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# Sedimental symbols

«In our description of nature, the purpose is not to disclose the real essence of phenomena but only to track down as far as possible relations between the multifold aspects of our experience.»
Niels Bohr, 1939

Physicists refer to properties of Nature, alias observables, by means of symbols which differ from purely mathematical notations since they must be experimentally validated, i.e. a definite procedure exists that assigns numbers to them. Nevertheless, phenomena may sometimes appear to be quite counterintuitive or rest on pure abstraction. Let it suffice to quote the well celebrated case of the elusive wave-particle dualism.
In Hamlet and his problems, Eliot wrote: «The only way of expressing emotion in the form of art is by finding an “objective correlative”; in other words, a set of objects, a situation, a chain of events which shall be the formula of that particular emotion; such that when the external facts, which must terminate in sensory experience, are given, the emotion is immediately evoked.»[1] Laying the scene thus addresses the expectation of immediacy and concreteness of emotion.
It may be stimulating to rephrase this “chain of events” in terms of the quantum principle of superposition: «the way of expressing any state of any microscopic system in the form of quantum physics is by finding a “superposition state”; in other words, a set of complex numbers and a maximal set of commuting observables which, combined together, shall conveniently define the formula of any state of that particular system; such that when the external devices, which must terminate in an experimental result, are given, the microscopic system is immediately set into a particular state». While not actually solving any potential excess of abstraction, it may help bridge the gap between symbols and “reality”.
Several maximal sets of observables often exist, each giving a complete, yet qualitatively different, description of the system. This demands a subtler philosophical attitude, not to ascribe what would appear as an unacceptable paradox to the lack of accuracy in the measuring apparatus. Misconceptions especially occur when phenomena exceed conventional logic and the will of figuring everything out in the usual terms is confronted with the fairly non-pictorial character of modern physics.
Classical and quantum approach are, to a certain degree, mutually exclusive conceptual structures. Each observation involves two confronting systems at least: the phenomenon under study and the measuring devices, including investigators. According to classical physics, a distinction between the two terms can always be drawn and any reciprocal interference be confined to our accuracy. It thus makes perfectly sense to think that the contents of outer experiences are recorded rather than determined. Quite the opposite, as expressed by Bohr, «the quantum theory is characterized by the acknowledgement of a fundamental limitation in the classical physical ideas when applied to atomic phenomena. […] Its essence may be expressed in the so-called quantum postulate, which attributes to any atomic process an essential discontinuity, or rather individuality, completely foreign to classical theories and symbolized by Planck’s quantum of action. […] The quantum postulate implies that any observation of atomic phenomena will involve an interaction with the agency of observation not to be neglected. Accordingly, an independent reality in the ordinary physical sense can neither be ascribed to the phenomena nor to the agencies of observation. After all, the concept of observation is in so far arbitrary as it depends upon which objects are included in the system to be observed. Ultimately, every observation can, of course, be reduced to our sense perceptions»[2].
As a first example, let electrons, coming from very far away, pass through a diffraction slit, whose size is comparable with their de Broglie wavelength $\lambda=h/p$ [3]. Several difficulties arise in the interpretation of the experimental results. By shooting them one at a time and allowing for a certain delay between two shots, the position of each one is collected on a screen past the slit. The diffraction pattern which goes forming point by point is identical to that obtained in the case of several electrons shot simultaneously. In addition to the principal maximum at the centre, a number of secondary maxima appear, whose separation depends on the size of the slit and on the electron’s wavelength. A similar pattern is obtained if photons are used in place of electrons, i.e. electromagnetic waves.
If electrons strictly behaved as classical particles fired in a straight line, a pattern, roughly corresponding to the size and shape of the slit, would be expected on the screen, not a diffraction-blurred one. It would be incorrect to interpret the results by depicting each electron as a tiny wave of energy split/spread by the slit, with one part going in one direction and the other(s) in a different one. Each point collected on the screen corresponds to one whole electron: points are recorded one at a time.
In the case of many electrons, a statistical interpretation of the solution $\Psi$ of Schrödinger’s equation would naturally lead to treat $|\Psi|^2$ as the distribution function depicting the observed diffraction. The number of recorded hits is maximum where $|\Psi|^2$ attains its maximum value and minimum correspondingly.
Were the experiment repeated on a macroscopic scale, using gum bullets for instance, each hit scored might be accidental, hence resulting in a Gaussian distribution with only one maximum at the centre, seemingly without fringes. Wavelengths of macroscopic objects are so negligible owing to their mass, that diffraction curves are smoothed as a consequence of some kind of average performed over finest fringes by the recording device.
How to interpret the cumulative effect of the electrons individually fired? Does each electron know in advance where it should go if it were part of a quantum-mechanical ensemble?
Classical physics postulates that each elementary particle can, at any instant of time, be located at some definite point x along a certain direction labelled as x-axis and have a definite velocity $v_x$, hence a definite momentum $p_x$ along the same direction. Provided that the external forces are known, its trajectory of motion can be exactly and uniquely predicted, given certain initial conditions. According to quantum mechanics, only the probability that the particle will move in a certain direction with a certain velocity/momentum can instead be calculated, the accuracy of this prediction being limited by Heisenberg’s uncertainty relation: $\Delta p_x\Delta x \approx h$[4]. Since the numerical value of h is considerably small, these effects become noticeable only on an atomic scale.
Since $p_x$ and x are canonical conjugate variables, their quantum-mechanical operators do not commute. Bohr’s complementarity principle (1928) postulates that two such variables form a complementary pair, i.e. either they are both undetermined or one of them is “perfectly” determined only. A trade-off between conjugate quantities must be faced: each act of measurement selects reality. Any experiment designed to measure one variable will inevitably destroy information about the other. «Within the scope of classical physics, all characteristic properties of a given object can in principle be ascertained by a single experimental arrangement, although in practice various arrangements are often convenient for the study of different aspects of the phenomena. In fact, data obtained in such a way simply supplement each other and can be combined into a consistent picture of the behaviour of the object under investigation. In quantum mechanics, however, evidence about atomic objects obtained by different experimental arrangements exhibits a novel kind of complementary relationship. Indeed, it must be recognized that such evidence which appears contradictory when combination into a single picture is attempted, exhaust all conceivable knowledge about the object. Far from restricting our efforts to put questions to nature in the form of experiments, the notion of complementarity simply characterizes the answers we can receive by such inquiry, whenever the interaction between the measuring instruments and the objects form an integral part of the phenomena.»[5]
Let the vertical direction in the above-mentioned experiment be the x-axis and the slit be A wide. After being fired, each electron travels horizontally, along a straight line, towards the target with a certain momentum. If initially $p_x=0$ “exactly” ($\Delta p_x \rightarrow 0$), nothing ($\Delta x \rightarrow \infty$) about the vertical position x is known: the particle will either hit somewhere about the slit, or pass through it and reach the screen. As soon as a score is registered, then an electron must have passed through the slit and its position, at some instant of time, must, in principle, have been known with uncertainty $\pm A/2$. Out of Heisenberg’s relation: $\Delta x \approx A/2 \rightarrow \Delta p_x \approx 2h/A$. Owing to this amount of information about the particle’s vertical position, the original accuracy ($\Delta p_x \rightarrow 0$) on px is lost and since $\Delta p_x \approx 2h/A \ne 0$, the trajectory has spread out. So, the smaller A, the wider the diffraction pattern. Physicists refer to this phenomenon as the chance of a particle to travel from the source to the screen, passing through the slit.
If Heisenberg’s principle were not at work in the Universe, the fact that the detected electron has passed through the slit (i.e. its position has become known with a given uncertainty at that time) would not effect its momentum and the scores would approximately collect within an A-wide range. «Sometimes people say quantum mechanics is all wrong. When the particle arrived from the left, its vertical momentum was zero. And now that it has gone through the slit, its position is known. Both position and momentum seem to be known with arbitrary accuracy. It is quite true that we can receive a particle, and on reception determine what its position is and what its momentum would have had to have been to have gotten there. That is true, but that is not what the uncertainty relation refers to. It refers to the predictability of a situation, not remarks about the past. It does no good to say “I knew what the momentum was before it went through the slit, and now I know the position,” because now the momentum knowledge is lost. The fact that it went through the slit no longer permits us to predict the vertical momentum. We are talking about a predictive theory, not just measurements after the fact. So we must talk about what we can predict.»[6]
The whole set of observables any phenomenon consists of cannot be investigated as such to an arbitrary degree of accuracy. Contrary classical expectations, objects do not posses definite properties at all times. The maximal knowledge of all of its parts is not allowed. As previously discussed in the case of x and $p_x$, canonical conjugate variables affect each other. This is a fact in itself not a technological factor. Symbolism serves to elaborate predictions, but the conventional picture of reality as a still bedrock at full disposal cannot be restored and falls beyond language.
Space-time offers the natural framework for all the relationships scaling up from atoms to planets. But how to visualise it? «Someone might call it ether, but for you/ the light at the end of the tunnel is never quite air,// and breath is a shape that sails out over the rooftops,/ into the lights off the quay and the tethered yawls.»[7] Again, the drawback of any visualisation resides in the peculiarity of the terms under comparison. The joyful idea of electrons spinning like toy tops is, in many ways, misleading and so was it for the atomic planetary model. An accurate approach requires to get acquainted with abstract signs and to manipulate them.
To illustrate this point with a last example, let a proton and an electron be in the neighbourhood of each other. The relative orientations of their spins, which can be either up or down, are usually described with the help of the handy ket notation: $|+ +>$; (electron up, proton up), $|+ ->$; (electron up, proton down) and so forth. These symbols act as a base in the vector space of the states of the system, hence any state can be described as a linear combination of them. Once the properties of vector spaces are known and the existence of an entity called spin is verified, complex problems may be rationalised and worked out without even the idea of how to depict the spin actually. Kets do not stand for shortened sentences or words. They extend language at a level at which phenomena cannot be figured out otherwise. Operating with kets serves to decompose systems into fewer elements to be experimentally questioned. «The conception of the objective reality of the particles has thus evaporated in a curious way, not into the fog of some new, obscure, or not yet understood reality concept, but into the transparent clarity of a mathematics that represents no longer the behaviour of the elementary particles but rather our knowledge of this behaviour.»[8]
Do physics and poetry disturb the Universe that tacitly works? Their repeated attempts to trace and record numbers and words are beside the point, for, while doing so, they are part of the whole work.
«If you think of a poem, you can think of it as the most super-charged kind of language; the way you compress meaning into a very, very brief area on the page. Dirac was producing equations that had that kind of concision. And you can then unpack them just as you re-read a Shakespeare sonnet, see more and more in it, see more and more elegance. Same with the Dirac equation. You find an equation there, and you keep finding things that were not obvious on first reading. In fact Dirac actually said that the equation was smarter than he was, because it actually gave more stuff out than he put into it.»[9]

Federico Federici

Footnotes

[1] Eliot, Thomas Stearns: Hamlet and his problems, in: The Sacred Woods: essays on poetry and criticism, London 1920, p. 92.
[2] Bohr, Niels: Atomic theory and the description of nature, Cambridge 1934, p. 54.
[3] $h=6.6\times 10^{-34} \ Js$ is Planck’s constant; $p=\gamma mv$ is the momentum, $\gamma\rightarrow 1$ if $v/c \rightarrow 0$, being $c=3.0\times 10^8 \ m/s$ the speed of light in vacuum.
[4] $\Delta p_x$ and $\Delta x$: uncertainty in momentum and position along x-axis respectively.
[5] Bohr, Niels: Essays 1958-1962 on atomic physics and human knowledge, New York 1963, p. 4.
[6] Feynman, Richard: The Feynman Lectures on Physics vol. 3, New York 2010, p. 2-5.
[7] Burnside, John: The hunt in the forest, London 2009, p. 3.
[8] Heisenberg, Werner: The representation of nature in contemporary physics, in: Deadalus n.87, Cambridge 1958, p. 95.
[9] Farmelo, Graham: The strangest man: the hidden life of Paul Dirac, quantum genius, London 2009.

