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DIIS and the CC Iterative Procedure

In an earlier project, we examined how DIIS (“direct inversion in the iterative subspace”) can reduce substantially the number of iterations required to converge the self-consistent field. In this project, we examine how DIIS can similarly speed convergence in the CC amplitude equations. This approach was first used by Scuseria, Lee, and Schaefer in 1986 [“Accelerating the conference of the coupled-cluster approach. The use of the DIIS method”, Chem. Phys. Lett. 130, 236 (1986)].

Error Vectors

The DIIS approach requires a well-defined “error vector”, a quantity that approaches zero as the equations converge. In the SCF procedure, we use the atomic-orbital representation of the occupied-virtual block of the Fock matrix. In the CC method, we could choose the difference between successive sets of cluster amplitudes:

where ${\mathbf T}_i$ represents a vector containing all the cluster amplitudes for the i-th iteration.

There are three important points to note about this choice of error vector:

• Defining the error vectors as differences between successive sets of amplitudes implies that one cannot begin the DIIS extrapolation until at least three iterations are complete.
• Just as in the SCF DIIS procedure, error vectors should only be computed using non-extrapolated sets of cluster amplitudes.
• The set of cluster amplitudes can require substantial memory for larger molecules; hence, the choice of the number of error vectors used in the CC DIIS procedure is potentially dependent on the available storage (disk and memory).

Extrapolation

Given the above definition of the error vectors, the set of linear equations to be solved is the same as that for the SCF DIIS procedure:

where lambda is a Lagrangian multiplier and the elements Bij are computed as dot products of error matrices:

A new set of cluster amplitudes is then obtained as a linear combinations of older amplitudes using the coefficients from the linear equations above:

Again: The extrapolated cluster amplitudes should be used only in the CC amplitude equations, not to compute subsequent error vectors.

Results

Once the procedure is working, you should observe a considerable reduction in the number of iterations required to converge the CC amplitude equations to a given tolerance. For example, without DIIS extrapolation, the STO-3G H2O test case from Project #5 converges in 38 iterations to a precision of 10-12:

iter =  1  Ecc =      -0.062758205955
iter =  2  Ecc =      -0.067396582597
iter =  3  Ecc =      -0.069224536410
iter =  4  Ecc =      -0.070007757556
iter =  5  Ecc =      -0.070360041902
iter =  6  Ecc =      -0.070523820218
iter =  7  Ecc =      -0.070602032617
iter =  8  Ecc =      -0.070640293027
iter =  9  Ecc =      -0.070659428829
iter = 10  Ecc =      -0.070669194426
iter = 11  Ecc =      -0.070674268048
iter = 12  Ecc =      -0.070676944995
iter = 13  Ecc =      -0.070678375859
iter = 14  Ecc =      -0.070679148887
iter = 15  Ecc =      -0.070679570139
iter = 16  Ecc =      -0.070679801279
iter = 17  Ecc =      -0.070679928796
iter = 18  Ecc =      -0.070679999445
iter = 19  Ecc =      -0.070680038717
iter = 20  Ecc =      -0.070680060604
iter = 21  Ecc =      -0.070680072825
iter = 22  Ecc =      -0.070680079661
iter = 23  Ecc =      -0.070680083488
iter = 24  Ecc =      -0.070680085633
iter = 25  Ecc =      -0.070680086836
iter = 26  Ecc =      -0.070680087511
iter = 27  Ecc =      -0.070680087891
iter = 28  Ecc =      -0.070680088103
iter = 29  Ecc =      -0.070680088223
iter = 30  Ecc =      -0.070680088290
iter = 31  Ecc =      -0.070680088328
iter = 32  Ecc =      -0.070680088349
iter = 33  Ecc =      -0.070680088361
iter = 34  Ecc =      -0.070680088368
iter = 35  Ecc =      -0.070680088372
iter = 36  Ecc =      -0.070680088374
iter = 37  Ecc =      -0.070680088375
iter = 38  Ecc =      -0.070680088376

The DIIS extrapolation with eight error vectors reduces the number of iterations to just 16:

iter =  1  Ecc =      -0.062758205955
iter =  2  Ecc =      -0.067396582597
iter =  3  Ecc =      -0.070325967097
iter =  4  Ecc =      -0.070642541851
iter =  5  Ecc =      -0.070681441999
iter =  6  Ecc =      -0.070679216733
iter =  7  Ecc =      -0.070680252965
iter =  8  Ecc =      -0.070680084062
iter =  9  Ecc =      -0.070680084577
iter = 10  Ecc =      -0.070680089084
iter = 11  Ecc =      -0.070680088015
iter = 12  Ecc =      -0.070680088239
iter = 13  Ecc =      -0.070680088359
iter = 14  Ecc =      -0.070680088381
iter = 15  Ecc =      -0.070680088377
iter = 16  Ecc =      -0.070680088376

Additional Reading

• G.E. Scuseria, T.J. Lee, and H.F. Schaefer, “Accelerating the conference of the coupled-cluster approach. The use of the DIIS method”, Chem. Phys. Lett. 130, 236 (1986).
• G.D. Purvis and R.J. Bartlett, “The reduced linear equation method in coupled cluster theory”, J. Chem. Phys. 75, 1284 (1981).
crawdad/programming/project10.txt · Last modified: 2011/07/05 21:39 by crawdad

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# Prove the Reduction Forumla

1. Homework Statement
a)Prove the reduction formula:

$$\int\ tan^n\ x\ dx\ =\ \frac{1}{n-1}tan^{n-1}\ x\ -\int\ tan^{n-2}\ x\ dx$$

Hint: first write $$tan^n x$$ as $$tan^{n-2} \ x\ tan^2\ x$$ and the rewrite using $$tan^2\ x+1=sec^2\ x$$.

b) Use the formula twice to find $$\int\ tan^4\ dx$$

3. The Attempt at a Solution

I not sure what they're asking for when they say "prove". How should I begin with this one? Thanks

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
Vid
Use methods of integration to show the left side equals the right. From the form of the right side, it should be pretty obvious which method to use.


