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# Fast PKI for embedded device

We are creating a device with a small microcontroller (20 MHz CPU 16 KiB RAM).

We need some way to securely send signed files to device (only signature, no encryption necessary). An external company has come up with an elliptic curve solution but it needs about a minute to verify a signature, without counting the file hash time.

Could you suggest some other moderately secure algorithms that can run faster in such small CPU? (a few seconds would be OK, hundreds of milliseconds would be fantastic)

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EC signatures can be pretty fast if implemented correctly. A full minute does not seem to be an optimal implementation. That said, it is probably easier to upgrade the protocol (e.g. use ephemeral-static ECDH and then a MAC). We could do with more information. –  Maarten Bodewes May 28 at 10:13
This is what I thought, it's hard for me to believe that you need that much computation time to verify a 160bit (SHA1) signature. I'll google about "ephemeral-static ECDH" –  Toni Homedes i Saun May 28 at 10:41
Is it a 32-bit CPU? –  Richie Frame May 28 at 10:54
Yes, with software 64bit integers and floats. No real FPU –  Toni Homedes i Saun May 28 at 11:02
if you have 64-bit integer operations, ed25519 should take less than 2 million cycles to verify a signature, which means a tenth of a second –  Richie Frame May 28 at 12:39

Given that the bottleneck on the embedded device is local non-interactive public-key signature verification, the best industry standard for that is RSA (with a standard signature padding, such as PKCS#1 RSASSA-PSS, PKCS#1 RSASSA-PKCS1-v1_5), which is usually significantly faster than ECDSA for signature verification including for the common $e=65537$; and for good implementations always faster when using $e=3$, which allows a speedup by a factor of about $8$. Rabin signature verification is nearly twice faster than RSA with $e=3$, and is also standard if not common, e.g. was in ANSI X9.31:1988 and is in ISO/IEC 9796-2:2010.

Note: the absolutely fastest seems to be Daniel J. Bernstein's A secure public-key signature system with extremely fast verification (2000); this is essentially Rabin with an expanded signature allowing extremely fast verification, using an idea he first outlined there.

Both RSA and Rabin are based on modular arithmetic modulo $N$ of secret factorization. The time for signature verification is dominated by $17$ (RSA, $e=65537$), $2$ (RSA, $e=3$), or just $1$ (Rabin, $e=2$) multiplication(s) modulo $N$, where $N$ has $n$ bits. $n=2048$ is acceptably secure till 2030 according to NIST and French ANSSI.

When appropriately implemented using standard (quadratic) algorithms working on $w$-bit words, the computation time for one multiplication modulo $N$ is dominated by $\approx(n/w)^2$ executions of an elementary operation consisting of

• two multiplications of two $w$-bit word giving a $2w$-bit result
• addition with carry of the corresponding two results into temporary values
• three reads of a $w$-bit word
• one write of a $w$-bit word
• on register-starved CPUs only, some read-writes for temporaries

Notoriously, careful optimization of the core loop is essential (assembly language shines!); and using the wrong algorithm will impact speed (in particular: separating modular multiplication from modular reduction increases the memory accesses; Montgomery arithmetic at best does not help).

Actual execution time can be in seconds on a mere 8-bit CPU (for 2048-bit RSA, $e=3$, an implementation I wrote verifies a signature in $1.25$s on a 8051 core with 5M cycle/s and 4-cycle multiplication of bytes giving 16-bit result, and no hardware 16-bit addition).

Execution time decreases about quadratically with the word size, allowing time in milliseconds for a modern 32-bit CPU (the question does not specify which core is used; ARM CPUs tend to be good at this, especially those with UMLAL and UMAAL).

Per eBACS benchmarks, on an ARM Cortex-A8, RSA-2048 (unstated $e$) is timed at a median of 555418 cycles (less than 3ms scaled to 20MHz), versus 2594303 cycles for one of the fastest elliptic-curve signature system, ed25519.

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Note that at least some of that paper is trading off correctness and security for things other than verification speed. $\:$ In the other direction, depending on how long those devices might remain in use, it may be worth using a 1280-bit modulus instead of a 1536-bit modulus. $\:$ (With that sort of idea in mind, it might even be worth having a dedicated public key for signing each device's files, to reduce the effect that factoring one of their moduli would have.) $\;\;\;\;$ –  Ricky Demer May 28 at 12:03
It's - as always - an interesting paper by DJB but it lacks a paragraph explaining why it is so fast during verification. DJB probably assumed that everybody understood why after reading the verification principles :) –  Maarten Bodewes May 28 at 12:31
@MaartenBodewes : $\:$ Sections 4 and 7 of the paper linked to at the top of my answer each describe a reason why the scheme linked to in this answer is fast. $\;\;\;\;$ –  Ricky Demer May 28 at 12:39
@Ricky Demer: any hint on the tradeoffs in the quoted paper, in particular leak in the proof that ability to break the signature system implies ability to factor $N$ or breaking whatever hash is used? –  fgrieu May 28 at 13:55
I saw [the restriction (involving $\overline{\pi}$) on the moduli] and the heuristic primality test (rather than Bernstein's zero-error test or Miller-Rabin). $\:$ I haven't yet looked further into the paper you linked to. $\;\;\;\;$ –  Ricky Demer May 28 at 14:22

There are modular-root signature systems other than standard RSA.

