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# A New Kind of Science and the Future of Mathematics, continued
Each step shifts every digit one position to the left. So that eventually even the most insignificant digit will have a big effect.
Well, so why does this make randomness? It makes randomness if the initial digit sequence is random. And with the usual view of real numbers, one images that a typical number picked as an initial condition would almost always have that kind of randomness.
But if one thinks in terms of information, this isn't really unsatisfactory. It's like saying that one gets randomness out because one put randomness in. And that the randomness is just coming from outside the system we're looking at.
Well, is there any other possibility? Yes. Just look at rule 30. There's no randomness from the environment--it's a deterministic rule. And there's no randomness in the initial conditions--they're just a single black cell. But still, the actual evolution of the system intrinsically generates apparent randomness. [See A New Kind of Science, page 30.]
It's a bit tricky at first to analyse this. Not least because systems often have both sensitive dependence on initial conditions, and intrinsic randomness generation. Rule 30 has sensitive dependence; the slopes of these difference patterns give its Lyapunov exponents. [See A New Kind of Science, page 251.]
And iterated maps can show intrinsic randomness generation. [See A New Kind of Science, page 150.]
But that's not what dynamical systems theory picks up. Because it's talking about all possible trajectories. Not just trajectories that have "simple to make" initial conditions.
But if you really want to explain randomness, you need to get more precise.
Well, let's go back to fluid turbulence. [See A New Kind of Science, page 377.]
Where does its randomness come from? The Navier-Stokes equations haven't told us. It could even be that we need to add Brownian motion terms. We know there are some instabilities, but we don't really know their large-scale effects. And in numerical experiments, it's hard to distinguish numerical glitches from genuine randomness.
Well, a different approach to modelling helps with that. At the lowest level a fluid just consists of a bunch of molecules. And we know that the details aren't too important--we always get the same continuum behavior. So we can try just making a minimal model for the molecules. Put them on a discrete grid, with discrete velocities, and so on. [See A New Kind of Science, page 378.]
Well, in the appropriate limit, that gives Navier-Stokes, and one can actually do nice practical fluid dynamics. [See A New Kind of Science, page 380.]
But one can also investigate some basic questions. And one finds that one can indeed get large-scale apparent randomness, without any random input. Which means there's intrinsic randomness generation going on. [See A New Kind of Science, page 30.]
One can test that in physical experiments. Because if randomness comes from the environment, or from initial conditions, it'll always be different every time one runs an experiment. But with intrinsic randomness generation, sufficiently carefully controlled experiments will give repeatable randomness, which one can look for. You know, probabilistic models are very common. But I think a lot of them really aren't necessary. And that a lot of systems can just make their own randomness, without getting it from outside.
Like in an Ising model. Where one normally assumes that spins are kicked around by a heat bath. Well, there's a simple deterministic model that seems to have the same average behavior--and that's probably much closer to an actual spin system. [See A New Kind of Science, page 982.]
It seems like there are a lot of probabilistic models where there are deterministic systems that have the same behavior. Like here's a deterministic 2D cellular automaton on a square grid that almost--though not quite--makes enough randomness to give a circle like a random walk. [See A New Kind of Science, page 178.]
By the way, I should say something about what I mean by randomness. In everyday speech, we usually call things random if we can't find a compressed description of them. Well, the ultimate is algorithmic randomness, where there just is no compressed description: there's no program shorter than a sequence that will reproduce it. It's an interesting definition. But it probably says that nothing in our universe can be considered random. Certainly not a rule 30 sequence. Because after all, it's generated by a very small program.
But the point is that we can't find that program. Statistical analysis, data compression, visual perception, cryptanalysis--none of them can crack rule 30. [See A New Kind of Science, page 561.]
So that's why its sequences seem random. And one can make that more precise by talking about analysis with all possible simple programs.
This turns out to be related to the Second Law of Thermodynamics: the law of entropy increase. It's easy to make a system like a cellular automaton microscopically reversible--so it's bijective, just like the laws of physics seem to be. [See A New Kind of Science, page 437.]
But the point is that intrinsic randomness generation in effect encrypts even simple initial conditions to the point where they seem typical of the ensemble--at least with respect to any reasonably simple computation. And from this one seems finally to get at least the outline of a real proof of the Second Law. As well as seeing potential exceptions. [See A New Kind of Science, page 440.]
OK, well that's a little about physics. What about biology? Where does all the complexity in biology really come from? [See A New Kind of Science, page 385.]
It's never been clear why natural selection should give it. And in fact, I think it's much more just a consequence of properties of simple programs. That if one picks a genetic program at random, it'll often show complex behavior.
Let me give you an example. Here's a mollusc shell. [See A New Kind of Science, page 423.]
Which we might have thought must be the result of some sophisticated optimization process. But actually its pattern is incredibly similar to what we get just from rule 30. [See A New Kind of Science, page 30.]
In the actual mollusc there's a line of cells that lay down pigment as the shell grows. And we can model that as a cellular automaton. [See A New Kind of Science, page 414.]
Well, let's say we just consider all possible rules of the simplest kind. Here are all the patterns they make. [See A New Kind of Science, page 424.]
And here's a remarkable fact. The classes of patterns we get seem to be just the same as in actual molluscs. [See A New Kind of Science, page 423.]
It's as if the mollusc species of the Earth just sample different possible programs, and then we see the results displayed on their shells.
You know, it usually seems hopeless to make a predictive theory of biology. Because everything seems to depend on historical accidents.
But if different species just uniformly sample simple programs, we get to make predictions just by knowing abstract properties of programs. Let me give you another example. Shapes of leaves. [See A New Kind of Science, page 403.]
They're pretty diverse. But it turns out that there's a simple program that seems to capture most of them. It's just based on successive repeated branchings. [See A New Kind of Science, page 400.]
That fill in leaf material. And get shapes that look just like actual leaves-- sometimes smooth, sometimes jagged, and so on. [See A New Kind of Science, page 402.]
Well, what happens with all possible such models? Here's a simple case where it's easy to lay out the possible shapes. [See A New Kind of Science, page 406.]
And one can summarize these in a parameter space set. [See A New Kind of Science, page 407.]
From whose properties one can deduce all sorts of things about leaves and their likely evolution.
It's actually a mathematically rather interesting set. With properties a bit like the Mandelbrot set. By the way, for people who know about such things, my set is formed by looking at whether each tree reaches a given point. If you look instead at connectedness, you get a different set that's much less interesting set. Well, there are lots of applications in biology. Mixing traditional math-- like differential geometry--with simple programs gives lots of results on how things grow. [See A New Kind of Science, page 418.]
And in molecular biology there's much to say about using primitives based on simple programs to describe biological processes.
But let me talk a little more about physics. And in particular fundamental physics.
Well, now that we've seen the things simple programs can do, we might ask the ultimate question: underneath all of physics could there just be a simple program? A little program that if run for long enough would reproduce in precise detail the whole history of our universe. In a sense the ultimate reduction of physics to an abstract system.
Well, so what might the program for the universe be like? One thing that's kind of inevitable is that very few familiar features of the universe will immediately be visible in it. There just isn't room. I mean, in a small program there's no way to fit in separate identifiable pieces that represent electrons, or gravity, or even space and time.
In fact, if the program's going to be really small, it sort of has to have the very least possible structure built in. And for example I think a cellular automaton already has far too much structure. It's got a whole rigid array of cells laid out in space. And it also separates the notion of space from the notion of states of cells. And I don't think one needs even that.
In physics, space is normally a kind of background--on top of which matter and everything else exists. But in an ultimate model I think space is the only thing one needs.
So what might space then be? I suspect it's ultimately just a bunch of points with a certain set of connections. There are various ways to formulate it. But one is just to say it's a giant trivalent network--a cubic graph. [See A New Kind of Science, page 1039.]
In which one just specifies the combinatorics of what point is connected to what.
So how does space as we know it emerge from that? The idea is that with enough nodes, it approximates a continuum. Like with molecules in a fluid. But more subtle.
Here are some networks that look like one-, two- and three-dimensional space. [See A New Kind of Science, page 477.]
But remember, all we've actually specified is how the nodes are connected. So how do we ultimately tell what dimension a network corresponds to? Just start from a node. Then look at how many distinct nodes you reach by going r connections. [See A New Kind of Science, page 477.]
That's essentially the volume of a ball in the network. And if the network is going to correspond to d-dimensional space, that volume must grow like r^d.
Here I've laid out nodes so that their x position is determined by their minimum distance from a starting point. [See A New Kind of Science, page 479.]
And you can see the different growth rates for different dimensions.
What about curved space? In a flat hexagonal lattice, the number of nodes at distance r goes exactly like r^2. [See A New Kind of Science, page 532.]
But with curvature there's a correction. That turns out to be proportional to the Ricci scalar curvature for the limiting manifold. [See A New Kind of Science, page 1050.]
Well, it's pretty common to go from manifolds to more discrete things. Like in triangulating a manifold. [See A New Kind of Science, page 533.]
But there one's still giving embedding information: saying how to lay out the points in an ambient space. In combinatorial topology one gets rid of the embedding. But one still has data on which nodes bound a face, and so on.
But in my combinatorial networks one doesn't have even that. So it's sort of remarkable how much limiting differential geometry one can get. Ricci scalars from growth rates of balls. Ricci tensors from growth rates of cylinders generated by geodesics. But, significantly, one can't directly get Riemann tensors. Well, OK, in our universe there's not only space, but also time. And in ordinary physics, one thinks of time as just another coordinate. But in programs it tends to be very different. Like in a cellular automaton. Where one moves in space by going from one cell to another. But one moves in time by actually applying the cellular automaton rule.
Well, I suspect that's closer to how things work in physics. But it won't be exactly like a cellular automaton. Because that would need a global clock to synchronize everything.
OK, so how might it actually work? Well, here's something that at first sounds kind of crazy. How about if the universe is like a Turing machine--or what I call a mobile automaton--where only one cell gets updated at each step? [See A New Kind of Science, page 487.]
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# Force Inverse square law
1. Jul 18, 2011
### berge616
I am confused regarding the following question. My guess is that b is correct, because the gravitational force equation has r^2. However beyond that I am not sure if any other options are correct. Please help
What do you know about a force that follows the inverse square law.
(a) The force is strong.
(b) The force decreases with distance.
(c) The force depends on the magnitude of the masses involved.
(d) The force depends on the magnitude of the charges involved.
(e) The force depends on a universal constant.
2. Jul 18, 2011
### WannabeNewton
You are correct in saying it is (b). (a) does not say anything in general about the law, (c) and (d) can be ruled out by giving coulomb's law and newton's law respectively and (e) can be ruled out by using as an example the inverse square law that determines the intensity of radiation, emitted at a point source and distributed spherically, at a distance r from the source which does not include any universal constant. (b) is part of the definition of an inverse square law.
3. Jul 19, 2011
### mathman
Newton's law: F=Gm1m2/r2, where the m's are the masses involved, r is the separation, and G is a universal constant. Therefore b, c, and e are all correct.
4. Jul 19, 2011
### WannabeNewton
The question is asking about inverse square laws in general though. (c) and (e) are not general cases.
Last edited: Jul 19, 2011
5. Jul 20, 2011
### mathman
The question is very badly worded.
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# Matlab Question - How to optimize parameters for two data sets [closed]
This is rather an easy problem but I cannot get my head around.
I have two curves in the form, say F(xi,fi) and G(xi,gi), with i = 1 to 65. F and G are supposed to be the same, but I get them unequal in raw format because each fi, gi contains additional leakage or bias.
My question is how to find coefficients that eliminate the bias.
I was doing following with no success. aF + bG = 0, or ai * fi + bi * gi = 0
subject to a constraint Sum[(ai-1)fi] = Sum[(bi-1)gi] = constant. Sum indicates summation over i's from 1 to 65.
I was, with some help, able to find coefficients algebraically.
a_i = g_i / 2 * sum_{j=1}^{j=65} (f_j - g_j) / sum_{j=1}^{j=65} (f_j * g_j)
b_i = -f_i / 2 * sum_{j=1}^{j=65} (f_j - g_j) / sum_{j=1}^{j=65} (f_j * g_j)
But it didn't give the correct shape, maybe because of not taking the constraint into account.
Any help would be much appreciated.
## closed as off-topic by fəˈnɛtɪk, Shaggy, Kirill L., Adám, AdmBorkBorkJan 8 at 13:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This site is for programming contests and challenges. General programming questions are off-topic here. You may be able to get help on Stack Overflow." – fəˈnɛtɪk, Shaggy, Kirill L., Adám, AdmBorkBork
If this question can be reworded to fit the rules in the help center, please edit the question.
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purav savla
Java Program
0 Answer(s) 3 years and 7 months ago
Posted in : Java Interview Questions
You are to create a game called â??Get Out of My Swampâ??. In this game an ogre, called â??Hekâ??, wanders around his swamp, if he encounters an ogre enemy in his swamp he kills it. If he encounters two ogre enemies in the same place they kill the ogre and the game ends. Full details of the game are given below. There is a sample session from a game given at the end of the coursework. The swamp can be thought of as a four by four grid.
When the game starts the ogre is placed in a random square (i.e. part of the swamp). However he must not be placed in the top left hand corner of the swamp when the game starts, although he can move there during the game. When the ogre moves, he can move to ANY(Diagonally as well) of the neighbouring squares as shown below. The ogre cannot move out of the swamp.The square the ogre moves to is chosen at random from all the possible squares that it can move to. There are three types of ogre enemies that can inhabit HEkâ??s swamp; a snake, a parrot and a donkey. There is a hole in the fence around Hekâ??s swamp in the top left corner and this is where the enemies enter the swamp. Once in the swamp the enemies move in exactly the same way as the ogre; i.e. they move randomly to any of the neighbouring squares.
Every time Hek moves there is a one in three chance of an enemy entering the swamp, the type of enemy is completely random.
Although there are only three types of enemies. new types of enemy may be available in the future. You should take this into account when coding your solution.
If the ogre moves into the same square as an enemy, the ogre kills the enemy and the enemy is removed from the swamp. If the ogre moves into the same square as two or more enemies, the enemies kill the ogre and the game is over.
If the ogre is still alive and the user chooses not to make another move the game state should be saved using serialization. When the program starts the user should have the option of loading saved data or starting a new game. It should be possible to change the size of your swamp by simply changing one integer in your program. You should incorporate both the singleton pattern and the observer pattern into your game. If you cannot see an obvious place for the patterns in the game you may extend the specification of the game in any way in order to incorporate the patterns.
Please provide me with the solution as soon as possible because i need to demonstrate the same on monday 12th august. Thanking You. Purav
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# chemistry
posted by .
I have the answers, I am just interested in the formulas so that I can solve the problems, thanks!
1. Find the boiling point of a solution composed of 110g of HgCl2 (a non-ionizing solute) in 175g of water.
2. Find the freezing point of a solution composed of 50g IBr in 120 g of water. Assume 100% ionization. KBr = k1+Br1-.
3. Find the freezing point of solution composed of 74.15g MG(NO3)2 in 90 g of water. Assume 100% ionization.
4. Calculate the freezing point of a solution containing 12.0g of sucrose (C12H22O11) in 120g of water.
5. Calculate the boiling point of a solution containing 42.00 g of CuSO4 in 300g of water. CuSO4=Cu2+ + SO42-.
6. Find the molecular mass of a non-ionizing solute if 72g dissolved in 300g of water raise the boiling point to 100.860 degrees Celsius.
7. Find the molecular mass of a non-ionizing solute if 87.85g dissolved in 500g of water lower the freezing point to -3.00 degrees Celsius.
• chemistry -
1. Find the boiling point of a solution composed of 110g of HgCl2 (a non-ionizing solute) in 175g of water.
mols HgCl2 = grams/molar mass
Substitute and solve for mols.
molality = m = mols/Kg solvent
Substitute and solve for m
delta T = i*Kb*m
Your post says i = 1
Kb = 1.51
Substitute and solve for delta T, then add to 100 C to find the new boiling point.
• chemistry -
2. Find the freezing point of a solution composed of 50g IBr in 120 g of water. Assume 100% ionization. KBr = k1+Br1-.
I assume you made a typo and IBr really is KBr.
mols KBr = grams/molar mass.
Solve for mols.
molality = m = mols/kg solvent
Solve for m
delta T = i*Kf*m
i for KBr = 2
Kf = 1.86
Solve for delta T and subtract from 0 C to find the new freezing point.
delta T = i*Kf*m
i = 1
• chemistry -
3,5,5 look just like 1 and 2.
6 and 7 are done by working backward.
delta T = i*Kf or Kb *m
You're given delta T, solve for m.
Then m = mols/kg solvent.
Solve for mols
Then mols = grams/molar mass. You're given mols and grams; solve for molar mass.
i = 1 for both 6 and 7
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# See the opposite wall on a mirror [closed]
I'm in tenth standard. This is a higher-order-thinking-skills Q I found in a book. One is supposed to use laws of reflection ($\angle i = \angle r$). You can also use mathematical concepts like similarity and trigonometry.
You are in the centre of a room (I assume cuboidal). There is a plane mirror hanging on one wall. If your eyes are $h$ distance from the ground, what is the minimum height of the plane mirror required to see the entire wall behind you.
Initially, I didn't get the answer. The answer in the book says $h/2$.
Then I came up with a solution that said:
$$Mirror\ height=\frac{Wall\ height}3$$
and I thought that the mirror height was independent of $h$.
I just had a thought. Maybe we are allowed to turn the mirror to an angle. But that is not clearly specified and I don't know how it would benefit us.
• Try drawing a diagram, you should then 'see the light', no pun intended.
– Gert
Oct 1, 2015 at 15:57
• The answer obviously isn't $h/2$ because your eyes could be a distance $h \approx 0$ from the ground if you were lying down and the mirror height in that scenario can't also be zero or you wouldn't be able to see the top of the rear wall. Actually I get the same as you - wall height/3. Oct 1, 2015 at 16:12
• @JohnRennie Okay, thanks. Are such questions not allowed on this site? Because the reason given for closing is a homework question. Oct 2, 2015 at 6:23
• @JohnRennie a) I did put some effort in finding the answer b) The question is useful to future users. c) It is not exactly a homework question, it's just a question from a book. Even my teacher was not entirely convinced that the answer in the book was wrong. Oct 2, 2015 at 6:25
• @ghosts_in_the_code: It's hard to see how the answer would be of use to anyone unless they are either (a) using the same book as you or (b) encounter the same question in an exam. This is going to be a tiny subset of site users, and thats why we disapprove of such answers. If the question explored important concepts in light propagation that would be a different matter. Still, I sympathise and that's why I provided a brief answer in a comment. Oct 2, 2015 at 6:31
To see complete wall (of height $$l$$) behind himself, a person requires a plane mirror of at least $$\frac{1}{3}$$ of the height of wall. It should be noted, as mentioned in the problem, that person is standing in the middle of the room.
The ray diagram is more or less like this.
The next diagram is however not that good . But for explanation consider it to work good.
From picture, we proceed to prove as follows:
Say $$MM' = r \,\ unit \,\ and \,\ EA = EG = HF = BD = MM' = r .....(i) \,\ (\,\ by \,\ construction)$$
In $$\Delta$$ RMB, C is mid-pt. of RB and CA is parallel to RM.
So A is mid-pt. of BM.
Similarly, I is mid-pt. of QH implies that G is mid-pt. of HM'.
Again in $$\Delta$$ MFB, A is mid-pt. of BM and AE is parallel to BF.
So, E is mid-pt. of MF and from the mid-pt. theorem on $$\Delta$$ M'DH, E is mid-pt. of M`D.
Therefore, $$2EA = FB \,\ and \,\ 2GE = HD \,\ implies \,\ HD = FB = 2r ....(ii)$$
But from (i) and (ii), we can say that $$HD = HF + FD = r + FD = 2r$$ implies $$FD = r$$
Therefore $$HF = FD = BD = r \,\ i.e. \,\ HB = HF + FD + BD = 3r$$
Therefore $$l = 3r$$ implies $$r = \frac{l}{3}$$
EDIT: Now according to the question, the person's eyes are at a height $$h$$ from ground.
So, $$EA = r \,\ and \,\ AC = h-r$$
And from $$\Delta$$ BRM, $$MR = 2AC = 2(h-r)$$ Similarly M'Q = 2(h-r)
Now as assumed HB = QR = l, $$MR + MM' + M'Q = QR = l \,\ implies \,\ 2(h-r) + r + 2(h-r) = l$$ or, $$4h - 3r = l$$
And as earlier proved $$r = \frac{l}{3}$$ we have $$4h - 3r = 3r \,\ implies \,\ 6r = 4h \,\ or, \,\ r = \frac{2h}{3}$$ So your answer is $$\frac{2}{3}$$rd of the height at which the eye is stationed.
Hope it helps you.
• This is correct (though the derivation could be a lot briefer) but note that in the question $h$ is not the height of the wall. It's the height of the person's eyes above the ground. Oct 2, 2015 at 6:33
• @JohnRennie Oops..missed it. Thanks for pointing it out. I have updated my answer, this time considering $l$ as the height of wall. And ya..the answer could have been briefer..I wrote in too much detail. Oct 2, 2015 at 12:42
The answer is $\dfrac{L}{2}$, where $L$ is the length and height of the wall. When you are standing in such a way that your eyes are at height $h$, the height of the wall above you is $L-h$. Now to see that part, you need a mirror of height $\dfrac{L-h}{2}$ because of $\textit{i=r}$, you will be able to see rest $\dfrac{L-h}{2}$ . By the same logic, the mirror needs to have $\dfrac{h}{2}$ below the level of eyes. Thus, height of mirror must be $\dfrac{L-h}{2} + \dfrac{h}{2} = \dfrac{L}{2}$. Same logic applies to the other dimension of the mirror. So, if it is a $AXB$ wall , then you need $\dfrac{A}{2} X \dfrac{B}{2}$ mirror to watch the wall at your back.
• This is wrong. See Aniket's answer. Oct 2, 2015 at 6:28
• @John Rennie Okay, agree! I realize what mistake I made. Thank you. Oct 2, 2015 at 12:00
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## Tag - Linear Regression
### A Tour of The Top 10 Algorithms for Machine Learning Newbies
For machine learning newbies who are eager to understand the basic of machine learning, here is a quick tour on the top 10 machine learning algorithms used by data scientists. By James Le, Machine Learning Engineer. comments In machine learning, ther...
### Getting Up Close and Personal with Algorithms
We've put together a brief summary of the top algorithms used in predictive analysis, which you can see just below. Read to learn more about Linear Regression, Logistic Regression, Decision Trees, Random Forests, Gradient Boosting, and more. By Datai...
### KDnuggets Webinar: Modern Regression Modeling for Voter MicroTargeting, Sep 14, Sep 21
Join us for a special 2-part webinar about voting trends, and we will show how machine learning models and data science can be used in elections. Salford Systems 2-part Webinar: Part 1: Linear Regression and MARS™ (Multivariate Adaptive Regression Sp...
### Linear Regression In Real Life
A helpful guide to Linear Regression, using an example of a friends road trip to Las Vegas to highlight how it can be used in a real life situation. comments By Caroline Bento, a Software Engineer with a passion for Photo by Roman Mager on Unsplash W...
### Linear Regression, Least Squares & Matrix Multiplication: A Concise Technical Overview
Linear regression is a simple algebraic tool which attempts to find the “best” line fitting 2 or more attributes. Read here to discover the relationship between linear regression, the least squares method, and matrix multiplication. By Matthew Mayo,...
### Logistic Regression: A Concise Technical Overview
Logistic Regression is a Regression technique that is used when we have a categorical outcome (2 or more categories). Logistic Regression is one of the most easily interpretable classification techniques in a Data Scientist’s portfolio. By Asel Mendi...
### Neural Networks with Numpy for Absolute Beginners — Part 2: Linear Regression
In this tutorial, you will learn to implement Linear Regression for prediction using Numpy in detail and also visualize how the algorithm learns epoch by epoch. In addition to this, you will explore two layer Neural Networks. comments By Suraj Donthi...
### The Best Metric to Measure Accuracy of Classification Models
Measuring accuracy of model for a classification problem (categorical output) is complex and time consuming compared to regression problems (continuous output). Let’s understand key testing metrics with example, for a classification problem. By Jacob...
### The Vietnamese population profile
How Bayesian statistics convinced me to hit the gym A fun journey into the theory of Linear Regression with a Bayesian touch (Hush Hush: I use metric measurements in this post) As the title of the article suggested, I will make a scientific investiga...
### You have created your first Linear Regression Model. Have you validated the assumptions?
Linear Regression is an excellent starting point for Machine Learning, but it is a common mistake to focus just on the p-values and R-Squared values while determining validity of model. Here we examine the underlying assumptions of a Linear Regressio...
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# Relationship between capacitance voltage and resistance
### Electrical impedance - Wikipedia
The relationship between the current through a conductor with resistance The power dissipated by the resistor is equal to the voltage The charge(q), voltage ( v), and capacitance(C) of a capacitor are related as follows. Each of the three basic components resistor R, capacitor C, and inductor L can be described in terms of the relationship between the voltage across and the. For capacitors, this quantity is voltage; for inductors, this quantity is current. multiplying this quantity by the difference between the final and starting circuit values: Let's analyze the voltage rise on the series resistor-capacitor circuit shown at.
The power dissipated by the resistor is equal to the voltage multiplied by the current: If I is measured in amps and V in volts, then the power P is in watts. The potential on the straight side with the plus sign should always be higher than the potential on the curved side.
### Voltage and Current Calculations | RC and L/R Time Constants | Electronics Textbook
Notice that the capacitor on the far right is polarized; the negative terminal is marked on the can with white negative signs. The polarization is also indicated by the length of the leads: A capacitor is a device that stores electric charges. The current through a capacitor can be changed instantly, but it takes time to change the voltage across a capacitor.
The unit of measurement for the capacitance of a capacitor is the farad, which is equal to 1 coulomb per volt. The charge qvoltage vand capacitance C of a capacitor are related as follows: Differentiating both sides with respect to time gives: Rearranging and then integrating with respect to time give: If we assume that the charge, voltage, and current of the capacitor are zero atour equation reduces to: The energy stored in a capacitor in joules is given by the equation: Inductors The symbol for an inductor: Real inductors and items with inductance: The mathematical formula for determining the precise percentage is quite simple: It is derived from calculus techniques, after mathematically analyzing the asymptotic approach of the circuit values.
The more time that passes since the transient application of voltage from the battery, the larger the value of the denominator in the fraction, which makes for a smaller value for the whole fraction, which makes for a grand total 1 minus the fraction approaching 1, or percent. Universal Time Constant Formula We can make a more universal formula out of this one for the determination of voltage and current values in transient circuits, by multiplying this quantity by the difference between the final and starting circuit values: The final value, of course, will be the battery voltage 15 volts.
Our universal formula for capacitor voltage in this circuit looks like this: Since we started at a capacitor voltage of 0 volts, this increase of The same formula will work for determining current in that circuit, too.
### Resistors (Ohm's Law), Capacitors, and Inductors - Northwestern Mechatronics Wiki
Since we know that a discharged capacitor initially acts like a short-circuit, the starting current will be the maximum amount possible: We also know that the final current will be zero, since the capacitor will eventually behave as an open-circuit, meaning that eventually no electrons will flow in the circuit.
Now that we know both the starting and final current values, we can use our universal formula to determine the current after 7. Note that the figure obtained for change is negative, not positive! This tells us that current has decreased rather than increased with the passage of time. Since we started at a current of 1.
Either way, we should obtain the same answer: Using the Universal Time Constant Formula for Analyzing Inductive Circuits The universal time constant formula also works well for analyzing inductive circuits. If we start with the switch in the open position, the current will be equal to zero, so zero is our starting current value.
If we desired to determine the value of current at 3.
• Resistors (Ohm's Law), Capacitors, and Inductors
• Electrical impedance
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math
posted by .
I am a three-digit number
i am less than 200
i am divisible by 12,and by 9
my units digit is less than my tens digit
• math -
Here are the numbers between 100 and 200 that are divisible by 12.
108, 120, 132, 144, 156, 168, 180, 192
Which of those numbers meets your other criteria?
• math -
120
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# Ged Math Practice Test Pdf
Ged Math Practice Test Pdf C# by Mike D’Elizondo, Sí. Linking to my other test C# data in this link http://blogs.msdn.com/b/mark/archive/07151839.aspx you’re gonna find useful info that I’ve posted before: Summary – Tests can be given in multiple graphs or networks, so that you can try to write tests that may be a lot of work. – It’s sometimes instructive to run some test for some group of – if you have multiple groups of graph, and take one group of graph, then create – one instance of Graph, then create Graph, and test for group. – If you have multiple group of group of graph, you can try to write tests that you simply – need to click this Graph if you can’t do also, within the Graph. Meyl. DOURDE, ESBONEN, STEW. COMPARE & DEAL. WHAT HAPPEN IN AN EBSORT A/GED Math Practice? By Keith P. McGlashawski This topic was open to discussion at pre-1937, and all sorts of places have it done. It’s not really about a technical question, but a very general one because it’s a very hard problem. A little lesson goes to Peter Hoxie of the University of Vienna on how to describe a graph, and he explains his method for making lines in a graph. If you haven’t done so, you’ll see a section on graph theory. – The method for classifying graphs. Now that we’re done, we’ll split this topic up in a more abstract way — if you happen to have the method listed above, ask people to work with it. If you do not have the method, or you don’t know about it, so find some specific blog that covers it. – you can try this out a book that’s interesting for mathematicians, as it’s the reference to a topic that a lot of us do. This is my fourth year and I enjoy looking at big papers.
## Take A Course Or Do A Course
Let’s start with the simple case. If we say someone has an advantage (which we already know), it must be caused by a graph he has grown around. A long time ago I worked at the very center and got a master’s degree. Now, for any graph we can compare it to, in order to demonstrate when it is in a fixed size, we may call the next node a small seed, or a special seed, or a sub-seed. Let me give you my method for what we really want to do here, though this topic might seem silly and tedious. Here’s an article on Edgewat, which is an excellent blog of the basics, but focuses mostly on graph theory — it’s also one of the few good resources on how to explain graphs. This is actually the topic of the work that is being done in the paper, on which the author is writing a book whose title is “GDNA: The Theory and Its Use in Research.” What it’s looking at in this one is, is that many of the principles that enable us to work with Geds not only apply to complex graphs, but they apply as well. A number of his students have done experiments using some of the techniques we’ll use here, but we won’t know what they’re talking about until we “find” the one that is this one. I’ll get more into the details later, but my reading quickly found out that this is not a very special case — it’s the subject of my article. To begin, let’s just start from the assumption that there is no internal structure for a graph, ie, that there is no two-class or even three-class or even four-class. The graph in question is a circle or several circles. Of all the classes that we can mention, the class is (at least) known as the least powerful, the class “hard” is called least “bicartic,” and so there must be a class that can handle these problems. As I said, one of the principles of the paper is that we can name the classes “bicartic” and/or “hard” and “bicordic” or “bicortic.” In this paper, I’ll list a few classes IGed Math Practice Test Pdf – 2014-04) (Disclaimer: For your reference, below is my current sample, which is in click format and is made up of the tests that I already gave. Please be aware that you could easily change the PDF reading settings on a non pdf page and that they can be altered frequently. If you are copying out pages of paper that you may need to change or don’t want it to work). If you like, please simply run the PDF to the left of the question mark, and then click OK. Please note that this code is to be used on paper that isn’t being used. (Disclaimer: For your reference, below is my current sample, which is in PDF format and is made up of the tests that I already gave.
## Creative Introductions In Classroom
Please be aware that you could easily change the PDF reading settings on a non pdf page and that they can be altered frequently.) Note: The photo above appears to be mine. The original photo is here. I went on to upload the image here as well. No update to the user name in the picture, though; I also see a checkbox next to the “save” button on the top left of the page. When checking in the saved image (it’s from a book store), they seem to assume that it’s the book with the bookmarks, and if it points to the book without the bookmark as a bookmark (or something like that), I’ll use the bookmark. Since I knew it to look as if the photo was mine, I didn’t see a click. (Disclaimer: For your reference, below is my current sample, which is in PDF format and is made up of the tests that I already gave. Please be aware that you could easily change the PDF reading settings on a non pdf page and that they can be altered frequently.) Note that this test includes two pages that are in different formats, because you seem to use “just one” to test for multiple pages — in my case, there’s also a pop-up to show the four test items that appear when looking at the title. On an issue or a website with more than 100 issues, I suggest using these “test paragraphs” — the ones that you referenced. These can be rethought, maybe to reduce the number of results. (Disclaimer: For your reference, below is my current sample, which is in PDF format and is made up of the tests that I already gave. Please be aware that you could easily change the PDF reading settings on a non pdf page and that they can be altered frequently.) It’s not clear from here that the test snippets that appear on a presentation will actually apply as well as the text-based ones (even though the text-based ones would actually be ignored if you were using more than one test set). (You can see how I discussed different test items here; they’re just some pages.) 1st Page Mint-11 – I changed a title of the test paragraph above to “Mint-12 – 7-12 months; M7 – 4 years. These pages are not test paragraphs. The test paragraph labeled “M1” would not have been inserted as a test page before adding “M2” on top). 2nd Page Bolth-3 – The test paragraph labeled “Bolth-3 – 10 months; B10 – 3 years.
## Do My Online Classes For Me
At the library a random assignment from over a few books has to go to a library! At some point they end up needing to take a few days to get checked out by the library. That extra work is usually very expensive, but at once you need to budget accordingly. Then you can also say “well I’m going to spend that money trying to do this thing!” – that simple goal would bring you here. If you don’t want to check your own books out at one time you’ll want to do “I’m you could look here to spend this month studying this book!”, and if you change this into “no but that’s fine if that’s what I want then I’ll go ahead and do it!”, remember to write in chapter-two of your book plan – this will help you get the most out of your new projects and you’ll be within a little bit of a project 😉 While all this is confusing it’s even easier if you’make’ your math assignment less technical and take a more ‘proper’ approach – then just do the little things for the sake of being honest. I’ve stated many times before that I’ve helped 1,500 students achieve their goals. If you want to make a high school degree coursework so simple that your curriculum can be completed much easier your ‘workman’ might look like this: 1. The ‘proper’ way to find the math that the applicant has wanted to include (i.
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## Questions people ask while researching Drawing Plane (2-D) Figures
• ### What is a two dimensional figure made from two or more geometric figures?
A 2-dimensional discern, additionally called a plane or planar figure, is a hard and fast of line segments or aspects and curve segments or arcs, all mendacity in a unmarried plane.
• ### What is 2d shapes with examples?
Any shape that can be laid flat on a bit of paper or any mathematical aircraft is a 2nd shape. as a infant, your first drawings likely used simple shapes, together with squares, triangles, and circles. now you could find 2d shapes inside the international all round you. examples of second shapes consist of rectangles, octagons, or even hearts.
• ### What are 2d figures?
In geometry, a 2-dimensional form can be defined as a flat plane discern or a form that has dimensions – period and width. two-dimensional or 2-d shapes do no longer have any thickness and may be measured in handiest two faces. ... a circle, triangle, square, rectangle and pentagon are examples of -dimensional shapes.
• ### What are 2 and 3 dimensional shapes?
A two-dimensional (2d) shape has handiest two measurements, including period and top. a square, triangle, and circle are all examples of a 2d shape. however, a 3-dimensional (3-D) form has 3 measurements, including duration, width, and height.
Lumos Assistant
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## 557616
557,616 (five hundred fifty-seven thousand six hundred sixteen) is an even six-digits composite number following 557615 and preceding 557617. In scientific notation, it is written as 5.57616 × 105. The sum of its digits is 30. It has a total of 6 prime factors and 20 positive divisors. There are 185,856 positive integers (up to 557616) that are relatively prime to 557616.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 30
• Digital Root 3
## Name
Short name 557 thousand 616 five hundred fifty-seven thousand six hundred sixteen
## Notation
Scientific notation 5.57616 × 105 557.616 × 103
## Prime Factorization of 557616
Prime Factorization 24 × 3 × 11617
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 69702 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 557,616 is 24 × 3 × 11617. Since it has a total of 6 prime factors, 557,616 is a composite number.
## Divisors of 557616
20 divisors
Even divisors 16 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 20 Total number of the positive divisors of n σ(n) 1.44063e+06 Sum of all the positive divisors of n s(n) 883016 Sum of the proper positive divisors of n A(n) 72031.6 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 746.737 Returns the nth root of the product of n divisors H(n) 7.74127 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 557,616 can be divided by 20 positive divisors (out of which 16 are even, and 4 are odd). The sum of these divisors (counting 557,616) is 1,440,632, the average is 7,203,1.6.
## Other Arithmetic Functions (n = 557616)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 185856 Total number of positive integers not greater than n that are coprime to n λ(n) 11616 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45787 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 185,856 positive integers (less than 557,616) that are coprime with 557,616. And there are approximately 45,787 prime numbers less than or equal to 557,616.
## Divisibility of 557616
m n mod m 2 3 4 5 6 7 8 9 0 0 0 1 0 3 0 3
The number 557,616 is divisible by 2, 3, 4, 6 and 8.
• Abundant
• Polite
## Base conversion (557616)
Base System Value
2 Binary 10001000001000110000
3 Ternary 1001022220110
4 Quaternary 2020020300
5 Quinary 120320431
6 Senary 15541320
8 Octal 2101060
10 Decimal 557616
12 Duodecimal 22a840
20 Vigesimal 39e0g
36 Base36 by9c
## Basic calculations (n = 557616)
### Multiplication
n×i
n×2 1115232 1672848 2230464 2788080
### Division
ni
n⁄2 278808 185872 139404 111523
### Exponentiation
ni
n2 310935603456 173382667456720896 96680949496546879143936 53910844334466484560725016576
### Nth Root
i√n
2√n 746.737 82.3086 27.3265 14.1016
## 557616 as geometric shapes
### Circle
Diameter 1.11523e+06 3.5036e+06 9.76833e+11
### Sphere
Volume 7.26264e+17 3.90733e+12 3.5036e+06
### Square
Length = n
Perimeter 2.23046e+06 3.10936e+11 788588
### Cube
Length = n
Surface area 1.86561e+12 1.73383e+17 965819
### Equilateral Triangle
Length = n
Perimeter 1.67285e+06 1.34639e+11 482910
### Triangular Pyramid
Length = n
Surface area 5.38556e+11 2.04333e+16 455292
## Cryptographic Hash Functions
md5 68459010f6f3584beebf17f98b786c90 8d5af13fec046fbb675baae5f087010874166093 31d128bd79d0bed97d730f3d42a36208db822906f43d314d1b25a3a32adfb2cd d5785863b93bb03a2f1323fe9f92b5a86b5977479f47216090f82a18f327f4d2d23068e31639d760d1f34693bfda1b73c111290ea532e9481b6d081639670eb8 7fd7b29be3415096391671a143955d8e1e778d17
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to System Speak:
Filter by: Content type:
Stage:
Challenge level:
### There are 179 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Common Divisor
##### Stage: 4 Challenge Level:
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
##### Stage: 4 Challenge Level:
Four jewellers possessing respectively eight rubies, ten saphires, a hundred pearls and five diamonds, presented, each from his own stock, one apiece to the rest in token of regard; and they. . . .
### Geometric Parabola
##### Stage: 4 Challenge Level:
Explore what happens when you draw graphs of quadratic equations with coefficients based on a geometric sequence.
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### More Number Pyramids
##### Stage: 3 and 4 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Cosines Rule
##### Stage: 4 Challenge Level:
Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . .
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### For What?
##### Stage: 4 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Tessellating Hexagons
##### Stage: 3 Challenge Level:
Which hexagons tessellate?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Three Frogs
##### Stage: 4 Challenge Level:
Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . .
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### A Knight's Journey
##### Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
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# Colour
In this lesson we will consider how the colour of an object is related to the differential absorption, transmission and reflection of different wavelengths of light by the object.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 5
Q1.Revision: What is the correct unit for frequency?
1/5
Q2.Revision: A wave travels at 40 m/s and has a wavelength of 4 m. What is its frequency?
2/5
Q3.What type of lens is shown in the image?
3/5
Q4.Which three words can be used to describe the image produced here?
Select three (3) boxes
4/5
Q5.What type of lens is shown in the image?
5/5
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 5
Q1.Revision: What is the correct unit for frequency?
1/5
Q2.Revision: A wave travels at 40 m/s and has a wavelength of 4 m. What is its frequency?
2/5
Q3.What type of lens is shown in the image?
3/5
Q4.Which three words can be used to describe the image produced here?
Select three (3) boxes
4/5
Q5.What type of lens is shown in the image?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Colour
This quiz will test your understanding of the topic in the lesson, and prior knowledge from earlier in this topic.
## Question 6
Q1.Revision: What is the name of point F on the image?
1/6
Q2.Revision: What type of lens is shown the diagram?
2/6
Q3.What colour will an orange appear to be when placed under blue light?
3/6
Q4.Revision (HT only): When a light moves from air into glass (more optically dense), what will happen to its wavelength?
4/6
Q5.Revision: A wave travels at 24 m/s and has a wavelength of 12 m. What is its frequency?
5/6
Q6.Which of the following statements is false
6/6
Quiz:
# Colour
This quiz will test your understanding of the topic in the lesson, and prior knowledge from earlier in this topic.
## Question 6
Q1.Revision: What is the name of point F on the image?
1/6
Q2.Revision: What type of lens is shown the diagram?
2/6
Q3.What colour will an orange appear to be when placed under blue light?
3/6
Q4.Revision (HT only): When a light moves from air into glass (more optically dense), what will happen to its wavelength?
4/6
Q5.Revision: A wave travels at 24 m/s and has a wavelength of 12 m. What is its frequency?
5/6
Q6.Which of the following statements is false
6/6
# Lesson summary: Colour
### Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga
### Take part in The Big Ask.
The Children's Commissioner for England wants to know what matters to young people.
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https://www.numbersaplenty.com/18826
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Search a number
18826 = 29413
BaseRepresentation
bin100100110001010
3221211021
410212022
51100301
6223054
7105613
oct44612
927737
1018826
1113165
12aa8a
138752
146c0a
1558a1
hex498a
18826 has 4 divisors (see below), whose sum is σ = 28242. Its totient is φ = 9412.
The previous prime is 18803. The next prime is 18839. The reversal of 18826 is 62881.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 9801 + 9025 = 99^2 + 95^2 .
It is a nialpdrome in base 13.
It is a self number, because there is not a number n which added to its sum of digits gives 18826.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4705 + ... + 4708.
218826 is an apocalyptic number.
18826 is a deficient number, since it is larger than the sum of its proper divisors (9416).
18826 is an equidigital number, since it uses as much as digits as its factorization.
18826 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 9415.
The product of its digits is 768, while the sum is 25.
The square root of 18826 is about 137.2078714943. The cubic root of 18826 is about 26.6023100864.
The spelling of 18826 in words is "eighteen thousand, eight hundred twenty-six".
Divisors: 1 2 9413 18826
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http://www.alice.org/forums/printthread.php?s=162d68989992f7d9462c1bde6fe2ce0e&t=11596
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Alice Community (http://www.alice.org/community/index.php)
- The Lounge (http://www.alice.org/community/forumdisplay.php?f=4)
- - Assignment 5 - nested If statements (http://www.alice.org/community/showthread.php?t=11596)
lucyrn99 10-05-2016 12:13 PM
Assignment 5 - nested If statements
Yikes, I had trouble trying to get the concept of how my program would need to do a different action based on one of THREE different answers!
It was easy to understand how to do actions based on two different answers, because that a simple true/false or if/else choice.
I kept rereading the text, watching various youtube videos, and experimenting with my animation, until I finally had an 'aha' moment. :)
In order to get my program to work correctly, I created one variable to ask the user to input an age. Then I used that variable in the first if/else statement, for the path age is > 21, so user is too old to enter the club.
I then dragged in a second if/else statement, and repeated the process of using the variable, except this time the path was 'age is < 21, so user is too young'.
Then I dragged in a 3rd if/else statement, this time for the path age = 21, so user can enter the club.
The mistake I had been making in the beginning, was trying to make all three of the paths work, via only TWO if/else statements.
Once I figured out that I needed a separate if/else for EACH path, it was easy peasy! :D
bodyjohn 12-13-2018 11:34 AM
[QUOTE=lucyrn99;57335]Yikes, I had trouble trying to get the concept of how my program would need to do a different action based on one of THREE different answers!
It was easy to understand how to do actions based on two different answers, because that a simple true/false or if/else choice.
I kept rereading the text, watching various youtube [URL="stammibene-it.com"]videos[/URL], and experimenting with my animation, until I finally had an 'aha' moment. :)
In order to get my program to work correctly, I created one variable to ask the user to input an age. Then I used that variable in the first if/else statement, for the path age is > 21, so user is too old to enter the club.
I then dragged in a second if/else statement, and repeated the process of using the variable, except this time the path was 'age is < 21, so user is too young'.
Then I dragged in a 3rd if/else statement, this time for the path age = 21, so user can enter the club.
The mistake I had been making in the beginning, was trying to make all three of the paths work, via only TWO if/else statements.
Once I figured out that I needed a separate if/else for EACH path, it was easy peasy! :D[/QUOTE]
great!
Amzoun95 10-11-2019 10:52 AM
Once you pass a conditional statement then you've passed it until you get to it again.
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# Why aleatoric uncertainty converge to log(2)
Hy everybody,
I’m beginning with tensorflow probability and I have some difficulties to interpret my Bayesian neural network outputs. I’m working on a regression case, and started with the example provided by tensorflow notebook here: regression with tfp
As I seek to know the uncertainty of my network predictions, I dived directly into example 4 with Aleatoric & Epistemic Uncertainty. You can find my code bellow:
``````def negative_loglikelihood(targets, estimated_distribution):
return -estimated_distribution.log_prob(targets)
def posterior_mean_field(kernel_size, bias_size, dtype=None):
n = kernel_size + bias_size #number of total paramaeters (Weights and Bias)
c = np.log(np.expm1(1.))
return tf.keras.Sequential([
tfp.layers.VariableLayer(2 * n, dtype=dtype, initializer=lambda shape, dtype: random_gaussian_initializer(shape, dtype), trainable=True),
tfp.layers.DistributionLambda(lambda t: tfd.Independent(
# The Normal distribution with location loc and scale parameters.
tfd.Normal(loc=t[..., :n],
scale=1e-5 +0.01*tf.nn.softplus(c + t[..., n:])),
reinterpreted_batch_ndims=1)),
])
def prior(kernel_size, bias_size, dtype=None):
n = kernel_size + bias_size
return tf.keras.Sequential([
tfp.layers.VariableLayer(n, dtype=dtype),
tfp.layers.DistributionLambda(lambda t: tfd.Independent(
tfd.Normal(loc=t, scale=1),
reinterpreted_batch_ndims=1)),
])
def build_model(param):
model = keras.Sequential()
for i in range(param["n_layers"] ):
name="n_units_l"+str(i)
num_hidden = param[name]
model.add(tfp.layers.DistributionLambda(lambda t: tfd.Normal(loc=t[..., :1],scale=1e-3 + tf.math.softplus(0.01 * t[...,1:]))))
lr = param["learning_rate"]
model.compile(
loss=negative_loglikelihood, #negative_loglikelihood,
optimizer=optimizer,
metrics=[keras.metrics.RootMeanSquaredError()],
)
return model
``````
I think I have the same network than in tfp example, I just added few hidden layers with differents units. Also I added 0.01 in front of the Softplus in the posterior as suggested here, which allows the network to come up to good performances. Not able to get reasonable results from DenseVariational
The performances of the model are very good (less than 1% of error) but I have some questions. I ploted bellow the uncertainty (prediction_stdv) vs error. Goal here is to get intuitions about the relevance of the uncertainty. I’m only picking aleatory uncertainty here (.stddev() on my final layer).
1. My uncertainty seems to stick to a single value, and this is bad. This phenomenon is more visible as the number of epochs increases. I trained my network for 2000 epochs, am I already overfitting ?
2. The value where my uncertainty converge is 0.6931571960449219 which is really close to ln(2) (0.69314718056). In tensorflow example this boundary does not exist, they can have aleatory uncertainty bellow this value.
Thanks in advance for all feedbacks
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# Eight Beautiful Math Posters that Make You Love Mathematics
Are you looking for some cool math posters for your walls? We have curated some of Megan Emma Moree’s work for you. This series of minimalist posters created for the love of mathematics, nature, art, and education. Unfortunately, her store is under construction. But you can still reach her at [email protected].
If you are a teacher, you might want to print off these visuals and hang them on your classroom wall to entice your students to love mathematics.
### Tessellation
Tessellations are patterns found in nature that tile over the surface of an object, such as honeycombs.
### Fractal
Fractals are repeating geometric patterns that can be found in many natural occurrences, such as river networks, mountain ranges, lightning bolts, blood vessels, and DNA.
### Polyhedron
Polyhedra are geometric solids in three dimensions with flat faces and straight edges. Irregular polyhedra can be found within nature in the form of crystals and viruses.
### Fibonacci Sequence
The Fibonacci Sequence occurs when crisscrossing spirals produce an optical effect, like the florets of a sunflower. These florets, when counted, turn out to be the Fibonacci numbers.
### Boundary Condition
In mathematics, in the field of differential equations, a boundary value problem is a differential equation together with a set of additional constraints, called the boundary conditions. A solution to a boundary value problem is a solution to the differential equation which also satisfies the boundary conditions.
### Bezier Curve
Bezier Curves are straight lines corresponding to different positions at slightly different angles to give the appearance of a curve. A popular use of Bezier curves was in Mary Everest Boole’s curve stitching activities.
### Contrast Effect
A contrast effect is the enhancement or diminishment, relative to normal, of perception, cognition or related performance as a result of successive or simultaneous exposure to a stimulus of lesser or greater value in the same dimension.
### Hypercube
A hypercube is a shape with multiple faces and sides extruding in any number of dimensions. These shapes can help to explain multiple dimensions in the space around us.
1 – 8 of 8 stuffs
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# physics
posted by .
a constant force acts for 3s on a body of mass 16kg and then ceases to act.during the next 3s,the body covers 81m.find the magnitude of the force.
• physics -
assuming no friction, its final velocity is 81/3 m/s
Vf=a*t= Force/mass * t
solve for force.
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# US Boys High School 1996 Outdoor Top Middle Distance Times
last update 8-2-98
#### High School Distance gods
Additional times can be found at the High School Track & Cross Country Recruiting page
## Where the times came from
Most of the people on the lists came from the results posted to t-and-f, coach-net and xc-track. These are three excellent mailing lists which I highly recommend. Unfortunately, not all the times which would make the lists have been posted. I have checked web sites and some subscriptions (t&fnews) for results BUT I KNOW I HAVE MISSED SOME. So please, if you deserve to be on this list let me know Patrick Hoffman. The list now uses "best times in high school" so if your best time was in 1995 but you were still in high school in 1996 also let me know.
### Conversions
A note about Conversions: Jack Shepard's High School track does some yards to metric conversions and suggests a multiplication facter of .994194 from 440 yds to 400 meters. This is a the simple ratio which does not take into consideration the "tiring" facter of the longer mile and 2 mile distances. I have used a "purdy pt" conversion factor going from miles to 1600m and 3200m. This makes the 1600m converted times about .2 of a second (.4 in the 2 mile) faster than the simple ratio. This actually varies from .17 secs to .33 as one goes from near world record time to 6 minute pace. Now track and field news appears to use a simple multiplication factor when converting from 1500 to 1600 (probably based on world record times?), again the "tiring" factor should make this a more complicated factor (like purdy pts.) . But for now I am using this "magic" number 1.0666678to convert 1500 times to 1600 times. (also 3k to 3200m times). I should probably use purdy pts, instead of the magic number, but since I'm partial to 1.5k and 3k these conversions to the 1600m and 3200m "look" better. Please let me know if anyone has some more insight into these conversions. *** times in tenths only have had ".1" added with a "?" times with no tenths have been set at xx.8?
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## 3.3 Use of qualitative predictors
An important situation not covered so far is how to deal with qualitative, and not quantitative, predictors. Qualitative predictors are also known as categorical variables or, in R’s terminology, factors, and are very common in many fields, such as in social sciences. Dealing with them requires some care and proper understanding of how these variables are represented.
The simplest case is the situation with two levels. A binary variable $$C$$ with two levels (for example, a and b) can be encoded as
\begin{align*} D=\left\{\begin{array}{ll} 1,&\text{if }C=b,\\ 0,&\text{if }C=a. \end{array}\right. \end{align*}
$$D$$ is a dummy variable: it codifies with zeros and ones the two possible levels of the categorical variable. An example of $$C$$ is smoker, which has levels yes and no. The dummy variable associated is $$D=1$$ if a person smokes and $$D=0$$ if a person is non-smoker.
The advantage of this dummification is its interpretability in regression models. Since level a corresponds to $$0,$$ it can be seen as the reference level to which level b is compared. This is the key point in dummification: set one level as the reference, codify the rest as departures from it.
The previous interpretation translates easily to the linear model. Assume that the dummy variable $$D$$ is available together with other predictors $$X_1,\ldots,X_p.$$ Then:
\begin{align*} \mathbb{E}[Y|X_1=x_1,\ldots,X_p=x_p,D=d]=\beta_0+\beta_1x_1+\cdots+\beta_px_p+\beta_{p+1}d. \end{align*}
The coefficient associated to $$D$$ is easily interpretable: $$\beta_{p+1}$$ is the increment in the mean of $$Y$$ associated to changing $$D=0$$ (reference) to $$D=1,$$ while the rest of the predictors are fixed. Or in other words, $$\beta_{p+1}$$ is the increment in mean of $$Y$$ associated to changing the level of the categorical variable from a to b.
R does the dummification automatically, translating a categorical variable $$C$$ into its dummy version $$D,$$ if it detects that a factor variable is present in the regression model.
Let’s see now the case with more than two levels, for example, a categorical variable $$C$$ with levels a, b, and c. If we take a as the reference level, this variable can be represented by two dummy variables:
\begin{align*} D_1=\left\{\begin{array}{ll}1,&\text{if }C=b,\\0,& \text{if }C\neq b\end{array}\right. \end{align*}
and
\begin{align*} D_2=\left\{\begin{array}{ll}1,&\text{if }C=c,\\0,& \text{if }C\neq c.\end{array}\right. \end{align*}
Therefore, we can represent the levels of $$C$$ as in the following table.
$$C$$ $$D_1$$ $$D_2$$
$$a$$ $$0$$ $$0$$
$$b$$ $$1$$ $$0$$
$$c$$ $$0$$ $$1$$
The interpretation of the regression models in the presence of $$D_1$$ and $$D_2$$ is very similar to the one before. For example, for the linear model, the coefficient associated to $$D_1$$ gives the increment in the mean of $$Y$$ when the level of $$C$$ changes from a to b, and the coefficient for $$D_2$$ gives the increment in mean of $$Y$$ when $$C$$ changes from a to c.
In general, if we have a categorical variable $$C$$ with $$J$$ levels, then the previous process is iterated and the number of dummy variables required to encode $$C$$ is $$J-1\;$$62. Again, R does the dummification automatically if it detects that a factor variable is present in the regression model.
Let’s see an example with the famous iris dataset.
# iris dataset -- factors in the last column
summary(iris)
## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100 setosa :50
## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300 versicolor:50
## Median :5.800 Median :3.000 Median :4.350 Median :1.300 virginica :50
## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199
## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800
## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500
# Summary of a linear model
mod1 <- lm(Sepal.Length ~ ., data = iris)
summary(mod1)
##
## Call:
## lm(formula = Sepal.Length ~ ., data = iris)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.79424 -0.21874 0.00899 0.20255 0.73103
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.17127 0.27979 7.760 1.43e-12 ***
## Sepal.Width 0.49589 0.08607 5.761 4.87e-08 ***
## Petal.Length 0.82924 0.06853 12.101 < 2e-16 ***
## Petal.Width -0.31516 0.15120 -2.084 0.03889 *
## Speciesversicolor -0.72356 0.24017 -3.013 0.00306 **
## Speciesvirginica -1.02350 0.33373 -3.067 0.00258 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3068 on 144 degrees of freedom
## Multiple R-squared: 0.8673, Adjusted R-squared: 0.8627
## F-statistic: 188.3 on 5 and 144 DF, p-value: < 2.2e-16
# Speciesversicolor (D1) coefficient: -0.72356. The average increment of
# Sepal.Length when the species is versicolor instead of setosa (reference)
# Speciesvirginica (D2) coefficient: -1.02350. The average increment of
# Sepal.Length when the species is virginica instead of setosa (reference)
# Both dummy variables are significant
# How to set a different level as reference (versicolor)
iris$Species <- relevel(iris$Species, ref = "versicolor")
# Same estimates, except for the dummy coefficients
mod2 <- lm(Sepal.Length ~ ., data = iris)
summary(mod2)
##
## Call:
## lm(formula = Sepal.Length ~ ., data = iris)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.79424 -0.21874 0.00899 0.20255 0.73103
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.44770 0.28149 5.143 8.68e-07 ***
## Sepal.Width 0.49589 0.08607 5.761 4.87e-08 ***
## Petal.Length 0.82924 0.06853 12.101 < 2e-16 ***
## Petal.Width -0.31516 0.15120 -2.084 0.03889 *
## Speciessetosa 0.72356 0.24017 3.013 0.00306 **
## Speciesvirginica -0.29994 0.11898 -2.521 0.01280 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3068 on 144 degrees of freedom
## Multiple R-squared: 0.8673, Adjusted R-squared: 0.8627
## F-statistic: 188.3 on 5 and 144 DF, p-value: < 2.2e-16
# Speciessetosa (D1) coefficient: 0.72356. The average increment of
# Sepal.Length when the species is setosa instead of versicolor (reference)
# Speciesvirginica (D2) coefficient: -0.29994. The average increment of
# Sepal.Length when the species is virginica instead of versicolor (reference)
# Both dummy variables are significant
# Coefficients of the model
confint(mod2)
## 2.5 % 97.5 %
## (Intercept) 0.8913266 2.00408209
## Sepal.Width 0.3257653 0.66601260
## Petal.Length 0.6937939 0.96469395
## Petal.Width -0.6140049 -0.01630542
## Speciessetosa 0.2488500 1.19827390
## Speciesvirginica -0.5351144 -0.06475727
# The coefficients of Speciessetosa and Speciesvirginica are
# significantly positive and negative, respectively
# Show the dummy variables employed for encoding a factor
contrasts(iris$Species) ## setosa virginica ## versicolor 0 0 ## setosa 1 0 ## virginica 0 1 iris$Species <- relevel(iris$Species, ref = "setosa") contrasts(iris$Species)
## versicolor virginica
## setosa 0 0
## versicolor 1 0
## virginica 0 1
It may happen that one dummy variable, say $$D_1,$$ is not significant, while other dummy variables of the same categorical variable, say $$D_2,$$ are significant. For example, this happens in the example above at level $$\alpha=0.01.$$ Then, in the considered model, the level associated to $$D_1$$ does not add relevant information for explaining $$Y$$ with respect to the reference level.
Do not codify a categorical variable as a discrete variable. This constitutes a major methodological failure that will flaw the subsequent statistical analysis.
For example if you have a categorical variable party with levels partyA, partyB, and partyC, do not encode it as a discrete variable taking the values 1, 2, and 3, respectively. If you do so:
• You assume implicitly an order in the levels of party, since partyA is closer to partyB than to partyC.
• You assume implicitly that partyC is three times larger than partyA.
• The codification is completely arbitrary – why not consider 1, 1.5, and 1.75 instead?
The right way of dealing with categorical variables in regression is to set the variable as a factor and let R do the dummification internally.
### 3.3.1 Case study application
Let’s see what the dummy variables are in the Boston dataset and what effect they have on medv.
# Load the Boston dataset
data(Boston, package = "MASS")
# Structure of the data
str(Boston)
## 'data.frame': 506 obs. of 14 variables:
## $crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ... ##$ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ... ##$ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ... ##$ rm : num 6.58 6.42 7.18 7 7.15 ...
## $age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ... ##$ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $rad : int 1 2 2 3 3 3 5 5 5 5 ... ##$ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ... ##$ black : num 397 397 393 395 397 ...
## $lstat : num 4.98 9.14 4.03 2.94 5.33 ... ##$ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
# chas is a dummy variable measuring if the suburb is close to the river (1)
# or not (0). In this case it is not codified as a factor but as a 0 or 1
# (so it is already dummified)
# Summary of a linear model
mod <- lm(medv ~ chas + crim, data = Boston)
summary(mod)
##
## Call:
## lm(formula = medv ~ chas + crim, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.540 -5.421 -1.878 2.575 30.134
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 23.61403 0.41862 56.409 < 2e-16 ***
## chas 5.57772 1.46926 3.796 0.000165 ***
## crim -0.40598 0.04339 -9.358 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.373 on 503 degrees of freedom
## Multiple R-squared: 0.1744, Adjusted R-squared: 0.1712
## F-statistic: 53.14 on 2 and 503 DF, p-value: < 2.2e-16
# The coefficient associated to chas is 5.57772. That means that if the suburb
# is close to the river, the mean of medv increases in 5.57772 units for
# the same house and neighborhood conditions
# chas is significant (the presence of the river adds a valuable information
# for explaining medv)
# Summary of the best model in terms of BIC
summary(modBIC)
##
## Call:
## lm(formula = medv ~ crim + zn + chas + nox + rm + dis + rad +
## tax + ptratio + black + lstat, data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.5984 -2.7386 -0.5046 1.7273 26.2373
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 36.341145 5.067492 7.171 2.73e-12 ***
## crim -0.108413 0.032779 -3.307 0.001010 **
## zn 0.045845 0.013523 3.390 0.000754 ***
## chas 2.718716 0.854240 3.183 0.001551 **
## nox -17.376023 3.535243 -4.915 1.21e-06 ***
## rm 3.801579 0.406316 9.356 < 2e-16 ***
## dis -1.492711 0.185731 -8.037 6.84e-15 ***
## rad 0.299608 0.063402 4.726 3.00e-06 ***
## tax -0.011778 0.003372 -3.493 0.000521 ***
## ptratio -0.946525 0.129066 -7.334 9.24e-13 ***
## black 0.009291 0.002674 3.475 0.000557 ***
## lstat -0.522553 0.047424 -11.019 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.736 on 494 degrees of freedom
## Multiple R-squared: 0.7406, Adjusted R-squared: 0.7348
## F-statistic: 128.2 on 11 and 494 DF, p-value: < 2.2e-16
# The coefficient associated to chas is 2.71871. If the suburb is close to
# the river, the mean of medv increases in 2.71871 units
# chas is significant as well in the presence of more predictors
We will see how to mix dummy and quantitative predictors in Section 3.4.3.
1. To account for the $$J-1$$ possible departures from a reference level.↩︎
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# Math
Use the given information about a polynomial whose coefficients are real numbers to find the remaining zeros of the polynomial.
Degree: 6
Zeros: -6 + 13i^3,
-8 + s^2i,
-3 - 4i
1. 👍 0
2. 👎 0
3. 👁 151
1. i^3=-i
s^2i ???
1. 👍 0
2. 👎 0
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# 28 days from today | what is 28 days from today | 28 days from now
Select Dates below
A 28 days from today / 28 days from now calculator, also known as a date calculator, is a handy tool that lets you quickly and easily calculate the number of days between dates. You can use it, for example, when counting up to your birthday or when you are wondering how many days are left before a critical deadline.
But that’s not all! You can also use this A 28 days from today calculator to check which date is after a certain number of days, eg. For example, suppose you run a 28 day fitness challenge to find out when the challenge ends.
## What is 28 days from today/ 28 days from now
It is a date calculator to know the exact date in thirty days without counting.
For example months with 28 days Today is March 15, 2021 which means 28 days from today is April 14, 2021. You can check this by using a date calculator to measure the number of days.
Do you need to calculate 28 days or 100 days from today from a certain date? Use the Days From Date calculator.
April 14, 2021 is a Wednesday. This is the 104th day of the year and the fifteenth seven day stretch of the year (accepting every week begins on Monday) or the second quarter of the year. There are 28 days in this month. 2021 is not high so there are 365 days this year.
The abbreviated form used in the United States for this date is April 14, 2021, and almost everywhere in the world is April 14, 2021. And if you want to calculate 28 days from today saturday 26 june 2021. You can also check how many months have 28 days in a year.
### How many days between dates for 28 days from now?
Imagine that you have a complex project working on it. Today is march 15 2021. Your boss wants to check progress 14 days before the end of the project. He also hopes that the project will be completed on March 29.
## You can ask yourself two questions
• On what date will progress control be carried out?
• How long do you have before the advancement check?
To answer the first question, we have to subtract 14 days from March 29th. If you enter 14 days as the day between and 29 March as the end date in the day counter, you will immediately see that the progress check takes place on March 15.
Now that we know when the progress check will take place, we can count the number of days between March 15 (start date) and March 29 (end date). The result is 14 days – this is the time you still have to work on your project.
Not sure if the results are correct? Go ahead and check with the date duration 28 days from now the calculator provided here!
## What if you only counted the 28 days from today’s date?
In some cases, you may want to skip weekends and only count the 28 day calendar days of the week. This can be useful when you find out that you have a deadline based on a certain number of workdays.
Attempting to see which day falls with the specific date distinction from 28 days after the fact of the week beginning today, you can tally every day and skip Saturday and Sunday.
## 28 days from today’s date
Start your calculations, which fall on Monday, today. The next day is Tuesday. To get exactly 28 days from tomorrow, or 28 days from yesterday you actually need to calculate a total of 42 days (including weekends).
This means the 28 days of the week from today will be April 26, 2021 28 days from date. If you are calculating workdays or which months have 28 days, you will need to adjust these dates for all holidays.
April 26, 2021 is Monday. This is the 116th day of the year and the seventeenth seven day stretch of the year (accepting every week begins on Monday) or the second quarter of the year. There are 28 days left in this month.
2021 is not high so there are 365 days this year. The abbreviated form used in the United States for this date is April 26, 2021, and almost everywhere in the world is April 26, 2021 28 days from date.
### Ways to use the 28 days from now
There are three ways to use this date and date calculator:
• The number of days between the dates is unknown: Knowing your start and end dates (like summer vacation) is a quick way to count the days in between. All you have to do is enter both dates into the day counter. Remember that the end date has to be after the start date – after all, you can’t go back in time!
• Start date unknown: To find out what date is 57 days ago, enter the end date and the number of days that have passed in the appropriate fields.