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# Rotation of a Disk

1. Dec 8, 2006

### the_dialogue

1. The problem statement, all variables and given/known data

Suppose a disk of radius 'r' is rotation on a surface. If the center G moves a distance 'd', then what is the distance traveled by a point on the top of the disk (on its edge or circumference).

2. Relevant equations

theta = s / r ; where theta is the rotation in rad, s is the arc length, r is the radius

3. The attempt at a solution
I know that if G moves a distance 'd', then the entire circle rotates 'theta'=d/r. But I'm not sure how to make this a general case.

A thought: Can i treat the point of contact between the disk and ground as a n "origin" and then state that a point directly above it on the edge of the disk moves '2r*theta' ?

Thank you,
Alex.

-------

EDIT:

I hope you don't mind if I make the problem a bit more specific. Suppose a gear of radius r_o is moving with an inner hub of radius r_i. If I know the origin moves 'd', then how far does a point on the circumference of the inner hub move?

Last edited: Dec 8, 2006
2. Dec 8, 2006

### OlderDan

It is not clear to me what you are asking. First, are you talking about a disk that is rolling in a straight line on it edge? I think so, but need to be sure. Are you looking for the net displacement of the point on the edge, or its actual path length? Are you looking at all points, or only the point that started at the top. By changing the problem with your edit, you seem to be generalizing to any point on the wheel. The answer is not the same for all points, whether you are talking about displacements or path lengths. Please restate the problem being specific about the orientation of the disk and what exactly you are trying to calculate.

Last edited: Dec 8, 2006
3. Dec 8, 2006

### BobG

Are you talking about a wheel rolling along the ground?

You want the center of the wheel to be your origin. You find the horizontal component (x-component) of the point on the circumference relative to the origin. It would be best to pick the trailing edge of the wheel as your point of interest. That way, it's start point would be -r.

As the wheel rotates, find the point's x component relative to the center. You should be able to find a general equation that would handle any location around the circumference, plus be correct for your start position. Your equation $$\theta = \frac{s}{r}$$ is on the right track, but you need to find out just the horizontal displacement.

Add in the distance that the center of the wheel moved.

Subtract your start position from the above sum.


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### Distributed spanner construction in doubling metric spaces.

This paper presents a distributed algorithm that runs on an n-node unit ball graph (UBG) G residing in a metric space of constant doubling dimension, and constructs, for any " > 0, a (1 + ")-spanner H of G with maximum degree bounded above by a constant. In addition, we show that H is \lightweight", in the following sense. Let ¢ denote the aspect ratio of G, that is, the ratio of the length of a longest edge in G to the length of a shortest edge in G. The total weight of H is bounded above by O(log¢)¢wt(MST), where MST denotes a minimum spanning tree of the metric space. Finally, we show that H satis¯es the so called leapfrog property, an immediate implication being that, for the special case of Euclidean metric spaces with ¯xed dimension, the weight of H is bounded above by O(wt(MST)). Thus, the current result subsumes the results of the authors in PODC 2006 that apply to Euclidean metric spaces, and extends these results to metric spaces with constant doubling dimension.

Main Author: Damian, Mirela. Pandit, Saurav., Pemmaraju, Sriram. English 2006 http://ezproxy.villanova.edu/login?url=https://digital.library.villanova.edu/Item/vudl:175665


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at FinTECHTalents 2018 on Oct 30, 2018 [reveal]
Neil D. Lawrence, Amazon and University of Sheffield

#### Abstract

What is the nature of machine intelligence and how does it differ from humans? In this talk we introduce embodiment factors. They represent the extent to which our intelligence is locked inside us. The locked in nature of our intelligence makes us fundamentally different from the machine intelligences we are creating around us. Having summarized these differences we consider the Three Ds of machine learning system design: a set of considerations to take into acount when building machine intelligences.

## The Diving Bell and the Butterfly

The Diving Bell and the Butterfly is the autobiography of Jean-Dominique Bauby.

In 1995, when he was editor-in-chief of the French Elle magazine, he suffered a stroke, which destroyed his brainstem. He became almost totally physically paralyzed, but was still mentally active. He acquired what is known as locked-in syndrome.

Incredibly, Bauby wrote his memoir after he became paralyzed.

His left eye was the only muscle he could voluntarily move, and he wrote the entire book by winking it.

How could he do that? Well, first, they set up a mechanism where he could scan across letters and blink at the letter he wanted to use. In this way, he was able to write each letter.

It took him 10 months of four hours a day to write the book. Each word took two minutes to write.

Imagine doing all that thinking, but so little speaking, having all those thoughts and so little ability to communicate.

The idea behind this talk is that we are all in that situation. While not as extreme as for Bauby, we all have somewhat of a locked in intelligence.

## Embodiment Factors

Let me explain what I mean. Claude Shannon introduced a mathematical concept of information for the purposes of understanding telephone exchanges.

Information has many meanings, but mathematically, Shannon defined a bit of information to be the amount of information you get from tossing a coin.

If I toss a coin, and look at it, I know the answer. You don’t. But if I now tell you the answer I communicate to you 1 bit of information. Shannon defined this as the fundamental unit of information.

If I toss the coin twice, and tell you the result of both tosses, I give you two bits of information. Information is additive.

Shannon also estimated the average information associated with the English language. He estimated that the average information in any word is 12 bits, equivalent to twelve coin tosses.

So every two minutes Bauby was able to communicate 12 bits, or six bits per minute.

This is the information transfer rate he was limited to, the rate at which he could communicate.

Compare this to me, talking now. The average speaker for TEDX speaks around 160 words per minute. That’s 320 times faster than Bauby or around a 2000 bits per minute. 2000 coin tosses per minute.

But, just think how much thought Bauby was putting into every sentence. Imagine how carefully chosen each of his words was. Because he was communication constrained he could put more thought into each of his words. Into thinking about his audience.

So, his intelligence became locked in. He thinks as fast as any of us, but can communicate slower. Like the tree falling in the woods with no one there to hear it, his intelligence is embedded inside him.

Two thousand coin tosses per minute sounds pretty impressive, but this talk is not just about us, it’s about our computers, and the type of intelligence we are creating within them.