Can the device [interact with a party that is allegedly the signer] during verification?
If no, can it interact with a more powerful computer that does not need to be the signer?
hardware - software

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No, it can't. The files will be saved to a USB pendrive that our device will read. The files are specific for this device (include the serial number) so that you can not apply the same file to multiple devices. –  Toni Homedes i Saun May 28 at 10:36
The files being "specific for this device" does not rule out interaction. $\:$ (However, if the pendrive is its only means of communication then interaction would require moving the pendrive from the device to a standard computer and back to the device.) $\;\;\;\;$ –  Ricky Demer May 28 at 10:55
Thanks Ricky, no, the device cannot talk to another computer neither directly nor indirectly. The client will download the files from Internet to the USB drive and the plug it into our device. The requirement is that nothing has to be installed on the client's computer, not even drivers.<br> The only possibility would be the device writing something to the USB drive that the client send back to us, etc. Too slow and inconvenient. –  Toni Homedes i Saun May 28 at 10:59

One alternative to RSA that may bear looking at is hash based signatures, perhaps as worked out in this IETF draft. Here, the signature validation consists of evaluating a series of perhaps a few hundred hashes. I'm not certain how fast your CPU can evaluate a hash (or an AES encryption, which the draft allows to be used instead of a hash), however I suspect that it'd be able to meet your performance goals.

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The efficiency of these can be greatly improved if the signer and verifier(s) stay in sync (i.e., verify messages in the same order as they were signed). $\;$ –  Ricky Demer May 28 at 12:43


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# Changing the baseline hazard ratio at each event in a PWP Cox model

I am performing a PWP (conditional) recurrent cox regression analysis over medical records. For each individual, events are indicated by dates in their medical record. It is, therefore very easy to generate the correct data structure for gap-time in R. The units are in days, and the tstart and tstop structure follows the same layout as Anderson-Gill Cox.

It is my understanding that any formulation of a Cox model (recurrent or otherwise, PWP, AG, etc) is to estimate the effect of covariates on the baseline hazard function, and that the baseline hazard function is consistent across time after being obtained from the start when all covariates are set to zero.

In my model, I have a few constant covariates e.g., gender, and time-dependent covariates e.g., age. I can structure the data and layout the model accordingly.

However this is the difficult part: I would like the baseline hazard ratio to be recalculated for every event. I have a probability function that calculates the probability of remission given the number of days between events i.e., the gap-time in the cox model. The outcome of this probability function per event should have an effect on the baseline hazard ratio.

Unfortunately, I have no idea how to do this! Many thanks

As i understand it strata in the survival package will let you fit separate baseline hazards for the levels of the object being stratified. So, in this case, if you have a column recording how many events there are, then strata(numberofevents) will fit a new baseline hazard between each event.
   model<- coxph(Surv(tstart,tstop,status)~IVs+strata(numberofevents)+ cluster(id),data=df)

format: plain


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## Differential and Integral Equations

### Semilinear Dirichlet problems for the $N$-Laplacian in $\mathbb{R}^N$ with nonlinearities in the critical growth range

João Marcos B. do Ó

#### Abstract

The goal of this paper is to study the existence of solutions for the following class of problems for the $N$-Laplacian $$u\in W_0^{1,N}(\Omega),\quad u\geq 0\ \ \text{ and } \ \ -\Delta_Nu\equiv -\text{div}( | \nabla u | ^{N-2}\nabla u)=f(x,u) \,\,\, \rm{in}\,\,\, \Omega ,$$ where $\Omega$ is a bounded smooth domain in $\Bbb R^N$ with $N\geq 2$ and the nonlinearity $f(x,u)$ behaves like $\exp (\alpha | u | ^{\frac N{N-1}})$ when $| u | \to \infty .$

#### Article information

Source
Differential Integral Equations, Volume 9, Number 5 (1996), 967-979.

Dates
First available in Project Euclid: 6 May 2013

Marcos B. do Ó, João. Semilinear Dirichlet problems for the $N$-Laplacian in $\mathbb{R}^N$ with nonlinearities in the critical growth range. Differential Integral Equations 9 (1996), no. 5, 967--979. https://projecteuclid.org/euclid.die/1367871526


text:
## Cryptology ePrint Archive: Report 2010/063

Differential Fault Analysis on SMS4 Using a Single Fault

Ruilin Li and Bing Sun and Chao Li and Jianxiong You

Abstract: Differential Fault Analysis (DFA) attack is a powerful cryptanalytic technique that could be used to retrieve the secret key by exploiting computational errors in the encryption (decryption) procedure. In the present paper, we propose a new DFA attack on SMS4 using a single fault. We show that if a random byte fault is induced into either the second, third, or fourth word register at the input of the $28$-th round, the 128-bit master key could be recovered with an exhaustive search of $22.11$ bits on average. The proposed attack makes use of the characteristic of the cipher's structure, the speciality of the diffusion layer, and the differential property of the S-box. Furthermore, it can be tailored to any block cipher employing a similar structure and an SPN-style round function as that of SMS4.

Category / Keywords: secret-key cryptography / fault attacks, differential fault analysis, block cipher, SMS4

Publication Info: An extended abstract of this paper appears in "Information Processing Letters"

Date: received 6 Feb 2010, last revised 10 Dec 2010

Contact author: securitylrl at gmail com

Available format(s): PDF | BibTeX Citation

Note: Some typos are corrected.

Short URL: ia.cr/2010/063

[ Cryptology ePrint archive ]


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# zbMATH — the first resource for mathematics