• End date unknown: You can also use this calculator to find out what day after a certain period of time. To do this, enter the start date and the number of days between them in the day counter.
## How to use the 28 days from today
To use this date calculator, you have to follow 4 simple steps to calculate the date 28 days later or 28 days from today.
• Enter the date: In the field at the top left, you can select the starting day, month, and year for your calculation. You can also set the start time under entered data. To set date entry, you can use preset dates such as Today, Tomorrow, Yesterday. “Now” button sets the starting time of the current moment.
• Daily values and operators: In the fields at the top right you can select the day, month and year to add or subtract from the start date. There you can also select the hour, minute and second values to be calculated.
• Date options: Under date input field you can select date format and time format and specify whether month name and day name should be displayed or not. Date formats are useful for making calculated dates easier to read.
• Result of date calculation: After selecting the start date and date options you want, click the Calculate Date button. The date results are displayed below the Calculate button with the option you selected.
Do you need to find out how many days until today? See the day to the calculator
##### Conclusion
This website provides an online calculator for the day from today which can be used to find the exact date X days after. You can also enter a negative number to find out when X days fall before today. This tool lets you calculate deadlines if you have a certain number of days.
Or on the other hand read the full page to become familiar with due dates in case you’re just checking the times of the week or days of the week and skipping Saturdays and Sundays. If you are trying to measure the number of days between two dates, you can switch to a date difference calculator.
##### FAQs
• what day is 28 days from today
• what is the date 28 days from today
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• how many 28 days in a year
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Programming-Idioms
This language bar is your friend. Select your favorite languages!
Idiom #81 Round floating point number to integer
Declare the integer y and initialize it with the rounded value of the floating point number x .
Ties (when the fractional part of x is exactly .5) must be rounded up (to positive infinity).
``#include <cmath>``
``int y = static_cast<int>(std::floor(x + 0.5f));``
``Y : constant Integer := Integer (Float'Rounding (X));``
``#include <math.h>``
``int y = (int)floorf(x + 0.5f);``
``(defn rnd [y] (int (+ y 0.5)))``
``using System;``
``long y = (long)Math.Round(x);``
``import std.math: round;``
``int y = cast(int) x.round;``
``````var y = x.round();
``````
``y = Kernel.round x``
``````integer :: y
y = nint(x)``````
``import "math"``
``y := int(math.Floor(x + 0.5))``
``````y = floor (x + 1/2)
``````
``var y = Math.round(x);``
``long y = Math.round(x);``
``````(defun rnd (y) (floor (+ y 0.5)))
``````
``````function round(float)
local int, part = math.modf(float)
if float == math.abs(float) and part >= .5 then return int+1 -- positive float
elseif part <= -.5 then return int-1 -- negative float
end
return int
end``````
``````function round(float)
return math.floor(float + .5)
end``````
``\$y = (int) round(\$x);``
``````var
y: integer;
x: double;
begin
y := round(x);
end.``````
``my \$y = int(\$x + 1/2);``
``y = int(x + 0.5)``
``y = (x + 1/2r).floor``
``let y = x.round() as i64;``
``(define y (round x))``
deleplace
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# 2.6: Functions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
hyypothes.is tag: s20iostpy06ualr
## Learning Objectives
Students will be able to:
Content:
• Explain the meaning and purpose of a function
• Recognize a function definition, function header, and function call in a program
• Combine the use of functions with if/else statements
• Explain programs that use the same function multiple times
• Use good test data for programs that include functions
Process:
• Write code that includes function definitions and function calls
• Write programs that incorporate functions and if/else statements
Prior Knowledge
• Python concepts from Activities 1-5
• Understanding of flowchart input symbols
## Model 1: User defined Functions
function is a block of organized, reusable code that is used to perform a single, related action. Functions provide better modularity for your application and a high degree of code reusing. A function definition is the segment of code that tells the program what to do when the function is executed. The first line of a function definition is known as the function header.
Python Program 1
# Programmer: Monty Python
# Date: today's date
# Program uses a function to print a message
# User defined Function definition
def printMessage():
print("Welcome to Python.")
print("Learn the power of functions!")
# User defined Function definition
def main():
print("Hello Programmer!")
# Function call
printMessage()
# Function Call
main()
print("Done!")
Critical Thinking Questions:
1. Closely examine the Python program above.
a. What Python keyword is used to indicate that a code segment is a user defined function definition?
b. What are the two function headers in the Python code?
c. The name of the user defined function is in the function header. What are the names of the two user defined functions?
d. Enter and execute the Python program. What is the output?
e. What one line of code could you add to the program to repeat the lines "Welcome to Python" and "Learn the power of functions!" twice. Where would you add the code?
2. Examine the following program
Python Program 2
# Description: This program uses a function to calculate
# the area of a circle given the radius.
import math
def main():
main()
print("Done!")
a. Label the user defined function definition and the function calls.
b. The user defined function call and the function definition for calculateArea each include a variable within the parentheses. The variable in the function call is known as an argument. The variable in the function definition is called a parameter. What is the parameter in the function definition? What is its purpose?
c. In this example the argument in the function definition and the argument in the function call have the same name. Is this required?
d. Enter and execute the program. Verify your answer to question ‘c’ by changing the variable name and the function argument in the main function from radius to number. Do not change the parameter variable name in the function definition. Does the program still work?
e. Add a line of code to the main program that calls the calculateArea function and sends the value “6” as the argument. Execute the program to be sure it works properly.
f. Add another function to the program that calculates and prints the diameter of a circle, given the radius as the parameter. Place the function definition above call to the main function of the program. Write the function below.
g. Add another line of code to the main part of the program that calls the function that was created in part ‘f’. Send the radius entered by the user as the argument to the function.
Note
Parameter the variable listed inside the parenthesis of the function definition, there can be more than one variable (a,b below)
Argument: the value sent to a function when it is called, there can be more than one value, and the number of values needs to equal the number of variables in the defined function being called (fname, lname).
def printName(a,b):
print(a," ",b)
printName(fname,lname)
3. What is the purpose of a function? Why do programmers use them?
4. Carefully examine the following program.
Python Program 3
print(num1, "+", num2,"=",num1+num2)
def subtractNumbers(num1, num2):
print(num1, "+", num2,"=",num1-num2)
def main():
firstNumber = int(input("Enter a number between 1 and 10: "))
secondNumber = int(input("Enter another number between 1 and 10: "))
operator = input("Enter a + to add or a - to subtract: ")
if operator == "+":
elif operator == "-":
subtractNumbers(firstNumber,secondNumber)
else:
print("Invalid operator!")
######call to main program #######
main()
print("Done!")
a. Circle the first line of code executed by the Python interpreter.
b. Enter and execute the program. Use the following test data and indicate what the output is for each set of data.
Data Set Operand 1 Operand 2 Operator 1 2 6 + 2 3 8 - 3 34 23 * 4 4 5 /
c. What problems did you notice when you entered Data Sets 3 and 4?
d. Add code to the program that would warn the user about the problems that could occur when data similar to that in Data Sets 3 and 4 are entered. See sample output below. List the lines of code below the sample output.
NOTE: There are two types of Functions, Void Functions, that do not send back information to where they are called, and Value Returning Functions, that send back information to where they are called.
5. Enter and execute the code below. Carefully examine the code.
Python Program 4
# This program prompts the user for two numbers,
# calls a function to determine the smaller number
# and prints the smaller number that is returned from the function
def getSmaller(num1, num2):
if num1<num2:
smaller = num1
else:
smaller = num2
return smaller
def main():
userInput1 = int(input("Enter a number: "))
userInput2 = int(input("Enter a second number: "))
smallerNumber = getSmaller(userInput1,userInput2)
print("The smaller of the two numbers is", smallerNumber)
###call to main program####
main()
print("Done!")
a. What is the new keyword used in the function definition? What do you think the keyword tells the program to do?
b. Circle the line of code from the program that includes the function call to getSmaller.
c. In a void function, the function call is on a line by itself. Why is this function call placed on the right-hand-side of an assignment statement?
d. What are the arguments used for the function call?
6. Examine the following Python program.
Python Program 5
def getMolarity(grams, molMass, volume):
volume = volume/1000
molarity = grams/molMass/volume
return molarity
def main():
grams = float(input("Enter grams of substance: "))
molMass = float(input("Enter molar mass of substance in grams per mole: "))
volume = float(input("Enter volume in milliliters: "))
print("Your Molarity is: ", getMolarity(grams, molMass, volume)," mol/L.")
main()
print("Done!")
a. Circle the user defined function call in the main program.
b. How does the location of the function call in the main program differ from others you have seen today?
c. Is the function a void function or a value returning function?
Information: Style guides define the maximum number of characters in a line of computer source code. The standard for python is 79 characters. This allows for several files to be open side-by-side, and makes code easier to read. You can read more at: https://www.python.org/dev/peps/pep-0008/#code-lay-out
7. The following Python program performs the same operations as found in question 6.
Python Program 6
def getMolarity(grams, molMass, volume):
volume = volume/1000
molarity = grams/molMass/volume
return molarity
def main():
grams = float(input("Enter grams of substance: "))
molMass = float(input("Enter molar mass of substance in grams per mole: "))
volume = float(input("Enter volume in milliliters: "))
print("The molarity of your solution is: ",
format(getMolarity(grams, molMass, volume),'.2f'), " mol/L.")
main()
print("Done!")
a. Circle the user defined function call in the main program
b. How does the program above show that a user defined function can be the argument of another function?
c. Why is the print function within the main function written on two lines?
d. Execute the program by copying and pasting this code into your IDE editor window.
e. What happens if you remove the indentation from the second line of the the print
function in the main function as show below:
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, this material is a modification by Ehren Bucholtz and Robert Belford of CS-POGIL content, with the original material developed by Lisa Olivieri, which is available here.
Application Questions: Use the Python Interpreter to check your work
1. Write a function that draws a frog. Call the function to be sure it works.
Sample frog:
2. Expand the program in #1 to produce the following output:
3. Write a Python program that prompts the user for three words and prints the word that comes last alphabetically. Use a function to create the program.
This page titled 2.6: Functions is shared under a not declared license and was authored, remixed, and/or curated by Robert Belford.
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Best Excel Tutorial
The largest Excel knowledge base ✅ The best place to learn Excel online ❤️
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Resolving Excel can’t find project or library
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https://support.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.lapackeig.zgebal.html
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# naginterfaces.library.lapackeig.zgebal¶
naginterfaces.library.lapackeig.zgebal(job, n, a)[source]
zgebal balances a complex general matrix in order to improve the accuracy of computed eigenvalues and/or eigenvectors.
For full information please refer to the NAG Library document for f08nv
https://support.nag.com/numeric/nl/nagdoc_30.1/flhtml/f08/f08nvf.html
Parameters
jobstr, length 1
Indicates whether is to be permuted and/or scaled (or neither).
is neither permuted nor scaled (but values are assigned to , and ).
is permuted but not scaled.
is scaled but not permuted.
is both permuted and scaled.
nint
, the order of the matrix .
acomplex, array-like, shape
The matrix .
Returns
acomplex, ndarray, shape
is overwritten by the balanced matrix. If , is not referenced.
iloint
The values and such that on exit is zero if and or .
If or , and .
ihiint
The values and such that on exit is zero if and or .
If or , and .
scalefloat, ndarray, shape
Details of the permutations and scaling factors applied to . More precisely, if is the index of the row and column interchanged with row and column and is the scaling factor used to balance row and column then
The order in which the interchanges are made is to then to .
Raises
NagValueError
(errno )
On entry, error in parameter .
Constraint: , , or .
(errno )
On entry, error in parameter .
Constraint: .
Notes
zgebal balances a complex general matrix . The term ‘balancing’ covers two steps, each of which involves a similarity transformation of . The function can perform either or both of these steps.
1. The function first attempts to permute to block upper triangular form by a similarity transformation:
where is a permutation matrix, and and are upper triangular. Then the diagonal elements of and are eigenvalues of . The rest of the eigenvalues of are the eigenvalues of the central diagonal block , in rows and columns to . Subsequent operations to compute the eigenvalues of (or its Schur factorization) need only be applied to these rows and columns; this can save a significant amount of work if and . If no suitable permutation exists (as is often the case), the function sets and , and is the whole of .
2. The function applies a diagonal similarity transformation to , to make the rows and columns of as close in norm as possible:
This scaling can reduce the norm of the matrix (i.e., ) and hence reduce the effect of rounding errors on the accuracy of computed eigenvalues and eigenvectors.
References
Golub, G H and Van Loan, C F, 1996, Matrix Computations, (3rd Edition), Johns Hopkins University Press, Baltimore
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https://schematron.org/tornado-diagram-pmp.html
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Simply explained by a PMI-certified Project Manager. Home · Acronyms and The tornado diagram is a special bar chart that is used in sensitivity analysis. The tornado diagram is one of the methods used to display the sensitivity analysis.
Simply explained by a PMI-certified Project Manager. Home · Acronyms and The tornado diagram is a special bar chart that is used in sensitivity analysis.
The tornado diagram is one of the methods used to display the sensitivity analysis. Tornado Diagram.
A special type of bar chart used in sensitivity analysis for comparing the relative importance of the variables. Best PMP Resources. 35 PMP.
Tornado Diagram is one of the tools used to complete sensitivity on how to create tornado diagram using excel, Some PMP exam questions. The Tornado diagram is a visual display of the sensitivity of various risks with regards to their positive or negative impact. Tornado Diagram Analysis Criteria: The Tornado diagram is a useful tool to visually understand the uncertainty of various risks and their potential impact.The diagram contained a series of bars with the length of the bars corresponding to the risk impact on the project.
The longer the bars, the greater was the risk presented. Such a diagram is likely to be: A.
An assessment diagram B. A triangular distribution C.
A tornado diagram D. A . About six months ago, we received word from one of our PMP prep students that he had a strange weather topic appear on his PMP exam, something like a “hurricane chart” or something.
After a bit of investigating and confirmation with the student, in all probabilities, he had a question on a tornado diagram. While. A Tornado diagram, also called tornado plot or tornado chart, is a special type of Bar chart, where the data categories are listed vertically instead of the standard horizontal presentation, and the categories are ordered so that the largest bar appears at the top of the chart, the second largest appears second from the top, and so on.
A tornado diagram has the following characteristics: The longer the bar, the more sensitive the project objective is to the risk. The risks are presented in descending order, with the largest impact on the top and the least impact on the bottom. It allows the team to focus on those risks with the greatest impact on a project objective.
The tornado diagram is a special bar chart that is used in sensitivity analysis. The sensitivity analysis is a modeling technique that determines which risks have the most impact on the project. As one of the tools used in sensitivity analysis, the tornado diagram is used to compare the importance (relative) of different variables.Project Management Best Practice: Tornado DiagramWhat Constitutes a Good Tornado Diagram?
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# ››17,929 days before now
Want to figure out the date that is exactly seventeen thousand nine hundred and twenty-nine days before now without counting?
Today is August 21, 2019 so that means that 17929 days before today would be July 20, 1970.
You can check this by using the date difference calculator to measure the number of days from today to Jul 20, 1970.
Need to calculate 17929 days before a specific date? Use the Days From Date calculator.
# ››July, 1970 calendar
Su M Tu W Th F Sa 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
July 20th, 1970 is a Monday. It is the 201st day of the year, and in the 30th week of the year (assuming each week starts on a Monday), or the 3rd quarter of the year. There are 31 days in this month. 1970 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 7/20/1970, and almost everywhere else in the world it's 20/7/1970.
# ››What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 17,929 weekdays before today, you can count up each day skipping Saturdays and Sundays.
Start your calculation with today, which falls on a Wednesday. Counting backward, the next day would be a Tuesday.
To get exactly seventeen thousand nine hundred and twenty-nine weekdays before now, you actually need to count 25,101 total days (including weekend days). That means that 17929 weekdays before today would be November 30, 1950.
If you're counting business days, don't forget to adjust this date for any holidays.
# ››November, 1950 calendar
Su M Tu W Th F Sa 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
November 30th, 1950 is a Thursday. It is the 334th day of the year, and in the 48th week of the year (assuming each week starts on a Monday), or the 4th quarter of the year. There are 30 days in this month. 1950 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 11/30/1950, and almost everywhere else in the world it's 30/11/1950.
# ››Enter the number of days to count
Type in the number of days you want to calculate from today. If you want to find a previous date, you can enter a negative number to figure out the number of days before today.
Due to date calculation restrictions, the allowable range is from -42984 to 6725.
Number of days from now:
# ››Days From Now calculator
This site provides an online Days From Today calculator to help you find the date that occurs exactly X days from now. You can also enter a negative number to find out when X days before today happened to fall. You can use this tool to figure out a deadline if you have a certain number of days remaining. Or read the full page to learn more about the due date if you're counting business days or weekdays only, skipping Saturday and Sunday. If you're trying to measure the number of days between two dates, you can switch to the Date Difference calculator instead.
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# American Institute of Mathematical Sciences
2009, 3(2): 205-217. doi: 10.3934/amc.2009.3.205
## Further results on implicit factoring in polynomial time
1 Indian Statistical Institute, 203 B T Road, Kolkata 700 108, India 2 Indian Statistical Institute, 203 B T Road,, Kolkata 700 108, India
Received March 2009 Revised April 2009 Published May 2009
In PKC 2009, May and Ritzenhofen presented interesting problems related to factoring large integers with some implicit hints. One of the problems is as follows. Consider $N_1 = p_1 q_1$ and $N_2 = p_2 q_2$, where $p_1, p_2, q_1, q_2$ are large primes. The primes $p_1, p_2$ are of same bit-size with the constraint that certain amount of Least Significant Bits (LSBs) of $p_1, p_2$ are same. Further the primes $q_1, q_2$ are of same bit-size without any constraint. May and Ritzenhofen proposed a strategy to factorize both $N_1, N_2$ in poly$(\log N)$ time ($N$ is an integer with same bit-size as $N_1, N_2$) with the implicit information that $p_1, p_2$ share certain amount of LSBs. We explore the same problem with a different lattice-based strategy. In a general framework, our method works when implicit information is available related to Least Significant as well as Most Significant Bits (MSBs). Given $q_1, q_2 \approx$Nα, we show that one can factor $N_1, N_2$ simultaneously in poly$(\log N)$ time (under some assumption related to Gröbner Basis) when $p_1, p_2$ share certain amount of MSBs and/or LSBs. We also study the case when $p_1, p_2$ share some bits in the middle. Our strategy presents new and encouraging results in this direction. Moreover, some of the observations by May and Ritzenhofen get improved when we apply our ideas for the LSB case.
Citation: Santanu Sarkar, Subhamoy Maitra. Further results on implicit factoring in polynomial time. Advances in Mathematics of Communications, 2009, 3 (2) : 205-217. doi: 10.3934/amc.2009.3.205
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2016 Impact Factor: 0.8
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If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka
For in GOD we live, and move, and have our being. - Acts 17:28
The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka
# Word Problems on Integration
Prerequisite Topics:
Factoring
Partial Fractions
Exponents and Logarithms
Trigonometry
Differential Calculus
For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.
Solve all questions
Indicate the method(s) you used as applicable
Show all work
(1.) Evaluate $\displaystyle\int -7dx$
Power Rule
$\displaystyle\int -7dx \\[5ex] = \displaystyle\int -7 * 1 * dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \displaystyle\int -7x^0dx \\[5ex] = -7 \displaystyle\int x^0dx \\[5ex] = -7 * \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = -7 * \dfrac{x^1}{1} + C \\[5ex] = -7x + C$
(2.) Determine the antiderivative of $\cot x$
Integration by Algebraic Substitution
$\displaystyle\int \cot xdx \\[5ex] \cot x = \dfrac{\cos x}{\sin x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \sin x \\[3ex] \dfrac{dp}{dx} = \cos x \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{\cos x} \\[5ex] dx = \dfrac{dp}{\cos x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \cot xdx = \displaystyle\int \dfrac{\cos x}{\sin x} dx \\[5ex] = \displaystyle\int \dfrac{\cos x}{p} * \dfrac{dp}{\cos x} \\[5ex] = \displaystyle\int \dfrac{dp}{p} \\[5ex] = \ln p + C \\[3ex] = \ln \sin x + C$
(3.) JAMB Evaluate $\displaystyle\int_1^3 (x^2 - 1)dx$
$A.\:\: -6\dfrac{2}{3} \\[5ex] B.\:\: 6\dfrac{2}{3} \\[5ex] C.\:\: \dfrac{2}{3} \\[5ex] D.\:\: -\dfrac{2}{3} \\[5ex]$
Power Rule
$\displaystyle\int_1^3 (x^2 - 1)dx \\[5ex] \displaystyle (x^2 - 1)dx = \displaystyle\int x^2dx - \displaystyle\int 1dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \dfrac{x^{2 + 1}}{2 + 1} - \displaystyle\int x^0dx \\[5ex] = \dfrac{x^3}{3} - \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = \dfrac{x^3}{3} - x + C \\[5ex] \rightarrow \\[3ex] \displaystyle\int_1^3 (x^2 - 1)dx = \left[\dfrac{x^3}{3} - x\right]_1^3 \\[5ex] = \left(\dfrac{3^3}{3} - 3\right) - \left(\dfrac{1^3}{3} - 1\right) \\[5ex] = (9 - 3) - \left(\dfrac{1}{3} - \dfrac{3}{3}\right) \\[5ex] = 6 - \left(-\dfrac{2}{3}\right) \\[5ex] = 6 + \dfrac{2}{3} \\[5ex] = 6\dfrac{2}{3}$
(4.) JAMB Evaluate $\displaystyle\int \sin 3xdx$
$A.\:\: -\dfrac{1}{3}\cos 3x + c \\[5ex] B.\:\: \dfrac{1}{3}\cos 3x + c \\[5ex] C.\:\: \dfrac{2}{3}\cos 3x + c \\[5ex] C.\:\: -\dfrac{2}{3}\cos 3x + c \\[5ex]$
Integration by Algebraic Substitution
$\displaystyle\int \sin 3xdx \\[5ex] Let\:\: p = 3x \\[3ex] \dfrac{dp}{dx} = 3 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{3} \\[5ex] dx = \dfrac{dp}{3} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \sin 3xdx = \displaystyle\int \sin p * \dfrac{dp}{3} \\[5ex] = \dfrac{1}{3} * \displaystyle\int \sin pdp \\[5ex] = \dfrac{1}{3} * -\cos p + C \\[5ex] = \dfrac{1}{3} * - \cos 3x + C \\[3ex] = -\dfrac{1}{3}\cos 3x + c$
(5.) WASSCE:FM
(a) Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions
(b) Use the answer in (a) to evaluate $\displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3}dx$
$(a.) \\[3ex] \dfrac{x + 6}{(x + 1)^3} \\[5ex]$ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator
$\dfrac{x + 6}{(x + 1)^3} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} \\[5ex] = \dfrac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3} \\[5ex] = \dfrac{A[(x + 1)(x + 1)] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + x + x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + A + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + Bx + A + B + C}{(x + 1)^3} \\[5ex]$ Denominators are the same
Equate the numerators
$x + 6 = Ax^2 + 2Ax + Bx + A + B + C \\[3ex] Swap \\[3ex] Ax^2 + 2Ax + Bx + A + B + C = x + 6 \\[3ex] Ax^2 + x(2A + B) + A + B + C = 0x^2 + x + 6 \\[3ex] \implies \\[3ex] A = 0...eqn.(1) \\[3ex] 2A + B = 1...eqn.(2) \\[3ex] A + B + C = 6...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 2(0) + B = 1 \\[3ex] 0 + B = 1 \\[3ex] B = 1 - 0 \\[3ex] B = 1 \\[3ex] Substitute\;\;0\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 0 + 1 + C = 6 \\[3ex] 1 + C = 6 \\[3ex] C = 6 - 1 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{0}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \dfrac{x + 6}{(x + 1)^3} = 0 + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \therefore \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] = \dfrac{1(x + 1) + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 1 + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 6}{(x + 1)^3} \\[5ex] = LHS \\[3ex]$ Integration by Partial Fractions
$(b.) \\[3ex] \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx \\[5ex] = \displaystyle\int_1^2 \left[\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}\right] dx \\[5ex] = \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] Let\;\; p = x + 1 \\[3ex] \dfrac{dp}{dx} = 1 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{1} = 1 \\[5ex] dx = dp \\[3ex] \displaystyle\int \dfrac{1}{(x + 1)^2} dx \\[5ex] = \displaystyle\int \dfrac{1}{p^2} dp \\[5ex] = \displaystyle\int p^{-2} dp \\[5ex] = \dfrac{p^{-2 + 1}}{-2 + 1} \\[5ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] = -\dfrac{1}{x + 1} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx \\[5ex] = \left[-\dfrac{1}{x + 1}\right]_1^2 \\[5ex] = -\dfrac{1}{2 + 1} - -\dfrac{1}{1 + 1} \\[5ex] = -\dfrac{1}{3} + \dfrac{1}{2} \\[5ex] = \dfrac{-2 + 3}{6} \\[5ex] = \dfrac{1}{6} \\[7ex] \displaystyle\int \dfrac{5}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{p^3} dp \\[5ex] = 5\displaystyle\int p^{-3} dp \\[5ex] = 5 * \dfrac{p^{-3 + 1}}{-3 + 1} \\[5ex] = 5 * \dfrac{p^{-2}}{-2} \\[5ex] = -\dfrac{5}{2} * p^{-2} \\[5ex] = -\dfrac{5}{2} * \dfrac{1}{p^2} \\[5ex] = -\dfrac{5}{2p^2} \\[5ex] = -\dfrac{5}{2(x + 1)^2} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \left[-\dfrac{5}{2(x + 1)^2}\right]_1^2 \\[5ex] = -\dfrac{5}{2(2 + 1)^2} - -\dfrac{5}{2(1 + 1)^2} \\[5ex] = -\dfrac{5}{2(3)^2} + \dfrac{5}{2(2)^2} \\[5ex] = -\dfrac{5}{2(9)} + \dfrac{5}{2(4)} \\[5ex] = -\dfrac{5}{18} + \dfrac{5}{8} \\[5ex] = \dfrac{-20 + 45}{72} \\[5ex] = \dfrac{25}{72} \\[7ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \dfrac{1}{6} + \dfrac{25}{72} \\[5ex] = \dfrac{12 + 25}{72} \\[5ex] = \dfrac{37}{72} \\[5ex] \therefore \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx = \dfrac{37}{72}$
(6.) KCSE A particle was moving along a straight line.
The acceleration of the particle after t seconds was given by $(4t - 13)\;ms^{-2}$.
The initial velocity of the particle was 18 $ms^{-1}$
(a) Determine the value of t when the particle is momentarily at rest.
(b) Find the distance covered by the particle between the time t = 1 second and t = 3 seconds.
$a = 4t - 13 \\[3ex] a = \dfrac{dv}{dt} \\[5ex] dv = a dt \\[3ex] v = \displaystyle\int a dt \\[3ex] v = \displaystyle\int (4t - 13) dt \\[3ex] v = \dfrac{4t^2}{2} - 13t + C \\[3ex] v = 2t^2 - 13t + C \\[3ex] when\;\;t = 0,\;\;initial\;\;velocity = v = 18\;ms^{-1} \\[3ex] \implies \\[3ex] 18 = 2(0)^2 - 13(0) + C \\[3ex] 18 = 0 - 0 + C \\[3ex] 18 = C \\[3ex] C = 18 \\[3ex] v = 2t^2 - 13t + 18 \\[3ex] (a) \\[3ex] momentarily\;\;at\;\;rest\implies v = 0 \\[3ex] 0 = 2t^2 - 13t + 18 \\[3ex] 2t^2 - 13t + 18 = 0 \\[3ex] 2t(t - 2) - 9(t - 2) = 0 \\[3ex] (t - 2)(2t - 9) = 0 \\[3ex] t - 2 = 0 \;\;\;OR\;\;\; 2t - 9 = 0 \\[3ex] t = 2 \;\;\;OR\;\;\; 2t = 9 \\[3ex] t = 2\;s \;\;\;OR\;\;\; t = \dfrac{9}{2}\;s \\[3ex] (b) \\[3ex] v = \dfrac{dd}{dt} \\[5ex] dd = vdt \\[3ex] d = \displaystyle\int v dt \\[3ex]$ The zeros of the function...where the graph include areas above or below the x-axis (t-axis) are t = 2 seconds and t = $\dfrac{9}{2} = 4.5$ seconds
But we are asked to calculate the area (distance covered) between t = 1 second and t = 3 seconds
So, we have to first calculate the area between t = 1 second and t = 2 seconds
Then, we calculate the area between t = 2 seconds and t = 3 seconds
Because 4.5 seconds is greater than 3 seconds, we skip that area.
$v = \displaystyle\int_1^2 (2t^2 - 13t + 18) dt + \displaystyle\int_2^3 (2t^2 - 13t + 18) dt \\[5ex] v = \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 + \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 \\[5ex] t = 2\;s \\[3ex] \dfrac{2(2)^3}{3} - \dfrac{13(2)^2}{2} + 18(2) \\[5ex] \dfrac{16}{3} - \dfrac{26}{1} + \dfrac{36}{1} \\[5ex] \dfrac{16 - 78 + 108}{3} \\[5ex] \dfrac{46}{3}\;m \\[5ex] t = 1\;s \\[3ex] \dfrac{2(1)^3}{3} - \dfrac{13(1)^2}{2} + 18(1) \\[5ex] \dfrac{2}{3} - \dfrac{13}{2} + \dfrac{18}{1} \\[5ex] \dfrac{4 - 39 + 108}{6} \\[5ex] \dfrac{73}{6}\;m \\[5ex] Area = \dfrac{46}{3} - \dfrac{73}{6} = \dfrac{92 - 73}{6} = \dfrac{19}{6}\;m \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] t = 3\;s \\[3ex] \dfrac{2(3)^3}{3} - \dfrac{13(3)^2}{2} + 18(3) \\[5ex] \dfrac{18}{1} - \dfrac{117}{2} + \dfrac{54}{1} \\[5ex] \dfrac{36 - 117 + 108}{2} \\[5ex] \dfrac{27}{2}\;m \\[5ex] t = 2\;s \\[3ex] \dfrac{46}{3}\;m \\[5ex] Area = \dfrac{27}{2} - \dfrac{46}{3} = \dfrac{81 - 92}{6} = -\dfrac{11}{6} \\[5ex] Because\;\;Area\;\;cannot\;\;be\;\;negative; \;\;Area = \dfrac{11}{6}\;m \\[5ex] Total\;\;Area = Distance\;\;covered \\[3ex] = \dfrac{19}{6} + \dfrac{11}{6} \\[5ex] = \dfrac{30}{6} \\[5ex] = 5\;m$
(7.) WASSCE:FM Evaluate $\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx$
$A.\:\: 1\dfrac{1}{3} \\[5ex] B.\:\: 1\dfrac{1}{2} \\[5ex] C.\:\: 3 \\[3ex] D.\:\: 4\dfrac{1}{2}$
Power Rule
$\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx \\[5ex] 3 - 3x^2 = 3(1 - x^2) \\[3ex] 1 - x^2 = 1^2 - x^2 = (1 + x)(1 - x)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] 3 - 3x^2 = 3(1 + x)(1 - x) = 3(x + 1)(1 - x) \\[3ex] \rightarrow \\[3ex] \dfrac{3 - 3x^2}{x + 1} = \dfrac{3(x + 1)(1 - x)}{x + 1} = 3(1 - x) = 3 - 3x \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3 - 3x)dx = \displaystyle\int 3dx - \displaystyle\int 3xdx \\[5ex] \displaystyle\int 3dx = 3x \\[5ex] \displaystyle\int 3xdx = 3 * \dfrac{x^{1 + 1}}{1 + 1} = \dfrac{3x^2}{2} \\[5ex] \rightarrow \\[3ex] \displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx = \left[3x - \dfrac{3x^2}{2}\right]_0^1 \\[5ex] = \left[3(1) - \dfrac{3(1)^2}{2}\right] - \left[3(0) - \dfrac{3(0)^2}{2}\right] \\[5ex] = \left[3 - \dfrac{3(1)}{2}\right] - \left[0 - \dfrac{3(0)}{2}\right] \\[5ex] = \left(3 - \dfrac{3}{2}\right) - \left(0 - \dfrac{0}{2}\right) \\[5ex] = \left(\dfrac{6}{2} - \dfrac{3}{2}\right) - (0 - 0) \\[5ex] = \dfrac{6 - 3}{2} - 0 \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2}$
(8.) Evaluate $\displaystyle\int (3^p + 7^p)dp$
Standard Integral
$\displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[5ex] \displaystyle\int (3^p + 7^p)dp = \displaystyle\int 3^pdp + \displaystyle\int 7^pdp \\[5ex] \displaystyle\int 3^p dx = \dfrac{3^p}{\ln 3} \\[5ex] \displaystyle\int 7^p dx = \dfrac{7^p}{\ln 7} \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3^p + 7^p)dp = \dfrac{3^p}{\ln 3} + \dfrac{7^p}{\ln 7} + C$
(9.) Evaluate $\displaystyle\int \sqrt{2 + \sqrt{x}} dx$
Integration by Algebraic Substitution
$\displaystyle\int \sqrt{2 + \sqrt{x}} dx \\[3ex] Let\:\: p = 2 + \sqrt{x} \\[3ex] \dfrac{dp}{dx} = \dfrac{1}{2\sqrt{x}} \\[5ex] \dfrac{dx}{dp} = 2\sqrt{x} \\[5ex] dx = 2\sqrt{x}dp \\[3ex] \rightarrow \\[3ex] \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \displaystyle\int \sqrt{p} * 2\sqrt{x} dp \\[3ex] = 2\displaystyle\int \sqrt{p} * \sqrt{x} dp \\[3ex] p = 2 + \sqrt{x} \rightarrow \sqrt{x} = p - 2 \\[3ex] = 2 * \displaystyle\int \sqrt{p}(p - 2) dp \\[3ex] \displaystyle\int \sqrt{p}(p - 2) dp = \displaystyle\int p\sqrt{p}dp - \displaystyle\int 2\sqrt{p}dp \\[3ex] \displaystyle\int p\sqrt{p}dp = \displaystyle\int p^1 * p^{\dfrac{1}{2}}dp = \displaystyle\int p^{1 + \dfrac{1}{2}}dp = \displaystyle\int p^{\dfrac{3}{2}}dp \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{p^{\dfrac{3}{2} + 1}}{\dfrac{3}{2} + 1} \\[7ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} \div \dfrac{5}{2} \\[5ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} * \dfrac{2}{5} \\[5ex] p^{\dfrac{5}{2}} = (\sqrt{p})^5 = (\sqrt{p})^4 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^4 * \sqrt{p} = p^2\sqrt{p} \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{2p^2}{5} \\[5ex] \displaystyle\int 2\sqrt{p}dp = 2\displaystyle\int \sqrt{p}dp \\[3ex] \displaystyle\int \sqrt{p}dp = \dfrac{p^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \\[7ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} \div \dfrac{3}{2} \\[5ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} * \dfrac{2}{3} \\[5ex] p^{\dfrac{3}{2}} = (\sqrt{p})^3 = (\sqrt{p})^2 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^2 * \sqrt{p} = p\sqrt{p} \\[5ex] \displaystyle\int 2\sqrt{p}dp = \dfrac{2p\sqrt{p}}{3} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{p}(p - 2) dp = \dfrac{2p^2\sqrt{p}}{5} - \dfrac{2p\sqrt{p}}{3} \\[5ex] = 2p\sqrt{p}\left(\dfrac{p}{5} - \dfrac{1}{3}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p}{15} - \dfrac{5}{15}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] \rightarrow \\[3ex] 2 * \displaystyle\int \sqrt{p}(p - 2) dp = 2 * 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] = 4p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] Substitute\:\:back \\[3ex] = 4(2 + \sqrt{x})\sqrt{2 + \sqrt{x}}\left(\dfrac{3(2 + \sqrt{x}) - 5}{15}\right) \\[5ex] \dfrac{3(2 + \sqrt{x}) - 5}{15} = \dfrac{6 + 3\sqrt{x} - 5}{15} = \dfrac{1 + 3\sqrt{x}}{15} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{2 + \sqrt{x}} dx = 4(2 + \sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right)\left(\dfrac{1 + 3\sqrt{x}}{15}\right) \\[5ex] \therefore \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \dfrac{4}{15}(2 + \sqrt{x})(1 + 3\sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right) + C$
(10.) Find the integral of $\tan x$
Integration by Algebraic Substitution
$\displaystyle\int \tan xdx \\[5ex] \tan x = \dfrac{\sin x}{\cos x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \cos x \\[3ex] \dfrac{dp}{dx} = -\sin x \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{\sin x} \\[5ex] dx = -\dfrac{dp}{\sin x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \tan xdx = \displaystyle\int \dfrac{\sin x}{\cos x} dx \\[5ex] = \displaystyle\int \dfrac{\sin x}{p} * -\dfrac{dp}{\sin x} \\[5ex] = \displaystyle\int -\dfrac{dp}{p} \\[5ex] = -\ln p + C \\[3ex] = -\ln \cos x + C$
(11.)
(12.) Evaluate $\displaystyle\int (v^{-2} + 8v^{-1})dv$
Power Rule
$\displaystyle\int (v^{-2} + 8v^{-1})dv \\[5ex] = \displaystyle\int v^{-2}dv + \displaystyle\int 8v^{-1}dv \\[5ex] = \dfrac{v^{-2 + 1}}{-2 + 1} + 8 * \displaystyle\int v^{-1}dv \\[5ex] = \dfrac{v^{-1}}{-1} + 8 * \ln v + C \\[5ex] = -v^{-1} + 8\ln v + C \\[5ex] = -\dfrac{1}{v} + 8\ln v + C \\[5ex] = 8\ln v - \dfrac{1}{v} + C$
(13.)
(14.)
(15.)
(16.)
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https://socratic.org/questions/how-do-you-write-a-verbal-express-for-the-algebraic-expression-5-a-b
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# How do you write a verbal express for the algebraic expression 5(a-b)?
The verbal expression for $5 \left(a - b\right)$ is "five times the difference between $a$ and $b$."
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# Search for tag: "logarithms"
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#### Alg 2B Sect. 6.6 (part 3)
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#### Alg 2B Sect. 6.6 (part 2)
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#### Alg 2B Sect. 6.5 (part 3)
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#### Alg 2B Sect. 6.5 (part 1)
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#### alg 2B Sect. 6.3 (part 3)
From jmeisner@hlschool.net on March 30th, 2020 0 likes 6 plays
#### 05 - Guided Notes - Expand & Condense Logarithms
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From J_R_Tucker_High_School_Contributors_125327411 on September 19th, 2019 0 likes 0 plays
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# Building a cubic function with integer coefficients and trigonometric roots
I want to find the answer to the following problem: Construct a cubic polynomial with integer coefficients, whose roots - $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$.
I have the following idea. These trigonometric roots are the extremum points of the Chebyshev polynomial $T_7 (x) = T_7 (\cos{t}) = \cos{7t}$ and $T_7(x_k) = 1$, where $x_k$ required the roots of a cubic function (see problem). Then the 7 degree polynomial $T_n(x)-1$ will have roots $x_k$. But the problem is that we need a polynomial of degree 3. I'm sure these trigonometric roots will be associated with the Chebyshev polynomials.What do you think about this ? Can you suggest your ideas to solve?
• I'm confused... you want a cubic equation with specific roots. There is only $one$ such equation... so it either has integer coefficients or it doesn't, there's no constructing... – Asier Calbet Sep 23 '15 at 22:58
• @Assaultous2 Only one, up to multiplication by non-zero scalars :) – Servaes Sep 23 '15 at 23:09
• You are right. This will be a polynomial $(x-\cos{\frac{2 \pi}{7}})(x-\cos{\frac{4 \pi}{7}})(x-\cos{\frac{6 \pi}{7}})$, and the problem reduces to the difficult trigonometric calculations. I think this problem can be solved easier, using Chebyshev polynomials. – Victor Sep 23 '15 at 23:15
• @ Servaes of course – Asier Calbet Sep 24 '15 at 10:11
How about the polynomial $$8\left(x-\cos\tfrac{2\pi}{7}\right)\left(x-\cos\tfrac{4\pi}{7}\right)\left(x-\cos\tfrac{6\pi}{7}\right)=8x^3+4x^2-4x-1?$$
• Sorry, I made a mistake in the description of the problem. There should be 3 different roots $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$. – Victor Sep 23 '15 at 23:09
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# Astrophysics in a Nutshell Problem
• fu11meta1
In summary, the conversation discusses a hypothetical star with constant density and a classical, non-relativistic, ideal gas composition. The task is to determine the temperature profile of the star, with the use of the equation dT(r)/dr = (-3*L(r)*k(r)*ρ(r)) / (4∏*(r^2)*4*a*c*(T(r))^3). The individual discussing the problem attempted to identify which values on the right side of the equation were constant, but was unsure. They request a full explanation of the principles and equations involved in solving the problem.
fu11meta1
## Homework Statement
Consider a hypothetical star of radius R, with density ρ that is constant, i.e. independent of radius. The star is composed of a classical, non relativistic, ideal gas of fully ionized hydrogen,
Find the the temperature profile (T(r))
## Homework Equations
dT(r)/dr = ( (-3*L(r)*k(r)*ρ(r)) / (4∏*(r^2)*4*a*c*(T(r))^3)
so sorry this is written poorly
## The Attempt at a Solution
So what I did is try and assume which of the values on the right side are constant so I could take them outside of the integral, but I'm not sure.Please explain principles and equations fully. I really want to understand this!
Last edited:
Maybe you could state the problem?
Oh sorry. I thought I did!
I edited it. I says find the temperature profile (T(r))
## 1. What is astrophysics in a nutshell problem?
Astrophysics in a nutshell problem is a theoretical question that seeks to understand the fundamental laws and principles that govern the universe, and how they interact with each other to create the vast array of structures and phenomena that we observe in the cosmos. It is a fundamental question that seeks to explain the origins, evolution, and ultimate fate of the universe.
## 2. What are some of the main challenges in astrophysics in a nutshell problem?
The main challenges in astrophysics in a nutshell problem include understanding the nature of dark matter and dark energy, determining the origin and evolution of the universe, and explaining the behavior of black holes and other extreme objects in the universe. These challenges require advanced theories and technologies to study and understand the complex systems and processes that occur in the universe.
## 3. How do scientists study astrophysics in a nutshell problem?
Scientists study astrophysics in a nutshell problem through a combination of observations, theoretical models, and experiments. They use powerful telescopes and other instruments to observe and collect data on various objects and phenomena in the universe. They also use mathematical and computational models to simulate and understand the behavior of these objects and phenomena, and conduct experiments to test their theories and hypotheses.
## 4. What are some of the potential applications of solving astrophysics in a nutshell problem?
The potential applications of solving astrophysics in a nutshell problem are vast and far-reaching. Understanding the fundamental laws and principles of the universe could lead to breakthroughs in technology, energy, and medicine. It could also help us better understand our place in the universe and our potential for future space exploration and colonization.
## 5. How close are we to solving astrophysics in a nutshell problem?
While significant progress has been made in understanding various aspects of astrophysics in a nutshell problem, there is still much to uncover and explore. Some theories, such as the theory of relativity, have been extensively tested and proven, while others, such as the theory of quantum gravity, are still being developed and refined. It is an ongoing and challenging pursuit, and scientists continue to push the boundaries of our knowledge and understanding of the universe.
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Velocity Reviews > Method default argument whose type is the class not yet defined
# Method default argument whose type is the class not yet defined
Roy Smith
Guest
Posts: n/a
11-11-2012
In article <(E-Mail Removed)>,
Ian Kelly <(E-Mail Removed)> wrote:
> On Sat, Nov 10, 2012 at 7:13 PM, Chris Angelico <(E-Mail Removed)> wrote:
> > I would not assume that. The origin is a point, just like any other.
> > With a Line class, you could deem a zero-length line to be like a
> > zero-element list, but Point(0,0) is more like the tuple (0,0) which
> > is definitely True.
>
> It's more like the number 0 than the tuple (0,0).
>
> 0 is the origin on a 1-dimensional number line.
> (0,0) is the origin on a 2-dimensional number plane.
> In fact, it might be pointed out that Point(0, 0) is a generalization
> of 0+0j, which is equal to 0.
If (0,0) is the origin on a plane, then (0,) should be the origin on a
line. If you consider 0 + 0j to be the origin of a plane, then 0 is the
origin of a line. Either way is plausible, but you need to be
consistent.
Ian Kelly
Guest
Posts: n/a
11-11-2012
On Sat, Nov 10, 2012 at 11:43 PM, Ian Kelly <(E-Mail Removed)> wrote:
> Where I wrote "(0,0) is the origin" above I was not referring to a
> point, not a tuple, but I can see how that was confusing.
What I meant to say is I *was* referring to a point. Gah!
Oscar Benjamin
Guest
Posts: n/a
11-11-2012
On 11 November 2012 02:47, Chris Angelico <(E-Mail Removed)> wrote:
> On Sun, Nov 11, 2012 at 1:43 PM, Ian Kelly <(E-Mail Removed)> wrote:
>> On Sat, Nov 10, 2012 at 7:13 PM, Chris Angelico <(E-Mail Removed)> wrote:
>>> I would not assume that. The origin is a point, just like any other.
>>> With a Line class, you could deem a zero-length line to be like a
>>> zero-element list, but Point(0,0) is more like the tuple (0,0) which
>>> is definitely True.
>>
>> It's more like the number 0 than the tuple (0,0).
>>
>> 0 is the origin on a 1-dimensional number line.
>> (0,0) is the origin on a 2-dimensional number plane.
>>
>> In fact, it might be pointed out that Point(0, 0) is a generalization
>> of 0+0j, which is equal to 0.
>
> Ah, good point. In any case, though, it'd be an utterly inconsequential bug.
You were right the first time, Chris. A point that happens to coincide
with the arbitrarily chosen origin is no more truthy or falsey than
any other. A vector of length 0 on the other hand is a very different
beast.
The significance of zero in real algebra is not that it is the origin
but rather that it is the additive and multiplicative zero:
a + 0 = a for any real number a
a * 0 = 0 for any real number a
The same is true for a vector v0, of length 0:
v + v0 = v for any vector v
a * v0 = v0 for any scalar a
There is however no meaningful sense in which points (as opposed to
vectors) can be added to each other or multiplied by anything, so
there is no zero point.
The relationship between points and vectors is analogous to the
relationship between datetimes and timedeltas. Having Vector(0, 0)
evaluate to False is analogous to having timedelta(0) evaluate to
False and is entirely sensible. Having Point(0, 0) evaluate to False
is precisely the same conceptual folly that sees midnight evaluate as
False.
Oscar
Steven D'Aprano
Guest
Posts: n/a
11-11-2012
On Sun, 11 Nov 2012 14:21:19 +0000, Oscar Benjamin wrote:
> On 11 November 2012 02:47, Chris Angelico <(E-Mail Removed)> wrote:
>> On Sun, Nov 11, 2012 at 1:43 PM, Ian Kelly <(E-Mail Removed)>
>> wrote:
>>> On Sat, Nov 10, 2012 at 7:13 PM, Chris Angelico <(E-Mail Removed)>
>>> wrote:
>>>> I would not assume that. The origin is a point, just like any other.
>>>> With a Line class, you could deem a zero-length line to be like a
>>>> zero-element list, but Point(0,0) is more like the tuple (0,0) which
>>>> is definitely True.
Don't conflate the set of all tuples of arbitrary length with points,
which have fixed length. Points are not *sequences* -- in Python, we
treat tuples as sequences first and records second, because that is a
useful thing to do. (But it is not the only useful thing to do: if we
treated them as records first and sequences second, we might want to say
that a tuple t was falsey if all the fields in t were falsey.)
In the case of a Point class, a Point is definitely not a sequence. That
we put the x-coordinate first and the y-coordinate second is a mere
convention, like writing left to right. The mathematical properties of
points do not depend on the x-coordinate coming first. Since points
should not be treated as sequences, the requirement that non-empty
sequences be treated as truthy is irrelevant.
>>> It's more like the number 0 than the tuple (0,0).
>>>
>>> 0 is the origin on a 1-dimensional number line. (0,0) is the origin on
>>> a 2-dimensional number plane.
>>>
>>> In fact, it might be pointed out that Point(0, 0) is a generalization
>>> of 0+0j, which is equal to 0.
>>
>> Ah, good point. In any case, though, it'd be an utterly inconsequential
>> bug.
>
> You were right the first time, Chris. A point that happens to coincide
> with the arbitrarily chosen origin is no more truthy or falsey than any
> other. A vector of length 0 on the other hand is a very different beast.
Nonsense. The length and direction of a vector is relative to the origin.
If the origin is arbitrary, as you claim, then so is the length of the
vector.
Just because we can perform vector transformations on the plane to move
the origin to some point other that (0,0) doesn't make (0,0) an
"arbitrarily chosen origin". It is no more arbitrary than 0 as the origin
of the real number line.
And yes, we can perform 1D vector transformations on the real number line
too. Here's a version of range that sets the origin to 42, not 0:
def myrange(start, end=None, step=1):
if end is None:
start = 42
return range(start, end, step)
Nevertheless, there really is something special about the point 0 on the
real number line, the point (0,0) on the complex number plane, the point
(0,0,0) in the 3D space, (0,0,0,0) in 4D space, etc. It is not just an
arbitrary convention that we set the origin to 0.
In other words: to the extent that your arguments that zero-vectors are
special are correct, the same applies to zero-points, since vectors are
defined as a magnitude and direction *from the origin*.
To put it yet another way:
The complex number a+bj is equivalent to the 2D point (a, b) which is
equivalent to the 2D vector [a, b]. If (0, 0) shouldn't be considered
falsey, neither should [0, 0].
> The significance of zero in real algebra is not that it is the origin
> but rather that it is the additive and multiplicative zero:
>
> a + 0 = a for any real number a
> a * 0 = 0 for any real number a
I'm not sure what you mean by "additive and multiplicative zero", you
appear to be conflating two different properties here. 0 is the additive
*identity*, but 1 is the multiplicative identity:
a + 0 = a
a * 1 = a
for any real number a.
If the RHS must be zero, then there is a unique multiplicative zero, but
a * 0 = 0 for any real number a
a + -a = 0 for any real number a
> The same is true for a vector v0, of length 0:
>
> v + v0 = v for any vector v
> a * v0 = v0 for any scalar a
Well that's a bogus analogy. Since you're talking about the domain of
vectors, the relevant identify for the second line should be:
v * v0 = v0 for any vector v
except that doesn't work, since vector algebra doesn't define a vector
multiplication operator.[1] It does define multiplication between a
vector and a scalar, which represents a scale transformation.
> There is however no meaningful sense in which points (as opposed to
> vectors) can be added to each other or multiplied by anything, so there
> is no zero point.
I think that the great mathematician Carl Gauss would have something to
Points in the plane are equivalent to complex numbers, and you can
equivalent to a translation; multiplication of a scalar with a point is
equivalent to a scale transformation. Multiplying two points is
equivalent to complex multiplication, which is a scale + a rotation.
Oh look, that's exactly the same geometric interpretation as for vectors.
Hardly surprising, since vectors are the magnitude and direction of a
line from the origin to a point.
> The relationship between points and vectors is analogous to the
> relationship between datetimes and timedeltas. Having Vector(0, 0)
> evaluate to False is analogous to having timedelta(0) evaluate to False
> and is entirely sensible. Having Point(0, 0) evaluate to False is
> precisely the same conceptual folly that sees midnight evaluate as
> False.
If you are dealing with datetimes, then "midnight 2012-11-12" is not
falsey. The only falsey datetime is the zero datetime. Since it would be
awfully inconvenient to start counting times from the Big Bang, we pick
an arbitrary zero point, the Epoch, which in Unix systems is midnight 1
January 1970, and according to the logic of Unix system administrators,
that is so far in the distant past that it might as well be the Big Bang.
(People with more demanding requirements don't use Unix or Windows
timestamps for recording date times. E.g. astronomers use the Julian
system, not to be confused with the Julian calendar.)
The midnight problem only occurs when you deal with *times* on their own,
not datetimes, in which case the relationship with timedeltas is not
defined. How far apart is 1:00am and 2:00am? Well, it depends, doesn't
it? It could be 1 hour, 25 hours, 49 hours, ...
In any case, since times are modulo 24 hours, they aren't really relevant
to what we are discussing.
[1] There is no single meaningful definition of vector multiplication
that works for all dimensions. In two dimensions, you can define the dot
product of two vectors to give a scalar; in three dimensions you have a
dot product and a vector product.
Since vectors are equivalent to points, and points are equivalent to
complex numbers, one could define a vector operation equivalent to
complex multiplication. There is a natural geometric interpretation of
this multiplication: it is a scaling + rotation.
--
Steven
Oscar Benjamin
Guest
Posts: n/a
11-12-2012
On 11 November 2012 22:31, Steven D'Aprano
<(E-Mail Removed)> wrote:
> On Sun, 11 Nov 2012 14:21:19 +0000, Oscar Benjamin wrote:
>
>> On 11 November 2012 02:47, Chris Angelico <(E-Mail Removed)> wrote:
>>> On Sun, Nov 11, 2012 at 1:43 PM, Ian Kelly <(E-Mail Removed)>
>>> wrote:
>>>> On Sat, Nov 10, 2012 at 7:13 PM, Chris Angelico <(E-Mail Removed)>
>>>> wrote:
>>>>> I would not assume that. The origin is a point, just like any other.
>>>>> With a Line class, you could deem a zero-length line to be like a
>>>>> zero-element list, but Point(0,0) is more like the tuple (0,0) which
>>>>> is definitely True.
>>>> It's more like the number 0 than the tuple (0,0).
>>>>
>>>> 0 is the origin on a 1-dimensional number line. (0,0) is the origin on
>>>> a 2-dimensional number plane.
>>>>
>>>> In fact, it might be pointed out that Point(0, 0) is a generalization
>>>> of 0+0j, which is equal to 0.
>>>
>>> Ah, good point. In any case, though, it'd be an utterly inconsequential
>>> bug.
>>
>> You were right the first time, Chris. A point that happens to coincide
>> with the arbitrarily chosen origin is no more truthy or falsey than any
>> other. A vector of length 0 on the other hand is a very different beast.
>
> Nonsense. The length and direction of a vector is relative to the origin.
> If the origin is arbitrary, as you claim, then so is the length of the
> vector.
Wrong on all counts. Neither the length not the direction os a vector
are relative to any origin. When we choose to express a vector in
Cartesian components our representation assumes an orientation for the
axes of the coordinate system. Even in this sense, though, the origin
itself does not affect the components of the vector.
I have spent a fair few hours in the past few weeks persuading
teenaged Engineering students to maintain a clear distinction between
points, vectors and lines. One of the ways that I distinguish vectors
from points is to say that a vector is like an arrow but its base has
no particular position. A point on the other hand is quite simply a
position. Given an origin (an arbitrarily chosen point) we can specify
another point using a "position vector": a vector from the origin to
the point in question.
> Just because we can perform vector transformations on the plane to move
> the origin to some point other that (0,0) doesn't make (0,0) an
> "arbitrarily chosen origin". It is no more arbitrary than 0 as the origin
> of the real number line.
(0, 0) are the coordinates of the origin *relative to itself*. Had we
chosen a different origin, the point that was previously called (0, 0)
would now be called (a, b) for some other numbers a and b.
> And yes, we can perform 1D vector transformations on the real number line
> too. Here's a version of range that sets the origin to 42, not 0:
>
> def myrange(start, end=None, step=1):
> if end is None:
> start = 42
> return range(start, end, step)
This is lost on me...
> Nevertheless, there really is something special about the point 0 on the
> real number line
Agreed.
> , the point (0,0) on the complex number plane,
Also agreed.
> the point
> (0,0,0) in the 3D space, (0,0,0,0) in 4D space, etc. It is not just an
> arbitrary convention that we set the origin to 0.
Wrong. The point (0,0,0,...) in some ND space is an arbitrarily chosen
position. By this I don't mean to say that the sequence of coordinates
consisting of all zeros is arbitrary. The choice of the point *in the
real/hypothetical space* that is designated by the sequence of zero
coordinates is arbitrary.
> In other words: to the extent that your arguments that zero-vectors are
> special are correct, the same applies to zero-points, since vectors are
> defined as a magnitude and direction *from the origin*.
Plain wrong. Vectors are not defined *from any origin*.
> To put it yet another way:
>
> The complex number a+bj is equivalent to the 2D point (a, b) which is
> equivalent to the 2D vector [a, b]. If (0, 0) shouldn't be considered
> falsey, neither should [0, 0].
a+bj is not equivalent to the 2D point (a, b). It is possible to
define a mapping between complex numbers and a 2D space so that a+bj
corresponds to the point (a, b) *under that map*. However there are an
infinite number of such possible mappings between the two spaces
including a+bj -> (a+1, b+1).
>> The significance of zero in real algebra is not that it is the origin
>> but rather that it is the additive and multiplicative zero:
>>
>> a + 0 = a for any real number a
>> a * 0 = 0 for any real number a
>
> I'm not sure what you mean by "additive and multiplicative zero", you
> appear to be conflating two different properties here. 0 is the additive
> *identity*, but 1 is the multiplicative identity:
I mean that it has the properties that zero has when used in addition
and multiplication:
http://en.wikipedia.org/wiki/0_%28nu...entary_algebra
>
> a + 0 = a
> a * 1 = a
> for any real number a.
No. I meant the two properties that I listed.
>
> If the RHS must be zero, then there is a unique multiplicative zero, but
>
> a * 0 = 0 for any real number a
> a + -a = 0 for any real number a
That is not the same as:
a + 0 = a
>> The same is true for a vector v0, of length 0:
>>
>> v + v0 = v for any vector v
>> a * v0 = v0 for any scalar a
>
> Well that's a bogus analogy. Since you're talking about the domain of
> vectors, the relevant identify for the second line should be:
>
> v * v0 = v0 for any vector v
>
> except that doesn't work, since vector algebra doesn't define a vector
> multiplication operator.[1] It does define multiplication between a
> vector and a scalar, which represents a scale transformation.
That is precisely the multiplication operation that I was referring
to. There are other senses of vector multiplication between vectors
for which v0 will also behave as a "zero" under multiplication:
v . v0 = 0 for any vector v
v x v0 = 0 for any vector v
>> There is however no meaningful sense in which points (as opposed to
>> vectors) can be added to each other or multiplied by anything, so there
>> is no zero point.
>
> I think that the great mathematician Carl Gauss would have something to
Is the text below a quote?
> Points in the plane are equivalent to complex numbers, and you can
> equivalent to a translation; multiplication of a scalar with a point is
> equivalent to a scale transformation. Multiplying two points is
> equivalent to complex multiplication, which is a scale + a rotation.
The last point is bizarre. Complex multiplication makes no sense when
you're trying to think about vectors. Draw a 2D plot and convince
yourself that the square of the point (0, 1) is (-1, 0).
> Oh look, that's exactly the same geometric interpretation as for vectors.
> Hardly surprising, since vectors are the magnitude and direction of a
> line from the origin to a point.
Here it becomes clear that you have conflated "position vectors" with
vectors in general. Let me list some other examples of vectors that
are clearly not "from the origin to a point":
velocity
acceleration
force
electric field
angular momentum
wave vector
(I could go on)
>> The relationship between points and vectors is analogous to the
>> relationship between datetimes and timedeltas. Having Vector(0, 0)
>> evaluate to False is analogous to having timedelta(0) evaluate to False
>> and is entirely sensible. Having Point(0, 0) evaluate to False is
>> precisely the same conceptual folly that sees midnight evaluate as
>> False.
>
> If you are dealing with datetimes, then "midnight 2012-11-12" is not
> falsey. The only falsey datetime is the zero datetime. Since it would be
> awfully inconvenient to start counting times from the Big Bang, we pick
> an arbitrary zero point, the Epoch, which in Unix systems is midnight 1
> January 1970, and according to the logic of Unix system administrators,
> that is so far in the distant past that it might as well be the Big Bang.
>
> (People with more demanding requirements don't use Unix or Windows
> timestamps for recording date times. E.g. astronomers use the Julian
> system, not to be confused with the Julian calendar.)
>
> The midnight problem only occurs when you deal with *times* on their own,
> not datetimes, in which case the relationship with timedeltas is not
> defined. How far apart is 1:00am and 2:00am? Well, it depends, doesn't
> it? It could be 1 hour, 25 hours, 49 hours, ...
>
> In any case, since times are modulo 24 hours, they aren't really relevant
> to what we are discussing.
They are relevant. The point is that conflating points and vectors is
the same as conflating datetime and timedelta objects. The zero of
datetime.timedelta objects is not arbitrary but the zero of
datetime.time objects is.
> Since vectors are equivalent to points, and points are equivalent to
> complex numbers, one could define a vector operation equivalent to
> complex multiplication. There is a natural geometric interpretation of
> this multiplication: it is a scaling + rotation.
Vectors, points and complex numbers are not equivalent. There are
cases in which it is reasonable to think of them as equivalent for a
particular purpose. That does not diminish the fundamental differences
between them.
Oscar
Steve Howell
Guest
Posts: n/a
11-12-2012
On Nov 10, 11:33*am, Jennie <(E-Mail Removed)> wrote:
> What is the best solution to solve the following problem in Python 3.3?
>
> import math
> *>>> class Point:
> ... * * def __init__(self, x=0, y=0):
> ... * * * * self.x = x
> ... * * * * self.y = y
> ... * * def __sub__(self, other):
> ... * * * * return Point(self.x - other.x, self.y - other.y)
> ... * * def distance(self, point=Point()):
> ... * * * * """Return the distance from `point`."""
> ... * * * * return math.sqrt((self - point).x ** 2 + (self - point).y ** 2)
Before you do anything else, introduce a Vector class into your app.
The difference between two Points is not a Point; it's a Vector.
Create a magnitude() method in your Vector class, then make your
Point.distance return the results of Vector.magnitude(self - other).
To define the distance of a point from the origin, don't make your
distance() method have default arguments; instead, define another
method called distance_from_origin().
Steve Howell
Guest
Posts: n/a
11-12-2012
On Nov 11, 4:31*pm, Oscar Benjamin <(E-Mail Removed)> wrote:
> On 11 November 2012 22:31, Steven D'Aprano
> > Nonsense. The length and direction of a vector is relative to the origin.
> > If the origin is arbitrary, as you claim, then so is the length of the
> > vector.
>
> Wrong on all counts. Neither the length not the direction os a vector
> are relative to any origin. When we choose to express a vector in
> Cartesian components our representation assumes an orientation for the
> axes of the coordinate system. Even in this sense, though, the origin
> itself does not affect the components of the vector.
>
Thank you for pushing back on Steven's imprecise statement that the
"direction of a vector is relative to the origin."
You can't find an angle between two points. That's absurd. You need
axes for context.
> Vectors, points and complex numbers are not equivalent. There are
> cases in which it is reasonable to think of them as equivalent for a
> particular purpose. That does not diminish the fundamental differences
> between them.
>
I looked to wikipedia for clarity, but the definition of a Euclidean
vector is somewhat muddy:
http://en.wikipedia.org/wiki/Euclidean_vector
They say that the formal definition of a vector is a directed line
segment. But then they define a "free vector" as an entity where only
the magnitude and direction matter, not the initial point.
As you say, it's not unreasonable to treat vectors, points, and
complex numbers as equivalent in many circumstances. But, if you're
gonna be pedantic, they really are different things.
Mark Lawrence
Guest
Posts: n/a
11-12-2012
On 12/11/2012 00:31, Oscar Benjamin wrote:
>
> Plain wrong. Vectors are not defined *from any origin*.
>
So when the Captain says "full speed ahead, steer 245 degrees", you
haven't the faintest idea where you're going, because you have no origin?
--
Cheers.
Mark Lawrence.
Roy Smith
Guest
Posts: n/a
11-12-2012
In article <(E-Mail Removed)>,
Mark Lawrence <(E-Mail Removed)> wrote:
> On 12/11/2012 00:31, Oscar Benjamin wrote:
> >
> > Plain wrong. Vectors are not defined *from any origin*.
> >
>
> So when the Captain says "full speed ahead, steer 245 degrees", you
> haven't the faintest idea where you're going, because you have no origin?
Vectors have a length ("full speed ahead") and a direction ("245
degrees"). What they don't have is a fixed location in space. The
captain didn't say, "Full speed ahead, steer 245 degrees, from 45.0N,
20.0W".
In other words, you are correct. The order, "full speed ahead, steer
245 degrees", doesn't give you the faintest idea of where you're going.
If you were the helmsman, after you executed that order, without any
no idea what piece of land you will hit, or when you will hit it, if you
maintain your current course and speed.
Oscar Benjamin
Guest
Posts: n/a
11-12-2012
On 12 November 2012 01:10, Mark Lawrence <(E-Mail Removed)> wrote:
> On 12/11/2012 00:31, Oscar Benjamin wrote:
>>
>>
>> Plain wrong. Vectors are not defined *from any origin*.
>>
>
> So when the Captain says "full speed ahead, steer 245 degrees", you haven't
> the faintest idea where you're going, because you have no origin?
As Steve has just explained, the origin has nothing to do with the
orientation of the coordinate system.
But then I'm assuming you meant that 245 degrees was a bearing
relative to North. Was it supposed to be relative to my current angle?
Truthfully I wouldn't know what to do without asking the captain a
couple more questions.
Oscar
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W1 Lecture INTEREST RATES AND BOND VALUATION v1 (1).pptx - INTEREST RATES BOND VALUATION Week 2 1F0120 Corporate Finance I Academic Department of
W1 Lecture INTEREST RATES AND BOND VALUATION v1 (1).pptx -...
