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# American Institute of Mathematical Sciences

May  2009, 3(2): 199-210. doi: 10.3934/ipi.2009.3.199

## On a Hybrid method for shape reconstruction of a buried object in an elastostatic half plane

 1 Faculty of Applied Mathematics and Computer Science, Ivan Franko National University of Lviv, 79000 Lviv

Received  October 2008 Revised  March 2009 Published  May 2009

We apply the hybrid method for determining the shape of a bounded object in the elastostatic half plane from given Cauchy data on the boundary. The identifiability of the shape is investigated. For the integral representation of the function and the traction on the boundaries, the Green's function approach based on Kelvin's fundamental solution is used. The approximation of the integral operators with various singularities is made by trigonometrical and sinc quadratures. The presented numerical experiments exhibit the feasibility of the hybrid method for the system of differential equations and its stability in the case of noisy data.
Citation: Roman Chapko. On a Hybrid method for shape reconstruction of a buried object in an elastostatic half plane. Inverse Problems & Imaging, 2009, 3 (2) : 199-210. doi: 10.3934/ipi.2009.3.199
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# Fundamentals of Quantum Electrodynamics

In quantum electrodynamics, the classical Hamiltonian is obtained from the classical electromagnetic Lagrangian. Then the classical electric and magnetic fields are promoted to operators, as is the classical 4-vector potential $A_{\mu}$. The appropriate commutation relations are expected between the fields and their conjugate momenta.

Now, my question is, do the principles of quantum electrodynamics follow as a consequence of the fact that the charged particle producing the field is a quantum particle which must follow the principles of quantum mechanics?

Let me give a specific example. Consider a slow moving(for simplicity) free electron moving with a constant velocity initially.

Now, classically, the magnetic field at a point $P$ would be given by a function $\vec{B} = \vec{f}(\vec{r},\vec{x},\vec{p})$, where $\vec{r}$ is the position vector of the point at which the field is being 'measured' and $x$ and $p$ are the position and momenta of the charged particle evaluated at the retarded time.

Now, supposing I apply the principles of quantum mechanics to this electron and promote the above mentioned expression for the magnetic field at point $P$ to an operator by the usual quantum mechanical prescription. Would this prescription yield the correct values for the measured magnetic field at point $P$? Why? or Why not?

The bottom line of my entire question is whether the quantum field theory of an electron is a direct consequence of the fact that the particle producing the field is a quantum particle (and not a classical one) or does it involve much more than that?

EDIT: Thank you for your responses. I would also like to know if the above mentioned prescription for obtaining the magnetic field would yield accurate results for slow moving electrons(non-relativistic)?

The bottom line of my entire question is whether the quantum field theory of an electron is a direct consequence of the fact that the particle producing the field is a quantum particle (and not a classical one) or does it involve much more than that?

It involves "much more than that":

If I understand correctly, you're taking the classical expression for, say, the Coulomb field resulting from a source charge, or a magnetic field resulting from a current element and then saying that, since the position/momentum of the sources are quantized, they become operators and in this way the field becomes an operator since it's a function of those positions/momenta.

In QED, it's possible to describe a freely propagating field quantum (e.g. a photon). Freely propagating means that once it's been produced, its existence is now independent of any source. I don't see how this is possible in the scheme where you just quantize the source. Any time dependence of the source would always be immediately transferred to the electromagnetic field.

In QED, you quantize the electromagnetic field and the electron/positron fields independently. Neither is in any sense more fundamental than the other. One can act as a source for the other only after you've introduced an interaction term in the theory. So the source/field relationship isn't the basis of the quantization.

Also one of the key features of quantum field theory which distinguishes it from quantum mechanics is that it offers a mechanism to create and destroy particles. This would not be possible with a prescription such as the one you describe. Even your electron description is still a single particle one.

• Thank you for the informative post. In addition, I would like to know if the procedure outlined by me to calculate the magnetic field of the electron at the point P, would give accurate results for a slow moving(v<<c) electron? – guru Jun 24 '13 at 16:59

In Quantum mechanics , you have operators $X(t)$, where $t$ is a parameter, and $X$ is the operator.

What happens in Quantum Field theory ?

If we take a real scalar field, you have operators $\Phi(x,t)$, where $x$ and $t$ are parameters, and $\Phi$ is the operator.

In Quantum Electrodynamics, the photonic field is represented by operators $A_{\mu}(x,t)$, where $x$ and $t$ are parameters, and $A_{\mu}$ is the operator. The electron/positron field is represented by operators $\Psi(x,t)$, where $x$ and $t$ are parameters, and $\Psi$ is the operator.

So, in Quantum Field Theory, $x$ is not an operator, it is a (space) parameter. So, you have not the right to "mix" the 2 formalisms.

Especially, you have not the right to think about something like $\vec B=f(\vec X,t)$, where $\vec B$ and $\vec X$ would be operators representing magnetic field and position, because that will be a incoherent mixing of the 2 formalisms of Quantum Mechanics and Quantum Field Theory.

• Sure you can. – Michael Brown Jun 24 '13 at 8:51
• @MichaelBrown : Very Interesting paper, like this one, but the idea in these papers, if I correctly understand, is that a $1$st quantized formalism would be equivalent to a $2$nd quantized formalism. But there is no mixing of the 2 formalisms. – Trimok Jun 24 '13 at 16:22

Now, my question is, do the principles of quantum electrodynamics follow as a consequence of the fact that the charged particle producing the field is a quantum particle which must follow the principles of quantum mechanics?

First, a semantic issue. A principle is a starting point so, if an alleged principle follows from something else, can it in fact be called a principle?

Anyhow, what are the principles of QED or, more generally, QFT?

Certainly, one principle is the identification of fundamental "particles" as the quanta of quantized field modes. Each mode of a (free) field obeys a quantum harmonic oscillator equation and thus, each mode has associated ladder operators that create and destroy quanta, i.e., particles.

So, an electron (or positron) with definite momentum $k$ is identified as a quantum of the $k$th mode of Dirac "field".

And, a photon with definite momentum $k$ is identified as a quantum of the $k$th mode of the vector potential "field".

But, importantly, the Dirac field is not the source of the vector potential field in this picture. In fact, the vector potential is seen as a gauge field that is required by the principle of local gauge invariance.

So, it seems to be the case that the answer to your question must be no.

format: plain


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Complex Power Series Radius of Convergence Proof

1. jsi

24
1. The problem statement, all variables and given/known data

If f(z) = $\sum$ an(z-z0)n has radius of convergence R > 0 and if f(z) = 0 for all z, |z - z0| < r ≤ R, show that a0 = a1 = ... = 0.

2. Relevant equations

3. The attempt at a solution

I know it is a power series and because R is positive I know it converges. And if f(z) = 0 then the sum itself would be 0 which would mean that each term must add to 0. This doesn't necessarliy imply that each coefficient ai is 0 though because they could alternate a1 = 1, a2 = -1, a3 = 2, a4 = -2 etc... Am I right in thinking this? And if so, I'm not sure how to go from there... Unless f(z) = 0 for all z implies that |z - z0| ≠ 0 somehow then for f(z) to equal 0, the ai would have to equal 0? Any help would be appreciated, thanks!

2. Some Pig

38
Use Open Mapping Theorem.

3. Dick

25,853
It's pretty easy to see that all of the derivatives of f(z) are zero at z0, isn't it? Try to use derivatives to say something about the coefficients.

4. jsi

24
oh ok, so if say, since f(z) = 0, d/dz = 0, then taking the derivative of every term of the series yields d/dz (an(z - z0)n) = (nan(z - z0)n-1) = (nan(z - z0)n)/(z - z0)1 and then that means (z - z0) can't = 0 so the ai must be 0? Is that right, or sort of right, or totally wrong?

5. Dick

25,853
Take it one step at a time. The series is f(z)=a0+a1(z-z0)+a2(z-z0)^2+... Since f(z0)=0 that means a0=0. f'(z)=a1+2*a2(z-z0)+... What does f'(z0)=0 tell you? Now use the second derivative to say something about a2. Etc.

6. jsi

24
I think I got it; f'(0) = 0 implies that a1 = 0 and f''(0) = 0 implies that a2 = 0, so f to the nth derivative of 0 = 0 implies that an = 0, so all an must be 0?

7. Dick

25,853
Yes, that's it. Except you want to evaluate all of the derivatives at z=z0. Not z=0.

8. jsi

24
great, thank you!


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Math Help - Numerical ODE question

1. Numerical ODE question

hi, I'm wondering how to find the approximate solution of
dy/dx= e^x y y(0)=1

at x = 0.1 using the Taylor Series method. The expansion should include the
first four non-zero terms. Work to six decimal places accuracy.

Here is what I did but I am unsure which one is correct :
first attempt:
y'=e^x y y'(0)=1
y''=e^x y' y''(0)=1
y'''=e^x y'' y'''(0)=1
sub them into Taylor series method:
y(x0+h)=y(0+0.1)=y(0.1)= 1+h+h^2 /2 + h^3 /6

second attempt:
used product rule to differentiate:
y''=e^x y' + e^x y y''(0)=1+1=2
y'''=e^x y'' + e^x y' + e^x y' +e^x y y'''(0)=2+1+1+1=5
sub them into Taylor:
y(0.1)=1+1*0.1+ 2*(0.1)^2 /2 + 5*(0.1)^3 /6

thank you

2. $\frac{dy}{dx}= e^xy$

=> $\frac{dy}{y}=e^xdx$

Integrating, we have

=> $ln (y)=e^x+c$ ( c is a constant)

Now from the condition $y(0)=1$, $c$ comes out to be $-1$

So $y=e^{e^x-1}$

Why are you using Taylor's method?

3. because the question is from numerical methods class. cheers

4. Originally Posted by hazeleyes
hi, I'm wondering how to find the approximate solution of
dy/dx= e^x y y(0)=1

at x = 0.1 using the Taylor Series method. The expansion should include the
first four non-zero terms. Work to six decimal places accuracy.