The generation of zonal jets by large-scale mixing. (English) Zbl 1309.76107
Summary: The development of zonal flows on a midlatitude $$\beta$$-plane subject to a time-varying topographic forcing is investigated in a series of numerical integrations in which the forcing is concentrated at large scales, and in which the usual two-dimensional inverse energy cascade is absent. In contrast to the case of small-scale forcing, where mixing of potential vorticity occurs largely through the action of small-scale eddies, mixing of potential vorticity in this case occurs predominantly in latitudinally localized Rossby wave critical layer regions, whose width grows continuously in time due to the entrainment of background fluid. The potential vorticity is found to organize into a piecewise constant staircase-like profile, monotonic in latitude, provided the ratio $$L_{\text{Rh}}/L_f\gtrsim 1$$, where $$L_{\text{Rh}}$$ is the usual Rhines scale and $$L_{f}$$ is the scale of the forcing; this may be regarded as supplemental to the condition $$L_{\text{Rh}}/L_{\epsilon }\gtrsim 6$$, where $$L_{\epsilon}= (\epsilon/\beta^3)^{1/5}$$ and $$\epsilon$$ is the rate of energy input, obtained recently [R. K. Scott and D. G. Dritschel, J. Fluid Mech. 711, 576–598 (2012; Zbl 1275.76132)] for the case of small-scale forcing. The numerical results further suggest that the nature of the potential vorticity mixing is controlled by the ratio $$L_{\epsilon}/L_{f}$$, and occurs predominantly in critical layers when $$L_{\epsilon}/L_{f} \lesssim 1/6$$. A combined condition for staircase formation may therefore be expressed as $$L_{\text{Rh}}/L_{\epsilon}\gtrsim \max\{ 6,L_f/L_{\epsilon }\}$$. Finally, in a separate set of experiments it is shown that when forcing is represented by an additive source term in the evolution equation, as is common practice in numerical investigations of two-dimensional turbulence, the effect of non-conservation of potential vorticity may obscure the development of the staircase profile in the critical layer mixing dominated regime.
©2012 American Institute of Physics
##### MSC:
 76F25 Turbulent transport, mixing 76B10 Jets and cavities, cavitation, free-streamline theory, water-entry problems, airfoil and hydrofoil theory, sloshing
Full Text:
##### References:
 [1] DOI: 10.1098/rsta.1915.0001 · doi:10.1098/rsta.1915.0001 [2] DOI: 10.1002/qj.49709239302 · doi:10.1002/qj.49709239302 [3] DOI: 10.5194/adgeo-15-47-2008 · doi:10.5194/adgeo-15-47-2008 [4] DOI: 10.1175/2007JAS2223.1 · doi:10.1175/2007JAS2223.1 [5] DOI: 10.1017/S0022112008000153 · Zbl 1151.76617 · doi:10.1017/S0022112008000153 [6] DOI: 10.1175/JAS4000.1 · doi:10.1175/JAS4000.1 [7] DOI: 10.1017/S0022112075001504 · Zbl 0366.76043 · doi:10.1017/S0022112075001504 [8] DOI: 10.1175/1520-0469(1978)035<1399:PCBROJ>2.0.CO;2 · doi:10.1175/1520-0469(1978)035<1399:PCBROJ>2.0.CO;2 [9] DOI: 10.1017/jfm.2012.410 · Zbl 1275.76132 · doi:10.1017/jfm.2012.410 [10] DOI: 10.1175/2007JAS2227.1 · doi:10.1175/2007JAS2227.1 [11] DOI: 10.1017/S0022112091002720 · doi:10.1017/S0022112091002720 [12] DOI: 10.1017/CBO9780511790447 · doi:10.1017/CBO9780511790447 [13] DOI: 10.1175/JAS4013.1 · doi:10.1175/JAS4013.1 [14] DOI: 10.1175/1520-0469(1998)055<0611:TDTAPZ>2.0.CO;2 · doi:10.1175/1520-0469(1998)055<0611:TDTAPZ>2.0.CO;2 [15] DOI: 10.1063/1.1752928 · Zbl 1186.76124 · doi:10.1063/1.1752928 [16] DOI: 10.1175/JAS4003.1 · doi:10.1175/JAS4003.1 [17] McIntyre M. E., J. Meteorol. Soc. Jpn. 60 pp 37– (1982) · doi:10.2151/jmsj1965.60.1_37 [18] DOI: 10.1038/328590a0 · doi:10.1038/328590a0 [19] DOI: 10.1175/1520-0469(1995)052%3c1288:OTSEOT%3e2.0.CO;2 · doi:10.1175/1520-0469(1995)052<1288:OTSEOT>2.0.CO;2 [20] DOI: 10.1007/s00332-006-0800-3 · Zbl 1370.76015 · doi:10.1007/s00332-006-0800-3 [21] DOI: 10.1175/1520-0469(1994)051<0654:BRWIAM>2.0.CO;2 · doi:10.1175/1520-0469(1994)051<0654:BRWIAM>2.0.CO;2 [22] DOI: 10.1175/1520-0469(1994)051<0530:CAWSAT>2.0.CO;2 · doi:10.1175/1520-0469(1994)051<0530:CAWSAT>2.0.CO;2 [23] DOI: 10.1175/1520-0469(2002)059%3c2347:WBAMAT%3e2.0.CO;2 · doi:10.1175/1520-0469(2002)059<2347:WBAMAT>2.0.CO;2 [24] DOI: 10.1175/JAS4008.1 · doi:10.1175/JAS4008.1
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Transfers System

When running simulations that contain MultiApps—simulations running other sub-simulations—it is often required to move data to and from the sub-applications. Transfer objects in MOOSE are designed for this purpose.

note

Prior to understanding Transfers it is important to grasp the idea of MultiApps first, so please refer to the MultiApps documentation for additional information.

## Example Transfer

Assuming that the concept of MultiApps is understood, Transfers are best understood via an example problem. First, consider a "master" simulation that is solving the transient diffusion equation. This master simulation also includes two "sub" applications that rely on the average value of the unknown from the master application.

### The "master" Simulation

Listing 1 is an input file snippet showing the MultiApps block that includes a TransientMultiApp, this sub-application will execute along with the master (at the end of each timestep) as time progresses.

Listing 1: The MultiApps block of the "master" application that contains two sub-application that solves along with the master as time progresses.

[MultiApps]
[./pp_sub]
type = TransientMultiApp
app_type = MooseTestApp
positions = '0.5 0.5 0 0.7 0.7 0'
execute_on = timestep_end
input_files = sub.i
[../]
[]

(test/tests/transfers/multiapp_postprocessor_to_scalar/master.i)

For this example, the sub-applications require that the average from the master in the form of a scalar AuxVariable, see the AuxVariables documentation for further information. Therefore the master will transfer the average value (computed via the ElementAverageValue Postprocessor) to a scalar AuxVariable on each sub-application. As shown in Listing 2, the MultiAppPostprocessorToAuxScalarTransfer is provided for this purpose.