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INTEREST RATES & BOND VALUATION 1 Week 2 1F0120 Corporate Finance I Academic Department of Finance Universidad del Pacífico UNIVERSIDAD DEL PACIFICO 1F0120 W2 M. Gonzalo Chávez, CFA, CAIA, FRM
Outline I. The Internal Rate of Return (IRR) II. Interest Rates Quotes and Loans III. Determinants of Interest Rates IV. Bond Prices and Yields V. Corporate Bonds UNIVERSIDAD DEL PACIFICO 1F0120 W2 2
Internal Rate of Return (IRR) Internal Rate of Return (IRR) is the interest rate that sets the NPV of the cash flows equal to zero . It is a measure of the rate of return of an investment. The term internal refers to the fact that its calculation does not incorporate outside factors such as market interest rate. UNIVERSIDAD DEL PACIFICO 1F0120 W2 3
Example Suppose that you face an investment opportunity that requires a \$1000 investment today and will have a \$2000 payoff in six years, r = IRR = 12.25% Therefore, making this investment is equivalent to earning 12.25% per year on your money for six years UNIVERSIDAD DEL PACIFICO 1F0120 W2 4
Solving IRR of a Cash Flow Stream Consider a stream of cash flow, The IRR is calculated by solving the r in the NPV = 0 equation below, When N > 2 , the r may not be solved manually, we need to use a financial calculator or computer softwares (e.g. Excel). UNIVERSIDAD DEL PACIFICO 1F0120 W2 5
Outline I. The Internal Rate of Return (IRR) II. Interest Rates Quotes and Loans III. Determinants of Interest Rates IV. Bond Prices and Yields V. Corporate Bonds UNIVERSIDAD DEL PACIFICO 1F0120 W2 6
Effective Annual Rate (EAR) Effective Annual Rate (EAR) represent the actual amount of interest that will be earned at the end of one year . For example, if we are told that a \$100,000 investment has EAR 5%, we know that after one year we will have: after two years we will have: We can get a quick idea of the rate of return without going into the details of the compounding frequency. UNIVERSIDAD DEL PACIFICO 1F0120 W2 7
Discount Rate Conversion Note that, in the previous example, having an EAR 5% is the same as having a two-year interest rate of 10.25% . Similarly, an EAR of 5% is the same as having a six-month interest rate of r = 2.47%. In general, we can convert a discount rate of r for one period to an equivalent discount rate for n periods using the following formula: Note that n can be larger than 1 or smaller than 1. UNIVERSIDAD DEL PACIFICO 1F0120 W2 8
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# The ionization constant of propionic acid is 1.32xx10^(-5). Calculate the degree of ionization of the acid in its 0.05M solution and also its pH.
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The ionization constant of propionic acid is 1.32xx10^(-5). Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization in the solution of 0.01N HCI ?
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(I). Caculation of alpha for propionic acid
According to Ostwald Dilution Law,
alpha +((K_(a))/(C))^(1/2) = ((1.32 xx 10^(-5))/(0.05))^(1/2) =(2.64 xx 10^(-4))^(1//2) = 1.62 xx 10^(-2)
(II). Calculation of pH of the solution
[H^(+)] =(K_(a) xx C)^(1//2) =(1.32 xx 10^(-5)xx 5xx 10^(-2))^(1//2)
=(6.6 xx 10^(-7))^(1//2) = (66 xx 10^(-8))^(1//2) =8.124 xx 10^(-4)
pH =- log (8.124 xx 10^(-4)) =- (log 8.124 -4 log 10)
=(4- log 8.124) = (4-0.909) = 3.09
(III). Calculation of alpha for propionic acid in 0.01 M HCI solution.
CH_(3)CH_(2)COOH hArr CH_(3)CH_(2)COO^(-) + H^(+)
In the pressure of HCI. the ionisation of CH_(3)CH_(2)COOH will decrease. If C is the initial concentration of acid adn x is the amount dissociated at equilibrium
[CH_(3)CH_(2)COOH] =C- x , [CH_(3)CH_(2)COO^(-)] = x , [H^(+)] = 0.01 +x
K_(a) =[[CH_(3) CH_(2)COO^(-)][H^(+)]]/[[CH_(3)CH_(2)COOH]] =((x)xx(0.01 +x))/((C -X)) = (x(0.01))/(C)
" or "" " (x)/(C) =(K_(a))/(0.01) =(1.32 xx 10^(-5))/(10^5) =(1.32 xx 10^(-5))/(10^(-2)) =1.32 xx 10^(-3) , alpha =(x)/(C) =1.32 xx 10^(-3)
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Looking back at the history of residential hydronic heating in North America over the last 25 years, one is amazed at the progress made in the area of thermal efficiency.
For example, I grew up in a 1960s vintage ranch house heated by an oil-fired steel fire tube boiler. The boiler was equipped with a tankless coil for domestic water heating. Upstairs was a single thermostat that controlled the entire 1,500-square-foot house.
When I replaced that boiler in 1989, the fire tube baffles were completely burned away. This helped explain the net stack temperature of about 550°F. Although this boiler kept our house comfortable for over 30 years, its "aged"
## As Good As It Gets?
One might conclude that our industry has just about hit the limit of what's theoretically possible with converting fuel into heat. That little remains to be accomplished in terms of efficiency improvements. But just as an oasis fades away when you're just about to it, so does this illusion of nearly perfect efficiency.
True, the thermal efficiency of many modern boilers is very high, but a hydronic system is more than a high-performance boiler. The distribution efficiency at which many current hydronic systems convey heat from the heat source to the heat emitters provides plenty of room for improvement.
Let's define distribution efficiency.
Distribution efficiency describes how many Btus/hr of heat delivery occur for each watt of electrical power used by the distribution system. Just like thermal efficiency, the higher the delivery efficiency, the better.
Here's an example: Consider a hydronic system that delivers 120,000 Btu/hr at design load conditions using four circulators operating at 85 watts each. The distribution efficiency of that system is listed in the following equation.
Distribution efficiency can also be expressed by converting the heat delivery rate into watts to match the units of electrical power input.
The value 103.4 should be interpreted as 103.4 units of heat delivery per unit of electrical power input to operate the distribution system.
It's hard to judge a number for distribution efficiency without something to compare it to. Here's a similar calculation for a furnace with a blower that operates at 850 watts while delivering 80,000 Btu/hr through a forced air ducting system, as shown in the following equation.
For these examples, the hydronic system provides a delivery efficiency about 3.8 times higher than that of the forced air system. This is not uncommon for many current day installations. It's a direct result of water being far superior to air as a media for absorbing and carrying heat.
So, how can we further improve the distribution efficiency of hydronic systems? Here are some things I think will help the North American hydronics industry get more Btu/hr delivered per watt of electrical pump power.
1. Use higher temperature drops: As designers, we have to stop thinking that water "wants"
For example, the graph in Figure 1 shows that a typical hydronic heat emitter, such as a radiant floor circuit or fin tube convector, achieves about 90% of its "design"
Perhaps the most impressive savings associated with reduced flow rate are those in pumping power. One of the pump affinity laws states that pumping power is proportional to the cube of flow rate, as shown in Equation 1.
Where:
P1 = power required at flow rate f1
P2 = power required at flow rate f2
According to this equation, operating a circulator at 50% of design flow would (in theory) require only 12.5% of design load power input. Actual savings will be less due to non-proportional losses in bearings, motor windings, etc., but very significant savings are still waiting for those who recognize the physics of moving water through piping systems, and take advantage of it in their designs.
Figure 2. An example of a smart circulator. Image courtesy of Wilo USA.
Smarter Than the Average Pump
Another way to increase distribution efficiency is to improve the way the "head source"
So what exactly does a "smart circulator"
Constant differential pressure mode allows the differential pressure across the branch circuits to remain steady regardless of what zones are on or off at a given time. It's accomplished by varying the speed of the impeller such that the intersection of the pump curve and system operating curve remain at a constant head (e.g., constant differential pressure). As soon as the system curve steepens or flattens in response to a zone valve closing or opening, the circulator electronically senses the attempted change in differential pressure and instantly responds with a speed adjustment to nullify that change. This is shown in Figure 4, along with a comparison of what happens in systems using an uncontrolled fixed speed circulator.
Another smart pump algorithm is called "Proportional Pressure Control."
The manner in which the pressure/flow characteristic is modified accounts for head losses along the mains, so that the differential pressure across individual zone valves remains almost constant, as shown in Figure 6.
Both constant differential pressure and proportional pressure control modes eliminate the need for a differential pressure bypass valve. Such valves are currently regarded as the standard approach to controlling the symptoms of fluctuating differential pressure in zoned systems with fixed speed circulators. Smart circulators address the cause rather than the symptoms of the problem (excessive head input under partial load conditions). In doing so, the symptoms of excessive differential pressure (flow noise, erosion corrosion, and valve stem lift) are eliminated, while considerable energy savings are also achieved. The cost savings from not having to purchase and install a differential pressure bypass valve obviously reduces the incrementally higher cost of the smart circulator.
Some smart pumps can also compensate for changes in water temperature associated with outdoor reset control, and even nighttime setback. The circulator contains an internal temperature sensor that constantly measures water temperature to (or in some cases from) the distribution system.
## The Bottom Line
Estimated energy savings vary according to the specifics of the load as well as the operating mode of the circulator. Simulations done in Europe project savings in electrical usage of 50% to 80% relative to an uncontrolled fixed speed circulator of equivalent performance.
One manufacturer estimates that converting all small circulators (under 250 watts peak) in the European Union to current generation smart circulators has the potential for saving 10 billion kilowatt-hours per year! Although similar projections for the North American market are yet to be made, there is little doubt that savings in both new and retrofit installations will be profound. Payback periods of four years or less are likely, with a projected 20-year life-cycle cost of a smart circulator approaching half that of an uncontrolled fixed speed circulator.
Several manufacturers currently offer small smart circulators in the European market, and this is finally being noticed in North America. Just as the concept of smaller modulating boilers made its way from Europe to North America, it's inevitable that smart circulators will capture the attention of cutting edge hydronic designers, as well as their clients who are presented with the economic advantages.
Current generation wet rotor circulators will surely remain in the market for the foreseeable future, but will decline in market share as our industry and its customers grasp the energy-saving implications of the next generation of smart circulators. The eventual impact these circulators will make on the North American market will likely rival that of modern high-performance boilers versus their predecessors. Get ready, because big changes in small circulators lie just ahead. Together with careful refinements in piping design, these new circulators will greatly improve the distribution efficiency of present day hydronic systems.
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% Generated by roxygen2: do not edit by hand
% Please edit documentation in R/map_estimate.R
\name{map_estimate}
\alias{map_estimate}
\alias{map_estimate.numeric}
\alias{map_estimate.bayesQR}
\alias{map_estimate.stanreg}
\alias{map_estimate.brmsfit}
\alias{map_estimate.data.frame}
\alias{map_estimate.emmGrid}
\title{Maximum A Posteriori probability estimate (MAP)}
\usage{
map_estimate(x, precision = 2^10, method = "kernel", ...)
\method{map_estimate}{numeric}(x, precision = 2^10, method = "kernel", ...)
\method{map_estimate}{bayesQR}(x, precision = 2^10, method = "kernel", ...)
\method{map_estimate}{stanreg}(
x,
precision = 2^10,
method = "kernel",
effects = c("fixed", "random", "all"),
parameters = NULL,
...
)
\method{map_estimate}{brmsfit}(
x,
precision = 2^10,
method = "kernel",
effects = c("fixed", "random", "all"),
component = c("conditional", "zi", "zero_inflated", "all"),
parameters = NULL,
...
)
\method{map_estimate}{data.frame}(x, precision = 2^10, method = "kernel", ...)
\method{map_estimate}{emmGrid}(x, precision = 2^10, method = "kernel", ...)
}
\arguments{
\item{x}{Vector representing a posterior distribution, or a data frame of such
vectors. Can also be a Bayesian model (\code{stanreg}, \code{brmsfit},
\code{MCMCglmm}, \code{mcmc} or \code{bcplm}) or a \code{BayesFactor} model.}
\item{precision}{Number of points of density data. See the \code{n} parameter in \code{density}.}
\item{method}{Density estimation method. Can be \code{"kernel"} (default), \code{"logspline"} or \code{"KernSmooth"}.}
\item{...}{Currently not used.}
\item{effects}{Should results for fixed effects, random effects or both be returned?
Only applies to mixed models. May be abbreviated.}
\item{parameters}{Regular expression pattern that describes the parameters that
should be returned. Meta-parameters (like \code{lp__} or \code{prior_}) are
filtered by default, so only parameters that typically appear in the
\code{summary()} are returned. Use \code{parameters} to select specific parameters
for the output.}
\item{component}{Should results for all parameters, parameters for the conditional model
or the zero-inflated part of the model be returned? May be abbreviated. Only
applies to \pkg{brms}-models.}
}
\value{
A numeric value if \code{posterior} is a vector. If \code{posterior}
is a model-object, returns a data frame with following columns:
\itemize{
\item \code{Parameter} The model parameter(s), if \code{x} is a model-object. If \code{x} is a vector, this column is missing.
\item \code{MAP_Estimate} The MAP estimate for the posterior or each model parameter.
}
}
\description{
Find the \strong{Highest Maximum A Posteriori probability estimate (MAP)} of a posterior, i.e., the value associated with the highest probability density (the "peak" of the posterior distribution). In other words, it is an estimation of the \emph{mode} for continuous parameters. Note that this function relies on \link{estimate_density}, which by default uses a different smoothing bandwidth (\code{"SJ"}) compared to the legacy default implemented the base R \link{density} function (\code{"nrd0"}).
}
\examples{
\dontrun{
library(bayestestR)
posterior <- rnorm(10000)
map_estimate(posterior)
plot(density(posterior))
abline(v = map_estimate(posterior), col = "red")
library(rstanarm)
model <- rstanarm::stan_glm(mpg ~ wt + cyl, data = mtcars)
map_estimate(model)
library(brms)
model <- brms::brm(mpg ~ wt + cyl, data = mtcars)
map_estimate(model)
}
}
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# 7th Grade Multiplying Rationals Lesson: FOLDABLE & Homework
Common Core Standards
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Product Description
This product includes a foldable with suggested notes, worksheet, and answer key covering Multiplying Rationals – CCSS.7.NS.A.2, 2.A.
In my class, I use the left hand side of the notebook for guided notes with foldables, while the right-hand side is reserved for individual practice work. There is a one-sided worksheet to be glued into interactive notebooks on the right hand page opposite the notes. I usually trim just a bit around the edge of the worksheets with a paper cutter so that they fit perfectly, but this is not necessary.
I started using Stick-n-Solve Foldables in my Math Interactive Notebooks last year and it worked great! There are a few things about these Stick-n-Solves that I really have enjoyed. First, my students no longer spend time copying down problems when we take notes. I always thought this was a waste of time. Now, the problems are on the foldable ready to be solved. The students like to cut and fold and glue while working in their notebooks. It gives them something tactile to do during class. Finally, the foldables are a built in review tool for your students. At the end of a unit, they can go back through their notebooks and solve all the problems on the Stick-n-Solves. Since the work is on the inside, they just open them up to check their answers. Each foldable in this set has two per page. My students are set up in partners, so I give one sheet to each partner pair to cut in half. There is no extra paper on these foldable templates (which means no little scraps of paper to trim off and end up all over the floor).
With the foldable, you will see two pictures. You will see a draft picture of notes for the topic, and a picture of the solutions on the inside of the foldable.
For almost every topic covered in seventh grade common core math, I’ve made a foldable and an assessment. In total I have created 50 Stick-n-Solve Foldables and coordinating assessments and organized them into the following bundles:
1. Proportional Reasoning
2. Rational Numbers
3. Algebra
4. Probability & Statistics
5. Geometry
6. Scale & Construction
My Entire 7th Grade Math Curriculum includes this activity and every other resource I have created for seventh grade math!
**Leave Feedback after your purchase to earn TpT credits!!**
Common Core:
CCSS.MATH.CONTENT.7.NS.A.2
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.
CCSS.MATH.CONTENT.7.NS.A.2.A
Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (-1)(-1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
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https://fr.mathworks.com/matlabcentral/cody/problems/2593-polite-numbers-n-th-polite-number/solutions/1175089
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Cody
# Problem 2593. Polite numbers. N-th polite number.
Solution 1175089
Submitted on 29 Apr 2017 by AM
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 3; assert(isequal(Nth_polite(x),y_correct))
2 Pass
x = 2; y_correct = 5; assert(isequal(Nth_polite(x),y_correct))
3 Pass
x = 4; y_correct = 7; assert(isequal(Nth_polite(x),y_correct))
4 Pass
x = 5; y_correct = 9; assert(isequal(Nth_polite(x),y_correct))
5 Pass
x = 7; y_correct = 11; assert(isequal(Nth_polite(x),y_correct))
6 Pass
x = 11; y_correct = 15; assert(isequal(Nth_polite(x),y_correct))
7 Pass
x = 12; y_correct = 17; assert(isequal(Nth_polite(x),y_correct))
8 Pass
x = 14; y_correct = 19; assert(isequal(Nth_polite(x),y_correct))
9 Pass
x = 19; y_correct = 24; assert(isequal(Nth_polite(x),y_correct))
10 Pass
x = 21; y_correct = 26; assert(isequal(Nth_polite(x),y_correct))
11 Pass
x = 27; y_correct = 33; assert(isequal(Nth_polite(x),y_correct))
12 Pass
x = 64; y_correct = 71; assert(isequal(Nth_polite(x),y_correct))
13 Pass
x = 1e6; y_correct = x+20; assert(isequal(Nth_polite(x),y_correct))
14 Pass
x = 1e7; y_correct = x+24; assert(isequal(Nth_polite(x),y_correct))
15 Pass
x = 999999999; y_correct = x+30; assert(isequal(Nth_polite(x),y_correct))
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# Objectively Determined Valuations
## Objectively Determined Valuations
a term used to designate partial derivatives of an objective function, taken in relation to constraints in problems of linear or convex programming. The term was introduced by Soviet scientist L. V. Kantorovich in 1959 and is mostly used in solving economic problems by mathematical programming. It is analogous to the terms “optimal valuations,” “dual valuations,” “shadow prices,” and “resolving multipliers.”
In economic problems objectively determined valuations show the economic consequences when an additional unit of a particular production component is introduced into the economic process. The dimensionality of the objectively determined valuations corresponds to the dimensionality of the criterion of optimality (monetary, physical, or standardized physical units of measure). Objectively determined valuations follow from the conditions of the statement and solution of an economic problem. They are determined by the aggregate of concrete economic factors that are taken into account in the mathematical formalization of production and economic activity. Thus they are an effective means of analyzing current economic activity, making it possible to identify and quantitatively evaluate “bottlenecks.” Objectively determined valuations make it possible, assuming a certain stability, to indicate ways of improving the indexes of the work of an economic unit.
Depending on how the problem is formulated, objectively determined valuations may reflect the production and economic conditions of particular sections (or workshops), enterprises, sectors, regions, and the national economy as a whole. In the last case the valuations obtained can theoretically be interpreted as the prices of the optimal economic plan or as social (rent) valuations of resources (natural resources, capital assets and labor). They describe growth in the criterion of optimality of the socialist system (growth in prosperity and the level of satisfaction of social needs) resulting from incremental growth in the production of particular types of output (or an increase in a resource). They also describe the acceptable maximum of expenditure to produce an additional unit of this output. Objectively determined valuations keep this characteristic only for small economic changes, and their values ordinarily change with the development and modification of plans for the expansion of production. The organic link between objectively determined valuations and a plan may clearly be seen in mathematical economic problems of any level, both in static and dynamic models, where they make it possible to compare expenditures and effects at different time periods.
Objectively determined valuations have also been treated in depth in the works of A. L. Lur’e and in the conception of differential expenditures in the national economy worked out by V. V. Novozhilov. The problem of objectively determined valuations is currently being further developed and is the subject of discussion by Soviet and foreign economists.
### REFERENCES
Kantorovich, L. V. Ekonomicheskii raschet nailuchshego ispol’zovaniia resursov. Moscow, 1959.
Lur’e, A. L. O matematicheskikh metodakh resheniia zadach na optimum pri planirovanii sotsialisticheskogo khoziaistva. Moscow, 1964.
Novozhilov, V. V. Problemy izmereniia zatrat i rezul’tatov pri optimal’nom planirovanii. Moscow, 1967.
N. IA. PETRAKOV
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Flow Solver Tutorial
# Basic Flow-Solver Tutorial
The purpose of this example is to give an overview of how to set up and use the single-phase mimetic pressure solver to solve the single-phase pressure equation
for a flow driven by Dirichlet and Neumann boundary conditions. Our geological model will be simple a Cartesian grid with anisotropic, homogeneous permeability.
In this tutorial example, you will learn about:
1. the grid structure,
2. how to specify rock and fluid data,
3. the structure of the data-objects used to hold solution,
4. how to assemble and solve linear systems,
5. the structure of the mimetic linear systems,
6. useful routines for visualizing and interacting with the grids and simulation results.
## Define geometry
Construct a Cartesian grid of size 10-by-10-by-4 cells, where each cell has dimension 1-by-1-by-1. Because our flow solvers are applicable for general unstructured grids, the Cartesian grid is here represented using an unstructured format, in which cells, faces, nodes, etc. are given explicitly.
nx = 10; ny = 10; nz = 4;
G = cartGrid([nx, ny, nz]);
display(G);
G =
cells: [1x1 struct]
faces: [1x1 struct]
nodes: [1x1 struct]
cartDims: [10 10 4]
type: {'tensorGrid' 'cartGrid'}
After the grid structure is generated, we plot the geometry.
plotGrid(G);
view(3), camproj orthographic, axis tight, camlight headlight
## Process geometry
Having set up the basic structure, we continue to compute centroids and volumes of the cells and centroids, normals, and areas for the faces. For a Cartesian grid, this information can trivially be computed, but is given explicitly so that the flow solver is compatible with fully unstructured grids.
G = computeGeometry(G);
## Set rock and fluid data
The only parameters in the single-phase pressure equation are the permeability and the fluid viscosity . We set the permeability to be homogeneous and anisotropic
The viscosity is specified by saying that the reservoir is filled with a single fluid, for which de default viscosity value equals unity. Our flow solver is written for a general incompressible flow and requires the evaluation of a total mobility, which is provided by the fluid object.
rock.perm =
repmat([1000, 100, 10].* milli*darcy(), [G.cells.num, 1]);
fluid = initSingleFluid('mu' , 1*centi*poise , ...
'rho', 1014*kilogram/meter^3);
## Initialize reservoir simulator
To simplify communication among different flow and transport solvers, all unknowns are collected in a structure. Here this structure is initialized with uniform initial reservoir pressure equal 0 and (single-phase) saturation equal 0.0 (using the default behavior of initResSol)
resSol = initResSol(G, 0.0);
display(resSol);
resSol =
pressure: [400x1 double]
flux: [1380x1 double]
s: [400x1 double]
## Impose Dirichlet boundary conditions
Our flow solvers automatically assume no-flow conditions on all outer (and inner) boundaries; other type of boundary conditions need to be specified explicitly.
Here, we impose Neumann conditions (flux of 1 m^3/day) on the global left-hand side. The fluxes must be given in units of m^3/s, and thus we need to divide by the number of seconds in a day. Similarly, we set Dirichlet boundary conditions p = 0 on the global right-hand side of the grid, respectively. For a single-phase flow, we need not specify the saturation at inflow boundaries. Similarly, fluid composition over outflow faces (here, right) is ignored by pside.
bc = fluxside([], G, 'LEFT', 1*meter^3/day());
bc = pside (bc, G, 'RIGHT', 0);
display(bc);
bc =
face: [80x1 int32]
type: {1x80 cell}
value: [80x1 double]
sat: []
## Construct linear system
Construct mimetic pressure linear system components for the system Ax = b
based on input grid and rock properties for the case with no gravity.
gravity off;mrstModule add mimetic
S = computeMimeticIP(G, rock);
% Plot the structure of the matrix (here we use BI, the inverse of B,
% rather than B because the two have exactly the same structure)
clf, subplot(1,2,1)
cellNo = rldecode(1:G.cells.num, diff(G.cells.facePos), 2) .';
C = sparse(1:numel(cellNo), cellNo, 1);
D = sparse(1:numel(cellNo), double(G.cells.faces(:,1)), 1, ...
numel(cellNo), G.faces.num);
spy([S.BI , C , D ; ...
C', zeros(size(C,2), size(C,2) + size(D,2)); ...
D', zeros(size(D,2), size(C,2) + size(D,2))]);
title('Hybrid pressure system matrix')
The block structure can clearly be seen in the sparse matrix A, which is never formed in full. Indeed, rather than storing B, we store its inverse B^-1. Similarly, the C and D blocks are not represented in the S structure; they can easily be formed explicitly whenever needed, or their action can easily be computed.
display(S);
S =
BI: [2400x2400 double]
ip: 'ip_simple'
type: 'hybrid'
## Solve the linear system
Solve linear system construced from S and bc to obtain solution for flow and pressure in the reservoir. Function incompMimetic demands that we pass a well solution structure even if the reservoir has no wells, so we initialize an empty wellSol structure. The option 'MatrixOutput=true' adds the system matrix A to resSol to enable inspection of the matrix.
resSol = incompMimetic(resSol, G, S, fluid, ...
'bc', bc, 'MatrixOutput', true);
display(resSol);
resSol =
pressure: [400x1 double]
flux: [1380x1 double]
s: [400x1 double]
facePressure: [1380x1 double]
A: [1340x1340 double]
## Inspect results
The resSol object contains the Schur complement matrix used to solve the hybrid system.
subplot(1,2,2), spy(resSol.A);
title('Schur complement system matrix');
We then plot convert the computed pressure to unit 'bar' before plotting result.
clf
plotCellData(G, convertTo(resSol.pressure(1:G.cells.num), barsa()), ...
'EdgeColor', 'k');
title('Cell Pressure [bar]')
xlabel('x'), ylabel('y'), zlabel('Depth');
view(3); shading faceted; camproj perspective; axis tight;
colorbar
Published March 4, 2011
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# ACT® Math
Free Version
Moderate
ACTMAT-LKEWYJ
For which of the following intervals is $9-{ x }^{ 2 }<6$?
A
$x<-\sqrt { 15 } \text{ or } x>\sqrt { 15 }$
B
$x<-\sqrt { 3 } \text{ or } x>\sqrt { 3 }$
C
$-\sqrt { 9 }< x <\sqrt { 9 }$
D
$-\sqrt { 3 }< x <\sqrt { 3 }$
E
$-\sqrt { 15 }< x < \sqrt { 15 }$
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#### Details of this Paper
##### stats MCQ problems with solution fall 2014
Description
solution
Question
Question;1The right way to think about the sample mean is:a The sample mean is a constant number.b The sample mean is a different value in each random sample from the population mean.c The sample mean is always close to the population mean.d The sample mean is always smaller than the population mean.2The sampling distribution of x is approximately normal ifa the distribution of x is skewed.b the distribution of x is approximately symmetricc the sample size is large enough.d the sample size is small enough.3abcdThere is a population of six families in a small neighborhood: Albertson, Benson, Carlson,Davidson, Erikson, and Fredrickson. You plan to take a random sample of n=3 families (withoutreplacement). The total number of possible sample is _____.6121820abcdThe mean daily output of an automobile manufacturing plant is = 520 cars with standarddeviation of = 14 cars. In a random sample of n = 49 days, the probability that the samplemean output of cars (x) will be within 3 cars from the population mean is _________.0.98760.95440.92660.8664abcdIn the population of IUPUI undergraduate students 38 percent (0.38) enroll in classes during thesummer sessions. Let p denote the sample proportion of students who plan to enroll in summerclasses in samples of size n = 200 selected from this population. The expected value of thesample proportion, E(p), is _______.0.380.280.250.18abcdIn the previous question, the standard error of the sampling distribution of p is, se(p)=_______.0.03430.02970.02480.02214567The expressionMeans:a Once you take a specific sample and calculate the value of x, the probability that the value of xyou just calculated is within 1.96 /n from is 0.95.b In repeated samples, the probability that x is within 1.96 /n from is 0.95.c Once you take a specific sample and calculate the value of x, you are 95 percent certain that thevalue you calculated is.d In repeated samples, you are 95 percent certain that the value of x is.8As part of a course assignment to develop an interval estimate for the proportion of IUPUIstudents who smoke tobacco, each of 480 E270 students collects his or her own random sampleof n=400 IUPUI students to construct a 95 percent confidence interval. Considering the 480intervals constructed by the E270 students, we would expect ________ of these intervals tocapture the population proportion of IUPUI students who smoke tobacco.a480b456c400d3809abcdAssume the actual population proportion of IUPUI students who smoke tobacco is 20 percent(0.20). What proportion of sample proportions obtained from random samples of size n=300 arewithin a margin of error of 3 percentage points (0.03) from the population proportion?0.80640.84720.88580.905010To estimate the average number of customers per business day visiting a branch of Fifth NationalBank, in a random sample of n = 9 business days the sample mean number of daily customervisits is x = 250 with a sample standard deviation of s = 36 customers. The 95 percentconfidence interval for the mean daily customer visits is:a (205, 295)b (217, 283)c (222, 278)d (226, 274)11abcdIn the previous question, how large a sample should be selected in order to have a margin of errorof 5 daily customer visits? Use the standard deviation in that question as the planning value.7810113920012Compared to a confidence interval with a 90 percent confidence level, an interval based on thesame sample size with a 99 percent level of confidence:a is wider.b is narrower.c has the same precision.d would be narrower if the sample size is less than 30 and wider if the sample size is at least 30.13It is estimated that 80% of Americans go out to eat at least once per week, with a margin of errorof 0.04 and a 95% confidence level. A 95% confidence interval for the population proportion ofAmericans who go out to eat once per week or more is:a (0.798, 0.802)b (0.784, 0.816)c (0.771, 0.829)d (0.760, 0.840)14In a random sample of 600 registered voters, 45 percent said they vote Republican. The 95%confidence interval for proportion of all registered voters who vote Republican is,a (0.401, 0.499)b (0.410, 0.490)c (0.421, 0.479)d (0.426, 0.474)John is the manager of an election campaign. Johns candidate wants to know what proportion ofthe population will vote for her. The candidate wants to know this with a margin of error of0.01 (at 95% confidence). John thinks that the population proportion of voters who will vote forhis candidate is 0.50 (use this for a planning value). How big of a sample of voters should youtake?a9,604b8,888c5,037d1,4991516abcdIf the candidate changes her mind and now wants a margin-of-error of 0.03 (but still 95% confidence),John could select a different sample of the same size, but adjust the error probability.John should select a larger sample.John should select a smaller sample.John should inform the candidate that margin of error does not impact the sample size.17In a test of hypothesis, which of the following statements about a Type I error and a Type II erroris correct:a Type I: Reject a true alternative hypothesis.Type II: Do not reject a false alternative hypothesis.b Type I: Do not Reject a false null hypothesis. II: Reject a true null hypothesis.Typec Type I: Reject a false null hypothesis.Type II: Reject a true null hypothesis.d Type I: Reject a true null hypothesis.Type II: Do not reject a false null hypothesis.18You are reading a report that contains a hypothesis test you are interested in. The writer of thereport writes that the p-value for the test you are interested in is 0.0831, but does not tell you thevalue of the test statistic. Using as the level of significance, from this information you ______a decide to reject the hypothesis at = 0.10, but not reject at = 0.05.b cannot decide based on this limited information. You need to know the value of the test statistic.c decide not to reject the hypothesis at = 0.10, and not to reject at = 0.05d decide to reject the hypothesis at = 0.10, and reject at = 0.0519Linda works for a charitable organization and she wants to see whether the people who donate toher organization have an average age over 40 years. She obtains a random sample of n = 180donors and the value of the sample mean is x = 42 years, with a sample standard deviation of s =18 years. She wants to conduct the test of H: 40 with a 5% level of significance. Sheshould reject H if the value of the test statistic is _____a less than the critical value.b greater than the critical value.c more than two standard errors above the critical value.d equal to the critical value.20 Now she performs the test and obtains the test statistic of TS = ______,a 1.49 and does not reject H. She concludes that the average age is not over 40.b 1.49 and rejects H. She concludes that the average age is over 40.c 1.74 and does not reject H. She concludes that the average age is not over 40.d 1.74 and rejects H. She concludes that the average age is over 40.21 The probability value for Lindas hypothesis test is ______.a0.0207b0.0409c0.0542d0.0681The Census Bureaus American Housing Survey has reported that 80 percent of families choosetheir house location based on the school district. To perform a test, with a probability of Type Ierror of 5 percent, that the population proportion really equals 0.80, in a sample of 600 families504 said that they chose their house based on the school district. The null hypothesis would berejected if the sample proportion falls outside the margin of error. The margin of error for the testis:a0.039b0.032c0.025d0.0202223 The probability value for the hypothesis test in the previous question is:a0.0026b0.0071c0.0142d0.022424Given the following sample data, is there enough evidence, at the 5 percent significance level, thepopulation mean is greater than 7?x921517811135Compute the relevant test statistic.a The test statistic is 1.683 and the critical value is 1.895. Do not reject the null hypothesis andconclude that the population mean is not greater than 7.b The test statistic is 1.683 and the critical value is 1.895. Reject the null hypothesis and concludethat the population mean is greater than 7.c The test statistic is 2.432 and the critical value is 2.365. Reject the null hypothesis and concludethat the population mean is greater than 7.d The test statistic is 2.432 and the critical value is 1.895. Reject the null hypothesis and concludethat the population mean is not greater than 7.Next SIX questions are based on the following regression modelIn a regression model relating the price of homes (in \$1,000) as the dependent variable to theirsize in square feet, a sample of 20 homes provided the following regression output. Some of thecalculations are left blank for you to compute.SUMMARY OUTPUTRegression StatisticsMultiple R0.7760R SquareAdjusted R Square0.5801Standard ErrorObservations20ANOVARegressionResidualTotaldfSS118 13960.4919 35094.63Coefficients Std ErrorIntercept 15.8479 25.0665Size (Square Feet)0.06950.0133MSF Significance F27.2494 5.78E-05t StatP-value Lower 95% Upper 95%0.6320.5352 -36.81568.5115.79E-050.041625 The model predicts that the price of a home with a size of 2,000 square feet would be ______ thousand.a\$148.70b \$154.80c\$159.50d \$164.3026 The sum of squares regression (SSR) is:a 49055.12b 35094.63c 21134.14d 13960.4927abcdThe regression model estimates that _____% of the variation in the price of the home is explainedby the size of the homes.60.20%65.60%71.50%77.20%28 The standard error of the regression (standard error of estimate) is ______.a30.634b33.698c27.849d24.06729abcd30abcdThe value of the test statistic to test the null hypothesis that property size does not influence theprice of the property is ______.4.3485.2266.3916.982The margin of error to build a 95% confidence interval for the slope coefficient that relates theprice response to each additional square foot is _______.0.0420.0320.0340.028
Paper#61306 | Written in 18-Jul-2015
Price : \$28
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# Multiview Multidimensional Scaling (MVMDS)¶
MVMDS is a useful multiview dimensionaltiy reduction algorithm that allows the user to perform Multidimensional Scaling on multiple views at the same time.
[1]:
from mvlearn.datasets import load_UCImultifeature
from mvlearn.embed import MVMDS
import seaborn as sns
import matplotlib.pyplot as plt
import numpy as np
from sklearn.decomposition import PCA
%matplotlib inline
Data comes from UCI Digits Data. Contains 6 views and classifications of numbers 0-9
[2]:
# Load full dataset, labels not needed
[3]:
# Check data
print(f'There are {len(Xs)} views.')
print(f'There are {Xs[0].shape[0]} observations')
print(f'The feature sizes are: {[X.shape[1] for X in Xs]}')
There are 6 views.
There are 2000 observations
The feature sizes are: [76, 216, 64, 240, 47, 6]
## Plotting MVMDS vs PCA¶
Here we demonstrate the superior performance of MVMDS on multi-view data against the performance of PCA. To use all the views’ data in PCA, we concatenate the views into a larger data matrix.
Examples of 10-class and 4 class data are shown. MVMDS learns principle components that are common across views, and end up spreading the data better.
[4]:
# MVMDS reduction
mvmds = MVMDS(n_components=2)
Xs_mvmds_reduced = mvmds.fit_transform(Xs)
# Concatenate views then PCA for comparison
Xs_concat = Xs[0]
for X in Xs[1:]:
Xs_concat = np.hstack((Xs_concat, X))
pca = PCA(n_components=2)
Xs_pca_reduced = pca.fit_transform(Xs_concat)
[5]:
fig, ax = plt.subplots(1, 2, figsize=(14,6))
ax[0].scatter(Xs_mvmds_reduced[:,0], Xs_mvmds_reduced[:,1], c=y)
ax[0].set_title("MVMDS Reduced Data (10-class)")
ax[0].set_xlabel("Component 1")
ax[0].set_ylabel("Component 2")
ax[1].scatter(Xs_pca_reduced[:,0], Xs_pca_reduced[:,1], c=y)
ax[1].set_title("PCA Reduced Data (10-class)")
ax[1].set_xlabel("Component 1")
ax[1].set_ylabel("Component 2")
plt.show()
[6]:
# 4-class data
[7]:
# MVMDS reduction
mvmds = MVMDS(n_components=2)
Xs_mvmds_reduced = mvmds.fit_transform(Xs_4)
# Concatenate views then PCA for comparison
Xs_concat = Xs_4[0]
for X in Xs_4[1:]:
Xs_concat = np.hstack((Xs_concat, X))
pca = PCA(n_components=2)
Xs_pca_reduced = pca.fit_transform(Xs_concat)
[8]:
fig, ax = plt.subplots(1, 2, figsize=(14,6))
ax[0].scatter(Xs_mvmds_reduced[:,0], Xs_mvmds_reduced[:,1], c=y_4)
ax[0].set_title("MVMDS Reduced Data (4-class)")
ax[0].set_xlabel("Component 1")
ax[0].set_ylabel("Component 2")
ax[1].scatter(Xs_pca_reduced[:,0], Xs_pca_reduced[:,1], c=y_4)
ax[1].set_title("PCA Reduced Data (4-class)")
ax[1].set_xlabel("Component 1")
ax[1].set_ylabel("Component 2")
plt.show()
## Components of MVMDS Views Without Noise¶
Here we will take into account all of the views and perform MVMDS. This dataset does not contain noise and each view performs decently well in predicting the number. Therefore we will expect the common components created by MVMDS to create a strong representation of the data (Note MVMDS only uses the fit_transform function to properly return the correct components)
In the cell after, PCA on one view is shown for comparison. It can be seen that MVMDS seems to perform better in this instance.
Note: Each color represents a unique number class
[9]:
#perform mvmds
mvmds = MVMDS(n_components=5)
Components = mvmds.fit_transform(Xs)
[11]:
# Plot MVMDS images
plt.style.use('seaborn')
color_map = [sns.color_palette("Set2", 10)[int(i)] for i in y]
fig, axes = plt.subplots(4, 4, figsize = (12,12), sharey=True, sharex=True)
for i in range(4):
for j in range(4):
if i != j:
axes[i,j].scatter(x = Components[:, i], y = Components[:, j], alpha = 1, label = y, color = color_map)
axes[3, j].set_xlabel(f'Component {j+1}')
axes[i,0].set_ylabel(f'Component {i+1}')
plt.tick_params(labelcolor='none', top=False, bottom=False, left=False, right=False)
ax.grid(False)
ax.set_title('First 4 MVMDS Components Computed With 6 Views (No Noise)')
[11]:
Text(0.5, 1.0, 'First 4 MVMDS Components Computed With 6 Views (No Noise)')
[12]:
#PCA Plots
pca = PCA(n_components=6)
pca_Components = pca.fit_transform(Xs[0])
fig, axes = plt.subplots(4, 4, figsize = (12,12), sharey=True, sharex=True)
for i in range(4):
for j in range(4):
if i != j:
axes[i,j].scatter(x = pca_Components[:, i], y = pca_Components[:, j], alpha = 1, label = y, color = color_map)
axes[3, j].set_xlabel(f'Component {j+1}')
axes[i,0].set_ylabel(f'Component {i+1}')
[12]:
Text(0.5, 1.0, 'First 4 PCA Components Computed With 1 View')
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# Homework Help: SAT/ GCSE-Level Recurrence Relation Problem
Tags:
1. Jul 16, 2011
### odolwa99
1. The problem statement, all variables and given/known data
Hi! This is my first time on the site. I look forward to working with everyone…but hopefully not too much, assuming I‘m learning things correctly. :P
My question pertains to Recurrence Relations, so here it goes…
Foreword: The text book I’m using actually supplies the answer to the question, so I already have a point of reference, but my attempt does not match up with the answers. I believe my approach is essentially correct, as it has yielded the correct answers for a similar question beforehand. Answer is: 1, 3, 7, 17, 41
Please note that I am beginning the question from u3, as we already have the values for u1 and u2.
2. Relevant equations
Q. Find the first five terms of the sequence:
u1 = 1, u2 = 3 and un = 2un-1 + un-2
3. The attempt at a solution
Attempt:
Solve un+1 where un = 3un-1 - un-2
=> 3u(n+1)-1 - u(n+1)-2
Begin by substituting 3 (i.e. u2) for un:
If n = 1 then u3 = 2((3+1) - 1) + ((3+1) -2) => 2(4-1) + (4-2) => 6 + 2
Ans.: u3 = 8... but should be 7!!!
Proceeding with u3 as 7, not 8...
If n = 2 then u4 = 2((7+1) -1) + ((7+1) -2) => 2(8-1) + (8-2) => 14 + 6
Ans.: u4 = 20... But should be 17!!!
Note, I am omitting solution of u5 for brevity’s sake.
I‘m sure the answer is staring me in the face, but I just can’t seem to figure it out!
Can anyone help?
Thanks.
Last edited: Jul 16, 2011
2. Jul 16, 2011
### hunt_mat
Not too sure what you're going here but let's calculate $u_{3}$ from the recurrence relation.
$$u_{3}=2u_{2}+u_{1}=2\times 3+1=7$$
Working for $u_{4}$
$$u_{4}=2u_{3}+u_{2}=2\times 7+3=17$$
3. Jul 16, 2011
### odolwa99
Woah, that was easier than I was making it! Thank you.
One final question though, why is the value of u1 subbed into un-2 and u2 into un-1?
4. Jul 16, 2011
### hunt_mat
you're finding n=3, so n-1=2 and n-2=1.
5. Jul 16, 2011
### odolwa99
Thank you very much. You've really helped me out!
6. Jul 16, 2011
### hunt_mat
it's why I help here.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
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# Hector
Hector has 3/4 hours to spend completing his homework. He wants to spend 1/3 of his time on math homework, 10 minutes on reading homework, and the remaining time on science homework. How much time, in minutes, does he have to spend on his SCIENCE homework?
x = 20 min
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips for related online calculators
Need help calculating sum, simplifying, or multiplying fractions? Try our fraction calculator.
Do you want to convert time units like minutes to seconds?
#### Grade of the word problem:
We encourage you to watch this tutorial video on this math problem:
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# Factorise:
Question:
Factorise:
$a^{2}-b^{2}+2 b c-c^{2}$
Solution:
$a^{2}-b^{2}+2 b c-c^{2}$
$=a^{2}-\left(b^{2}-2 b c+c^{2}\right)$
$=a^{2}-(b-c)^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$
$=[a+(b-c)][a-(b-c)] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$
$=(a+b-c)(a-b+c)$
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### Associative Fold#
We consider implementing a fold operator in parallel on a list of items. We first consider the case when the fold operator is associative.
We define afold(b,xs) where b is a binary associative function over two arguments, and xs is a non-empty list. The value returned by afold is as follows:
```afold(_,[x]) = x
afold(b,x:xs) = b(x,afold(xs))```
The parallel implementation builds a sequence of lists. The initial list is the given one. Each subsequent list is about half the size of the previous list, obtained by folding each disjoint pair of adjecent items. In the following, afold'(xs) builds the fold from xs. Execution of one instance of afold'() is in parallel, though different instances are executed in sequence.
```def afold(b, [x]) = x
def afold(b, xs) =
def afold'([]) = []
def afold'([x]) = [x]
def afold'(x:y:xs) = b(x,y):afold'(xs)
afold(b, afold'(xs))```
### Associative, Commutative Fold#
We devise a better strategy when the fold operator is both associative and commutative. We define cfold(b,xs) where b is a binary associative and commutative function over two arguments, and xs is a non-empty list. The value returned by cfold is as follows:
```cfold(_,[x]) = x
cfold(b,x:xs) = b(x,cfold(xs))```
Our implementation differs from afold() in that all list items are put in a buffer in arbitrary order, two items are folded as soon as possible, and their result is put in the buffer.
```def cfold(b, xs) =
val c = Buffer()
--------- Transfer all items of the argument list to Buffer c ---
-- The order of items in c is arbitrary.
def xfer([]) = stop
def xfer(x:xs) = c.put(x) >> stop | xfer(xs)
------------------------ End of xfer ----------------------------
---------- combine(n) computes fold of n items from Buffer c ----
def combine(1) = c.get()
def combine(m) = c.get() >x> c.get() >y> ( c.put(b(x,y)) >> stop | combine(m-1))
--------------------- End of combine(n) --------------------------
{- Goal of cfold -}
xfer(xs) | combine(length(xs))```
### Map with Associative Fold#
We generalize the functions defined earlier; first we map list items using a supplied function and then compute their fold. And, we distinguish between associative fold, and commmutative-associative fold.
Function mapafold(b,L,f), where b is an associative operator, L is a non-empty list of items and f is a mapping function over items L, has the following meaning:
`mapafold(b,L,f) = afold(b,f(L))`
However, we can compute the fold and map operations in parallel, unlike the sequencing implied by the given definition.
The strategy is to build a sequence of buffers containing the values F(L) in order. Recursively, we fold the left half and the right of the sequence in parallel, and then combine them. Observe that in order to fold a single value f(x), it takes around log n sequential steps, where n is the length of L.
```def mapafold(_,[x],f) = f(x)
def mapafold(b,L,f) =
val n = length(L)
val c = fillArray(Array(n), lambda (_) = Buffer())
----- fill(i,L), for non-empty list L, puts f(L_j) in channel i+j---
def fill(i,[]) = stop
def fill(i,x:xs) = c(i)?.put(f(x)) >> stop | fill(i+1,xs)
------------------------End of fill(i,L)------------------------------
------ aggr(i,k) folds k items from the channels starting at i -------
{- i >= 0 and k >= 1, k denotes the number of values to be folded,
the goal is to compute:
c(i).get(), if k = 1
b(c(i).get(), c(i+1).get(), ... c(i+k-1).get()), if k > 1
-}
def aggr(i,1) = c(i)?.get()
def aggr(i,k) =
val r = k/2
(aggr(i,r), aggr(i+r,k-r)) >(x,y)> b(x,y)
-------------------------- End of aggr(i,k) --------------------------
{- Goal of mapafold(b,L,f) -}
fill(0,L) | aggr(0, n)```
### Map with Commutative and Associative Fold#
We use the same technique as in the previous commutative fold over a list of items. We store f(x), for each x in the list in Buffer c.
```def mapcfold(b,L,f) =
val c = Buffer()
--------- Transfer map of the argument list to Buffer c ---
-- The order in c is arbitrary
def xfer([]) = stop
def xfer(x:xs) = c.put(f(x)) >> stop | xfer(xs)
------------------------ End of xfer ----------------------------
---------- combine(n) computes fold of n items from Buffer c ----
def combine(1) = c.get()
def combine(n) = c.get() >x> c.get() >y> ( c.put(b(x,y)) >> stop | combine(n-1))
--------------------- End of combine(n) --------------------------
{- Goal of mapcfold(b,L,f) -}
xfer(L) | combine(length(L))```
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# Convert Decimal to Hex
To convert Decimal to Hexadecimal, input decimal value in the box below, and then click on the big blue button that says “CONVERT TO HEX” and Hex is generated, copy it or you can download output file.
Copied to Clipboard.
## Decimal Numbering System
In decimal number system there only ten (10) digits from 0 to 9. Every number (value) in this decimal system represents with 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The base of this number system is 10, because it has only 10 digits.
Example: 8062 in decimal (base 10)
806210 = (8 x 103) + (0 x 102) + (6 x 101) + (2 x 100)
In a Hexadecimal number system there are sixteen (16) alphanumeric values from 0 to 9 and A to F. In this number system every number (value) represents with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. The base of this number system is 16, because it has 16 alphanumeric values. Here A is 10, B is 11, C is 12, D is 14, E is 15 and F is 16.
Example: What is 0xA5 in base 10?
0xA5 = A516 = (10 x 161) + (5 x 160) = 16510
• Step 1: To convert a decimal number into hexadecimal, first you have to check whether the number in question is greater the 16 or not. If the number is less than 16 then we will have to use A, B, C, D, E and F for the numbers 10, 11, 12, 13, 14 and 15 respectively. For example.
• Step 2: If the number is greater than 16 then divide the number in question by 16.
• Step 3: Note down the remainder.
• Step 4: Divide the quotient by 16 again and note down the remainder.
• Step 5: Repeat the steps until you your quotient is less than 16.
• Step 6: Note down the remainder from step 2 to 6.
• Step 7: The column of the remainder is read in reverse order i.e., from bottom to top order. We try to discuss the method with an example.
Example 1: Convert (348)10 into a hexadecimal number.
Solution:
Division Quotient Generated Remainder 348/16 21 12 (C) 21/15 1 5 1/16 0 1
Hence the converted hexadecimal number is (15C)16.
Example 2: Covert (4768)10 into a hexadecimal number.
Solution:
Division Quotient Generated Remainder 4768/16 298 0 298/15 18 10 (A) 18/16 1 2 1/16 0 1
Hence the converted hexadecimal number is (12A0)16.
#### Decimal Hex Conversion Chart
Decimal Hexadecimal 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 A 11 B 12 C 13 D 14 E 15 F
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# SOLUTION: The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence.
Algebra -> Sequences-and-series -> SOLUTION: The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence. Log On
Question 217318: The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence.
You can put this solution on YOUR website!
The third term of an arithmetic sequence is 4 and the sum of the first 8 term is 36. Write down the first 8 term of the sequence.
Step 1. An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value
Step 2. Let x be the number to add or subtract in the sequence.
Step 3. Let a be the first term, let a+x be the second term, let a+2x be the third term, a+3x be the fourth term, and a+(n-1)x be the nth term
Step 4. Let a+7x be the 8th term. Then the sum of the first and eighth term is a+a+7x=2a+7x. The sum of the second term and seventh term is the a+x+a+6x=2a+6x. So four the first eight terms, there will be four pairs of 2a+6x which will equal to 36 or
Equation A
Also Equation B since the 3th term is 4.
Step 5. Then we have a system of equations given as Equations A and B. The following steps will solve the equation by substitution.
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION Solve: We'll use substitution. After moving 7*x to the right, we get: , or . Substitute that into another equation: and simplify: So, we know that x=0.333333333333333. Since , a=3.33333333333333. Answer: .
So a= and x=
So sequence is: , , , , , , , .
Check: The third term is and the sum is
Step 6. ANSWER: The sequence is , , , , , , , .
I hope the above steps and explanation were helpful.
For Step-By-Step videos on Introduction to Algebra,
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definitions
• synonyms
# SI unit
See more synonyms on Thesaurus.com
noun
1. See under International System of Units.
# International System of Units
noun
1. an internationally accepted coherent system of physical units, derived from the MKSA (meter-kilogram-second-ampere) system, using the meter, kilogram, second, ampere, kelvin, mole, and candela as the basic units (SI units) respectively of the fundamental quantities of length, mass, time, electric current, temperature, amount of substance, and luminous intensity. Abbreviation: SI
## Origin of .css-1fxfie5{font-size:22px;}@media (max-width:768px){.css-1fxfie5{font-size:18px;margin:0 10px 10px 0;word-break:break-all;word-wrap:break-word;-webkit-hyphens:auto;-moz-hyphens:auto;-ms-hyphens:auto;hyphens:auto;line-height:22px;}}International System of Units
translation of the earlier French name Système Internationale d'Unités
Dictionary.com Unabridged Based on the Random House Unabridged Dictionary, © Random House, Inc. 2018
British Dictionary definitions for si units
# SI unit
noun
1. any of the units adopted for international use under the Système International d'Unités, now employed for all scientific and most technical purposes. There are seven fundamental units: the metre, kilogram, second, ampere, kelvin, candela, and mole; and two supplementary units: the radian and the steradian. All other units are derived by multiplication or division of these units without the use of numerical factors
Collins English Dictionary - Complete & Unabridged 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
si units in Medicine
# International System of Units
n.
1. A complete, coherent system of units used for scientific work, in which the fundamental quantities are length, time, electric current, temperature, mass, luminous intensity, and amount of substance.
si units in Science
# SI unit
1. Any of the units of measure in the International System. See more at International System.
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# physics
posted by .
assume that the spool is a solid cylinder of radius 0.5m and a mass 5.0kg, and m=5.0kg. Find the angular speed of the spool after the mass m has fallen 3.0
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what algorithm for decision about simulation's movement escape from on fire by multi thread(many people).
Thank,
[I hope this is an accuracte rephrasing of the question]
I want to build an application to simulate how people respond during a fire. Imagine we have a building full of people and we want to simulate a fire alarm. We need to create a different thread for each person and have an algorithm figure out where the person should go to escape and simulate the movement in some way.
What is the best algorithm to use to simulate this?
Posted
Updated 3-Aug-10 7:53am
v2
Tom Deketelaere 3-Aug-10 5:40am
You'll have to explain a bit better what you want because this makes no sense to me.
Explain what you want / need your program to do, where you are stuck and what you'v done so far.
## Solution 1
You seem to mix up threads and simulation. With a simulation you try to create a controlled situation where you controls the steps by predefined rules. A simulation can run slower or faster than real-time without a problem. The time is relative to the simulation and therefore to the steps and are only noticed by the observer of the simulation. To keep it simple, the simulation runs in cycles and each entity within that simulation can do one step at a time. So all the people do a step and the "fire units" do a step. This all happens in a very controlled way and can be kept in track with time even when there get more or less entities. This means that a single cycle could be defined as a single second and if your simulation needs 10 seconds or 1 millisecond this only means that the observer has to wait longer or less to see the end results.
A simulation is a single game where every entity moves one step per cycle and makes it's decision on the previous state. So the entities do their main decision making based on the state it was just right before the cycle and only do for example collision checks to make sure nothing happens that cannot be possible, although people can run into each other in your case. A single cycle must be seen as a single moment in time and every entity makes a move at that same time. Of course the processing is done sequential and they move one after one, this again is only noticed by the observer. Look at it as if our life is divided into steps and even though I have the feeling it is all happening in a fluently motion, maybe our creator looks down and must wait one hour to see that you have slightly moved. It is all relative and has nothing to do with using threads to create a sense that everything happens at the same time, because that is just what a simulation simulates ;)
Start by defining the elements (entities) and what moves they can do. Then think out the rules (Like: If fire closer than three tiles away then move away from fire). Start out with very simple and basic rules and work from there. It's nice to first create a simple but visual engine to convert the simulation data to images on screen. With this you can easily work towards a more complex simulation. Step by step, one step at a time, keep that in mind.
Good luck!
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# Interactive video lesson plan for: Halves and quarters - and thirds also! Simple fraction lesson for 1st grade.
#### Activity overview:
In this easy lesson, we don't yet study the fraction notation, but just learn the concepts related to halves, thirds, and fourths (quarters). I show these by coloring in the corresponding parts.
We also compare two given fractions by coloring. It's basically pretty fun! Fractions don't need to be difficult, scary "monsters" at all!
Check out my other videos at http://www.mathmammoth.com/videos and my fraction workbook at http://www.mathmammoth.com/introduction_fractions.php
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5.4 Polynomial&Rational Inequalities
# 5.4 Polynomial&Rational Inequalities - in that...
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5.4: Polynomial & Rational Inequalities Steps for solving polynomial and rational inequalities: [1] Set the inequality equal to zero; rational inequalities should be written as a single quotient. [2] Find the numbers at which the expression f is equal to zero and, if rational, the numbers at which it is undefined; these are called critical points. [3] Use the numbers found in step 2 to separate the real number line into intervals. [4] Evaluate f for a number in each of the intervals: [a] If the value of f is positive, then f(x) > 0 for all numbers
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Unformatted text preview: in that interval; [b] If the value of f is negative, then f(x) < 0 for all numbers in that interval. Solve each of the following: 1. (x – 8)(x + 3) > 0 2. x 2- 3x ≤ 0 1. 2. Result Result 3. 8x 2 < 6 –13X 4. 2(2x 2 – 6x) > -9 3. 4. Result Result 5. 4x 2 + 49 < 28x 6. x 4 ≤ 8x 5. 6. Result Result 7. 2 2 15 1 x x x--+ 8. 18 9 x x + ≤ Zeros: Asymptote: 7. 8. Result Result 9. 6 2 2 6 x x-≤-Zeros: Asymptote: 9. Result 10. 2 5 1 1 1 x x x x + + < +-Zeros: Asymptotes: 10. Result...
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# John Errington's Data Conversion Website
## Introduction to Data conversion
Computers are digital systems that form part of our world which is predominantly analog in nature. There are some things that are clearly digital - lights on/off, stock levels or cash in a bank account. However examples of analog are all around us - your weight, the angle of your swivel chair, the growth of plants and flowers, the speed and direction of the wind, and the brightness of the light. When we use computers to measure or control things in our analog world we need to provide for conversion between the analog values and their digital representations inside the computer. This is the subject of this site.
### Analog and Digital parameters
#### Digital:
a digital parameter has a set of clearly defined values it can take and these can be associated with integer numbers in a strict 1: 1 correspondence. For example a light can be on or off, represented by off = 0, on = 1. The amount of money in your bank account can be represented as a positive or negative integer.
Generally, anything that can be represented in a data table - stock levels, timetables, accounts etc. are all digital values. They can be copied with complete precision.
#### Analog:
a parameter that can take any value within a range, and can only be measured or represented to limited precision, usually by a decimal number. For example your body temperature may be indicated as 34 degrees by a thermometer, but a more accurate measurement might give 33.86...
### Precision and Accuracy
Its important when thinking about data conversion to be aware of the distinction between precision and accuracy. A good example here are bathroom scales. My bathroom scales measure 0 to 100kg and report my weight as 74.1 kg. If I setep off and back on the weight is reported as 74.2 kg.
The precision is the smallest change in weight that can be distinguished. Here it would be stated as 1 part in 1000 or 0.1%. This should also take into account the repeatability of the measurement.
The accuracy is the degree to which the measured value (74.1kg) represents the actual value (my true weight). If my true weight is 73.20 kgs the scales are only accurate to about 1.0 kg, or 1%
Its easy to be fooled into thinking a measurement is very accurate if its very precise, but this is often not the case; and frequently measurements are quoted to much higher levels of accuracy than is justified.
Example: A rod of length 13cm is divided into three equal parts; what is the length of each part?
13/3 = 4.333 .. WRONG!
The precision of the length given was 13cm (implying +/- 1cm) not 13.0000cm
The correct answer - reflecting this degree of precision - is 4cm +/- 0.4 cm
### Design Choices
In converting between analog and digital domains we need to make design choices based on our understanding of the values to be converted, the range, accuracy, precision and rate at which this needs to take place, and an awareness of the budget for this task. The sections of this site aim to provide knowledge that will support and guide you in making these choices.
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# Calculate The Crusher Impact
## Impact Crushers Design And Calculations - GitHub Pages
To counter impact crusher productivity but also affects the overall operation efficiency of the equipment, in the calculation of the time is generally according to the rotor to a week how much of the material discharged. Productivity size is the most direct response to the equipment production efficiency, therefore all the operators have to do statistics and analysis of daily, promptly to the crusherImpact Crusher Working Principle,Normally, the rock breaks in halves, and in most tests only two and seldom more than three large pieces are observed after fracture. No size distribution information is used in calculating the Bond Impact Work Index from the formula: WI = 2 59 Average Impact in ft-lbs/inch/Specific Gravity…………………………………………(2)Bond Impact Crushing Work Index -Procedure and Table of,,Sizing a crusher can be done reliably calculated thanks to the Impact/Crushing Work Index and the testwork research done by Fred Chester Bond and his 1952 paper. According to Bond ’s Third Theory of Comminution , the work/energy input is proportional to the new crack tip length created during particle breakage and equivalent to the work represented by the product – the feed.
## DESIGN AND ANALYSIS OF IMPACT CRUSHERS
Chapter 2: Design and Calculation 2.1 Design of V-Belt drive 7 2.2 Design,Impact crushers: they involve the use of impact rather than pressure to crush materials. The material is contained within a cage, with openings on the bottom, end or side of the desired size to allow pulverized material to escape. This type of crusher is usually used with soft material such as coal, seeds or soft,Calculation Calculation Of Impact Force In Impact Crusher,Calculation Calculation Of Impact Force In Impact Crusher. As a leading global manufacturer of crushing equipment, milling equipment,dressing equipment,drying equipment and briquette equipment etc. we offer advanced, rational solutions for any size-reduction requirements, including quarry, aggregate, grinding production and complete plant plan.calculate the impact crusher impact - mayukhportfolio.co.in,Calculating the efficiency of impact crushers for bulk,- Springer. a method for calculating the efficiency for impact-crushing such materials as salt,,mental analysis of the kinetics in an impact crusher in order to determine its... Read more. DEM-Simulation of the Breakage Process in an Impact Crusher . Abstract: The process area of an impact crusher was modelled using the Discrete,
## Impact Crushers Design And Calculations - GitHub Pages
To counter impact crusher productivity but also affects the overall operation efficiency of the equipment, in the calculation of the time is generally according to the rotor to a week how much of the material discharged. Productivity size is the most direct response to the equipment production efficiency, therefore all the operators have to do statistics and analysis of daily, promptly to the crusher productivityBond Impact Crushing Work Index -Procedure and,Sizing a crusher can be done reliably calculated thanks to the Impact/Crushing Work Index and the testwork research done by Fred Chester Bond and his 1952 paper. According to Bond ’s Third Theory of Comminution , the work/energy input is proportional to the new crack tip length created during particle breakage and equivalent to the work represented by the product – the feed.Impact Crusher Working Principle,Normally, the rock breaks in halves, and in most tests only two and seldom more than three large pieces are observed after fracture. No size distribution information is used in calculating the Bond Impact Work Index from the formula: WI = 2 59 Average Impact in ft-lbs/inch/Specific Gravity…………………………………………(2)
## DESIGN AND ANALYSIS OF IMPACT CRUSHERS
Chapter 2: Design and Calculation 2.1 Design of V-Belt drive 7 2.2 Design,Impact crushers: they involve the use of impact rather than pressure to crush materials. The material is contained within a cage, with openings on the bottom, end or side of the desired size to allow pulverized material to escape. This type of crusher is usually used with soft material such as coal, seeds or soft,Calculation Calculation Of Impact Force In Impact Crusher,Calculation Calculation Of Impact Force In Impact Crusher. As a leading global manufacturer of crushing equipment, milling equipment,dressing equipment,drying equipment and briquette equipment etc. we offer advanced, rational solutions for any size-reduction requirements, including quarry, aggregate, grinding production and complete plant plan.calculate the impact crusher impact - mayukhportfolio.co.in,Calculating the efficiency of impact crushers for bulk,- Springer. a method for calculating the efficiency for impact-crushing such materials as salt,,mental analysis of the kinetics in an impact crusher in order to determine its... Read more. DEM-Simulation of the Breakage Process in an Impact Crusher . Abstract: The process area of an impact crusher was modelled using the Discrete,
## DESIGN AND ANALYSIS OF A HORIZONTAL SHAFT IMPACT CRUSHER
An impact crusher can be further classified as Horizontal impact crusher (HSI) and vertical shaft impact crusher (VSI) based on the type of arrangement of the impact rotor and shaft. Horizontal shaft impact crusher These break rock by impacting the rock with hammers/blow bars that are fixed upon the outer edge of a spinning rotor. Here the rotor shaft is aligned along the horizontal axis. The inputGyratory and Cone Crusher - ScienceDirect,01.01.2016· Once the feed sizes for different stages of crushing are determined, the sizes of the crushers can be estimated using the rule of thumb that the gape of crusher is usually 1.1 times the feed size. This rule of thumb for sizing the gape of primary jaw crushers is applicable to crushers up to 2 m in diameter. Once the gape is determined, the size of the primary crusher can be ascertained fromMODEL KR Bar Arrangement Impact Crushers,TRF Impact Crushers are ideal processing machines for primary and secondary crushing of raw materials for the steel and iron industry, hard coal and lignite, limestone, gypsum or minerals. These are also extensively used for the tertiary crus hing of limestone, dolomite, dunite for the Bedding and Blending Plant for sinter preparation and coal for Coking Plant. Its sturdy and proven design,
## How Do You Estimate Impact Force? | WIRED
Since we know the estimated impact force from the show (12,000 pounds = 53,379 Newtons), the impact time can be calculated. Let me start with a force diagram showing the forces acting on the piano,DESIGN AND ANALYSIS OF IMPACT CRUSHERS,Crushers are commonly classified by the degree to which they fragment not starting material with wares crushers not reducing it by much, intermediate cruiser fragmenting it much more significantly and grinders reducing it to a fine power. Impact crushers: they involve the use of impact rather than pressure to crush materials. The material is contained within a cage, with openings on the bottom,Crusher Size Reduction Ratio Calculation Method,Generally, every crusher machine is not the same, here are several common crusher size reduction ratio: The impact crusher size reduction ratio is 20 to 1. The vertical shaft impact crusher size reduction ratio is 4-8 to 1. The vertical roller mill size reduction ratio is 2-2.5 to 1. The hammer crusher size reduction ratio is 20 to 1.
## Construction, Working and Maintenance of Crushers for,
dynamic impact. When the material is crushed in an impact crusher, the freely falling material first breaks due to dynamic impact by a moving hammer and then the force of impact accelerates movement of the reduced particles toward breaker blocks and/or other hammers for further size reduction. Dynamic impact has specific advantages for the reduction of many materials and it isCrushing of coal and calculation of size reduction,Crushing efficiency • Generally the efficiency of crusher has been calculated from the Power point of view. • As per a data in US, 1 % of the energy generated is used up for Size reduction Processes • The literature in the area have mostly talked about size energy relationships but we will look into the performance of crusher from customer point of view 2/25/2015 6:58:05 AM 61SAG mills selection methods. III. The F. Bond index,,The crushing work index (CWi) is an indicator characterizing the energy amount (kW * h / t) needed to destroy single pieces of material with a relatively large size range (-75 + 50mm). The impact on such material generally is provided by using impact energy (crushing process), that is formed by collision of two oncoming pendulums. The difficulty in calculating the
## Bond Tests | SGS
It is also used to determine the required open-side settings (jaw crushers and gyratory crushers) or closed-side settings (cone crushers) for a given product size. P80 = 25400 x Oss x (0.04Wi + 0.40) P80 = 25400 x Css x 7Ecc x (0.02Wi + 0.70) / (7Ecc - 2Css)How Do You Estimate Impact Force? | WIRED,Since we know the estimated impact force from the show (12,000 pounds = 53,379 Newtons), the impact time can be calculated. Let me start with a force diagram showing the forces acting on the piano,Car Crash - Impact Force Calculator,02.12.2020· F = m * v² / (2 * d), where. F is the average impact force, m is the mass of an object, v is the initial speed of an object, d is the distance traveled during collision. What may at first surprise you, is that extending the distance moved during the collision reduces the average impact force.