Here is what I did but I am unsure which one is correct :
first attempt:
y'=e^x y y'(0)=1
y''=e^x y' y''(0)=1
y'''=e^x y'' y'''(0)=1
sub them into Taylor series method:
y(x0+h)=y(0+0.1)=y(0.1)= 1+h+h^2 /2 + h^3 /6

second attempt:
used product rule to differentiate:
y''=e^x y' + e^x y y''(0)=1+1=2
y'''=e^x y'' + e^x y' + e^x y' +e^x y y'''(0)=2+1+1+1=5
sub them into Taylor:
y(0.1)=1+1*0.1+ 2*(0.1)^2 /2 + 5*(0.1)^3 /6

thank you
Because of the fact that the initial data is given at 0 your first method is the best idea. so you know that

$y(0)=1$
$y'=e^{x}y(x) \implies y'(0)=e^{0}y(0)=1$
$y''='e^{x}y(x)+e^{x}y'(x) \implies y''(0)=2$

Note in general if you nave two functions (and in your cases $n \ge 2$)

General Leibniz rule - Wikipedia, the free encyclopedia

you get that
$\displaystyle \frac{d^n}{dx^n}=y^{(n)}(x)=\sum_{k=0}^{n-1}\binom{n-1}{k}\left( \frac{d^k}{dx^k}y \right) \left( \frac{d^{n-1-k}}{dx^{n-1-k}}e^{x}\right)=\sum_{k=0}^{n-1}\binom{n-1}{k}e^{x}y^{k}$

Putting this together you get that the value of the nth derivative is
(The n-1 comes from the fact we didn't need to use the product rule to take the first derivative it was given by the ODE)
$\displaystyle y^{n}(0)=\sum_{k=0}^{n-1}\binom{n-1}{k}y^{k}(0)$

You can use this to get as many terms as you need the series look like this

After cranking out a few terms I get that the series looks like

$y=1+x+x^2+\frac{5}{6}x^3+\frac{5}{8}x^4+\frac{13}{ 30}x^5$

Now use Taylors remainder theorem to show the error bound is where you want it.


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### Home > CALC > Chapter 4 > Lesson 4.1.3 > Problem4-39

4-39.

Given the function $f ( x ) = \left\{ \begin{array} { l l } { 2 x ^ { 2 } - 4 \text { for } x \leq 3 } \\ { - 2 x - 5 \text { for } x > 3 } \end{array} \right.$ find:

1. $\lim\limits_ { x \rightarrow 3 ^ { + } } f ( x )$

Notice that the boundary point is $x = 3$. Which piece is to the right of that boundary point?

$\lim\limits_ { x \rightarrow 3 ^ { - } } f ( x )$

Which piece of $f(x)$ is to the left of the boundary point?

1. What do your results above tell you about $f(x)$?

Do not claim continuity (or discontinuity) without accounting for all three conditions.


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# Using RSA keys for Broadcast Encryption?

What is the best way of using both broadcast encryption and RSA private keys? I'm particularly interested in Broadcast Encryption as implemented by BGW 1, but I guess that the question stands for any broadcast encryption system.

My confusion comes from the fact that the initial phase of Broadcast Encryption takes in the number of users and creates one global Public Key and a Private Key for each user. But suppose that the users in the system already had RSA private keys assigned. Is there a way to mitigate the fact that we need to maintain two private keys per each user (one from RSA and one from Broadcast Encryption) ?

1 Boneh, Dan, Craig Gentry, and Brent Waters. "Collusion resistant broadcast encryption with short ciphertexts and private keys." Annual International Cryptology Conference. Springer Berlin Heidelberg, 2005.

• Considering that ciphertext size in this scheme is $O(\sqrt{n})$ for $n$ listeners I'd say you can just as well encrypt for everyone separately and save yourself the hassle. Apr 27 '17 at 13:39
• I think that's a valid point for small n. I would guess that the benefit of a BE system is that it scales better than encrypting separately for large number of users (>1M). May 2 '17 at 9:03

The Broadcast encryption scheme you link to uses elliptic curve groups based on the bilinear map (bilinear pairing). RSA is not elliptic curve based.

Is there a way to mitigate the fact that we need to maintain two private keys per each user (one from RSA and one from Broadcast Encryption) ?

This is not trivially possible. You would have to devise a scheme that combines those two primitives.

The only scheme I know of that combines RSA with bilinear pairing is Comparison-based encryption for fine-grained access control in clouds by Zhu et al. (DOI: 10.1145/2133601.2133614)

You might want to check that to see if you can salvage some of that, but I doubt that you will be successful.

### Why integrating them is not a good idea in your case:

If your users already have their RSA private keys, then those RSA private keys are not linked in mathematically aside from following the same key generation process. In order to get the ciphertext size improvement of broadcast encryption (BE), you'd need to find some hidden property between all of the RSA private keys just by looking at the RSA public keys of your users. For that, you will either have to break RSA (not an easy feat) or you need to ask the users for their RSA private keys. If you ask for RSA private keys, you might as well just replace them with the BE private keys and skip the integration.

• Replacing the RSA keys with BE private keys sounds tempting. However, I need the functionality of users signing stuff with their private keys.... which was the justification of using the RSA keys. Could then Elliptic Curve keys be used for both Broadcast Encryption and signing by users ? May 2 '17 at 8:53
• Search for "broadcast signcryption" in your favorite search engine and you will probably find what you're looking for. It was not clear from your question that you wanted the private key holders to sign stuff. If this doesn't pan out, you can still ask a new question with a clear description of you what you want and what you literature study showed. May 3 '17 at 18:31


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# GLM Model checking Plots - Quasi Poisson - Poisson

I wonder whether accounting for overdispersion in a GLM (Quasi - Poisson instead of Poisson family) has an effect on the model checking plots (plot of residuals against fitted values, a scale–location plot of residuals against fitted values, a normal QQ plot and a plot of residuals against leverages). I tested this and apart from the scale in the plots nothing changes.

Is this hazard or is this always the case?

• It's always the case -- quasi Poisson is Poisson with an additional scale factor. – Glen_b May 1 '15 at 10:34
• is the scale factor the variance (which is assumed to be a linear function of the mean for quasi-poisson)? – kalakaru May 1 '15 at 10:40
• see the edits to my answer – Glen_b May 1 '15 at 10:55

It's always the case -- quasi Poisson is effectively Poisson with an additional scale factor.

The dispersion parameter corresponds to rescaling a Poisson variable by that parameter.

To be more specific, the scale factor I referred above to is the dispersion parameter.

In a quasi-Poisson GLM, $E(Y)=\mu\,$, $Var(Y)=\phi.\mu$.

You can get that by taking a Poisson($\lambda$) and multiplying by $\phi$, such that $\mu=\lambda.\phi$. e.g. $Z\sim \text{Pois}(\lambda)$ and $Y=\phi.Z$; then $Y$ has the required properties for a quasi-Poisson. (To my recollection it's the only member of the natural exponential family that has the right variance function.)

(People using them to model overdispersed counts don't worry too much about that since they look on it as a quasi-likelihood model.)

For example, Clark and Thayer [1] say

The Over-Dispersed Poisson distribution is a generalization of the Poisson, in which the range is a constant $\phi$ times the positive integers. That is, the variable $Y$ can take on values ${0,1\phi, 2\phi, 3\phi, 4\phi,...}$. It has a variance equal to $\phi$ times the mean.

[1]: Clark D.R. and Thayer, C.A. (2004),
A Primer on the Exponential Family of Distributions,
CAS 2004 Discussion Papers 117-148,
Casualty Actuarial Society. Arlington, Virginia

http://www.citeulike.org/user/yaroslavvb/article/331414


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# 10.1: The Nature of Monopoly

Learning Objectives

1. Define monopoly and the relationship between price setting and monopoly power.
2. List and explain the sources of monopoly power and how they can change over time.
3. Define what is meant by a natural monopoly.

Monopoly is at the opposite end of the spectrum of market models from perfect competition. A monopoly firm has no rivals. It is the only firm in its industry. There are no close substitutes for the good or service a monopoly produces. Not only does a monopoly firm have the market to itself, but it also need not worry about other firms entering. In the case of monopoly, entry by potential rivals is prohibitively difficult.

A monopoly does not take the market price as given; it determines its own price. It selects from its demand curve the price that corresponds to the quantity the firm has chosen to produce in order to earn the maximum profit possible. The entry of new firms, which eliminates profit in the long run in a competitive market, cannot occur in the monopoly model.

A firm that sets or picks price based on its output decision is called a price setter. A firm that acts as a price setter possesses monopoly power. We shall see in the next chapter that monopolies are not the only firms that have this power; however, the absence of rivals in monopoly gives it much more price-setting power.

As was the case when we discussed perfect competition in the previous chapter, the assumptions of the monopoly model are rather strong. In assuming there is one firm in a market, we assume there are no other firms producing goods or services that could be considered part of the same market as that of the monopoly firm. In assuming blocked entry, we assume, for reasons we will discuss below, that no other firm can enter that market. Such conditions are rare in the real world. As always with models, we make the assumptions that define monopoly in order to simplify our analysis, not to describe the real world. The result is a model that gives us important insights into the nature of the choices of firms and their impact on the economy.

## Sources of Monopoly Power

Why are some markets dominated by single firms? What are the sources of monopoly power? Economists have identified a number of conditions that, individually or in combination, can lead to domination of a market by a single firm and create barriers that prevent the entry of new firms.

Barriers to entry are characteristics of a particular market that block new firms from entering it. They include economies of scale, special advantages of location, high sunk costs, a dominant position in the ownership of some of the inputs required to produce the good, and government restrictions. These barriers may be interrelated, making entry that much more formidable. Although these barriers might allow one firm to gain and hold monopoly control over a market, there are often forces at work that can erode this control.

## Economies of Scale

Scale economies and diseconomies define the shape of a firm’s long-run average cost (LRAC) curve as it increases its output. If long-run average cost declines as the level of production increases, a firm is said to experience economies of scale.

A firm that confronts economies of scale over the entire range of outputs demanded in its industry is a natural monopoly. Utilities that distribute electricity, water, and natural gas to some markets are examples. In a natural monopoly, the LRAC of any one firm intersects the market demand curve where long-run average costs are falling or are at a minimum. If this is the case, one firm in the industry will expand to exploit the economies of scale available to it. Because this firm will have lower unit costs than its rivals, it can drive them out of the market and gain monopoly control over the industry.

Suppose there are 12 firms, each operating at the scale shown by ATC1 (average total cost) in Figure 10.1 “Economies of Scale Lead to Natural Monopoly”. A firm that expanded its scale of operation to achieve an average total cost curve such as ATC2 could produce 240 units of output at a lower cost than could the smaller firms producing 20 units each. By cutting its price below the minimum average total cost of the smaller plants, the larger firm could drive the smaller ones out of business. In this situation, the industry demand is not large enough to support more than one firm. If another firm attempted to enter the industry, the natural monopolist would always be able to undersell it.