## 5 thoughts on “Sedimental symbols”

1. Thank you Emma for your precious feedback. I am really happy that I could see you through my travel from physics to poetry (and back)…

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2. emma1951.wordpress.com says:

Excellent reading and expression… I especially understand your philosophies on poetry in amongst your thinking – “meaning compression” (powerful) – the strength of poetry. And I’m not a physical science major!

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3. I have browsed some Online Etymological Dictionary and they all state that the term quark was coined by Gell-Mann himself: not any hint at some other usage of the term.
Finally, I have found this, which may glue the puzzle:
“In 1963, when I assigned the name “quark” to the fundamental constituents of the nucleon, I had the sound first, without the spelling, which could have been “kwork”. Then, in one of my occasional perusals of Finnegans Wake, by James Joyce, I came across the word “quark” in the phrase “Three quarks for Muster Mark”. Since “quark” (meaning, for one thing, the cry of the gull) was clearly intended to rhyme with “Mark”, as well as “bark” and other such words, I had to find an excuse to pronounce it as “kwork”. But the book represents the dream of a publican named Humphrey Chimpden Earwicker. Words in the text are typically drawn from several sources at once, like the “portmanteau” words in “Through the Looking-Glass”. From time to time, phrases occur in the book that are partially determined by calls for drinks at the bar. I argued, therefore, that perhaps one of the multiple sources of the cry “Three quarks for Muster Mark” might be “Three quarts for Mister Mark”, in which case the pronunciation “kwork” would not be totally unjustified. In any case, the number three fitted perfectly the way quarks occur in nature.”
Can you quote some other 19C writer who used “quark” as the sound of birds? I would like to embody these information in the second darft of my paper.
F.

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4. Nice annotation! I will correct the second line in my citation according to what you suggest and will explore the reasons why Gell-Mann chose that name (and why he eventually tricked the true explanation…).
In my Oxford Dictionary there’s no trace of any old meaning of the term “quark”. I am going to check it in some ethimological one.
I have much appreciated your contribution.
Thanks again
F.

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5. Vincent Deane says:

Contrary to Gell-Mann’s amusing but idiosyncratic view, ‘quark’ is not a Joycean invention, but an imitative English word, used by 19C writers for the cry of an animal (most of the OED citations give birds). The second line of your citation should read ‘Sure he hasn’t got much of a bark’. For what it’s worth, I should add that the Penguin edition retains all the worst errors of the pre-corrected 1939 first printing of FW, some of which were subsequently corrected in the Faber/Viking editions of 1950 and 1964.

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# convergence of a series using the root test

I want to find whether this series diverges or converges using the root test $$\sum_{n=1}^\infty \arccos^n \left(\frac{1}{n^2}\right)$$ $$\sqrt[n] {\arccos^n \left(\frac{1}{n^2}\right)}$$ $$\arccos \sqrt[n] {\left(\frac{1}{n^2}\right)}$$ $$\arccos 1 = 0$$ and since its 0 the series should converge but I am not sure about the calculations there.

• The third line is wrong. – Claude Leibovici Oct 30 '20 at 14:10
• @ClaudeLeibovici I don't know how to contact you in a polite way. Raffaele's hypothesis is false :( so I deleted my answer. $341, 561, 645$ are counterexamples in the first thousand... – Raffaele Oct 30 '20 at 14:16
• $\arccos\frac{1}{n^2}\to 1$ – Raffaele Oct 30 '20 at 14:17
• I still dont understand fully how is arcos 1/n^2 equal to one if we take the limit – GregoryStory16 Oct 30 '20 at 14:22
• @GregoryStory16. $\frac \pi 2$ – Claude Leibovici Oct 30 '20 at 14:26

The series diverges. Indeed $$\arccos^n\left(\frac{1}{n^2}\right)\sim \left(\frac{\pi }{2}\right)^n+O\left(\frac{1}{n^2}\right);\;n\to\infty$$ The first $$50$$ terms sum is about $$1.7\times 10^{10}$$.

Recall that

$$\arccos(x) = \frac{\pi}{2} - \arcsin(x) =\frac{\pi}{2}-x+O(x^2) \to \frac \pi 2$$

therefore the series diverges since $$a_n \not \to 0$$.

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## Reduced density matrix and partial trace

The reduced matrix is defined as the partial trace of the density matrix. In linear algebra and functional analysis, the partial trace is a generalization of the trace. Whereas the trace is a scalar valued function on operators, the partial trace is an operator-valued function. The partial trace has applications in quantum information and decoherence which is relevant for quantum measurement and thereby to the decoherent approaches to interpretations of quantum mechanics, including consistent histories and the relative state interpretation.

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Poisson random measures, Lévy processes, generalized stochastic calculus, continuous state branching processes
 发布时间： 2017-10-13     10:22   【返回上一页】 发布人：Etienne Pardoux教授

The Poisson distribution appears very naturally as the law of a random cloud of points, under mild conditions. We will define and study Poisson Random measures of points. We shall then show how to use them in order to define all Lévy processes other than the Brownian motion. This requires defining stochastic integrals of deterministic functions with respect to a compensated Poisson random measure. We shall also integrate predictable random processes, establish a generalized Itô formula, solve SDEs directed by a Lévy process, and prove the Markov property of the solution. Finally we shall study Dawson-Li type SDEs, whose solutions are general continuous state branching processes. We shall see that the Lamperti transform changes those into Lévy processes.

1016日下午15:30-17:301018日下午14:00-17:001020日上午8:40-11:401022日上午8:40-11:40


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# More than 99% of groups of order less than 2000 are of order 1024?

In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024.

Is this for real? How can one deduce this result? Is there a nice way or do we just check all finite groups up to isomorphism?

Thanks!

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There are $49487365422$ of order 1024 –  PAD Nov 20 '12 at 15:32
This was a result of Besche, Eick and O'Brien. Note that O'Brien $\Rightarrow$ computer usage. See mathscinet. Basically, the paper is a survey discussing methods and algorithms used to construct (small) groups. –  user1729 Nov 20 '12 at 15:34
No doubt this about isomorphism classes of groups. –  Marc van Leeuwen Nov 20 '12 at 17:28
Is there a rough estimate (say one significant digit) of the number of groups of order 2048? –  yatima2975 Nov 20 '12 at 23:06
@yatima2975: From this article : "gnu(2048) is still not precisely known, but it strictly exceeds 1774274116992170, which is the exact number of groups of order 2048 that have exponent-2 class 2, and can confidently be expected to agree with that number in its first 3 digits." –  Mikko Korhonen Nov 21 '12 at 10:59

Here is a list of the number of groups of order $n$ for $n=1,\ldots,2015$. If you add up the number of groups of order other than $1024$, you get $423{,}164{,}062$. There are $49{,}487{,}365{,}422$ groups of order $1024$, so you can see the assertion is true. (In fact the percentage is about $99.15\%$.)

As far as I know there is no reasonable way to deduce a priori the number of isomorphism classes of groups of a given order, though I believe that combinatorial group theory has some methods for specific cases. A general rule of thumb is that there are a ton of $2$-groups, and in fact I have heard it said that "almost all finite groups are $2$-groups" (though I cannot cite a reference for this statement).

EDIT: As pointed out in the comments, "almost all finite groups are $2$-groups" is still a conjecture. There is an asymptotic bound on the number of $p$-groups of order $p^n$, however. Denoting by $\mu(p,n)$ the number of groups of order $p^n$, $$\mu(p,n)=p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3},$$ which is proven here. This colossal growth along with the results of Besche, Eick & O'Brien seem to be what primarily motivated the conjecture.

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A while ago I tried to find a reference for this "almost all..." result. I think it is just a folklore statement, with the paper which is the subject of this thread proffered as evidence. –  user1729 Nov 20 '12 at 15:42
(I wonder if there are more groups of order $3^{10}$ than of order $2^{10}$? Genericity proofs are...unsavoury...at least to my pallet...) –  user1729 Nov 20 '12 at 15:44
Of course it's possible to deduce the number of isomorphism classes of groups of (finite) order $n$: write down all possible $n$ by $n$ multiplication tables, check which satisfy the group axioms, check every bijection between each pair to see if it's a group isomorphism. Since everything is finite, this can all be computed in finite time. The hard part is finding ways to do it in a sane amount of time. –  Chris Eagle Nov 20 '12 at 15:55
According to the list linked in the answer, there are 504 groups of order $3^6=729$ and 267 groups of order $2^6=64$. There are 15 groups of order $5^4=625$ and also of order $3^4=81$ and 14 groups of order $2^4=16$. –  Mark Bennet Nov 20 '12 at 16:40
Almost all groups are infinite. –  Marc van Leeuwen Nov 20 '12 at 17:29

This is true. The amount of groups of order at most 2000 (up to isomorphism) was calculated precisely for the first time in 2001 by Besche, Eick and O'Brien. Here is the announcement of their result:

We announce the construction up to isomorphism of the $49 910 529 484$ groups of order at most $2000$.

In table 1 the number of groups of order $1024$ is given, it is $49 487 365 422$. Hence ~99.2% of all groups of order at most $2000$ have order $1024$.

