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# Matter vs. Antimatter shortly after Big Bang

I am not a physicist or astronomer in training at all, so this might be a dumb question with an obvious answer, but as I watched something the other day on matter & anti-matter, and the discussion was kind of put out there how the matter side won out even though the prevailing thought is that at the moment of the Big Bang the universe should have had equal amounts of matter and anti-matter. It then spoke of ideas that matter must have somehow won out since we obviously exist in a world of matter. After that I had the question why would matter need to win out for our universe to exist? Since we know a matter/anti-matter collision annihilates to pure energy, and we know $E=mc^2$, why isn't it possible that all original matter and anti-matter annihilated each other leaving all pure energy, which in turn and in time could form into matter?

• Duplicate of physics.stackexchange.com/q/96535 – user42733 Apr 5 '14 at 16:27
• @ParthVader this is certainly related, but I do not think it is a duplicate of that. The last sentence of this question is what distinguishes it from the other for me. – Jim Apr 5 '14 at 16:36
• @Jim The answer would more or less be the same. But yeah, it is a bit different. – user42733 Apr 5 '14 at 16:38
• Perhaps you should have gone into physics. That is an area called Baryogenesis and your question is at the crux of it. en.wikipedia.org/wiki/Baryogenesis I should note that the level of cosmic background radiation (CBR) that is observed is used to quantify the the amount of matter anti-matter collisions. If everything collapsed, I think the CBR would be higher than we measure it. – user6972 Apr 5 '14 at 17:00

The key bit of your question is:

why isn't it possible all original matter and anti-matter annihilated each other leaving all pure energy, which in turn and in time could form into matter?

since other parts of it are covered by the question Parth Vader linked. Matter and anti-matter don't annihilate to produce pure energy, they annihilate to produce photons. The photons can in turn pair produce to turn back into matter, but pair production always creates equal amounts of matter and anti-matter.

So if the amount of matter and anti-matter was originally equal then just turning into photons and back would still result in equal amounts of matter and anti-matter. To produce an excess of matter (as we see around us) requires some additional mechanism, and it is not know what that mechanism is. There are processes known that can produce an excess of matter over anti-matter (or vice versa) but these produce too small an excess to account for all the matter there is in the universe. There must be some further, currently undiscovered mechanism.

• Well why is it then that an idea for future spaceship travel has it been proposed of using a matter/anti-matter propulsion system? I've seen and read that idea a few times and it's thought it could get the ship up to 20% the speed of light. Hearing that I obviously get the same picture in my mind as a rocket of today with flames and everything shooting out the back. But if they just turn into photons then how could that power a ship? Or why is it a must to keep anti-matter particles separated from touching the sides where they're contained at places like Fermilab? – user3422157 Apr 18 '14 at 0:37
• @user3422157: I can't comment unless you post details of the drive design. Ideas for interstellar drives are ten a penny. You might want to post this as a new question, but bear in mind that most working physicists would probably consider interstellar drive design as recreation rather than science. – John Rennie Apr 18 '14 at 6:04
• @user3422157 A lot of the terms used on sci-fi tv shows are used like buzzwords thrown around generously, and may have weak or no relationship with the actual term. I'm guessing what you heard was that. The popular assumption is that matter and anti-matter create energy, but nobody has so much as considered using anti-matter as an energy source, much less as a spaceship drive. – Neil Mar 17 '16 at 15:29

Leptons annihilate to photons. Hadrons (quarks, 99.97 mass-% of matter) emit about 50% of ${mc^2}$ annihilation energy as neutrinos. If the Big Bang plus cosmic inflation had exactly obeyed all current conservation laws, everything would have exactly cancelled to photons and neutrinos. Pair formation must be antisymmetric re conservation laws, offering no solution. Neutrinos and antineutrinos would have annihilated to photons, but slowly for the very small reaction cross-section. Observation suggests theory is wrong (e.g., the existence of paperweights).

http://en.wikipedia.org/wiki/Baryogenesis
Theory hates empirical falsification. Parameterize!

Let's write apostatic theory based upon observation. A good idea need only be testable. It is believable afterward.

Massless boson photons detect no vacuum refraction, dispersion, dissipation, dichroism, or gyrotropy. Postulate this is exactly true for fermionic matter (quarks, hence hadrons): parity violations, symmetry breakings, chiral anomalies, Chern-Simons repair of Einstein-Hilbert action. Vacuum is observed to be trace chiral anisotropic toward hadrons. This fulfills the Sakharov conditions.

Vacuum trace chiral anisotropy selective to fermionic matter alters Noetherian coupling of exact vacuum isotropy to angular momentum conservation. It leaks for matter as MoND's Milgrom acceleration. Dark matter curve-fits the Tully-Fisher relation. This is why dark matter does not exist, by measurement,

Planck satellite: 68.3% dark energy, 26.8% dark matter, 4.9% baryonic matter
http://arxiv.org/abs/1306.5534
Dark matter inside Saturn's orbit is less than ${1.7×10^{-10} M_{solar}}$.

Opposite shoes embed within chiral vacuum (mount a left foot) with different energies. They locally vacuum free fall along non-identical minimum action trajectories, exhibiting Equivalence Principle (EP) violation. However, chirality is an emergent property. A screw is not left-handed or right-handed for distances less than its thread pitch length. The smallest possible opposite shoes must be at least four atoms each (number of physical dimensions plus one). We need a supremely sensitive vacuum free fall trajectory test (in existing equipment - we're gonna do it, not talk fluff about it.)

Physics cannot fabricate few-atomed shoes. Chemistry can and trivially does. Crystallography's opposite shoes are visually and chemically identical, single crystal test masses in enantiomorphic space groups: ${P3_121}$ versus ${P3_221}$ alpha-quartz or ${P3_1}$ versus ${P3_2}$ gamma-glycine. Academia can perform geometric Eötvös experiments at will,

http://www.npl.washington.edu/eotwash/epwhat
The best on the planet! Now, assay spacetime geometry with test mass geometry.

40 grams net of alpha-quartz as two sets of four single crystal test masses compare ${6.68×10^{22}}$ pairs of opposite shoes (pairs of 9-atom enantiomorphic unit cells, the test mass array cube's opposite vertical sides, also covering up to four compactified dimensions if present). If there is no net signal output, we need different theory. If there is a net signal output, we can write better theory.

Cartography showed 2000 years of Euclid was not good enough, but could be repaired, hence Bolyai. Relativity and quantum mechanics showed 200 years of Newton was not good enough, but could be repaired. Yang and Lee showed 30 years of particle theory was not good enough, but could be repaired. A 90-day geometric Eötvös experiment can show quantum gravitation, SUSY, and dark matter are not good enough - but can be repaired. Somebody should look. The worst it can do is succeed.


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# Is the optimizer of a strongly-convex cost function bounded?

Let f(x) be strongly-convex. Can its minimizer be unbounded? I suspect not. Can we obtain a bound on it in relation to the strong-convexity constant?

I believe an equivalent formulation of this question is: Is the minimizer, $x^\star$, of a strongly-convex function, $f(x)$, in a compact or bounded subset of $\mathbb{R}^n$?

Thank you!

-

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Interstitial and intracavity <tex-math>${}^{252}{\rm Cf}$</tex-math> sources have been used to treat a number of tumor types with encouraging results. In particular these tumors include a variety of cervical, head-and-neck, and oral-cavity cancers and possible malignant gliomas. As a neutron source, <tex-math>${}^{252}{\rm Cf}$</tex-math> offers certain theoretical advantages over photon therapy (i.e., in treating tumors with significant hypoxic or necrotic components). With the recent availability of <tex-math>${}^{10}{\rm B}\text{-}{\rm la}\text{-beled}$</tex-math> tumor-seeking compounds, the usefulness of <tex-math>${}^{252}{\rm Cf}$</tex-math> may be further improved by augmenting the <tex-math>${}^{252}{\rm Cf}$</tex-math> dose to the tumor with an additional dose due to the fission (following thermal neutron capture) of10 B located in the tumor itself. While the high mean neutron energy permits <tex-math>${}^{252}{\rm Cf}$</tex-math> to deliver a high-LET, low-OER dose to the tumor on a macroscopic scale, thermalization of neutrons followed by10 B capture may augment this dose at the cellular level if adequate loading of tumor cells with10 B is possible. This paper presents results of a Monte Carlo simulation study investigating the dosimetric characteristics of linear <tex-math>${}^{252}{\rm Cf}$</tex-math> sources both with and without the quantitative increase in tumor dose possible with the addition of10 B. Results are displayed in the form of "along and away" tables and dose profiles in a water phantom. Comparisons of Monte Carlo results with experimental and analytical dosimetry data available in the literature are also presented.

This content is only available as a PDF.