Figure 10.1 Economies of Scale Lead to Natural Monopoly

A firm with falling LRAC throughout the range of outputs relevant to existing demand (D) will monopolize the industry. Here, one firm operating with a large plant (ATC2) produces 240 units of output at a lower cost than the $7 cost per unit of the 12 firms operating at a smaller scale (ATC1), and producing 20 units of output each. ## Location Sometimes monopoly power is the result of location. For example, sellers in markets isolated by distance from their nearest rivals have a degree of monopoly power. The local movie theater in a small town has a monopoly in showing first-run movies. Doctors, dentists, and mechanics in isolated towns may also be monopolists. ## Sunk Costs The greater the cost of establishing a new business in an industry, the more difficult it is to enter that industry. That cost will, in turn, be greater if the outlays required to start a business are unlikely to be recovered if the business should fail. Suppose, for example, that entry into a particular industry requires extensive advertising to make consumers aware of the new brand. Should the effort fail, there is no way to recover the expenditures for such advertising. An expenditure that has already been made and that cannot be recovered is called a sunk cost. If a substantial fraction of a firm’s initial outlays will be lost upon exit from the industry, exit will be costly. Difficulty of exit can make for difficulty of entry. The more firms have to lose from an unsuccessful effort to penetrate a particular market, the less likely they are to try. The potential for high sunk costs could thus contribute to the monopoly power of an established firm by making entry by other firms more difficult. ## Restricted Ownership of Raw Materials and Inputs In very few cases the source of monopoly power is the ownership of strategic inputs. If a particular firm owns all of an input required for the production of a particular good or service, then it could emerge as the only producer of that good or service. The Aluminum Company of America (ALCOA) gained monopoly power through its ownership of virtually all the bauxite mines in the world (bauxite is the source of aluminum). The International Nickel Company of Canada at one time owned virtually all the world’s nickel. De Beers acquired rights to nearly all the world’s diamond production, giving it enormous power in the market for diamonds. With new diamond supplies in Canada, Australia, and Russia being developed and sold independently of DeBeers, however, this power has declined, and today DeBeers controls a substantially smaller percentage of the world’s supply. ## Government Restrictions Another important basis for monopoly power consists of special privileges granted to some business firms by government agencies. State and local governments have commonly assigned exclusive franchises—rights to conduct business in a specific market—to taxi and bus companies, to cable television companies, and to providers of telephone services, electricity, natural gas, and water, although the trend in recent years has been to encourage competition for many of these services. Governments might also regulate entry into an industry or a profession through licensing and certification requirements. Governments also provide patent protection to inventors of new products or production methods in order to encourage innovation; these patents may afford their holders a degree of monopoly power during the 17-year life of the patent. Patents can take on extra importance when network effects are present. Network effects arise in situations where products become more useful the larger the number of users of the product. For example, one advantage of using the Windows computer operating system is that so many other people use it. That has advantages in terms of sharing files and other information. ### Key Takeaways • An industry with a single firm, in which entry is blocked, is called a monopoly. • A firm that sets or picks price depending on its output decision is called a price setter. A price setter possesses monopoly power. • The sources of monopoly power include economies of scale, locational advantages, high sunk costs associated with entry, restricted ownership of key inputs, and government restrictions, such as exclusive franchises, licensing and certification requirements, and patents. • A firm that confronts economies of scale over the entire range of output demanded in an industry is a natural monopoly. ### Try It! What is the source of monopoly power—if any—in each of the following situations? 1. The U.S. Food and Drug Administration granted Burroughs Wellcome exclusive rights until 2005 to manufacture and distribute AZT, a drug used in the treatment of AIDS. 2. John and Mary Doe run the only shoe repair shop in town. 3. One utility company distributes residential electricity in your town. 4. The widespread use of automatic teller machines (ATMs) has proven a boon to Diebold, the principal manufacturer of the machines. #### Case in Point: The Ambassador Bridge Fights to Maintain Its Monopoly Figure 10.2 Irf – ambassador bridge, detroit, mi. – CC BY-NC-ND 2.0. Matty Moroun was quietly enjoying his monopoly power. He is the owner of the 75-year-old Ambassador Bridge, a suspension bridge that is the only connection between Detroit, Michigan and Windsor, Ontario. He purchased the bridge from Warren Buffet in 1974 for$30 million. Forbes estimates that it is worth more than $500 million today. Mr. Moroun now oversees the artery over which$100 billion of goods—one quarter of U.S. trade with Canada and 40% of all truck shipments from the U.S.—make their way between the two countries.

Despite complaints of high and rising tolls—he has more than doubled fares for cars and tripled fares for trucks—Mr. Moroun has so far held on. Kenneth Davies, a lawyer who often battles Mr. Moroun in court, is a grudging admirer. “He’s very intelligent and very aggressive. His avarice and greed are just American capitalism at work,” he told Forbes.

What are the sources of his monopoly power? With the closest alternative bridge across the Detroit River two hours away, location is a big plus. In addition, the cost of creating a new transportation link is high. A group that is considering converting an old train tunnel to truck use and boring a new train tunnel some distance away is facing a $600 million price tag for the project. In addition to having entry by potential competitors blocked , he has a status not shared by most other monopolists. The Michigan Supreme Court ruled in 2008 that the city of Detroit cannot regulate his business because of the bridge’s international nature. Canadian courts have barred any effort by Canadian authorities to regulate him. He will not even allow inspectors from the government of the United States to set foot on his bridge. Increased security since 9/11 has caused delays, but Mr. Moroun has eased these by increasing his own spending on security to$50,000 a week and by building additional inspection stations and gifting them to the U.S. inspection agency, the General Services Administration. Even a monopolist understands the importance of keeping his customers content!

Because of the terrorist attacks on 9/11 and the concern about vulnerability and security, calls to deal with this monopoly have increased. Some people argue that the government should buy what is the most important single international arterial in North America, while others have called for more regulatory oversight. Canadian groups are exploring the development of alternative means of bringing traffic between the United States and Canada. Time will tell whether Mr. Moroun can hold onto what Forbes writers Stephane Fitch and Joann Muller dubbed “the best monopoly you never heard of.”

Sources: Stephane Fitch and Joann Muller, “The Troll Under the Bridge,” Forbes 174:10 (November 15 2004): 134–139; John Gallagher, “Plan Uses Parkway to Ease Ambassador Bridge Traffic,” Detroit Free Press, May 1, 2008; and “State Supreme Court Sides with Ambassador Bridge in Dispute,” Detroit News, May 7, 2008.

### Answers to Try It! Problems

1. The government’s grant of an exclusive franchise to the drug gave the firm monopoly power.
2. While John and Mary have the only shop in town, this is an easy entry business. Further, there may be competitors in the nearby town. John and Mary probably have monopoly power, but they do not have a monopoly.
3. Natural monopoly
4. Patent with strong network effects


text:
Keywords = integral group ring
##### Torsion units for some projected special linear groups

Volume 6, Issue 1, March 2017, Pages 37-53

Joe Gildea

##### Sylow like theorems for $V(\mathbb{Z}G)$

Volume 4, Issue 4, December 2015, Pages 49-59

Wolfgang Kimmerle

##### The prime graph conjecture for integral group rings of some alternatings groups

Volume 2, Issue 1, March 2013, Pages 175-185

Mohamed Salim


text:
You are currently browsing the category archive for the ‘linear programming’ category.

Sydney Afriat arrived in Purdue in the late 60’s with a Bentley in tow. Mort Kamien described him as having walked out of the pages of an Ian Flemming novel. Why he brought the Bentley was a puzzle, as there were no qualified mechanics as far as the eye could see. In Indiana, that is a long way. Afriat would take his Bentley on long drives only to be interrupted by mechanical difficulties that necessitated the Bentley being towed to wait for parts or specialized help.

I came upon Afriat when I learnt about the problem of rationalizability.  One has a model of choice and a collection of observations about what an agent selected. Can one rationalize the observed choices by the given model of choice? In Afriat’s seminal paper on the subject, the observations consisted of price-quantity pairs for a vector of goods and a budget. The goal was to determine if the observed choices were consistent with an agent maximizing a concave utility function subject to the budget constraint. Afriat’s paper has prompted many other papers asking the same question for different models of choice. There is an aspect of these papers, including Afriat’s, that I find puzzling.

To illustrate, consider rationalizing expected utility (Eran Shmaya suggested that expected consumption’ might be more accurate). Let ${S = \{1,2 \ldots, n\}}$ be the set of possible states. We are given a sequence of observations ${\{x^{i},p^{i}\}_{i=1}^{m}}$ and a single budget ${b}$. Here ${x^i_j}$ represents consumption in state ${j}$ and ${p^i_j}$ is the unit price of consumption in state ${j}$ in observation ${i}$. We want to know if there is a probability distribution over states, ${v=(v_{1},...,v_{n})}$, such that each ${(x^i, p^i)}$ maximizes expected utility. In other words, ${(x^i, p^i)}$ solves

$\displaystyle \max \sum_{j=1}^{n}v_{j}x^i_{j}$

subject to

$\displaystyle \sum_{j=1}^{n}p^i_{j}x_{j}\leq b$

$\displaystyle x^i_{j}\geq 0\,\,\forall j \in S$

The solution to the above program is obvious. Identify the variable with the largest objective coefficient to constraint ratio and make it as large as possible. It is immediate that a collection of observations ${\{x^{i},p^{i}\}_{i=1}^{m}}$ can be rationalized by a suitable set ${\{v_{j}\} _{j=1}^{n}}$ of non-zero and nonnegative ${v_{j}}$‘s if the following system has a feasible solution:

$\displaystyle \frac{v_{r}}{p^i_r}\geq \frac{v_{j}}{p^i_{j}} \,\,\forall j, \,\, x^i_r> 0$

$\displaystyle \sum_{j \in S}v_{j}=1$

$\displaystyle v_{j}\geq 0\,\,\forall j \in S$

This completes the task as formulated by Afriat. A system of inequalities has been identified, that if feasible means the given observations can be rationalized. How hard is this to do in other cases? As long as the model of choice involves optimization and the optimization problem is well behaved in that first order conditions, say, suffice to characterize optimality, its a homework exercise. One can do this all day, thanks to Afriat; concave, additively separable concave, etc. etc.

Interestingly, no rationalizability paper stops at the point of identifying the inequalities. Even Afriat’s paper goes a step farther and proceeds to characterize’ when the observations can be rationalized. But, feasibility of the inequalities themselves is just such a characterization. What more is needed?