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If Black-Scholes model assumptions hold, then a derivative payoff can always be replicated in such a way it would never provide return more than risk free interest rate, otherwise it will lead to arbitrage opportunities. But assumption never ... -3 I've been collecting daily option prices for all symbols/strikes since beginning 2014. If anyone interested, let me know (not free) 0 Probably Option Metrics which you can find on WRDS is what you are looking for. 1 Note that, for K_1 < K < K_2, \begin{align*} -(K_2-K_1) \le (S_T-K_2)^+ - (S_T-K_1)^+ \le 0. \end{align*} Taking the conditional expectation with respect to information set \mathcal{F}_t, \begin{align*} -(K_2-K_1)B_t(T) \le C(T, K_2, S, t) - C(T, K_1, S, t) \le 0. \end{align*} That is, \begin{align*} -B_t(T) \le \frac{C(T, K_2, S, t) - C(T, K_1, ... 3 For the case where \sum_{i=1}^n \lambda_i =1, you need only note that the payoff \begin{align*} (x-K)^+ \end{align*} is a convex function in x. That is, \begin{align*} \Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+ \le \sum_{i=1}^n\lambda_i(S_i-K)^+. \end{align*} Then \begin{align*} e^{-rT}E\left(\Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+\right) \le ... 3 you just add in any auxiliary variables accumulated along the path that determine the pay-off to the regression variables. So path-dependence is not a problem. If you have previous decisions, you may need to do different regressions based on their possible values or make them into a continuous variables that can be used for regression. 2 The easiest way to think of this is as follows: Settlement Price - Price at which the exchange margins all accounts for those options. Closing Price - Mid/Bid/Ask of Active Market at the exchanges last trade time. E.g. for TY Contracts this is at 5pm EST vs. a Settle Time of 3pm EST. Last Trade Price - Not all options trade every day. This is the price ... 0 One example could be someone using option strategies and its underlying dividends. In these cases, the trader could use early excersise to capture the dividend value. Google it for more information. 3 Options are in zero net supply (like futures and other derivatives), so for evry long there is a short and for every short there is a long. The open interest is the sum of the longs which also equals the sum of the short positions. 2 You should not use the Feller condition as a constraint. In many cases its violation will be required for a good fit to the market data. 0 There was a recent thread discussing this historic options data That data source mentioned there is commercial, though reasonably priced. Their data in CSV format readily ingestible by Excel. One note of caution is options' daily closing quotes are not always accurate. 1 it is not a "newbie question". actually it is very deep question, especially if you are trying to trade options in some way. all Greeks are about change in premium of the option. Premium consists of two components: time value + intrinsic. Greeks cannot tell you anything about this decomposition, they deal only with total sum (i.e. with overall premium). So ... 0 From e.g. the CBOE web site you are able to download a full table of option quotes (delayed) for a specific stock (menu quotes - delayed quotes). Take it from there, appply data manipulation (Excel VBA, R, Python..) to the extent needed and then you should be be able to end up with the desired result. This wouldn't be a simple copy & paste procedure; ... 5 If you want to know what Greeks the market assigns to an option, i.e. the market implied Greeks, then you would use the implied volatility. And that is what traders like to look at. 0 Check out BarchartOnDemand. They have CME EOD data as an API...it may not be advertised but I know is available via their API (http://www.barchartondemand.com/api.php ). Pricing is not published but should be reasonable for an individual just looking for EOD. You can also get the same data through QuantGo.com but have to use it on your own computer instance ... 0 To get IV I do the following: 1) change sig many times and calculate C in BS formula every time. That can be done with OIC calculator All other parameters are kept constant in BS call price calculations. The sig that corresponds to C value closest to the call market value is probably right. 2) without OIC calculator for every chosen sig I am using old ... 2 You can guesstimate by vega weighted implied vol. This is why: Say that you have a portfolio of options with prices P_j. Each one of them has a different pricing function f_j (as function of vol) and a different implied vol \sigma_j. For each option f_j(\sigma_j)=P_j. Now you put them together in a single product. If the implied vol of the product ... 3 For the US market nearly all options on securities are american whereas the options on indexes are european. What you can do is to use a database such as OptionMetrics which adjusts the stock american options to european options by taking into account the early exercise premium. 3 Using the answer from: Chris Taylor, on math stackexchange (link): Let the price of an option at strike K be given by V(K). To say that the price is convex in the strike means thatV(K-\delta) + V(K+\delta) > 2 V(K) for all $K>0$ and $\delta>0$. Let's assume that the opposite is true, i.e. that there exist tradeable option contracts ...

Top 50 recent answers are included


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# If Line k in the xy-plane has equation y = mx + b, where m

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02 Sep 2010, 11:52
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7).
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7)

I have a doubt in stmnt (2). K intersects the line with equation y=2(x)+3 at point (2,7) means for the line y=2(x)+3 slope is 2,but with this we cannot able to find slope of the line K. is my reasoning is right. please explain
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02 Sep 2010, 12:19
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

$$y=mx+b$$ is called point-intercept form of equation of a line. Where: $$m$$ is the slope of the line; $$b$$ is the y-intercept of the line; $$x$$ is the independent variable of the function $$y$$.

So we are asked to find the value of $$m$$.

(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is $$1-m$$, so $$1-m=m$$ --> $$m=\frac{1}{2}$$. Sufficient.

(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> $$7=2m+b$$ --> can not solve for $$m$$. Not sufficient.

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26 Mar 2012, 13:14
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is
the slope of k?
(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7)

I have a doubt in stmnt (2). K intersects the line with equation y=2(x)+3 at point (2,7) means for the line y=2(x)+3 slope is 2,but with this we cannot able to find slope of the line K. is my reasoning is right. please explain

My 2 cents:

Questions is asking you what is the value of M

so

1) Since k and this line with slope (1-m) are parallel

then m = 1 - m or m = 1/2

thus it is sufficient

2) we know that the point 2, 7 is on the line line y = mx + b

so 7 = 2m + b

But we know nothing else so we cannot solve for m thus it is insufficient
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06 Jun 2012, 23:54
For line K, y=mx+b implies slope=m

From Statement-1,

As slopes of parallel lines are equal, m = (1-m)

m=1/2
Statement-1 is sufficient.

From statement-2,

line K passes through (2,7)
7=2.m + b
m = (7-b)/2
Not sufficient.

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text:
# Variance of Logistic Distribution/Proof 2

## Theorem

Let $X$ be a continuous random variable which satisfies the logistic distribution:

$X \sim \map {\operatorname {Logistic} } {\mu, s}$

The variance of $X$ is given by:

$\var X = \dfrac {s^2 \pi^2} 3$

## Proof

By Moment Generating Function of Logistic Distribution, the moment generating function of $X$ is given by:

$\ds \map {M_X} t = \map \exp {\mu t} \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u$

for $\size t < \dfrac 1 s$.

$\ds \var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Logistic Distribution, we have:

$\expect X = \mu$
$\ds \map { {M_X}''} t = \map \exp {\mu t} \paren {\mu^2 \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u}$

Hence setting $t = 0$:

 $\ds \map { {M_X}''} 0$ $=$ $\ds \mu^2 \int_{\to 0}^{\to 1} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u$ $\ds$ $=$ $\ds \mu^2 - 2 s \mu \paren {0 } + s^2 \paren {\dfrac {\pi^2} 3}$ Definite Integral of Constant, Expectation of Logistic Distribution/Lemma 3 and Lemma 4 $\ds$ $=$ $\ds \mu^2 + \dfrac {s^2 \pi^2} 3$

So:

 $\ds \var X$ $=$ $\ds \mu^2 + \dfrac {s^2 \pi^2} 3 - \mu^2$ $\ds$ $=$ $\ds \dfrac {s^2 \pi^2} 3$

$\blacksquare$


text:
# How do you divide (x^4+x^2-3x-3)/(5x-2)?

Jan 19, 2016

Long divide the coefficients of the polynomials to find:

$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$

$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$

#### Explanation:

I like to divide polynomials by long dividing their coefficients, including $0$'s for any missing powers of $x$ ...

This is similar to long division of numbers.

Write the dividend $1 , 0 , 1 , - 3 , - 3$ under the bar and the divisor $5 , - 2$ to the left of the bar.

Write each term of the quotient in turn, chosen to match the leading term of the running remainder when multiplied by the divisor.

So we write $\textcolor{b l u e}{\frac{1}{5}}$ which multiplied by $5 , - 2$ results in $1 , - \frac{2}{5}$, matching the leading term $1$ of the dividend.

Subtract the product from the dividend and bring down the next term of the dividend alongside the result as the running remainder.

Choose the next term $\textcolor{b l u e}{\frac{2}{25}}$ to match the leading term of this running remainder and subtract the product, etc.

Repeat until the running remainder is shorter than the divisor and there is no remaining term to bring down from the dividend. This is the final remainder.

In this particular example we find that the quotient polynomial is:

$\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}$

with remainder $- \frac{2509}{625}$

That is:

$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$

$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$

or if you prefer:

${x}^{4} + {x}^{2} - 3 x - 3$

$= \left(5 x - 2\right) \left(\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}\right) - \frac{2509}{625}$


text:
find the basis for a Jordan algebra of matrices - Maple Programming Help

Home : Support : Online Help : Mathematics : DifferentialGeometry : LieAlgebras : DifferentialGeometry/LieAlgebras/JordanMatrices

LieAlgebras[JordanMatrices] - find the basis for a Jordan algebra of matrices

LieAlgebras[JordanProduct] - find the Jordan product of two Jordan matrices

Calling Sequences

JordanMatrices(n, alg, option )

JordanProduct(A, B)

Parameters

n        - an integer

alg      - a name or string of an initialized algebra, the string "R" or the string "C"

option   - the keyword argument signature = [p, q], where p and q are integers and p + q = n

A, B     - square matrices

Description

 • Let be the algebra of real numbers, the complex numbers, the quaternions, the octonions, or one of their split versions. A Jordan matrix is a square matrix with entries in which is Hermitian with respect to the conjugation in the algebra, that is,  More generally, if is the diagonal matrix ${I}_{\mathrm{pq}}=\left[\begin{array}{rr}{I}_{p}& 0\\ 0& -{I}_{q}\end{array}\right]$ and then $J$ is called a Jordan matrix with respect to . The set of such matrices is always a real vector space.
 • The command JordanMatrix(n, alg) returns a list of matrices which form a basis for the real vector space of  square matrices with entries in . With the keyword argument signature = [p, q] a basis for the Jordan matrices with respect to ${I}_{\mathrm{pq}}$ is determined.
 • The Jordan product of 2 Jordan matrices and is the symmetric product . The set of Jordan matrices with Jordan product is an algebra which is denoted by $\mathrm{𝕁}\left(n\mathit{,}\mathrm{𝔸}\right)$or .
 • The structure equations for a general Jordan algebra can be calculated with the command AlgebraData. The structure equations for a few low dimensional Jordan algebras are also available through the command AlgebraLibraryData.

Examples

 > with(DifferentialGeometry): with(LieAlgebras): with(Tools):

Example 1.

In this example we construct a basis for the Jordan algebra of matrices over the quaternions. The first step is to use the command AlgebraLibraryData to retrieve the structure equations for the quaternions.

