots ,\phi ({\overline {y_{m}}}))=\phi (g)}$ since φ is an algebra homomorphism. Also, ${\displaystyle f^{\#a}=({\overline {y_{1}}}\circ f,\dots ,{\overline {y_{m}}}\circ f)=f.}$
5. ^ Proof: Let A be the coordinate ring of such an affine neighborhood of x. If f = g/h with some g in A and some nonzero h in A, then f is in A[h−1] = k[D(h)]; that is, f is a regular function on D(h).
6. ^ Hartshorne, Ch. I, § 3.
7. ^ Vakil, Foundations of algebraic geometry, Proposition 6.5.7.
8. ^ Hartshorne, Ch. I,Theorem 4.4.
9. ^ Hartshorne, Ch. I, Proposition 6.8.
10. ^ Proof: it's enough to consider the case when the variety is affine and then use the fact that a Noetherian integrally closed domain is the intersection of all the localizations at height-one prime ideals.
11. ^ Mumford, Ch. I, § 8. Theorems 2, 3.
12. ^ Fulton, Example 18.3.9.
## References
• William Fulton, Intersection theory 2nd edition
• Robin Hartshorne (1997). Algebraic Geometry. Springer-Verlag. ISBN 0-387-90244-9.
• Milne, Algebraic geometry, old version v. 5.xx.
• Mumford, David (1999). The Red Book of Varieties and Schemes: Includes the Michigan Lectures (1974) on Curves and Their Jacobians (2nd ed.). Springer-Verlag. doi:10.1007/b62130. ISBN 354063293X.
• Igor Shafarevich (1995). Basic Algebraic Geometry I: Varieties in Projective Space (2nd ed.). Springer-Verlag. ISBN 0-387-54812-2.
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2021-01-22 11:05:53
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https://chemistry.stackexchange.com/questions/80961/why-do-polarizable-nucleophiles-favor-sn2-over-e2/80963
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# Why do polarizable nucleophiles favor SN2 over E2?
Weakly basic ions and polarizable bases favor $$\mathrm{S_{N}2}$$ over $$\mathrm{E2}$$. Why? The argument used that polarizable nucleophiles form bonds earlier, stabilizing the transition state, can be applied to either $$\mathrm{S_{N}2}$$ or $$\mathrm{E2}$$.
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2022-01-19 05:51:35
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-3-solving-multi-step-equations-practice-and-problem-solving-exercises-page-99/56
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## Algebra 1
$$x=\frac{13}{3}$$
$$8(\frac{3x}{8}-1)=8(\frac{5}{8})$$ the problem is in this step, the 8 needs to be distributed to the 1 as well $$3x-8=5$$ $$3x=13$$ $$x=\frac{13}{3}$$
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2019-09-20 20:34:39
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https://mathspace.co/textbooks/syllabuses/Syllabus-406/topics/Topic-7189/subtopics/Subtopic-96060/?activeTab=interactive
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# Use the distributive property
## Interactive practice questions
Which of the following area models match the statement $6\times35$6×35?
Select all the correct options.
$30$30 $3$3 $6$6
A
$30$30 $5$5 $6$6
B
$32$32 $3$3 $6$6
C
$30$30 $5$5 $8$8
D
$30$30 $3$3 $6$6
A
$30$30 $5$5 $6$6
B
$32$32 $3$3 $6$6
C
$30$30 $5$5 $8$8
D
Easy
Less than a minute
We want to find $2\times45$2×45.
We want to find $9\times27$9×27.
We want to find $5\times76$5×76.
### Outcomes
#### NA3-1
Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.
#### NA3-7
Generalise the properties of addition and subtraction with whole numbers
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2022-01-27 17:14:11
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https://blog.bi0s.in/2019/03/18/Crypto/Block-Ciphers/confidenceteaser19-countmein/
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# Count me in - Teaser CONFidence CTF 2019
tl;dr Known plaintext attack on a multithreaded AES-CTR
Challenge Points: 59
Challenge Solves:
Solved by: v3ct0r
This challenge is a customised version of AES-CTR. Here is the challenge script:
import multiprocessing
from Crypto.Cipher import AES
from secret import key, flag
counter = 0
aes = AES.new(key, AES.MODE_ECB)
def chunk(input_data, size):
return [input_data[i:i + size] for i in range(0, len(input_data), size)]
def xor(*t):
from functools import reduce
from operator import xor
return [reduce(xor, x, 0) for x in zip(*t)]
def xor_string(t1, t2):
t1 = map(ord, t1)
t2 = map(ord, t2)
return "".join(map(chr, xor(t1, t2)))
pad_byte = 16 - len(data) % 16
def worker_function(block):
global counter
result = xor_string(block, key_stream)
counter += 1
return result
def distribute_work(worker, data_list, processes=8):
pool = multiprocessing.Pool(processes=processes)
result = pool.map(worker, data_list)
pool.close()
return result
def encrypt_parallel(plaintext, workers_number):
results = distribute_work(worker_function, chunks, workers_number)
return "".join(results)
def main():
plaintext = """The Song of the Count
You know that I am called the Count
Because I really love to count
I could sit and count all day
Sometimes I get carried away
I count slowly, slowly, slowly getting faster
Once I've started counting it's really hard to stop
Faster, faster. It is so exciting!
I could count forever, count until I drop
1! 2! 3! 4!
1-2-3-4, 1-2-3-4,
1-2, i love couning whatever the ammount haha!
1-2-3-4, heyyayayay heyayayay that's the sound of the count
I count the spiders on the wall...
I count the cobwebs in the hall...
I count the candles on the shelf...
When I'm alone, I count myself!
I count slowly, slowly, slowly getting faster
Once I've started counting it's really hard to stop
Faster, faster. It is so exciting!
I could count forever, count until I drop
1! 2! 3! 4
1-2-3-4, 1-2-3-4, 1,
2 I love counting whatever the
ammount! 1-2-3-4 heyayayay heayayay 1-2-3-4
That's the song of the Count!
""" + flag
encrypted = encrypt_parallel(plaintext, 32)
print(encrypted.encode("hex"))
if __name__ == '__main__':
multiprocessing.freeze_support()
main()
After observing the script carefully, we can see that the each block has been parallelly encrypted using multithreading , but here lies the vulnerability!
Since multithreading is used, a few set of blocks are encrypted with the same nonce. Here comes the use of
the given plaintext, since we have many plaintext and ciphertext block pairs we can recover the encrypted
nonce.
It is basically a xor challenge where the encrypted nonces are the keys which when XORed with the plaintext
gives us the ciphertext. So basically the part of the flag maybe XORed with the same key block with which
some of the previous blocks have been XORed. But we don’t which of these is used so we try all of them.
There are two steps involved in solving this challenge:
1. Recovering the key sets from the known plaintext-ciphertext pairs
We can get that by XORing the pt-ct pairs and dividing it into chunks of 16 since each block is 16.
2. Getting the flag
Find which key set used to XOR with the flag. Just try all of them and there is nothing else to do but getting
the flag.
Here is the exploit script:
from Crypto.Cipher import AES
import string
from count import plaintext as pt
chars = string.ascii_lowercase+string.digits+"{}_"
def chunk(input_data, size):
return [input_data[i:i + size] for i in range(0, len(input_data), size)]
def xor(a,b):
from itertools import cycle
return ''.join(chr(ord(i)^ord(j)) for i,j in zip(a,cycle(b)))
# Possible Keys
k = xor(ct,pt)[:len(pt)]
keys = chunk(k,16)
ctflag = ct[-64:]
flag = ''
for key in set(keys):
out = chunk(xor(ctflag,key),16)
for i in out:
if all(char in chars for char in i[:8]):
flag+=i
print flag
Running the above script gives out the flag as:
p4{at_the_end_of_the_day_you_can_only_count_on_yourself}!
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2020-05-30 02:03:09
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http://isotopicmaps.org/pipermail/tmql-wg/2004-March/000202.html
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# [tmql-wg] Result set requirements
Robert Barta [email protected]
Wed, 3 Mar 2004 07:25:23 +1000
On Tue, Mar 02, 2004 at 12:17:13PM +0100, Rani Pinchuk wrote:
> Maybe I should emphasis that I speak about complexity in learning the
> language - so in the syntax of the language:
>
> If the query returns ALWAYS one list of strings, there is no use to
> include that fact in the query. The query language should not have at
> all the select clause (which makes its syntax simpler).
....
> If the query include apart from that also the ability to return XML,
> the select clause should include extra syntax to support also returning
> of XML (like DTD or like the way you do it).
Rani,
Additional syntax makes it easier to learn the language? I am not sure
whether a
SELECT <<ATTENTION, ATTENTION HERE NOW COMES XML>>
<xml-here/>
compared to
return
<xml-here/>
is simpler to learn :-)
And - if we ever dare to walk into the snakepit of XML schema
languages - we should be as agnostic as possible. I would find it
great if we could exclude DTDs and friends completely.
> With algorithms I mean that the data from the query can still go
> through an algorithm that is written in the query language (which
> obviously becomes very generic language at that point).
So you would not include functions implementing algorithms in a
query language to keep it simple?
If so, then SQL is not simple, because it has functions:
SELECT 1+2*3;
> So for example, if the data from the query is list of words, the
> algorithm can be an algorithm that returns instead the types of the
> words (verb/noun/adj/adv etc). To have that, you must program a nice
> algorithm that identify the words into their groups.
Hmmm, if something COULD be done with the query language, why MUST
I tell the developer that he SHOULD NOT do it within the query language,
but with some external program?
> > In case of AsTMa we DO NOT expand strings. It is true that we write
> > the query as a string (well, of course), but the XML (as well as the
> > list and Topic Map constructor) are no text templates, but
> > internalized.
> I am not sure I understood that. What I meant is that the syntax of
> AsTMa suggests that the {$a} and {$b} are strings.
Expressions like {$a} or {$a/bn} are interpreted in the context. If
I use it like so
<coffee id="{$a/bn}">need more</coffee> then obviously I want a string to be inserted. If {$a/bn} can be
evaluated to a string, then AsTMa? can do that. If I simply ask for
{$a/bn} then the context could also say "return it as information item" (I have not implemented this yet because I never needed it). Why use a different syntax to hammer home such a simple feature? So in case of <coffee id="{$a/bn}">need more</coffee>
this looks as if this is like a template expansion, because XML is
string-oriented. But
({$a/bn}, {$a/oc(homepage)}, ...)
is handled differently, and so it is the case with creating a TM:
# stupid way to copy all coffee-cup topics
# get a list
function addict () as list return {
forall $c [ * (coffee-cups) return {$c}
}
# get a map, consisting of these topics only
function addict () as map return {
forall $c [ * (coffee-cups) return {$c}
}
> So the way it is implemented is not that relevant for my
> argument. To put it more clear - could the very exact syntax of
> AsTMa be implemented using templates and expanding strings?
Hmm, _everything_ can be implemented expanding strings. Every machine
in the Chomsky hierarchy expands strings. I am not sure where you
are going with this.
Fact is, that
return
<coffee id="{$a/bn}">I really need more now</coffee> CAN BE implemented using internal data structures (DOM fragments, whatever), so that I DO NOT HAVE TO do it with "string handling" (which is much slower). It also means that the application gets this in a already parsed form. Which is a hell of a difference to templating. I found the design of XQuery quite instructive, btw. > > Sure, sure. But how much have we actually gained and how much lost? > We gained the separation between the languages. You don't have the mix. > It is readable and maintainable. We lost the same we loose when we use > for example OO - some speed in development (at least in the first > phase), and some speed in performance. What you suggest (creating templates for designers or to encapsulate XML snippets) may be all very well. I have seen very stupid designers and I have worked with VERY talented ones. So I would assume for some projects massive templating is ok, for others may be not. To hard-wire the "you-must-factor-everything-into-some-template" would mean to patronize. I would not like to go that path. If someone wants templates, then he/she should pick one of the 10000000000 templating packages and put it an top of TMQL. Others will choose not to use templates at all, but will use TMQL at some higher abstraction level. These people will generate abstraction layers and will offer objects and object classes to the application. These people will not be interested in lists or XML, they will use map results to create their abstraction, I would assume. > > But isn't that just an XMLish notation of AsTMa? itself? And what happens > > now in case of nested queries? > > > > <albums>{ > > forall$t [ $a (album) > > bn:$bn ] in $m > > return > > <album id="{$a}">{$bn} > > { > > forall [ (is-producer-of) > > album:$a
> > producer: $p ] in$m
> > return
> > <producer>{$p/bn}</producer> > > } > > </album> > > } > > </albums> > > You can implement it as follows: > > <albums> > <while condition="loop_over_albums"> > <album id="$album_id">$album_name > <while condition="loop_over_producer_of_albun"> > <producer>$producer_name</producer>
> </while>
> </album>
> </while>
> </albums>
>
> In your program you query for all the albums ids.
>
> loop_over_albums contains code that gets the next album id, and run a
> query over all the producers of that album.
How does the outer loop know what the values are to iterate over? How
does the inner loop know what the values are to iterate over? Is this
somehow communicated behind the scenes? And what happens when the
template-engineer changes the above to
<albums>
<x:while condition="loop_over_albums">
<album id="$album_id">$album_name
<x:while condition="loop_over_producer_of_albun">
<producer>$producer_name</producer> </x:while> <x:while condition="loop_over_producer_of_albun"> <producer>$producer_name</producer>
</x:while>
</album>
</x:while>
</albums>
Would the functions realize what is going on?
Making this separation systematical you will have to introduce A LOT
of communication between parts of the template AND between the
template and the functions in the background doing the "business
logic" stuff.
Introducing MORE communication patterns does NOT MAKE the language
simpler, IMHO. If you can isolate particular use cases and create
organisation with your particular programmers, that is ok.
As a general approach to a language I am not convinced.
> I guess the above looks a bit strange when you are not used to it.
> But when your code becomes bigger, this separation makes it much more
> clear.
I have used and programmed XML taglibs. They work very well for a
defined set of tags, a defined set of parameters and for things which
are good to isolate from the environment. For instance
some other XML stuff here
<my:weather zip="12345" country="australia"/>
some other XML stuff here
They are a pain in the ... earlobe when you start to connect things
and pass around values.
> Try to think about a scenario when you want to make changes after the
> application is already done (and big) - how do you change the XML?
Change the XML. :-)
> How do you change the queries (suppose the topic map structure is
> changed).
Change the queries. :-)
> In the separation approach this will be much easier because there is
> no mix in the languages and it is all more readable.
Again, if someone wants to have this separation, then this can be
implemented on top of TMQL. If someone else knows what he is doing,
then we should not patronize him.
> > > So I don't try to avoid running the same query. I try to avoid
> > > hard-coding the same query.
> >
> > Isn't that exactly what I said above? That 'some pattern P' should
> > be in one place and is reused with different constructors?
> Yes, but for me the sentence "I try to avoid mixing those two" is
> more important. Exactly like with error messages in English. It is
> better to put them outside of your code
It may depend. If someone writes a single script of 30 lines and
starts pulling error message codes from external files then I would
seriously ask some questions (after shooting of course :-)
In a big system where maybe internationalization is an issue, then
this might be standard procedure. Still, there are many different
approaches. None of them has been built into, say, Oracle. It can be
built on-top of it, though.
\rho
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2018-11-17 03:00:16
|
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|
https://undergroundmathematics.org/trigonometry-compound-angles/r8525/solution
|
Review question
# Can we prove $\sin a \sin b \leq \sin^2\frac{1}{2}(a+b)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8525
## Solution
Prove that, for all values of $a$ and $b$, $\sin a \sin b \le \sin^2\frac{1}{2}(a+b).$
We know that $\cos (x+y) = \cos x\cos y-\sin x\sin y \quad \text{and} \quad \cos (x-y) = \cos x\cos y+\sin x\sin y.$
Thus $\sin x\sin y = \dfrac{1}{2}(\cos (x-y)-\cos (x+y)).$
This means $\sin \left(\dfrac{a+b}{2}\right)\sin \left(\dfrac{a+b}{2}\right) = \dfrac{1}{2}(1-\cos (a+b)).$
We also have $\sin a\sin b = \dfrac{1}{2}(\cos (a-b)-\cos (a+b)).$
Since $\cos(a-b) \leq 1$, on comparing the last two equations we see $\sin a\sin b \leq \sin^2 \left(\dfrac{a+b}{2}\right)$.
Show further that, if $a$, $b$, $c$ and $d$ all lie between $0$ and $\pi$, then $\sin a \sin b \sin c \sin d \leq \left( \sin\frac{1}{4}(a+b+c+d)\right)^4;$
If $0 \leq a, b, c, d \leq \pi,$ then
$0 \leq \sin a, \sin b, \sin c, \sin d \leq 1.$
Therefore
$0 \leq \sin a \sin b \leq \sin^2{\frac{1}{2}(a+b)}$
and
$0 \leq \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)}.$
Because the left hand side of each inequality is greater than or equal to zero, we can multiply the inequalities to get
$$$\sin a \sin b \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)} \sin^2{\frac{1}{2}(c+d)}. \label{eq:1}$$$ Using the inequality from a), we also have \begin{align} \sin{\frac{1}{2}(a+b)} \sin{\frac{1}{2}(c+d)} &\leq \sin^2{\frac{1}{2}\bigl[\frac{1}{2}(a+b)+\frac{1}{2}(c+d)\bigr]} \notag \\ \quad &= \sin^2{\frac{1}{4}(a + b + c +d)}. \label{eq:2} \end{align}
We know
$0 \leq \frac{1}{2}(a+ b), \frac{1}{2}(c+d) \leq \pi,$
and so
$0 \leq \sin \frac{1}{2}(a+b) \sin \frac{1}{2}(c+d),$
which means we can square $\eqref{eq:2}$ to get
$$$\sin^2{\frac{1}{2}(a+b)} \sin^2 \frac{1}{2}(c+d) \leq \sin^4{\frac{1}{4}(a+b+c+d)}. \label{eq:3}$$$
Combining $\eqref{eq:1}$ and $\eqref{eq:3}$ we get
\begin{align*} \sin a \sin b \sin c \sin d &\leq \sin^4\frac{1}{4}(a+b+c+d) \\ & \quad = \bigl(\sin \frac{1}{4}(a+b+c+d)\bigr)^4, \end{align*}
as required.
… and, by writing $d=\frac{1}{3}(a+b+c)$, deduce that $\sin a \sin b \sin c \le \left( \sin\frac{1}{3}(a+b+c)\right)^3.$
Setting $d = \frac{1}{3}(a+b+c),$ we see that
$a + b + c + d = \frac{4}{3}(a+b+c),$
and we can deduce that
$$$\sin a \sin b \sin c \sin \frac{1}{3}(a+b+c) \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^4. \label{eq:4}$$$
From $0 \leq a, b, c \leq \pi$, we deduce that $0 \leq \frac{1}{3}(a+b+c) \leq \pi$ and, except in the cases $a,b,c=0$ and $a,b,c=\pi$, we have the strict inequality $0 < \frac{1}{3}(a+b+c) < \pi$, and so
$\sin \frac{1}{3}(a+b+c) > 0.$
Given this, we can divide $\eqref{eq:4}$ by $\sin \frac{1}{3}(a+b+c)$ to get
$\sin a \sin b \sin c \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^3,$
as required.
The final task is to consider the cases $a = b = c = 0$ and $a = b = c = \pi.$ In both of these cases, both sides of the inequality are $0$ (as $\sin 0 = \sin \pi = 0$), and so the inequality still holds.
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2018-01-17 10:58:59
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http://mathoverflow.net/questions/53514/seiberg-witten-equation-on-s2-times-s1
|
Seiberg-Witten equation on S^2\times S^1
What are the irreducible solutions of Seiberg-Witten equation on S^2\times S^1? Thanks.
-
The equations depends on a Riemannian metric and on a perturbation term (a closed 2-form). The usual metric on $S^2\times S^1$ has positive scalar curvature, which implies that there are no irreducible solutions when the 2-form vanishes.
This is a consequence of the Weitzenboeck formula for the Dirac operator, as was observed by Witten in his paper Monopoles and four-manifolds. (The 3-dimensional SW equations are the 4-dimensional SW equations on the product with $S^1$, with the additional condition of translation-invariance in the circle-direction.)
|
2015-09-03 15:24:03
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https://indico.cern.ch/event/344485/contributions/1744597/
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# ICRC2015
July 29, 2015 to August 6, 2015
World Forum
Europe/Amsterdam timezone
## Reacceleration of electrons in supernova remnants
Aug 4, 2015, 4:00 PM
1h
Amazon Foyer (World Forum)
### Amazon Foyer
#### World Forum
Churchillplein 10 2517 JW Den Haag The Netherlands
Board: 176
Poster contribution CR-TH
### Speaker
Prof. Martin Pohl (DESY)
### Description
The radio spectra of many shell-type supernova remnants show deviations from those expected on theoretical grounds.}{In this paper we determine the effect of stochastic reacceleration on the spectra of electrons in the GeV band and at lower energies, and we investigate whether reacceleration can explain the observed variation in radio spectral indices. We explicitely calculated the momentum diffusion coefficient for 3 types of turbulence expected downstream of the forward shock: fast-mode waves, small-scale non-resonant modes, and large-scale modes arising from turbulent dynamo activity. After noting that low-energy particles are efficiently coupled to the quasi-thermal plasma, a simplified cosmic-ray transport equation can be formulated and is numerically solved. Only fast-mode waves can provide momentum diffusion fast enough to significantly modify the spectra of particles. Using a synchrotron emissivity that accurately reflects a highly turbulent magnetic field, we calculated the radio spectral index and find that soft spectra with index $\alpha \le -0.6$ can be maintained over more than 2 decades in radio frequency, even if the electrons experience reacceleration for only one acceleration time. A spectral hardening is possible but considerably more frequency-dependent. The spectral modification imposed by stochastic reacceleration downstream of the forward shock depends only weakly on the initial spectrum provided by, e.g., diffusive shock acceleration at the shock itself.
Registration number following "ICRC2015-I/" 82 -- not specified --
### Primary author
Prof. Martin Pohl (DESY)
### Presentation materials
There are no materials yet.
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2023-03-21 18:02:09
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http://math.stackexchange.com/questions/88341/branch-cut-and-logz-derivative
|
# Branch cut and $\log(z)$ derivative
I'm a little confused about the branch cut thing. Given an entire functions $f(z),g(z),h(z)$, $z\in \mathbb C$, such that $f(x)=g(x)+h(x)$ for all $x\in \mathbb R$, $f$ and $g+h$ doesn't vanish on $\mathbb R$ . I take the $\log$ for both sides then differentiate and get:
$$\log f(x)=\log(g(x)+h(x))$$ $$\frac{f'(x)}{f(x)}=\frac{g'(x)+h'(x)}{g(x)+h(x)}$$
I thought this is correct, but I was asked about "what is the branch cut I used here?", and I don't know what does this mean, and what is the answer for this question! Any help!?
Edit: I also have the same problem with the following case:
If $f(x)=e^{g(x)}$ then $\log f(x)=g(x)$, and $\frac{f'(x)}{f(x)}=g'(x)$, also a branch cut issue!
-
If $f=k$ and $f$ doesn't vanish then it is true that $f'/f = k'/k$, simply because $f=k$ and $f'=k'$; no logs needed. It is also true that whenever a logarithm of $f$ exists, i.e. a function $p$ such that $\exp(p)=f$, then it is true that $p'=f'/f$, a consequence of the chain rule. I'm not sure exactly what the question is here. Maybe you should ask the person who asked you for clarification of what they are asking. They might just be hinting at the fact that logarithms don't always exist globally and aren't unique, so care should be taken in what "log" means. – Jonas Meyer Dec 4 '11 at 20:33
One thing to note: There is no continuous logarithm defined on the range of a nonconstant entire function. However, for every nonvanishing analytic $f:U\to \mathbb C$, where $U$ is simply connected, there exists an analytic $p:U\to \mathbb C$ such that $\exp(p)=f$; such a $p$ is called a logarithm of $f$. – Jonas Meyer Dec 4 '11 at 20:46
Sometimes, problems arise when trying to extend continuous functions from $\mathbb{R}$ to $\mathbb{C}$. For example, consider the complex-valued function $f(z) = \sqrt{z}$. Consider the following (poorly-drawn) diagram:
If your domain $D$ sweeps out a quarter-circle of radius $r$, the the image $f(D)$ sweeps out half the angle (an eighth-circle) with radius $\sqrt{r}$. Suppose then that your domain $D$ was an entire circle in the domain.
Consider the blue-red interface in the domain. As the image suggests, $\sqrt{z}$ is discontinuous there, because as you transition across the line $x = 0$ in the domain, you jump half-way across the circle in the image. To deal with this, we extend $\sqrt{z}$ to a new domain: two complex planes "glued" together. It looks something like this.
Now the function is continuous. Instead of transitioning directly from (1) to (2) in the image plane, we "move through the cut" to the blue (3) on the second complex plane, so that $\sqrt{z}$ is continuous. Then we travel around the second plane, jumping from the red (4) to the red (2) in the first complex. We have gone around twice in the domain, and once in the image, and our issues with continuity are resolved. To be more precise about the gluing, the domain actually looks like this:
Now, as you can expect, you need to glue three complex planes together to make $\sqrt[3]{z}$ continuous, and so on, etc. Unfortunately, for the complex logarithm, no number of complex planes in the domain will make the image continuous. Thus, you need to define the complex logarithm on an infinite collection of complex planes, "spiraling" away in either direction. It looks like this.
Recall the example with the square root earlier. We had two copies of the complex plane, and their images under $\sqrt{z}$ were different. In the first copy, the image was the right semi-circle, and in the second copy, it was the left semicircle. Thus, when working with the complex square root, one needs to specify which copy of the complex plane one wishes to work with. Each such copy is called a branch (symmetrically, you can think of it as two copies of $\sqrt{z}$ defined on the same complex plane). Similarly, when working with the complex log, you need to talk about which of the infinitely many complex planes in the domain you wish to work with, and so you must specify which branch you are using. The popular choice is the so-called "principal branch":
Log$(z) = ln|z| + i$Arg$(z)$.
-
...very nice pictures, btw – Grigory M Dec 4 '11 at 21:19
Those are nice Riemann surfaces you've shown... :) – Ｊ. Ｍ. Dec 5 '11 at 1:10
Wow! Nice pictures indeed. – Nick Strehlke Feb 1 '12 at 7:42
The problem here is that $\log(z)$ is a multi-valued function on the complex plane. How can you see this? First of all recall that $\log(z)$ is usually defined to be an inverse of $e^{z}$ (i.e. $log(e^{z})=z$). Working out what this means shows that one way of defining the $\log$ is to take $$\log(z)=\ln|z| +iArg(z)$$ but then you run into the problem that $Arg(z)$ has an infinite number of values, just add or subtract a multiple of $2 \pi$. Now consider $\log(1)$. We want $\log$ to be continuous (and even analytic), so it should satisfy $$\lim_{y \to 0^{+}} \log(1+iy) =\lim_{\theta \to 2\pi^{-}} \log(1e^{i\theta}) = 0$$ The first one appears to be true, but the above definition of log shows that in the second limit $Arg(z)$ approaches $2\pi$ so that $$\lim_{\theta \to 2\pi^{-}} \log(1e^{i\theta}) = 2\pi i$$ The problem here is that the domain of the log function is not simply connected. A way to fix this problem is to remove all of the negative real axis and zero so that you can't go all the way around $0$ and force 2 different values of $Arg(z)$. Removing these points gives you an analytic function on $\mathbb{C}\setminus \{x\in \mathbb{R}|x\leq 0\}$ and this is called a branch cut because it singles out a single valued function of $\log$ from an infinite number of choices for the range of $Arg(z)$. One can also make a branch cut by removing any ray that connects zero and infinity.
In your problem, we are not sure of what domain your final statement is true on as we dont know what branch of the $\log$ function you have chosen.
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2014-03-07 11:17:46
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https://bradzzz.gitbooks.io/ga-seattle-dsi/content/dsi/dsi_06_trees_methods/5.2-lesson/readme.html
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# Communicating results
Week 6| Lesson 5.2
### LEARNING OBJECTIVES
After this lesson, you will be able to:
• Analyze costs and benefits when presenting data
• Decide how to present it based on audience
• Find common mistakes in visualizations
• Correct such mistakes
• Visualize results in tableau
### STUDENT PRE-WORK
Before this lesson, you should already be able to:
• Demonstrate ability to perform simple EDA using Tableau
• Recall the defining criteria of a confusion matrix
• Illustrate the principles of a strong classification report
### INSTRUCTOR PREP
Before this lesson, instructors will need to:
• Read in / Review any dataset(s) & starter/solution code
• Generate a brief slide deck
• Prepare any specific materials
• Provide students with additional resources
### LESSON GUIDE
TIMING TYPE TOPIC
5 min Opening Opening:
15 min Review Review: Model Quality
10 min Discussion Discussion: Cost Benefit Analysis
5 min Introduction Intro: The criteria for a good visualization
10 min Guided-practice Guided practice: Find the error
10 min Guided-practice Guided practice: Correct the error
25 min Ind-practice Independent practice: Visualizing model results in Tableau
5 min Conclusion Conclusion
## Opening: (5 min)
In Week 4, Lesson 5.1 we discussed how to present data to stakeholders that are less technical than we are. We will extend that topic now considering a few additional tools and in particular how to communicate results when using ensemble models.
Check: Do you expect huge differences in how to handle presentation of results for ensemble models?
Answer: not really, presentation of results should be almost model independent
## Review: Model Quality (15 min)
Let's review the methods we have learned so far to assess model quality.
### Classification problems
Check: Can you name few ways to judge the quality of a classification model?
Accuracy Score, Precision, Recall, F1-Score, Confusion Matrix
Check: For each of these ways, can you give the definition?
Instructor note: help them give definition, then write definition on whiteboard. You can review definitions here
Check: Can you think of situations you have encountered where you would choose one or the other?
Debate: Finesse VS communicability
### Regression problems
Check: Can you name few ways to judge the quality of a classification model?
RMSE, R2
Check: For each of these ways, can you give the definition?
Instructor note: help them give definition, then write definition on whiteboard. You can review definitions here
Check: Can you think of situations you have encountered where you would choose one or the other?
Debate: Finesse VS communicability
Check: What graphical tools have we encountered to explain model quality?
Learning curves, Validation curves
## Discussion: Cost Benefit Analysis (10 min)
One tool that complements the confusion matrix is cost-benefit analysis, where you attach a value to correctly and incorrectly predicted data.
Like the Precision-Recall trade off, there is a balancing point to the probabilities of a given position in the confusion matrix, and the cost or benefit to that position. This approach allows you to not only add a weighting system to your confusion matrix, but also to speak the language of your business stakeholders (i.e. communicate your values in dollars!).
Consider the following marketing problem: As a data scientist working on marketing spend, you've build a model that reduces user churn -- the number of users who decide to stop paying for a product -- through a marketing campaign. Your model generates a confusion matrix with the following probabilities (these probabilities are calculated as the value in that position over the sum of the sample):
TP: 0.2 FP: 0.2
FN: 0.1 TN: 0.5
In this case:
• The benefit of a true positive $B_{TP}$ is the retention of a user ($10 for the month) • The cost of a false positive$C_{FP}$is the spend of the campaign per user (-$0.05)
• The cost of a false negative $C_{FN}$ (someone who could have retained if sent the campaign) is, effectively, 0 (we didn't send it... but we certainly didn't benefit!)
• The benefit of a true negative $B_{TN}$ is 0: No spend on users who would have never retained.
To calculate Cost-Benefit, we'll use this following function:
$$p(TP) B{TP} + p(TN) B{TN} + p(FP) C{FP} + p(FN) C{FN}$$ which for our marketing problem, comes out to this:
$$(0.2 \times 10) + (0.5 \times 0) - (0.2 \times 0.05) - (0.1 \times 0)$$ or \$1.99 per user targeted.
Instructor note: this could be done in small groups or pairs or with the whole class.
Think about precision, recall, and cost benefit analysis from the above problem to answer the following questions:
• How would you rephrase the business problem if your model was optimizing toward precision? i.e., How might the model behave differently, and what effect would it have?
• How would you rephrase the business problem if your model was optimizing toward recall?
• What would the most ideal model look like in this case?
• Can you think of business situations where different stakeholders would take different decisions on what metric to optimize? For example, stakeholders with competing interests may decide to weigh false positives or false negatives differently. Can you think of a concrete example?
Answer: E.g. model to predict cancer: Health insurance would like to minimize false positives, patient would like to minimize false negatives.
## Intro: The criteria for a good visualization (5 min)
As we have learned in week 4, visualizations and presentations should be:
• Simplified
• Easy to interpret
• Clearly Labeled
Check: Can you give a few examples of good/poor visualizations?
• Simplified: At most, you'll want to include figures that either explain a variable on its own (explaining the sample or population), or explain that variable's relationship with a target. If your model used a data transformation (like the natural log of a variable), just visualize the original data, as log functions involve an additional layer of explanation.
• Easily interpretable: Any stakeholder looking at the figure should be seeing the exact same thing you're seeing. A good test for this: share the visual with others less familiar with the data, and see if they came to same conclusion. How long did it take them?
• Clearly Labeled: Take the time to clearly label your axis, title the plot, and double check your scales - especially if the figures should be comparable. If you're showing two graphs side by side, they should follow the same Y axis.
When building visuals for another audience, ask yourself these questions:
• Who: Who is my target audience for this visual?
• How: How does my project affect this audience? How might they interpret (or misinterpret) the data?
## Guided practice: Find the error (10 min)
WTFViz is a hilarious collection of visualizations done wrong. For the next activity, form groups of 3-4 people and go to the WTFVIZ website. Use the first 5 minutes to find 3 visualizations that you find particularly intriguing. Look for visualizations that are:
• Hilarious: find at least one visualization that is so wrong you can't stop laughing
• Tricky: find at least one visualization that is almost right and where it's hard to spot why it's wrong
• Boring: find a visualization that is not particularly exciting
In the following 5 minutes each group will present to the rest of the class its 3 visualizations together with their explanations for choosing them.
Instructor note: have each group name stand up and come out to present. Have your laptop ready on wtfviz website so they can browse it there.
## Guided practice: Correct the Errors (10 min)
Back in your groups, discuss how to amend the error: work in your groups to either actually correct or break down the steps of what you would do to correct the problems identified in each of their 3 selections. How could you have presented the results in a more compelling and clearer way?
Communicate a few suggestions to the rest of the class.
Instructor note: Give them a few minutes to discuss and then a few minutes fow presenting their corrections to the class.
## Independent practice: Visualizing model results in Tableau (25 min)
Tableau is a very powerful and easy to use tool that allows us to create nice visualizations. In the assets folder are some of the results of the labs you've been conducting in the past week. See if you can reproduce the visualizations below.
In addition create at least one more visualization using the data you can find in the asset folder and share your attempts within your team of the previous activity.
## Conclusion (5 min)
In this class we learned how the stakeholders should always be central when considering which visualization to produce.
Check: What did you learn? Share the things you discovered today.
Telling a sound story is as important as building an effective model, and crafting a nice story requires time and attention to detail.
Check: how would you allocate the time in a data science project between the following phases:
• data collection and cleaning
• model building
• visualization
Answer: it depends, but 1. and 3. should take more time than 2. (although it's hardly ever the case).
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2020-09-26 11:57:27
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https://proxieslive.com/existence-and-uniqueness-of-entropy-solutions-for-a-scalar-conservation-law/
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# Existence and uniqueness of entropy solutions for a scalar conservation law
Consider the conservation law
$$(\ast) \qquad u_t + \partial_x(u^\alpha) = 0$$ where $$\alpha > 0$$.
For what values of $$\alpha$$ is it known that there exists a (unique) entropy solution for the initial value problem associated to $$(\ast)$$?
In particular, I’m interested in knowing what happens in the case $$\alpha \in (1,3]$$.
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2020-01-22 12:49:45
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https://elteoremadecuales.com/removable-singularity/
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# Removable singularity
For instance, the (unnormalized) sinc function {displaystyle {text{sinc}}(z)={frac {sin z}{z}}} has a singularity at z = 0. This singularity can be removed by defining {displaystyle {text{sinc}}(0):=1,} which is the limit of sinc as z tends to 0. The resulting function is holomorphic. In this case the problem was caused by sinc being given an indeterminate form. Taking a power series expansion for {textstyle {frac {sin(z)}{z}}} around the singular point shows that {displaystyle {text{sinc}}(z)={frac {1}{z}}left(sum _{k=0}^{infty }{frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}right)=sum _{k=0}^{infty }{frac {(-1)^{k}z^{2k}}{(2k+1)!}}=1-{frac {z^{2}}{3!}}+{frac {z^{4}}{5!}}-{frac {z^{6}}{7!}}+cdots .} Formally, if {displaystyle Usubset mathbb {C} } is an open subset of the complex plane {displaystyle mathbb {C} } , {displaystyle ain U} a point of {displaystyle U} , and {displaystyle f:Usetminus {a}rightarrow mathbb {C} } is a holomorphic function, then {displaystyle a} is called a removable singularity for {displaystyle f} if there exists a holomorphic function {displaystyle g:Urightarrow mathbb {C} } which coincides with {displaystyle f} on {displaystyle Usetminus {a}} . We say {displaystyle f} is holomorphically extendable over {displaystyle U} if such a {displaystyle g} exists.
Contents 1 Riemann's theorem 2 Other kinds of singularities 3 See also 4 External links Riemann's theorem Riemann's theorem on removable singularities is as follows: Theorem — Let {displaystyle Dsubset mathbb {C} } be an open subset of the complex plane, {displaystyle ain D} a point of {displaystyle D} and {displaystyle f} a holomorphic function defined on the set {displaystyle Dsetminus {a}} . The following are equivalent: {displaystyle f} is holomorphically extendable over {displaystyle a} . {displaystyle f} is continuously extendable over {displaystyle a} . There exists a neighborhood of {displaystyle a} on which {displaystyle f} is bounded. {displaystyle lim _{zto a}(z-a)f(z)=0} .
The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 are trivial. To prove 4 ⇒ 1, we first recall that the holomorphy of a function at {displaystyle a} is equivalent to it being analytic at {displaystyle a} (proof), i.e. having a power series representation. Define {displaystyle h(z)={begin{cases}(z-a)^{2}f(z)&zneq a,\0&z=a.end{cases}}} Clearly, h is holomorphic on {displaystyle Dsetminus {a}} , and there exists {displaystyle h'(a)=lim _{zto a}{frac {(z-a)^{2}f(z)-0}{z-a}}=lim _{zto a}(z-a)f(z)=0} by 4, hence h is holomorphic on D and has a Taylor series about a: {displaystyle h(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+cdots ,.} We have c0 = h(a) = 0 and c1 = h'(a) = 0; therefore {displaystyle h(z)=c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+cdots ,.} Hence, where z ≠ a, we have: {displaystyle f(z)={frac {h(z)}{(z-a)^{2}}}=c_{2}+c_{3}(z-a)+cdots ,.} However, {displaystyle g(z)=c_{2}+c_{3}(z-a)+cdots ,.} is holomorphic on D, thus an extension of f.
Other kinds of singularities Unlike functions of a real variable, holomorphic functions are sufficiently rigid that their isolated singularities can be completely classified. A holomorphic function's singularity is either not really a singularity at all, i.e. a removable singularity, or one of the following two types: In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number {displaystyle m} such that {displaystyle lim _{zrightarrow a}(z-a)^{m+1}f(z)=0} . If so, {displaystyle a} is called a pole of {displaystyle f} and the smallest such {displaystyle m} is the order of {displaystyle a} . So removable singularities are precisely the poles of order 0. A holomorphic function blows up uniformly near its other poles. If an isolated singularity {displaystyle a} of {displaystyle f} is neither removable nor a pole, it is called an essential singularity. The Great Picard Theorem shows that such an {displaystyle f} maps every punctured open neighborhood {displaystyle Usetminus {a}} to the entire complex plane, with the possible exception of at most one point. See also Analytic capacity Removable discontinuity External links Removable singular point at Encyclopedia of Mathematics Categories: Analytic functionsMeromorphic functionsBernhard Riemann
Si quieres conocer otros artículos parecidos a Removable singularity puedes visitar la categoría Analytic functions.
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2023-03-28 10:56:47
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http://www.ck12.org/algebra/Order-of-Operations/lesson/Evaluate-Numerical-and-Variable-Expressions-Using-the-Order-of-Operations-MSM8/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Order of Operations
## The order of operations outlines the order by which certain operatives should take place in a mathematical expression for it to remain true
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Evaluate Numerical and Variable Expressions Using the Order of Operations
### [Figure1] License: CC BY-NC 3.0
Jeb enters his name into a draw to win a handheld game system from his local electronics megastore. A week later he visits the store for the draw and his name is picked! In order to claim the prize, he must correctly answer a skill testing question.
The question is:
Jeb has to answer the skill testing question without the use of technology. How can Jeb answer this question correctly to claim the prize?
In this concept, you will learn to use the order of operations to solve numerical and variable expressions.
### Order of Operations
In mathematics, you will often hear the word evaluate. Before you begin, it is important for you to understand what the word evaluate means. When you evaluate a mathematical sentence, you figure out the value of the number sentence. Often times you think of evaluating as solving, and it can be that, but more specifically, evaluating is figuring out the value of a sentence.
In mathematics, you can evaluate different types of number sentences. Sometimes you will be working with equations and other times you will be working with expressions. First you need to know the difference between an equation and an expression.
An equation is a statement that two mathematical expressions have the same value. An equation has an equal sign such that the quantity on the left side of the equal sign is equal to the quantity on the right side of the equal sign. This means both sides of the equation stand for the same number. is an equation. It states that have the same value.
An expression is a general term in mathematics for a group of numbers, symbols and variables representing numbers and operations. You evaluate an expression to figure out the value of the mathematical statement itself, you are not trying to make one side equal another, as with an equation. is an expression. There is no equal sign.
Two eighth grade math students evaluated the expression . Macy’s answer was ten and Cole’s answer was eight. The students you are asked to write their step by step solutions on the board.
Macy‘s Solution:
Cole‘s Solution:
It appears that each student performed the indicated operations in different orders. Of course, there cannot be two correct solutions for the same expression. There is actually a specific order in which operations must be performed.
The order of operations is a rule that tells you which operation you need to perform and the order in which it must be done to achieve the correct answer. The order of operations is often called PEMDAS and each of the letters represents one part of the rule. P: parenthesis and grouping symbols; E: exponents; M: multiplication; D: division; A: addition; S: subtraction. MD (multiplication and division) are performed in the order they appear in the expression from left to right. AS (addition and subtraction) are also performed in the order they appear in the expression from left to right.
Looking at the two solutions for evaluating the expression who has the correct answer?
Macy
Cole
Macy evaluated the expression by performing the addition first, followed by multiplication and division. Cole performed the multiplication and division first, then addition. Macy simply completed the operations as they appeared from left to right. Cole completed the multiplication and division as they appeared from left to right and then performed the addition as his final step. Cole used the order of operations rule, PEMDAS, and his answer is correct.
This is called evaluating or simplifying a numerical expression. A numerical expression is an expression made up only of numbers and operations.
In addition to numbers, expressions can also have letters. The letters in an expression are called variables. These variables represent an unknown quantity. When an expression is written with a variable in it, you call it a variable expression.
A variable expression is evaluated using the order of operations in the same way as a numerical expression is evaluated. In this Concept there will be a value given for the variable and you will substitute it into the variable expression before evaluating the expression.
Evaluate the variable expression:
First, substitute the given value for in the expression:
Remember that, used this way, parenthesis mean you are multiplying 2 times (5).
Now apply the order of operations (PEMDAS) and continue to evaluate the expression.
First, divide: and write the new expression.
Next, multiply: since this is the first multiplication from left to right.
Next, multiply: and write the new expression.
Next, add: and write the new expression.
Then, subtract:
Now let’s add in the grouping symbols. The grouping symbols that you will be working with are brackets [ ] and parenthesis ( ). According to the order of operations (PEMDAS), you perform all operations inside the grouping symbols BEFORE any other operation in the list.
Evaluate the numerical expression
First, perform the operation in the parenthesis: and write the new expression.
Next, multiply to clear the parenthesis and write the new expression.
Next, add: and write the new expression.
Then, subtract:
Brackets can be used to group more than one operation. When you see a set of brackets, remember that brackets are a way of grouping numbers and operations with other groups already in parenthesis. Always evaluate grouping symbols from the innermost to the outermost.
Evaluate the numerical expression.
First, evaluate the innermost grouping symbols: the parenthesis.
Multiply: and write the new expression.
Next, perform the operation inside the brackets.
Add: and write the new expression.
Next, add: and write the new expression.
Then, subtract:
### Examples
#### Example 1
Earlier, you were given a problem about Jeb and his (almost) prize.
Remember, when solving a problem like this skill testing question, you need to complete the indicated operations in the order of PEMDAS.
Notice there are parenthesis within the brackets. You need to evaluate from the inside out. So start with the parenthesis. According to PEMDAS multiplication comes before subtraction, and this is just as true within a group as without.
First, multiply: and write the new expression.
Next, subtract: and write the new expression.
Next, subtract: and write the new expression.
Next, multiply: and write the new expression.
Then, divide:
Jeb needs to answer 20 to claim his prize.
#### Example 2
Evaluate the expression
First, multiply and write down the new expression.
Next, divide and write down the new expression.
Next add and write down the new expression.
#### Example 3
Evaluate the variable expression
First, substitute into the expression.
Next, perform the operation inside the parenthesis.
Multiply: and write the new expression.
Next, multiply: to clear the parenthesis and write the new expression.
Next add: and write the new expression.
Then, subtract:
#### Example 4
Evaluate the numerical expression.
Remember to evaluate the grouping symbols from the innermost to the outermost.
First, perform the operation inside the parenthesis.
Multiply: and write the new expression.
Next, perform the operation inside the brackets.
Add: and write the new expression.
Then multiply
Next, add: and write the new expression.
Finally, subtract
#### Example 5
Evaluate the variable expression.
First substitute into the expression.
First, multiply: and write the new expression.
Next, divide: and write the new expression.
Then, multiply again: and write another new expression.
Next, add: and write the new expression.
Finally, subtract :
### Review
Evaluate each numerical expression using the order of operations.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Evaluate each variable expression. Remember to use PEMDAS when necessary.
11.
12.
13.
14.
15.
To see the Review answers, open this PDF file and look for section 1.4.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Brackets
Brackets [ ], are symbols that are used to group numbers in mathematics. Brackets are the 'second level' of grouping symbols, used to enclose items already in parentheses.
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Grouping Symbols
Grouping symbols are parentheses or brackets used to group numbers and operations.
nested parentheses
Nested parentheses describe groups of terms inside of other groups. By convention, nested parentheses may be identified with other grouping symbols, such as the braces "{}" and brackets "[]" in the expression $\{ 3 + [ 2 - ( 5 + 4 ) ] \}$. Always evaluate parentheses from the innermost set outward.
Numerical expression
A numerical expression is a group of numbers and operations used to represent a quantity.
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
PEMDAS
PEMDAS (Please Excuse My Daring Aunt Sally) is a mnemonic device used to help remember the order of operations: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.
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2016-12-09 05:19:24
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https://mathoverflow.net/questions/281095/formulation-of-contour-variational-problem
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# Formulation of contour variational problem
I am having difficulty formulating a problem, which involves optimizing a contour shape, into a well-posed variational form that would give a reasonable answer.
Within a bounded region on the $xy$ plane, say $x\in[-x_{0},x_{0}], y\in[-y_{0},y_{0}]$, we have a continuous scalar field $H=H(x,y)$. Both the field and the geometry of the problem exhibit no variations in the $z$ direction (i.e. $\partial/\partial z=0)$. On the plane within the specified region, there exists a closed loop (contour) $g(x,y)=0$ that encloses and defines a planar area $A$ that is penetrated by the field $H$. It is known from the physics of the problem that varying the shape of the contour $g$ can result in extremising the functional
$$J(y):=\iint_{A} H(x,y)dA$$
and I need to find the optimal shape $g$ of the contour, for a given $H$ function that is nontrivial ($\neq 0)$ and captured area $A$ that is nonzero. In writing $J$ here I assumed that $x$ is the independent variable and $y$ is dependent on it, to draw the contour shape.
Anticipating that the closed contour function will be most likely expressible in parameteric form $(x(t),y(t))$, and since classical variational formulations I am familiar with usually deal with paths rather than areas, I tried to write the functional in terms of the contour (instead of the area) as follows, using Green's theorem:
$$J=\iint_{A} H(x,y)dA=\iint_{A} \left(\frac{\partial F_{y}}{\partial x} -\frac{\partial F_{x}}{\partial y}\right)dA=\oint_{g}(F_{x}dx+F_{y}dy) =\int_{t=0}^{2\pi}(F_{x}\dot{x}+F_{y}\dot{y})dt$$
where $\boldsymbol{F}=F_{x}\hat{i}+F_{y}\hat{j}$ is some vector field whose curl may be defined to give $H$ (assuming we can find such field), dotted symbols like $\dot{x}$ denoting derivate in parameter $t\in[0,2\pi]$, and $\oint_{g}$ denoting integral around closed contour $g$. So, we can think of the Lagrangian of this problem as $L(x,y,\dot{x},\dot{y}):=F_{x}\dot{x}+F_{y}\dot{y}$.
The problem now is, if I don't impose any further contraints, the two Euler-Lagrange equation here (in $t$ now as independent variable and both $x$ and $y$ as dependents) give the same result (instead of two independent answers), which says that $H=0$. I plugged in different test fields $H$, and this is always the answer.
If I try to improve the formulation by imposing a constraint that $\iint_{A}dA=A_{0}$ to make the area nonzero constant, thus:
$$\frac{1}{2}\int_{0}^{2\pi}(x\dot{y}-y\dot{x})dt=A_{0} \Rightarrow \int_{0}^{2\pi}\left[ \frac{x\dot{y}}{2}-\frac{y\dot{x}}{2}-\frac{A_{0}}{2\pi} \right]dt=0,$$
giving a new (constrained) Lagrangian as $L(x,y,\dot{x},\dot{y},\lambda):=F_{x}\dot{x}+F_{y}\dot{y}+\frac{\lambda}{2}(x\dot{y}-y\dot{x}-\frac{A_{0}}{2\pi}),$
then, again, the two Euler-Lagrange equations in $x$ and $y$ give the same answer, basically that $H=\lambda$, where $\lambda$ is Lagrange's multiplier for this contraint.
What is wrong with my formulation, and how do I make it well posed for this problem so I can proceed?
• What is the "physics of the problem"? – Keith McClary Sep 15 '17 at 4:51
• @KeithMcClary [1/2] The problem represents a case of resonance, where two types of energies become equal at resonance, and they are (each) defined as integrals: $\iint_{A_{1}}W_{1}dA$ and $\iint_{A_{1}}W_{2}dA$, where $A_{1}$ is the initial domain. So, resonance will occur when their difference is zero, hence the functional $\iint_{A_{1}}WdA$, where $W=W_{1}-W_{2}$. We know already the solution of $W$ to this problem of domain $A_{1}$. The issue now comes when this field is faced with a new geometric feature inserted in the domain, perturbing it slightly but not sufficiently to change ... – user135626 Sep 15 '17 at 23:17
• @KeithMcClary [2/3] ... its eigenfunctions, and hence $W$ is assumed to be unchanged. The inserted feature is a "hole" of region $A\subset A_{1}$ and thus the new domain is now $A_{1}-A$, giving new functional as $\iint_{A_{1}-A}WdA=\iint_{A_{1}}WdA-\iint_{A}WdA=0-\iint_{A}WdA$. It is observed by physical experiment that, although W (related to the resonant eigenfunctions) is approximatly fixed, the eignevalues (wavelength) will shift, subject to choice of SHAPE of the contour enclosing this "hole", which may be tweaked to reach a resonant point same as the original wavelength (eigenvalue)... – user135626 Sep 15 '17 at 23:24
• @KeithMcClary [3/3] ... even if the area $A$ is fixed (say $A_{0}$). We are thus seeking the optimal shape of the contour enclosing this "hole" region, which will restore the functional to zero. Clearly, this contour shape will reflect the way the present fields $W_{1}, W_{2}$ (or their difference $W$) vary in space. So, I tried to attack the problem as I wrote in the question above, hoping to reach an answer. I also tried plugging in example values for $W_{1},W_{2}$, such as the simple case of $B \sin k x$ and $C \cos k x$, but no reasonable progress is reached, as explained above. – user135626 Sep 15 '17 at 23:31
• What exactly do you mean by the extremum of a vector-valued function? (Or is $H$ actually real-valued despite you call it a "field"?). From what you wrote in comments, it seems like you actually have a scalar function $H$ and just want a domain over which the integral is $0$ but then you start talking about "eigenfunctions that do not change" and that makes me totally confused... – fedja Sep 18 '17 at 8:12
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2019-10-14 14:26:11
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http://math.stackexchange.com/questions/134574/carmichael-numbers
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# Carmichael numbers?
$a^{p-1} \equiv 1 \pmod p$
Why do Carmichael numbers prevent Fermat's Little Theorem from being a guaranteed test of primality? Fermat' Little theorem works for any $a$ such that $1≤a\lt p$, where $p$ is a prime number. Carmichael numbers only work for $a$'s coprime to $N$ (where $N$ is the modulus). Doesn't this mean that for some non-coprime $a$ the Carmichael number will fail the test? Therefore if every $a$ is tested, a Carmichael number wouldn't pass.
-
Google (weak) probable prime and pseudoprime and then let us know if you stll have a question. – Bill Dubuque Apr 20 '12 at 23:15
@Farhad Yusufali: Perfectly correct. But if we view the matter probabilistically, it may be that numbers between $1$ and $m-1$ that are not relatively prime to $m$ are rare, so a test that relied on bumping into such a number would be impractical. – André Nicolas Apr 20 '12 at 23:20
If $a$ is composite and not Carmichael then the proportion of $a \bmod m$ such that $a^{m-1} \not\equiv 1 \bmod m$ is at least $50\%$, which is substantial. But if $a$ is composite and Carmichael then the proportion of $a \bmod m$ such that $a^{m-1} \not\equiv 1 \bmod m$ can be very small; in fact the only such $a$ are precisely those sharing a common factor greater than 1 with $m$, so finding them randomly amounts to finding a factor of $m$ randomly. For instance, if $m = 294409$ then the proportion of $a \bmod m$ such that $a^{m-1} \not\equiv 1 \bmod m$ is $4.9\%$. – KCd Apr 21 '12 at 1:59
If you think testing every $a$ is a good idea, then you can ignore all nonobvious math and just do trial division. – KCd Apr 21 '12 at 2:00
@KCd - I am obviously not going to test every $a$ - that would be trivial. I was just curious why the test would fail IF you tested every $a$ – user26649 Apr 21 '12 at 16:26
If the goal of the Fermat Primality test is to guarantee that a number is prime, then testing against all possible $a$ is no better than simply trying to divide our number by all primes.
In particular, if it were easy to find a number that is not coprime to $n$, then it's easy to factor $n$ and so we wouldn't need to use any sort of Fermat primality.
But you are correct. Notably, if $p | n$, then doing the test$\mod p$ would yield $0$.
In case you looked at the Wikipedia article on Carmichael numbers, your question may have resulted from the sentence "Since Carmichael numbers exist, [the Fermat] primality test cannot be relied upon to prove the primality of a number". This is a bad formulation, since the Fermat primality test isn't meant to be used as proof of the primality of a number, but as a probabilistic test that is very likely to prove the compositeness of any composite number. It's that latter use that Carmichael numbers interfere with. As the article on the Fermat primality test shows, for numbers $n$ other than primes and Carmichael numbers, at least half of all numbers coprime to $n$ are Fermat witnesses, i.e. let the test prove compositeness. Thus the Fermat primality test serves its function well for non-Carmichael numbers, whereas for Carmichael numbers with relatively high prime factors, such as $8911=7\cdot19\cdot67$, the probability of proving compositeness with a randomly chosen number $\lt n$ is significantly reduced (roughly from $1$ in $2$ to in this case $1$ in $7$ per test).
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2015-07-30 18:43:37
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http://golem.ph.utexas.edu/~distler/blog/archives/2010_06.shtml
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## June 27, 2010
### Crib Notes
It is difficult to get a man to understand something when his livelihood depends on him not understanding it.
— Upton Sinclair
You probably don’t want to read this post. It has an intended audience of one — my erstwhile coauthor, Skip Garibaldi.
Skip and I wrote a paper, last year, which proved that Garrett Lisi’s “Theory of Everything” (or any $E_8$-based variant thereof) could not yield chiral fermions (much less 3 Standard Model generations worth of fermions). Anyone with training in high energy theory instantly apprehends the consequence that this “theory” cannot, therefore, have anything remotely to do with the real world. Unfortunately, if your PhD is in pure mathematics (or, apparently, in hydrodynamics), this may not be immediately obvious to you.
Skip has the unenviable task of lecturing on our paper at a workshop, next week, with Garrett in attendance. (Well, OK, the workshop is in lovely Banff Alberta, so perhaps some envy is warranted.) This post is designed to help him fill in the dots. It contains only material which — to someone schooled in high energy theory — is of an embarrassingly elementary nature.
You have been warned!
Posted by distler at 9:17 PM | Permalink | Followups (13)
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2014-04-24 10:59:46
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https://www.physicsforums.com/threads/srednicki-problem-3-4.728480/
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# Homework Help: Srednicki Problem 3.4
1. Dec 15, 2013
### WannabeNewton
1. The problem statement, all variables and given/known data
Recall that $T(a)^{-1}\varphi(x)T(a) = \varphi(x - a)$ where $T(a) = e^{-iP^{\mu}a_{\mu}}$ is the space-time translation operator and $P^{\mu}$ is the 4-momentum operator.
(a) Let $a^{\mu}$ be infinitesimal and derive an expression for $[P^{\mu},\varphi]$.
(b) Show that the time component of your result is equivalent to the Heisenberg equation of motion $\dot{\varphi} = i[H,\varphi]$.
(c) For a free field, use the Heisenberg equation to derive the Klein-Gordon equation.
(d) Define a spatial momentum operator $\mathbf{P} = -\int d^{3}x (\pi \nabla\varphi)$. Use the canonical commutation relations to show that $\mathbf{P}$ obeys the relation you derived in part (a).
(e) Express $\mathbf{P}$ in terms of $a(\mathbf{k})$ and $a^{\dagger}(\mathbf{k})$.
2. Relevant equations
The Hamiltonian density is given by $\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla \varphi)^2 + \frac{1}{2}m^2 \varphi^2$ where $\pi$ is the field conjugate momentum as usual.
The equal time commutation relations for the field and the field conjugate momentum are $[\varphi(x,t),\varphi(x',t)] = 0$, $[\pi(x,t),\pi(x',t)] = 0$, and $[\varphi(x,t),\pi(x',t)] = i\delta^{3}(\mathbf{x} - \mathbf{x'})$.
You probably already noticed but $\varphi$ is a real field. Also I'll use the notation $\varphi' \equiv \varphi(x',t)$ with $\pi'$ and $\nabla'$ defined similarly just to make things cleaner.
3. The attempt at a solution
(a) $\varphi(x^{\mu} - a^{\mu}) = \varphi(x^{\mu}) - a_{\mu}\partial^{\mu}\varphi +O(a^2)$ and $T(a^{\mu}) = I - ia_{\mu}P^{\mu}+O(a^2)$ so $T^{-1}\varphi T = (I + ia_{\mu}P^{\mu})(\varphi - ia_{\mu}\varphi P^{\mu}) = \varphi(x^{\mu}) + ia_{\mu}[P^{\mu},\varphi]+O(a^2)$ hence $i[P^{\mu},\varphi] = -\partial^{\mu}\varphi$ since $a_{\mu}$ was arbitrary.
(b) $\mu = 0$ gives $i[H,\varphi] = -\partial^{0}\varphi = \dot{\varphi}$
(c) We have $2[\varphi,H] = [\varphi,\int d^3{x'}(\pi'^2 + (\nabla'\varphi')^2 + m^2 \varphi'^2)] = \int d^{3}x'([\varphi,\pi'^2] + [\varphi,(\nabla'\varphi')^2] + m^2[\varphi,\varphi'^2])$; the equal time commutator can be pulled into the integral since the commutator is evaluated at the field point $x$ whereas the integral is over $x'$.
Now $[\varphi,\nabla'\varphi'] = \nabla'[\varphi,\varphi'] = 0$ hence $[\varphi,(\nabla'\varphi')^2] = 0$. Similarly $[\varphi,\varphi'^2] = 0$. This leaves us with $2[\varphi,H] = \int d^{3}x'[\varphi,\pi'^2]$.
We have $[\varphi,\pi'^2]= i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi' + (\pi' \varphi)\pi' - (\pi' \varphi) \pi' + i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi' = 2i\delta^{3}(\mathbf{x} - \mathbf{x}')\pi'$ hence $[\varphi,H] = i\int d^{3}x' \delta^{3}(\mathbf{x} - \mathbf{x}')\pi' = i\pi$. The Heisenberg equation of motion thus gives $\dot{\varphi} = \pi$.
Similarly, $2[\pi,H] = \int d^{3}x'([\pi,\pi'^2] + [\pi,(\nabla'\varphi')^2] + m^2[\pi,\varphi'^2])$; $[\pi,\pi'] = 0$ so $[\pi,\pi'^2] = 0$ and $[\pi,\varphi'^2] = - 2i\delta^{3}(\mathbf{x} - \mathbf{x'})\varphi'$.
Furthermore $[\pi,(\nabla'\varphi')^2] = (\pi \nabla'\varphi')\cdot\nabla'\varphi' - \nabla'\varphi'\cdot((\nabla'\varphi')\pi) \\= (\nabla'\varphi')\cdot\pi (\nabla'\varphi') - i(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi' - (\nabla'\varphi')\cdot\pi(\nabla'\varphi') - i\nabla'\varphi'\cdot(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'})) \\= -2i(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi'$
This leaves us with $[\pi,H] = -i\int d^{3}x'\delta^{3}(\mathbf{x} - \mathbf{x'})\varphi' - i\int d^{3}x'(\nabla'\delta^{3}(\mathbf{x} - \mathbf{x'}))\cdot\nabla'\varphi' = i(\nabla^2\varphi - m^2\varphi)$.
So now here's the part that I'm not sure about. The Heisenberg equation of motion derived above is valid for any scalar quantum field $\varphi$ right? In the derivation itself I think all I used was the fact that $\varphi$ is an arbitrary scalar field, making no reference to a specific scalar field. So it works for the field conjugate momentum $\pi$ as well? If so then we get $i[\pi,H] = -\dot{\pi} = -\ddot{\varphi}$ and at the same time $i[\pi,H] = -(\nabla^2\varphi - m^2\varphi)$ so that $\ddot{\varphi} - \nabla^2\varphi + m^2\varphi = -\partial^{\mu}\partial_{\mu}\varphi+m^2\varphi = 0$ as desired.
Does this all check out i.e. are all my calculations sound? I just want to make sure the calculations are sound so that I can move on to parts (d) and (e). Also if you know of a faster calculation to get the KG equation from the Heisenberg equation of motion then I'd appreciate it if you could show it. Thanks in advance!
2. Dec 15, 2013
### WannabeNewton
Actually part (d) was really short so I may as well post that too; hopefully it checks out :)
We have by definition $P^{i} = -\int d^{3}x'(\pi' \partial'^{i}\varphi')$ and from part (a) above we also have $i[P^{\mu},\varphi] = -\partial^{\mu}\varphi$.
Now $[\pi' \partial'^{i}\varphi',\varphi] \\= \pi'((\partial'^{i}\varphi') \varphi) - (\varphi\pi')\partial'^{i}\varphi' \\= \pi'\partial'^{i}(\varphi'\varphi)-\pi'(\varphi (\partial'^{i}\varphi')) - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' \\= \pi'\partial'^{i}[\varphi',\varphi] - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' = - i\delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi'$
hence $i[P^{i},\varphi] = -i\int d^{3}x'[(\pi' \partial'^{i}\varphi'),\varphi] = -\int d^{3}x' \delta^{3}(\mathbf{x} - \mathbf{x'})\partial'^{i}\varphi' = -\partial^{i}\varphi$ as desired.
3. Dec 15, 2013
### WannabeNewton
Also I'm stuck on part (e) so if someone could help me out with part (e) that would be swell :)
The free scalar field solution is $\varphi(x) =\int d\tilde{k} [a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]$ where $d\tilde{k} = \frac{d^3 k}{(2\pi)^3 2\omega}$ is the Lorentz invariant measure.