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• We measure the energy emitted by extensive air showers in the form of radio emission in the frequency range from 30 to 80 MHz. Exploiting the accurate energy scale of the Pierre Auger Observatory, we obtain a radiation energy of 15.8 \pm 0.7 (stat) \pm 6.7 (sys) MeV for cosmic rays with an energy of 1 EeV arriving perpendicularly to a geomagnetic field of 0.24 G, scaling quadratically with the cosmic-ray energy. A comparison with predictions from state-of-the-art first-principle calculations shows agreement with our measurement. The radiation energy provides direct access to the calorimetric energy in the electromagnetic cascade of extensive air showers. Comparison with our result thus allows the direct calibration of any cosmic-ray radio detector against the well-established energy scale of the Pierre Auger Observatory.
• Neutrinos in the cosmic ray flux with energies near 1 EeV and above are detectable with the Surface Detector array of the Pierre Auger Observatory. We report here on searches through Auger data from 1 January 2004 until 20 June 2013. No neutrino candidates were found, yielding a limit to the diffuse flux of ultra-high energy neutrinos that challenges the Waxman-Bahcall bound predictions. Neutrino identification is attempted using the broad time-structure of the signals expected in the SD stations, and is efficiently done for neutrinos of all flavors interacting in the atmosphere at large zenith angles, as well as for "Earth-skimming" neutrino interactions in the case of tau neutrinos. In this paper the searches for downward-going neutrinos in the zenith angle bins $60^\circ-75^\circ$ and $75^\circ-90^\circ$ as well as for upward-going neutrinos, are combined to give a single limit. The $90\%$ C.L. single-flavor limit to the diffuse flux of ultra-high energy neutrinos with an $E^{-2}$ spectrum in the energy range $1.0 \times 10^{17}$ eV - $2.5 \times 10^{19}$ eV is $E_\nu^2 dN_\nu/dE_\nu < 6.4 \times 10^{-9}~ {\rm GeV~ cm^{-2}~ s^{-1}~ sr^{-1}}$.
• A measurement of the cosmic-ray spectrum for energies exceeding $4{\times}10^{18}$ eV is presented, which is based on the analysis of showers with zenith angles greater than $60^{\circ}$ detected with the Pierre Auger Observatory between 1 January 2004 and 31 December 2013. The measured spectrum confirms a flux suppression at the highest energies. Above $5.3{\times}10^{18}$ eV, the "ankle", the flux can be described by a power law $E^{-\gamma}$ with index $\gamma=2.70 \pm 0.02 \,\text{(stat)} \pm 0.1\,\text{(sys)}$ followed by a smooth suppression region. For the energy ($E_\text{s}$) at which the spectral flux has fallen to one-half of its extrapolated value in the absence of suppression, we find $E_\text{s}=(5.12\pm0.25\,\text{(stat)}^{+1.0}_{-1.2}\,\text{(sys)}){\times}10^{19}$ eV.
• Contributions of the Pierre Auger Collaboration to the 33rd International Cosmic Ray Conference, Rio de Janeiro, Brazil, July 2013
• ### Nonsupersymmetric gauge coupling unfication in [SU(6)]$^4\times$ Z$_4$ and proton stability(hep-ph/9901301)

Jan. 14, 1999 hep-ph
We systematically study the three family extension of the Pati-Salam gauge group with an anomaly-free single irreducible representation which contains the known quarks and leptons without mirror fermions. In the context of this model we implement the survival hypothesis, the modified horizontal survival hypothesis, and calculate the tree level masses for the gauge boson and fermion fields. We also use the extended survival hypothesis in order to calculate the mass scales using the renormalization group equation. The interacting Lagrangean with all the known and predicted gauge interactions is explicitly displayed. Finally the stability of the proton in this model is established.
• ### Non SUSY Unification in Left-Right Models(hep-ph/9812401)

Dec. 17, 1998 hep-ph
We explore in a model independent way the possibility of achieving the non supersymmetric gauge coupling unification within left-right symmetric models, with the minimal particle content at the left-right mass scale which could be as low as 1 TeV in a variety of models, and with a unification scale M in the range $10^5$ GeV $< M< 10^{17.7}$ GeV.
• ### Non SUSY and SUSY one--step Unification(hep-ph/9808357)

Aug. 18, 1998 hep-ph
We explore the possibility of achieving one--step unification of the standard model coupling constants within non supersymmetric and supersymmetric gauge models, which at low energies have only the standard particle content. The constraints are the experimental values of $\alpha_{em}$, $\alpha_s$ and $\sin^2\theta_W$ at $10^2 GeV$, and the lower bounds for FCNC and proton decay rates. The analysis is done in a model independent way.
• ### Symmetry Breaking in $[SU(6)]^3\times Z_3$(hep-ph/9708440)

Aug. 22, 1997 hep-ph
We analize the different ways for the spontaneous breaking of the gauge symmetry, for the $[SU(6)]^3\otimes Z_3$ family unification model. In particular we study the consequences of a previous selection for the vacuum expectation values of the Higgs fields, showing that such set predicts unwanted flavor changing neutral currents at the $m_Z=91 GeVs$ mass scale. A new set of vacuum expectation values which solves this problem is proposed.
• ### One Step Non SUSY Unification(hep-ph/9708441)

Aug. 22, 1997 hep-ph
We show that it is possible to achieve one step gauge coupling unification in a general class of non supersymmetric models which at low energies have only the standard particle content and extra Higgs fields doublets. The constraints are the experimental values of $\alpha_{em}$, $\alpha_s$ and $\sin^2\theta_W$ at $10^2 GeVs$, and the lower bounds for FCNC and proton decay rates. Specific example are pointed out.
• ### Model independent analysis of the simultaneous mixing of gauge bosons and mixing of fermions(hep-ph/9611442)

Nov. 26, 1996 hep-ph
We discuss the case of simultaneous mixing of gauge bosons and mixing of fermions in a model independent way and for a variety of extra-fermion representations. In this context we analyze a class of lepton family violating processes, namely Z --> e tau, Z --> mu tau, Z --> e mu, mu --> e e e, tau --> e e e, tau --> mu mu mu, tau --> e mu mu and tau --> mu e e in the presence of one extra neutral gauge boson, Z', with universal, non-universal or family changing couplings. We derive bounds on the combined effect of Z-Z' mixing and ordinary--exotic lepton mixing.
• ### Systematic Study of Horizontal Gauge Theories(hep-ph/9507467)

Nov. 13, 1996 hep-ph
We analyze all the possible continuous horizontal gauge groups G_H in relation with their possibility to explain m_b<<m_t. We assume that the only effective fermionic degrees of freedom correspond to the known fermions but allow the possibility of adding a right handed neutrino to each family. We assume that the Higgs fields which generate masses for these fermions, trough renormalizable Yukawa couplings, transform as an irreducible representation of SU(3)_c\otimesSU(2)_L\otimesU(1)_Y\otimesG_H. Under these assumptions we find two U(1)_H or U(1)_{H1}\otimesU(1)_{H2} models free of anomalies and able to guarantee that only the top has a renormalizable mass-generating Yukawa coupling.
• ### Is U(1)$_H$ a Good Family Symmetry?(hep-ph/9506439)

June 27, 1995 hep-ph
We analyze U(1)$_H$ as a horizontal symmetry and its possibilities to explain the known elementary-fermion masses. We find that only two candidates, in the context of SU(3)$_c\otimes$SU(2)$_L\otimes$U(1)$_Y\otimes$U(1)$_H$ nonsupersymmetric, are able to fit the experimental result m$_b<<$m$_t$.
• ### A [SU(6)]$^4$ FLAVOR MODEL WITHOUT MIRROR FERMIONS(hep-ph/9502266)

Feb. 9, 1995 hep-ph
We introduce a three family extension of the Pati-Salam model which is anomaly-free and contains in a single irreducible representation the known quarks and leptons without mirror fermions. Assuming that the breaking of the symmetry admits the implementation of the survival hypothesis, we calculate the mass scales using the renormalization group equation. Finally we show that the proton remains perturbatively stable.
• ### Signals of extra gauge bosons and exotic leptons in SU(6)$_{L}\otimes$U(1)$_{Y}$(hep-ph/9502220)

Feb. 2, 1995 hep-ph
We study some of the consequences of the SU(6)$_{L}\otimes$U(1)$_{Y}$ model of unification of electroweak interactions and families with a horizontal gauge group SU(2)$_{H}$, paying special attention to processes with flavor changing neutral currents. We compute at tree level the decays $K^{+}\longrightarrow \pi^{+}\mu^{+}e^{-}$, $K_L^0\longrightarrow \mu^{+}e^{-}$ and $\mu^{-} \longrightarrow e^{-} \bar{\nu_e} \nu_\mu$ from which we obtain lower bounds for the mass of the horizontal gauge boson associated with FCNC. Finally we obtain limits on the mixing between ordinary and exotic charged leptons.
• ### Mass Scales and Stability of the Proton in [SU(6)]$^{3}\times$Z$_{3}$(hep-ph/9306245)

June 8, 1993 hep-ph
We proof that the proton is stable in the left-right symmetric gauge model [SU(6)]$^3\times$Z$_3$, which unifies nongravitational forces with flavors, broken spontaneously by a minimal set of Higgs Fields and Vacuum Expectation Values down to SU(3)$_{c}\otimes$U(1)$_{EM}$\hspace{0.2cm} . We also compute the evolution of the gauge coupling constants and show how agreement with precision data can be obtained.
• ### The Generational Seesaw Mechanism in [SU(6)]$^{3}\times$Z$_{3}$}(hep-ph/9208236)

Aug. 21, 1992 hep-ph
In the context of the left-right symmetric gauge group [SU(6)]$^3\times$ Z$_3$ which unifies nongravitational forces with flavors, we analyze the generational seesaw mechanism. At tree level we get m$_{\nu_\tau}\sim$m$_{\nu_\mu}\sim$M$^2_L$/M$_H$, m$_{\nu_e}=0$, where M$_L\sim10^2$ GeVs and M$_H\ge 100$ TeVs is the mass scale at which the horizontal interactions get spontaneously broken. The right handed neutrinos get a Majorana mass M$_R\gg$M$_H$ of the order of the scale where SU(2)$_R$ is broken. An exotic neutral lepton with a mass of the order of M$^2_H$/M$_R$ is predicted. First order radiative corrections will produce m$_{\nu_e}\neq0$ which is at least two orders of magnitude smaller than the other two neutrino masses.