 ${\mathrm{AD}}{:=}\left[{{\mathrm{e1}}}^{{2}}{=}{\mathrm{e1}}{,}{\mathrm{e1}}{.}{\mathrm{e2}}{=}{\mathrm{e2}}{,}{\mathrm{e1}}{.}{\mathrm{e3}}{=}{\mathrm{e3}}{,}{\mathrm{e1}}{.}{\mathrm{e4}}{=}{\mathrm{e4}}{,}{\mathrm{e2}}{.}{\mathrm{e1}}{=}{\mathrm{e2}}{,}{{\mathrm{e2}}}^{{2}}{=}{-}{\mathrm{e1}}{,}{\mathrm{e2}}{.}{\mathrm{e3}}{=}{\mathrm{e4}}{,}{\mathrm{e2}}{.}{\mathrm{e4}}{=}{-}{\mathrm{e3}}{,}{\mathrm{e3}}{.}{\mathrm{e1}}{=}{\mathrm{e3}}{,}{\mathrm{e3}}{.}{\mathrm{e2}}{=}{-}{\mathrm{e4}}{,}{{\mathrm{e3}}}^{{2}}{=}{-}{\mathrm{e1}}{,}{\mathrm{e3}}{.}{\mathrm{e4}}{=}{\mathrm{e2}}{,}{\mathrm{e4}}{.}{\mathrm{e1}}{=}{\mathrm{e4}}{,}{\mathrm{e4}}{.}{\mathrm{e2}}{=}{\mathrm{e3}}{,}{\mathrm{e4}}{.}{\mathrm{e3}}{=}{-}{\mathrm{e2}}{,}{{\mathrm{e4}}}^{{2}}{=}{-}{\mathrm{e1}}\right]$ (2.1)

Initialize this algebra.

 > DGsetup(AD, '[e, i, j, k]', '[omega]');
 ${\mathrm{algebra name: Q}}$ (2.2)

Generate a basis for the Jordan algebra of matrices over the quaternions.

 > M := JordanMatrices(3, Q);
 ${M}{:=}\left[\left[\begin{array}{ccc}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {e}& {0}{}{e}\\ {e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {e}\\ {0}{}{e}& {e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {i}& {0}{}{e}\\ {-}{i}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {j}& {0}{}{e}\\ {-}{j}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {k}& {0}{}{e}\\ {-}{k}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {i}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {-}{i}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {j}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {-}{j}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {k}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {-}{k}& {0}{}{e}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {i}\\ {0}{}{e}& {-}{i}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {j}\\ {0}{}{e}& {-}{j}& {0}{}{e}\end{array}\right]{,}\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {k}\\ {0}{}{e}& {-}{k}& {0}{}{e}\end{array}\right]\right]$ (2.3)

We form the general element of $\mathrm{𝕁}\left(\mathit{3}\mathit{,}\mathrm{ℚ}\right)$ and check it is Hermitian.

 Q > C := [seq(c||n, n = 1 .. 15)];
 ${C}{:=}\left[{\mathrm{c1}}{,}{\mathrm{c2}}{,}{\mathrm{c3}}{,}{\mathrm{c4}}{,}{\mathrm{c5}}{,}{\mathrm{c6}}{,}{\mathrm{c7}}{,}{\mathrm{c8}}{,}{\mathrm{c9}}{,}{\mathrm{c10}}{,}{\mathrm{c11}}{,}{\mathrm{c12}}{,}{\mathrm{c13}}{,}{\mathrm{c14}}{,}{\mathrm{c15}}\right]$ (2.4)
 Q > J := evalDG(DGzip(C, M, "plus"));
 ${J}{:=}\left[\begin{array}{ccc}{\mathrm{c1}}{}{e}& {\mathrm{c4}}{}{e}{+}{\mathrm{c7}}{}{i}{+}{\mathrm{c8}}{}{j}{+}{\mathrm{c9}}{}{k}& {\mathrm{c5}}{}{e}{+}{\mathrm{c10}}{}{i}{+}{\mathrm{c11}}{}{j}{+}{\mathrm{c12}}{}{k}\\ {\mathrm{c4}}{}{e}{-}{\mathrm{c7}}{}{i}{-}{\mathrm{c8}}{}{j}{-}{\mathrm{c9}}{}{k}& {\mathrm{c2}}{}{e}& {\mathrm{c6}}{}{e}{+}{\mathrm{c13}}{}{i}{+}{\mathrm{c14}}{}{j}{+}{\mathrm{c15}}{}{k}\\ {\mathrm{c5}}{}{e}{-}{\mathrm{c10}}{}{i}{-}{\mathrm{c11}}{}{j}{-}{\mathrm{c12}}{}{k}& {\mathrm{c6}}{}{e}{-}{\mathrm{c13}}{}{i}{-}{\mathrm{c14}}{}{j}{-}{\mathrm{c15}}{}{k}& {\mathrm{c3}}{}{e}\end{array}\right]$ (2.5)

Here is the conjugate transpose of J.

 Q > Jdagger := DGconjugate(J)^+;
 ${\mathrm{Jdagger}}{:=}\left[\begin{array}{ccc}{\mathrm{c1}}{}{e}& {\mathrm{c4}}{}{e}{+}{\mathrm{c7}}{}{i}{+}{\mathrm{c8}}{}{j}{+}{\mathrm{c9}}{}{k}& {\mathrm{c5}}{}{e}{+}{\mathrm{c10}}{}{i}{+}{\mathrm{c11}}{}{j}{+}{\mathrm{c12}}{}{k}\\ {\mathrm{c4}}{}{e}{-}{\mathrm{c7}}{}{i}{-}{\mathrm{c8}}{}{j}{-}{\mathrm{c9}}{}{k}& {\mathrm{c2}}{}{e}& {\mathrm{c6}}{}{e}{+}{\mathrm{c13}}{}{i}{+}{\mathrm{c14}}{}{j}{+}{\mathrm{c15}}{}{k}\\ {\mathrm{c5}}{}{e}{-}{\mathrm{c10}}{}{i}{-}{\mathrm{c11}}{}{j}{-}{\mathrm{c12}}{}{k}& {\mathrm{c6}}{}{e}{-}{\mathrm{c13}}{}{i}{-}{\mathrm{c14}}{}{j}{-}{\mathrm{c15}}{}{k}& {\mathrm{c3}}{}{e}\end{array}\right]$ (2.6)

We see that J is Hermitian.

 Q > J &MatrixMinus Jdagger;
 $\left[\begin{array}{ccc}{0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\\ {0}{}{e}& {0}{}{e}& {0}{}{e}\end{array}\right]$ (2.7)

Now define two elements of $\mathrm{𝕁}\left(\mathit{3}\mathit{,}\mathrm{ℚ}\right)$ and calculate their Jordan product.

 Q > A := evalDG(M[8] + M[12]);
 ${A}{:=}\left[\begin{array}{ccc}{0}{}{e}& {j}& {k}\\ {-}{j}& {0}{}{e}& {0}{}{e}\\ {-}{k}& {0}{}{e}& {0}{}{e}\end{array}\right]$ (2.8)
 Q > B := evalDG(M[7] + M[15]);
 ${B}{:=}\left[\begin{array}{ccc}{0}{}{e}& {i}& {0}{}{e}\\ {-}{i}& {0}{}{e}& {k}\\ {0}{}{e}& {-}{k}& {0}{}{e}\end{array}\right]$ (2.9)
 Q > JordanProduct(A, B);
 $\left[\begin{array}{ccc}{0}{}{e}& \frac{{1}}{{2}}{}{e}& \frac{{1}}{{2}}{}{i}\\ \frac{{1}}{{2}}{}{e}& {0}{}{e}& \frac{{1}}{{2}}{}{j}\\ {-}\frac{{1}}{{2}}{}{i}& {-}\frac{{1}}{{2}}{}{j}& {0}{}{e}\end{array}\right]$ (2.10)

Example 2.

In this example we construct a basis for the Jordan matrices over the split octonions with respect to the inner product . First we retrieve the structure equations for the split octonions and initialize.