Then $\pi = \dot{\varphi} = \int d\tilde{k} [ -a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]i\omega$
and $\nabla \varphi = \int d\tilde{k} [a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} - a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}]i\mathbf{k}$
so we have $\mathbf{P} = \int d\tilde{k}d\tilde{k}' d^{3}x[ -a(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} + a^{\dagger}(\mathbf{k})e^{i(-\mathbf{k}\cdot \mathbf{x} + \omega t)}] [a(\mathbf{k'})e^{i(\mathbf{k'}\cdot \mathbf{x} - \omega' t)} - a^{\dagger}(\mathbf{k'})e^{i(-\mathbf{k'}\cdot \mathbf{x} + \omega' t)}]\omega\mathbf{k'}\\ = \int d\tilde{k}d\tilde{k}' d^{3}x[-a(\mathbf{k})a(\mathbf{k'})e^{i(\mathbf{k}+\mathbf{k'})\cdot \mathbf{x}} e^{-i(\omega + \omega') t} + a^{\dagger}(\mathbf{k})a(\mathbf{k'})e^{i(\mathbf{k'}-\mathbf{k})\cdot \mathbf{x}} e^{-i(\omega - \omega') t} \\+a(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{i(\mathbf{k}-\mathbf{k'})\cdot \mathbf{x}} e^{-i(\omega' - \omega) t} - a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{-i(\mathbf{k}+\mathbf{k'})\cdot \mathbf{x}} e^{i(\omega + \omega') t}]\omega\mathbf{k'}\\ = -(2\pi)^3 \int d\tilde{k}d\tilde{k}'\delta^{3}(\mathbf{k} +\mathbf{k'})[a(\mathbf{k})a(\mathbf{k'})e^{-i(\omega + \omega') t}+ a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{i(\omega + \omega') t}]\omega\mathbf{k'}\\+(2\pi)^3 \int d\tilde{k}d\tilde{k}'\delta^{3}(\mathbf{k} - \mathbf{k'})[a^{\dagger}(\mathbf{k})a(\mathbf{k'})e^{-i(\omega - \omega') t} +a(\mathbf{k})a^{\dagger}(\mathbf{k'})e^{-i(\omega' - \omega) t}]\omega\mathbf{k'}\\ = -\frac{1}{2} \int d\tilde{k}[a(\mathbf{k})a(\mathbf{-k})e^{-2i\omega t}+ a^{\dagger}(\mathbf{k})a^{\dagger}(\mathbf{-k})e^{2i\omega t}]\mathbf{k}\\+\frac{1}{2} \int d\tilde{k}[a^{\dagger}(\mathbf{k})a(\mathbf{k}) +a(\mathbf{k})a^{\dagger}(\mathbf{k})]\mathbf{k}$
I don't know how to simplify this any further. The second term in the final equality looks like it should stay intact since the Hamiltonian for the free scalar field is given by $H = \frac{1}{2} \int d\tilde{k}[a^{\dagger}(\mathbf{k})a(\mathbf{k}) +a(\mathbf{k})a^{\dagger}(\mathbf{k})]\omega$.
I'm guessing it's the first term in the final equality that has to vanish somehow. However I can't seem to make it vanish. I tried using a parity inversion $\mathbf{k}\rightarrow -\mathbf{k}$ and the fact that $[a(\mathbf{k}),a(\mathbf{-k})] = [a^{\dagger}(\mathbf{k}),a^{\dagger}(\mathbf{-k})] = 0$ but the lone $\mathbf{k}$ in the integrand picks up an overall sign that cancels out with the overall sign picked up by the Lorentz invariant measure $d\tilde{k}$, giving me back the original integral even though what I need is an overall negative of the original integral in order for it to vanish. Thanks in advance for any help.
EDIT: Oops nevermind! I forgot that the bounds of the integral also flip under the parity inversion so I have to flip them back and that gives me the overall negative sign that I need to make the first term vanish.
Last edited: Dec 15, 2013
4. Dec 15, 2013
Thanks wbn!
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2018-07-21 05:01:24
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https://galotib.palmolive2day.com/ra-milikans-photoelectric-experiments-that-supported-einsteins-theory-77867bb.html
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# Ra milikans photoelectric experiments that supported einsteins theory
Today, the accepted value is 1. Working with Fletcher, Millikan showed that the charge of the droplets were always a whole number multiple of 1. Fortunately, these electrons aren't moving at relativistic speeds! The fact that 16 years went by before Einstein won a Nobel Prize for his ground-breaking work on the photoelectric effect, reflects how long it took the scientific world to accept that radiant energy is quantized.
I have in recent years been subjecting this experiment to some searching experimental tests from a variety of viewpoints, and have been led to the conclusion that, whatever its origin, it actually represents very accurately the behavior… of all the substances with which I have worked.
And while I wasn't annoyed enough to clear my schedule of more important day job stuff in order to respond in blog time, I do want to say something to highlight Millikan's other great work.
It is an extremely important and unjustly overlooked bit of work, though, which cleared up some controversies about earlier measurements, and played an important role in locking up the case for quantum theory. This was consistent with his hypothesis that the charge on the electron was the same as that found in hydrogen.
InMillikan wrote: Robert Millikan, the very man who showed that the equation really did work, would have nothing to do with its physical interpretation.
Some of the electrons would be shaken so hard, the theory went, that eventually they'd be tossed out altogether. If the body is charged to a positive potential V0 and is surrounded by conductors at zero potential, and if V0 is just large enough to prevent the loss of electricity, it follows that where e is the electronic charge All the light of a particular frequency comes in little bullets of the same energy, equal to the frequency multiplied by Planck's constantand that's the key to understanding the photoelectric effect.
In his mid-twenties Compton hatched a theory of the intensity of X-ray reflection from crystals that gave a powerful tool for studying the crystallographic arrangement of electrons and atoms in a substance.
The glass vessel shown below was evacuated; one of the metal samples mounted on the wheel W at the center was rotated to bring it next to the knife K, which scraped it clean; then the sample was rotated degrees to face light shining through aperture O.
Hertz found that by shining ultraviolet light onto metal electrodeshe could lower the voltage needed to make sparks hop between the electrodes. In this figure from Lenard's paper, the potential voltage is plotted on the horizontal axis, and the current on the vertical axis.
The key breakthrough came in and was published the following year. If, as physicists believed, the photoelectric effect followed from an interaction between electrons and electromagnetic waves, then intensifying the radiation ought to shake the electrons in the metal surface harder and so shoot them out with more energy.
This seems like it ought to be easy to explain, but classical physics turned out not to do a very good job of it, as I explained back in the pre-ScienceBlogs days of this blog. Yet, incredibly, Einstein never won a Nobel Prize for relativity. One of the great improvements was the use of oil drops instead of the cloud of water drops that Thomson used.
Yet the semi-corpuscular theory by which Einstein arrived at his equation seems at present to be wholly untenable. Light from source L shines onto plate U. Hertz was creating and detecting waves using spark gaps-- discharging a current through a small gap to make a spark in the air produced a blast of radiation, which he then detected by its ability to trigger a spark between another pair of electrodes some distance away.
As Einstein said in"There are Robert Millikan, the very man who showed that the equation really did work, would have nothing to do with its physical interpretation. In the conclusion of the March paper, he opens by saying: Kadisch, and he thanks Walter Whitney for doing spectroscopy to determine the wavelengths of light from his source.
He also used optical filters to eliminate the effects of stray light at wavelengths other than the one he was trying to test-- a precaution that anybody who's ever tried this in a modern physics course can tell you is absolutely essential-- and investigated the effects of contacts between the metal being studied and the different wires used to connect to the measuring apparatus in great detail.
NO electrons are produced if the frequency of the light waves is below a critical value. Having established that the model worked, Millikan then put it to use, in the March paper.
By modern standards, those people would probably be co-authors. Contributors Overview Although Hertz discovered the photoelectron init was not until that a theory was proposed that explained the effect completely. Thus, only electrons ejected with enough kinetic energy to get up this potential hill would contribute to the current.
Experimentalists railed at the prospect of what Einstein's equation of the photoelectric effect implied. Light could manifest itself as particles. I have in recent years been subjecting this experiment to some searching experimental tests from a variety of viewpoints, and have been led to the conclusion that, whatever its origin, it actually represents very accurately the behavior… of all the substances with which I have worked.Foremost among them was the photoelectric effect: the ejection of electrons from a metal when irradiated by light.
The wave theory of light could not yield a satisfactory account of this, since the energy of a wave is spread over its entire surface. Dec 17, · Einstein’s theory of photoelectric effect. Photoelectric effect: The process in which visible light, x rays, or gamma rays incident on matter cause an electron to be ejected.
The ejected electron is called a photoelectron. This led to the two papers above, and contributed to Nobel Prizes for both Einstein (the photoelectric effect is the only specific theory mentioned in his prize citation) and Millikan. What was measured in Millikan's study of the photoelectric effect?
David R. Lloyd Citation: American Journal of Physics 83, (); doi: / unambiguous experimental support for Einstein’s theory of the photoelectric effect. His graphical about what Millikan’s experiment does.
Although Millikan’s.
Einstein's simple explanation (Equation \ref{Eq1}) completely accounted for the observed phenomena in Lenard's and Millikan's experiments (Figure $$\PageIndex{4}$$) and began an investigation into the field we now call quantum mechanics.
This new field seeks to provide a quantum explanation for classical mechanics and create a more unified theory of physics and thermodynamics. The photoelectric effect posed a significant challenge to the study of optics in the latter portion of the s.
It challenged the classical wave theory of light, which was the prevailing theory of the time. It was the solution to this physics dilemma that catapulted Einstein into prominence in the physics community, ultimately earning him the Nobel Prize.
Ra milikans photoelectric experiments that supported einsteins theory
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2019-08-25 15:09:24
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https://discuss.codechef.com/questions/98089/mxmedian-editorial
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×
# MXMEDIAN - Editorial
Practice
Contest
Author: Praveen Dhinwa
Tester: Pawel Kacprzak and Misha Chorniy
Editorialist: Bhuvnesh Jain
EASY
# Prerequisites
Greedy, Sorting, Median
# Problem
You are given an array $A$ of size $2 * N$ consisting of positive integers, where $N$ is an odd number. You can construct an array $B$ from $A$ as follows, $B[i] = max(A[2 * i - 1], A[2 * i])$. Consider all permutations of $A$. Print any one of them which maximises the median of $B$
# Explanation
The larger elements we have in the array $B$, the larger will be its median. Since, array $B$ has $n$ elements only, we would like to have the largest $n$ elements of $A$ somehow, go into the array $B$. Let us assume we have a black box which permutes array $A$ in some manner such that the largest $n$ elements go into array $B$. Now, what will be the median of array $B$ in such a case? It will be simply the middle element once the array $B$ is sorted.
Now, let us describe the black box now. We see that elements of $B$ are from adjacent elements from $A$. i.e. they are independent of each other in their selection. Thus, we simply put all the highest $n$ elements in either odd or even positions in array $A$. This will ensure that are always selected into array $B$. So, the only requirement is to sort the array $A$ to get the highest $n$ elements. The sorting can be done using mergesort, randomised quick-sort or any inbuilt sort function available.
The above is just one of the methods to construct the array and solve the problem. Multiple solutions to the problem might exist. Feel free to discuss them below.
# Time Complexity
$O(n \log{n})$ per test case
Setter's solution
Tester's solution
Editorialist solution
This question is marked "community wiki".
6★likecs
3.7k2481
accept rate: 9%
19.8k350498541
0 My solution is almost similar to the Editorialist solution but i got WA where am i wrong https://www.codechef.com/viewsolution/13499478 answered 18 May '17, 17:15 4★divik544 525●1●1●10 accept rate: 10% Your answer is correct, but the array required to give that answer is is incorrect, after sorting your array would look lie a[0],a[1],a[2],a[3],.......a[2n-1] but the array which gives required answer will look like a[0],a[n],a[1],a[n+1].....a[n-1=,a[2n-1]. (18 May '17, 17:53)
0 @divik544 You sorted the initial array, but you forgot to permute it! Remember, you need max elements in each adjacent pair of elements. If the array is a0, a1, a2, a4, ... a(2n-1) after sorting. A valid permutation would be: a0, an, a1, a(n+1), ... a(n-1), a(2n-1) after which the B array becomes an, a(n+1), a(n+2), ... a(2n-1) and the median would be a[(3n-1)/2] that is, [(n) + (2n-1)]/2. Here's my solution answered 18 May '17, 17:39 63●2 accept rate: 7%
0 ******************************************/ #include using namespace std; typedef long long LL; typedef long double LD; typedef pair pii; typedef pair pll; int main() { #ifndef ONLINE_JUDGE freopen("inp.txt", "r", stdin); #endif Could someone please explain how we are saving time through this in programs? As I am new to programming. answered 10 Jun '17, 14:21 1 accept rate: 0% 15.5k●1●20●66 How do you edit code to preview in readable format @vijju123 ? (10 Jun '17, 14:48) 1 Copy paste the entire code, then select it, and THEN click "Enter code" (Make sure to give atleast 1 space to code from any sentence before or after the code, else it wont have any effect) Eg- Have a look at my code #include.... //This is wrong Have a look at my code #include...//this is right (10 Jun '17, 14:51) Thanks mate ! ^-^ (10 Jun '17, 14:58) @adiaspirant, we just saving some tying time by these statements and nothing else. The ifndef statement just helps us to execute the code directly from sublime text without going to the terminal again and again. The ifndef statement will only work when the "ONLINE_JUDGE" flag is defined while compilation which can be seen on the specific websites. (17 Jun '17, 14:03) likecs6★
0 Setter & Tester solution links not yet updated :( answered 10 Jun '17, 14:35 419●2●10 accept rate: 7%
0 please check this code https://www.codechef.com/viewsolution/14267962 answered 17 Jun '17, 10:19 1 accept rate: 0%
0 Simple and Elegant : Code Time Complexity: O(2NLOGN) Space Complexity: O(2N) answered 28 Oct '17, 12:46 0●1 accept rate: 0%
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question asked: 18 May '17, 13:46
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2019-03-21 18:20:25
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http://grephysics.net/ans/9677/56
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GR 8677927796770177 | # Login | Register
GR9677 #56
Problem
GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Optics$\Rightarrow$}Total Internal Reflections
Total internal reflection is when one has a beam of light having all of the incident wave reflected. Going through a bit of formalism in electromagnetism one can derive Snell's Law for Total Internal Reflection,
$n_{inside} \sin\theta = n_{outside},
$
where $n_{inside}=1.33$, and one assumes that the surface has $n_{outside}=1$ for air.
One must solve the equation $\theta = \sin^{-1}(1/1.33)$. One can immediately throw out choices (A) and (E). From the unit circle, one recalls that $\sin(30^{\deg})=1/2$ and $\sin(60^{\deg})=1.7/2=0.85$. Since $1/1.33 \approx 0.7$, one deduces that the angle must be choice (C).
Alternate Solutions
calcuttj
2014-09-20 06:48:40
1/1.33 -> 3/4 -> (3/2)/2 -> 1.5/2
sin(45) = $\sqrt{2}$/2, $\sqrt{2}$ is about 1.4
so our angle is slightly above 45
tensorwhat
2009-04-03 07:42:49
For a piece of plastic or glass with $n=1.5$ you have a critical angle of $41.5^o$ (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about $50^o$
ramparts2009-11-03 18:06:23 Yeah. I'll be sure to look that one up on the test. Thanks a lot.
calcuttj
2014-09-20 06:48:40
1/1.33 -> 3/4 -> (3/2)/2 -> 1.5/2
sin(45) = $\sqrt{2}$/2, $\sqrt{2}$ is about 1.4
so our angle is slightly above 45
walczyk
2012-09-29 21:53:57
Just figured out a good approximate expansion people might want to memorize, arcsin in degrees: arcsin(z) ~ 60z + 10z^3, you get about 49.2 for this answer. You just have to be quick with your fractions.
mistaj
2011-08-25 11:19:41
This is Brewster's Law: $\rm tan\theta = \frac{n_t}{n_i}$ where t is the transmission medium (air n = 1.00) and i is the incident medium (water n = 1.33). Dividing 1 by 1.33, you get about 0.7. Now, $\rm tan\theta \simeq \theta$ (which is really good for this problem). So, in radians we have $\theta = 0.7$. Now, we can get an idea of this in terms of $\pi$ by solving $\frac{\pi}{x} = 0.7$ for $x$ which is roughly $x = \frac{\pi}{4}$ which is roughly half of 90 degrees. Choice C is the only option.
mistaj2011-08-25 11:38:57 Woops, misunderstood this. It still works though! But forget Brewster's Law and use sin instead of tan.
walczyk
2011-02-25 16:42:21
so here's how i figured out the hard part, arcsin(1/1.33): 1/1.33 is close to 3/4 (remember the inverse is 1.33!!) . sin(45) is 1/sqrt(2) so its bigger than 45 degrees (sqrt(2) is like 1.414.. so its bigger than 1.33!!) now sin(60) is sqrt(3)/2, which is like .8 or something so its less than 60 degrees. the only option left is 50 degrees, and we're done. the first time i did it i used the fact that (sqrt(3)/2)^2 is 3/4 so its obviously greater than 3/4.
torturedbabycow
2010-03-27 19:54:37
As far as solving that annoying inverse sine, I think the easiest way by far is to just draw out the triangle - one side is 1, and the hypotenuse is 1.33. So, since 1.33 is pretty close to $\sqrt{2}$, the angle should be pretty close to 45 degrees. Stare at the triangle a few more seconds, and it becomes obvious (at least to me) that it should be a little more than 45, so voila, answer (C)!
jmason86
2009-10-01 19:33:10
This is probably one to just have memorized. Stupid ETS trying to make us solve that inverse sign of 1/1.33. I hate 'em.
lrichey2011-11-04 17:40:57 I agree... but here is a way I figured out.... 1.33~1 1/3, hence 4/3 1/(4/3)=3/4. 3/4 is the square of a 30-60-90 triangle relation, giving inversesine(3/4)~ 60. Which is closer to 50 degrees than 75 degrees don't know if this works or helps at all
f4hy
2009-04-03 17:32:47
I only got this one because I remembered doing this exact problem in an optics class and knew that for water the angle is 50 degrees
tensorwhat
2009-04-03 07:42:49
For a piece of plastic or glass with $n=1.5$ you have a critical angle of $41.5^o$ (look it up), so by decreasing the index of refraction (eg. water 1.33) you would slightly increase the critical angle so there about $50^o$
ramparts2009-11-03 18:06:23 Yeah. I'll be sure to look that one up on the test. Thanks a lot.
LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$
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2023-03-22 19:27:45
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https://math.stackexchange.com/questions/154985/do-omega-omega-2-aleph-0-aleph-1
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# Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
As we know, $2^{\aleph_0}$ is a cardinal number, so it is a limit ordinal number. However, it must not be $2^\omega$, since $2^\omega=\sup\{2^\alpha|\alpha<\omega\}=\omega=\aleph_0<2^{\aleph_0}$, and even not be $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$. Since $\|\sum_{i = n<\omega}^{0}\omega^i\cdot a_i\| \le \aleph_0$ for all of them.
Besides, $\sup\{\sum_{i = n<\omega}^{0}\omega^i\cdot a_i|\forall i \le n(a_i \in \omega)\}=\omega^\omega$, and $\|\omega^\omega\|=2^{\aleph_0}$ since every element in there can be wrote as $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$ and actually $\aleph_{0}^{\aleph_0}=2^{\aleph_0}$ many.
Therefore $\omega^\omega$ is the least ordinal number such that has cardinality $2^{\aleph_0}$, and all ordinal numbers below it has at most cardinality $\aleph_0$. Hence $\omega^\omega=2^{\aleph_0}=\aleph_1$?
• $2^{\aleph_0} = \aleph_1$ is the Continuum Hypothesis, which is independent of ZFC. Jun 7, 2012 at 6:27
• Cardinality of the ordinal number $\omega^\omega$ is $\aleph_0$, see e.g. here or here. Jun 7, 2012 at 6:30
• Yes, you and Scott are correct, thank you. Jun 7, 2012 at 8:21
• You need to be careful with the different kinds of exponentiation. In some set theory books you can see (on different pages) $\omega=\aleph_0$, $2^\omega=\omega$, $2^{\aleph_0}\gt \aleph_0$. Sep 17, 2012 at 17:45
Your notation confuses cardinal and ordinal exponentiation, which are two very different things. If you’re doing cardinal exponentiation, $2^\omega$ is exactly the same thing as $2^{\aleph_0}$, just expressed in a different notation, because $\omega=\aleph_0$. If you’re doing ordinal exponentiation, then as you say, $2^\omega=\omega$.
But if you’re doing ordinal exponentiation, then $$\omega^\omega=\sup_{n\in\omega}\omega^n=\bigcup_{n\in\omega}\omega^n\;,$$ which is a countable union of countable sets and is therefore still countable; it doesn’t begin to reach $\omega_1$. Similarly, still with ordinal exponentiation, $\omega^{\omega^\omega}$ is countable, $\omega^{\omega^{\omega^\omega}}$ is countable, and so on. The limit of these ordinals, known as $\epsilon_0$, is again countable, being the limit of a countable sequence of countable ordinals, and so is smaller than $\omega_1$. (It’s the smallest ordinal $\epsilon$ such that $\omega^\epsilon=\epsilon$.)
Now back to cardinal exponentiation: for that operation you have $2^\omega\le\omega^\omega\le(2^\omega)^\omega=2^{\omega\cdot\omega}=2^\omega$, where $\omega\cdot\omega$ in the exponent is cardinal multiplication, and therefore $2^\omega=\omega^\omega$ by the Cantor-Schröder-Bernstein theorem. The statement that this ordinal is equal to $\omega_1$ is known as the continuum hypothesis; it is both consistent with and independent of the other axioms of set theory.
There are very different definitions for cardinal and ordinal exponentiation. Ordinal exponentiation is defined in a way which allows us to generate well-orderings of a particular set; where as cardinal exponentiation strips out the ordering and deals with cardinality of all functions from one set to another.
This is why some authors differentiate the two by using $^\omega\omega$ for cardinal exponentiation and $\omega^\omega$ for ordinal exponentiation (at least where context is ambiguous). Personally I am not a big fan of this approach, despite the fact it may clear some possible confusion.
Lastly, as commented, $2^{\aleph_0}$ need not be equal to $\aleph_1$. This is known as The Continuum Hypothesis which was proved unprovable from ZFC.
To add on the confusion, let me give a short list of some common uses for $\omega^\omega$:
• The first limit ordinal which is a limit of limit ordinals each a limit of limit ordinals which are not limits of limit ordinals.
• The set of all sequences of natural numbers (which also form the underlying set for the following uses).
• The Baire space.
• The real numbers (in some contexts).
• I’m not a fan of it either, since for me $^\omega\omega$ is the set of functions from $\omega$ to $\omega$, not the cardinality of that set. Jun 7, 2012 at 6:56
• Brian Scott says $\omega^{\omega}$ is countable while you seem to imply it's isomorphic to the reals. How do these two reconcile? Aug 16, 2018 at 8:40
• @Robert: Ordinal arithmetic vs. cardinal arithmetic. Aug 16, 2018 at 8:41
• So people using the form you describe here are doing cardinal arithmetic with what appear to be ordinal numbers? Looks like something I'll need to watch out for. Aug 16, 2018 at 8:41
• @Robert: People, myself included, use $A^B$ to denote the set of all functions from $B$ to $A$. Brian prefers ${}^BA$ for that set. The reason being that the set of all functions from $\omega$ to $\omega$ is denoted $\omega^\omega$, but so does the ordinal exponentiation of $\omega^\omega$. This ends up being confusing, so people have two different notations. I just prefer clarifying the context. Aug 16, 2018 at 8:44
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2022-05-24 18:22:14
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http://physics.stackexchange.com/questions/79522/given-wave-function-at-t-0-what-is-the-process-of-deriving-time-dependent-wav
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# Given wave function at $t=0$, what is the process of deriving time dependent wave equation? [closed]
Suppose $$\Psi (x, t=0)=Ae^{i\alpha _1}\psi _1(x)+Be^{i\alpha _2}\psi_2(x)+Ce^{i\alpha _3}\psi_3(x).$$ If $\psi _n$ are the energy eigenfunctions how would I derive $\Psi (x,t)$?
I am having trouble with the $\psi_n$ for which I don't know how to deal with when I use Fourier transform
$$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \left(Ae^{i\alpha _1}\psi _1(x)+Be^{i\alpha _2}\psi_2(x)+Ce^{i\alpha _3}\psi_3(x)\right)\, dx$$
Any hint would be appreciated.
-
## closed as off-topic by Colin McFaul, Brandon Enright, Chris White, DavePhD, Kyle OmanJun 16 '14 at 16:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Colin McFaul, Brandon Enright, DavePhD, Kyle Oman
If this question can be reworded to fit the rules in the help center, please edit the question.
The time dependence is given by $$\psi(x,t) = \psi_n(x) e^{-iE_nt/\hbar}$$ where $E_n$ is the energy of the system.
I would put subscript $n$ on your $\psi$s to indicate it is only valid when $\psi$ is an energy eigenfunction. – BMS Oct 3 '13 at 18:53
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2015-08-01 12:06:06
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https://cooking.stackexchange.com/questions/42569/how-can-i-reduce-the-heat-output-of-this-heating-element/42571
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# How can I reduce the heat output of this heating element?
I'd like to add a Brinkmann 1500 watt element to my Bradley smoker to bring my temps up in cold weather.
I'd like to limit the amount of heat it puts out, is there a voltage regulator or resistor I can use to drop the heat? Something variable maybe?
## migrated from diy.stackexchange.comMar 6 '14 at 19:39
This question came from our site for contractors and serious DIYers.
• Can you use a grill blanket (I am not making that up) to help conserve the heat you already have? Most smokers are pretty much not insulated at all. – SAJ14SAJ Mar 6 '14 at 20:07
• Your going to need a high wattage resistor. -expensive. I suppose you could run your 1500 watt heater in series with an 800 watt heater, outside the grill. A Variac with that power output will cost an arm and a leg. Maybe a switching power supply, but solid state at that wattage will not come cheap. I'd go with a smaller space heater in series outside the unit for a cheap effective solution. V=IR and W = VI should tell you the wattage you'll need. – Wayfaring Stranger Nov 18 '17 at 5:30
First, you will want to check the wiring and fuses in your current smoker to make sure it can safely handle the 1500w upgrade.
Assuming that your wiring and fuses can handle the 1500w element, keep in mind how the system works in the first place. The heating element is controlled by the thermal measurement unit. A more powerful element would still turn off once the set temperature at the measurement unit is reached. The difference would be how fast it reaches the temperature and more hot spots closer to the element.
For the "variable" options you can alter the voltage or the current running to the heater. One issue to be aware of is that you would need to keep the power supply to the smoker's control board separated and unaltered.
# Altering the voltage:
There is a power supply called a variac or variable transformer.
Pros:
• Since this plugs into the a home receptacle and has a standard receptacle coming out of it, it would be safer from a wire splicing perspective. Keep in mind, you'd need to acquire a variac that can the wattage.
• You will look and feel like a cold war submarine commander when adjusting the voltage on this kind of equipment so this is a high testosterone solution.
• There's conversational value here for parties.
Cons:
• A potentially fiery death. Playing with high wattage hacks requires strict attention to detail and close supervision of the device.
• Cost. The price ranges for these seem to be $180-$400ish in the 2000W range. This could significantly increase the overall cost of your smoker setup.
• The control board of the smoker can certainly cope with some range of voltage under the usual 120V, but it may not operate at the lower voltage
• Even after everything is working you probably don't want to invite your fire insurance provider over for dinner if you like your current insurance rates.
# Altering the current (by adding resistance):
You can add a variable resistor. Just like the above solution you'd need a higher wattage rated device.
Pros:
• It would indeed adjustably control the amount of current flowing through your element.
Cons:
• A potentially fiery death.
• This resistor would be handling some serious current and it doesn't have any housing as-is. You wouldn't want your cat to brush against this.
• This solution require wire splicing and would be inherently less safe for anyone not very familiar with electric systems.
• Cost. This is in the $350ish range • This could heat up depending on the setting you choose. # Summary: Both of these solutions are costly compared to most smoker's retail value. These higher power devices require alloys that are expensive to produce. The variac would be my preferred cooler and safer solution out of the two, but keep in mind that the smoker control board wants plain old 120 line voltage. It will not operate very far out of the 120 volt range and, even worse, could malfunction and not turn off the element. Regardless of any solution,If you do proceed you will be hacking a 1000+ Watt device please be very careful. • Who the heck beside me has a variac ? – blacksmith37 Nov 17 '17 at 20:57 You could use a Sous Vide controller. Normally these are intended to drive a heater such as a crock pot, but I don't see why it wouldn't work here. Their website says it will output up to 15amps at 120V which is 1800W. This drives the element by having it be on at 100% for only some partial amount time, rather than driving it slowly at all times, but you should be able to achieve your desired result. Dork Food Sous Vide Temperature Controller • The temperature range of a Sous Vide controller might not go high enough to be useful to smoke. That model you link to will only go up to 200˚F. But maybe some others are more adjustable? – Hank Mar 6 '14 at 16:50 A couple low-tech options: • A long extension cord will drop the voltage at the element and therefore its power output. E.g. a 50ft 18-gauge cord will reduce the voltage ~10% when used with a 15A element. You could chain a few together to get a larger drop. • You could get a mechanical light timer (make sure you get one rated for 15A), which will can be set to turn the element on and off. Many timers have a resolution of 15 minutes which may be too long for you, but I have one that has 5-minute increments. • For the extension cord trick, if you're not careful you could end up smoking your house or garage as well as your food. – The Photon Mar 6 '14 at 17:55 • Yeah, if the OP wants to use extension cords to drop the voltage, I would recommend 1) not coiling hundreds of feet up in a giant hot ball, 2) not leaving the contraption unattended. But let's put this in perspective: the heating element itself is clearly a lot more dangerous than the extension cord. – Hank Mar 6 '14 at 18:09 You can find Variac's at ham radio shows or electronic surplus places, most are bulk or without a case. Check a place called "Circuit specialist" They have models built in a case which is a safer unit, keep it out of rain. One I saw a 20 amp for$89. The element is an inductive load, you can google for a 1500-2000 watt electronic controller. Another way to look at this problem is thermostat the element with a temp controller thermostat one rated at 1500 Watt. Ebay has electronic temp controllers digital read out for \$15, it controls 10 amps, add a relay maybe 39 amp relay. This way the element will come on when its needed, it don't matter if it is 1500 watt, it will turn on and off frequently but only if needed.
• That's awful cheap for a 20 amp Variac. I'd carefully check reviews before buying. – Wayfaring Stranger Nov 18 '17 at 5:31
The device you want to use, IF no thermostatic control is desired, is a burst control dimmer rated for the power. A phase angle dimmer at that wattage will be an EMC liability, a resistor will be very energy inefficient, huge and a fire risk, and a variac is bulky and expensive.
Note that common household light dimmers are a) phase angle designs and b) too weak.
However, using some kind of thermostat that switches in the heating if more heat is needed and switches it off when it is getting too hot might be the more desirable solution.
For a VERY simple power dropper (also inefficient, but safer than a resistor) put very high powered incandescent lights (maybe a few of them paralleled), like construction lights, in series with the heating element. Set them up so they operate safely, these lights get hot too!
Some kind of overtemperature cutoff device in the smoker would be very well advised unless the whole smoker design is inherently, absolutely impossible to set on fire, even if the element was running at full power - which is a possible failure mode in case your method of heat control fails - and it will if you expect it not to.
If you have any doubt about your design being safe, consult a professional electrician.
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2019-11-22 00:07:50
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http://www.go4expert.com/forums/makefile-gmake-t21405/
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# Makefile gmake
Light Poster
Hi!
Can someone tell me what this means in line makefile:
include \$ (DVD_MAKE) / generic.mak
Thanks!
0
Light Poster
Quote:
Originally Posted by irdeto
Hi!
Can someone tell me what this means in line makefile:
include \$ (DVD_MAKE) / generic.mak
Thanks!
What means here sign of dollar \$.
0
Mentor
It expands the symbol DVD_MAKE, so instead of trying to include the file DVD_MAKE/generic.mak, if DVD_MAKE is set to /usr/home/me/dvdstuff then it will include /usr/home/me/dvdstuff/generic.mak.
0
Light Poster
Quote:
Originally Posted by xpi0t0s
It expands the symbol DVD_MAKE, so instead of trying to include the file DVD_MAKE/generic.mak, if DVD_MAKE is set to /usr/home/me/dvdstuff then it will include /usr/home/me/dvdstuff/generic.mak.
Thank you.
I tried to compile some file ,with gmake and compiler, when started gmake ,I get error cant't find generic.mak , generic.mak is in different branches in the SW tree.
Do you know why I get error?
0
Light Poster
This is my SW tree:
/ xxxx\gamke > makefile
Root
\ yyyy\generic.mak
See above post:
0
Light Poster
set DVD_MAKE=%1/make
what mean %1
thanks!
0
Mentor
I don't think I can explain why you get the error. make itself has told you why you got the error ("can't find generic.mak") and I can't think of any way to explain "can't find generic.mak" without actually saying "can't find generic.mak".
So either generic.mak isn't on your computer, or make is looking in the wrong place.
%1 is the first parameter. So for the command "dir *.txt /s", dir would see *.txt as %1 and /s as %2.
If generic.mak is in \yyyy then probably you need to set DVD_MAKE=\yyyy.
irdeto like this
0
Light Poster
I found what is wrong ,think, batch file don't
Quote:
"set DVD_MAKE=%1/make"
that is why what gmake and make file can't recognize generic.mak.
Don't know why.
Names of folder and file is OK.
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2014-10-24 13:44:02
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https://codereview.stackexchange.com/questions/217150/database-model-security-for-book-inventory
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# Database Model Security For Book Inventory
This code is one of the models developed for the Book Inventory application that is open for review. It is also based on a comment on this answer to a C# question. Based on the comment I did some research and found this stackoverflow question.
The code presented for review is the database model for a series of books in the Book Inventory. A series of books has an author and a title. The author is represented by a key into the author table. To add a series to the database the user has to select the author from a list of authors already in the database. They then have to add the title or name of the series.
Below I present a before and after, the before is one function that was used in the answer above, the after is the entire code to be reviewed including the refactored function.
Questions:
Is there anything else I can do to prevent SQL Injection attacks?
What else can I do to improve the code, keep in mind I've been on this site for a while, I'm not asking "Does my code suck?", all code sucks, I'm asking how does my code suck.
Before
private DataRow GetRawSeriesData(uint seriesId)
{
DataRow rawData = null;
if (seriesId > 0)
{
try
{
using (MySqlConnection conn = new MySqlConnection(_dbConnectionString))
{
string queryString = "SELECT * FROM series WHERE idSeries = '" + seriesId.ToString() + "';";
int ResultCount = 0;
DataTable Dt = new DataTable();
conn.Open();
using (MySqlCommand cmd = new MySqlCommand())
{
cmd.Connection = conn;
cmd.CommandType = CommandType.Text;
cmd.CommandText = queryString;
ResultCount = sda.Fill(Dt);
if (ResultCount > 0)
{
rawData = Dt.Rows[0];
}
}
}
}
catch (Exception ex)
{
string errorMsg = "Database Error: " + ex.Message;
MessageBox.Show(errorMsg);
}
}
return rawData;
}
After: Code to be Reviewed: SeriesTableModel.cs
using System;
using System.Collections.Generic;
using System.Data;
using System.Windows;
using MySql.Data.MySqlClient;
namespace pacsw.BookInventory.Models
{
public class SeriesTableModel : CDataTableModel
{
private int seriesTitleIndex;
private int seriesKeyIndex;
private int seriesAuthorKeyIndex;
public SeriesTableModel() : base("series", "getAllSeriesData", "addAuthorSeries")
{
seriesTitleIndex = GetDBColumnData("SeriesName").IndexBasedOnOrdinal;
seriesKeyIndex = GetDBColumnData("idSeries").IndexBasedOnOrdinal;
seriesAuthorKeyIndex = GetDBColumnData("AuthorOfSeries").IndexBasedOnOrdinal;
}
public DataTable Series { get { return DataTable; } }
{
SeriesModel seriesModel = (SeriesModel)iSeriesData;
}
{
}
public List<string> SeriesSelectionListCreator(AuthorModel author)
{
List<string> seriesSelectionList = new List<string>();
if (author != null && author.IsValid)
{
DataTable currentSeriesList = Series;
string filterString = "LastName = '" + author.LastName + "' AND FirstName = '" + author.FirstName + "'";
DataRow[] seriesTitleList = currentSeriesList.Select(filterString);
foreach (DataRow row in seriesTitleList)
{
}
}
return seriesSelectionList;
}
public uint GetSeriesKey(AuthorModel author, string seriesTitle)
{
uint key = 0;
if (author != null && author.IsValid)
{
string SqlQuery = "SELECT series.idSeries FROM series WHERE series.SeriesName = @title AND series.AuthorOfSeries = @authorid;";
using (MySqlConnection conn = new MySqlConnection(_dbConnectionString))
{
int ResultCount = 0;
DataTable Dt = new DataTable();
try
{
conn.Open();
using (MySqlCommand cmd = new MySqlCommand())
{
cmd.Connection = conn;
cmd.CommandType = CommandType.Text;
cmd.CommandText = SqlQuery;
cmd.Parameters["@title"].Value = seriesTitle;
cmd.Parameters["@authorid"].Value = author.AuthorId;
cmd.ExecuteNonQuery();
ResultCount = sda.Fill(Dt);
if (ResultCount > 0)
{
key = Dt.Rows[0].Field<uint>(0);
}
}
}
catch (Exception ex)
{
string errorMsg = "Database Error: " + ex.Message;
MessageBox.Show(errorMsg);
key = 0;
}
}
}
return key;
}
public string GetSeriesTitle(uint seriesId)
{
string title = string.Empty;
if (seriesId > 0)
{
string SqlQuery = "SELECT series.SeriesName FROM series WHERE series.idSeries = @seriesid;";
using (MySqlConnection conn = new MySqlConnection(_dbConnectionString))
{
int ResultCount = 0;
DataTable Dt = new DataTable();
try
{
conn.Open();
using (MySqlCommand cmd = new MySqlCommand())
{
cmd.Connection = conn;
cmd.CommandType = CommandType.Text;
cmd.CommandText = SqlQuery;
cmd.Parameters["@seriesid"].Value = seriesId;
cmd.ExecuteNonQuery();
ResultCount = sda.Fill(Dt);
if (ResultCount > 0)
{
title = Dt.Rows[0].Field<string>(0);
}
}
}
catch (Exception ex)
{
string errorMsg = "Database Error: " + ex.Message;
MessageBox.Show(errorMsg);
}
}
}
return title;
}
public SeriesModel GetSeriesModel(uint seriesId)
{
SeriesModel seriesData = null;
DataRow rawSeriesData = GetRawSeriesData(seriesId);
if (rawSeriesData != null)
{
seriesData = ConvertDataRowToSeriesModel(rawSeriesData);
}
return seriesData;
}
protected override void InitializeSqlCommandParameters()
{
AuthorTableModel authorTable = ((App)Application.Current).Model.AuthorTable;
}
private SeriesModel ConvertDataRowToSeriesModel(DataRow rawSeriesData)
{
uint authorId;
uint.TryParse(rawSeriesData[seriesAuthorKeyIndex].ToString(), out authorId);
string title = rawSeriesData[seriesTitleIndex].ToString();
AuthorModel author = ((App)Application.Current).Model.AuthorTable.GetAuthorFromId(authorId);
SeriesModel seriesModel = new SeriesModel(author, title);
return seriesModel;
}
private DataRow GetRawSeriesData(uint seriesId)
{
DataRow rawData = null;
if (seriesId > 0)
{
try
{
using (MySqlConnection conn = new MySqlConnection(_dbConnectionString))
{
string queryString = "SELECT * FROM series WHERE idSeries = @seriesid;";
int ResultCount = 0;
DataTable Dt = new DataTable();
conn.Open();
using (MySqlCommand cmd = new MySqlCommand())
{
cmd.Connection = conn;
cmd.CommandType = CommandType.Text;
cmd.CommandText = queryString;
cmd.Parameters["@seriesid"].Value = seriesId;
ResultCount = sda.Fill(Dt);
if (ResultCount > 0)
{
rawData = Dt.Rows[0];
}
}
}
}
catch (Exception ex)
{
string errorMsg = "Database Error: " + ex.Message;
MessageBox.Show(errorMsg);
}
}
return rawData;
}
}
}
Anytime I see something like these I wonder "is there a better way we can do this?":
cmd.Parameters.Add("@title", MySqlDbType.String);
cmd.Parameters["@title"].Value = seriesTitle;
cmd.Parameters["@authorid"].Value = author.AuthorId;
Well, in this case, I'm going to suggest a slightly-functional approach to this process.
C# has a (mostly) handy feature called "Extension Methods". In general, I try to avoid them unless there's a good application for them, such as here.
I would define an extension method here quick:
public static MySqlCommand AddParameter<T>(this MySqlCommand command, string name, MySqlDbType type, T value) {
command.Parameters[name].Value = value;
return command;
}
Now, here, I returned the MySqlCommand on purpose, because I want to chain this:
command.AddParameter("@title", MySqlDbType.String, seriesTitle)
Since you aren't using an ORM (and I suspect you have reasons for that, I definitely won't try to persuade you to use one) I would do a couple of these small things to make life a little easier on myself. It's pretty cheap, and it allows us to work our code more easily.
I would take some of these shorter functions and apply some expression-bodied members:
public DataTable Series { get { return DataTable; } }
{
SeriesModel seriesModel = (SeriesModel)iSeriesData;
}
{
}
To:
public DataTable Series => DataTable;
For simple functions like those, it's trivial to do and saves you some vertical space.
• If I implement the first suggestion in the abstract class CDataTableModel in my original question will it work in all the inherited DataTable models classes? – pacmaninbw Apr 9 '19 at 18:34
• No, you will need to make a new public static class for the extension methods, but then as long as the namespace containing that public static class is available, it will work in any class. – Der Kommissar Apr 9 '19 at 18:37
• Usually it's a better idea to cast a class into an interface rather the the other way around as in your last two lines but I guess this is just a blind copy/paste of OP's code and turning the methods into expression-bodied ones but in this case the overload taking a SeriesModel is no longer necessary. – t3chb0t Apr 10 '19 at 10:35
• @t3chb0t Yeah I didn't look at any of the dependency trees, just went off what was in the question. – Der Kommissar Apr 10 '19 at 11:37
public bool AddSeries(ISeriesModel iSeriesData)
{
SeriesModel seriesModel = (SeriesModel)iSeriesData;
}
If SeriesModel implements ISeriesModel interface (which I suppose it does) then you need only the overload taking the interface.
Also casting an interface into a class is rarely a good idea because we then loose the advantages of having the interface in the first place like using a mock type.
|
2020-02-25 07:20:00
|
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|
http://mathoverflow.net/questions/6538/generalizations-of-boolean-posets-lattices/6566
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# Generalizations of Boolean posets/lattices
A Boolean lattice has a number of rather nice properties which give it a central role in many parts of combinatorics. For instance, it's a lattice, it can be augmented with a ring structure, it can also be augmented with an associative algebra structure, it has a complementation operation, it can be "identified" in different ways with things like the hypercube or a powerset, it satisfies the Stone representability theorem, etc.
It's kind of obvious that Boolean lattices are pretty closely related to the number 2. Like, they have a duality, finite ones all have order a power of 2, etc. One way to see this is to look at a (finite) Boolean lattice as the set of all functions from a set $S \rightarrow \{0, 1\}$, with the lattice and complementation structures acting pointwise.
To what extent can you get an analogue of Boolean lattices (or even posets) with some other natural number k taking the place of 2? You could consider the set of all functions from $S \rightarrow \{0, 1, ..., k-1\}$, which again gives you a poset with a lattice structure, but we don't get a complementation map (although I guess we do get some sort of "k-ality.")
There are a number of questions that this raises: Is this the proper generalization of Boolean lattices in this direction? Does some weird analogue of the Stone theorem hold? What's true for Boolean lattices but isn't true for these guys? What are the right notions to replace "hypercube" and "power set?" Etc.
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Sort of the driving question here, FWIW, is the relationship between extremal combinatorics of set systems and Ramsey theory, in particular (and most concretely) the point of view in which we take (density) Hales-Jewett to be a generalization of Sperner's theorem. – Harrison Brown Nov 23 '09 at 6:58
I also don't understand what you mean by an analogue of Stone's theorem. What part of it do you care about? – Qiaochu Yuan Nov 23 '09 at 23:28
I guess I really don't care that much about it, since I'm really mostly concerned with finite sets. It would be interesting to know if there's some way of looking at these things as sets of a topological space, though. – Harrison Brown Nov 24 '09 at 2:55
A standard extention is to the lattice of subspaces of an n dimensional space over a field with q elements. Many nice properties of the Boolean lattice extend while others do not. Many extremal combinatorics result extend and some are even simpler.
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Hm, interesting. Is there a name for this other than just "lattice of subspaces?" – Harrison Brown Nov 24 '09 at 2:59
Here is a link to a paper springerlink.com/content/v141451p51820482 which studies simultaneuosly a basic questions regarding ranks of incidence matrices for the Boolean lattice and for the lattice of subspaces over a field with q elements. – Gil Kalai Nov 24 '09 at 15:28
This lattice coincides with the (incidence geometry) of the building of type $A_{n-1}$ over the field $\mathbb{F}_q$. The points of this geometry are the 1-dimensional subspaces, the lines the 2-dimensional subspaces etc. Buildings in general are indeed known to admit a rich theory and countless deep and beautiful properties, applications and interpretations, in many fields of mathematics, such as (finite and infinite) group theory, representation theory, combinatorics, geometry, ... – Max Horn Jul 1 '11 at 8:44
The generalization you seek exists when k itself is a power of 2 (but gives no additional examples). This is because, as Q Yuan points out, the important properties of 2 that you seem to require are that is it a "truth" poset 2={false,true}, and when k=2^m, then there is a Boolean algebra of size k that can serve a similar purpose.
That is, the suggestion is that you should replace 2 with an arbitrary Boolean algebra B. For example, if you look at functions f:S to B, you can still perform lattice operations and complements pointwise. Perhaps this is the generalization you seek.
But you won't get any new examples this way, since they will still just be (larger) Boolean algebras.
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What he said. ☼ – Jon Awbrey Dec 18 '09 at 13:02
Harrison, I've been meaning to blog a bit about this somewhere, but I might as well put something down here in case it's useful to you.
As you know, there are lots of ways of defining Boolean algebras. I'm going to focus on one (due in large part to Lawvere) that may open one's eyes to some not-too-well-known possibilities. It starts by observing that for any finitary algebraic theory $T$, the category of algebras is equivalent to the category of product-preserving functors
$$Kl(T)^{op} \to Set$$
where $Kl(T)$, aka the Kleisli category, is the category of finitely generated free algebras. (The morphisms here are of course natural transformations.) This is "Lawvere Theories 101" (the Lawvere theory of an algebraic theory or monad $T$ being defined as $Kl(T)^{op}$). In the case we are concerned with, the finitely generated free Boolean algebras are finite Boolean algebras of cardinality $2^{2^n}$, and by a baby form of Stone duality, the opposite of the category of finitely generated free Boolean algebras is equivalent to the category of finite sets having cardinalities of the form $2^n$.
Let's call this category $Fin_{2^{-}}$. Thus the category of Boolean algebras is equivalent to the category of product-preserving functors
$$Fin_{2^{-}} \to Set$$
But we can say it more nicely than that. A functor $F: C \to Set$ extends uniquely (up to isomorphism) to the so-called Cauchy completion of $C$, aka the Karoubi envelope or idempotent-splitting completion of $C$, which I'll denote as $\bar{C}$. Moreover, if $C$ has finite products and $F$ preserves them, then it's an easy exercise that $\bar{C}$ acquires finite products and the extension $\bar{F}: \bar{C} \to Set$ preserves them. In the present case, it is easy to see that the idempotent-splitting completion of $C = Fin_{2^-}$ is the category $Fin$ of all finite sets, basically because any (nonempty) finite set is a retract of a finite set of cardinality $2^n$.
Putting all this together, we obtain what I think is a pretty description of the category of Boolean algebras: it is equivalent to the category of product-preserving functors
$$Fin \to Set$$
(which I believe Lawvere and Schanuel have taken to calling "distributions".) One of the beauties of this description is that it is totally unbiased: there is no special bias toward finite sets of cardinality $2^n$.
In fact, all this shows that we could, if we want, change our bias to, say, finite sets of cardinality $3^n$. In other words, the category of such sets is also a category with finite products, and its Cauchy completion is again $Fin$, and therefore we are within our rights to describe the category of Boolean algebras as equivalent to the category of product-preserving functors of the form
$$Fin_{3^-} \to Set$$
The category $Fin_{3^-}$ a perfectly legitimate single-sorted Lawvere theory $T'$; the generic object is the finite set $3^1$, of which all other objects in $T'$ are finite products. Using this to extract an alternative operations-and-equations presentation of the theory of Boolean algebras is an interesting exercise, but maybe I'll confine myself to some additional remarks that bear a bit on Stone duality.
From Lawvere Theories 101, we know that the free $T'$-algebra on $n$-generators, $F(n)$, is in this case given by the representable (note that representable functors automatically preserve products)
$$Fin_{3^-}(3^n, -) = Fin(3^n, -): Fin_{3^-} \to Set$$
If $A$ is a $T'$-algebra, then the underlying set is naturally identified as
$$U(A) = Set(1, U(A)) \cong T'\text{-Alg}(F(1), A) \cong Nat(Fin(3^1, -), A-)$$
(in the last step identifying $T'$-algebras with product-preserving functors $A$). For example, if $A = F(n) = Fin(3^n, -)$, then we have
$$U(A) \cong Nat(Fin(3^1, -), Fin(3^n, -)) \cong Fin(3^n, 3)$$
by the Yoneda lemma, whence the underlying set of the $T'$-algebra $F(n)$ has $3^{3^n}$ elements, in perfect analogy with the standard description of the free Boolean algebra on $n$ generators having $2^{2^n}$ elements. (Obviously I shouldn't say "the" underlying set; one of the morals here is that there can be many underlying-set functors which are monadic for a given variety of algebras.)
Next, we can analogize baby Stone duality for finite Boolean algebras, which says that homming into the free Boolean algebra on 0 generators (the Boolean algebra with two elements) induces an equivalence
$$Bool_{fp}^{op} \to Fin$$
where $Bool_{fp}$ stands for the category of finitely presented Boolean algebras. Of course, notions like "finitely presented Boolean algebras" and the "free Boolean algebra on 0 generators" have perfectly invariant unbiased descriptions, but if we allow ourselves to be biased toward $T'$-algebras, where the free $T'$-algebra has $3^{3^0} = 3$ elements, then baby Stone duality reads
• The functor $T'\text{-Alg}(-, 3): T'\text{-Alg}(-, 3): T'\text{-Alg}_{fp}^{op} \to Fin$ is an equivalence.
The other direction of the equivalence is the obvious functor $Fin(-, 3): Fin^{op} \to T'-\text{Alg}_{fp}$, which sends a finite set $X$ to $3^X$, with the pointwise-defined $T'$-algebra operations induced by the $T'$-algebra structure on $3$.
Similarly, an ultrafilter on a set $X$ (or an ultrafilter in a $T'$-algebra $A$) may be defined as a $T'$-algebra map $3^X \to 3$ (a $T'$-algebra map $A \to 3$, resp.). Thus the general Stone duality may be lifted to the context of $T'$-algebras.
These observations (which I discovered for myself only recently, but which are undoubtedly known to people like Lawvere and Schanuel) may help shed some light on an observation made by Lawvere which might be a bit mysterious otherwise. Namely, if we consider the subtheory of $T'$ generated by the unary operations, noticing that unary operations under composition form a monoid which is isomorphic to the monoid of endomorphisms $M = Fin(3, 3)$ (again by the Yoneda lemma), then we get a forgetful functor
$$Bool \simeq T'\text{-Alg} \to Set^M$$
where $Set^M$ denotes the category of sets equipped with an $M$-action, or $M$-sets for short. It turns out that this is a full embedding; in other words, if a function between $T'$-algebras preserves just the unary operations, then it preserves all the $T'$-operations. (This is a nice exercise.) Therefore, we may identify an ultrafilter on a set $X$ as being essentially the same as a homomorphism of $M$-sets
$$3^X \to 3$$
(This doesn't work if $3$ is replaced by $2$!) And similarly if $3$ is replaced by any $n \geq 3$. (Lawvere goes on to say that countably complete ultrafilters may be equivalently defined as functions $\mathbb{N}^X \to \mathbb{N}$ which preserve the evident actions by the monoid of endofunctions on $\mathbb{N}$. Thus, the canonical map
$$prin_X: X \to Set^{End(\mathbb{N})}(\mathbb{N}^X, \mathbb{N})$$
which is the unit of an evident monad
$$Set \stackrel{\mathbb{N}^-}{\to} Set^{End(\mathbb{N})} \stackrel{\hom(-, \mathbb{N})}{\to} Set$$
on $Set$, fails to be an isomorphism if and only if there exists a measurable cardinal.
Thanks to a comment of Gerhard Paseman on my answer to this MO question, I recently learned that the algebras of the Lawvere theory $Fin_{3^-}$ are better known as $3$-valued Post algebras, and there is a similar notion of $n$-valued Post algebra. That gives perhaps a quicker answer to Harrison's question.
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One can consider the set of functions $S \to P$ where $P$ is any poset with $k$ elements. The reason the Boolean lattice is special is that the "truth poset" happens to lie at the intersection of a lot of important ideas (logic, set theory, universal algebra, etc.) and I don't think the same can be said about just any old poset.
If you want to generalize most of the important properties of the Boolean lattices I'm not convinced cardinality is the right think to be looking at.
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Well, you'd probably be right, except that I have a well-defined goal in mind, and cardinality's gotta come into play somehow... – Harrison Brown Nov 23 '09 at 23:16
I think you could look at MV-algebras. They were introduced in the 50's by Chang in order to prove completness of the Lukasziewicz propositional many valued (MV) logic. From what I know, this class of algebras is the most special proper generalization of Boolean algebras that makes sense.
Every MV-algebra is a subdirect product of totally ordered MV-algebras; this is a direct generalization of the algebraic version of Stone's theorem. In the finite case, this reduces to the fact that every finite MV-algebra is isomorphic to a direct product of finite chains and vice versa. Even some of the more elaborate parts of the theory of Boolean algebras, like the Loomis-Sikorski theorem have their MV-algebraic versions.
As proved by Mundici in the 80's, MV-algebras are categorically equivalent to lattice ordered abelian groups with strong order unit.
The standard reference book for MV-algebras is this one.
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Alfred Foster worked on "n-ality" in some structures. While not quite a generalization of Boolean algebras, he did work on structures with a notion that generalized duality. In one of his papers, he found (something like, I am operating with faulty memory here) a substructure that resembled a ring of idempotents that helped carry the notion. He was also interested in generalizations of the Chinese Remainder Theorem for structures.
That and n-valued logics are the closest I can think of toward the "n" part of your question.
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2015-03-03 00:52:43
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https://socratic.org/questions/how-do-you-differentiate-g-y-x-2-1-x-2-2-3-2-using-the-product-rule
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# How do you differentiate g(x) =(x^2 - 1) (x^2 - 2)^(3/2 using the product rule?
Nov 11, 2017
$g ' \left(x\right) = x {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left(5 {x}^{2} - 7\right)$
#### Explanation:
$\text{given "g(x)=f(x)h(x)" then}$
$g ' \left(x\right) = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$
$f \left(x\right) = {x}^{2} - 1 \Rightarrow f ' \left(x\right) = 2 x$
$h \left(x\right) = {\left({x}^{2} - 2\right)}^{\frac{3}{2}}$
$\text{differentiate using the "color(blue)"chain rule}$
$\Rightarrow h ' \left(x\right) = \frac{3}{2} {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \times 2 x = 3 x {\left({x}^{2} - 2\right)}^{\frac{1}{2}}$
$\Rightarrow g ' \left(x\right) = 3 x \left({x}^{2} - 1\right) {\left({x}^{2} - 2\right)}^{\frac{1}{2}} + 2 x {\left({x}^{2} - 2\right)}^{\frac{3}{2}}$
$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = x {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left[3 \left({x}^{2} - 1\right) + 2 \left({x}^{2} - 2\right)\right]$
$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = x {\left({x}^{2} - 2\right)}^{\frac{1}{2}} \left(5 {x}^{2} - 7\right)$
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2019-04-22 22:24:57
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https://www.mfmat.org/cours_giorgio.html
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GdT Scattering des EDP non-lineaires
Scattering theory for Schrodinger equations (Giorgio Velo)
Basic ideas in scattering theory will be introduced and applied to the nonlinear Schroedinger equation (NLS) and to the Hartree equation (HE). In situations of short range the Cauchy problem with infinite initial time will be presented and asymptotic completeness will be discussed. The boundary between short and long range will be illustrated in detail. Some general information on the Cauchy problem with infinite initial time for the long range HE will be given.
Cours 1
• Introduction au "Scattering" des equations de Schroedinger non-lineaires; inegalites de Strichartz.
Cours 2
• Problemes de Cauchy avec donnees initiales a l'infini: $$L^{2}(\mathbb{R}^{n})$$, $$H^{1}(\mathbb{R}^{n})$$.
Cours 3
• Problemes de Cauchy en $$Y^{1}(I)$$ ; loi de transformation pseudoconformes; completude asymptotique.
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2022-08-14 14:46:00
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http://cfgt.net/blog/2012/11/10/when-to-spend-more
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# When To Spend More
Buying things seems to get more complicated every year. This increased complexity can include opportunity costs, the choices and sometimes, even the price. Then at some point in your life, you get a job and now you have more disposable income to spend. It's a lot easier to make a decision when you have a small budget. Good things at low prices generally don't come in huge varieties – take headphones for example. There is the Koss KSC75 at $15 - and then a huge gaping chasm in quality until you can afford at least a pair of Grado SR60s, which come in at around$80. When you have more to spend, you have a lot more choice. At the $100 -$150 price range, you have an abundance of excellent choices. In fact, you can now choose based on other criteria, like comfort, open or closed, size and portability. Unfortunately, the inverse is also true – there are also few choices at the high end of the market. If you are in the market for a new car that costs more than $2 million dollars, you have little choice besides a Bugatti Veyron Super Sports. Needless to say, this leaves us with a conundrum most of the time. When should you splurge for the better product? There are probably a couple of factors to consider, some more obvious than others. Is the more expensive product better? This could be an objective scale or a subjective scale. Sometimes, the comparison is simple: 3-ply toilet paper is strictly better than 1-ply toilet paper (maybe, in the case where you aren't using for its intended purpose, you could make an argument otherwise). Other times, like choosing between the iPhone 5 and the Samsung Galaxy S III, you can have read countless comparisons and reviews and not come to a good conclusion. Can you tell the difference? Don't just read what you can find on the Internet and take it as gospel. Sure, a 24k gold plated HDMI cable from Monster Cable is probably strictly better than a$2 cable from Monoprice; but you won't be able to tell the difference. After all, the signals are digital and the cable is going to be hidden behind your television. Nobody (except, perhaps you) would ever care how much you spent on your damn cable.
Are you buying it because of the brand? A lot of geeks and nerds look down on this pattern of buying. Why? This behaviour means that you might be buying a product (and paying more) based simply on marketing – not its actual merit. A good example of this are Beats headphones; they aren't actually bad, but they sure are expensive. Most Beats headphones have excellent Sennheiser alternatives for less than half the price - so if you own Beats headphones, it's pretty clear your intention sways more towards flaunting your purchasing power rather proving than your audiophile cred.
Now, I'm giving pretty specific examples with clear cut differences. Really, the product you can comparing can come with numerous differences that have their tradeoffs – and this can include things like after sales service, included accessories, aesthetics and even seemingly tiny items like a special hinge on a laptop or an additional button on a mouse.
So when should you spend more? It's really all up to you. Buy what's best for you. Who cares what angry people on the Internet think?
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2018-09-25 16:52:04
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https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_24&diff=cur&oldid=71436
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# Difference between revisions of "2001 AMC 10 Problems/Problem 24"
## Problem
In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB, and $AD=7$. What is $AB\cdot CD$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$
## Solution
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((0.2,4.92), 1.3)); draw(circle((1.04,1.58), 2.14)); draw((-1.1,4.92)--(0.2,4.92)); draw((0.2,4.92)--(1.04,1.58)); draw((1.04,1.58)--(-1.1,1.58)); draw((-1.1,1.58)--(-1.1,4.92)); /* dots and labels */ dot((-1.1,4.92),dotstyle); label("A", (-1.02,5.12), NE * labelscalefactor); dot((0.2,4.92),dotstyle); label("B", (0.28,5.12), NE * labelscalefactor); dot((-1.1,1.58),dotstyle); label("D", (-1.02,1.78), NE * labelscalefactor); dot((1.04,1.58),dotstyle); label("C", (1.12,1.78), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
If $AB=x$ and $CD=y$, then $BC=x+y$. By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49.$ Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$.
## Solution 2
Simpler is just drawing the trapezoid and then using what is given to solve. Draw a line parallel to $\overline{AD}$ that connects the longer side to the corner of the shorter side. Name the bottom part $x$ and top part $a$. By the Pythagorean theorem, it is obvious that $a^{2} + 49 = (2x+a)^{2}$ (the RHS is the fact the two sides added together equals that). Then, we get $a^2 + 49 = 4x^2 + 4ax + a^2$, cancel out and factor and we get $49 = 4x(x+a)$. Notice that $x(x+a)$ is what the question asks, so the answer is $\boxed{\textbf{(B)}\ 12.25}$.
Solution by IronicNinja
## Solution 3
We know it is a trapezoid and that $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$. If they are perpendicular to $\overline{AD}$ that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know $\overline{AD}$ is $7$. We can then set the length of $\overline{AB}$ to be $x$ and the length of $\overline{DC}$ to be $y$. $\overline{BC}$ would then be $x+y$. Let's draw a straight line down from point $B$ which is perpendicular to $\overline{DC}$ and parallel to $\overline{AD}$. Let's name this line $M$. Then let's name the point at which line $M$ intersects $\overline{DC}$ point $E$. Line $M$ partitions the trapezoid into rectangle $ADEB$ and triangle $BEC$. We will use the triangle to solve for $x \times y$ using the Pythagorean theorem. The line segment $\overline{EC}$ would be $y-x$ because $\overline{DC}$ is $y$ and $\overline{DE}$ is $x$. $\overline{DE}$ is $x$ because it is parallel to $\overline{AB}$ and both are of equal length. Because of the Pythagorean theorem, we know that $(EC)^2+(BE)^2=(BC)^2$. Substituting the values we have we get $(y-x)^2+(7)^2=(x+y)^2$. Simplifying this we get $(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)$. Now we get rid of the $x^2$ and $y^2$ terms from both sides to get $(-2xy)+(49)=(2xy)$. Combining like terms we get $(49)=(4xy)$. Then we divide by $4$ to get $(12.25)=(xy)$. Now we know that $x \times y$ (same thing as $xy$) is equal to $12.25$ which is answer choice $\boxed{\textbf{(B)}\ 12.25}$.
Solution By: MATHCOUNTSCMS25
P.S. I Don't Know How to Format It Properly Using $\LaTeX$ So Could Someone Please Fix It
EDIT: Fixed! (As much as my ability can) - Mliu630XYZ
EDIT: Fixed Completely! - palaashgang
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2021-10-18 19:20:41
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http://blog.omega-prime.co.uk/2012/12/06/rpath-emulation-absolute-dll-references-on-windows/
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When creating an executable or shared library on Linux, it’s possible to include an ELF RPATH header which tells the dynamic linker where to search for the any shared libraries that you reference. This is a pretty handy feature because it can be used to nail down exactly which shared library you will link against, without leaving anything up to chance at runtime.
Unfortunately, Windows does not have an equivalent feature. However, it does have an undocumented feature which may be enough to replace your use of rpath if you are porting software from Linux.
Executables or DLLs or Windows always reference any DLLs that they import by name only. So, the import table for an executable will refer to kernel32.dll rather than C:\Windows\kernel32.dll. Window’s dynamic loader will look for a file with the appropriate name in the DLL search path as usual. (For full details on DLL import tables and more, you can check out my previous in depth post.)
However, Window’s dynamic loader will, as a completely undocumented (and presumably unsupported) feature, also accept absolute paths in the import table. This is game-changing because it means that you can hard-code exactly which DLL you want to refer to, just like you would be able to with rpath on Linux.