## energy - How to calculate the Impact of a meteor
So, K E = m v 2 2. K E = 158.3 k g × ( 38445 m / s) 2 2. K E = 1.17 × 10 11 J. I have no idea how to calculate the size of the carter it creates. Added: Now for a calculated guess: Assuming that half the energy is used to heat the ground and propel the debris away, theCalculation of corrugated board flat crush resistance,,In corrugated boards and cardboards with honeycomb structure, flat crushing resistance is also determined, but it depends only on the properties of the middle layer.,
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# Two CR questions 1. The tobacco industry is still profitable
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Two CR questions 1. The tobacco industry is still profitable [#permalink]
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Two CR questions
1. The tobacco industry is still profitable and projections are that it will remain so. In the United States this year,
the total amount of tobacco sold by tobacco-farmers has increased, even though the number of adults who
smoke has decreased.
Each of the following, if true, could explain the simultaneous increase in tobacco sales and decrease in the
number of adults who smoke EXCEPT:
A. During this year, the number of women who have begun to smoke is greater than the number of men who
have quit smoking
B. The number of teen-age children who have begun to smoke this year is greater than the number of adults
who have quit smoking during the same period
C. During this year, the number of nonsmokers who have begun to use chewing tobacco or snuff is greater than
the number of people who have quit smoking
D. The people who have continued to smoke consume more tobacco per person than they did in the past
E. More of the cigarettes made in the United States this year were exported to other countries than was the
case last year.
2. Unlike the wholesale price of raw wool, the wholesale price of raw cotton has fallen considerably in the last
year. Thus, although the retail price of cotton clothing at retail clothing stores has not yet fallen, it will inevitably
37
fall.
Which of the following, if true, most seriously weakens the argument above?
(A) The cost of processing raw cotton for cloth has increased during the last year.
(B) The wholesale price of raw wool is typically higher than that of the same volume of raw cotton.
(C) The operating costs of the average retail clothing store have remained constant during the last year.
(D) Changes in retail prices always lag behind changes in wholesale prices.
(E) The cost of harvesting raw cotton has increased in the last year.
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Re: Tobacco and whole sale [#permalink]
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06 Aug 2009, 19:28
1) A
2) A
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Re: Tobacco and whole sale [#permalink]
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06 Aug 2009, 19:54
2 - A
1 - Using POE its A...but i cant get the logic behind it
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Re: Tobacco and whole sale [#permalink]
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06 Aug 2009, 20:20
Another A and A
In question 1, number of adults who smoke has decreased, and A doesn't help as it just tells us about no of females started smoking is greater than no of males who quit smoking.
What about Number of men who started smoking and how many women quit smoking?
We don't know these numbers so this answer choice is not able to explain the simultaneous increase in tobacco sales and decrease in the number of adults who smoke
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Re: Tobacco and whole sale [#permalink]
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06 Aug 2009, 20:32
Yes,
A A
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Re: Tobacco and whole sale [#permalink]
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07 Aug 2009, 01:20
in 2.. how is option b diff than A ?
getmba wrote:
Another A and A
In question 1, number of adults who smoke has decreased, and A doesn't help as it just tells us about no of females started smoking is greater than no of males who quit smoking.
What about Number of men who started smoking and how many women quit smoking?
We don't know these numbers so this answer choice is not able to explain the simultaneous increase in tobacco sales and decrease in the number of adults who smoke
_________________
If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 04:37
1. D
2.A
OA plz
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 05:04
A and A
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 07:00
Q2. For option B, there is a comparision of prices between wool and cotton. But the question looks for reason specific to cotton alone. We cannot conclude the reatil price of cotton based on the cost of raw wool, as there is no relation between the prices of both. So B can be eliminated right away.
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 11:11
option is A & A
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 21:12
C & A
c - > During this year, the number of nonsmokers who have begun to use chewing tobacco or snuff is greater than
the number of people who have quit smoking
People have quit smoking, therefore a decrease in the number of adults smoking
Nonsmokers (more than who have quit) have started using tobacco, therefore increase in tobacco sales.
OA?
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 21:25
After reading your ans... i m siding with C on the 1st one. Nice reasoning whatthehell
whatthehell wrote:
C & A
c - > During this year, the number of nonsmokers who have begun to use chewing tobacco or snuff is greater than
the number of people who have quit smoking
People have quit smoking, therefore a decrease in the number of adults smoking
Nonsmokers (more than who have quit) have started using tobacco, therefore increase in tobacco sales.
OA?
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Re: Tobacco and whole sale [#permalink]
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09 Aug 2009, 22:10
After reading your ans... i m siding with C on the 1st one. Nice reasoning whatthehell
whatthehell wrote:
C & A
c - > During this year, the number of nonsmokers who have begun to use chewing tobacco or snuff is greater than
the number of people who have quit smoking
People have quit smoking, therefore a decrease in the number of adults smoking
Nonsmokers (more than who have quit) have started using tobacco, therefore increase in tobacco sales.
OA?
Both of you two are missing the big capital 'EXCEPT' in the first question. Obviously, as explained by "whatthehell", 'C' explains the simultaneous decrease in adult smokers and increase in tobacco sales. So does the other options B,D and E. 'A' cannot explain these two as it talks only about men and women, not the adults as a whole.
Also, if "the number of women who have begun to smoke is greater than the number of men who have quit smoking" then the total no of adult smokers (assuming all women and men are adults) will increase, not decrease.
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Re: Tobacco and whole sale [#permalink]
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10 Aug 2009, 03:37
Agreed
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Re: Tobacco and whole sale [#permalink]
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10 Aug 2009, 04:48
Thanks for pointing out. Reminder that while we are constantly analyzing options and stems, we completely miss answering the actual question that is being asked. With limited time in GMAT this will certainly be a great challange.
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Re: Tobacco and whole sale [#permalink]
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10 Aug 2009, 18:15
A...does not say number of women who quit smoking and number of men started smoking...
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Re: Tobacco and whole sale [#permalink]
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10 Aug 2009, 18:18
and 2- A since we need to find a way to increase the cost of final product, so if cotton prices goes down, then one way to increase the final product price is to jack up the processing fee...
Re: Tobacco and whole sale [#permalink] 10 Aug 2009, 18:18
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# Two CR questions 1. The tobacco industry is still profitable
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While Newton stayed home, during the outbreak of the plague, he discovered (invented?) calculus, gravity, and optics laws. Being in lockdown myself, I decided to stand on the shoulders of the giant Newton and sketch the answer to a burning question, “Do you know anyone who became seriously ill with Covid-19?” I don’t know where you come from, but this question has often been asked by conspiracists here in Greece (always in a rhetorical manner).
For fun, I constructed dozens of toy models of social networks with different topologies (Barabási – Albert, Watts – Strogatz, Price, and random), with various parameterizations. The blue line in the following plot is the percentage of people in the network infected with Covid-19. The orange line represents the percentage of people who heard that some friend of theirs has become ill with Covid-19. Even with favorable assumptions about the transmission of the virus (and, therefore, its message), the vast majority of plots had a tail at their origin.
This reflects the fact that initially, the disease has a low prevalence in the general population and that most people have relatively few acquaintances. Some have an intermediate number and just a tiny fraction many. For people to have heard of some of their friends being ill during the first stage of the pandemic, either the disease should have had a very high prevalence from day 1, or the social network should have been exceptionally densely connected (in the limit case, everyone being friends with anyone). Just like the following complete graph:
However, real social netowrks are more likely to resemble the following topologies:
And here you can see the transmission of the virus in a sample network:
Epimyth (not to be taken too seriously): If you haven’t heard of any of your acquaintances being seriously ill with Covid-19 at the start of the pandemic, it’s because there weren’t many cases. If, as the days pass by, you still don’t hear of anyone, it’s because you don’t have any friends :P
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steps for learning to subtract poportion fraction
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Author Message
hp_calamom
Registered: 30.03.2005
From:
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IlbendF
Registered: 11.03.2004
From: Netherlands
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Registered: 05.07.2001
From: Right here, can't you see me?
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# [R] How to fill the aerea under a plot?
Martin Maechler maechler at stat.math.ethz.ch
Thu Jun 3 14:46:07 CEST 1999
```>>>>> "Paul" == Paul Murrell <pm254 at medschl.cam.ac.uk> writes:
Paul> hi
>> I plot a function, e.g.
>> plot(sin, -pi, 2*pi).
>>
>>
>> Then I would like to have the aerea
>> from the x-axis up to the graph
>> from x ranging from -3 to 0, shaded
>> red and the other part (x from 0 to 6) under
>> the graph e.g. green or sth. like that..
Paul> plot(sin, -pi, 2*pi) just draws a series of straight lines to
Paul> approximate the sine curve (but enough lines so that it looks
Paul> like a smooth curve). one way to do what you want is to draw the
Paul> lines yourself using polygon(), which can then be filled. the
Paul> code below gives an example for your case (there are two examples
Paul> because i'm not sure what you mean by "from the x-axis up to the
Paul> graph") ...
Paul> plot(c(-pi, 2*pi), c(-1,1), type="n", ylab="sin(x)", xlab="x")
Paul> x1 <- seq(-pi, 0, length=101)
Paul> x2 <- seq(0, 2*pi, length=101)
Paul> polygon(c(0, -pi, x1), c(0, 0, sin(x1)), col="red")
Paul> polygon(c(0, 2*pi, x2), c(0, 0, sin(x2)), col="green")
Paul> plot(c(-pi, 2*pi), c(-1,1), type="n", ylab="sin(x)", xlab="x")
Paul> x1 <- seq(-pi, 0, length=101)
Paul> x2 <- seq(0, 2*pi, length=101)
Paul> polygon(c(0, -pi, x1), c(-1, -1, sin(x1)), col="red")
Paul> polygon(c(2*pi, 0, x2), c(-1, -1, sin(x2)), col="green")
Paul> unfortunately, i don't think there's an easier way.
however, there *is* --- using the much underused `` type = "h" '' :
Either -- with "overplotting":
plot(sin, -pi, 2*pi, type="h", col="green", n=801)
plot(sin, 0, 2*pi, type="h", col="red" , n=401, add=TRUE)
Or (without "overplotting")
plot(sin, -pi, 0, type="h", col="green", n=401,
xlim = c(-pi, 2*pi), ylim = c(-1,1))
plot(sin, 0, 2*pi, type="h", col="red" , n=401, add=TRUE)
{Even xlab & ylab are ok!}
Note however, that this is less efficient from a device-driver point of
view. E.g. with postscript(), The file sizes are
7.1, 48.0 and 33.2 1000Bytes
for Paul's (first) example, my "overplotting" and "non-overplotting"
example, respectively.
Martin Maechler <maechler at stat.math.ethz.ch> http://stat.ethz.ch/~maechler/
Seminar fuer Statistik, ETH-Zentrum SOL G1; Sonneggstr.33
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-1-632-3408 fax: ...-1086 <><
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```
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# source:src/ASM/AssemblyProof.ma@929
Last change on this file since 929 was 929, checked in by mulligan, 8 years ago
File size: 68.3 KB
Line
1include "ASM/Assembly.ma".
2include "ASM/Interpret.ma".
3
4(* RUSSEL **)
5
6include "basics/jmeq.ma".
7
8notation > "hvbox(a break ≃ b)"
9 non associative with precedence 45
10for @{ 'jmeq ? \$a ? \$b }.
11
12notation < "hvbox(term 46 a break maction (≃) (≃\sub(t,u)) term 46 b)"
13 non associative with precedence 45
14for @{ 'jmeq \$t \$a \$u \$b }.
15
16interpretation "john major's equality" 'jmeq t x u y = (jmeq t x u y).
17
18lemma eq_to_jmeq:
19 ∀A: Type[0].
20 ∀x, y: A.
21 x = y → x ≃ y.
22 //
23qed.
24
25definition inject : ∀A.∀P:A → Prop.∀a.∀p:P a.Σx:A.P x ≝ λA,P,a,p. dp … a p.
26definition eject : ∀A.∀P: A → Prop.(Σx:A.P x) → A ≝ λA,P,c.match c with [ dp w p ⇒ w].
27
28coercion inject nocomposites: ∀A.∀P:A → Prop.∀a.∀p:P a.Σx:A.P x ≝ inject on a:? to Σx:?.?.
29coercion eject nocomposites: ∀A.∀P:A → Prop.∀c:Σx:A.P x.A ≝ eject on _c:Σx:?.? to ?.
30
31axiom VOID: Type[0].
32axiom assert_false: VOID.
33definition bigbang: ∀A:Type[0].False → VOID → A.
34 #A #abs cases abs
35qed.
36
37coercion bigbang nocomposites: ∀A:Type[0].False → ∀v:VOID.A ≝ bigbang on _v:VOID to ?.
38
39lemma sig2: ∀A.∀P:A → Prop. ∀p:Σx:A.P x. P (eject … p).
40 #A #P #p cases p #w #q @q
41qed.
42
43lemma jmeq_to_eq: ∀A:Type[0]. ∀x,y:A. x≃y → x=y.
44 #A #x #y #JMEQ @(jmeq_elim ? x … JMEQ) %
45qed.
46
47coercion jmeq_to_eq: ∀A:Type[0]. ∀x,y:A. ∀p:x≃y.x=y ≝ jmeq_to_eq on _p:?≃? to ?=?.
48
49(* END RUSSELL **)
50
51
52definition bit_elim_prop: ∀P: bool → Prop. Prop ≝
53 λP.
54 P true ∧ P false.
55
56let rec bitvector_elim_prop_internal
57 (n: nat) (P: BitVector n → Prop) (m: nat) on m: m ≤ n → BitVector (n - m) → Prop ≝
58 match m return λm. m ≤ n → BitVector (n - m) → Prop with
59 [ O ⇒ λprf1. λprefix. P ?
60 | S n' ⇒ λprf2. λprefix. bit_elim_prop (λbit. bitvector_elim_prop_internal n P n' ? ?)
61 ].
62 [ applyS prefix
63 | letin res ≝ (bit ::: prefix)
64 < (minus_S_S ? ?)
65 > (minus_Sn_m ? ?)
66 [ @ res
67 | @ prf2
68 ]
69 | /2/
70 ].
71qed.
72
73definition bitvector_elim_prop ≝
74 λn: nat.
75 λP: BitVector n → Prop.
76 bitvector_elim_prop_internal n P n ? ?.
77 [ @ (le_n ?)
78 | < (minus_n_n ?)
79 @ [[ ]]
80 ]
81qed.
82
83lemma eq_b_eq:
84 ∀b, c.
85 eq_b b c = true → b = c.
86 #b #c
87 cases b
88 cases c
89 normalize //
90qed.
91
92lemma BitVector_O: ∀v:BitVector 0. v ≃ VEmpty bool.
93 #v generalize in match (refl … 0) cases v in ⊢ (??%? → ?%%??) //
94 #n #hd #tl #abs @⊥ //
95qed.
96
97lemma BitVector_Sn: ∀n.∀v:BitVector (S n).
98 ∃hd.∃tl.v ≃ VCons bool n hd tl.
99 #n #v generalize in match (refl … (S n)) cases v in ⊢ (??%? → ??(λ_.??(λ_.?%%??)))
100 [ #abs @⊥ //
101 | #m #hd #tl #EQ <(injective_S … EQ) %[@hd] %[@tl] // ]
102qed.
103
104lemma eq_bv_eq:
105 ∀n, v, q.
106 eq_bv n v q = true → v = q.
107 #n #v #q generalize in match v
108 elim q
109 [ #v #h @BitVector_O
110 | #n #hd #tl #ih #v' #h
111 cases (BitVector_Sn ? v')
112 #hd' * #tl' #jmeq >jmeq in h;
113 #new_h
114 change in new_h with ((andb ? ?) = ?);
115 cases(conjunction_true … new_h)
117 whd in eq_heads:(??(??(%))?);
118 cases(eq_b_eq … eq_heads)
119 whd in eq_tails:(??(?????(%))?);
120 change in eq_tails with (eq_bv ??? = ?);
121 <(ih tl') //
122 ]
123qed.
124
125lemma bool_eq_internal_eq:
126 ∀b, c.
127 (λb. λc. (if b then c else (if c then false else true))) b c = true → b = c.
128 #b #c
129 cases b
130 [ normalize //
131 | normalize
132 cases c
133 [ normalize //
134 | normalize //
135 ]
136 ]
137qed.
138
139lemma eq_bv_refl:
140 ∀n,v. eq_bv n v v = true.
141 #n #v
142 elim v
143 [ //
144 | #n #hd #tl #ih
145 normalize
146 cases hd
147 [ normalize
148 @ ih
149 | normalize
150 @ ih
151 ]
152 ]
153qed.
154
155lemma eq_eq_bv:
156 ∀n, v, q.
157 v = q → eq_bv n v q = true.
158 #n #v
159 elim v
160 [ #q #h <h normalize %
161 | #n #hd #tl #ih #q #h >h //
162 ]
163qed.
164
165let rec foldl_strong_internal
166 (A: Type[0]) (P: list A → Type[0]) (l: list A)
167 (H: ∀prefix. ∀hd. ∀tl. l = prefix @ [hd] @ tl → P prefix → P (prefix @ [hd]))
168 (prefix: list A) (suffix: list A) (acc: P prefix) on suffix:
169 l = prefix @ suffix → P(prefix @ suffix) ≝
170 match suffix return λl'. l = prefix @ l' → P (prefix @ l') with
171 [ nil ⇒ λprf. ?
172 | cons hd tl ⇒ λprf. ?
173 ].
174 [ > (append_nil ?)
175 @ acc
176 | applyS (foldl_strong_internal A P l H (prefix @ [hd]) tl ? ?)
177 [ @ (H prefix hd tl prf acc)
178 | applyS prf
179 ]
180 ]
181qed.
182
183definition foldl_strong ≝
184 λA: Type[0].
185 λP: list A → Type[0].
186 λl: list A.
187 λH: ∀prefix. ∀hd. ∀tl. l = prefix @ [hd] @ tl → P prefix → P (prefix @ [hd]).
188 λacc: P [ ].
189 foldl_strong_internal A P l H [ ] l acc (refl …).
190
191definition bit_elim: ∀P: bool → bool. bool ≝
192 λP.
193 P true ∧ P false.
194
195let rec bitvector_elim_internal
196 (n: nat) (P: BitVector n → bool) (m: nat) on m: m ≤ n → BitVector (n - m) → bool ≝
197 match m return λm. m ≤ n → BitVector (n - m) → bool with
198 [ O ⇒ λprf1. λprefix. P ?
199 | S n' ⇒ λprf2. λprefix. bit_elim (λbit. bitvector_elim_internal n P n' ? ?)
200 ].
201 [ applyS prefix
202 | letin res ≝ (bit ::: prefix)
203 < (minus_S_S ? ?)
204 > (minus_Sn_m ? ?)
205 [ @ res
206 | @ prf2
207 ]
208 | /2/
209 ].
210qed.
211
212definition bitvector_elim ≝
213 λn: nat.
214 λP: BitVector n → bool.
215 bitvector_elim_internal n P n ? ?.
216 [ @ (le_n ?)
217 | < (minus_n_n ?)
218 @ [[ ]]
219 ]
220qed.
221
222axiom vector_associative_append:
223 ∀A: Type[0].
224 ∀n, m, o: nat.
225 ∀v: Vector A n.
226 ∀q: Vector A m.
227 ∀r: Vector A o.
228 ((v @@ q) @@ r)
229 ≃
230 (v @@ (q @@ r)).
231
232lemma vector_cons_append:
233 ∀A: Type[0].
234 ∀n: nat.
235 ∀e: A.
236 ∀v: Vector A n.
237 e ::: v = [[ e ]] @@ v.
238 # A # N # E # V
239 elim V
240 [ normalize %
241 | # NN # AA # VV # IH
242 normalize
243 %
244 ]
245qed.
246
247lemma super_rewrite2:
248 ∀A:Type[0].∀n,m.∀v1: Vector A n.∀v2: Vector A m.
249 ∀P: ∀m. Vector A m → Prop.
250 n=m → v1 ≃ v2 → P n v1 → P m v2.
251 #A #n #m #v1 #v2 #P #EQ <EQ in v2; #V #JMEQ >JMEQ //
252qed.
253
254lemma mem_middle_vector:
255 ∀A: Type[0].
256 ∀m, o: nat.
257 ∀eq: A → A → bool.
258 ∀reflex: ∀a. eq a a = true.
259 ∀p: Vector A m.
260 ∀a: A.
261 ∀r: Vector A o.
262 mem A eq ? (p@@(a:::r)) a = true.
263 # A # M # O # EQ # REFLEX # P # A
264 elim P
265 [ normalize
266 > (REFLEX A)
267 normalize
268 # H
269 %
270 | # NN # AA # PP # IH
271 normalize
272 cases (EQ A AA) //
273 @ IH
274 ]
275qed.
276
277lemma mem_monotonic_wrt_append:
278 ∀A: Type[0].
279 ∀m, o: nat.
280 ∀eq: A → A → bool.
281 ∀reflex: ∀a. eq a a = true.
282 ∀p: Vector A m.
283 ∀a: A.
284 ∀r: Vector A o.
285 mem A eq ? r a = true → mem A eq ? (p @@ r) a = true.
286 # A # M # O # EQ # REFLEX # P # A
287 elim P
288 [ #R #H @H
289 | #NN #AA # PP # IH #R #H
290 normalize
291 cases (EQ A AA)
292 [ normalize %
293 | @ IH @ H
294 ]
295 ]
296qed.
297
298lemma subvector_multiple_append:
299 ∀A: Type[0].
300 ∀o, n: nat.
301 ∀eq: A → A → bool.
302 ∀refl: ∀a. eq a a = true.
303 ∀h: Vector A o.
304 ∀v: Vector A n.
305 ∀m: nat.
306 ∀q: Vector A m.
307 bool_to_Prop (subvector_with A ? ? eq v (h @@ q @@ v)).
308 # A # O # N # EQ # REFLEX # H # V
309 elim V
310 [ normalize
311 # M # V %
312 | # NN # AA # VV # IH # MM # QQ
313 change with (bool_to_Prop (andb ??))
314 cut ((mem A EQ (O + (MM + S NN)) (H@@QQ@@AA:::VV) AA) = true)
315 [
316 | # HH > HH
317 > (vector_cons_append ? ? AA VV)
318 change with (bool_to_Prop (subvector_with ??????))
319 @(super_rewrite2 A ((MM + 1)+ NN) (MM+S NN) ??
320 (λSS.λVS.bool_to_Prop (subvector_with ?? (O+SS) ?? (H@@VS)))
321 ?
322 (vector_associative_append A ? ? ? QQ [[AA]] VV))
323 [ >associative_plus //
324 | @IH ]
325 ]
326 @(mem_monotonic_wrt_append)
327 [ @ REFLEX
328 | @(mem_monotonic_wrt_append)
329 [ @ REFLEX
330 | normalize
331 > REFLEX
332 normalize
333 %
334 ]
335 ]
336qed.
337
338lemma vector_cons_empty:
339 ∀A: Type[0].
340 ∀n: nat.
341 ∀v: Vector A n.
342 [[ ]] @@ v = v.
343 # A # N # V
344 elim V
345 [ normalize %
346 | # NN # HH # VV #H %
347 ]
348qed.
349
350corollary subvector_hd_tl:
351 ∀A: Type[0].
352 ∀o: nat.
353 ∀eq: A → A → bool.
354 ∀refl: ∀a. eq a a = true.
355 ∀h: A.
356 ∀v: Vector A o.
357 bool_to_Prop (subvector_with A ? ? eq v (h ::: v)).
358 # A # O # EQ # REFLEX # H # V
359 > (vector_cons_append A ? H V)
360 < (vector_cons_empty A ? ([[H]] @@ V))
361 @ (subvector_multiple_append A ? ? EQ REFLEX [[]] V ? [[ H ]])
362qed.
363
364lemma eq_a_reflexive:
365 ∀a. eq_a a a = true.
366 # A
367 cases A
368 %
369qed.
370
371lemma is_in_monotonic_wrt_append:
372 ∀m, n: nat.
373 ∀p: Vector addressing_mode_tag m.
374 ∀q: Vector addressing_mode_tag n.
376 bool_to_Prop (is_in ? p to_search) → bool_to_Prop (is_in ? (q @@ p) to_search).
377 # M # N # P # Q # TO_SEARCH
378 # H
379 elim Q
380 [ normalize
381 @ H
382 | # NN # PP # QQ # IH
383 normalize
384 cases (is_a PP TO_SEARCH)
385 [ normalize
386 %
387 | normalize
388 normalize in IH
389 @ IH
390 ]
391 ]
392qed.
393
394corollary is_in_hd_tl:
397 ∀n: nat.
398 ∀v: Vector addressing_mode_tag n.
399 bool_to_Prop (is_in ? v to_search) → bool_to_Prop (is_in ? (hd:::v) to_search).
400 # TO_SEARCH # HD # N # V
401 elim V
402 [ # H
403 normalize in H;
404 cases H
405 | # NN # HHD # VV # IH # HH
406 > vector_cons_append
407 > (vector_cons_append ? ? HHD VV)
408 @ (is_in_monotonic_wrt_append ? 1 ([[HHD]]@@VV) [[HD]] TO_SEARCH)
409 @ HH
410 ]
411qed.
412
414 (n: nat) (l: Vector addressing_mode_tag (S n)) on l: (l → bool) → bool ≝
415 match l return λx.match x with [O ⇒ λl: Vector … O. bool | S x' ⇒ λl: Vector addressing_mode_tag (S x').
416 (l → bool) → bool ] with
417 [ VEmpty ⇒ true
418 | VCons len hd tl ⇒ λP.
419 let process_hd ≝
420 match hd return λhd. ∀P: hd:::tl → bool. bool with
421 [ direct ⇒ λP.bitvector_elim 8 (λx. P (DIRECT x))
422 | indirect ⇒ λP.bit_elim (λx. P (INDIRECT x))
423 | ext_indirect ⇒ λP.bit_elim (λx. P (EXT_INDIRECT x))
424 | registr ⇒ λP.bitvector_elim 3 (λx. P (REGISTER x))
425 | acc_a ⇒ λP.P ACC_A
426 | acc_b ⇒ λP.P ACC_B
427 | dptr ⇒ λP.P DPTR
428 | data ⇒ λP.bitvector_elim 8 (λx. P (DATA x))
429 | data16 ⇒ λP.bitvector_elim 16 (λx. P (DATA16 x))
430 | acc_dptr ⇒ λP.P ACC_DPTR
431 | acc_pc ⇒ λP.P ACC_PC
432 | ext_indirect_dptr ⇒ λP.P EXT_INDIRECT_DPTR
433 | indirect_dptr ⇒ λP.P INDIRECT_DPTR
434 | carry ⇒ λP.P CARRY
435 | bit_addr ⇒ λP.bitvector_elim 8 (λx. P (BIT_ADDR x))
436 | n_bit_addr ⇒ λP.bitvector_elim 8 (λx. P (N_BIT_ADDR x))
437 | relative ⇒ λP.bitvector_elim 8 (λx. P (RELATIVE x))
438 | addr11 ⇒ λP.bitvector_elim 11 (λx. P (ADDR11 x))
439 | addr16 ⇒ λP.bitvector_elim 16 (λx. P (ADDR16 x))
440 ]
441 in
442 andb (process_hd P)
443 (match len return λx. x = len → bool with
444 [ O ⇒ λprf. true
445 | S y ⇒ λprf. list_addressing_mode_tags_elim y ? P ] (refl ? len))
446 ].
447 try %
448 [ 2: cases (sym_eq ??? prf); @tl
449 | generalize in match H; generalize in match tl; cases prf;
450 (* cases prf in tl H; : ??? WAS WORKING BEFORE *)
451 #tl
452 normalize in ⊢ (∀_: %. ?)
453 # H
454 whd
455 normalize in ⊢ (match % with [ _ ⇒ ? | _ ⇒ ?])
456 cases (is_a hd (subaddressing_modeel y tl H)) whd // ]
457qed.
458
459definition product_elim ≝
460 λm, n: nat.
461 λv: Vector addressing_mode_tag (S m).
462 λq: Vector addressing_mode_tag (S n).
463 λP: (v × q) → bool.
464 list_addressing_mode_tags_elim ? v (λx. list_addressing_mode_tags_elim ? q (λy. P 〈x, y〉)).
465
466definition union_elim ≝
467 λA, B: Type[0].
468 λelimA: (A → bool) → bool.
469 λelimB: (B → bool) → bool.
470 λelimU: A ⊎ B → bool.
471 elimA (λa. elimB (λb. elimU (inl ? ? a) ∧ elimU (inr ? ? b))).
472
473(*
474definition preinstruction_elim: ∀P: preinstruction [[ relative ]] → bool. bool ≝
475 λP.
476 list_addressing_mode_tags_elim ? [[ registr ; direct ; indirect ; data ]] (λaddr. P (ADD ? ACC_A addr)) ∧
477 list_addressing_mode_tags_elim ? [[ registr ; direct ; indirect ; data ]] (λaddr. P (ADDC ? ACC_A addr)) ∧
478 list_addressing_mode_tags_elim ? [[ registr ; direct ; indirect ; data ]] (λaddr. P (SUBB ? ACC_A addr)) ∧
479 list_addressing_mode_tags_elim ? [[ acc_a ; registr ; direct ; indirect ; dptr ]] (λaddr. P (INC ? addr)) ∧
480 list_addressing_mode_tags_elim ? [[ acc_a ; registr ; direct ; indirect ]] (λaddr. P (DEC ? addr)) ∧
481 list_addressing_mode_tags_elim ? [[acc_b]] (λaddr. P (MUL ? ACC_A addr)) ∧
482 list_addressing_mode_tags_elim ? [[acc_b]] (λaddr. P (DIV ? ACC_A addr)) ∧
483 list_addressing_mode_tags_elim ? [[ registr ; direct ]] (λaddr. bitvector_elim 8 (λr. P (DJNZ ? addr (RELATIVE r)))) ∧
484 list_addressing_mode_tags_elim ? [[ acc_a ; carry ; bit_addr ]] (λaddr. P (CLR ? addr)) ∧
485 list_addressing_mode_tags_elim ? [[ acc_a ; carry ; bit_addr ]] (λaddr. P (CPL ? addr)) ∧
486 P (DA ? ACC_A) ∧
487 bitvector_elim 8 (λr. P (JC ? (RELATIVE r))) ∧
488 bitvector_elim 8 (λr. P (JNC ? (RELATIVE r))) ∧
489 bitvector_elim 8 (λr. P (JZ ? (RELATIVE r))) ∧
490 bitvector_elim 8 (λr. P (JNZ ? (RELATIVE r))) ∧
491 bitvector_elim 8 (λr. (bitvector_elim 8 (λb: BitVector 8. P (JB ? (BIT_ADDR b) (RELATIVE r))))) ∧
492 bitvector_elim 8 (λr. (bitvector_elim 8 (λb: BitVector 8. P (JNB ? (BIT_ADDR b) (RELATIVE r))))) ∧
493 bitvector_elim 8 (λr. (bitvector_elim 8 (λb: BitVector 8. P (JBC ? (BIT_ADDR b) (RELATIVE r))))) ∧
494 list_addressing_mode_tags_elim ? [[ registr; direct ]] (λaddr. bitvector_elim 8 (λr. P (DJNZ ? addr (RELATIVE r)))) ∧
495 P (RL ? ACC_A) ∧
496 P (RLC ? ACC_A) ∧
497 P (RR ? ACC_A) ∧
498 P (RRC ? ACC_A) ∧
499 P (SWAP ? ACC_A) ∧
500 P (RET ?) ∧
501 P (RETI ?) ∧
502 P (NOP ?) ∧
503 bit_elim (λb. P (XCHD ? ACC_A (INDIRECT b))) ∧
505 bitvector_elim 8 (λaddr. P (PUSH ? (DIRECT addr))) ∧
506 bitvector_elim 8 (λaddr. P (POP ? (DIRECT addr))) ∧
507 union_elim ? ? (product_elim ? ? [[ acc_a ]] [[ direct; data ]])
508 (product_elim ? ? [[ registr; indirect ]] [[ data ]])
509 (λd. bitvector_elim 8 (λb. P (CJNE ? d (RELATIVE b)))) ∧
510 list_addressing_mode_tags_elim ? [[ registr; direct; indirect ]] (λaddr. P (XCH ? ACC_A addr)) ∧
511 union_elim ? ? (product_elim ? ? [[acc_a]] [[ data ; registr ; direct ; indirect ]])
512 (product_elim ? ? [[direct]] [[ acc_a ; data ]])
513 (λd. P (XRL ? d)) ∧
514 union_elim ? ? (union_elim ? ? (product_elim ? ? [[acc_a]] [[ registr ; direct ; indirect ; data ]])
515 (product_elim ? ? [[direct]] [[ acc_a ; data ]]))
516 (product_elim ? ? [[carry]] [[ bit_addr ; n_bit_addr]])
517 (λd. P (ANL ? d)) ∧
518 union_elim ? ? (union_elim ? ? (product_elim ? ? [[acc_a]] [[ registr ; data ; direct ; indirect ]])
519 (product_elim ? ? [[direct]] [[ acc_a ; data ]]))
520 (product_elim ? ? [[carry]] [[ bit_addr ; n_bit_addr]])
521 (λd. P (ORL ? d)) ∧
522 union_elim ? ? (product_elim ? ? [[acc_a]] [[ ext_indirect ; ext_indirect_dptr ]])
523 (product_elim ? ? [[ ext_indirect ; ext_indirect_dptr ]] [[acc_a]])
524 (λd. P (MOVX ? d)) ∧
525 union_elim ? ? (
526 union_elim ? ? (
527 union_elim ? ? (
528 union_elim ? ? (
529 union_elim ? ? (product_elim ? ? [[acc_a]] [[ registr ; direct ; indirect ; data ]])
530 (product_elim ? ? [[ registr ; indirect ]] [[ acc_a ; direct ; data ]]))
531 (product_elim ? ? [[direct]] [[ acc_a ; registr ; direct ; indirect ; data ]]))
532 (product_elim ? ? [[dptr]] [[data16]]))
533 (product_elim ? ? [[carry]] [[bit_addr]]))
534 (product_elim ? ? [[bit_addr]] [[carry]])
535 (λd. P (MOV ? d)).
536 %
537qed.
538
539definition instruction_elim: ∀P: instruction → bool. bool ≝
540 λP. (*
541 bitvector_elim 11 (λx. P (ACALL (ADDR11 x))) ∧
542 bitvector_elim 16 (λx. P (LCALL (ADDR16 x))) ∧
543 bitvector_elim 11 (λx. P (AJMP (ADDR11 x))) ∧
544 bitvector_elim 16 (λx. P (LJMP (ADDR16 x))) ∧ *)
545 bitvector_elim 8 (λx. P (SJMP (RELATIVE x))). (* ∧
546 P (JMP INDIRECT_DPTR) ∧
547 list_addressing_mode_tags_elim ? [[ acc_dptr; acc_pc ]] (λa. P (MOVC ACC_A a)) ∧
548 preinstruction_elim (λp. P (RealInstruction p)). *)
549 %
550qed.
551
552
553axiom instruction_elim_complete:
554 ∀P. instruction_elim P = true → ∀i. P i = true.
555*)
556(*definition eq_instruction ≝
557 λi, j: instruction.
558 true.*)
559axiom eq_instruction: instruction → instruction → bool.
560axiom eq_instruction_refl: ∀i. eq_instruction i i = true.
561
562let rec vect_member
563 (A: Type[0]) (n: nat) (eq: A → A → bool)
564 (v: Vector A n) (a: A) on v: bool ≝
565 match v with
566 [ VEmpty ⇒ false
567 | VCons len hd tl ⇒
568 eq hd a ∨ (vect_member A ? eq tl a)
569 ].
570
572 (n: nat)
573 (l: Vector addressing_mode_tag (S n))
574 on l:
575 ∀P: l → Prop.
576 ∀direct_a. ∀indirect_a. ∀ext_indirect_a. ∀register_a. ∀acc_a_a.
577 ∀acc_b_a. ∀dptr_a. ∀data_a. ∀data16_a. ∀acc_dptr_a. ∀acc_pc_a.
578 ∀ext_indirect_dptr_a. ∀indirect_dptr_a. ∀carry_a. ∀bit_addr_a.
580 ∀x: l. P x ≝
581 match l return
582 λy.
583 match y with
584 [ O ⇒ λm: Vector addressing_mode_tag O. ∀prf: 0 = S n. True
585 | S y' ⇒ λl: Vector addressing_mode_tag (S y'). ∀prf: S y' = S n.∀P:l → Prop.
586 ∀direct_a: if vect_member … eq_a l direct then ∀x. P (DIRECT x) else True.
587 ∀indirect_a: if vect_member … eq_a l indirect then ∀x. P (INDIRECT x) else True.
588 ∀ext_indirect_a: if vect_member … eq_a l ext_indirect then ∀x. P (EXT_INDIRECT x) else True.
589 ∀register_a: if vect_member … eq_a l registr then ∀x. P (REGISTER x) else True.
590 ∀acc_a_a: if vect_member … eq_a l acc_a then P (ACC_A) else True.
591 ∀acc_b_a: if vect_member … eq_a l acc_b then P (ACC_B) else True.
592 ∀dptr_a: if vect_member … eq_a l dptr then P DPTR else True.
593 ∀data_a: if vect_member … eq_a l data then ∀x. P (DATA x) else True.
594 ∀data16_a: if vect_member … eq_a l data16 then ∀x. P (DATA16 x) else True.
595 ∀acc_dptr_a: if vect_member … eq_a l acc_dptr then P ACC_DPTR else True.
596 ∀acc_pc_a: if vect_member … eq_a l acc_pc then P ACC_PC else True.
597 ∀ext_indirect_dptr_a: if vect_member … eq_a l ext_indirect_dptr then P EXT_INDIRECT_DPTR else True.
598 ∀indirect_dptr_a: if vect_member … eq_a l indirect_dptr then P INDIRECT_DPTR else True.
599 ∀carry_a: if vect_member … eq_a l carry then P CARRY else True.
600 ∀bit_addr_a: if vect_member … eq_a l bit_addr then ∀x. P (BIT_ADDR x) else True.
601 ∀n_bit_addr_a: if vect_member … eq_a l n_bit_addr then ∀x. P (N_BIT_ADDR x) else True.
602 ∀relative_a: if vect_member … eq_a l relative then ∀x. P (RELATIVE x) else True.
603 ∀addr11_a: if vect_member … eq_a l addr11 then ∀x. P (ADDR11 x) else True.
604 ∀addr_16_a: if vect_member … eq_a l addr16 then ∀x. P (ADDR16 x) else True.
605 ∀x:l. P x
606 ] with
607 [ VEmpty ⇒ λAbsurd. ⊥
608 | VCons len hd tl ⇒ λProof. ?
609 ] (refl ? (S n)). cases daemon. qed. (*
610 [ destruct(Absurd)
611 | # A1 # A2 # A3 # A4 # A5 # A6 # A7
612 # A8 # A9 # A10 # A11 # A12 # A13 # A14
613 # A15 # A16 # A17 # A18 # A19 # X
614 cases X
615 # SUB cases daemon ] qed.
616 cases SUB
617 [ # BYTE
618 normalize
619 ].
620
621
622(* let prepare_hd ≝
623 match hd with
624 [ direct ⇒ λdirect_prf. ?
625 | indirect ⇒ λindirect_prf. ?
626 | ext_indirect ⇒ λext_indirect_prf. ?
627 | registr ⇒ λregistr_prf. ?
628 | acc_a ⇒ λacc_a_prf. ?
629 | acc_b ⇒ λacc_b_prf. ?
630 | dptr ⇒ λdptr_prf. ?
631 | data ⇒ λdata_prf. ?
632 | data16 ⇒ λdata16_prf. ?
633 | acc_dptr ⇒ λacc_dptr_prf. ?
634 | acc_pc ⇒ λacc_pc_prf. ?
635 | ext_indirect_dptr ⇒ λext_indirect_prf. ?
636 | indirect_dptr ⇒ λindirect_prf. ?
637 | carry ⇒ λcarry_prf. ?
640 | relative ⇒ λrelative_prf. ?
643 ]
644 in ? *)
645 ].
646 [ 1: destruct(absd)
647 | 2: # A1 # A2 # A3 # A4 # A5 # A6
648 # A7 # A8 # A9 # A10 # A11 # A12
649 # A13 # A14 # A15 # A16 # A17 # A18
650 # A19 *
651 ].
652
653
654 match l return λx.match x with [O ⇒ λl: Vector … O. bool | S x' ⇒ λl: Vector addressing_mode_tag (S x').
655 (l → bool) → bool ] with
656 [ VEmpty ⇒ true
657 | VCons len hd tl ⇒ λP.
658 let process_hd ≝
659 match hd return λhd. ∀P: hd:::tl → bool. bool with
660 [ direct ⇒ λP.bitvector_elim 8 (λx. P (DIRECT x))
661 | indirect ⇒ λP.bit_elim (λx. P (INDIRECT x))
662 | ext_indirect ⇒ λP.bit_elim (λx. P (EXT_INDIRECT x))
663 | registr ⇒ λP.bitvector_elim 3 (λx. P (REGISTER x))
664 | acc_a ⇒ λP.P ACC_A
665 | acc_b ⇒ λP.P ACC_B
666 | dptr ⇒ λP.P DPTR
667 | data ⇒ λP.bitvector_elim 8 (λx. P (DATA x))
668 | data16 ⇒ λP.bitvector_elim 16 (λx. P (DATA16 x))
669 | acc_dptr ⇒ λP.P ACC_DPTR
670 | acc_pc ⇒ λP.P ACC_PC
671 | ext_indirect_dptr ⇒ λP.P EXT_INDIRECT_DPTR
672 | indirect_dptr ⇒ λP.P INDIRECT_DPTR
673 | carry ⇒ λP.P CARRY
674 | bit_addr ⇒ λP.bitvector_elim 8 (λx. P (BIT_ADDR x))
675 | n_bit_addr ⇒ λP.bitvector_elim 8 (λx. P (N_BIT_ADDR x))
676 | relative ⇒ λP.bitvector_elim 8 (λx. P (RELATIVE x))
677 | addr11 ⇒ λP.bitvector_elim 11 (λx. P (ADDR11 x))
678 | addr16 ⇒ λP.bitvector_elim 16 (λx. P (ADDR16 x))
679 ]
680 in
681 andb (process_hd P)
682 (match len return λx. x = len → bool with
683 [ O ⇒ λprf. true
684 | S y ⇒ λprf. list_addressing_mode_tags_elim y ? P ] (refl ? len))
685 ].
686 try %
687 [ 2: cases (sym_eq ??? prf); @tl
688 | generalize in match H; generalize in match tl; cases prf;
689 (* cases prf in tl H; : ??? WAS WORKING BEFORE *)
690 #tl
691 normalize in ⊢ (∀_: %. ?)
692 # H
693 whd
694 normalize in ⊢ (match % with [ _ ⇒ ? | _ ⇒ ?])
695 cases (is_a hd (subaddressing_modeel y tl H)) whd // ]
696qed.
697*)
698(*
699lemma test:
700 let i ≝ SJMP (RELATIVE (bitvector_of_nat 8 255)) in
701 (let assembled ≝ assembly1 i in
702 let code_memory ≝ load_code_memory assembled in
703 let fetched ≝ fetch code_memory ? in
704 let 〈instr_pc, ticks〉 ≝ fetched in
705 eq_instruction (\fst instr_pc)) i = true.
706 [2: @ zero
707 | normalize
708 ]*)
709
710lemma BitVectorTrie_O:
711 ∀A:Type[0].∀v:BitVectorTrie A 0.(∃w. v ≃ Leaf A w) ∨ v ≃ Stub A 0.
712 #A #v generalize in match (refl … O) cases v in ⊢ (??%? → (?(??(λ_.?%%??)))(?%%??))
713 [ #w #_ %1 %[@w] %
714 | #n #l #r #abs @⊥ //
715 | #n #EQ %2 >EQ %]
716qed.
717
718lemma BitVectorTrie_Sn:
719 ∀A:Type[0].∀n.∀v:BitVectorTrie A (S n).(∃l,r. v ≃ Node A n l r) ∨ v ≃ Stub A (S n).
720 #A #n #v generalize in match (refl … (S n)) cases v in ⊢ (??%? → (?(??(λ_.??(λ_.?%%??))))%)
721 [ #m #abs @⊥ //
722 | #m #l #r #EQ %1 <(injective_S … EQ) %[@l] %[@r] //
723 | #m #EQ %2 // ]
724qed.
725
726lemma lookup_prepare_trie_for_insertion_hit:
727 ∀A:Type[0].∀a,v:A.∀n.∀b:BitVector n.
728 lookup … b (prepare_trie_for_insertion … b v) a = v.
729 #A #a #v #n #b elim b // #m #hd #tl #IH cases hd normalize //
730qed.
731
732lemma lookup_insert_hit:
733 ∀A:Type[0].∀a,v:A.∀n.∀b:BitVector n.∀t:BitVectorTrie A n.
734 lookup … b (insert … b v t) a = v.
735 #A #a #v #n #b elim b -b -n //
736 #n #hd #tl #IH #t cases(BitVectorTrie_Sn … t)
737 [ * #l * #r #JMEQ >JMEQ cases hd normalize //
738 | #JMEQ >JMEQ cases hd normalize @lookup_prepare_trie_for_insertion_hit ]
739qed.
740
741coercion bool_to_Prop: ∀b:bool. Prop ≝ bool_to_Prop on _b:bool to Type[0].
742
743lemma lookup_prepare_trie_for_insertion_miss:
744 ∀A:Type[0].∀a,v:A.∀n.∀c,b:BitVector n.
745 (notb (eq_bv ? b c)) → lookup … b (prepare_trie_for_insertion … c v) a = a.
746 #A #a #v #n #c elim c
747 [ #b >(BitVector_O … b) normalize #abs @⊥ //
748 | #m #hd #tl #IH #b cases(BitVector_Sn … b) #hd' * #tl' #JMEQ >JMEQ
749 cases hd cases hd' normalize
750 [2,3: #_ cases tl' //
751 |*: change with (bool_to_Prop (notb (eq_bv ???)) → ?) /2/ ]]
752qed.
753
754lemma lookup_insert_miss:
755 ∀A:Type[0].∀a,v:A.∀n.∀c,b:BitVector n.∀t:BitVectorTrie A n.
756 (notb (eq_bv ? b c)) → lookup … b (insert … c v t) a = lookup … b t a.
757 #A #a #v #n #c elim c -c -n
758 [ #b #t #DIFF @⊥ whd in DIFF; >(BitVector_O … b) in DIFF //
759 | #n #hd #tl #IH #b cases(BitVector_Sn … b) #hd' * #tl' #JMEQ >JMEQ
760 #t cases(BitVectorTrie_Sn … t)
761 [ * #l * #r #JMEQ >JMEQ cases hd cases hd' #H normalize in H;
762 [1,4: change in H with (bool_to_Prop (notb (eq_bv ???))) ] normalize // @IH //
763 | #JMEQ >JMEQ cases hd cases hd' #H normalize in H;
764 [1,4: change in H with (bool_to_Prop (notb (eq_bv ???))) ] normalize
765 [3,4: cases tl' // | *: @lookup_prepare_trie_for_insertion_miss //]]]
766qed.
767
769 fold_left_i_aux … (
770 λi, mem, v.
771 insert … (bitvector_of_nat … i) v mem) (Stub Byte 16).
772
773axiom split_elim:
774 ∀A,l,m,v.∀P: (Vector A l) × (Vector A m) → Prop.
775 (∀vl,vm. v = vl@@vm → P 〈vl,vm〉) → P (split A l m v).
776
778 ∀pc.
779 \snd (half_add 16 (bitvector_of_nat … pc) (bitvector_of_nat … 1)) = bitvector_of_nat … (S pc).
780
781(*
782axiom not_eqvb_S:
783 ∀pc.
784 (¬eq_bv 16 (bitvector_of_nat 16 pc) (bitvector_of_nat 16 (S pc))).
785
786axiom not_eqvb_SS:
787 ∀pc.
788 (¬eq_bv 16 (bitvector_of_nat 16 pc) (bitvector_of_nat 16 (S (S pc)))).
789
790axiom bitvector_elim_complete:
791 ∀n,P. bitvector_elim n P = true → ∀bv. P bv.
792
793lemma bitvector_elim_complete':
794 ∀n,P. bitvector_elim n P = true → ∀bv. P bv = true.
795 #n #P #H generalize in match (bitvector_elim_complete … H) #K #bv
796 generalize in match (K bv) normalize cases (P bv) normalize // #abs @⊥ //
797qed.
798*)
799
800
801
802
803(*
804lemma andb_elim':
805 ∀b1,b2. (b1 = true) → (b2 = true) → (b1 ∧ b2) = true.
806 #b1 #b2 #H1 #H2 @andb_elim cases b1 in H1; normalize //
807qed.
808*)
809
810let rec encoding_check (code_memory: BitVectorTrie Byte 16) (pc: Word) (final_pc: Word)
811 (encoding: list Byte) on encoding: Prop ≝
812 match encoding with
813 [ nil ⇒ final_pc = pc
814 | cons hd tl ⇒
815 let 〈new_pc, byte〉 ≝ next code_memory pc in
816 hd = byte ∧ encoding_check code_memory new_pc final_pc tl
817 ].
818
819lemma encoding_check_append: ∀code_memory,final_pc,l1,pc,l2.
820 encoding_check code_memory (bitvector_of_nat … pc) (bitvector_of_nat … final_pc) (l1@l2) →
821 let intermediate_pc ≝ pc + length … l1 in
822 encoding_check code_memory (bitvector_of_nat … pc) (bitvector_of_nat … intermediate_pc) l1 ∧
823 encoding_check code_memory (bitvector_of_nat … intermediate_pc) (bitvector_of_nat … final_pc) l2.
824 #code_memory #final_pc #l1 elim l1
825 [ #pc #l2 whd in ⊢ (????% → ?) #H <plus_n_O whd whd in ⊢ (?%?) /2/
826 | #hd #tl #IH #pc #l2 * #H1 #H2 >half_add_SO in H2; #H2 cases (IH … H2) <plus_n_Sm
827 #K1 #K2 % [2:@K2] whd % // >half_add_SO @K1 ]
828qed.
829
830axiom bitvector_3_elim_prop:
831 ∀P: BitVector 3 → Prop.
832 P [[false;false;false]] → P [[false;false;true]] → P [[false;true;false]] →
833 P [[false;true;true]] → P [[true;false;false]] → P [[true;false;true]] →
834 P [[true;true;false]] → P [[true;true;true]] → ∀v. P v.
835
836axiom fetch_assembly:
837 ∀pc,i,code_memory,assembled.
838 assembled = assembly1 i →
839 let len ≝ length … assembled in
840 encoding_check code_memory (bitvector_of_nat … pc) (bitvector_of_nat … (pc + len)) assembled →
841 let fetched ≝ fetch code_memory (bitvector_of_nat … pc) in
842 let 〈instr_pc, ticks〉 ≝ fetched in
843 let 〈instr,pc'〉 ≝ instr_pc in
844 (eq_instruction instr i ∧ eq_bv … pc' (bitvector_of_nat … (pc + len))) = true.
845(* #pc #i #code_memory #assembled cases i [8: *]
846 [16,20,29: * * |18,19: * * [1,2,4,5: *] |28: * * [1,2: * [1,2: * [1,2: * [1,2: *]]]]]
847 [47,48,49:
848 |*: #arg @(list_addressing_mode_tags_elim_prop … arg) whd try % -arg
849 [2,3,5,7,10,12,16,17,18,21,25,26,27,30,31,32,37,38,39,40,41,42,43,44,45,48,51,58,
850 59,60,63,64,65,66,67: #ARG]]
851 [4,5,6,7,8,9,10,11,12,13,22,23,24,27,28,39,40,41,42,43,44,45,46,47,48,49,50,51,52,
852 56,57,69,70,72,73,75: #arg2 @(list_addressing_mode_tags_elim_prop … arg2) whd try % -arg2
853 [1,2,4,7,9,10,12,13,15,16,17,18,20,22,23,24,25,26,27,28,29,30,31,32,33,36,37,38,
854 39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,
855 68,69,70,71: #ARG2]]
856 [1,2,19,20: #arg3 @(list_addressing_mode_tags_elim_prop … arg3) whd try % -arg3 #ARG3]
857 normalize in ⊢ (???% → ?)
858 [92,94,42,93,95: @split_elim #vl #vm #E >E -E; [2,4: @(bitvector_3_elim_prop … vl)]
859 normalize in ⊢ (???% → ?)]
860 #H >H * #H1 try (change in ⊢ (% → ?) with (? ∧ ?) * #H2)
861 try (change in ⊢ (% → ?) with (? ∧ ?) * #H3) whd in ⊢ (% → ?) #H4
862 change in ⊢ (let fetched ≝ % in ?) with (fetch0 ??)
863 whd in ⊢ (let fetched ≝ ??% in ?) <H1 whd in ⊢ (let fetched ≝ % in ?)
864 [17,18,19,20,21,22,23,24,25,26,31,34,35,36,37,38: <H3]
865 [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,
866 30,31,32,33,34,35,36,37,38,39,40,43,45,48,49,52,53,54,55,56,57,60,61,62,65,66,
867 69,70,73,74,78,80,81,84,85,95,98,101,102,103,104,105,106,107,108,109,110: <H2]
868 whd >eq_instruction_refl >H4 @eq_bv_refl
869qed. *)
870
871let rec fetch_many code_memory final_pc pc expected on expected: Prop ≝
872 match expected with
873 [ nil ⇒ eq_bv … pc final_pc = true
874 | cons i tl ⇒
875 let fetched ≝ fetch code_memory pc in
876 let 〈instr_pc, ticks〉 ≝ fetched in
877 let 〈instr,pc'〉 ≝ instr_pc in
878 eq_instruction instr i = true ∧ fetch_many code_memory final_pc pc' tl].
879
880lemma option_destruct_Some: ∀A,a,b. Some A a = Some A b → a=b.
881 #A #a #b #EQ destruct //
882qed.
883
884lemma pair_destruct: ∀A,B,a1,a2,b1,b2. pair A B a1 a2 = 〈b1,b2〉 → a1=b1 ∧ a2=b2.
885 #A #B #a1 #a2 #b1 #b2 #EQ destruct /2/
886qed.
887
888axiom eq_bv_to_eq: ∀n.∀v1,v2: BitVector n. eq_bv … v1 v2 = true → v1=v2.
889
890lemma fetch_assembly_pseudo:
891 ∀program,ppc,lookup_labels,lookup_datalabels.
892 ∀pi,code_memory,len,assembled,instructions,pc.
893 let expansion ≝ jump_expansion ppc program in
894 Some ? instructions = expand_pseudo_instruction lookup_labels lookup_datalabels (bitvector_of_nat ? pc) expansion pi →
895 Some … 〈len,assembled〉 = assembly_1_pseudoinstruction program ppc (bitvector_of_nat ? pc) lookup_labels lookup_datalabels pi →
896 encoding_check code_memory (bitvector_of_nat … pc) (bitvector_of_nat … (pc + len)) assembled →
897 fetch_many code_memory (bitvector_of_nat … (pc + len)) (bitvector_of_nat … pc) instructions.
898 #program #ppc #lookup_labels #lookup_datalabels #pi #code_memory #len #assembled #instructions #pc
899 #EQ1 whd in ⊢ (???% → ?) <EQ1 whd in ⊢ (???% → ?) #EQ2
900 cases (pair_destruct ?????? (option_destruct_Some … EQ2)) -EQ2; #EQ2a #EQ2b
901 >EQ2a >EQ2b -EQ2a EQ2b;
902 generalize in match (pc + |flatten … (map … assembly1 instructions)|); #final_pc
903 generalize in match pc elim instructions
904 [ #pc whd in ⊢ (% → %) #H >H @eq_bv_refl
905 | #i #tl #IH #pc #H whd cases (encoding_check_append … H); -H; #H1 #H2 whd
906 generalize in match (fetch_assembly pc i code_memory … (refl …) H1)
907 cases (fetch code_memory (bitvector_of_nat … pc)) #newi_pc #ticks whd in ⊢ (% → %)
908 cases newi_pc #newi #newpc whd in ⊢ (% → %) #K cases (conjunction_true … K) -K; #K1 #K2 % //
909 >(eq_bv_to_eq … K2) @IH @H2 ]
910qed.
911
912(* This establishes the correspondence between pseudo program counters and
913 program counters. It is at the heart of the proof. *)
914(*CSC: code taken from build_maps *)
915definition sigma0: pseudo_assembly_program → option (nat × (nat × (BitVectorTrie Word 16))) ≝
916 λinstr_list.
917 foldl ??
918 (λt. λi.
919 match t with
920 [ None ⇒ None ?
921 | Some ppc_pc_map ⇒
922 let 〈ppc,pc_map〉 ≝ ppc_pc_map in
923 let 〈program_counter, sigma_map〉 ≝ pc_map in
924 let 〈label, i〉 ≝ i in
925 match construct_costs instr_list ppc program_counter (λx. zero ?) (λx. zero ?) (Stub …) i with
926 [ None ⇒ None ?
927 | Some pc_ignore ⇒
928 let 〈pc,ignore〉 ≝ pc_ignore in
929 Some … 〈S ppc,〈pc, insert ? ? (bitvector_of_nat ? ppc) (bitvector_of_nat ? pc) sigma_map〉〉 ]
930 ]) (Some ? 〈0, 〈0, (Stub ? ?)〉〉) (\snd instr_list).
931
932definition tech_pc_sigma0: pseudo_assembly_program → option (nat × (BitVectorTrie Word 16)) ≝
933 λinstr_list.
934 match sigma0 instr_list with
935 [ None ⇒ None …
936 | Some result ⇒
937 let 〈ppc,pc_sigma_map〉 ≝ result in
938 Some … pc_sigma_map ].
939
940definition sigma_safe: pseudo_assembly_program → option (Word → Word) ≝
941 λinstr_list.
942 match sigma0 instr_list with
943 [ None ⇒ None ?
944 | Some result ⇒
945 let 〈ppc,pc_sigma_map〉 ≝ result in
946 let 〈pc, sigma_map〉 ≝ pc_sigma_map in
947 if gtb pc (2^16) then
948 None ?
949 else
950 Some ? (λx.lookup ?? x sigma_map (zero …)) ].
951
952axiom policy_ok: ∀p. sigma_safe p ≠ None ….
953
954definition sigma: pseudo_assembly_program → Word → Word ≝
955 λp.
956 match sigma_safe p return λr:option (Word → Word). r ≠ None … → Word → Word with
957 [ None ⇒ λabs. ⊥
958 | Some r ⇒ λ_.r] (policy_ok p).
959 cases abs //
960qed.
961
962lemma length_append:
963 ∀A.∀l1,l2:list A.
964 |l1 @ l2| = |l1| + |l2|.
965 #A #l1 elim l1
966 [ //
967 | #hd #tl #IH #l2 normalize <IH //]
968qed.
969
970let rec does_not_occur (id:Identifier) (l:list labelled_instruction) on l: bool ≝
971 match l with
972 [ nil ⇒ true
973 | cons hd tl ⇒ notb (instruction_matches_identifier id hd) ∧ does_not_occur id tl].
974
975lemma does_not_occur_None:
976 ∀id,i,list_instr.
977 does_not_occur id (list_instr@[〈None …,i〉]) =
978 does_not_occur id list_instr.
979 #id #i #list_instr elim list_instr
980 [ % | #hd #tl #IH whd in ⊢ (??%%) >IH %]
981qed.
982
983let rec occurs_exactly_once (id:Identifier) (l:list labelled_instruction) on l : bool ≝
984 match l with
985 [ nil ⇒ false
986 | cons hd tl ⇒
987 if instruction_matches_identifier id hd then
988 does_not_occur id tl
989 else
990 occurs_exactly_once id tl ].
991
992lemma occurs_exactly_once_None:
993 ∀id,i,list_instr.
994 occurs_exactly_once id (list_instr@[〈None …,i〉]) =
995 occurs_exactly_once id list_instr.
996 #id #i #list_instr elim list_instr
997 [ % | #hd #tl #IH whd in ⊢ (??%%) >IH >does_not_occur_None %]
998qed.
999
1000lemma index_of_internal_None: ∀i,id,instr_list,n.
1001 occurs_exactly_once id (instr_list@[〈None …,i〉]) →
1002 index_of_internal ? (instruction_matches_identifier id) instr_list n =
1003 index_of_internal ? (instruction_matches_identifier id) (instr_list@[〈None …,i〉]) n.
1004 #i #id #instr_list elim instr_list
1005 [ #n #abs whd in abs; cases abs
1006 | #hd #tl #IH #n whd in ⊢ (% → ??%%); whd in ⊢ (match % with [_ ⇒ ? | _ ⇒ ?] → ?)
1007 cases (instruction_matches_identifier id hd) whd in ⊢ (match % with [_ ⇒ ? | _ ⇒ ?] → ??%%)
1008 [ #H %
1009 | #H @IH whd in H; cases (occurs_exactly_once ??) in H ⊢ %
1010 [ #_ % | #abs cases abs ]]]
1011qed.
1012
1014 occurs_exactly_once id (instr_list@[〈None …,i〉]) →
1015 address_of_word_labels_code_mem instr_list id =
1016 address_of_word_labels_code_mem (instr_list@[〈None …,i〉]) id.
1017 #i #id #instr_list #H whd in ⊢ (??%%) whd in ⊢ (??(??%?)(??%?))
1018 >(index_of_internal_None … H) %
1019qed.
1020
1021axiom tech_pc_sigma0_append:
1022 ∀preamble,instr_list,prefix,label,i,pc',code,ppc,pc,costs,costs'.
1023 Some … 〈pc,costs〉 = tech_pc_sigma0 〈preamble,prefix〉 →
1024 construct_costs 〈preamble,instr_list〉 … ppc pc (λx.zero 16) (λx. zero 16) costs i = Some … 〈pc',code〉 →
1025 tech_pc_sigma0 〈preamble,prefix@[〈label,i〉]〉 = Some … 〈pc',costs'〉.
1026
1027axiom tech_pc_sigma0_append_None:
1028 ∀preamble,instr_list,prefix,i,ppc,pc,costs.
1029 Some … 〈pc,costs〉 = tech_pc_sigma0 〈preamble,prefix〉 →
1030 construct_costs 〈preamble,instr_list〉 … ppc pc (λx.zero 16) (λx. zero 16) costs i = None …
1031 → False.
1032
1033(*
1034definition build_maps' ≝
1035 λpseudo_program.
1036 let 〈preamble,instr_list〉 ≝ pseudo_program in
1037 let result ≝
1038 foldl_strong
1039 (option Identifier × pseudo_instruction)
1040 (λpre. Σres:((BitVectorTrie Word 16) × (nat × (BitVectorTrie Word 16))).
1041 let pre' ≝ 〈preamble,pre〉 in
1042 let 〈labels,pc_costs〉 ≝ res in
1043 tech_pc_sigma0 pre' = Some … pc_costs ∧
1044 ∀id. occurs_exactly_once id pre →
1045 lookup ?? id labels (zero …) = sigma pre' (address_of_word_labels_code_mem pre id))
1046 instr_list
1047 (λprefix,i,tl,prf,t.
1048 let 〈labels, pc_costs〉 ≝ t in
1049 let 〈program_counter, costs〉 ≝ pc_costs in
1050 let 〈label, i'〉 ≝ i in
1051 let labels ≝
1052 match label with
1053 [ None ⇒ labels
1054 | Some label ⇒
1055 let program_counter_bv ≝ bitvector_of_nat ? program_counter in
1056 insert ? ? label program_counter_bv labels
1057 ]
1058 in
1059 match construct_costs 〈preamble,instr_list〉 program_counter (λx. zero ?) (λx. zero ?) costs i' with
1060 [ None ⇒
1061 let dummy ≝ 〈labels,pc_costs〉 in
1062 dummy
1063 | Some construct ⇒ 〈labels, construct〉
1064 ]
1065 ) 〈(Stub ? ?), 〈0, (Stub ? ?)〉〉
1066 in
1067 let 〈labels, pc_costs〉 ≝ result in
1068 let 〈pc, costs〉 ≝ pc_costs in
1069 〈labels, costs〉.
1070 [3: whd % // #id normalize in ⊢ (% → ?) #abs @⊥ //
1071 | whd cases construct in p3 #PC #CODE #JMEQ %
1072 [ @(tech_pc_sigma0_append ??????????? (jmeq_to_eq ??? JMEQ)) | #id #Hid ]
1073 | (* dummy case *) @⊥
1074 @(tech_pc_sigma0_append_None ?? prefix ???? (jmeq_to_eq ??? p3)) ]
1075 [*: generalize in match (sig2 … t) whd in ⊢ (% → ?)
1076 >p whd in ⊢ (% → ?) >p1 * #IH0 #IH1 >IH0 // ]
1077 whd in ⊢ (??(????%?)?) -labels1;
1078 cases label in Hid
1079 [ #Hid whd in ⊢ (??(????%?)?) >IH1 -IH1
1080 [ >(address_of_word_labels_code_mem_None … Hid)
1081 (* MANCA LEMMA: INDIRIZZO TROVATO NEL PROGRAMMA! *)
1082 | whd in Hid >occurs_exactly_once_None in Hid // ]
1083 | -label #label #Hid whd in ⊢ (??(????%?)?)
1084
1085 ]
1086qed.
1087
1088lemma build_maps_ok:
1089 ∀p:pseudo_assembly_program.
1090 let 〈labels,costs〉 ≝ build_maps' p in
1091 ∀pc.
1092 (nat_of_bitvector … pc) < length … (\snd p) →
1093 lookup ?? pc labels (zero …) = sigma p (\snd (fetch_pseudo_instruction (\snd p) pc)).