 Q > AD := AlgebraLibraryData("Octonions", Os, type = "Split");
 ${\mathrm{AD}}{:=}\left[{{\mathrm{e1}}}^{{2}}{=}{\mathrm{e1}}{,}{\mathrm{e1}}{.}{\mathrm{e2}}{=}{\mathrm{e2}}{,}{\mathrm{e1}}{.}{\mathrm{e3}}{=}{\mathrm{e3}}{,}{\mathrm{e1}}{.}{\mathrm{e4}}{=}{\mathrm{e4}}{,}{\mathrm{e1}}{.}{\mathrm{e5}}{=}{\mathrm{e5}}{,}{\mathrm{e1}}{.}{\mathrm{e6}}{=}{\mathrm{e6}}{,}{\mathrm{e1}}{.}{\mathrm{e7}}{=}{\mathrm{e7}}{,}{\mathrm{e1}}{.}{\mathrm{e8}}{=}{\mathrm{e8}}{,}{\mathrm{e2}}{.}{\mathrm{e1}}{=}{\mathrm{e2}}{,}{{\mathrm{e2}}}^{{2}}{=}{-}{\mathrm{e1}}{,}{\mathrm{e2}}{.}{\mathrm{e3}}{=}{\mathrm{e4}}{,}{\mathrm{e2}}{.}{\mathrm{e4}}{=}{-}{\mathrm{e3}}{,}{\mathrm{e2}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e6}}{,}{\mathrm{e2}}{.}{\mathrm{e6}}{=}{\mathrm{e5}}{,}{\mathrm{e2}}{.}{\mathrm{e7}}{=}{-}{\mathrm{e8}}{,}{\mathrm{e2}}{.}{\mathrm{e8}}{=}{\mathrm{e7}}{,}{\mathrm{e3}}{.}{\mathrm{e1}}{=}{\mathrm{e3}}{,}{\mathrm{e3}}{.}{\mathrm{e2}}{=}{-}{\mathrm{e4}}{,}{{\mathrm{e3}}}^{{2}}{=}{-}{\mathrm{e1}}{,}{\mathrm{e3}}{.}{\mathrm{e4}}{=}{\mathrm{e2}}{,}{\mathrm{e3}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e7}}{,}{\mathrm{e3}}{.}{\mathrm{e6}}{=}{\mathrm{e8}}{,}{\mathrm{e3}}{.}{\mathrm{e7}}{=}{\mathrm{e5}}{,}{\mathrm{e3}}{.}{\mathrm{e8}}{=}{-}{\mathrm{e6}}{,}{\mathrm{e4}}{.}{\mathrm{e1}}{=}{\mathrm{e4}}{,}{\mathrm{e4}}{.}{\mathrm{e2}}{=}{\mathrm{e3}}{,}{\mathrm{e4}}{.}{\mathrm{e3}}{=}{-}{\mathrm{e2}}{,}{{\mathrm{e4}}}^{{2}}{=}{-}{\mathrm{e1}}{,}{\mathrm{e4}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e8}}{,}{\mathrm{e4}}{.}{\mathrm{e6}}{=}{-}{\mathrm{e7}}{,}{\mathrm{e4}}{.}{\mathrm{e7}}{=}{\mathrm{e6}}{,}{\mathrm{e4}}{.}{\mathrm{e8}}{=}{\mathrm{e5}}{,}{\mathrm{e5}}{.}{\mathrm{e1}}{=}{\mathrm{e5}}{,}{\mathrm{e5}}{.}{\mathrm{e2}}{=}{\mathrm{e6}}{,}{\mathrm{e5}}{.}{\mathrm{e3}}{=}{\mathrm{e7}}{,}{\mathrm{e5}}{.}{\mathrm{e4}}{=}{\mathrm{e8}}{,}{{\mathrm{e5}}}^{{2}}{=}{\mathrm{e1}}{,}{\mathrm{e5}}{.}{\mathrm{e6}}{=}{\mathrm{e2}}{,}{\mathrm{e5}}{.}{\mathrm{e7}}{=}{\mathrm{e3}}{,}{\mathrm{e5}}{.}{\mathrm{e8}}{=}{\mathrm{e4}}{,}{\mathrm{e6}}{.}{\mathrm{e1}}{=}{\mathrm{e6}}{,}{\mathrm{e6}}{.}{\mathrm{e2}}{=}{-}{\mathrm{e5}}{,}{\mathrm{e6}}{.}{\mathrm{e3}}{=}{-}{\mathrm{e8}}{,}{\mathrm{e6}}{.}{\mathrm{e4}}{=}{\mathrm{e7}}{,}{\mathrm{e6}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e2}}{,}{{\mathrm{e6}}}^{{2}}{=}{\mathrm{e1}}{,}{\mathrm{e6}}{.}{\mathrm{e7}}{=}{\mathrm{e4}}{,}{\mathrm{e6}}{.}{\mathrm{e8}}{=}{-}{\mathrm{e3}}{,}{\mathrm{e7}}{.}{\mathrm{e1}}{=}{\mathrm{e7}}{,}{\mathrm{e7}}{.}{\mathrm{e2}}{=}{\mathrm{e8}}{,}{\mathrm{e7}}{.}{\mathrm{e3}}{=}{-}{\mathrm{e5}}{,}{\mathrm{e7}}{.}{\mathrm{e4}}{=}{-}{\mathrm{e6}}{,}{\mathrm{e7}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e3}}{,}{\mathrm{e7}}{.}{\mathrm{e6}}{=}{-}{\mathrm{e4}}{,}{{\mathrm{e7}}}^{{2}}{=}{\mathrm{e1}}{,}{\mathrm{e7}}{.}{\mathrm{e8}}{=}{\mathrm{e2}}{,}{\mathrm{e8}}{.}{\mathrm{e1}}{=}{\mathrm{e8}}{,}{\mathrm{e8}}{.}{\mathrm{e2}}{=}{-}{\mathrm{e7}}{,}{\mathrm{e8}}{.}{\mathrm{e3}}{=}{\mathrm{e6}}{,}{\mathrm{e8}}{.}{\mathrm{e4}}{=}{-}{\mathrm{e5}}{,}{\mathrm{e8}}{.}{\mathrm{e5}}{=}{-}{\mathrm{e4}}{,}{\mathrm{e8}}{.}{\mathrm{e6}}{=}{\mathrm{e3}}{,}{\mathrm{e8}}{.}{\mathrm{e7}}{=}{-}{\mathrm{e2}}{,}{{\mathrm{e8}}}^{{2}}{=}{\mathrm{e1}}\right]$ (2.11)
 ${\mathrm{algebra name: Os}}$ (2.12)

Here are the Jordan matrices we seek.

 Os > M := JordanMatrices(2, Os, signature = [1,1]);
 ${M}{:=}\left[\left[\begin{array}{cc}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\\ {0}{}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e1}}\\ {-}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e2}}\\ {\mathrm{e2}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e3}}\\ {\mathrm{e3}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e4}}\\ {\mathrm{e4}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e5}}\\ {\mathrm{e5}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e6}}\\ {\mathrm{e6}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e7}}\\ {\mathrm{e7}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e8}}\\ {\mathrm{e8}}& {0}{}{\mathrm{e1}}\end{array}\right]{,}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\\ {0}{}{\mathrm{e1}}& {\mathrm{e1}}\end{array}\right]\right]$ (2.13)

We form the general element of and check that it is Hermitian.

 Q > C := [seq(c||n, n = 1 .. 10)];
 ${C}{:=}\left[{\mathrm{c1}}{,}{\mathrm{c2}}{,}{\mathrm{c3}}{,}{\mathrm{c4}}{,}{\mathrm{c5}}{,}{\mathrm{c6}}{,}{\mathrm{c7}}{,}{\mathrm{c8}}{,}{\mathrm{c9}}{,}{\mathrm{c10}}\right]$ (2.14)
 Q > J := evalDG(DGzip(C, M, "plus"));
 ${J}{:=}\left[\begin{array}{cc}{\mathrm{c1}}{}{\mathrm{e1}}& {\mathrm{c2}}{}{\mathrm{e1}}{+}{\mathrm{c3}}{}{\mathrm{e2}}{+}{\mathrm{c4}}{}{\mathrm{e3}}{+}{\mathrm{c5}}{}{\mathrm{e4}}{+}{\mathrm{c6}}{}{\mathrm{e5}}{+}{\mathrm{c7}}{}{\mathrm{e6}}{+}{\mathrm{c8}}{}{\mathrm{e7}}{+}{\mathrm{c9}}{}{\mathrm{e8}}\\ {-}{\mathrm{c2}}{}{\mathrm{e1}}{+}{\mathrm{c3}}{}{\mathrm{e2}}{+}{\mathrm{c4}}{}{\mathrm{e3}}{+}{\mathrm{c5}}{}{\mathrm{e4}}{+}{\mathrm{c6}}{}{\mathrm{e5}}{+}{\mathrm{c7}}{}{\mathrm{e6}}{+}{\mathrm{c8}}{}{\mathrm{e7}}{+}{\mathrm{c9}}{}{\mathrm{e8}}& {\mathrm{c10}}{}{\mathrm{e1}}\end{array}\right]$ (2.15)

Here is the conjugate transpose of J.

 Q > Jdagger := DGconjugate(J)^+;
 ${\mathrm{Jdagger}}{:=}\left[\begin{array}{cc}{\mathrm{c1}}{}{\mathrm{e1}}& {-}{\mathrm{c2}}{}{\mathrm{e1}}{-}{\mathrm{c3}}{}{\mathrm{e2}}{-}{\mathrm{c4}}{}{\mathrm{e3}}{-}{\mathrm{c5}}{}{\mathrm{e4}}{-}{\mathrm{c6}}{}{\mathrm{e5}}{-}{\mathrm{c7}}{}{\mathrm{e6}}{-}{\mathrm{c8}}{}{\mathrm{e7}}{-}{\mathrm{c9}}{}{\mathrm{e8}}\\ {\mathrm{c2}}{}{\mathrm{e1}}{-}{\mathrm{c3}}{}{\mathrm{e2}}{-}{\mathrm{c4}}{}{\mathrm{e3}}{-}{\mathrm{c5}}{}{\mathrm{e4}}{-}{\mathrm{c6}}{}{\mathrm{e5}}{-}{\mathrm{c7}}{}{\mathrm{e6}}{-}{\mathrm{c8}}{}{\mathrm{e7}}{-}{\mathrm{c9}}{}{\mathrm{e8}}& {\mathrm{c10}}{}{\mathrm{e1}}\end{array}\right]$ (2.16)
 Os > I22 := Matrix([[1, 0], [0, -1]]);
 ${\mathrm{I22}}{:=}\left[\begin{array}{rr}{1}& {0}\\ {0}& {-}{1}\end{array}\right]$ (2.17)
 Os > evalDG(I22.J) &MatrixMinus evalDG(Jdagger.I22);
 $\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\\ {0}{}{\mathrm{e1}}& {0}{}{\mathrm{e1}}\end{array}\right]$ (2.18)

Now define two elements of $\mathrm{𝕁}\left(\mathit{2}\mathit{,}\mathrm{ℚs}\right)$ and calculate their Jordan product.

 Q > A := evalDG(M[8] + M[10]);
 ${A}{:=}\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& {\mathrm{e7}}\\ {\mathrm{e7}}& {\mathrm{e1}}\end{array}\right]$ (2.19)
 Q > B := evalDG(M[1] + M[4]);
 ${B}{:=}\left[\begin{array}{cc}{\mathrm{e1}}& {\mathrm{e3}}\\ {\mathrm{e3}}& {0}{}{\mathrm{e1}}\end{array}\right]$ (2.20)
 Q > JordanProduct(A, B);
 $\left[\begin{array}{cc}{0}{}{\mathrm{e1}}& \frac{{1}}{{2}}{}{\mathrm{e3}}{+}\frac{{1}}{{2}}{}{\mathrm{e7}}\\ \frac{{1}}{{2}}{}{\mathrm{e3}}{+}\frac{{1}}{{2}}{}{\mathrm{e7}}& {0}{}{\mathrm{e1}}\end{array}\right]$ (2.21)


text:
System of Equations Solution Method

Supposing I had two equations: $\alpha/\beta=\mu$ and $\alpha/\beta^{2}=\sigma^{2}$, where $\alpha$, $\beta$, $\sigma$ and $\mu \in \mathbb R$. How can I solve for $\alpha$ and $\beta$?

Furthermore, is there a general method for this kind of system?

-
Divide. We get $\beta=\frac{\sigma^2}{\mu}$. Square the first expression and divide. We get $\alpha=\frac{\sigma^2}{\mu^2}$. –  André Nicolas Oct 9 '12 at 14:24

Solving one equation for $\alpha$ gives p. e. $\alpha = \beta\mu$. Plug this into the other equation, giving $\beta\mu/\beta^2 = \sigma^2 \iff \beta = \mu/\sigma^2$. So $\alpha = \mu^2/\sigma^2$.


text:
# 8.1 Mining and mineral processing  (Page 7/7)

 Page 7 / 7

## Phosphates

Rock phosphate [Ca ${}_{10}$ (PO ${}_{4}$ ) ${}_{6}$ F ${}_{2}$ ], mined from open pit mines at Phalaborwa, is an important raw material in the production of fertilisers. The following two reactions are used to transform rock phosphate into water soluble phosphates:

• Ca ${}_{10}$ (PO ${}_{4}$ ) ${}_{6}$ F ${}_{2}$ + 7X + 3H ${}_{2}$ O $\to$ 3Ca(H ${}_{2}$ PO ${}_{4}$ ) ${}_{2}$ H ${}_{2}$ O + 2HF + 7CaSO ${}_{4}$
• Ca ${}_{10}$ (PO ${}_{4}$ ) ${}_{6}$ F ${}_{2}$ + 14Y + 10H ${}_{2}$ O $\to$ 10Ca(H ${}_{2}$ PO ${}_{4}$ ) ${}_{2}$ H ${}_{2}$ O + 2HF
1. Identify the acids represented by X and Y.
2. Despite similar molecular formulae, the products Ca(H ${}_{2}$ PO ${}_{4}$ ) ${}_{2}$ formed in the two reactions have different common names. Write down the names for each of these products in reactions A and B.
3. Refer to the products in reactions A and B and write down TWO advantages of reaction B over reaction A.
4. Why is rock phosphate unsuitable as fertiliser?