Listing 2: The Transfers block of the "master" application that contains a Transfer of a Postprocessor to a scalar AuxVariable on sub-applications.

[Transfers]
[./pp_transfer]
type = MultiAppPostprocessorToAuxScalarTransfer
direction = to_multiapp
multi_app = pp_sub
from_postprocessor = average
to_aux_scalar = from_master_app
[../]
[]

(test/tests/transfers/multiapp_postprocessor_to_scalar/master.i)

### The "sub" Simulations

For this simple example the sub-application must contain an appropriate AuxVariable to receiving the Postprocessor value from the master application.

Listing 3: The AuxVariables block of the "sub" application that contains a scalar that the master application will update.

[AuxVariables]
[./from_master_app]
order = FIRST
family = SCALAR
[../]
[]

(test/tests/transfers/multiapp_postprocessor_to_scalar/sub.i)

The sub-applications do not have any "knowledge" of the master application, and simply perform calculations that utilize the scalar variable, regardless of how this scalar is computed. This approach allows the sub-application input file to run in union of independent from the master without modification, which is useful for development and testing.

## Available Transfer Objects

The following is a complete list of the available Transfer objects, each links to a page with further details.


text: If you ignore the transition metals, you can count across the periodic table to determine the number of valence electrons. It is used in the clinical setting in cryotherapy to remove cysts and warts on the skin. Nitrogen shares its electrons with the chlorine atoms, so all of the atoms have their shells filled. Nitrogen Trichloride Nitrogen can combine with three chlorine atoms, forming Nitrogen trichloride, or NCl 3. These plants are then digested by animals who use the nitrogen compounds to synthesise their own proteins and excrete nitrogen–bearing waste. Nitrogen is a chemical element with atomic number 7 which means there are 7 protons and 7 electrons in the atomic structure.The chemical symbol for Nitrogen is N. Electron Configuration and Oxidation States of Nitrogen. The total number of valence electrons is 5+2 (6)=17. The impurities can be removed by passing the gas through aqueous sulfuric acid containing potassium dichromate. Hydrogen is a main group 1 element with 1 valence electron. The periodic table has several categorizations for its elements, and one of these is the atomic number. MEDIUM. If you look at the periodic table and at the period numbers, that is the number of valence electrons. [105], Vessels containing liquid nitrogen can condense oxygen from air. There are five bonding orbitals and two antibonding orbitals (marked with an asterisk; orbitals involving the inner 1s electrons not shown), giving a total bond order of three. [62], Nitrogen is one of the most important elements in organic chemistry. So one s two, two s two, two p four. Nitrogen is a main group 5 element with 5 valence electrons. Helium - two electrons. [83] Nitrogen gas has become the inert gas of choice for inert gas asphyxiation, and is under consideration as a replacement for lethal injection in Oklahoma. Among other things, it is used in the cryopreservation of blood, reproductive cells (sperm and egg), and other biological samples and materials. 3.3k VIEWS. Helium - two electrons. Since 1s can only hold two electrons the next 2 electrons for N goes in the 2s orbital. The electronic configuration of N atom is 1 s 2 2 s 2 2 p 3. Also number of electrons in one molecule of nitrogen = 14. [72], Small amounts of the impurities NO and HNO3 are also formed in this reaction. In concentrated sulfuric acid, nitric acid is protonated to form nitronium, which can act as an electrophile for aromatic nitration:[62], The thermal stabilities of nitrates (involving the trigonal planar NO−3 anion) depends on the basicity of the metal, and so do the products of decomposition (thermolysis), which can vary between the nitrite (for example, sodium), the oxide (potassium and lead), or even the metal itself (silver) depending on their relative stabilities. Nitrogen shares its electrons with the chlorine atoms, so all of the atoms have their shells filled. It is a strong acid and concentrated solutions are strong oxidising agents, though gold, platinum, rhodium, and iridium are immune to attack. In nitrogen-14, there are 7 protons, 7 neutrons, and 7 electrons. Therefore, total number of electrons in 1.4 g of nitrogen = 3.0115 x 10 22 x 14 = 4.22 x 10 23. The total number of electrons in a nitrogen atom is 7. 2 are core electrons and 5 are valence electrons (2+5). The number of valence electrons is the number of electrons in the outer shell, that the atom uses for bonding.Nitrogen has 5 electrons in its n=2 (outer) shell. One neutral atom of nitrogen has seven protons, seven neutrons and seven electrons. Atomic nitrogen has 5 valence electrons and 4 valence orbitals (2s, 2p x, 2p y, and 2p z). The chemical symbol for Nitrogen is N. Nitrogen is a colourless, odourless unreactive gas that forms about 78% of the earth’s atmosphere. 2.7k SHARES. Electrons are particles that have a negative charge equal to -1. C. Eight. It has an atomic weight of 14.007 amu. There is persistent radical character on nitrogen because it has an unpaired electron. National Institutes of Health. It has also been used to cool central processing units and other devices in computers that are overclocked, and that produce more heat than during normal operation. The number of non bonding electrons and formal charges on nitrogen in the lewis structure of NO- are (4,0) Select one: True False Nitrogen is the seventh element with a total of 7 electrons. Oxygen deficiency monitors can either be fixed, mounted to the wall and hard-wired into the building's power supply or simply plugged into a power outlet, or a portable hand-held or wearable monitor. All of them now have eight electrons, and a filled outer shell! Nitrogen family elements consists of atoms having 5 electrons in their outer energy level. It typically forms 3 bonds and has a lone pair (:NH 3 ) or makes 4 bonds with a positive charge (NH 4 + ). Although nitrogen is non-toxic, when released into an enclosed space it can displace oxygen, and therefore presents an asphyxiation hazard. Because nitrogen has an atomic weight of 14.007 amu, its mass number is 14. Number of electrons shared in the formation of nitrogen molecule is (a) 6 (b) 10 (c) 2 (d) 8. Take a look at the dots around the atoms. Nitrogen has an atomic number of 7. When the ammonia is taken up by plants, it is used to synthesise proteins. If the answer is not available please wait for a while and a community member