Perhaps, the characterization involving inequalities lacks interpretation’. Or, if the given system for a set of observations was infeasible, we may be interested in the obstacle to feasibility. Afriat’s paper gave a characterization in terms of the strong axiom of revealed preference, i.e., an absence of cycles of certain kinds. But that is precisely the Farkas alternative to the system of inequalities identified in Afriat. The absence of cycles condition follows from the fact that the initial set of inequalities is associated with the problem of finding a shortest path (see the chapter on rationalizability in my mechanism design book). Let me illustrate with the example above. It is equivalent to finding a non-negative and non trivial solution to

$\displaystyle \frac{v_{r}}{v_j}\geq \frac{p^i_{r}}{p^i_{j}} \,\,\forall j, \,\, x^i_r> 0$

Take logs:

$\displaystyle \ln{v_r} - \ln{v_j} \geq \ln{\frac{p^i_{r}}{p^i_{j}}} \,\,\forall j, \,\, x^i_r> 0$

This is exactly the dual to the problem of finding a shortest path in a suitable network (I believe that Afriat has a paper, that I’ve not found, which focuses on systems of the form $b_{rs} >$ $x_s - x_r$ ).The cycle characterization would involve products of terms like ${\frac{p^i_{r}}{p^i_{j}}}$ being less than 1 (or greater than 1 depending on convention). So, what would this add?

In my salad days, school masters would assign boys returning from the summer hols an essay: What I did during the summer’. Yes, masters and boys. I served a portion of my youth in a misbegotten penal colony upon a wind blasted heath’. The only females present were master’s wives, matrons and the French mistress. No, not that kind, the kind that offers instruction in French. As you can see, to the lascivious minds of boys, there was no end to the double entendres. However, I digress.

Over the summer Thanh Nguyen and myself completed a paper about stable matchings. The abstract is reproduced below.

The National Resident Matching program strives for a stable matching of medical students to teaching hospitals. With the presence of couples, stable matchings need not exist. For any student preferences, we show that each instance of a stable matching problem has a nearby’ instance with a stable matching. The nearby instance is obtained by perturbing the capacities of the hospitals. Specifically, given a reported capacity $k_h$ for each hospital $h$, we find a redistribution of the slot capacities $k'_h$ satisfying $|k_h-k'_h|\le 4$ for all hospitals $h$ and $\sum_h k_h\le \sum k'_h \le \sum_h k_h+9$, such that a stable matching exists with respect to $k'$. Our approach is general and applies to other type of complementarities, as well as matchings with side constraints and contracts.

In other words, with the addition of at most 9 additional slots, one can guarantee the existence of a stable matchings. This is independent of the size of the market or doctors preferences (it does assume responsive preferences on the part of hospitals). The key tool  is Scarf’s lemma which is a wonderful device for converting results about cardinal matching problems into results about ordinal matching problems. For more on this, consult the paper by Kiralyi and Pap, who should be credited with a formulation of Scarf’s lemma that makes its usefulness evident.

The last of the trio, Harold Kuhn, passed away on July 2nd, 2014. Upon hearing the news, I was moved to dig up some old lecture notes of Kuhn’s in which KTK is stated an proved. I’ve been carrying them around with me since 1981. From the condition they are in, this must have been the last time I looked at them. With good reason, for as I re-read them, it dawned upon me how much of them I had absorbed and taken to be my own thoughts. Kuhn motivates the KTK theorem by replacing the non-linear functions by their first order Taylor approximations. This turns the exercise into a linear program. The LP duality theorem suggests the theorem to be proved, and the separating hyperplane theorem does the rest. For details see the relevant chapter of my book. The notes go on to describe Kuhn and Tucker’s excitement and subsequent despair as they uncover a counterexample and the need for a constraint qualification.

William Karush, who passed in 1997, had arrived at the same theorem many years earlier in his 1939 University of Chicago Masters Thesis (Kuhn-Tucker is 1951). When Kuhn learned of Karush’s contribution through a reading of Takayama’s book on Mathematical Economics. Upon doing so he wrote Karush:

In March I am talking at an AMS Symposium on “Nonlinear Programming – A Historical View.” Last summer I learned through reading Takayama’s Mathematical Economics of your 1939 Master’s Thesis and have obtained a copy. First, let me say that you have clear priority on the results known as the Kuhn–Tucker conditions (including the constraint qualification). I intend to set the record as straight as I can in my talk.

The missive closes with this paragraph:

Dick Cottle, who organized the session, has been told of my plans to rewrite history and says you must be a saint’ not to complain about the absence of recognition. Al Tucker remembers you from RAND, wonders why you never called this to his attention and sends his best regards.

Karush’s reply, 6 days later, equally gracious:

Thank you for your most gracious letter. I appreciate your thoughtfulness in wanting to draw attention to my early work. If you ask why I did not bring up the matter of priority before, perhaps the answer lies in what is now happening – I am not only going to get credit for my work, but I am going to crowned a “saint”.

This is the second posts about stability and equilibrium in trading networks. The first post may be found here. In it I assumed a single homogenous divisble good being traded by agents with quasi-linear preferences. Each agent was associated with a vertex of a graph and the edges represented pairs of agents who were permitted to trade with each other. Furthermore, each agent was designated the role of buyer and seller. Buyers could only trade with sellers and vice-versa. This meant the underlying graph was bipartite. In this post I will drop the assumption that they trade a homogenous good and that it is dvisible.

As noted in the earlier post, when the utility of buyers and the costs of sellers are linear in quantity, total unimodularity of the underlying constraint matrix is sufficient to ensure the existence of Walrasian prices that support an efficient allocation that is integral. If buyer utility functions become concave or sellers’s cost functions become convex, this is no longer true.

Suppose instead that the buyers utility functions are M${^{\#}}$-concave and seller’s cost functions are M${^{\#}}$-convex (object to be defined soon), then, the efficient allocation (the precise formulation will appear below) is indeed integer valued and supporting Walrasian prices exist. The corresponding allocations and prices can be interpreted as stable outcomes which cannot be blocked. This result was established by Murota and Tamura (2001).

I’ll define M${^{\#}}$-concavity. For any integer valued vector ${x}$ in ${\Re^n}$ let ${supp^+(x) = \{i: x_i > 0\}}$ and ${supp^-(x) = \{i : x_i < 0\}}$. A real valued function ${f}$ defined on the integer vectors of ${\Re^n}$ (with non-empty domain) is M${^{\#}}$ -concave if for all ${x, y}$ in its domain and any index ${u \in supp^+(x-y)}$ there exists an index ${v}$ in ${\{0\} \cup supp^{-}(x-y)}$ such that

$\displaystyle f(x) + f(y) \leq f(x - \chi_u + \chi_v) + f(y + \chi_u - \chi_v).$

Here ${\chi_i}$ is the 0-1 vector with a zero in every component except component ${i}$. I describe three different ways one might interpret this notion.

First, one can think of M${^{\#}}$-concavity as trying to capture an essential feature of the basis exchange property of matroids. If you don’t know what a matroid is, skip to the next paragraph. Think of the vectors ${x}$ and ${y}$ as being 0-1 and representing subsets of columns of some given matrix, ${A}$, say. Define ${f(x)}$ to be the rank of the columns in the subset associated with ${x}$. What the definition of M${^{\#}}$-concavity captures is this: for any column, ${u}$ I remove from ${x}$, that is not in ${y}$ and add it to ${y}$, I can find (possibly) a column from ${y}$ (not in ${x}$) to add to ${x}$ and not diminish the sum of the ranks of the two sets. The argument is straightforward.

A second interpretation is to be had by comparing the definition of M${^{\#}}$-concavity to the following way of stating concavity of a function:

$\displaystyle f(x) + f(y) \leq f(x - \frac{x-y}{2}) + f(y + \frac{x-y}{2})$

If ${x}$ and ${y}$ were numbers with ${x > y}$, then, the first term on the right represents a step from ${x}$ closer to ${y}$ and the second term is a step from ${y}$ closer to ${x}$. M${^{\#}}$-concavity can be interpreted as a discrete analogy of this. Think of ${x}$ being above and to the left of ${y}$. A step closer to ${y}$ would means one step down in the ${u}$ direction and a step to the right in the ${v}$ direction., i.e. ${x - \chi_u + \chi_v}$. The vector ${y + \chi_u - \chi_v}$ has a similar interpretation.

The third interpretation comes from an equivalent definition of M${^{\#}}$-concavity, one that shows that it is identical to the notion of gross substitutes proposed by Kelso and Crawford (1982). Let ${p \in \Re^n}$ be a non-negative price vector. Let ${x}$ be a non-negative integer that maximizes ${f(x) - p \cdot x}$, i.e., a utility maximizing bundle. Consider a new price vector ${p' \geq p}$. Then there exists a new utility maximizing bundle ${y}$ such that ${y_i \geq x_i}$ for all ${i}$ such that ${p_i = q_i}$. In words, the demand for goods whose price did not change cannot go down.

The key result of Murota and Tamura is that the following linear program has an integer optimal solution when the ${U_i}$‘s are M${^{\#}}$-concave and the ${C_j}$‘s are M${^{\#}}$-convex.

$\displaystyle \max \sum_{i \in B}U_i(x_{ij}:\{j: (i,j) \in E\}) - \sum_{j \in S}C_j(x_{ij}: \{i:(i,j) \in E\})$

subject to

$\displaystyle 0 \leq x_{ij} \leq d_{ij}\,\, \forall (i,j) \in E$

Notice, that buyer’s utilities do not necessarily depend on the total amount consumed. They can depend on the vector of inflows from the various sellers (similarly for the sellers). We can interpret this to mean that the sellers sell different goods which are substitutes for each other and there is an upper limit, ${d_{ij}}$ on the amount of seller ${j}$‘s good that can go to seller ${i}$. Such a bound is necessary to make sure the problem is well defined. Consider the case when utilities and costs are linear in quantity. The problem could be unbounded then. A second interpretation is that seller’s sell the same good but under different contractual terms (payment plan, delivery date etc.) which make them, in the buyer’s eyes, imperfect substitutes. The only element of the contract not fixed is the price and that is what is determined in equilibrium.

In 1937, representatives of the Plywood trust called upon Comrade Professor Leonid Vitalievich Kantorovich with a problem. The trust produced 5 varieties of plywood using 8 different machines. How, they asked, should they allocate their limited supply of raw materials to the various machines so as to produce the maximum output of plywood in the required proportions? As problems go, it was, from this remove unremarkable. Remarkable is that the Comrade Professor agreed to take it on. The so called representatives might have been NKVD. Why? Uncle Joe’s first act upon taking power in 1929 was to purge the economists, or more precisely the Jewish ones. This was well before the purge of the communist party in 1936. Why the economists? They complained about waste in a planned economy dizzy with success.’ Yet, here were the apparatchiks of the Trust asking the Comrade Professor to reduce waste.

Kantorovich writes, that at the time he was burnt out by pure mathematics. Combined with a concern at the rise of Hitler, he felt compelled to do something practical. And, so he turned his mind to the problem of the Plywood Trust. Frances Spufford, in his delightful work of faction’ called Red Plenty, imagines what Kantorovich might have been thinking.