(DoE Exemplar Paper, Grade 11, 2007)

## Case study : controversy on the wild coast - titanium mining

Read the extract below, which has been adapted from an article that appeared in the Mail and Guardian on 4th May 2007, and then answer the questions that follow.

A potentially violent backlash looms in Pondoland over efforts by an Australian company to persuade villagers to back controversial plans to mine an environmentally sensitive strip of the Wild Coast. The mining will take place in the Xolobeni dunes, south of Port Edward. The application has outraged environmental groups, largely because the proposed mining areas form part of the Pondoland centre of endemism, which has more species than the United Kingdom, some of which are endemic and facing extinction.

Exploratory drilling revealed Xolobeni has the world's 10th largest titanium deposit, worth about R11 billion. The amount of money that will be spent over the mine's 22 years, including a smelter, is estimated at R1.4 billion. The Australian mining company predicts that 570 direct jobs will be created.

But at least two communities fiercely oppose the mining plans. Some opponents are former miners who fear Gauteng's mine dumps and compounds will be replicated on the Wild Coast. Others are employees of failed ecotourism ventures, who blame the mining company for their situation. Many are suspicious of outsiders. The villagers have also complained that some of the structures within the mining company are controlled by business leaders with political connections, who are in it for their own gain. Intimidation of people who oppose the mining has also been alleged. Headman Mandoda Ndovela was shot dead after his outspoken criticism of the mining.

Mzamo Dlamini, a youth living in one of the villages that will be affected by the mining, said 10% of the Amadiba 'who were promised riches by the mining company' support mining. 'The rest say if people bring those machines, we will fight.'

1. Explain what the following words means:
1. endemic
2. smelter
3. ecotourism
2. What kinds of 'riches' do you think the Amadiba people have been promised by the mining company?
3. In two columns, list the potential advantages and disadvantages of mining in this area.
4. Imagine that you were one of the villagers in this area. Write down three questions that you would want the mining company to answer before you made a decision about whether to oppose the mining or not. Share your ideas with the rest of the class.
5. Imagine that you are an environmentalist. What would your main concerns be about the proposed mining project? Again share your answers with the rest of the class.

How do hydrogen and chlorine atoms bond covalently in a molecule of hydrogen chloride?
what is the parallel circuit
how many moles of H2O can be formed if 12,5 moi CH4 reacts with sufficient NH3 and O2
what atoms form covalent bond
The combination of two non metals
Ndumiso
combined two non-metals 😊
Trudy
what is a mole
a mole is a measure of large quantities of small entities such as atoms, molecules etc
DINEO
what does STP stand for?
Deeco
STP stands for Standard Temperature and Pressure.
Dolly
A car drives straight off the edge of a cliff that is 54m high. The police at the scene of the accident observed that the point of the impact is 130m from the base of the clif. Calculate the initial velocity of the car when it went ovet the clif.
wat happens to current if resistors are in parallel connection
more current flows from the source that would flow for any of them individually,so the total total resistance is lower
Theoo
In the parallel circuit the current is divided among the resistors
Ndumiso
what are the isotopes
are different types of the same elemant but with different mass or atomic no.
Ayanda
i sotopes are different type of element with same atomic number but different mass number
Tareskay
yes i forgot some details😁
Ayanda
next quetion plz
Tareskay
why don't we insert the negative sign for 5 × 10 - 9 when substituting
Why does an enclosed gas exert pressure on the walls of a container
State the gay lussacs law
what is the coefficient of Na in order to balanced the equation?_Na + MgCl2=2NaCl+Mg?
2
Okuhle
2
Tlotlisang
2
Praise
2
Anna
2
Anelehr
2
Liyema
the combining power of an element, especially as measured by the number of hydrogen atoms it can displace or combine with.
Thanks
REJOYCE
Dealon
what are orbitals
examples of atoms whose Valence energy levels are not full and more to bond and become more stable
Sphe
What is a valency
REJOYCE
What a lone pairs...
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Valency is the number of electrons than an atom must gain, lose or share to achieve noble gas configuration.
Mcebisi
what is vacuum
Vacuum, space in which there is no matter or in which the pressure is so low that any particles in the space do not affect any processes being carried on there. It is a condition well below normal atmospheric pressure and is measured in units of pressure (the pascal).
Khayalethu


text:
# Finding pair of sum in sorted array

In a sorted array, I am trying to find pairs that sum up to a certain value. I was wondering if anyone could help me improve my code in performance or memory. I believe my code is O(n). If there are other method for finding pairs the sum up to a certain value that is much more efficient, I am open to these methods!

static int[] intArray = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};

public static ArrayList<String> pairSum(int[] j, int sum) {
int len = j.length;
int pointer1;
int pointer2;
ArrayList<String> storage = new ArrayList<String>();

for (int i = 0; i < len; i++) {
pointer1 = j[i];
for (int k = len - 1; k >= 0; k--) {
pointer2 = j[k];
if (pointer1 + pointer2 == sum && pointer1 != pointer2) {
i++;
String pair = Integer.toString(pointer1) + " and " + Integer.toString(pointer2) ;
}
}
}
return storage;
}

If I ran this code, say pairSum(intArray,16), I would get:

[0 and 16, 2 and 14, 4 and 12, 6 and 10, 9 and 7, 11 and 5, 13 and 3, 15 and 1]

If I remove the i++, the output would be:

[0 and 16, 1 and 15, 2 and 14, 3 and 13, 4 and 12, 5 and 11, 6 and 10, 7 and 9, 9 and 7, 10 and 6, 11 and 5, 12 and 4, 13 and 3, 14 and 2, 15 and 1, 16 and 0]
-
The i++; produces the wrong result. I'd recommend you remove it from your question, since the code under review out to run and produce correct results. Why is it there? –  Wayne Conrad Jan 10 at 13:22

EDIT: Please see update at end because there is an improved solution that is inspired by Mike Chamberlain's answer...

There are two solutions which come to mind for your problem, but first, let's look through your code:

• you have the variables pointer1 and pointer2. These are very descriptive names, which is great, but unfortunately they are not pointers. They are the values at the i and k positions in the array. In other words, i and k are the actual pointers. I would call pointer1 and pointer2 something like val1 and val2. But, in fact, I would drop the pointers entirely, and just use an index in to the data (why copy the data when you don't need to....? )
• In most languages the variable name i is 'reserved' for indexed looping through data, just like you have for (int i = 0; i < len; i++) {...}. It is standard to use j and k as the variable names for any inner/nested loops you need. i is the top level, j is the first nested level, k is the subsequent level, and, if you need more than that, you should be calling a function/method (Never use l as a variable name, it is too easy to confuse with 1). You are breaking this convention by using j as an array, and going immediately to k inside your loop. My suggestion is to rename j to be data, and to rename k to be j.
• as a general rule, when you have a loop constrained by an index, like you have for (int i = 0; i < len; i++) { ..., you should never modify the index inside the loop. I cannot think of a reason why it should ever be necessary, and, certainly, it should never be conditionally modified (inside an if-statement inside another loop). This will lead to bugs. In fact, the simple fact that you have that inner i++ tells me you do not understand your own code's behaviour.
• Your method should return the interface type List<String> instead of the concrete class ArrayList<String> since there is no reason for outsiders to know the physical implementation.
• storage is not a great name for a variable.... everything could be called storage since all variables store things... I would choose something like: results.
• Your inner if condition makes sure that the sums are a match, but also checks pointer1 != pointer2. This second check is a problem, because it should not be testing the values, but the indexes.... For example what if the data is [4,4] and the target sum is 8?
• To get your result pair as a String you do: String pair = Integer.toString(pointer1) + " and " + Integer.toString(pointer2).... this is serious overkill, try this instead: String pair = pointer1 + " and " + pointer2. In Java the + String operator will implicitly convert the integers to a String... no need to do it yourself.

So, you have some style problems, and some concerns about your inner workings.

Now, let's talk about your algorithm..... It loops through each value. For each value, it then compares it against every other value (except itself). So, you loop n times, and, for each n you do m checks. Your complexity is O(n x m), but, in reality, n and m are the same, so your complexity is O(n2)

Now, about that i++ inside the k loop.... the reason you need it is because your k loop is starting from the wrong place. your algorithm is wrong. Let me explain.

If you have the data [ 0, 1, 2, 3, 4, 5 ] , you are using your i index to loop through all the data. Your first data will be 0. You sum this with 5 (Note, you have summed 0 + 5 now), and check the result, then with 4, and so on. When you are done, you move i to the value 1, and you keep moving i until you get to the value 5 for i. You then compare **this value 5 to all the other values, including 0. This is *the second time you sum the items 0 + 5 (but it is now 5 + 0) *.

You need to change the algorithm to only scan those values that have not yet been scanned. The easiest way to do this is to modify your inner/nested loop. Consider your code:

for (int i = 0; i < len; i++) {
...
for (int k = len - 1; k >= 0; k--) {
...

Now, we change the inner loop as follows:

for (int i = 0; i < len; i++) {
....
for (int j = i + 1; j < len; j++) {

(note, I have renamed k to be j). We start our j index at i + 1, because we don't need to sum values before i since that sum was done already when i was smaller.....

Now, our loops only calculate the sum value once for each pair of values.

Unfortunately, our complexity is still O(n2) because even though our inner loop is only testing half (on average) the values, we have not actually changed the way the algorithm scales (if you double the input data, you with quadruple the execution time).

Still, even with this algorithm change, and some variable renaming, and some bug fixing, your code will look like:

public static List<String> pairSum(int[] data, int sum) {
int len = data.length;
int val1;
int val2;
List<String> results = new ArrayList<String>();

for (int i = 0; i < len; i++) {
val1 = data[i];
for (int j = i + 1; j < len; j++) {
val2 = data[j];
if (val1 + val2 == sum) {
String pair = val1 + " and " + val2;
}
}
}
return results;
}

This is OK, but my preference would be to strip all the val1 variables entirely, and you can simplify some other things too. Consider the following more 'compact' code:

public static List<String> pairSum(int[] data, int sum) {
List<String> results = new ArrayList<String>();

for (int i = 0; i < data.length; i++) {
for (int j = i + 1; j < data.length; j++) {
if (data[i] + data[j] == sum) {
results.add(data[i] + " and " + data[j]);
}
}
}

return results;
}

That is what I would recommend for a simple, easy-to-read option that is O(n2). For small data sets (say less than 50 or so), this solution will be great.