# Demonstration
To demonstrate this technique, we’re going to need code for a DLL and a referring EXE:
If we were building a DLL and EXE normally, we would do this:
gcc -c library.c
gcc -shared -o library.dll library.o
gcc -o rpath rpath.c -L./ -llibrary
This all works fine:
$./rpath Made library call! However, as you would expect, if you move library.dll elsewhere, the EXE will fail to start: $ mv library.dll C:/library.dll
$./rpath /home/Max/rpath/rpath.exe: error while loading shared libraries: library.dll: cannot open shared object file: No such file or directory Now let’s work some magic! If we open up rpath.exe in a hex editor, we see something like this: Let’s just tweak that a bit to change the relative path to library.dll to an absolute path. Luckily there is enough padding to make it fit: The EXE will now work perfectly! $ ./rpath
# In practice
Knowing that this feature exists is one thing. Actually making use of it in a reliable way is another. The problem is that to my knowledge no linkers are capable of creating a DLL or EXE which include an absolute path in their import tables. Sometimes we will be lucky enough that the linker creates an EXE or DLL with enough padding in it for us to manually edit in an absolute path, but with the method above there is no guarantee that this will be possible.
In order to exploit this technique robustly, we’re going to use a little trick with import libraries. Instead of using GCC’s ability to link directly to a DLL, we will generate an import library for the DLL, which we will call library.lib:
$dlltool --output-lib library.lib --dllname veryverylongdllname.dll library.o When you use dlltool you either need to write a .def file for the DLL you are creating an import library for, or you need to supply all the object files that were used to create the DLL. I’ve taken the second route here and just told dlltool that the our DLL was built from library.o. Now we have an import library, we can do our hex-editing trick again, but this time on the library. Before: And after (note that I have null-terminated the new absolute path): The beauty of editing the import library rather than the output of the linker is that using the --dllname option we can ensure that the import library contains as much space as we need to fit the entire absolute path of the DLL, no matter how long it may be. This is the key to making robust use of absolute paths in DLL loading, even if linkers don’t support them! Now we have the import library, we can link rpath.exe again, but this time using the import library rather than library.dll: $ gcc -o rpath rpath.c library.lib
$./rpath Made library call! Yes, it really is using the DLL on the C: drive: $ mv C:/library.dll C:/foo.dll
\$ ./rpath
/home/Max/rpath/rpath.exe: error while loading shared libraries: C:\library.dll: cannot open shared object file: No such file or directory
# Conclusion
I haven’t seen this technique for using absolute paths for DLL references anywhere on the web, so it doesn’t seem to be widely known. However, it works beautifully on Windows 7 and probably on all other versions of Windows as well.
I may apply these techniques to the Glasgow Haskell Compiler in order to improve the support for Haskell shared objects on Windows: more information on this topic can be found on the GHC wiki.
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2018-08-17 01:54:44
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https://colinsalmcorner.com/install-and-configure-sql-server-using-powershell-dsc/
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# Install and Configure SQL Server using PowerShell DSC
I’m well into my journey of discovering the capabilities of PowerShell DSC and Release Management’s DSC feature (See my previous posts: PowerShell DSC: Configuring a Remote Node to “Reboot If Needed”, Using PowerShell DSC in Release Management: The Hidden Manual and More DSC Release Management Goodness: Readying a Webserver for Deployment). I’ve managed to work out how to use Release Management to run DSC scripts on nodes. Now I am trying to construct a couple of scripts that I can use to deploy applications to servers – including, of course, configuring the servers – using DSC. (All scripts for this post are available for download here).
## SQL Server Installation
To install SQL Server via a script, there are two prerequisites: the SQL install sources and a silent (or unattended) installation command.
Fortunately the SQL server installer takes care of the install command – you run the install wizard manually, specifying your installation options as you go. On the last page, just before clicking “Install”, you’ll see a path to the ini conifguration file. I saved the configuration file and cancelled the install. Then I opened the config file and tweaked it slightly (see this post and this post on some tweaking ideas)– till I could run the installer from the command line (using the /configurationFile switch). That takes care of the install command itself.
There are many ways to make the SQL installation sources available to the target node. I chose to copy the ISO to the node (using the File DSC resource) from a network share, and then use a Script resource to mount the iso. Once it’s mounted, I can run the setup command using the ini file.
SQL Server requires .NET 3.5 to be installed on the target node, so I’ve added that into the script using the WindowsFeature resource. Here’s the final script:
Configuration SQLInstall
{
param (
[Parameter(Mandatory=$true)] [ValidateNotNullOrEmpty()] [String]$PackagePath,
[Parameter(Mandatory=$true)] [ValidateNotNullOrEmpty()] [String]$WinSources
)
Node $AllNodes.where{$_.Role.Contains("SqlServer") }.NodeName
{
Log ParamLog
{
Message = "Running SQLInstall. PackagePath = $PackagePath" } WindowsFeature NetFramework35Core { Name = "NET-Framework-Core" Ensure = "Present" Source =$WinSources
}
WindowsFeature NetFramework45Core
{
Name = "NET-Framework-45-Core"
Ensure = "Present"
Source = $WinSources } # copy the sqlserver iso File SQLServerIso { SourcePath = "$PackagePath\en_sql_server_2012_developer_edition_x86_x64_dvd_813280.iso"
DestinationPath = "c:\temp\SQLServer.iso"
Type = "File"
Ensure = "Present"
}
# copy the ini file to the temp folder
File SQLServerIniFile
{
SourcePath = "$PackagePath\ConfigurationFile.ini" DestinationPath = "c:\temp" Type = "File" Ensure = "Present" DependsOn = "[File]SQLServerIso" } # # Install SqlServer using ini file # Script InstallSQLServer { GetScript = {$sqlInstances = gwmi win32_service -computerName localhost | ? { $_.Name -match "mssql*" -and$_.PathName -match "sqlservr.exe" } | % { $_.Caption }$res = $sqlInstances -ne$null -and $sqlInstances -gt 0$vals = @{
Installed = $res; InstanceCount =$sqlInstances.count
}
$vals } SetScript = { # mount the iso$setupDriveLetter = (Mount-DiskImage -ImagePath c:\temp\SQLServer.iso -PassThru | Get-Volume).DriveLetter + ":"
if ($setupDriveLetter -eq$null) {
throw "Could not mount SQL install iso"
}
Write-Verbose "Drive letter for iso is: $setupDriveLetter" # run the installer using the ini file$cmd = "$setupDriveLetter\Setup.exe /ConfigurationFile=c:\temp\ConfigurationFile.ini /SQLSVCPASSWORD=P2ssw0rd /AGTSVCPASSWORD=P2ssw0rd /SAPWD=P2ssw0rd" Write-Verbose "Running SQL Install - check %programfiles%\Microsoft SQL Server\120\Setup Bootstrap\Log\ for logs..." Invoke-Expression$cmd | Write-Verbose
}
TestScript =
{
$sqlInstances = gwmi win32_service -computerName localhost | ? {$_.Name -match "mssql*" -and $_.PathName -match "sqlservr.exe" } | % {$_.Caption }
$res =$sqlInstances -ne $null -and$sqlInstances -gt 0
if ($res) { Write-Verbose "SQL Server is already installed" } else { Write-Verbose "SQL Server is not installed" }$res
}
}
}
}
# command for RM
#SQLInstall -ConfigurationData $configData -PackagePath "\\rmserver\Assets" -WinSources "d:\sources\sxs" # test from command line SQLInstall -ConfigurationData configData.psd1 -PackagePath "\\rmserver\Assets" -WinSources "d:\sources\sxs" Start-DscConfiguration -Path .\SQLInstall -Verbose -Wait -Force Here’s some analysis: • (Line 7 / 12) The config takes in 2 parameters:$PackagePath (location of SQL ISO and config ini file) and $WinSources (Path to windows sources). • (Line 15) I changed my config data so that I can specify a comma-separated list of roles (since a node might be a SQLServer and a WebServer) so I’ve made the comparer a “contains” rather than an equals (as I’ve had in my previous scripts) – see the config script below. • (Line 22 / 29) Configure .NET 3.5 and .NET 4.5 Windows features, using the$WinSources path if the sources are required
• (Line 37) Copy the SQL iso to the target node from the $PackagePath folder • (Line 46) Copy the ini file to the target node from the$PackagePath folder
• (Line 58) Begins the Script to install SQL server
• The Get-Script does a check to see if there is a SQL server service running. If there is, it returns the SQL instance count for the machine.
• The Set-Script mounts the iso, saving the drive letter to a variable. Then I invoke the setup script (passing in the config file and required passwords) writing the output to Write-Verbose, which will appear on the DSC invoking machine as the script executes.
• The Test-Script does the same basic “is there a SQL server service running” check. If there is, skip the install – else run the install. Of course this could be refined to ensure each and every component is installed, but I didn’t want to get that granular.
• The last couple of lines of the script show the command for Release Management (commented out) as well as the command to run the script manually from a PowerShell prompt.
Here’s my DSC config script:
#$configData = @{ @{ AllNodes = @( @{ NodeName = "*" PSDscAllowPlainTextPassword =$true
},
@{
NodeName = "fabfiberserver"
Role = "WebServer,SqlServer"
}
);
}
# Note: different 1st line for RM or command line invocation
# use \$configData = @{ for RM
# use @{ for running from command line
You can download the above scripts (and my SQL configuration ini file for reference) here.
## What’s Next
After running this script, I have a server with SQL Server installed and configured according to my preferences (which are contained in the ini file). From here, I can run restores or dacpac deployments and so on. Of course this is going to be executed from within Release Management as part of the release pipeline.
Next up will be the full WebServer DSC script – and then we’ll be ready to tackle the actual application deployment, since we’ll have servers ready to host our applications.
Until then, happy releasing!
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2022-08-08 07:18:33
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https://practice.geeksforgeeks.org/problems/special-series-sum0903/1
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Special Series Sum
Basic Accuracy: 75.79% Submissions: 2049 Points: 1
Given the value of n, we need to find the sum of the series where i-th term is sum of first i natural numbers.
NOTE: Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + …… + (1+2+3+4+…+n)
Example 1:
Input: n = 5
Output: 35
Explanation: 1 + (1+2) + (1+2+3).. = 35
Hence sum of the series is 35.
Example 2:
Input: n = 10
Output: 220
Explanation: 1 + (1+2) + (1+2+3) +
(1+2+3+4) + (1+2+3+4+5)... = 210.
Hence sum of the series is 210.
You dont need to read input or print anything. Complete the function sumOfTheSeries() which takes n as input parameter and returns the sum of the series of n terms.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1<= n <=1000
We are replacing the old Disqus forum with the new Discussions section given below.
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2021-09-17 04:05:38
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https://programmathically.com/introduction-to-the-hypothesis-space-and-the-bias-variance-tradeoff-in-machine-learning/
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# Introduction to the Hypothesis Space and the Bias-Variance Tradeoff in Machine Learning
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In this post, we introduce the hypothesis space and discuss how machine learning models function as hypotheses. Furthermore, we discuss the challenges encountered when choosing an appropriate machine learning hypothesis and building a model, such as overfitting, underfitting, and the bias-variance tradeoff.
The hypothesis space in machine learning is a set of all possible models that can be used to explain a data distribution given the limitations of that space. A linear hypothesis space is limited to the set of all linear models. If the data distribution follows a non-linear distribution, the linear hypothesis space might not contain a model that is appropriate for our needs.
To understand the concept of a hypothesis space, we need to learn to think of machine learning models as hypotheses.
## The Machine Learning Model as Hypothesis
Generally speaking, a hypothesis is a potential explanation for an outcome or a phenomenon. In scientific inquiry, we test hypotheses to figure out how well and if at all they explain an outcome.
In supervised machine learning, we are concerned with finding a function that maps from inputs to outputs.
But machine learning is inherently probabilistic. It is the art and science of deriving useful hypotheses from limited or incomplete data. Our functions are not axioms that explain the data perfectly, and for most real-life problems, we will never have all the data that exists. Accordingly, we will not find the one true function that perfectly describes the data.
Instead, we find a function through training a model to map from known training input to known training output. This way, the model gradually approximates the assumed true function that describes the distribution of the data.
So we treat our model as a hypothesis that needs to be tested as to how well it explains the output from a given input. We do this using a test or validation data set.
### The Hypothesis Space
During the training process, we select a model from a hypothesis space that is subject to our constraints. For example, a linear hypothesis space only provides linear models. We can approximate data that follows a quadratic distribution using a model from the linear hypothesis space.
Of course, a linear model will never have the same predictive performance as a quadratic model, so we can adjust our hypothesis space to also include non-linear models or at least quadratic models.
## The Data Generating Process
The data generating process describes a hypothetical process subject to some assumptions that make training a machine learning model possible. We need to assume that the data points are from the same distribution but are independent of each other. When these requirements are met, we say that the data is independent and identically distributed (i.i.d.).
### Independent and Identically Distributed Data
How can we assume that a model trained on a training set will perform better than random guessing on new and previously unseen data?
First of all, the training data needs to come from the same or at least a similar problem domain. If you want your model to predict stock prices, you need to train the model on stock price data or data that is similarly distributed. It wouldn’t make much sense to train it on whether data.
Statistically, this means the data is identically distributed.
But if data comes from the same problem, training data and test data might not be completely independent. To account for this, we need to make sure that the test data is not in any way influenced by the training data or vice versa. If you use a subset of the training data as your test set, the test data evidently is not independent of the training data. Statistically, we say the data must be independently distributed.
## Overfitting and Underfitting
We want to select a model from the hypothesis space that explains the data sufficiently well. During training, we can make a model so complex that it perfectly fits every data point in the training dataset. But ultimately, the model should be able to predict outputs on previously unseen input data. The ability to do well when predicting outputs on previously unseen data is also known as generalization.
There is an inherent conflict between those two requirements.
If we make the model so complex that it fits every point in the training data, it will pick up lots of noise and random variation specific to the training set, which might obscure the larger underlying patterns. As a result, it will be more sensitive to random fluctuations in new data and predict values that are far off. A model with this problem is said to overfit the training data and, as a result, to suffer from high variance.
To avoid the problem of overfitting, we can choose a simpler model or use regularization techniques to prevent the model from fitting the training data too closely. The model should then be less influenced by random fluctuations and instead, focus on the larger underlying patterns in the data. The patterns are expected to be found in any dataset that comes from the same distribution. As a consequence, the model should generalize better on previously unseen data.
But if we go too far, the model might become too simple or too constrained by regularization to accurately capture the patterns in the data. Then the model will neither generalize well nor fit the training data well. A model that exhibits this problem is said to underfit the data and to suffer from high bias. If the model is too simple to accurately capture the patterns in the data (for example, when using a linear model to fit non-linear data), its capacity is insufficient for the task at hand.
When training neural networks, for example, we go through multiple iterations of training in which the model learns to fit an increasingly complex function to the data.
Typically, your training error will decrease during learning the more complex your model becomes and the better it learns to fit the data. In the beginning, the training error decreases rapidly. In later training iterations, it typically flattens out as it approaches the minimum possible error.
Your test or generalization error should initially decrease as well, albeit likely at a slower pace than the training error. As long as the generalization error is decreasing, your model is underfitting because it doesn’t live up to its full capacity. After a number of training iterations, the generalization error will likely reach a trough and start to increase again. Once it starts to increase, your model is overfitting, and it is time to stop training.
Ideally, you should stop training once your model reaches the lowest point of the generalization error. The gap between the minimum generalization error and no error at all is an irreducible error term known as the Bayes error that we won’t be able to completely get rid of in a probabilistic setting. But if the error term seems too large, you might be able to reduce it further by collecting more data, manipulating your model’s hyperparameters, or altogether picking a different model.
We’ve talked about bias and variance in the previous section. Now it is time to clarify what we actually mean by these terms.
### Understanding Bias and Variance
In a nutshell, bias measures if there is any systematic deviation from the correct value in a specific direction. If we could repeat the same process of constructing a model several times over, and the results predicted by our model always deviate in a certain direction, we would call the result biased.
Variance measures how much the results vary between model predictions. If you repeat the modeling process several times over and the results are scattered all across the board, the model exhibits high variance.
In their book “Noise” Daniel Kahnemann and his co-authors provide an intuitive example that helps understand the concept of bias and variance.
Imagine you have four teams at the shooting range.
Team B is biased because the shots of its team members all deviate in a certain direction from the center. Team B also exhibits low variance because the shots of all the team members are relatively concentrated in one location.
Team C has the opposite problem. The shots are scattered across the target with no discernible bias in a certain direction.
Team D is both biased and has high variance.
Team A would be the equivalent of a good model. The shots are in the center with little bias in one direction and little variance between the team members.
Generally speaking, linear models such as linear regression exhibit high bias and low variance.
Nonlinear algorithms such as decision trees are more prone to overfitting the training data and thus exhibit high variance and low bias.
A linear model used with non-linear data would exhibit a bias to predict data points along a straight line instead of accomodating the curves. But they are not as susceptible to random fluctuations in the data.
A nonlinear algorithm that is trained on noisy data with lots of deviations would be more capable of avoiding bias but more prone to incorporate the noise into its predictions. As a result, a small deviation in the test data might lead to very different predictions.
To get our model to learn the patterns in data, we need to reduce the training error while at the same time reducing the gap between the training and the testing error. In other words, we want to reduce both bias and variance. To a certain extent, we can reduce both by picking an appropriate model, collecting enough training data, selecting appropriate training features and hyperparameter values.
At some point, we have to trade-off between minimizing bias and minimizing variance. How you balance this trade-off is up to you.
### The Bias Variance Decomposition
Mathematically, the total error can be decomposed into the bias and the variance according to the following formula.
Total\; Error = Bias^2 + Variance + Bayes\; Error
Remember that Bayes’ error is an error that cannot be eliminated.
Our machine learning model represents an estimating function \hat f(X) for the true data generating function f(X) where X represents the predictors and y the output values.
Now the mean squared error of our model is the expected value of the squared difference of the output produced by the estimating function \hat f(X) and the true output Y.
MSE(X) = E[(\hat f(X) - Y)^2]
The bias is a systematic deviation from the true value. We can measure it as the squared difference between the expected value produced by the estimating function (the model) and the values produced by the true data-generating function.
Bias(X) = (E[(\hat f(X)] - f(X))^2
Of course, we don’t know the true data generating function, but we do know the observed outputs Y, which correspond to the values generated by f(x) plus an error term.
Y = f(X) + \epsilon
The variance of the model is the squared difference between the expected value and the actual values of the model.
Var(X) = E[(\hat f(X) - E[\hat f(X)])^2 ]
Now that we have the bias and the variance, we can add them up along with the irreducible error to get the total error.
Err(X) = (E[(\hat f(X)] - f(X))^2 +E[(\hat f(X) - E[\hat f(X)])^2 ] + \sigma^2
## Summary
A machine learning model represents an approximation to the hypothesized function that generated the data. The chosen model is a hypothesis since we hypothesize that this model represents the true data generating function.
We choose the hypothesis from a hypothesis space that may be subject to certain constraints. For example, we can constrain the hypothesis space to the set of linear models.
When choosing a model, we aim to reduce the bias and the variance to prevent our model from either overfitting or underfitting the data. In the real world, we cannot completely eliminate bias and variance, and we have to trade-off between them.
The total error produced by a model can be decomposed into the bias, the variance, and irreducible (Bayes) error.
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2022-08-14 20:43:54
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https://en.m.wikisource.org/wiki/1911_Encyclop%C3%A6dia_Britannica/Infinitesimal_Calculus/Nature_of_the_Calculus
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# 1911 Encyclopædia Britannica/Infinitesimal Calculus/Nature of the Calculus
Infinitesimal Calculus (§1) Infinitesimal CalculusI. Nature of the Calculus (§2-12) II. History (§13-22)
I. Nature of the Calculus.
2. The guise in which variable quantities presented themselves to the mathematicians of the 17th century was that of the lengths of variable lines. This method of representing variable quantities dates from the 14th century, when it was employed by Nicole Oresme, who studied Geometrical representation of Variable Quantities. and afterwards taught at the Collège de Navarre in Paris from 1348 to 1361. He represented one of two variable quantities, e.g. the time that has elapsed since some epoch, by a length, called the “longitude,” measured along a particular line; and he represented the other of the two quantities, e.g. the temperature at the instant, by a length, called the “latitude,” measured at right angles to this line. He recognized that the variation of the temperature with the time was represented by the line, straight or curved, which joined the ends of all the lines of “latitude.” Oresme’s longitude and latitude were what we should now call the abscissa and ordinate. The same method was used later by many writers, among whom Johannes Kepler and Galileo Galilei may be mentioned. In Galileo’s investigation of the motion of falling bodies (1638) the abscissa OA represents the time during which a body has been falling, and the ordinate AB represents the velocity acquired during that time (see fig. 1). The velocity being proportional to the time, the “curve” obtained is a straight line OB, and Galileo showed that the distance through which the body has fallen is represented by the area of the triangle OAB.
Fig. 1.
The most prominent problems in regard to a curve were the problem of finding the points at which the ordinate is a maximum or a minimum, the problem of drawing a tangent to the curve at an assigned point, and the problem of determining the area of the curve. The relation of The problems of Maxima and Minima, Tangents, and Quadratures. the problem of maxima and minima to the problem of tangents was understood in the sense that maxima or minima arise when a certain equation has equal roots, and, when this is the case, the curves by which the problem is to be solved touch each other. The reduction of problems of maxima and minima to problems of contact was known to Pappus. The problem of finding the area of a curve was usually presented in a particular form in which it is called the “problem of quadratures.” It was sought to determine the area contained between the curve, the axis of abscissae and two ordinates, of which one was regarded as fixed and the other as variable. Galileo’s investigation may serve as an example. In that example the fixed ordinate vanishes. From this investigation it may be seen that before the invention of the infinitesimal calculus the introduction of a curve into discussions of the course of any phenomenon, and the problem of quadratures for that curve, were not exclusively of geometrical import; the purpose for which the area of a curve was sought was often to find something which is not an area—for instance, a length, or a volume or a centre of gravity.
3. The Greek geometers made little progress with the problem of tangents, but they devised methods for investigating the problem of quadratures. One of these methods was afterwards called the “method of exhaustions,” and the principle on which it is based was laid down in the Greek methods. lemma prefixed to the 12th book of Euclid’s Elements as follows: “If from the greater of two magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there will at length remain a magnitude less than the smaller of the proposed magnitudes.” The method adopted by Archimedes was more general. It may be described as the enclosure of the magnitude to be evaluated between two others which can be brought by a definite process to differ from each other by less than any assigned magnitude. A simple example of its application is the 6th proposition of Archimedes’ treatise On the Sphere and Cylinder, in which it is proved that the area contained between a regular polygon inscribed in a circle and a similar polygon circumscribed to the same circle can be made less than any assigned area by increasing the number of sides of the polygon. The methods of Euclid and Archimedes were specimens of rigorous limiting processes (see Function). The new problems presented by the analytical geometry and natural philosophy of the 17th century led to new limiting processes.
4. In the problem of tangents the new process may be described as follows. Let P, P′ be two points of a curve (see fig. 2). Let x, y be the coordinates of P, and x+Δx, y+Δy those of P′. The symbol Δx means “the difference of two x’s” and there is a like meaning for the symbol Δy.Differentiation.
Fig. 2.
The fraction Δy/Δx is the trigonometrical tangent of the angle which the secant PP′ makes with the axis of x. Now let Δx be continually diminished towards zero, so that P′ continually approaches P. If the curve has a tangent at P the secant PP′ approaches a limiting position (see § 33 below). When this is the case the fraction Δy/Δx tends to a limit, and this limit is the trigonometrical tangent of the angle which the tangent at P to the curve makes with the axis of x. The limit is denoted by
${\displaystyle {\frac {dy}{dx}}}$.
If the equation of the curve is of the form y=ƒ(x) where ƒ is a functional symbol (see Function), then
${\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}}$.
and
${\displaystyle {\frac {dy}{dx}}=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}$.
The limit expressed by the right-hand member of this defining equation is often written
${\displaystyle f^{\prime }(x)\,}$
and is called the “derived function” of ƒ(x), sometimes the “derivative” or “derivate” of ƒ(x). When the function ƒ(x) is a rational integral function, the division by Δx can be performed, and the limit is found by substituting zero for Δx in the quotient. For example, if ƒ(x) = x2, we have
${\displaystyle {\frac {f(x+\Delta x)-f(x)}{\Delta x}}={\frac {f(x+\Delta x)^{2}-x^{2}}{\Delta x}}={\frac {2x\Delta x+(\Delta x)^{2}}{\Delta x}}=2x+\Delta x}$
and
${\displaystyle f^{\prime }(x)=2x\,}$
The process of forming the derived function of a given function is called differentiation. The fraction Δy/Δx is called the “quotient of differences,” and its limit dy/dx is called the “differential coefficient of y with respect to x.” The rules for forming differential coefficients constitute the differential calculus.
The problem of tangents is solved at one stroke by the formation of the differential coefficient; and the problem of maxima and minima is solved, apart from the discrimination of maxima from minima and some further refinements, by equating the differential coefficient to zero (see Maxima and Minima).
5. The problem of quadratures leads to a type of limiting process which may be described as follows: Let yx be the equation of a curve, and let AC and BD be the ordinates of the points C and D (see fig. 3). Let a, b be the abscissae of these points. Let the segment AB be divided into a numberIntegration. of segments by means of intermediate points such as M, and let MN be one such segment. Let PM and QN be those ordinates of the curve which have M and N as their feet. On MN as base describe
Fig. 3.
two rectangles, of which the heights are the greatest and least values of y which correspond to points on the arc PQ of the curve. In fig. 3 these are the rectangles RM, SN. Let the sum of the areas of such rectangles as RM be formed, and likewise the sum of the areas of such rectangles as SN. When the number of the points such as M is increased without limit, and the lengths of all the segments such as MN are diminished without limit, these two sums of areas tend to limits. When they tend to the same limit the curvilinear figure ACDB has an area, and the limit is the measure of this area (see § 33 below). The limit in question is the same whatever law may be adopted for inserting the points such as M between A and B, and for diminishing the lengths of the segments such as MN. Further, if P′ is any point on the arc PQ, and P′M′ is the ordinate of P′, we may construct a rectangle of which the height is P′M′ and the base is MN, and the limit of the sum of the areas of all such rectangles is the area of the figure as before. If x is the abscissa of P, x+Δx that of Q, x′ that of P′, the limit in question might be written
${\displaystyle lim.{\begin{matrix}\sum _{a}^{b}f(x^{\prime })\Delta x\end{matrix}}}$
where the letters a, b written below and above the sign of summation Σ indicate the extreme values of x. This limit is called “the definite integral of ƒ(x) between the limits a and b,” and the notation for it is
${\displaystyle \int _{a}^{b}f(x)\,dx}$
The germs of this method of formulating the problem of quadratures are found in the writings of Archimedes. The method leads to a definition of a definite integral, but the direct application of it to the evaluation of integrals is in general difficult. Any process for evaluating a definite integral is a process of integration, and the rules for evaluating integrals constitute the integral calculus.
6. The chief of these rules is obtained by regarding the extreme ordinate BD as variable. Let ξ now denote the abscissa of B. The area A of the figure ACDB is represented by the integral ${\displaystyle \int _{a}^{\xi }f(x)\,dx}$, and it is a function of ξ. Let BD be displaced to B′D′ so that becomes ${\displaystyle \xi +\Delta \xi }$ (seeTheorem of Inversion. fig. 4). The area of the figure ACD′B′ is represented by the integral ${\displaystyle \int _{a}^{\xi +\Delta \xi }f(x)\,dx}$ and the increment ΔA is given by the formula:
${\displaystyle \Delta A=\int _{\xi }^{\xi +\Delta \xi }f(x)\,dx}$
which represents the area BDD′B′.
Fig. 4.
This area is intermediate between those of two rectangles, having as a common base the segment BB′, and as heights the greatest and least ordinates of points on the arc DD′ of the curve. Let these heights be H and h. Then ΔA is intermediate between HΔξ and hΔξ, and the quotient of differences ΔA/Δξ is intermediate between H and h. If the function ƒ(x) is continuous at B (see Function), then, as Δξ is diminished without limit, H and h tend to BD, or ƒ(ξ), as a limit, and we have:
${\displaystyle {\frac {d{\text{A}}}{d\xi }}=f(\xi )}$
The introduction of the process of differentiation, together with the theorem here proved, placed the solution of the problem of quadratures on a new basis. It appears that we can always find the area A if we know a function F(x) which has ƒ(x) as its differential coefficient. If ƒ(x) is continuous between a and b, we can prove that
${\displaystyle A=\int _{a}^{b}f(x)\,dx=F(b)-F(a)}$
When we recognize a function F(x) which has the property expressed by the equation
${\displaystyle {\frac {dF(x)}{dx}}=f(x)}$,
we are said to integrate the function ƒ(x), and F(x) is called the indefinite integral of ƒ(x) with respect to x, and is written
${\displaystyle \int f(x)\,dx}$
7. In the process of § 4 the increment Δy is not in general equal to the product of the increment Δx and the derived function ƒ′(x). In general we can write down an equation of the formDifferentials.
${\displaystyle \Delta y=f^{\prime }(x)\Delta x+{\text{R}}}$,
in which R is different from zero when Δx is different from zero; and then we have not only
${\displaystyle \lim _{\Delta x\to 0}{\text{R}}=0}$,
but also
${\displaystyle \lim _{\Delta x\to 0}{\frac {\text{R}}{\Delta x}}=0}$
We may separate Δy into two parts: the part ƒ′(x)Δx and the part R. The part ƒ′(x)Δx alone is useful for forming the differential coefficient, and it is convenient to give it a name. It is called the differential of ƒ(x), and is written dƒ(x), or dy when y is written for ƒ(x). When this notation is adopted dx is written instead of Δx, and is called the “differential of x,” so that we have
${\displaystyle df(x)=f^{\prime }(x)dx.}$
Thus the differential of an independent variable such as x is a finite difference; in other words it is any number we please. The differential of a dependent variable such as y, or of a function of the independent variable x, is the product of the differential of x and the differential coefficient or derived function. It is important to observe that the differential coefficient is not to be defined as the ratio of differentials, but the ratio of differentials is to be defined as the previously introduced differential coefficient. The differentials are either finite differences, or are so much of certain finite differences as are useful for forming differential coefficients.
Again let F(x) be the indefinite integral of a continuous function ƒ(x), so that we have
${\displaystyle {\frac {dF(x)}{dx}}=f(x),\int _{a}^{b}f(x)\,dx=F(b)-F(a)}$
When the points M of the process explained in § 5 are inserted between the points whose abscissae are a and b, we may take them to be n − 1 in number, so that the segment AB is divided into n segments. Let x1, x2, ... xn−1 be the abscissae of the points in order. The integral is the limit of the sum
${\displaystyle f(a)(x_{1}-a)+f(x_{1})(x_{2}-x_{1})+\ldots +f(x_{r})(x_{r+1}-x_{r})+\ldots +f(x_{n-1})(b-x_{n-1})}$
every term of which is a differential of the form ƒ(x)dx. Further the integral is equal to the sum of differences
${\displaystyle {F(x_{1})-F(a)}+{F(x_{2})-F(x_{1})}+\ldots +{F(x_{r+1})-F(x_{r})}+\ldots +{F(b)-F(x_{n-1})}}$
for this sum is F(b) − F(a). Now the difference F(xr+1) − F(xr) is not equal to the differential ƒ(xr) (xr+1xr), but the sum of the differences is equal to the limit of the sum of these differentials. The differential may be regarded as so much of the difference as is required to form the integral. From this point of view a differential is called a differential element of an integral, and the integral is the limit of the sum of differential elements. In like manner the differential element ydx of the area of a curve (§ 5) is not the area of the portion contained between two ordinates, however near together, but is so much of this area as need be retained for the purpose of finding the area of the curve by the limiting process described.
8. The notation of the infinitesimal calculus is intimately bound up with the notions of differentials and sums of elements. The letter “d ” is the initial letter of the word differentia (difference) and the symbol “∫” is a conventionally written “S”, the initial letter of the word summa Notation.(sum or whole). The notation was introduced by Leibnitz (see §§ 25-27, below).
9. The fundamental artifice of the calculus is the artifice of forming differentials without first forming differential coefficients. From an equation containing x and y we can deduce a new equation, containing also Δx and Δy, by substituting x+Δx for x and y+Δy for y. If there is a differential coefficient Fundamental Artifice.of y with respect to x, then Δy can be expressed in the form φ.Δx + R, where lim.Δx = 0 (R/Δx) = 0, as in § 7 above. The artifice consists in rejecting ab initio all terms of the equation which belong to R. We do not form R at all, but only φ.Δx, or φ.dx, which is the differential dy. In the same way, in all applications of the integral calculus to geometry or mechanics we form the element of an integral in the same way as the element of area y.dx is formed. In fig. 3 of § 5 the element of area y.dx is the area of the rectangle RM. The actual area of the curvilinear figure PQNM is greater than the area of this rectangle by the area of the curvilinear figure PQR; but the excess is less than the area of the rectangle PRQS, which is measured by the product of the numerical measures of MN and QR, and we have
${\displaystyle {\begin{matrix}\lim _{MN\to 0}{\frac {MN.QR}{MN}}\end{matrix}}=0.}$
Thus the artifice by which differential elements of integrals are formed is in principle the same as that by which differentials are formed without first forming differential coefficients.
10. This principle is usually expressed by introducing the notion of orders of small quantities. If x, y are two variable numbers which are connected together by any relation, and if when x tends to zero y also tends to zero, the fraction y/x may tend to a finite limit. In this case x and y are said to be “of the Orders of small quantities.same order.” When this is not the case we may have either
${\displaystyle {\begin{matrix}\lim _{x\to 0}{\frac {x}{y}}\end{matrix}}=0,}$
or
${\displaystyle {\begin{matrix}\lim _{x\to 0}{\frac {y}{x}}\end{matrix}}=0.}$
In the former case y is said to be “of a lower order” than x; in the latter case y is said to be “of a higher order” than x. In accordance with this notion we may say that the fundamental artifice of the infinitesimal calculus consists in the rejection of small quantities of an unnecessarily high order. This artifice is now merely an incident in the conduct of a limiting process, but in the 17th century, when limiting processes other than the Greek methods for quadratures were new, the introduction of the artifice was a great advance.
11. By the aid of this artifice, or directly by carrying out the appropriate limiting processes, we may obtain the rules by which differential coefficients are formed. These rules may be classified as “formal rules” and “particular results.” The formal rules may be stated as Rules of Differentiation. follows:—
(i.) The differential coefficient of a constant is zero
(ii.) For a sum u+v+ . . ., where u,v,... are functions of x,
${\displaystyle {\frac {d(u+v+\ldots +z)}{dx}}={\frac {du}{dx}}+{\frac {dv}{dx}}+\ldots +{\frac {dz}{dx}}}$
(iii.) For a product uv
${\displaystyle {\frac {d(uv)}{dx}}=u{\frac {dv}{dx}}+v{\frac {du}{dx}}.}$
(iv.) For a quotient u/v
${\displaystyle {\frac {d(u/v)}{dx}}=\left(v{\frac {du}{dx}}-u{\frac {dv}{dx}}\right){\Bigg /}v^{2}.}$
(v.) For a function of a function, that is to say, for a function y expressed in terms of a variable z, which is itself expressed as a function of x,
${\displaystyle {\frac {dy}{dx}}={\frac {dy}{dz}}\cdot {\frac {dz}{dx}}.}$
In addition to these formal rules we have particular results as to the differentiation of simple functions. The most important results are written down in the following table:—
${\displaystyle y\,}$ ${\displaystyle {\frac {dy}{dx}}}$ ${\displaystyle x^{n}\,}$ ${\displaystyle {\begin{matrix}nx^{n-1}\\{\mbox{ for all values of }}n\end{matrix}}}$ ${\displaystyle \log _{a}x\,}$ ${\displaystyle x^{-1}log_{a}e\,}$ ${\displaystyle a^{x}}$ ${\displaystyle a^{x}log_{e}a\,}$ ${\displaystyle \sin {x}\,}$ ${\displaystyle \cos {x}\,}$ ${\displaystyle \cos {x}\,}$ ${\displaystyle -\sin {x}\,}$ ${\displaystyle \sin ^{-1}x\,}$ ${\displaystyle (1-x^{2})^{-{\tfrac {1}{2}}}\,}$ ${\displaystyle \tan ^{-1}x\,}$ ${\displaystyle (1+x^{2})^{-1}\,}$
Each of the formal rules, and each of the particular results in the table, is a theorem of the differential calculus. All functions (or rather expressions) which can be made up from those in the table by a finite number of operations of addition, subtraction, multiplication or division can be differentiated by the formal rules. All such functions are called explicit functions. In addition to these we have implicit functions, or such as are determined by an equation containing two variables when the equation cannot be solved so as to exhibit the one variable expressed in terms of the other. We have also functions of several variables. Further, since the derived function of a given function is itself a function, we may seek to differentiate it, and thus there arise the second and higher differential coefficients. We postpone for the present the problems of differential calculus which arise from these considerations. Again, we may have explicit functions which are expressed as the results of limiting operations, or by the limits of the results obtained by performing an infinite number of algebraic operations upon the simple functions. For the problem of differentiating such functions reference may be made to Function.
12. The processes of the integral calculus consist largely in transformations of the functions to be integrated into such forms that they can be recognized as differential coefficients of functions which have previously been differentiated. Corresponding to the results in the table of § 11 we Indefinite Integrals. have those in the following table:—
${\displaystyle f(x)\,}$ ${\displaystyle \int f(x)dx}$ ${\displaystyle x^{n}\,}$ ${\displaystyle {\begin{matrix}{\frac {x^{n+1}}{n+1}}\\{\mbox{ for all values of }}n{\mbox{ except }}-1\end{matrix}}}$ ${\displaystyle {\frac {1}{x}}\,}$ ${\displaystyle \log _{e}x\,}$ ${\displaystyle e^{ax}\,}$ ${\displaystyle a^{-1}e^{ax}\,}$ ${\displaystyle \cos {x}\,}$ ${\displaystyle \sin {x}\,}$ ${\displaystyle \sin {x}\,}$ ${\displaystyle -\cos {x}\,}$ ${\displaystyle (a^{2}-x^{2})^{-{\frac {1}{2}}}\,}$ ${\displaystyle \sin ^{-1}{\frac {x}{a}}\,}$ ${\displaystyle {\frac {1}{a^{2}+x^{2}}}}$ ${\displaystyle {\frac {1}{a}}\tan ^{-1}{\frac {x}{a}}}$
The formal rules of § 11 give us means for the transformation of integrals into recognizable forms. For example, the rule (ii.) for a sum leads to the result that the integral of a sum of a finite number of terms is the sum of the integrals of the several terms. The rule (iii.) for a product leads to the method of integration by parts. The rule (v.) for a function of a function leads to the method of substitution (see § 48 below).
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2022-07-02 23:26:20
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https://leanprover-community.github.io/archive/stream/113489-new-members/topic/Negated.20proof.20on.20else-branch.html
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## Stream: new members
### Topic: Negated proof on else-branch
#### Marcus Rossel (Feb 24 2021 at 20:27):
If-then-else expressions give us the option of obtaining a proof of the condition we're checking on the then-branch:
if h : my_condition
then use_condition h
else ...
Is there a way of obtaining a proof of ¬h on the else-branch?
#### Alex J. Best (Feb 24 2021 at 20:29):
In the else branch the identifier h should refer to a term of type ¬my_condition
Last updated: May 13 2021 at 21:12 UTC
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2021-05-13 21:48:04
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http://mathoverflow.net/questions/5865/classical-calculi-as-universal-quotients?sort=votes
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# Classical Calculi as Universal Quotients
As is well known, every differential calculus $(\Omega,d)$ over an algebra $A$ is a quotient of the universal calculus $(\Omega_A,d)$, by some ideal $I$. In the classical case, when $A$ is the coordinate ring of a variety $V(J)$ (for some ideal of polynomials $J$), and $(\Omega,d)$ is its ordinary calculus, how is $I$ related to $J$?
-
In the classical case, if $\Omega(A)$ is the kernel of the multiplication map $m:A\otimes A\to A$, then—since $A$ is commutative, so that $m$ is not only a map of $A$-bimodules but also a morphism of $k$-algebras,—it turns out that $\Omega(A)$ is an ideal of $A\otimes A$, not only a sub-$A$-bimodule. In particular, you can compute its square $(\Omega(A))^2$. Then the classical module of Kähler differentials $\Omega^1_{A/k}$ is the quotient $\Omega(A)/\Omega(A)^2$.
One way to think of this is that $\omega(A)$ is universal for derivations into arbitrary $A$-bimodules; the "classical" Kähler module $\omega(A)/\omeg(A)^2$ is universal for derivations into symmetric (a.k.a. commutative) $A$-bimodules. So the change of category means one works with a different notion of "universal derivation". – Yemon Choi Nov 18 '09 at 4:08
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2015-11-25 08:32:25
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https://math.stackexchange.com/questions/401707/true-or-false-statements-odd-integers
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# True or false statements odd integers
One is false one is true write the negation for the false one and prove both of them
a) for all integers a, there exists an integer b so that a+b=odd
b) there exists an integer b such that for all integers a, a+b=odd
I said (a) is false and (b) is true. I wrote the negation for (a) then I split it in two cases. Case 1: b is even. Let b=2k and let a=0 then a+b=2k which is even. Case 2: b is odd. Let b=2k+1 and let a=-1 then a+b=2k
Then for (b) I set up two cases again. Case 1: a is even. A=2k and b=1 then a+b=2k+1 which is odd Case 2: a is odd, let a=2k+1 and b=0 then a+b=2k+1 which is odd
Input in whether I'm right or not would be great thanks
I’m afraid that you got them exactly backwards. Let’s take a look.
(a) For each integer $a$ there is an integer $b$ such that $a+b$ is odd.
Think of this in terms of a game. I give you an integer $a$, and you win if you can find an integer $b$ such that $a+b$ is odd; if you cannot find such a $b$, I win. The assertion (a) says that you can always win, no matter how I choose my $a$.
Suppose I give you the integer $101$; can you find an integer to add to it to make an odd integer? Sure: it’s already odd, so just add $0$. Suppose instead that I give you the integer $100$; can you find an integer to add to it to make an odd integer? $0$ won’t work, but $1$ will: $100+1=101$, which is certainly odd. Was there anything very special here about $101$ and $100$? No: all I used was the fact that $101$ is odd, so that adding $0$ was bound to give me an odd total, and the fact that $100$ is even, so that adding $1$ would give me an odd total. If I give you any odd integer $a$, you can use $b=0$ and be sure that $a+b=a+0=a$ will be odd. And if I give you any even integer $a$, you can use $b=1$ and be sure that $a+b=a+1$ is odd. Every integer is either even or odd, so no matter what integer $a$ I give you, you’re covered: you know how to pick a $b$ such that $a+b$ is odd. In other words, (a) is true: you do have a winning strategy.
(Of course there are other choices that work besides the ones that I’ve mentioned; mine are just the simplest.)
(b) There is an integer $b$ such that for all integers $a$, $a+b$ is odd.
Again you can think of this in terms of a game, with you picking $b$ and me picking $a$. The difference is that in this game you have to play first: you pick some integer $b$, then I pick my $a$, knowing what your $b$ is. You win if $a+b$ is odd, you lose if it isn’t, and (b) says that you have a winning strategy: there is some $b$ that you can pick that will make $a+b$ odd no matter what integer I pick for $a$.
But that’s clearly not true: if your $b$ is even, I’ll just let $a=0$, so that $a+b=b$ is even, and you lose. And if your $b$ is odd, I’ll pick $a=1$, so that $a+b=1+b$ is even, and again you lose. No matter how you play — no matter what integer you choose for your $b$ — I can beat you by choosing $a$ to make $a+b$ even.
Here the first true,because if $a=2k$( is even) choose $b=1$ if $a=2k+1$(odd) choose $b=2$ to complete proving the first fact.
In your proof i guess you have missed out the critical point "there exists".
• The first statement is not the negation of the second: the negation of the second is for each integer $b$ there is an integer $a$ such that $a+b$ is even. May 25, 2013 at 3:39
• Isnt this the contrapositive of the 2nd statement???? I am a bit confused. May 25, 2013 at 3:42
• Only implications have contrapositives. Neither statement is of the form $\varphi\to\psi$, so neither has a contrapositive. (And negation and contrapositive are completely different things.) May 25, 2013 at 3:47
• Thanks @BrianM.Scott for clarifying . May 25, 2013 at 3:56
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2022-12-03 04:41:58
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http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P3869&l=en
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Magyar Information Contest Journal Articles
# Problem P. 3869. (February 2006)
P. 3869. A sample of ideal gas is taken through the cyclic process shown in the figure. The temperature of the gas at state A is TA=200 K. At states B and C the temperature of the gas is the same.
a) What is the greatest temperature of the gas during the cyclic process?
b) Graph the cyclic process in the T-V diagram.
(4 pont)
Deadline expired on 13 March 2006.
### Statistics:
125 students sent a solution. 4 points: 54 students. 3 points: 30 students. 2 points: 2 students. 1 point: 22 students. 0 point: 8 students. Unfair, not evaluated: 9 solutions.
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2017-12-13 03:29:33
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https://socratic.org/questions/what-is-half-of-1-1-2-cups
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What is half of 1 1/2 cups?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Explanation
Explain in detail...
Explanation:
I want someone to double check my answer
149
Mar 9, 2016
$\frac{3}{4}$ cup
Explanation:
$1 \frac{1}{2}$ can be written as an improper fraction $\frac{3}{2}$
$\frac{1}{2}$ of anything means $\frac{1}{2} \times$ anything
So
$\frac{1}{2}$ of $1 \frac{1}{2}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2} \times \frac{3}{2}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{1 \times 3}{2 \times 2}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{3}{4}$
or 0.75
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Explanation
Explain in detail...
Explanation:
I want someone to double check my answer
37
Oct 19, 2016
$\frac{3}{4}$ cup
Explanation:
to answer this question you first change $1 \frac{1}{2}$ into a improper fraction.
$\frac{3}{2}$
then multiply it by $\frac{1}{2}$
$\frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Explanation
Explain in detail...
Explanation:
I want someone to double check my answer
15
Aug 18, 2016
$\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$
Explanation:
We can work this out with a bit of thinking and without having to use the fraction operations which students find difficult.
Half of $1 \frac{1}{2}$ cups can be thought of as:
half of one cup and half of half a cup.
Half of 1 is $\frac{1}{2}$
Half of $\frac{1}{2} i s \frac{1}{4}$
Now we have $\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$
This can be nicely shown by using the measuring cups of different sizes used in baking.
Nice introduction to fractions from a practical point of view.
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2018-02-18 17:58:35
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https://www.transtutors.com/questions/identify-two-disadvantages-of-using-the-payback-period-for-compa-256214.htm
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# Identify two disadvantages of using the payback period for compa
Identify two disadvantages of using the payback period for comparing investments.
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2019-04-20 12:52:04
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https://electronics.stackexchange.com/questions/420108/is-c0g-a-synonym-for-np0
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# Is C0G a synonym for NP0?
This question is in the context of a production environment where part numbers and change notices are strictly controlled, and any alternate manufacturer parts under an internal part number must be equivalent components.
For Class 2 Dielectrics, the rating codes are well defined. When requesting a new part number X5R and X7R are not equivalent are are given different internal part numbers (even if they are valid substitutes in a given design).
For Class 1 Dielectrics the picture to me is a little fuzzy. I recognize that there are a few different dielectric codes associated with Class 1 parts.However, C0G and NP0 specifically are used by some manufacturers interchangeably. Also, some distributor search engines (e.g. Digikey) will give option as NP0/C0G when filtering capacitors.
As an example there are two equivalent parts from TDK, an 0603 1000pF Capacitor. One is sold under NP0 designation the other under C0G, their spec sheets are identical...
C1608NP01H102J080AA
C1608C0G1H102J080AA
My question is, when applying the most stringent view of production Change Control and Controlled Part Numbers, Could NP0 and C0G be considered equivalent or synonymous?
• I don't know how responsive TDK is to questions, but given they're apparently publishing two different datasheets, you might be able to reach out to a sales engineer or equivalent and see what they say. Feb 2, 2019 at 10:36
Yes they are equivalent.
C0G is the material class that gives an NP0 (NP-zero ) tempco. But it is slang (NP-oh) where the 0 looks like O.
Beware that it is often spelt both ways with oh or zero, but it is intended to be a zero in both cases.
They also have P100 for +100 ppm/‘C and N100, N200 etc for -PPM over a greater range since ceramics tend to be offset this way. The ceramic composition is varied for analog applications requiring this .
C0G: Normally used for low C values. It has a low dielectric constant, but highest level of stability vs T and Vdc
X7R: Used for high C values with much higher dielectric constant than COG, but a lower stability.
Z5U: Used for even higher C values but has a lower stability than either COG or X7R, thus lower costs
The other tempco ratings that are below NP0 have different alpha-numeric codes. eg. P2H R2H S2H etc
see Murata specs
• I know NP0 is always a zero, but is it COG (with the letter) or C0G (with the number)? Feb 2, 2019 at 14:31
• It is both with zeroes, like Canadian postal codes, there are other codes in Murata’s link for different tempco that indicates the chemistry thermal properties besides 0 which each have a 50ppm tolerance. It’s just faster to read as “oh” Feb 2, 2019 at 16:35
• Also notice only the “Code” font shows the zero best. Feb 2, 2019 at 17:27
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2022-11-27 04:51:08
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https://gmatclub.com/forum/m21-184291.html
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# M21-37
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16 Sep 2014, 00:15
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Question Stats:
37% (01:12) correct 63% (01:23) wrong based on 134 sessions
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If for any positive integer $$x$$, $$d[x]$$ denotes its smallest positive odd divisor and $$D[x]$$ denotes its largest odd divisor, is $$x$$ even?
(1) $$D[x] - d[x] = 0$$
(2) $$D[3x] = 3$$
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Joined: 02 Sep 2009
Posts: 53066
### Show Tags
16 Sep 2014, 00:15
2
1
Official Solution:
If for any positive integer $$x$$, $$d[x]$$ denotes its smallest positive odd divisor and $$D[x]$$ denotes its largest odd divisor, is $$x$$ even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus $$d[x]=1$$ for any positive integer $$x$$.
(1) $$D[x] - d[x] = 0$$. Since $$d[x]=1$$, then $$D[x] - 1 = 0$$, which gives $$D[x] = 1$$. $$x$$ can be 1, so odd or $$2^n$$, (2, 4, 8, ...), so even. Not sufficient.
(2) $$D[3x] = 3$$. Again $$x$$ can be 1, so odd, as the largest odd divisor of $$3x=3$$ is 3 or $$x$$ can be $$2^n$$ (2, 4, 8, ...), so even, as the largest odd divisor of $$3*2=6$$ or $$2*4=12$$ is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that $$x$$ can be either 1, so odd or $$2^n$$, so even. Not sufficient.
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Joined: 24 Jun 2015
Posts: 46
### Show Tags
02 Jul 2015, 04:20
Hi,
Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?
Thanks a lot.
Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 53066
### Show Tags
02 Jul 2015, 04:37
2
luisnavarro wrote:
Hi,
Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?
Thanks a lot.
Luis Navarro
Looking for 700
If for any positive integer x, d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus $$d[x]=1$$ for any x.
(1) D[x] - d[x] = 0 --> $$D[x] - 1 = 0$$ --> $$D[x] = 1$$ --> x can be 1, so odd or $$2^n$$, (2, 4, 8, ...), so even. Not sufficient.
(2) D[3x] = 3 --> again x can be 1, so odd, as the largest odd divisor of $$3x=3$$ is 3 or x can be $$2^n$$ (2, 4, 8, ...), so even, as the largest odd divisor of 3*2=6 or 2*4=12 is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that x can be either 1, so odd or 2^n, so even. Not sufficient.
_________________
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Joined: 24 Jun 2015
Posts: 46
### Show Tags
02 Jul 2015, 05:21
2
Bunuel wrote:
luisnavarro wrote:
Hi,
Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?
Thanks a lot.
Luis Navarro
Looking for 700
If for any positive integer x, d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus $$d[x]=1$$ for any x.
(1) D[x] - d[x] = 0 --> $$D[x] - 1 = 0$$ --> $$D[x] = 1$$ --> x can be 1, so odd or $$2^n$$, (2, 4, 8, ...), so even. Not sufficient.
(2) D[3x] = 3 --> again x can be 1, so odd, as the largest odd divisor of $$3x=3$$ is 3 or x can be $$2^n$$ (2, 4, 8, ...), so even, as the largest odd divisor of 3*2=6 or 2*4=12 is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that x can be either 1, so odd or 2^n, so even. Not sufficient.
Very interesting, great¡¡¡ Actually in my analysis I had not view the 2^n posibilities, and now it is clear to me: For all 2^n the smallest and greatest odd divisor are the same = 1. It is easy to understand when you actually see explanations like yours but, its hard to do "alone".
Thanks.
Luis Navarro
Looking for 700
Intern
Joined: 29 Jun 2017
Posts: 28
### Show Tags
29 Jan 2019, 00:33
[quote="Bunuel"]Official Solution:
If for any positive integer $$x$$, $$d[x]$$ denotes its smallest positive odd divisor and $$D[x]$$ denotes its largest odd divisor, is $$x$$ even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus $$d[x]=1$$ for any positive integer $$x$$.
(1) $$D[x] - d[x] = 0$$. Since $$d[x]=1$$, then $$D[x] - 1 = 0$$, which gives $$D[x] = 1$$. $$x$$ can be 1, so odd or $$2^n$$, (2, 4, 8, ...), so even. Not sufficient.
(2) $$D[3x] = 3$$. Again $$x$$ can be 1, so odd, as the largest odd divisor of $$3x=3$$ is 3 or $$x$$ can be $$2^n$$ (2, 4, 8, ...), so even, as the largest odd divisor of $$3*2=6$$ or $$2*4=12$$ is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that $$x$$ can be either 1, so odd or $$2^n$$, so even. Not sufficient.
Hi Bunuel,
Can you please explain why we need to check for 2^n? Doesn't the problem end at x=1?
Thanks,
Megha.
Math Expert
Joined: 02 Sep 2009
Posts: 53066
### Show Tags
29 Jan 2019, 00:37
Megha1119 wrote:
Bunuel wrote:
Official Solution:
If for any positive integer $$x$$, $$d[x]$$ denotes its smallest positive odd divisor and $$D[x]$$ denotes its largest odd divisor, is $$x$$ even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus $$d[x]=1$$ for any positive integer $$x$$.
(1) $$D[x] - d[x] = 0$$. Since $$d[x]=1$$, then $$D[x] - 1 = 0$$, which gives $$D[x] = 1$$. $$x$$ can be 1, so odd or $$2^n$$, (2, 4, 8, ...), so even. Not sufficient.
(2) $$D[3x] = 3$$. Again $$x$$ can be 1, so odd, as the largest odd divisor of $$3x=3$$ is 3 or $$x$$ can be $$2^n$$ (2, 4, 8, ...), so even, as the largest odd divisor of $$3*2=6$$ or $$2*4=12$$ is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that $$x$$ can be either 1, so odd or $$2^n$$, so even. Not sufficient.
Hi Bunuel,
Can you please explain why we need to check for 2^n? Doesn't the problem end at x=1?
Thanks,
Megha.
No. x can be 1 or 2, 4, 8, 16, 32, ... Notice that for all those numbers both statements hold true. For example, for all of them both smallest and largest odd divisors is 1.
_________________
Re: M21-37 [#permalink] 29 Jan 2019, 00:37
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# M21-37
Moderators: chetan2u, Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
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2019-02-22 12:25:40
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https://www.scilag.net/problem/P-180707.1
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# Boundary layers in periodic homogenization of Dirichlet problem
Partially SolvedYear of origin: 1978
Posted online: 2018-07-07 19:13:48Z by Hayk Aleksanyan151
Cite as: P-180707.1
• Analysis of PDEs
### Problem's Description
Consider the homogenization problem of the elliptic system $$- \nabla \cdot A \left( \frac{x}{\varepsilon} \right) \nabla u (x) = 0, \ \ x \in D, \tag{1}$$ in a domain $D\subset \mathbb{R}^d$, ($d\geq 2$), and with oscillating Dirichlet boundary data $$u(x) = g \left(x , \frac{x}{\varepsilon} \right), \ \ x \in \partial D. \tag{2}$$
Here $\varepsilon> 0$ is a small parameter, and $A= A^{\alpha \beta } (x) \in M_N(\mathbb{R})$, $x\in \mathbb{R}^d$ is a family of functions indexed by $1\leq \alpha, \beta \leq d$ and with values in the set of matrices $M_N( \mathbb{R})$. For each $\varepsilon>0$ let $\mathcal{L}_\varepsilon$ be the differential operator in question, i.e. the $i$-th component of its action on a vector function $u=(u_1,...,u_N)$ is defined as $$(\mathcal{L}_\varepsilon u)_i (x)= - \left( \nabla \cdot A \left( \frac{\cdot}{\varepsilon} \right) \nabla u \right)_{i} (x) = -\partial_{x_\alpha} \left[ A^{\alpha \beta }_{ij} \left( \frac{\cdot}{\varepsilon} \right) \partial_{x_\beta} u_j \right],$$ where $1\leq i \leq N$ (the case $N=1$ corresponds to scalar equations). We impose the following conditions on the problem $(1)$-$(2)$:
(Ellipticity) there exists a constant $\lambda>0$ such that for any $x\in \mathbb{R}^d$, and any $\xi=(\xi^\alpha_i)\in \mathbb{R}^{dN}$ one has $$\lambda \xi^\alpha_i \xi^\alpha_i \leq A^{\alpha \beta}_{ij}(x) \xi^\alpha_i \xi^\beta_j \leq \frac{1}{\lambda} \xi^\alpha_i \xi^\alpha_i.$$
(Periodicity) $A$ and, $g$ in its second variable, are both $\mathbb{Z}^d$-periodic, i.e. $A(y+h) = A(y)$, and $g(x, y + h) = g(y)$ for all $x\in \partial D$, $y\in \mathbb{R}^d$, and $h\in \mathbb{Z}^d$.
We skip further assumptions on the smoothness and geometry of the domain, these are not relevant for introducing the general problem.
For each $\varepsilon > 0$ let $u_\varepsilon$ be the unique solution to $(1)$-$(2)$.
The main question, in general terms, asks: $$\textbf{What can we say about the limit of } u_\varepsilon \textbf{ as } \varepsilon \to 0 \ ?$$ In other words does $(1)$-$(2)$ have a homogenized limit and if so what properties does it have?
Following is a brief discussion based on [1] (see also [2]-[4] for details) on how one may naturally appear at $(1)$-$(2)$.
Consider problem $(1)$ but with a fixed boundary data $g$, i.e. $$-\nabla \cdot A\left( \frac{x}{\varepsilon} \right) \nabla u (x) = 0 , \ x\in D \qquad \text{and} \qquad u= g(x), \ x\in \partial D.$$ By Lax-Milgram for each $\varepsilon>0$ this problem has a unique weak solution $u_\varepsilon \in H^1(D; \mathbb{R}^N)$, which converges weakly in $H^1(D)$ to solution $u_0$ of the homogenized problem $$-\nabla \cdot A^0 \nabla u_0 (x) = 0 , \ x\in D \qquad \text{ and } \qquad u_0= g(x), \ x\in \partial D.$$ Here $A^0$ is the homogenized coefficient tensor and is defined via the solutions of the cell-problem, namely for $1\leq \gamma \leq d$ define $\chi = \chi^\gamma (y) \in M_N(\mathbb{R})$ to be the periodic solution of the problem $$-\partial_\alpha [ A^{\alpha \beta}(y) \partial_\beta \chi^\gamma(y) ] = \partial_\alpha A^{\alpha \gamma } (y) \text{ in } \mathbb{T}^d \text{ and } \int_{\mathbb{T}^d} \chi^\gamma(y) dy = 0 ,$$ where $\mathbb{T}^d$ is the unit torus, and we adopted the summation convention of repeated indices. The homogenized coefficients are defined by $$A^{0, \alpha \beta} = \int_{ \mathbb{T}^d } A^{\alpha \beta} dy + \int_{\mathbb{T}^d} A^{\alpha \gamma} \partial_{\gamma } \chi^{\beta} dy.$$
Setting $u_1 (x,y) = - \chi^{\alpha} (y) \partial_{\alpha} u_0(x)$ one obtains $$u_\varepsilon(x) = u_0(x) + \varepsilon u_1 \left( x, \frac{x}{\varepsilon} \right) + O( \varepsilon^{1/2} ) \text{ in } H^1(D).$$ The latter is the justification that the first two terms of the formal two-scale expansion of $u_\varepsilon$ are correct. Now observe that there is a mismatch of the boundary data in the left and right hand sides of the last expansion. The profile $u_1$ being periodic in its second variable oscillates near the boundary, giving rise to the so-called boundary layer phenomenon, which is responsible for the $\varepsilon^{1/2}$ loss in the approximation of $u_\varepsilon$. Quoting [1],
Of particular importance is the analysis of the behavior of solutions near boundaries and, possibly, any associated boundary layers. Relatively little seems to be known about this problem.
Indeed, correcting the boundary data by $u_{1,\varepsilon }^{bl}$ defined as $$\mathcal{L}_\varepsilon u_{1, \varepsilon }^{bl} = 0 \text{ in } D \qquad \text{and} \qquad u_{1, \varepsilon }^{bl} = -u_1 \left( x, \frac{x}{\varepsilon} \right) \text{ on } \partial D,$$ we get $$u_\varepsilon(x) = u_0(x) + \varepsilon u_1 \left( x, \frac{x}{\varepsilon} \right) + \varepsilon u_{1, \varepsilon}^{bl} (x) + O( \varepsilon ) \text{ in } H^1(D).$$ While this expansion shows that correcting the boundary data in the two-scale expansion of $u_\varepsilon$ gives a better approximation, it is of little use as long as we do not understand the behavior of $u_{1, \varepsilon}^{bl}$ as $\varepsilon \to 0$. The last question is precisely a partial case of $(1)$-$(2)$, with boundary data $g(x,y) = - \chi^{\alpha} (y) \partial_{\alpha} u_0(x)$.
1. ## BookIs an originAsymptotic analysis for periodic structures
year of publication: 1978
2. ## Article Homogenization and boundary layers
Acta Mathematica 209 (1), 133-178, 2012
3. ## Article Recent progress in the theory of homogenization with oscillating Dirichlet data
pp. 1-13, year of publication: 2013arXiv
4. ## Article Asymptotic analysis of boundary layer correctors in periodic homogenization
SIAM Journal on Mathematical Analysis 45 (1), 345-387, 2013
No remarks yet
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2019-05-25 07:41:49
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https://math.stackexchange.com/questions/1846769/eigenvalues-of-product-of-a-matrix-with-a-diagonal-matrix?noredirect=1
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# Eigenvalues of product of a matrix with a diagonal matrix
I have got a question and I would appreciate if one could help with it.
Assume $S$ is a diagonal matrix (with diagonal entries $s_1, s_2, \cdots$) and $M$ is a positive symmetric matrix with eigen value decomposition as follows:
$$\mathrm{eig}(M) = ULU^T$$
where $U^T$ means the transpose of $U$. I am trying to find out about the eigenvalues of $SM$. In other words, is there any relation between the eigenvalues of the matrix $S$, the matrix $M$ and their product $SM?$
any help/hint is appreciated
• $\mathrm{eig}(U^T M U) = L$ and $\mathrm{eig}(U^T S U)$ is another positive symmetric matrix. The $\mathrm{eig}(U^T S U U^T M U) = SM$. I think I do not understand what you mean, or I am missing something. Do you mean that since $S = \mathrm{eig}(U^T S U)$ and $\mathrm{eig}(M) = \mathrm{eig}(U^T M U)$, we can write $\mathrm{eig}(SM) = \mathrm{eig}(U^T S U U^T M U)$? If this is the case, I doubt to agree. I just checked it with matlab and this does not seem to be correct – M.X Jul 2 '16 at 14:05
• If the entries of $S$ are permuted, say to $S_1$, which preserves the eigenvalues of $S$, then in general, the eigenvalues of $S_1 M$ are not equal to the eigenvalues of $S M$. So clearly, order counts for $S$, not just its eigenvalues. – Mark L. Stone Jul 2 '16 at 15:33
Take any unit vector $u$ in the plane, and let $u'$ be 90 degrees counterclockwise from $u$. And let $Q$ be the matrix whose rows are $u$ and $u'$. Then for any diagonal matrix $D$, $Q^t D Q$ represents stretching along $u$ by $d_1$ and along $u'$ by $d_2$ (where these are the diagonal entries of $D$). Hold that thought.