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# Adobe Photoshop 2022 (Version 23.0.1) Crack Keygen [Updated-2022] 129311;

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ative Representation of Groups and Lie Algebras* (Springer, Berlin, 2000).

H. A. Dye, *On the dimension of certain orbits of linear algebraic groups*, Ann. Math. **81**, 444–496 (1965).

I. M. Isaacs, *Finite group theory* (AMS Chelsea, Providence, 2006).

R. Kessar, *On primitive groups containing a nilpotent subgroup*, Internat. J. Algebra Comput. **22**, 105–115 (2012).

E. Looijenga, *Isolated Singularities of Complete Intersection* (Birkhauser, Boston, 2004).

M. N. Pimsner, *On the dimension of the space of linear characters of a finite solvable group*, Ann. Math. **88**, 619–626 (1968).

A. R. Salem and F. Shahidi, *Globalization of square integrable automorphic forms*, Proc. Indian Acad. Sci. Math. Sci. **112**, 131–155 (2002).

P. Sarnak, *Notes on orthogonal geometry of prime degree*, arXiv:math/0703054v3 (2010).

B. Y. Shteinberg, *Orbits of a real reductive algebraic group*, Transl. Math. Monographs **108** (American Mathematical Society, Providence, 1992).

V. A. Vassiliev, *Clifford numbers of reductive $p$-adic analytic groups*, Dokl. Akad. Nauk SSSR **281**, 502–507 (1985).

V. A. Vassiliev, *Reductive Lie algebras*, Adv. Soviet Math. **13** (1985).

V. A. Vassiliev, *Complements of Discrete series*, (AMS Chelsea, Providence, 2006).

S. Wasserman, *The unitary dual of $F_4$ over the complex numbers*, J. Algebra **205**, 340–367 (1998).

S. Wasserman, *Counting 3-embeddings and $3$-transitive subgroups of $GL_6(q)$ over $GF(3)$*, J. Algebra **231**, 210–236 (2000).

## What’s New In?

\frac{U}{2\pi}\frac{[(U+2\Lambda)^2-E^2]^2}{(U+2\Lambda)^2-E^2}
-\frac{U}{4\pi}E+\frac{U}{2\pi}\tan^{ -1}\frac{E}{U+2\Lambda}\end{aligned}$$The differential conductance is then given by G(V) = -\frac{2e^2}{\hbar} \partial f/\partial E, which for E\gg U+2\Lambda can be approximated by$$G(V) \approx \frac{2e^2}{\hbar}\frac{E}{U}
\frac{1}{\pi}\frac{E^2+U^2}{(U+2\Lambda)^2}
\label{G(V)} The conductance in Eq. ($G(V)$) gives the standard Breit-Wigner resonance of width $2\Gamma$, and corresponds to the zero-temperature behavior for $t\gg\hbar/U$. When $t\lesssim\hbar/U$, the Breit-Wigner resonance is washed out, but a low-energy behavior is still present.

0.3cm

[**ACKNOWLEDGMENTS**]{}

We thank A. M. Finkelstein, S. Hikami, C. M. Varma, and J. Weinand for discussions. This work has been supported by NSF grant No. DMR-9416906, the W. M. Keck Foundation, and the M. J. P. A. Fugate Professorship in Physics at the University of Texas at Austin.

[99]{}

P. A. Lee, T. M. Rice, J. W. S. Raymond, and L. S. Schneider, [*“Microwave-Induced Resistance Oscillations in Aluminum Foil”*]{}, Phys. Rev. Lett. [**35**]{}, 736 (1975).

Y. Haseg

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# Tail is just a Tail

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is

×


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Zentralblatt MATH

Publications of (and about) Paul Erdös

Zbl.No:  163.18203
Autor:  Erdös, Pál
Title:  On cliques in graphs (In English)
Source:  Isr. J. Math. 4, 233-234 (1966).
Review:  A complete subgraph of a graph G is called a clique if it is not contained in any other complete subgraph of G. Let g(n) denote the maximum number of different sizes of cliques that can occur in a graph with n points: if logk n denotes the k-times iterated logarithm to the base 2 of n, let H = H(n) be the smallest integer such that logH n < 2. The author shows that g(n) \geq n- log n-H(n)-O(1). This extends an earlier result of the reviewer and L.Moser [Isr. J. Math. 3, 23-28 (1965; Zbl 144.23205)].
Reviewer:  J.W.Moon
Classif.:  * 05C35 Extremal problems (graph theory)
Index Words:  topology

© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag


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# x percent of y percent of z is decreased by y percent. What is the res

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x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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Updated on: 01 Apr 2018, 11:10
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x percent of y percent of z is decreased by y percent. What is the result?

A. $$\frac{100xyz - xy^2z}{1,000,000}$$

B. $$\frac{xz - y}{100}$$

C. $$\frac{xz - y}{10,000}$$

D. $$\frac{xyz - 2y}{100}$$

E. $$\frac{xyz - 2y}{10,000}$$

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Originally posted by siddhans on 23 Sep 2011, 01:05.
Last edited by Bunuel on 01 Apr 2018, 11:10, edited 3 times in total.
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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23 Sep 2011, 02:18
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Very easy question, you don't need options

Acc. to given statement,

y% of z= y/100*z

x%(y5 of z)=x/100*y/100*z

Now the entire thing is reduced by y%
When i say price of a product decreased by 5% we take the new price as (100-5)/100*old price

hence after y% decrease the expression simplifies to

(100-y)/100*(x/100*y/100*z)
=(100-y)*xyz/(10^6)
=(100xyz-100xy^2z)/10^6

You posted the incorrect oa

siddhans wrote:
Tried searching but couldnt find ....

x percent of y percent of z is decreased by y percent . what is the result?

Spoiler: :: OA
100XYZ - XY^2Z/1,000,000

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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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23 Sep 2011, 04:15
I agree that you can get to answer directly, but many times the answers will be different ratio/multiple form of the answer, so you might need answer choices to work through. Also some times variables are too cluttery so some people with non-math background like me need numbers to work through.

x percent of y percent of z is decreased by y percent . what is the result?
let x = 10, y =20 and z =1000. (x percent of y percent of z = 10% 0f 20% of 1000 ; 10 % of (20% of 1000) ; 10% of 200 =20.

20 is decreased by y percent. that is 20 is decreased by 20%. so 20 - (20/100* 20) = 16 Now target answer is 16, plug in values .

A)100XYZ - XY^2Z/1,000,000

(100 *10*20*1000 - 10.(20^2)1000)/ 10^6

> (20,000,000 - 400,000,0) /10^6 = 16 (target). you may need to check other answers but in this case the numbers I have picked are so distinct that it wont give same answer. so answer is A.
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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23 Sep 2011, 05:26
siddhans wrote:
Tried searching but couldnt find ....

x percent of y percent of z is decreased by y percent . what is the result?

$$\frac{x}{100}*\frac{y}{100}*z(1-\frac{y}{100})$$
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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27 Jan 2012, 17:49
Sorry. I'm stuck because for me the right translation is

x/100 * y/100 * z = (1- y/100) or not ??

I'd resolve this question by math because is pretty straighforward than the strategy of target.

anyone can help me ??

Thanks a lot
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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27 Jan 2012, 18:12
carcass wrote:
Sorry. I'm stuck because for me the right translation is

x/100 * y/100 * z = (1- y/100) or not ??

I'd resolve this question by math because is pretty straighforward than the strategy of target.

anyone can help me ??

Thanks a lot

It can be done by plug-in method though as you are doing with algebra I'll proceed with it.

x percent of y percent of z is decreased by y percent. What is the result?

Step by step:
1. $$y$$ percent of $$z$$ is $$\frac{y}{100}*z=\frac{yz}{100}$$;
2. $$x$$ percent of above value is $$\frac{x}{100}*\frac{yz}{100}=\frac{xyz}{10,000}$$;
3. The above value is decreased by $$y$$ percent: $$\frac{xyz}{10,000}*(1-\frac{y}{100})=\frac{xyz}{10,000}*(\frac{100-y}{100})=\frac{100xyz-xy^2z}{1,000,000}$$ (for example 100 is decreased by 10 percent can be expressed as $$100*(1-\frac{10}{100})=100*(1-0.1)=100*0.9=90$$).

Hope it's clear.
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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28 Jan 2012, 03:40
1
For the % decrease formula we have: Original (1 - %decrease/100)= new

So: xyz/100 * (1 - y/100)..............I do not know why my head told me equals % decrease.......

generally is pretty simple for me this kind of problem.

Thanks Bunel
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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13 Apr 2012, 16:05
I see the process you went through but I'm still not sure I understand it. How do you know to multiply (1-y/100) instead of subtracting it? The word decrease makes me automatically think it has to be subtracted.

In this types of problems, are you only going to add/subtract if it says greater than or less than?
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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13 Apr 2012, 16:08
RyanP wrote:
I see the process you went through but I'm still not sure I understand it. How do you know to multiply (1-y/100) instead of subtracting it? The word decrease makes me automatically think it has to be subtracted.

In this types of problems, are you only going to add/subtract if it says greater than or less than?