But, if you have more data, a better solution would be more complicated, but faster.

In your example code, you have an input test array that is all positive, sorted, and the values are unique. Let's assume that all the data will be this way (we can modify this assumption later).

Consider the algorithm that does:

take all values that are less than half the target sum.
find a value that is the difference

It would look like:

public static List<String> pairSumFast(int[] data, int sum) {
List<String> results = new ArrayList<String>();

int limit = sum / 2;
for (int i = 0; i < data.length && data[i] <= limit; i++) {
int match = Arrays.binarySearch(data, i + 1, data.length, sum - data[i]);
if (match >= 0) {
results.add(data[i] + " and " + data[match]);
}
}

return results;
}

This has the complexity of O( nlog(n) ) for the loop, and the binary-search respectively.

If your data set is large, and the values are unique, but the data is unsorted, then I would recommend the change:

public static List<String> pairSumFast(int[] indata, int sum) {
List<String> results = new ArrayList<String>();

int[] data = Arrays.copyOf(indata, indata.length);
Arrays.sort(data);

int limit = sum / 2;
for (int i = 0; i < data.length && data[i] <= limit; i++) {
int match = Arrays.binarySearch(data, i + 1, data.length, sum - data[i]);
if (match >= 0) {
results.add(data[i] + " and " + data[match]);
}
}

return results;
}

This is O(n log(n) ) for the complexity still... (it scales this way, even through it does more computation....)

Finally, if the data is not unique, then you will get duplicate output values. To accommodate that, I would add an inner loop to the match....:

public static List<String> pairSumFast(int[] data, int sum) {
List<String> results = new ArrayList<String>();

int limit = sum / 2;
for (int i = 0; i < data.length && data[i] <= limit; i++) {
int match = Arrays.binarySearch(data, i + 1, data.length, sum - data[i]);
if (match >= 0) {
String pair = data[i] + " and " + data[match];
// binary search will return the index of one match, look around to find others.
int more = match - 1;
while (more >=0 && data[more] == data[match]) {
more--;
}
more = match + 1;
while (more < data.length && data[more] == data[match]) {
more++;
}
}
}

return results;
}

## EDIT: A better solution...

Mike Chamberlain suggests using a Map to preserve the values you have seen in order to match the pairs in a single O(n) iteration of the list.

This single-iteration made me realize that there is a much simpler solution to this problem... iterate from both sides of the data.... consider the following:

public static List<String> pairSumFastest(int[] data, int sum) {
ArrayList<String> results = new ArrayList<String>();

int low = 0;
int high = data.length - 1;
while (low < high) {
while (low < high && data[low] + data[high] < sum) {
low++;
}
while (low < high && data[low] + data[high] > sum) {
high--;
}
if (low < high) {
results.add( data[low] + " and " + data[high]);
low++;
high--;
}
}

return results;
}

This has true O(n) complexity, and has no memory overhead either.

-
Much more comprehensive than my answer. I hope the OP gives this one the checkmark. –  Wayne Conrad Jan 10 at 14:08
Yes thank you so much for taking the time to write this ! I really appreciate it! I learn a lot!! @WayneConrad thank you for your answer as well! –  Liondancer Jan 10 at 14:10
Thanks! I was 2-thirds through the answer when I saw your's pop up .... and I figured I covered things you had not, so I posted.... That's why there's a lot of duplication between our answers, sorry ;-) –  rolfl Jan 10 at 14:10
@rolfi, One of the things I like about codereview is that the fastest gun doesn't always win. I've been out of the Java game for a long time, so I'm glad someone who knows it well was able to cover what I missed. –  Wayne Conrad Jan 10 at 14:16
A few things I saw, the results declaration should be List<String>, int more can be directly set and there's a stray space at more --. –  Bobby Jan 10 at 14:30

# Code improvements

Before addressing efficiency, let's see if there are any improvements that could be made to the code.

I suspect some of the names could be improved.

• pointer1 -> value1
• pointer2 -> value2
• j -> values
• storage -> value_pairs

I would consider getting rid of this temporary:

int len = j.length;

as it's not carrying its weight. Using j.length inline is not burdensome, and the compiler/runtime are likely to produce adequately fast code without it.

Lastly, I wonder about the decision to return the pairs as strings. This depends upon how the function is being used, but I would often rather return the pairs of numbers, and let the caller worry about formatting. This decouples the function from I/O.

# A bug?

This term in the if:

pointer1 != pointer2

Might be a bug. If the list is guaranteed to contain no duplicates, it will work fine (but might be more communicative as i != k). If, however, the list may contain duplicates, and it is the intent of the code to return the resulting duplicate pairs, then this must be i != k.

# Efficiency.

In order to be O(n), the algorithm must visit each element a constant number of times. We typically think of a O(n) algorithm as one that visits each element once. In this algorithm, both the outer and inner loop visit each element. That makes this an O(n^2) algorithm.

A google search for "efficiently finding pairs that add up to a sum" finds, not surprisingly, several Stack Overflow questions. Here's one of them that may be useful to you: http://stackoverflow.com/questions/1494130/design-an-algorithm-to-find-all-pairs-of-integers-within-an-array-which-sum-to-a

-
Thanks for this analysis! My code is O(n^2) didnt realize this! –  Liondancer Jan 10 at 13:54
Make that i != k, as j currently refers to the array. –  nmclean Jan 10 at 13:55
@mmclean Thanks. Fixed! –  Wayne Conrad Jan 10 at 14:15

I believe you can do this in O(n).

The algorithm below loops through the list only once. It keeps track of the numbers it's seen using a HashSet. For each number, it works out its required complement. If it's seen the complement already then it's found a match. If not, it adds the number to the set - we may see its complement later.

This actually works for unordered as well as ordered data, with the same performance characteristics. It also works for non-unique values (though we get n-1 repeated pairs of x for n repeated values of x - possibly not what you want, but a final stage could be to remove the duplicates if you so wished).

Note that the order of the pairs returned, and indeed the order of two numbers in a pair, may not be the same as your original algorithm. It wasn't specified if this was important, but I'm guessing not.

This reduced complexity comes at the expense of memory, as it must keep track of the numbers its seen in a separate data structure.

One thing I would point out is that the algorithm is more flexible if it returns raw data, rather than formatted strings. Pretty-printing can be done by the caller.

I've done this in C# because that's what I'm familiar with. I guess it should be trivial to convert to Java.

using System;
using System.Collections.Generic;
using System.Linq;
using NUnit.Framework;

namespace FindSummingPairs
{
public static class Algorithm
{
public static Tuple<int, int>[] FindSummingPairs(int[] input, int sumTarget)
{
var result = new List<Tuple<int, int>>();

// holds the numbers we have seen so far
var seenValues = new HashSet<int>();

// loop through our input
foreach(var number in input)
{
// work out the required value to make our target sum
var required = sumTarget - number;

// have we seen our required value in the list already?
if (seenValues.Contains(required))
{
// score!
}
else
{
}
}

return result.ToArray();
}
}

[TestFixture]
internal class Tests
{
[Test]
public void EmptyInput()
{
var input = new int[0];
var result = Algorithm.FindSummingPairs(input, 0);
Assert.False(result.Any());
}

[Test]
public void SingleNumberAsInput()
{
var input = new[] {1};
var result = Algorithm.FindSummingPairs(input, 2);
Assert.IsTrue(result.Length == 0);
}

[Test]
public void TwoNumbersThatDontSum()
{
var input = new[] {1, 2};
var result = Algorithm.FindSummingPairs(input, 2);
Assert.IsTrue(result.Length == 0);
}

[Test]
public void TwoNumbersThatDoSum()
{
var input = new[] {1, 2};
var result = Algorithm.FindSummingPairs(input, 3);
var expected = new[] { new Tuple<int, int>(1, 2) };
Assert.IsTrue(AreEquivalent(expected, result));
}

[Test]
public void LotsOfNumbers()
{
var input = new[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
var result = Algorithm.FindSummingPairs(input, 10);
var expected = new[]
{
new Tuple<int, int>(1, 9),
new Tuple<int, int>(2, 8),
new Tuple<int, int>(3, 7),
new Tuple<int, int>(4, 6)
};
Assert.IsTrue(AreEquivalent(expected, result));
}

[Test]
public void LotsOfNumbersThatDontSum()
{
var input = new[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
var result = Algorithm.FindSummingPairs(input, 20);
Assert.IsTrue(result.Length == 0);
}

[Test]
public void TestCaseFromQuestion()
{
var input = new[] {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};

var result = Algorithm.FindSummingPairs(input, 16);
var expected = new[]
{
new Tuple<int, int>(0, 16),
new Tuple<int, int>(1, 15),
new Tuple<int, int>(2, 14),
new Tuple<int, int>(3, 13),
new Tuple<int, int>(4, 12),
new Tuple<int, int>(5, 11),
new Tuple<int, int>(6, 10),
new Tuple<int, int>(7, 9)
};
Assert.IsTrue(AreEquivalent(expected, result));
}

// calculates whether two pair arrays are equivalent, without respect to the order of the pairs
private bool AreEquivalent(Tuple<int, int>[] e1, Tuple<int, int>[] e2)
{
if (e1.Length != e2.Length)
return false;

foreach(var pair1 in e1)
{
foreach (var pair2 in e2)
{
if (AreEquivalent(pair1, pair2))
return true;
}
}
return false;
}

// calculates whether 2 pairs are equivalent, without respect to the order of the numbers
private bool AreEquivalent(Tuple<int, int> pair1, Tuple<int, int> pair2)
{
return (pair1.Item1 == pair2.Item1 && pair1.Item2 == pair2.Item2)
|| (pair1.Item2 == pair2.Item1 && pair1.Item2 == pair2.Item1);
}
}
}
-


text:
# Non-integrable differential equation and non-holonomic contraints

From the constraint $$v=a\dot{\phi}$$ of a rolling disk over a plane, where $$a$$ is the radius of the disk we can derive these two equations:

we have two differential equations of constraint:

$$dx=asin\theta d\phi$$

$$dy=-acos\theta d\phi$$

Can you rigorously explain to me why can't I integrate these functions? Is it their non-integrability that makes these constraints non-holonomic?