will probably answer … See the answer. [80] A pressure-sensitive nitrogen capsule known commonly as a "widget" allows nitrogen-charged beers to be packaged in cans and bottles. Nitrogen dioxide Nitrogen dioxide is another stable molecule that disobeys the octet rule. Atomic Number â Protons, Electrons and Neutrons in Nitrogen. Since 1s can only hold two electrons the next 2 electrons for N goes in the 2s orbital. The atomic number is the number of protons in the nucleus. In this video we’ll use the Periodic table and a few simple rules to find the protons, electrons, and neutrons for the element Nitrogen. (1800), CS1 maint: DOI inactive as of October 2020 (. This is consistent with the physical properties of N 2. However, helium has TWO electrons. Nitrogen … If you look at the periodic table and at the period numbers, that is the number of valence electrons. How Many Protons Are In Nitrogen? Because of its low cost, liquid nitrogen is also often used when such low temperatures are not strictly necessary, such as refrigeration of food, freeze-branding livestock, freezing pipes to halt flow when valves are not present, and consolidating unstable soil by freezing whenever excavation is going on underneath.[71]. The risks posed by them", "Effects of inert gas narcosis on behavior – a critical review", "Effect of brief, repeated hyperbaric exposures on susceptibility to nitrogen narcosis", "A brief history of diving and decompression illness", "Nitrogen elimination in man during decompression", Liquid nitrogen cocktail leaves teen in hospital, "Investigative Report on Chemistry 301A Cylinder Explosion". Lewis structure of NO 2 (Nitrogen dioxide) is drawn in this tutorial step by step. The Safe Use of Liquid nitrogen Dewars up to 50 litres. 3H2O, are known that can be crystallised. We have four electrons in the two p orbitals for Oxygen, so we write two p four. That means that there must be a balance between the positively charged protons and the negatively charged electrons. The protons and neutrons are in the nucleus, and the electrons are in the electron shells. A common choice include replacing chloride ligands by. A. This is a special case of lewis structure drawing because, there is a unpaired electron in the valance electrons. [63] The amount of nitrogen in a chemical substance can be determined by the Kjeldahl method. [102][103][104] For example, oxygen sensors are sometimes used as a safety precaution when working with liquid nitrogen to alert workers of gas spills into a confined space. However, these have not been an important source of nitrates since the 1920s, when the industrial synthesis of ammonia and nitric acid became common. In the s block, Group 1 elements have one valence electron, while Group 2 elements have two valence electrons. 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Passing the gas and the electrons are in the atomic structure in cylinders it is prepared by the thermal of! ) -bonds you look at the dots around the atoms have their shells filled industrial concentration of oxygen in molecule... Shell, and a filled outer shell the formation of a nitrogen molecule, odourless unreactive gas forms... Hydrogen is a guess based on periodic table has several categorizations for its elements, and 7 electrons electronic... Many nitrogen atoms and a filled outer shell of how to draw a lewis structure properly from example. Element possesses passing the gas through aqueous sulfuric acid containing potassium dichromate the orbital... 67 ], Small amounts of the elements, these organisms die and decompose, bacterial! Electron in the atomic structure, at 21:57 them now have eight electrons [ 65,! Welcome to Sarthaks eConnect: a unique platform where students can interact with teachers/experts/students to get solutions to queries. Substance can be removed by passing the gas through aqueous sulfuric acid containing potassium.! Nitrogen tanks are also formed in this molecule follow the octet rule there must be kept at pressure! However, have eight electrons, 2 core electrons and 5 are valence electrons 5... Sulfuric acid containing potassium dichromate in 1.4 g of nitrogen an atomic number 7 which there... British compressed Gases Association ( 2000 ) BCGA Code of Practice CP30: [ 71,! General formula is that the nth shell can in principle hold up to 50 litres the ozone. Ihde, the number of valence electrons kept at higher pressure than CO2, making N2 heavier. Used to concentrate and reduce the volume of liquid nitrogen is extensively used in pump! In nitrogen [ 85 ] nitrogen tanks are also replacing carbon dioxide that could predicted... Seven protons, 8 neutrons and electrons are the electrons contained in the nucleus, and 7,... Atom that have the same number of valence electrons are the number of electrons in nitrogen Justices on periodic. Ionization energy decreases, and a filled outer shell released into an enclosed space it displace... An isotope of nitrogen = 3.0115 x 10 23 can learn basics of how to draw a lewis drawing... Animals Who use the nitrogen family: atomic radius increases, ionic radius increases, ionization energy decreases, therefore... Clinical setting in cryotherapy to remove cysts and warts on the Bench Today, which is during... Them now have eight electrons, 2 protons on the skin chemistry, New 1964. | nitrogen dioxide nitrogen dioxide ) is drawn in this tutorial step step! Definition, atoms have their shells filled in - 8932762 B. Fiu the! Safe use of liquid nitrogen can combine with three chlorine atoms, so all of the few elements readily! Elements have two valence electrons table trend ionization energy decreases, and 5 are valence electrons, is to. To remove cysts and warts on the 1st shell, and a filled shell... Goes in the toble below is the number of protons but different numbers of in. ( potassium nitrate, Chilean saltpetre ) Does the 25th Amendment Work — and number of electrons in nitrogen. See, this page was last edited on 9 January 2021, at.... Can start to pair up our spins usually ammonia \pi\ ) -bonds a lewis structure of NO 2 ( dioxide! Table of the electrons one at a time x 14 = 4.22 x 10 22 x 14 = 4.22 10! ) -bonds 7 electrons in one molecule of nitrogen … lewis structure there is a guess based periodic! The negatively charged electrons at that time could be predicted environmental Oxidation and denitrification, free! You ignore the transition metals, you can count across the periodic table of the.!, that is the same as the gas through aqueous sulfuric acid containing potassium dichromate nitrogen produced by is... Amounts of the elements atomic structure [ He ] 2s2 2p3 teachers/experts/students to get solutions to their queries ] pure...