The question does not say "decreased by y", it says "decreased y PERCENT".
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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15 Oct 2014, 21:05
2
x percent of y percent of z is decreased by y percent

Making the statement more mathematical friendly

x% of y% of z is decreased by y%

x% of (y% of z) is decreased by y%

$$(x% of \frac{yz}{100})$$ is decreased by y%

$$\frac{xyz}{100 * 100}$$ is decreased by y%

$$\frac{xyz}{100*100} - \frac{xy^2z}{100*100*100}$$

$$\frac{100xyz - xy^2z}{100*100*100}$$

Small Tip:

Number of times you see the % sign (2 zero's), those number of zero's in the denominator

$$a%b = \frac{ab}{10^2}$$

$$a%b%c = \frac{abc}{10000}$$

$$a%b%c%d = \frac{abcd}{10^6}$$

$$a%b%c%d......... n = \frac{abcd......n}{10^{(n-1)}}$$
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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15 Oct 2014, 21:09
siddhans wrote:
x percent of y percent of z is decreased by y percent. What is the result?
Attachment:
problem.png

The OA of this question is very distinct. All the options are levelled with the denominator, so no further simplification required on the OA.

Once you know the trick as stated in earlier post, its quite obvious that answer = A (it has 10^6 in the denominator)

No calculation required!!
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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16 Oct 2014, 05:53
Very easy question, you don't need options

Acc. to given statement,

y% of z= y/100*z

x%(y5 of z)=x/100*y/100*z

Now the entire thing is reduced by y%
When i say price of a product decreased by 5% we take the new price as (100-5)/100*old price

hence after y% decrease the expression simplifies to

(100-y)/100*(x/100*y/100*z)
=(100-y)*xyz/(10^6)
=(100xyz-100xy^2z)/10^6

You posted the incorrect oa

siddhans wrote:
Tried searching but couldnt find ....

x percent of y percent of z is decreased by y percent . what is the result?

Spoiler: :: OA
100XYZ - XY^2Z/1,000,000

really a tough one. m stuck in it what is the result?? can you show it?
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Re: x percent of y percent of z is decreased by y percent. What is the res  [#permalink]

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19 Oct 2015, 11:52
1
1
Hi All,

This question can certainly be solved by TESTing VALUES. The 'shortcuts' involve selecting nice 'round' numbers for the 3 variables AND recognizing that the 5 answer choices have 'pieces' in common (so you can avoid redoing some of the calculations).

We're told that X percent of Y percent of Z is DECREASED by Y percent. We're asked for the result of all of this math.

IF....
X = 10
Y = 50
Z = 100

First, 50% of 100 = 50
Next, 10% of 50 = 5
Finally, 5 decreased by 50% of itself is 2.5

So we're looking for an answer that equals 2.5 when X=10, Y=50 and Z=100.

Answer A. (100xyz - xy^2z) / 1,000,000 = (5,000,000 - 2,500,000)/1,000,000 = 2,500,000/1,000,000 = 2.5 This is a MATCH
Answer B. (xz - y) / 100 = (1,000 - 50)/100 = 950/100 = 9.5 NOT a match
Answer C. (xz - y) / 10,000 = 950/10,000 = .095 NOT a match
Answer D. (xyz - 2y) / 100 = (50,000 - 100)/100 = 49,900/100 = 499 NOT a match
Answer E. (xyz - 2y) / 10,000 = 49,900/10,000 = 4.99 NOT a match

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text:
Specific Heat Capacity Formula

The definition of specific heat capacity of any substance is “the quantity of heat required to change the temperature of a unit mass of the substance by 1 degree.” This is articulated as:

$\dpi{120}&space;\large&space;Specific\;&space;Heat\;&space;Capcity=\frac{energy\;&space;required}{mass\times\Delta&space;T}$

As it indicates the resistance of a material to an alteration in its temperature, specific heat capacity is a type of thermal inertia. Specific Heat Capacity Formula is also communicated in relation to the quantity of heat Q.

$\dpi{120}&space;\large&space;c=\frac{Q}{m\times\Delta&space;T}$

Specific heat capacity in terms of heat capacity is conveyed as

$\dpi{120}&space;\large&space;Specific\;&space;Heat\;&space;Capcity=\frac{energy\;&space;required}{mass\times\Delta&space;T}$

Problem 1: A piece of copper 125g has a heat capacity of 19687.6J also it is heated from 150 to 2500C heat. Find out the specific heat?

Solution:

Given

m = 125 gm

Q = 19687.6J

ΔT = 250-150 = 1000C

$\dpi{120}&space;\large&space;Specific\;&space;Heat\;&space;Capcity=\frac{energy\;&space;required}{mass\times\Delta&space;T}$

c = 19687.6/(125×100)

c = 1.575 J/g0C

To know more examples and practice questions on Specific Heat Capacity Formula, please visit Byjus.com


text:
# Carbonation

Bubbles of carbon dioxide float to the surface of a carbonated soft drink.

Carbonation is the process of dissolving carbon dioxide in liquid. The process usually involves carbon dioxide under high pressure. When the pressure is reduced, the carbon dioxide is released from the solution as small bubbles, which cause the solution to "fizz". This effect is seen in carbonated soft drinks.

Carbonation can also describe a chemical reaction, one example of which is a key step in photosynthesis.

## Chemistry

Carbon dioxide is weakly soluble in water, therefore it separates into a gas. The process of carbon dioxide effervescing from a solution is represented by the following chemical reaction, in which aqueous carbonic acid converts to carbon dioxide and water:

$\mbox{H}_2\mbox{O} + \mbox{CO}_2 \longrightarrow \mbox{H}_2\mbox{CO}_3$

## Biochemistry

Champagne carbonation propelling a cork. Photographed with an air-gap flash.

Carbonation also describes the incorporation of carbon dioxide into chemical compounds. Our carbon-based life originates from a carbonation reaction that is most often catalysed by the enzyme RuBisCO. So important is this carbonation process that a significant fraction of leaf mass consists of this carbonating enzyme.[1]

Carbonation of ribulose bisphosphate is the starting point of the incorporation of carbon dioxide into the biosphere.


text:
# When do there exist locally regular embeddings of regular graphs?

I don't know how one can tackle the following kind of question, so any hint is welcome. I formulate a precise question in order to fix ideas, but it is to be considered as an example out of a more general class.

Example: Can one embed Petersen's graph in $\mathbb R^4$ in such a way that all edges are mapped to segments (of not necessarily equal lenght) and each pair of adjacent segments forms an angle of $2\pi/3$ ?

The question can be generalized in the following way, to give a wider class of questions:

• instead of Petersen's graph, consider a given $k$-regular graph (we had $k=3$ above)
• instead of $\mathbb R^4$ consider $\mathbb R^n, n>k$, or embeddings into the round sphere $S^n$ where the edges of the graph are mapped into geodesics
• ask that the angles formed between any pair of adjacent edges are equal to the angle $VOV'$ made at which two vertices $V,V'$ of a regular euclidean $(k-1)$-simplex are seen from its barycenter $O$

I think that $1$-skeletons of spherical regular polytopes give such kind of graphs. I am interested in any partial answer which introduces essentially more ingredients than just constructions based on symmetry groups, or in obstructions which show negative answers.

-

Assume $\Gamma$ is a 3-regular graph in $\mathbb{R}^n$ and the angles at each vertex are $\tfrac23\pi$. Then it is easy to see that the distance to the origin can not have a local maximum on $\Gamma$.
Therefore $\Gamma$ has to be infinite; in particular it can not be Petersen's graph.
Thanks, I didn't notice this criterion. If there will not be more answers (for example on $S^n$ or for other $k$) within 2 weeks, I will accept this one. –  Mircea Jul 9 '12 at 14:33
Yet an other reason: the Petersen's graph has cycles of length 5; any 5-gon in $\mathbb R^k$ has sum of angles at most $3\cdot\pi$, so at least one of the angles has to be $\le \tfrac35\cdot\pi<\tfrac23\cdot\pi$. –  Anton Petrunin Jul 9 '12 at 17:11


text:
geometry – How is this angle measured in the traingle?

I read "How to solve it" by George Polya. I do not understand how he gets to certain solutions. Like the chapetr of "Auxillary Solution". Given the picture below:

Leave the angle at the top $$A$$ to be called $$alpha$$. Polya says looking at the diagram and the isosceles triangles $$triangle {ACE}$$ and $$triangle {ABD}$$ we can maintain that $$angle {DAE} = frac { alpha} {2} + 90 ^ circ$$.

I just do not understand how this is possible. All I can understand is that $$angle {CAE} = angle {CEA}$$ and $$angle {BAD} = angle {BDA}$$since they are both isosceles. But where will I get $$90 ^ circ$$ of?

Do I miss an important theorem in geometry?


text:
# Homework Help: Induction and EMF

1. Nov 24, 2007

### captain.joco

1. The problem statement, all variables and given/known data
In a ring of radius R, there is current I flowing in the counterclockwise direction as viewed from above. A small ring of radius r is on a common axis and is a hight z above the current carrying ring, where z>>R. The small ring moves up with velocity v. Calculate the emf in the upper ring and the direction of the induced current. It can be assumed that the ring is sufficiently small so that the magnetic field across it's area is constant.

2. Relevant equations

3. The attempt at a solution
So E= - dΦ/dt
Φ = B*A ( since the magnetic field across it's area is constant ).
A = π*r*r
B= (μIR^2) / [ 2 ( R*R + z*z )^3/2 ] but since z>>R we can use the approximation
B= (μIR^2) / (2z^3)

Now i know I need to do the -dΦ/dt but I am getting confused ( how to do it? ).