For $u_1 = e_1$ and $(d_1, d_2) = (1, 4)$, for instance, we get a matrix $M$ that's just diagonal, with $1$ and $4$ on the diagonal. If $u$ is a unit vector in the $45$-degree direction, we get something else, but again with eigenvalues $1$ and $4$. For each possible angle $\theta$, let $$M_\theta = Q^t \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}Q$$ where $Q$ is the the matrix made from $u$ and $u'$, and $u$ is the vector $$\begin{bmatrix} \cos u \\ \sin u \end{bmatrix}.$$
So $M_\theta$ has eigenvalues $1$ and $4$. Let $$S = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$$ Then for $\theta = 0$, we have $SM_\theta$ is a diagonal matrix with eigenvalues 2 and 12, the product of smallest and product of largest eigenvalues. But for $\theta = \pi/2$, the product has eigenvalues $8 = 2 \cdot 4$ and $3 = 1 \cdot 3$, the "middle" two products of the eigenvalues of the two original matrices. For intermediate values of $\theta$, you get other possible eigenvalues. This shows that the eigenvalues of the product can range all the way from the smallest possible (the product two smallest evals of the factors) to the largest possible.
Of course, the PRODUCT of the eigenvalues of $SM$ is the products of those of $S$ with those of $M$, by determinants. So that's another samll constraint on them.
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2019-09-24 09:07:31
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http://liorpachter.wordpress.com/
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Nature Publishing Group claims on its website that it is committed to publishing “original research” that is “of the highest quality and impact”. But when exactly is research “original”? This is a question with a complicated answer. A recent blog post by senior editor Dorothy Clyde at Nature Protocols provides insight into the difficulties Nature faces in detecting plagiarism, and identifies the issue of self plagiarism as particularly problematic. The journal tries to avoid publishing the work of authors who have previously published the same work or a minor variant thereof. I imagine this is partly in the interests of fairness, a service to the scientific community to ensure that researchers don’t have to sift through numerous variants of a single research project in the literature, and a personal interest of the journal in its aim to publish only the highest level of scholarship.
On the other hand, there is also a rationale for individual researchers to revisit their own previously published work. Sometimes results can be recast in a way that makes them accessible to different communities, and rethinking of ideas frequently leads to a better understanding, and therefore a better exposition. The mathematician Gian-Carlo Rota made the case for enlightened self-plagiarism in one of his ten lessons he wished he had been taught when he was younger:
3. Publish the same result several times
After getting my degree, I worked for a few years in functional analysis. I bought a copy of Frederick Riesz’ Collected Papers as soon as the big thick heavy oversize volume was published. However, as I began to leaf through, I could not help but notice that the pages were extra thick, almost like cardboard. Strangely, each of Riesz’ publications had been reset in exceptionally large type. I was fond of Riesz’ papers, which were invariably beautifully written and gave the reader a feeling of definitiveness.
As I looked through his Collected Papers however, another picture emerged. The editors had gone out of their way to publish every little scrap Riesz had ever published. It was clear that Riesz’ publications were few. What is more surprising is that the papers had been published several times. Riesz would publish the first rough version of an idea in some obscure Hungarian journal. A few years later, he would send a series of notes to the French Academy’s Comptes Rendus in which the same material was further elaborated. A few more years would pass, and he would publish the definitive paper, either in French or in English. Adam Koranyi, who took courses with Frederick Riesz, told me that Riesz would lecture on the same subject year after year, while meditating on the definitive version to be written. No wonder the final version was perfect.
Riesz’ example is worth following. The mathematical community is split into small groups, each one with its own customs, notation and terminology. It may soon be indispensable to present the same result in several versions, each one accessible to a specific group; the price one might have to pay otherwise is to have our work rediscovered by someone who uses a different language and notation, and who will rightly claim it as his own.
The question is: where does one draw the line?
I was recently forced to confront this question when reading an interesting paper about a statistical approach to utilizing controls in large-scale genomics experiments:
J.A. Gagnon-Bartsch and T.P. Speed, Using control genes to corrected for unwanted variation in microarray dataBiostatistics, 2012.
A cornerstone in the logic and methodology of biology is the notion of a “control”. For example, when testing the result of a drug on patients, a subset of individuals will be given a placebo. This is done to literally control for effects that might be measured in patients taking the drug, but that are not inherent to the drug itself. By examining patients on the placebo, it is possible to essentially cancel out uninteresting effects that are not specific to the drug. In modern genomics experiments that involve thousands, or even hundreds of thousands of measurements, there is a biological question of how to design suitable controls, and a statistical question of how to exploit large numbers of controls to “normalize” (i.e. remove unwanted variation) from the high-dimensional measurements.
Formally, one framework for thinking about this is a linear model for gene expression. Using the notation of Gagnon-Bartsch & Speed, we have an expression matrix $Y$ of size $m \times n$ (samples and genes) modeled as
$Y_{m \times n} = X_{m \times p}\beta_{p \times n} + Z_{m \times q}\gamma_{q \times n} + W_{m \times k} \alpha_{k \times n} + \epsilon_{m \times n}$.
Here is a matrix describing various conditions (also called factors) and associated to it is the parameter matrix $\beta$ that records the contribution, or influence, of each factor on each gene. $\beta$ is the primary parameter of interest to be estimated from the data Y. The $\epsilon$ are random noise, and finally and are observed and unobserved covariates respectively. For example Z might encode factors for covariates such as gender, whereas W would encode factors that are hidden, or unobserved. A crucial point is that the number of hidden factors in W, namely k, is not known. The matrices $\gamma$ and $\alpha$ record the contributions of the Z and W factors on gene expression, and must also be estimated. It should be noted that X may be the logarithm of expression levels from a microarray experiment, or the analogous quantity from an RNA-Seq experiment (e.g. log of abundance in FPKM units).
Linear models have been applied to gene expression analysis for a very long time; I can think of papers going back 15 years. But They became central to all analysis about a decade ago, specifically popularized with the Limma package for microarray data analysis. In an important paper in 2007, Leek and Storey focused explicitly on the identification of hidden factors and estimation of their influence, using a method called SVA (Surrogate Variable Analysis). Mathematically, they described a procedure for estimating k and W and the parameters $\alpha$. I will not delve into the details of SVA in this post, except to say that the overall idea is to first perform linear regression (assuming no hidden factors) to identify the parameters $\beta$ and to then perform singular value decomposition (SVD) on the residuals to identify hidden factors (details omitted here). The resulting identified hidden factors (and associated influence parameters) are then used in a more general model for gene expression in subsequent analysis.
Gagnon-Bartsch and Speed refine this idea by suggesting that it is better to infer W from controls. For example, house-keeping genes that are unlikely to correlate with the conditions being tested, can be used to first estimate W, and then subsequently all the parameters of the model can be estimated by linear regression. They term this two-step process RUV-2 (acronym for Remote Unwanted Variation) where the “2” designates that the procedure is a two-step procedure. As with SVA, the key to inferring W from the controls is to perform singular value decomposition (or more generally factor analysis). This is actually clear from the probabilistic interpretation of PCA and the observation that what it means to be a in the set of “control genes” C in a setting where there are no observed factors Z, is that
$Y_C = W \alpha_C + \epsilon_C$.
That is, for such control genes the corresponding $\beta$ parameters are zero. This is a simple but powerful observation, because the explicit designation of control genes in the procedure makes it clear how to estimate W, and therefore the procedure becomes conceptually compelling and practically simple to implement. Thus, even though the model being used is the same as that of Leek & Storey, there is a novel idea in the paper that makes the procedure “cleaner”. Indeed, Gagnon-Bartsch & Speed provide experimental results in their paper showing that RUV-2 outperforms SVA. Even more convincing, is the use of RUV-2 by others. For example, in a paper on “The functional consequences of variation in transcription factor binding” by Cusanovitch et al., PLoS Genetics 2014, RUV-2 is shown to work well, and the authors explain how it helps them to take advantage of the controls in experimental design they created.
There is a tech report and also a preprint that follow up on the Gagnon-Bartsch & Speed paper; the tech report extends RUV-2 to a four step method RUV-4 (it also provides a very clear exposition of the statistics), and separately the preprint describes an extension to RUV-2 for the case where the factor of interest is also unknown. Both of these papers build on the original paper in significant ways and are important work, that to return to the original question in the post, certainly are on the right side of “the line”
The wrong side of the line?
The development of RUV-2 and SVA occurred in the context of microarrays, and it is natural to ask whether the details are really different for RNA-Seq (spoiler: they aren’t). In a book chapter published earlier this year:
D. Risso, J. Ngai, T.P. Speed, S. Dudoit, The role of spike-in standards in the normalization of RNA-Seq, in Statistical Analysis of Next Generation Sequencing Data (2014), 169-190.
the authors replace “log expression levels” from microarrays with “log counts” from RNA-Seq and the linear regression performed with Limma for RUV-2 with a Poisson regression (this involves one different R command). They call the new method RUV, which is the same as the previously published RUV, a naming convention that makes sense since the paper has no new method. In fact, the mathematical formulas describing the method are identical (and even in almost identical notation!) with the exception that the book chapter ignores altogether, and replaces $\epsilon$ with O.
To be fair, there is one added highlight in the book chapter, namely the observation that spike-ins can be used in lieu of housekeeping (or other control) genes. The method is unchanged, of course. It is just that the spike-ins are used to estimate W. Although spike-ins were not mentioned in the original Gagnon-Bartsch paper, there is no reason not to use them with arrays as well; they are standard with Affymetrix arrays.
My one critique of the chapter is that it doesn’t make sense to me that counts are used in the procedure. I think it would be better to use abundance estimates, and in fact I believe that Jeff Leek has already investigated the possibility in a preprint that appears to be an update to his original SVA work. That issue aside, the book chapter does provide concrete evidence using a Zebrafish experiment that RUV-2 is relevant and works for RNA-Seq data.
The story should end here (and this blog post would not have been written if it had) but two weeks ago, among five RNA-Seq papers published in Nature Biotechnology (I have yet to read the others), I found the following publication:
D. Risso, J. Ngai, T.P. Speed, S. Dudoit, Normalization of RNA-Seq data using factor analysis of control genes or samples, Nature Biotechnology 32 (2014), 896-902.
This paper has the same authors as the book chapter (with the exception that Sandrine Dudoit is now a co-corresponding author with Davide Risso, who was the sole corresponding author on the first publication), and, it turns out, it is basically the same paper… in fact in many parts it is the identical paper. It looks like the Nature Biotechnology paper is an edited and polished version of the book chapter, with a handful of additional figures (based on the same data) and better graphics. I thought that Nature journals publish original and reproducible research papers. I guess I didn’t realize that for some people “reproducible” means “reproduce your own previous research and republish it”.
At this point, before drawing attention to some comparisons between the papers, I’d like to point out that the book chapter was refereed. This is clear from the fact that it is described as such in both corresponding authors’ CVs (Davide Risso CV and Sandrine Dudoit CV).
How similar are the two papers?
Final paragraph of paper in the book:
Internal and external controls are essential for the analysis of high-throughput data and spike-in sequences have the potential to help researchers better adjust for unwanted technical effects. With the advent of single-cell sequencing [35], the role of spike-in standards should become even more important, both to account for technical variability [6] and to allow the move from relative to absolute RNA expression quantification. It is therefore essential to ensure that spike-in standards behave as expected and to develop a set of controls that are stable enough across replicate libraries and robust to both differences in library composition and library preparation protocols.
Final paragraph of paper in Nature Biotechnology:
Internal and external controls are essential for the analysis of high-throughput data and spike-in sequences have the potential to help researchers better adjust for unwanted technical factors. With the advent of single-cell sequencing27, the role of spike-in standards should become even more important, both to account for technical variability28 and to allow the move from relative to absolute RNA expression quantification. It is therefore essential to ensure that spike- in standards behave as expected and to develop a set of controls that are stable enough across replicate libraries and robust to both differences in library composition and library preparation protocols.
Abstract of paper in the book:
Normalization of RNA-seq data is essential to ensure accurate inference of expression levels, by adjusting for sequencing depth and other more complex nuisance effects, both within and between samples. Recently, the External RNA Control Consortium (ERCC) developed a set of 92 synthetic spike-in standards that are commercially available and relatively easy to add to a typical library preparation. In this chapter, we compare the performance of several state-of-the-art normalization methods, including adaptations that directly use spike-in sequences as controls. We show that although the ERCC spike-ins could in principle be valuable for assessing accuracy in RNA-seq experiments, their read counts are not stable enough to be used for normalization purposes. We propose a novel approach to normalization that can successfully make use of control sequences to remove unwanted effects and lead to accurate estimation of expression fold-changes and tests of differential expression.
Abstract of paper in Nature Biotechnology:
Normalization of RNA-sequencing (RNA-seq) data has proven essential to ensure accurate inference of expression levels. Here, we show that usual normalization approaches mostly account for sequencing depth and fail to correct for library preparation and other more complex unwanted technical effects. We evaluate the performance of the External RNA Control Consortium (ERCC) spike-in controls and investigate the possibility of using them directly for normalization. We show that the spike-ins are not reliable enough to be used in standard global-scaling or regression-based normalization procedures. We propose a normalization strategy, called remove unwanted variation (RUV), that adjusts for nuisance technical effects by performing factor analysis on suitable sets of control genes (e.g., ERCC spike-ins) or samples (e.g., replicate libraries). Our approach leads to more accurate estimates of expression fold-changes and tests of differential expression compared to state-of-the-art normalization methods. In particular, RUV promises to be valuable for large collaborative projects involving multiple laboratories, technicians, and/or sequencing platforms.
Abstract of Gagnon-Bartsch & Speed paper that already took credit for a “new” method called RUV:
Microarray expression studies suffer from the problem of batch effects and other unwanted variation. Many methods have been proposed to adjust microarray data to mitigate the problems of unwanted variation. Several of these methods rely on factor analysis to infer the unwanted variation from the data. A central problem with this approach is the difficulty in discerning the unwanted variation from the biological variation that is of interest to the researcher. We present a new method, intended for use in differential expression studies, that attempts to overcome this problem by restricting the factor analysis to negative control genes. Negative control genes are genes known a priori not to be differentially expressed with respect to the biological factor of interest. Variation in the expression levels of these genes can therefore be assumed to be unwanted variation. We name this method “Remove Unwanted Variation, 2-step” (RUV-2). We discuss various techniques for assessing the performance of an adjustment method and compare the performance of RUV-2 with that of other commonly used adjustment methods such as Combat and Surrogate Variable Analysis (SVA). We present several example studies, each concerning genes differentially expressed with respect to gender in the brain and find that RUV-2 performs as well or better than other methods. Finally, we discuss the possibility of adapting RUV-2 for use in studies not concerned with differential expression and conclude that there may be promise but substantial challenges remain.
Many figures are also the same (except one that appears to have been fixed in the Nature Biotechnology paper– I leave the discovery of the figure as an exercise to the reader). Here is Figure 9.2 in the book:
The two panels appears as (b) and (c) in Figure 4 in the Nature Biotechnology paper (albeit transformed via a 90 degree rotation and reflection from the dihedral group):
Basically the whole of the book chapter and the Nature Biotechnology paper are essentially the same, down to the math notation, which even two papers removed is just a rehashing of the RUV method of Gagnon-Bartsch & Speed. A complete diff of the papers is beyond the scope of this blog post and technically not trivial to perform, but examination by eye reveals one to be a draft of the other.
Although it is acceptable in the academic community to draw on material from published research articles for expository book chapters (with permission), and conversely to publish preprints, including conference proceedings, in journals, this case is different. (a) the book chapter was refereed, exactly like a journal publication (b) the material in the chapter is not expository; it is research, (c) it was published before the Nature Biotechnology article, and presumably prepared long before, (d) the book chapter cites the Nature Biotechnology article but not vice versa and (e) the book chapter is not a particularly innovative piece of work to begin with. The method it describes and claims to be “novel”, namely RUV, was already published by Gagnon-Bartsch & Speed.
Below is a musical rendition of what has happened here:
“An entertaining freshness… Tic Tac!” This is Ferrero‘s tag line for its most successful product, the ubiquitous Tic Tac. And the line has stuck. As WikiHow points out in how to make your breath freshfirst buy some mints, then brush your teeth.
One of the amazing things about Tic Tacs is that they are sugar free. Well… almost not. As the label explains, a single serving (one single Tic Tac) contains 0g of sugar (to be precise, less than 0.5g, as explained in a footnote). In what could initially be assumed to be a mere coincidence, the weight of a single serving is 0.49g. It did not escape my attention that 0.50-0.49=0.01. Why?
To understand it helps to look at the labeling rules of the FDA. I’ve reproduced the relevant section (Title 21) below, and boldfaced the relevant parts:
TITLE 21–FOOD AND DRUGS
CHAPTER I–FOOD AND DRUG ADMINISTRATION DEPARTMENT OF HEALTH AND HUMAN SERVICES
SUBCHAPTER B–FOOD FOR HUMAN CONSUMPTION
(c) Sugar content claims –(1) Use of terms such as “sugar free,” “free of sugar,” “no sugar,” “zero sugar,” “without sugar,” “sugarless,” “trivial source of sugar,” “negligible source of sugar,” or “dietarily insignificant source of sugar.” Consumers may reasonably be expected to regard terms that represent that the food contains no sugars or sweeteners e.g., “sugar free,” or “no sugar,” as indicating a product which is low in calories or significantly reduced in calories. Consequently, except as provided in paragraph (c)(2) of this section, a food may not be labeled with such terms unless:
(i) The food contains less than 0.5 g of sugars, as defined in 101.9(c)(6)(ii), per reference amount customarily consumed and per labeled serving or, in the case of a meal product or main dish product, less than 0.5 g of sugars per labeled serving; and
(ii) The food contains no ingredient that is a sugar or that is generally understood by consumers to contain sugars unless the listing of the ingredient in the ingredient statement is followed by an asterisk that refers to the statement below the list of ingredients, which states “adds a trivial amount of sugar,” “adds a negligible amount of sugar,” or “adds a dietarily insignificant amount of sugar;” and
(iii)(A) It is labeled “low calorie” or “reduced calorie” or bears a relative claim of special dietary usefulness labeled in compliance with paragraphs (b)(2), (b)(3), (b)(4), or (b)(5) of this section, or, if a dietary supplement, it meets the definition in paragraph (b)(2) of this section for “low calorie” but is prohibited by 101.13(b)(5) and 101.60(a)(4) from bearing the claim; or
(B) Such term is immediately accompanied, each time it is used, by either the statement “not a reduced calorie food,” “not a low calorie food,” or “not for weight control.”
It turns out that Tic Tacs are in fact almost pure sugar. Its easy to figure this out by looking at the number of calories per serving (1.9) and multiplying the number of calories per gram of sugar (3.8) by 0.49 => 1.862 calories. 98% sugar! Ferrero basically admits this in their FAQ. Acting completely within the bounds of the law, they have simply exploited an arbitrary threshold of the FDA. Arbitrary thresholds are always problematic; not only can they have unintended consequences, but they can be manipulated to engineer desired outcomes. In computational biology they have become ubiquitous, frequently being described as “filters” or “pre-processing steps”. Regardless of how they are justified, thresholds are thresholds are thresholds. They can sometimes be beneficial, but they are dangerous when wielded indiscriminately.
There is one type of thresholding/filtering in used in RNA-Seq that my postdoc Bo Li and I have been thinking about a bit this year. It consists of removing duplicate reads, i.e. reads that map to the same position in a transcriptome. The motivation behind such filtering is to reduce or eliminate amplification bias, and it is based on the intuition that it is unlikely that lightning strikes the same spot multiple times. That is, it is improbable that many reads would map to the exact same location assuming a model for sequencing that posits selecting fragments from transcripts uniformly. The idea is also called de-duplication or digital normalization.
Digital normalization is obviously problematic for high abundance transcripts. Consider, for example, a transcripts that is so abundant that it is extremely likely that at least one read will start at every site (ignoring the ends, which for the purposes of the thought experiment are not relevant). This would also be the case if the transcript was twice as abundant, and so digital normalization would prevent the possibility for estimating the difference. This issue was noted in a paper published earlier this year by Zhou et al. The authors investigate in some detail the implications of this problem, and quantify the bias it introduces in a number of data sets. But a key question not answered in the paper is what does digital normalization actually do?
To answer the question, it is helpful to consider how one might estimate the abundance of a transcript after digital normalization. One naive approach is to just count the number of reads after de-duplication, followed by normalization for the length of the transcript and the number of reads sequenced. Specifically if there are sites where a read might start, and of the sites had at least one read, then the naive approach would be to use the estimate $\frac{k}{n}$ suitably normalized for the total number of reads in the experiment. This is exactly what is done in standard de-duplication pipelines, or in digital normalization as described in the preprint by Brown et al. However assuming a simple model for sequencing, namely that every read is selected by first choosing a transcript according to a multinomial distribution and then choosing a location on it uniformly at random from among the sites, a different formula emerges.
Let be a random variable that denotes the number of sites on a transcript of length n that are covered in a random sequencing experiment, where the number of reads starting at each site of the transcript is Poisson distributed with parameter c (i.e., the average coverage of the transcript is c). Note that
$Pr(X \geq 1) = 1-Pr(X=0) = 1-e^{-c}$.
The maximum likelihood estimate for can also be obtained by the method of moments, which is to set
$\frac{k}{n} = 1-e^{-c}$
from which it is easy to see that
$c = -log(1-\frac{k}{n})$.
This is the same as the (derivation of the) Jukes-Cantor correction in phylogenetics where the method of moments equation is replaced by $\frac{4}{3}\frac{k}{n} = 1-e^{-\frac{4}{3}c}$ yielding $D_{JC} = -\frac{3}{4}log(1-\frac{4}{3}\frac{k}{n})$, but I’ll leave an extended discussion of the Jukes-Cantor model and correction for a future post.
The point here, as noticed by Bo Li, is that since $log(1-x) \approx -x$ by Taylor approximation, it follows that the average coverage can be estimated by $c \approx \frac{k}{n}$. This is exactly the naive estimate of de-duplication or digital normalization, and the fact that $\frac{k}{n} \rightarrow 1$ as $k \rightarrow n$ means that $-log(1-\frac{k}{n})$ blows up, at high coverage hence the results of Zhou et al.
Digital normalization as proposed by Brown et al. involves possibly thresholding at more than one read per site (for example choosing a threshold C and removing all but at most C reads at every site). But even this modified heuristic fails to adequately relate to a probabilistic model of sequencing. One interesting and easy exercise is to consider the second or higher order Taylor approximations. But a more interesting approach to dealing with amplification bias is to avoid thresholding per se, and to instead identify outliers among duplicate reads and to adjust them according to an estimated distribution of coverage. This is the approach of Hashimoto et al. in a the paper “Universal count correction for high-throughput sequencing” published in March in PLoS One. There are undoubtedly other approaches as well, and in my opinion the issue will received renewed attention in the coming year as the removal of amplification biases in single-cell transcriptome experiments becomes a priority.
As mentioned above, digital normalization/de-duplication is just one of many thresholds applied in a typical RNA-Seq “pipeline”. To get a sense of the extent of thresholding, one need only scan the (supplementary?) methods section of any genomics paper. For example, the GEUVADIS RNA-Seq consortium describe their analysis pipeline as follows:
“We employed the JIP pipeline (T.G. & M.S., data not shown) to map mRNA-seq reads and to quantify mRNA transcripts. For alignment to the human reference genome sequence (GRCh37, autosomes + X + Y + M), we used the GEM mapping suite24 (v1.349 which corresponds to publicly available pre-release 2) to first map (max. mismatches = 4%, max. edit distance = 20%, min. decoded strata = 2 and strata after best = 1) and subsequently to split-map (max.mismatches = 4%, Gencode v12 and de novo junctions) all reads that did not map entirely. Both mapping steps are repeated for reads trimmed 20 nucleotides from their 3′-end, and then for reads trimmed 5 nucleotides from their 5′-end in addition to earlier 3′-trimming—each time considering exclusively reads that have not been mapped in earlier iterations. Finally, all read mappings were assessed with respect to the mate pair information: valid mapping pairs are formed up to a maximum insert size of 100,000 bp, extension trigger = 0.999 and minimum decoded strata = 1. The mapping pipeline and settings are described below and can also be found in https://github.com/gemtools, where the code as well as an example pipeline are hosted.”
This is not a bad pipeline- the paper shows it was carefully evaluated– and it may have been a practical approach to dealing with the large amount of RNA-Seq data in the project. But even the first and seemingly innocuous thresholding to trim low quality bases from the ends of reads is controversial and potentially problematic. In a careful analysis published earlier this year, Matthew MacManes looked carefully at the effect of trimming in RNA-Seq, and concluded that aggressive trimming of bases below Q20, a standard that is frequently employed in pipelines, is problematic. I think his Figure 3, which I’ve reproduced below, is very convincing:
It certainly appears that some mild trimming can be beneficial, but a threshold that is optimal (and more importantly not detrimental) depends on the specifics of the dataset and is difficult or impossible to determine a priori. MacManes’ view (for more see his blog post on the topic) is consistent with another paper by Del Fabbro et al. that while seemingly positive about trimming in the abstract, actually concludes that “In the specific case of RNA-Seq, the tradeoff between sensitivity (number of aligned reads) and specificity (number of correctly aligned reads) seems to be always detrimental when trimming the datasets (Figure S2); in such a case, the modern aligners, like Tophat, seem to be able to overcome low quality issues, therefore making trimming unnecessary.”
Alas, Tic Tac thresholds are everywhere. My advice is: brush your teeth first.
When I was a teenager I broke all the rules on Friday night. After dinner I would watch Louis Rukeyser’s Wall Street Week at 8:30pm, and I would be in bed an hour later. On new year’s eve, he had a special “year-end review”, during which he hosted “financial experts” who would opine on the stock market and make predictions for the coming year.
What I learned from Louis Rukeyser was:
1. Never trust men in suits (or tuxedos).
2. It’s easier to perpetrate the 1024 scam than one might think!
Here are the experts in 1999 all predicting increases for the stock market in 2000:
As it turned out, the NASDAQ peaked on March 10, 2000, and within a week and a half had dropped 10%. By the end of the year the dot-com bubble had completely burst and a few years later the market had lost almost 80% of its value.
Predictions on the last day of the 20th century represented a spectacular failure for the “pundits”, but by then I had already witnessed many failures on the show. I’d also noted that almost all the invited “experts” were men. Of course correlation does not imply causation, but I remember having a hard time dispelling the notion that the guests were wrong because they were men. I never wanted to be sexist, but Louis Rukeyser made it very difficult for me!
Gender issues aside, the main lesson I learned from Louis Rukeyser’s show is that it’s easy to perpetrate the 1024 scam. The scam goes something like this: a scammer sends out 1024 emails to individuals that are unlikely to know each other, with each email making a prediction about the performance of the stock market in the coming week. For half the people (512), she predicts the stock market will go up, and for the other half, that it will go down. The next week, she has obviously sent a correct prediction of the market to half the people (this assumes the market is never unchanged after a week). She ignores the 512 people who have received an incorrect prediction, dividing those who received the correct prediction into two halves (256 each). Again, she predicts the performance of the market in the coming week, sending 256 individuals a prediction that the market will go up, and the other 256 a prediction that it will go down. She continues this divide-and-conquer for 10 weeks, at which time there is one individual that has received correct predictions about the movement of the stock market for 2.5 months! This person may believe that the scammer has the ability to predict the market; after all, $(\frac{1}{2})^{10} = 0.00098$ which looks like a very significant p-value. This is when the scammer asks for a “large investment”. Of course what is missing is knowledge of the other prediction emails sent out, or in other words the multiple testing problem.
The Wall Street Week guest panels essentially provided a perfect setting in which to perpetrate this scam. “Experts” that would err would be unlikely to be invited back. Whereas regular winners would be back for another chance at guessing. This is a situation very similar to the mutual fund management market, where managers are sacked when they have a bad year, only to have large firms with hundreds of funds on the books highlight funds that have performed well for 10 years in a row in their annual glossy brochures. But that is not the subject matter of this blog post. Rather, it’s the blog itself.
I wrote and posted my first blog entry (Genesis of *Seq) exactly a year ago. I began writing it for two reasons. First, I thought it could be a convenient and useful forum for discussion of technical developments in computational biology. I was motivated partly by the seqanswers website, which allows users to share information and experience in dealing with high-throughput sequence data. But I was also inspired by the What’s New Blog that has created numerous bridges in the mathematics community via highly technical yet accessible posts that have democratized mathematics. Second, I had noticed an extraordinary abuse of multiple testing in computational biology, and I was desperate for a forum where I could bring the issue to peoples attention. My initial frustration with outlandish claims in papers based on weak statistics had also grown over time to encompass a general concern for lack of rigor in computational biology papers. None of us are perfect but there is a wide gap between perfect and wrong. Computational biology is a field that is now an amalgamation of many subjects and I hoped that a blog would be able to reach the different silos more effectively than publications.
And thus this blog was born on August 19th 2013. I started without a preconception of how it would turn out over time, and I’m happy to say I’ve been surprised by its impact, most notably on myself. I’ve learned an enormous amount from reader feedback, in part via comments on individual posts, but also from private emails to me and in personal conversations. For this (selfish) reason alone, I will keep blogging. I have also been asked by many of you to keep posting, and I’m listening. When I have nothing left to say, I promise I will quit. But for now I have a backlog of posts, and after a break this summer, I am ready to return to the keyboard. Besides, since starting to blog I still haven’t been to Las Vegas.
There has recently been something of an uproar over the new book A Troublesome Inheritance by Nicholas Wade, with much of the criticism centering on Wade’s claim that race is a meaningful biological category. This subject is one with which I1 have some personal connection since as a child growing up in South Africa in the 1980s, I was myself categorized very neatly by the Office for Race Classification: 10. A simple pair of digits that conferred on me numerous rights and privileges denied to the majority of the population.
Explanation of identity numbers assigned to citizens by the South African government during apartheid.
And yet the system behind those digits was anything but simple. The group to which an individual was assigned could be based on not only their skin color but also their employment, eating and drinking habits, and indeed explicitly social factors as related by Muriel Horrell of the South African Institute of Race Relations: “Should a man who is initially classified white have a number of coloured friends and spend many of his leisure hours in their company, he stands to risk being re-classified as coloured.”
With these memories in mind, I found Wade’s concept of race as a biological category quite confusing, a confusion which only deepened when I discovered that he identifies not the eight races designated by the South African Population Registration Act of 1950, but rather five, none of which was the Griqua! With the full force of modern science on his side2, it seemed unlikely that these disparities represented an error on Wade’s part. And so I was left with a perplexing question: how could it be that the South African apartheid regime — racists par excellence — had failed to institutionalize their racism correctly? How had Wade gotten it right when Hendrik Verwoerd had gone awry?
Eventually I realized that A Troublesome Inheritance itself might contain the answer to this conundrum. Institutions, Wade explains, are genetic: “they grow out of instinctual social behaviors” and “one indication of such a genetic effect is that, if institutions were purely cultural, it should be easy to transfer an institution from one society to another.”3 So perhaps it is Wade’s genetic instincts as a Briton that explain how he has navigated these waters more skillfully than the Dutch-descended Afrikaners who designed the institutions of apartheid.
One might initially be inclined to scoff at such a suggestion or even to find it offensive. However, we must press boldly on in the name of truth and try to explain why this hypothesis might be true. Again, A Troublesome Inheritance comes to our aid. There, Wade discusses the decline in English interest rates between 1400 and 1850. This is the result, we learn, of rich English people producing more children than the poor and thereby genetically propagating those qualities which the rich are so famous for possessing: “less impulsive, more patient, and more willing to save.”4 However this period of time saw not only falling interest rates but also the rise of the British Empire. It was a period when Englishmen not only built steam engines and textile mills, but also trafficked in slaves by the millions and colonized countries whose people lacked their imperial genes. These latter activities, with an obvious appeal to the more racially minded among England’s population, could bring great wealth to those who engaged in them and so perhaps the greater reproductive fitness of England’s economic elite propagated not only patience but a predisposition to racism. This would explain, for example, the ability of John Hanning Speke to sniff out “the best blood of Abyssinia” when distinguishing the Tutsi from their Hutu neighbors.
Some might be tempted to speculate that Wade is himself a racist. While Wade — who freely speculates about billions of human beings — would no doubt support such an activity, those who engage in such speculation should perhaps not judge him too harshly. After all, racism may simply be Wade’s own troublesome inheritance.
#### Footnotes
1. In the spirit of authorship designation as discussed in this post, we describe the author contributions as follows: the recollections of South Africa are those of Lior Pachter, who distinctly remembers his classification as “white”. Nicolas Bray conceived and composed the post with input from LP. LP discloses no conflicts of interest. NB discloses being of British ancestry.
2. Perhaps not quite the full force, given the reception his book has received from actual scientists.
3. While this post is satirical, it should be noted for clarity that, improbably, this is an actual quote from Wade’s book.
4. Again, not satire.
In the Jeopardy! game show contestants are presented with questions formulated as answers that require answers in the form questions. For example, if a contestant selects “Normality for $200″ she might be shown the following clue: “The average $\frac{x_1+x_2+\cdots + x_n}{n}$,” to which she would reply “What is the maximum likelihood estimate for the mean of independent identically distributed Gaussian random variables from which samples $x_1,x_2,\ldots,x_n$ have been obtained?” Host Alex Trebek would immediately exclaim “That is the correct answer for$200!”
The process of doing mathematics involves repeatedly playing Jeopardy! with oneself in an unending quest to understand everything just a little bit better. The purpose of this blog post is to provide an exposition of how this works for understanding principal component analysis (PCA): I present four Jeopardy clues in the “Normality” category that all share the same answer: “What is principal component analysis?” The post was motivated by a conversation I recently had with a well-known population geneticist at a conference I was attending. I mentioned to him that I would be saying something about PCA in my talk, and that he might find what I have to say interesting because I knew he had used the method in many of his papers. Without hesitation he replied that he was well aware that PCA was not a statistical method and merely a heuristic visualization tool.
The problem, of course, is that PCA does have a statistical interpretation and is not at all an ad-hoc heuristic. Unfortunately, the previously mentioned population geneticist is not alone; there is a lot of confusion about what PCA is really about. For example, in one textbook it is stated that “PCA is not a statistical method to infer parameters or test hypotheses. Instead, it provides a method to reduce a complex dataset to lower dimension to reveal sometimes hidden, simplified structure that often underlie it.” In another one finds out that “PCA is a statistical method routinely used to analyze interrelationships among large numbers of objects.” In a highly cited review on gene expression analysis PCA is described as “more useful as a visualization technique than as an analytical method” but then in a paper by Markus Ringnér titled the same as this post, i.e. What is principal component analysis? in Nature Biotechnology, 2008, the author writes that “Principal component analysis (PCA) is a mathematical algorithm that reduces the dimensionality of the data while retaining most of the variation in the data set” (the author then avoids going into the details because “understanding the details underlying PCA requires knowledge of linear algebra”). All of these statements are both correct and incorrect and confusing. A major issue is that the description by Ringnér of PCA in terms of the procedure for computing it (singular value decomposition) is common and unfortunately does not shed light on when it should be used. But knowing when to use a method is far more important than knowing how to do it.
I therefore offer four Jeopardy! clues for principal component analysis that I think help to understand both when and how to use the method:
1. An affine subspace closest to a set of points.
Suppose we are given numbers $x_1,\ldots,x_n$ as in the initial example above. We are interested in finding the “closest” number to these numbers. By “closest” we mean in the sense of total squared difference. That is, we are looking for a number $m$ such that $\sum_{i=1}^n (m-x_i)^2$ is minimized.
This is a (straightforward) calculus problem, solved by taking the derivative of the function above and setting it equal to zero. If we let $f(m) = \sum_{i=1}^n (m-x_i)^2$ then $f'(m) = 2 \cdot \sum_{i=1}^n (m-x_i)$ and setting $f'(m)=0$ we can solve for $m$ to obtain $m = \frac{1}{n} \sum_{i=1}^n x_i$.
The right hand side of the equation is just the average of the n numbers and the optimization problem provides an interpretation of it as the number minimizing the total squared difference with the given numbers (note that one can replace squared difference by absolute value, i.e. minimization of $\sum_{i=1}^n |m-x_i|$, in which case the solution for is the median; we return to this point and its implications for PCA later).
Suppose that instead of n numbers, one is given n points in $\mathbb{R}^p$. That is, point is ${\bf x}^i = (x^i_1,\ldots,x^i_p)$. We can now ask for a point ${\bf m}$ with the property that the squared distance of ${\bf m}$ to the n points is minimized. This is asking for $min_{\bf m} \sum_{i=1}^n ||{\bf m}-{\bf x}^i||_2$.
The solution for $m$ can be obtained by minimizing each coordinate independently, thereby reducing the problem to the simpler version of numbers above, and it follows that ${\bf m} = \frac{1}{n} \sum_{i=1}^n {\bf x}^i$.
This is 0-dimensional PCA, i.e., PCA of a set of points onto a single point, and it is the centroid of the points. The generalization of this concept provides a definition for PCA:
Definition: Given n points in $\mathbb{R}^p$, principal components analysis consists of choosing a dimension $k < p$ and then finding the affine space of dimension with the property that the squared distance of the points to their orthogonal projection onto the space is minimized.
This definition can be thought of as a generalization of the centroid (or average) of the points. To understand this generalization, it is useful to think of the simplest case that is not 0-dimensional PCA, namely 1-dimensional PCA of a set of points in two dimensions:
In this case the 1-dimensional PCA subspace can be thought of as the line that best represents the average of the points. The blue points are the orthogonal projections of the points onto the “average line” (see, e.g., the red point projected orthogonally), which minimizes the squared lengths of the dashed lines. In higher dimensions line is replaced by affine subspace, and the orthogonal projections are to points on that subspace. There are a few properties of the PCA affine subspaces that are worth noting:
1. The set of PCA subspaces (translated to the origin) form a flagThis means that the PCA subspace of dimension k is contained in the PCA subspace of dimension k+1. For example, all PCA subspaces contain the centroid of the points (in the figure above the centroid is the green point). This follows from the fact that the PCA subspaces can be incrementally constructed by building a basis from eigenvectors of a single matrix, a point we will return to later.
2. The PCA subspaces are not scale invariant. For example, if the points are scaled by multiplying one of the coordinates by a constant, then the PCA subspaces change. This is obvious because the centroid of the points will change. For this reason, when PCA is applied to data obtained from heterogeneous measurements, the units matter. One can form a “common” set of units by scaling the values in each coordinate to have the same variance.
3. If the data points are represented in matrix form as an $n \times p$ matrix $X$, and the points orthogonally projected onto the PCA subspace of dimension are represented as in the ambient p dimensional space by a matrix $\tilde{X}$, then $\tilde{X} = argmin_{M:rk(M)=k} ||X-M||_2$. That is, $\tilde{X}$ is the matrix of rank k with the property that the Frobenius norm $||X-\tilde{X}||_2$ is minimized. This is just a rephrasing in linear algebra of the definition of PCA given above.
At this point it is useful to mention some terminology confusion associated with PCA. Unfortunately there is no standard for describing the various parts of an analysis. What I have called the “PCA subspaces” are also sometimes called “principal axes”. The orthogonal vectors forming the flag mentioned above are called “weight vectors”, or “loadings”. Sometimes they are called “principal components”, although that term is sometimes used to refer to points projected onto a principal axis. In this post I stick to “PCA subspaces” and “PCA points” to avoid confusion.
Returning to Jeopardy!, we have “Normality for $400″ with the answer “An affine subspace closest to a set of points” and the question “What is PCA?”. One question at this point is why the Jeopardy! question just asked is in the category “Normality”. After all, the normal distribution does not seem to be related to the optimization problem just discussed. The connection is as follows: 2. A generalization of linear regression in which the Gaussian noise is isotropic. PCA has an interpretation as the maximum likelihood parameter of a linear Gaussian model, a point that is crucial in understanding the scope of its application. To explain this point of view, we begin by elaborating on the opening Jeopardy! question about Normality for$200:
The point of the question was that the average of n numbers can be interpreted as a maximum likelihood estimation of the mean of a Gaussian. The Gaussian distribution is
$f(x,\mu,\sigma) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$. Given the numbers $x_1,\ldots,x_n$, the likelihood function is therefore
$L(\mu,\sigma) = \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i-\mu)^2}{2\sigma^2}}$. The maximum of this function is the same as the maximum of its logarithm, which is
$log L(\mu,\sigma) = \sum_{i=1}^n \left( log \frac{1}{\sqrt{2 \pi \sigma^2}} -\frac{(x_i-\mu)^2}{2\sigma^2} \right)$. Therefore the problem of finding the maximum likelihood estimate for the mean is equivalent to that of finding the minimum of the function
$S(\mu) = \sum_{i=1}^n (x_i-\mu)^2$. This is exactly the optimization problem solved by 0-dimensional PCA, as we saw above. With this calculation at hand, we turn to the statistical interpretation of least squares:
Given n points $\{(x_i,y_i)\}_{i=1}^n$ in the plane (see figure above), the least squares line $y=mx+b$ (purple in figure) is the one that minimizes the sum of the squares $\sum_{i=1}^n \left( (mx_i+b) - y_i \right)^2$. That is, the least squares line is the one minimizing the sum of the squared vertical distances to the points. As with the average of numbers, the least squares line has a statistical interpretation: Suppose that there is some line $y=m^{*}x+b^{*}$ (black line in figure) that is unknown, but that “generated” the observed points, in the sense that each observed point was obtained by perturbing the point $m^{*}x_i +b^{*}$ vertically by a random amount from a single Gaussian distribution with mean 0 and variance $\sigma^2$. In the figure, an example is shown where the blue point on the unknown line “generates” the observed red point; the Gaussian is indicated with the blue streak around the point. Note that the model specified so far is not fully generative, as it depends on the hidden points $m^{*}x_i +b^{*}$ and there is no procedure given to generate the $x_i$. This can be done by positing that the $x_i$ are generated from a Gaussian distribution along the line $y=m^{*}x+b$ (followed by the points $y_i$ generated by Gaussian perturbation of the y coordinate on the line). The coordinates $x_i$ can then be deduced directly from the observed points as the Gaussian perturbations are all vertical. The relationship between the statistical model just described and least squares is made precise by a theorem (which we state informally, but is a special case of the Gauss-Markov theorem):
Theorem (Gauss-Markov): The maximum likelihood estimate for the line (the parameters and b) in the model described above correspond to the least squares line.
The proof is analogous to the argument given for the average of numbers above so we omit it. It can be generalized to higher dimensions where it forms the basis of what is known as linear regression. In regression, the $x_i$ are known as independent variables and $y$ the dependent variable. The generative model provides an interpretation of the independent variables as fixed measured quantities, whereas the dependent variable is a linear combination of the independent variables with added noise. It is important to note that the origins of linear regression are in physics, specifically in work of Legendre (1805) and Gauss (1809) who applied least squares to the astronomical problem of calculating the orbits of comets around the sun. In their application, the independent variables were time (for which accurate measurements were possible with clocks; by 1800 clocks were accurate to less than 0.15 seconds per day) and the (noisy) dependent variable the measurement of location. Linear regression has become one of the most (if not the most) widely used statistical tools but as we now explain, PCA (and its generalization factor analysis), with a statistical interpretation that includes noise in the $x_i$ variables, seems better suited for biological data.
The statistical interpretation of least squares can be extended to a similar framework for PCA. Recall that we first considered a statistical interpretation for least squares where an unknown line $y=m^{*}x+b^{*}$ “generated” the observed points, in the sense that each observed point was obtained by perturbing the point $m^{*}x_i +b^{*}$ vertically by a random amount from a single Gaussian distribution with mean 0 and variance $\sigma^2$. PCA can be understood analogously by replacing vertically by orthogonally (this is the probabilistic model of Collins et al., NIPS 2001 for PCA). However this approach is not completely satisfactory as the orthogonality of the perturbation is is not readily interpretable. Stated differently, it is not obvious what physical processes would generate points orthogonal to a linear affine subspace by perturbations that are always orthogonal to the subspace. In the case of least squares, the “vertical” perturbation corresponds to noise in one measurement (represented by one coordinate). The problem is in naturally interpreting orthogonal perturbations in terms of a noise model for measurements. This difficulty is resolved by a model called probabilistic PCA (pPCA), first proposed by Tipping and Bishop in a Tech Report in 1997, and published in the J. of the Royal Statistical Society B 2002, and independently by Sam Roweis, NIPS 1998, that is illustrated visually in the figure below, and that we now explain:
In the pPCA model there is an (unknown) line (affine space in higher dimension) on which (hidden) points (blue) are generated at random according to a Gaussian distribution (represented by gray streak in the figure above, where the mean of the Gaussian is the green point). Observed points (red) are then generated from the hidden points by addition of isotropic Gaussian noise (blue smear), meaning that the Gaussian has a diagonal covariance matrix with equal entries. Formally, in the notation of Tipping and Bishop, this is a linear Gaussian model described as follows:
Observed random variables are given by $t = Wx + \mu + \epsilon$ where x are latent (hidden) random variables, W is a matrix describing a subspace and $Wx+\mu$ are the latent points on an affine subspace ($\mu$ corresponds to a translation). Finally, $\epsilon$ is an error term, given by a Gaussian random variable with mean 0 and covariance matrix $\sigma^2 I$. The parameters of the model are $W,\mu$ and $\sigma^2$. Equivalently, the observed random variables are themselves Gaussian, described by the distribution $t \sim \mathcal{N}(\mu,WW^T + \psi)$ where $\psi \sim \mathcal{N}(0,\sigma^2I)$. Tipping and Bishop prove an analogy of the Gauss-Markov theorem, namely that the affine subspace given by the maximum likelihood estimates of $W$ and $\mu$ is the PCA subspace (the proof is not difficult but I omit it and refer interested readers to their paper, or Bishop’s Pattern Recognition and Machine Learning book).
It is important to note that although the maximum likelihood estimates of $W,\mu$ in the pPCA model correspond to the PCA subspace, only posterior distributions can be obtained for the latent data (points on the subspace). Neither the mode nor the mean of those distributions corresponds to the PCA points (orthogonal projections of the observations onto the subspace). However what is true, is that the posterior distributions converge to the PCA points as $\sigma^2 \rightarrow 0$. In other words, the relationship between pPCA and PCA is a bit more subtle than that between least squares and regression.
The relationship between regression and (p)PCA is shown in the figure below:
In the figure, points have been generated randomly according to the pPCA model. the black smear shows the affine space on which the points were generated, with the smear indicating the Gaussian distribution used. Subsequently the latent points (light blue on the gray line) were used to make observed points (red) by the addition of isotropic Gaussian noise. The green line is the maximum likelihood estimate for the space, or equivalently by the theorem of Tipping and Bishop the PCA subspace. The projection of the observed points onto the PCA subspace (blue) are the PCA points. The purple line is the least squares line, or equivalently the affine space obtained by regression (y observed as a noisy function of x). The pink line is also a regression line, except where is observed as a noisy function of y.
A natural question to ask is why the probabilistic interpretation of PCA (pPCA) is useful or necessary? One reason it is beneficial is that maximum likelihood inference for pPCA involves hidden random variables, and therefore the EM algorithm immediately comes to mind as a solution (the strategy was suggested by both Tipping & Bishop and Roweis). I have not yet discussed how to find the PCA subspace, and the EM algorithm provides an intuitive and direct way to see how it can be done, without the need for writing down any linear algebra:
The exact version of the EM shown above is due to Roweis. In it, one begins with a random affine subspace passing through the centroid of the points. The “E” step (expectation) consists of projecting the points to the subspace. The projected points are considered fixed to the subspace. The “M” step (maximization) then consists of rotating the space so that the total squared distance of the fixed points on the subspace to the observed points is minimized. This is repeated until convergence. Roweis points out that this approach to finding the PCA subspace is equivalent to power iteration for (efficiently) finding eigenvalues of the the sample covariance matrix without computing it directly. This is our first use of the word eigenvalue in describing PCA, and we elaborate on it, and the linear algebra of computing PCA subspaces later in the post.
Another point of note is that pPCA can be viewed as a special case of factor analysis, and this connection provides an immediate starting point for thinking about generalizations of PCA. Specifically, factor analysis corresponds to the model $t \sim \mathcal{N}(\mu,WW^T + \psi)$ where the covariance matrix $\psi$ is less constrained, and only required to be diagonal. This is connected to a comment made above about when the PCA subspace might be more useful as a linear fit to data than regression. To reiterate, unlike physics, where some coordinate measurements have very little noise in comparison to others, biological measurements are frequently noisy in all coordinates. In such settings factor analysis is preferable, as the variance in each coordinate is estimated as part of the model. PCA is perhaps a good compromise, as PCA subspaces are easier to find than parameters for factor analysis, yet PCA, via its pPCA interpretation, accounts for noise in all coordinates.
A final comment about pPCA is that it provides a natural framework for thinking about hypothesis testing. The book Statistical Methods: A Geometric Approach by Saville and Wood is essentially about (the geometry of) pPCA and its connection to hypothesis testing. The authors do not use the term pPCA but their starting point is exactly the linear Gaussian model of Tipping and Bishop. The idea is to consider single samples from n independent identically distributed independent Gaussian random variables as one single sample from a high-dimensional multivariate linear Gaussian model with isotropic noise. From that point of view pPCA provides an interpretation for Bessel’s correction. The details are interesting but tangential to our focus on PCA.
We are therefore ready to return to Jeopardy!, where we have “Normality for $600″ with the answer “A generalization of linear regression in which the Gaussian noise is isotropic” and the question “What is PCA?” 3. An orthogonal projection of points onto an affine space that maximizes the retained sample variance. In the previous two interpretations of PCA, the focus was on the PCA affine subspace. However in many uses of PCA the output of interest is the projection of the given points onto the PCA affine space. The projected points have three useful related interpretations: 1. As seen in in section 1, the (orthogonally) projected points (red -> blue) are those whose total squared distance to the observed points is minimized. 2. What we focus on in this section, is the interpretation that the PCA subspace is the one onto which the (orthogonally) projected points maximize the retained sample variance. 3. The topic of the next section, namely that the squared distances between the (orthogonally) projected points are on average (in the $l_2$ metric) closest to the original distances between the points. The sample variance of a set of points is the average squared distance from each point to the centroid. Mathematically, if the observed points are translated so that their centroid is at zero (known as zero-centering), and then represented by an $n \times p$ matrix X, then the sample covariance matrix is given by $\frac{1}{n-1}X^tX$ and the sample variance is given by the trace of the matrix. The point is that the jth diagonal entry of $\frac{1}{n-1}X^tX$ is just $\frac{1}{n-1}\sum_{i=1}^n (x^i_j)^2$, which is the sample variance of the jth variable. The PCA subspace can be viewed as that subspace with the property that the sample variance of the projections of the observed points onto the subspace is maximized. This is easy to see from the figure above. For each point (blue), Pythagoras’ theorem implies that $d(red,blue)^2+d(blue,green)^2 = d(red,green)^2$. Since the PCA subspace is the one minimizing the total squared red-blue distances, and since the solid black lines (red-green distances) are fixed, it follows that the PCA subspace also maximizes the total squared green-blue distances. In other words, PCA maximizes the retained sample variance. The explanation above is informal, and uses a 1-dimensional PCA subspace in dimension 2 to make the argument. However the argument extends easily to higher dimension, which is typically the setting where PCA is used. In fact, PCA is typically used to “visualize” high dimensional points by projection into dimensions two or three, precisely because of the interpretation provided above, namely that it retains the sample variance. I put visualize in quotes because intuition in two or three dimensions does not always hold in high dimensions. However PCA can be useful for visualization, and one of my favorite examples is the evidence for genes mirroring geography in humans. This was first alluded to by Cavalli-Sforza, but definitively shown by Lao et al., 2008, who analyzed 2541 individuals and showed that PCA of the SNP matrix (approximately) recapitulates geography: Genes mirror geography from Lao et al. 2008: (Left) PCA of the SNP matrix (2541 individuals x 309,790 SNPs) showing a density map of projected points. (Right) Map of Europe showing locations of the populations . In the picture above, it is useful to keep in mind that the emergence of geography is occurring in that projection in which the sample variance is maximized. As far as interpretation goes, it is useful to look back at Cavalli-Sforza’s work. He and collaborators who worked on the problem in the 1970s, were unable to obtain a dense SNP matrix due to limited technology of the time. Instead, in Menozzi et al., 1978 they performed PCA of an allele-frequency matrix, i.e. a matrix indexed by populations and allele frequencies instead of individuals and genotypes. Unfortunately they fell into the trap of misinterpreting the biological meaning of the eigenvectors in PCA. Specifically, they inferred migration patterns from contour plots in geographic space obtained by plotting the relative contributions from the eigenvectors, but the effects they observed turned out to be an artifact of PCA. However as we discussed above, PCA can be used quantitatively via the stochastic process for which it solves maximum likelihood inference. It just has to be properly understood. To conclude this section in Jeopardy! language, we have “Normality for$800″ with the answer “A set of points in an affine space obtained via projection of a set of given points so that the sample variance of the projected points is maximized” and the question “What is PCA?”
4. Principal component analysis of Euclidean distance matrices.
In the preceding interpretations of PCA, I have focused on what happens to individual points when projected to a lower dimensional subspace, but it is also interesting to consider what happens to pairs of points. One thing that is clear is that if a pair of points are projected orthogonally onto a low-dimensional affine subspace then the distance between the points in the projection is smaller than the original distance between the points. This is clear because of Pythagoras’ theorem, which implies that the squared distance will shrink unless the points are parallel to the subspace in which case the distance remains the same. An interesting observation is that in fact the PCA subspace is the one with the property where the average (or total) squared distances between the points is maximized. To see this it again suffices to consider only projections onto one dimension (the general case follows by Pythagoras’ theorem). The following lemma, discussed in my previous blog post, makes the connection to the previous discussion:
Lemma: Let $x_1,\ldots,x_n$ be numbers with mean $\overline{x} = \frac{1}{n}\sum_i x_i$. If the average squared distance between pairs of points is denoted $D = \frac{1}{n^2}\sum_{i,j} (x_i-x_j)^2$ and the variance is denoted $V=\frac{1}{n}\sum_i (x_i-\overline{x})^2$ then $V=\frac{1}{2}D$.
What the lemma says is that the sample variance is equal to the average squared difference between the numbers (i.e. it is a scalar multiple that does not depend on the numbers). I have already discussed that the PCA subspace maximizes the retained variance, and it therefore follows that it also maximizes the average (or total) projected squared distance between the points. Alternately, PCA can be interpreted as minimizing the total (squared) distance that is lost, i,e. if the original distances between the points are given by a distance matrix $D$ and the projected distances are given by $\tilde{D}$, then the PCA subspace minimizes $\sum_{ij} (D^2_{ij} - \tilde{D}^2_{ij})$, where each term in the sum is positive as discussed above.
This interpretation of PCA leads to an interesting application of the method to (Euclidean) distance matrices rather than points. The idea is based on a theorem of Isaac Schoenberg that characterizes Euclidean distance matrices and provides a method for realizing them. The theorem is well-known to structural biologists who work with NMR, because it is one of the foundations used to reconstruct coordinates of structures from distance measurements. It requires a bit of notation: is a distance matrix with entries $d_{ij}$ and $\Delta$ is the matrix with entries $-\frac{1}{2}d^2_{ij}$. ${\bf 1}$ denotes the vector of all ones, and ${\bf s}$ denotes a vector.
Theorem (Schoenberg, 1938): A matrix D is a Euclidean distance matrix if and only if the matrix $B=(I-{\bf 1}{\bf s}')\Delta(I-{\bf s}{\bf 1}')$ is positive semi-definite where ${\bf s}'{\bf 1} = 1$.
For the case when ${\bf s}$ is chosen to be a unit vector, i.e. all entries are zero except one of them equal to 1, the matrix B can be viewed as the Gromov transform (known as the Farris transform in phylogenetics) of the matrix with entries $d^2_{ij}$. Since the matrix is positive semidefinite it can be written as $B=XX^t$, where the matrix X provides coordinates for points that realize D. At this point PCA can be applied resulting in a principal subspace and points on it (the orthogonal projections of X). A point of note is that eigenvectors of $XX^t$ can be computed directly, avoiding the need to compute $X^tX$ which may be a larger matrix if $n < p$.
The procedure just described is called classic multidimensional scaling (MDS) and it returns a set of points on a Euclidean subspace with distance matrix $\tilde{D}$ that best represent the original distance matrix D in the sense that $\sum_{ij} (D^2_{ij} - \tilde{D}^2_{ij})$ is minimized. The term multidimensional scaling without the “classic” has taken on an expanded meaning, namely it encapsulates all methods that seek to approximately realize a distance matrix by points in a low dimensional Euclidean space. Such methods are generally not related to PCA, but classic multidimensional scaling is PCA. This is a general source of confusion and error on the internet. In fact, most articles and course notes I found online describing the connection between MDS and PCA are incorrect. In any case classic multidimensional scaling is a very useful instance of PCA, because it extends the utility of the method to cases where points are not available but distances between them are.
Now we return to Jeopardy! one final time with the final question in the category: “Normality for $1000″. The answer is “Principal component analysis of Euclidean distance matrices” and the question is “What is classic multidimensional scaling?” An example To illustrate the interpretations of PCA I have highlighted, I’m including an example in R inspired by an example from another blog post (all commands can be directly pasted into an R console). I’m also providing the example because missing in the discussion above is a description of how to compute PCA subspaces and the projections of points onto them. I therefore explain some of this math in the course of working out the example: First, I generate a set of points (in $\mathbb{R}^2$). I’ve chosen a low dimension so that pictures can be drawn that are compatible with some of the examples above. Comments following commands appear after the # character. set.seed(2) #sets the seed for random number generation. x <- 1:100 #creates a vector x with numbers from 1 to 100 ex <- rnorm(100, 0, 30) #100 normally distributed rand. nos. w/ mean=0, s.d.=30 ey <- rnorm(100, 0, 30) # " " y <- 30 + 2 * x #sets y to be a vector that is a linear function of x x_obs <- x + ex #adds "noise" to x y_obs <- y + ey #adds "noise" to y P <- cbind(x_obs,y_obs) #places points in matrix plot(P,asp=1,col=1) #plot points points(mean(x_obs),mean(y_obs),col=3, pch=19) #show center At this point a full PCA analysis can be undertaken in R using the command “prcomp”, but in order to illustrate the algorithm I show all the steps below: M <- cbind(x_obs-mean(x_obs),y_obs-mean(y_obs))#centered matrix MCov <- cov(M) #creates covariance matrix Note that the covariance matrix is proportional to the matrix$M^tM$. Next I turn to computation of the principal axes: eigenValues <- eigen(MCov)$values #compute eigenvalues
eigenVectors <- eigen(MCov)$vectors #compute eigenvectors The eigenvectors of the covariance matrix provide the principal axes, and the eigenvalues quantify the fraction of variance explained in each component. This math is explained in many papers and books so we omit it here, except to say that the fact that eigenvalues of the sample covariance matrix are the principal axes follows from recasting the PCA optimization problem as maximization of the Raleigh quotient. A key point is that although I’ve computed the sample covariance matrix explicitly in this example, it is not necessary to do so in practice in order to obtain its eigenvectors. In fact, it is inadvisable to do so. Instead, it is computationally more efficient, and also more stable to directly compute the singular value decomposition of M. The singular value decomposition of M decomposes it into $M=UDV^t$ where is a diagonal matrix and both $U$ and $V^t$ are orthogonal matrices. I will also not explain in detail the linear algebra of singular value decomposition and its relationship to eigenvectors of the sample covariance matrix (there is plenty of material elsewhere), and only show how to compute it in R: d <- svd(M)$d #the singular values
v <- svd(M)\$v #the right singular vectors
The right singular vectors are the eigenvectors of $M^tM$. Next I plot the principal axes:
lines(x_obs,eigenVectors[2,1]/eigenVectors[1,1]*M[x]+mean(y_obs),col=8)
This shows the first principal axis. Note that it passes through the mean as expected. The ratio of the eigenvectors gives the slope of the axis. Next
lines(x_obs,eigenVectors[2,2]/eigenVectors[1,2]*M[x]+mean(y_obs),col=8)
shows the second principal axis, which is orthogonal to the first (recall that the matrix $V^t$ in the singular value decomposition is orthogonal). This can be checked by noting that the second principal axis is also
lines(x_obs,-1/(eigenVectors[2,1]/eigenVectors[1,1])*M[x]+mean(y_obs),col=8)
as the product of orthogonal slopes is -1. Next, I plot the projections of the points onto the first principal component:
trans <- (M%*%v[,1])%*%v[,1] #compute projections of points
P_proj <- scale(trans, center=-cbind(mean(x_obs),mean(y_obs)), scale=FALSE)
points(P_proj, col=4,pch=19,cex=0.5) #plot projections
segments(x_obs,y_obs,P_proj[,1],P_proj[,2],col=4,lty=2) #connect to points
The linear algebra of the projection is simply a rotation followed by a projection (and an extra step to recenter to the coordinates of the original points). Formally, the matrix M of points is rotated by the matrix of eigenvectors W to produce $T=MW$. This is the rotation that has all the optimality properties described above. The matrix T is sometimes called the PCA score matrix. All of the above code produces the following figure, which should be compared to those shown above:
There are many generalizations and modifications to PCA that go far beyond what has been presented here. The first step in generalizing probabilistic PCA is factor analysis, which includes estimation of variance parameters in each coordinate. Since it is rare that “noise” in data will be the same in each coordinate, factor analysis is almost always a better idea than PCA (although the numerical algorithms are more complicated). In other words, I just explained PCA in detail, now I’m saying don’t use it! There are other aspects that have been generalized and extended. For example the Gaussian assumption can be relaxed to other members of the exponential family, an important idea if the data is discrete (as in genetics). Yang et al. 2012 exploit this idea by replacing PCA with logistic PCA for analysis of genotypes. There are also many constrained and regularized versions of PCA, all improving on the basic algorithm to deal with numerous issues and difficulties. Perhaps more importantly, there are issues in using PCA that I have not discussed. A big one is how to choose the PCA dimension to project to in analysis of high-dimensional data. But I am stopping here as I am certain no one is reading at this far into the post anyway…
The take-home message about PCA? Always be thinking when using it!
Acknowledgment: The exposition of PCA in this post began with notes I compiled for my course MCB/Math 239: 14 Lessons in Computational Genomics taught in the Spring of 2013. I thank students in that class for their questions and feedback. None of the material presented in class was new, but the exposition was intended to clarify when PCA ought to be used, and how. I was inspired by the papers of Tipping, Bishop and Roweis on probabilistic PCA in the late 1990s that provided the needed statistical framework for its understanding. Following the class I taught, I benefited greatly from conversations with Nicolas Bray, Brielin Brown, Isaac Joseph and Shannon McCurdy who helped me to further frame PCA in the way presented in this post.
The Habsburg rulership of Spain ended with an inbreeding coefficient of F=0.254. The last king, Charles II (1661-1700), suffered an unenviable life. He was unable to chew. His tongue was so large he could not speak clearly, and he constantly drooled. Sadly, his mouth was the least of his problems. He suffered seizures, had intellectual disabilities, and was frequently vomiting. He was also impotent and infertile, which meant that even his death was a curse in that his lack of heirs led to a war.
None of these problems prevented him from being married (twice). His first wife, princess Henrietta of England, died at age 26 after becoming deeply depressed having being married to the man for a decade. Only a year later, he married another princess, 23 year old Maria Anna of Neuberg. To put it mildly, his wives did not end up living the charmed life of Disney princesses, nor were they presumably smitten by young Charles II who apparently aged prematurely and looked the part of his horrific homozygosity. The princesses married Charles II because they were forced to. Royals organized marriages to protect and expand their power, money and influence. Coupled to this were primogeniture rules which ensured that the sons of kings, their own flesh and blood and therefore presumably the best-suited to be in power, would indeed have the opportunity to succeed their fathers. The family tree of Charles II shows how this worked in Spain:
It is believed that the inbreeding in Charles II’s family led to two genetic disorders, combined pituitary hormone deficiency and distal renal tubular acidosis, that explained many of his physical and mental problems. In other words, genetic diversity is important, and the point of this blog post is to highlight the fact that diversity is important in education as well.