• Note that this question is Derivation 4 in Chapter 1 of Goldstein's Classical Mechanics, from which your example, figure, and notation are taken. – Michael Seifert May 28 '19 at 20:02

HINT: Rewrite the first constraint as $$f \left[ dx +(- a \sin \theta) d\phi + (0) d\theta \right] = 0.$$ where $$f(x, \theta, \phi)$$ is some unknown integrating function. We want to know whether this can be written as $$dg = \frac{\partial g}{\partial x} dx + \frac{\partial g}{\partial \phi} d\phi+ \frac{\partial g}{\partial \theta} d\theta$$ Assuming that $$g$$ is a "nicely-behaved" function of the coordinates, its mixed partial derivatives are independent of the order of differentiation. Using this fact, can you show that this implies $$f = 0$$?
To answer your second question: there are different definitions of "non-holonomic" constraints used by different authors. If you're following Goldstein, any constraint that cannot be written in the form $$f(x_i) = 0$$ is non-holonomic; but there are also non-holonomic constraints that can only be written as equalities among the higher derivatives, or as inequalities.


text:
# SuanShu, a Java numerical and statistical library

com.numericalmethod.suanshu.stats.test.distribution.kolmogorov

## Class KolmogorovSmirnov

• Direct Known Subclasses:
KolmogorovSmirnov1Sample, KolmogorovSmirnov2Samples

public abstract class KolmogorovSmirnov
extends HypothesisTest
The Kolmogorov-Smirnov test (KS test) compares a sample with a reference probability distribution (one-sample KS test), or to compare two samples (two-sample KS test). The Kolmogorov-Smirnov statistic quantifies the distance between the empirical distribution function of the sample and the cumulative distribution function of the reference distribution (one-sample KS test), or between the empirical distribution functions of two samples (two-sample KS test). The null distribution of this statistic is calculated under the null hypothesis that the sample is drawn from the reference distribution (in the one-sample case), or that the samples are drawn from the same distribution (in the two-sample case). In each case, the distributions considered under the null hypothesis are continuous distributions but are otherwise unrestricted. The test-statistics is either Dn, Dn+, or Dn-, depending on the side to compute.

By R convention, when the exact p-value is not available, we use the limiting distribution. The exact p-value is not available in the case of ties for the one-sample case.

The R equivalent function is ks.test.

• ### Nested Class Summary

Nested Classes
Modifier and Type Class and Description
static class  KolmogorovSmirnov.Side
the available types of the Kolmogorov-Smirnov statistic
static class  KolmogorovSmirnov.Type
the available types of the Kolmogorov-Smirnov tests
• ### Method Summary

All Methods
Modifier and Type Method and Description
String getAlternativeHypothesis()
Get the description of the alternative hypothesis.
String getNullHypothesis()
Get a description of the null hypothesis.
• ### Methods inherited from class com.numericalmethod.suanshu.stats.test.HypothesisTest

isNullRejected, nGroups, nObs, oneSidedPvalue, pValue, statistics
• ### Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Method Detail

• #### getNullHypothesis

public String getNullHypothesis()
Description copied from class: HypothesisTest
Get a description of the null hypothesis.
Specified by:
getNullHypothesis in class HypothesisTest
Returns:
the null hypothesis description
Wikipedia: Null hypothesis
• #### getAlternativeHypothesis

public String getAlternativeHypothesis()
Description copied from class: HypothesisTest
Get the description of the alternative hypothesis.
Specified by:
getAlternativeHypothesis in class HypothesisTest
Returns:
the alternative hypothesis description


text:
# what the best way to calculate Rg gate driver for Mosfet

I need to drive a MOSFET IPW90R120C3 from Infineon here the Specification of MOSFET

VDS @ TJ=25°C 900 V
Rdson @ TJ=25°C   : 0.12ohm
Qg = 270nC

Specification of driver IR2110

Isource: 2A
Isink : 2A
VOUT 10 - 20V
ton/off (typ.) 120 & 94 ns
Driver supply : 15V

The switch frequency Fsw= 50KH

what the best way to calculate the resistance Rg to drive proprely the MOSFET ? best regards

• Probably include the datasheets of the MOSFETs. – Bradman175 Feb 20 '17 at 22:00
• Datasheet of Mosfet : infineon.com/dgdl/… – user3212448 Feb 20 '17 at 22:06
• I bet it is the maximal gate current. The MOSFET gate is capacitor like, therefore it will sink/source current $I_g=\dfrac{V_g}{R_g+R_{g_{internal}}}$ – Marko Buršič Feb 20 '17 at 22:19
• UPDATE: The maximal current of the gate driver on the picture. – Marko Buršič Feb 20 '17 at 22:24
• @ Marko Buršič, Yes it is the maximal gate current. I searched and read many Application notes from supplier , but did not find a specific answer or a genaralist formula to calculate the correct value for a Rg resistor for a MOSFET's. the right formula must includ the Gate Charge Qg, and the swithing time turn ON - Turn OFF, Or by using the Output voltage slop dVout/dt – user3212448 Feb 20 '17 at 22:29

"best way" comes downto what you are trying to achieve?
Switching is all about charge transfer. You need to transfer 270nC worth of charge in (or out) of the gate region to turn the device on (or off). It is also about charging up the relevant capacitance.

$Q_{gs} \Rightarrow Q_{gd} \Rightarrow Q_g$

The turn on of the device can be split onto two region's

1. Period 1. Time to charge the gate to the threshold. This is the delay and is influenced by the input capacitance $C_{iss}$ (or $C_{gd} + C_{gs}$)
2. Period 2. Time to saturate the device, This is the rise time & is governed by the remaining gate charge.

The actual turn on characteristics (and equally turn-off) is split into three regions, (charging to threshold, charging through the millar, charging to saturation) but without a detailed charge plot of the millar plateau it is more of a talking point

$\Delta V = (V_{end} - V_{start})(1 - \frac{1}{e^{\frac{t}{\tau}}})$

$t = RC ln(1 - \frac{\Delta V}{ V_{end} - V_{start}})$

Period 1

$V_{gg}$ is 10V = $V_{end}$

$V_{0}$ is 0V = $V_{start}$

$V_{th}$ is 3V = $\Delta V$

C = 6.8nF (from $C_{iss}$)

From this a 1st pass approximation of $R_g$ can be done based upon a required delay time

Period 2

$V_{gg}$ is 10V = $V_{end}$

$V_{th}$ is 3V = $V_{start}$

$V_{gg} -V_{th}$ is 7V = $\Delta V$

$Q_g$ is 270nC -> $C_g = Q_g/Vg = 27nF$

C = 20.2nF (from $C_g - C_{iss}$)

This will facilitate deriving a gate resistance for a given rise time. This value is of greater importance than delay time.

There are three other considerations with regards to gate drive

Power

$P_{drv} = Q_g f_{in} \Delta V_g$

The faster you want to switch, the more power it will take

Current

$\hat{I_{out}} = \frac{\Delta V_g}{R_{g\_min}}$

The lowest gate resistor is governed by the output current capability of your driver

Stability

$R_{g\_min} = 2 \sqrt(\frac{L_g}{Cg})$

To minimise creating a pierce oscillator, the damping factor associated with the L-C circuit must fulfil the damping condition

• the stability is something new to me , so i got curious , by inductance of the gate is you mean : lead mosfet package and PCB layout , right ?? – ElectronS Feb 24 '17 at 15:16
• yup. The "gate resistance" is all the resistance from the rail to the gate (driver, purpose resistor, traces, chip resistance) and the "gate lead inductance" is the stray inductance due to any leads in the package, traces etc. The number of gate-drive designs that have inadvertently created a pearce oscillator simply because of poor layout and a very low resistor is just silly..,. – JonRB Feb 24 '17 at 15:20
• i get the resistance part and it is straight forward , but the inductance part is a bit difficult to measure , right ? and what are the clues that the gate driver has created a Pearce oscillator ? is ringing on the gate a symptom ? – ElectronS Feb 24 '17 at 20:11

You should look on datasheet of the gate driver circuit, then look for max. current and $R_{ds_{ON}}$, look for internal gate resistance of the MOSFET, too. $$I_g=\dfrac{V_{driver}}{R_{ds_{ON}}+R_g+R_{g_{MOSFET}}}$$

The gate driver you have chosen is not a good fit. The Source/Sink Current is 2A, but if you calculate the Ig based on Turn On Time (90 nS) and required gate electric charge, which is 270 nC: Ig= dQ/dt= 270/90= 3A (at least this 3A is required, I would go for 5A)

The best design solution is explained by @JonRB Thank you for your complete response, but I have some concerns about it:

By using the datasheet parameters, for the first period of Ton (delay), I calculated Rg1=~29 ohms. For the second period, the rise time, you mentioned: Vgg is 10V = Vend Vth is 3V = Vstart Vgg−Vth is 7V = ΔV By substituting these values we get Ln (1-7/7) = Ln (0), which is not defined. If we say: Vgg is 10V = Vend Vth is 0V = Vstart Vgg−Vth is 7V = ΔV Then, we get Rg2=~ 1 ohm. However, this doesn't agree with what you said before: "This will facilitate deriving a gate resistance for a given rise time. This value is of greater importance than delay time." Please let me know where I make a mistake.


text:
# Rounding error?

Logic Level 3

Consider the following scenario.

A group of $$n$$ friends together gather an amount of money, contributing $$a_1, a_2, \ldots, a_n$$ of the total. Collectively, they manage to buy $$m$$ horses. They now want to distribute these horses according to the money they contribute. Normally, the $$i$$-th person should get $$m a_i$$ horses, but since horses cannot be divided, these numbers need to be rounded. The $$i$$-th person starts by getting $$\lfloor m a_i \rfloor$$ horses. At the end, some horses will remain; these will be given in order to the people with the highest fractional part of $$m a_i$$ until none is left. (In case of a tie, begin with the person with the lowest index.)

For example, consider $$a_1 = 0.5, a_2 = 0.25, a_3 = 0.25$$; that is, the first person contributes twice as much as each of the others. They manage to buy $$m = 5$$ horses. They are now distributed: the first person should get $$m a_1 = 2.5$$ horses, the second and third $$m a_2 = m a_3 = 1.25$$. We first ignore the fractions, so the first person gets 2, and the other two 1 each. Now, the highest fractional part comes from the first person, so the first person gets an additional horse; that's 3 for the first person and 1 for the other two.

With $$m = 6$$, the second person would get an additional horse (to 2), and with $$m = 7$$, the third person would also get one more (to 2 as well).

The above example shows that increasing $$m$$ wouldn't make someone to get less horses. Is this true? That is, for any fixed $$a_1, \ldots, a_n$$, is it true that if we raise $$m$$, nobody will lose horses?

×


text:
Free Version
Easy

# Coefficient in $(5x-2y)^7$

COMBIN-XIIW4G

What is the coefficient of $x^3y^4$ in the expansion of $(5x-2y)^7$?

A

$725$

B

$32$

C

$126$

D

$252$

E

$1024$