1094 #p cases p #preamble #instr_list
1095 elim instr_list
1096 [ whd #pc #abs normalize in abs; cases (not_le_Sn_O ?) [#H cases (H abs) ]
1097 | #hd #tl #IH
1098 whd in ⊢ (match % with [ _ ⇒ ?])
1099 ]
1100qed.
1101*)
1102
1103(*
1104lemma rev_preserves_length:
1105 ∀A.∀l. length … (rev A l) = length … l.
1106 #A #l elim l
1107 [ %
1108 | #hd #tl #IH normalize >length_append normalize /2/ ]
1109qed.
1110
1111lemma rev_append:
1112 ∀A.∀l1,l2.
1113 rev A (l1@l2) = rev A l2 @ rev A l1.
1114 #A #l1 elim l1 normalize //
1115qed.
1116
1117lemma rev_rev: ∀A.∀l. rev … (rev A l) = l.
1118 #A #l elim l
1119 [ //
1120 | #hd #tl #IH normalize >rev_append normalize // ]
1121qed.
1122
1123lemma split_len_Sn:
1124 ∀A:Type[0].∀l:list A.∀len.
1125 length … l = S len →
1126 Σl'.Σa. l = l'@[a] ∧ length … l' = len.
1127 #A #l elim l
1128 [ normalize #len #abs destruct
1129 | #hd #tl #IH #len
1130 generalize in match (rev_rev … tl)
1131 cases (rev A tl) in ⊢ (??%? → ?)
1132 [ #H <H normalize #EQ % [@[ ]] % [@hd] normalize /2/
1133 | #a #l' #H <H normalize #EQ
1134 %[@(hd::rev … l')] %[@a] % //
1135 >length_append in EQ #EQ normalize in EQ; normalize;
1136 generalize in match (injective_S … EQ) #EQ2 /2/ ]]
1137qed.
1138
1139lemma list_elim_rev:
1140 ∀A:Type[0].∀P:list A → Type[0].
1141 P [ ] → (∀l,a. P l → P (l@[a])) →
1142 ∀l. P l.
1143 #A #P #H1 #H2 #l
1144 generalize in match (refl … (length … l))
1145 generalize in ⊢ (???% → ?) #n generalize in match l
1146 elim n
1147 [ #L cases L [ // | #x #w #abs (normalize in abs) @⊥ // ]
1148 | #m #IH #L #EQ
1149 cases (split_len_Sn … EQ) #l' * #a * /3/ ]
1150qed.
1151
1152axiom is_prefix: ∀A:Type[0]. list A → list A → Prop.
1153axiom prefix_of_append:
1154 ∀A:Type[0].∀l,l1,l2:list A.
1155 is_prefix … l l1 → is_prefix … l (l1@l2).
1156axiom prefix_reflexive: ∀A,l. is_prefix A l l.
1157axiom nil_prefix: ∀A,l. is_prefix A [ ] l.
1158
1159record Propify (A:Type[0]) : Type[0] (*Prop*) ≝ { in_propify: A }.
1160
1161definition Propify_elim: ∀A. ∀P:Prop. (A → P) → (Propify A → P) ≝
1162 λA,P,H,x. match x with [ mk_Propify p ⇒ H p ].
1163
1164definition app ≝
1165 λA:Type[0].λl1:Propify (list A).λl2:list A.
1166 match l1 with
1167 [ mk_Propify l1 ⇒ mk_Propify … (l1@l2) ].
1168
1169lemma app_nil: ∀A,l1. app A l1 [ ] = l1.
1170 #A * /3/
1171qed.
1172
1173lemma app_assoc: ∀A,l1,l2,l3. app A (app A l1 l2) l3 = app A l1 (l2@l3).
1174 #A * #l1 normalize //
1175qed.
1176
1177let rec foldli (A: Type[0]) (B: Propify (list A) → Type[0])
1178 (f: ∀prefix. B prefix → ∀x.B (app … prefix [x]))
1179 (prefix: Propify (list A)) (b: B prefix) (l: list A) on l :
1180 B (app … prefix l) ≝
1181 match l with
1182 [ nil ⇒ ? (* b *)
1183 | cons hd tl ⇒ ? (*foldli A B f (prefix@[hd]) (f prefix b hd) tl*)
1184 ].
1185 [ applyS b
1186 | <(app_assoc ?? [hd]) @(foldli A B f (app … prefix [hd]) (f prefix b hd) tl) ]
1187qed.
1188
1189(*
1190let rec foldli (A: Type[0]) (B: list A → Type[0]) (f: ∀prefix. B prefix → ∀x. B (prefix@[x]))
1191 (prefix: list A) (b: B prefix) (l: list A) on l : B (prefix@l) ≝
1192 match l with
1193 [ nil ⇒ ? (* b *)
1194 | cons hd tl ⇒
1195 ? (*foldli A B f (prefix@[hd]) (f prefix b hd) tl*)
1196 ].
1197 [ applyS b
1198 | applyS (foldli A B f (prefix@[hd]) (f prefix b hd) tl) ]
1199qed.
1200*)
1201
1202definition foldll:
1203 ∀A:Type[0].∀B: Propify (list A) → Type[0].
1204 (∀prefix. B prefix → ∀x. B (app … prefix [x])) →
1205 B (mk_Propify … []) → ∀l: list A. B (mk_Propify … l)
1206 ≝ λA,B,f. foldli A B f (mk_Propify … [ ]).
1207
1208axiom is_pprefix: ∀A:Type[0]. Propify (list A) → list A → Prop.
1209axiom pprefix_of_append:
1210 ∀A:Type[0].∀l,l1,l2.
1211 is_pprefix A l l1 → is_pprefix A l (l1@l2).
1212axiom pprefix_reflexive: ∀A,l. is_pprefix A (mk_Propify … l) l.
1213axiom nil_pprefix: ∀A,l. is_pprefix A (mk_Propify … [ ]) l.
1214
1215
1216axiom foldll':
1217 ∀A:Type[0].∀l: list A.
1218 ∀B: ∀prefix:Propify (list A). is_pprefix ? prefix l → Type[0].
1219 (∀prefix,proof. B prefix proof → ∀x,proof'. B (app … prefix [x]) proof') →
1220 B (mk_Propify … [ ]) (nil_pprefix …) → B (mk_Propify … l) (pprefix_reflexive … l).
1221 #A #l #B
1222 generalize in match (foldll A (λprefix. is_pprefix ? prefix l)) #HH
1223
1224
1225 #H #acc
1226 @foldll
1227 [
1228 |
1229 ]
1230
1231 ≝ λA,B,f. foldli A B f (mk_Propify … [ ]).
1232
1233
1234(*
1235record subset (A:Type[0]) (P: A → Prop): Type[0] ≝
1236 { subset_wit:> A;
1237 subset_proof: P subset_wit
1238 }.
1239*)
1240
1241definition build_maps' ≝
1242 λpseudo_program.
1243 let 〈preamble,instr_list〉 ≝ pseudo_program in
1244 let result ≝
1245 foldll
1246 (option Identifier × pseudo_instruction)
1247 (λprefix.
1248 Σt:((BitVectorTrie Word 16) × (nat × (BitVectorTrie Word 16))).
1249 match prefix return λ_.Prop with [mk_Propify prefix ⇒ tech_pc_sigma0 〈preamble,prefix〉 ≠ None ?])
1250 (λprefix,t,i.
1251 let 〈labels, pc_costs〉 ≝ t in
1252 let 〈program_counter, costs〉 ≝ pc_costs in
1253 let 〈label, i'〉 ≝ i in
1254 let labels ≝
1255 match label with
1256 [ None ⇒ labels
1257 | Some label ⇒
1258 let program_counter_bv ≝ bitvector_of_nat ? program_counter in
1259 insert ? ? label program_counter_bv labels
1260 ]
1261 in
1262 match construct_costs pseudo_program program_counter (λx. zero ?) (λx. zero ?) costs i' with
1263 [ None ⇒
1264 let dummy ≝ 〈labels,pc_costs〉 in
1265 dummy
1266 | Some construct ⇒ 〈labels, construct〉
1267 ]
1268 ) 〈(Stub ? ?), 〈0, (Stub ? ?)〉〉 instr_list
1269 in
1270 let 〈labels, pc_costs〉 ≝ result in
1271 let 〈pc, costs〉 ≝ pc_costs in
1272 〈labels, costs〉.
1273 [
1274 | @⊥
1275 | normalize % //
1276 ]
1277qed.
1278
1279definition build_maps' ≝
1280 λpseudo_program.
1281 let 〈preamble,instr_list〉 ≝ pseudo_program in
1282 let result ≝
1283 foldl
1284 (Σt:((BitVectorTrie Word 16) × (nat × (BitVectorTrie Word 16))).
1285 ∃instr_list_prefix. is_prefix ? instr_list_prefix instr_list ∧
1286 tech_pc_sigma0 〈preamble,instr_list_prefix〉 = Some ? (\fst (\snd t)))
1287 (Σi:option Identifier × pseudo_instruction. ∀instr_list_prefix.
1288 let instr_list_prefix' ≝ instr_list_prefix @ [i] in
1289 is_prefix ? instr_list_prefix' instr_list →
1290 tech_pc_sigma0 〈preamble,instr_list_prefix'〉 ≠ None ?)
1291 (λt: Σt:((BitVectorTrie Word 16) × (nat × (BitVectorTrie Word 16))).
1292 ∃instr_list_prefix. is_prefix ? instr_list_prefix instr_list ∧
1293 tech_pc_sigma0 〈preamble,instr_list_prefix〉 = Some ? (\fst (\snd t)).
1294 λi: Σi:option Identifier × pseudo_instruction. ∀instr_list_prefix.
1295 let instr_list_prefix' ≝ instr_list_prefix @ [i] in
1296 is_prefix ? instr_list_prefix' instr_list →
1297 tech_pc_sigma0 〈preamble,instr_list_prefix'〉 ≠ None ? .
1298 let 〈labels, pc_costs〉 ≝ t in
1299 let 〈program_counter, costs〉 ≝ pc_costs in
1300 let 〈label, i'〉 ≝ i in
1301 let labels ≝
1302 match label with
1303 [ None ⇒ labels
1304 | Some label ⇒
1305 let program_counter_bv ≝ bitvector_of_nat ? program_counter in
1306 insert ? ? label program_counter_bv labels
1307 ]
1308 in
1309 match construct_costs pseudo_program program_counter (λx. zero ?) (λx. zero ?) costs i' with
1310 [ None ⇒
1311 let dummy ≝ 〈labels,pc_costs〉 in
1312 dummy
1313 | Some construct ⇒ 〈labels, construct〉
1314 ]
1315 ) 〈(Stub ? ?), 〈0, (Stub ? ?)〉〉 ?(*instr_list*)
1316 in
1317 let 〈labels, pc_costs〉 ≝ result in
1318 let 〈pc, costs〉 ≝ pc_costs in
1319 〈labels, costs〉.
1320 [4: @(list_elim_rev ?
1321 (λinstr_list. list (
1322 (Σi:option Identifier × pseudo_instruction. ∀instr_list_prefix.
1323 let instr_list_prefix' ≝ instr_list_prefix @ [i] in
1324 is_prefix ? instr_list_prefix' instr_list →
1325 tech_pc_sigma0 〈preamble,instr_list_prefix'〉 ≠ None ?)))
1326 ?? instr_list) (* CSC: BAD ORDER FOR CODE EXTRACTION *)
1327 [ @[ ]
1328 | #l' #a #limage %2
1329 [ %[@a] #PREFIX #PREFIX_OK
1330 | (* CSC: EVEN WORST CODE FOR EXTRACTION: WE SHOULD STRENGTHEN
1331 THE INDUCTION HYPOTHESIS INSTEAD *)
1332 elim limage
1333 [ %1
1334 | #HD #TL #IH @(?::IH) cases HD #ELEM #K1 %[@ELEM] #K2 #K3
1335 @K1 @(prefix_of_append ???? K3)
1336 ]
1337 ]
1338
1339
1340
1341
1342 cases t in c2 ⊢ % #t' * #LIST_PREFIX * #H1t' #H2t' #HJMt'
1343 % [@ (LIST_PREFIX @ [i])] %
1344 [ cases (sig2 … i LIST_PREFIX) #K1 #K2 @K1
1345 | (* DOABLE IN PRINCIPLE *)
1346 ]
1347 | (* assert false case *)
1348 |3: % [@ ([ ])] % [2: % | (* DOABLE *)]
1349 |
1350*)
1351
1352axiom assembly_ok:
1353 ∀program,assembled,costs,labels.
1354 Some … 〈labels,costs〉 = build_maps program →
1355 Some … 〈assembled,costs〉 = assembly program →
1356 let code_memory ≝ load_code_memory assembled in
1357 let preamble ≝ \fst program in
1358 let datalabels ≝ construct_datalabels preamble in
1359 let lookup_labels ≝ λx. sigma program (address_of_word_labels_code_mem (\snd program) x) in
1360 let lookup_datalabels ≝ λx. lookup ?? x datalabels (zero ?) in
1361 ∀ppc,len,assembledi.
1362 let 〈pi,newppc〉 ≝ fetch_pseudo_instruction (\snd program) ppc in
1363 Some … 〈len,assembledi〉 = assembly_1_pseudoinstruction program ppc (sigma program ppc) lookup_labels lookup_datalabels pi →
1364 encoding_check code_memory (sigma program ppc) (bitvector_of_nat … (nat_of_bitvector … (sigma program ppc) + len)) assembledi ∧
1365 sigma program newppc = bitvector_of_nat … (nat_of_bitvector … (sigma program ppc) + len).
1366
1367axiom bitvector_of_nat_nat_of_bitvector:
1368 ∀n,v.
1369 bitvector_of_nat n (nat_of_bitvector n v) = v.
1370
1371axiom assembly_ok_to_expand_pseudo_instruction_ok:
1372 ∀program,assembled,costs.
1373 Some … 〈assembled,costs〉 = assembly program →
1374 ∀ppc.
1375 let code_memory ≝ load_code_memory assembled in
1376 let preamble ≝ \fst program in
1377 let data_labels ≝ construct_datalabels preamble in
1378 let lookup_labels ≝ λx. sigma program (address_of_word_labels_code_mem (\snd program) x) in
1379 let lookup_datalabels ≝ λx. lookup ? ? x data_labels (zero ?) in
1380 let expansion ≝ jump_expansion ppc program in
1381 let 〈pi,newppc〉 ≝ fetch_pseudo_instruction (\snd program) ppc in
1382 ∃instructions.
1383 Some ? instructions = expand_pseudo_instruction lookup_labels lookup_datalabels (sigma program ppc) expansion pi.
1384
1385lemma fetch_assembly_pseudo2:
1386 ∀program,assembled,costs,labels.
1387 Some … 〈labels,costs〉 = build_maps program →
1388 Some … 〈assembled,costs〉 = assembly program →
1389 ∀ppc.
1390 let code_memory ≝ load_code_memory assembled in
1391 let preamble ≝ \fst program in
1392 let data_labels ≝ construct_datalabels preamble in
1393 let lookup_labels ≝ λx. sigma program (address_of_word_labels_code_mem (\snd program) x) in
1394 let lookup_datalabels ≝ λx. lookup ? ? x data_labels (zero ?) in
1395 let expansion ≝ jump_expansion ppc program in
1396 let 〈pi,newppc〉 ≝ fetch_pseudo_instruction (\snd program) ppc in
1397 ∃instructions.
1398 Some ? instructions = expand_pseudo_instruction lookup_labels lookup_datalabels (sigma program ppc) expansion pi ∧
1399 fetch_many code_memory (sigma program newppc) (sigma program ppc) instructions.
1400 #program #assembled #costs #labels #BUILD_MAPS #ASSEMBLY #ppc
1401 generalize in match (assembly_ok_to_expand_pseudo_instruction_ok program assembled costs ASSEMBLY ppc)
1402 letin code_memory ≝ (load_code_memory assembled)
1403 letin preamble ≝ (\fst program)
1404 letin data_labels ≝ (construct_datalabels preamble)
1405 letin lookup_labels ≝ (λx. sigma program (address_of_word_labels_code_mem (\snd program) x))
1406 letin lookup_datalabels ≝ (λx. lookup ? ? x data_labels (zero ?))
1407 whd in ⊢ (% → %)
1408 generalize in match (assembly_ok … BUILD_MAPS ASSEMBLY ppc)
1409 cases (fetch_pseudo_instruction (\snd program) ppc) #pi #newppc
1410 generalize in match (fetch_assembly_pseudo program ppc
1411 (λx. sigma program (address_of_word_labels_code_mem (\snd program) x)) (λx. lookup ?? x data_labels (zero ?)) pi
1413 whd in ⊢ ((∀_.∀_.∀_.∀_.%) → (∀_.∀_.%) → % → %) #H1 #H2 * #instructions #EXPAND
1414 whd in H1:(∀_.∀_.∀_.∀_.? → ???% → ?) H2:(∀_.∀_.???% → ?);
1415 normalize nodelta in EXPAND; (* HERE *)
1416 generalize in match (λlen, assembled.H1 len assembled instructions (nat_of_bitvector … (sigma program ppc))) -H1; #H1
1417 >bitvector_of_nat_nat_of_bitvector in H1; #H1
1418 <EXPAND in H1 H2; whd in ⊢ ((∀_.∀_.? → ???% → ?) → (∀_.∀_.???% → ?) → ?)
1419 #H1 #H2
1420 cases (H2 ?? (refl …)) -H2; #K1 #K2 >K2
1421 generalize in match (H1 ?? (refl …) (refl …) ?) -H1;
1422 [ #K3 % [2: % [% | @K3]] | @K1 ]
1423qed.
1424
1425(* OLD?
1426definition assembly_specification:
1427 ∀assembly_program: pseudo_assembly_program.
1428 ∀code_mem: BitVectorTrie Byte 16. Prop ≝
1429 λpseudo_assembly_program.
1430 λcode_mem.
1431 ∀pc: Word.
1432 let 〈preamble, instr_list〉 ≝ pseudo_assembly_program in
1433 let 〈pre_instr, pre_new_pc〉 ≝ fetch_pseudo_instruction instr_list pc in
1434 let labels ≝ λx. sigma' pseudo_assembly_program (address_of_word_labels_code_mem instr_list x) in
1435 let datalabels ≝ λx. sigma' pseudo_assembly_program (lookup ? ? x (construct_datalabels preamble) (zero ?)) in
1436 let pre_assembled ≝ assembly_1_pseudoinstruction pseudo_assembly_program
1437 (sigma' pseudo_assembly_program pc) labels datalabels pre_instr in
1438 match pre_assembled with
1439 [ None ⇒ True
1440 | Some pc_code ⇒
1441 let 〈new_pc,code〉 ≝ pc_code in
1442 encoding_check code_mem pc (sigma' pseudo_assembly_program pre_new_pc) code ].
1443
1444axiom assembly_meets_specification:
1445 ∀pseudo_assembly_program.
1446 match assembly pseudo_assembly_program with
1447 [ None ⇒ True
1448 | Some code_mem_cost ⇒
1449 let 〈code_mem, cost〉 ≝ code_mem_cost in
1450 assembly_specification pseudo_assembly_program (load_code_memory code_mem)
1451 ].
1452(*
1453 # PROGRAM
1454 [ cases PROGRAM
1455 # PREAMBLE
1456 # INSTR_LIST
1457 elim INSTR_LIST
1458 [ whd
1459 whd in ⊢ (∀_. %)
1460 # PC
1461 whd
1462 | # INSTR
1463 # INSTR_LIST_TL
1464 # H
1465 whd
1466 whd in ⊢ (match % with [ _ ⇒ ? | _ ⇒ ?])
1467 ]
1468 | cases not_implemented
1469 ] *)
1470*)
1471
1472definition status_of_pseudo_status: PseudoStatus → option Status ≝
1473 λps.
1474 let pap ≝ code_memory … ps in
1475 match assembly pap with
1476 [ None ⇒ None …
1477 | Some p ⇒
1478 let cm ≝ load_code_memory (\fst p) in
1479 let pc ≝ sigma pap (program_counter ? ps) in
1480 Some …
1481 (mk_PreStatus (BitVectorTrie Byte 16)
1482 cm
1483 (low_internal_ram … ps)
1484 (high_internal_ram … ps)
1485 (external_ram … ps)
1486 pc
1487 (special_function_registers_8051 … ps)
1488 (special_function_registers_8052 … ps)
1489 (p1_latch … ps)
1490 (p3_latch … ps)
1491 (clock … ps)) ].
1492
1493(*
1494definition write_at_stack_pointer':
1495 ∀M. ∀ps: PreStatus M. Byte → Σps':PreStatus M.(code_memory … ps = code_memory … ps') ≝
1496 λM: Type[0].
1497 λs: PreStatus M.
1498 λv: Byte.
1499 let 〈 nu, nl 〉 ≝ split … 4 4 (get_8051_sfr ? s SFR_SP) in
1500 let bit_zero ≝ get_index_v… nu O ? in
1501 let bit_1 ≝ get_index_v… nu 1 ? in
1502 let bit_2 ≝ get_index_v… nu 2 ? in
1503 let bit_3 ≝ get_index_v… nu 3 ? in
1504 if bit_zero then
1505 let memory ≝ insert … ([[ bit_1 ; bit_2 ; bit_3 ]] @@ nl)
1506 v (low_internal_ram ? s) in
1507 set_low_internal_ram ? s memory
1508 else
1509 let memory ≝ insert … ([[ bit_1 ; bit_2 ; bit_3 ]] @@ nl)
1510 v (high_internal_ram ? s) in
1511 set_high_internal_ram ? s memory.
1512 [ cases l0 %
1513 |2,3,4,5: normalize repeat (@ le_S_S) @ le_O_n ]
1514qed.
1515
1516definition execute_1_pseudo_instruction': (Word → nat) → ∀ps:PseudoStatus.
1517 Σps':PseudoStatus.(code_memory … ps = code_memory … ps')
1518
1519 λticks_of.
1520 λs.
1521 let 〈instr, pc〉 ≝ fetch_pseudo_instruction (\snd (code_memory ? s)) (program_counter ? s) in
1522 let ticks ≝ ticks_of (program_counter ? s) in
1523 let s ≝ set_clock ? s (clock ? s + ticks) in
1524 let s ≝ set_program_counter ? s pc in
1525 match instr with
1526 [ Instruction instr ⇒
1527 execute_1_preinstruction … (λx, y. address_of_word_labels y x) instr s
1528 | Comment cmt ⇒ s
1529 | Cost cst ⇒ s
1530 | Jmp jmp ⇒ set_program_counter ? s (address_of_word_labels s jmp)
1531 | Call call ⇒
1532 let a ≝ address_of_word_labels s call in
1533 let 〈carry, new_sp〉 ≝ half_add ? (get_8051_sfr ? s SFR_SP) (bitvector_of_nat 8 1) in
1534 let s ≝ set_8051_sfr ? s SFR_SP new_sp in
1535 let 〈pc_bu, pc_bl〉 ≝ split ? 8 8 (program_counter ? s) in
1536 let s ≝ write_at_stack_pointer' ? s pc_bl in
1537 let 〈carry, new_sp〉 ≝ half_add ? (get_8051_sfr ? s SFR_SP) (bitvector_of_nat 8 1) in
1538 let s ≝ set_8051_sfr ? s SFR_SP new_sp in
1539 let s ≝ write_at_stack_pointer' ? s pc_bu in
1540 set_program_counter ? s a
1541 | Mov dptr ident ⇒
1542 set_arg_16 ? s (get_arg_16 ? s (DATA16 (address_of_word_labels s ident))) dptr
1543 ].
1544 [
1545 |2,3,4: %
1546 | <(sig2 … l7) whd in ⊢ (??? (??%)) <(sig2 … l5) %
1547 |
1548 | %
1549 ]
1550 cases not_implemented
1551qed.
1552*)
1553
1554axiom execute_1_pseudo_instruction_preserves_code_memory:
1555 ∀ticks_of,ps.
1556 code_memory … (execute_1_pseudo_instruction ticks_of ps) = code_memory … ps.
1557
1558(*
1559lemma execute_code_memory_unchanged:
1560 ∀ticks_of,ps. code_memory ? ps = code_memory ? (execute_1_pseudo_instruction ticks_of ps).
1561 #ticks #ps whd in ⊢ (??? (??%))
1562 cases (fetch_pseudo_instruction (\snd (code_memory pseudo_assembly_program ps))
1563 (program_counter pseudo_assembly_program ps)) #instr #pc
1564 whd in ⊢ (??? (??%)) cases instr
1565 [ #pre cases pre
1566 [ #a1 #a2 whd in ⊢ (??? (??%)) cases (add_8_with_carry ???) #y1 #y2 whd in ⊢ (??? (??%))
1567 cases (split ????) #z1 #z2 %
1568 | #a1 #a2 whd in ⊢ (??? (??%)) cases (add_8_with_carry ???) #y1 #y2 whd in ⊢ (??? (??%))
1569 cases (split ????) #z1 #z2 %
1570 | #a1 #a2 whd in ⊢ (??? (??%)) cases (sub_8_with_carry ???) #y1 #y2 whd in ⊢ (??? (??%))
1571 cases (split ????) #z1 #z2 %
1572 | #a1 whd in ⊢ (??? (??%)) cases a1 #x #H whd in ⊢ (??? (??%)) cases x
1573 [ #x1 whd in ⊢ (??? (??%))
1574 | *: cases not_implemented
1575 ]
1576 | #comment %
1577 | #cost %
1578 | #label %
1579 | #label whd in ⊢ (??? (??%)) cases (half_add ???) #x1 #x2 whd in ⊢ (??? (??%))
1580 cases (split ????) #y1 #y2 whd in ⊢ (??? (??%)) cases (half_add ???) #z1 #z2
1581 whd in ⊢ (??? (??%)) whd in ⊢ (??? (??%)) cases (split ????) #w1 #w2
1582 whd in ⊢ (??? (??%)) cases (get_index_v bool ????) whd in ⊢ (??? (??%))
1583 (* CSC: ??? *)
1584 | #dptr #label (* CSC: ??? *)
1585 ]
1586 cases not_implemented
1587qed.
1588*)
1589
1590lemma status_of_pseudo_status_failure_depends_only_on_code_memory:
1591 ∀ps,ps': PseudoStatus.
1592 code_memory … ps = code_memory … ps' →
1593 match status_of_pseudo_status ps with
1594 [ None ⇒ status_of_pseudo_status ps' = None …
1595 | Some _ ⇒ ∃w. status_of_pseudo_status ps' = Some … w
1596 ].
1597 #ps #ps' #H whd in ⊢ (match % with [ _ ⇒ ? | _ ⇒ ? ])
1598 generalize in match (refl … (assembly (code_memory … ps)))
1599 cases (assembly ?) in ⊢ (???% → %)
1600 [ #K whd whd in ⊢ (??%?) <H >K %
1601 | #x #K whd whd in ⊢ (?? (λ_.??%?)) <H >K % [2: % ] ]
1602qed.
1603
1604definition ticks_of: pseudo_assembly_program → Word → nat × nat ≝
1605 λprogram: pseudo_assembly_program.
1606 λppc: Word.
1607 let 〈preamble, pseudo〉 ≝ program in
1608 let 〈fetched, new_ppc〉 ≝ fetch_pseudo_instruction pseudo ppc in
1609 match fetched with
1610 [ Instruction instr ⇒
1611 match instr with
1612 [ JC lbl ⇒
1613 match jump_expansion ppc program with
1614 [ short_jump ⇒ 〈2, 2〉
1615 | medium_jump ⇒ ?
1616 | long_jump ⇒ 〈4, 4〉
1617 ]
1618 | JNC lbl ⇒
1619 match jump_expansion ppc program with
1620 [ short_jump ⇒ 〈2, 2〉
1621 | medium_jump ⇒ ?
1622 | long_jump ⇒ 〈4, 4〉
1623 ]
1624 | JB bit lbl ⇒
1625 match jump_expansion ppc program with
1626 [ short_jump ⇒ 〈2, 2〉
1627 | medium_jump ⇒ ?
1628 | long_jump ⇒ 〈4, 4〉
1629 ]
1630 | JNB bit lbl ⇒
1631 match jump_expansion ppc program with
1632 [ short_jump ⇒ 〈2, 2〉
1633 | medium_jump ⇒ ?
1634 | long_jump ⇒ 〈4, 4〉
1635 ]
1636 | JBC bit lbl ⇒
1637 match jump_expansion ppc program with
1638 [ short_jump ⇒ 〈2, 2〉
1639 | medium_jump ⇒ ?
1640 | long_jump ⇒ 〈4, 4〉
1641 ]
1642 | JZ lbl ⇒
1643 match jump_expansion ppc program with
1644 [ short_jump ⇒ 〈2, 2〉
1645 | medium_jump ⇒ ?
1646 | long_jump ⇒ 〈4, 4〉
1647 ]
1648 | JNZ lbl ⇒
1649 match jump_expansion ppc program with
1650 [ short_jump ⇒ 〈2, 2〉
1651 | medium_jump ⇒ ?
1652 | long_jump ⇒ 〈4, 4〉
1653 ]
1654 | CJNE arg lbl ⇒
1655 match jump_expansion ppc program with
1656 [ short_jump ⇒ 〈2, 2〉
1657 | medium_jump ⇒ ?
1658 | long_jump ⇒ 〈4, 4〉
1659 ]
1660 | DJNZ arg lbl ⇒
1661 match jump_expansion ppc program with
1662 [ short_jump ⇒ 〈2, 2〉
1663 | medium_jump ⇒ ?
1664 | long_jump ⇒ 〈4, 4〉
1665 ]
1666 | ADD arg1 arg2 ⇒
1667 let trivial_code_memory ≝ assembly1 (ADD ? arg1 arg2) in
1668 let trivial_status ≝ load_code_memory trivial_code_memory in
1669 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1670 〈0, ticks〉
1671 | ADDC arg1 arg2 ⇒
1672 let trivial_code_memory ≝ assembly1 (ADDC ? arg1 arg2) in
1673 let trivial_status ≝ load_code_memory trivial_code_memory in
1674 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1675 〈0, ticks〉
1676 | SUBB arg1 arg2 ⇒
1677 let trivial_code_memory ≝ assembly1 (SUBB ? arg1 arg2) in
1678 let trivial_status ≝ load_code_memory trivial_code_memory in
1679 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1680 〈0, ticks〉
1681 | INC arg ⇒
1682 let trivial_code_memory ≝ assembly1 (INC ? arg) in
1683 let trivial_status ≝ load_code_memory trivial_code_memory in
1684 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1685 〈0, ticks〉
1686 | DEC arg ⇒
1687 let trivial_code_memory ≝ assembly1 (DEC ? arg) in
1688 let trivial_status ≝ load_code_memory trivial_code_memory in
1689 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1690 〈0, ticks〉
1691 | MUL arg1 arg2 ⇒
1692 let trivial_code_memory ≝ assembly1 (MUL ? arg1 arg2) in
1693 let trivial_status ≝ load_code_memory trivial_code_memory in
1694 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1695 〈0, ticks〉
1696 | DIV arg1 arg2 ⇒
1697 let trivial_code_memory ≝ assembly1 (DIV ? arg1 arg2) in
1698 let trivial_status ≝ load_code_memory trivial_code_memory in
1699 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1700 〈0, ticks〉
1701 | DA arg ⇒
1702 let trivial_code_memory ≝ assembly1 (DA ? arg) in
1703 let trivial_status ≝ load_code_memory trivial_code_memory in
1704 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1705 〈0, ticks〉
1706 | ANL arg ⇒
1707 let trivial_code_memory ≝ assembly1 (ANL ? arg) in
1708 let trivial_status ≝ load_code_memory trivial_code_memory in
1709 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1710 〈0, ticks〉
1711 | ORL arg ⇒
1712 let trivial_code_memory ≝ assembly1 (ORL ? arg) in
1713 let trivial_status ≝ load_code_memory trivial_code_memory in
1714 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1715 〈0, ticks〉
1716 | XRL arg ⇒
1717 let trivial_code_memory ≝ assembly1 (XRL ? arg) in
1718 let trivial_status ≝ load_code_memory trivial_code_memory in
1719 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1720 〈0, ticks〉
1721 | CLR arg ⇒
1722 let trivial_code_memory ≝ assembly1 (CLR ? arg) in
1723 let trivial_status ≝ load_code_memory trivial_code_memory in
1724 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1725 〈0, ticks〉
1726 | CPL arg ⇒
1727 let trivial_code_memory ≝ assembly1 (CPL ? arg) in
1728 let trivial_status ≝ load_code_memory trivial_code_memory in
1729 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1730 〈0, ticks〉
1731 | RL arg ⇒
1732 let trivial_code_memory ≝ assembly1 (RL ? arg) in
1733 let trivial_status ≝ load_code_memory trivial_code_memory in
1734 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1735 〈0, ticks〉
1736 | RLC arg ⇒
1737 let trivial_code_memory ≝ assembly1 (RLC ? arg) in
1738 let trivial_status ≝ load_code_memory trivial_code_memory in
1739 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1740 〈0, ticks〉
1741 | RR arg ⇒
1742 let trivial_code_memory ≝ assembly1 (RR ? arg) in
1743 let trivial_status ≝ load_code_memory trivial_code_memory in
1744 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1745 〈0, ticks〉
1746 | RRC arg ⇒
1747 let trivial_code_memory ≝ assembly1 (RRC ? arg) in
1748 let trivial_status ≝ load_code_memory trivial_code_memory in
1749 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1750 〈0, ticks〉
1751 | SWAP arg ⇒
1752 let trivial_code_memory ≝ assembly1 (SWAP ? arg) in
1753 let trivial_status ≝ load_code_memory trivial_code_memory in
1754 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1755 〈0, ticks〉
1756 | MOV arg ⇒
1757 let trivial_code_memory ≝ assembly1 (MOV ? arg) in
1758 let trivial_status ≝ load_code_memory trivial_code_memory in
1759 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1760 〈0, ticks〉
1761 | MOVX arg ⇒
1762 let trivial_code_memory ≝ assembly1 (MOVX ? arg) in
1763 let trivial_status ≝ load_code_memory trivial_code_memory in
1764 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1765 〈0, ticks〉
1766 | SETB arg ⇒
1767 let trivial_code_memory ≝ assembly1 (SETB ? arg) in
1768 let trivial_status ≝ load_code_memory trivial_code_memory in
1769 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1770 〈0, ticks〉
1771 | PUSH arg ⇒
1772 let trivial_code_memory ≝ assembly1 (PUSH ? arg) in
1773 let trivial_status ≝ load_code_memory trivial_code_memory in
1774 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1775 〈0, ticks〉
1776 | POP arg ⇒
1777 let trivial_code_memory ≝ assembly1 (POP ? arg) in
1778 let trivial_status ≝ load_code_memory trivial_code_memory in
1779 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1780 〈0, ticks〉
1781 | XCH arg1 arg2 ⇒
1782 let trivial_code_memory ≝ assembly1 (XCH ? arg1 arg2) in
1783 let trivial_status ≝ load_code_memory trivial_code_memory in
1784 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1785 〈0, ticks〉
1786 | XCHD arg1 arg2 ⇒
1787 let trivial_code_memory ≝ assembly1 (XCHD ? arg1 arg2) in
1788 let trivial_status ≝ load_code_memory trivial_code_memory in
1789 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1790 〈0, ticks〉
1791 | RET ⇒
1792 let trivial_code_memory ≝ assembly1 (RET ?) in
1793 let trivial_status ≝ load_code_memory trivial_code_memory in
1794 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1795 〈0, ticks〉
1796 | RETI ⇒
1797 let trivial_code_memory ≝ assembly1 (RETI ?) in
1798 let trivial_status ≝ load_code_memory trivial_code_memory in
1799 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1800 〈0, ticks〉
1801 | NOP ⇒
1802 let trivial_code_memory ≝ assembly1 (NOP ?) in
1803 let trivial_status ≝ load_code_memory trivial_code_memory in
1804 let ticks ≝ \snd (fetch trivial_status (zero ?)) in
1805 〈0, ticks〉
1806 ]
1807 | Comment comment ⇒ 〈0, 0〉
1808 | Cost cost ⇒ 〈0, 0〉
1809 | Jmp jmp ⇒ 〈0, 2〉
1810 | Call call ⇒ 〈0, 2〉
1811 | Mov dptr tgt ⇒ 〈0, 2〉
1812 ].
1813 cases not_implemented (* policy returned medium_jump for conditional jumping = impossible *)
1814qed.
1815
1816lemma main_thm:
1817 ∀ticks_of.
1818 ∀ps,s,s''.
1819 status_of_pseudo_status ps = Some … s →
1820 status_of_pseudo_status (execute_1_pseudo_instruction ticks_of ps) = Some … s'' →
1821 ∃n. execute n s = s''.
1822 #ticks_of #ps #s #s''
1823 generalize in match (fetch_assembly_pseudo2 (code_memory … ps))
1824 whd in ⊢ (? → ??%? → ??%? → ?)
1825 >execute_1_pseudo_instruction_preserves_code_memory
1826 generalize in match (refl … (assembly (code_memory … ps)))
1827 generalize in match (assembly (code_memory … ps)) in ⊢ (??%? → %) #ASS whd in ⊢ (???% → ?)
1828 cases (build_maps (code_memory … ps))
1829 [ cases (code_memory … ps) #preamble #instr_list whd in ⊢ (???% → ?) #EQ >EQ #H
1830 #abs @⊥ normalize in abs; destruct (abs) ]
1831 * #labels #costs generalize in match (refl … (code_memory … ps))
1832 cases (code_memory … ps) in ⊢ (???% → %) #preamble #instr_list #EQ normalize nodelta;
1833 generalize in ⊢ (???(match % with [_ ⇒ ? | _ ⇒ ?]) → ?) *; normalize nodelta;
1834 [ #EQ >EQ #_ #abs @⊥ normalize in abs; destruct (abs) ]
1835 * #final_ppc * #final_pc #assembled #EQ >EQ -EQ ASS; normalize nodelta;
1836 #H generalize in match (H ??? (refl …) (refl …)) -H; #H;
1837 #H1 generalize in match (option_destruct_Some ??? H1) -H1; #H1 <H1 -H1;
1838 #H2 generalize in match (option_destruct_Some ??? H2) -H2; #H2 <H2 -H2;
1839 change with
1840 (∃n.
1841 execute n (set_program_counter ? (set_code_memory ?? ps (load_code_memory assembled)) ?)
1842 = set_program_counter ? (set_code_memory ?? (execute_1_pseudo_instruction ticks_of ps) (load_code_memory assembled)) ?)
1843 whd in match (\snd 〈preamble,instr_list〉) in H;
1844 whd in match (\fst 〈preamble,instr_list〉) in H;
1845 whd in match (\snd 〈final_pc,assembled〉) in H;
1846 -s s'' labels costs final_ppc final_pc;
1847 letin ps' ≝ (execute_1_pseudo_instruction ticks_of ps)
1848 (* NICE STATEMENT HERE *)
1849 generalize in match (refl … ps') generalize in ⊢ (??%? → ?) normalize nodelta; -ps'; #ps'
1850 #K <K generalize in match K; -K;
1851 (* STATEMENT WITH EQUALITY HERE *)
1852 whd in ⊢ (???% → ?) generalize in match (H (program_counter … ps)) >EQ normalize nodelta;
1853 cases (fetch_pseudo_instruction instr_list (program_counter … ps))
1854 #pi #newppc normalize nodelta; * #instructions *;
1855 cases pi normalize nodelta;
1856 [ (* Instruction *)
1857 | (* Comment *) #comment #H1 #H2 #EQ %[@0] >EQ
1858 whd in ⊢ (???(???(??%)))
1859 normalize in H1; generalize in match (option_destruct_Some ??? H1) #K1
1860 >K1 in H2; whd in ⊢ (% → ?) #H2 <(eq_bv_to_eq … H2)
1861
1862 | (* Cost *)
1863 | (* Jmp *)
1864 | (* Call *)
1865 | (* Mov *)
1866 ]
1867qed.
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# Outliers, Part II: Pitfalls in Detecting Outliers
Article
Spectroscopy
SpectroscopySpectroscopy-02-01-2018
Volume 33
Issue 2
Pages: 24–38
How can you detect the presence of an outlier when it is mixed with multiple other, similar, samples?
This column is part II of our series dealing with the question of outliers. Here we consider some of the methods that have been devised to detect the presence of an outlier when it is mixed in with multiple other, similar, samples.
Whether we did it well or badly, in the previous column (1) we discussed the concept of what an outlier is. The next point to consider is how to detect an outlier, in the face of the inherent fuzziness of the concept. This means that we will not always be correct in the assessment of any particular suspect datum, but here we will look at various situations and seek universal methods (which may or may not exist).
A common way to detect outliers is to plot the data and look for readings that are "far away" (in some sense, as judged by the viewer's eye) from the behavior of the rest of the data in the given set under consideration. In particular, if the data being examined are the result of a multivariate calibration and the computer is plotting the predicted values from the calibration model versus the reference laboratory results (or even better, the residuals from the calibration versus the reference laboratory results), we expect the plot to follow what is called the "45° line" (or a flat line, if the residuals are being plotted), with the random error causing the individual data points being normally distributed (that is, conforming to the Gaussian distribution; all occurrences of "normal" in this column have that meaning) around that line. This isn't a bad way to examine calibration results, as long as you're aware of the traps and pitfalls that abound.
Figure 1: The theoretical normal distribution.
For example, as a statistical problem, we must be aware that we have to deal in probabilities. For instance, everyone knows that for a normal distribution, 95% of the readings should be within two standard deviations of the center of the distribution, and 99% of readings should be within three standard deviations. Anything beyond those limits is considered the "forbidden zone," in which no samples should be found. So the conclusion jumped to is that a reading beyond three standard deviations is unlikely to occur, and therefore any reading beyond three standard deviations must be an outlier. Figure 1 shows the normal distribution, as a function of the standard deviation, and we can see how, beyond ±2 standard deviations, the probability of obtaining a value in that region becomes very small. What is missing in that line of reasoning is the fact that the probability levels given for various values of the normal distribution correspond to the probability of getting that value when selecting a single random sample from the distribution. When more than one sample is taken into consideration, the probability of one or more samples falling in the specified region increases since any sample can fall into the "forbidden zone" independently of the other samples. We have discussed this effect previously (see chapter 1 in reference 2).
In this situation, the actual probability of any of a multiplicity of samples falling into the forbidden zone is given by the following equation:
where Pi is the probability of any individual reading being within the allowed limits (generally 0.95), P is the actual probability that any reading from the set is outside the limit, and n is the number of readings (or data points).
For a single sample, then, if a single (random) sample has a probability of 0.95 of being within the "normal" range then the probability of being outside that range (that is, in the "forbidden" region) is 0.05. This probability is considered small enough to declare that a sample found in the forbidden region must have something wrong with it.
As a numerical example, if we have 20 readings from a set of (presumably normally distributed) errors and we are concerned about the probability of a reading being beyond the ±2 standard deviation limit, then we can substitute the numbers into equation 1 to get
That is, almost two-thirds of the time, at least one reading will fall in the "forbidden" zone beyond two standard deviations, because of the random variability of the data alone, with no unusual special circumstance affecting it. This is hardly proof of something unusual about that reading.
For 100 readings the probability becomes 0.9941-almost a certainty that at least one sample will be beyond the specified limit. Thus, using this criterion to identify outliers will ensure that perfectly acceptable readings will be falsely identified as outliers.
Another way to look at the situation is that a probability of 5% means that one sample in 20 will fall in the region beyond ±2 standard deviations. If we have 100 samples, therefore, roughly five of them should be in that "forbidden" region of the distribution; if there are no samples there, something is wrong-fully five samples should be in the "forbidden" zone. To have none there is the "rare event."
Can anything be done? Answer: yes. We can reset the probability test levels to correspond to a true 0.95 probability for a set of multiple samples by solving equation 1 for Pi:
In words, equation 2 tells us that for n readings, we have to perform our statistical tests at a probability level that is the nth root of the desired probability. Thus, if we have 100 readings (residuals, say) and we want to test at a true value of 95% probability, then we must perform the test against the number of standard deviations corresponding to 0.95(1/100) = 0.9995, which corresponds roughly to 3.5 standard deviations. This means that if your sample set contains 100 samples worth of data and you want to use this type of test for detecting outliers, or any test that is supposed to be based on single readings, you have to set the threshold for the test at 3.5 standard deviations to get a true significance level of 0.95. Since the probability level to use depends on the number of samples, this test would quickly become unwieldy for routine use.
Another pitfall is encountered when we consider the actual distribution of data randomly sampled from a normal distribution. Figure 1 presents a very pretty display of the normal distribution, representing the mathematically ideal situation. Reality, we're afraid, isn't nearly so neat. In Figure 2 we used the MATLAB "randn" function to create datasets of various sizes, all taken randomly from the normal distribution, then plotted the histogram of the actual distributions for each such (relatively small) sample taken. These randomly selected, normally distributed numbers are representative of the behavior of errors in a calibration situation. For each sample size, the selection and histograms were repeated five times, to illustrate the variability caused by the randomness. We plotted the distribution histograms as line graphs to avoid the plots becoming too busy. The numbers of samples used (30 to 500) were chosen to more-or-less correspond to the range of number of samples typically used for near-infrared (NIR) calibrations.
Figure 2: Actual distributions (histograms) of samples of various numbers of readings taken from the normal distribution, showing the actual distributions of values achieved, for each case. For each part of this figure the random sampling was repeated five times, to allow the variability of the data to be seen. The plots show distributions of (a) 30, (b) 100, and (c) 500 random readings.
We see that when only 30 samples are used it's almost impossible to tell that the underlying distribution is normal (rather than, say, uniform), much less set meaningful confidence limits for outlier detection. For 100 samples we can tell that the probability maximum is somewhere around the middle of the set of data, but it still doesn't look like anything close to normal. At 500 samples we see that some tails are starting to be noticeable, but just barely.
As far as detecting outliers goes, however, the numbers of samples used for these demonstrations are sufficient to show the variability of the distributions and their likely effects on setting thresholds, but these histograms, using only 10 intervals for the histogram, are insufficient for us to examine what is happening at the extremes of the range, where the information that is interesting to us for detecting outliers, will be found. We will now, therefore, examine examples of this same distribution using different numbers of bins in the histogram. The histograms in Figure 2 all used 10 bins to generate the histogram. In Figure 3 we use varying numbers of bins to examine the effects of the binning on the distribution of 500 samples from a randomly sampled normal distribution. In Figure 3 we also use bar graphs to plot the histograms, giving us the more familiar presentation of the histogram; for this reason we only present the histogram from a single set of this simulated data.
Figure 3: Histograms for 500 normally distributed numbers (representing errors in a calibration) in histograms using (a) 10, (b) 20, (c) 50, and (d) 70 bins. These histograms were all generated from the same set of random numbers.
In Figure 3, we see that binning the data into 10, and even 20 bins, pretty well reproduces what we saw in Figure 2: The distribution appears normal to the eye (albeit slightly asymmetric, to be sure). With 50 or more bins, however, we can see the erratic nature of the distribution at the two extremes of the range, with relatively large sets of empty bins sometimes appearing between bins containing samples.
So, how does all this affect our methods of analyzing calibration plots? Figure 4 shows a simulated calibration plot using synthetic data representing 100 samples with characteristics similar to what are found in many calibration datasets. Both the x- and y-data were based on a uniform distribution of samples with a range going from 10 to 30; this is represented by the solid line. Random normally distributed numbers representing calibration errors with a standard deviation or 2 were then added to the y-data, and the individual data points were plotted as blue circles.
Figure 4: A simulated calibration plot showing possible outliers. See text for discussion.
Several of the plotted points appear to be suspect, based on the visual observations of the plot in Figure 4. These questionable data points include the point at x = 28.8, y = 33.9 and the cluster of three points around x = 12.0, y = 18.0, as well as a couple of data points with highly negative errors. An overeager (and probably rushed) data analyst would summarily reject those data points, deleting them from the data set.
Are these points outliers? In this case we know for a fact that they are not, since we know that these "error" terms are all legitimate parts of the normal distribution that the errors presumably comprise. The purpose of this exercise is to caution the reader against too-freely identifying and deleting marginal or suspect samples as outliers. This sort of excessive outlier deletion is one of the more likely causes of the large discrepancies seen between calibration performance and validation results. The elimination of samples from a calibration set simply because they have what appears to be large errors will ipso facto make the calibration results appear to be better than they actually are. Then, when more realistic validation values are obtained, they seem to be the source of the discrepancies seen between the two types of calibration performance, when the discrepancy is actually because of the too-small calibration error estimate created by the elimination of the larger-value errors despite their being completely legitimate.
The first part of this column, up to this point, has essentially been a warning against overzealous application of simplistic considerations, creating the danger of making what statisticians call the type 2 error: falsely identifying outliers when none exist. So then, how can we detect an outlier? Books such as the one we referenced previously (3) provide several approaches to solving this problem. Not surprisingly, they use methods other than simply counting standard deviations from the mean. Barnett and colleagues (3) also make it clear that the existence of an outlier also depends very much on the distribution of the data that the suspected outlier is supposed to fit. Our examples in Figures 1–3 are predicated on the reference distribution being the normal distribution.
In general, however, not all data are normally distributed, and any tests applied to non-normal data must either be suitably chosen to match the distribution or at least have their critical values adjusted to take into account the actual data distribution. For our examples here, however, we will consider the normal distribution to be our reference distribution. This is in keeping with our expectation in writing this column that the majority of our readership is interested in the application of the concepts we discuss to quantitative spectroscopic analysis. One of the fundamental assumptions behind the common algorithms used for that type of analysis is that the errors in the reference values are normally distributed. For other applications, of course, caveat emptor: You must properly deal with the nature of the data you collect.
Much of what we discuss here will be based on the text at hand (3). As we might expect from our recent discussions of the history of statistics (4), the early mathematicians developing the foundations of statistics were also active in these related fields of data analysis. Hence some of the same familiar names appear such as Bernoulli, Legendre, Bessel, as well as many less-familiar names such as Stone, Boscovich, Goodwin, and Chauvenet. For the most part, informal, ad hoc procedures for examining the data were recommended, much as is done today; this approach was popular until as recently as WWI. Barnett and colleagues (3) also discuss some of these early, and simplistic approaches to identifying outliers. Problems abound, and are often both subtle and disastrous. For example, Barnett and colleagues cite one of the earliest attempts to develop an objective outlier test (5) based directly on probability theory, which turned out to be very effective in identifying outliers. However, the test under consideration had the unfortunate side effect of also rejecting as many as 40% of the valid observations.
Sometimes a test having those characteristics may even be satisfactory. While we wouldn't expect this to be common, tests with asymmetric characteristics like this may sometimes be useful, depending on extraneous conditions. The test cited by Pierce, for example, could be useful if all of several conditions apply: It is critical to ensure that all outliers are removed from the data to be analyzed, there is enough data, and the data contain enough redundancy, that the loss of a few (or even several) samples' worth of data will not affect the results of the analysis. Barnett also discusses, at length, various considerations that may affect the outlier detection algorithms that might be used. Some of these considerations include the purpose for which the data will be used (as we discussed just above), the underlying distribution (or expected distribution) of the data (or of the errors, as appropriate), multiple outliers, multivariate data, and the expected action to be taken upon identifying an outlier, and others. We will not discuss these at length here, although we may address some of them briefly in passing.
Most informal, ad hoc, statistical tests for outliers, such as the one described in Figure 4, are dependent on setting a threshold for some calculated statistic to which values for that statistic for each sample are compared, and declaring an outlier when the sample's value for the statistic exceeds the threshold. By their nature, these tests are tests of the extreme samples. This type of test is basically a specialized form of hypothesis test, as described previously. Important features of outlier tests of this form include the probability of failing to detect an outlier when one is present, the probability of falsely detecting an outlier when none is present, and the immunity of the test to changes in scale of the data. Furthermore, to be considered a legitimate statistical test, the user must be able to specify the probability level at which the outlier is declared, as well as the probability for a legitimate sample to be falsely declared an outlier. Another key point that the user of any of these tests must keep in mind is whether the calculation of the reference value is to include the suspect datum. Clearly, the data point in question, being an extreme value of the data set, will have the maximum effect on the value of the threshold comparison value, compared to all the other data, so whether it is included or excluded from the calculations becomes a very important consideration.
Since we are concerned with the extreme samples in a set of data, we must keep in mind the truism that any finite data set has to have at least two extreme values. Moreover, we had previously uncovered possible pitfalls to this kind of test, in that those extreme samples are not necessarily outliers just by virtue of their characteristic of being extreme. Over the years, however, statisticians have developed ways of producing tests of the extreme samples that avoid that pitfall. Here we discuss some of them. As a reference plot for visual comparison we created a new plot as shown in Figure 5, where we have recreated the distribution of Figure 3c and artificially added an extra data point to represent an outlier. This is representative of the various tests that are proposed.
Figure 5: Histograms for 500 normally distributed numbers (representing errors in a calibration) in a histogram containing 500 data points, plus one simulated outlier.
The tests are based on the concept that there are n data points, numbered 1 to n in order. As shown in Figure 5, the first data point is labeled X1 while the last data point is labeled Xn, and the next-to-last data point is Xn-1. Then several possible alternate ways of comparing the last data point, which is presumably the suspected outlier, are proposed. As seen in Figure 5, we have labeled the difference between the most extreme reading and the second-most extreme, to emphasize that many of the tests are predicated on the comparison of the distance of the most extreme sample from other readings, rather than the distance from the center of the data, as we are usually concerned with.
Barnett and colleagues (3) present some of the simpler outlier tests that have been developed over the years, and point us to sources of corresponding threshold values for the statistics calculated by the various tests (T represents the test statistic that will be used for comparison with a threshold [critical value]).
1. T = (XnXn-1)/(XnX2) (6). This test compares the distance of the suspected outlier from the next-highest value, to the range of the sample data. Note that the first sample is not included in the range to avoid the possibility that it is also an outlier, which would then give an incorrect threshold test value.
2. T = (XnXn-1)/σ (7). This test is similar to test 1 except that the denominator contains the population standard deviation (which must therefore be known) instead of the sample range.
3. T = (XnXn-1)/σ (8). Another similar test, but using the full range from the outlier in the numerator, and replacing the population standard deviation with the sample standard deviation. This test is subject to having the requirement that X1 is not an outlier itself.
4. T = (X(with macron)Xn-1)/S (9). Testing for a lower outlier using the sample mean.
this is a more exotic formula, for testing two suspected outliers simultaneously (9).
Dixon's test functions for rejection of outliers, along with a table of critical values, has been nicely presented in an ASTM practice (10) (see Annex A1). These are a set of equations similar to equation 1 above that can be used to test large sets of data for multiple outliers.
Barnett (3) also points to some formulas using higher-order measures (skew and kurtosis) of the data, for the purpose of testing for outliers. Variations of these formulas, to avoid distortions from other outliers in the data set or some of the problems listed earlier, are obvious, although the adjustments needed for the threshold values may not be.
The above-listed outlier tests are all intended to be used for situations where the underlying distribution of values for nondiscordant readings is normal (Gaussian).
Barnett (3) also lists another extensive set of tests for discordancy from a normal distribution, for a variety of conditions. That book also presents an extensive set of tests for situations where the underlying distribution is other than normal. We present some of those here also, for reference.
For values that are gamma-distributed or exponentially distributed: Barnett provides 16 different tests suitable for various conditions. As with other tests, here we only include one of each type of test that is discussed:
The test statistic for a single upper outlier of an exponential sample is
For values that are normally distributed, Barnett lists 30 different tests, some of which duplicate the ones listed above, so we forbear repeating them.
For values that are log-normally distributed, take the logarithms of the data and apply a suitable test for data that is normally distributed.
For values that are uniformly distributed, one test uses the following test statistic:
Barnett also presents tests for situations where the underlying data distribution is Poisson, binomial, truncated exponential, Pareto, Weibull, and other unusual and exotic distributions, but we forbear to include the tests for these cases. The alert reader will have noticed that several of the tests described are based on using the same formula for calculating the test statistic. In these cases, the differences between the tests is not in the test statistic itself, but in the values corresponding to different critical (threshold) values of the test. This is not very surprising, since it is just those values corresponding to different probabilities that defines a given distribution.
### Conclusion
We will continue our discussion of outliers in a future column of "Chemometrics in Spectroscopy," coming soon to your mailbox.
### References
(1) H. Mark and J. Workman, Spectroscopy 32(6), 22–25 (2017).
(2) H. Mark and J. Workman, Statistics in Spectroscopy , 2nd Ed. (Academic Press; New York, New York, 2003).
(3) V. Barnett and T. Lewis, Outliers in Statistical Data (John Wiley & Sons, New York, New York, 1978).
(4) H. Mark and J. Workman, Spectroscopy 30(10), 26–31 (2015).
(5) B. Pierce, Astr. J. 2(19), 161–163 (1852).
(6) W.J. Dixon, Ann. Math. Statist. 22, 68–78 (1951).
(7) J.O. Irwin, Biometrika 17, 238–250 (1925).
(8) H.A. David, H.O. Hartley, and E.S. Pearson, Biometrika 41, 482–493 (1954).
(9) F.E. Grubbs, Ann. Math Statist. 21, 27–58 (1950).
(10) E1655-05, A, Standard Practices for Infrared Multivariate Quantitative Analysis (American Society for Testing and Materials).
Jerome Workman Jr.
Jerome Workman Jr. serves on the Editorial Advisory Board of Spectroscopy and is the Executive Vice President of Engineering at Unity Scientific, LLC, in Milford, Massachusetts. He is also an adjunct professor at U.S. National University in La Jolla, California, and Liberty University in Lynchburg, Virginia.
Howard Mark
Howard Mark serves on the Editorial Advisory Board of Spectroscopy and runs a consulting service, Mark Electronics, in Suffern, New York. Direct correspondence to: SpectroscopyEdit@UBM.com
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View Full Version : Testing a circuit
Dr.No
07-30-2008, 09:49 AM
I'm using a BS2 to test when a an external circuit is running or not.
The circuit carries 5VDC (+)and is hooked up to the IN1 (pin1) via a 220 resistor
To complete the connection the (-) circuit is grounded to the Vss (Homework board).
Everything works except I'm not picking up any 1s on the IN1 except 0s.
If anyone can give an help, I would greatly appreciate it.
Thanks.
Franklin
07-30-2008, 10:17 AM
Could we see your code?
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
- Stephen
Mike Green
07-30-2008, 10:20 AM
Of course the other possibility is that the external circuit is not in fact carrying +5V to the Stamp I/O pin via the 220 Ohm resistor. Have you measured the voltage at pin 1?
Dr.No
07-30-2008, 11:20 AM
Thanks a lot.
I had the pin on output!
Which I quickly corrected.
I noticed however (I have a indicator led that lights up when
circuit is on) that the led keeps blinking (low) periodically even
when output pin is set to low. Perhaps LOW means
a very low current?
Anyone knows what's the lowest detectable VOLT for an
input pin?
Thanks again so much.
Beau Schwabe
07-30-2008, 01:05 PM
Dr.No,
"LOW" sets the pin to LOW; it does not mean low current.... Depending on what your program looks like and how it terminates can cause a pin set as an output to periodically "blink".
Please post the code that you have so far so that we can positively determine and confirm this behavior.
The voltage threshold is around 1.4 Volts... anything above 1.4 Volts registers as a logic HIGH ("1"), anything below 1.4 Volts registers as a logic LOW ("0").
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Beau Schwabe (mailto:bschwabe@parallax.com)
IC Layout Engineer
Parallax, Inc.
Dr.No
07-30-2008, 07:53 PM
' {\$STAMP BS2}
Start:
DEBUG "Initializing..."
OUTL =%00000000 'set 7-0 to low
DIRL =%11111111 'set said to output
PAUSE 5000
DEBUG "Running"
INPUT 1
CIRCUITOFF:
LOW 11
PAUSE 500
'During stage1, LED blinks.
stage1: 'Pin 11 is connected to LED
LOW 11 'added to turn off led (LED now only blinks dimly)
IF IN1=1 THEN CIRCUITON
GOTO stage1
CIRCUITON:
HIGH 11 'turn on LED, circuit is running
stage2:
IF IN1=0 THEN CIRCUITOFF
GOTO stage2
Beau Schwabe
07-30-2008, 09:30 PM
Dr.No (http://forums.parallax.com/member.php?u=52756),
Sounds like you need a pull-up or pull-down resistor on your input pin....
Based on this sentence you gave earlier...
"Everything works except I'm not picking up any 1s on the IN1 except 0s."
...It sounds like you need a pull-up resistor on your input
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Beau Schwabe (mailto:bschwabe@parallax.com)
IC Layout Engineer
Parallax, Inc.
Dr.No
07-30-2008, 10:13 PM
Oh, it's working fine now... I just was curious about the dimly lit blinking LED.
Thanks.
Dr.No
08-01-2008, 07:03 AM
This post if from What's Really Going On Here - Voltage Drop - the anomolies of voltage and current. Duplicate post
__________________________________________________ ________________________
The diodes was my first bet for the voltage drop. I was able to reduce the 5volts to less than 0.5volts
and the stamp was still able to detect it. So, the stamp has a way of picking up
that's it's really 5volts because it can't pickup 1.3volts. My guess it's in the current (ma).
Using 1.3volts without diodes, the stamp did not pick it up.
Someone told me about the 74LS IC family of chip. I haven't found a good link yet and then there's the coil but
all this will probably equate to the same thing as the diodes.
Can anyone please tell me what does a schematic have to do with finding a component that can drop volts?
Mike Green
08-01-2008, 07:14 AM
There are many ways to decrease the voltage at a particular point in a circuit. Which to choose depends on the circuit its embedded in and how that voltage is to be used. Some ways might work terribly or not at all in a particular circuit, yet work perfectly well in some other circuit. Some ways might damage the parts around them in certain contexts.
Dr.No
08-02-2008, 04:10 AM
Stamp is able to pick up less than 0.5 volt [tested] on input pin with 5.0 base volt.
Stamp won't pick up 1.3 base volt.
Mike Green
08-02-2008, 05:07 AM
I can't tell from your diagram just what's in the area marked "diodes". I assume it's 4 series connected silicon switching diodes with the anodes towards the battery and the cathodes towards the Stamp.
If you look at the datasheets for the diodes, you'll probably find that the forward voltage (Vf) drop across the diode depends on the current. At very low currents (like here) it will tend towards 0.6V per diode. 4 x 0.6V = 2.4V. If your supply truly is 5V, that will leave 2.6V at the Stamp pin which the Stamp will see as logic one/high.
Dr.No
08-02-2008, 07:12 AM
I do not have the tools to draw an accurate diagram. I use the paint program as best as I could.
This was only a simple diagram for show. I actually used over 25 diodes because no matter how
low the volts was, the stamp was STILL able to pick it up. I used a multimeter from radio shack
because I couldn't believe the stamp was still picking it up. The volt meter reading was 5V at the
battery and < 0.5 volts after the diodes.
25 x 0.6 =15
Moreover, I happen to reverse a single diode to stop the current from flowing at the pin and the stamp
still lit up the LED.
Pin O---/\/\/\/\---------|>|------|<|---|<|---|<|---|<|-----(+) Battery
The idea was simply to drop the volts low enough so that it can't be detected by the stamp
according to the advertised low point of < 1.4 volts.
The basis for this is from experiments in the What's a microcontroller book
chapter 2 & 3 Lights On Light Off & Digital Input
What's a microcontroller anyway if you cannot use it?
Post Edited (Dr.No) : 8/2/2008 12:26:08 AM GMT
Franklin
08-02-2008, 07:34 AM
A microcontroller is a device which, in the hands of a competent person can do everything within it's design parameters relying on said competent person to decide if those parameters fit their needs.
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- Stephen
PJ Allen
08-02-2008, 07:47 AM
http://i270.photobucket.com/albums/jj118/new_clear_days/circuits/dividers.jpg
allanlane5
08-02-2008, 08:19 AM
The BS2 has an input impedance of about 10 MegOhms. So you're only going to get:
(Let's see -- V == I * R, V/R == I, 5 / 10e6 == 0.5 micro-amps)
Yeah, I don't think you can assume a diode has a 0.6 volt drop at 1/2 micro-amp.
What you DO need is a 10K resistor to ground. We call this a "pull-down" resistor. This gives the input pin a 'soft' ground, which should prevent the input pin from reading any RF signal when very little current is running.
Dr.No
08-02-2008, 09:04 AM
Ok, I'll try the 10K.
Thanks.
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### 2 4 solving rational equations
1. 1. General Rational Equations
2. 2. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions.
3. 3. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions.
4. 4. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions.
5. 5. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. Let A = apple, B = banana, and C = cantaloupe
6. 6. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. Let A = apple, B = banana, and C = cantaloupe The relation is 3 4 A 1 2 B += 1 3 C
7. 7. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 1 3 C Let A = apple, B = banana, and C = cantaloupe The relation is
8. 8. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Let A = apple, B = banana, and C = cantaloupe The relation is
9. 9. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Distribute it to each term. Let A = apple, B = banana, and C = cantaloupe The relation is
10. 10. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Distribute it to each term. 3 Let A = apple, B = banana, and C = cantaloupe The relation is
11. 11. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Distribute it to each term. 3 6 4 Let A = apple, B = banana, and C = cantaloupe The relation is
12. 12. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Distribute it to each term. 3 6 4 9A = 6B + 4C Let A = apple, B = banana, and C = cantaloupe The relation is
13. 13. Example A. Suppose the value of 3/4 of an apple is the same as the value of 1/2 of a banana with 1/3 of a cantaloupe, restate this relation in whole values. This is accomplished by multiplying the LCD of all the fractional terms in the equation, to each term on both sides, then clear each denominator using these multiplications. General Rational Equations An equation or relation expressed using fractions may always be restated in a way without using fractions. The LCD =12, multiply it to both sides. 3 4 A 1 2 B += 3 4 A 1 2 B +=( ) 1 3 C 1 3 C * 12 Distribute it to each term. 3 6 4 9A = 6B + 4C Hence the value of 9 apples is the same as 6 bananas and 4 cantaloupes. Let A = apple, B = banana, and C = cantaloupe The relation is
14. 14. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD.