The problem of inbreeding in academia has been studied previously, albeit to a limited extent. One interesting article is Navel Grazing: Academic Inbreeding and Scientific Productivity by Horta et al published in 2010 (my own experience with an inbred academic from a department where 39% of the faculty are self-hires anecdotally confirms the claims made in the paper). But here I focus on the downsides of inbreeding of ideas rather than of faculty. For example home-schooling, the educational equivalent of primogeniture, can be fantastic if the parents happen to be good teachers, but can fail miserably if they are not. One thing that is guaranteed in a school or university setting is that learning happens by exposure to many teachers (different faculty, students, tutors, the internet, etc.) Students frequently complain when there is high variance in teaching quality, but one thing such variance ensures is that is is very unlikely that any student is exposed only to bad teachers. Diversity in teaching also helps to foster the development of new ideas. Different teachers, by virtue of insight or error, will occasionally “mutate” ideas or concepts for better or for worse. In other words, one does not have to fully embrace the theory of memes to acknowledge that there are benefits to variance in teaching styles, methods and pedagogy. Conversely, there is danger in homogeneity.
This brings me to MOOCs. One of the great things about MOOCs is that they reach millions of people. Udacity claims it has 1.6 million “users” (students?). Coursera claims 7.1 million. These companies are greatly expanding the accessibility of education. Starving children in India can now take courses in mathematical methods for quantitative finance, and for the first time in history, a president of the United States can discreetly take a freshman course on economics together with its high school algebra prerequisites (highly recommended). But when I am asked whether I would be interested in offering a MOOC I hesitate, paralyzed at the thought that any error I make would immediately be embedded in the brains of millions of innocent victims. My concern is this: MOOCs can greatly reduce the variance in education. For example, Coursera currently offers 641 courses, which means that each courses is or has been taught to over 11,000 students. Many college courses may have less than a few dozen students, and even large college courses rarely have more than a few hundred students. This means that on average, through MOOCs, individual professors reach many more (2 orders of magnitude!) students. A great lecture can end up positively impacting a large number of individuals, but at the same time, a MOOC can be a vehicle for infecting the brains of millions of people with nonsense. If that nonsense is then propagated and reaffirmed via the interactions of the people who have learned it from the same source, then the inbreeding of ideas has occurred.
I mention MOOCs because I was recently thinking about intuition behind Bessel’s correction replacing n with n-1 in the formula for sample variance. Formally, Bessel’s correction replaces the biased formula
$s^2_n = \frac{1}{n} \sum_{i=1}^n (x_i - \overline{x})^2$
for estimating the variance of a random variable from samples $x_1,\ldots,x_n$ with
$s^2_{n-1} = \frac{1}{n-1} \sum_{i=1}^n (x_i-\overline{x})^2$.
The switch from to n-1 is a bit mysterious and surprising, and in introductory statistics classes it is frequently just presented as a “fact”. When an explanation is provided, it is usually in the form of algebraic manipulation that establishes the result. The issue came up as a result of a blog post I’m writing about principal components analysis (PCA), and I thought I would check for an intuitive explanation online. I googled “intuition sample variance” and the top link was a MOOC from the Khan Academy:
The video has over 51,000 views with over 100 “likes” and only 6 “dislikes”. Unfortunately, in this case, popularity is not a good proxy for quality. Despite the title promising “review” and “intuition” for “why we divide by n-1 for the unbiased sample variance” there is no specific reason given why is replaced by n-1 (as opposed to another correction). Furthermore, the intuition provided has to do with the fact that $x_i-\overline{x}$ underestimates $x_i-\mu$ (where $\mu$ is the mean of the random variable and $\overline{x}$ is the sample mean) but the explanation is confusing and not quantitative (which it can easily be). In fact, the wikipedia page for Bessel’s correction provides three different mathematical explanations for the correction together with the intuition that motivates them, but it is difficult to find with Google unless one knows that the correction is called “Bessel’s correction”.
Wikipedia is also not perfect, and this example is a good one for why teaching by humans is important. Among the three alternative derivations, I think that one stands out as “better” but one would not know by just looking at the wikipedia page. Specifically, I refer to “Alternate 1″ on the wikipedia page, that is essentially explaining that variance can be rewritten as a double sum corresponding to the average squared distance between points and the diagonal terms of the sum are zero in expectation. An explanation of why this fact leads to the n-1 in the unbiased estimator is as follows:
The first step is to notice that the variance of a random variable is equal to half of the expected squared difference of two independent identically distributed random variables of that type. Specifically, the definition of variance is:
$var(X) = \mathbb{E}(X - \mu)^2$ where $\mu = \mathbb{E}(X)$. Equivalently, $var(X) = \mathbb{E}(X^2) -\mu^2$. Now suppose that Y is another random variable identically distributed to X and with X,Y independent. Then $\mathbb{E}(X-Y)^2 = 2 var(X)$. This is easy to see by using the fact that
$\mathbb{E}(X-Y)^2 = \mathbb{E}(X^2) + \mathbb{E}(Y^2) - 2\mathbb{E}(X)\mathbb{E}(Y) = 2\mathbb{E}(X^2)-2\mu^2$.
This identity motivates a rewriting of the (uncorrected) sample variance $s_n$ in a way that is computationally less efficient, but mathematically more insightful:
$s_n = \frac{1}{2n^2} \sum_{i,j=1}^n (x_i-x_j)^2$.
Of note is that in this summation exactly n of the terms are zero, namely the terms when i=j. These terms are zero independently of the original distribution, and remain so in expectation thereby biasing the estimate of the variance, specifically leading to an underestimate. Removing them fixes the estimate and produces
$s_{n-1}^2 = \frac{1}{2n(n-1)} \sum_{i,j=1, i \neq j}^n (x_i-x_j)^2$.
It is easy to see that this is indeed Bessel’s correction. In other words, the correction boils down to the fact that $n^2-n = n(n-1)$, hence the appearance of n-1.
Why do I like this particular derivation of Bessel’s correction? There are two reasons: first, n-1 emerges naturally and obviously from the derivation. The denominator in $s_{n-1}^2$ matches exactly the number of terms being summed, so that it can be understood as a true average (this is not apparent in its standard form as $s_{n-1}^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i-\overline{x})^2$. There is really nothing mysterious anymore, its just that some terms having been omitted from the sum because they were non-inofrmative. Second, as I will show in my forthcoming blog post on PCA, the fact that the variance of a random variable is half of the expectation of the squared difference of two instances, is key to understanding the connection between multi-dimensional scaling (MDS) and PCA. In other words, as my student Nicolas Bray is fond of saying, although most people think a proof is either right or wrong, in fact some proofs are more right than others. The connection between Bessel’s correction and PCA goes even deeper: as explained by Saville and Wood in their book Statistical Methods: A Geometric Approach n-1 can be understood to be a reduction in one dimension from the point of view of probabilistic PCA (Saville and Wood do not explicitly use the term probabilistic PCA but as I will explain in my PCA post it is implicit in their book). Finally, there are many subtleties to Bessel’s correction, for example it is an unbiased estimator for variance and not standard deviation. These issues ought to be mentioned in a good lecture about the topic. In other words, the Khan lecture is neither necessary nor sufficient, but unlike a standard lecture where the damage is limited to a small audience of students, it has been viewed more than 50,000 times and those views cannot be unviewed.
In writing this blog post I pondered the irony of my call for added diversity in teaching while I preach my own idea (this post) to a large number of readers via a medium designed for maximal outreach. I can only ask that others blog as well to offer alternative points of view :) and that readers inform themselves on the issues I raise by fact-checking elsewhere. As far as the statistics goes, if someone finds the post confusing, they should go and register for one of the many fantastic MOOCs on statistics! But I reiterate that in the rush to MOOCdom, providers must offer diversity in their offerings (even multiple lectures on the same topic) to ensure a healthy population of memes. This is especially true in Spain, where already inbred faculty are now inbreeding what they teach by MOOCing via Miriada X. Half of the MOOCs being offered in Spain originate from just 3 universities, while the number of potential viewers is enormous as Spanish is now the second most spoken language in the world (thanks to Charles II’s great-great-grandfather, Charles I).
May Charles II rest in peace.
One of my distinct memories from elementary school is going to “library class” to learn about the Dewey decimal classification and how to use a card catalog to find books. Searching for books efficiently was possible because cards in the catalog were sorted lexicographically.
It didn’t occur to me at the time, but the system required authors of books to be totally ordered. Without an ordering of authors in a book with multiple authors, there would be no way to decide where to place the card for the book in a catalog searchable by author. The practice of ordering authors on publications is evident in the oldest printed texts and has persisted to this day. I have never thought that it could be any other way.
However this past Wednesday I was visiting the University of Washington to deliver a seminar, and among the highlights of the visit was my meeting with the graduate students. I met 12 for lunch and two more came for dinner. Meeting with students is always my favorite part of a visit to a university. They have original and creative ideas, and most importantly, are not bound in their thought by archaic tradition. They frequently don’t know what one is supposed to think and how one is supposed to say it. They just think and speak!
One of the students I met on Wednesday was Vanessa Gray, a student of Doug Fowler, who in a conversation on authorship practices suggested to me the radical and brilliant idea that papers should be published without an ordering of authors.
Many journals now have a section called “Author contributions” where roles of individuals in collaborative projects can be described (many journals now require such descriptions). So why bother ordering the authors for a list underneath the title? As far as indexing and searching goes, Google and other search engines require only a set of authors, and not a specific ordering.
I agree with Vanessa that ending author ordering on publications would greatly improve fairness in the biological sciences, where many current projects involve complex assemblies of teams with complementary skills. “First authorship” is not well-defined when one author performed a large number of difficult experiments, and another developed novel algorithms and wrote complex software for analyzing the experiments. Similarly, “last authorship” fails as a concept when students are co-advised, or one principal investigator provides substantial funding on a project, while another is participating in doing the work. And recently, large consortium projects have completely destroyed any meaning of “author” by having hundreds, or even thousands of authors on projects. Even when there are relatively few authors people rarely credit anyone except the first and last authors, even if others did substantial work. In the recent ENCODE paper published in PNAS with 30 authors, it appears to me from the responses to my previous blog post about the paper that the 5th and 6th authors did a lot (majority?) of the work in putting together figures and results, yet I suspect the “credit” for the paper will go to the first author (the flip side in that case is that the first author is where blame is assigned as well).
There is also a theoretical justification for not ordering authors. Ordering of authors on a publication can be thought of as a ranking produced by “votes” of the participants in the project. Of course in practice not all votes are equal. In what is called dictatorship in social choice theory, PIs frequently make the decisions independently of how specific authors feel they may have contributed. This may work on a paper where there is a single PI (although it may be considered unfair by the graduate students and postdocs), however dictatorship as a system for determining authorship certainly breaks down when multiple PIs collaborate on a project. Arrow’s impossibility theorem explains that in the absence of dictatorship, there is a problem in producing a single ordering satisfying two other seemingly basic and essential fairness criteria. Informally, the theorem states that there is no authorship ordering system based on voting of contributing authors that can satisfy the following three criteria:
• If every author thinks that X should be ordered before Y, then the author list should have X placed before Y.
• For a fixed list of voting preferences regarding the ordering of X vs. Y, the ordering between X and Y in the author list will remain unchanged even if does not depend on the ordering of other pairs such as X and Z, Y and Z, or Z and W.
• There is no “dictator”, i.e. no single author possesses the power to determine the author ordering.
Authors frequently have differing opinions about the impact of their own contribution to a publication, and therefore their preferences (votes) for author ordering are discordant. This means that any system for ordering authors will not satisfy everyone’s preferences, and in the sense of Arrow’s impossibility theorem will be unfair. One way around Arrow’s impossibility theorem is to specify authorship order without regard to authors’ preferences, for example by always ordering authors alphabetically (the Hardy-Littlewood rule). This method, usually the one used in the mathematical sciences, is also fraught with problems. Of course, listing author contributions for what they are is not entirely trivial. For example, different authors may have conflicting views about what it means to have “written the text of the paper”. But using words to describe contributions allow for much more detail about what each author did, and allows for nuanced contributions to be described (e.g., John and Jane were in the room when the initial idea for the project was discussed, but did not contribute anything afterwards).
To summarize, in the modern era of electronic publishing ordering of authors is unnecessary, and if it is unnecessary, then why confront Arrow’s theorem and inevitably produce orderings unfairly? Publications should just explain the author contributions. Time to end ordered authorship.
The card catalog at Yale University’s Sterling Memorial Library (from Wikipedia).
I was recently reading the latest ENCODE paper published in PNAS when a sentence in the caption of Figure 2 caught my attention:
“Depending on the total amount of RNA in a cell, one transcript copy per cell corresponds to between 0.5 and 5 FPKM in PolyA+ whole-cell samples according to current estimates (with the upper end of that range corresponding to small cells with little RNA and vice versa).”
Although very few people actually care about ENCODE, many people do care about the interpretation of RNA-Seq FPKM measurements and to them this is likely to be a sentence of interest. In fact, there have been a number of attempts to provide intuitive meaning for RPKM (and FPKM) in terms of copy numbers of transcripts per cell. Even though the ENCODE PNAS paper provides no citation for the statement (or methods section explaining the derivation), I believe its source is the RNA-Seq paper by Mortazavi et al. In that paper, the authors write that
“…absolute transcript levels per cell can also be calculated. For example, on the basis of literature values for the mRNA content of a liver cell [Galau et al. 1977] and the RNA standards, we estimated that 3 RPKM corresponds to about one transcript per liver cell. For C2C12 tissue culture cells, for which we know the starting cell number and RNA preparation yields needed to make the calculation, a transcript of 1 RPKM corresponds to approximately one transcript per cell. “
This statement has been picked up on in a number of publications (e.g., Hebenstreit et al., 2011, van Bakel et al., 2011). However the inference of transcript copies per cell directly from RPKM or FPKM estimates is not possible and conversion factors such as 1 RPKM = 1 transcript per cell are incoherent. At the same time, the estimates of Mortazavi et al. and the range provided in the ENCODE PNAS paper are informative. The “incoherence” stems from a subtle issue in the normalization of RPKM/FPKM that I have discussed in a talk I gave at CSHL, and is the reason why TPM is a better unit for RNA abundance. Still, the estimates turn out to be “informative”, in the sense that the effect of (lack of) normalization appears to be smaller than variability in the amount of RNA per cell. I explain these issues below:
Why is the sentence incoherent?
RNA-Seq can be used to estimate transcript abundances in an RNA sample. Formally, a sample consists of n distinct types of transcripts, and each occurs with different multiplicity (copy number), so that transcript appears $m_i$ times in the sample. By “abundance” we mean the relative amounts $\rho_1,\ldots,\rho_n$ where $\rho_i = \frac{m_i}{\sum_{i=1}^n m_i}$. Note that $0 \leq \rho_i \leq 1$ and $\sum_{i=1}^n \rho_i = 1$. Suppose that $m_j=1$ for some j. The corresponding $\rho_j$ is therefore $\rho_j = \frac{1}{M}$ where $M = \sum_{i=1}^n m_i$. The question is what does this $\rho$ value correspond to in RPKM (or FPKM).
RPKM stands for “reads per kilobase of transcript per million reads mapped” and FPKM is the same except with “fragment” replacing read (initially reads were not paired-end, but with the advent of paired-end sequencing it makes more sense to speak of fragments, and hence FPKM). As a unit of measurement for an estimate, what FPKM really refers to is the expected number of fragments per kilboase of transcript per million reads. Formally, if we let $l_i$ be the length of transcript and define $\alpha_i = \frac{\rho_i l_i}{\sum_{j=1}^n \rho_j l_j}$ then abundance in FPKM for transcript is abundance measured as $FPKM_i = \frac{\alpha_i \cdot 10^{6}}{l_i/(10^3)}$. In terms of $\rho$, we obtain that
$FPKM_i = \frac{\rho_i \cdot 10^9}{\sum_{j=1}^n \rho_j l_j}$.
The term in the denominator can be considered a kind of normalization factor, that while identical for each transcript, depends on the abundances of each transcript (unless all lengths are equal). It is in essence an average of lengths of transcripts weighted by abundance. Moreover, the length of each transcript should be taken to be taken to be its “effective” length, i.e. the length with respect to fragment lengths, or equivalently, the number of positions where fragments can start.
The implication for finding a relationship between FPKM and relative abundance constituting one transcript copy per cell is that one cannot. Mathematically, the latter is equivalent to setting $\rho_i = \frac{1}{M}$ in the formula above and then trying to determine $FPKM_i$. Unfortunately, all the remaining $\rho$ are still in the formula, and must be known in order to calculate the corresponding FPKM value.
The argument above makes clear that it does not make sense to estimate transcript copy counts per cell in terms of RPKM or FPKM. Measurements in RPKM or FPKM units depend on the abundances of transcripts in the specific sample being considered, and therefore the connection to copy counts is incoherent. The obvious and correct solution is to work directly with the $\rho$. This is the rationale of TPM (transcripts per million) used by Bo Li and Colin Dewey in the RSEM paper (the argument for TPM is also made in Wagner et al. 2012).
Why is the sentence informative?
Even though incoherent, it turns out there is some truth to the ranges and estimates of copy count per cell in terms of RPKM and FPKM that have been circulated. To understand why requires noting that there are in fact two factors that come into play in estimating the FPKM corresponding to abundance of one transcript copy per cell. First, is M as defined above to be the total number of transcripts in a cell. This depends on the amount of RNA in a cell. Second are the relative abundances of all transcripts and their contribution to the denominator in the $FPKM_i$ formula.
The best paper to date on the connection between transcript copy numbers and RNA-Seq measurements is the careful work of Marinov et al. in “From single-cell to cell-pool transcriptomes: stochasticity in gene expression and RNA splicing” published in Genome Research earlier this year. First of all, the paper describes careful estimates of RNA quantities in different cells, and concludes that (at least for the cells studied in the paper) amounts vary by approximately one order of magnitude. Incidentally, the estimates in Marinov et al. confirm and are consistent with rough estimates of Galau et al. from 1977, of 300,000 transcripts per cell. Marinov et al. also use spike-in measurements are used to conclude that in “GM12878 single cells, one transcript copy corresponds to ∼10 FPKM on average.”. The main value of the paper lies in its confirmation that RNA quantities can vary by an order of magnitude, and I am guessing this factor of 10 is the basis for the range provided in the ENCODE PNAS paper (0.5 to 5 FPKM).
In order to determine the relative importance of the denominator in $FPKM_i$ I looked at a few RNA-Seq datasets we are currently examining. In the GEUVADIS data, the weighted average can vary by as much as 20% between samples. In a rat RNA-Seq dataset we are analyzing, the difference is a factor of two (and interestingly very dependent on the exact annotation used for quantification). The point here is that even the denominator in $FPKM_i$ does vary, but less, it seems, than the variability in RNA quantity. In other words, the estimate of 0.5 to 5 FPKM corresponding to one transcript per cell is incoherent albeit probably not too far off.
One consequence of all of the above discussion is that while differential analysis of experiments can be performed based on FPKM units (as done for example in Cufflinks, where the normalization factors are appropriately accounted for), it does not make sense to compare raw FPKM values across experiments. This is precisely what is done in Figure 2 of the ENCODE PNAS paper. What the analysis above shows, is that actual abundances may be off by amounts much larger than the differences shown in the figure. In other words, while the caption turns out to to containing an interesting comment the overall figure doesn’t really make sense. Specifically, I’m not sure the relative RPKM values shown in the figure deliver the correct relative amounts, an issue that ENCODE can and should check. Which brings me to the last part of this post…
What is ENCODE doing?
Having realized the possible issue with RPKM comparisons in Figure 2, I took a look at Figure 3 to try to understand whether there were potential implications for it as well. That exercise took me to a whole other level of ENCODEness. To begin with, I was trying to make sense of the x-axis, which is labeled “biochemical signal strength (log10)” when I realized that the different curves on the plot all come from different, completely unrelated x-axes. If this sounds confusing, it is. The green curves are showing graphs of functions whose domain is in log 10 RPKM units. However the histone modification curves are in log (-10 log p), where p is a p-value that has been computed. I’ve never seen anyone plot log(log(p-values)); what does it mean?! Nor do I understand how such graphs can be placed on a common x-axis (?!). What is “biochemical signal strength” (?) Why in the bottom panel is the grey H3K9me3 showing %nucleotides conserved decreasing as “biochemical strength” is increasing (?!) Why is the green RNA curves showing conservation below genome average for low expressed transcripts (?!) and why in the top panel is the red H3K4me3 an “M” shape (?!) What does any of this mean (?!) What I’m supposed to understand from it, or frankly, what is going on at all ??? I know many of the authors of this ENCODE PNAS paper and I simply cannot believe they saw and approved this figure. It is truly beyond belief… see below:
All of these figures are of course to support the main point of the paper. Which is that even though 80% of the genome is functional it is also true that this is not what was meant to be said , and that what is true is that “survey of biochemical activity led to a significant increase in genome coverage and thus accentuated the discrepancy between biochemical and evolutionary estimates… where function is ascertained independently of cellular state but is dependent on environment and evolutionary niche therefore resulting in estimates that differ widely in their false-positive and false-negative rates and the resolution with which elements can be defined… [unlike] genetic approaches that rely on sequence alterations to establish the biological relevance of a DNA segment and are often considered a gold standard for defining function.”
The ENCODE PNAS paper was first published behind a paywall. However after some public criticism, the authors relented and paid for it to be open access. This was a mistake. Had it remained behind a paywall not only would the consortium have saved money, I and others might have been spared the experience of reading the paper. I hope the consortium will afford me the courtesy of paywall next time.
In reading the news yesterday I came across multiple reports claiming that even casually smoking marijuana can change your brain. I usually don’t pay much attention to such articles; I’ve never smoked a joint in my life. In fact, I’ve never even smoked a cigarette. So even though as a scientist I’ve been interested in cannabis from the molecular biology point of view, and as a citizen from a legal point of view, the issues have not been personal. However reading a USA Today article about the paper, I noticed that the principal investigator Hans Breiter was claiming to be a psychiatrist and mathematician. That is an unusual combination so I decided to take a closer look. I immediately found out the claim was a lie. In fact, the totality of math credentials of Hans Breiter consist of some logic/philosophy courses during a year abroad at St. Andrews while he was a pre-med student at Northwestern. Even being an undergraduate major in mathematics does not make one a mathematician, just as being an undergraduate major in biology does not makes one a doctor. Thus, with his outlandish claim, Hans Breiter had succeeded in personally offending me! So, I decided to take a look at his paper underlying the multiple news reports:
This is quite possibly the worst paper I’ve read all year (as some of my previous blog posts show I am saying something with this statement). Here is a breakdown of some of the issues with the paper:
### 1. Study design
First of all, the study has a very small sample size, with only 20 “cases” (marijuana users), a fact that is important to keep in mind in what follows. The title uses the term “recreational users” to describe them, and in the press release accompanying the article Breiter says that “Some of these people only used marijuana to get high once or twice a week. People think a little recreational use shouldn’t cause a problem, if someone is doing OK with work or school. Our data directly says this is not the case.” In fact, the majority of users in the study were smoking more than 10 joints per week. There is even a person in the study smoking more than 30 joints per week (as disclosed above, I’m not an expert on this stuff but if 30 joints per week is “recreation” then it seems to me that person is having a lot of fun). More importantly, Breiter’s statement in the press release is a lie. There is no evidence in the paper whatsoever, not even a tiny shred, that the users who were getting high once or twice a week were having any problems. There are also other issues with the study design. For example, the paper claims the users are not “abusing” other drugs, but it is quite possible that they are getting high on cocaine, heroin, or ??? as well, an issue that could quite possibly affect the study. The experiment consisted of an MRI scan of each user/control, but only a single scan was done. Given the variability in MRI scans this also seems problematic.
### 2. Multiple testing
The study looked at three aspects of brain morphometry in the study participants: gray matter density, volume and shape. Each of these morphometric analyses constituted multiple tests. In the case of gray matter density, estimates were based on small clusters of voxels, resulting in 123 tests (association of each voxel cluster with marijuana use). Volumes were estimated for four regions: left and right nucleus accumbens and amygdala. Shape was also tested in the same four regions. What the authors should have done is to correct the p-values computed for each of these tests by accounting for the total number of tests performed. Instead, (Bonferroni) corrections were performed separately for each type of analysis. For example, in the volume analysis p-values were required to be less than 0.0125 = 0.05/4. In other words, the extent of testing was not properly accounted for. Even so, many of the results were not significant. For example, the volume analysis showed no significant association for any of the four tested regions. The best case was the left nucleus accumbens (Figure 1C) with a corrected p-value of 0.015 which is over the authors’ own stated required threshold of 0.0125 (see caption). They use the language “The association with drug use, after correcting for 4 comparisons, was determined to be a trend toward significance” to describe this non-effect. It is worth noting that the removal of the outlier at a volume of over $800 mm^3$ would almost certainly flatten the line altogether and remove even the slight effect. It would have been nice to test this hypothesis but the authors did not release any of their data.
Figure 1c.
In the Fox News article about the paper, Breiter is quoted saying ““For the NAC [nucleus accumbens], all three measures were abnormal, and they were abnormal in a dose-dependent way, meaning the changes were greater with the amount of marijuana used,” Breiter said. “The amygdala had abnormalities for shape and density, and only volume correlated with use. But if you looked at all three types of measures, it showed the relationships between them were quite abnormal in the marijuana users, compared to the normal controls.” The result above shows this to be a lie. Volume did not significantly correlate with use.
This is all very bad, but things get uglier the more one looks at the paper. In the tables reporting the p-values, the authors do something I have never seen before in a published paper. They report the uncorrected p-values, indicating those that are significant (prior to correction) in boldface, and then put an asterisk next to those that are significant after their (incomplete) correction. I realize my own use of boldface is controversial… but what they are doing is truly insane. The fact that they put an asterisk next to the values significant after correction indicates they are aware that multiple testing is required. So why bother boldfacing p-values that they know are not significant? The overall effect is an impression that more tests are significant than is actually the case. See for yourself in their Table 4:
Table 4.
The fact that there are multiple columns is also problematic. Separate tests were performed for smoking occasions per day, joints per occasion, joints per week and smoking days per week. These measures are highly correlated, but even so multiply testing them requires multiple test correction. The authors simply didn’t perform it. They say “We did not correct for the number of drug use measures because these measures tend not be independent of each other”. In other words, they multiplied the number of tests by four, and chose to not worry about that. Unbelievable.
Then there is Table 5, where the authors did not report the p-values at all, only whether they were significant or not… without correction:
Table 5.
### 3. Correlation vs. causation
This issue is one of the oldest in the book. There is even a wikipedia entry about itCorrelation does not imply causation. Yet despite the fact the every result in the paper is directed at testing for association, in the last sentence of the abstract they say “These data suggest that marijuana exposure, even in young recreational users, is associated with exposure-dependent alterations of the neural matrix of core reward structures and is consistent with animal studies of changes in dendritic arborization.” At a minimum, such a result would require doing a longitudinal study. Breiter takes this language to an extreme in the press release accompanying the article. I repeat the statement he made that I quoted above where I boldface the causal claim: “”Some of these people only used marijuana to get high once or twice a week. People think a little recreational use shouldn’t cause a problem, if someone is doing OK with work or school. Our data directly says this is not the case.” I believe that scientists should be sanctioned for making public statements that directly contradict the content of their papers, as appears to be the case here. There is precedent for this.
A few years ago after the birth of our second daughter and in anticipation of our third, I started designing a one-room addition for our house. One of the problems I faced was figuring out the shape of the roof. I learned of the concept of the straight skeleton of a polygon, first defined by Oswin Aichholzer and Franz Aurenhammer in a book chapter “Straight Skeletons for General Polygonal Figures in the Plane” in 1996. Wikipedia provides an intuitive definition:
“The straight skeleton of a polygon is defined by a continuous shrinking process in which the edges of the polygon are moved inwards parallel to themselves at a constant speed. As the edges move in this way, the vertices where pairs of edges meet also move, at speeds that depend on the angle of the vertex. If one of these moving vertices collides with a nonadjacent edge, the polygon is split in two by the collision, and the process continues in each part. The straight skeleton is the set of curves traced out by the moving vertices in this process. In the illustration the top figure shows the shrinking process and the middle figure depicts the straight skeleton in blue.”
but the concept is best understood by picture:
The fact that straight skeletons fit “symmetrically” into the polygons that generated them, made me think about whether they could constitute aesthetic representations of phylogenetic trees. So I asked the inverse question: given a phylogenetic tree, i.e. a graph that is a tree with weighted edges, together with a cyclic orientation on its vertices, is there a convex polygon such that the tree is the straight skeleton of that polygon? A few google searches didn’t reveal anything, but fortunately and coincidentally, Satyan Devadoss, who is a topologist and computational geometer was visiting my group on his sabbatical (2009–2010).
Now, a few years later, Satyan and coauthors have written a paper providing a partial answer to my question. Their paper is about to appear in the next issue of Discrete Applied Mathematics:
The main theorem is formally about ribbon trees:
Definition. A ribbon tree is a tree (a connected graph with no cycles) for which each edge is assigned a nonnegative length, each internal vertex has degree at least three, and the edges incident to each vertex are cyclically ordered.
The authors prove the interesting result that there exists only a finite set of planar embeddings of a tree appearing as straight skeletons of convex polygons. Specifically, they show that:
Theorem. A ribbon tree with n leaves has at most 2n−5 suitable convex polygons.
Its fun to work out by hand the case of a star tree with three leaves:
The algebra works out to solving a cubic system of three equations that can be seen to have one unique positive solution (Lemma 5 in the paper).
The proof of the main theorem relies on some elementary trigonometry and algebra, as well as an interesting analogy of Cauchy’s “arm lemma” (if one increases one of the angles of a convex polygonal chain, then the distance between the endpoints will only increase). Furthermore, a few interesting cases and connections are pointed out along the way. For example, some ribbons, even caterpillar ribbons, are not realized by any polygon. There is also a related conference publication by many of the same authors, and in addition including Aichholzer himself, that provides some interesting constructions in special cases:
• Oswin Aichholzer, Howard Cheng, Satyan L. Devadoss, Thomas Hackl, Stefan Huber, Brian Li, Andrej Risteski, What makes a tree a straight skeleton?, Canadian Conference on Computational Geometry, 2012.
Straight skeletons may yet find application in drawing phylogenetic trees, but for now the best out there are radial or circular representations optimizing various layout considerations.
My addition is now complete and the roof is absolutely beautiful.
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2014-09-19 13:49:04
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http://dekalogblog.blogspot.com/
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Sunday, 1 March 2015
Particle Swarm Optimisation, Part 2.
Following on from my last post, here is an Octave .oct file implementation of the one dimensional Particle swarm optimisation routine, with one slight twist: instead of using a for loop I've implemented it within a while loop with a stopping condition that the algorithm should cease once there has been no improvement in the global_best value for 25 iterations.
DEFUN_DLD ( pso_conversion_code, args, nargout,
"-*- texinfo -*-\n\
@deftypefn {Function File} {} pso_conversion_code (@var{target_val,max_lambda})\n\
The output of this test function should be half the input target_val.\n\
@end deftypefn" )
{
octave_value_list retval_list ;
int nargin = args.length () ;
// check the number of input arguments
if ( nargin != 2 )
{
error ( "Invalid number of arguments." ) ;
return retval_list ;
}
if ( args(0).length () != 1 )
{
error ( "Invalid target_val. Should be a single value." ) ;
return retval_list ;
}
if ( args(1).length () != 1 )
{
error ( "Invalid max_lambda value. Should be a single value for the maximum 'guess'." ) ;
return retval_list ;
}
if (error_state)
{
error ( "Invalid arguments." ) ;
return retval_list ;
}
// end of input checking
double target_val = args(0).double_value() ;
double max_lambda = args(1).double_value() ;
double loocv_value ;
// the the pso algorithm
int no_iterations_until_cease = 25 ;
int no_of_particles = 100 ;
double global_best = std::numeric_limits::infinity() ;
double global_best_lambda = 0.0 ; // initially set to unregularised
ColumnVector local_best( no_of_particles , 1 ) ; local_best.fill( global_best ) ;
ColumnVector local( no_of_particles , 1 ) ;
ColumnVector local_best_so_far( no_of_particles , 1 ) ;
ColumnVector velocity( no_of_particles , 1 ) ; velocity.fill( 0.0 ) ; // particle velocity vector
// A Mersenne Twister random number generator can be declared with a simple from the #include "MersenneTwister.h"
MTRand mtrand1 ;
// values for the random updating process
double r1 ;
double r2 ;
// an inertial constant. Good values are usually slightly less than 1. Or it could be randomly initialized for each particle.
double w_ic = 0.9 ;
// c1 and c2 are constants that say how much the particle is directed towards good positions. They represent a "cognitive" and a "social" component, respectively,
// in that they affect how much the particle's personal best and the global best (respectively) influence its movement. Usually we take c_1, c_2 approx = 2.
// Or they could be randomly initialized for each particle.
double c1 = 2.0 ;
double c2 = 2.0 ;
// fill the local vector with initial random values < max_lambda, temporarily using r1
for ( octave_idx_type ii (0) ; ii < no_of_particles ; ii++ )
{
r1 = mtrand1.randDblExc () ;
local(ii) = r1 * max_lambda ;
}
int while_counter = 0 ;
while ( while_counter < no_iterations_until_cease )
{
// loop once over local_best and local vectors
for ( octave_idx_type jj (0) ; jj < no_of_particles ; jj++ )
{
// Replace this code box as necessary
//*************************************************************//
// //
// fitness function evaluation //
loocv_value = local(jj) * local(jj) - target_val * local(jj) ; //
// //
//*************************************************************//
// check if the local_best has improved
if ( loocv_value < local_best(jj) )
{
// update local_best and local_best_so_far vector if it has
local_best(jj) = loocv_value ;
local_best_so_far(jj) = local(jj) ;
}
// check if the above local_best has also improved the global_best
if ( local_best(jj) < global_best )
{
// update global_best and global_best_lambda if it has
global_best = local_best(jj) ;
global_best_lambda = local_best_so_far(jj) ;
while_counter = 0 ;
}
} // end of loop once over local_best and local vectors
// now update the particle velocity and position vectors
for ( octave_idx_type jj (0) ; jj < no_of_particles ; jj++ )
{
r1 = mtrand1.randDblExc () ;
r2 = mtrand1.randDblExc () ;
velocity(jj) = w_ic * velocity(jj) + c1 * r1 * ( local_best_so_far(jj) - local(jj) ) + c2 * r2 * ( global_best - local(jj) ) ;
local(jj) = local(jj) + velocity(jj) ;
} // end of particle velocity and position vectors updates loop
while_counter += 1 ;
} // end of main while loop
retval_list(1) = global_best ;
retval_list(0) = global_best_lambda ;
return retval_list ;
} // end of function
In the "commented" function section there is the same test function as in the vectorised code in my previous post. In real life application, of course, this code would be replaced - the above is just a test of my conversion of the pso algorithm from Octave to C++ code.
I've been looking at pso because I think I can easily use it as a simple Hyperparameter optimisation tool to tune the regularisation of the weights in my proposed neural net trading system. The reason I chose the fitness function I did in the above code is simply that it has a global minimum, which is what neural net training is all about - minimisation of an error function.
Friday, 20 February 2015
Particle Swarm Optimisation
Having decided that I'm going to use my mfe_mae indicator as a target for my neural net, over the last couple of months I've been doing some research on what would make good features for this target. In the course of this I've also decided that Particle swarm optimization would be a useful tool to use.
Thanks to the pseudo-code on this wiki and the code on in this stackoverflow thread I've been able to write some Octave code to perform pso over one dimension, which is shown in the code box below:
clear all
function [out] = evaluate_fitness( R , val )
out = R .* R .- val .* R ;
end
val = input( 'Enter test val: ' ) ;
% Initialize the particle positions and their velocities
n_particles = 100 ;
upper_limit = val ;
n_iterations = 50 ;
w = 0.9 .* ones( n_particles , 1 ) ; % is an inertial constant. Good values are usually slightly less than 1. Or it could be randomly initialized for each particle.
% c1 and c2 are constants that say how much the particle is directed towards good positions. They represent a "cognitive" and a "social" component, respectively,
% in that they affect how much the particle's personal best and the global best (respectively) influence its movement. Usually we take c_1, c_2 approx = 2.
% Or they could be randomly initialized for each particle.
C1 = 2.0 ;
C2 = 2.0 ;
X = upper_limit * rand( n_particles , 1 ) ; % == particle position vector containing lambda values
V = zeros( size(X) ) ; % particle velocity vector
X_lbest = X ; % == particle position vector containing lambda values for local best
% Initialize the global and local fitness to the worst possible. Fitness == LOOCV "press" statistic
fitness_gbest = Inf ; % _gbest == global best
fitness_lbest = fitness_gbest * ones( n_particles , 1 ) ; % _lbest == local best
% Loop until convergence, in this example a finite number of iterations chosen
for ii = 1 : n_iterations
% evaluate the fitness of each particle, i.e. do the linear regression and get
% the LOOCV statistic
fitness_X = evaluate_fitness( X , val ) ;
% Update the local bests and their fitness
ix = find( fitness_X < fitness_lbest ) ; % if local LOOCV "press" statistic improves
fitness_lbest( ix ) = fitness_X( ix ) ; % record this better LOOCV statistic value
X_lbest( ix ) = X( ix ) ; % and the lambda value that gave rise to it
% Update the global best and its fitness
[ min_fitness min_fitness_index ] = min( fitness_X ) ;
if ( min_fitness < fitness_gbest ) % if global LOOCV "press" statistic improves
fitness_gbest = min_fitness ; % record this better LOOCV statistic value
X_gbest = X( min_fitness_index ) ; % and the lambda value that gave rise to it
end % end if
% Update the particle velocity and position vectors
R1 = rand( n_particles , 1 ) ;
R2 = rand( n_particles , 1 ) ;
V = w .* V + C1 .* R1 .* ( X_lbest .- X ) .+ C2 .* R2 .* ( X_gbest .* ones( n_particles , 1 ) .- X ) ;
X = X .+ V ;
end % end main ii loop
and which is vectorised as much as I can make it at the moment. The evaluate_fitness function I've chosen to use is a Quadratic function of the form
f(x) = x^2 - bx
which, when a positive value for "val" is input as the test value, ensures the function curve goes through the origin and has a minimum y-axis value at a point on the x-axis that is half the input test value. This make it easy to quickly verify that the pso code is performing as expected, with the global minimum x_axis value found by the algorithm being given by the variable X_gbest. My reasons for choosing a test function of this form, and for looking at pso in general, will be given in my next post.
Thursday, 11 December 2014
MFE/MAE Indicator Test Results
Following on from the previous post, the test I outlined in that post wasn't very satisfactory, which I put down to the fact that the Sigmoid transformation of the raw MFE/MAE indicator values is not amenable to the application of standard deviation as a meaningful measure. Instead, I changed the test to one based on the standard error of the mean, an example screen shot of which is shown below:-
The top pane shows the the long version of the indicator and the bottom pane the short version. In each there are upper and lower limits of the sample standard error of the mean above and below the population mean (mean of all values of the indicator) along with the cumulative mean value of the top N matches as shown on the x-axis. In this particular example it can be seen that around the 170-180 samples mark the cumulative mean moves inside the standard error limits, never to leave them again. The meaning I ascribe to this is that there is no value to be gained from using more than approximately 180 samples for machine learning purposes, for this example, as to use more samples would be akin to training on all available data, which makes the use of my Cauchy Schwarz matching algo superfluous. I repeated the above on all instances of sigmoid transformed and untransformed MFE/MAE indicator values to get an average of 325 samples for transformed, and an average of 446 samples for the untransformed indicator values across the 4 major forex pairs. Based on this, I have decided to use the top 450 Cauchy Schwarz matches for training purposes, which has ramifications for model complexity will be discussed shortly.
Returning to the above screen shot, the figure 2 inset shows the price bars that immediately follow the price bar for which the main screen shows the top N matches. Looking at the extreme left of the main screen it can be seen that the lower pane, short indicator has an almost maximum reading of 1 whilst the upper pane, long indicator shows a value of approx. 2.7, which is not much above the global minimum for this indicator and well below the 0.5 neutral level. This strongly suggests a short position, and looking at the inset figure it can be seen that over the 3 days following the extreme left matched bar a short position was indeed the best position to hold. This is a pattern that seems to frequently present itself during visual inspection of charts, although I am unable to quantify this in any way.
On the matter of model complexity alluded to above, I found the Learning From Data course I have recently completed on the edX platform to be very enlightening, particularly the concept of the VC dimension, which is nicely explained in the Learning From Data Video library. I'll leave it to interested readers to follow the links, but the big take away for me is that using 450 samples as described above implies that my final machine learning model must have an upper bound of approximately 45 on the VC dimension, which in turn implies a maximum of 45 weights in the neural net. This is a design constraint that I will discuss in a future post.
Wednesday, 26 November 2014
Test of the MFE/MAE Indicator
Continuing from my last post, wherein I stated I was going to conduct a more pertinent statistical test of the returns of the bars(s) immediately following the N best, Cauchy Schwarz matching algorithm matched bars in the price history, readers may recall that the basic premise behind this algorithm is that by matching current price action to the N best matches, the price action after these matches can be used to infer what will occur after the current price action. However, rather than test the price action directly I have decided to apply the test to the MFE/MAE indicator. There are several reasons for this, which are enumerated below.
1. I intend to use the indicator as the target function for future Neural net training
2. the indicator represents a reward to risk ratio, which indirectly reflects price action itself, but without the noise of said action
3. this reward to risk ratio is of much more direct concern, from a trading perspective, than accurately predicting price
4. since the indicator is now included as a feature in the matching algorithm, testing the indicator is, very indirectly, a test of the matching algorithm too
The test I have in mind is a hybrid of a hypothesis test, Cross validation and the application of Statistical process control. Consider the chart below:
This shows two sampling distributions of the mean for Long MFE/MAE indicator values > 0.5, the upper pane for sample sizes of 20 and the lower pane for 75. For simplicity I shall only discuss the Long > 0.5 version of the indicator, but everything that follows applies equally to the Short version. As expected the upper pane shows greater variance, and for the envisioned test a whole series of these sampling distributions will be produced for different sampling rates. The way I intend it to work is as follows:
• take a single bar in the history and see what the value of the MFE/MAE indicator value is 3 bars later (assume > 0.5 for this exposition, so we compare to long sampling distributions only)
• get the top 20 matched bars for the above selected bar and the corresponding 20 indicator values for 3 bars later and take the mean of these 20 indicator values
• check if this mean falls within the sampling distribution of the mean of 20, as shown in the upper pane above by the vertical black line at 0.8 on the x axis. If it does fall with the sampling distribution, we accept the null hypothesis that the 20 best matches in history future indicator values and the value of the indicator after the bar to be matched come from the same distribution
• repeat the immediately preceding step for means of 21, 22, ... etc until such time as the null hypothesis can be rejected, shown in the lower pane above. At this point, we then then declare an upper bound on the historical number of matches for the bar to be predicted
For any single bar to be predicted we can then produce the following chart, which is completely artificial and just for illustrative purposes:
where the cyan and red lines are the +/- 2 standard deviations above/below a notional mean value for the whole distribution of approximately 0.85, and the chart can be considered to be a type of control chart. The upper and lower control lines converge towards the right, reflecting the decreasing variance of increasingly large N sample means, as shown in the first chart above. The green line represents the cumulative N sample mean of the best N historical matches' future values. I have shown it as decreasing as it is to be expected that as more N matches are included, the greater the chance that incorrect matches, unexpected price reversals etc. will be caught up in this mean calculation, resulting in the mean value moving into the left tail of the sampling distribution. This effect combines with the shrinking variance to reach a critical point (rejection of the null hypothesis) at which the green line exits below the lower control line.
The purpose of all the above is provide a principled manner to choose the number N matches from the Cauchy-Schwarz matching algorithm to supply instances of training data to the envisioned neural net training. An incidental benefit of this approach is that it is indirectly a hypothesis test of the fundamental assumption underlying the matching algorithm; namely that past price action has predictive ability for future price action, and furthermore, it is a test of the MFE/MAE indicator. Discussion of the results of these tests in a future post.
Wednesday, 12 November 2014
First Use for the MFE/MAE Indicator
This first use is as an input to my Cauchy-Schwarz matching algorithm, previous posts about which can be read herehere and here. The screen shot below shows what I would characterise as a "good" set of matches:
The top left pane shows the original section of the price series to be matched, and the panes labelled #1, #5, etc. are the best match, 5th best match and so on respectively. The last 3 rightmost bars in each pane are "future" price bars, i.e. the 4th bar in from the right is the target bar that is being matched, matched over all the bars to the left or in the past of this target bar.
I consider the above to be a set of "good" matches because, for the #1 through #25 matches for "future" bars:
• if one considers the logic of the mfe/mae indicator each pane gives indicator readings of "long," which all agree with the original "future" bars
• similarly the mae (maximum adverse excursion) occurs on the day immediately following the matched day
• the mfe (maximum favourable excursion) occurs on the 3rd "future" bar, with the slight exception of pane #10
• the marked to market returns of an entry at the open of the 1st "future" bar to the close of the 3rd "future" bar all show a profit, as does the original pane
However, it can be seen that the above noted "goodness" breaks down for panes #25 and #30, which leads me to postulate that there is an upper bound on the number of matches for which there is predictive ability for "future" returns.
In the above linked posts the test statistic used to judge the predictive efficacy of the matching algorithm was effect size. However, I think a more pertinent test statistic to use would be the average bar return over the bars immediately following a matched bar, and a discussion of this will be the subject of my next post.
Wednesday, 5 November 2014
A New MFE/MAE Indicator.
After stopping my investigation of tools for spectral analysis over the last few weeks I have been doing another mooc, this time Learning from Data, and also working on the idea of one of my earlier posts.
In the above linked post there is a video showing the idea as a "paint bar" study. However, I thought it would be a good idea to render it as an indicator, the C++ Octave .oct code for which is shown in the code box below.
DEFUN_DLD ( adjustable_mfe_mae_from_open_indicator, args, nargout,
"-*- texinfo -*-\n\
@deftypefn {Function File} {} adjustable_mfe_mae_from_open_indicator (@var{open,high,low,close,lookback_length})\n\
This function takes four input series for the OHLC and a value for lookback length. The main outputs are\n\
two indicators, long and short, that show the ratio of the MFE over the MAE from the open of the specified\n\
lookback in the past. The indicators are normalised to the range 0 to 1 by a sigmoid function and a MFE/MAE\n\
ratio of 1:1 is shifted in the sigmoid function to give a 'neutral' indicator reading of 0.5. A third output\n\
is the max high - min low range over the lookback_length normalised by the range of the daily support and\n\
resistance levels S1 and R1 calculated for the first bar of the lookback period. This is also normalised to\n\
give a reading of 0.5 in the sigmoid function if the ratio is 1:1. The point of this third output is to give\n\
some relative scale to the unitless MFE/MAE ratio and to act as a measure of strength or importance of the\n\
MFE/MAE ratio.\n\
@end deftypefn" )
{
octave_value_list retval_list ;
int nargin = args.length () ;
// check the input arguments
if ( nargin != 5 )
{
error ( "Invalid arguments. Arguments are price series for open, high, low and close and value for lookback length." ) ;
return retval_list ;
}
if ( args(4).length () != 1 )
{
error ( "Invalid argument. Argument 5 is a scalar value for the lookback length." ) ;
return retval_list ;
}
int lookback_length = args(4).int_value() ;
if ( args(0).length () < lookback_length )
{
error ( "Invalid argument lengths. Argument lengths for open, high, low and close vectors should be >= lookback length." ) ;
return retval_list ;
}
if ( args(1).length () != args(0).length () )
{
error ( "Invalid argument lengths. Argument lengths for open, high, low and close vectors should be equal." ) ;
return retval_list ;
}
if ( args(2).length () != args(0).length () )
{
error ( "Invalid argument lengths. Argument lengths for open, high, low and close vectors should be equal." ) ;
return retval_list ;
}
if ( args(3).length () != args(0).length () )
{
error ( "Invalid argument lengths. Argument lengths for open, high, low and close vectors should be equal." ) ;
return retval_list ;
}
if (error_state)
{
error ( "Invalid arguments. Arguments are price series for open, high, low and close and value for lookback length." ) ;
return retval_list ;
}
// end of input checking
// inputs
ColumnVector open = args(0).column_vector_value () ;
ColumnVector high = args(1).column_vector_value () ;
ColumnVector low = args(2).column_vector_value () ;
ColumnVector close = args(3).column_vector_value () ;
// outputs
ColumnVector long_mfe_mae = args(0).column_vector_value () ;
ColumnVector short_mfe_mae = args(0).column_vector_value () ;
ColumnVector range = args(0).column_vector_value () ;
// variables
double max_high = *std::max_element( &high(0), &high( lookback_length ) ) ;
double min_low = *std::min_element( &low(0), &low( lookback_length ) ) ;
double pivot_point = ( high(0) + low(0) + close(0) ) / 3.0 ;
double s1 = 2.0 * pivot_point - high(0) ;
double r1 = 2.0 * pivot_point - low(0) ;
for ( octave_idx_type ii (0) ; ii < lookback_length ; ii++ ) // initial ii loop
{
// long_mfe_mae
if ( open(0) > min_low ) // the "normal" situation
{
long_mfe_mae(ii) = 1.0 / ( 1.0 + exp( -( ( max_high - open(0) ) / ( open(0) - min_low ) - 1.0 ) ) ) ;
}
else if ( open(0) == min_low )
{
long_mfe_mae(ii) = 1.0 ;
}
else
{
long_mfe_mae(ii) = 0.5 ;
}
// short_mfe_mae
if ( open(0) < max_high ) // the "normal" situation
{
short_mfe_mae(ii) = 1.0 / ( 1.0 + exp( -( ( open(0) - min_low ) / ( max_high - open(0) ) - 1.0 ) ) ) ;
}
else if ( open(0) == max_high )
{
short_mfe_mae(ii) = 1.0 ;
}
else
{
short_mfe_mae(ii) = 0.5 ;
}
range(ii) = 1.0 / ( 1.0 + exp( -( ( max_high - min_low ) / ( r1 - s1 ) - 1.0 ) ) ) ;
} // end of initial ii loop
for ( octave_idx_type ii ( lookback_length ) ; ii < args(0).length() ; ii++ ) // main ii loop
{
// assign variable values
max_high = *std::max_element( &high( ii - lookback_length + 1 ), &high( ii + 1 ) ) ;
min_low = *std::min_element( &low( ii - lookback_length + 1 ), &low( ii + 1 ) ) ;
pivot_point = ( high(ii-lookback_length) + low(ii-lookback_length) + close(ii-lookback_length) ) / 3.0 ;
s1 = 2.0 * pivot_point - high(ii-lookback_length) ;
r1 = 2.0 * pivot_point - low(ii-lookback_length) ;
// long_mfe_mae
if ( open( ii - lookback_length + 1 ) > min_low && open( ii - lookback_length + 1 ) < max_high ) // the "normal" situation
{
long_mfe_mae(ii) = 1.0 / ( 1.0 + exp( -( ( max_high - open( ii - lookback_length + 1 ) ) / ( open( ii - lookback_length + 1 ) - min_low ) - 1.0 ) ) ) ;
}
else if ( open( ii - lookback_length + 1 ) == min_low )
{
long_mfe_mae(ii) = 1.0 ;
}
else
{
long_mfe_mae(ii) = 0.0 ;
}
// short_mfe_mae
if ( open( ii - lookback_length + 1 ) > min_low && open( ii - lookback_length + 1 ) < max_high ) // the "normal" situation
{
short_mfe_mae(ii) = 1.0 / ( 1.0 + exp( -( ( open( ii - lookback_length + 1 ) - min_low ) / ( max_high - open( ii - lookback_length + 1 ) ) - 1.0 ) ) ) ;
}
else if ( open( ii - lookback_length + 1 ) == max_high )
{
short_mfe_mae(ii) = 1.0 ;
}
else
{
short_mfe_mae(ii) = 0.0 ;
}
range(ii) = 1.0 / ( 1.0 + exp( -( ( max_high - min_low ) / ( r1 - s1 ) - 1.0 ) ) ) ;
} // end of main ii loop
retval_list(2) = range ;
retval_list(1) = short_mfe_mae ;
retval_list(0) = long_mfe_mae ;
return retval_list ;
} // end of function
The way to interpret this is as follows:
• if the "long" indicator reading is above 0.5, go long
• if the "short" is above 0.5, go short
• if both are below 0.5, go flat
An alternative, if the indicator reading is flat, is to maintain any previous non flat position. I won't show a chart of the indicator itself as it just looks like a very noisy oscillator, but the equity curve(s) of it, without the benefit of foresight, on the EURUSD forex pair are shown below.
The yellow equity curve is the cumulative, close to close, tick returns of a buy and hold strategy, the blue is the return going flat when indicated, and the red maintaining the previous position when flat is indicated. Not much to write home about. However, this second chart shows the return when one has the benefit of the "peek into the future" as discussed in my earlier post.
The colour of the curves are as before except for the addition of the green equity curve, which is the cumulative, vwap value to vwap value tick returns, a simple representation of what an equity curve with realistic slippage might look like. This second set of equity curves shows the promise of what could be achievable if a neural net to accurately predict future values of the above indicator can be trained. More in an upcoming post.
Tuesday, 23 September 2014
High Resolution Tools for Spectral Analysis - Update
Following on from my initial enthusiasm for the code on the High Resolution Tools for Spectral Analysis page, I have say that I have been unable to get the code performing as I would like it for my intended application to price time series.
My original intent was to use the zero crossing period estimation function, the subject of my last few posts, to get a rough idea of the dominant cycle period and then use the most recent data in a rolling window of this length as input to the high resolution code. This approach, however, ran into problems.
Firstly, windows of just the dominant cycle length (approximately 10 to 30 data points only) would lead to all sorts of errors being thrown from the toolkit functions as well as core Octave functions, such as divide by zero warnings and cryptic error messages that even now I don't understand. My best guess here is that the amount of data available in such short windows is simply insufficient for the algorithm to work, in much the same way as the Fast Fourier transform may fail to work if given too little data that is not a power of 2 in length. It might be possible to substantially rewrite the relevant functions, but my understanding of the algorithm and the inner workings of Octave means this is well beyond my pay grade.
My second approach was to simply extend the amount of data available by using the Octave repmat function on the windowed data so that all the above errors disappeared. This had very hit and miss results - sometimes they were very accurate, other times just fair to middling, and on occasion just way off target. I suspect here that the problem is the introduction of signal artifacts via the use of the repmat function, which results in Aliasing of the underlying signal.
As a result I shall not continue with investigating this toolbox for now. Although I only investigated the use of the me.m function (there are other functions available) I feel that my time at the moment can be used more productively.
|
2015-03-29 06:04:19
|
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|
https://robotics.stackexchange.com/questions/19713/number-of-pixel-comparison-needed-to-establish-correspondence/19715
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# Number of pixel comparison needed to establish correspondence
I am reading the paper titled Variable Baseline/Resolution Stereo David Gallup, Jan-Michael Frahm, Philippos Mordohai, Marc Pollefeys
Section 3 of the paper talks about the time complexity of the standard stereo, within it, subsection for fixed baseline case says the following
In stereo, each pixel must be tested against the pixels along the corresponding epipolar line within the disparity range of the scene. Because the depthrange is defined by the scene, the disparity range is some fraction of the image width, and thus increases with resolution. Letting $$D$$ be the ratio of the disparity range to the image width ($$w$$), the number of pixel comparisons needed is
$$T_{fixed} = D~w^2~h = \frac{D~w^3}{a}$$
Here, symbol $$a$$ is the aspect ratio and $$h$$ is the height of the image ($$h = {w}/{a}$$).
We know that the search for correspondence is restricted to epipolar line and thus actual number of comparisons should be way lower than the number given above.
OR
I am missing something?
## 1 Answer
I am missing something?
Indeed, the context is standard stereo and (all) pixel comparisons ($$w\times h$$).
In a rectified image, the epipolar line can be typically be about the width of the image. Thus the search range would be $$D\times w$$.
Thus total number of comparisons = $$D\times w \times wh$$.
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2020-07-08 11:29:14
|
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https://mathspace.co/textbooks/syllabuses/Syllabus-914/topics/Topic-19403/subtopics/Subtopic-259833/
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# 4.01 Review: Laws of exponents
Lesson
Let's review the laws of exponents. It's important to remember the order of operations when we're simplifying these expressions.
Laws of exponents
• The product of powers property: $a^m\times a^n=a^{m+n}$am×an=am+n
• The quotient of powers property: $a^m\div a^n=a^{m-n}$am÷an=amn
• The zero exponent property: $a^0=1$a0=1
• The power of a power property: $\left(a^m\right)^n=a^{mn}$(am)n=amn
• The negative exponent definition: $a^{-m}=\frac{1}{a^m}$am=1am
A question may have any combination of laws of exponents. We just need to simplify it step by step, making sure we follow the order of operations.
#### Worked examples
##### Question 1
Simplify: $p^7\div p^3\times p^5$p7÷p3×p5
Think: We need to apply the exponent division and exponent multiplication laws.
Do:
$p^7\div p^3\times p^5$p7÷p3×p5 $=$= $p^{7-3+5}$p7−3+5 $=$= $p^9$p9
Reflect: We can choose to do this in more steps by first doing $p^{7-3}\times p^5=p^4\times p^5$p73×p5=p4×p5 and then getting our final answer of $p^9$p9
##### Question 2
Simplify: $\frac{\left(u^{x+3}\right)^3}{u^{x+1}}$(ux+3)3ux+1
Think: We need to simplify the numerator using the power of a power property, then apply the quotient property.
Do:
$\frac{\left(u^{x+3}\right)^3}{u^{x+1}}$(ux+3)3ux+1 $=$= $\frac{u^{3\left(x+3\right)}}{u^{x+1}}$u3(x+3)ux+1 Simplify the numerator using the power of a power property $=$= $\frac{u^{3x+9}}{u^{x+1}}$u3x+9ux+1 Apply the distributive property $=$= $u^{3x+9-\left(x+1\right)}$u3x+9−(x+1) Use the quotient property and subtract the powers $=$= $u^{3x+9-x-1}$u3x+9−x−1 Simplify by collecting the like terms $=$= $u^{2x+8}$u2x+8
##### Question 3
Express $\left(4^p\right)^4$(4p)4 with a prime number base in exponential form.
Think: We could express $4$4 as $2^2$22 which has a prime number base.
Do:
$\left(4^p\right)^4$(4p)4 $=$= $4^{4p}$44p Use the power of a power property $=$= $\left(2^2\right)^{4p}$(22)4p Use the fact that $4=2^2$4=22 $=$= $2^{8p}$28p Use the power of a power property
Reflect: This skill will become increasingly important as we look at simplifying expressions with related bases such as $2^{3p}\times\left(4^p\right)^4$23p×(4p)4.
##### Question 4
Simplify $20m^6\div5m^{13}\times9m^2$20m6÷5m13×9m2, expressing your answer in positive exponential form.
Think: Let's express this as a fraction so the powers are on the numerator and the denominator for easy comparison.
Do:
$\frac{20m^6}{5m^{13}}\times9m^2$20m65m13×9m2 $=$= $\frac{4}{m^7}\times9m^2$4m7×9m2 Simplify the fraction using the quotient property $=$= $\frac{36m^2}{m^7}$36m2m7 Simplify the multiplication $=$= $36m^{-5}$36m−5 Use the quotient property - this step is sometimes omitted $=$= $\frac{36}{m^5}$36m5 Write as a positive exponent
#### Practice questions
##### Question 5
Simplify $\frac{\left(x^2\right)^6}{\left(x^2\right)^2}$(x2)6(x2)2
##### Question 6
Simplify $\left(u^9\times u^5\div u^{19}\right)^2$(u9×u5÷u19)2, expressing your answer in positive exponential form.
##### Question 7
Express $\left(5y^3\right)^{-3}$(5y3)3 with a positive exponent.
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2022-01-26 07:34:01
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https://socratic.org/questions/how-do-you-evaluate-27-8-4-3
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# How do you evaluate (-27/8)^(4/3)?
${\left(- \frac{27}{8}\right)}^{\frac{4}{3}}$
$= {\left({\left(- \frac{27}{8}\right)}^{\frac{1}{3}}\right)}^{4}$
$= {\left(\frac{\sqrt[3]{- 27}}{\sqrt[3]{8}}\right)}^{4}$
$= {\left(- \frac{3}{2}\right)}^{4}$
$= \frac{81}{16}$
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2022-05-25 14:21:04
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https://proxies-free.com/numerical-integration-nintegrate-and-maxrecursion/
|
# numerical integration – NIntegrate and MaxRecursion
I’m using `NIntegrate` to get the value of a very complicated function `f[k,T,M]`, where `T` and `M` are some parameters:
``````S[T_, M_, MaxRec_]:= NIntegrate[f[k, T, M], {k,0,Infinity}, PrecisionGoal->10, MaxRecursion->MaxRec]
``````
What puzzles me is the fact that, for the same parameters, I get different values depending on `MaxRec`:
``````S[10^12, 10^12, 10]
``````
``````1.30494*10^30
``````
versus
``````S[10^12, 10^12, 50]
``````
``````-1.161*10^45
``````
How should I interpret these results? I could provide the explicit form of `f` if needed. Thanks a lot!
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2021-09-28 00:54:47
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https://www.electro-tech-online.com/threads/hacking-steering-wheel-controls-to-radio.140501/page-2
|
# Hacking steering wheel controls to radio
#### alec_t
##### Well-Known Member
In the LM3914's 'dot' mode there is a slight overlap of the respective voltage levels for the dots. Is this likely to be a problem?
#### JimB
##### Super Moderator
For information, please see the attachment which is a scan of the schematic sketch which I used to build my converter.
Some aspects of it may be a bit cryptic.
Overlap on the dots was not a problem for me.
I did have to use two of the 3914 outputs for one switch position to ensure correct operation under all conditions.
Despite my best efforts at juggling the 390 ohm potential divider resistance, there was always on switch case which fell very near one of the 3914 switching thresholds, so I used both outputs above and below the threshold.
JimB
JimB:
Clever.
#### therossinator
##### New Member
The chips arrived today and I have everything I think I will need. I calculated my resistor network based on the second page of the data sheet which uses a 6.8-18v input and a 0-5v signal. The couple things I noticed were that the 3.83k and 1.21k ohm resistors called for in the diagram don't really exist in the 100 count multipack I picked up from Radio Shack, secondly the formula provided results in 5.206v for the ref out. These seem like very specific resistances to end up with a formula that doesn't equal 5.000. That being said, on to the steering wheel switches.
I chose these values, voltages are based off 12-14v
R5 (to ground)- 150
R1 (switch)- 15k, 1.01-1.27v
Phantom R (pressing the following 2 simultaneously)- 800, 1.89-2.21v
R2 (switch)- 470, 2.90-3.39v
R3 (switch)- 330, 3.75-4.38v
R4 (switch)- 248 (390 and 680 in parallel), 4.52-5.28v
I figured I could use 2 outputs per switch to make 5 steps instead of 10 to ensure the correct command is registered. I will use three leds for R1 and one for R3 and R4(they overlap in the 7th band). Or I could use a ref out of 5 exactly and have steps of 0.5v which work nicely if 5.28v would still register and send a signal.
Any thoughts?
#### dougy83
##### Well-Known Member
You could just use four 100 ohm resistors on the switches and a 220 ohm from Vin to Gnd. That gives a voltage of 0.69, 0.52, 0.42 and 0.35 (as a fraction of RHI), which will operate four contiguous outputs.
Alternately, the E12 standard values 120, 56, 91, 150 (and 220 from Vin to gnd) provide fractional voltages of 0.65, 0.55, 0.45, 0.35 (of RHI), which fall right in the middle of each comparator window.
The positive wire going to the switches should come from whatever you're feeding into the LM3914 RHI pin.
Last edited:
#### therossinator
##### New Member
The positive wire going to the switches should come from whatever you're feeding into the LM3914 VHI pin.
Why? I understand that its convenient to do so but had I, I would not have been able to create a different function from pressing 2 buttons at once. If the consensus is that my method will not work then yes, I'm going straight to your solution, but I like my 5th function and I already did the maths.
#### therossinator
##### New Member
I could rehash it using the Vhi as a source but I still want to keep the switches in parallel for the sake of the 5th function.
#### dougy83
##### Well-Known Member
Ok, sorry, I didn't notice the extra function. The reason to use the same reference as the LM3914 is to make it irrelevant what the supply voltage is. If you have values that work, then go for it.
#### dougy83
##### Well-Known Member
These are some values for parallel resistors that give you 7 outputs from 4 switches if you use combinations of two at a time.