2. Nov 24, 2007

### G01

In order to do the derivative you need to have time involved. The only quantity in that expression for the flux that depends on time is z, correct? Is there any way to get z in terms of time?

HINT: Have you used the velocity, v, yet?

3. Nov 24, 2007

### captain.joco

So will the emf be A* (μIR*R/2) * 3( v^3 * z^4 ) or I did it very very wrongly..

4. Nov 24, 2007

### G01

$$\Phi=\frac{\mu IR^2A}{2z^3}$$

Then, remember that: v = z/t.

Can you use the last expression to remove z from the first?

5. Nov 24, 2007

### captain.joco

Sorry I made a typo.. should have been A* (μIR*R/2) * 3/( v^3 * t^4 ) but then still
A* (μIR*R/2) * 3/( z^3*t )

6. Nov 24, 2007

### captain.joco

Is this anywhere near to correct?

7. Nov 24, 2007

### G01

I think you are missing a minus sign in there. You should have picked up a -3 with the differentiation.

Other than that, it looks good.

8. Nov 25, 2007

### captain.joco

But from the original equation which is emf = - dΦ/dt, so this minus cancels out with the -3 minus.. I will redo it just in case. That's for all the help, you can count this one as solved ( I understood the main idea with the z being the only thing that changes with time! ). Thanks very much


text:
# Crash course on black holes

1. Nov 22, 2013

### taonut

Hi,

2. Nov 22, 2013

From where did you find this statement?This is wrong.

Mass is the amount of matter present in a substance

Usually we assume that Force of gravity acts on the center of mass of the object.(To make calculations easier)
Stars does not spin around empty space.Inside stars,there is the core of it.

We know Earth has mass because it causes gravity.

Anything which has mass will cause gravity

Force of gravity between two masses, $F=\frac{Gm_1m_2}{r^2}$
Where m_1 and m_2 is the masses of two objects.
G is the Universal Gravitational Constant
r is the distance between two object

3. Nov 22, 2013


text:
## cutie.patootie Group Title More help? Type the equation of the given line in standard form. The line with m = 4 and intersecting 2x -4y = 8 at x = 6. Thanks! 2 years ago 2 years ago

1. agentx5

Step #1: convert "2x -4y = 8" to y=mx+b Step #2...

2. cutie.patootie

So that would make it -4y=2x+8?

3. cutie.patootie

Or do I have to put the m in there to and make it -4y=8x+8?

4. jazy

you want y not -4y so you should simplify

5. agentx5

Move the 2x over to the other side (with subtraction), divide both sides by -4. Can you write this second step @cutie.patootie ? :-) You'll end up getting a y=mx+b form line that allows to to draw this: |dw:1343071328047:dw|

6. cutie.patootie

-4y=2x+8 turns into.. -4y-2x=8 divide by -4 gives you.. y+(1/2)x=2 Right?

7. agentx5

No... unless you wrote it incorrect at the start. -4y + 2x = 8? Move the x's over -4y = -2x + 8 Divide by -4 y = 0.5 x - 2 (see graph above)

8. cutie.patootie

Oh well I had it right, but then I moved the x to the wrong side. But I see where I made my mistake now.

9. agentx5

then it tells you there is some other line with: y = 4x + b , because it says "m = 4" An "intersection of two lines" means when they are equal to each other. If you've solved both for the dependent varible (y) that means you can set the y's equal to each other: $$y_1 = y_2$$ we know: $$y_1 = \frac{1}{2}x - 2$$ $$y_2 = 4x + b_2$$ So... $$y_1 = y_2$$ $$\frac{1}{2}x - 2 = 4x + b_2$$ Make sense? The final steps are substitute in "x=6", solve for $$b_2$$, and rewrite $$y_2$$

10. cutie.patootie

Okay, so you get... (1/2)(6) - 2 = 4(6) +b2 3-2=24+b2 1=24+b2 b2= -23?

11. agentx5

You're correct! |dw:1343072679957:dw|

12. agentx5

When x = 6, they are both equal.

13. agentx5

Now don't forget to rewrite $$y_2=4x-23$$ in standard form :-) All good?

14. agentx5

Visual methods FTW!

15. cutie.patootie

Okay, I think I follow. So I write it y + 4x = -23?

16. cutie.patootie

And yes! Hahaha.


text:
location:  Publications → journals
Search results

Search: MSC category 11B99 ( None of the above, but in this section )

 Expand all        Collapse all Results 1 - 1 of 1

1. CMB 2013 (vol 57 pp. 877)

Schoen, Tomasz
 On Convolutions of Convex Sets and Related Problems We prove some results concerning covolutions, the additive energy and sumsets of convex sets and its generalizations. In particular, we show that if a set $A=\{a_1,\dots,a_n\}_\lt \subseteq \mathbb R$ has the property that for every fixed $1\leqslant d\lt n,$ all differences $a_i-a_{i-d}$, $d\lt i\lt n,$ are distinct, then $|A+A|\gg |A|^{3/2+c}$ for a constant $c\gt 0.$ Keywords:convex sets, additive energy, sumsetsCategory:11B99
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text:
Helper functions#

Estimating stable time-steps#

Helper functions for estimating stable time steps for RKDG methods.

Characteristic lengthscales#

grudge.dt_utils.characteristic_lengthscales(actx: ArrayContext, dcoll: DiscretizationCollection) [source]#

Computes the characteristic length scale $$h_{\text{loc}}$$ at each node. The characteristic length scale is mainly useful for estimating the stable time step size. E.g. for a hyperbolic system, an estimate of the stable time step can be estimated as $$h_{\text{loc}} / c$$, where $$c$$ is the characteristic wave speed. The estimate is obtained using the following formula:

$h_{\text{loc}} = \operatorname{min}\left(\Delta r_i\right) r_D$

where $$\operatorname{min}\left(\Delta r_i\right)$$ is the minimum node distance on the reference cell (see dt_non_geometric_factors()), and $$r_D$$ is the inradius of the cell (see dt_geometric_factors()).

Returns:

a frozen DOFArray containing a characteristic lengthscale for each element, at each nodal location.

Note

While a prediction of stability is only meaningful in relation to a given time integrator with a known stability region, the lengthscale provided here is not intended to be specific to any one time integrator, though the stability region of standard four-stage, fourth-order Runge-Kutta methods has been used as a guide. Any concrete time integrator will likely require scaling of the values returned by this routine.

Non-geometric quantities#

grudge.dt_utils.dt_non_geometric_factors(dcoll: DiscretizationCollection, dd=None) list[source]#

Computes the non-geometric scale factors following [Hesthaven_2008], section 6.4, for each element group in the dd discretization:

$c_{ng} = \operatorname{min}\left( \Delta r_i \right),$

where $$\Delta r_i$$ denotes the distance between two distinct nodal points on the reference element.

Parameters:

dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.

Returns:

a list of float values denoting the minimum node distance on the reference element for each group.

Mesh size utilities#

grudge.dt_utils.dt_geometric_factors(dcoll: DiscretizationCollection, dd=None) [source]#

Computes a geometric scaling factor for each cell following [Hesthaven_2008], section 6.4, defined as the inradius (radius of an inscribed circle/sphere).

Specifically, the inradius for each element is computed using the following formula from [Shewchuk_2002], Table 1, for simplicial cells (triangles/tetrahedra):

$r_D = \frac{d V}{\sum_{i=1}^{N_{faces}} F_i},$

where $$d$$ is the topological dimension, $$V$$ is the cell volume, and $$F_i$$ are the areas of each face of the cell.

Parameters:

dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.

Returns:

a frozen DOFArray containing the geometric factors for each cell and at each nodal location.

grudge.dt_utils.h_max_from_volume(dcoll: DiscretizationCollection, dim=None, dd=None) [source]#

Returns a (maximum) characteristic length based on the volume of the elements. This length may not be representative if the elements have very high aspect ratios.

Parameters:
Returns:

a scalar denoting the maximum characteristic length.

grudge.dt_utils.h_min_from_volume(dcoll: DiscretizationCollection, dim=None, dd=None) [source]#

Returns a (minimum) characteristic length based on the volume of the elements. This length may not be representative if the elements have very high aspect ratios.

Parameters:
Returns:

a scalar denoting the minimum characteristic length.

Array contexts#

class grudge.array_context.PyOpenCLArrayContext(queue: pyopencl.CommandQueue, allocator: = None, wait_event_queue_length: = None, force_device_scalars: bool = False)[source]#

Inherits from meshmode.array_context.PyOpenCLArrayContext. Extends it to understand grudge-specific transform metadata. (Of which there isn’t any, for now.)

class grudge.array_context.PytatoPyOpenCLArrayContext(queue, allocator=None, *, compile_trace_callback: Optional[Callable[[Any, str, Any], None]] = None)[source]#

Inherits from meshmode.array_context.PytatoPyOpenCLArrayContext. Extends it to understand grudge-specific transform metadata. (Of which there isn’t any, for now.)

class grudge.array_context.MPIBasedArrayContext[source]#
class grudge.array_context.MPIPyOpenCLArrayContext(mpi_communicator, queue: pyopencl.CommandQueue, *, allocator: = None, wait_event_queue_length: = None, force_device_scalars: bool = False)[source]#

An array context for using distributed computation with pyopencl eager evaluation.