He had thought about ways to distinguish between better answers and worse answers to questions which had no right answer. He had seen a method which could do what the detective work of conventional algebra could not, in situations like the one the Plywood Trust described, and would trick impossibility into disclosing useful knowledge. The method depended on measuring each machine’s output of one plywood in terms of all the other plywoods it could have made.

If he was right — and he was sure he was, in essentials — then anyone applying the new method to any production situation in the huge family of situations resembling the one at the Plywood Trust, should be able to count on a measureable percentage improvement in the quantity of product they got from a given amount of raw materials. Or you could put that the other way around: they would make a measureable percentage saving on the raw materials they needed to make a given amount of product.

He didn’t know yet what sort of percentage he was talking about, but just suppose it was 3%. It might not sound like much, only a marginal gain, an abstemious eking out of a little bit more from the production process, at a time when all the newspapers showed miners ripping into fat mountains of solid metal, and the output of plants booming 50%, 75%, 150%. But it was predictable. You could count on the extra 3% year after year. Above all it was free. It would come merely by organising a little differently the tasks people were doing already. It was 3% of extra order snatched out of the grasp of entropy. In the face of the patched and mended cosmos, always crumbling of its own accord, always trying to fall down, it built; it gained 3% more of what humanity wanted, free and clear, just as a reward for thought. Moreover, he thought, its applications did not stop with individual factories, with getting 3% more plywood, or 3% more gun barrels, or 3% more wardrobes. If you could maximise, minimise, optimise the collection of machines at the Plywood Trust, why couldn’t you optimise a collection of factories, treating each of them, one level further up, as an equation? You could tune a factory, then tune a group of factories, till they hummed, till they purred. And that meant –

An english description of Kantorovich’s appeared in the July 1960 issue of Management Science. The opening line of the paper is:

The immense tasks laid down in the plan for the third Five Year Plan period require that we achieve the highest possible production on the basis of the optimum utilization of the existing reserves of industry: materials, labor and equipment.

The paper contains a formulation of the Plywood Trust’s problem as a linear program. A recognition of the existence of an optimal solution at an extreme point as well as the hopelessness of enumerating extreme as a solution method. Kantorovich then goes on to propose his method, which he calls the method of resolving multipliers. Essentially, Kantorovich proposes that one solve the dual and then use complementary slackness to recover the primal. One might wonder how Kantorovich’s contribution differs from the contributions of Koopmans and Dantzig. That is another story and as fair a description of the issues as I know can be found in Roy Gardner’s 1990 piece in the Journal of Economic Literature. I reproduce one choice remark:

Thus, the situation of Kantorovich is rather like that of the discoverer Columbus. He really never touched the American mainland, and he didn’t give its name, but he was the first one in the area.

As an aside, David Gale is the one often forgotten in this discussion. If the Nobel committee has awarded the prize for Linear Programming, Dantzig and Gale would have been included. Had Gale lived long enough, he might have won it again for matching making him the third to have won the prize twice in the same subject. The others are John Bardeen and Frederick Sanger.

Continuing with Spufford’s imaginings:

– and that meant that you could surely apply the method to the entire Soviet economy, he thought. He could see that this would not be possible under capitalism, where all the factories had separate owners, locked in wasteful competition with one another. There, nobody was in a position to think systematically. The capitalists would not be willing to share information about their operations; what would be in it for them? That was why capitalism was blind, why it groped and blundered. It was like an organism without a brain. But here it was possible to plan for the whole system at once. The economy was a clean sheet of paper on which reason was writing. So why not optimise it? All he would have to do was to persuade the appropriate authorities to listen.

Implementation of Kantorovich’s solution at the Plywood trust led to success. Inspired, Kantorovich sent a letter to Gosplan urging adoption of his methods. Here the fact that Kantorovich solved the dual first rather than the primal is important. Kantorovich interpreted his resolving multipliers (shadow prices today) as objectively determined prices. Kantorovich’s letter to Gosplan urged a replacement of the price system in place by his resolving multipliers. Kantorovich intended to implement optimal production plans through appropriate pieces. Gosplan, responded that reform was unecessary. Kantorovich narrowly missed a trip to the Gulag and stopped practicing Economics, for a while. Readers wanting a fuller sense of what mathematical life was like in this period should consult this piece by G. G. Lorentz.

After the war, Kantorovich took up linear programming again. At Lenningrad, he headed a team to reduce scrap metal produced at the Egorov railroad-car plant. The resulting reduction in waste reduced the supply of scrap iron for steel mills disrupting their production! Kantorovich escaped punishment by the Leningrad regional party because of his work on atomic reactors.

Kantorovich’s interpretation of resolving multipliers which he renamed as objectively determined valuations put him at odds with the prevailing labor theory of value. In the post Stalin era, he was criticized for being under the sway of Bohm-Bawerk, author of the notion of subjective utility. Aron Katsenelinboigen, relates a joke played by one of these critics on Kantorovich. A production problem was presented to Kantorovich where the labor supply constraint would be slack at optimality. Its objectively determined valuation’ was therefore zero, contradicting the labor theory of value.

Nevertheless, Kantorovich survived. This last verse from the Ballard of L. V. Kantorvich authored by Josph Lakhman explains why:

Then came a big scholar with a solution.
Alas, too clever a solution.
`Objectively determined valuations’-
That’s the panacea for each and every doubt!
Truth be told, the scholar got his knukcles rapped
Slightly rapped


text:
Note

This documentation is for a development version. Click here for the latest stable release (v0.2.2).

# The Lorenz Chaotic Attractor¶

This example show how a classical chaotic dynamical system (the Lorenz “butterfly” attractor) can be implemented in a neural population. The dynamical equations for this attractor are:

\begin{split}\begin{align} \dot{x}_0 &= \sigma(x_1 - x_0) \\ \dot{x}_1 &= x_0(\rho - x_2) - x_1 \\ \dot{x}_2 &= x_0 x_1 - \beta x_2 \end{align}\end{split}

Since $$x_2$$ is approximately centered around $$\rho$$, and because NEF ensembles are typically optimized to represent values within a pre-defined radius of the origin, we substitute $$x'_2 = x_2 - \rho$$, yielding these equations:

\begin{split}\begin{align} \dot{x}_0 &= \sigma(x_1 - x_0) \\ \dot{x}_1 &= -x_0 x'_2 - x_1 \\ \dot{x}'_2 &= x_0 x_1 - \beta(x'_2 + \rho) - \rho \end{align}\end{split}

As with the previous oscillator examples, an exponential synapse is being used to implement these dynamics, meaning that the neural feedback matrix is:

$\mathbf{A}' = \tau\mathbf{A} + \mathbf{I}$

Because the Lorenz attractor has no input signal $$u$$, this in turn means that the neural feedback function can be computed as:

\begin{split}\begin{align} f_{fdbk}(x) &= \mathbf{A}' x \\ &= \tau\mathbf{A}x + x \\ &= \tau\dot{x} + x \\ &= \begin{bmatrix} \tau\dot{x}_0 + x_0 \\ \tau\dot{x}_1 + x_1 \\ \tau\dot{x}'_2 + x_2 \end{bmatrix} \end{align}\end{split}

## Step 1: Set up the Python Imports¶

[1]:

%matplotlib inline
import matplotlib.pyplot as plt
from IPython.display import HTML

import nengo

from mpl_toolkits.mplot3d import Axes3D

from anim_utils import make_anim_chaotic

import nengo_fpga
from nengo_fpga.networks import FpgaPesEnsembleNetwork

/home/travis-ci/tmp/nengo-fpga-993.3/miniconda/envs/travis-ci-993.3/lib/python3.6/site-packages/paramiko/transport.py:33: CryptographyDeprecationWarning: Python 3.6 is no longer supported by the Python core team. Therefore, support for it is deprecated in cryptography and will be removed in a future release.
from cryptography.hazmat.backends import default_backend
/home/travis-ci/tmp/nengo-fpga-993.3/miniconda/envs/travis-ci-993.3/lib/python3.6/site-packages/paramiko/transport.py:236: CryptographyDeprecationWarning: Blowfish has been deprecated
"class": algorithms.Blowfish,

## Step 2: Choose an FPGA Device¶

Define the FPGA device on which the remote FpgaPesEnsembleNetwork will run. This name corresponds with the name in your fpga_config file. Recall that in the fpga_config file, device names are identified by the square brackets (e.g., [de1] or [pynq]). The names defined in your configuration file might differ from the example below. Here, the device de1 is being used.

[2]:

board = "de1"  # Change this to your desired device name

## Step 3: Create the FPGA Ensemble¶

Create a remote FPGA neural ensemble (FpgaPesEnsembleNetwork) using the board defined above, 2000 neurons, 3 dimensions, and with no learning rate. We will also specify the recurrent connection here.

[3]:

tau = 0.1  # Synaptic time constant

# Lorenz attractor parameters
sigma = 10
beta = 8.0 / 3.0
rho = 28

with nengo.Network(label="Lorenz Attractor") as model:
# Remote FPGA neural ensemble
fpga_ens = FpgaPesEnsembleNetwork(
board,  # The board to use (from above)
n_neurons=2000,  # The number of neurons to use in the ensemble
dimensions=3,  # 3 dimensions
learning_rate=0,  # No learning for this example
feedback=1,  # Activate the recurrent connection
)

# The representational space of the Lorenz attractor is pretty big,

# Setting feedback=1 creates a feedback connection object
# with the identity transform. To implement the oscillator, it
# is necessary to set the transform on the feedback connection
# using .transform.
fpga_ens.feedback.synapse = tau  # nengo.LowPass(tau)

# Define the feedback function
def func_fdbk(x):
# These are the three variables represented by the ensemble
x0, x1, x2 = x

dx0 = sigma * (x1 - x0)
dx1 = -x0 * x2 - x1
dx2 = x0 * x1 - beta * (x2 + rho) - rho

return [tau * dx0 + x0, tau * dx1 + x1, tau * dx2 + x2]

# Assign the feedback function to the feedback connection
fpga_ens.feedback.function = func_fdbk

## Step 4: Add Probes to Collect Data¶

[4]:

with model:
# The output from the FPGA ensemble
# (filtered with a 10ms post-synaptic filter)
output_p = nengo.Probe(fpga_ens.output, synapse=0.01)

## Step 5: Run the Model¶

[5]:

with nengo_fpga.Simulator(model) as sim:
sim.run(20)

## Step 8: Plot the Results¶

[6]:

def plot_results():
plt.figure(figsize=(8, 12))
plt.subplot(211)
plt.plot(sim.trange(), sim.data[output_p])
plt.legend(["$x_0$", "$x_1$", "$x_2$"], loc="upper right")
plt.xlabel("Time (s)")

# Make 3D plot
ax = plt.subplot(212, projection=Axes3D.name)
ax.plot(*sim.data[output_p].T)
ax.set_title("3D Plot")

plt.show()

plot_results()

The plots above show the decoded output of the oscillator exhibiting the dynamics of a Lorenz “butterfly” attractor. The 3D plot illustrates why this attractor is called a “butterfly” attractor.