So how does two thousand compare to our digital companions? When computers talk to each other, they do so with billions of coin tosses per minute.

Let’s imagine for a moment, that instead of talking about communication of information, we are actually talking about money. Bauby would have 6 dollars. I would have 2000 dollars, and my computer has billions of dollars.

The internet has interconnected computers and equipped them with extremely high transfer rates.

However, by our very best estimates, computers actually think slower than us.

How can that be? You might ask, computers calculate much faster than me. That’s true, but underlying your conscious thoughts there are a lot of calculations going on.

Each thought involves many thousands, millions or billions of calculations. How many exactly, we don’t know yet, because we don’t know how the brain turns calculations into thoughts.

Our best estimates suggest that to simulate your brain a computer would have to be as large as the UK Met Office machine here in Exeter. That’s a 250 million pound machine, the fastest in the UK. It can do 16 billion billon calculations per second.

It simulates the weather across the word every day, that’s how much power we think we need to simulate our brains.

So, in terms of our computational power we are extraordinary, but in terms of our ability to explain ourselves, just like Bauby, we are locked in.

For a typical computer, to communicate everything it computes in one second, it would only take it a couple of minutes. For us to do the same would take 15 billion years.

If intelligence is fundamentally about processing and sharing of information. This gives us a fundamental constraint on human intelligence that dictates its nature.

I call this ratio between the time it takes to compute something, and the time it takes to say it, the embodiment factor (Lawrence 2017). Because it reflects how embodied our cognition is.

If it takes you two minutes to say the thing you have thought in a second, then you are a computer. If it takes you 15 billion years, then you are a human.

So when it comes to our ability to compute we are extraordinary, not compute in our conscious mind, but the underlying neuron firings that underpin both our consciousness, our subconsciousness as well as our motor control etc.

If we think of ourselves as vehicles, then we are massively overpowered. Our ability to generate derived information from raw fuel is extraordinary. Intellectually we have formula one engines.

But in terms of our ability to deploy that computation in actual use, to share the results of what we have inferred, we are very limited. So when you imagine the F1 car that represents a psyche, think of an F1 car with bicycle wheels.

Just think of the control a driver would have to have to deploy such power through such a narrow channel of traction. That is the beauty and the skill of the human mind.

In contrast, our computers are more like go-karts. Underpowered, but with well-matched tires. They can communicate far more fluidly. They are more efficient, but somehow less extraordinary, less beautiful.

## Human Communication

For human conversation to work, we require an internal model of who we are speaking to. We model each other, and combine our sense of who they are, who they think we are, and what has been said. This is our approach to dealing with the limited bandwidth connection we have. Empathy and understanding of intent. Mental dispositional concepts are used to augment our limited communication bandwidth.

Fritz Heider referred to the important point of a conversation as being that they are happenings that are “psychologically represented in each of the participants” (his emphasis) (Heider 1958)

### Bandwidth Constrained Conversations

import pods
from ipywidgets import IntSlider
pods.notebook.display_plots('anne-bob-conversation{sample:0>3}.svg',
'../slides/diagrams',  sample=IntSlider(0, 0, 7, 1))

Embodiment factors imply that, in our communication between humans, what is not said is, perhaps, more important than what is said. To communicate with each other we need to have a model of who each of us are.

To aid this, in society, we are required to perform roles. Whether as a parent, a teacher, an employee or a boss. Each of these roles requires that we conform to certain standards of behaviour to facilitate communication between ourselves.

Control of self is vitally important to these communications.

The high availability of data available to humans undermines human-to-human communication channels by providing new routes to undermining our control of self.

### A Six Word Novel

But this is a very different kind of intelligence than ours. A computer cannot understand the depth of the Ernest Hemingway’s apocryphal six word novel: “For Sale, Baby Shoes, Never worn”, because it isn’t equipped with that ability to model the complexity of humanity that underlies that statement.

## Computer Conversations

import pods
from ipywidgets import IntSlider
pods.notebook.display_plots('anne-bob-conversation{sample:0>3}.svg',
'../slides/diagrams',  sample=IntSlider(0, 0, 7, 1))

Similarly, we find it difficult to comprehend how computers are making decisions. Because they do so with more data than we can possibly imagine.

In many respects, this is not a problem, it’s a good thing. Computers and us are good at different things. But when we interact with a computer, when it acts in a different way to us, we need to remember why.

Just as the first step to getting along with other humans is understanding other humans, so it needs to be with getting along with our computers.

Embodiment factors explain why, at the same time, computers are so impressive in simulating our weather, but so poor at predicting our moods. Our complexity is greater than that of our weather, and each of us is tuned to read and respond to one another.

Their intelligence is different. It is based on very large quantities of data that we cannot absorb. Our computers don’t have a complex internal model of who we are. They don’t understand the human condition. They are not tuned to respond to us as we are to each other.

Embodiment factors encapsulate a profound thing about the nature of humans. Our locked in intelligence means that we are striving to communicate, so we put a lot of thought into what we’re communicating with. And if we’re communicating with something complex, we naturally anthropomorphize them.

We give our dogs, our cats and our cars human motivations. We do the same with our computers. We anthropomorphize them. We assume that they have the same objectives as us and the same constraints. They don’t.

This means, that when we worry about artificial intelligence, we worry about the wrong things. We fear computers that behave like more powerful versions of ourselves that will struggle to outcompete us.

In reality, the challenge is that our computers cannot be human enough. They cannot understand us with the depth we understand one another. They drop below our cognitive radar and operate outside our mental models.

The real danger is that computers don’t anthropomorphize. They’ll make decisions in isolation from us without our supervision, because they can’t communicate truly and deeply with us.

# Evolved Relationship with Information

The high bandwidth of computers has resulted in a close relationship between the computer and data. Large amounts of information can flow between the two. The degree to which the computer is mediating our relationship with data means that we should consider it an intermediary.

Originaly our low bandwith relationship with data was affected by two characteristics. Firstly, our tendency to over-interpret driven by our need to extract as much knowledge from our low bandwidth information channel as possible. Secondly, by our improved understanding of the domain of mathematical statistics and how our cognitive biases can mislead us.