Code:
Rgnd: 220
SW R Vo
SW1 120 0.65
SW2 270 0.45
SW3 680 0.24
SW4 1200 0.15
SW1/2 83.07 0.73
SW2/3 193.26 0.53
SW3/4 434.04 0.34
The resistors are 120, 270, 680, 1200 ohms (220 RHI to gnd). This should activate outputs 6, 4, 2 and 1 for single switch presses, and outputs 7, 5 and 3 for combination switch presses.
You may have a problem using combinations of buttons because you won't be able to press or release them at exactly the same time, and therefore you will have unwanted outputs when you try.
#### therossinator
##### New Member
These are some values for parallel resistors that give you 7 outputs from 4 switches if you use combinations of two at a time.
Code:
Rgnd: 220
SW R Vo
SW1 120 0.65
SW2 270 0.45
SW3 680 0.24
SW4 1200 0.15
SW1/2 83.07 0.73
SW2/3 193.26 0.53
SW3/4 434.04 0.34
The resistors are 120, 270, 680, 1200 ohms (220 RHI to gnd). This should activate outputs 6, 4, 2 and 1 for single switch presses, and outputs 7, 5 and 3 for combination switch presses.
You may have a problem using combinations of buttons because you won't be able to press or release them at exactly the same time, and therefore you will have unwanted outputs when you try.
You, sir, are a god among men. Thank you!
Is the Vref a function of the Vin. That is, if Vin goes from 12v to 14 would the Vref adjust proportionally?
#### dougy83
##### Well-Known Member
The Vref provides a constant voltage regardless of the supply voltage (provided V+ is large enough). So if you set Vref to 5V, it will be 5V if V+ is 8V and Vref will still be 5V when V+ is 15V.
I just read in the datasheet that RHI should be < V+ - 1.5V, so you'd want to use Vref or a voltage dropper. I've attached the schem. if using Vref. The transistor is just to allow sufficient current to go to the resistors without affecting the LED current too much (the the LED current is proportional to the current flowing out of Vref). You may wish to put a capacitor between RHI and GND if there are noise issues, however there may not be any...
#### therossinator
##### New Member
why did you use an 8k2 resistor for R2? I like using an Vref of 5V so I'll be able to use 1/4 watt resistors. If I followed the diagram on the data sheet and use an 3k83 resistor would I still need the transistor?
#### dougy83
##### Well-Known Member
R2 can be changed to any value you want, but you should keep the transistor. Because the LED current is 10x the current drawn from the VREF-OUT pin, it is important to buffer the reference with the transistor, or your LED current will skyrocket when you press a switch.
#### therossinator
##### New Member
I put in a 3k9 and hooked everything up on my breadboard, I get 4.9V for my Vhi. Now the problem is that the resistors are not tripping the correct step on the chip. I replaced the resistor for a potentiometer and while it can turn a little over a 180 degrees the lights only start illuminating when on the final 45 degrees of rotation. I double checked my Vlo and it is properly grounded.
#### dougy83
##### Well-Known Member
If you have a linear pot between RHI and RLO, with the wiper to VIN, the LEDs should come on with roughly equal rotation of the pot. If you're using a log pot then the changes of the LEDs lighting will be bunched up toward one end of the pot.
This is assuming that you haven't connected the steering wheel resistors or the 220R resistor between VIN and GND.
#### therossinator
##### New Member
I have everything hooked up and it works great. Thanks for all your help. If anyone wants more details on what I did please just post a reply to the thread. I will be subscribed to this until the end of time.
#### JimB
##### Super Moderator
A success story, sounds good to me.
JimB
#### SyndromeOCZ
##### New Member
Sorry about resurrecting an old post, I came across this thread via a google search and I believe you've done exactly what I am looking to do with steering wheel control buttons.
I'm not the most educated when it comes to knowing what components to use to get my project accomplished, but I know how to run a DMM, soldering, and most of the simple steps. Not so much an engineer.
My car is a 2001 Buick Regal LS, It has 8 buttons on the steering wheel, I am planning on getting a bluetooth remote and having the steering wheel buttons operate it. I wont be running a standard headunit so I can't use a typical steering wheel adapter.
Did your steering wheel buttons already have the resistors attached to them so you are getting different voltage out of each button when pressing it, or did you have to install the resistors onto each button? When I checked my buttons they seemed to have the same voltage no matter which button I pressed. But this is when my battery was weak from 2 days of working on the vehicle, battery is currently charging so I can test at full capacity.
#### KeepItSimpleStupid
##### Well-Known Member
SyndromeOCZ said:
Did your steering wheel buttons already have the resistors attached to them so you are getting different voltage out of each button when pressing it, or did you have to install the resistors onto each button? When I checked my buttons they seemed to have the same voltage no matter which button I pressed. But this is when my battery was weak from 2 days of working on the vehicle, battery is currently charging so I can test at full capacity.
If you look at post #10, you will find 12 V somehow connected to the network. If you measured the "voltage" with this 12V connected, you would get nearly 12 V for all combinations of buttons. If there was a "device" connected, you should read different voltages.
The point is, the resistance should change when the network is potential free.
If your resistances turn out nearly the same, then each output from the bar graph IC could drive a FET optocoupler. This OPTOCOUPLER can be placed across pushbuttons on your remote.
Last edited:
#### ANAS90
##### New Member
hi
i used the same schematic for the Mazda, 1*LM3914 2* 7414 2*4066. my problem is everything its okay when the jack is not plug in the stereo, when i plug it i got just 2 buttons correct the others are completely wrong, any idea please, i used i diode still got the problem
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2018-09-26 06:27:39
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https://cadet.github.io/master/modelling/unit_operations/inlet.html
|
You're reading the documentation for a development version. For the latest released version, please have a look at v4.3.0.
# Inlet¶
A system inlet unit operation is a pseudo unit operation since there is no physical correspondence. The inlet serves as a mass source in the network of unit operations. Consequently, it only possesses an outlet port and no inlet port. Note that an inlet unit operation can provide arbitrary many components and there can be arbitrary many inlet unit operations in a network.
An inlet unit operation provides a feed in which the concentration of each component is given by a profile. The most common profile is a piecewise cubic polynomial, which can both represent discontinuous signals (e.g., pulse or step) and smooth $$C^2$$ signals (cubic spline):
\begin{aligned} c_i(t) = \sum_{k = 1}^{N_{\text{sect}}} \mathbb{R}_{\left[t_k, t_{k+1} \right)}(t) \left[ a_{k,i} \left( t - t_k \right)^3 + b_{k,i} \left( t - t_k \right)^2 + d_{k,i} \left( t - t_k \right) + f_{k,i} \right], \end{aligned}
where $$0 \leq t_1 < t_2 < \dots < t_{N_{\text{sect}} + 1} \leq T_{\text{sim}}$$ is a decomposition of the simulation time interval $$\left[0, T_{\text{sim}}\right]$$ into pieces $$\left[t_k, t_{k+1} \right)$$. On each piece, the profile is given by a cubic (fourth order) polynomial shifted to the beginning $$t_k$$ of the piece.
For information on model parameters see Inlet.
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2022-05-17 06:59:59
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http://en.wikipedia.org/wiki/Binary_classification
|
# Binary classification
Binary or binomial classification is the task of classifying the elements of a given set into two groups on the basis of a Classification rule. Some typical binary classification tasks are
• medical testing to determine if a patient has certain disease or not (the classification property is the presence of the disease)
• quality control in factories; i.e. deciding if a new product is good enough to be sold, or if it should be discarded (the classification property is being good enough)
• deciding whether a page or an article should be in the result set of a search or not (the classification property is the relevance of the article, or the usefulness to the user)
Statistical classification in general is one of the problems studied in computer science, in order to automatically learn classification systems; some methods suitable for learning binary classifiers include the decision trees, Bayesian networks, support vector machines, neural networks, probit regression, and logit regression.
Sometimes, classification tasks are trivial. Given 100 balls, some of them red and some blue, a human with normal color vision can easily separate them into red ones and blue ones. However, some tasks, like those in practical medicine, and those interesting from the computer science point-of-view, are far from trivial, and may produce faulty results if executed imprecisely.
## Evaluation of binary classifiers
true positive (TP) eqv. with hit true negative (TN) false positive (FP) eqv. with false alarm, Type I error false negative (FN) eqv. with miss, Type II error sensitivity or true positive rate (TPR) eqv. with hit rate, recall $\mathit{TPR} = \mathit{TP} / P = \mathit{TP} / (\mathit{TP}+\mathit{FN})$ specificity (SPC) or True Negative Rate $\mathit{SPC} = \mathit{TN} / N = \mathit{TN} / (\mathit{FP} + \mathit{TN})$ precision or positive predictive value (PPV) $\mathit{PPV} = \mathit{TP} / (\mathit{TP} + \mathit{FP})$ negative predictive value (NPV) $\mathit{NPV} = \mathit{TN} / (\mathit{TN} + \mathit{FN})$ fall-out or false positive rate (FPR) $\mathit{FPR} = \mathit{FP} / N = \mathit{FP} / (\mathit{FP} + \mathit{TN})$ false discovery rate (FDR) $\mathit{FDR} = \mathit{FP} / (\mathit{FP} + \mathit{TP}) = 1 - \mathit{PPV}$ Miss Rate or False Negative Rate (FNR) $\mathit{FNR} = \mathit{FN} / (\mathit{FN} + \mathit{TP})$ accuracy (ACC) $\mathit{ACC} = (\mathit{TP} + \mathit{TN}) / (P + N)$ F1 score is the harmonic mean of precision and sensitivity $\mathit{F1} = 2 \mathit{TP} / (2 \mathit{TP} + \mathit{FP} + \mathit{FN})$ Matthews correlation coefficient (MCC) $\frac{ TP \times TN - FP \times FN } {\sqrt{ (TP+FP) ( TP + FN ) ( TN + FP ) ( TN + FN ) } }$ Uncertainty coefficient, aka Proficiency \begin{align} L &= (P+N)\times \log(P+N) \\ LTP &= TP \times \log\frac{TP}{(TP+FP)(TP+FN)} \\ LFP &= FP \times \log\frac{FP}{(FP+TP)(FP+TN)} \\ LFN &= FN \times \log\frac{FN}{(FN+TP)(FN+TN)} \\ LTN &= TN \times \log\frac{TN}{(TN+FP)(TN+FN)} \\ LP &= P \times \log \frac{P}{P+N} \\ LN &= N \times \log\frac{N}{P+N} \\ UC &= \frac{L + LTP + LFP + LFN + LTN}{L + LP + LN} \end{align} Informedness = Sensitivity + Specificity - 1 Markedness = Precision + NPV - 1 Source: Fawcett (2006).
From the confusion matrix you can derive four basic measures
To measure the performance of a classifier or predictor there are several values that can be used. Different fields have preferences for specific metric due to the known biases that are accepted. For example, in medicine the concepts sensitivity and specificity are often used. Say we test some people for the presence of a disease. Some of these people have the disease, and our test says they are positive. They are called true positives (TP). Some have the disease, but the test claims they don't. They are called false negatives (FN). Some don't have the disease, and the test says they don't - true negatives (TN). Finally, there might be healthy people who have a positive test result - false positives (FP). Thus, the number of true positives, false negatives, true negatives, and false positives add up to 100% of the set.
Let us define an experiment from P positive instances and N negative instances for some known condition. The four outcomes can be formulated in a 2×2 contingency table or confusion matrix, as follows:
Condition (as determined by "Gold standard") Condition positive Condition negative Test outcome Test outcome positive True positive False positive (Type I error) Precision = Σ True positive Σ Test outcome positive Test outcome negative False negative (Type II error) True negative Negative predictive value = Σ True negative Σ Test outcome negative Sensitivity = Σ True positive Σ Condition positive Specificity = Σ True negative Σ Condition negative Accuracy
Specificity (TNR) is the proportion of people that tested negative (TN) of all the people that actually are negative (TN+FP). As with sensitivity, it can be looked at as the probability that the test result is negative given that the patient is not sick. With higher specificity, fewer healthy people are labeled as sick (or, in the factory case, the less money the factory loses by discarding good products instead of selling them).
Sensitivity (TPR), also known as recall, is the proportion of people that tested positive (TP) of all the people that actually are positive (TP+FN). It can be seen as the probability that the test is positive given that the patient is sick. With higher sensitivity, fewer actual cases of disease go undetected (or, in the case of the factory quality control, the fewer faulty products go to the market).
The relationship between sensitivity and specificity, as well as the performance of the classifier, can be visualized and studied using the ROC curve.
In theory, sensitivity and specificity are independent in the sense that it is possible to achieve 100% in both (such as in the red/blue ball example given above). In more practical, less contrived instances, however, there is usually a trade-off, such that they are inversely proportional to one another to some extent. This is because we rarely measure the actual thing we would like to classify; rather, we generally measure an indicator of the thing we would like to classify, referred to as a surrogate marker. The reason why 100% is achievable in the ball example is because redness and blueness is determined by directly detecting redness and blueness. However, indicators are sometimes compromised, such as when non-indicators mimic indicators or when indicators are time-dependent, only becoming evident after a certain lag time. The following example of a pregnancy test will make use of such an indicator.
Modern pregnancy tests do not use the pregnancy itself to determine pregnancy status; rather, human chorionic gonadotropin is used, or hCG, present in the urine of gravid females, as a surrogate marker to indicate that a woman is pregnant. Because hCG can also be produced by a tumor, the specificity of modern pregnancy tests cannot be 100% (in that false positives are possible). Also, because hCG is present in the urine in such small concentrations after fertilization and early embryogenesis, the sensitivity of modern pregnancy tests cannot be 100% (in that false negatives are possible).
In addition to sensitivity and specificity, the performance of a binary classification test can be measured with positive predictive value (PPV), also known as precision, and negative predictive value (NPV). The positive prediction value answers the question "If the test result is positive, how well does that predict an actual presence of disease?". It is calculated as (true positives) / (true positives + false positives); that is, it is the proportion of true positives out of all positive results. (The negative prediction value is the same, but for negatives, naturally.)
accuracy measures the fraction of all instances that are correctly categorized; it is the ratio of the number of correct classifications to the total number of correct or incorrect classifications.
The F1 score is a measure of a test's performance when a single value is wanted. It considers both the precision and the recall of the test to compute the score. The traditional or balanced F-score is the harmonic mean of precision and recall:
$F_1 = 2 \cdot \frac{\mathrm{precision} \cdot \mathrm{recall}}{\mathrm{precision} + \mathrm{recall}}$.
Note, however, that the F-scores do not take the true negative rate into account, and that measures such as the Phi coefficient, Matthews correlation coefficient, Informedness or Cohen's kappa may be preferable to assess the performance of a binary classifier.[1] As a correlation coefficient, the Matthews correlation coefficient is the geometric mean of the regression coefficients of the problem and its dual. The component regression coefficients of the Matthews correlation coefficient are markedness (deltap) and informedness (deltap').[2]
### Example
As an example, suppose there is a test for a disease with 99% sensitivity and 99% specificity. If 2000 people are tested, 1000 of them are sick and 1000 of them are healthy. About 990 true positives 990 true negatives are likely, with 10 false positives and 10 false negatives. The positive and negative prediction values would be 99%, so there can be high confidence in the result.
However, if of the 2000 people only 100 are really sick: the likely result is 99 true positives, 1 false negative, 1881 true negatives and 19 false positives. Of the 19+99 people tested positive, only 99 really have the disease - that means, intuitively, that given that a patient's test result is positive, there is only 84% chance that he or she really has the disease. On the other hand, given that the patient's test result is negative, there is only 1 chance in 1882, or 0.05% probability, that the patient has the disease despite the test result.
## Converting continuous values to binary
Tests whose results are of continuous values, such as most blood values, can artificially be made binary by defining a cutoff value, with test results being designated as positive or negative depending on whether the resultant value is higher or lower than the cutoff.
However, such conversion causes a loss of information, as the resultant binary classification does not tell how much above or below the cutoff a value is. As a result, when converting a continuous value that is close to the cutoff to a binary one, the resultant positive or negative predictive value is generally higher than the predictive value given directly from the continuous value. In such cases, the designation of the test of being either positive or negative gives the appearance of an inappropriately high certainty, while the value is in fact in an interval of uncertainty. For example, with the urine concentration of hCG as a continuous value, a urine pregnancy test that measured 52 mIU/ml of hCG may show as "positive" with 50 mIU/ml as cutoff, but is in fact in an interval of uncertainty, which may be apparent only by knowing the original continuous value. On the other hand, a test result very far from the cutoff generally has a resultant positive or negative predictive value that is lower than the predictive value given from the continuous value. For example, a urine hCG value of 200,000 mIU/ml confers a very high probability of pregnancy, but conversion to binary values results in that it shows just as "positive" as the one of 52 mIU/ml.
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2014-04-20 19:55:31
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https://answers.opencv.org/questions/220383/revisions/
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Revision history [back]
cv2.VideoCapture() importing file seq. with numbers on the name. Is this a bug?
I was trying to open a tiff file seq. with cv2.VideoCapture(), because it's easier to handle then using cv2.imread.
It works ok in general, but if the file name have numbers in the basename, (before the number padding) it only opens the first frame. looks like its messing with the padding numbers.
If the filename as numbers on the base name ( eg: test123a.000001.tif) it ignores the sequence, and reads 1st frame only.
Steps to reproduce:
Create a file sequence (eg: test123a.000001.tif, test123a.000002.tif, test123a.000003.tif, etc) or download from my share here
import cv2
import numpy
cap = cv2.VideoCapture('C:/images/5frames/test123a.' + '{:06d}'.format(1, ) + '.jpg', cv2.CAP_IMAGES)
cap.set(1, 2)# 2nd number is the frame to show
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2020-06-04 05:59:55
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https://brianbuccola.com/lenovo-x140e-and-arch-linux/
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# Lenovo X140e and (Arch) Linux
January 23, 2015
My previous laptop was an Asus that ran Linux beautifully, until, after a few years, several pieces of hardware stopped working (first battery, then wifi, then monitor). I had heard great things about Lenovo, specifically how nicely they play with Linux, so last summer I bought myself a Lenovo X140e. On the whole, I’m quite happy with it, but to my surprise (and sadness), it didn’t work out of the box like my Asus did. In this post I’ll detail some of the issues I had and my workarounds for dealing with them.
The main issues that I’ve managed to solve are:
There’s one issue that I haven’t manage to solve:
• jittery touchpad, but only while on AC power
Basically, there’s no issue while on battery power; but on AC power, touching the touchpad, even without moving your finger around, causes the cursor to jut around back and forth really fast. This (annoying) problem has been confirmed elsewhere, but I haven’t found a solution to it.
If you have a solution, please email me!
## WiFi
By far the most serious (but luckily, easiest to solve) issue is the Broadcom wifi card BCM43228:
$lspci -vnn | grep Broadcom 01:00.0 Network controller [0280]: Broadcom Corporation BCM43228 802.11a/b/g/n [14e4:4359] This card was not supported under the Linux kernel until kernel version 3.17. (We’re now on 3.18, as of this writing; when I bought the laptop, we were at 3.16.) The workaround for kernels below 3.17 is to use the AUR package broadcom-wl. However, even once we arrived at 3.17, I found the native support (with the b43 driver and firmware) to be lacking: weak wifi connections, constant dropping, etc. So I still use broadcom-wl. Here’s how it works. First, download broadcom-wl with your favorite AUR helper, and install it. For example: $ cower -d broadcom-wl
$cd ~/aur/broadcom-wl$ makepkg -csi
Second, restart computer.
That’s all! Well, almost. Now, every time you update your kernel, you need to rebuild and reinstall broadcom-wl. For example:
$cd ~/aur/broadcom-wl$ makepkg -csif
The -f flag forces a rebuild and overwrites the current .pkg.tar.xz file.
(Another option is to use broadcom-wl-dkms, which automatically rebuilds itself after a kernel update.)
(This solution is obviously specific to Arch Linux, but most major distros should have some analog of the broadcom-wl package available, which should likewise solve the issue.)
## Brightness keys
You’re supposed to be able to change the brightness with <Fn-F8> and <Fn-F9> (that is, the function key together with F8 or F9). For me, this works fine in console, but not in X11, where most people (including me) spend most of their time. I’ve read that a BIOS upgrade fixes this, but I haven’t tried that.
My workaround was to write a simple bash script, brt. The idea:
• brt (no argument): output current brightness level (0–255).
• brt [n]: set brightness level to n (0–255).
• brt down: decrease brightness level by 20 (bind this to <Fn-F8>).
• brt up: increase brightness level by 20 (bind this to <Fn-F9>).
I’ll illustrate how the main part of it works, the brt_change() function (the rest of the script should be pretty self-explanatory):
#!/bin/bash
max_brightness=255
min_brightness=5
current_brightness=$(cat "$brightness_file")
up_amt=20
down_amt=20
brt_change() {
echo "$1" | sudo tee "$brightness_file"
}
brt_up() {
local new_brightness=$(($current_brightness + $up_amt)) if [[$new_brightness -le $max_brightness ]]; then brt_change "$new_brightness"
else
brt_change "$max_brightness" fi } brt_down() { local new_brightness=$(($current_brightness -$up_amt))
if [[ $new_brightness -ge$min_brightness ]]; then
brt_change "$new_brightness" else brt_change "$min_brightness"
fi
}
if [[ $# -eq 1 ]]; then case "$1" in
[0-9]*) brt_change "$1" && exit 0;; up) brt_up && exit 0;; down) brt_down && exit 0;; *) echo "Error: invalid argument. Pick a brightness level ($min_brightness-$max_brightness), or say 'up' or 'down'." && exit 1;; esac elif [[$# -eq 0 ]]; then
echo "\$current_brightness"
exit 0
else
echo "Error: too many arguments."
exit 1
fi
The file /sys/class/backlight/radeon_bl0/brightness contains the current brightness level, which for my Lenovo X140e is between 0 and 255. To change the brightness, you just change this file. The problem is that since it’s located in /sys/..., you need root permission to change it. That means that
echo "100" > /sys/...
won’t work, but neither will
sudo echo "100" > /sys/...
The reason is because in the latter, sudo is only operating on the echo command. It’s like saying, run echo as root, and now (not as root) append the output to /sys/.... To solve this, we use tee, which allows piping from stdin to a file, as root:
echo "100" | sudo tee /sys/...
This command will successfully set the brightness level to 100, and that’s the crux of the script.
There’s one remaining issue, though: we don’t want to run this script in a terminal; we want to bind it to a key. But the script uses sudo, which requires a password to be typed, which you can’t really do outside of a terminal. The solution is to allow tee to be run as root without a password. To do this, you need to change the sudoers file by running visudo (as root) and adding this line:
# Add tee' to list of commands that user brian' can run without password
brian ALL = NOPASSWD: /usr/bin/tee
What this does is allow the user “brian” (that’s me) to run tee as root (sudo tee) without a password.1 /usr/bin/tee is of course the full path to tee. To find that out on your system, run which tee in a terminal.
Now the script can be run effectively. Just bind brt down and brt up to <Fn-F8> and <Fn-F9> (or whatever you want) in whatever way is required by your desktop environment or window manager. (For me, I bind keys in xmonad.hs since I use xmonad.)
## No insert key
The laptop keyboard does not come with any physical Insert key. I guess that’s because most people nowadays don’t use it very often. But I do. One of the best features of Linux (well, X11) is the X clipboard: whenever you highlight something, it gets added to the X clipboard (no need to <Ctrl-C>), and you can paste it with <Shift-Insert>. (I also use Insert to go into ignore-mode in Vimperator for Firefox.)
What I did was bind the Windows key (which was serving no purpose) to Insert. Here’s how:
xmodmap -e "keycode 133 = Insert" # map windows button to insert
You can find the keycode of a key by running xev (X event program) from a terminal, typing the key, and looking for “keycode” in the output. (Hit <Ctrl-C> to exit xev.)
Now you can run this xmodmap command in a terminal, and it should work. But the best solution is to include it in your .xinitrc file so that it’s run every time X starts. (I have a whole keyboard-adjust script that adds the Dvorak layout, switches caps and control lock, etc., maps the Windows button to insert, etc.; I then call the whole script from my .xinitrc file.)
1. NB: This is a (potential) security risk, as it allows “brian”, or anyone logged in as “brian”, to overwrite any file! (Example: echo "all gone" | sudo tee /path/to/important/file.) Only do this if you’re the only one with access to this user profile.
|
2019-02-21 05:50:32
|
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|
http://codeforces.com/problemset/problem/497/E
|
Reminder: in case of any technical issues, you can use the lightweight website m1.codeforces.com, m2.codeforces.com, m3.codeforces.com. ×
E. Subsequences Return
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Assume that sk(n) equals the sum of digits of number n in the k-based notation. For example, s2(5) = s2(1012) = 1 + 0 + 1 = 2, s3(14) = s3(1123) = 1 + 1 + 2 = 4.
The sequence of integers a0, ..., an - 1 is defined as . Your task is to calculate the number of distinct subsequences of sequence a0, ..., an - 1. Calculate the answer modulo 109 + 7.
Sequence a1, ..., ak is called to be a subsequence of sequence b1, ..., bl, if there is a sequence of indices 1 ≤ i1 < ... < ik ≤ l, such that a1 = bi1, ..., ak = bik. In particular, an empty sequence (i.e. the sequence consisting of zero elements) is a subsequence of any sequence.
Input
The first line contains two space-separated numbers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 30).
Output
In a single line print the answer to the problem modulo 109 + 7.
Examples
Input
4 2
Output
11
Input
7 7
Output
128
Note
In the first sample the sequence ai looks as follows: (0, 1, 1, 0). All the possible subsequences are:
(), (0), (0, 0), (0, 1), (0, 1, 0), (0, 1, 1), (0, 1, 1, 0), (1), (1, 0), (1, 1), (1, 1, 0).
In the second sample the sequence ai looks as follows: (0, 1, 2, 3, 4, 5, 6). The subsequences of this sequence are exactly all increasing sequences formed from numbers from 0 to 6. It is easy to see that there are 27 = 128 such sequences.
|
2020-04-08 15:55:12
|
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|
http://documenta.sagemath.org/vol-20/30.html
|
#### DOCUMENTA MATHEMATICA, Vol. 20 (2015), 1055-1069
Christian Bopp
Syzygies of 5-Gonal Canonical Curves
We show that for 5-gonal curves of odd genus $g≥ 13$ and even genus $g≥ 28$ the $\lceil \frac{g-1}2\rceil$-th syzygy module of the curve is not determined by the syzygies of the scroll swept out by the special pencil of degree 5.
2010 Mathematics Subject Classification: 13D02,14H51
Keywords and Phrases: syzygies, canonical curves, gonality, special divisor
Full text: dvi.gz 36 k, dvi 103 k, ps.gz 466 k, pdf 238 k.
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2017-06-23 13:38:35
|
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|
http://ieeexplore.ieee.org/xpl/tocresult.jsp?reload=true&isnumber=6195043
|
By Topic
# IEEE Transactions on Nanotechnology
## Filter Results
Displaying Results 1 - 25 of 37
• ### [Front cover]
Publication Year: 2012, Page(s): C1
| PDF (182 KB)
• ### IEEE Transactions on Nanotechnology publication information
Publication Year: 2012, Page(s): C2
| PDF (40 KB)
Publication Year: 2012, Page(s):429 - 430
| PDF (190 KB)
• ### A Low-Power 40-Gb/s 1:2 Demultiplexer IC Based on a Resonant Tunneling Diode
Publication Year: 2012, Page(s):431 - 434
Cited by: Papers (6)
| | PDF (526 KB) | HTML
A low-power 1:2 demultiplexer (DEMUX) IC based on a resonant tunneling diode (RTD) is proposed. In order to achieve low-power consumption, the unique negative differential resistance (NDR) characteristics arising from the quantum effect of the RTD are exploited. The proposed DEMUX IC consists of an return to zero (RZ)-mode 1:2 demultiplexing block and an RZ-to-nonreturn to zero converting block, w... View full abstract»
• ### High Current Gain Microwave Performance of Organic Metal-Base Transistor
Publication Year: 2012, Page(s):435 - 436
| | PDF (316 KB) | HTML
We report the realization of 140 nm emitter TPD/CuPc (N,N'-diphenyl-N,N'-bis(3-methylphenyl)-(1,1'-biphenyl)-4,4'-diamine/copper phthalocyanine) with cutoff frequency fT = 300 kHz. Our devices were grown on high-resistivity p-type float-zone silicon collector and implemented with an aluminum grid base feature a high and stable current gain well above 400. The present transistors are the fir... View full abstract»
• ### Electrical and Magnetic Properties of Higher Manganese Silicide Nanostructures
Publication Year: 2012, Page(s):437 - 440
Cited by: Papers (3)
| | PDF (406 KB) | HTML
Higher manganese silicide, Mn15Si26, nanostructures were grown using CVD using a coordination compound precursor. These nanostructures exhibit p-type semiconducting behavior. They also exhibit a nonzero magnetic moment even at room temperature and the magnetic transition temperature appears to be near 330 K. View full abstract»
• ### PolyMethyl Methacrylate Thin-Film-Based Field Emission Microscope
Publication Year: 2012, Page(s):441 - 443
Cited by: Papers (4)
| | PDF (271 KB) | HTML
A field emission microscope (FEM) is a useful tool for investigating molecular surface structures. Conventional FEMs suffer from poor image contrast level and low sensitivities when low-energy electron beams are applied. In this article, a new anode material is employed to improve the FEM imaging performance. We demonstrate that the device has the capability of clearly capturing images of facet bo... View full abstract»
• ### Fabrication of optical device arrays using patterned growth of ZnO nanostructures
Publication Year: 2012, Page(s):444 - 447
Cited by: Papers (3)
| | PDF (443 KB) | HTML
An on-site catalyst-free and seedless synthesis method is presented for patterned growth of ZnO nanostructures and easy integration into arrayed microdevices. The optoelectronic ZnO devices, exhibiting Schottky diode behavior, are highly sensitive to UV illumination. We demonstrate that the patterned synthesis method combined with conventional lithography provides a facile way of fabricating array... View full abstract»
• ### Single Mesowire Transistor From Perylene Tetracarboxylic Diimide
Publication Year: 2012, Page(s):448 - 450
| | PDF (461 KB) | HTML
We have recently fabricated new nano/mesowires of perylene tetracarboxylic diimide (PTCDI) without side chains by self-assembling them from a gas phase. In this letter, we discuss a single PTCDI mesowire transistor that was successfully fabricated, characterized, and modeled. This organic n-channel field effect transistor shows good output and transfer characteristics. Our transistor model include... View full abstract»
• ### A Scalable Memory-Based Reconfigurable Computing Framework for Nanoscale Crossbar
Publication Year: 2012, Page(s):451 - 462
Cited by: Papers (16)
| | PDF (1309 KB) | HTML
Nanoscale molecular electronic devices amenable to bottom-up self-assembly into a crossbar structure have emerged as a promising candidate for future electronic systems. To address some of the design challenges in molecular crossbar, we propose “memory-based architecture for reconfigurable computing” (MBARC), where memory, instead of switch-based logic functions, is used as the compu... View full abstract»
• ### Theoretical Investigation of Traveling-Wave Amplification in Metallic Carbon Nanotubes Biased by a DC Field
Publication Year: 2012, Page(s):463 - 471
Cited by: Papers (5)
| | PDF (570 KB) | HTML
Traveling-wave amplification along a carbon nanotube (CNT) under dc-ac fields is theoretically investigated. The ac conductivity of a metallic CNT is found with respect to the applied dc bias. For this purpose, the Boltzmann transport equation (BTE) is solved within the relaxation time approximation (RTA) by separating the ac and dc distributions. The problem is solved both exactly and approximate... View full abstract»
• ### Work Function Engineering With Linearly Graded Binary Metal Alloy Gate Electrode for Short-Channel SOI MOSFET
Publication Year: 2012, Page(s):472 - 478
Cited by: Papers (17)
| | PDF (426 KB) | HTML
Over the last few decades, silicon-on-insulator (SOI) technology has been identified as one possible solution for enhancing the performance of CMOS because of its numerous advantages over conventional bulk CMOS technology. One of the primary drawbacks of short-channel SOI MOSFET is the degradation of device threshold voltage with decreasing channel length. Drain-induced barrier-lowering (DIBL) eff... View full abstract»
• ### MOSFETs Made From GaN Nanowires With Fully Conformal Cylindrical Gates
Publication Year: 2012, Page(s):479 - 482
Cited by: Papers (11)
| | PDF (396 KB) | HTML
We report novel metal-oxide-semiconductor field effect transistors (MOSFETs) based on individual gallium nitride (GaN) nanowires with fully conformal cylindrical gates. The W/Al2O3 gates were deposited by atomic layer deposition. Reverse-bias breakdown voltages exceeded the largest gate voltage tested (-35 V). The nanowire MOSFETs showed complete pinchoff, with threshold volt... View full abstract»
• ### MOS Devices With High-κ (ZrO$_2$) $_x$(La $_2$O$_3$) $_{1-x}$ Alloy as Gate Dielectric Formed by Depositing ZrO $_2$/La$_2$O $_3$/ZrO$_2$ Laminate and Annealing
Publication Year: 2012, Page(s):483 - 491
Cited by: Papers (3)
| | PDF (781 KB) | HTML
An amorphous (ZrO2)x(La2O3)1-x alloy formed by depositing a ZrO2/La2O3/ZrO2 laminate and a subsequent annealing was employed as the gate dielectric for metal-oxide-semiconductor (MOS) devices. The (ZrO2)x(La2O3)1-x alloy is found to have a high permi... View full abstract»
• ### Quantum Transport and Current Distribution at Radio Frequency in Multiwall Carbon Nanotubes
Publication Year: 2012, Page(s):492 - 500
Cited by: Papers (4)
| | PDF (746 KB) | HTML
Multiwall carbon nanotubes represent a low-dimensional material that could serve as building blocks for future carbon-based nanoelectronics. The understanding of the electromagnetic performances at radio frequency of these materials for use in nanointerconnects is strictly related to the analysis of their transport properties as function of the working conditions. In this paper, we present an expl... View full abstract»
• ### Performance Analysis of Si Nanowire Biosensor by Numerical Modeling for Charge Sensing
Publication Year: 2012, Page(s):501 - 512
Cited by: Papers (21)
| | PDF (1135 KB) | HTML
A numerical study on the operation of Si nanowire (NW) biosensors in charge-based sensing is presented. The simulation is built on physical models that, upon numerical convergence, coherently account for Fermi-Dirac, Poisson-Boltzman, site-binding and Gouy-Chapman-Stern theories in self-consistent manner. The method enables us to disentangle the impact of key design and experimental setup factors ... View full abstract»
• ### Properties of Metal–Graphene Contacts
Publication Year: 2012, Page(s):513 - 519
Cited by: Papers (26)
| | PDF (627 KB) | HTML
We present a study on the metal-graphene contact properties. Utilizing a dual-gate field-effect transistor device, an energetic separation between the Fermi level and the Dirac point in the contact areas can be adjusted deliberately by applying an appropriate front-gate voltage that acts only on the channel. This front-gate voltage is compensated by an opposite large-area back-gate voltage, thereb... View full abstract»
• ### A Flexible ZnO Nanowire-Based Humidity Sensor
Publication Year: 2012, Page(s):520 - 525
Cited by: Papers (13)
| | PDF (779 KB) | HTML
In this paper, the authors report the direct growth of ZnO nanowires (NWs) on a flexible substrate by the hydrothermal process and the fabrication of ZnO NW-based humidity sensor. It was found that average length and diameter of the ZnO NWs were 0.6 μm and 50 nm, respectively. It was also found that resistance of the ZnO NWs decreased by 45% as we increased the relative humidity from 52% to... View full abstract»
• ### Influence of Channel and Underlap Engineering on the High-Frequency and Switching Performance of CNTFETs
Publication Year: 2012, Page(s):526 - 533
Cited by: Papers (15)
| | PDF (814 KB) | HTML
We have comprehensively studied the influence of the channel and underlap engineering on the switching and high-frequency performance of carbon nanotube field-effect transistors (CNTFETs). Various source/drain and channel-engineered CNTFETs have been investigated and optimized architectures have been concluded from simulations. Performance parameters such as switching time τ and cutoff freq... View full abstract»
• ### Tuning Electronic Structure of Graphene: A First-Principles Study
Publication Year: 2012, Page(s):534 - 541
Cited by: Papers (15)
| | PDF (712 KB) | HTML
Based on first-principles study, tuning of electronic structure of graphene is reported. The emergence of band gap in this semimetal can be accomplished through different mechanisms. In this study, we have reported on the band gap modulations in graphene through chemical functionalization with oxygen, under the application of external stress, and through the creation of vacancies. Our study sugges... View full abstract»
• ### Orientation Modulated Epitaxy of Cu Nanorods on Si(1 0 0) Substrate
Publication Year: 2012, Page(s):542 - 545
Cited by: Papers (5)
| | PDF (359 KB) | HTML
Epitaxial growth of Cu nanorod films on hydrogen-terminated Si(100) substrates by oblique angle deposition (OAD) was investigated. It is found out that the crystallographic orientation of Cu nanorod films exhibits a 45° in-plane rotation with respect to the substrate. When the incident angle of deposition was increased from normal to 80°, the distribution of Cu(111) poles is observed... View full abstract»
• ### PLGA Micro- and Nanoparticles Loaded Into Gelatin Scaffold for Controlled Drug Release
Publication Year: 2012, Page(s):546 - 553
Cited by: Papers (8)
| | PDF (637 KB) | HTML
Curcumin and bovine serum albumin (BSA) were used as model drugs and loaded into microand nanoparticles of biodegradable poly(lactic-co-glycolic acid) (PLGA). The PLGA was incorporated into hydrophilic and biocompatible gelatin scaffolds to design a controlled drug release system. The gelatin scaffolds were cross-linked using glutaraldehyde. The controlled delivery of drugs from biologically activ... View full abstract»
• ### Transmission-Line Model for Multiwall Carbon Nanotubes With Intershell Tunneling
Publication Year: 2012, Page(s):554 - 564
Cited by: Papers (17)
| | PDF (619 KB) | HTML
The electromagnetic behavior of multiwall carbon nanotubes (MWCNTs), in the frequency range where only intraband transitions are allowed, depends on the combinations of different aspects: the number of effective conducting channels of each shell, the electron tunneling between adjacent shells, and the electromagnetic interaction between shells and the environment. This paper proposes a general tra... View full abstract»
• ### Electrical Properties of Silicon Nanowire Fabricated by Patterning and Oxidation Process
Publication Year: 2012, Page(s):565 - 569
Cited by: Papers (2)
| | PDF (393 KB) | HTML
We are reporting electrical properties of Si nanowire field-effect transistors with a Schottky barrier formed at the electrodes. The channel widths are varied using a top-down process of electron-beam patterning followed by surface oxidation from a few micrometers to the sub-10-nm level. The field-effect mobility increases gradually with decreasing channel width to 20 nm. On the other hand, the mo... View full abstract»
• ### Joint Energy Harvesting and Communication Analysis for Perpetual Wireless Nanosensor Networks in the Terahertz Band
Publication Year: 2012, Page(s):570 - 580
Cited by: Papers (65)
| | PDF (710 KB) | HTML
Wireless nanosensor networks (WNSNs) consist of nanosized communicating devices, which can detect and measure new types of events at the nanoscale. WNSNs are the enabling technology for unique applications such as intrabody drug delivery systems or surveillance networks for chemical attack prevention. One of the major bottlenecks in WNSNs is posed by the very limited energy that can be stored in a... View full abstract»
## Aims & Scope
The IEEE Transactions on Nanotechnology is devoted to the publication of manuscripts of archival value in the general area of nanotechnology, which is rapidly emerging as one of the fastest growing and most promising new technological developments for the next generation and beyond.
Full Aims & Scope
## Meet Our Editors
Editor-in-Chief
Fabrizio Lombardi
Dept. of ECE
Northeastern Univ.
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2017-07-27 00:04:27
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|
https://labs.tib.eu/arxiv/?author=91405%20Orsay%20Cedex
|
• The flux of very high-energy neutrinos produced in our Galaxy by the interaction of accelerated cosmic rays with the interstellar medium is not yet determined. The characterization of this flux will shed light on Galactic accelerator features, gas distribution morphology and Galactic cosmic ray transport. The central Galactic plane can be the site of an enhanced neutrino production, thus leading to anisotropies in the extraterrestrial neutrino signal as measured by the IceCube Collaboration. The ANTARES neutrino telescope, located in the Mediterranean Sea, offers a favourable view on this part of the sky, thereby allowing for a contribution to the determination of this flux. The expected diffuse Galactic neutrino emission can be obtained linking a model of generation and propagation of cosmic rays with the morphology of the gas distribution in the Milky Way. In this paper, the so-called "Gamma model" introduced recently to explain the high-energy gamma ray diffuse Galactic emission, is assumed as reference. The neutrino flux predicted by the "Gamma model" depends of the assumed primary cosmic ray spectrum cut-off. Considering a radially-dependent diffusion coefficient, this proposed scenario is able to account for the local cosmic ray measurements, as well as for the Galactic gamma ray observations. Nine years of ANTARES data are used in this work to search for a possible Galactic contribution according to this scenario. All flavour neutrino interactions are considered. No excess of events is observed and an upper limit is set on the neutrino flux of $1.1$ ($1.2$) times the prediction of the "Gamma model" assuming the primary cosmic ray spectrum cut-off at 5 (50) PeV. This limit excludes the diffuse Galactic neutrino emission as the major cause of the "spectral anomaly" between the two hemispheres measured by IceCube.
• The z=0.89 molecular absorber toward PKS1830-211 provides us with the opportunity to probe the chemical and physical properties of the interstellar medium in the disk of a galaxy at a look-back time of half the present age of the Universe. Recent ALMA observations of hydrides have unveiled the multi-phase composition of this source's interstellar medium along two absorbing sightlines. Here, we report ALMA observations of CH+ and SH+, and of their 13C- and 34S- isotopologues, as potential tracers of energetic processes in the interstellar medium. CH+ and 13CH+ are detected toward both images of PKS1830-211, CH+ showing the deepest and broadest absorption among all species observed so far. The [CH+]/[13CH+] abundance ratio is ~100 in the south-west line of sight. [...] Toward the north-east image, we find an even larger value of [CH+]/[13CH+], 146 +/- 43, although with a large uncertainty. This sightline intercepts the absorber at a larger galactocentric radius than the southwestern one, where material might be less processed in stellar nucleosynthesis. In contrast to CH+ and its 13C isotopologue, SH+ and 34SH+ are only detected on the south-west sightline. These are the first detections of extragalactic SH+ and interstellar 34SH+. The spectroscopic parameters of SH+ are reevaluated and improved rest frequencies of 34SH+ are obtained. The [CH+]/[SH+] column density ratios show a large difference between the two lines of sight: ~25 and >600 toward the SW and NE image, respectively. We are not able to shed light on the formation process of CH+ and SH+ with these data, but the differences in the two sightlines toward PKS1830-211 suggest that their absorptions arise from gas with molecular fraction gtrsim 10%, with SH+ tracing significantly higher molecular fractions than CH+.
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2020-11-29 15:08:56
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https://codegolf.stackexchange.com/questions/220833/highest-single-turn-score-in-match-land
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# Highest single-turn score in Match Land
## Background
Match Land is a mobile game that falls into the Match-3 genre (think Bejeweled or Candy Crush Saga series): swap two orthogonally adjacent pieces to make a 3-in-a-row or longer. However, Match Land has an additional rule that makes the game much more interesting.
Once you make a valid match-3 move, the matched blocks are not removed immediately; instead, you get a small time window where you can create more/longer match-3 chains by swapping not-yet-matched blocks. If you make a wrong swap (that does not create a new match or extend existing matches), your turn ends immediately.
Exact rules for this challenge are as follows (I don't play this game right now, so the details might be different from the actual game. And the details not relevant to this challenge are omitted):
• The game is played on a rectangular board. It is turn-based, and the player can make one or more moves in a turn.
• The board is filled with tiles, each of which has one of six different types (denoted 1 to 6 in the examples later).
• The player makes a "match" (three or more same tiles in a row, horizontally or vertically) by swapping two orthogonally (horizontally or vertically) adjacent tiles. The action of swapping two tiles is called a "move".
• After the initial match, the player can make an additional move if the move extends the length of an existing match or it creates at least one new match (3-in-a-row).
In the following example, B4-C4 and B3-B4 are valid moves but A3-B3 is not (simply moving the 4 adjacent to an existing match without creating a 3-in-a-row is not valid):
| 1 2 3 4
--+------------
A | 2 1 4 3
B | 1 2 3 4
C |(4)(4)(4) 5
• The player cannot move already matched tiles.
For example, consider the following 5x7 board (with coordinates for ease of explanation):
| 1 2 3 4 5 6 7
--+---------------------
A | 2 4 4 3 5 2 4
B | 3 2 1 4 1 3 5
C | 4 2 4 4 3 1 4
D | 2 4 3 1 4 2 3
E | 2 4 2 2 3 3 4
The player can make a move at A3-A4 to match three 4's:
| 1 2 3 4 5 6 7
--+---------------------
A | 2 4 3 (4) 5 2 4
B | 3 2 1 (4) 1 3 5
C | 4 2 4 (4) 3 1 4
D | 2 4 3 1 4 2 3
E | 2 4 2 2 3 3 4
Then D7-E7 to match some 3's:
| 1 2 3 4 5 6 7
--+---------------------
A | 2 4 3 (4) 5 2 4
B | 3 2 1 (4) 1 3 5
C | 4 2 4 (4) 3 1 4
D | 2 4 3 1 4 2 4
E | 2 4 2 2 (3)(3)(3)
Then C5-D5 (note that a match is extended only if the new tile to be matched is aligned with the existing match, so the 3 moving into D5 is not matched with the existing E5-E7):
| 1 2 3 4 5 6 7
--+---------------------
A | 2 4 3 (4) 5 2 4
B | 3 2 1 (4) 1 3 5
C | 4 2 (4)(4)(4) 1 4
D | 2 4 3 1 3 2 4
E | 2 4 2 2 (3)(3)(3)
You can continue this until you run out of possible moves.
I found a sequence of 9 moves from the initial state that matches 26 tiles in total (not confirmed yet if it is optimal):
C1-C2 B1-B2 A2-B2 C5-D5 D6-E6 E5-E6 D3-D4 B6-C6 B3-B4
| 1 2 3 4 5 6 7
--+---------------------
A |(2) 3 (4) 3 5 2 4
B |(2)(4)(4)(1)(1)(1) 5
C |(2)(4)(4)(4)(4)(3) 4
D |(2)(4) 1 (3)(3)(3)(3)
E |(2)(4)(2)(2)(2)(3) 4
## Challenge
Given a board state in Match Land, output the maximum number of tiles that can be matched in a single turn.
The input is a 2D array of integers from 1 to 6 inclusive. You can use any equivalent representation of a rectangular 2D array in your language, and any 6 distinct values in place of the integers. You can further assume that the input does not already have any 3-in-a-rows (as is the case in the actual game), and it has at least one valid initial move.
Standard rules apply. Shortest code in bytes wins. Imaginary brownie points for a solution that is provably correct and works reasonably fast in practice.
## Test cases
Input:
1 1 2 1 1
Output: 4
Input:
1 1 2
2 2 3
3 3 1
Output: 9
Input:
2 2 1 2 1
2 1 2 1 2
3 2 1 2 1
Output: 14 (A3-A4, B4-B5, B2-B3, B3-B4, C3-C4)
Input:
3 2 1 2 2
2 1 2 1 2
3 2 1 2 1
Output: 12 (B4-C4, B2-C2, A2-A3)
(I believe matching 13 tiles is impossible without breaking existing matches)
• "works reasonably fast in practice" - Do you know the lowest time complexity possible? – Jonah Mar 18 at 1:52
• @Jonah No, I don't. But I know that naive backtracking would definitely explode exponentially :) – Bubbler Mar 18 at 1:58
• @Arnauld The latter is correct. One way to think of it is to ignore the word "extend" -- every move should create at least one new 3-in-a-row. (In your example, if you make D2-D3 move, a new vertical match of B2-C2-D2 is created in this sense.) – Bubbler Mar 18 at 7:58
• I think this might've been cooler as a fastest code challenge than code-golf. – Manish Kundu Mar 18 at 8:02
• Does the final configuration need to be a single connected region? – Jonah Mar 18 at 18:56
# JavaScript (ES6), 315 307 303 bytes
Brute force, but quite fast on the test cases.
f=(m,T=M=0,C=c=>m.map((r,y)=>r.map((_,x)=>c(x,y,1,0)|c(x,y,0,1))))=>C((x,y,X,Y,r=m[y],v=r[x],R=m[y+Y]||0,V=R[X+=x])=>V<8&v<8&&(Y=m.map(r=>[...r]),r[x]=V,R[X]=v,a=[n=0,1,2],C((x,y,w,z)=>a.some(p=i=>p-(p=(m[y+z*i]||0)[x+w*i]&7))||a.map(i=>n+=(R=m[y+z*i])[i=x+w*i]<(R[i]|=8)))|n&&f(m,T+n),m=Y),M=T>M?T:M)|M
Try it online!
### Commented and speed-optimized
By adding a cache, we can make it fast enough to solve the 7x5 example given in the challenge in a few seconds.
A solution for 30 tiles in 11 moves is:
A4-B4 B5-C5 C4-D4 C3-D3 B3-C3 B1-B2 D7-E7 A7-B7 E1-E2 D1-E1 C1-C2
f = (
// m[] is the input grid
m,
// T is the current score, M is the best score
T = M = 0,
// ungolfed version only: keep track of the moves and use a cache
P = [],
S = new Set,
// C is a helper function that walks through all cells and invokes a callback
// function with the position (x, y) and a vector (+1, 0) or (0, +1)
C = c => m.map((r, y) => r.map((_, x) => c(x, y, 1, 0) | c(x, y, 0, 1))),
B
) =>
// cache hit? (ungolfed version only)
( S.has(key = JSON.stringify(m))
||
// look for all pairs of cells that can be swapped
C((x, y, X, Y, r = m[y], v = r[x], R = m[y + Y] || 0, V = R[X += x]) =>
// valid only if none of them already belongs to a match
// ungolfed version only: also make sure they're not equal
V < 8 && v < 8 && v ^ V && (
// make a backup of the grid
B = m.map(r => [...r]),
// swap the cells
r[x] = V,
R[X] = v,
// check the grid
a = [n = 0, 1, 2],
C((x, y, w, z) =>
// 3 cells in a row?
a.some(p = i =>
p - (p = (m[y + z * i] || 0)[x + w * i] & 7)
) ||
// if yes, mark them as matched and increment n for each new cell
a.map(i =>
n += (R = m[y + z * i])[i = x + w * i] < (R[i] |= 8)
)
) | n &&
// do a recursive call if at least one new cell was added
f(m, T + n, [...P, [[x, y], [X, y + Y]]], S),
// restore the grid
m = B
),
// update M to max(M, T)
M = T > M ? (best = P, T) : M,
// update the cache (ungolfed version only)
),
[ best, M ]
)
Try it online!
# R, 239 bytes
f=function(m,x=!m,w=nrow(m),+=sum,/=c){for(i in seq(m))for(j in 0:1){l=m
k=i--(i%%w&!j)-j*w*(i>w)
l[i/k]=l[k/i]
n=0
for(d in 1:2)n=t(n|apply(l,d,function(v,r=rle(v)$l)unlist(Map(rep,r>2,r)))&!+x[i/k]) if(+n>+x)F=max(F,+n,f(l,t(n)))} F} Try it online! ### Semi-ungolfed # m - input matrix # x - matrix of existing matches initialized with FALSEs of the shape of m # w - number of rows f = function(m, x=!m, w=nrow(m)) { # loop through each cell in m twice: # once for a vertical match and once for a horizontal one for(i in seq(m)) for(j in 0:1) { # Work on a copy of m l = m # The index of the cell to swap: # one cell down from i (except the last row), or # one cell left from i (except the first column) k = i + (i%%w&!j) - j*w*(i>w) # Swap ith and kth cell l[c(i,k)] = l[c(k,i)] # Record new matches, start with 0 n = 0 # If neither i, nor k was matched earlier, check both dimensions: if(!any(x[c(i,k)])) for(d in 1:2) { # Find runs of 3-in-a-row across the given dimension: # take Run Length Encoding of a row/column, # expand it so that runs longer than 2 get mapped to TRUE # e.g.: 1 2 2 2 3 3 => F T T T F F, # then logical OR with previous n n = n | apply(l, d, function(v, r=rle(v)$l)unlist(Map(rep, r>2, r)))
# Apply always collects the results columnwise, so that
# a transpose is necessary to make the results conformable
n = t(n)
}
# If there are new matches compared to previous iteration
if(sum(n)>sum(x))
# Call itself recursively with the modified input l, and
# n transposed to original shape, as new existing matches
# The maximum of all iterations will be the final result
F = max(F, sum(n), f(l,t(n)))
}
F
}
|
2021-05-16 18:04:47
|
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|
https://voer.edu.vn/c/photon-momentum/0e60bfc6/da278131
|
Giáo trình
College Physics
Science and Technology
Photon Momentum
Tác giả: OpenStaxCollege
Measuring Photon Momentum
The quantum of EM radiation we call a photon has properties analogous to those of particles we can see, such as grains of sand. A photon interacts as a unit in collisions or when absorbed, rather than as an extensive wave. Massive quanta, like electrons, also act like macroscopic particles—something we expect, because they are the smallest units of matter. Particles carry momentum as well as energy. Despite photons having no mass, there has long been evidence that EM radiation carries momentum. (Maxwell and others who studied EM waves predicted that they would carry momentum.) It is now a well-established fact that photons do have momentum. In fact, photon momentum is suggested by the photoelectric effect, where photons knock electrons out of a substance. [link] shows macroscopic evidence of photon momentum.
[link] shows a comet with two prominent tails. What most people do not know about the tails is that they always point away from the Sun rather than trailing behind the comet (like the tail of Bo Peep’s sheep). Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. The dust particles recoil away from the Sun when photons scatter from them. Evidently, photons carry momentum in the direction of their motion (away from the Sun), and some of this momentum is transferred to dust particles in collisions. Gas atoms and molecules in the blue tail are most affected by other particles of radiation, such as protons and electrons emanating from the Sun, rather than by the momentum of photons.
Momentum is conserved in quantum mechanics just as it is in relativity and classical physics. Some of the earliest direct experimental evidence of this came from scattering of x-ray photons by electrons in substances, named Compton scattering after the American physicist Arthur H. Compton (1892–1962). Around 1923, Compton observed that x rays scattered from materials had a decreased energy and correctly analyzed this as being due to the scattering of photons from electrons. This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. Energy and momentum are conserved in the collision. (See [link]) He won a Nobel Prize in 1929 for the discovery of this scattering, now called the Compton effect, because it helped prove that photon momentum is given by
$p=\frac{h}{\lambda },$
where $h$ is Planck’s constant and $\lambda$ is the photon wavelength. (Note that relativistic momentum given as $p=\gamma \text{mu}$ is valid only for particles having mass.)
We can see that photon momentum is small, since $p=h/\lambda$ and $h$ is very small. It is for this reason that we do not ordinarily observe photon momentum. Our mirrors do not recoil when light reflects from them (except perhaps in cartoons). Compton saw the effects of photon momentum because he was observing x rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron.
Electron and Photon Momentum Compared
(a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum. (c) What is the energy of the electron, and how does it compare with the energy of the photon?
Strategy
Finding the photon momentum is a straightforward application of its definition: $p=\frac{h}{\lambda }$. If we find the photon momentum is small, then we can assume that an electron with the same momentum will be nonrelativistic, making it easy to find its velocity and kinetic energy from the classical formulas.
Solution for (a)
Photon momentum is given by the equation:
$p=\frac{h}{\lambda }.$
Entering the given photon wavelength yields
$p=\frac{6\text{.}\text{63}×{\text{10}}^{\text{–34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}}{\text{500}×{\text{10}}^{\text{–9}}\phantom{\rule{0.25em}{0ex}}\text{m}}=\text{1}\text{.}\text{33}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}.$
Solution for (b)
Since this momentum is indeed small, we will use the classical expression $p=\text{mv}$ to find the velocity of an electron with this momentum. Solving for $v$ and using the known value for the mass of an electron gives
$v=\frac{p}{m}=\frac{1\text{.}\text{33}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}}{9\text{.}\text{11}×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}}=\text{1460 m/s}\approx \text{1460 m/s}.$
Solution for (c)
The electron has kinetic energy, which is classically given by
${\text{KE}}_{e}=\frac{1}{2}{\text{mv}}^{2}.$
Thus,
${\text{KE}}_{e}=\frac{1}{2}\left(9\text{.}\text{11}×{\text{10}}^{\text{–3}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{1455 m/s}{\right)}^{2}=\text{9.64}×{\text{10}}^{\text{–25}}\phantom{\rule{0.25em}{0ex}}\text{J}.$
Converting this to eV by multiplying by $\left(\text{1 eV}\right)/\left(1\text{.}\text{602}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}J\right)$ yields
${\text{KE}}_{e}=\text{6.02}×{\text{10}}^{\text{–6}}\phantom{\rule{0.25em}{0ex}}\text{eV}.$
The photon energy $E$ is
$E=\frac{\text{hc}}{\lambda }=\frac{\text{1240 eV}\cdot \text{nm}}{\text{500 nm}}=2\text{.}\text{48 eV},$
which is about five orders of magnitude greater.
Discussion
Photon momentum is indeed small. Even if we have huge numbers of them, the total momentum they carry is small. An electron with the same momentum has a 1460 m/s velocity, which is clearly nonrelativistic. A more massive particle with the same momentum would have an even smaller velocity. This is borne out by the fact that it takes far less energy to give an electron the same momentum as a photon. But on a quantum-mechanical scale, especially for high-energy photons interacting with small masses, photon momentum is significant. Even on a large scale, photon momentum can have an effect if there are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually take spacecraft from place to place in the solar system. (See [link].)
Relativistic Photon Momentum
There is a relationship between photon momentum $p$ and photon energy $E$ that is consistent with the relation given previously for the relativistic total energy of a particle as ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$. We know $m$ is zero for a photon, but $p$ is not, so that ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$ becomes
$E=\text{pc},$
or
$p=\frac{E}{c}\left(photons\right).$
To check the validity of this relation, note that $E=\text{hc}/\lambda$ for a photon. Substituting this into $p=E/c$ yields
$p=\left(\text{hc}/\lambda \right)/c=\frac{h}{\lambda },$
as determined experimentally and discussed above. Thus, $p=E/c$ is equivalent to Compton’s result $p=h/\lambda$. For a further verification of the relationship between photon energy and momentum, see [link].
Photon Energy and Momentum
Show that $p=E/c$ for the photon considered in the [link].
Strategy
We will take the energy $E$ found in [link], divide it by the speed of light, and see if the same momentum is obtained as before.
Solution
Given that the energy of the photon is 2.48 eV and converting this to joules, we get
$p=\frac{E}{c}=\frac{\left(2.48 eV\right)\left(1\text{.}\text{60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J/eV}\right)}{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=\text{1}\text{.}\text{33}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}.$
Discussion
This value for momentum is the same as found before (note that unrounded values are used in all calculations to avoid even small rounding errors), an expected verification of the relationship $p=E/c$. This also means the relationship between energy, momentum, and mass given by ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$ applies to both matter and photons. Once again, note that $p$ is not zero, even when $m$ is.
Section Summary
• Photons have momentum, given by $p=\frac{h}{\lambda }$, where $\lambda$ is the photon wavelength.
• Photon energy and momentum are related by $p=\frac{E}{c}$, where $E=\text{hf}=\text{hc}/\lambda$ for a photon.
Conceptual Questions
Which formula may be used for the momentum of all particles, with or without mass?
Is there any measurable difference between the momentum of a photon and the momentum of matter?
Why don’t we feel the momentum of sunlight when we are on the beach?
Problems & Exercises
(a) Find the momentum of a 4.00-cm-wavelength microwave photon. (b) Discuss why you expect the answer to (a) to be very small.
(a) $\text{1.66}×{\text{10}}^{-\text{32}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$
(b) The wavelength of microwave photons is large, so the momentum they carry is very small.
(a) What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?
(a) What is the wavelength of a photon that has a momentum of $5\text{.}\text{00}×{\text{10}}^{-\text{29}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$? (b) Find its energy in eV.
(a) 13.3 μm
(b) $9\text{.}\text{38}×{\text{10}}^{-2}$ eV
(a) A $\gamma$-ray photon has a momentum of $8\text{.}\text{00}×{\text{10}}^{-\text{21}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$. What is its wavelength? (b) Calculate its energy in MeV.
(a) Calculate the momentum of a photon having a wavelength of $2\text{.}\text{50 μm}$. (b) Find the velocity of an electron having the same momentum. (c) What is the kinetic energy of the electron, and how does it compare with that of the photon?
(a) $2\text{.}\text{65}×{\text{10}}^{-\text{28}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$
(b) 291 m/s
(c) electron $3\text{.}\text{86}×{\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}\text{J}$, photon $7\text{.}\text{96}×{\text{10}}^{-\text{20}}\phantom{\rule{0.25em}{0ex}}\text{J}$, ratio $2\text{.}\text{06}×{\text{10}}^{6}$
Repeat the previous problem for a 10.0-nm-wavelength photon.
(a) Calculate the wavelength of a photon that has the same momentum as a proton moving at 1.00% of the speed of light. (b) What is the energy of the photon in MeV? (c) What is the kinetic energy of the proton in MeV?
(a) $1\text{.}\text{32}×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{m}$
(b) 9.39 MeV
(c) $4.70×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{MeV}$
(a) Find the momentum of a 100-keV x-ray photon. (b) Find the equivalent velocity of a neutron with the same momentum. (c) What is the neutron’s kinetic energy in keV?
Take the ratio of relativistic rest energy, $E={\mathrm{\gamma mc}}^{2}$, to relativistic momentum, $p=\gamma \text{mu}$, and show that in the limit that mass approaches zero, you find $E/p=c$.
$E={\mathrm{\gamma mc}}^{2}$ and $P=\mathrm{\gamma mu}$, so
$\frac{E}{P}=\frac{{\text{γmc}}^{2}}{\text{γmu}}=\frac{{\text{c}}^{2}}{\text{u}}.$
As the mass of particle approaches zero, its velocity $u$ will approach $c$, so that the ratio of energy to momentum in this limit is
${lim}_{m\to 0}\frac{E}{P}=\frac{{c}^{2}}{c}=c$
which is consistent with the equation for photon energy.
Consider a space sail such as mentioned in [link]. Construct a problem in which you calculate the light pressure on the sail in ${\text{N/m}}^{2}$ produced by reflecting sunlight. Also calculate the force that could be produced and how much effect that would have on a spacecraft. Among the things to be considered are the intensity of sunlight, its average wavelength, the number of photons per square meter this implies, the area of the space sail, and the mass of the system being accelerated.
A car feels a small force due to the light it sends out from its headlights, equal to the momentum of the light divided by the time in which it is emitted. (a) Calculate the power of each headlight, if they exert a total force of $2\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{N}$ backward on the car. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
(a) $3\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{W}$
|
2019-11-22 13:50:39
|
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|
https://bdsmlog.mistresstissa.com/whats-your-number/
|
## 4 thoughts on “What’s Your Number?”
1. Jack says:
What a fun survey to take! The images make is much easier that the usual on a scale of 1 – 10!
Thanks for the post, was fun making up my mind picturing which image I would most enjoy.
Jack
1. Mistress Tissa says:
Thanks for your comment, Jack. Glad you enjoyed pondering where you fell on the chart!
2. john davis says:
Thought I was a real wimpy at 3 but results say different.Almost 43 % go with 3 much higher than I thought for a low number.I find it best to rely on your judgement,you always get it right.
1. Mistress Tissa says:
The results haven’t surprised me, actually. I think most people are generally in the middle of things, with fewer on either of the two ends. So somewhere around 3 or 4 makes sense to me.
|
2022-01-29 08:31:26
|
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https://www.freemathhelp.com/forum/threads/simplify-into-double-angle-formula.117487/
|
# Simplify into Double Angle Formula
#### KatieJM
##### New member
Hi there,
I need some help with this answer ASAP, please and thank you!
Simplify the expression by using a double-angle formula:
(I am attaching a screen shot of the formula because I can't find how to type the symbols).
Thanks!
Katie
#### Subhotosh Khan
##### Super Moderator
Staff member
Hi there,
I need some help with this answer ASAP, please and thank you!
Simplify the expression by using a double-angle formula:
View attachment 13240
(I am attaching a screen shot of the formula because I can't find how to type the symbols).