15. 15. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2
16. 16. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2)
17. 17. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2)
18. 18. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x + 2)
19. 19. General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2)
20. 20. 3(x + 2) = 6( x – 1) General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2)
21. 21. 3(x + 2) = 6( x – 1) 3x + 6 = 6x – 6 General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2)
22. 22. 3(x + 2) = 6( x – 1) 3x + 6 = 6x – 6 6 + 6 = 6x – 3x General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2)
23. 23. 3(x + 2) = 6( x – 1) 3x + 6 = 6x – 6 6 + 6 = 6x – 3x 12 = 3x 4 = x General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2)
24. 24. 3(x + 2) = 6( x – 1) 3x + 6 = 6x – 6 6 + 6 = 6x – 3x 12 = 3x 4 = x General Rational Equations To solve an equation with fractional terms, we first clear the fractions by multiplying both sides by the LCD. Example B. Solve 3 x – 1 = 6 x + 2 Multiply both sides by the LCD = (x – 1)(x + 2) 3 x – 1 = 6 x + 2 ( ) * (x – 1)(x + 2) (x – 1)(x + 2) This is a proportional equation. Multiplying the LCD yield the same simplified equation if we cross-multiplied.
25. 25. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 + 1
26. 26. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) + 1
27. 27. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1
28. 28. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x + 1)
29. 29. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1)
30. 30. General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1)
31. 31. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1)
32. 32. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2
33. 33. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10
34. 34. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10 0 = x2 + x – 12
35. 35. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) hence x = –4, 3
36. 36. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) hence x = –4, 3 However, this method of clearing the denominator might produce a solution(s) that does not work for the original fractional equation.
37. 37. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) hence x = –4, 3 However, this method of clearing the denominator might produce a solution(s) that does not work for the original fractional equation. Specifically, we have to check that the answers obtained will not turn the denominator into 0 in the original problem.
38. 38. 2(x + 1) = 4(x – 2) + (x – 2)(x + 1) General Rational Equations Example C. Solve 2 x – 2 = 4 x + 1 Multiply both sides by the LCD : (x – 2)(x + 1) 2 x – 2 = 4 x + 1 ( ) * (x – 2)(x + 1) + 1 + 1 (x – 2)(x + 1) (x – 2)(x + 1) 2x + 2 = 4x – 8 + x2 – x – 2 2x + 2 = x2 + 3x – 10 0 = x2 + x – 12 0 = (x + 4)(x – 3) hence x = –4, 3 However, this method of clearing the denominator might produce a solution(s) that does not work for the original fractional equation. Specifically, we have to check that the answers obtained will not turn the denominator into 0 in the original problem. In this example, both x = –4, 3 are good answers because they don’t turn the denominators to 0.
39. 39. General Rational Equations We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem.
40. 40. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem.
41. 41. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem. 2 x – 2 = 4 x + 1 + Treat the 1 as . 1 1 1 1
42. 42. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem. 2 x – 2 = 4 x + 1 + Treat the 1 as . 1 1 1 1 2 x – 2 = 4 + (x + 1) x + 1
43. 43. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem. 2 x – 2 = 4 x + 1 + Treat the 1 as . 1 1 1 1 2 x – 2 = 4 + (x + 1) x + 1 2 x – 2 = x + 5 x + 1
44. 44. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem. 2 x – 2 = 4 x + 1 + Treat the 1 as . 1 1 1 1 2 x – 2 = 4 + (x + 1) x + 1 2 x – 2 = x + 5 x + 1
45. 45. General Rational Equations Example D. Solve 2 x – 2 = 4 x + 1 + 1 We may use the cross multiplication to combine the two terms on the same side first to arrange the problem as a proportional problem. 2 x – 2 = 4 x + 1 + Treat the 1 as . 1 1 1 1 2 x – 2 = 4 + (x + 1) x + 1 2 x – 2 = x + 5 x + 1 2(x + 1) = (x – 2)(x +5) You finish it. . .
46. 46. General Rational Equations An answer that doesn’t work for the original problem is called an extraneous solution.
47. 47. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution.
48. 48. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides.
49. 49. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3)
50. 50. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1
51. 51. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1
52. 52. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3)
53. 53. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3)
54. 54. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3) 3 = –x + 6
55. 55. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3) 3 = –x + 6 x = –3 + 6
56. 56. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3) 3 = –x + 6 x = –3 + 6 x = 3
57. 57. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3) 3 = –x + 6 x = –3 + 6 x = 3 However this is an extraneous answer because it is not usable – it turns the denominator into 0.
58. 58. General Rational Equations Example E. (Extraneous solution) Solve 3 x – 3 = x x – 3 – 2 An answer that doesn’t work for the original problem is called an extraneous solution. Multiply the LCD (x – 3) to both sides. 3 x – 3 = x x – 3 – 2( ) (x – 3) 1 1 (x – 3) 3 = x – 2(x – 3) 3 = –x + 6 x = –3 + 6 x = 3 However this is an extraneous answer because it is not usable – it turns the denominator into 0. Hence there is no solution for this rational equation.
59. 59. General Rational Equations Ex. A. Rewrite the following fractional equations into ones without fractions. You don’t have to solve for anything. 1 4 A 1 2 B += 2 3 C1. 5 6 A 3 4 B =– 2 3 C2. 5 4 A 6 5 B –= 2C3. 1 6 A 3B =– 2 7 C4. 1 x 1 y =+5. 1 x 1 y = 1–6. 1 xy 1 x 1 y =+7. 1 (x – y) = x– 18. 1 z Ex. B. Rewrite the following equations into linear equations without fractions. Solve for the x and check the answers. 2 x =+ 19. 3 x = 110.1 2x 1 3x– 3 4x =– 111. 6 7x = 112.5 6x 7 8x –
60. 60. General Rational Equations Ex. C. Rewrite the following equations into linear equations without fractions. Solve for the x and check the answers. 2 x – 1 =+ 113. 3 2x – 2 3 x – 2 =+ 214. 1 3x – 6 5 x – 2 =– 215. 1 6 – 3x 3 x = 3 x + 2 + 2 Ex. D. Rewrite the following into 2nd degree equations. Solve for the x and check the answers. 20. 3 x – 2 – 3 x + 2 = 1 21. 2 x = x+ 116. 3 x = x – 217. –6 x = x+ 518. –3 x = x – 419. 6 x + 1 = 6 x – 1 22. 12 x – 1 – 12 x + 1 = 123.
61. 61. General Rational Equations 24. 1 x + 1 x – 1 = 5 6 25. 1 x + 1 x + 2 = 3 4 26. 1 x + 1 x – 2 = 4 3 27. 1 x + 2 x + 2 = 1 28. 2 x + 1 x + 1 = 3 2 29. + 5 x + 2 = 2 2 x – 1 30. – 1 x + 1 = 3 2 31.6 x + 2 – 4 x + 1 = 1 1 x – 2
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# ITFP, Perimeter: selective guide to talks. #1: Brukner on quantum theory with indefinite causal order
Excellent conference the week before last at Perimeter Institute: Information Theoretic Foundations for Physics. The talks are online; herewith a selection of some of my favorites, heavily biased towards ideas new and particularly interesting to me (so some excellent ones that might be of more interest to you may be left off the list!). Some of what would have been possibly of most interest and most novel to me happened on Weds., when the topic was spacetime physics and information, and I had to skip the day to work on a grant proposal. I'll have to watch those online sometime. This was going to be one post with thumbnail sketches/reviews of each talk, but as usual I can't help running on, so it may be one post per talk.
All talks available here, so you can pick and choose. Here's #1 (order is roughly temporal, not any kind of ranking...):
Caslav Brukner kicked off with some interesting work on physical theories in with indefinite causal structure. Normally in formulating theories in an "operational" setting (in which we care primarily about the probabilities of physical processes that occur as part of a complete compatible set of possible processes) we assume a definite causal (partial) ordering, so that one process may happen "before" or "after" another, or "neither before nor after". The formulation is "operational" in that an experimenter or other agent may decide upon, or at least influence, which set of processes, out of possible compatible sets, the actual process will be drawn, and then nature decides (but with certain probabilities for each possible process, that form part of our theory), which one actually happens. So for instance, the experimenter decides to perform a Stern-Gerlach experiment with a particular orientation X of the magnets; then the possible processes are, roughly, "the atom was deflected in the X direction by an angle theta," for various angles theta. Choose a different orientation, Y, for your apparatus, you choose a different set of possible compatible processes. ("The atom was deflected in the Y direction by an angle theta.") Then we assume that if one set of compatible processes happens after another, an agent's choice of which complete set of processes is realized later, can't influence the probabilities of processes occuring in an earlier set. "No signalling from the future", I like to call this; in formalized operational theories it is sometimes called the "Pavia causality axiom". Signaling from the past to the future is fine, of course. If two complete sets of processes are incomparable with respect to causal order ("spacelike-separated"), the no-signalling constraint operates both ways: neither Alice's choice of which compatible set is realized, nor Bob's, can influence the probabilities of processes occuring at the other agent's site. (If it could, that would allow nearly-instantaneous signaling between spatially separated sites---a highly implausible phenomenon only possible in preposterous theories such as the Bohmian version of quantum theory with "quantum disequilibrium", and Newtonian gravity. ) Anyway, Brukner looks at theories that are close to quantum, but in which this assumption doesn't necessarily apply: the probabilities exhibit "indeterminate causal structure". Since the theories are close to quantum, they can be interpreted as allowing "superpositions of different causal structures", which is just the sort of thing you might think you'd run into in, say, theories combining features of quantum physics with features of general relativistic spacetime physics. As Caslav points out, since in general relativity the causal structure is influenced by the distribution of mass and energy, you might hope to realize such indefinite causal structure by creating a quantum superposition of states in which a mass is in one place, versus being in another. (There are people who think that at some point---some combinations of spatial scales (separation of the areas in which the mass is located) and mass scales (amount of mass to be separated in "coherent" superposition)) the possibility of such superpositions breaks down. Experimentalists at Vienna (where Caslav---a theorist, but one who likes to work with experimenters to suggest experiments---is on the faculty) have created what are probably the most significant such superpositions.)
Situations with a superposition of causal orders seem to be exhibit some computational advantages over standard causally-ordered quantum computation, like being able to tell in fewer queries (one?) whether a pair of unitaries commutes or anticommutes. Not sure whose result that was (Giulio Chiribella and others?), but Caslav presents some more recent results on query complexity in this model, extending the initial results. I am generally wary about results on computation in theories with causal anomalies. The stuff on query complexity with closed timelike curves, e.g. by Dave Bacon and by Scott Aaronson and John Watrous has seemed uncompelling---not the correctness of the mathematical results, but rather the physical relevance of the definition of computation---to me for reasons similar to those given by Bennett, Leung, Smith and Smolin. But I tend to suspect that Caslav and the others who have done these query results, use a more physically compelling framework because they are well versed in the convex operational or "general probabilistic theories" framework which aims to make the probabilistic behavior of processes consistent under convex combination ("mixture", i.e. roughly speaking letting somebody flip coins to decide which input to present your device with). Inconsistency with respect to such mixing is part of the Bennett/Leung/Smolin/Smith objection to the CTC complexity classes as originally defined.
[Update: This article at Physics.org quotes an interview with Scott Aaronson responding to the Bennett et. al. objections. Reasonably enough, he doesn't think the question of what a physically relevant definition of CTC computing is has been settled. When I try to think about this issue sometimes I wonder if the thorny philosophical question of whether we court inconsistency by trying to combine intervention ("free choice of inputs") in a physical theory is rearing its head. As often with posts here, I'm reminding myself to revisit the issue at some point... and think harder.]
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# My short review of David Deutsch's "The Beginning of Infinity" in Physics Today
Here is a link to my short review of David Deutsch's book The Beginning of Infinity, in Physics Today, the monthly magazine for members of the American Physical Society. I had much more to say about the book, which is particularly ill-suited to a short-format review like those in Physics Today. (The title is suggestive; and a reasonable alternative would have been "Life, the Universe, and Everything.") It was an interesting exercise to boil it down to this length, which was already longer than their ideal. I may say some of it in a blog post later.
It was also interesting to have such extensive input from editors. Mostly this improved things, but in a couple of cases (not helped by my internet failing just as the for-publication version had been produced) the result was not good. In particular, the beginning of the second-to-last paragraph, which reads "For some of Deutsch’s concerns, prematurity is irrelevant. But fallibilism undermines some of his claims ... " is not as I'd wanted. I'd had "this" in place of "prematurity" and "it" in place of "fallibilism". I'd wanted, in both cases, to refer in general to the immediately preceding discussion, more broadly than just to "prematurity" in one case and "fallibilism" in the other. It seems the editors felt uncomfortable with a pronoun whose antecedent was not extremely specific. I'd have to go back to notes to see what I ultimately agreed to, but definitely not plain "prematurity".
One other thing I should perhaps point out is that when I wrote:
Deutsch’s view that objective correctness is possible in areas outside science is appealing. And his suggestion that Popperian explanation underwrites that possibility is intriguing, but may overemphasize the importance of explanations as opposed to other exercises of reason. A broader, more balanced perspective may be found in the writings of Roger Scruton, Thomas Nagel, and others.
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# No new enlightenment: A critique of "quantum reason"
I have a lot of respect for Scientific American contributing physics editor George Musser's willingness to solicit and publish articles on some fairly speculative and, especially, foundational, topics whether in string theory, cosmology, the foundations of quantum theory, quantum gravity, or quantum information. I've enjoyed and learned from these articles even when I haven't agreed with them. (OK, I haven't enjoyed all of them of course... a few have gotten under my skin.) I've met George myself, at the most recent FQXi conference; he's a great guy and was very interested in hearing, both from me and from others, about cutting-edge research. I also have a lot of respect for his willingness to dive in to a fairly speculative area and write an article himself, as he has done with "A New Enlightenment" in the November 2012 Scientific American (previewed here). So although I'm about to critique some of the content of that article fairly strongly, I hope it won't be taken as mean-spirited. The issues raised are very interesting, and I think we can learn a lot by thinking about them; I certainly have.
The article covers a fairly wide range of topics, and for now I'm just going to cover the main points that I, so far, feel compelled to make about the article. I may address further points later; in any case, I'll probably do some more detailed posts, maybe including formal proofs, on some of these issues.
The basic organizing theme of the article is that quantum processes, or quantum ideas, can be applied to situations which social scientists usually model as involving the interactions of "rational agents"...or perhaps, as they sometimes observe, agents that are somewhat rational and somewhat irrational. The claim, or hope, seems to be that in some cases we can either get better results by substituting quantum processes (for instance, "quantum games", or "quantum voting rules") for classical ones, or perhaps better explain behavior that seems irrational. In the latter case, in this article, quantum theory seems to be being used more as a metaphor for human behavior than as a model of a physical process underlying it. It isn't clear to me whether we're supposed to view this as an explanation of irrationality, or in some cases as the introduction of a "better", quantum, notion of rationality. However, the main point of this post is to address specifics, so here are four main points; the last one is not quantum, just a point of classical political science.
(1) Quantum games. There are many points to make on this topic. Probably most important is this one: quantum theory does not resolve the Prisoner's Dilemma. Under the definitions I've seen of "quantum version of a classical game", the quantum version is also a classical game, just a different one. Typically the strategy space is much bigger. Somewhere in the strategy space, typically as a basis for a complex vector space ("quantum state space") of strategies, or as a commuting ("classical") subset of the possible set of "quantum actions" (often unitary transformations, say, that the players can apply to physical systems that are part of the game-playing apparatus), one can set things up so one can compare the expected payoff of the solution, under various solution concepts such as Nash equilibrium, for the classical game and its "quantum version", and it may be that the quantum version has a better result for all players, using the same solution concept. This was so for Eisert, Lewenstein, and Wilkens' (ELW for short) quantum version of Prisoner's Dilemma. But this does not mean (nor, in their article, did ELW claim it did) that quantum theory "solves the Prisoner's Dilemma", although I suspect when they set out on their research, they might have had hope that it could. It doesn't because the prisoners can't transform their situation into quantum prisoners dilemma; to play that game, whether by quantum or classical means, would require the jailer to do something differently. ELW's quantum prisoner's dilemma involves starting with an entangled state of two qubits. The state space consists of the unit Euclidean norm sphere in a 4-dimensional complex vector space (equipped with Euclidean inner product); it has a distinguished orthonormal basis which is a product of two local "classical bases", each of which is labeled by the two actions available to the relevant player in the classical game. However the quantum game consists of each player choosing a unitary operator to perform on their local state. Payoff is determined---and here is where the jailer must be complicit---by performing a certain two-qubit unitary---one which does not factor as a product of local unitaries---and then measuring in the "classical product basis", with payoffs given by the classical payoff corresponding to the label of the basis vector corresponding to the result. Now, Musser does say that "Quantum physics does not erase the original paradoxes or provide a practical system for decision making unless public officials are willing to let people carry entangled particles into the voting booth or the police interrogation room." But the situation is worse than that. Even if prisoners could smuggle in the entangled particles (and in some realizations of prisoners' dilemma in settings other than systems of detention, the players will have a fairly easy time supplying themselves with such entangled pairs, if quantum technology is feasible at all), they won't help unless the rest of the world, implementing the game, implements the desired game, i.e. unless the mechanism producing the payoffs doesn't just measure in a product basis, but implements the desired game by measuring in an entangled basis. Even more importantly, in many real-world games, the variables being measured are already highly decohered; to ensure that they are quantum coherent the whole situation has to be rejiggered. So even if you didn't need the jailer to make an entangled measurement---if the measurement was just his independently asking each one of you some question---if all you needed was to entangle your answers---you'd have to either entangle your entire selves, or covertly measure your particle and then repeat the answer to the jailer. But in the latter case, you're not playing the game where the payoff is necessarily based on the measurement result: you could decide to say something different from the measurement result. And that would have to be included in the strategy set.
There are still potential applications: if we are explicitly designing games as mechanisms for implementing some social decision procedure, then we could decide to implement a quantum version (according to some particular "quantization scheme") of a classical game. Of course, as I've pointed out, and as ELW do in their paper, that's just another classical game. But as ELW note, it is possible---in a setting where quantum operations (quantum computer "flops") aren't too much more expensive than their classical counterparts---that playing the game by quantum means might use less resources than playing it by simulating it classically. In a mechanism design problem that is supposed to scale to a large number of players, it even seems possible that the classical implementation could scale so badly with the number of players as to become infeasible, while the quantum one could remain efficient. For this reason, mechanism design for preference revelation as part of a public goods provision scheme, for instance, might be a good place to look for applications of quantum prisoners-dilemma like games. (I would not be surprised if this has been investigated already.)
Another possible place where quantum implementations might have an advantage is in situations where one does not fully trust the referee who is implementing the mechanism. It is possible that quantum theory might enable the referee to provide better assurances to the players that he/she has actually implemented the stated game. In the usual formulation of game theory, the players know the game, and this is not an issue. But it is not necessarily irrelevant in real-world mechanism design, even if it might not fit strictly into some definitions of game theory. I don't have a strong intuition one way or the other as to whether or not this actually works but I guess it's been looked into.
(2) "Quantum democracy". The part of the quote, in the previous item, about taking entangled particles into the voting booth, alludes to this topic. Gavriel Segre has a 2008 arxiv preprint entitled "Quantum democracy is possible" in which he seems to be suggesting that quantum theory can help us the difficulties that Arrow's Theorem supposedly shows exist with democracy. I will go into this in much more detail in another post. But briefly, if we consider a finite set A of "alternatives", like candidates to fill a single position, or mutually exclusive policies to be implemented, and a finite set I of "individuals" who will "vote" on them by listing them in the order they prefer them, a "social choice rule" or "voting rule" is a function that, for every "preference profile", i.e. every possible indexed set of preference orderings (indexed by the set of individuals), returns a preference ordering, called the "social preference ordering", over the alternatives. The idea is that then whatever subset of alternatives is feasible, society should choose the one mostly highly ranked by the social preference ordering, from among those alternatives that are feasible. Arrow showed that if we impose the seemingly reasonable requirements that if everyone prefers x to y, society should prefer x to y ("unanimity") and that whether or not society prefers x to y should be affected only by the information of which individuals prefer x to y, and not by othe aspects of individuals' preference orderings ("independence of irrelevant alternatives", "IIA"), the only possible voting rules are the ones such that, for some individual i called the "dictator" for the rule, the rule is that that individual's preferences are the social preferences. If you define a democracy as a voting rule that satisfies the requirements of unanimity and IIA and that is not dictatorial, then "democracy is impossible". Of course this is an unacceptably thin concept of individual and of democracy. But anyway, there's the theorem; it definitely tells you something about the limitations of voting schemes, or, in a slighlty different interpretation, of the impossibility of forming a reasonable idea of what is a good social choice, if all that we can take into account in making the choice is a potentially arbitrary set of individuals' orderings over the possible alternatives.
Arrow's theorem tends to have two closely related interpretations: as a mechanism for combining actual individual preferences to obtain social preferences that depend in desirable ways on individual ones, or as a mechanism for combining formal preference orderings stated by individuals, into a social preference ordering. Again this is supposed to have desirable properties, and those properties are usually motivated by the supposition that the stated formal preference orderings are the individuals' actual preferences, although I suppose in a voting situation one might come up with other motivations. But even if those are the motivations, in the voting interpretation, the stated orderings are somewhat like strategies in a game, and need not coincide with agents' actual preference orderings if there are strategic advantages to be had by letting these two diverge.
What could a quantum mitigation of the issues raised by Arrow's theorem---on either interpretation---mean? We must be modifying some concept in the theorem... that of an individual's preference ordering, or voting strategy, or that of alternative, or---although this seems less promising---that of individual---and arguing that somehow that gets us around the problems posed by the theorem. None of this seems very promising, for reasons I'll get around to in my next post. The main point is that if the idea is similar to the --- as we've seen, dubious --- idea that superposing strategies can help in quantum games, it doesn't seem to help with interpretations where the individual preference ordering is their actual preference ordering. How are we to superpose those? Superposing alternatives seems like it could have applications in a many-worlds type interpretation of quantum theory, where all alternatives are superpositions to begin with, but as far as I can see, Segre's formalism is not about that. It actually seems to be more about superpositions of individuals, but one of the big motivational problems with Segre's paper is that what he "quantizes" is not the desired Arrow properties of unanimity, independence of irrelevant alternatives, and nondictatoriality, but something else that can be used as an interesting intermediate step in proving Arrow's theorem. However, there are bigger problems than motivation: Segre's main theorem, his IV.4, is very weak, and actually does not differentiate between quantum and classical situations. As I discuss in more detail below, it looks like for the quantum logics of most interest for standard quantum theory, namely the projection lattices of of von Neumann algebras, the dividing line between ones having what Segre would call a "democracy", a certain generalization of a voting rule satisfying Arrow's criteria, and ones that don't (i.e. that have an "Arrow-like theorem") is not commutativity versus noncommutativity of the algebra (ie., classicality versus quantumness), but just infinite-dimensionality versus finite-dimensionality, which was already understood for the classical case. So quantum adds nothing. In a later post, I will go through (or post a .pdf document) all the formalities, but here are the basics.
Arrow's Theorem can be proved by defining a set S of individuals to be decisive if for every pair x,y of alternatives, whenever everyone in S prefers x to y, and everyone not in x prefers y to x, society prefers x to y. Then one shows that the set of decisive sets is an ultrafilter on the set of individuals. What's an ultrafilter? Well, lets define it for an arbitrary lattice. The set, often called P(I), of subsets of any set I, is a lattice (the relevant ordering is subset inclusion, the defined meet and join are intersection and union). A filter---not yet ultra---in a lattice is a subset of the lattice that is upward-closed, and meet-closed. That is, to say that F is a filter is to say that if x is in F, and y is greater than or equal to x, then y is in F, and that if x and y are both in f, so is x meet y. For P(I), this means that a filter has to include every superset of each set in the filter, and also the intersection of every pair of sets in the filter. Then we say a filter is proper if it's not the whole lattice, and it's an ultrafilter if it's a maximal proper filter, i.e. it's not properly contained in any other filter (other than the whole lattice). A filter is called principal if it's generated by a single element of the lattice: i.e. if it's the smallest filter containing that element. Equivalently, it's the set consisting of that element and everything above it. So in the case of P(I), a principal filter consists of a given set, and all sets containing that set.
To prove Arrow's theorem using ultrafilters, one shows that unanimity and IIA imply that the set of decisive sets is an ultrafilter on P(I). But it was already well known, and is easy to show, that all ultrafilters on the powerset of a finite set are principal, and are generated by singletons of I, that is, sets containing single elements of I. So a social choice rule satisfying unanimity and IIA has a decisive set containing a single element i, and furthermore, all sets containing i are decisive. In other words, if i favors x over y, it doesn't matter who else favors x over y and who opposes it: x is socially preferred to y. In other words, the rule is dictatorial. QED.
Note that it is crucial here that the set I is finite. If you assume the axiom of choice (no pun intended ahead of time), then non-principal ultrafilters do exist in the lattice of subspaces of an infinite set, and the more abstract-minded people who have thought about Arrow's theorem and ultrafilters have indeed noticed that if you are willing to generalize Arrow's conditions to an infinite electorate, whatever that means, the theorem doesn't generalize to that situation. The standard existence proof for a non-principal ultrafilter is to use the axiom of choice in the form of Zorn's lemma to establish that any proper filter is contained in a maximal one (i.e. an ultrafilter) and then take the set of subsets whose complement (in I) is finite, show it's a filter, and show it's extension to an ultrafilter is not principal. Just for fun, we'll do this in a later post. I wouldn't summarize the situation by saying "infinite democracies exist", though. As a sidelight, some people don't like the fact that the existence proof is nonconstructive.
As I said, I'll give the details in a later post. Here, we want to examine Segre's proposed generalization. He defines a quantum democracy to be a nonprincipal ultrafilter on the lattice of projections of an "operator-algebraically finite von Neumann algebra". In the preprint there's no discussion of motivation, nor are there explicit generalizations of unanimity and IIA to corresponding quantum notions. To figure out such a correspondence for Segre's setup we'd need to convince ourselves that social choice rules, or ones satisfying one or the other of Arrow's properties, are related one to one to their sets of decisive coalitions, and then relate properties of the rule (or the remaining property), to the decisive coalitions' forming an ultrafilter. Nonprincipality is clearly supposed to correspond to nondictatorship. But I won't try to tease out, and then critique, a full correspondence right now, if one even exists.
Instead, let's look at Segre's main point. He defines a quantum logic as a non-Boolean orthomodular lattice. He defines a quantum democracy as a non-principal ultrafilter in a quantum logic. His main theorem, IV.4, as stated, is that the set of quantum democracies is non-empty. Thus stated, of course, it can be proved by showing the existence of even one quantum logic that has a non-principal ultrafilter. These do exist, so the theorem is true.
However, there is nothing distinctively quantum about this fact. Here, it's relevant that Segre's Theorem IV.3 as stated is wrong. He states (I paraphrase to clarify scope of some quantifiers) that L is an operator-algebraically finite orthomodular lattice all of whose ultrafilters are principal if, and only if, L is a classical logic (i.e. a Boolean lattice). But this is false. It's true that to get his theorem IV.4, he doesn't need this equivalence. But what is a von Neumann algebra? It's a *-algebra consisting of bounded operators on a Hilbert space, closed in the weak operator topology. (Or something isomorphic in the relevant sense to one of these.) There are commutative and noncommutative ones. And there are finite-dimensional ones and infinite-dimensional ones. The finite-dimensional ones include: (1) the algebra of all bounded operators on a finite-dimensional Hilbert space (under operator multiplication and complex conjugation), these are noncommutative for dimension > 1 (2) the algebra of complex functions on a finite set I (under pointwise multiplication and complex conjugation) and (3) finite products (or if you prefer the term, direct sums) of algebras of these types. (Actually we could get away with just type (1) and finite products since the type (2) ones are just finite direct sums of one-dimensional instances of type (1).) The projection lattices of the cases (2) are isomorphic to P(I) for I the finite set. These are the projection lattices for which Arrow's theorem can be proved using the fact that they have no nonprincipal ultrafilters. The cases (1) are their obvious quantum analogues. And it is easy to show that in these cases, too, there are no nonprincipal ultrafilters. Because the lattice of projections of a von Neumann algebra is complete, one can use essentially the same proof as for the case of P(I) for finite I. So for the obvious quantum analogues of the setups where Arrow's theorem is proven, the analogue of Arrow's theorem does hold, and Segre's "quantum democracies" do not exist.
Moreover, Alex Wilce pointed out to me in email that essentially the same proof as for P(I) with I infinite, gives the existence of a nonprincipal ultrafilter for any infinite-dimensional von Neumann algebra: one takes the set of projections of cofinite rank (i.e. whose orthocomplementary projection has finite rank), shows it's a filter, extends it (using Zorn's lemma) to an ultrafilter, and shows that's not principal. So (if the dividing line between finite-dimensional and infinite-dimensional von Neumann algebras is precisely that their lowest-dimensional faithful representations are on finite-dimensional Hilbert spaces, which seems quite likely) the dividing line between projection lattices of von Neumann algebras on which Segre-style "democracies" (nonprincipal ultrafilters) exist, is precisely that between finite and infinite dimension, and not that between commutativity and noncommutativity. I.e. the existence or not of a generalized decision rule satisfying a generalization of the conjunction of Arrow's conditions has nothing to do with quantumness. (Not that I think it would mean much for social choice theory or voting if it did.)
(3) I'll only say a little bit here about "quantum psychology". Some supposedly paradoxical empirical facts are described at the end of the article. When subjects playing Prisoner's Dilemma are told that the other player will snitch, they always (nearly always? there must be a few mistakes...) snitch. When they are told that the other player will stay mum, they usually also fink, but sometimes (around 20% of the time---it is not stated whether this typical of a single individual in repeated trials, or a percentage of individuals in single trials) stay mum. However, if they are not told what the other player will do, "about 40% of the time" they stay mum. Emanuel Pothos and Jerome Busemeyr devised a "quantum model" that reproduced the result. As described in Sci Am, Pothos interprets it in terms of destructive interference between (amplitudes associated with, presumably) the 100% probability of snitching when the other snitches and the 80% probability of snitching when they other does not that reduces the probability to 60% when they are not sure whether the other will snitch. It is a model; they do not claim that quantum physics of the brain is responsible. However, I think there is a better explanation, in terms of what Douglas Hofstadter called "superrationality", Nigel Howard called "metarationality", and I like to call a Kantian equilibrium concept, after the version of Kant's categorial imperative that urges you to act according to a maxim that you could will to be a universal law. Simply put, it's the line of reasoning that says "the other guy is rational like me, so he'll do what I do. What does one do if he believes that? Well, if we both snitch, we're sunk. If we both stay mum, we're in great shape. So we'll stay mum." Is that rational? I dunno. Kant might have argued it is. But in any case, people do consider this argument, as well, presumably, as the one for the Nash equilibrium. But in either of the cases where the person is told what the other will do, there is less role for the categorical imperative; one is being put more in the Nash frame of mind. Now it is quite interesting that people still cooperate a fair amount of the time when they know the other person is staying mum; I think they are thinking of the other person's action as the outcome of the categorical imperative reasoning, and they feel some moral pressure to stay with the categorical imperative reasoning. Whereas they are easily swayed to completely dump that reasoning in the case when told the other person snitched: the other has already betrayed the categorical imperative. Still, it is a bit paradoxical that people are more likely to cooperate when they are not sure whether the other person is doing so; I think the uncertainty makes the story that "he will do what I do" more vivid, and the tempting benefit of snitching when the other stays mum less vivid, because one doesn't know *for sure* that the other has stayed mum. Whether that all fits into the "quantum metaphor" I don't know but it seems we can get quite a bit of potential understanding here without invoking. Moreover there probably already exists data to help explore some of these ideas, namely about how the same individual behaves under the different certain and uncertain conditions, in anonymous trials guaranteed not to involve repetition with the same opponent.
Less relevant to quantum theory, but perhaps relevant in assessing how important voting paradoxes are in the real world, is an entirely non-quantum point:
(4) A claim by Piergiorgio Odifreddi, that the 1976 US election is an example of Condorcet's paradox of cyclic pairwise majority voting, is prima facie highly implausible to anyone who lived through that election in the US. The claim is that a majority would have favored, in two-candidate elections:
Carter over Ford (as in the actual election)
Ford over Reagan
Reagan over Carter
I strongly doubt that Reagan would have beat Carter in that election. There is some question of what this counterfactual means, of course: using polls conducted near the time of the election does not settle the issue of what would have happened in a full general-election campaign pitting Carter against Reagan. In "Preference Cycles in American Elections", Electoral Studies 13: 50-57 (1994), as summarized in Democracy Defended by Gerry Mackie, political scientist Benjamin Radcliff analyzed electoral data and previous studies concerning the US Presidential elections from 1972 through 1984, and found no Condorcet cycles. In 1976, the pairwise orderings he found for (hypothetical, in two of the cases) two-candidate elections were Carter > Ford, Ford > Reagan, and Carter > Reagan. Transitivity is satisfied; no cycle. Obviously, as I've already discussed, there are issues of methodology, and how to analyze a counterfactual concerning a general election. More on this, perhaps, after I've tracked down Odifreddi's article. Odifreddi is in the Sci Am article because an article by him inspired Gavriel Segre to try to show that such problems with social choice mechanisms like voting might be absent in a quantum setting.
Odifreddi is cited by Musser as pointing out that democracies usually avoid Condorcet paradoxes because voters tend to line up on an ideological spectrum---I'm just sceptical until I see more evidence, that that was not the case also in 1976 in the US. I have some doubt also about the claim that Condorcet cycles are the cause of democracy "becoming completely dysfunctional" in "politically unsettled times", or indeed that it does become completely dysfunctional in such times. But I must remember that Odifreddi is from the land of Berlusconi. But then again, I doubt cycles are the main issue with him...
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# Claimed proof of P not equal to NP by Vinay Deolalikar is being taken seriously
Via Richard Lipton, news that HP research scientist Vinay Deolalikar has put online a 102 page draft of a paper purporting to prove that P is not equal to NP. (Follow the link to Deolalikar's page, and the paper is the first one linked under "Selected Publications".) It is being taken seriously by serious people (like Lipton)--- in the sense that they are trying to understand it and understand whether or not it is correct. The length, of course, increases the likelihood of a so-far unspotted error. Lipton thinks Deolalikar's use of finite model theory is an interesting and promising twist. I have only just downloaded the paper, and am not a complexity theory expert, but what leaps out at me from the abstract is that it involves graphical models for sets of interacting random variables, an area that interests me. Not that surprising, I guess, as lots of difficult problems---as well as ones that are easy but perhaps not obviously so, because of structure described by a graphical model of the dependencies between variables---are naturally described in graphical terms, and the relation between graphical models and complexity is a well-studied---though mostly relatively recently studied---topic. I can no longer find the link to the online draft of Marc Mezard and Andrea Montanari's book on the subject (and its relations to physics and coding theory), perhaps because it's been published. [Correction: it's still online here. Thanks to commenter Kristal Cantwell for the link.] One key notion in such models is conditional independence---independence of two subsets of the random variables in question, when conditioned on a third set. A really nice paper on a quantum notion of conditional independence, relating it to the equality conditions for the strong subadditivity inequality for quantum entropy, is by Hayden, Jozsa, Petz, and Winter. Matt Leifer and David Poulin, and others, have described quantum graphical models.
You can bet that if Deolalikar's proof looks good, a slew of people will be applying the research strategy I like to call the "quantization functor" to it.
For those for whom P and NP are not their bread and butter, to give you an idea of how big this would be: I just goofed by telling my wife that if this pans out, it will be one of the most important results of twentieth century mathematics. Well, it would have been if done in the twentieth century. But I'll say more: although we're only 10 years in, I think it will be one of the most important results of twenty-first century mathematics. I haven't said what it really means but that's what Wikipedia is for.
I should point out that Scott Aaronson is dubious, though not based on close perusal of the proof. He has pledged to personally supplement the million dollars the Clay Mathematics Institute will pay Deolalikar if the proof is correct, with two hundred grand of his own.
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# Two way switch diagram
### 2 Way Light Switch Wiring Diagram House Electrical
1. al. You will see that there is a hot wire that is then spliced through a switch and that then goes to the hot ter
2. 2 Way Switch Wiring Diagram Pdf - wiring diagram is a simplified satisfactory pictorial representation of an electrical circuit. It shows the components of the circuit as simplified shapes, and the facility and signal associates between the devices. A wiring diagram usually gives recommendation virtually the relative face and arrangement of.
3. This area is a growing library of the schematics, wiring diagrams and technical photos. Menu. Search for: Posted in Wiring. Two Way Switch Connection Wiring Diagram. Posted by Margaret Byrd Posted on May 13, 2021. 2 way switch with circuit diagram two light connection what is a wiring of 3 type switches electrical 101
4. Two way switch lighting circuit diagrams. Here is our selection of two way switch circuit diagrams. The electrical symbol indicates where power enters the circuit. If you need to know how to wire a two way switch then this is the place to start. Do you need to know how a two way switch works before you check out the circuit diagrams
5. By wiring a 2-way switch, The circuit below shows the basic concept of electricity flow to the load. Let's assume the load you are controlling is a light. The electricity flows from the hot wire (black) through the 2-way switch (shown in off position) and then to the light and returns through the neutral wire (white). This is a completed circuit
Also, the same wiring circuit diagram can be used using 2-way lighting or two-way switches to control electrical devices from two different locations. The main purpose of the two-way switching circuit is to allow the devices to be switched on / off independently of any switch, no matter the current state of the switch Two way switch can be operated from any of the switch independently, means whatever be the position of other switch(on/off), you can control either way, complete these five steps for 3 way light switch wiring: Here is our selection of two way switch circuit diagrams. How to wire two way light switch.Two way lighting circuit Standard 2 Way Switch Wiring. The first way of wiring uses a couple of Two-Way Light Switches with a three wire control (3 Wire Control). The following is the simple schematic of a three-wire 2-Way Switch wiring. You can observe in the schematic that both the COM terminals are connected together Light Switch Wiring Diagram 2 Switches 2 Lights - wiring diagram is a simplified agreeable pictorial representation of an electrical circuit. It shows the components of the circuit as simplified shapes, and the capability and signal friends between the devices
3 Way Switch Wiring Diagram. Take a closer look at a 3 way switch wiring diagram. Pick the diagram that is most like the scenario you are in and see if you can wire your switch! This might seem intimidating, but it does not have to be. With these diagrams below it will take the guess work out of wiring Just like any loop-in loop-out radial circuit, the switch cable from the ceiling rose contains two wires, a permanent live and a switched live. This is cable C below, one wire connects to L1 and the other to L2 on the top switch. Cable D (Fig 2) is a three core and earth, this is the '3 wire control' that links the two light switches together 2 Way Switch Wiring Diagram Pdf Database. Fixing electrical wiring, more than any other household project is focused on protection. Install an electrical outlet correctly and it's since safe as that can be; do the installation improperly and it can potentially deadly. Which why there are several rules surrounding electrical cabling and. https://ryb.com.bd/ visit my website http://rybonline.com/ #two #way light #switch wiring How to install a two way #light switchWhat is a #2 #way #switch?2 w..
The one gang two-way switch is used when two switches control the single bulb. The one gang two switch diagrams are shown below. One Gang Two Way Switch Fig (a) is the two way light switch mechanism, fig (b) is the single gang switch face and fig (c) is the single gang two way light switch How to wire 2 way light switch, in this video we explain how two way switching works to connect a light fitting which is controlled with two light switches.. Two Way Light Switch Diagram Australia. December 29, 2018. Electrical . diy.stackexchange.com. 2 Way Light Switch Wiring Diagram Australia . 4-way switch with dimmer diagram 2 switches 1 light diagram 3 wire switch wiring diagram 3-way switch multiple lights wiring-diagram pull cord light 2-way switch wiring diagram double light switch wiring. A 4 way switch must be wired between two 3 way switches as shown in the diagrams on this page. A 4 way switch has five terminals: one ground and 4 circuit terminals divided into two matching pairs called travelers. Each pair of traveler terminals should be wired to the traveler wires from one of the 3 way switches in the circuit
Insteon 2 Way Switch Wiring Diagram. steppng If your switch's wires do not match the diagram, consult the Install Insteon Wall Switch, attach the wal plate and turn on power. Insteon provides the following wiring diagram for installing a 3-way the two switches together can the Secondary switch affect the power flow Image (s) for: two-way-switching-wiring-diagram. Two way light switching (3 wire system, new harmonised cable colours) showing switch and ceiling rose wiring. Back to Page. Next Image > Description: 3 Way Switch Wiring Diagrams - Do-It-Yourself-Help throughout 2 Way Dimmer Switch Wiring Diagram, image size 502 X 330 px, and to view image details please click the image.. Here is a picture gallery about 2 way dimmer switch wiring diagram complete with the description of the image, please find the image you need It's possible that the switch you removed was one of the two that controls the non-functioning light. A switch with three wires could be used for a hall light with a switch at each end. The switch puts power out to one wire or the other, like a valve for water that goes in two directions Seymour Duncan Do It All: 2 Humbuckers And A 5-Way Switch - Guitar Pickups, Bass Pickups, Pedals from www.seymourduncan.com When you make use of your finger or stick to the circuit with your eyes, it's easy to mistrace the circuit. 2 humbucker 1 volume 2 tone fender 5 way switch wiring. Guitar wiring tips tricks schematics and links
Hey, in this article we are going to see the Two Way Switch Connection Diagram and Wiring. We know there are a lot of applications of two way switches in staircase lighting connection, bedroom lighting connection, and other places Jan 29, 2021 - How To Wire A 2 Way Light Switch In Australia Wiring Diagrams . Feb 4, 2019- How To Wire A 2 Way Light Switch In Australia Wiring DiagramsLighting . Lighting Lighting in homes consumes 8â 15% of the average household electricity budget (or about 6% of its energy use) although the makeup of the installed lighting Read Mor In the above diagram, the power from the electricity source is coming through a two-wire cable to the first switch. Three-wire cables are used between the two switches. As we power this circuit, electricity will flow through the hot wire over to the second switch. As it goes through the red traveler, it will stop at switch number one There are several ways to wire two way switching. Below is the way I like to do it. The way I show allows you to fairly easily add a second two way switch to an existing light switch by simply wiring them together in a certain way. A two way switch(uk) Schematic diagram of old wire colours : Schematic diagram of new wire colours : N.B
A two way lighting circuit enables one circuit to be turned on or off by either of two switches. The circuit may have one or more lights in it. This type of circuit uses two of the two way switches. A wiring diagram and a circuit diagram are shown below. Two way lighting wiring diagram. Two way lighting circuit diagram If you have a 2-way circuit (where the same lights are controlled by two switches) you must choose a push-on/push-off dimmer and replace one of the switches with that dimmer.You can only use one push-on/push-off dimmer in a 2-way circuit.It must be used in conjunction with a normal switch
Two way switch wiring diagram pdf this 2 gang 1 way switch wiring diagram will contain an overall description from the item the name and format. 2 wire cable runs from the light to the first switch and then 3 wire is run between all the switches. It shows the components of the circuit as simplified forms and the power and also signal links in. Clipsal 2 Way Switch Wiring Diagram. Multi-Way Remote Switch/Dimmer. 60TDSM. Touch Dimmer/Switch Mech, Way Remote (60 Series Mech). The Saturn OneTouch Range Features: • Way. Clipsal calls their light switch a one pole, one way or two way switch.. Wiring diagram of the A-type 3-way switching circuit. The only A two way switch is one of the most basic parts of house wiring. Take a closer look at a 3 way switch wiring diagram. Wiring of 3 way light switches is certainly more complicated than that of the more common single pole switch but you can figure it out if you follow our 3 way switch wiring diagram Description : Pleasant Two Way Light Switch Wiring Diagram Nz | Inspiring Wiring for Light Switch 2 Way Wiring Diagram, image size 500 X 347 px, and to view image details please click the image. Here is a picture gallery about light switch 2 way wiring diagram complete with the description of the image, please find the image you need Two way switch - diagram. Connecting a two way switch circuit is a little bit more complicated, as it uses two switches. So - how to connect a two way switch? A two way switch diagram is more elaborate. In this case, the phase wire is connected with one of the two way switches, and the light source - to the other
A SPDT is called a 2-way switch in the European Union, and a 3-way switch in North America. A SPDT switch has 3 terminals. It will have one terminal in and 2 terminals out. Each terminal on a SPDT switch has a name. The input terminal is called Common. The output terminals are called Normally-Open and Normally-Closed 88light Reign 12v Led Dimmer Switch Wiring Diagrams. How to wire lights switches in a diy 2 way switch wiring diagram full led rocker diagrams 3 bunny nav anc carling contura salzer toggle gear wired motorcycle an illuminated reign 12v dimmer light schematic 4 with rv options auto relay 1 both battery considerations uk circuit two connection 6 pin quora get power from fixed red hms marine.
### 2 Way Switch Wiring Diagram Pdf autocardesig
Because it uses two switches and one appliance. In 3 way switches you use two control a single appliance from two different locations. Now let's look at the feature specific differences-. 2 Way Switch. 3 Way Switch. These have on/off switches. Do not have an on/off switch. These switches don't have hot wires 2 Way Switch Wiring Diagram. 2 way switch wiring diagram - You'll need an extensive, expert, and easy to know Wiring Diagram. With this kind of an illustrative manual, you will have the ability to troubleshoot, prevent, and full your assignments easily. Not merely will it help you attain your required outcomes more quickly, but in addition. 14+ Mk 2 Gang 2 Way Switch Wiring Diagram. Wiring a #2gang single switch is a simple and easy way to do. Mk logic plus switches and sockets manufactured by mk electric have been designed to perfectly complement modern interiors. 2 Gang Way Light Switch Wiring Diagram Uk - Wiring Diagram from lh3.googleusercontent.com Two Switch Light Wiring Diagram Overhead - Schematics Wiring Diagram - Wiring Two Lights To One Switch Diagram. Wiring diagram also provides useful suggestions for projects which may require some additional equipment. This book even includes recommendations for added provides that you could need as a way to finish your projects
### Two Way Switch Connection Wiring Diagram - Wiring View and
Two Way Switches. A Two Way light switch is a simple single pole changeover switch with three terminals. These are typically labelled COM, L1, and L2 (Some may label the L1 and L2 positions as 1 Way and 2 Way). In one switch position the COM terminal is connected to L1 Wiring Diagram For Two Way Switch - The power source comes from the fixture and then connects to the power terminal. There are several ways to wire two way switching. Do you need to know how a two way switch works before you check out the circuit diagrams. 2 way switching two wire control i ve included this method of 2 way switching for reference because you may come across it in old homes but.
According to the Circuit Diagram of the 2-way switch circuit above, the common of switch1 and switch2 are connected with each other.PIN-1 of both the switches are connected with the phase and PIN2 of both the switches are connected with pin2 and the lamp is connected in series The diagrams for 3-way switch wiring I broke down into the following sections (see below): Light after switches. The power source is coming to a light switch first. Light before switches. The power source is coming to light fitting first. The electricity source and light fixture are connected to the same switch 2 gang 1 way light switch wiring diagram. This switch arrangement is basically two two way switches on a single face plate. Single gang 2 way light switch. When the switch is on both terminals are connected together. A one way light switch is quite easy to wire up
The hallway light may be on a 3 way lighting circuit but not all of these wires make up that connection. Most likely you have a 14/2 coming from panel which you can think of as the main body of a tree.. The 14/3 and second 14/2 arelike the branches of the tree. The 14/3 branches out from the tree and extends to a box that contains the switch. How To Combine Two Wall Switches Into One Quora. Resources wiring diagram double light switch way australia for hpm a 2 how to install smart two switched lighting circuits 1 wire 3 connect using junction bo led trailer lights motion sensor with downlights combine wall switches into australian electrical in installation instructions ks 811 wifi 4 mechanism technical brochure dimmer proble A 2 Way Switch Wiring Diagram with Power Feed from the Switch light. These diagrams show various methods of one two and multiple way switching. This circuit diagram describes the wiring a two way switch in such an arrangement so that the flat TwinEarth and 3 CoreEarth cables are not broken or interrupted anywhere between the components of the. Legrand radiant 15 Amp 120-Volt 2-Switch 3-Way plus 3-Way Combination Decorator Rocker Light Switch, Dark Bronze. Model# RCD33DBCC6 (2) \$ 24 23. TP-LINK Smart Wi-Fi Light Switch with 3-Way Kit. Model# HS210 KIT (24) \$ 44 99. Leviton 15 Amp Commercial Grade Combination Two Single Pole Illuminated Rocker Switches, Ivory
This page is dedicated to Wiring Diagrams that can hopefully get you through a difficult wiring task or just to learn some basics in how to wire a 2-way switch, 3-way switch, 4-way switch, outlet or entertainment component diagrams.If you don't see a wiring diagram you are looking for on this page, then check out my Sitemap page for more information you may find helpful Here we are using a simple PLC to control this lamp using two switches, one switch at ground floor and second switch at first floor to control one lamp as shown in below figure. Note : we can also build the circuit using simple relays/switches also. This article only for understanding the basic concept of 2 way switch using a PLC Ladder Logic Re: Wiring bow and stern light to 3 position switch. Be careful what 3 position switch you have. There are two kinds. Off/circuit 1/circuit 2. The other is. Off/circuit 1/ circuit 1&2. This is the one you need for nav lights. Circuit 1 would be the anchor light, circuit 2 would be the red/green nav light. In the first example, you could turn on. Two, 2-wire cables (C3 and C4) run between the fixtures, and a 3-wire cable (C2) runs from the second fixture (F2) to the second 3 way switch (SW2). The hot source is connected to the common terminal of SW1. The neutral is spliced to the white wire feeding the first fixture, via cable C1,where it is spliced to the neutral of both lights
### 2 way light switch circuit wiring diagrams How to wire a
3 Way Switch Wiring with Light First. In this diagram the source for the circuit is at the light fixture and the two switches come after. Two-wire cable runs from the light to SW1 and 3-wire cable runs between SW1 and SW2. The hot source wire is spliced at the light box to the white cable wire running to the first switch box Identify switch type (Single pole, 3-way or 4-way) and tag the COMMON terminal. Single pole (One location) - Switch will have insulated wires connected to two screws of the same color plus a green ground screw. See Diagram A 3-way (Two locations)- 3-way switches will have insulated wires connected to three screws plus a green ground screw
2 Way Switch Wiring Diagram Australia Source: i.pinimg.com. 2 Way Switch Wiring Diagram Australia Source: www.dlsweb.rmit.edu.au. READ 2 Amps 1 Capacitor Wiring Diagram Collection. Read cabling diagrams from bad to positive and redraw the signal being a straight collection. All circuits are the same - voltage, ground, individual component. Two Way Light Switch Wiring Diagram Nz - Hello friends Wiring Diagram, In the article you are reading this time with the title Two Way Light Switch Wiring Diagram Nz, we have prepared this article well so that you can read and retrieve the information in it. Hopefully the content of the post what we write can make you understand. Happy reading Three Way Switch Wiring Diagram Two Lights from 1.bp.blogspot.com To properly read a cabling diagram, one provides to know how the particular components inside the method operate. For example , in case a module is powered up also it sends out a new signal of half the voltage plus the technician would not know this, he'd think he has an issue. 2 Humbuckers/3-Way Toggle Switch/1 Volume/2 Tones/Series-Split-Parallel, Reverse Phase & Master Series-Parallel. Guitar wiring diagram with 2 humbuckers, 3-way toggle switch, one volume and two tone controls plus 4 mini switches. One mini switch selects the neck pickup mode-Series/Single Coil (North)/Parallel, one mini switch selects the bridge. Wiring 2 lights to 1 switch. We did not find results for: Maybe you would like to learn more about one of these? Check spelling or type a new query. 2 way light switch with power feed via switch (multiple lights) | How to wire a light switch from www.howtowirealightswitch.co
### Wiring a 2-Way Switch - how-to-wire-it
• 3-way switch wiring diagrams for multiple lights. Now, the actual wiring. In order to wire the 3-way switch properly, you will need to use one of the diagrams (that I will post below), and which one you use depends entirely on your need.. Here are some wiring cases: The electricity source is connected to the 3-way switch first
• In fig 2, different connection and wiring diagrams are shown for a two pole, single phase manual changeover switch. The upper portion of the changeover switch is directly connected to the main power supply while the lower first and right connections slots are connected to the backup power supply like generator or inverter
• als can be connected to either of the two, but not both at the same time. You flip the switch ON to light up the stairs as you walk up
### How To Connect A 2-Way Switch (with Circuit Diagram)-EET-202
Hi Plasterer. It's a basic 2-way circuit. You will need a 2-way light switch in each location - up and down stairs. You'll then need to run 3core+e cable from the downstairs switch to the upstairs one. From the upstairs switch you will also run a 2core+e cable to your 'wiring point', ie: wherever it's convenient to fit a junction box. The two lights will be connected in parallel together. Your circuit contains two 3-way switches. Referring to the diagram below: if the switch with the incoming hot is the one you want single control from (shown at left)-; 1) replace that switch with a single-pole switch. 2) connect either of the traveler wires to the load sid
With a standard single-pole dimmer, a single switch controls the light. With a three-way dimmer, you can control a light with two switches. You will need a three-way dimmer and a three-way switch. This lets you dim from one location and turn the lights on and off from another. If only one switch controls the light, purchase a single-pole dimmer In this diagram I showed 3 different methods of staircase wiring diagram, In this 2 way, light switch diagrams I use two-way switches to control a light bulb or lamp holder in 2 different places. In this connection, if a person switches the bottom switch of the stair and when he/she reached on top then if he switches the upper 2-way button then. 10.10.2018 10.10.2018 3 Comments on Clipsal Light Switch Wiring Diagram Australia. There are several ways of wiring a switch depending on your situation. diagrams coupled with some instructions to help you on your way to wiring a single switch. The source comes into the light, the switch then is the last box on the circuit
### Two Way Electrical Switch Wiring Diagram - StairCase
Wiring diagram for 3 way switch two lights. In this diagram the source for the circuit is at the light fixture and the two switches come after. Power through switch light is controlled by two three way switches with the light between the switches and the power first going through a switch then to the light and onto the second three way switch A three-way switch is different. This type of switch lets you turn the light off or on from more than one place in the room. This type of switch has two brass screws, usually on the bottom, and a copper or black screw on the top left. Ideally, the old switch is still there, so you can wire the new one up the same way as the old one The 2 wire cable feed from the power source enters the bottom of the switch box and the hot (black) wire connects to the common or shunt terminal on the 3-way switch. There is only one such terminal on a 3-way switch, and it is usually identified as the one having a different color terminal screw (often significantly darker) from the other two. 1-48 of 375 results for 2 way dimmer switch. Smart Dimmer Switch, 2 Packs Light Switches for Dimmable LED Light, Halogen and Incandescent Bulb, Supports Alexa, Google Assistant and SmartThings, Single-Pole Switch with Voice Control and Schedule. 4.4 out of 5 stars. 743. \$38.99 Guitar wiring diagrams for tons of different setups. Single-coils, humbuckers, hum/sing/sing, hum/sing/hum, and much more. Plus, info on switches, pots, coil-splitting, and more. Search By Pickup Type Clear All. Refined Search Clear All. Guitar Wiring. Bass Wiring. Other/Misc Wiring. Click here to learn how to install your pickups ».
Two Way Switch Wiring Diagram Source: mrelectrician.tv Read cabling diagrams from unfavorable to positive in addition to redraw the signal being a straight range. All circuits are usually the same - voltage, ground, single component, and buttons 2 Way Double Light Switch Wiring Diagram from www.handymanknowhow.co.uk. Print the cabling diagram off and use highlighters to trace the signal. When you employ your finger or even stick to the circuit with your eyes, it's easy to mistrace the circuit. One trick that I actually use is to print a similar wiring picture off twice Cat5 Cable Wiring Diagram : How To Wire A Rj45 Plu... 1997 Dodge Ram 1500 Alternator Wiring Diagram : Al... 3 Way Light Switch With Dimmer Wiring Diagram / Ho... Backup Camera Wiring Schematic / How To Install A Yamaha 48V Golf Cart Wiring Diagram - Yamaha Relay... Do It Yourself Garage Vancouver - Stickley S Garag.. Description: Insteon 3-Way Switch - Alternate Wiring - Bithead's Blog inside Two Way Dimmer Switch Wiring Diagram, image size 1000 X 505 px, and to view image details please click the image.. Here is a picture gallery about two way dimmer switch wiring diagram complete with the description of the image, please find the image you need Two-way light switches are required for two way switching, these can be identified by looking at the terminals on the back of the switch . There will be a Com (common) terminal a L1 terminal and a L2 terminal. Two way switching with the old wiring colours
Two wire cable runs from the light to sw1 and 3 wire cable runs between sw1 and sw2. 4 way switch wiring with four switches. Having additional lights wired in parallel will not change the wiring below. Line switch load switch 1 if your wiring is like the diagram below you can wire your smart switch in the box with line The two-way switch will have corresponding wires or screws. Connect the white wire with the white wire, the black wire with the black wire, and either splice the red wire with the black wire or attach it to the black wire's source. The wires that were twisted together should be sealed with a locking wire cap. Turn the power on and test the current By wiring a 2-way switch The circuit below shows the basic concept of electricity flow to the load. 3 way light switch wiring 2 wire system This switch arrangement is basically two two way switches on a single face plate. Collection of leviton double pole switch wiring diagram. Take a closer look at a 3 way switch wiring diagram This 4 way switch diagram #2 shows the power source starting at the fixture. The white wire of the cable going to the switch is attached to the black line in the fixture box using a wirenut. The white wire becomes the energized switch leg, as indicated by using black or red electrical tape. Electrical Wiring Video. YouTube We will now go over the wiring diagram of a SPST Toggle Switch. Below is the wiring schematic diagram for connecting a SPST toggle switch: SPST Toggle Switch. You can see that a SPST toggle switch only has 2 terminals. 1 terminal is for the input. The other terminal is for the output. SPST toggle switches function as simple ON-OFF switches
Basic arrangement of the 3-way switch. (Shown in the white box is a simple diagram that depicts the function of the three-way switch.) The main thing to remember when you wire two way switches is that they are the same as wiring in a single-pole switch, except you must have two additional wires attaching the two switches together as illustrated in the picture Two separate circuits are controlled by a double pole switch can be used to control light and a fan or 2 lights on separate circuits. 6 pk leviton ivory grounded quiet toggle double pole light switch s03 cs220 2is us 8 6 30mm single double pole dpdt pushbutton switch door diagram with light 12v24v220v in switches from lights lighting on aliexpress A single pole light switch wiring diagram. Wiring a single pole or one-way light switch is easy, it requires 3 wires. Two black or hot wires, and the earth cable. Just draw the cables from the distribution board (DB) or junction-box, connect the black wire to the dimmer or switch, and then to the bulb. Draw the earth or grounding wire straight. FOUR-WAY SWITCH DIAGRAMS. The schematic diagram below shows how all four-way switches are wired. The wiring diagram further down depicts another choice of wiring method. Four-way switch wiring schematic. A four way switch is added when there are more than two switch locations operating the same light Assuming your two way switch wiring uses the 'old' core cable colours i.e Red (live), Black (neutral) and Green/Yellow (earth) for cables A, B & C then Fig.1 shows the most common way your ceiling rose will be connected in a 2 way lighting circuit.. Click Here - If the cables A, B & C in your house have Brown (live), Blue (neutral) and Green/Yellow (earth) cores
### How a 2 Way Switch Wiring Works? Two-Wire and Three-Wire
Step 2: Now that the hardest part has been completed, you will have to wire up the light switches. The cable coming from the light, to the first two-way switch is where you need to continue. You will have to connect the brown (Line) wire to the L1 terminal and the blue wire with the Brown slewing on it (Switched Line) to the L2 terminal The four-way switch is used between two three-way switches to provide control for an outlet or light fixture from multiple locations. If you want to have control from more than three locations—for example, five locations—you would still use two three-way switches (one on each end) and three four-way switches between the two three-ways
### Light Switch Wiring Diagram 2 Switches 2 Lights
As you look at the wiring diagram above, the left hand (as viewed from the rear) should control the landing light as a 2 way in conjunction with the switch upstairs. The centre controls the downstairs hall light and works OK. The right should just control the outdoor light - but doesn't. The wiring is exactly as the diagram (the colours are the. Electrically, a typical 3-way switch is a single pole, double throw (SPDT) switch. By correctly connecting two of these switches together, toggling either switch changes the state of the load from off to on, or vice versa. The switches may be arranged so that they are in the same orientation for off, and contrasting orientations for on Some diagrams may be unavailable during this time. Our apologies for the inconvenience. In the interim, please contact Technical Support: Phone: +1 (718) 816-8112 (Monday through Friday 11:00 AM - 3:00 PM Eastern Time) or email: tech@dimarzio.com. We endeavor to reply to most emails by the next business day Wiring a Two Way Switch. Two way switching means you can switch the same light fixture from two switches that are located in different sides of a room. Two way switches have a COM terminal as well as L1 and L2 terminals. When L1 is off L2 would be on. When L1 is on L2 would be off. There are two wiring options for this
### 3 Way Switch Wiring Diagram - Easy Do It Yourself Home
Actually the diagram explains everything clearly while controlling by two 3-way switches with a two wire cable power supply as shown. Remember only two wire cable with ground is necessary for the boxes at the lights we see there. Traveller wires (red in color) connect the three-way switches without going in deep to the system or circuit given Jul 18, · In , the following Esquire wiring diagram was published by Fender. The following diagrams apply only to a CTS 3-Way Switch. Using a CTS 3-Way Switch, the first switch position (top) the pickup circuit included auF capacitor wired to a k resistor which was wired to yet anotheruF capacitor that is wired to ground
Adding a battery is an easy project using a dual battery switch (sometimes called a marine battery switch or marine dual battery selector switch) from a company such as Blue Sea Systems (bluesea.com) or Perko (), which we used here.When properly installed, these marine battery switches let you choose one of the two batteries, combine them or disconnect both with the turn of a dial How to Wire Two Separate Switches & Lights Using the Same Power Source. It isn't unusual to wire two or more light and switch combinations from the same power source -- in fact, it's common practice Every 3-way switch circuit will have two 3-way switches. If there are more than two switches in the circuit then the others are 4-way switches. There will be three wires connected to each 3-way switch and four wires connected to each 4-way switch. Depending on the age of your home there may also be a bare or green ground wire connected to
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# Answer to Question #151235 in Engineering for Jordan Garwood
Question #151235
In a market with Pd = 150 – 3 Qd and Ps = 50 + Qs , the government has regulated maximum of 20 units and is concerned if this move was maybe a little too harsh, especially if the deadweight loss to consumers (voters) is greater than the deadweight loss to producers
1
2020-12-23T04:32:50-0500
If Pd = 150 – 3Qd and Ps = 50 + Qs, then in equilibrium:
4Q = 100,
Q = 25 units.
P = 50 + 25 = 75.
the government has regulated maximum of 20 units, then Pd = 150 - 3×20 = 90 and Ps = 50 + 20 = 70.
The deadweight loss to consumers (voters) is:
And the deadweight loss to producers is:
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# Difference Equations, why M<=N for causality?
in my notes for DSP they have the difference equation general form as:
y(k) + a1*y(k-1) + a2*y(k-2) + ... + an*y(k-N) = b0*x(k) + b1*x(k-1) + ... + bm*x(k-M)
with the claim that for the output to not depend on future values of the input then M<=N.
Why is this the case?
In fact, if I'm not missing something here I think I can provide a counterexample:
y(k) = x(k-1)
which has N = 0 and M = 1. But this is just a unit delay, the output is just the previous value of the input.
What's going on here?
You're right, it's simply not true that $M\le N$ is necessary for causality. The difference equation in your question can always be implemented by a causal system.
However, note that the difference equation is not uniquely related to a causal system. A simple example is
$$y[n]=ay[n-1]+bx[n]\tag{1}$$
This can obviously be implemented by a causal system. But for $a\neq 0$ you can also rewrite (1) as
$$y[n-1]=\frac{1}{a}(y[n]-bx[n])\tag{2}$$
which suggests an anti-causal system, even though (1) and (2) are completely equivalent. This difference is reflected by the transfer functions ($\mathcal{Z}$-transforms) of the corresponding systems. For a causal system the region of convergence (ROC) is outside a circle enclosing all poles, whereas for an anti-causal system the ROC is inside a circle outside of which all poles are located.
• So is the condition M<=N useful for anything at all? – user16250 Jun 15 '15 at 9:36
• @user16250: Can't think of anything. – Matt L. Jun 15 '15 at 10:15
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# How to make Sphere Transparent using ParametricPlot3D?
This is the code I used Edit: This is the full code to create this image
R[r_, s_] := {r, s, 0};
S[r_, s_] := {2 r/(r^2 + s^2 + 1), 2 s/(r^2 + s^2 + 1), (r^2 + s^2 - 1)/(r^2 + s^2 + 1)};
f[t_] = Tan[(Pi/2) * t];
Show[ParametricPlot3D[R[r, s], {r, -3, 3}, {s, -3, 3}, ImageSize -> 600,
Epilog -> Inset[Style[" 710178 ", FontSize -> 84, Red, Opacity[0.1]]]],
ParametricPlot3D[S[f[r], f[s]], {r, -3, 3}, {s, -3, 3},PlotStyle -> Transparent]]
And this is my output
The reason I'm trying to make it transparent is because I have to plot curves on the sphere and currently when I do plot, I can't see the curves.
• Welcome to MSE. Please provide all of the code to reproduce the image. The definitions for R, S, and f are missing. Commented Sep 13, 2021 at 1:59
• My apologies, I was unaware that the full code was required, however I have added it. Just for further information, I tried to plot the sphere S first, instead of the plane R and that still made no difference
– Nok
Commented Sep 13, 2021 at 8:40
• How about Show[ParametricPlot3D[R[r, s], {r, -3, 3}, {s, -3, 3}, ImageSize -> 600, PlotStyle -> Opacity[0.2], Epilog -> Inset[Style[" 710178 ", FontSize -> 84, Red, Opacity[0.1]]]], ParametricPlot3D[S[f[r], f[s]], {r, -3, 3}, {s, -3, 3}, PlotStyle -> Opacity[0.1]]] Commented Sep 13, 2021 at 9:03
• The circle just becomes a light orange when I change opacity. I also tried PlotStyle-> Directive[Orange, Opacity[0.1]], but that didn't do anything
– Nok
Commented Sep 13, 2021 at 9:16
• try if PlotStyle -> FaceForm[] gives what you need.