While a static plots are informative, an animated figure can be used to show how the Lorenz attractor evolves over time. It may take a moment to generate the animation.

[7]:

def make_anim():
_, _, anim = make_anim_chaotic(
sim.data[output_p][:, 0],
sim.data[output_p][:, 1],
sim.data[output_p][:, 2],
ms_per_frame=20,
)
return anim

HTML(make_anim().to_html5_video())

[7]:

## Step 9: Spiking Neurons¶

The plots above show the output of the Lorenz attractor network implemented with non-spiking rectified linear neurons. The network can also be simulated using spiking neurons to illustrate the similarities and differences between a spiking and a non-spiking network.

Below, we configure the FPGA neural ensemble to use spiking neurons, run the simulation, and plot the results.

[8]:

with model:
fpga_ens.ensemble.neuron_type = nengo.SpikingRectifiedLinear()

with nengo_fpga.Simulator(model) as sim:
sim.run(20)
plot_results()

[9]:

HTML(make_anim().to_html5_video())

[9]:

The plots above show that with spiking neurons, the output of the network is, expectedly, more noisy (less precise) than the results of the non-spiking network. However, despite this, the chaotic attractor network in its current configuration is stable, and exhibits the expected behaviour of the Lorenz attractor.

It should be noted that the output of the spiking Lorenz attractor network differs from that of the non-spiking network because of two factors. First, every time the Nengo simulator is created, the network parameters are regenerated (randomized). Second, because the dynamics of the Lorenz attractor are extremely sensitive to the initial conditions of the network (which is what makes it chaotic), even if the spiking and non-spiking networks were generated with the same parameters, the additional noise introduced by the spikes cause the spiking output to diverge from the trajectory of the non-spiking network.

## Step 10: Experiment!¶

Try playing around with the number of neurons in the FPGA ensemble as well as the synaptic time constant (tau) to see how it effects performance (e.g., observe how changing these numbers affect the stability of the oscillator)! Additionally, modify the parameters of the Lorenz attractor to see what effect it has on the shape of it. Be sure to run the simulation multiple times and observe if you can see any identical patterns (because the attractor is chaotic, every run should be different).

Explore the deterministic (yet chaotic) nature of the Lorenz attractor network, by constructing the FpgaPesEnsembleNetwork with the seed parameter, and re-running this notebook. You should observe because the Lorenz attractor is deterministic, with a pre-set seed, every run of the notebook will produce identical results. However, because the attractor is chaotic (very sensitive to the recurrent activity in the network), the additional noise introduced by the spikes cause it to follow a different trajectory from the non-spiking network.


text:
Page content

A little while back, I offered up a one-liner to scan your SQL Server instances and report which ones are out of date. But what if you need to take the next step, determining which updates need to be downloaded? That’s exactly what Josh asked on the SQL Community Slack recently.

Can Test-DbaBuild also bring back the KB number? From reading the docs it looks possible as it does return the compliant version form the .json file, or at least I think it does 😀. The intention is if the KB is returned, then one could pipe it through to Get-DbaKbUpdate and/or Save-DbaKbUpdate to have the KB’s automatically downloaded to the desired path.

I found the question interesting, so I decided to dig in a bit.

Let’s rewind to that previous post. But this time, we’ll add three more parameters to the output - KBLevel, BuildLevel and BuildTarget

The KBLevel property of Test-DbaBuild’s output tells you the KB corresponding to the current service pack or cumulative update you’re running. Not terribly useful if you need to sort out which updates need to be downloaded - it’s telling us where we are, not where we need to be. BuildTarget, on the other hand, is where we want to be.

 1 2 3 Test-DbaBuild -SqlInstance $Localinstances -Update -MaxBehind 0CU  | Select-Object SqlInstance, NameLevel, KBLevel, SPLevel, SPTarget, CULevel, CUTarget, BuildLevel, BuildTarget, Compliant  | Format-Table -AutoSize; SqlInstance NameLevel KBLevel SPLevel SPTarget CULevel CUTarget BuildLevel BuildTarget Compliant FLEXO\SQL17 2017 4527377 RTM RTM CU18 CU21 14.0.3257 14.0.3335 False FLEXO\SQL19 2019 4527376 RTM RTM CU1 CU5 15.0.4003 15.0.4043 False We can hop over to SQLServerUpdates.com to check those build and KB numbers against the CU numbers. ## What Next? We can use Get-DbaBuildReference to sort out what KB we need to look up and download. Get-DbaBuildReference scans through the build information that dbatools has and can report build numbers, KB number, names, etc. using build number, KB number, CU or SP name as input. Let’s see what it knows about build 14.0.3335, which is the target for the SQL Sever 2017 instance.  1 Get-DbaBuildReference -Build 14.0.3335 | Format-Table -AutoSize; Build BuildLevel CULevel KBLevel MatchType NameLevel SPLevel SupportedUntil Warnings 14.0.3335 14.0.3335 CU21 4557397 Exact 2017 RTM 2027-10-12 00:00:00 OK, now that’s something we can work with. We’ve got the KB number corresponding to CU21. ## Download Via Browser? Nope! Another of Chrissy LeMaire’s (blog | twitter) projects is KBUpdate. Compared to dbatools it’s a pretty compact module, but it’s incredibly useful - it’ll seek out information about KB updates and even download them for you! She’s also rolled these functions into dbatools for convenience, so we don’t need to install or import that module separately. Get-DbaKbUpdate provides a wealth of information about the KB number that’s passed into it.  1 Get-DbaKbUpdate 4557397 Title : SQL Server 2017 RTM Cumulative Update (CU) 21 KB4557397 Description : CU21 for SQL Server 2017 RTM upgraded all SQL Server 2017 RTM instances and components installed through the SQL Server setup. CU21 can upgrade all editions and servicing levels of SQL Server 2017 RTM to the CU21 level. For customers in need of additional installation options, please visit the Microsoft Download Center to download the latest Cumulative Update (https://support.microsoft.com/en-us/kb/957826). To learn more about SQL Server 2017 RTM CU21, please visit the Microsoft Support (http://support.microsoft.com) Knowledge Base article KB4557397. Architecture : NameLevel : 2017 SPLevel : RTM KBLevel : 4557397 CULevel : CU21 BuildLevel : 14.0.3335 SupportedUntil : 2027-10-12 00:00:00 Language : Classification : Updates SupportedProducts : Microsoft SQL Server 2017 MSRCNumber : n/a MSRCSeverity : n/a Hotfix : false Size : 533.3 MB UpdateId : 6bc92812-80ea-4c05-833c-433d60457e64 RebootBehavior : Can request restart RequestsUserInput : No ExclusiveInstall : NetworkRequired : No UninstallNotes : SQL Server 2017 RTM CU21 may be removed by selecting KB4557397 from "View Installed Updates" in the Programs and Features section of the Control Panel, under SQL Server 2017. UninstallSteps : n/a SupersededBy : n/a Supersedes : {SQL Server 2017 RTM Cumulative Update (CU) 1 KB4038634, SQL Server 2017 RTM Cumulative Update (CU) 2 KB4052574, SQL Server 2017 RTM Cumulative Update (CU) 3 KB4052987, SQL Server 2017 RTM Cumulative Update (CU) 4 KB4056498…} LastModified : 7/1/2020 Link : {http://download.windowsupdate.com/d/msdownload/update/others/2020/07/32055076\_0e9fcf41f92570c69b61 5407ef8cae1083e1600c.cab, http://download.windowsupdate.com/d/msdownload/update/software/updt/2020/ 07/sqlserver2017-kb4557397-x64\_e91bfa33a34accf82a0c374c9e8b7d0ce3b7ce05.exe} Great, but I need to download the installer. It’s covered by Save-DbaKbUpdate. Save-DbaKbUpdate -Name 4557397 -Path C:\Users\Andy\Downloads\ Wait a few seconds, and it’s downloaded. ## Putting it all together Recapping real quick, here’s what we can do with the snippets above: • Scan SQL Server instances to find out if they’re running the latest update • Find the KB number for the latest build of each version of SQL Server • Find and download the update installer based on the KB number Great! Let’s glue the pieces together.$DownloadPath = "C:\Users\Andy\Downloads\SQLUpdates";
$LocalInstances = @('flexo\sql17', 'flexo\sql19');$BuildTargets = Test-DbaBuild -SqlInstance $LocalInstances -MaxBehind 0CU -Update | Where-Object { !$PSItem.Compliant } | Select-Object -ExpandProperty BuildTarget -Unique;

Get-DbaBuildReference -Build $BuildTargets | ForEach-Object { Save-DbaKBUpdate -Path$DownloadPath -Name $PSItem.KBLevel }; ### First Line • Test-DbaBuild - I covered this in the aforementioned blog post • Where-Object {!$PSItem.Compliant} - Only pick out the instances which are not compliant with my MaxBehind check. I care about the compliant instances, I’m just not interested in them for this exercise.
• Select-Obect BuildTarget -Unique - I might have multiple instances that are returned, but I only want one copy of each BuildTarget (target build number).

### Second Line

• Get-DbaBuildReference -Build \$BuildTargets - Look through the build reference and get the corresponding KB numbers. While this function does take a -Update parameter to update the build reference file, we already did that on the previous line with Test-DbaBuild so it’s unnecessary here.


text:
# How to return a value from a function in LaTeX3

I have this MWE that tries to return a value from \l_example_foo_int from a function:

\documentclass[a4paper]{article}

\usepackage{expl3}
\usepackage{xparse}

\ExplSyntaxOn

\NewDocumentCommand{\returnvalue}{}{
\int_new:N \l_example_foo_int
% return value
\l_example_foo_int
}

\begin{document}

Should print value here:
\returnvalue

\int_new:N \l_example_bar_int

Should print value here:
\l_example_bar_int

\end{document}

It throws an error however.

Wondering how to get it so it can return a value from a function and that value can be typeset or \typeout or the like.

• Again you have a \...new instruction inside of NewDocumentCommand.. for your real issue: \int_use:N \foo_int could help.. but using int_add will make the whole thing unexpandable – user31729 Mar 26 '18 at 17:46
• If I move \int_new outside of the function I still get an error ! Missing number, treated as zero.. – Lance Pollard Mar 26 '18 at 17:48
• The \int_new is not the cause of your problem. . ... and I am typing on my smartphone therefore no answer – user31729 Mar 26 '18 at 17:50
• Seeing now the real problem (being at my computer at home: \returnvalue is unexpandable and can't be fed to \int_add:Nn again – user31729 Mar 26 '18 at 18:04
• You might also be interested in tex.stackexchange.com/q/56294/86 – Loop Space Mar 26 '18 at 21:06

\int_add:Nn is not expandable, i.e. it will not be fed correctly to \int_add:Nn again.