With this new set up there is a potential for assimilating far more information via the computer, but the computer can present this to us in various ways. If it’s motives are not aligned with ours then it can misrepresent the information. This needn’t be nefarious it can be simply as a result of the computer pursuing a different objective from us. For example, if the computer is aiming to maximize our interaction time that may be a different objective from ours which may be to summarize information in a representative manner in the shortest possible length of time.

For example, for me, it was a common experience to pick up my telephone with the intention of checking when my next appointment was, but to soon find myself distracted by another application on the phone, and end up reading something on the internet. By the time I’d finished reading, I would often have forgotten the reason I picked up my phone in the first place.

There are great benefits to be had from the huge amount of information we can unlock from this evolved relationship between us and data. In biology, large scale data sharing has been driven by a revolution in genomic, transcriptomic and epigenomic measurement. The improved inferences that that can be drawn through summarizing data by computer have fundamentally changed the nature of biological science, now this phenomenon is also infuencing us in our daily lives as data measured by happenstance is increasingly used to characterize us.

Better mediation of this flow actually requires a better understanding of human-computer interaction. This in turn involves understanding our own intelligence better, what its cognitive biases are and how these might mislead us.

For further thoughts see Guardian article on marketing in the internet era from 2015.

You can also check my blog post on System Zero..

## Boulton and Watt’s Steam Engine

James Watt’s steam engine contained an early machine learning device. In the same way that modern systems are component based, his engine was composed of components. One of which is a speed regulator sometimes known as Watt’s governor. The two balls in the center of the image, when spun fast, rise, and through a linkage mechanism.

The centrifugal governor was made famous by Boulton and Watt when it was deployed in the steam engine. Studying stability in the governor is the main subject of James Clerk Maxwell’s paper on the theoretical analysis of governors (Maxwell 1867). This paper is a founding paper of control theory. In an acknowledgment of its influence, Wiener used the name cybernetics to describe the field of control and communication in animals and the machine (Wiener 1948). Cybernetics is the Greek word for governor, which comes from the latin for helmsman.

A governor is one of the simplest artificial intelligence systems. It senses the speed of an engine, and acts to change the position of the valve on the engine to slow it down.

Although it’s a mechanical system a governor can be seen as automating a role that a human would have traditionally played. It is an early example of artificial intelligence.

The centrifugal governor has several parameters, the weight of the balls used, the length of the linkages and the limits on the balls movement.

Two principle differences exist between the centrifugal governor and artificial intelligence systems of today.

1. The centrifugal governor is a physical system and it is an integral part of a wider physical system that it regulates (the engine).
2. The parameters of the governor were set by hand, our modern artificial intelligence systems have their parameters set by data.

This has the basic components of sense and act that we expect in an intelligent system, and this system saved the need for a human operator to manually adjust the system in the case of overspeed. Overspeed has the potential to destroy an engine, so the governor operates as a safety device.

The first wave of automation did bring about sabotoage as a worker’s response. But if machinery was sabotaged, for example, if the linkage between sensor (the spinning balls) and action (the valve closure) was broken, this would be obvious to the engine operator at start up time. The machine could be repaired before operation.

## What is Machine Learning?

Machine learning allows us to extract knowledge from data to form a prediction.

$$\text{data} + \text{model} \xrightarrow{\text{compute}} \text{prediction}$$

A machine learning prediction is made by combining a model with data to form the prediction. The manner in which this is done gives us the machine learning algorithm.

Machine learning models are mathematical models which make weak assumptions about data, e.g. smoothness assumptions. By combining these assumptions with the data, we observe we can interpolate between data points or, occasionally, extrapolate into the future.

Machine learning is a technology which strongly overlaps with the methodology of statistics. From a historical/philosophical view point, machine learning differs from statistics in that the focus in the machine learning community has been primarily on accuracy of prediction, whereas the focus in statistics is typically on the interpretability of a model and/or validating a hypothesis through data collection.

The rapid increase in the availability of compute and data has led to the increased prominence of machine learning. This prominence is surfacing in two different but overlapping domains: data science and artificial intelligence.

## From Model to Decision

The real challenge, however, is end-to-end decision making. Taking information from the environment and using it to drive decision making to achieve goals.

# Deep Learning

### DeepFace

The DeepFace architecture (Taigman et al. 2014) consists of layers that deal with translation and rotational invariances. These layers are followed by three locally-connected layers and two fully-connected layers. Color illustrates feature maps produced at each layer. The neural network includes more than 120 million parameters, where more than 95% come from the local and fully connected layers.

### Deep Learning as Pinball

Sometimes deep learning models are described as being like the brain, or too complex to understand, but one analogy I find useful to help the gist of these models is to think of them as being similar to early pin ball machines.

In a deep neural network, we input a number (or numbers), whereas in pinball, we input a ball.

Think of the location of the ball on the left-right axis as a single number. Our simple pinball machine can only take one number at a time. As the ball falls through the machine, each layer of pins can be thought of as a different layer of ‘neurons’. Each layer acts to move the ball from left to right.

In a pinball machine, when the ball gets to the bottom it might fall into a hole defining a score, in a neural network, that is equivalent to the decision: a classification of the input object.

An image has more than one number associated with it, so it is like playing pinball in a hyper-space.

import pods
from ipywidgets import IntSlider
pods.notebook.display_plots('pinball{sample:0>3}.svg',
'../slides/diagrams',
sample=IntSlider(1, 1, 2, 1))

Learning involves moving all the pins to be in the correct position, so that the ball ends up in the right place when it’s fallen through the machine. But moving all these pins in hyperspace can be difficult.

In a hyper-space you have to put a lot of data through the machine for to explore the positions of all the pins. Even when you feed many millions of data points through the machine, there are likely to be regions in the hyper-space where no ball has passed. When future test data passes through the machine in a new route unusual things can happen.

Adversarial examples exploit this high dimensional space. If you have access to the pinball machine, you can use gradient methods to find a position for the ball in the hyper space where the image looks like one thing, but will be classified as another.