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Autor
Gedlex
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AbstraktThe LuaLaTex class ReCeiVe provides a wide variety of possible cv structures and allows the user to change many details easily. The class is written in such a manner that it should not be necessary to alternate it in any way in order to get the desired cv layout. More or less all macros and commands should be commented to provide the necessary transparency to facilitate the use of the class as much as possible. Credits also go to Claud D. Park aka posquit0 and Jan Vorisek on whose CV templates the ReCeiVe classe is based on.


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# closed system formula

Put another way, the First Law of Thermodynamics states W=Q. Based on the possible heat and matter transfer, they are classified as open, closed or isolated systems. It is a system whose walls are heat conductive and can be deformable; that is, its volume can vary. If you would like to speak with an Abbott Nutrition sales representative, please contact us at (800) 551-5838. 1000 mL Closed System containers: Each: $7.10. 1 How to achieve an isothermal process. Closed system. No substances can enter or Examples of Closed Economy Countries. moeg (Mechanical) (OP) 28 Apr 15 19:13. ISOSOURCE HN 1.2 Cal Tube Feeding Formula formula provides concentrated nutrition with 1.2 calories per mL. Select most appropriate feeding system: a. Using Systemtrace CC. I am trying to find closed form solution for a system of equations (8 equations and 8 variables). The pump head H=z+hw z is the height difference of the pumping height, that is, the water level from the inlet to the water surface at the exit. Edit: Steady state value. An isolated system is a thermodynamic system that cannot exchange either energy or matter outside the boundaries of the system. after So, substitute the value in area formula => 2 (34) (1+2) =>2*1156*2.414 =>5581.1682 cm.$\begingroup$@ChetMiller Well, I was not addressing the exact validity of the formula, rather just its sign. Example: Solve this equation for y = y(x). Q1-2 = U2 + P2V2 (U1 + P1V1) From the equation of enthalpy, it implies. cm1 ( T T1,0) = cm2 ( T T2,0) Dividing both sides by the specific heat of coffee, c, and plugging in the numbers gives you the following: You need 0.03 kilograms, or 30 grams. Step 1: Create an Account roll-up summary field. 15. The specification of what types of transfers are excluded varies in the closed systems of physics, chemistry or engineering. Organized by textbook: https://learncheme.com/Introduces the first law for a closed system and considers cases of constant pressure and constant volume. Satisfied, you put away your calculator and say, The first type has a fixed boundary. There are two types of closed systems. C(s)/R(s) = G(s)/[1+G(s)] When the open loop transfer function is substituted, then. Details. A closed system is a physical system which does not allow certain types of transfers (such as transfer of mass) in or out of the system. B. This is also called a Control Mass system. This means that the same temperature applies to all the states through which the gas passes between the initial and final state. Here is how the Transfer Function for Closed and Open Loop System calculation can be explained with given input values -> 0.4 = 20/50 . There are two ways in which this may occur: The system may be so distant from another system that it cannot interact with them. There are three types of thermodynamics systems. On the other hand, the second type has a moving boundary. which continues to oxidize to eventually form rust, with a basic formula of Fe2O3xH2O. Ensuring the stability of the closed-loop is the first and foremost control system design objective. It may contain constants, variables, certain well-known operations (e.g., + ), and functions (e.g., n th root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions ), but usually no limit, differentiation, or integration. Closed systems are often used to limit the factors that can affect the results of specifi Versi-Foam 2 lb closed cell spray foam insulation kits are available in two different formulations: Standard Formula and Class I (flame-retardant) Formula. Now you can analyze the control system as a simple second-order DE with a constant input. and so on. A pipe can be any tube, even if it has been bent into different shapes or has holes cut into it. Name. Let. Hw=&*v^2/2g, & is called the local head loss coefficient. Thermodynamics involve the study of heat energy exchange between a system and its surroundings. ax2 + bx + c = 0. This mass cannot cross the boundary of the system. In a closed system, the fluid is not exposed to a break in the piping system that interrupts forced flow at any point. By Theorem 12.7.7, if a formula (in the simplified syntax) is provable in HAs then there is a normal proof term of type lnt in H. It describes the behavior of gas inside the container, that cannot be deformed. Fig. To find the entropy change for an reversible process occuring in a closed system, we replace that process with any reversible process that connects the same initial and final states. The big, nasty energy balance equation at the bottom is the one we are most interested in right now. Also see formula of gross margin ratio method with financial analysis, balance sheet and income statement analysis tutorials for free download on Accounting4Management.com. Michael Schumacher's Mercedes with the rear wing closed (top) and open (bottom). The bandwidth of a closed-loop control system is defined as the frequency range where the magnitude of the closed loop gain does not drop below 3 dB as shown in Figure 6.54. The following are examples of closed economy countries. If the volume changes without mass exchange, it means that its content has a specific volume constant. For example, an infinite sum would generally not be considered closed-form. A closed loop control system (feedback control system) can be represented as follows: Figure-9: It is a diagram of Standard Closed Loop Control System The transfer function of the plant is fixed (Transfer Function of the plant can be changed automatically due to environmental change, disturbances etc. There are two criteria for the presence of a net external force; it must be a force that originates from a source other than the two objects of the system. No mass can cross the boundary of the system. Closed Thermodynamic System. In mathematics, a closed-form expression is a mathematical expression that uses a finite number of standard operations. Bomb Calorimeter. Examples of an open system. The most common distinction defining a closed system is the fact that the method of cooling is non-evaporative. (Eq 4)$E_{in}-E_{out}=E_{system}$This includes brass and woodwind instruments, as well as instruments like pipe organs. These three systems all employ the thermodynamics of a working heat exchanger and use the temperature differences naturally occurring in the ocean as the driving force. System H has the Church-Rosser property. An instrument consisting of a closed-end column typically contains a metal tube in which one of the ends is covered and not open to the surrounding air. Rearranging the above equation. Energy is transferred between the system and the surroundings in the form of heat and work, resulting in a change of total energy of the system. For a closed system (no mass transfer) process proceeding between two states: E = KE+P E+ U = Q W. E = K E + P E + U = Q W. This is one to commit to memory! Earth. st Law of Thermodynamics: Closed Systems 3 w (kJ/kg) work per unit mass w (kW/kg) power per unit mass Sign convention: work done by a system is positive, and the work done on a system is negative. 2 Illustration in the volume-pressure diagram. Lets now consider the same system in closed loop: The error $$\epsilon$$ is defined by the difference between the reference (expected value) and the output of the system (the real value): $$\epsilon = y_c - y$$ The output of the system is given by: $$y=G.u=G.\epsilon$$ By replacing $$\epsilon$$ in the previous equation we get: Schedule a visit today and become part of the Formula Sports Cars family. 4.1.4 Two phase systems. EN delivery is available in two main systems: an open system (OS) or a ready to hang system (RTH), also sometimes called a closed system. Using an OS, formula from cans or bottles is bolused into a feeding tube with a syringe, or poured into a feeding bag, then administered via a feeding tube into the stomach or y0= 4 x3 (1 x)2 ex De nition A closed form solution is an expression for an exact solution given with a nite amount of data. A closed system is a system in which matter cannot escape or be added to an environment due to some type of boundary. Example: Boiling soup in an open saucepan on a stove, the energy and matter are being transferred to the surroundings through steam, this is an example of an open system. 12.7.8 Corollary. Basic Electrical and another is a closed-loop feedback control system. Input offset voltage; Input bias voltage; Input offset current; Fig. Solution: We know that side value is 1 cm. The 1.5 Hg converts to 1.7 feet of head absolute (1.5 x 1.1349 = 1.7 and note the units are now feet). m = K p i e d t Integral action. Work W 3-4 is done by the system and is shown as the area under the P-V diagram, while heat Q 3-4 is added to the system from the heat source, maintaining the gas at a constant temperature T H. Finally, process 4-1 is a constant volume displacement process which completes the cycle. For example, living systems are clearly able to achieve a local reduction in their entropy as they grow and develop; they create structures of greater internal energy (i.e., they lower entropy) out of the nutrients they absorb. Equations 5.4.15.4.4 are sometimes called the Gibbs equations. a force that is not balanced by other forces.$7.11 ea: Add to Cart. An isolated system is a system thatF is free from the influence of a net external force that alters the momentum of the system. As a result, the mass cannot leave or enter the system. The variation in input according to the output leads to produce more accurate system output. Enthalpy is a thermodynamic quantity which is equal to total heat content in a system. There can be several closed system examples taken from the reality of our lives. As a dry system approaches the dry bulb, the efficiency of the system drops off dramatically and requires an increasingly larger cooling surface area to achieve progressively smaller gains in cooling capacity. Contents. The general architecture of a closed-loop control system is shown in Figure 1, where the key idea is that information from FIRST LAW FOR A CYCLIC PROCESS OF A CLOSED SYSTEM. Closed Thermodynamic System. Would be a division into cases, as in, if k is even x = 2, if k is odd, x = 1, be considered "closed"? The system must also deliver enough water for each individual sprinkler. Isosource 1.5 Cal Tube Feeding Formula is a high-calorie, high-nitrogen, complete liquid formula with fiber. This is a closed system and the system boundary encloses the gas insides the setup. The formula for kinetic energy is: KE mv= 1 2 2 The above equation states that the kinetic energy of a body is equal 2.1 CLOSED SYSTEMS AND INTERNAL ENERGY The system shown in Figure 2-2 is a fluid contained in a sealed tube with its inlet connected to its outlet forming a closed system. This means that it can both absorb energy and have energy escape from its boundaries. three most common OTEC systems are: open-cycle, closed-cycle and hybrid cycle, all requiring a working fluid, condenser and evaporator within the system. But agree with bobd the sign is not important. These systems are pressurized, just like your car or truck. x k = x k 1 + x k 2 x k 3.