I suggest to use two separate commands, one that adds a value and another one, that is a wrapper for \l_example_foo_int.

However, the real use case is unclear, so the whole wrapping in \returnvalue is not really necessary, most likely.

Don't use a \...new:N call inside of a document level macro unless really checking. The next call of the original \returnvalue would try to define \int_new:N ... again, which is an error.

\documentclass[a4paper]{article}

\usepackage{expl3}
\usepackage{xparse}

\ExplSyntaxOn

\int_new:N \l_example_foo_int
\int_new:N \l_example_bar_int

}

\NewDocumentCommand{\returnvalue}{}{
\int_use:N \l_example_foo_int
}

\begin{document}

Should print value here:
\returnvalue

Should print value here:
\int_use:N \l_example_bar_int

\end{document}

• So this means all variables need to be declared globally then is what im understanding – Lance Pollard Mar 26 '18 at 18:13
• @LancePollard: In most cases, yes. If you want to have them local, use \int_set:N etc. in a group – user31729 Mar 26 '18 at 18:14
• I was hoping to have the function behave like in a normal programming language where you do (a better name instead of returnvalue would be initializeint lets say for this example) initializeint and the output is the result of the function. I would like for it to work like that if at all possible. – Lance Pollard Mar 26 '18 at 18:15
• @LancePollard: You fell into a 'normal' programmer trap however. That's what I had to learn some years ago as well: TeX etc. is different. As soon as you perform an assignment the whole thing is unexpandable... – user31729 Mar 26 '18 at 18:17
• @ChristianHupfer \int_new:N creates always a global variable. – Ulrike Fischer Mar 26 '18 at 18:53

\int_add:Nn is not expandable (it has no red star in the documentation). So it won't work like this. But you can use \int_eval:Nn:

\documentclass[a4paper]{article}

\usepackage{xparse}

\begin{document}

\ExplSyntaxOn
\int_new:N \l_example_foo_int
\int_new:N  \l_example_bar_int

\NewDocumentCommand\returnvalue {} {
\int_eval:n { \l_example_foo_int + 10}
}

\returnvalue\par

• Would be helpful to know (1) why you moved \int_new:N out of the function, (2) why the need for \int_use:N (I read in the documentation that it "recovers the content and places it in the input stream" but I don't know what that means or where it was before the input stream, and (3) if \int_eval:n is good-practice (since eval is avoided in some programming languages). – Lance Pollard Mar 26 '18 at 18:10
• \int_new:N isn't expandable either. If you want a command that can be used in a calculation it shouldn't contain this stuff. – Ulrike Fischer Mar 26 '18 at 18:47


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 10.6: The Discriminant

Difficulty Level: At Grade Created by: CK-12

You have seen parabolas that intersect the \begin{align*}x-\end{align*}axis twice, once, or not at all. There is a relationship between the number of real \begin{align*}x-\end{align*}intercepts and the Quadratic Formula.

Case 1: The parabola has two \begin{align*}x-\end{align*}intercepts. This situation has two possible solutions for \begin{align*}x\end{align*}, because the value inside the square root is positive. Using the Quadratic Formula, the solutions are \begin{align*}x=\frac{-b+\sqrt{b^2-4ac}}{2a}\end{align*} and \begin{align*}x=\frac{-b-\sqrt{b^2-4ac}}{2a}\end{align*}.

Case 2: The parabola has one \begin{align*}x-\end{align*}intercept. This situation occurs when the vertex of the parabola just touches the \begin{align*}x-\end{align*}axis. This is called a repeated root, or double root. The value inside the square root is zero. Using the Quadratic Formula, the solution is \begin{align*}x=\frac{-b}{2a}\end{align*}.

Case 3: The parabola has no \begin{align*}x-\end{align*}intercept. This situation occurs when the parabola does not cross the \begin{align*}x-\end{align*}axis. The value inside the square root is negative, therefore there are no real roots. The solutions to this type of situation are imaginary, which you will learn more about in a later textbook.

The value inside the square root of the Quadratic Formula is called the discriminant. It is symbolized by \begin{align*}D\end{align*}. It dictates the number of real solutions the quadratic equation has. This can be summarized with the Discriminant Theorem.

• If \begin{align*}D>0\end{align*}, the parabola will have two \begin{align*}x-\end{align*}intercepts. The quadratic equation will have two real solutions.
• If \begin{align*}D=0\end{align*}, the parabola will have one \begin{align*}x-\end{align*}intercept. The quadratic equation will have one real solution.
• If \begin{align*}D<0\end{align*}, the parabola will have no \begin{align*}x-\end{align*}intercepts. The quadratic equation will have zero real solutions.

Example 1: Determine the number of real solutions to \begin{align*}-3x^2+4x+1=0\end{align*}.

Solution: By finding the value of its discriminant, you can determine the number of \begin{align*}x-\end{align*}intercepts the parabola has and thus the number of real solutions.

\begin{align*}D &= b^2-4(a)(c)\\ D &= (4)^2-4(-3)(1)\\ D &= 16+12=28\end{align*}

Because the discriminant is positive, the parabola has two real \begin{align*}x-\end{align*}intercepts and thus two real solutions.

Example: Determine the number of solutions to \begin{align*}-2x^2+x=4\end{align*}.

Solution: Before we can find its discriminant, we must write the equation in standard form \begin{align*}ax^2+bx+c=0\end{align*}.

Subtract 4 from each side of the equation: \begin{align*}-2x^2+x-4=0\end{align*}.

\begin{align*}\text{Find the discriminant.} && D &= (1)^2-4(-2)(-4)\\ && D &= 1-32=-31\end{align*}

The value of the discriminant is negative; there are no real solutions to this quadratic equation. The parabola does not cross the \begin{align*}x-\end{align*}axis.

Example 2: Emma and Bradon own a factory that produces bike helmets. Their accountant says that their profit per year is given by the function \begin{align*}P=0.003x^2+12x+27,760\end{align*}, where \begin{align*}x\end{align*} represents the number of helmets produced. Their goal is to make a profit of 40,000 this year. Is this possible? Solution: The equation we are using is \begin{align*}40,000=0.003x^2+12x+27,760\end{align*}. By finding the value of its discriminant, you can determine if the profit is possible. Begin by writing this equation in standard form: \begin{align*}0 &= 0.003x^2+12x-12,240\\ D &= b^2-4(a)(c)\\ D &= (12)^2-4(0.003)(-12,240)\\ D &= 144+146.88=290.88\end{align*} Because the discriminant is positive, the parabola has two real solutions. Yes, the profit of40,000 is possible.

Multimedia Link: This http://sciencestage.com/v/20592/a-level-maths-:-roots-of-a-quadratic-equation-:-discriminant-:-examsolutions.html - video, presented by Science Stage, helps further explain the discriminant using the Quadratic Formula.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. What is a discriminant? What does it do?
2. What is the formula for the discriminant?
3. Can you find the discriminant of a linear equation? Explain your reasoning.
4. Suppose \begin{align*}D=0\end{align*}. Draw a sketch of this graph and determine the number of real solutions.
5. \begin{align*}D=-2.85\end{align*}. Draw a possible sketch of this parabola. What is the number of real solutions to this quadratic equation.
6. \begin{align*}D>0\end{align*}. Draw a sketch of this parabola and determine the number of real solutions.

Find the discriminant of each quadratic equation.

1. \begin{align*}2x^2-4x+5=0\end{align*}
2. \begin{align*}x^2-5x=8\end{align*}
3. \begin{align*}4x^2-12x+9=0\end{align*}
4. \begin{align*}x^2+3x+2=0\end{align*}
5. \begin{align*}x^2-16x=32\end{align*}
6. \begin{align*}-5x^2+5x-6=0\end{align*}

Determine the nature of the solutions of each quadratic equation.

1. \begin{align*}-x^2+3x-6=0\end{align*}
2. \begin{align*}5x^2=6x\end{align*}
3. \begin{align*}41x^2-31x-52=0\end{align*}
4. \begin{align*}x^2-8x+16=0\end{align*}
5. \begin{align*}-x^2+3x-10=0\end{align*}
6. \begin{align*}x^2-64=0\end{align*}

A solution to a quadratic equation will be irrational if the discriminant is not a perfect square. If the discriminant is a perfect square, then the solutions will be rational numbers. Using the discriminant, determine whether the solutions will be rational or irrational.

1. \begin{align*}x^2=-4x+20\end{align*}
2. \begin{align*}x^2+2x-3=0\end{align*}
3. \begin{align*}3x^2-11x=10\end{align*}
4. \begin{align*}\frac{1}{2}x^2+2x+\frac{2}{3}=0\end{align*}
5. \begin{align*}x^2-10x+25=0\end{align*}
6. \begin{align*}x^2=5x\end{align*}
7. Marty is outside his apartment building. He needs to give Yolanda her cell phone but he does not have time to run upstairs to the third floor to give it to her. He throws it straight up with a vertical velocity of 55 feet/second. Will the phone reach her if she is 36 feet up? (Hint: The equation for the height is given by \begin{align*}y=-32t^2+55t+4\end{align*}.)
8. Bryson owns a business that manufactures and sells tires. The revenue from selling the tires in the month of July is given by the function \begin{align*}R=x(200-0.4x)\end{align*} where \begin{align*}x\end{align*} is the number of tires sold. Can Bryson’s business generate revenue of \$20,000 in the month of July?
9. Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation \begin{align*}y=-\frac{32}{6400}x^2+x\end{align*}, where \begin{align*}y\end{align*} is the height and \begin{align*}x\end{align*} is the horizontal distance the ball travels. We want to know if Marcus kicked the ball hard enough to go over the goal post, which is 10 feet high.

Mixed Review

1. Factor \begin{align*}6x^2-x-12\end{align*}.
2. Find the vertex of \begin{align*}y=-\frac{1}{4} x^2-3x-12=y\end{align*} by completing the square.
3. Solve using the Quadratic Formula: \begin{align*}-4x^2-15=-4x\end{align*}.
4. How many centimeters are in four fathoms? (Hint: 1 fathom = 6 feet)
5. Graph the solution to \begin{align*}\begin{cases} 3x+2y \le -4\\ x-y>-3 \end{cases}\end{align*}.
6. How many ways can 3 toppings be chosen from 7 options?