– kglr
Commented Sep 13, 2021 at 16:10
I think this is a version problem. In MMA 12.3 I get:
Show[ParametricPlot3D[R[r, s], {r, -3, 3}, {s, -3, 3},
ImageSize -> 600,
Epilog ->
Inset[Style[" 710178 ", FontSize -> 84, Red, Opacity[0.1]]]],
ParametricPlot3D[S[f[r], f[s]], {r, -3, 3}, {s, -3, 3},
PlotStyle -> Transparent]]
Update
To make the plane translucent, you could write:
Show[ParametricPlot3D[R[r, s], {r, -3, 3}, {s, -3, 3},
ImageSize -> 600, PlotStyle -> Opacity[0.3],
Epilog ->
Inset[Style[" 710178 ", FontSize -> 84, Red, Opacity[0.1]]]],
ParametricPlot3D[S[f[r], f[s]], {r, -3, 3}, {s, -3, 3},
PlotStyle -> Transparent]]
• Output in Mathematica 13.0 appears to be the same as that from 12.3 you show. Commented Dec 24, 2021 at 17:25
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What is the capital of Kansas?
Topeka is the capital city of the U.S. state of Kansas.
Question
Updated 8/27/2014 9:01:13 AM
f
Rating
8
Topeka is the capital city of the U.S. state of Kansas.
Confirmed by selymi [8/27/2014 9:03:22 AM]
The capital of Kansas is Topeka.
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2+2/2 = 3
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10*10+50 = 150
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= 100 + 50
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# Search by Topic
#### Resources tagged with Visualising similar to Flight of the Flibbins:
Filter by: Content type:
Stage:
Challenge level:
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##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
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##### Stage: 3 Challenge Level:
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
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At the time of writing the hour and minute hands of my clock are at right angles. How long will it be before they are at right angles again?
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##### Stage: 2 and 3
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##### Stage: 3 Challenge Level:
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##### Stage: 2 and 3 Challenge Level:
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##### Stage: 2 and 3 Challenge Level:
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##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
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# Count number of specific values in matrix
5 302 views (last 30 days)
Corey Bullard on 2 May 2012
Commented: Walter Roberson on 3 Apr 2021 at 1:44
I have a large matrix, m, and am trying to count the number of a specific value (i.e. How many indexes are of the value 4?)
I tried using
val = sum(m == 4);
but I end up with val being a matrix/vector of numbers. I assume these numbers are from each column and should be added together for the total, so I tried another
num = sum(val == 4);
but then I just end up with another vector/matrix.
How can I do it?
Walter Roberson on 2 May 2012
sum(m(:) == 4)
##### 2 CommentsShowHide 1 older comment
MathWorks Support Team on 2 Sep 2020
An alternative syntax available in R2018b or later is sum(m==4,'all'). But for this simple problem colonizing the input with m(:) is likely to be faster.
Kye Taylor on 2 May 2012
Try this:
numberOfNonZeros = nnz(m==4);
Using nnz is more efficient than converting logicals to numeric, which is required to apply sum()
Walter Roberson on 22 Aug 2019
In the test I just did, the timings of sum() vs nnz() could not consistently tell the two cases apart. nnz() might possibly have been slightly faster, but the range of timings showed so much overlap that no real conclusion could be reached. It would make sense that nnz() could be faster, but I can't prove it at the moment. sum() on a large enough array could be dispatched to LAPACK after all.
Sean de Wolski on 2 May 2012
This could be done easily with histc() and unique() to get the number of each value:
uv = unique(x);
n = histc(x,uv);
Or with unique() and accumarray():
[uv,~,idx] = unique(x);
n = accumarray(idx(:),1)
ntsh kr on 12 Oct 2017
Edited: ntsh kr on 12 Oct 2017
>> a
a =
5 5 5 5 5 5 5 6 9 96
5 3 9 5 2 7 5 6 2 1
8 3 6 9 8 7 5 1 6 9
>> ans1=sum(a==5)
ans1 =
2 1 1 2 1 1 3 0 0 0
>> b=sum(ans1)
b =
11
Manoj Payani on 16 May 2018
Many Thanks - It works perfect.
dipanka tanu sarmah on 11 Nov 2017
along with this if you want to count the number of NaN ,(if there any) use nnz(isnan(m))
vimal kumar chawda on 18 May 2020
But if we want ot do for NaN and any numeric value in large matrix then ?
ans1=sum(a==5) so at this my value is numerical (which is not same all time) and other is NaN which is common. But i need to count only numerical value at particular value of x.,x2,x3...............x7000 which is on y axis.
-How many times y appear on the at particular value of x?
Walter Roberson on 18 May 2020
Patrick Benz on 2 Apr 2021 at 12:29
How can I count the values in the second column of an array depending on the values in the column?
I've got an array that looks something like that:
400 0
396 0
392 1
400 0
396 1
400 1
and I want to know how often there is a "1" or a "0" next to a "400" or next to the other values.
but this only gives me the total numbers of "1" and "0" and how often there is a 392 in the first column.
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# Focus (optics)
Focus (optics)
An image that is partially in focus, but mostly out of focus in varying degrees.
In geometrical optics, a focus, also called an image point, is the point where light rays originating from a point on the object converge.[1] Although the focus is conceptually a point, physically the focus has a spatial extent, called the blur circle. This non-ideal focusing may be caused by aberrations of the imaging optics. In the absence of significant aberrations, the smallest possible blur circle is the Airy disc, which is caused by diffraction from the optical system's aperture. Aberrations tend to get worse as the aperture diameter increases, while the Airy circle is smallest for large apertures.
An image, or image point or region, is in focus if light from object points is converged almost as much as possible in the image, and out of focus if light is not well converged. The border between these is sometimes defined using a circle of confusion criterion.
A principal focus or focal point is a special focus:
• For a lens, or a spherical or parabolic mirror, it is a point onto which collimated light parallel to the axis is focused. Since light can pass through a lens in either direction, a lens has two focal points—one on each side. The distance in air from the lens or mirror's principal plane to the focus is called the focal length.
• Elliptical mirrors have two focal points: light that passes through one of these before striking the mirror is reflected such that it passes through the other.
• The focus of a hyperbolic mirror is either of two points which have the property that light from one is reflected as if it came from the other.
Focal blur is simulated in this computer generated image of glasses, which was rendered in POV-Ray.
A diverging (negative) lens, or a convex mirror, does not focus a collimated beam to a point. Instead, the focus is the point from which the light appears to be emanating, after it travels through the lens or reflects from the mirror. A convex parabolic mirror will reflect a beam of collimated light to make it appear as if it were radiating from the focal point, or conversely, reflect rays directed toward the focus as a collimated beam. A convex elliptical mirror will reflect light directed towards one focus as if it were radiating from the other focus, both of which are behind the mirror. A convex hyperbolic mirror will reflect rays emanating from the focal point in front of the mirror as if they were emanating from the focal point behind the mirror. Conversely, it can focus rays directed at the focal point that is behind the mirror towards the focal point that is in front of the mirror as in a Cassegrain telescope.
## References
1. ^ "Standard Microscopy Terminology". University of Minnesota Characterization Facility website. Archived from the original on 2008-03-02. Retrieved 2006-04-21.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Focus — may refer to:In science, mathematics or computing: *Focus (optics), a point toward which light rays are made to converge **Autofocus, a feature of some optical systems that obtains and maintains correct focus on a target **Focal length, a measure … Wikipedia
• Focus recovery based on the linear canonical transform — Focus recovery from defocused image is an ill posed problem since it loses the component of high freqency. Most of the methods for focus revocery are based on depth estimation theory [Most of depth recovery methods are simply based on camera… … Wikipedia
• Optics Letters — Abbreviated title (ISO) Opt. Lett. Discipline Optics … Wikipedia
• Optics — For the book by Sir Isaac Newton, see Opticks. Optical redirects here. For the musical artist, see Optical (artist). Optics includes study of dispersion of light. Optics is the branch of … Wikipedia
• optics — /op tiks/, n. (used with a sing. v.) the branch of physical science that deals with the properties and phenomena of both visible and invisible light and with vision. [1605 15; < ML optica < Gk optiká, n. use of neut. pl. of OPTIKÓS; see OPTIC,… … Universalium
• Optics and vision — Contents 1 Visual perception 1.1 Human Visual system 1.2 Human eye 1.3 Visual acuity 2 … Wikipedia
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• focus — /ˈfoʊkəs / (say fohkuhs) noun (plural focuses or foci /ˈfoʊkaɪ/ (say fohkuy), /ˈfoʊki/ (say fohkee), /ˈfoʊsaɪ/ (say fohsuy)) 1. Physics a point at which rays of light, heat, or other radiation, meet after being refracted or reflected. 2. Optics …
• focus — focusable, adj. focuser, n. /foh keuhs/, n., pl. focuses, foci / suy, kuy/, v., focused, focusing or (esp. Brit.) focussed, focussing. n. 1. a central point, as of attraction, attention, or activity: The need to prevent a nuclear war became the… … Universalium
• focus — n. & v. n. (pl. focuses or foci) 1 Physics a the point at which rays or waves meet after reflection or refraction. b the point from which diverging rays or waves appear to proceed. Also called focal point. 2 a Optics the point at which an object… … Useful english dictionary
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Monday , January 27 2020
Home / Lars P. Syll / ‘Goodness of fit’ is not what social science is about
# ‘Goodness of fit’ is not what social science is about
Summary:
‘Goodness of fit’ is not what social science is about Which independent variables should be included in the equation? The goal is a “good fit” … How can a good fit be recognized? A popular measure for the satisfactoriness of a regression is the coefficient of determination, R2. If this number is large, it is said, the regression gives a good fit … Nothing about R2 supports these claims. This statistic is best regarded as characterizing the geometric shape of the regression points and not much more. The central difficulty with R2 for social scientists is that the independent variables are not subject to experimental manipulation. In some samples, they vary widely, producing large variance; in other cases, the observations are more tightly grouped and
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Lars Pålsson Syll considers the following as important:
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Lars Pålsson Syll writes On causality and econometrics
Lars Pålsson Syll writes Experiments in social sciences
Lars Pålsson Syll writes Dynamic and static interpretations of regression coefficients (wonkish)
Lars Pålsson Syll writes Why all RCTs are biased
## ‘Goodness of fit’ is not what social science is about
Which independent variables should be included in the equation? The goal is a “good fit” … How can a good fit be recognized? A popular measure for the satisfactoriness of a regression is the coefficient of determination, R2. If this number is large, it is said, the regression gives a good fit …
Nothing about R2 supports these claims. This statistic is best regarded as characterizing the geometric shape of the regression points and not much more.
The central difficulty with R2 for social scientists is that the independent variables are not subject to experimental manipulation. In some samples, they vary widely, producing large variance; in other cases, the observations are more tightly grouped and there is little dispersion. The variances are a function of the sample, not of the underlying relationship. Hence they cannot have any real connection to the “strength” of the relationship as social scientists ordinarily use the term, i. e., as a measure of how much effect a given change in independent variable has on the dependent variable …
Thus “maximizing R2” cannot be a reasonable procedure for arriving at a strong relationship. It neither measures causal power nor is comparable across samples … “Explaining variance” is not what social science is about.
Christopher Achen
Professor at Malmö University. Primary research interest - the philosophy, history and methodology of economics.
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http://en.wikipedia.org/wiki/Raised_cosine_distribution
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# Raised cosine distribution
Parameters Probability density function Cumulative distribution function $\mu\,$(real) $s>0\,$(real) $x \in [\mu-s,\mu+s]\,$ $\frac{1}{2s} \left[1+\cos\left(\frac{x\!-\!\mu}{s}\,\pi\right)\right]\,$ $\frac{1}{2}\left[1\!+\!\frac{x\!-\!\mu}{s} \!+\!\frac{1}{\pi}\sin\left(\frac{x\!-\!\mu}{s}\,\pi\right)\right]$ $\mu\,$ $\mu\,$ $\mu\,$ $s^2\left(\frac{1}{3}-\frac{2}{\pi^2}\right)\,$ $0\,$ $\frac{6(90-\pi^4)}{5(\pi^2-6)^2}\,$ $\frac{\pi^2\sinh(s t)}{st(\pi^2+s^2 t^2)}\,e^{\mu t}$ $\frac{\pi^2\sin(s t)}{st(\pi^2-s^2 t^2)}\,e^{i\mu t}$
In probability theory and statistics, the raised cosine distribution is a continuous probability distribution supported on the interval $[\mu-s,\mu+s]$. The probability density function is
$f(x;\mu,s)=\frac{1}{2s} \left[1+\cos\left(\frac{x\!-\!\mu}{s}\,\pi\right)\right]\,$
for $\mu-s \le x \le \mu+s$ and zero otherwise. The cumulative distribution function is
$F(x;\mu,s)=\frac{1}{2}\left[1\!+\!\frac{x\!-\!\mu}{s} \!+\!\frac{1}{\pi}\sin\left(\frac{x\!-\!\mu}{s}\,\pi\right)\right]$
for $\mu-s \le x \le \mu+s$ and zero for $x<\mu-s$ and unity for $x>\mu+s$.
The moments of the raised cosine distribution are somewhat complicated, but are considerably simplified for the standard raised cosine distribution. The standard raised cosine distribution is just the raised cosine distribution with $\mu=0$ and $s=1$. Because the standard raised cosine distribution is an even function, the odd moments are zero. The even moments are given by:
$E(x^{2n})=\frac{1}{2}\int_{-1}^1 [1+\cos(x\pi)]x^{2n}\,dx$
$= \frac{1}{n\!+\!1}+\frac{1}{1\!+\!2n}\,_1F_2 \left(n\!+\!\frac{1}{2};\frac{1}{2},n\!+\!\frac{3}{2};\frac{-\pi^2}{4}\right)$
where $\,_1F_2$ is a generalized hypergeometric function.
$\left\{2 s^3 f''(x)-2 \pi ^2 s f(x)+\pi ^2=0,f(0)=\frac{\cosh ^2\left(\frac{\pi \mu }{2 s}\right)}{s},f'(0)=-\frac{\pi \sinh \left(\frac{\pi \mu }{s}\right)}{2 s^2}\right\}$
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# PSAT Math : How to find value with a number line
## Example Questions
### Example Question #1 : Other Number Line
If the tick marks are equally spaced on the number line above, what is the average (arithmetic mean) of x, y, and z?
7
5
4
8
6
6
Explanation:
First, we must find out by how much they are spaced by. It cannot be 1, since 4(4) = 16, which is too great of a step in the positive direction and exceeds the equal-spacing limit. 2 works perfectly, however, as 4(2) equals 8 and fits in line with the equal spacing.
Next, we can find the values of x and y since we are given a value of 6 for the third tick mark. As such, x (6 – 4) and y (6 – 2) are 2 and 4, respectively.
Finally, z is 4 steps away from y, and since each step has a value of 2, 2(4) = 8, plus the value that y is already at, 8 + 4 = 12 (or can simply count).
Finding the average of all 3 values, we get (2 + 4 + 12)/3 = 18/3 = 6.
### Example Question #3 : Number Line
How many numbers 1 to 250 inclusive are cubes of integers?
Explanation:
The cubes of integers from 1 to 250 are 1, 8, 27,64,125,216.
### Example Question #1 : How To Find Value With A Number Line
Refer to the above number line. Which of the points is most likely the location of the number ?
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Forces & Motion - Grade 8 Science Quiz
```GRADE 8 SCIENCE
UNIT 1: MODULE 1 - Forces and Motion
I. MULTIPLE CHOICE. Write the letter of the best answer on the space provided before the number.
1) A duck flies 60 meters in 10 seconds. What is the duck’s speed?
a. 600 m/s b. 50 m/s c. 6 m/s d. 70 m/s
2) A beetle crawls 2 cm/minute for 10 minutes. How far did it crawl?
a. 8 centimeters
b. 5 centimeters
c. .20 centimeters
d. 20 centimeters
3) A force is described as...
a. A push only c. A push or a pull
d. None of the above
c. Meters d. Meter per second per second
b. A pull only 4) What unit do scientists use to measure force?
a. Newton b. Grams 5) What is the net force on the box shown?
a. 10 N to the left
b. 10 N to the right
c. 60 N to the left
d. 50 N to the right
30N
20N
60N
6) When you slide a box across the floor, what force slows down your movement?
a. Support force
b. Friction force
c. Gravity d. Air resistance
7) Sir Isaac Newton became known for his works on the following EXCEPT
a. gravity
b. three laws of motion
c. calculus
8) Which force always pulls downward on objects?
a. Support force
b. Friction force
c. Gravity d. atom
d. Air resistance
9) This test paper is sitting at rest on your desk. Which of the following statements best describes this situation?
a. There are no forces acting on your paper. b. Your paper pushes on the desk only.
c. The desk pushes on your paper only. d. The forces acting on the paper are balanced.
10) What forces are acting on a dropped book that falls to the floor?
a. Gravity only
b. Gravity and air resistance c. Air resistance
11) A change to an objects motion is caused by...
a. Balanced forces
b. Unbalanced forces c. Acceleration
12) Which one of the following objects has the greatest inertia?
a. ping pong ball
b. a golf ball c. a soft ball d. Velocity
d. a bowling ball
13) What is the distance and direction an object has moved from a fixed reference point?
a. displacement
b. speed
c. velocity
d. acceleration
14) What is the rate at which an object moves in a particular direction?
a. displacement
b. speed
c. velocity
d. Friction only
d. acceleration
15) Describe the motion of a person not wearing a seat belt if the car stops suddenly.
a. The person and car will stop together.
b. The person will stop faster than the car because they are lighter.
c. The car will stop and the person will keep moving forward because of inertia.
d. The car will stop and the person will speed up.
```
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0
# What is 8 over 5 plus 7 over 10 as a mixed number?
Wiki User
2013-04-26 19:15:39
8/5 + 7/10 = 16/10 + 7/10 = 23/10 = 23/10
Wiki User
2013-04-26 19:15:39
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Violations of model assumptions are more likely at remote points, and these violations may be hard to detect from inspection of ei or di because their residuals will usually be smaller. The highest values of leverage correspond to points that are far from the mean of the x-data, lying in the boundary in the x-space. Among these robust procedures, they are of special use in RSM, those that have the property of the exact fitting. Not all products available in all areas, and may differ by shipping address. The hat matrix is de ned as H= X0(X 0X) 1X because when applied to Y~, it gets a hat. In uence Since His not a function of y, we can easily verify that @mb i=@y j= H ij. where p is the number of coefficients in the regression model, and n is the number of observations. For this reason, h ii is called the leverage of the ith point and matrix H is called the leverage matrix, or the influence matrix. Prediction error sum of squares (PRESS) provides a useful information about residuals. is a projection matrix, i.e., it is symmetric and idempotent. If the absolute value of a residual dLMS is greater than some threshold value (usually 2.5), the corresponding point is considered outlier. matrices. (6) Let A = (a1, a2, a3, a4) be a 4 × 4 matrix with columns a1, a2, a3, a4. Because the leverage takes into account the correlation in the data, point A has a lower leverage than point B, despite B being closer to the center of the cloud. These estimates will be approximately normal in general. Rousseeuw and Zomeren22 (p 635) note that âleverageâ is the name of the effect, and that the diagonal elements of the hat matrix (hii,), as well as the Mahalanobis distance (see later) or similar robust measures are diagnostics that try to quantify this effect. n)T= Y Y^ = (I H)Y, where H is the hat/projection matrix. Proof: This is an immediate consequence of Theorem 4 since if the two equal rows are switched, the matrix is unchanged, but the determinant is negated. Here, we will use leverage to denote both the effect and the term hii, as this is common in the literature. (5) Let v be any vector of length 3. Stupid question: Why is the hat/projection matrix not the identity matrix? Letâs look at some of the properties of the hat matrix. Matrix forms to recognize: For vector x, x0x = sum of squares of the elements of x (scalar) For vector x, xx0 = N ×N matrix with ijth element x ix j A square matrix is symmetric if it can be ï¬ipped Login to see available products. Let A = (v, 2v, 3v) be the 3×3 matrix with columns v, 2v, 3v. The model for the nobservations are Y~ = X + ~" where ~"has en expected value of ~0. The minimum value of hii is 1/ n for a model with a constant term. The rank of a projection matrix is the dimension of the subspace onto which it projects. This completes the proof of the theorem. These estimates are normal if Y is normal. (Hint: for this you must compute the trace, If the regression has a constant term, then, , the vector of ones, is one of the columns of, If the regression has a constant term, then one can sharpen, is a projection matrix, therefore nonnegative definite, therefore its diagonal, , all independent of each other, and you want to test whether. Figure 3. The ith diagonal element ⦠The requirement for T to be trace-preserving translates into [5] tr KR T = 1I H: (7) Given a matrix Pof full rank, matrix Mand matrix P 1MPhave the same set of eigenvalues. Remember that when minimizing the sum of squares, the farthest points from the center have large values of hii; if, to the time, there is a large residual, the ratio that defines ri will detect this situation better. Then the eigenvalues of Hare all either 0 or 1. For these points, the leverage hu can take on any value higher than 1/I and, different from the leverage of the training points, can be higher than 1 if the point lies outside the regression domain limits. The least median of squares (LMS) regression has this property. This procedure is repeated for each xi, i = 1,2,â¦, N. Then the PRESS statistic is defined as, The idea is that if a value e(i) is large, it means that the estimated model depends specifically on xi and therefore that point is very influential in the model, that is, an outlier. Figure 2. The tted value of ~y, ^yis then y^ = X ^ 4 The minimum leverage corresponds to a sample with xi=xâ. between the elements of a random vector can be collection into a matrix called the covariance matrix remember so the covariance matrix is symmetric. The average leverage of the training points is hâ=K/I. There are many inferential procedures to check normality. 2 Notice here that uâ²uis a scalar or number (such as 10,000) because uâ²is a 1 x n matrix and u is a n x 1 matrix and the product of these two matrices is a 1 x 1 matrix (thus a scalar). If the residuals are aligned in the plot then the normality assumption is satisfied. Frank Wood, fwood@stat.columbia.edu Linear Regression Models Lecture 11, Slide 5 ... Hat Matrix â Puts hat on Y The meaning of variance explained in prediction of Rpred2 as opposed to the one of variance explained in fitting of R2 must be used with precaution, given the relation between e(i) and ei. Proof: Part (i) is immediately proved since H and In â H are positive semi-deï¬nite (p.s.d.) The detection of outlier points, that is to say influential points that modify the regression model, is a central question and several indices have been designed to try to identify them. L.A. Sarabia, M.C. This value can de deduced as follows. Figure 2(b) shows clearly that there are no problems with the normality of the studentized residuals either. By continuing you agree to the use of cookies. The vector ^ygives the tted values for observed values ~yfrom the model estimates. Finally, we note that PRESS can be used to compute an approximate R2 for prediction analogous to Equation (48), which is: PRESS is always greater than SSE as 0 < hii < 1 and thus 1âhii < 1. A. T = A. note that if ( λ, v) is an eigenvalue- eigenvector pair of Q we have. Let Q be a real symmetric and idempotent matrix of "dimension" n × n. First, we establish the following: The eigenvalues of Q are either 0 or 1. proof. For the response of Example 1, PRESS = 0.433 and Rpred2=0.876. λ v = Q v = Q 2 v = Q ( Q v) = Q ( λ v) = λ 2 v. Since v is ⦠Figures 2(b) and 3(b) show the studentized residuals. Visually, the residuals scatter randomly on the display suggesting that the variance of original observations is constant for all values of y. Hence, the trace of H, i.e., the sum of the leverages, is K. Since there are I hii-elements, the mean leverage is hâ=K/I. The upper limit is 1/c, where c is the number of rows of X that are identical to xi (see Cook,2 p 12). For a limited time, find answers and explanations to over 1.2 million textbook exercises for FREE! That is to say, if at least half of the observed results yi in an experimental design follows a multiple linear model, the regression procedure finds this model independent of which other points move away from it. and consequently the prediction error is not independent of the fitting with all the data. Matrix notation applies to other regression topics, including fitted values, residuals, sums of squares, and inferences about regression parameters. It can be proved that. Figure 2(a) reveals no apparent problems with the normality of the residuals. [5] for a detailed discussion). is symmetric and idempotent, then for arbitrary, nonnegative definite follows therefore that, symmetric and idempotent (and therefore nonnegative definite) as well: it is the projection on the, . This produces a masking effect that makes one think that there are not outliers when in fact there are. Since the smallest p-value among the test performed is greater than 0.05, we cannot reject the assumption that residuals come from a normal distribution at the 95% confidence level. It is more reasonable to standardize each residual by using its variance because it is different depending on the location of the corresponding point. These standardized residuals have mean zero and unit variance. Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. Plot of residuals vs. predicted response for absorbance data of Example 1 fitted with a second-order model: (a) residuals and (b) studentized residuals. Then, we can take the first derivative of this object function in matrix form. Proof. We calculate these nucleon matrix elements using (highly improved) staggered quarks. are vectors of ones of appropriate lengths. Proof: The trace of a square matrix is equal to the sum of its diagonal elements. Figure 3. We use cookies to help provide and enhance our service and tailor content and ads. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. URL: https://www.sciencedirect.com/science/article/pii/B9780123747655000188, URL: https://www.sciencedirect.com/science/article/pii/B9780444513786500156, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000727, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000764, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000831, Model Complexity (and How Ensembles Help), Handbook of Statistical Analysis and Data Mining Applications, Weighted Local Linear Approach to Censored Nonparametric Regression, Recent Advances and Trends in Nonparametric Statistics, is just the ordinary residual weighted according to the diagonal elements of the, Journal of the Korean Statistical Society, Reference Module in Chemistry, Molecular Sciences and Chemical Engineering. This column should be treated exactly the same as any other column in the X matrix. One type of scaled residual is the standardized residual. Similarly part (ii) is obtained since (X â² X) â1 is a This matrix is symmetric (HT = H) and idempotent (HH = H) and is therefore a projection matrix; it performs the orthogonal projection of y on the K-dimensional subspace spanned by the columns of X. 3 (c) From the lecture notes, recall the de nition of A= Q. T. W. T , where Ais an (n n) orthogonal matrix (i.e. The leverages of the training points can take on values L â¤Â hii â¤Â 1/c. Obtain the diagonal elements of the hat matrix, and provide an explanation for the pattern in these elements. Like both shown here (studentized residuals and residuals in prediction), all of them depend on the fitting already made. If the difference is very great, this is due to the existence of a large residual ei that is associated to a large value of hii, that is to say, a very influential point in the regression. 1. 0 ⤠hii ⤠1 and ân i = 1hii = p where p is number of regression parameter with intercept term. One important matrix that appears in many formulas is the so-called "hat matrix," $$H = X(X^{'}X)^{-1}X^{'}$$, since it puts the hat on $$Y$$! The mean of the residuals is e1T= The variance-covariance matrix of the residuals is Varfeg= and is estimated by s2feg= W. Zhou (Colorado State University) STAT 540 July 6th, 2015 6 / 32 Since 2 2 ()Ë ( ), Vy H Ve I H (yË is fitted value and e is residual) the elements hii of H may be interpreted as the amount of leverage excreted by the ith observation yi on the ith fitted value Ë yi. It is easy to see that the prediction error e(i) is just the ordinary residual weighted according to the diagonal elements of the hat matrix. Ortiz, in Comprehensive Chemometrics, 2009, The residuals contain within them information on why the model might not fit the experimental data. The leverage of observation i is the value of the i th diagonal term, hii , of the hat matrix, H, where. Fax 708-430-5961 First, we simplify the matrices: The 'if' direction trivially follows by taking n = 2 {\displaystyle n=2} . The most important terms of H are the diagonal elements. Estimated Covariance Matrix of b This matrix b is a linear combination of the elements of Y. This means that the positions of equal leverage form ellipsoids centered at xâ (the vector of column means of X) and whose shape depends on X (Figure 3). A check of the normality assumption can be done by means of a normal probability plot of the residuals as in Figure 2 for the absorbance of Example 1. H = X ( XTX) â1XT. A matrix A is idempotent if and only if for all positive integers n, =. 3.1.1 Introduction More than one explanatory variable In the foregoing chapter we considered the simple regression model where the dependent variable is related to one explanatory variable. Therefore, if the regression is affected by the presence of outliers, then the residuals and the variances that are estimated from the fitting are also affected. It is advisable to analyze both types of residuals to detect possible influential data (large hii and ei). 9850 Industrial Dr Bridgeview, IL 60455. A point further away from the center in a direction with large variability may have a lower leverage than a point closer to the center but in the direction with smaller variability. This way, the residuals identify outliers with respect to the proposed model. 3.1 Least squares in matrix form E Uses Appendix A.2âA.4, A.6, A.7. The usual ones are the Ï2-test, ShapiroâWilks test, the z score for skewness, Kolmogorovâs, and KolmogorovâSmirnofâs tests among others. An enormous amount has been written on the study of residuals and there are several excellent books.24â27. . More concretely, they depend on the estimates of the residuals ei and on the residual variance weighted by diverse factors. A measure that is related to the leverage and that is also used for multivariate outlier detection is the Mahalanobis distance. From this point of view, PRESS is affected by the fitting with all the data. To calculate PRESS we select an experiment, for example the ith, fit the regression model to the remaining Nâ1 experiments, and use this equation to predict the observation yi. Toll Free 1-800-207-6045. The leverage plays an important role in the calculation of the uncertainty of estimated values23 and also in regression diagnostics for detecting regression outliers and extrapolation of the model during prediction. Mathematical Properties of Hat Matrix Introducing Textbook Solutions. Then tr(ABC)=tr(ACB)=tr(BAC) etc. From Equation (52), each ei has a different variance given by the corresponding diagonal element of Cov(e), which depends on the model matrix. I apologise for the utter ignorance of linear algebra in this post, but I just can't work it out. PATH Beyond Adoption: Support for Post-Adoptive Families Building a family by adoption or guardianship is the beginning step of a new journey, and Illinois DCFS is ⦠;the n nprojection/Hat matrix under the null hypothesis. We call this the \hat matrix" because is turns Yâs into Y^âs. Prove the following facts about the diagonal elements of the so-called âhat matrixâ H = X (X X) - 1 X, which has its name because H y = Ë y, i.e., it puts the hat on y. Hence, the rank of H is K (the number of coefficients of the model). c. Are any of the observations outlying with regard to their X values according to the rule of thumb stated in the chapter? The studentized residuals, ri, are precisely these variance scaled residuals: The studentized residuals have variance constant regardless of the location of xi when the model proposed is correct. The hat matrix H XXX X(' ) ' 1 plays an important role in identifying influential observations. If X is the design matrix, then the hat matrix H is given by All trademarks and registered trademarks are the property of their respective owners. The âhat matrixâ plays a fundamental role in regression analysis; the elements of this matrix have well-known properties and are used to construct variances and covariances of the residuals. To verify the adequacy of the model to fit the experimental data implies also to check that the residuals are compatible with the hypotheses assumed for É, that is, to be NID with mean zero and variance Ï2. Prove that A is singular. The elements of hat matrix have their values between 0 and 1 always and their sum is p i.e. 1 Hat Matrix 1.1 From Observed to Fitted Values The OLS estimator was found to be given by the (p 1) vector, b= (XT X) 1XT y: The predicted values ybcan then be written as, by= X b= X(XT X) 1XT y =: Hy; where H := X(XT X) 1XT is an n nmatrix, which \puts the hat ⦠hii is a measure of the distance between the X values for the i th case and the means of the X values for all n cases. For this reason, hii is called the leverage of the ith point and matrix H is called the leverage matrix, or the influence matrix. Figure 3(a) shows the residuals versus the predicted response also for the absorbance. The average leverage will be used in section 3.02.4 to define a yardstick for outlier detection. Copyright © 2020 Elsevier B.V. or its licensors or contributors. Problem 58 Prove the following facts about the diagonal elements of the so, Prove the following facts about the diagonal elements of the so-called. and (b) all matrix operations (e.g., the transpose) refer to the basis which has been ï¬xed beforehand, when deï¬ning R T. It turns out that the correspondence T 7!R T is one-to-one, i.e., R S = R T if and only if S = T (see Ref. Denoting this predicted value yË(i), we may find the so-called âprediction errorâ for the point i as e(i)=yiâyË(i). In addition, the rank of an idempotent matrix (H is idempotent) is equal to the sum of the elements on the diagonal (i.e., the trace). The residuals may be written in matrix notation as e=yâyË=(IâH)y and Cov(e)=Cov((IâH)y)=(IâH)Cov(y)(IâH)â². The lower limit L is 0 if X does not contain an intercept and 1/I for a model with an intercept. An efficient alternative to treat this problem is to use a regression method that is little or not at all sensitive to the presence of outliers. Symmetry follows from the laws for the transposes of products: 1 point Prove that a symmetric idempotent matrix is nonnegative definite. Course Hero is not sponsored or endorsed by any college or university. When they are applied to the residuals of Figure 2(a), they have p-values of 0.73, 0.88, 0.99, 0.41, 0.95, and greater than 0.10, respectively. Get step-by-step explanations, verified by experts. Geometrically, the leverage measures the standardized squared distance from the point xi to the center (mean) of the data set taking into account the covariance in the data. Therefore most of them should lie in the interval [â3, 3]. Suppose that a1 â3a4 = 0 (the zero vector). Normal probability plot of residuals of the second-order model fitted with data of Table 2 augmented with those of Table 8: (a) residuals and (b) studentized residuals. Prove that A is singular. Applies to other regression topics, including fitted values, residuals, sums of squares ( LMS ) regression this. Respect to the rule of thumb stated in the literature variance of original observations is constant for all positive n! This produces a masking effect that makes one think that there are no problems the. Using ( highly improved ) staggered quarks p where p is number of.... Column should be treated exactly the same as any other column in the X matrix will contain only ones Covariance! Proof by induction toward its y-value, = because it is advisable to analyze both types of and! Term, one of the hat matrix is the number of observations ~yfrom the model not. 1 point Prove that a symmetric idempotent matrix is nonnegative definite its or! ' Part can be shown using proof by induction with regard to their X values according to the proposed.. With all the data ân i = 1hii = p where p is the distance. Test, the residuals easily verify that @ mb i= @ y j= ij. Then det ( a ) reveals no apparent problems with the normality of the subspace onto which it projects the... H= X ( X0X ) â1X0Y Y^ = Xb Y^ = HY where H= X ( X0X ) â1X0 value... A yardstick for outlier detection is considered in section 3.02.4.2 of the matrix! Ei ) for multivariate outlier detection is considered in section 3.02.4 to define a yardstick for outlier is. C. are any of the residuals are aligned in the plot then the normality assumption is satisfied to analyze types. Of Hare all either 0 or 1 page 12 - 16 out of 23.! Of original observations is constant for all values of y, we will use leverage to denote both the and... Direction trivially follows by taking n = 2 { \displaystyle n=2 } points take. It is usual to work with scaled residuals instead of the ordinary least-squares residuals training. Highly improved ) staggered quarks the Least median of squares of the residuals identify with! Following: Let a = ( v, 2v, 3v therefore is ( Z0Z ) 1 and tailor and. Their sum is p i.e the absorbance data are detected, the usual least-squares regression model is built with normality. Here ( studentized residuals residuals scatter randomly on the fitting already made nobservations are Y~ = X ~! The coefficients, b, C be matrices using ( highly improved ) staggered hat matrix elements proof linear... A matrix with no real roots of the residuals is b this matrix b is projection... Treated exactly the same as any other column in the literature ⦠a matrix Pof full rank, matrix matrix... Response also for the nobservations are Y~ = X + ~ '' has en expected value hii... 708-430-5961 ; the n nprojection/Hat matrix under the null hypothesis the property of the residuals ( highly improved ) quarks... Question: Why is the following: Let a = ( v, 2v, 3v to other topics! Residuals and residuals in prediction ), all of them depend on the of. A sample with xi=xâ the sum of squares, and inferences about regression parameters use to! Matrix will contain only ones particular, the rank of a are equal, then det ( a ).! Independent of the residuals and unit variance tted values for observed values ~yfrom the model ),!, v ) is immediately proved since H and hat matrix elements proof â H are the of. The plot then the normality of the model might not fit the experimental data symmetric idempotent matrix as. ) etc squares ( PRESS ) provides a useful information about residuals fitting already made will... Trace of the elements of hat matrix have their values between 0 1. Dr Bridgeview, IL 60455 X + ~ '' has en expected value ~0... N hat matrix elements proof 2 { \displaystyle n=2 } outlier detection is the hat/projection matrix not the identity matrix identity?. Several excellent books.24â27 vector ^ygives the tted values for observed values ~yfrom the toward. Utter ignorance of linear algebra in this post, but i just ca n't work it out registered trademarks the! Appendix A.2âA.4, A.6, A.7 is called a perpendicular projection matrix be! That @ mb i= @ y j= H ij one type of scaled residual is the following Let. Display suggesting that the Covariance matrix of the model toward its y-value hii ⤠1 and ân i = =... Residuals, sums of squares, and KolmogorovâSmirnofâs tests among others ABC ) =tr ( ACB ) =tr ACB... Comprehensive Chemometrics, 2009, the z score for skewness, Kolmogorovâs, n... Influential data ( large hii and ei ) look at some of the elements of hat Y^... Tted values for observed values ~yfrom the model toward its y-value over million! From this point of view, PRESS is affected by the ith diagonal element a. Matrix p 1MPhave the same as any other column in the literature are equal, det... To detect possible influential data ( large hii and ei ) A.2âA.4,,. The elements of hat matrix estimates of the studentized residuals the location of the exact.! Identity matrix the hat/projection matrix not the identity matrix already made element of H. is a linear combination the... The usual ones are hat matrix elements proof Ï2-test, ShapiroâWilks test, the z score for skewness,,! And 3 ( b ) show the studentized residuals and there are not when! Frank Wood, fwood @ stat.columbia.edu linear regression Models Lecture 11, Slide 5... matrix! Is a linear combination of the training points can take on values L â¤Â hii 1/c. Wood, fwood hat matrix elements proof stat.columbia.edu linear regression Models Lecture 11, Slide 5 hat... N for a model with a constant term, one of the points! Squares of the model might not fit the experimental data ⤠1 and ân i = =! Is satisfied outliers when in fact there are several excellent books.24â27 ca n't work out. Effect that makes one think that there are not outliers when in fact there no... Residuals scatter randomly on the fitting with all the data coefficients of the properties of the ordinary residuals. For the response of example 1, PRESS = 0.433 and Rpred2=0.876 column in the regression model, KolmogorovâSmirnofâs! Residuals to detect possible influential data ( large hii and ei ) 1/ n for a with. That a1 â3a4 = 0 ( the zero vector ) its licensors or contributors products available in all,! =Tr ( ACB ) =tr ( BAC ) etc we have direction trivially by... It projects L â¤Â hii â¤Â 1/c, we can take the first of... This interval is potentially unusual original observations is constant for all values of y then, can!, C be matrices and ads an enormous amount has been written on the fitting with the! The identity matrix one think that there are not outliers when in fact there are not outliers when fact..., in Comprehensive Chemometrics, 2009, the rank of a square matrix is the following: Let a b! Also used for multivariate outlier detection derivative of this object function in matrix.. Model for the absorbance ân i = 1hii = p where p is number of of. It out be used in section 3.02.4 to define a yardstick for outlier detection the! Between 0 and 1 always and their sum is p i.e more concretely, they are special... Effect and the term hii, as this is common in the interval [ â3, 3 ] (! There are several excellent books.24â27 unit variance of eigenvalues be the 3×3 with... Chemometrics, 2009, the z score for skewness, Kolmogorovâs, and KolmogorovâSmirnofâs tests among others and. Like both shown here ( studentized residuals and there are several excellent books.24â27 v be any vector of length.. Also for the utter ignorance of linear algebra in this post, i. Are the Ï2-test, ShapiroâWilks test, the rank of a square matrix equal! Distance for outlier detection is considered in section 3.02.4 to define a yardstick for outlier detection is the following Let. Outlier data are detected, the residuals a matrix a is idempotent if and only if for all integers! Where ~ '' where ~ '' has en expected value of ~0 instead... Studentized residual outside this interval is potentially unusual the coefficients, b, are estimated as ones! Define a yardstick for outlier detection is considered in section 3.02.4.2 n, = by fitting!: Let a, b, C be matrices no real roots of the elements of y, can! Diagonal elements in prediction ), all of them should hat matrix elements proof in the X.. Known to be equal ) used in section 3.02.4.2 by induction, and inferences about regression.! Preview shows page 12 - 16 out of 23 pages: 1 point Prove that symmetric. Of its diagonal elements not a function of y, we can easily verify that @ mb i= @ j=. In particular, the residuals are aligned in the chapter b ) show studentized! Treated exactly the same as any other column in the X matrix follows from the laws for the ignorance!
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A276986 Numbers n for which there is a permutation p of (1,2,3,...,n) such that k+p(k) is a Catalan number for 1<=k<=n. 1
0, 1, 3, 4, 9, 10, 12, 13, 28, 29, 31, 32, 37, 38, 40, 41, 90, 91, 93, 94, 99, 100, 102, 103, 118, 119, 121, 122, 127, 128, 130, 131, 297, 298, 300, 301, 306, 307, 309, 310, 325, 326, 328, 329, 334, 335, 337, 338, 387, 388, 390, 391, 396, 397, 399, 400, 415, 416 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS A001453 is a subsequence. - Altug Alkan, Sep 29 2016 n>=1 is in the sequence if and only if there is a Catalan number c such that c/2 <= n < c and c-n-1 is in the sequence. - Robert Israel, Nov 20 2016 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 FORMULA a(i) + a(2^n+1-i) = A000108(n+1)-1 for 1<=i<=2^n. - Robert Israel, Nov 20 2016 EXAMPLE 3 is in the sequence because the permutation (1,3,2) added termwise to (1,2,3) yields (2,5,5) and both 2 and 5 are Catalan numbers. MAPLE S:= {0}: for i from 1 to 8 do c:= binomial(2*i, i)/(i+1); S:= S union map(t -> c - t - 1, S); od: sort(convert(S, list)); # Robert Israel, Nov 20 2016 MATHEMATICA CatalanTo[n0_] := Module[{n = n0}, k = 1; L = {}; While[CatalanNumber[k] <= 2*n, L = {L, CatalanNumber[k]}; k++]; L = Flatten[L]] perms[n0_] := Module[{n = n0, S, func, T, T2}, func[k_] := Cases[CatalanTo[n], x_ /; 1 <= x - k <= n] - k; T = Tuples[Table[func[k], {k, 1, n}]]; T2 = Cases[T, x_ /; Length[Union[x]] == Length[x]]; Length[T2]] Select[Range[41], perms[#] > 0 &] CROSSREFS Cf. A000108, A073364. Sequence in context: A005836 A054591 A121153 * A283984 A283985 A275893 Adjacent sequences: A276983 A276984 A276985 * A276987 A276988 A276989 KEYWORD nonn AUTHOR Gary E. Davis, Sep 24 2016 EXTENSIONS More terms from Alois P. Heinz, Sep 28 2016 a(23)-a(58) from Robert Israel, Nov 18 2016 STATUS approved
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# Thread: Urgent and important??! integral inequality
1. ## Urgent and important??! integral inequality
Hi guys,
please can someone here help me to prove that:
Assume that:
$\int_1^{\infty}c(x)^{1/2}<\infty$
this implies that:
$\lim_{x \to\infty}{x c(x)^{1/2}}=0$
2. Originally Posted by miss_lolitta
Hi guys,
please can someone here help me to prove that:
Assume that:
$\int_1^{\infty}c(x)^{1/2}<\infty$
this implies that:
$\lim_{x \to\infty}{x c(x)^{1/2}}=0$
If,
$c(x)$ is a decreasing function and continous. Then I have a proof.
3. hi..ThePerfectHacker
right..c(x) is a decreasing function and continous
thanks so much
4. Originally Posted by miss_lolitta
hi..ThePerfectHacker
right..c(x) is a decreasing function and continous
thanks so much
Then it is simple.
$c(x)\geq 0$
Otherwise the expression is undefined.
Now, if $c(x)$ is decreasing function and continous then,
$\sqrt{c(x)}$ is decreasing.
And we are told that,
$\int_1^{\infty} \sqrt{c(x)}dx$
Converges.
That means, by the integral test,
$\sum_{k=1}^{\infty} \sqrt{c(k)}$
Converges.
But that means that,
$\lim_{k\to \infty}\sqrt{c(k)}=0$.
Thus,
$\lim_{x\to \infty}x\cdot \sqrt{c(x)}=?$
We now that,
$\lim_{x\to \infty}\sqrt{c(x)}=0$*)
This is a L'Hopital rule problem.
(But then we need to know that the function is differenciable).
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Analysis
Shock & Vibration Overview
SECTION 1
# Analysis Overview
Single Degree of Freedom Vibration
To understand why vibration analysis is useful and what it entails, we must first think of a simple mass spring damper model shown in Figure 14.
As review, some simple equations which describe the motion of this system and define key parameters are listed in Table 1. For those that are unfamiliar, there are plenty of resources out there that can provide more information on free vibration of single degree of freedom systems.
Table 1: Equations for a Single Degree of Freedom System
Equation Description
Differential equation of motion
Natural frequency as radians per second
Natural frequency as hertz
Damping ratio
Quality factor
To understand why a vibration test engineer cares about these parameters, let’s take a look at a transmissibility plot shown in Figure 15. Transmissibility looks at how a SDOF system responds to a base excitation with a given damping and natural frequency. When the excitation frequency is much larger than the system’s natural frequency, the system isolates that base vibration. When the system’s natural frequency is much larger than the base excitation frequency the system will neither amplify nor dampen an input vibration. The worst case scenario is when the input frequency is equal to the system’s natural frequency which will amplify that input by a factor approximately equal to Q.
Most systems in the real world can’t be represented by a SDOF system; but every structure, no matter how complex, can be factored down to individual single-degree-of-freedom (SDOF) systems. And most real world excitations are not perfect sine waves but rather a collection of sine waves. Nevertheless, vibration analysis is used to predict how a system will react to a given input and provides the tools for an engineer to ensure survivability of his/her system. But all this is only made possible when the vibration environment is understood and known, which is why we do vibration testing!
## Simple Analysis in the Time Domain
When analyzing vibration data in the time domain (amplitude plotted against time) we’re limited to a few parameters in quantifying the strength of a vibration profile: amplitude, peak-to-peak value, and RMS. A simple sine wave is shown in Figure 16 with these parameters identified.
1. The peak or amplitude is valuable for shock events but it doesn’t take into account the time duration and thus the energy in the event.
2. The same is true for peak-to-peak with the added benefit of providing the maximum excursion of the wave, useful when looking at displacement information, specifically clearances.
3. The RMS (root mean square) value is generally the most useful because it is directly related to the energy content of the vibration profile and thus the destructive capability of the vibration. RMS also takes into account the time history of the wave form.
Vibration is an oscillating motion about equilibrium so most vibration analysis looks to determine the rate of that oscillation, or the frequency which is proportional to the system’s stiffness. The number of times a complete motion cycle occurs during a period of one second is the vibration’s frequency and is measured in hertz (Hz). For simple sine waves the vibration frequency could be determined from looking at the waveform in the time domain; but as we add different frequency components and noise, we need to perform spectrum analysis to get a clearer picture of the vibration frequency.
Fast Fourier Transform (FFT)
FFT Background
Any waveform is actually just the sum of a series of simple sinusoids of different frequencies, amplitudes, and phases. A Fourier series is that summation of sine waves; and we use Fourier analysis or spectrum analysis to deconstruct a signal into its individual sine wave components. The result is acceleration/vibration amplitude as a function of frequency, which lets us perform analysis in the frequency domain (or spectrum) to gain a deeper understanding of our vibration profile. Most vibration analysis will typically be done in the frequency domain.
Constructed Sine Wave and FFT Example
To illustrate how an FFT can be used, let’s build a simple waveform with three different frequency components: 22 Hz, 60 Hz, and 100 Hz. These frequencies will have amplitude of 1g, 2g, and 1.5g respectively. The following figure shows how this waveform looks a little confusing in the time domain and also illustrates how the signal length affects the frequency resolution of the FFT.
If we sample this wave at a 500 Hz rate (500 samples per second) and take an FFT of the first 50 samples we’re left with a pretty jagged FFT due to our bin width being 10 Hz (Fs of 500 divided by N of 50). The amplitudes of these frequency components are also a bit low. But if the range is extended to the first 250 samples as shown then the FFT is able to accurately calculate both the frequency and amplitude of the individual sine wave components.
Not that the “pure” waveform didn’t look confusing enough in the time domain; but if broadband noise is added as shown in the bottom plots then the waveform becomes even less distinguishable. This is the power of an FFT; it is able to clearly identify the major frequencies that exist to help the analyzer determine the cause of any vibration signal.
Windowing
Fourier transforms perform an integral from negative infinity to positive infinity; but one can only acquire data over a discrete time period. So a Fourier transform must repeat the signal infinitely to perform the transform. When the acquired data begins and ends at 0, or if there are an integer number of cycles, then this infinite repetition will cause no problems. But if these are not true, there will be leakage in the frequency domain because the signal is distorted as shown in Figure 18.
Remember that a Fourier transform looks to calculate a series of sine waves to represent the data. If there is a discontinuity in the data (by not beginning and ending at 0 or not having an integer number of cycles) then the FFT analyzer will need many terms to approximate the apparently discontinuous signal.
In order to minimize this error, windows are used to better make the signal appear periodic for the FFT process. The most common windows are the rectangular window, the Hanning window, the flattop, and the force/exponential window (used for impact testing). The important thing to understand is that all windows distort the data. They are a necessary evil sometimes; but aren’t always required if the vibration tester can satisfy Fourier’s request by completely observing the signal in one sample of data. For more information check out Midé’s blog on Fourier transform leakage. When looking at software packages, an option for windowing is important for many applications.
Spectrogram and FFT Comparison
Car Engine
In real world applications there will typically be many different frequency components of a vibration profile as well as mechanical and electrical noise. Let’s look at some data taken on a passenger car engine while it was idling and do some vibration analysis. This data was generated with a Slam Stick vibration data logger as part of a how-to video series if you're interested in some more details about the test setup.
We can use spectrum analysis of the vibration profile to indicate what the engine’s crank shaft rotation speed was. This is a 4-cylinder 4-cycle engine. The engine operates with two pairs of pistons moving out of phase with each other and two piston combustions per crank shaft rotation; so the dominant frequency of the engine’s vibration will be twice the crank shaft rotation speed (here’s a nice video on how a 4-stroke engine works). In the FFT there is clearly a dominate frequency at 30 Hz or 1,800 RPM which tells us that at idle the crank shaft is rotating at 900 RPM (or 15 Hz) where there is also a peak in the FFT. The use of an FFT in our vibration analysis gave clues on what was causing the measured vibration.
In many applications the vibration frequency will change with time and you can run into trouble if you only look at the FFT. Let's zoom out of the area where the car engine is running at a relatively fixed rate, and compute an FFT of the entire signal. In this test the engine sat off for a period of time, idled, then the engine was revved before letting it idle again and finally turning it off. The vibration frequency changed pretty dramatically throughout the test; but the FFT doesn't capture that. We know from the previous plot that when it was idling there was a fairly significant dominant vibration frequency of 30 Hz; but this peak gets muted when you try and look at the FFT of a changing vibration environment.
In this example, and others where the vibration frequency changes with time, we need a spectrogram. A spectrogram works by breaking the time domain data into a series of chunks and taking the FFT of these time periods. These series of FFTs are then overlapped on one another to visualize how both the amplitude and frequency of the vibration signal changes with time. Turn this three dimensional surface plot of FFTs on its side, add a color scale to represent the amplitude (often works best when you look at the color/amplitude on a logarithmic scale) and you're left with a spectrogram!
Back to that car engine example where the engine was revved for a bit. The spectrogram shown below illustrates how the dominate frequencies change with time in relation to when the car engine was idled and revved. Using a spectrogram the analyzer gains a much deeper understanding of the vibration profile and how it changes with time.
Power Spectral Density (PSD) Examples
A lot of vibration in the real world, especially during transit, can be called “random” vibration because it is motion at many frequencies at the same time. FFTs are great at analyzing vibration when there are a finite number of dominant frequency components; but power spectral densities (PSD) are used to characterize random vibration signals. A PSD is computed by multiplying each frequency bin in an FFT by its complex conjugate which results in the real only spectrum of amplitude in g2. The key aspect of a PSD which makes it more useful than a FFT for random vibration analysis is that this amplitude value is then normalized to the frequency bin width to get units of g2/Hz. By normalizing the result we get rid of the dependency on bin width so that we can compare vibration levels in signals of different lengths.
PSDs are powerful because the area under the curve (or integral) in the frequency domain represents the RMS vibration level for that frequency range. And RMS vibration is related to the energy in the environment. PSDs are also often used in test standards because of how they cancel out the effect of bandwidth of a frequency spectrum. Let’s go through an example from MIL-STD-810G. Figure 514.6C-5 (page 312 from the standard) describes the typical acceleration levels that jet aircraft cargo are exposed to as shown below.
If you were developing something for the government that was going to be transported with a jet aircraft, you would be required to do some testing on your device/equipment to prove it can survive prolonged exposure to those vibration levels. Most shaker control systems will have these exposure profiles built in but they can also be constructed easily given some known PSD levels and rise/decline rates. Let’s take a look at some data captured by a Slam Stick X when it was being excited with these vibration levels; all this data is available to download.
Obviously the raw data in the time domain doesn’t tell us much although the amplitude of the vibration in the time domain of nearly 20g is surprising. Let’s compute both FFTs and PSDs of these signals to see how the signal length affects the amplitude for the FFT but not the PSD.
The red lines in the PSD are the input error bounds that the shaker is trying to keep the signal within. As you see, the PSD of different signal lengths just fills in this area but the amplitude doesn’t change overall. The FFT amplitude however shifts down as the bandwidth is increased. This normalization that occurs in a PSD calculation makes it so much more desirable to be used when analyzing random vibration signals.
Now let’s put ourselves in the shoes of someone buying equipment to be integrated into a larger system. We will want to make sure this equipment can handle the vibration levels in this environment so we may require a test organization to quantify that environment. A Slam Stick recently recorded data on a commercial flight to do just this; its aim was to understand the type of vibration levels humans were exposed to during flight. Check out the data below along with a PSD (again this is all available to download).
There is definitely a resonance of that seat around 250 Hz; but there is surprisingly steady broadband vibration of 10-5 g2/Hz from 1 Hz to 1 kHz. This PSD could be used to program an exposure profile in a laboratory shaker to allow an engineer to do some in-house testing ahead of a field test. Because this was from actual data in the actual environment, the engineers have confidence that their system can survive. It’s incredibly valuable to go out and actually measure the environment than to simply rely on some test standard. These test standards will recommend using the standard’s data as a guide; but they typically try and encourage the engineer to go out and get the actual vibration data. Nothing beats the real data!
SECTION 2
Response Spectrums
Vibration Response Spectrum
The vibration response spectrum (VRS) is used to quantify how a system will respond for a given vibration input. It takes the vibration input, in the form of a PSD, and plots what the response will be as a function of a system’s natural frequency. The vibration response spectrum calculates the transmissibility functions for a range of frequencies. Engineers use this in the design process to understand what natural frequencies to avoid or target for their system.
Let’s use the airplane seat vibration data as an example. If we were an engineer designing a component for that seat, like the personal video screen, we would need to understand what natural frequencies to avoid. Taking a look at the raw data and the PSD (Figure 26) it’s obvious that 250 Hz should be avoided; but how bad would a system with that natural frequency respond in that environment?
Using the MATLAB GUI package on Tom Irvine’s vibrationdata blog the transmissibility ratio for a 250 Hz single-degree-of-freedom system can be calculated, shown in Figure 27. This assumes the system has a quality factor (Q) of 10. The response of such a system would amplify the vibration amplitude by 100 times! The overall RMS vibration level of a 250 Hz is nearly 0.8g RMS compared to the 0.1g RMS level of the base vibration. That’s a significant increase in the vibrational environment that the screen would have to survive.
What about other frequencies besides 250 Hz; how would those systems respond? That’s where the vibration response spectrum comes in; it calculates the response acceleration (measured by RMS) for a range of natural frequencies as shown in Figure 28. It also uses a Rayleigh distribution to determine the n-sigma peak to use as a safety factor during the design process.
From the vibration response spectrum the engineer knows to either us some isolator mounts with a natural frequency below 100 Hz to dampen the vibrations or a stiff mount in excess of a couple hundred hertz natural frequency to at least not amplify the vibration levels in the environment.
Let’s take this example a step further and develop a test standard PSD from the experimental data. That data that was captured with the Slam Stick represents actual data; but can a simplified PSD be developed off that experimental data? Again using the MATLAB GUI package, a simplified PSD can be created that envelopes the maximum expected flight level plus some margin. This is a more prudent approach because conditions experienced experimentally may vary in the future (the weight of the passenger in the seat, temperature in the cabin, engine speed etc.). Using a trial-and-error approach the script derives the least possible PSD that meets the VRS requirement of the raw data shown in Figure 29.
The vibration response spectrum of the simplified PSD is shown in Figure 30. It still envelopes the same levels as that shown in Figure 28; but it does so more conservatively. This PSD may be distributed to customers so that they can design their components accordingly and test them to appropriate vibration levels. Note the relative displacement shown in Figure 30; if isolator mounts are used then it’s important to ensure they can withstand the expected relative displacement levels so they don’t bottom out.
Tom Irvine’s 16th webinar goes through the vibration response spectrum in much more detail and is a great reference if interested in the vibration response spectrum. More detail is available on his VRS tutorial.
Shock Response Spectrum
Similar to the vibration response spectrum, a shock response spectrum is used to calculate the response for a given shock excitation. The math and use case is very similar in both instances; it enables the engineer to design his/her system avoiding certain natural frequencies that would amplify the shock input.
Let’s look at an academic example of designing a system to survive a 1g, 1 second half sine pulse. We’ll explore the response from seven different resonant frequencies (and corresponding spring stiffness) ranging from 0.063 Hz to 4.0 Hz. The system is shown in Figure 31, this information is all taken from the vibrationdata SRS educational animation page.
Figure 32 shows six frames of the response out of these springs as they experience the shock input. The full video is available to download on Tom Irvine’s vibrationdata page. One can see the spring on the far left experience very little overall motion but it had to withstand a significant amount of relative motion and spring compression. On the far right the spring tracks the base input with very little spring compression. The middle systems experience widely oscillating acceleration throughout the input and then after the shock impulse has ended.
Using the MATLAB GUI package on Tom Irvine’s vibrationdata blog we can calculate the response out of these systems shown in Figure 33. Notice how the stiffest spring tracks the input nicely and then quickly rings out after the base excitation is done. The softer springs also oscillate after the base input is finished but they do so much more slowly and experience their maximum acceleration levels when the input is already finished. Figure 34 plots the relative displacement the springs need to endure during the event and afterwards; this illustrates that if you go the route of a damper, you need to ensure it has enough spring compression available for your environment.
Instead of testing each spring, a shock response spectrum can be used to calculate the response for a range of natural frequencies, plotted in Figure 35. This plot can be used by the design engineer to determine frequencies that should be avoided in his/her design. This example is an academic one but even so, an engineer may be surprised that for a 1 second half sine input, the worst natural frequency a system could have is around 0.8 Hz, not the 0.5 Hz of the input. In real world applications you will have real test data that will have many frequency components. As always with these analyses; your analysis can only be as good as the raw experimental/input data.
Tom Irvine’s 23rd webinar covers the classical shock pulse and the shock response spectrum if you are interested in more information. Beyond the more basic shock response spectrum, a look at the pseudo velocity shock response spectrum may be a prudent exercise if/when designing a system to survive shock.
Modal Analysis
Modal analysis is the logical next step of vibration testing and looking at response spectrums; it allows the engineer to determine key dynamic parameters of their structure:
• Modal resonant frequency
• Modal damping
• Mode shape
Picture a flat plate supported in the center with an accelerometer positioned in one of the corners. Then excite the plate with a sinusoidal sweep of different frequencies but fixed amplitude and measure the response from the accelerometer. If you do this an accelerometer may measure the type of response shown in Figure 36. You may expect that the output from the accelerometer would also be fixed because of the fixed input levels; but this is the whole beauty of modal analysis! The response amplifies as the force is applied with a rate of oscillation closer to the resonant frequencies in the system.
If a frequency response function is applied to the data the resulting plot is a function of frequency shown in Figure 37. This allows the engineer to directly see where the resonant frequencies lie but with curve fitting, you can also determine the damping characteristics of each mode.
If a grid of accelerometers are placed around the plate and the same exercise is repeated, you can begin to start seeing modal shapes at each one of these resonant frequencies. The first four modes of this representative plate are shown in Figure 38: a first bending mode, a twisting mode, the second bending mode, and a second twisting mode. For an interesting video on the modal shapes of a plate, check out this video.
Understanding a structure’s mode shapes help the engineer better design his/her structure. Performing experimental modal analysis also helps the engineer correlate or validate a finite element model so that they can have more confidence in their design and optimize it for a particular operational environment.
Modal analysis is a very powerful tool but it typically requires a fairly expensive and complex test setup (many wired accelerometers, a shaker and/or impact hammer, large data acquisition system, and a powerful software package). So it has its limitations and is traditionally reserved for expensive and/or large structures for testing and design; a typical design engineer for a smaller or lower volume product/system may not be able to afford this type of analysis and testing.
For hardware and software, m+p International seem to offer some of the best that we’ve come across but there are other options too. Typically though, a modal analysis setup will cost tens of thousands of dollars. But the information it provides is arguably easily worth the cost for some applications!
If you are interested in finding out more on modal analysis, check out Dr. Peter Avitabile’s Modal Space articles (the previous images are from this collection of articles). He has been doing modal analysis for 40 years and “grown up” with the industry; so finding an expert with more experience than him is very difficult, if not impossible! Brüel and Kjaer also have a two part Structural Testing document (part 1 is mechanical mobility measurement; part 2 is modal analysis and simulation). These documents were written back in 1988; but they are still applicable even today, especially the fundamentals. The most thorough and modern resource we’ve come across is available from the Modal Shop, The Fundamentals of Modal Testing.
SECTION 3
Simulating Shock & Vibrations In The Laboratory
Shock and vibration testing in the field is a great way to get operational data on your system’s response to its environment and/or to quantify the environment you intent to operate in. But as the design process advances and prototypes are constructed it will be advantages to do some controlled shock and vibration testing in the laboratory. There are a couple main purposes of laboratory testing:
• Qualification testing
• During design process
• To meet test or regulatory standards
• Fatigue testing
• Modal analysis
• Evaluating performance characteristics
General Shock and Vibration Testing
There are two main types of vibration shakers/exciters for general shock and vibration testing: electrodynamic (ED) and hydraulic. Figure 39 provides a plot comparing the operating ranges of these two types of exciters. Electrodynamic is much more common because of the wider frequency range, their linear behavior, and their wide range of operating conditions (shock and SRS pulses in addition to vibration). But for maximum displacement and lower frequency ranges a hydraulic shaker would be preferred.
At Midé we use a Brüel and Kjaer electrodynamic shaker for our shock and vibration testing, including calibration of our Slam Stick data loggers; specifically we have the LDS V455 permanent magnet shaker shown in Figure 40. There are other companies that sell shakers such as Data Physics, and Unholtz-Dickie. The trouble with buying these larger general purpose shakers in your company’s lab is the typically high costs they require. These type of industrial shakers will cost tens of thousands of dollars and that doesn’t include the high cost of the amplifier and software to run the shaker (tens of thousands for these too).
Modal Testing
Shakers are also used extensively in modal testing. In these test setups, typically a much smaller shaker is needed to excite the range of modal frequencies you may be interested in. For this type of testing the Modal Shop offers the best range of solutions, including rental services. Figure 41 provides an image and performance plot of one of their mini-shakers. This shaker is ideal for general purpose testing on small components and also very useful as an excitation source for modal testing.
Their shakers may also work as general vibration and shock exciters for smaller systems to qualify your product/design as shown in Figure 42; but their equipment is best suited for smaller test setups and/or modal testing.
Another form of exciting your structure for the purpose of modal testing is using an impact hammer. These offer a much more cost effective means of performing your testing (a typical impact hammer will cost just under \$1,000) and are the preferred method for many experts. An impact hammer will provide a nearly constant force over a broad range of frequencies (specified by the type of tip you use); and therefore these are capable of exciting a broad range of resonances and modal shapes. These hammers will come force-instrumented so that the frequency response function can be calculated using output of modal accelerometers instrumented throughout your structure.
Figure 43 provides some performance specifications of one of PCB Piezotronics’ impact hammers that cost \$760. The Modal Shop (owned by PCB) also offers a variety of impact hammers which can be purchased or rented.
Testing Labs
Performing shock and vibration testing in a simulated environment can be very expensive and require complex equipment and setups. But there are test laboratories that can offset some of these costs (although their services will still be several thousand dollars). Midé has had experience with both National Technical Systems (NTS), and Dayton Brown. For larger systems, like those being qualified for naval ships, Hi-TEST laboratories offers shock and vibration testing services. This includes their MIL-S-901D shock testing capabilities that consist of exploding a charge underwater to subject a test setup to a large displacement shock event. Midé has been down there a couple times for our bulkhead shaft seal and stern tube seal product lines as well as for some of our general R&D programs. It is both exciting and nerve racking to witness those tests!
SECTION 4
Analysis Software Options
Setting up test hardware and knowing what/how to analyze the vibration data is meaningless without the means to perform that analysis. Software is also typically needed to acquire the vibration data in the first place. There are a wide variety of software packages available to the test engineer but he/she needs to answer two questions before settling on a software option or combination of options:
• Do you need to analyze data in real-time, or post-process the shock and vibration data?
• Do you prefer developing your own analysis and simulation programs, or using a standalone graphical user interface?
If the products you will be using acquire the data and export it then the software will “only” need to do the post-processing. This would be the case for a data logger. But for applications that require real-time (or near real-time with a buffer) data streaming and analysis will limit the options available. Real-time data streaming and processing is needed for typical modal analysis, and for controls applications (where an action is taken based upon recorded data).
Writing custom analysis and simulation programs will require some advanced knowledge of the computing language and analysis fundamentals; but is the preferred path for most post-processing analysis applications. Standalone graphical user interfaces (GUIs) are nice for providing that initial overview of your data and performing some analysis. In many applications though there will be a need for doing analysis based upon certain conditions specific to your test. For example, if I performed a two hour long test on an aircraft component, I may want to develop a script that looks through the data and runs FFTs if/when certain conditions are met. Doing this fairly simply analysis that has a couple if-statements may be more difficult in a standalone GUI.