Probabilistic methods explore more of the space by considering a range of possible paths for the ball through the machine. This helps to make them more data efficient and gives some robustness to adversarial examples.

# References

Heider, Fritz. 1958. The Psychology of Interpersonal Relations. John Wiley.

Hilbert, Martin, and Priscila López. 2011. “The World’s Technological Capcity to Store, Communicate and Compute Information.” Science 332 (6025): 60–65. https://doi.org/10.1126/science.1200970.

Lawrence, Neil D. 2017. “Living Together: Mind and Machine Intelligence.” arXiv. https://arxiv.org/abs/1705.07996.

Maxwell, James Clerk. 1867. “On Governors.” Proceedings of the Royal Society of London 16. The Royal Society: 270–83. http://www.jstor.org/stable/112510.

Taigman, Yaniv, Ming Yang, Marc’Aurelio Ranzato, and Lior Wolf. 2014. “DeepFace: Closing the Gap to Human-Level Performance in Face Verification.” In Proceedings of the IEEE Computer Society Conference on Computer Vision and Pattern Recognition. https://doi.org/10.1109/CVPR.2014.220.

Wiener, Norbert. 1948. Cybernetics: Control and Communication in the Animal and the Machine. Cambridge, MA: MIT Press.


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# Help: Functional Equation 3

$f\left( x \right) f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2$

If $$f\left( x \right)$$ is a polynomial satisfying the equation above for all $$x$$ and $$y$$ and $$f\left( 2 \right)=5$$, then compute $$\displaystyle \lim _{ x\rightarrow 2 }{ f^{ ' }\left( x \right) }$$.

If anyone solves this one, please do tell me what is the error in this one

1 year, 4 months ago

Sort by:

Hey there!
We have,
$$f(x)f(y)-f(x)-f(y)+1\quad =\quad f(xy)-1\\ (f(x)-1)(f(y)-1)\quad =\quad f(xy)-1\quad \\ g(x)=f(x)-1 \\ \Rightarrow g(x)g(y) = g(xy) \\$$

This is famous, see here.
We then have that,
$$g(x) = x ^ t, t = 2. \\ g(x) = x^2 \\ f(x) = x^2 + 1 \\ f'(x) = 2x \\ \Rightarrow \lim _{ x\rightarrow 2 }{ f^{ ' }(x) } = 4$$ · 1 year, 4 months ago

Put x and y as 2 which will give you the value of f(4) as 17

Put x as 4 and y as 2 and you get f(8) as 65. Similarly you can check with other values and this leads to the conclusion that f(x) is one greater than the square of x.

Hence the limit computation will be 4

Please excuse me if you were looking for a formal proof or evaluation of the same. · 1 year, 4 months ago