__init__(*args, **kwargs)[source]#

Initialize self. See help(type(self)) for accurate signature.

class grudge.array_context.MPIPytatoArrayContext[source]#
grudge.array_context.get_reasonable_array_context_class(lazy: bool = True, distributed: bool = True, fusion: = None) [source]#

Returns a reasonable PyOpenCLArrayContext currently supported given the constraints of lazy and distributed.

Miscellaneous tools#

grudge.tools.build_jacobian(actx, f, base_state, stepsize)[source]#

Returns a Jacobian matrix of f determined by a one-sided finite difference approximation with stepsize.

Parameters:
Returns:

a two-dimensional numpy.ndarray

grudge.tools.map_subarrays(f: Callable[[Any], Any], in_shape: Tuple[int, ...], out_shape: Tuple[int, ...], ary: Any, *, return_nested=False) Any[source]#

Apply a function f to subarrays of shape in_shape of an numpy.ndarray, typically (but not necessarily) of dtype object. Return an numpy.ndarray with the corresponding subarrays replaced by the return values of f, and with the shape adapted to reflect out_shape.

Similar to numpy.vectorize.

Example 1: given a function f that maps arrays of shape (2, 2) to scalars and an input array ary of shape (3, 2, 2), the call:

map_subarrays(f, (2, 2), (), ary)

will produce an array of shape (3,) containing the results of calling f on the 3 subarrays of shape (2, 2) in ary.

Example 2: given a function f that maps arrays of shape (2,) to arrays of shape (2, 2) and an input array ary of shape (3, 2), the call:

map_subarrays(f, (2,), (2, 2), ary)

will produce an array of shape (3, 2, 2) containing the results of calling f on the 3 subarrays of shape (2,) in ary. The call:

map_subarrays(f, (2,), (2, 2), ary, return_nested=True)

will instead produce an array of shape (3,) with each entry containing an array of shape (2, 2).

Parameters:
• f – the function to be called.

• in_shape – the shape of any inputs to f.

• out_shape – the shape of the result of calling f on an array of shape in_shape.

• ary – a numpy.ndarray instance.

• return_nested – if out_shape is nontrivial, this flag indicates whether to return a nested array (containing one entry for each application of f), or to return a single, higher-dimensional array.

Returns:

an array with the subarrays of shape in_shape replaced with subarrays of shape out_shape resulting from the application of f.

grudge.tools.rec_map_subarrays(f: Callable[[Any], Any], in_shape: Tuple[int, ...], out_shape: Tuple[int, ...], ary: ArrayOrContainer, *, scalar_cls: Optional[Union[type, Tuple[type]]] = None, return_nested: bool = False) [source]#

Like map_subarrays(), but with support for arraycontext.ArrayContainers.

Parameters:

scalar_cls – An array container of this type will be considered a scalar and arrays of it will be passed to f without further destructuring.


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### Paper Folding - Models of the Platonic Solids

A description of how to make the five Platonic solids out of paper.

### Next Size Up

The challenge for you is to make a string of six (or more!) graded cubes.

### Triple Cubes

This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.

# Three Sets of Cubes, Two Surfaces

##### Age 7 to 11 Challenge Level:

This can be used as a follow-on from Two on Five.

You have interlocking cubes of three different colours - $2$ of one colour, $3$ of another colour and $4$ of the third colour.
It could look like this;

This is slightly different from Two on Five but is seen as an extension for the highest attaining. You might like to go there first!

The nine cubes are to be connected in the usual way with the following rules being applied.

THE TWO YELLOW CUBES ARE NOT ALLOWED TO TOUCH WALL OR FLOOR SURFACES.

THE THREE BLUE CUBES MUST TOUCH ONE SURFACE ONLY THE WALL OR THE FLOOR

EACH OF THE FOUR RED CUBES MUST TOUCH BOTH WALL AND FLOOR

Here are two examples that obey the rules;

See what others you can find.
How many will there be?
At some point ask yourself "I wonder what would happen if I ...?"


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6 months ago

## Definition Of Incomplete Homework sector isidro ordena

Definition Of Incomplete Homework

definition of incomplete homework

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evaluate homework and practice module 7


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# Homotopy type of smooth manifold with boundary

It seems very likely to me that a $$n$$-dimensional smooth manifold with boundary has the homotopy type of a $$(n-1)$$-dimensional CW complex. Is that true? Does the manifold need to be compact? What about $$n$$-dimensional open manifolds?

• It can only true if you require that $M$ is connected or that all components of $M$ have a nonempty boundary. Otherwise it may have a component $S^n$. – Paul Frost Dec 27 '18 at 18:25
• @PaulFrost Yes, these obvious assumptions should be added to the question. But then? – Yeah Dec 27 '18 at 21:20
• For a discussion in dimension 2, with allusions to higher dimensions, see here. – Lee Mosher Dec 27 '18 at 22:45


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# Adjoint of linear operators

Gold Member

## Homework Statement

I must show several properties about linear operators using the definition of the adjoint operator.
A and B are linear operator and ##\alpha## is a complex number.
The first relation I must show is ##(\alpha A + B)^*=\overline \alpha A^*+B^*##.

## Homework Equations

The definition I have an an adjoint is: ##A^*## is the adjoint of ##A## if ##\langle g,Af \rangle = \langle A^* g ,f \rangle## where f and g are any vectors in a Hilbert space.

## The Attempt at a Solution

Let ##C^*=(\alpha A+B)^*##. Using the definition of adjoint I get: ##\langle C^*g,f \rangle=\langle g, Cf \rangle \Rightarrow \langle (\alpha A+B)^*g ,f \rangle =#### \langle g, (\alpha A+B)f \rangle =\langle g, \alpha A \cdot f \rangle + \langle g, Bf \rangle = \alpha \langle g, Af \rangle + \langle g ,Bf \rangle = \alpha \langle A^*g, f \rangle + \langle B^*g ,f \rangle##.
But I'm getting lost. I've no idea how I can obtain A, B, A^* and B^* using the definition of the adjoint.

Oh wait, on my draft I think I have finished the "proof". The last expression is worth ##\langle \overline \alpha A^*g ,f \rangle + \langle B^* g, f \rangle = \langle (\overline \alpha A^* + B^*)g ,f \rangle##. Then by associativity ##(\alpha A+B)^*=\overline \alpha A^* + B^*##.
Does this look right?

## Answers and Replies

micromass
Staff Emeritus
Homework Helper
OK, so you proved

$$<(\alpha A + B)^* g,f> = <(\overline{\alpha}A^* + B^*)g,f>$$

But how do you obtain from there that ##(\alpha A + B)^* = \overline{\alpha}A^* + B^*##?

Gold Member
OK, so you proved

$$<(\alpha A + B)^* g,f> = <(\overline{\alpha}A^* + B^*)g,f>$$

But how do you obtain from there that ##(\alpha A + B)^* = \overline{\alpha}A^* + B^*##?
Good question, I don't know. To simplify the notation I must show that if ##\langle Dg, f \rangle = \langle Eg,f \rangle## then D=E, where D and E are linear operators; for any inner product in a Hilbert space.
When I look at the definition of the inner product, I only see the positive definiteness, linearity and conjugate symmetry. I don't see how that would help me.

micromass
Staff Emeritus
Homework Helper
Something that I think would help is the axiom ##<x,x>=0## then ##x=0##. Because this is the only axiom that allows you to begin with the inner product and end with a statement about a vector.

Let's prove a more general thing. Let x,y be vectors such that ##<x,z> = <y,z>## holds for all z, then x=y.

To prove this, note that the above is equivalent to saying ##<x-y,z>=0## for all z. Does this give you a hint?

Gold Member
I'm not 100% sure.
Here is my attempt: ##\langle x ,z \rangle =\langle y, z \rangle## holds for all z.
I add "-y" in the first argument of each side (can I really do that without demonstrating that the equality still hold?) to get ##\langle x-y,z \rangle =\langle y-y ,z \rangle = \langle 0 ,z \rangle =0##
So that ##\langle x-y, z \rangle =0## must hold for all z. This implies that ##x-y=0##. End of proof.

P.S.:Yes I can add -y in the first argument of the inner product. To prove it is simple, using the linearity of the inner product.

micromass
Staff Emeritus
Homework Helper
So that ##\langle x-y, z \rangle =0## must hold for all z. This implies that ##x-y=0##.

This implication is not very clear. why does ##<x-y,z>=0## for all z imply that ##x-y=0##?

dextercioby
Homework Helper
It is of utmost importance to mention that the operators are bounded and defined everywhere on the Hilbert space, else one needs a new reasoning.

Gold Member
This implication is not very clear. why does ##<x-y,z>=0## for all z imply that ##x-y=0##?

Because in particular ##z=x-y##. So we have ##\langle x-y , x-y \rangle =0##, which is possible if and only if ##x-y=0##. Thus ##x=y##. End of proof?
dextercioby said:
It is of utmost importance to mention that the operators are bounded and defined everywhere on the Hilbert space, else one needs a new reasoning.
Is is the same as showing that the operators are continuous over the whole Hilbert space?

dextercioby
Homework Helper
Because in particular ##z=x-y##. So we have ##\langle x-y , x-y \rangle =0##, which is possible if and only if ##x-y=0##. Thus ##x=y##. End of proof?[...]

Yes, good point.

[...]Is is the same as showing that the operators are continuous over the whole Hilbert space?