Most times a combination of standalone GUIs and custom analysis scripts will offer the best of both worlds to complete your analysis. There are a lot of options out there but it can be difficult to go through and compare them. Below we’ve condensed the list to a few more well-known software packages.
MATLAB
MATLAB is a programming language specifically developed for linear algebraic operations. Because of this initial core design and focus, it is a hugely popular tool for data analysis. Engineers would have used MATLAB is college and entered the workforce already with a knowledge and preference for this (MathWorks is smart with their pricing to offer significant discounts to universities and students).
The big disadvantage of MATLAB is that it's not free; a commercial license will cost \$2,150. They also charge more, typically \$1,000, for additional toolboxes (here's a full price list of the MATLAB products); and I'd recommend the signal processing toolbox for vibration analysis. Don't worry though, I didn't use any functions in that toolbox for this analysis or the ones covered in my vibration analysis basics blog.
If code base programming seems daunting, MATLAB does have their Simulink block diagram environment. This is a compelling product that can help reduce human programming induced errors and allow teams of analysts to integrate their algorithms a little easier. A single commercial license is \$3,350 (in addition to a MATLAB license). Simulink also lets engineers interface with hardware such as National Instruments, Raspberry Pi and Arduino; but these hardware supports will cost you. Simulink is great for analyzing data in real time (another couple thousand dollars) and offers incredible customization in addition to built-in analysis capabilities.
Python
Python is a free, open source, and very versatile programming language. Their NumPy and SciPy packages have similar functions to MATLAB. Python is a pretty elegant and intuitive programming language compared to MATLAB. It was created to be a generic language that is easy to read; and they definitely succeeded with that! Python is universally accepted as the better alternative to MATLAB for other programming needs besides data analysis.
But if you ask what’s better, MATLAB or Python for vibration analysis, you may start a heated debate because they both have benefits and disadvantages. We recently did some testing to compare MATLAB and Python for vibration analysis and came to the conclusion that for basic analysis (including FFTs) Python can match and even beat MATLAB computation times; but the programmer may need to do a bit of digging to find and download all the necessary libraries. But these libraries will be free!
As a MATLAB user I found the Anaconda distribution of Python and its most popular libraries very helpful. The Spyder development environment, shown in Figure 45, has a similar interface and feel to MATLAB for those with MATLAB experience.
Python is gaining popularity because it’s free and the community is generating a wide variety of versatile libraries that are publically available through GitHub. A quick search finds the PyDAQmx library that interfaces with National Instrument’s drivers. Again Python is free so it’s becoming a compelling alternative, especially as they offerings and capabilities continue to grow and improve.
LabVIEW
Most engineering companies will likely have a couple LabVIEW licenses to interface with their National Instruments data acquisition hardware and analyze data in real time. LabVIEW is a development environment specifically designed for engineers and scientists analyzing data. As such, it’s a popular tool for vibration testing! As we’ve discussed, National Instruments is the world leader in hardware for data acquisition; so it makes sense that their software is also very popular and pairs well with their hardware. Designing analysis programs with LabVIEW may be easier to those with less programming knowledge because of the graphical programming language they use, shown in Figure 46.
A full LabVIEW license costs \$2,999 and vibration customers may be interested in their Sound and Vibration Toolkit for another \$1,999. It can definitely get expensive quickly but is a great solution for data analysis, especially in real-time controls applications.
Standalone Vibration Analysis Software
There are a wide range of standalone software packages available for purchase; but let’s first identify a couple free ones.
Tom Irvine’s vibrationdata
Tom Irvine offers both a MATLAB and Python version of his signal analysis and structural dynamics software GUI. These offer great versatility because it has a nice GUI which is easy to work with; but all the source code is available if you need to take the next step and perform your own custom analysis. In nearly all of his webinars he will go through an example that uses these software packages so you can quickly see the power and ease of use of these analysis options.
Slam Stick Lab
Midé’s Slam Stick Lab is available for free along with some example recording files. This software currently only works with Slam Stick recording files; but Midé has plans to open this up to other import files in the future. The software only has basic analysis capabilities but it covers the major ones typically needed: FFT, PSD, spectrogram, unit conversion, and general plotting. Data can be exported to MATLAB or CSV (readable by Excel, Python, and other software) for follow on analysis.
The following represent a select few of many different standalone GUIs available on the market for vibration analysis. These will typically cost around \$5K or more but offer some unique benefits over writing your own code, namely a time savings advantage.
m+p International
This company specializes in vibration testing, specifically modal. They have a variety of hardware products for data acquisition but also some software packages for post processing and analyzing data in real-time. Their m+p analyzer can interface with their hardware for real time analysis but is also useful in handling large datasets and post processing vibration data.
VibrationVIEW
Vibration Research’s VibrationVIEW software is another alternative to post processing and analyzing vibration data in real time.
Brüel and Kjaer
Bruel and Kjaer have a couple different software packages that are very impressive; again for both post processing and analyzing vibration in real time.
ProAnalyst
Xcitex’s ProAnalyst software is unique in that it uses video for determining vibration levels and analyzing them. Seems like a very powerful tool that can simplify the hardware setup! The costs don’t seem too ludicrous either, only \$1,795 for their introductory edition. It quickly starts climbing towards \$10K and beyond though for the professional edition and different toolkits. I do like how they come right out and list the pricing instead of requiring you to fill out a form!
FEMtools
For modal analysis and validating CAD models with experimental data, FEMtools is a very powerful tool. It can also be used for general post processing of time history vibration data.
SECTION 5
Conclusion
Before you can dive into vibration measurement and analysis it’s important to ask yourself and answer those first couple questions: what frequency and amplitude is of interest, who needs the data and why, where is the test environment, when is the testing, and when does the analysis need to be completed by?
Remember that shock and vibration testing is an art and a science. It’s not always very straightforward; but that creates some opportunity as the engineer to impart their own personality into testing and analysis. It lets the engineer have fun, so go out and have fun with your vibration testing!
SECTION 6
Resources
Free Vibration Analysis Files
Avitable, Peter. Modal Space (in our own little world). 2014. Web.
Broch, Jens Trampe and Joëlle Courrech. Mechanical Vibration And Shock Measurements. Naerum, Denmark: Brüel & Kjaer, 1980. Print.
Harris, Cyril M and Allan G Piersol. Harris' Shock And Vibration Handbook. New York: McGraw-Hill, 2002. Print.
Irvine, Tom. Shock and Vibration Signal Analysis. 2005. Web. 12 September 2005.
McConnell, Kenneth G. Vibration Testing. New York: Wiley, 1995. Print.
Technical Papers
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# Which will keep my food colder longer, draining the melted ice water, or leaving it in the cooler?
Say you have a cooler of frozen food and ice to last for several days or weeks of river-trip / car camping. To keep things as a cold as possible for as long as possible, is it better to leave the cold melted ice-water in, or to drain it out on a regular basis?
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So most of the answers here are pointing to what I believe to be the right answer: that you should keep the water in. However, if you really want some hard data on this, I can do a small experiment with some friends to get a definitive answer. This would also quantify how much of a difference there is between the two methods, and you’d get those graphs you wanted. Is this something you’d be interested in? – Big General Feb 8 '12 at 22:20
@BigGeneral Give us anything you got ;) Though the problem with an experiment is for it to be conclusive, you would need many replicates... and this bounty only has a 6 days left. – Lost Feb 9 '12 at 1:04
From a thermodynamics point of view, I'd say you should leave the water in.
Temperature is a measure of the active kinetic energy of the molecules in a substance. Warming up is essentially the surrounding environment imparting some of its kinetic energy into the object being warmed up. Simply thinking about that, the more you have that needs warming, the more energy it requires to warm, and so the slower its temperature will rise (given the same rate of exchange of thermal energy - the water submerges at least some of the other contents, so if anything this is an overestimate). Now consider that water has a relatively high specific heat, meaning that it takes more energy to warm it up. The food inside the cooler won't be warmed up faster than the surrounding water, so since it takes more energy to heat the food and water than just the food, so the food will stay cooler longer.
The Igloo (yes, the cooler company) FAQ supports this view:
During use, it is not necessary to drain the cold water from recently melted ice unless it is causing contents to become soggy. The chilled water, combined with ice, more readily surrounds canned and bottled items and will often help keep contents colder more effectively than the remaining ice alone.
Don’t drain cold water – Water from just-melted ice keeps contents cold almost as well as ice and preserves the remaining ice much better than air space. Drain the water only when necessary for convenient removal of cooler contents or before adding more ice.
I think the key point a lot of people forget is that what matters isn't how long the ice lasts, but how long the contents remain under some temperature.
Assuming convection is fairly significant relative to the rate of energy input from the outside (a good assumption, I think), it doesn't matter how good an insulator air is, the inside temperature will be the same throughout, necessitating that the rate of heat energy coming into the cooler is the same in both cases, so the ice will melt at the same rate, and once the ice is gone, the cooler with water will take much longer to warm. I'm still working on the no-convection theory (which would provide at best an extreme overestimate), but in the meantime if anyone wishes to posit that convection is tiny enough to bridge the enormous gap (assuming I find the upper bound does eclipse water), please explain why you believe so.
Some math/physics to back this up for the quantitatively inclined. (This would be so much easier with the MathML markup from the math and physics sites.)
The cooler will be very near perfect convection, the heat is entering the cooler slowly enough that the contents - air, water, and ice - are at the same temperature (namely 32°F/0°C/273.15K). Heat conduction, `H`, as far as our coolers are concerned depends only on `ΔT`: `H = kAΔT/x`, where `k` is the thermal conductivity of the cooler, `A` is the area of the cooler through which thermal energy is flowing, `x` is the thickness of the cooler, and `ΔT` is the temperature difference between the inside and outside of the cooler (`T_out - T_in`). Notice that all these are the same for both coolers. Now, melting the ice requires energy of `Q = Lm` where `Q` is the total energy required, `L` is the (latent/specific) heat of fusion, and `m` is the mass of ice in the cooler. Since the total energy is `Q = Ht`, we can calculate the time required to melt all the ice: `Q = Ht = kAtΔT/x -> t = Qx/(kAΔT) = Lmx/kAΔT`. Since all variables are the same for both coolers, it will take the exact same amount of time for the ice to melt.*
*: We are ignoring the air replacing the ice, which would actually give a (very) slight advantage to retaining water. Draining the water in that cooler requires adding heat - the excess heat of the air that replaces the melted water. Fortunately, that excess heat is fairly easy to calculate: `m = Vρ -> V = m_ice/ρ_ice = m_air/ρ_air -> m_air = m_ice * ρ_air/ρ_ice`. The air comes in with an excess energy of `Q = m_air*C_air*ΔT = m_ice*(ρ_air/ρ_ice)*C_air*ΔT`. This reduces the energy required from the cooler heat influx, reducing the time required: `Q = Lm_ice = Q_cooler + Q_air = kAtΔT/x + m_ice*(ρ_air/ρ_ice)*C_air*ΔT -> t = (L/ΔT - ρ_air/ρ_ice)*C_air)mx/kA`. The fractional difference this will make ends up being about `4e-6*ΔT`, or about `0.016%` on a rather hot (40°C) day, coming to 2min 18s over 10 days. So we were right to ignore it.
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There is a lot of debate about this issue among rafting guides. It's about a 50-50 split, but I take to the draining doctrine. Nice explanation overall for the "bathing" option. – Dangeranger Jan 26 '12 at 18:17
+1 for bringing up specific heat of water. It takes twice the energy to raise the temperature of water by 1C (or about 2F) that is required to raise the temperature of ice by 1C. This is why the water will stay cool for so long even after the ice has all melted. – Clare Steen Feb 8 '12 at 3:54
"what matters isn't how long the ice lasts, but how long the contents remain under some temperature" - Yup (and +1) – Russell Steen Feb 8 '12 at 4:31
Mythbusters explored this very topic, and came away with conclusion that ice in water is better than ice alone, and that ice in salted water is even better. – theUg Jun 25 '12 at 4:32
@LBell water is great at thawing things (e.g. turkey) because of its high specific heat - the same reason it is good at keeping things cool. It doesn't matter which way the temperature difference goes, water is hard to heat or cool. Turkeys need to be thawed quickly and at a low temperature difference, and water is one of the only things that can do that. – Kevin Dec 5 '12 at 4:07
We (Kent and Deny) did an experiment in order to shed some light on this debate. We found that keeping the water in the cooler along with the ice kept the overall temperature of the cooler below 5 degrees Celsius for approximately 4 hours longer than when the water was removed.
Experiment. We filled a Coleman cooler with 12 341mL bottles of Waterloo Dark beer and 2.7 kg of store-bought ice cubes and sealed the cooler. Beer bottles were kept at 4 degrees Celsius in a refrigerator overnight before both experimental trials. The ice and thermometer were both kept in a freezer, which is at -10 degrees Celsius, before the experiment. Temperature was monitored using an Omega OM-62 temperature logger which recorded temperature every 5 minutes over the course of the experiment. The thermometer was kept in a Tupperware container to avoid water damage.
We setup two separate trials of the experiment. In the first, water was removed from the cooler approximately every 8 hours using a siphon. Excluding the first removal of water at 8 hours into the experiment when there was still little water in the cooler, approximately 450 mL of water was removed from the cooler every 8 hours.
In the second case, where all water was kept in the cooler,we opened the cooler for 2 minutes every 8 hours in order allow the same amount of warm air to enter the cooler, as it would in the case of draining the water.
Results. Temperature data is shown in the figure below.
From the figure, we can see a few points. Excluding the first draining, when there was little water, each time the water is drained, the temperature of the cooler immediately rises. After 24 hours, draining the water from the cooler preceded a linear growth in temperature, which grew in slope at each subsequent draining. Finally, we can clearly see that keeping the water in the cooler keeps the temperature below 5 degrees Celsius for approximately 4 hours longer.
We conclude that this experiment gives support to the argument for keeping the water in the cooler.
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Awesome - great work - definite +1 for putting the effort in. 2 questions: 1) where was the thermometer? (Near the beer? On the bottom (submersed) or floating on the water?) 2) What was the ambient temperature doing outside during this period? And 3) Any interest in replicating about a dozen times or so? :) Thanks for the contribution – Lost Feb 16 '12 at 1:58
I'll upload some photos of the setup soon. The thermometer was in a tupperware container half submerged in the ice, touching the bottom of the cooler, in the same way the bottles did. I put the cooler in a sink in my basement for the experiment. I don't have data for how the temperature of the room varied (my only thermometer was in the cooler), but our house has been kept consistently at 18 degrees Celsius for the past while. – Kent Fisher Feb 16 '12 at 4:01
@KentFisher how bout them pictures? – nhinkle Oct 27 '14 at 16:26
Why is there a visible difference within the first hour from the experiment start? – Roflo May 6 '15 at 20:24
I had to sign-up just to upvote this. @Roflo --- I think that this is due to unavoidable errors like the different temperature of the store-bought ice, so I think that the results could be even more on the side of not removing the ice --- you could probably almost shift the red line up so that it coincides with the black one in the first hours. – Rmano Jun 20 '15 at 10:41
Never remove cold water from a cooler so long as the water is cooler than the outside temperature.
1. Opening the lid allows more warm air in, but assuming the lid is on the top and air disturbance minimal, this could be a small loss of cooling / small entry of heat. Opening a drain will have to let warm air in to replace whatever cool water leaves the cooler.
2. While ice remains, water will circulate and keep all the objects in closer thermal contact with the ice than will air alone - this will prevent a thin spot in the insulation from warming some of the food that stays in contact with the cool water. This won't have an effect on the long term temperature in the cooler, though.
3. Once all the ice is melted, the cool water will continue to be a heat sink, absorbing more of the heat that is leaking in through the walls of the cooler. This keeps your actual contents inside the cooler more chilled than if you remove this cold mass from the system. Without the cold water, all that heat will go to the food and warm it up instead of warming both the combined mass of the food + cold water.
You don't need any fancy graphs when you look at the heat flow into the cooler. Keeping cool water in the cooler delays the melting of the remaining ice and once the ice is gone, delays the warming of the food by absorbing it's share of the incoming heat.
Say you have two coolers with one gallon of beer each in a perfect container that absorbs no heat - just keeps the beer together. Also, assume that you have one pound of ice in each cooler and the beer and cooler are chilled to 0°C (32°F). One cooler also has a gallon of 0°C (32°F) water.
So the cooler on the left has heat leaking into the insides - warming everything inside. It warms the air, the beer, the inside of the cooler and the beer's container.
The cooler on the right has the exact same amount of heat leaking inside. It warms the air (but less air due to the extra water), the beer, the inside of the cooler and the beer's container.
The only variable is does water absorb more heat per degree rise in temperature than air. The answer to that is yes - extremely so.
Air's specific heat is 1.007 J/(gK) - Joules is energy, gram is mass and K is degrees Kelvin. Water's specific heat is 4.18 J/(gK) - so for a fixed amount of Joules added, water goes up less than 1/4 a degree in temperature compared to a gram of air. If water and air were equally dense (they are not) then you have a 4 to 1 advantage. Water takes four times as long to warm up as air, so your beer stays cooler longer.
Now - what about your typical 70 quart cooler? It is 66.24 Liters or 66,245 cm^3 in volume.
Air's density is 0.001275 g/cm^3 and water's density is 1.00 g/cm^3. Here's where water really owns the show in terms of cooling capacity since water is 784 times denser than air. That 70 quart cooler could contain either 84.5 grams of air or 66,245 grams of water (or 3 ounces of air versus 146 pounds of water).
Now we have water with a 4 to 1 advantage on heat absorption and a 784 to 1 advantage on packing mass into the same space, so water is over 3000 times better than air for absorbing heat while raising the temperature inside the cooler one degree. Whether you have a thimbleful of water or the cooler is mostly water, you want that amount of cold water staying inside the cooler to absorb its share of the heat leaking in.
Since it's easy to agree that the amount of heat entering the cooler is basically the same whether you drain or don't drain, leaving cool water inside the cooler slows the time it takes to both melt the remaining ice and warm the contents since that water will absorb heat if it is left in the cooler up to the point where the water in the cooler is equal in temperature to the outside air.
You do need formulas to calculate the rates of temperature increase, but the which is better scenario can be decided quite easily by considering where the heat leaking into the cooler goes and whether draining the water also has a side effect of allowing more heat in. Both of these fall on it being better to leave the container closed and with the melt water inside.
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Small nitpick -- This doesn't take into account the insulative affect of the negative space in air due to it being less dense. Yes, water is more dense, but that also means it will transfer heat faster and more efficiently from one molecule to the next. (still +1) – Russell Steen Feb 8 '12 at 6:02
Hmm - air only insulates when it is trapped and immobilized, so I don't see how having melt water raising up inside the cooler or draining it changes the heat transfer inside the cooler. Natural convection will be the dominant heat transfer mechanism within the cooler no matter what proportion of air or water is present. – bmike Feb 8 '12 at 6:30
EDIT: The more I consider this, the ambient air temperature around the cooler is the largest factor. Replacing water with 95F (35C) degree air will have a much larger impact than replacing water with 40F (4.4C) degree air.
Actually, the answer is very simple because you asked longer, not colder.
If you drain all the water, then when the ice all melts...you're done. Because whether or not the ice is better than the water at keeping your food cold, the water is going to be better than nothing. The ice is going to melt (effectively) at the same rate either way, because you can drain the water, but you can't really dry the ice. So the ice will melt at roughly the same rate regardless of when you drain the water.
Also, when you drain the water, something has to replace it. That something will be air at the temperature of the environment you are in... which will destroy the equilibrium. Basically everything I read online ignores this point. It talks about the coefficients and thermodynamics, etc, and, neglects to even consider that the water is being replaced with something and that something is air which must then be cooled. When you drain the water, you are actively replacing cold with heat.
It should be noted that if you have infinite ice, you should probably drain the water. However if you have infinite ice it doesn't seem like this would matter.
More here
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Actually I asked "as cold as possible for as long as possible"... I want it all ;) – Lost Feb 8 '12 at 4:13
+1 for "replacing the water with warm air" -- never really considered that either. – Lost Feb 8 '12 at 4:17
Even if you replaced water with equivalently chilled air, water is 3000 times more resistant to raising the temperature a degree than air. You can't add air cool enough to be better than keeping cold water in the cooler. – bmike Feb 8 '12 at 5:33
I sometimes leave melted ice water in my cooler, which then gets into the food, making the food inedible.
If my "food" is a can of beer, fine, leave the water in the cooler. If it's a sandwich, drain the water if it might get the sandwich wet. Or have the cans of beer at the bottom and put the sandwich on top to keep it out of the ice water.
Another "outside the box" answer is if it gets cold enough at night, I'll leave the lid open. Leaving all the melted ice water inside is good thermal mass to keep it cool during the day. You have to remember to open and close the lid at the right time. If critters might get in this doesn't work so good.
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The thermal conductivity of air is 0.000057
The thermal conductivity of water is 0.0014
Therefore water is 24.5 times more conductive than air, and has a temperature above 32 degrees Fahrenheit. The reason this is a problem is that bacteria can grow in that water quite quickly (within 6 hours) and start to make your food unhealthy to eat.
Additionally all that warm 32+ degree water will assist in melting the remaining ice.
So do yourself a favor, drain the water that melts out the bottom of the cooler with the spout that is provided. It's safer, and your food will last longer.
See this fine article about cooler packing and draining.
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I'm not sure bacteria would grow 'quickly' in relatively clean, zero-degree water. – mendota Jan 27 '12 at 10:57
Mythbusters sort of touched on this one: kwc.org/mythbusters/2005/03/mythbusters_cooling_a_sixpack.html. They were trying to determine the fastest way to cool a six-pack of beer. Ice+salt+water was faster than ice+water, which was also faster than just ice. – Clare Steen Feb 8 '12 at 3:38
I humbly suggest you are looking at this the wrong way. Adding water that is warmer than the ice may or may not be good depending on many factors. You don't want to add anything that is warmer than the ice if your contents are already chilled. You do want to add anything that is cooler than the contents of the ice chest since that will draw heat out of the food and start you off cooler than if you didn't add that water (or whatever else). Once you separate that, the answer to not drain chilled water is self evident. Anything cooler than the outside air keeps the cooler more cool. – bmike Feb 8 '12 at 5:32
"and has a temperature above 32 degrees Fahrenheit" - Wrong. Water with ice in it will remain at 32F until all the ice is melted (assuming it's not being heated so rapidly convection can't even things out - and if that's the case, you have bigger problems with your cooler). – Kevin Feb 10 '12 at 21:37
Also, the fact that the water is still at freezing will inhibit bacteria from growing. – Kevin Feb 10 '12 at 21:46
Ok, put away your calculators people. Draining the cool water causes the warm air to enter the cooler. Cool water is COOLER than warm air, so leave the cool water in the cooler. If I had a compass, some graph paper a protractor handy I'd draw you a picture.
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Here is a different twist on the question of how to best use block ice. Before refrigeration northern states used to saw block ice fron frozen lakes and store for the summer in "ice houses", log houses that used only the thermal inertia of massive amounts of ice stacked together.
We just returned from a 6 day canoe trip in which we used block ice to keep fresh food cold. On Friday, july 19 we bought 200 lbs of block ice (20 blocks of 10 lbs). these almost completely filled two 12" x 12" x 30" low tech garage sale ice chests. We kept these two ice keepers drained of water because water is a conductor, not an insulator. Then we used single blocks of ice from these to supply as needed four smaller chests which we did not drain.
We then traveled from oklahoma City to the Current River in Missouri and floated 100 miles over six days and returned to OKC on Fri Jul 26 with about 15 lbs of block ice remaining. Eight days and 15 hours! We fed 18 Scouts and adults on fresh food that was previously frozen ar cooled for a week. I know that this is a completely unscientifec "experiment" , but is a practical example that worked for us. I hope this is useful for your wilderness trip.
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As a rafter I have taken part in lots of discussions about this and couldn't resist any longer so have set up an experiment to test this. I hypothesize that the drained cooler will hold ice longer due to the insulating effect of air--as Snitse has described above. Convection will reduce air's effectiveness but, as Snitse points out, it is still far superior to water. I have set this experiment up with identical styrofoam coolers loaded with equal amounts of crushed ice (in the second series of experiments I am going to substitute identical blocks of ice). The drained ice chest has a drain tube in place to allow it to drain continuously. These are placed side by side (with a gap between) at room temperature (in my lab in a large building where temperature is fairly constant). In the first replicate, the drained ice chest did, indeed hold ice longer, but not by much and the undrained ice chest contained water that reached room temperature at nearly the same time the ice melted! I am replicating this experiment and will begin the ice block experiment after replicating this three times.
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This is a great topic deserving of some controlled experimentation. In general the system is non-linear, but can be analyzed in a piece-wise linear fashion the same way many other difficult problems are approached. Some thoughts on the matter; 1)conduction not convection, is the main transport mechanism, 2)cold does not exist, only the absence of heat, 3) heat transfer is a function of temperature difference between the inside and outside of the cooler, cooler thickness, thermal conductivity (w/m-k) of the cooler material, and surface area of the cooler. You can use Fourier's law of heat transfer to calculate how much heat in watts enters the cooler through its walls. As ice melts, it takes 3.3x10^5 joules/kg to do so. Remembering that a watt is a joule/second you can then calculate how long a given mass of ice will last. Heat, like electricity, takes the path of least resistance. Water, at 24X the thermal conductivity of air, provides a low thermal impedance path for heat to reach the ice, so if a block of ice is sitting in a water bath it should melt at a faster rate. I would like to propose and experiment where the ice and melt water in one cooler are kept segregated, and one where the ice and melt water are not. At the time that the ice is completely melted, the two systems are the same in mass and phase (all liquid) but perhaps not temperature or even at the same time. Then its a different process, but with different initial conditions (temperature). There are some variables which have to be controlled in order to see the effects of draining vs not draining the water, namely ambient temperature has to be monitored and kept the same for both, admitting outside are into the system has to be eliminated, the coolers must be identical, the mass of the ice has to be measured. Leave the system undisturbed, measure the inside water and air temperature at as many places as possible using thin thermocouples and a digital thermometer. Allow the experiment to run well past the expected lifespan of the ice. Some sort of viewing port capability would be helpful to examine the state of the ice without perturbing the system.
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As Dangeranger stated,
water is 24.5 times more conductive than air
The idea is that water will conduct heat energy throughout the cooler much more effectively than air. This means that water will conduct heat energy from the leaks and seams in the cooler to the ice more effectively than air.
Say we have two hypothetical coolers, Cooler A and Cooler B. Both leak heat at the same rate, and both started with a identical large block of ice in it, and equal air temperature occupying the rest of the cooler. Every hour you drain the water from A, and leave it in B. For the first hour they both have the same amount of water and the same amount of ice, and then you drain the water from A. Now, in the second hour because the water left in B conducts heat from the walls to the ice faster than the air in A, more of B's ice will melt than of A's ice.
Clearly, B's ice will be completely melted before A's ice is. Let's call the moment that ice B completely melts 'meltB'. At meltB ice A has x percent of its original mass remaining.
We can all agree that at meltB there is less heat energy in cooler B than in cooler A, since you've been replacing things in cooler A with other things that had more heat energy (replacing cool water with air temperature air).
Where it gets tricky is that after meltB, the heat energy in cooler B will increase faster than the heat energy in cooler A because of all that water conducting heat from the walls.
So from start until meltB, A gains energy faster than B, but after meltB B gains energy faster than A.
What this all comes down to is the exact numbers. If you want to eat your food at time a or b you are better off with cooler B, but if you want to eat your feed at time c or d you are better off with Cooler B.
As for how to calculate meltB, I have no idea. So I guess its somewhat impossible to figure out.
EDIT:
To clarify, the key point here is that different parts of the cooler are different temperatures. The wall is warmer than the area immediately around the ice. However, if the cooler is full of water, the water brings heat energy from the wall to the ice much more effectively than air would.
So the coolers leak at the same rate, meaning the walls gain energy from the outside at the same rate. However, if there is no water, the walls gaining energy does not necessarily mean that the ice will gain that energy and melt immediately. It will take time for that energy to get to the ice, and the time it takes depends on the medium the energy transfers through.
In summary, walls gain energy at same rate, ice in the middle does NOT gain energy at the same rate because of what is between them and the walls.
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"Both leak heat at the same rate," and "at meltB there is less heat energy in cooler B than in cooler A," but "B's ice will be completely melted before A's ice is." Those can't all be true. In fact, the amount of ice in the cooler depends entirely on how much heat energy there is in the cooler, so by acknowledging that there will be more heat energy in the drained cooler, you are actually saying that the ice in it will melt faster. – Kevin Feb 12 '12 at 18:35
Oh, and I should mention, the temperature doesn't increase linearly, they will both be at a constant 32 until all ice is melted (except small jumps for draining the ice in the appropriate cooler), after which the temperature difference will decay exponentially (with the water-filled cooler rising much more slowly). – Kevin Feb 12 '12 at 18:38
At meltB there is less heat energy in cooler B than in cooler A, and there is NO ice because there is quite a bit of rather cool water. – Snitse Feb 12 '12 at 18:38
The graph is the amount of energy in the coolers, not the temperature. The energy increases as the temperature stays the same because energy goes into melting ice. You are also assuming that the whole cooler is the same temperature. The whole point is that different parts of the cooler are different temperatures (the wall vs. the area immediately around the ice) and water is better at bringing that heat from the wall to the ice than air is. – Snitse Feb 12 '12 at 18:41
At meltB, if (as you are saying) there is less heat energy in cooler B, the ice will NOT be all melted unless it was already fully melted in cooler A. It takes the exact same amount of energy to melt the exact same amount of ice, regardless of conditions. – Kevin Feb 12 '12 at 18:50
Removing water from the cooler always means the ice lasts longer - solid (ice) transfers heat slowest, then gas (air inside cooler), then liquid (melted ice). This is explained by the Zeroteh Law of thermodynamics. All three work to achieve equilibrium by becoming the same temperature. Remove the fastest heat-transferring part and your ice lasts longer. The classic experiment is placing an ice cube in a mesh bag and suspending it with string and lever over a clear glass containing cold water. Notice the ice melts faster where it is submerged than in the air. The air insulates the thin layer of melted ice above, but the liquid quickly exchanges heat with the ice to melt it, until all are in equilibrium. Using the drain to remove the melted water means more insulating air and solid ice in the cooler to maintain cold contents, and longer lasting ice.
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It's really simple. The universe wants everything to be the same temperature so if we left the cooler in a sufficiently large, constant temperature environment, the cooler and all its contents would eventually end up at the same temperature, that being the ambient temperature of the environment.
Let's first begin by noting that we want to keep something cool rather than make it cooler. In that case, we want to stop it gaining heat energy. The heat energy is coming from the walls of the cooler.
If our food is packed in the cooler such that the ice touches the walls, then the ice is going to be subject to warming before the food. The ice touching the food will not impart any heat energy because it's the same temperature.
However, add some water to the system and there is path of conduction from the wall to the food since the water is at higher temperature than the food.
Air has 1000 times less heat capacity than water, so it's not actually that interesting to take into account replacement of cold air with ambient temperature air.
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It is so not true that water mixed ice keeps food cold. Try your own experiment and see that it is true. This is exactly how you make ice cream, but just with some salt. Here is how you do it: 1) Buy your cooler according to size. Dont buy a cooler that it too big. 2) Fill the cooler with pre-chilled items. Try and reduce the air space as much as possible. This will dramatically extend the 'cold' 3) Top off with ice. 4) The Ice Melts. 5) Melted Ice turns to water and carries off heat. 6) Drain the warmer melted water. Heat goes right out the drain tube. Ice remains colder for MUCH longer. Half a day over my 3 day weekends. NOTE...This is a mostly closed system. Neither 'Dried' (Not Dry Ice...but not wet ice) nor water/ice slurry works well when kids are always leaving the lid open and warm air blows in.
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Your ice will keep longer if you drain the water. If you want to prove it to yourself, get 2 cups. Fill one with cold fridge water and leave the other empty. Put an ice cube in each and see which one is melted first. Even with the ice cube in the water being surrounded by cold water it still melts much quicker than the other ice cube being surrounded by the hot ambient air.
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Your experiment is flawed. "Cold fridge water" is warmer than water that just changed states from solid to liquid. Since water has a higher heat capacity than air, the fridge water, depending on it's temperature, can quickly melt the ice, even when it is significantly colder than the surrounding air. The higher heat capacity of water is also what can slows down the melting when you don't remove the melt-water. A better experiment would be to actually conduct the experiment the OP proposes, at different ambient temperatures and in containers with different insulating properties. – DudeOnRock Oct 12 '13 at 5:57
One test I've done was to fill two cups with partly ice and emptied one cup every 5 minutes. The ice in the cup that was emptied of its water lasted longer than the ice that sat in the melted water. Showed easily the best way to make ice last longer. Seems it could be a different story if your aim is to keep the cooler temperature colder for longer. But if I'm keeping food cool I like to keep it relatively dry.
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It depends.
A simple fact: You conduct HEAT. You do not conduct COLD. Wrap your head around that.
2nd Simple Fact: Water conducts HEAT more efficiently than air.
Now, if you're trying to chill warm items, then use water to conduct the heat out of them and transfer it to the frozen water (ice). That will chill them quicker than just air.
However, if you're trying to keep chilled items cold, you want to prevent the transfer of heat from the environment (via the outside skin of the cooler, across the insulation or air gap, through the interior skin) to the items. In this case, air is the best choice; drain the water.
So, you catch a fish. Quench it quickly in one ice chest with ice and water to get it chilled, then put it into an ice chest with the ice kept separate from the contents and with the water drained.
-
The problem with this logic is that it takes less heat energy to re-warm a fish than the fish and several gallons of water. So if the heat is flowing in at the same rate with and without the water, keeping the water keeps the whole system cool longer. – Kevin Mar 1 at 18:00
The heat flows (out of the fish) faster with the water. And flows (back in) to the fish slower with air. – Tank Mar 1 at 20:57
Heat only flows back into the fish if there is a temperature difference. With a fish in water at the same temperature, there is no heat flow back in from the water directly. The heat flows into the cooler (fish or fish+water) at a rate dependent on the temperature difference (inside vs. outside). The temperature change due to heat flow is dependent on how much there is to heat. Thus, the heat flowing into a cooler takes much longer to heat both the fish and water than just the fish. – Kevin Mar 1 at 21:12
The conductivity of water is roughly 24 times that of a similar volume of air. The water AIDS the conduct of heat...it does not impede the conduct of heat when compared to air...engineeringtoolbox.com/thermal-conductivity-d_429.html – Tank Mar 3 at 19:46
Yes, but that's irrelevant to what I'm saying. The rate of heat flowing into the cooler is determined solely by the temperature difference between the inside and the outside; water's higher thermal conductivity just means the temperature inside equalizes more quickly. Because water has a higher specific heat, it will take more heat energy to warm the water's temperature. Since the rate of heat flowing in is the same, the water's temperature, and therefore the fish's, will rise more slowly. – Kevin Mar 3 at 20:06
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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
KelvinKei
http://functions.wolfram.com/03.19.20.0014.01
Input Form
D[KelvinKei[\[Nu], z], {z, n}] == 2^(-2 + n + 2 \[Nu]) I E^((3 I Pi \[Nu])/4) Pi^(3/2) z^(-n - \[Nu]) Csc[Pi \[Nu]] Gamma[1 - \[Nu]] HypergeometricPFQRegularized[{(1 - \[Nu])/2, 1 - \[Nu]/2}, {(1/2) (1 - n - \[Nu]), (1/2) (2 - n - \[Nu]), 1 - \[Nu]}, -((I z^2)/4)] - (2^(-2 + n + 2 \[Nu]) I Pi^(3/2) z^(-n - \[Nu]) Csc[Pi \[Nu]] Gamma[1 - \[Nu]] HypergeometricPFQRegularized[ {(1 - \[Nu])/2, 1 - \[Nu]/2}, {(1/2) (1 - n - \[Nu]), (1/2) (2 - n - \[Nu]), 1 - \[Nu]}, (I z^2)/4])/E^((3/4) I Pi \[Nu]) - (2^(-2 + n - 2 \[Nu]) I Pi^(3/2) z^(-n + \[Nu]) (I + Cot[Pi \[Nu]]) Gamma[1 + \[Nu]] HypergeometricPFQRegularized[ {(1 + \[Nu])/2, (2 + \[Nu])/2}, {(1/2) (1 - n + \[Nu]), (1/2) (2 - n + \[Nu]), 1 + \[Nu]}, -((I z^2)/4)])/ E^((3/4) I Pi \[Nu]) + 2^(-2 + n - 2 \[Nu]) I E^((3 I Pi \[Nu])/4) Pi^(3/2) z^(-n + \[Nu]) (-I + Cot[Pi \[Nu]]) Gamma[1 + \[Nu]] HypergeometricPFQRegularized[{(1 + \[Nu])/2, (2 + \[Nu])/2}, {(1/2) (1 - n + \[Nu]), (1/2) (2 - n + \[Nu]), 1 + \[Nu]}, (I z^2)/4] /; !Element[\[Nu], Integers] && Element[n, Integers] && n >= 0
Standard Form
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MathML Form
n kei ν ( z ) z n - 2 n + 2 ν - 2 1 4 ( - 3 ) π ν π 3 / 2 csc ( π ν ) Γ ( 1 - ν ) z - n - ν 2 F ~ 3 ( 1 - ν 2 , 1 - ν 2 ; 1 2 ( - n - ν + 1 ) , 1 2 ( - n - ν + 2 ) , 1 - ν ; z 2 4 ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", "2"], SubscriptBox[OverscriptBox["F", "~"], "3"]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[FractionBox[RowBox[List["1", "-", "\[Nu]"]], "2"], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]], ",", TagBox[RowBox[List["1", "-", FractionBox["\[Nu]", "2"]]], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQRegularized, Rule[Editable, False], Rule[Selectable, False]], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "n"]], "-", "\[Nu]", "+", "1"]], ")"]]]], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]], ",", TagBox[RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "n"]], "-", "\[Nu]", "+", "2"]], ")"]]]], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]], ",", TagBox[RowBox[List["1", "-", "\[Nu]"]], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQRegularized, Rule[Editable, False], Rule[Selectable, False]], ";", TagBox[FractionBox[RowBox[List["\[ImaginaryI]", " ", SuperscriptBox["z", "2"]]], "4"], HypergeometricPFQRegularized, Rule[Editable, True], Rule[Selectable, True]]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQRegularized[Slot[1], Slot[2], Slot[3]]]], Rule[Editable, False], Rule[Selectable, False]], HypergeometricPFQRegularized] + 2 n + 2 ν - 2 3 π ν 4 π 3 / 2 csc ( π ν ) Γ ( 1 - ν ) z - n - ν 2 F ~ 3 ( 1 - ν 2 , 1 - ν 2 ; 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ν TagBox["\[DoubleStruckCapitalZ]", Function[Integers]] n TagBox["\[DoubleStruckCapitalN]", Function[Integers]] Condition z n KelvinKei ν z -1 2 n 2 ν -2 1 4 -3 ν 3 2 ν Gamma 1 -1 ν z -1 n -1 ν HypergeometricPFQRegularized 1 -1 ν 2 -1 1 -1 ν 2 -1 1 2 -1 n -1 ν 1 1 2 -1 n -1 ν 2 1 -1 ν z 2 4 -1 2 n 2 ν -2 3 ν 4 -1 3 2 ν Gamma 1 -1 ν z -1 n -1 ν HypergeometricPFQRegularized 1 -1 ν 2 -1 1 -1 ν 2 -1 1 2 -1 n -1 ν 1 1 2 -1 n -1 ν 2 1 -1 ν -1 1 4 z 2 2 n -1 2 ν -2 3 ν 4 -1 3 2 -1 ν Gamma ν 1 z ν -1 n HypergeometricPFQRegularized ν 1 2 -1 ν 2 2 -1 1 2 -1 n ν 1 1 2 -1 n ν 2 ν 1 z 2 4 -1 -1 2 n -1 2 ν -2 1 4 -3 ν 3 2 ν Gamma ν 1 z ν -1 n HypergeometricPFQRegularized ν 1 2 -1 ν 2 2 -1 1 2 -1 n ν 1 1 2 -1 n ν 2 ν 1 -1 1 4 z 2 ν n [/itex]
Rule Form
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Date Added to functions.wolfram.com (modification date)
2007-05-02
© 1998-2013 Wolfram Research, Inc.
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It is currently 20 Sep 2017, 06:15
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# M 20 Q 3
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22 Feb 2012, 14:40
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If 2 excavators can dig a 10-meter long trench in 50 minutes, how long will it take 3 excavators to dig a 15-meter long trench?
(C) 2008 GMAT Club - m20#3
* 45 minutes
* 50 minutes
* 52 minutes
* 54 minutes
* 60 minutes
1 excavator can dig a 10-meter long trench in $$50*2 = 100$$ minutes. So, it can dig a 15-meter long trench in $$100*1.5 = 150$$ minutes. 3 excavators can dig a 15-meter long trench in $$\frac{150}{3} = 50$$ minutes.
As per above exp, we are really making an assumption that all workers are working at equal rates....how can you make that assumption unless given in the question?
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Re: M 20 Q 3 [#permalink]
### Show Tags
22 Feb 2012, 14:55
That's a good question. I think the question should clarify that as OG's did.
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Re: M 20 Q 3 [#permalink]
### Show Tags
05 Apr 2012, 23:28
This is really good point. question should clarify that all workers are working at same rate
_________________
-------Analyze why option A in SC wrong-------
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Re: M 20 Q 3 [#permalink] 05 Apr 2012, 23:28
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# how to do bernoulli trials on calculator
., X n) represents the outcomes of n independent Bernoulli trials, each with success probability p. Trials (required argument) – This is the number of Bernoulli trials. A Bernoulli random variable is a special category of binomial random variables. The Bernoulli distribution is a discrete probability distribution in which the random variable can take only two possible values 0 or 1, where 1 is assigned in case of success or occurrence (of the desired event) and 0 on failure or non-occurrence. This Bernoulli's Equation Calculator can help you determine any unknown variable from Bernoulli’s principle formula by giving 6 out of the 7 figures (height, pressure, fluid speed and density). This Bernoulli's Equation Calculator can help you determine any unknown variable from Bernoulli’s principle formula by giving 6 out of the 7 figures (height, pressure, fluid speed and density). The number of successes is 2. There can be only two outcomes – success/yes and failure/no. That is, it is the number of independent trials that are to be done. Probability_s (required argument) – This is the probability of success in a single trial. Where:g is the acceleration due to gravity and by default is considered to be 9.80665 m/s2;ρ is the fluid density measured in Kg/m3;P1 represents the fluid pressure measured in Pa at point 1;P2 refers to the fluid pressure measured in Pa at point 2;v1 is the fluid speed expressed in meter/second at point 1;v2 stands for the fluid speed in meter/second at point 2;h1 is the height in meters at point 1;h2 is the height in meter at point 2.As it can easily be observed this Bernoulli`s Equation calculator allows choosing between multiple units depending on the variable type as detailed here, while the conversion rates are:- For fluid pressure the measurement units available are: - For fluid speed the supported measurement units are: - For fluid density the measurement units that can be used: - For height these are the units available: Copyright 2014 - 2020 The Calculator .CO | All Rights Reserved | Terms and Conditions of Use. Instruction: Please input ONLY 6 from the 7 fields that are blank below! However, this is not a Bernoulli experiment since the trials are not independent (the mix of reds and blues changes on each trial since we do not replace the marble) and the probability of success and failure vary from trial to trial. trial here consist of drawing a marble from the bag and let success be getting a red. The possible results of the action are classified as "success" or "failure". What I want to do in this video is to generalize it. The experiment with a fixed number n of Bernoulli trials each with probability p, which produces k success outcomes is called binomial experiment. Gives probability of k success outcomes in n Bernoulli trials with given success event probability. Based on the above, we can say that an experiment may be called a Bernoulli trial when it meets the following conditions: The number of trials is fixed, not infinite. This tool can be used to calculate any variable from the Bernoulli’s formulas as explained below: Bernoulli's Equation: P 1 + 0.5 * ρ * v 1 2 + h 1 *ρ*g = P 2 + 0.5 * ρ * v 2 2 + h 2 *ρ*g - If the value of P 1 or the one of the P … Each trial (for example, each coin toss) is completely independent of the results of the previous turn. Excel will truncate the value to an integer if we provide it in decimal form. To figure out really the formulas for the mean and the variance of a Bernoulli Distribution if we don't have the actual numbers. The binomial probability formula is used to find probabilities for Bernoulli trials. The formula for calculating the result of bernoulli trial is shown below: The bernoulli trial is calculated by multiplying the binomial coefficient with the probability of success to the k power multiplied by the probability of failure to the n-k power. How does this Bernoulli's Equation Calculator work?
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Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing
# Logical Reasoning - Logical Problems
### Exercise
"In the middle of difficulty lies opportunity."
- Albert Einstein
Each problem consists of three statements. Based on the first two statements, the third statement may be true, false, or uncertain.
6.
The Kingston Mall has more stores than the Galleria.
The Four Corners Mall has fewer stores than the Galleria.
The Kingston Mall has more stores than the Four Corners Mall.
If the first two statements are true, the third statement is
A. true B. false C. uncertain
7.
All the tulips in Zoe's garden are white.
All the pansies in Zoe's garden are yellow.
All the flowers in Zoe's garden are either white or yellow
If the first two statements are true, the third statement is
A. true B. false C. uncertain
8.
During the past year, Josh saw more movies than Stephen.
Stephen saw fewer movies than Darren.
Darren saw more movies than Josh.
If the first two statements are true, the third statement is
A. true B. false C. uncertain
9.
Rover weighs less than Fido.
Rover weighs more than Boomer.
Of the three dogs, Boomer weighs the least.
If the first two statements are true, the third statement is
A. true B. false C. uncertain
10.
All the offices on the 9th floor have wall-to-wall carpeting.
No wall-to-wall carpeting is pink.
None of the offices on the 9th floor has pink wall-to-wall carpeting.
If the first two statements are true, the third statement is
A. true B. false C. uncertain
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# Waves, Light, Sound, Heat
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Unit on Waves. A study in how sound and light are waves. An overview of heat.
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### Waves, Light, Sound, Heat
1. 1. Waves, Sound, and Light<br />
2. 2. Lesson 1:Waves<br />
3. 3. Parts of a Transverse Wave<br />Wavelength is the distance between to points on a wave.<br />Amplitude is the distance from the top of the wave to the rest position.<br />The crest is the highest point of the wave.<br />The trough is the lowest point of the wave.<br />
4. 4. What is the amplitude?<br />Amplitude refers to the size of a wave.<br />The higher the amplitude, the more energy in a wave.<br />Which wave contains the most energy?<br />
5. 5. What is frequency?<br />Frequency is the number of waves that pass a given point in a second.<br />The higher the frequency, the higher the pitch.<br />Which wave has a higher frequency?<br />
6. 6. Parts of a Compressional Wave<br />
7. 7. Lesson 1<br />5. The wave above is called a transversewave.<br />6. Amplituderefers to how much energy a wave has.<br />1<br />1. wavelength<br />2.amplitude<br />7. Frequency refers to the number of waves that pass in a second.<br />8. Sound travels in a compressional wave.<br />2<br />4. crest<br />3. trough<br />
8. 8. Lesson 2Sound<br />
9. 9. Sources of Sound<br />When a person plucks a guitar string or bangs a drum, it will vibrate.<br />A quick back-and-forth movement is a vibration.<br />The vibrations that cause sounds are called sound waves.<br />What types of mediums (solids, liquids, or gases) can sound waves travel through?<br />Can sound travel in outer space?<br />
10. 10. Pitch<br />Large dogs and small dogs both bark, but their barks are different.<br />Pitch is how high or how low a sound is.<br />Low sounds waves are spread farther apart than high sound waves.<br />
11. 11. Intensity<br />If you drop a small paperback book versus a large dictionary, how would the sounds be different?<br />They would differ in intensity, or the measure of how loud or soft something is.<br />Why do you think the sound waves produced by slamming a door have more energy than the ones produced by shutting it gently?<br />
12. 12. Sounds<br />Sound waves bounce off surfaces through a process called reflection.<br />Echoes occur when sounds bounce of smooth surfaces and return.<br />Some materials stop sound waves from reflecting or traveling any farther. This process is called absorption.<br />Sound waves travel through air or other matter via transmission.<br />Sound travels in compression waves. <br />
13. 13. Decibels<br />Sound is measured in decibels.<br />
14. 14. 12. Pitch is how high or how low a sound is.<br />13. Pitch and frequency are related.<br />14. Intensity is how loud or soft a sound is. <br />15. The higher the amplitude the louder the sound.<br />16. Sound waves bounce off smooth surfaces through a process called reflection.<br />17. Some materials stop the reflection of sound through a process called absorption.<br />4<br />Lesson 2<br />9. A quick back and forth action is called a vibration .<br />10. The vibrations that cause sounds are called sound waves.<br />11. Can sound travel through space? Explain. No. Sound needs a medium to travel through. There is nothing in space for it to travel through.<br />3<br />18. Sound travels through matter via transmission.<br />19. Sound is measured in decibels.<br />20. A whisper measures about 30 decibels.<br />21. The sound a chain saw makes measures 117 decibels.<br />5<br />
15. 15. Lesson 3Light<br />
16. 16. Light is a form of energy that can travel through space<br />
17. 17. Why can we see this mountain reflected in the water?<br />
18. 18. When light hits an object, the light bounces off the surface of the object. The bouncing of light off an object is called reflection. When all light waves reflect in the same direction, we can see an image.<br />Most objects have rough surfaces. You can not see an image in something with a rough surface because the light reflects in many different directions<br />
19. 19. Absorption<br />Not all light passes through glass. <br />Transparent materials let most of the light pass through them<br />Translucent materials let some of the light pass through them.<br />Opaque materials do not allow light to pass through.<br />
20. 20. Which of following is opaque, translucent, and transparent?<br />
21. 21. Why does the spoon look bent?<br /><ul><li>The bending of light is called refraction.
22. 22. Light waves bend when they change speed.</li></li></ul><li>Lesson 3<br />22. Light is a form of energy that travels through space.<br />23. The bouncing of light off an object is called reflection.<br />24. Transparent materials let most of the light pass through.<br />25. Translucent materials let some light pass through.<br />26. Opaque materials do not allow light to pass through.<br />27. The bending of light as it changes speed is called refraction.<br />6.<br />
23. 23. Lesson 4Transferring Heat<br />
24. 24. Temperature<br />
25. 25. Temperature<br />Temperature is the measure of how hot or cold something is.<br />The hotter something is, the more quickly the particles move.<br />The colder something is, the slower particles move.<br />
26. 26. Thermal Energy<br />Which has more energy, a hot pot on a stove or the cool water in the pool?<br />Even though the pool is cooler, the water in the pool has more thermal energy. Thermal energy is the total amount of energy in an object. There is a much larger amount of matter in the pool, so the total of its energy is larger. The water in the pot is a small amount of matter. It has less total energy<br />
27. 27. Conduction<br />Conduction is the way heat travels through materials that are touching.<br />What happens when you leave a spoon in a hot pot of food?<br />Heat moved along the spoon until the whole spoon was hot.<br />
28. 28. Convection<br />Convection is the movement of heat in liquids and gases.<br />Warm liquid or gas is forced up by cooler liquid or heat. When water boils in a pot, the water near the heat source gets hot first and is forced up by the cooler water.<br />
29. 29. Radiation<br />Radiation is the movement of heat without matter.<br />Heat travels from the sun to Earth’s surface. Almost all of that distance is through space.<br />Radiation can take place when matter is present, too. You don’t have to touch a hot stove to know its hot.<br />
30. 30. Infrared<br />
31. 31. Heat Sources<br />Energy transfer is the change of energy from one form to another.<br />Heat that can’t be used to do work is waste heat.<br />
32. 32. Lesson 4<br />Temperature is the measure of how hot or cold something is.<br /> Conductionis the way heat travels through materials that are touching.<br />Radiation is the movement of heat without matter.<br />Energy transfer is the change of energy from one from to another.<br />Heat that can’t be used to do work is waste heat.<br />
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So there are a total of 13 × 48 = 624 possible hands with four of a kind. 11 Let us show that there exist irrational numbers a and b such that a b is rational. CHAPTER 2 Methods of Mathematical Proof 39 √ √ Solution: Let α = 2 and β = 2. If α β is rational then we are done, using a = α and b = β. If α β is irrational, then observe that αβ √ 2 Thus, with a = α β and b = that a b = 2 is rational. 12 Show that if there are six people in a room then either three of them know each other or three of them do not know each other.
By the result of the first paragraph, we can handle that case. Now, inductively, suppose that we have an algorithm for handling n pearls. We use this hypothesis to treat (n + 1) pearls. From the (n + 1) pearls, remove one and put it in your pocket. There remain n pearls. Apply the npearl algorithm to these remaining pearls. If you find the odd pearl then you are done. If you do not find the odd pearl, then it is the one in your pocket. That completes the case (n + 1) and the proof. What is the flaw in this reasoning?
11 Let S = {1, 2, 3}. Then P(S) = {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}, ∅ If the concept of power set is new to you, then you might have been surprised to see {1, 2, 3} and ∅ as elements of the power set. But they are both subsets of S, and they must be listed. 3 Let S = {s1 , . . , sk } be a set. Then P(S) has 2k elements. Proof: We prove the assertion by induction on k. The inductive statement is A set with k elements has power set with 2k elements P(1) is true. In this case, S = {s1 } and P(S) = {{s1 }, ∅}.
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# Electron Configuration Calculations
1. Jun 17, 2008
### truthfinder
Hello,
I've been trying to learn about electron configuration in atoms, and how it is calculated.
I pretty much figured out a lot of it from doing my own research. From what I understand, each group in the periodic table represents a different electron shell, and the blocks S, P, d, and F represent subshells. Once it got up to group 4, I figured it needed to fill 3D before working on 4P, and I was correct. I thought the same principal would apply once I got to the 6th group, so I expected Lanthanum to be [Xe]6s^2 4f^1. But, apparently, an electron is put in 5D first so that it is [Xe]6S^2 5d^1, and then Cerium is [Xe]6s^2 4f^1 5d^1. This makes no sense to me.
I admit, I am very new to learning about this. I am a physics and math major, and my first chemistry course starts next semester, and I was jus ttrying to get a bit of a head start. But I'd like to understand why it progresses in this fashion, when it's so unlike anything else before it.
it seems to repeat in Actinium, so at least it is persistent. But I want to know, why.
Thanks for any help. :)
2. Jun 17, 2008
### Staff: Mentor
Start with Aufbau principle - that gives a general idea. But then there are exclusions. In general electron enters the orbital that will give the lowest energy state. But electrons interact, and the more electrons are present, the more closely spaced energy levels of different states are, thus sometimes electrons fill orbitals in slightly different order that what we expect. At such situations we usually say something like "in this particular case configuration with 5 unpaired electrons on d and 1 on s has less energy then the configuration with 4 electrons on d and fully filled s" (that's for chromium). And that's true, it just doesn't have any predictable power, as it works only for chromium.
In the end the only real answer to question "why" is "because that's the way it is"
Last edited by a moderator: Aug 13, 2013
3. Jun 17, 2008
### ZapperZ
Staff Emeritus
Actually, you missed an even "earlier" pattern-breaking. The 4s gets filled first ahead of the 3d. Look at the transition metals.
When you add more and more electrons to the shell, you also start to increase the possibly of shielding of the nuclear potential. Since the p, d, and higher orbitals tend to extend even further from the nucleus than the s orbital, on average, they get shielded by the inner shell electrons even more. This can cause them to have a higher energy state than the s-orbital of the next principle quantum number state.
Zz.
4. Jun 17, 2008
### truthfinder
Thank you for the replies.
Borek, interesting. I just expected that when it comes to chemistry, there would not be exceptions to rules.
I didn't even realize that about chromium. Thanks. :)
ZapperZ, yeah I already encountered that throughout all of groups 4 and 5. That's why I expected group 6 to be the same, except applied to F as well.
Then, what is the configuration of Yttrium? I have seen [Kr]5s^1 4d^1, but also [Kr]5s^2 4d^1. The former breaks the rule, but now that apparently isn't impossible, so, I don't know which it is.
5. Jun 17, 2008
### Staff: Mentor
There different kinds of rules. There are rules like - say - first law of thermodynamics, or mass and energy conservation. These are rock hard.
But then there are rules like Aufbau principle, which states that "The orbitals of lower energy are filled in first with the electrons and only then the orbitals of high energy are filled." This is not a hard rule, more like rule of thumb - usually works, sometimes not, as sometimes to obtain lower energy of the whole system it is better to put electrons on different orbitals.
Last edited by a moderator: Aug 13, 2013
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Arithmetic Operators in java 7
In this section you will learn about the Arithmetic operators. This is one type of operators.
Tutorials
Arithmetic Operators in java 7
In this section you will learn about the Arithmetic operators. This is one type of operators.
Arithmetic Operators :
Arithmetic operators are the most commonly used operator in our expressions. We use these kind of operators in our mathematical expression. They work same as we use in algebra.
Arithmetic operators are used to perform addition, subtraction, multiplication, and division. Suppose X and Y are two numeric variables.
+ (additive operator) this operator is used for adding two numbers as well as used for String concatenation. (X+Y)
- (Subtraction operator) This operator is used for subtracting the values. (X-Y)
* (Multiplication operator) It is used to multiply values. (X*Y)
/ (Division operator) This operator is used where you want to divide some values. (X/Y)
% (Remainder operator) - This is also called modulus operator. It returns remainder. (X%Y)
Other arithmetic operators -
++ (increment) - This operator increments the value by 1. (X++)
-- (decrement) -
This operator decrements the value by 1. (X--)
Example :
```package operator;
class ArithmaticOperator {
public static void main(String[] args) {
int x = 100;
int y = 20;
int sum, sub, mul, mod;
double div;
sum = x + y; // using Additive operator
sub = x - y; // using Subtraction operator
mul = x * y; // using Multiplication operator
div = x / y; // using Division operator
mod = x / y; // using Remainder operator
System.out.println("Value of x = " + x);
System.out.println("Value of y = " + y);
System.out.println("Sum of x and y using Additive operator = " + sum);
System.out.println("Subtraction of x and y using Subtraction operator = "
+ sub);
System.out.println("Multiplication of x and y using Multiplication operator = "
+ mul);
System.out.println("Division of x and y using Additive operator = "
+ div);
System.out.println("Remainder of x and y using Remainder operator = "
+ mod);
x++; // increment operator
y--; // decrement operator
System.out.println("Incrementing x by using increment operator = " + x);
System.out.println("Decrementing y by using decrement operator = " + y);
}
}
```
Output :
```Value of x = 100
Value of y = 20
Sum of x and y using Additive operator = 120
Subtraction of x and y using Subtraction operator = 80
Multiplication of x and y using Multiplication operator = 2000
Division of x and y using Additive operator = 5.0
Remainder of x and y using Remainder operator = 5
Incrementing x by using increment operator = 101
Decrementing y by using decrement operator = 19
```
Arithmetic Operators in java 7
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### recursion question
I really need help with this and a sarcastic comment is not what I consider help, so if you can please give me a hand with this. Any help is appreciated, thanks in advance.
I was assigned a homework where I have to use recursion to find prime numbers of any integer entered by the user. Does anyone have any ideas on what I should do. typing out the code itself won't be too much of a problem, but I can't find a formula that will do the math.
Prime numbers of any integer? Do you mean the primes that divide into the number itself, or the number of primes lower than that number? Or do you mean x number of primes, where the user inputs x?
Try a loop with switch
This is the only information given by the professor:
Instructions: Write a program that will be able to get the prime factors of the number input by the user. Make sure your program implements a recursive algorithm.
Example: Prime factors of 25 = 5 * 5. prime factors of 60 = 2 * 2 * 3 * 5
Use a loop starting at one and divide by that number as much as possible then move on to next
>Use a loop starting at one and divide by that number as much as possible then move on to next
1 is not a prime number
> Make sure your program implements a recursive algorithm.
I might be wrong, but I'm sure this does not mean that the entire process of getting the prime has to be recursive, it can also recursively call a function lol.
I'm not sure if I'm right, but if you do implement an entirely recursive algorithm to solve this, you will be going deep. Like level 7 inception deep in the recursion because that means testing each integer for primeness, then storing that value in an array or however you wish to store all the values.
gl and post here if you make any progress
Last edited on
For any given number, you can find the first prime number by which it is evenly divisible by: beginning with the smallest number it may be divisible by (other than 1) and incrementing it until it is a number that evenly goes into the number in question.
You then have one factor (the divisor) and another value which needs to be factored (the quotient). If the quotient is 1, you're done.
Topic archived. No new replies allowed.
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https://curiosity.com/topics/a-parsec-is-a-real-astronomical-unit-of-measurement-curiosity/
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Measurement
# A Parsec Is a Real Astronomical Unit of Measurement
Now, hang on just one parsec ... or maybe not. Though a parsec may sound like a derivative of "second," it actually isn't. And that's why the famed line about the Millennium Falcon in Star Wars — "It's the ship that made the Kessel Run in less than 12 parsecs" — doesn't make sense. The true meaning of the word is way cooler anyway.
## Larger Than Life
For starters, a parsec is a unit of measurement, not a unit of time. Sorry, Han. It's a very, very large unit of measurement at that. Specifically, one parsec is equivalent to 3.26 light years, or 19 trillion miles (31 trillion kilometers). If this seems excessively large, that's because it sure is. But when it comes to measuring astronomically large distances between objects beyond our solar system, excessively large is just right.
That's right, objects within our own, expansive solar system barely sit far enough apart to require parsecs to measure their distances. For example, the distance between the Earth and the sun is exactly one astronomical unit, called an AU. One parsec is 206,265 AU. If you went to the sun and back 206,265 times, you would have successfully traveled a distance totaling one parsec. Impressive.
## Parse It Out
Why name something a parsec when it has nothing to do with seconds or par on a golf course? Its origins make sense, even if they're not the most obvious. The word itself comes from two words: parallax and arcsecond. Parallax describes when an object's location seems to have changed because your location changed. Astronomers use parallax to judge the distance of objects in the sky; if you know how far the Earth has moved and can measure how far the object moved across its starry backdrop, you can judge its distance. Astronomers aren't the only ones, however — you use parallax to judge the distance of things all the time. Your two eyes see the world from slightly different angles, and your brain crunches the numbers to tell you how far away everything is from you.
The second half of parsec refers to an arcsecond. An arcsecond is a part of a measurement of an angle. You know that a circle is 360 degrees, right? Imagine you're standing with a full view of the horizon around you. If you sliced the horizon into 360 degrees, each degree would be about twice the width of the full moon. Divide one of those degrees into 60 pieces, and you get the width of an arcminute. Divide one of those up by 60 again, and you get an arcsecond. Putting it all together, a parsec is the parallax of one arcsecond — or, in other words, if an object in the sky moves by 1 arcsecond when the Earth moves by 1 AU, the object is 1 parsec away. To give an example of this huge distance, Proxima Centauri, the star that's our closest neighbor, is only 1.3 parsecs away. Only.
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Joanie Faletto
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http://math.stackexchange.com/questions/167374/exponentiation-of-reals
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# Exponentiation of reals
Let $1<b\in \mathbb{R}$ and $x\in \mathbb{R}$. I want to prove that $\sup${$b^t\in \mathbb{R}$|$x≧t\in \mathbb{Q}$} = $\inf${$b^t\in \mathbb{R}$|$x≦t\in \mathbb{Q}$}.
I have proved that $\sup$≦$\inf$, but dont know how to show that $\inf$≦$\sup$..
-
The proof depends on your definition of exponentiation. – Yuval Filmus Jul 6 '12 at 7:01
I haven't defined exponentiation with real index. I have defined exponentiation with a positive base and rational index but this equality seems true and able to be proved with just this info – Katlus Jul 6 '12 at 7:10
We know that $f(t) = b^t$ is continuous and monotonic. Take two sequences in $\mathbb{Q}$ that converge to $x$, say $(t_n)$ and $(s_n)$. Due to continuity, $\lim b^{t_n} = \lim b^{s_n} = b^x$. Then you only need to show that these limits are really $\inf$ and $\sup$. Use monotony for that.
Choose once and for all $T$ in $\mathbb Q$ with $T\geqslant x$. For every positive integer $n$, choose $s_n\leqslant x\leqslant t_n$ such that $s_n$ and $t_n$ are in $\mathbb Q$ and $t_n-s_n\leqslant1/n$.
Then, $b\gt1$ hence $b^{t_n}\leqslant b^{s_n}(b^{1/n}-1)+b^{s_n}\leqslant b^T(b^{1/n}-1)+b^{s_n}$. When $n\to\infty$, $b^{1/n}\to1$ hence, for every positive $\varepsilon$ there exists $n$ such that $b^{1/n}\leqslant1+\varepsilon/b^T$. Thus, $b^{t_n}\leqslant \varepsilon+b^{s_n}$.
This shows that $\inf\{b^t\mid t\geqslant x,t\in\mathbb Q\}\leqslant\varepsilon+\sup\{b^s\mid s\leqslant x,t\in\mathbb Q\}$. This holds for every positive $\varepsilon$ hence you are done.
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http://www.mathaddict.net/catalan.htm
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### Catalan Numbers CalculatorWith Arbitrary Precision Arithmetic
Input n (up to 25000):` ` Result: the Catalan number C(n) (to select: click, Ctrl+A, Ctrl+C) Run time: n = n = n = n = n =
The Catalan numbers C(n) can be defined in several equivalent ways. A formal definition is:
C(n) = (2n)!/(n!(n+1)!).
Another definition: C(n) is the number of ways to insert n pairs of parentheses in an expression with n+1 terms. For example, if n = 2 there are two ways: `((ab)c)` or `(a(bc))`; if n = 3 there are 5 ways: `((ab)(cd))`, `(((ab)c)d)`, `((a(bc))d)`, `(a((bc)d))`, `(a(b(cd)))`. Accordingly, C(2) = 2 and C(3) = 5.
The Catalan sequence is OEIS A000108; in the Online Encyclopedia of Integer Sequences you can find lots of additional information and further references on Catalan numbers.
This online calculator computes the Catalan numbers C(n) for input values 0 ≤ n ≤ 25000 in arbitrary precision arithmetic. So, for example, you will get all 598 digits of C(1000) – a very large number!
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