@parv mor · 1 year, 4 months ago


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YouTuber MindYourDecisions adapted this mind-boggling math question from a similar one found on an elementary school student's homework in China. There is no need to download any app for these activities. If you choose to use a calculator, be sure it is permitted, is working on test day, and has reliable batteries. You're forgiven if you don't remember exactly how exponents work. Types of Questions Within the context of open-ended mathematical tasks, it is useful to group questions into four main categories (Badham, 1994). You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. These questions may be used in your classroom to help students feel comfortable with the types of assessment formats they will experience. If $$\Large a\ast b=2 \left(a+b\right)$$, then $$\Large 5\ast 2$$ is equal to: $$\Large a \ast b = 2 \left(a+b\right)$$, 8). Rely more on themselves to determine whether something is mathematically correct. Game. In order to solve this problem, you simply need to subtract the exponents (4-2) and solve for 32, which expands into 3 x 3 and equals 9. This elementary school-level problem is a little less problem solving and a little more memorization. Elementary Math; Elementary Math. Think back to that problem about the dogs at the dog show and use the same logic to solve this problem. Either way, very few individuals—even math teachers—have been able to find the correct answer to this problem. You will receive incredibly detailed scoring results at the end of your SSAT Elementary Level Math practice test to help you identify your strengths and weaknesses. E: This question requires that you consider the different ways in which a rectangle and a circle lying on a plane could overlap and determine the maximum number of points of intersection.It might be helpful to sketch various possibilities. Now you have 9 – 9 + 1, and from there you can simply work from left to right and get your final answer: 1. Since we know that there are two zombies for every three humans and that 2 + 3 = 5, we can divide 85 by 5 to figure out that in total, there are 17 groups of humans and zombies. The first kind is the content knowledge needed to do the work of the student curriculum, such as solving a math problem similar to one students would solve. Preschool. no. Experts are answering questions even after midnight Therefore, if you are stuck on a math problem or homework at the last minute or if it is 10 PM, or 11 PM and you have not found a solution to your math problems, you will get help immediately. This question was used in China to identify gifted 5th … If the rectangle is placed so that one side is tangent to the circle, there will be one point of intersection. Trust us, we get it. Most people automatically assume that half of the lake would be covered in half the time, but this assumption is wrong. (b) 836041 − 579406 = 256635 Here, 836041 is minuend, 579406 is subtrahend and the difference is 256635. But you're not done yet! Therefore, x has to be -3. Elementarymath.edc.org is dedicated to giving K–6 teachers, families, and researchers ideas and materials for: Thankfully, though, the New York Times explains step-by-step how to get to July 16, and you can read their detailed deduction here. Find resources to help prepare your child for elementary math tests, with study guides, practice tests and more. Instantly become team captain on trivia night. Take one of our many SSAT Elementary Level Math practice tests for a run-through of commonly asked questions. Other questions are meant to guide students to the heart of the mathematics. In order to do that, in case you forgot, you have to flip the fraction and switch from division to multiplication, thus getting 3 x 3 = 9. To answer these and many other common math questions, EDC has launched elementarymath.edc.org, which provides free access to a collection of EDC’s work in elementary mathematics. Here you'll find necessary information and be able to deepen your knowledge in all basic sections of elementary mathematics: arithmetic, algebra, geometry, trigonometry, functions and graphs, principles of analysis and so on. Once you've divided 192 by 3 to get 64, you can put the decimal place back where it belongs and get your final answer of 0.64. Every question has a worked solution. Basic Math Questions and Answers Test your understanding with practice problems and step-by-step solutions. Check out the following free elementary math tests below. If you live in this state, it's a possibility. first wrote about this deceiving question. All Rights Reserved. Fifth graders are also working on geometric understandings like area, perimeter, and three-dimensional shapes. Start your class with a question, … Conquer your inner couch potato once and for all. Then put your number crunching wisdom to the test with these tricky math questions taken straight from grade school tests and homework assignments and see for yourself. Believe it or not, this "math" question actually requires no math whatsoever. Solved mathematics questions with detailed explanations for easy understanding on various topics. Elementary school students are tested often in the various methods of mathematics, such as multiplication, division and algebra. Albert: Then I also know when Cheryl's birthday is. Both kinds of questions are necessary to develop mathematical proficiency. While logic might lead you to believe that your math skills have naturally gotten better as you've aged, the unfortunate reality is that, unless you've been solving algebra and geometry problems on a daily basis, the opposite is more likely the case. Let’s face it, math can get a little heavy sometimes so let’s lighten it up with a few quotes that have amused us over the years: An actual ACT Mathematics Test contains 60 questions to be answered in 60 minutes. Math symbols: Jokes: Forum: About us: Links: Contact us: Site map Program of Lessons Help Solution Answers Help Solution Answers Help Solution Answers. CBSE class 2 Maths question paper involves many elementary topics. Enter your email address to get the best tips and advice. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Jr. Mathletes is a math competition dedicated to encourage Elementary School students to learn more about math by solving problems and competing in fun competitions. 9). 7). Students in this grade often have a sophisticated number sense and are ready to do complex operations including multi-digit multiplication and long division. Another big part of fifth grade math is proportional reasoning, or gaining a better understanding of fractions, decimals, and percentages. This problem, mathematically speaking, is very similar to one of the other ones on this list. This test focuses on the essential content knowledge needed for teaching elementary mathematics. Don't let the fact that 8.563 has fewer numberrs than 4.8292 trip you up. To discover more amazing secrets about living your best life, click here to follow us on Instagram! Finding the answer to this final question will require using fractions. This problem might look easy, but a surprising number of adults are unable to solve it correctly. Not too bad, right? Since one measurement includes the cat's height and subtracts the turtle's and the other does the opposite, you can essentially just act like the two animals aren't there. This … This was one of the questions featured on the popular show Are You Smarter Than a 5th Grader? Don't believe us? All of the numbers in every row and column add up to 15! Your kid will identify quantities as fast as possible to serve hungry customers. It's unclear why Cheryl couldn't just tell both Albert and Bernard the month and day she was born, but that's irrelevant to solving this problem. Get answer explanations for each question and begin your preparation for the ACCUPLACER college placement test. You would study relentlessly for weeks for those silly standardized tests—and yet, come exam day, you'd still somehow have no idea what any of the equations or hard math problems were asking for. Of course, it's not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let's assume that it is. Primary Maths (Grades 4 and 5) - Free Questions and Problems With Answers Grade 4 and 5 maths questions and problems to test the understanding of maths concepts and procedures are presented. The note. When Best Life first wrote about this deceiving question, we had to ask a mathematician to explain the answer! Bestlifeonline.com is part of the Meredith Health Group. In order to solve this problem, you need to understand how the area of a parallelogram works. Unless you grew up to be an engineer, a banker, or an accountant, odds are that elementary and middle school math were the bane of your existence. Students. This problem comes straight from a standardized test given in New York in 2014. On the following pages you will find quizzes that cover several important elementary math topics. Unless you don't want your interviewee to work for you. When you add together the values given for S, B, and G, the sum comes out to 40, and making the missing letter J (which has a value of 14) makes the other diagonal's sum the same. Follow the question's "help" link, try your answer, then see our solution. This problem shouldn't be too difficult to solve if you play a lot of sudoku. Question: Find the area of the red triangle. If you're having trouble reading that, see here: "Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. A different question each day, from all subjects. Elementary Mathematics 5 Solution : (a) 650263 − 95742 = 554521 Here, 650263 is minuend, 95742 is subtrahend and the difference is 554521. If $$\Large \frac{2a+b}{a+4b}=3$$, then find the value of $$\Large \frac{a+b}{a+2b}$$. Questions are presented in multiple-choice format. This one comes from a first grader's homework. Try free elementary algebra questions for ACCUPLACER. (Also, 6 is the only number not represented out of numbers 1 through 9.). Start by solving the division part of the equation. 2nd Grade Math Tests . ACCUPLACER Math Study Guide – Free Algebra Questions. Then value of $$\Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) }$$ will be; 4). You'll never look at your canine companion the same again. © 2020 Galvanized Media. (PDF) document provides an overview of the characteristics of mathematics disability as well as information about accommodations, modifications, and assistive technology that can support a student with a disability in mathematics. From there, we can then multiply 17 by 2 and 3 and learn that there are 34 zombies and 51 humans respectively. Includes basic questions about math asked by elementary level students. Most Frequently Asked Questions for Class 2 Maths. And for more questions that will put your former education to the test, check out these 30 Questions You'd Need to Ace to Pass 6th Grade Geography. It was sneakily included in the legislation. If $$\Large a\ast b=a+b+ab$$, then $$\Large 3\ast 4-2\ast 3$$ is equal to: 10). Enclosed are samples of questions similar to those students will experience in the Elementary Math Assessment. If $$\Large x \circledast y = 3x+2y$$, then $$\Large 2 \circledast 3+3 \circledast 4$$ is equal to. This Sampler is not intended as a review – it is to familiarize students with format. Some people get stumped by the triangles hiding inside of the triangles and others forget to include the giant triangle housing all of the others. Don't worry, so was most of the world when this question, taken from a Singapore and Asian Schools Math Olympiad competition, went viral a few years ago. If $$\Large x = 7-4\sqrt{3}$$, then the value of $$\Large x+\frac{1}{x}$$. All you have to do is add a 0 to the end of 8.563 and then add like you normally would. … If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? Math) Includes problems and solutions for elementary level students. Using PEMDAS (an acronym laying out the order in which you solve it: "parenthesis, exponents, multiplication, division, addition, subtraction"), you would first solve the addition inside of the parentheses (1 + 2 = 3), and from there finish the equation as it's written from left to right. Following are the most frequently asked questions for class 2 maths … In examples no. In order to be a good instructor to your fifth graders, it is helpf… Adding two decimals together is easier than it looks. If $$\Large a\ast b = 2a-3b+ab$$, then $$\Large 3\ast 5+5\ast 3$$ is equal to: 2). 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# Absolute Values