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Homework Help: Integral variable substitution for removing singularity

1. Jan 10, 2012

Hokey

The problem statement, all variables and given/known data
Hi! In an assignment I have reached an integral that has the form:
$\int\frac{A+Bx+Cx^2}{Dx+Ex^2}$
where A-E are constants, the integration variable is x and the limits are 0 to 1. I'm supposed to remove the singularity at x=0 by substitution.

A-E have values but they're long and complicated and I hope they're not necessary to solve the problem. And sadly, no - I can't go backwards to complete squares or anything...

The attempt at a solution
This might be an easy question, but I really don't know what to substitute x for. I've tried squares, roots, inverted squares and roots, ln and exponential functions... but they all end up with the same singularity at 0. I don't know how to get around this. Is there any good way to figure out how to substitute in order to "remove" a singularity, in general?

Help and hints would be very much appreciated! Thank you!

/Jennifer

2. Jan 10, 2012

SammyS

Staff Emeritus
Try the following. Not sure how well it works for the whole problem.

Let u = 1/x . Of course that's equivalent to x = eu .

3. Jan 10, 2012

MednataMiza

Suggestion: Divide using long, polynomial division and rewrite the integral :)

4. Jan 10, 2012

SammyS

Staff Emeritus
That still leaves a singularity at x=0.

5. Jan 10, 2012

Hokey

Ahh. Thanks for your suggestions, they made me re-track my steps. And yep, I left out the part that the whole fraction is under a square root as well :tongue: That changes things a bit!

Substituting x with u2 seems to do the trick now! Phew


text:
My Math Forum Cube roots of -1

 Algebra Pre-Algebra and Basic Algebra Math Forum

December 30th, 2013, 05:16 AM   #11
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 234

Re: Cube roots of -1

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by agentredlum The result I'm looking for is ...
I'd represent it as

$\sqrt{3}= -ie^{i \pi/3} + ie^{-i\pi/3}$

And transform everything to $^{i}$ by the simple yet interesting identity

$e^\pi= i^{-2i}$

We can generalize this to any elementary number (i.e., an algebraic number with minimal polynomial having solvable Galois group), see Gauss sums and Kronecker-Weber theorem and the later for anybody who wants to dig deep in these things.
That's very nice, but I asked for a single term.

December 30th, 2013, 05:28 AM   #12
Math Team

Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
Thanks: 86

Math Focus: Number Theory
Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum That's very nice but i asked for a single term.
First thing I would have asked you, or any transcendental number theorist would, is what do you mean by a "single term".

But, for the sake of keeping everything fair, I'd rather not but simply say that yes, mine indeed is not representable in a single term (or apparently so). But if simplified, you get

$\sqrt{3}= i^{1/3} - i^{5/3}$

which is less "uglier" than that of the one you presented.

 December 30th, 2013, 05:43 AM #13 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: Cube roots of -1 Terms are separated by addition and/or subtraction, so, for instance, $x^2 \ + \ x \ - \ 5$ has 3 terms as written, and $3xy^2zw^3$ is a single term as written. So how would you represent 3 in terms of your beautiful method using $\ i \$?
December 30th, 2013, 05:44 AM   #14
Math Team

Joined: Apr 2010

Posts: 2,780
Thanks: 361

Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Bravo! I actually did it a little differently , $\Large i ^{i} \= \ \large e^{ \tiny{-} \large \frac{ \pi}{2}}$ Now, on RHS we want to change e to 3, get rid of $\ \pi \$, get rid of the negative and keep the 2 in the denominator for use as square root $$$\Large i^{ \Large i}$$^{ \Large { \large \frac{ \tiny{-} \ln 3}{ \pi}} \= \ $$\Large {e} ^{ \tiny{-} \frac{ \pi}{2}}$$^{ \tiny{-} \frac{ \ln 3}{ \pi} }$ And the result follows on the LHS.
Nice method!

Could this be what you're looking for:

$\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$

December 30th, 2013, 05:52 AM   #15
Math Team

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Math Focus: Number Theory
Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Terms are separated by addition and/or subtraction, so, for instance, $x^2 \ + \ x \ - \ 5$ has 3 terms as written, and $3xy^2zw^3$ is a single term as written. So how would you represent 3 in terms of your beautiful method using $\ i \$?
That is an ambiguous definition, you do understand that, no? Anyways, let's not discuss it here. I'll create a topic on NT section soon on some conjectures regarding this problem I have in my mind, so I hope to see you there.

As for representing 3, perhaps square it?

PS : My representation is nothing of much "beauty" although it is of interest, yes. Have you searched the keywords I have italicized?

December 30th, 2013, 05:54 AM   #16
Math Team

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Re: Cube roots of -1

Quote:
 Originally Posted by Hoempa $\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$
Hmm, interesting, but unfortunately I don't like this much Agent's and your previous representation makes thing so much interesting (topic coming up on NT)

December 30th, 2013, 05:56 AM   #17
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Re: Cube roots of -1

Quote:
Originally Posted by Hoempa
Quote:
 Originally Posted by agentredlum Bravo! I actually did it a little differently , $\Large i ^{i} \= \ \large e^{ \tiny{-} \large \frac{ \pi}{2}}$ Now , on RHS we want to change e to 3 , get rid of $\ \pi \$ , get rid of the negative and keep the 2 in the denominator for use as square root $$$\Large i^{ \Large i}$$^{ \Large { \large \frac{ \tiny{-} ln 3}{ \pi}} \= \ $$\Large {e} ^{ \tiny{-} \frac{ \pi}{2}}$$^{ \tiny{-} \frac{ ln 3}{ \pi} }$ And the result follows on the LHS.
Nice method!

Could this be what you're looking for:

$\sqrt{3}= \sqrt{\lfloor\cos(i \cdot i)^{\large i \cdot i} \cdot \cos(i \cdot i)^{\large i \cdot i} \rfloor}$
Very nice. It's in radians , right?

December 30th, 2013, 06:12 AM   #18
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Re: Cube roots of -1

Quote:
 Originally Posted by mathbalarka As for representing 3, perhaps square it?
Then you need to perform multiple additions and several multiplications using your result for $\ \sqrt{3}$

All i have to do is put a factor of 2 in front of the $\ iln3 \$ using my single term result for $\ \sqrt{3}$

Quote:
 Originally Posted by mathbalarka PS : My representation is nothing of much "beauty" although it is of interest, yes. Have you searched the keywords I have italicized?
Yes i looked at a couple of wiki articles , interesting.

December 30th, 2013, 06:16 AM   #19
Math Team

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From: India, West Bengal

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Math Focus: Number Theory
Re: Cube roots of -1

Quote:
 Originally Posted by agentredlum Then you need to perform multiple additions and several multiplications using your result for $\ \sqrt{3}$ All i have to do is put a factor of 2 in front of the $\ iln3 \$ using my single term result for $\ \sqrt{3}$
I understand you. I am posting the NT topic now, it will take 5-10 minutes to type it in. What we need is a better definition of "single term" or "multiple term" to work our way in and avoid ambiguity.

 December 30th, 2013, 06:47 AM #20 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Cube roots of -1 I have posted it here, feel free to add anything. I have to go now, bye!

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text:
clRNG An OpenCL library for random number generators (RNG)
clRNG.h File Reference

Library definitions common to all RNG's. More...

#include <CL/cl.h>

Go to the source code of this file.

## Typedefs

typedef enum clrngStatus_ clrngStatus
Error codes. More...

## Enumerations

enum  clrngStatus_
Error codes. More...

## Functions

const char * clrngGetErrorString ()
Retrieve the last error message. More...

const char * clrngGetLibraryDeviceIncludes (cl_int *err)
Generate an include option string for use with the OpenCL C compiler. More...

const char * clrngGetLibraryRoot ()
Retrieve the library installation path. More...

## Detailed Description

Library definitions common to all RNG's.

See also
clRNG_template.h

## Typedef Documentation

 typedef enum clrngStatus_ clrngStatus

Error codes.

Most library functions return an error status indicating the success or error state of the operation carried by the function. In case of success, the error status is set to CLRNG_SUCCESS. Otherwise, an error message can be retrieved by invoking clrngGetErrorString().

Note
In naming this type clrngStatus, we follow the convention from clFFT and clBLAS, where the homologous types are name clfftStatus and clblasStatus, respectively.

## Enumeration Type Documentation

 enum clrngStatus_

Error codes.

Most library functions return an error status indicating the success or error state of the operation carried by the function. In case of success, the error status is set to CLRNG_SUCCESS. Otherwise, an error message can be retrieved by invoking clrngGetErrorString().

Note
In naming this type clrngStatus, we follow the convention from clFFT and clBLAS, where the homologous types are name clfftStatus and clblasStatus, respectively.

## Function Documentation

 const char* clrngGetErrorString ( )

Retrieve the last error message.

The buffer containing the error message is internally allocated and must not be freed by the client.

Returns
Error message or NULL.
 const char* clrngGetLibraryDeviceIncludes ( cl_int * err )

Generate an include option string for use with the OpenCL C compiler.

Generate and return "-I${CLRNG_ROOT}/cl/include", where ${CLRNG_ROOT} is the value of the CLRNG_ROOT environment variable. This string is meant to be passed as an option to the OpenCL C compiler for programs that make use of the clRNG device-side headers. If the CLRNG_ROOT environment variable is not defined, the current directory of execution of the program is substituted for the library root path.

A static buffer is return and need not be released; it could change upon successive calls to the function.

An error is returned in err if the preallocated buffer is too small to contain the include string.

Parameters
 [out] err Error status variable, or NULL.
Returns
An OpenCL C compiler option to indicate where to find the device-side clRNG headers.
Examples:
MultiStream/multistream.c.
 const char* clrngGetLibraryRoot ( )

Retrieve the library installation path.

Returns
Value of the CLRNG_ROOT environment variable, if defined, else the current directory (.) of execution of the program.


text:
## Thomas' Calculus 13th Edition

a) For all (x,y,z) such that $x^2+y^2+z^2 \le 4$; b) For all $(x,y,z )$ such that $x^2+y^2+z^2 \ge 9$ except when $x^2+y^2+z^2=25$
a) There is no possible way to take the square root of the negative value.Therefore, $4-x^2-y^2-z^2 \gt 0$ Thus, for all (x,y,z) such that $x^2+y^2+z^2 \le 4$ b) There is no possible way to take the square root of the negative value. and there can be no zero in the denominator. hence,$x^2+y^2+z^2 \ge 9$ and also, $4- \sqrt {x^2+y^2+z^2-9} \gt 0$ Thus, For all $(x,y,z )$ such that $x^2+y^2+z^2 \ge 9$ except when $x^2+y^2+z^2=25$