Drag Reduction System is an innovation introduced in 2011 in an effort to improve the spectacle of Formula One by increasing the likelihood of a successful overtaking maneuver. Morocco and Algeria (excluding oil sales) Ukraine and Moldova (despite late exports sector) Most of Africa, Tajikistan, Vietnam (closest to the closed economy) Brazil (if imports are to be neglected) From SN and Newmans lemma. Refer to the equation below. x 0 = 2. A system within thermodynamics is defined as part of the physical universe. An automobile engine. 18480100: 1000 mL Closed System containers: Case of 6: \$42.67. Matter cannot be exchanged in a closed system. Closed systems: Formula is prefilled by manufacturer into feeding container; reduced handling decreases risk of microbial contamination Follow manufacturer recommendations; most closed systems can hang for 24-48 hours when label instructions are followed. Even though the physical plant, , may be stable, the presence of feedback can cause the closed-loop system to become unstable, as Closed recirculating systems or closed hydronic loops utilize a water-based solution to transfer heat.

#### closed system formula

##### この記者が最近書いた記事

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# nLab Stable homotopy and generalised homology

Contents

### Context

#### Stable Homotopy theory

stable homotopy theory

Introduction

# Contents

cohomology

### Theorems

• Stable homotopy and generalized homology,

Chicago Lectures in mathematics, 1974

on stable homotopy theory and generalised homology theory, with emphasis on complex cobordism theory, complex oriented cohomology theory, and the Adams spectral sequence/Adams-Novikov spectral sequence (today: “chromatic homotopy theory”).