In this lecture we shall discuss:

• Absolute values
• Inequalities involving absolute values
• Theorem 1.2.2 (√a2=|a|)
• Theorem inequality

### 1.2.1 Definition

The absolute value or magnitude of a real number a is denoted by |a| and is defined by

Example
|5|=5     Since 5>0
|-4/7|= -(-4/7) = 4/7   Since -4/7<0
|0|=0     Since 0≥0

Remark
|a| is a non-negative number for all values of a and
-|a|≤ a ≤ |a|
If a itself is negative, then -a is positive and +a is negative!!!

Example
Solve       |x-3|=4
Solution

 x-3= 4     x= 7 or -(x-3)= 4    x-3= -4        x= -1
The equation has 2 solutions: -1 and 7.

Example
Solve |3x-2|=|5x+4|

 3x-2   = 5x+43x-5x = 4+2    -2x = 6       x = -3 or 3x-2 = -(5x+4)    .    .       x = $-\frac{1}{4}$
The equation has 2 solutions: $-3$ and $-\frac{1}{4}$.

SQUARE ROOTS AND ABSOLUTE VALUES
b2 = a

(3)2 = 9
so b = 3
but!!!
(-3)2 = 9 so b = -3

The positive square root of the square of a number is equal to that number.

THEOREM 1.2.2
For any real number a
√a2 = |a|
e.g.
√(-4)2 = √16 = 4 = |-4|

1.2.3 THEOREM
If a and b are real numbers then,

1. |-a| = |a|    a number and its negative have the same absolute values.
2. |ab| = |a||b|    The absolute value of a product is the product of the absolute values.
3. |a/b| = |a|/|b|    The absolute value of a ratio is the ratio of the absolute values.

Proof
From theorem 1.2.2

(a)  |-a| = √(-a)2 = √a2 = |a|

(b)  |ab| = √(ab)2 = √a2b2 = √a2b2 = |a||b|

Examples

(a)  |-4| = |4|

(b)  |2.-3| = |-6| = 6 = |2|.|3| = 6

(c)  |5/4| = 5/4 = |5|/|4| = 5/4

The result (b) of above theorem can be extended to three or more factors.
For n-many real numbers
a1, a2, a3,...an

(a) |a1 a2 ...an| = |a1| |a2| ...|an|
(b) |an| = |a|n

Geometric interpretation of Absolute Value

Where A and B are points with coordinate a and b. The distance between A and B is

Theorem 1.2.4 (Distance formula)
If A and B are points on a coordinate line with coordinates a and b respectively, then the distance d between A and B
d = |b - a|

TABLE 1.2.2 (a)
|x-a| < k (k>0)

Alternative Form     -k < x-a < k
Solution Set           (a-k, a+k)

Example
The inequality
|x-3| < 4
rewritten as
-4 < x-3 < 4
-1 < x < 7
solution set (-1,7)

On real line

Example
Solve |x+4| ≥ 2
 x+4 ≤ -2x ≤ -6 x+4 ≥ 2x≥ -2
On combining these two sets
(-∞ , -6] ∪ [-2 , +∞ )

On real line

THE TRIANGLE INEQUALITY

It is not generally true that
|a+b| = |a| + |b|
e.g.
if a = 2 and b = -3, then a+b = -1 so that |a+b| = |-1| = 1
whereas
|a|+|b| = |2|+|-3| = 2+3 = 5 so |a+b| = |a|+|b|

1.2.5 THEOREM - (Triangle Inequality)
If   a  b  then |a+b| ≤ |a|+|b|
Proof
Since for any real number a and b, we know that
-|a| ≤ a ≤ |a|   and   -|b| ≤ b ≤ |b|
-|a| ≤ a ≤ |a|
+
-|b| ≤ b ≤ |b|
______________
= -|a| + -|b| ≤ a+b ≤ |a|+|b|
______________________________________________
Now we have two cases:

Case 1 where a+b ≥ 0
certainly a+b=|a+b|
Hence
|a+b| ≤ |a|+|b|

And

Case 2 where a+b < 0
|a+b| = -(a+b)
or
(a+b) = -|a+b|

Comparing with the intial inequality
-(|a|+|b|) ≤ -|a+b|
The result follows
_______________________________

Contact email:


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# American Institute of Mathematical Sciences

May  2013, 18(3): 601-641. doi: 10.3934/dcdsb.2013.18.601

## Mathematics of traveling waves in chemotaxis --Review paper--

 1 Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong

Received  November 2012 Revised  November 2012 Published  December 2012

This article surveys the mathematical aspects of traveling waves of a class of chemotaxis models with logarithmic sensitivity, which describe a variety of biological or medical phenomena including bacterial chemotactic motion, initiation of angiogenesis and reinforced random walks. The survey is focused on the existence, wave speed, asymptotic decay rates, stability and chemical diffusion limits of traveling wave solutions. The main approaches are reviewed and related analytical results are given with sketchy proofs. We also develop some new results with detailed proofs to fill the gap existing in the literature. The numerical simulations of steadily propagating waves will be presented along the study. Open problems are proposed for interested readers to pursue.
Citation: Zhi-An Wang. Mathematics of traveling waves in chemotaxis --Review paper--. Discrete & Continuous Dynamical Systems - B, 2013, 18 (3) : 601-641. doi: 10.3934/dcdsb.2013.18.601
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#### Volume 9, issue 4 (2005)

 Recent Issues
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Toward a general theory of linking invariants

### Vladimir V Chernov and Yuli B Rudyak

Geometry & Topology 9 (2005) 1881–1913
 arXiv: math.GT/0302295
##### Abstract

Let ${N}_{1},{N}_{2},M$ be smooth manifolds with $dim{N}_{1}+dim{N}_{2}+1=dimM$ and let ${\varphi }_{i}$, for $i=1,2$, be smooth mappings of ${N}_{i}$ to $M$ where $Im{\varphi }_{1}\cap Im{\varphi }_{2}=\varnothing$. The classical linking number $lk\left({\varphi }_{1},{\varphi }_{2}\right)$ is defined only when ${\varphi }_{1\ast }\left[{N}_{1}\right]={\varphi }_{2\ast }\left[{N}_{2}\right]=0\in {H}_{\ast }\left(M\right)$.