Thanks!
Katie
Can you expand
cos(2x)
in terms of cos(x)?
#### KatieJM
##### New member
Thank you, does this look correct?
#### ksdhart2
##### Senior Member
In the future, if you're ever unsure of an answer, you can always check it yourself. In this case, evaluating by hand would be next to impossible, so let's punch both the left-hand side and right-hand side into a calculator and see what we find:
Left-hand side: $$\cos(16^{\circ}) \approx 0.96126169 \implies 2 \cos^2(16^{\circ}) - 1 \approx 2 \cdot 0.96136169^2 - 1 \approx 0.84804807$$
Right-hand side: $$\cos(32^{\circ}) \approx 0.84804809$$
What say you? You may notice there's a very tiny discrepancy - a difference in the eighth decimal place. Why did this difference appear? What does it mean?
#### KatieJM
##### New member
Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?
Thank you,
Katie
#### pka
##### Elite Member
Thank you for that explanation!
I am stumped as to why the 8th decimal is different. What am I missing?
It is a difference in rounding, that is all.
#### ksdhart2
##### Senior Member
Consider a different equation altogether that's much simpler to solve: $$\displaystyle \frac{1}{3} \cdot \frac{1}{6}$$. What would happen if we used approximations instead of the real value?
$$\displaystyle \frac{1}{3} \approx 0.3333$$ and $$\displaystyle \frac{1}{6} \approx 0.1666$$
$$\displaystyle 0.3333 \cdot 0.1666 = 0.0555{\color{red}2778}$$
$$\displaystyle \frac{1}{18} = 0.0555{\color{red}5555}$$
Could we make the approximation tighter? What happens then?
$$\displaystyle \frac{1}{3} \approx 0.33333333$$ and $$\displaystyle \frac{1}{6} \approx 0.16666666$$
$$\displaystyle 0.33333333 \cdot 0.16666666 = 0.05555555{\color{red}27777778}$$
$$\displaystyle \frac{1}{18} = 0.05555555{\color{red}55555555}$$
What happens if we use the real, exact value?
$$\displaystyle \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}$$
#### KatieJM
##### New member
Consider a different equation altogether that's much simpler to solve: $$\displaystyle \frac{1}{3} \cdot \frac{1}{6}$$. What would happen if we used approximations instead of the real value?
$$\displaystyle \frac{1}{3} \approx 0.3333$$ and $$\displaystyle \frac{1}{6} \approx 0.1666$$
$$\displaystyle 0.3333 \cdot 0.1666 = 0.0555{\color{red}2778}$$
$$\displaystyle \frac{1}{18} = 0.0555{\color{red}5555}$$
Could we make the approximation tighter? What happens then?
$$\displaystyle \frac{1}{3} \approx 0.33333333$$ and $$\displaystyle \frac{1}{6} \approx 0.16666666$$
$$\displaystyle 0.33333333 \cdot 0.16666666 = 0.05555555{\color{red}27777778}$$
$$\displaystyle \frac{1}{18} = 0.05555555{\color{red}55555555}$$
What happens if we use the real, exact value?
$$\displaystyle \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18}$$
Got it, I was overthinking it, thank you!
|
2019-12-15 06:00:07
|
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|
https://gamedev.stackexchange.com/questions/9382/converting-to-and-from-local-and-world-3d-coordinate-spaces
|
# Converting to and from local and world 3D coordinate spaces?
I've been following a guide I found here (http://knol.google.com/k/matrices-for-3d-applications-view-transformation) on constructing a matrix that will allow me to 3D coordinates to an object's local coordinate space, and back again. I've tried to implement these two matrices using my object's look, side, up and location vectors and it seems to be working for the first three coordinates. I'm a little confused as to what I should expect for the w coordinate.
Here are couple of examples from the print outs I've made of the matricies that are constructed. I'm passing a test vector of [9, 8, 14, 1] each time to see if I can convert both ways:
Basic example:
localize matrix:
Matrix: 0.000000 -0.000000 1.000000 0.000000
0.000000 1.000000 0.000000 0.000000
1.000000 0.000000 0.000000 0.000000
5.237297 -45.530716 11.021271 1.000000
globalize matrix:
Matrix: 0.000000 0.000000 1.000000 0.000000
-0.000000 1.000000 0.000000 0.000000
1.000000 0.000000 0.000000 0.000000
-11.021271 -45.530716 -5.237297 1.000000
test:
Vector4f(9.000000, 8.000000, 14.000000, 1.000000)
localTest:
Vector4f(14.000000, 8.000000, 9.000000, -161.812256)
worldTest:
Vector4f(9.000000, 8.000000, 14.000000, -727.491455)
More complicated example:
localize matrix:
Matrix: 0.052504 -0.000689 -0.998258 0.000000
0.052431 0.998260 0.002068 0.000000
0.997241 -0.052486 0.052486 0.000000
58.806095 2.979346 -39.396252 1.000000
globalize matrix:
Matrix: 0.052504 0.052431 0.997241 0.000000
-0.000689 0.998260 -0.052486 0.000000
-0.998258 0.002068 0.052486 0.000000
-42.413120 5.975957 -56.419727 1.000000
test:
Vector4f(9.000000, 8.000000, 14.000000, 1.000000)
localTest:
Vector4f(-13.508600, 8.486917, 9.290090, 2.542114)
worldTest:
Vector4f(9.000190, 7.993863, 13.990230, 102.057129)
As you can see in the more complicated example, the coordinates after converting both ways loose some precision, but this isn't a problem. I'm just wondering how I should deal with the last (w) coordinate? Should I just set it to 1 after performing the matrix multiplication, or does it look like I've done something wrong?
It looks like you are confusing rows and columns in your matrices, either in the way your load or store them, or when you perform the matrix×vector multiplication. The w coordinate should always yield 1 with the matrices you are using.
|
2020-01-28 07:15:06
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|
https://www.xarg.org/puzzle/codingame/benfords-law/
|
# Codingame Solution: Benford's law
Original Problem
## Goal
In this puzzle that shows a real-life phenomenon, you have to detect whether a bank account have a high risk to contain fraudulent transactions.
The Benford law is used on real anti-fraud systems to detect frauds about government accounting fraud, fiscal fraud, or even election cheating.
Your program receives in input a list of transactions, like:
+1122.85 $-10.69$-21.79 $+12.08$
You have to count how much transactions start with "1", "2", ...
For example:
"+1122.85 $" starts with 1 "-0.50" starts with 5 "$ 242.00" starts with 2
And you must output "true" if the bank account seems fraudulent, or "false" if it seems regular.
For this puzzle, your program will naively output whether a bank account seems fraudulent or not. You will be given the Benford law percentages:
1: 30.1%2: 17.6%3: 12.5%4: 9.7%5: 7.9%6: 6.7%7: 5.8%8: 5.1%9: 4.6%
An account is considered regular if the account starting numbers percentage follow these percentages, with a margin of 10 percent. Which means that you may find:
transactions starting with 1: 30.1%, so between 20.1% and 40.1%
transactions starting with 2: 17.6%, so between 7.6% and 27.6%
...
If at least one of these percentage is outside the expected range, the account is considered as fraudulent, you must then output "true".
Note that transactions may be formatted like this:
-48.12$- 5,00 €+0.99350.10-25 €$ 500.0042 £
It can be any currency.
Input
Line 1: The number of transactions N.
Next N lines: The transaction. Can be formatted like "-48.12\$", "- 5,00 €", "+0.99", "350.10", "-25 €", ...
Output
One line: "true" if the account seems fraudulent, "false" if it seems regular.
Constraints
0 < N ≤ 1000
## Solution
The only problem on this task is the corrupted input format. I wanted to extract the first digit with $$d = \left\lfloor n \cdot 10^{-\lfloor\log_{10}(n)\rfloor}\right\rfloor$$, but the most reliable way is to search for the first digit in the string.
After solving the input, it's just a comparision if the frequecy is within the required 10% interval:
let res = false;
const N = +readline();
const cnt = new Uint32Array(10);
const bnf = [
.0,
.301,
.176,
.125,
.097,
.079,
.067,
.058,
.051,
.046
];
for (let i = 0; i < N; i++) {
cnt[readline().match(/\d/)[0]]++;
}
for (let i = 1; i < 10; i++) {
if (Math.abs(bnf[i] - cnt[i] / N) > .1) {
res = true;
break;
}
}
print(res);
« Back to problem overview
|
2021-04-16 03:13:28
|
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|
https://juanmarqz.wordpress.com/cucei-maths/multilinear-algebra-cucei-2009-b/multilineal-leccion-6/
|
# multilineal lección 6
### tensor fields on a surface and its covariant derivatives
the collection $T\Sigma$ of all the tangent spaces $T_p\Sigma$ is called the tangent bundle of the surface, i.e.
$T\Sigma=\bigsqcup_pT_p\Sigma$
A vector field in a surface is a mapping $X:\Sigma\to T\Sigma$ with the condition $p\mapsto X\in T_p\Sigma$ and since $\partial_1,\partial_2$ span $T_p\Sigma$ then
$X=X^s\partial_s$
This construction determines a contravariant tensor field of rank one, which is taken as the base to ask how other tensor fields -of any rank and any variance- vary.
A contravariant tensor field of rank two in a surface is a mapping $\Sigma\to T\Sigma\otimes T\Sigma$ which attach to each point of the surface a tensor of rank two
$p\mapsto B=B^{\mu\nu}\partial_{\mu}\otimes\partial_{\nu}\in T_p\Sigma\otimes T_p\Sigma$
where the scalars $B^{\mu\nu}$ are function around the point $p$, so if one is interested in the concept of how $B$ varies in the direction $X$ then one is compelled to attach a meaning to $\nabla_XB$ for which the most natural way is by means of the Leibniz’s rule:
$\nabla_XB=\nabla_X(B^{\mu\nu}\partial_{\mu}\otimes\partial_{\nu})$
$=X(B^{\mu\nu})\partial_{\mu}\otimes\partial_{\nu}+B^{\mu\nu}\nabla_X(\partial_{\mu}\otimes\partial_{\nu})$
where you have to remember that the directional derivative of a scalar is $Xf=\langle X,{\rm{grad}}f\rangle$ and also that we can abbreviate $\partial_kf=f_{,k}$ as well.
To simplify the grasping of the idea let us illustrate first by calculating the components of
$\nabla_{\partial_k}B$
$\nabla_{\partial_k}B=\nabla_{\partial_k}(B^{\mu\nu}\partial_{\mu}\otimes\partial_{\nu})$
$\nabla_{\partial_k}B=(B^{\mu\nu})_{,k}\partial_{\mu}\otimes\partial_{\nu}+B^{\mu\nu}(\nabla_{\partial_k}\partial_{\mu})\otimes\partial_{\nu}+B^{\mu\nu}\partial_{\mu}\otimes(\nabla_{\partial_k}\partial_{\nu})$
$\nabla_{\partial_k}B=(B^{\mu\nu})_{,k}\partial_{\mu}\otimes\partial_{\nu}+B^{\mu\nu}({\Gamma^s}_{k\mu}\partial_s)\otimes\partial_{\nu}+B^{\mu\nu}\partial_{\mu}\otimes({\Gamma^s}_{k\nu}\partial_s)$
$\nabla_{\partial_k}B=[(B^{\mu\nu})_{,k}+B^{s\nu}{\Gamma^{\mu}}_{ks}+B^{\mu s}{\Gamma^{\nu}}_{ks}]\partial_{\mu}\otimes\partial_{\nu}$
With this technique it is really easy to prove that for the metric tensor $g=g^{ij}\partial_i\otimes\partial_j$ we have:
$\nabla_{\partial_k}g=0$
that is: the metric tensor is covariantly constant. In the process you gonna need to use the formula
${\Gamma^i}_{jk}=\frac{1}{2}g^{is}[g_{sk,j}+g_{js,k}-g_{jk,s}]$
Another type of phenomena that you’ll gonna face is the problem of determine how is the behaviour of tensor quantities (of any rank and any variaance) defined in a surface it is with respect to change of parameterizations or change of coordiantes.
A surface can be GPS-ed with many diverse parameterization, zum beispiel for the two-sphere $S^2$ in $\mathbb{R}^3$, we show three:
• $\left(\!\!\begin{array}{c}v\\w\end{array}\!\!\right)\stackrel{\phi_1}\to\left(\!\!\begin{array}{c}v\\\\w\\\\\sqrt{1-v^2-w^2}\end{array}\!\!\right)$
• .
• $\left(\!\!\begin{array}{c}v\\w\end{array}\!\!\right)\stackrel{\phi_2}\to\left(\!\!\begin{array}{c}\cos v\cos w\\\\\cos v\sin w\\\\\sin v\end{array}\!\!\right)$
• .
• $\left(\!\!\begin{array}{c}v\\w\end{array}\!\!\right)\stackrel{\phi_3}\to\left(\begin{array}{c}\frac{2v}{1+v^2+w^2}\\\\\frac{2w}{1+v^2+w^2}\\\\\frac{1-v^2-w^2}{1+v^2+w^2}\end{array}\right)$
for these three maps one must choose a suitable domain for $v,w$ to ensure injectivity. The maps $\lambda_{ij}={\phi_j}^{-1}\cdot{\phi_i}$ are called change of coordinates for the surface. Note that for any two of them we have ${\phi_j}\cdot\lambda_{ij}=\phi_i$.
With a device like that it is possible to describe how the tangent coordinated basis change when different parameterization are taken…
A rank 2 tensor can constructed by asking how are the components of
$\nabla_{\partial/\partial x^k}X$
$\nabla_{\partial_k}X$
$\nabla_{\partial_{ x^k}}X$
$\nabla_{k}X$
the answer is by doing Leibniz in $\nabla_{\partial_k}(X^s\partial_s)$. So…
$\nabla_{\partial_k}(X^s\partial_s)=(\nabla_{\partial_k}X^s)\partial_s+X^s(\nabla_{\partial_k}\partial_s)$
$=(\partial_kX^s)\partial_s+X^s{\Gamma^t}_{ks}\partial_t=({X^s}_{,k}+X^t{\Gamma^s}_{kt})\partial_s$
so
${X^s}_{;k}={X^s}_{,k}+X^t{\Gamma^s}_{kt}$
are the components of this tensor.
This can be considered as a measure of what so far are each other:
• the std-calculus (${\Gamma^i}_{jk}=0$) and
• the with-curvature-calculus ${\Gamma^i}_{jk}\neq0$
The ${\Gamma^i}_{jk}$ they code or detect curvature
… continuará
but meanwhile, you would like to peer (to constrast and extra grasp) on Octavian lessons at manifolds 1 and manifolds 2
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2019-07-20 23:06:21
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https://plainmath.net/45781/hot-air-balloonist-rising-vertically-with-constant-velocity-magnitude
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# A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.
A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch ${a}_{y}-t,{v}_{y}t$, and y-t graphs for the motion.
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Fasaniu
Step 1
1Concepts and Principles
1- Particle under constant acceleration: If a particle moves in a straight line with a constant acceleration ${a}_{x}$, its motion is described by kinematics equations, from which we will use the following equation:
(1)${v}_{xf}={v}_{xi}+{a}_{y}t$
(2)${x}_{f}={x}_{i}+{v}_{\xi }t+\frac{1}{2}{a}_{x}{t}^{2}$
2- The sulution to a quadration in the from: $a{x}^{2}+bx+x=0$ is given by the quadrat formula:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Step 2
2 Given Data
Let the +y-direction be upwards and the initial position of the sand bag be at ${y}_{i}=0$.
${v}_{yi}\text{(initial velocity of the bag)}=5\frac{m}{s}$
${y}_{f}\text{(position of the bag when it reaches the ground)}=-40m$
${a}_{y}\text{(acceleration due to gravity)}=-9.8\frac{m}{{s}^{2}}$
Step 3
3 Required Data
In part (a), we are asked to find the position and the velocity of the sandbag at t=0.25 s and t=1 s.
In part(b), we are asked to find the time taken by the bag to reach the ground.
In part(c), we are asked to find the speed with which the bag strikes the ground.
In part(d), we are asked to find the maximum height the bag reaches above the ground.
In part(e), we are asked to sketch task graphs for the motion of the sand bag.
Step 4
4 Solution
(a)
Model the sand bag as a particle under constant acceleration and apply Equations (1) and (2) to find the final velocity and final position of the sand bag as a function of time:
${v}_{yf}={v}_{yi}+{a}_{y}t$
${y}_{f}={y}_{i}+{v}_{yi}t+\frac{1}{2}{a}_{y}{t}^{2}$
For t=0.25 s:
${v}_{yf}=5\frac{m}{s}+\left(-9.8\frac{m}{{s}^{2}}\right)\left(0.25s\right)=2.55\frac{m}{s}$
${y}_{f}=0+\left(5\frac{m}{s}\right)\left(0.25s\right)+\frac{1}{2}\left(-9.8\frac{m}{{s}^{2}}\right){\left(0.25s\right)}^{2}=0.944m$
Therefore, at t=0.25 s, the sand bag is moving with an upward velocity of magnitude 2.55 m/s and is 40 m+0.944 m=40.9 m above the ground.
For t=0.1 s:
${v}_{yf}=5\frac{m}{s}+\left(-9.8\frac{m}{{s}^{2}}\right)\left(1s\right)=-4.8\frac{m}{s}$
${y}_{f}=0+\left(5\frac{m}{s}\right)\left(1s\right)+\frac{1}{2}\left(-9.8\frac{m}{{s}^{2}}\right){\left(1s\right)}^{2}=0.1m$
Therefore, at t=1 s, the sand bag is moving with an downward velocity of magnitude 4.8 m/s and is 40 m+0.1 m=40.1 m above the ground.
Rita Miller
Step 5
(b)
The sand bag strikes the ground when its final position is ${y}_{f}=-40m$. From Equation (2):
${y}_{f}={y}_{i}+{v}_{yit}+\frac{1}{2}{a}_{y}{t}^{2}$
$-40m=0+\left(5\frac{m}{s}\right)t+\frac{1}{2}\left(-9.8\frac{m}{{s}^{2}}\right){t}^{2}$
$-40m=\left(5\frac{m}{s}\right)t-\left(4.9\frac{m}{{s}^{2}}\right){t}^{2}$
$\left(4.9m{s}^{2}\right){t}^{2}-\left(5\frac{m}{s}\right)t-40m=0$
Solve this equation for t using the quadratic formula from Equation (3), for :
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$t=\frac{-\left(-5\frac{m}{s}\right)±\sqrt{{\left(-5\frac{m}{s}\right)}^{2}-4\left(4.9\frac{m}{{s}^{2}}\right)\left(-40m\right)}}{2\left(4.9\frac{m}{{s}^{2}}\right)}$
=3.41 s or -2.39 s
A negative value for time is phisically unacceptable. Therefore, the time take for the sand bag to reach the ground is t=3.41 s.
Step 6
(c)
The velocity of the bag before it atrikes the ground is fround from Equation (l):
${v}_{yf}={v}_{yi}+{a}_{y}t$
Where t is the time taken for the sand bag to reach the ground which we found in part (b). Subsitute numerical values:
${v}_{yf}=5\frac{m}{s}+\left(-9.8\frac{m}{{s}^{2}}\right)\left(3.41s\right)=-28.4\frac{m}{s}$
The negative values indicates that the direction of the velocity ia downwads.
Step 7
(d)
The bag reaches maximum heightwhen its final velosity ${v}_{yf}=0$. From Equation (2), we find the time taken by the bag to reach this maximum height:
${v}_{yf}={v}_{yi}+{a}_{y}t$
$t=\frac{5\frac{m}{s}}{9.8\frac{m}{{s}^{2}}}=0.51s$
They by using Equation (2), we can find the maximum height:
$y={y}_{0}+{v}_{yi}t+\frac{1}{2}{a}_{y}{t}^{2}$
$y={0}_{5\frac{m}{s}}\left(0.51s\right)+\frac{1}{2}\left(-9.8\frac{m}{{s}^{2}}\right){\left(0.51s\right)}^{2}$
y=1.28 m
Therefore, the maximum height of the sand bag is 40 m+1.28 m=41.3 m.
Don Sumner
Step 8
(e)
The ${a}_{y}-t$ graph is a horithontal line since the acceleration is constant at ${a}_{y}=-9.8m/{s}^{2}$
Step 9
The velosity oa the sand bag is a function of time from Eguation (l) as follows:
${v}_{y}={v}_{yi}+{a}_{y}t=5m/s-\left(9.8m/{s}^{2}\right)t$
Therefore, the ${v}_{y}-t$ graphs looks like:
Step 10
The y - t graph look like (let y=0 at the ground level):
Result
(a)At t=0.25 s, the sand bag is moving with an upward velocity of magnitude 2.55 m/s and is 40.9 m above the ground. At t=1 s, the sand bag is moving with an downwoard velosity of magnitude 4.8 m/s and is 40.1 m above the ground.
(b)t=3.41 s
(c)${v}_{yt}=-28.4$ m/s
(d)The maximum height of the sand is 41.3 m.
(e)Click for graphs
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2022-09-30 18:52:36
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http://danielfurth.com/?q=node/7
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## Information transfer at multiple time-scales
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Summary:
Preliminary remark on the separate presentations held by Randy Gallistel, Greg Jensen, David Freestone during the last meetings of the Information, learning and temporal cognition seminars at Peter Balsam’s lab. I here argue that the concept of a trial is not a theoretical meaningful unit and can not even be sustained as a basis for methods of statistical evaluation.
An alternative 'psychophysical' approach is presented where time is conjectured as a fractal set. It can be shown that the scalar property of interval timing arise from properties of the wavelet function that represents physical time in the time-scale plane. Transfer entropy is introduced to define an asymmetric measurement of mutual information that deals with the third-variable problem, e.g. the influence of deprivation/satiation status.
Transfer entropy and wavelet analysis are combined to examine information transfer in a matching experiment. In agreement with previous studies on matching (Gallistel et al. (2007) 'Is Matching Innate?') it is shown that trial-by-trial (!) feed-back mechanisms are not necessary for matching to occur, and that the majority of information transfer flows from the response trajectory to the reinforcement trajectory at short to medium scales (trial-by-trial level) which is in line with a purely trial-by-trial feed-forward model that makes small-sample estimates of outcomes. It is shown that the sensitivity parameter of the matching law is determined by the extent that receiving a consecutive number of reinforcements (or lack thereof) cues the organism that a new schedule is introduced and that it is time to switch investment ratio (i.e. feedback mechanisms at higher time-scales)
# Introduction
The general question that has occured to most of us in these seminars is how to identify and characterize structure and hidden dependencies between components of a complex dynamical system such as the interaction between environment and the organisms behavior.
In it’s most general formulation the problem has been to characterize the set of functions that maps between the set of responses to that of the set of environmental events, such as a set of stimuli. Both the response and the stimulus is an event extended with time.
But what is an event? The answer to this question depend on whom you ask, the probability theorist will answer that an event is a set of outcomes (measurable subsets of the sample space, constituting a σ-algebra over the sample space itself) to which a probability is assigned. Of importance is that the σ-algebra is subset of the sample space that is closed under complementation and countable unions, an important topic I will return to in a later post (For now; note that a closed set restricts the semantics beyond what is reasonable to assume. The problem is that human everyday-reasoning on topics such as temporal contingency often violates the Law of Excluded Middle, $\large \dpi{80} {\color{Red} \lnot(p\wedge \lnot p)}$, as well as the Law of Double Negation $\large \dpi{80} {\color{Red} p = \lnot \lnot p}$. In future blog posts I will point to the epistemological problems associated with such a representation, especially if one aims at understanding memory).
The physicist may answer that an event indicates a physical situation or occurrence, located at a specific point in space and time (or space-time). The physicist may then go on to measure the distance between points in space and time through the use of appropriate metric tensor. Although, an important subject, physics provides little knowledge to deal with our problem. The physicist is interested in finding structure of our physical reality from measurement provided by our senses. The psychophysicist is interested in the opposite problem: to deduce the structure that constitute our sensory experience from physical measurements.
In my opinion the best answers are provided by philosophers. Bertrand Russell repeatedly pointed out an obvious problem for anyone who tries to build a theory of time perception grounded in the work of spatial perception. We begin with the simple observation that change is an ubiquitous property of our experience. Maybe one would like to claim that there can be change in things and change in events. Things change because they gain and lose properties, this is trivial. However, there cannot be change in events, because events have their properties essentially.
Objects are extended in time: if we encounter an object at a specific point in time we are not experiencing a temporal part of the object. In contrast, we often encounter spatial parts of very large objects as is the case when Crusoe is washed-up on the beach of an island. A process on the other hand is extended in time but we do not at any single moment encounter the whole process only parts of it. Since processes have parts that are themselves events it must be the case that events are characterized by change, in fact events are changes. It should thus be clear that asking if an event changes, that is often done in change-point analysis, is not always a meaningful question. An event ontology is crucial if we identify change as events. If changes are events then we can not include pure spatial variation as changes since we can not maintain the position of relationalism.
Within the framework of change-point analysis learning is conjectured merely as a process of change. The process is fully determined by its parts: events. These events occur in time. More importantly these events are thought to occur on a 'trial'. Therefore the concept of a trial needs to be investigated.
# Trials, sessions and the time-scale domain
The concept of a 'trial' is a convention made by the scientific community and by the scientist that designed the experiment. However, the ontological status of the 'trial' is comparable to any other social convention that splits continuous time into periods of repeated events like days, months, weeks etc. If we define the period of these repeated events by the unit of another temporal measurement we speak of different temporal frequencies.
The question then arise why one should settle for one specific temporal frequency over another. For example, sometimes learning theorists use the trial as the fundamental frequency of analysis and sometimes they use blocks of trials of any arbitrary length, mostly with the purpose to reduce what is regarded as high frequency 'noise' in the learning trajectory. Sometimes the training is clustered into the structure of trials embedded into sessions. The practice of arbitrary cluster trials together into blocks and sessions is thus highly prevalent in the literature, however, one could just as well argue that there exist some unmeasured period of high frequent events between trials, although this has never been put into practice by contemporary learning theorist because of the failure to see that the unit of a 'trial' is just one among an infinite number of possible frequencies of analysis. This fundamental problem has been acknowledge by Gallistel, and can best be expressed in my favorite passage from Memory and the Computational Brain "...life does not consist of trials" (Gallistel & King, 2009).
Nevertheless, what remains to be done is to take the leap of faith to get rid of the concept of a trail altogether, not just remove it as a unit in computational models of the organisms behavior but also ban the concept from the statistical methods by which we evaluate learning performance as well (e.g. change-point analysis). In what follows I will examine what exactly happens when we take this leap of faith, I assure you that the results will be quite informative.
Figure 1.A division of the time-scale plane by discrete-time wavelet series decomposition using ideal octave filter banks. The current time-scale plane, or temporal map, is of length 64 observations with 6 number of scales
# Wavelet filter banks, a non-probabilistic alternative to pacemaker models
All current models of time perception heavily rely on probability representations. However, as I remarked in the introduction probability representations (in fact all boolean algebras) are highly undesirable because of the structure of human contingency judgements. Therefore I will develop a completely deterministic account of time perception where probability theory have no room, n.b: boolean algebra is not (yet) rejected in the current model. We begin by defining the view that physical time is fractal by nature of the measurement procedure, a procedure that results in the time-scale space.
The time-scale space consists of a sequence of nested closed subspaces or time bundles that emit sensory signals $\large \dpi{80} \varphi$ for the brain to encode:
$\large \dpi{80} \cdots V_{2} \subset V_{1} \subset V_{0} \subset V_{-1} \subset V_{-2} \cdots$
With the following properties:
(iii) Time-scale invariance
$\large \dpi{80} f(t) \in V_{m} \rightarrow f(2^{m}t) \in V_{0}$
(iv) Translational ('shift') invariance
$\large \dpi{80} \all n \in \mathbb{Z}: f(t) \in V_{0} \rightarrow f(t-n) \in V_{0}$
(v) Normalized and orthogonal (orthonormal) basis:
$\large \dpi{80} \varphi(t) \in V_{0},\quad \{\varphi(t-n) | n \in \mathbb{Z} \}$
At present I will not comment on methods to verify or reject the empirical status of these axioms. However, the axiom of time-scale invariance is noteworthy. The axiom can be found in the introduction of John Gibbon's seminal paper 'Scalar Timing and Semi-Markov Chains in Free-Operant Avoidance' as equation 1a on page 111. Nevertheless, apart from that specific equation my approach and Gibbon's are very different. Gibbon conjectured time-scale invariance as a fundamental property of a probability distribution, hence randomness was an intricate part of the theory. In the theory that I will outline randomness is not a part of our perception of time. My final aim is a complete geometric representation and deterministic theory of time-perception, not a probabilistic one. At the heart of the theory is the concept of a temporal map represented as the time-scale plane in Figure 1.
The existence of an orthonormal basis in axiom (v) gives us the scaling function, $\large \dpi{80} \varphi(t)$. Together with time-scale invariance and translational invariance we can define the function or sensory time signal $\large \dpi{80} \varphi_{m, n}$ for any integers $\large \dpi{80} n$ and $\large \dpi{80} m$:
$\large \dpi{80} \varphi_{m, n}(t) = 2^{-\frac{m}{2}}\varphi(2^{-m}t-n),$
The factor $\large \dpi{80} 2^{-\frac{m}{2}}$ exists to normalize the norm $\large \dpi{80} \|\varphi_{m, n}\| = 1$ regardless of the choice of $\large \dpi{80} m$ and $\large \dpi{80} n$. It can be proven (Daubechies, ) that when (i)-(v) are satisfied for the sequence of subspaces (time bundles) there exists an orthonormal basis for $\large \dpi{80} L_{2}(\mathbb{R})$:
$\large \dpi{80} \psi_{m, n}(t) = 2^{-\frac{m}{2}}\psi(2^{-m}t-n),$
$\large \dpi{80} \{\psi_{m, n} \}$ serve as an orthonormal basis for another space: the psychometric time-scale space $\large \dpi{80} W_{m}$ which is the orthogonal complement of $\large \dpi{80} V_{m}$ in $\large \dpi{80} V_{m-1}$: $\large \dpi{80} V_{m-1} = V_{m} \oplus W_{m}$. The function $\large \dpi{80} \psi(t)$ is the wavelet function (or the brain's temporal encoding function). Because $\large \dpi{80} \varphi(t) \in V_{0} \subset V_{-1}$ and the functions of the form $\large \dpi{80} 2^{\frac{1}{2}}\varphi(2t-n)$ spans the space $\large \dpi{80} V_{m-1}$ the brain can decompose the sensory temporal signal into $\large \dpi{80} n$ number of components:
$\large \dpi{80} \varphi(t) =\sqrt{2} \displaystyle\sum_{n\in\mathbb{Z}}h_{0}(n) \varphi(2t-n),$
Where $\large \dpi{80} h_{0}(n)$ are normalized coefficients, $\large \dpi{80} \|h_{0}(n)\| = 1$. The preliminary representation has been outlined. next we will examine possible mechanism that could have physical parallels in the brain.
Down- and upsampling
The sensory signals that the brain receives contain too much information to be thoroughly sampled. Downsampling and upsampling are two signal processing techniques which are quite handy when a signal needs to be compressed but with out to much information loss. If we have a discrete input signal $\large \dpi{80} x_{n}$ then the downsampling of $\large \dpi{80} x_{n}$ results in a transformed signal $\large \dpi{80} y$. This can be written as:
$\large \dpi{80} y_{n} = x_{Nn}$
Where $\large \dpi{80} N$ is an integer. If we have a signal that transmit at a rate of 1000 Hz then if $\large \dpi{80} N = 1000$ we have downsampled the signal to 1 Hz. If the opposite is done we speak of upsampling the signal by inserting a bunch of zeros between the sampled intervals:
$\large \dpi{80} y_{n} = x_{n/M} n = kM$
Here $\large \dpi{80} k$ is an integer that gives the new sampling speed. Downsampling and upsampling is performed by a z-transformation of the original sequence of real or complex numbers into a complex frequency-domain representation. Because downsampling of a signal can induce aliasing it is recommended that a filter is used before the signal is downsampled. In a filter bank this is done by an array of different bandpass filters that separate different frequency components.
Two-channel filter bank
The two-channel filter bank transforms a signal, $\large \dpi{80} x_{n}$, into the complex frequency domain, $\large \dpi{80} X(z)$, and produces a perfect reconstruction $\large \dpi{80} \widehat{X}(z)$, where $\large \dpi{80} X(z)$ is the z-transform of $\large \dpi{80} x_{n}$ with $\large \dpi{80} z$ being a complex number. In matrix notation the perfect reconstruction of a two-channel filter bank is given by:
$\large \dpi{80} \hat{X}(z) = \frac{1}{2}\begin{bmatrix} F_{0}(z)\\ F_{1}(z) \end{bmatrix}^{T} \begin{bmatrix} H_{0}(z) & H_{0}(-z) \\ H_{1}(z)& H_{1}(-z) \end{bmatrix}\begin{bmatrix} X(z)\\ X(-z) \end{bmatrix}$
Where $\large \dpi{80} H_{1}(z)$ is a filter which passes only high frequency information further whereas $\large \dpi{80} H_{0}(z)$ is the corresponding low-pass filter. The lowest order high-pass filter gives us the threshold that establish a lower time-scale limit of our temporal perception, equivalent to the pacemaker rate in probabilistic models such as SET. The corresponding $\large \dpi{80} F(z)$ functions denotes the output functions of the system that reconstructs the signal after upsampling, analogous to the decision and memory components in SET. For a schematic of a multi-channel filter bank see Figure 2..
Figure 2.Schematic of a multi-channel filter bank (the neural machinery which can encode fractal time, V, into a wavelet representation in the time-scale plane, W, see Figure 1.). The filter bank can be extended by adding more and more components to filter the signal to the highest time-scale we can comprehend by experience. The mechanism by which the area of the time-scale plane (i.e. temporal map) is realized are open for discussion.
Scalar variability
SET is grounded on the observation that performance in interval timing obeys Weber's law, that the Just-noticeable difference of an interval is proportional to the interval's duration multiplied by a constant factor, the weber coefficient:
It was reasoned that a pacemaker timer with the arrival times drawn from a poisson distribution can not achieve the Weber law. It was hypothesized that small processing glitches in other components such as memory and decision making ad the desired normally distributed timing variability that follows Weber's law.
If one adopts a set of filter banks instead of a single pacemaker timer Weber's law follows naturally. Infact the channel capacity of the filter bank pacemaker system can be completely determined.
According to the axiom of translational scale invariance we see that the wavelet uses a varying width of the difference limen (i.e. the time window or j.n.d.), hence we have a fixed number of cycles $\large \dpi{80} \Delta fT=1$, increasing or decreasing the difference limen will affect the time resolution in the opposite direction in the time domain this is because $\large \dpi{80} \Delta f \Delta t$ is constant.
Therefore, $\large \dpi{80} \Delta f \propto f$ which gives us:
$\large \dpi{80} \frac{\Delta f}{f}=w$
Which is Weber's law in the frequency domain. It can further be shown (upon request I can show this in another blog post by the aid of the Hausdorff-Young inequality) that the channel capacity of the wavelet has the following limit:
$\large \dpi{80} \Delta f \Delta t \geq \frac{1}{2}$
This inequality establish why we experience instantaneous moments.
Inconclusion: the wavelet representation exhibit all the necessary psychophysical properties of interval timing and it can be physically realizable in the brain by, for example, biological mechanism that mimics the behavior of filter banks. Before implementing my view of temporal cognition to data I need to clearify why the concept of mutual information employed by Balsam and Gallistel is not a desirable measurement of contingency: since it is symmetric and subjected to the third-variable problem. This argument will be elaborated in the following section. In addition, the measurement of entropy transfer (Schreiber, 2000) is suggested to replace the use of mutual information.
# Information Transfer
For a discrete and stationary signal $\large \dpi{80} X(t)$, let $\large \dpi{80} p(x)$ be the time independent probability to observe symbol $\large \dpi{80} x$ from alphabet $\large \dpi{80} A$, e.g. $\large \dpi{80} x \in \{1, 2, ..., S\}$ where $\large \dpi{80} S$ denotes the number of symbols in the alphabet. Without taking possible correlations into account the average number of bits needed to optimally encode the signal $\large \dpi{80} X(t)$ is given by the Shannon entropy:
$\large \dpi{80} H_{X} = -\displaystyle\sum\limits_{x=1}^S p(x) \log_{2} p(x), \quad 0 \leq H_{X} \leq \log_{2}S,$
The Shannon entropy includes only the case where we have one possible probability distribution. In the settings of Pavlovian conditioning we however deals with multiple variables such as different stimuli or responses and their dependence. To resolve this we introduce the Kullback-Leibler divergence which is a measure of the expected number of extra bits required to code samples generated by $\large \dpi{80} p(\cdot )$ if we use $\large \dpi{80} q(\cdot )$ instead.
$\large \dpi{80} D_{KL}(P || Q) = \displaystyle\sum\limits_{x=1}^S p(x) \log_{2} \left( \frac{p(x)}{q(x)} \right),$
The Kullback-Leiber divergence resembles a metric distance since it is never negative, $\large \dpi{80} D_{KL}(P || Q) \geqslant 0$, nevertheless it can not be said to be a metric distance since it is asymmetric and does not obey the triangle inequality, properties which the norm derived metrics per definition obey. Hence forth I will simply refer to the Kullback-Leiber divergence as the relative entropy and we hereafter interpret the quantity as a measure of the information gain about the random variable $\large \dpi{80} X$ that is achieved if $\large \dpi{80} p$ can be used instead of $\large \dpi{80} q$.
We can use the relative entropy to measure how much information gain we can get about a random variable $\large \dpi{80} X$ (or random variable $\large \dpi{80} Y$) if we assume that $\large \dpi{80} X$ and $\large \dpi{80} Y$ are dependent instead of independent events.
$\large \dpi{80} I(X; Y) = \displaystyle\sum\limits_{x \in X}^S \displaystyle\sum\limits_{y \in Y}^S p(x, y) \log_{2} \left( \frac{p(x, y)}{p_{1}(x) p_{2}(y)} \right),$
Where $\large \dpi{80} p_{1}(x)$ and $\large \dpi{80} p_{2}(y)$ are the marginal distributions. If the random variables $\large \dpi{80} X$ and $\large \dpi{80} Y$ are independent then $\large \dpi{80} p_{1}(x) \cdot p_{2}(y) = p(x, y)$ and the ratio $\large \dpi{80} \log_{2}(p(x, y)/p_{1}(x) p_{2}(y))$ is 0, hence nothhing is gained by assuming that $\large \dpi{80} X$ and $\large \dpi{80} Y$ are dependent. The relative entropy function that measures the dependencies between random variables is known as the mutual information. One should note that the mutual information is a symmetric measure, it doesn't tell us if the information is gained from variable $\large \dpi{80} X$ or $\large \dpi{80} Y$.
Practically this means that if $\large \dpi{80} Y$ is some measurement of reinforcement delivery and $\large \dpi{80} X$ is some measurement of response elicitation then we cannot discriminate between if information flows from the delivery of reinforcement to the response elicitation (feedback system) or if the information flows purely from the response elicitation to the reinforcement (feedforward system). It has been suggested that these problems can be resolved by providing temporal structure to the measurement of mutual information in a rather ad hoc fashion by using the lagged mutual information;
$\large \dpi{80} I(X(t); Y(t-dt)) = \displaystyle\sum p(x_{t}, y_{t-dt}) \log_{2} \left( \frac{p(x_{t}, y_{t-dt})}{p(x) p(y)} \right),$
Where $\large \dpi{80} d$ denotes the time delay, or lag, between observing $\large \dpi{80} X(t)$ and $\large \dpi{80} Y(t)$. However, the lagged mutual information is not a suitable measurement for examining information flow between two or more time series in the context of learning. The reason is that the lagged mutual information is subjected to the third-variable problem, i.e. any possible relationship between X and Y is not due to the fact that Y provides input to X, or vice versa, but rather there exist a third, unmeasured variable, that affects the relationship between X and Y. Examples of these unmeasured variables could be the animal's status of satiation or deprivation in appetitive conditioning, or simply the general levels of arousal or attentional resources of the organism. In summary: all variables that clearly changes as a function of time into the conditioning protocol but are not necessarily related to the knowledge obtained by the organism are to be considered as 'third-variables'.
What we rather need is an asymmetric entropy measurement which can seperate statistical dependencies originating in the input signal Y from those dependencies which derives from the common history of both signals. This goal can be attained by considering the system as a k order Markov process:
$\large \dpi{80} p(x_{t+dt} | x_{1}, x_{2}, ... , x_{k}) = p(x_{t+dt} | x_{1}, x_{2}, ... , x_{k}, y_{1}, y_{2}, ... , y_{l})$
Where $\large \dpi{80} p( \cdot | \cdot )$ denotes the transition state probability and $\large \dpi{80} k$ and $\large \dpi{80} l$ denotes the length of the delay embedding vectors, or `block length' of the system's memory for Y and X, respectively. The average number of bits needed to encode one additional state of the system if all previous states are known can be formulated by the relative entropy between assuming either $\large \dpi{80} k +1$ or $\large \dpi{80} k$ dimensional delay vectors:
$\large \dpi{80} h_{X}(k) = -\sum p(x_{1}, x_{2}, ... , x_{k+1} ) \log_{2} \left( \frac{p(x_{1}, x_{2}, ... , x_{k}, x_{k+1} ) }{p(x_{1}, x_{2}, ... , x_{k} ) } \right)$
It is easy to see that this is merely the difference between the Shannon entropies of blocks with either length $\large \dpi{80} k +1$ or $\large \dpi{80} k$, that is to say the above equation can be rewritten as:
$\large \dpi{80} h_{X}(k) = H_{X} (k + 1) - H_{X} (k), \quad 0 \leq h_{x}(k) \leq H_{X}$
We will refer to $\large \dpi{80} h_{X}(k)$ as the entropy rate of the system. However, in the limit, as $\large \dpi{80} \lim_{k \to \infty} h_{X}(k) = h_{X}$, which I'll, for pedagogical reasons, refer to as the entropy of the source. In case of periodic and determinate signals $\large \dpi{80} h_{X} = 0$, and for purely stochastic signals $\large \dpi{80} h_{X} = H_{X}$. In practice, the block probabilities, e.g. $\large \dpi{80} p(x_{1}, x_{2}, ... , x_{k} )$, are estimated by the relative frequencies: $\large \dpi{80} p(x_{1}, x_{2}, ... , x_{k} ) = N^{-1} \cdot n(x_{1}, x_{2}, ... , x_{k} )$, where $\large \dpi{80} n(x_{1}, x_{2}, ... , x_{k} )$ denotes the number of occurrences of the sequence $\large \dpi{80} (x_{1}, x_{2}, ... , x_{k} )$ in the embedded series of length $\large \dpi{80} N$.
Obviously we have some severe estimation problems here since the limit $\large \dpi{80} k \rightarrow \infty$ is empirically impossible to achieve, hence we must search for a sufficiently large value of $\large \dpi{80} k$. However, as $\large \dpi{80} k$ increases the number of possible ordered samples with replacement (permutations with repetition) increases $\large \dpi{80} S^{k + l}$ where $\large \dpi{80} S$ is the number of symbols in the alphabet, if $\large \dpi{80} k$ is to large for the particular finite data set to support then the estimate of $\large \dpi{80} h_{X}$ will tend to 0 due to the finite sample effect. Hence, the amount of data needed is proportional to $\large \dpi{80} S^{k + l}$.
This estimation problem is the reason why the proposed method of analysis can not be used in small samples without careful adjustment of parameters. It is also the reason why my work on compiling this document got delayed so much. I would really appreciate any ideas or possible strategies (Bayesian?) to deal with the estimation problem in finite samples.
A simple, and rather ugly strategy, is to reduce the number of symbols, $\large \dpi{80} S$, in the alphabet by coarse graining the signal. One can also set the memory length of the hypothesized input signal $\large \dpi{80} l = 1$ and then choose $\large \dpi{80} k$ to be as large as possible. If this is done then we can estimate the entropy of the source extended to two time series, $\large \dpi{80} X(t)$ and $\large \dpi{80} Y(t)$. We will refer to this quantity as the {transfer entropy or information flow, $\large \dpi{80} T_{Y\rightarrow X}$. This quantity is asymmetric, i.e. in general $\large \dpi{80} T_{Y\rightarrow X} \neq T_{X\rightarrow Y}$. The transfer entropy $\large \dpi{80} T_{Y \rightarrow X}$, or information flow from $\large \dpi{80} Y$ into $\large \dpi{80} X$, is defined as the information gained about future observations of the dependent signal $\large \dpi{80} X(t + 1)$ from past joint observations of $\large \dpi{80} X$ and $\large \dpi{80} Y$ minus the information gained about future observations $\large \dpi{80} X(t+1)$ from past states of $\large \dpi{80} X$ only:
$\large \dpi{80} T_{Y \rightarrow X}(k, l) = \sum p(x_{1}, x_{2}, ... , x_{k+1}, y_{1}, ..., y_{l} ) \log_{2} \left( \frac{p( x_{k + 1} | x_{1}, x_{2}, ... , x_{k}, y_{1}, ..., y_{l}) }{p(x_{k + 1} | x_{1}, x_{2}, ... , x_{k},) } \right)$
Equation can be simplified to the block entropy form:
$\large \dpi{80} T_{Y \rightarrow X}(k, l) = h_{X}(k) - h_{XY}(k, l) , \quad 0 \leq T_{Y \rightarrow X} (k, l) \leq H_{X}$
One should be careful in choosing the value of $\large \dpi{80} k$, if it is too large to support the data then $\large \dpi{80} T_{Y \rightarrow X}$ is subjected to finite sample effects, if it is too small then we run the risk of misinterpret the information contained in past states of both variables as information flow from Y to X. Hence, if $\large \dpi{80} k$ is to small the transfer entropy measurement will not provide any additional information that the lagged mutual information not already reveals. In the next section we will combine what we now know about information transfer with our knowledge about wavelets to compute the information transfer at multiple time scales. In all analyses that follow the parameters are set to S = 3, k = 4, and l = 1. All wavelet representations are performed by using the Continious Wavelet Transform of the Morlet wavelet function with a central frequency of 5 radians per sample. The Morlet is used since it offers an excellent time versus frequency trade-off. When I am at Gallistel's lab I can demonstrate some principles and guidlines on how to estimate wavelet functions from data (a topic that I will not deal with here).
# Application to simulated data
Lets define an one-step autoregressive process, AR(1):
$\large \dpi{80} X(t + 1) = 0.7X(t) + 0.7\textup{cos}(0.3t) + \varepsilon_{X}$
First note that the prediction ahead combines 0.7 of the output from the past observation and combines it with a cosine frequency component together with signal specific random noise drawn from a normal distribution with mean zero and 0.2 in variance.
More importantly, let the AR(1) process $\large \dpi{80} X(t)$ serve as input to another AR(1) process $\large \dpi{80} Y(t)$:
$\large \dpi{80} Y(t + 1) = 0.7Y(t) + 0.7X(t) + 0.7\textup{sin}(0.06t) + \varepsilon_{Y}$
The reader should be comfortable with these two processes, their structure and dependencies otherwise the following simulation will not make sense. We begin by generating 5000 samples of the processes and then restrict analysis to the final 3000 observations. A small time window illustrating the processes can be seen in Figure 3., note that at this level we clearly see that some synchronicity is going on between the two trajectories, but we can not determine in which direction the coupling goes.
Figure 3.The two time-series X and Y which one is the input and which is the output?
The limitation of traditionally lagged measures such as the lagged mutual information can be seen in Figure 4.. The figure illustrates the Cross Correlation Function of the two time-series (a bivariate generalization of the autocorrelation function). Although we can clearly identify the phases where the processes synchronize we can not delineate in which direction they are coupled.
Figure 4.The failure of measures that try to recreate the asymmetry of temporal measurements by the use of ad hoc solutions can be illustrated by the cross correlation function (ccf) which is the bivariate generalization of the autocovariance function (ACF)
In contrast, the information transfer from Y to X (blue) and X to Y (black) depicted in Figure 5. clearly illustrate that the information flows from X into Y and not the other way around. In Figure 5. the transfer entropy is computed for different coupling parameters c between 0 and 1, notice that for the flow of information from X to Y the transmitted information increases monotonically until it reaches the channel's capacity. This simulation demonstrates that even for low coupled systems (c = 0.1~0.3) transfer entropy is sensitive enough.
Figure 5.Information transfer between the two time-series and as a function of the coupling parameter that directs the input of X to Y
However, both AR(1) processes contain frequency components. X includes a cosine wave with a frequency of 0.3 whereas the Y process both received a proportion of the lagged cosine term by it's coupling with X but also had a sine wave with a frequency of 0.6, a since wave that strictly originating from Y alone.
Any robust method of analysis not only needs to be able to extract information on coupled systems independent from a common source, but it also needs to able to decompose the information transfer into it's periodic elements. By using the Continious Wavelet Transform (CWT) of both time-series and it's representation into the time-scale plane on a total of 87 detail coefficients (see Figure 2. and Figure 1.) we can compute the information transfer between X and Y on each of the 87 time-scales. By definition an AR(1) process transfer information on a trial-by-trial basis (i.e. t+1), therefore we would expect a significant proportion of information transfer at the lowest time scale, 1.
In addition, because X with it's cosine 0.3 frequency oscillation partly serves as input into Y we should see information transfer at scales near the equivalent frequency of 0.3 (to understand how scales can be translated into frequencies the concept of central frequency needs to be understood, email me if you have any concerns about this). More importantly if multi-scale analysis of information transfer is sensitive we should not see any transfer around 0.06 which is the frequency of inherent oscillations of Y.
All of these three hypothesis, i.e. (1) transfer from X into Y at time-scale one (2) transfer of the cosine term at the 0.3 frequency (3) no transfer of the sine term around 0.06, can be seen in Figure 5 for the reader to verify. In Figure 5 information transfer is plotted as a function of time-scale represented on the abscissa, on the opposite axis the corresponding frequency is shown (note the logarithmic scale induced by the properties of the time-scale plane, compare with Figure 1.)
Figure 6.The wavelet representation of the information flow seen in Figure 5. Note that most information is transmitted in the lowest scale (1-5 Hz), more importantly we see a peak around 30 Hz which corresponds to the frequency of the cosine term in X. Lastly, note the absence of an equal amount of information transfered at the low frequency (0.06 Hz), hence the wavelet representation of information flow can distinguish between the relative contributions at different time-scales
# Implications for 'The Matching Law'
Since Richard Herrstein's original 1961 paper the attention received by what is today known as 'the matching law' has steadily increased, not just within the narrow field of behavioral analysis but within neuroscience and economics as well.
In 1974 Baum described the Generalized Matching Law that could incorporate deviations from strict matching by a sensitivity parameter, s, and a response bias, b. In the case of two concurrent operadum the Generalized Matching Law can be stated as follows:
$\large \dpi{80} \frac{B_{1}}{B_{2}}=b\left( \frac{R_{1}}{R_2}\right )^S$
In this section time-scale analysis of entropy transfer will be combined with multilevel linear modeling to examine the heterogeneity in matching seen among participants. Often this inter-individual variability is attributed to what is refered to as 'dynamical' or 'non-asymptotic' behavior and many studies of matching not interested in this 'dynamical' behavior therefore make certain that all data analyzed is obtained from sessions of well-trained organisms. This practice of ignoring valuable information is quite absurd, it is similar to standing at night under a street light with a flash light in your hand and directing the flash light's beam onto your own feet, a place that is obviously already illuminated. To their defence there are a souring interest in understanding this dynamical or non-asymptotic behavior. However, this approach is staggering with strange and silly concepts borrowed from physics/chemistry such as 'molecular', 'molar', 'momentum' together with a modeling jargong that compares animal behavior to the behavior of pistons, basins etc. as if an act of perception or remembering shared anything more than superficially to the mechanical behavior of a steam engine.
We begin by rearranging the behavioral ratio for the ith session with a specific concurrent schedule into it's log-transform. We then express the matching law as a function between random variables with Gaussian distributions. The mean is expressed as a linear function of a response bias parameter that is free to vary between individuals and the log-transformed reinforcement ratio multiplied by the sensitivity parameter (that also varies between individuals):
$\large \dpi{80} \log_{2} \left( \frac{[B_{1}]_{i}}{[B_{2}]_{i}} \right) \sim N\left( s_{j[i]} \log_{2}\left( \frac{[R_{1}]_{i}}{[R_2]_{i}}\right ) + \log_{2}(b)_{j[i]} \quad , \quad \sigma^{2} \right)$
As said before, the free parameters are allowed to vary between individuals j according to a normal distribution, formally:
$\large \dpi{80} \binom{\log_{2}(b)_{j}}{s_{j}} \sim N\left( \binom{\mu_{\log_{2}(b)}}{\mu_{s}}, \binom{\sigma^{2}_{\log_{2}(b)} \quad \rho \sigma_{\log_{2}(b)}\sigma_{s}}{\rho \sigma_{\log_{2}(b)}\sigma_{s} \quad \sigma^{2}_{s}} \right)$
The variability in parameters is determined by the between-individual correlation parameter $\large \dpi{80} \rho$. The above model will be applied to a instrumental social learning paradigm that mimics aspect of traditional free operant VI concurrent schedules.
The data to-be-analyzed is from an experiment on social dominance and aversive learning, see Figure 7.. In this simple paradigm the task is to participate in a ball thossing game, one can not win or loose. During the game if the participant throws the ball to one of the confederates there is a probability that the other confederate will disapprove of this and make a frown. At 1500 ms from the onset of the frown there is a possibility that a mild 100 ms electrotactile stimulation is delivered to the participant's armwrist. The Reinforcement Learning Module (RLM) of the paradigm is programmed by me for PC with the aid of DirectX, it was originally modelled after Eisenberger et al's (2009) paradigm for social exclusion.
For a short video that shows excerpt from the software (everything is made by me) see the following link: http://vimeo.com/21850959. Note that this is not a demonstration of the Reinforcement Learning Module (RLM) but of a paradigm that simulates social exclusion (see the article by Eisenberger, et al. 2003 in Science).
The current paradigm is far from an ideal paradigm to evaluate important aspects of matching since the paradigm comes with a rigid trial structure (one can not throw a ball that one does have). nevertheless, it is the only data set I have access to.
Figure 7.Example where the participant (here depicted as Paul Ekman) throws the ball to the male confederate, the female confederate frowns and 'delivers' a shock to the participant. The Reinforcement Learning Module (RLM), using frowning faces and electric shocks as US. The faces are animated from the Cohn-Kanade AU-Coded Facial Expression Database
All participants were exposed to the same order of programmed schedules occuring in blocks of 40 trials within in the same session without any interruption or cue. No change over delay (COD) was used since COD is not a natural part of the social interaction of interest. The probability schedule for a frowning face and electric shock was [ {0.7:0.3}, {0.3:0.7}, {0.9:0.1}, {0.15:0.85}, {0.85:0.15}, {0.1:0.9} ]. Each block consisted of 40 trials (throws made by the participant), resulting in 6 blocks with three ratios counter-balanced between confederates.
The matching plots from a sample of 25 individuals are shown in Figure 8. the reader can verify that some participants approximate the generalized matching law such that the fraction of allocated passes to a confederate is proportional to the fraction of obtained punishment delivered by that participant although note the variability in slopes (sensitivity parameter) between participants.
Figure 8.Sample of 25 individuals that participated in a matching law experiment on social dominance. Data points are obtained log-ratio of frequencies (behavior, B, or reinforcement, R) in each block, see text.
The statistical strategy is to enter the proportion of feedforward information transfer, $\large \dpi{80} \frac{{}T_{B_{1} \rightarrow R_{1}}}{T_{B_{1} \rightarrow R_{1}}+T_{R_{1} \rightarrow B_{1}}}$, as a subject-level predictor. $\large \dpi{80} {T_{B_{1} \rightarrow R_{1}}}$ denotes the entropy transfer from behavioral option 1 to reinforcement option 1, and $\large \dpi{80} T_{R_{1} \rightarrow B_{1}}$ denotes information flow in the opposite direction (reinforcement or feedback driven). By entering $\large \dpi{80} \frac{{}T_{B_{1} \rightarrow R_{1}}}{T_{B_{1} \rightarrow R_{1}}+T_{R_{1} \rightarrow B_{1}}}$, as a subject-level predictor. $\large \dpi{80} {T_{B_{1} \rightarrow R_{1}}}$ for each scale into the random effect model we can examine when the subject-predictor contributes significantly to the null model that has no information flow parameter that varies between individuals. The time-scale that reduces most variance in sensitivity parameters across participant is the time-scale that contributes the the hetereogenity seen in Figure 8. The null model has already been described, the alternative model that includes the subject-predictor for a specific time-scale is seen bellow:
$\large \dpi{80} \binom{\log_{2}(b)_{j}}{s_{j}} \sim N\left( \binom{\gamma^{log_{2}(b))}_{0}+\gamma^{log_{2}(b))}_{1}\mu_{j}}{\gamma^{s}_{0}+\gamma^{s}_{1}\mu_{j}}, \binom{\sigma^{2}_{\log_{2}(b)} \quad \rho \sigma_{\log_{2}(b)}\sigma_{s}}{\rho \sigma_{\log_{2}(b)}\sigma_{s} \quad \sigma^{2}_{s}} \right)$
The mean of the random effects have been replaced by a linear equation that describes the impact of on response bias $\large \dpi{80} \gamma^{log_{2}(b))}_{0}+\gamma^{log_{2}(b))}_{1}\mu_{j}$ and sensitivity $\large \dpi{80} \gamma^{s}+\gamma^{s}_{1}\mu_{j}$ across participants. Where $\large \dpi{80} \mu_{j} = \frac{{}T_{B_{1} \rightarrow R_{1}}}{T_{B_{1} \rightarrow R_{1}}+T_{R_{1} \rightarrow B_{1}}}$ for the jth participant.
Both the response onset and the delivery of an electric shock (option 1) were simultaneously sampled and offline downsampled for computational convenience to a sampling rate of 1 sample per 100 ms. The time-series were then transformed into time-scale space represented by 84 different time-scales. The resulting transfer entropy is seen in Figure 9.. The filled black circles and lines gives the transfer of information from the delivery of the US whereas the hollow circles gives the transfer of information from the response to the US.
Figure 9.Information transfer averaged across the 25 participants on each time-scale. Errorbars denotes +/-1 SEM.
Three properties are of importance in Figure 9.:
(1) The majority of information flows from the response to the US at small to mediate time-scales. This is equivalent to the behavior of a feed-forward model that makes episodic small-sample estimates (Gallistel et al., 2007)
(2) At time-scales 19 to 34 we see that the information that flows from the US to the response increases above that of the transfer from the response to the US. Scales 19 to 34 is equivalent the the average frequency of when the participant enters a new block and the schedule changes. The increase in information flow from the US to the response can be described by a feedback system at the time-scales of blocks where an apparent streak of shocks implicitly signals that the scheduled has changed.
(3) At higher time-scales the entropy transfer between the two series either converges or the total amount of information transmitted drops.
Before running the statistical model that will examine the time-scales that can explain variability in the sensitivity parameter, lets plot the information transfer in Figure 9. as a function of sensitivity/slope. This can be seen in the heatmap bellow:
Figure 10.Each thick on the y-axis denotes a single participant with it's corresponding sensitivity parameter. The x-axis shows the time-scales. The contours and colors in the heatmap shows the the proportion of information transfer that flows from the response to the US
On the y-axis each thick denotes a single participant with it's fitted sensitivity parameter (obtained by single subject OLS regression). The x-axis shows us the time-scale and the contours and colors in the heatmap shows us the measurement which I'll use as a between subject-predictor, i.e. the proportion of information transfer that flows from the response to the US.
In Figure 10. notice the following observations: (1) no systematic variability at short to medium time-scales between different slope values indicating that the episodic sampling behavior is not related to the sensitivity parameter. (2) at time-scales around 30 there seems to be the case that participants with a low sensitivity do not use the same amount of information from the US as highly sensitive individuals do, the dark blue area (3) green colors represents equilibrium (50-50).
Infact observation (2) is crucial. Because when we run our model we find that the majority of the variability in the sensitivity parameter can be accounted for by variability in information transfer at the rather narrow time scale 30 to 32, where it accounts for between 40-80% of the variability in slopes. It goes without saying that at these time-scales the model using information transfer as a fix effect is better than a model without, LL-ratio = 9.58, ChiSq = 9.59, df = 2, p< .01. The interaction of information transfer on sensitivity is depicted in Figure 11..
Figure 11.Individual participant's reinforcement sensitivity as a function of information transfer for time-scale 32. The error bars denotes individual subjects 95% CI. The regression lines CI is 95% as well
To grasp the temporal precision by which we can target the entropy transfer related to variability in reinforcement sensitivity we can superimpose these time-scales on the heatmap in Figure 11.. The three time-scales that contributes to the variability in the sensitivity parameter is highlighted as a short strip in the heatmap shown in Figure 12.:
Figure 12.The highlighted region shows the time-scale where variability in information transfer explains variability in reinforcement sensitivity
All scripts and functions used to compute information transfer will soon be available on this website, first for R then I will port the script to MatLab.
Yours Sincerely,
Daniel
### Blog
2012-10-26 10:34
It is very simple to inspect the source code of any command from a package in R by simply typing in the console the package...
2012-09-23 03:33
In this blog post I will especially look at the scaling problems that arise in the design of...
2011-04-12 20:51
Summary:
Preliminary remark on the separate presentations held by Randy Gallistel, Greg Jensen, David...
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2017-07-25 20:29:03
|
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https://puzzling.stackexchange.com/questions/12934/review-this-now/14397
|
# Review this, now!
What's happening is very strange,
I belong to another StackExchange!
I'm indeed off-topic here in Puzzling,
It isn't always clear what I'm saying.
Even though I'm often too broad,
Review it, something is flawed.
I'm just suggesting an edit,
A low quality job, ain't it?
For new users I'm a big trouble
A duplicate isn't always a double.
Your current work you shall reopen
Flag here, flag there, it's all broken!
Oh, don't forget to add a comment,
late it's not, takes just a moment!
What am I?
Hint #1:
I'm sometimes cause and consequence of inappropriate language"
Hint #2:
Upvoted or downvoted, doesn't matter, I generally appear in the bottom.
Hint #3:
I notify you to review something.
• sounds interesting !!!!.. . – coDe murDerer Apr 29 '15 at 12:38
• I'm voting to close this question because it's hilarious. – Ian MacDonald Apr 29 '15 at 14:12
• @IanMacDonald It's intended to be a funny "background", but this is a serious riddle, with a serious answer. – leoll2 Apr 29 '15 at 15:27
• @leoll2 I think someone is going around today giving out free down votes. Just ignore it as this seems to be a good riddle! – Mark N Apr 29 '15 at 16:01
• @leoll2 Mark N and i had the same problem on one of his questions. seems like someone is grumpy – Vincent Apr 29 '15 at 16:16
## 10 Answers
You are an
IDE Debugger
What's happening is very strange, I belong to another StackExchange!
Typically people would ask about problems found in the debugger on StackOverflow.
I'm indeed off-topic here in Puzzling, It isn't always clear what I'm saying.
Sometimes the error given is fairly vague and unhelpful.
Even though I'm often too broad, Review it, something is flawed.
The vague error message may be hard to interpret, but there is definitely something wrong that must be looked over.
I'm just suggesting an edit, A low quality job, ain't it?
The message is telling you to change something in the code.
For new users I'm a big trouble A duplicate isn't always a double.
Beginner programmers may have a hard time figuring out what the error is, especially when they've gotten the same error twice with a different cause.
Your current work you shall reopen Flag here, flag there, it's all broken!
Go back to your code, you need to check all the marked areas for bugs. Sometimes a single typo, like an extra or missing }, will cause huge amounts of code to be highlighted as 'broken'.
Oh, don't forget to add a comment, late it's not, takes just a moment!
When you fix the bug, add a comment to the code to make sure it is clear to others (or your future self) why the code is the way it is.
• Welcome to Puzzling! This is a very good answer for your first post. I hope that you can maintain a quality as high as this in many of your future posts here! – Mark N May 6 '15 at 12:57
• Oh oh! You're very close! The correct answer was actually "an error message", but you've interpreted all the lines correctly, so I accept this as best answer :-) – leoll2 May 6 '15 at 13:35
• @MarkN Thanks! I have some experience using StackOverflow from my work account and have been lurking around puzzling for a couple months now. – Root Vegetable May 6 '15 at 15:16
• @leoll2 That's pretty clever. I like the misdirection, making it seem like the answer is related to SE itself. I've been think everyone else's answers were so close for a week now. – Root Vegetable May 6 '15 at 15:17
• @RootVegetable That's exactly what I tried to do, I'm glad you appreciated it :) – leoll2 May 6 '15 at 15:25
You are a
Closed question
What's happening is very strange, I belong to another StackExchange!