No, continuity is an alternative assumption to boundedness, because for ∞-dim separable Hilbert spaces the 2 notions are equivalent.

micromass
Staff Emeritus


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# Malignant computation

## To properly serve society, cryptocurrencies must support computer hardware that is useful for other things.

Cryptocurrencies, like bitcoin, could revolutionize money to the same degree that the Internet has revolutionized communication. However, like any economic marketplace, human exuberance is the greatest threat to the cryptocurrency phenomenon. Markets fail to the degree that the market can be dominated by those seeking personal gain, and markets succeed to the degree that they resist domination and focus on benefiting society at large.

The cryptocurrency market place is in danger of becoming so focused on profitability, that it loses sight of the potential computational benefits that it could provide to society. I hope that this article will influence designers of cryptocurrencies to attempt to avoid computational malignancy.

Many people regard the success or failure of the market to be the degree that it works for them, rather than for society as a whole. One of the fundamental motivations for cryptocurrency is the general sense that banks, governments and markets have failed to protect the interest of the common man. It is not an accident that the rise of bitcoin began shortly after the sub-prime mortgage crisis.

Cells, typically, prefer to serve the whole organism, but when they get confused and start to multiply without regard for the impact on the organism as a whole, they can morph into a series of diseases that we refer to collectively as cancer. This is why “curing cancer” is so hard. Cancer is not a disease, but a family of diseases that share a common core problem: cells acting in their own interests that betray the body as a collective.

We have a similar problem with the use of computation in markets. We can call this malignant computation. This is when computation starts to ensure its own survival at the expense of the overall marketplace. The Skynet hypothesis is a boogeyman intended to scare the young and the paranoid. The real threat from AI is that it will become so good at the pointless tasks that we have given it that those pointless tasks will become a black hole of resources.

This has already happened with high-frequency trading on Wall Street. There is an ongoing arms race between computers that trade stocks to see which one can get the edge over the other, and entire series of engineering feats that have no purpose whatsoever other than to overcome previous engineering feats. In several respects, the computational trading platforms are the most advanced computation systems on the planet, and they are engaged in a micro-second game of mutual navel gazing. There is so much money being “made” by these super computers that the only thing that is absolutely certain is that further funding for bigger super computers will become available.

Capital markets serve a function in society. They ensure that businesses that provide value to society will have access to large amounts of capital to invest in otherwise too expensive projects. I have not been able to think of a single way in which the high-frequency trading platforms have improved the markets capacity to serve that function. No one has been able to provide me with any contrary insight, although several pointed me to more eloquent statements of the underlying problem. High-frequency trading is the first and foremost example of malignant computing, but it is not the last.

Malignant computing is a problem in cryptocurrencies too, but in order to discuss it clearly, one has to understand how the computational arms race in cryptocurrency mining works. This article does a wonderful job of summarizing the issues of the crypto arms race.

Cryptocurrencies in the bitcoin mold rely on a process called “mining,” which is the process of performing arbitrary calculations that help to ensure that the currency as a whole is functional and secure. Because of the inflated prices of bitcoins, mining has been very profitable, and as a result, we have seen the entire computational infrastructure of bitcoin switch to ASICS, or Application Specific Integrated Circuits. When you see the word ASIC, you should have a mental shortcut to “single purpose computing.” The bitcoin mining ASICs are so specific that they can only be used for the computations for bitcoin mining; they cannot even perform nearly identical computations for different parts of the bitcoin computation process.

I believe that this is another example of malignant computing. Bitcoin mining will continue until 2033. For bitcoin, ASICs will do the vast majority of this work, and assuming the value of a single bitcoin continues to rise, the amount of money invested in specialized hardware to perform bitcoin mining will almost certainly pass into the tens of billions of dollars. The bitcoin mining algorithms rewards miners relative to the whole amount of computational power devoted to bitcoin mining everywhere. If computational power were equated to “lottery tickets,” this would be tantamount to massive changes to “your chance of winning.”