The affine linking invariant $alk$ is a generalization of $lk$ to the case where ${\varphi }_{1\ast }\left[{N}_{1}\right]$ or ${\varphi }_{2\ast }\left[{N}_{2}\right]$ are not zero-homologous. In [?] we constructed the first examples of affine linking invariants of nonzero-homologous spheres in the spherical tangent bundle of a manifold, and showed that $alk$ is intimately related to the causality relation of wave fronts on manifolds. In this paper we develop the general theory.

The invariant $alk$ appears to be a universal Vassiliev–Goussarov invariant of order $\le 1$. In the case where ${\varphi }_{1\ast }\left[{N}_{1}\right]={\varphi }_{2\ast }\left[{N}_{2}\right]=0\in {H}_{\ast }\left(M\right)$, it is a splitting of the classical linking number into a collection of independent invariants.

To construct $alk$ we introduce a new pairing $\mu$ on the bordism groups of spaces of mappings of ${N}_{1}$ and ${N}_{2}$ into $M$, not necessarily under the restriction $dim{N}_{1}+dim{N}_{2}+1=dimM$. For the zero-dimensional bordism groups, $\mu$ can be related to the Hatcher–Quinn invariant. In the case ${N}_{1}={N}_{2}={S}^{1}$, it is related to the Chas–Sullivan string homology super Lie bracket, and to the Goldman Lie bracket of free loops on surfaces.

##### Keywords
linking invariants, winding numbers, Goldman bracket, wave fronts, causality, bordisms, intersections, isotopy, embeddings
##### Mathematical Subject Classification 2000
Primary: 57R19
Secondary: 14M07, 53Z05, 55N22, 55N45, 57M27, 57R40, 57R45, 57R52


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Tropical Geography ›› 2020, Vol. 40 ›› Issue (5): 881-892.

### Analysis of Mangrove Annual Changes in Guangdong Province during 1986—2018 Based on Google Earth Engine

Ziyu Wang1(), Kai Liu1,2(), Liheng Peng1, Jingjing Cao1,2, Yingxue Sun1, Yuxin Qian1, Shuyue Shi1

1. 1.School of Geography and Planning, Sun Yat-sen University//Provincial Engineering Research Center for Public Security and Disaster// Guangdong Key Laboratory for Urbanization and GeoSimulation, Guangzhou 510275, China
2.Southern Marine Science and Engineering Guangdong Laboratory (Zhuhai), Zhuhai 519000, China
• Received:2019-12-25 Revised:2020-07-15 Online:2020-09-28 Published:2020-10-10
• Contact: Kai Liu E-mail:wangzy38@mail2.sysu.edu.cn;liuk6@mail.sysu.edu.cn

Abstract:

Mangroves have critical ecological functions and social and economic value, and are an important target for protection in the coastal wetland ecosystem. By monitoring long-term dynamic changes in mangrove ecosystems, the overall change process can be systematically and accurately recorded, providing data support and a basis for decision-making on scientific protection and effective management of the ecosystem. This study focuses on mangrove forests in coastal areas of Guangdong Province. A map of the mangrove forest, from 1986 to 2018, was made using Landsat remote sensing image based on Google Earth Engine (GEE),which is a cloud computing platform. The Random Forest (RF) method was used to extract mangrove trees from 32 periods from 1986 to 2018, in Guangdong Province. The interannual variation in mangrove characteristics in coastal cities of Guangdong province were compared. In addition, the evolution characteristics of mangrove patches in Guangdong province were analyzed. The results show that 1) The computing capacity and massive data of the GEE cloud platform provide data support for analyzing the inter-annual evolution of mangroves in Guangdong province, which greatly improves the computing efficiency. From 1986 to 2018, the overall classification accuracy of mangrove remote sensing was higher than 90%, with high classification accuracy and reliable results. In general, the coastal mangrove area of Guangdong province first decreased and then increased, and the range of change gradually declined after 2014, remaining at about 11 000 hm2. Mangrove forests are unevenly distributed in the province and occur mainly in the west. 2) Concerning coastal cities, mangroves are distributed in 14 cities, among which Zhanjiang and Yangjiang have the largest mangrove area, which is about 70% of the mangrove area of Guangdong province. The mangrove area changes in each city fall under three categories: decreasing first and increasing later, increasing fluctuation, and no obvious change. 3) From 1986 to 2018, the overall number of patches in mangrove forests in Guangdong province showed a decreasing trend, but the average patch area (MPS) showedan increasing trend, and mangrove fragmentation was reduced. In 2018, the mangrove MPS was 4.11 hm2 in Guangdong province, and the total number of patches was 2 782. From 1986 to 2018, when the change trend of MPS in mangrove forests in Guangdong province was opposite to that of patch quantity, the changes of patches were mainly expansion and fragmentation. When MPS change trend was consistent with the change trend of plaque number, the increase and decrease in the change of plaque were dominant. Information on annual mangrove area distribution and structural changes can provide more detailed data and reference for the rational development and protection of mangroves and support ecological restoration and finely tuned mangrove management.

CLC Number:

• TP79


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# Awesome Fractions

$\frac{17820}{495} = 36$

The above is called an “awesome fraction” because it equals a positive integer and uses each of the digits 0 through 9 exactly once on either side of the equal sign ($$=$$).

What is the sum of all integers that can be expressed as awesome fractions?

Note: All integers must be expressed without leading zeros. You may make use of the coding environment below:

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