Questions can be closed if they belong to a different SE.
I'm indeed off-topic here in Puzzling,
Off-topic questions are closed.
It isn't always clear what I'm saying.
Questions that are unclear are closed.
Even though I'm often too broad,
Questions that are too broad are closed.
Review it, something is flawed.
If a question is closed, it can be flagged to be reopened, in which case it enters the review queue.
I'm just suggesting an edit,
Closed questions can be edited.
A low quality job, ain't it?
Low quality questions can be closed.
For new users I'm a big trouble
Closed questions can lead to a question ban, especially if you're a new user.
A duplicate isn't always a double.
Sometimes questions closed as duplicates aren't actually duplicates.
Your current work you shall reopen, Flag here, flag there, it's all broken!
If you can improve a closed question, you can flag to reopen it.
Oh, don't forget to add a comment, late it's not, takes just a moment!
People can still comment on closed questions.
• I think you nailed it :) – A.D. Apr 29 '15 at 14:58
• Nah, it's not a closed question! Good try! – leoll2 Apr 29 '15 at 15:25
• @leoll2, i was going to answer codereview.SE but then i saw pacoverflow's answer which i thought was the correct one, should i still bother with my answer? – Vincent Apr 29 '15 at 16:15
• @VincentAdvocaat Nah, codereview isn't correct! – leoll2 Apr 29 '15 at 16:16
• @leoll2 if this isn't the correct answer, your puzzle is too imprecise. Fix the puzzle, or mark this correct and make a new puzzle. – Ian MacDonald May 6 '15 at 12:10
I think I got it
you are a bug :)
What's happening is very strange,
I belong to another StackExchange!
when you have something weird on your code, you go on stackoverflow
I'm indeed off-topic here in Puzzling,
It isn't always clear what I'm saying.
you don't post questions about bugs here. And sometime, it is hard to explain something we don't understand.
Even though I'm often too broad,
Review it, something is flawed.
sometime the question is too vague to find it and usually we ask a review on a code to find it
I'm just suggesting an edit,
A low quality job, ain't it?
answers are usually simple edits because people do not proof read
For new users I'm a big trouble
A duplicate isn't always a double.
when you are new, it is hard to determine if the bug is a duplicate of something existing or not
Your current work you shall reopen
Flag here, flag there, it's all broken!
To isolate it, you have to open your source code and add markers to isolate it until you find the broken part
Oh, don't forget to add a comment,
late it's not, takes just a moment!
and always comment your bug so you don't do it again
Thanks to Spacemonkey, for inspiration
### Edit
If you are not a bug, you are a
bug report
• Very very close! The answer isn't a bug, but you're very close! – leoll2 Apr 30 '15 at 4:58
• @leoll2 I updated my answer with a new answer – A.D. Apr 30 '15 at 11:02
My answer is :
Coding (review)
What's happening is very strange,
I belong to another StackExchange!
Stackoverflowwwwwwwwww
I'm indeed off-topic here in Puzzling,
It isn't always clear what I'm saying.
Obviously doesn't belong in Puzzling and to most people here (maybe) asking code questions would be unclear
Even though I'm often too broad,
Review it, something is flawed.
Could refer to code or comments in code, allthough making it generic isnt bad, only in some cases.
I'm just suggesting an edit,
A low quality job, ain't it?
A lot of code problems come from low quality work
For new users I'm a big trouble
A duplicate isn't always a double.
Make sure you don't duplicate code. That being said a duplicate doesnt need to be an exact copy of something else (double), if it DOES the same thing, its a duplicate.
Your current work you shall reopen
Flag here, flag there, it's all broken!
If you did break code you'll have to go re-work it because it'll be breaking all over the place (flags representing errors in some coding environments)
Oh, don't forget to add a comment,
late it's not, takes just a moment!
The importance of commenting code.
• Very very good answer, still not the exact one! – leoll2 Apr 29 '15 at 19:28
• That's a good one. But it is probably "source code"! – A.D. Apr 29 '15 at 19:52
Let's try another answer (get some more activity back here). This is largely inspired by your recent post. Are you:
A Lateral-Thinking Question
What's happening is very strange,
I belong to another StackExchange!
There is some thought of moving lateral-thinking to Area 51.
I'm indeed off-topic here in Puzzling,
It isn't always clear what I'm saying.
Some consider lateral-thinking questions to be off topic as they aren't always good puzzles.
Even though I'm often too broad,
Review it, something is flawed.
The main issue with lateral-thinking is that they are often so broad that there are so many answers it isn't even funny.
I'm just suggesting an edit,
A low quality job, ain't it?
Most lateral-thinking could be improved with a nice edit to narrow down the possibilities.
For new users I'm a big trouble
A duplicate isn't always a double.
A lot of new users seem to try lateral-thinking and get some downvotes. Also, a lot of lateral-thinking puzzles seem to be similar.
Your current work you shall reopen
Flag here, flag there, it's all broken!
They can be reopened if they are fixed. A lot get flagged for deletion.
Oh, don't forget to add a comment,
late it's not, takes just a moment!
Most could be improved if they were commented on where the issue is.
• Good attempt, but no! I'll add another hint soon – leoll2 May 6 '15 at 11:24
You are:
The review queue!
I belong to another stack exchange
you belong to all the stack exchanges. They all have the review queue.
For everything else:
These are all line items or pieces of a review-able item.
• Nice try, but it's not! – leoll2 Apr 29 '15 at 15:56
• Darn, thought I had it. – Brian Robbins Apr 29 '15 at 15:57
Inspired by (and very close to) pacoverflow's answer, but hopefully different enough:
A flagged question
What's happening is very strange, I belong to another StackExchange!
Questions can be flagged to be migrated to the appropriate section.
I'm indeed off-topic here in Puzzling,
Off topic questions can be flagged for deletion. Also, many questions that would be considered "flaggable" (I take artistic license) on other sites would be fine here (answer known, not informative, etc.)
It isn't always clear what I'm saying.
Questions that are too broad are often flagged.
Even though I'm often too broad,
Same as before.
Review it, something is flawed.
Flagged questions are reviewed to address some problem.
I'm just suggesting an edit, A low quality job, ain't it?
Many flags are due to a need for an edit, possibly for low quality questions.
For new users I'm a big trouble
New users don't want to be flagged and kicked. Also, they can scare new users away.
A duplicate isn't always a double.
Sometimes flagged if it might be a duplicate, or to see if it is.
Your current work you shall reopen
Closed questions can be flagged for review and potential reopening.
Flag here, flag there, it's all broken!
Flags all around! You can only flag so many items in a time period.
Oh, don't forget to add a comment, late it's not, takes just a moment!
Users appreciate a comment as to why their questions are flagged, or you need to add a comment when you flag for what should be looked at.
• I like your plain answer, but the answer isn't this! Keep thinking! – leoll2 Apr 29 '15 at 18:00
Is it
a vote?
What's happening is very strange, I belong to another StackExchange! I'm indeed off-topic here in Puzzling, It isn't always clear what I'm saying.
In all of these cases you could down vote. Although, here also a flag would be the right answer.
Even though I'm often too broad, Review it, something is flawed. I'm just suggesting an edit, A low quality job, ain't it? For new users I'm a big trouble A duplicate isn't always a double.
Those definitely point more towards a flag. But a lot of people just down vote again.
Your current work you shall reopen Flag here, flag there, it's all broken!
As you mention a flag, it's not the flag. It's the vote to reopen a mistakenly flagged and closed question.
Oh, don't forget to add a comment, late it's not, takes just a moment!
If you vote (too often) without commenting, you are generally reminded by any SE site, that you should comment, too.
• Good attempt, but it's not a vote! – leoll2 May 6 '15 at 13:31
Since you are talking about all kinds of actions you can do here I'll take a guess and think you are
A question on the meta stackexchange http://meta.puzzling.stackexchange.com/
• Good attempt, but not the correct answer. There are many hints inside the text, find them! – leoll2 Apr 29 '15 at 13:19
EDITED ANSWER
Yet I'm not sure, but guess is
Spam Questions
?
What's happening is very strange, I belong to another StackExchange!
it's strange to identify it.
I'm indeed off-topic here in Puzzling, It isn't always clear what I'm saying.
Spam are off topics, which are not clear
Even though I'm often too broad, Review it, something is flawed.
spam message found broad sometimes, also flawed
For new users I'm a big trouble
its trouble for new users
Oh, don't forget to add a comment, late it's not, takes just a moment!
usually we add comment or flagged it as spam question.
• Good attempt, but it's not spam – leoll2 Apr 30 '15 at 5:02
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2020-02-18 04:49:30
|
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|
https://www.stellar.org/lumens/
|
One lumen (XLM) is a unit of digital currency, like a bitcoin.
## The Fundamentals
### What are lumens?
One lumen (XLM) is a unit of digital currency, like a bitcoin.
Lumens are the native asset of the Stellar network.
Native means that lumens are built into the network. Asset is how the network refers to an item of value that is stored on the ledger.
One lumen is a unit of digital currency, like a bitcoin.
While you can’t hold a lumen in your hand, they are essential to the Stellar network—they contribute to the ability to move money around the world and to conduct transactions between different currencies quickly and securely.
### Where did lumens come from?
In 2014 the Stellar network launched with 100 billion stellars, the original name of the network’s native asset.
In 2015, with the launch of the upgraded network, the name of the native asset changed from stellar to lumen to distinguish it from 1) the Stellar network itself and 2) Stellar.org, the nonprofit organization that contributes to development of the network.
### Why does the Stellar network need a native asset?
The Stellar network offers the innovative features of a shared public ledger on a distributed, global database—often referred to as blockchain technology. The Stellar network’s built-in currency, the lumen, serves two purposes:
First, lumens play a small anti-spam role.
Lumens are needed for transaction fees and minimum balances on accounts on the Stellar network in order to prevent people from overwhelming the network and to aid in prioritization.
Each transaction has a minor fee—0.00001 lumens—associated with it. This fee prevents users with malicious intentions from flooding the network (otherwise known as a DoS attack). Lumens serve as a security measure that mitigates DoS attacks that attempt to generate large numbers of transactions or consume large amounts of space in the ledger.
Similarly, the Stellar network requires all accounts to hold a minimum balance of 0.5 lumens. This requirement incentivizes users to declutter the ledger by eliminating abandoned accounts, thereby that ensuring that all accounts are likely to have economic utility on the network.
Second, lumens may facilitate multi-currency transactions.
Lumens sometimes facilitate trades between pairs of currencies between which there is not a large direct market, acting as a bridge. This function is possible when there is a liquid market between the lumen and each currency involved.
### What is XLM?
XLM is shorthand for lumen. Most currencies have 3-letter codes (USD, EUR, AUD, BTC) as an international standard.
The technical term for these shorthand codes is ISO 4217. Learn more about ISO codes.
### Who governs lumen creation?
Lumen supply is determined by fixed, protocol-level rules. The number of lumens created at genesis was 100 billion. Every year, there is a 1% inflation rate. New lumens cannot be generated arbitrarily by anyone.
While somebody could theoretically push a proposal to change the protocol to change the rule around lumen creation, validators on our network are very unlikely to accept and ratify it. The decentralization of the protocol prevents us from unilaterally controlling any changes — if we tried to force a change, people would lose trust in Stellar.org, refuse to accept the changes, and keep using the original protocol with the original rules. Therefore, we will never do it.
## Getting and Using Lumens
### Do I need lumens to use the Stellar network?
The Stellar network software is open-source and available to the general public.
Users will need to have lumens to pay transaction fees and make accounts on the live Stellar network.
There are two special values used to calculate fees:
1. The base fee (currently 100 stroops) is used in transaction fees.
2. The base reserve (currently 0.5 XLM) is used in minimum account balances.
Transaction Fee
The fee for a transaction is the number of operations the transaction contains multiplied by the base fee, which is 100 stroops (0.00001 XLM).
Transaction Fee = (# of operations × base fee)
For example, a transaction that allows trust on an account’s trustline (operation 1) and sends a payment to that account (operation 2) would have a fee of 2 × base fee = 200 stroops.
Extremely Low Cost
By design, transactions and accounts on Stellar are very low cost. As of April 2018 at $,$1.00 USD will cover about 400,000 transactions. Each account currently costs 0.5 lumens, or approximately \$0.15.
### Where can I get lumens?
Lumens are available on several licensed exchanges.
Please be aware of the risk associated with all digital currencies and make sure to safeguard your private keys to prevent loss or theft. While lumens will always have utility value for transactions and accounts in the Stellar network, the price of lumens denominated in fiat currencies may fluctuate and result in a complete loss of value.
Digital currencies are very innovative, but not insured by regulatory bodies such as the FDIC.
Before purchasing lumens, consult the Consumer Advisory brief by the Consumer Protection Finance Board (CPFB) on the potential risks associated with digital currencies. You should also read this educational document from the CFTC: “Customer Advisory: Understand the Risks of Virtual Currency Trading.”
### What is the price of lumens?
Please consult the available exchanges for the latest lumen price.
### What is the base fee? Who profits from those lumens?
There is a nominal fee, referred to as a base fee, associated with each operation in a transaction. The sender of the transaction incurs the fee.
The fee functions as a deterrent: Though nominal, it discourages users with malicious intentions from flooding the network (otherwise known as a DoS attack).
The base fee is currently set to .00001 XLM. The fee will increase if the system suspects an account is submitting transactions with the malicious intent to bring down the network.
No one profits from the base fee. The ledger collects those funds and redistributes them in the process of inflation (see the next question, below).
### How does inflation work with lumens? Why is there inflation?
The Stellar network has a built-in, fixed inflation mechanism. New lumens are added to the network at the rate of 1% each year. The network also collects a base fee for each operation in a transaction. The funds from base fees are added to the inflation pool.
As a balancing measure for the ecosystem, anyone who holds lumens can vote on where the funds in this pool go. Each week, the protocol distributes these lumens to any account that gets over .05% of the votes from other accounts on the network. Read more about inflation.
## Lumen Distribution and Giveaways
### How are lumens distributed?
Stellar.org intends to distribute the initial 100 billion lumens as follows:
• 50% given away to individuals:
• A small (50-300 XLM) amount of lumens will be given to each unique individual who signs up through an invitation link. You may get an invitation from one of our partners or by attending one of our meetups.
• 25% given away to partners:
• Lumens will be given to businesses, governments, institutions, or nonprofit organizations that contribute to the growth and adoption of the Stellar ecosystem.
• 20% given away to bitcoin and XRP holders:
• Stellar.org distributed lumens to holders of bitcoin and XRP (19% for bitcoin holders and 1% for XRP holders). This giveaway was completed in two rounds: one round was completed in October 2016 and another was completed in August 2017. Stellar.org does not intend to conduct another round of lumen giveaways for bitcoin or XRP holders.
• 5% reserved for Stellar.org operational expenses
For more information on this breakdown, please refer to Stellar.org’s mandate.
### Why is Stellar.org giving away lumens?
To achieve a more inclusive digital economy. Per our mission, Stellar.org works to connect people to low-cost financial services. Giving away lumens for free is an invitation to communities to design the services they need. Our hope is to eventually have global citizens own and use XLM, in both developing and developed economies.
To expand the reach of the network. The availability and active use of lumens on the network will increase the network’s utility by many orders of magnitude.
As of January 2018, we have now given away over 8 billion lumens.
### Why should I buy lumens if you are giving them away for free?
As an integrator or anchor (an integrator that is trusted to accept deposits and honor withdrawals, usually a licensed money service business), you may need a large quantity of lumens to cover base fees for transactions on the network, to create and use sophisticated smart contracts, and for account creation. As a money transfer operator or market maker, you may buy and use lumens as a bridge asset to facilitate trades of different asset trade pairs on the network.
As a developer or entrepreneur, you may buy and use lumens to learn about cryptocurrency, experiment with operations, and build innovative applications on the Stellar Network.
In the future, after we have given away all the lumens—which will happen over the next 10 years—everyone will need to procure lumens from third parties.
### What is the lumen auction and how does it work?
The Stellar.org mandate reserves 5% of the original 100 billion lumens to support the operations of Stellar.org.
Stellar.org covers its own operational costs via its own lumens in the following ways:
• Auction: We periodically offer portions of the reserved lumens at auction. We refer to this public process as the lumen auction. We will only use licensed intermediaries, e.g. ItBit, to conduct the auction. As an ethical safeguard, no one formally associated with Stellar.org—e.g., Stellar.org employees, consultants, or board members—will participate in the auction.
• Batches: We may sell larger batches of the reserved lumens to private parties interested in supporting the Stellar.org mission. We will use ItBit as an intermediary.
If you’re interested in acquiring a larger batch of lumens from the Stellar.org reserve, contact the foundation directly.
### How many lumens have been given away?
For up-to-date statistics on the number of lumens given away, see the Stellar.org dashboard.
|
2018-10-18 20:21:10
|
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https://www.actuaries.digital/2016/09/28/the-importance-of-projections-in-developing-retirement-strategies/
|
The Actuaries Institute’s Superannuation Projections and Disclosure (SPD) Subcommittee designed a projection model to estimate the income that assets would support during retirement for a number of case studies. Here, the SPD Subcommittee reports on its findings.
Experts are thick on the ground these days, sometimes with free advice that can prove costly later on. And it seems experts are particularly fond of advising retirees and those about to soak up the sun on weekdays about how and when to spend their money. One strand of free advice at the moment involves recommending that retirees should spend a bit more, or indeed a lot more, to secure a higher pension to take into account impending changes to the asset test.
Understandably, these changes from January 1 next year have many retirees and those close to retiring thinking hard about whether they should change their financial arrangements. To be more specific, after this date the age pension reduces by $78 per year for each$1,000 of non-home assets over certain thresholds. At first glance, this looks like you’d have to earn over 7.8% on the extra $1,000 or you’d be better off without the extra$1,000 of assets.
The Actuaries Institute cautions that retirees destined to live to a ripe old age should think twice before accepting some of the advice recently aired on this topic. Indeed, this advice ignores the fact that a partial age pension entitlement generally increases throughout retirement as assets reduce. The SPD Subcommittee have designed a projection model to estimate the income that assets would support during retirement for a number of case studies.
#### A Case Study
The SPD considered a number of scenarios. They were based on two single females (Anne and Barbara) who own their own homes. Their only asset, other than their home, was a balance in an allocated pension. It was assumed that the allocated pension was the only source of income for both women and that they continued to live in their own homes throughout their retirement. The modelling also assumed that the required level of income each year (the combination of the age pension and income from the allocated pension) would be equal to the annual expenditure of ASFA’s comfortable lifestyle for a single person indexed to CPI.
In this case study, we examine one of the scenarios considered.
This scenario assumes the two women plan to retire at age 65 on 1 January 2017 with potentially identical superannuation assets of $450,000. To highlight the long-term impact of spending some of the superannuation assets before retirement, we assumed that Anne increases her spending before 1 July 2017 so as to reduce her retirement assets and receive a higher age pension than Barbara, who decides to save her money. The additional spending was assumed to reduce Anne’s final retirement benefit available on 1 January 2017 to$250,000.
Chart 1 below provides a year-by-year projection of the incomes of these two individuals to age 100.
Chart 1 – Total income if retiring at age 65
Note: all projected values have been discounted to Today’s Dollars at the rate of Wage Inflation.
Assumptions Net investment return on allocated pension assets – 6.5% pa compound
Wage inflation – 3.5% pa compound
Price Inflation – 2.5% pa compound
Increase in desired income – Price inflation
Increase in age pension rate – Wage inflation
Increase in age pension asset test thresholds – Price inflation
The green and purple lines show the total income received in Today’s Dollars. The blue and red lines show the annual amount of age pension received.
It can be seen that the aged pension paid to Anne in the early years is higher because the pension assets she owns do not reduce her age pension. However, because Anne has less pension assets she exhausts her assets by age 84, after which she must live on the age pension or use her home to generate additional income.
Barbara, however, at age 84 still has pension assets and therefore receives a higher level of income than Anne for the rest of her retirement. Also Barbara’s total income received is equal to or greater than her desired income level throughout retirement. She will also maintain a balance in her allocated pension throughout retirement and can continue without resorting to using her own home to generate additional income.
An examination of the projected asset values is also instructive. Chart 2 below shows the value of their pension fund assets at the end of each year during retirement.
Chart 2 – Asset Values if retiring at age 65
Note: all projected values have been discounted to Today’s Dollars at the rate of Wage Inflation.
Barbara has significantly greater pension fund assets throughout retirement. This provides added flexibility in her spending pattern. It also allows for aged care costs or bequests in later age. The additional assets also provide a buffer if the net investment earnings are less than the 6.5% we have assumed. Importantly, the fact that Anne receives a larger age pension in the early retirement years does not indicate what strategy results in the best long-term outcome.
The example and related discussion above highlight the significant challenges involved in retirement income modelling and strategy choice. Such tasks cannot be properly addressed through conclusions based upon calculations of a retiree’s first year age pension and allocated pension income entitlements.
The interaction of the many pieces of Australia’s retirement income system is complex. It includes assets and income test rules for the pension, minimum superannuation assets withdrawal requirements and the interaction of other factors such as inflation and investment returns. Any conclusions based on only considering the income generated in the first year after retirement are liable to be incorrect. Only the output of a year-by-year projection can clearly show how these factors interact throughout a person’s retirement.
Retirees must make decisions about spending capital over time. Ideally, these should allow for a sensible assessment of future cash flow. Year-by- year projections throughout retirement are vital to capture the dynamic nature of the age pension rules as asset values change. However, this is just the start. Given each retiree has an unknown lifespan and faces unknown investment returns, people have valid concerns about outliving their capital. Models like this one can be extended to assess a full distribution of likely outcomes and take into account the retiree’s asset mix and even health status. This allows people to make informed decisions that meet their required levels of certainty.
A longer article which considers all the scenarios examined by the SPD Subcommittee is also available. If a copy of the longer article is required (or if there are any questions on the material contained in this article) please contact Andrew Boal, Convenor of the Institute’s Superannuation Practice Committee, or David Bell, Chief Executive Officer of the Actuaries Institute.
The Institue’s SPD Subcommittee authored this article. The members of the Subcommittee are:
Colin Grenfell, Convenor Glenn Langton
Thomas Sneddon, Secretary David Orford
Bill Buttler Richard Starkey
Esther Conway Ray Stevens
Ian Fryer Brnic Van Wyk
Jim Hennington
CPD: Actuaries Institute Members can claim two CPD points for every hour of reading articles on Actuaries Digital.
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2023-03-27 20:28:35
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https://en.wikipedia.org/wiki/Guarded_Command_Language
|
# Guarded Command Language
The Guarded Command Language (GCL) is a language defined by Edsger Dijkstra for predicate transformer semantics.[1] It combines programming concepts in a compact way, before the program is written in some practical programming language. Its simplicity makes proving the correctness of programs easier, using Hoare logic.
## Guarded command
The guarded command is the most important element of the guarded command language. In a guarded command, just as the name says, the command is "guarded". The guard is a proposition, which must be true before the statement is executed. At the start of that statement's execution, one may assume the guard to be true. Also, if the guard is false, the statement will not be executed. The use of guarded commands makes it easier to prove the program meets the specification. The statement is often another guarded command.
### Syntax
A guarded command is a statement of the form G ${\displaystyle \rightarrow }$ S, where
• G is a proposition, called the guard
• S is a statement
If G is true, the guarded command may be written simply S.
### Semantics
At the moment G is encountered in a calculation, it is evaluated.
• If G is true, execute S
• If G is false, look at the context to decide what to do (in any case, do not execute S)
## Skip and Abort
Skip and Abort are very simple as well as important statements in the guarded command language. Abort is the undefined instruction: do anything. The abort statement does not even need to terminate. It is used to describe the program when formulating a proof, in which case the proof usually fails. Skip is the empty instruction: do nothing. It is used in the program itself, when the syntax requires a statement, but the programmer does not want the machine to change states.
### Syntax
skip
abort
### Semantics
• Skip: do nothing
• Abort: do anything
## Assignment
Assigns values to variables.
### Syntax
v := E
or
v0, v1, ..., vn := E0, E1, ..., En
where
• v are program variables
• E are expressions of the same data type as their corresponding variables
## Concatenation
Statements are separated by one semicolon (;)
## Selection: if
The selection (often called the "conditional statement" or "if statement") is a list of guarded commands, of which one is chosen to execute. If more than one guard is true, one statement is nondeterministically chosen to be executed. If none of the guards are true, the result is undefined. Because at least one of the guards must be true, the empty statement "skip" is often needed.
### Syntax
if G0 ${\displaystyle \rightarrow }$ S0
| G1 ${\displaystyle \rightarrow }$ S1
...
| Gn ${\displaystyle \rightarrow }$ Sn
fi
### Semantics
Upon execution of a selection all guards are evaluated. If none of the guards evaluates to true then execution of the selection aborts, otherwise one of the guards that has the value true is chosen non-deterministically and the corresponding statement is executed.[2]
### Examples
#### Simple
In pseudocode:
if a < b then c := True
else c := False
In guarded command language:
if a < b ${\displaystyle \rightarrow }$ c := true
| a ≥ b ${\displaystyle \rightarrow }$ c := false
fi
#### Use of Skip
In pseudocode:
if error = True then x := 0
In guarded command language:
if error = true ${\displaystyle \rightarrow }$ x := 0
| error = false ${\displaystyle \rightarrow }$ skip
fi
If the second guard is omitted and error = False, the result is abort.
#### More guards true
if a ≥ b ${\displaystyle \rightarrow }$ max := a
| b ≥ a ${\displaystyle \rightarrow }$ max := b
fi
If a = b, either a or b is chosen as the new value for the maximum, with equal results. However, someone implementing this, may find that one is easier or faster than the other. Since there is no difference to the programmer, he is free to implement either way.
## Repetition: do
The repetition executes the guarded commands repeatedly until none of the guards are true. Usually there is only one guard.
### Syntax
do G0 ${\displaystyle \rightarrow }$ S0
| G1 ${\displaystyle \rightarrow }$ S1
...
| Gn ${\displaystyle \rightarrow }$ Sn
od
### Semantics
Upon execution of a repetition all guards are evaluated. If all guards evaluate to false then skip is executed. Otherwise one of the guards that has value true is chosen non-deterministically and the corresponding statement is executed after which the repetition is executed again.
### Examples
#### Original Euclidean algorithm
a, b := A, B;
do a < b ${\displaystyle \rightarrow }$ b := b - a
| b < a ${\displaystyle \rightarrow }$ a := a - b
od
This repetition ends when a = b, in which case a and b hold the greatest common divisor of A and B.
#### Extended Euclidean algorithm
a, b, x, y, u, v := A, B, 1, 0, 0, 1;
do b ≠ 0 ${\displaystyle \rightarrow }$
q, r := a div b, a mod b;
a, b, x, y, u, v := b, r, u, v, x - q*u, y - q*v
od
This repetition ends when b = 0, in which case the variables hold the solution to Bézout's identity: xA + yB = gcd(A,B) .
## Applications
### Programs correct by construction
Generalizing the observational congruence of Guarded Commands into a lattice has led to Refinement Calculus.[3] This has been mechanized in Formal Methods like B-Method that allow one to formally derive programs from their specifications.
### Asynchronous Circuits
Guarded commands are suitable for Quasi Delay Insensitive circuit design because the repetition allows arbitrary relative delays for the selection of different commands. In this application, a logic gate driving a node y in the circuit consists of two guarded commands, as follows:
PullDownGuard ${\displaystyle \rightarrow }$ y := 0
PullUpGuard ${\displaystyle \rightarrow }$ y := 1
PullDownGuard and PullUpGuard here are functions of the logic gate's inputs, which describe when the gate pulls the output down or up, respectively. Unlike classical circuit evaluation models, the repetition for a set of guarded commands (corresponding to an asynchronous circuit) can accurately describe all possible dynamic behaviors of that circuit. Depending on the model one is willing to live with for the electrical circuit elements, additional restrictions on the guarded commands may be necessary for a guarded-command description to be entirely satisfactory. Common restrictions include stability, non-interference, and absence of self-invalidating commands.[4]
### Model Checking
Guarded commands are used within the Promela programming language, which is used by the SPIN model checker. SPIN verifies correct operation of concurrent software applications.
### Other
The Perl module Commands::Guarded implements a deterministic, rectifying variant on Dijkstra's guarded commands.
## References
1. ^ Dijkstra, Edsger W. "EWD472: Guarded commands, non-determinacy and formal. derivation of programs." (PDF). Retrieved August 16, 2006.
2. ^ Anne Kaldewaij (1990), Programming: The Derivation of Algorithms, Prentice Hall
3. ^
4. ^ Martin, Alain J. "Synthesis of Asynchronous VLSI Circuits".
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2016-12-08 18:16:30
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|
https://math.stackexchange.com/questions/1904239/volume-form-on-a-complex-manifold-vs-volume-form-on-the-underlying-real-manifol
|
# Volume form on a complex manifold vs. volume form on the underlying real manifold
Consider a closed complex manifold $X=(M,J)$ of complex dimension $n$, where $M$ is a real smooth manifold of (real) dimension $2n$ and $J$ an integrable almost complex structure on it.
The holomorphic tangent bundle of $X$ is $T_X\cong T_X^{1,0}$, so a volume form for $X$ will be a nowhere vanishing section of $$\bigwedge^n T_X^*\cong\bigwedge^n(T_X^{1,0})^*,$$ i.e., a global section of the canonical bundle $\Omega^{n,0}_X=K_X$.
Up to this point everything seems to make sense, unless I'm making a mistake somewhere.
My confusion arises when we consider the underlying smooth manifold $M$. For $M$, the volume form is a section of $\Omega^{2n}_M$, which by Hodge decomposition is itself isomorphic to $\Omega^{n,n}_X$. Therefore, I conclude that a volume form of $M$ is a section of this vector bundle, which is clearly different from the other volume forms we got before.
I'm pretty sure the first approach I mention is the correct one, yet can't find the mistake in the second argument. Could anyone help me understand where am I going astray?
A nowhere zero section $\Omega$ of $K_X$ is called a holomorphic volume form, but it is not a volume form on the underlying smooth manifold $X$ precisely for the reason you point out. However, in any coordinate chart $(U, (z^1, \dots, z^n))$, $\Omega = fdz^1\wedge\dots\wedge dz^n$ for some nowhere-zero holomorphic function $f$ on $U$, so
As $f$ is nowhere-vanishing, $\Omega\wedge\overline{\Omega}$ is a nowhere vanishing $2n$-form and hence a volume form on the smooth $2n$-dimensional manifold $X$.
• Awesome, so just yo be completely sure: both my arguments are correct, but strictly speaking $\Omega$ is not a volume form, yet it gives rise to a legitimate volume form $\Omega\wedge\overline{\Omega}$. Aug 26, 2016 at 16:39
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2022-08-18 14:57:23
|
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https://dba.stackexchange.com/questions/41992/simple-model-database-transaction-log-full-checkpoint/41994
|
# Simple model database transaction log full 'CHECKPOINT'
I have a SQL Server 2012 SP1 database which has a problem that the transaction log is full. It consists of a single data file and a single log file, both about 800MB.
I had 5GB free disk space, and I've enlarged it to 20GB+, just in case there was a disk space problem.
The database is in SIMPLE recovery model.
I cannot do a full database backup - it fails with
transaction log is full, with wait_desc of 'CHECKPOINT'
MTA: Note - the wait_desc is CHECKPOINT (someone asked below about the wait desc strings)
I cannot add a log file - it fails with the same error.
MTA: I cannot change the growth model of the one existing log file - it fails with the same error.
MTA: The existing log file has +10% growth.
MTA: I also cannot increase the size of the existing log file - it fails with the same error.
I cannot change the recovery model to full or bulk-logged - it fails with the same error.
I cannot execute a manual CHECKPOINT - it fails with the same error.
DBCC LOGINFO (dbid) shows that there are a couple hundred VLFs, all in status 2.
I'm not using change tracking, mirroring, or transaction log shipping.
What can I do to resolve this?
(As recommended, moved from stackoverflow to dba.stackexchange)
MTA: More tech details here (but no answers at all): http://www.sqlservercentral.com/Forums/Topic1451836-2799-1.aspx
eg, "DBCC OPENTRAN gives me no open transactions"
• Can you increase the size of the log file? Or does that error out as well? – Cougar9000 May 13 '13 at 2:26
• Was the transaction log of your database only 800MB in size or was it in fact more (on disk) and was thus unable to grow? – John K. N. Apr 13 '17 at 11:47
It sounds like you've got a huge transaction that has remained open which has done a lot of work. Run DBCC OPENTRAN on the database and see how long the oldest transaction has been open for. You'll probably need to kill that transaction (or have the user commit if it's someone in Management Studio). Then the log will clear automatically.
Check the autogrowth setting of the log file. Make sure it is smaller than the max file size and can grow. You can use this query for that:
SELECT DB_NAME(mf.database_id) database_name,
mf.file_id,
mf.type_desc,
mf.name,
LTRIM(STR(vfs.size_on_disk_bytes/1024.0/1024.0,30,3)) size_mb,
CASE
WHEN mf.max_size = 0 OR mf.growth = 0 THEN '--'
WHEN mf.max_size = -1 THEN 'unlimited'
ELSE LTRIM(STR(mf.max_size*8192.0/1024.0/1024.0,30,3))
END max_size_mb,
CASE WHEN mf.max_size = 0 OR mf.growth = 0 THEN 'none'
ELSE
CASE mf.is_percent_growth
WHEN 0 THEN LTRIM(STR(mf.growth*8192.0/1024.0/1024.0,30,3)) +' mb'
ELSE LTRIM(STR(mf.growth,4,0)) +'%'
END
END growth,
mf.physical_name
FROM master.sys.master_files mf
CROSS APPLY sys.dm_io_virtual_file_stats(mf.database_id,mf.file_id) vfs
You can change these settings on the files tab in the database properties dialog:
To start with - check the values of log_reuse_wait and log_reuse_wait_desc columns in the sys.databases catalog view. You can find if anything is preventing log truncation and act based on that. You can find a list of available values and their meaning here:
http://msdn.microsoft.com/en-us/library/ms190925.aspx
I had exactly the same problem, on a test system luckily. I tried all the same steps to no avail, but eventually a full restore from backup worked for me. Replication kicked in, but because the log was not yet full, I could expand the log size and allow it to grow whilst replication caught up.
My problem was compounded by a broken replication setup on our development environments that filled the transaction log and then got stuck in this endless loop of checkpoint and log full errors.
|
2021-04-21 21:12:14
|
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|
http://mathematica.stackexchange.com/questions/15140/finitefields-package-is-very-slow-any-fast-substitute-for-mathematica?answertab=active
|
# FiniteFields package is very slow. Any fast substitute for Mathematica?
I want to compute the inverse of matrix, say with dimensions $100 \times 100$, defined over a large finite field extension such as $GF(2^{120})$. I am using the package FiniteFields, but Mathematica's computation time is exponential with respect to matrix dimensions. The following code illustrates the problem:
<< FiniteFields;
Table[
With[{ext = 12},
First@AbsoluteTiming@
Inverse[
Array[GF[2, ext][RandomInteger[{0, 1}, ext]] &, {n, n}]
]
],
{n, 1, 11}
]
I am using an Intel Xeon X5680 @ 3.33GHz (64-bit OS) and Mathematica v8.0.4.0. I have received the following timing results:
{0.0030, 0.0080, 0.0210, 0.0630, 0.1860, 0.5110, 1.3350, 3.3840, 8.9340, 23.0090, 57.4660}
I believe the source of the problem is that the FiniteFields package defines many UpValues, DownValues and SubValues of Times and Plus for head GF and, consequently, the pattern matching of arguments is degraded.
Does anyone know if a patch for the FiniteFields package or a faster substitute providing a similar interface?
Many thanks!
-
The textbook algorithm, that is, row reducing {A, I} to {I, A^(-1)} works.
n = 1000;
m = Array[RandomInteger[{0, 1}]&, {n, n}];
m // MatrixPlot
MatrixRank[m, Modulus -> 2] (* see if m is full rank *)
inverse = With[{n = Length@#},
RowReduce[MapThread[Join, {#, IdentityMatrix[n]}], Modulus -> 2][[;;,n+1;;]]]&;
inverseM = inverse[m]; // AbsoluteTiming
Mod[inverseM. m, 2] // MatrixPlot
On my laptop, computing inverse takes 15.262331 for a full rank matrix.
Often, you use the inverse matrix to solve A x == b, and Mathematica has fast LinearSolve in the FiniteFields package.
n = 1000;
LinearSolve[Array[RandomInteger[{0, 1}]&, {n, n}],
Array[RandomInteger[{0, 1}]&,n], Modulus -> 2] ;// AbsoluteTiming
On my laptop this last gives {5.110337, Null}
-
Works well over prime fields but not extensions thereof. – Daniel Lichtblau May 27 '13 at 19:46
This will not scale to dimension 100 but will be an improvement on what you now have. It is cribbed from the section "Linear Algebra over Galois Fields here as well as the section "Groebner bases over modules and related computations" in this notebook.
deg = 12;
flen = 3;
j = 0;
While[flen > 2 && j++ < 100,
defpoly =
x^deg + 1 + RandomInteger[{0, 1}, deg - 1].x^Range[1, deg - 1];
flen = Length[FactorList[defpoly, Modulus -> 2]]
]
dim = 20;
mat = Table[
RandomInteger[{0, 1}, deg].x^Range[0, deg - 1], {dim}, {dim}];
newvars = Array[z, 2*dim];
augmat = Transpose[
ArrayFlatten[{mat, IdentityMatrix[Length[mat]]}, 1]];
auxpolys = Union[Flatten[Outer[Times, newvars, newvars]]];
linpolys = Join[augmat.newvars, {defpoly}, auxpolys];
allvars = Append[newvars, x];
Here is the bulk of the computation.
Timing[gb = GroebnerBasis[linpolys, allvars, Modulus -> 2];]
(* {168.240000, Null} *)
Now extract the inverse matrix.
modulegb = Complement[gb, Join[auxpolys, {defpoly}]];
redmat = Reverse[Sort[Outer[D, modulegb, newvars]]];
invmat = Transpose[redmat[[All, dim + 1 ;; 2*dim]]];
We'll check the result.
PolynomialMod[invmat.mat, {defpoly, 2}] === IdentityMatrix[dim]
(* True *)
A serious bottleneck is the need, using this approach, to have $O(\tt{dim}^2)$ auxiliary polynomials.
One can improve on this by writing a direct row reduction using a list representation form the field elements. This would require working with undocumented functionality in the Algebra context, such as AlgebraPolynomialTimesModList. This TMJ article may give slightly more information should you opt to go that route.
--- edit ---
Here is an approach that will sometimes work. Treat the elements as algebraics over Q instead of GF[2]. Now you can use some built in functionality. If you are lucky you get substantially the same row reduction and can recover the mod-2 result. I'll show what I mean with the example above.
We first set up a matrix of AlgebraicNumber opjects.
alg = algnum[root[defpoly], CoefficientList[#1, x]] &;
mat2 = Map[alg, mat, {2}] /. x -> # /. root[a_] :> Root[a &, 1] /.
algnum -> AlgebraicNumber;
augmat = Transpose[
ArrayFlatten[{mat2, IdentityMatrix[Length[mat2]]}, 1]];
Now row reduce it and hope we do not get any denominators divisible by 2.
Timing[
rred = RowReduce[augmat, Method -> "OneStepRowReduction"];]
(* {36.53499999999994, Null} *)
Extract the inverse part.
invmat = Transpose[
rred[[All, dim + 1 ;; 2*dim]] /.
AlgebraicNumber[aa_, bb_] :>
PolynomialMod[bb, 2].x^Range[0, deg - 1]];
Check:
PolynomialMod[invmat.mat, {defpoly, 2}] === IdentityMatrix[dim]
(* True *)
--- end edit ---
-
Thank you for great references! It seems that I has to reimplement finite fields stuff in Mathematica to improve performance. This is bad news for me :( – Piotr Semenov Nov 25 '12 at 11:28
I think the key is AlgebraPolynomialExtendedGCDModList because it allows you to find field inverses. You invoke it using polynomials represented as their coefficient lists (constant term on the left and ascending from there). If defpoly is the list for your field defining polynomial, then to invert the field element r1 you would do r1inv=AlgebraPolynomialExtendedGCDModList[r1,defpoly,p][[2,1]]. In this notation p is the prime charatceristic, so it would be 2 for the cases of apparent interest. – Daniel Lichtblau Nov 26 '12 at 19:04
@Daniel Lichtblau Thank you for the point! I see, the easiest ways are to implement finite field extensions myself or do the patch for FiniteFields`. If I get some progress, I will share it in the community. – Piotr Semenov Nov 27 '12 at 10:11
Yes, please do share with community. Here and/or library.wolfram.com would be good places. Is your work only with linear algebra on field elements, or does it involve e.g. polynomials over extension fields? if the latter, there may be some existing functionality of interest I could locate. – Daniel Lichtblau Nov 27 '12 at 15:35
@DanielLichtblau My work includes both, the linear algebra and polynomials over extension fields. For example, I have to compute the determinant of big matrices over field extensions and to search roots of polynomials by their factorization. Sure, it will be great! Thanks. – Piotr Semenov Dec 3 '12 at 12:32
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2015-03-29 19:43:49
|
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http://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-1-page-17/17
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Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.1: 17
10 degrees Celsius
Work Step by Step
Use the formula $$C=\frac{5}{9}(F-32)$$ to convert 50 degrees Fahrenheit temperature to its equivalent temperature on the Celsius scale. Start by substituting 50 for F in the equation $$C=\frac{5}{9}(50-32)$$ complete grouped operations first $$C=\frac{5}{9}(18)$$ since the $\frac{5}{9}$ is a fraction, turn the 18 into a fraction (by putting it over 1) so that you can multiply the two fractions: $$C=\frac{5}{9}\times\frac{18}{1}$$ cross reduce the 9 with the 18 (both are divisible by 9) $$C=\frac{5}{1}\times\frac{2}{1}$$ $$C=\frac{10}{1}$$ $$C=10$$
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.
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2017-05-30 04:06:12
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https://im.kendallhunt.com/HS/students/4/2/20/index.html
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# Lesson 20
Interpreting Inequalities
• Let’s interpret the meaning of situations with phrases like “at least,” “at most,” and “up to.”
### 20.1: Math Talk: Solving Inequalities
Mentally solve for $$x$$.
• $$5x<10$$
• $$10>6x-2$$
• $$9x<5-23$$
• $$11(x-3)<46-2$$
### 20.2: Checking and Graphing Inequalities
1. $$\text-2x<4$$
1. Solve the inequality.
2. $$3x+5>6x-4$$
1. Solve the inequality.
3. $$\text-3(x+1)\geq13$$
1. Solve the inequality.
For each statement:
• Use a number line to show which values satisfy the inequality.
• Express the statement symbolically with an inequality.
1. The elevator can lift up to 1,200 pounds. Let $$x$$ represent the weight being lifted by the elevator.
2. Over the course of the senator's term, her approval rating was always around 53% ranging 3% above or below that value. Let $$x$$ represent the senator’s approval rating.
3. There's a minimum of 3 years of experience required. Let $$x$$ represent the years of experience a candidate has.
### 20.3: Card Sort: What’s the Situation?
Your teacher will give you a set of cards that show a graph, an inequality, or a situation. Sort the cards into groups of your choosing. Be prepared to explain the meaning of your categories. Then, sort the cards into groups in a different way. Be prepared to explain the meaning of your new categories.
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2023-02-08 14:45:42
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https://www.prepanywhere.com/prep/textbooks/advanced-functions-nelson/chapters/chapter-1-functions-characteristics-and-properties/materials/chapter-review
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Chapter Review
Chapter
Chapter 1
Section
Chapter Review
Solutions 50 Videos
Determine whether each relation is a function, and state its domain and range.
Q1a
Determine whether each relation is a function, and state its domain and range.
\displaystyle 3x^2 + 2y =6
Q1b
Determine whether each relation is a function, and state its domain and range.
Q1c
Determine whether each relation is a function, and state its domain and range.
\displaystyle x = 2^y
Q1d
A cell phone company charges a monthly fee of $30, plus$0.02 per minute of call time.
a) Write the monthly cost function, C(t), where t is the amount of time in minutes of call time during a month.
b) Find the domain and range of C.
Q2
Graph f(x) = 2|x + 3| -1, and state the domain and range.
Q3
Describe this interval using absolute value notation.
Q4
For the pair of functions, give a characteristic that the two functions have in common and a characteristic that distinguishes them.
\displaystyle f(x) =x^2 and \displaystyle g(x) = \sin x
Q5a
For the pair of functions, give a characteristic that the two functions have in common and a characteristic that distinguishes them.
\displaystyle f(x) =\frac{1}{x} and \displaystyle g(x) = x
Q5b
For the pair of functions, give a characteristic that the two functions have in common and a characteristic that distinguishes them.
\displaystyle f(x) =x^2 and \displaystyle g(x) = \sin x
Q5c
For the pair of functions, give a characteristic that the two functions have in common and a characteristic that distinguishes them.
\displaystyle f(x) =2^x and \displaystyle g(x) = x
Q5d
Identify the intervals of increase/decrease, the symmetry, and the domain and rage of each function.
\displaystyle f(x) = 3x
Q6a
Identify the intervals of increase/decrease, the symmetry, and the domain and range of each function.
\displaystyle f(x) = x^2+ 2
Q6b
For each of the following equations, state the parent function and the transformations that were applied. Graph the transformed function.
\displaystyle f(x) =|x + 1|
Q7a
For each of the following equations, state the parent function and the transformations that were applied. Graph the transformed function.
\displaystyle f(x) = -0.25\sqrt{3(x+ 7)}
Q7b
For each of the following equations, state the parent function and the transformations that were applied. Graph the transformed function.
\displaystyle f(x) = -2\sin(3x) + 1, 0 \leq x \leq 360^o
Q7c
For each of the following equations, state the parent function and the transformations that were applied. Graph the transformed function.
\displaystyle f(x) = 2^{-2x} -3
Q7d
The graph of y = x^2 is horizontally stretched by a factor of 2, reflected in the x—axis, and shifted 3 units down. Find the equation that results from the transformation, and graph it.
Q8
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y = -f(-x)+2
Q9a
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y =f(-2(x+ 9)) - 7
Q9b
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y =f(x -2) + 2
Q9c
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y = 0.3f(5(x-3))
Q9d
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y = 1 -f(1-x)
Q9e
(2, 1) is a point on the graph of y =f(x). Find the corresponding point on the graph of each of the following functions.
\displaystyle y = -f(2(x - 8))
Q9f
For the point on a function, state the corresponding point on the inverse relation.
(1, 2)
Q10a
For the point on a function, state the corresponding point on the inverse relation.
(-1, -9)
Q10b
For the point on a function, state the corresponding point on the inverse relation.
(0, 7)
Q10c
For the point on a function, state the corresponding point on the inverse relation.
f(5) = 7
Q10d
For the point on a function, state the corresponding point on the inverse relation.
g(0) = -3
Q10e
For the point on a function, state the corresponding point on the inverse relation.
h(1) = 10
Q10f
Given the domain and range of a function, state the domain and range of the inverse relation.
D = \{x\in \mathbb{R}\}, \displaystyle R = \{y \in \mathbb{R}, -2 < y < 2\}
Q11a
Given the domain and range of a function, state the domain and range of the inverse relation.
D = \{x\in \mathbb{R}, x \geq 7\}, \displaystyle R = \{y \in \mathbb{R}, y < 12\}
Q11b
Graph the function and its inverse relation on the same set of axes. Determine whether the inverse relation is a function.
f(x) = x^2 -4
Q12a
Graph the function and its inverse relation on the same set of axes. Determine whether the inverse relation is a function.
f(x) = 2^x
Q12b
Find the inverse of each function.
f(x) = 2x + 1
Q13a
Find the inverse of each function.
f(x) = x^3
Q13b
Graph the following function. Determine whether it is discontinuous and, if so, where. State the domain and the range of the function.
\displaystyle f(x) = \begin{cases} &2x, &\text{when } x < 1 \\ &x +1, &\text{when } x \geq 1 \end{cases}
Q14
Write the algebraic representation for the following piecewise function, using function notation.
Q15
If
\displaystyle f(x) = \begin{cases} &x^2 +1, &\text{when } x < 1 \\ &3x, &\text{when } x \geq 1 \end{cases}
is f(x) continuous at x =1? Explain.
Q16
A telephone company charges $30 a month and gives the customer 200 free call minutes. After the 200 min, the company charges$0.03 a minute.
a) Write the function using function notation.
b) Find the cost for talking 350 min in a month.
c) Find the cost for talking 180 min in a month.
Q17
Given f = \{(0, 6), (1, 3), (4, 7), (5, 8)\} and g= \{(-1, 2), (1, 5), (2, 3) ,(4, 8), (8, 9)\}, determine
f(x) + g(x)
Q18a
Given f = \{(0, 6), (1, 3), (4, 7), (5, 8)\} and g= \{(-1, 2), (1, 5), (2, 3) ,(4, 8), (8, 9)\}, determine
f(x)- g(x)
Q18b
Given f = \{(0, 6), (1, 3), (4, 7), (5, 8)\} and g= \{(-1, 2), (1, 5), (2, 3) ,(4, 8), (8, 9)\}, determine
f(x)\cdot g(x)
Q18c
Given f(x) = 2x^2 -2x, -2 \leq x \leq 3 and g(x) = -4x, -3 \leq x \leq 5, graph the following.
f
Q19a
Given f(x) = 2x^2 -2x, -2 \leq x \leq 3 and g(x) = -4x, -3 \leq x \leq 5, graph the following.
g
Q19b
Given f(x) = 2x^2 -2x, -2 \leq x \leq 3 and g(x) = -4x, -3 \leq x \leq 5, graph the following.
f +g
Q19c
Given f(x) = 2x^2 -2x, -2 \leq x \leq 3 and g(x) = -4x, -3 \leq x \leq 5, graph the following.
f -g
Q19d
Given f(x) = 2x^2 -2x, -2 \leq x \leq 3 and g(x) = -4x, -3 \leq x \leq 5, graph the following.
fg
Q19e
f(x) = x^2 +2x and g(x) = x + 1. Match the answer with the operation.
\displaystyle \begin{array}{llllllll} &(a) \phantom{.} x^3 + 3x^2 + 2x &A \phantom{.} f(x) + g(x) \\ &(b) \phantom{.} -x^2 -x + 1 &B \phantom{.} f(x) - g(x) \\ &(c) \phantom{.} x^2 +3x + 1 &C \phantom{.} g(x) - f(x) \\ &(d) \phantom{.} x^2 +x - 1 &D \phantom{.} g(x) \times f(x) \end{array}
Q20
f(x) = x^3 +2x^2 and g(x) = -x + 6.
a) Complete the table.
b) Use the table to graph f(x) and g(x) on the same axes.
c) Graph (f 1 g)(x) on the same axes as part b).
d) State the equation of (f + g)(x).
e) Verify the equation of (f + g)(x) using two of the ordered pairs in the table.
|
2021-04-16 17:19:57
|
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|
https://www.research-collection.ethz.ch/handle/20.500.11850/398970?show=full
|
dc.contributor.author
Basso, Giuliano
dc.contributor.supervisor
Lang, Urs
dc.contributor.supervisor
Lytchak, Alexander
dc.contributor.supervisor
Naor, Assaf
dc.date.accessioned
2020-02-17T07:56:46Z
dc.date.available
2020-02-12T16:28:44Z
dc.date.available
2020-02-13T10:26:37Z
dc.date.available
2020-02-16T10:19:00Z
dc.date.available
2020-02-17T07:56:46Z
dc.date.issued
2019-12
dc.identifier.uri
http://hdl.handle.net/20.500.11850/398970
dc.identifier.doi
10.3929/ethz-b-000398970
dc.description.abstract
The subject of this doctoral thesis is the class of barycentric metric spaces, which encompasses both Banach spaces and complete CAT(0) spaces. Encouraged by known results as well as open questions in the context of CAT(0) spaces, we study similar objectives in the framework of barycentric metric spaces. For example, we show that certain fixed point properties, which are given in CAT(0) spaces, do not hold for some barycentric metric spaces, and prove two fixed point results adapted to the new situation. These results are phrased for the class of metric spaces that allow a conical bicombing; this is no restriction, since the class of barycentric metric spaces agrees with this class. This equality leads to a variety of questions regarding the existence and uniqueness of certain classes of conical bicombings. In particular, we consider conical bicombings on open subsets of normed vector spaces and show that these bicombings are locally given by linear segments. This result implies that any open convex subset in a large class of Banach spaces possesses a unique consistent conical bicombing. Besides this, we consider various Lipschitz extension problems, where in some cases any complete barycentric metric space may appear as target space. One such Lipschitz extension problem involves the extension of a Lipschitz function to finitely many additional points. Our contribution consists of finding upper bounds for the distortion of the Lipschitz constant, and we construct examples which demonstrate that we found the best possible bounds in the case of an extension to one additional point. Many Lipschitz extension constants may be computed by solving an associated linear extension problem, which is why, in the last part, we turn our attention to absolute linear projection constants of real Banach spaces. We succeeded in finding a formula for the maximal linear projection constant amongst $$n$$-dimensional Banach spaces. By means of this formula, we give another proof of the Grünbaum conjecture, which was first proven by Chalmers and Lewicki in 2010.
en_US
dc.format
application/pdf
en_US
dc.language.iso
en
en_US
dc.publisher
ETH Zurich
en_US
dc.rights.uri
http://rightsstatements.org/page/InC-NC/1.0/
dc.subject
Metric spaces
en_US
dc.subject
Projection constants
en_US
dc.subject
Lipschitz maps
en_US
dc.subject
two-graphs
en_US
dc.subject
Non-positive curvature
en_US
dc.title
Fixed point and Lipschitz extension theorems for barycentric metric spaces
en_US
dc.type
Doctoral Thesis
In Copyright - Non-Commercial Use Permitted
dc.date.published
2020-02-13
ethz.size
101 p.
en_US
ethz.code.ddc
DDC - DDC::5 - Science::510 - Mathematics
en_US
ethz.identifier.diss
26486
en_US
ethz.publication.place
Zurich
en_US
ethz.publication.status
published
en_US
ethz.leitzahl
ETH Zürich::00002 - ETH Zürich::00012 - Lehre und Forschung::00007 - Departemente::02000 - Dep. Mathematik / Dep. of Mathematics::02003 - Mathematik Selbständige Professuren::03500 - Lang, Urs / Lang, Urs
en_US
ethz.date.deposited
2020-02-12T16:28:53Z
ethz.source
FORM
ethz.eth
yes
en_US
ethz.availability
Open access
en_US
ethz.rosetta.installDate
2020-02-13T10:26:48Z
ethz.rosetta.lastUpdated
2021-02-15T08:05:33Z
ethz.rosetta.versionExported
true
ethz.COinS
ctx_ver=Z39.88-2004&rft_val_fmt=info:ofi/fmt:kev:mtx:journal&rft.atitle=Fixed%20point%20and%20Lipschitz%20extension%20theorems%20for%20barycentric%20metric%20spaces&rft.date=2019-12&rft.au=Basso,%20Giuliano&rft.genre=unknown&rft.btitle=Fixed%20point%20and%20Lipschitz%20extension%20theorems%20for%20barycentric%20metric%20spaces
Search print copy at ETH Library
|
2021-10-23 05:40:58
|
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|
https://cstheory.stackexchange.com/questions/42492/3-dimensional-matching-shortest-solution-np-hard/42493
|
3 dimensional matching shortest solution NP-hard?
We have array of arbitrary number of elements - 3d vectors with positive integers components - for example
let a=[ [0,1,2], [1,0,2], [1,1,1], [1,2,0 ], [2,0,1 ], [2,1,0 ] ];
And we want to remove elements from that list, which have duplicate value on i-th index with other elements. This problem can have more than one solution:
• solution with 3 elements: [0,1,2],[1,2,0],[2,0,1]
• solution with 2 elements: [1,0,2],[2,1,0]
As you can see, the solution has this property that each solution element have unique value on i-th index (numbers on i-th position never duplicate) and if we add any other element from array a to that solution we loose this property.
Optimalization: from this wiki aricle I know that finding longest solution is NP-hard problem - but what about finding the shortest solution? Does exists algorithm which not use brute-force to find shortest solution (shortest 3-d matching)?
Here is interactive visualisation in javascript (works in browser, I put it also in wiki article). Here is implementation for find any solution in javascript.
Your problem is NP-complete, even in two dimensions. There is a straightforward reduction from MINIMUM MAXIMAL MATCHING in bipartite graphs:
MINIMUM MAXIMAL MATCHING in bipartite graphs
INSTANCE: a bipartite graph $$G=(V_1\cup V_2,E)$$ with $$E\subseteq V_1\times V_2$$; an integer $$k$$
QUESTION: Does $$G$$ possess a maximal matching $$E'$$ of cardinality at most $$k$$?
(A maximal matching is a subset $$E'\subseteq E$$ such that no two edges in $$E'$$ share a common endpoint, and such that every edge in $$E-E'$$ shares a common endpoint with some edge in $$E'$$.)
NP-hardness of MINIMUM MAXIMAL MATCHING in bipartite graphs has been shown in
M. Yannakakis, F. Gavril
Edge dominating sets in graphs.
SIAM J. Appl. Math. 38, 364–372 (1980)
https://epubs.siam.org/doi/10.1137/0138030
For the reduction, take the vertex set $$V_1\cup V_2$$ as ground set.
For every edge $$\{u,v\}\in E$$ with $$u\in V_1$$ and $$v\in V_2$$, create a corresponding vector $$[u,v]$$.
The equivalence to your problem is immediate.
(And if you insist on 3-dimensional vectors, then you should add a third component to these vectors and fill it with a lot of different values.)
• As is written in wikipedia on top of here and details here, the 2D version of this problem is NOT NP-hard. So the bug is in your answer, in wikipedia, or in my understanding? Please clarify this – Kamil Kiełczewski Mar 8 '19 at 8:42
• You are mixing up two very different questions: In the classical matching problem in the wikipedia, you want to find a matching of MAXIMUM cardinality. In the problem I am using, you want to find a matching that cannot be extended to a larger matching, and this matching should have minimum cardinality. – Gamow Mar 8 '19 at 10:37
• I found M. Yannakakis article here – Kamil Kiełczewski Mar 8 '19 at 11:29
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2020-08-09 11:40:00
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https://cms.math.ca/10.4153/CMB-2016-039-2
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location: Publications → journals → CMB
Abstract view
Zero-divisor Graphs of Ore Extensions over Reversible Rings
Published:2016-07-18
Printed: Dec 2016
• Ebrahim Hashemi,
Department of Mathematics, Shahrood University of Technology, , P.O.~Box: 316-3619995161, Shahrood, Iran
• R. Amirjan,
Department of Mathematics, Shahrood University of Technology, , P.O.~Box: 316-3619995161, Shahrood, Iran
Format: LaTeX MathJax PDF
Abstract
Let $R$ be an associative ring with identity. First we prove some results about zero-divisor graphs of reversible rings. Then we study the zero-divisors of the skew power series ring $R[[x;\alpha]]$, whenever $R$ is reversible and $\alpha$-compatible. Moreover, we compare the diameter and girth of the zero-divisor graphs of $\Gamma(R)$, $\Gamma(R[x;\alpha,\delta])$ and $\Gamma(R[[x;\alpha]])$, when $R$ is reversible and $(\alpha,\delta)$-compatible.
Keywords: zero-divisor graphs, reversible rings, McCoy rings, polynomial rings, power series rings
MSC Classifications: 13B25 - Polynomials over commutative rings [See also 11C08, 11T06, 13F20, 13M10] 05C12 - Distance in graphs 16S36 - Ordinary and skew polynomial rings and semigroup rings [See also 20M25]
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2018-03-19 01:07:07
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https://socratic.org/questions/57e4280f11ef6b05c44152fc
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# 2.55xx10^24 titanium atoms are cast into a cube that has a mass of 202.5*g. If the density of titanium metal is rho=4.60*g*cm^-3, what is the edge length of the cube?
Sep 22, 2016
The length of the metal cube $=$ ""^3sqrt(45.0*cm^3) $=$ ??cm
#### Explanation:
If there are $2.55 \times {10}^{24}$ $\text{titanium atoms}$, there are $\frac{2.55 \times {10}^{24}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1}$ $=$ $4.23 \cdot m o l$ $\text{titanium atoms}$.
And this has a mass of $4.23 \cdot m o l \times 47.87 \cdot g \cdot m o {l}^{-} 1$ $=$ $202.5 \cdot g$.
Now we have a cube that has a mass of $202.5 \cdot g$. But we know its $\text{density}$, which you have kindly provided, $\rho$ $=$ $4.50 \cdot g \cdot c {m}^{-} 3$.
$\rho$ $=$ $\text{Mass"/"Volume}$. Thus $\text{Volume}$ $=$ $\frac{\text{Mass}}{\rho}$ $=$ $\frac{202.5 \cdot g}{4.50 \cdot g \cdot c {m}^{-} 3}$ $=$ $45.0 \cdot c {m}^{3}$
And thus the side-length of the metal cube $=$ ""^3sqrt(45.0*cm^3) $=$ ??cm $\cong$ $3.5 \cdot c m$
Don't worry about making a mistake in the question; this happens. Mind you, I hope I have got your question right!
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2019-11-20 19:02:51
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https://robotics.stackexchange.com/questions/22247/looking-for-an-online-map-pgm-repository
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# looking for an online map pgm repository
I am currently writing a path planner. I wanted to test the planner for various scenarios.
I was wondering if there is a repository wherein I can find a number of pgm (+yaml) map files.
I came across this link but the two maps mentioned in the answer are very huge. Also, the corresponding yaml file seems to be missing as well.
Also, is there a way one can easily create 'small' maps so as to load them to the map_server ?
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2021-09-19 22:32:07
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https://www.physicsforums.com/threads/density-of-states-summation.174346/
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# Density of states summation?
1. Jun 18, 2007
### pivoxa15
If an infinite discrete sum is calculated via integrating over a density of states factor, is this integral an approximation to the discrete sum? i.e the discrete sums could be partition functions or Debye solids.
2. Jun 28, 2007
### Mr.Brown
you mean like when you have a huge amount of particles and instead of summing over all these particles to get the partition function you integrate?
Yes then it´s an approximation :)
3. Jun 29, 2007
### Manchot
Yes, it's an approximation, but it's a very, very good one. Suppose you were numerically integrating a slowly-varying function from 0 to 1e24. You'd probably do so using the definition of a Riemann integral, choosing a finite dx. That is, you'd partition the function into intervals, add their values together, and multiply by dx. So, if you chose dx=1e23, you'd add up the values of the function at 0, 1e23, 2e23, ..., 9e23 and multiply by 1e23. Of course, if you chose a smaller dx, (say 1e22) you'd get a better approximation to the integral. If you chose a dx=1, the error would be extremely small (differing by about 1e-24), and you'd get your answer by summing up the value at every integer and multiplying by 1. Of course, this means that the summation approximates the integral, and vice versa.
4. Aug 10, 2007
### Slaviks
Within the context your question referes to, it is an approximation which becomes exact in the thermodynamic limit (infinite system with fixed finite density). This is precisely the limit in which standard results of statistical mechanics make sense. Mathematically, by taking the thermodynamic limit the sum becomes an integral by the very definition of the latter.
It is also worth keeping in mind that caution is necessary in taking a sum to an integral when there are special modes like the condensate state in a bosonic system.
5. Aug 13, 2007
### quetzalcoatl9
remember that when the partition function was derived, the assumption of large N was made in order to invoke the sterling approximation:
$$\ln(N!) \approx N \ln N - N$$
numerical analysis shows that the approximation becomes valid very quickly. you can use the gamma function to investigate more thoroughly.
it is for this reason that we can get away with computer simulations of small particles and periodic boundaries (and still calculate meaningful averages). in fact, you can get reasonable results for a monoatomic gas of only 16 particles with periodic boundaries (at most state points, depending upon the potential).
Last edited: Aug 13, 2007
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2017-02-25 02:50:56
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https://brilliant.org/problems/nmtc-inter-level-problem-2/
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# Trio Of Primes
Number Theory Level 1
How many natural numbers $$n$$ exist such that the following are all primes?
$3n-4 \qquad 4n-5 \qquad 5n-3$
×
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2016-10-24 14:17:13
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