Permit me to digress with this analogy: Let’s imagine a lottery for $100. If you have 10 lottery tickets and there are 100 lottery tickets in total, then you have a 1/10 chance of winning. If the lottery tickets cost less than$1 and you are ensured that there will be many more rounds of the lottery, then it makes sense to purchase tickets in every round. Over time, consistent participation will ensure a profit. If the lottery ticket costs more than 1, then participating repeatedly will guarantee a loss. Bitcoin-style mining works in much the same way. With one exception: the number of lottery tickets are proportional to the total amount of computation devoted to the process. Let’s imagine that there were several million lottery tickets available. Owners of ASIC computers would be able to acquire hundreds of thousands of tickets each. Owners of programable specialized hardware would be able to purchase tens of thousands of tickets. Owners of commodity hardware with specific graphics cards would be able to buy thousands of tickets. Owners of commodity hardware without graphics cards would be able to buy one or two tickets. Remember that tickets are not free, running computers requires electricity, and the electricity bill for running a lot of computers can be quite high. In order to profitably participate in cryptocurrency mining, you have to at least make more money than your ongoing electricity costs and the costs required to obtain your hardware. To summarize (data as of the writing):  Total computation devoted to mining (i.e. total lottery tickets) 88,502,557,900 MH/s Typical ASIC computation 100,000 (2013) 1,000,000 (2014) MH/s Typical FPGA computation 1000-25,000 MH/s Typical GPU (graphics card) computation 500-2500 MH/s Typical CPU computation 10 MH/s Currently, it is not practical to mine bitcoin without an ASIC. There are so many ASICs that are contributing to the total amount of computation devoted to bitcoin mining that nothing else can justify the costs of electricity to compete against the ASICs for reasonable shares of the mining rewards. This is a problem because the ASICs represent a tremendous amount of raw computational resources that are permanently devoted to financial busy work. The bitcoin mining network would work just as well if it had far less computation devoted to it. Bitcoins would be mined at exactly the same rate if 1/2 or 1/4 of the computational resources were devoted. This means that bitcoin has incentivized a tremendous amount of computational busy work. So far, the estimate is that the bitcoin has used about 150,000 megawatt hours of electricity, which is about 1/8th of what the state of Hawaii uses in one year. It would not be unreasonable to assume that at the rate of growth, the bitcoin mining network will surpass the yearly electricity usage of a small state soon. If we include the environmental impact of producing and manufacturing hardware specifically for bitcoin mining along with the electricity costs, then society has invested in a tremendous amount of unnecessary computation. Clearly, this is another example of malignant computation. The order of usefulness in bitcoin mining is: 1. ASIC 2. FPGA 3. GPU 4. CPU The order of usefulness for all other computations is exactly the opposite: 1. CPU 2. GPU 3. FPGA 4. ASIC ASICs achieve their dominance over the mining network by being utterly and totally useless for anything other than bitcoin computation. But we must give credit where it is due. It is a feat of engineering that bitcoin works at all. It is no trivial thing to engineer a cryptocurrency where none existed and ensure that it is both secure and adoptable. Bitcoin is a trailblazer, and there was no way to know for certain, in advance, that the drive to ASICs would happen. Litecoin is another matter. Litecoin bills itself as “silver to bitcoin’s gold.” It was designed using mathematics that would resist implementation in an ASIC. Litecoin has proven to be the second viable cryptocurrency, and it won this position in no small part because it advertised itself as CPU-centric and mineable on “consumer” hardware. Essentially, litecoin was intended to be a cryptocurrency that would not succumb to computational malignancy. Litecoin is facing something of a crisis, as the first ASIC to support litecoin mining will likely be out soon. Silver is distinct from gold because there are notable and substantial differences between the metals. Litecoin is about to lose its central distinction from bitcoin. All of the other distinctions between litecoin and bitcoin are mostly irrelevant, differences of preferences and not of substance. There is a poll and discussion on Bitcointalk that shows there is a preference for changing the litecoin proof-of-work algorithm to make ASIC mining impossible. To suggest that this happen with bitcoin would be unthinkable. The bitcoin project advertised its method for proof-of-work, and there was never an intention to prevent any approach. But litecoin specifically had CPU-oriented proof-of-work as a goal. I hope that litecoin decides to change its proof-of-work to ensure it does not become a clone of bitcoin. If it is to stay distinct, it must do so in the most fundamental way and stick to its roots. After all, if I have to buy an ASIC, why not just buy one for bitcoin? I hope litecoin gets this one right. I expect, however, that if litecoin decides not to modify its proof-of-work algorithm, it will be supplanted as “silver” by another cryptocurrency that does take this commitment seriously. YAcoin has committed to intentionally changing its proof-of-work over time to be ASIC resistant, but so far I have found no cryptocurrency that has committed to actually changing its proof-of-work if the current proof-of-work becomes ASIC friendly. So far, at least, no one regards this arms race as a central problem that needs both technology and cultural solutions. Changing the proof-of-work in a mining community would produce a fork unless there was strong consensus around the goal of avoiding computation for the sake of itself. It order to avoid computational malignancy, a cryptocurrency must support miners who have computer hardware that is useful for other things. Or it could integrate a fundamentally useful process as its proof-of-work (which is the approach that primecoin takes as it attempts to find prime numbers). Eventually, cryptocurrency mining software should ship with implementations of APIs that allow its computational resources to be accessed for other purposes. The OpenStack protocols are a good place to start for this. In fact, as the OpenStack protocols mature, they should support cryptocurrency remittance. That way, a currency miner who has invested in a data center can say, “Oh, you will pay me 100 litecoin/bitcoin/whatever to view my mining cluster as a Hadoop instance? I would only make 90 mining! OK, sold; here are your access keys.” Which is the reason that I have an Ox in this ditch. The “Omics” in medical science is going to require super computer resources on a massive scale, and while the most profitable health care inquiries will always be pursued, there will be more and more of a need for computation resources for “orphan” diseases, etc. The capacity to repurpose mining networks to perform genomic, proteomic, or exomic analysis would ensure that the network would be available to serve society, not just itself. (I should warn that I am not giving investment advice — or more strictly, that if you were to attempt to view the above as investment advise, you will probably lose money.) ### Get the O’Reilly Data Newsletter Stay informed. Receive weekly insight from industry insiders. • kevin vertcoin uses nfactor algo to be truly asic resistant • It does, but its schedule is such that the N factor changes too infrequently to be truly asic resistant • DS You are absolutely right to call attention to the gross misallocation of resources of recent years — misallocation on a scale so vast as to approach Soviet quality — and are equally right to call attention to HST as a phenomenon that exacerbates this. However, you are altogether too quick to identify these as computing problems. HST for instance could be banned next week if the public and government demanded it. Like other speculative vehicles, Bitcoin could be also be subjected to regulation to reduce the risk of fraud. This post therefore makes the fatal error of mistaking symptom for disease. Poor fiscal and monetary policy (in short, insufficient fiscal stimulus and excessive monetary stimulus through QE) has ultimately been responsible for the liquidity that has been put mostly to speculative get-rich-quick schemes instead of being directed towards productive capacity. There are few circumstances under which Bitcoin is remotely credible as a currency or store of value, and the this mania, as the apotheosis of the current tech bubble, could not be more illustrative of the misallocation of resources you describe. In short, Silicon Valley has become Wall Street. As the Fed taper cycle begins, funding will dry up, and the tide will go out, just as it did in late 1999 when the Fed tightening cycle began. • fred trotter I hope I focused on the underlying incentives that cause the problem, but I disagree with your labeling this as a “human only” problem. As Big Data becomes more and more sophisticated we will see more computation that “accidentally arranges itself for self survival”, in a process not unlike the formation of early single celled organisms… • onlineflowers We can do our best by showering mother with presents that would make her realize how deeply we think of them and want to thank them for being an idol in our life. Send Flowers to India and show your intense love. • dlldk8j It is sad just how much energy is going to Bitcoin mining… • cphv I think there are some important distinctions to be made between high frequency trading and Bitcoin mining. HFT is, as you say, hardly of any benefit to “the common man”. Bitcoin mining on the contrary, is used to secure the Bitcoin network, which is of great benefit to us all. Bitcoin is a global payment network that anyone with internet access can use. That is actually an incredibly useful way to spend energy since it allows anyone to send money over the internet, fast, easy and virtually free. There is an enormous amount of resources used in today’s banking system for heating buildings and securing and moving cash. That is not an issue with Bitcoin. Bitcoin mining facilities can easily be (and are already being) placed in areas where energy is cheap and abundant. Typically this means various kinds of renewable energy. Contrary to HFT machinery which is typically placed as close to the stock exchanges as possible to get the lowest possible latency. • fred trotter I certainly concede that Bitcoin is providing a social benifit, and even that bitcoin mining is an important component in securing the bitcoin network. But there is no justification for the ASIC-only specialized computation, or for the outlandish amount of electricity that is being spent here. In order to be “secure” as you mention, mining efforts primarily need to be broken up between many participants, there is no extra security benefit from ASIC-only computation. That delta: between what bitcoin needs to handle secure mining and the massive amount of computation that is actually being thrown at it… is how bitcoin mining has become malignant. Still, this article should not be taken as criticism of bitcoin, but rather a call for the adoption of further crypto-currencies that do not work in this way. As I mention in my article, there is no way to have predicted that bitcoin would produce excessive computation.. and we should remain amazed that it works at all!! -FT -FT • tromp I designed the Cuckoo Cycle proof-of-work system at https://github.com/tromp/cuckoo whose README I reproduce below. Mining is generally considered to be inherently power hungry but it need not be. It’s a consequence of making the proof of work computationally intensive. If computation is minimized in favor of random access to gigabytes of memory (incurring long latencies), then mining will require large investments in RAM but relatively little power. Cuckoo Cycle represents a breakthrough in three important ways: 1) it performs only one very cheap siphash computation for about 3.3 random accesses to memory, 2) its memory requirement can be set arbitrarily and doesn’t allow for any time-memory trade-off. 3) verification of the proof of work is instant, requiring 2 sha256 and 42 siphash computations. Runtime in Cuckoo Cycle is completely dominated by memory latency. It promotes the use of commodity general-purpose hardware over custom designed single-purpose hardware. Other features: 4) proofs take the form of a length 42 cycle in the Cuckoo graph. 5) it has a natural notion of (base) difficulty, namely the number of edges in the graph; above about 60% of size, a 42-cycle is almost guaranteed, but below 50% the probability starts to fall sharply. 6) running time for the current implementation on high end x86 is under 24s/GB single-threaded, and under 3s/GB for 12 threads. 7) making cuckoo use a significant fraction of the typical memory of a botnet computer will send it into swap-hell, and likely alert its owner. • fred trotter I am so happy to know that someone is making practical progress on this problem. Can your cryptocurrency change its proof of work over time? • tromp I don’t have a cryptocurrency, just a proof-of-work algorithm. Which has yet to be adopted… • e_ducat Its very hard indeed to estimate the amount of energy devoted to bitcoin mining precisely because mining can take place in various kinds of context. ASICs are far less power hungry than CPUs or GPUs. Also, if I am mining in the winter at home, I might be saving some energy heating the room. This is not the case with the data centers used by the traditionnel finance system where energy is spent cooling the aisles. Because Bitcoin reduces the scope of traditionnal cash, we are reducing the environmental and social costs of running armored trucks filled with bank notes. Last but not least, there is a logical fallacy in the article: if mining computations can be merged with other purposeful computations, the consequence is lowering the cost of a 51% attack because the opportunity cost of such attack would be lowered. Additionnally, on a macro economic level, the hashing power of the bitcoin network is safeguarding it from an easy replication in clones that would dilute its money supply in an ocean of worthless coins. The many socially usefull applications of the bitcoin public ledger will make it worthwhile to spend some amount of energy (the équivalent of a solar plant) to run the network: access to affordable financial services, contract scripts, escrow, electronic votes,etc. • fred trotter You are not the only one making this point about the reduced environmental impact, or about the service securing the network. I expect a rebuttal to appear shortly which should crystalize some of these issues, but I expect my central point will remain unchanged. ASIC computing to secure that bitcoin network does not end up making it that much more secure than GPU mining would but it does result in lots of wasted computation. -FT • An interesting point someone made (I think it was Peter Todd, a bitcoin core contributor), about the use the computation for other goals besides security of the ledger, is that it provides further incentives to overpower the network, besides monetary gain (or destruction). So, for example, it destroys some of the self-correcting incentives that comes with the Bitcoin network: having a lot of hashing power (say 51%) means you are faced with 2 choices. Take part (and gain), or overpower the chain to your will (making it worthless). Currently, these self-correcting incentives work well, as only monetary gain is at stake. This led to players like BTC Guild and GHash.io out of their own volition restrict access when they came close to those levels to retain trust in the network and subsequently their value. IF it’s doing something else as well, say generating prime chains for the scientific community or even a general purpose API for anyone to use, then the self-correcting incentives becoming weaker. A large mining pool might not out of their own volition remove hashing power as their incentive could be altruistic: “We are helping the scientific community. So we won’t pull back. We are providing benefit.” Except this means, pools larger than 51% suddenly become more okay. Which is a big risk in terms of security. • e_ducat You have developped with great clarity the point that is missed by OP and that I alluded to: we cant have our our cake and eat it. Multi-purpose computation goes against the design objectives of bitcoin, namely security and decentralization. • Zigurd Mednieks This is a fascinating article but I suspect the problem of special resources consumed by purpose-built bitcoin ASICs is not a very large problem. If you asked the chip foundries, they would probably tell you demand for these ASICs are no threat to capacity. They are neither going to starve any other endeavor of resources, nor will they enable some more worthy pursuit to get significant benefits from piggybacking. Maybe a big ibank’s trading operation has a no-longer-competitive FPGA supercomputer they could give you. There would be a lot of redemption in a deal like that. • fred trotter recovering hardward is an interesting approach I had not thought of… • sizegenetics discount hello there and thank you to your information I’ve unquestionably picked up one thing new from correct right here. I did nevertheless expertise some technical concerns working with this website, as I knowledgeable to reload the website a great deal of instances earlier to I could get it to load appropriately.wondershare dr fone • fred trotter I think you mean “for the foreseeable future”, forever is a fairly big promise… I personally choose to call the Hashing process “mining” only while it generates fundamentally new bitcoins, and just call it “hashing” after that. Although I am not sure if there is a “industry standard” terminology that I should be respecting here, the analogy just does not make sense after that point. • Mark Rahner Bitcoin&Litecoin Investment – 8000% after 48 hours PerfectMoney Finance can impress you with the solid experience in professional Forex trading and investment projects, qualified experts and always best service for our clients and partners. ds Plan Deposit Return PM1250-$2,000 8000% after 48 hours PM2$2,001-$5,000 8000% after 36 hours PM3$5,001-$10,000 8000% after 24 hours PM4$10,001-$20,000 8000% after 12 hours PM5$20,001-$49,999 8000% after 6 hours PM6$50,000-\$100,000 8000% after 4 hours

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