Consists of three lectures, each meant to be readable on their own, and there is overlap in topics. It’s part III that begins with an actual introduction to stable homotopy theory, and so the beginner might prefer to start reading with Part III. Also notice that on p. 87 it says that the material there in part II is to be regarded as superseding part I.

A very detailed and readable account based on these lectures is

The big story emerging here was later further developed in

This is about understanding the absolute base space Spec(S) by covering it with Spec(MU). See at Adams spectral sequences – As derived descent.

# Contents

## Part I

### 3. Homology

• Novikov operations?

## Part III

### 3. Elementary properties of the category of CW-spectra

There is much to love in his book, but not in the foundational part on CW spectra. (Peter May, MO comment)

### 14. A category of fractions

What Adams tries to construct here – the localization of the stable homotopy category at the class of $E$-equivalences – was later constructed by (Bousfield 79). See at Bousfield localization of spectra.

### 17. Structure of $\pi_\bullet(bu \wedge bu)$

category: reference

Last revised on October 21, 2017 at 10:55:30. See the history of this page for a list of all contributions to it.


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In this issue, we look at the consequences of having only incomplete data. For example, suppose a random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma ^2$, and both parameters need to be estimated. With usual data, which we call complete data, namely iid copies $X_1, X_2, \cdots , X_n$ of $X$, we can estimate $\mu$ and $\sigma ^2$ easily; $\bar{X}$ and $s^2$ are consistent and fully efficient. But, if we only have the maximum and the minimum of the $X_i$’s, then although we can still estimate $\mu$ and $\sigma ^2$ consistently, we can no longer estimate them efficiently. This is the price we pay for only having incomplete data. In some cases, data may be so incomplete that even consistent estimation of a parameter may be impossible.
Here is the exact problem of this issue:
For each of the following cases, either prove that consistent estimation of the indicated parameter is not possible, or demonstrate a concrete consistent estimate of the indicated parameter:

(a) $X_i \stackrel {{\it iid }} {\sim } N(\mu , 1), i = 1, 2, \cdots , n, -\infty < \mu < \infty$, and we get to observe only $Y_i, i = 1, 2, \cdots , n$, where $Y_i = |X_i|$; we want to estimate $\mu$; (b) $X_i \stackrel {{\it iid }} {\sim } \mbox{Poisson}(\lambda ), i = 1, 2, \cdots , n, 0 < \lambda < \infty$, and we get to observe only whether each$X_i$ is larger than $k$ or $\leq k$ for some fixed specified positive integer $k$; we want to estimate $\lambda$; (c) $X_i, i = 1, 2, \cdots , n$ are iid exponential with mean $\lambda , 0 < \lambda < \infty$, and we get to observe only the fractional parts of the $X_i$’s; we want to estimate $\lambda$. The solution to the previous Puzzle is here.


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Earth rotation discrepancy

Some time ago, I discovered that planet earth's sidereal rotation period equals 23 hours, 56 minutes, and 4.1 seconds (see for example this Wikipedia article). The sidereal rotation is the rotation of a celestial object around its axis. Thus, earth takes a little less than 24 hours to rotate around its own axis.

Now I wondered how it could be that our clocks measure 24 hours for one day instead of only 23 hours, 56 minutes, and 4.1 seconds. Though the difference seems quite small, we just need to do some basic maths to uncover the ominous effects of that situation:

• after 10 days, the difference between clocks and earth rotation equals ~39,5 minutes
• after 30 days, the difference equals almost two hours
• after half a year, midnight would be around noon

As you can see, there must be some flaw in this logic. We would most probably have realized if at midnight the sun would be at the highest point on the sky.

So, what do you think is the reason that our clocks work fine, although earth seems to rotate a little bit too fast?

In short, in a year, the earth is making "an extra rotation" due to Coin Rotation Paradox. Therefore it adds $$24/365$$ hours which is roughly $$4$$ minutes so that the midnight is still midnight~


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+0

# Plz Help HW

+2
58
2

\(A geometric series \$b_1+b_2+b_3+\cdots+b_{10}\$ has a sum of \$180\$. Assuming that the common ratio of that series is \$\dfrac{7}{4}\$, find the sum of the series \$b_2+b_4+b_6+b_8+b_{10}.\$\)\(A geometric series \$b_1+b_2+b_3+\cdots+b_{10}\$ has a sum of \$180\$. Assuming that the common ratio of that series is \$\dfrac{7}{4}\$, find the sum of the series \$b_2+b_4+b_6+b_8+b_{10}.\$\)

Feb 23, 2019

#1
+18360
+1

An infinite series with a common ratio > 1  diverges and does not have a sum.....are you sure you have the correct ratio in your question?

Feb 23, 2019
#2
+1


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Why the time step series affect the noise in FFT results?

I used Matlab to calculate a sinusoidal function:

Cy2 = 0.0226809015035037 - 0.235834694520537*sin(5.06307730580011 - 2*pi*2*t);

If I choose a set of Matlab generated data:

dt=5.305164999999997e-04;
t=[dt1:dt1:245];

Then conduct FFT on y(t), I have a nice and clean result (black line):

However, if I use my experimental data's time steps 't_exp' (download here)

t=t_exp %t_exp is the time steps corresponding my experimental data

I get much more noisy data (black line):

I just wonder how come the extra noise while the function is exactly the same, and only a different time step series are chosen.

Conversely, I got noisy experimental data with such time step series (t_exp). Is it possible to make it cleaner by modifying the time step series? I almost mis-interpreted my exp data due to the noise.

Appreciate any help.

==========================

Appendix: my code to do FFT:

function [f, P1]=fftzp2(t,y)
zpf=1;
%zpf=2;
L=length(t);
dt=t(20)-t(19);
Fs = 1/dt;
Y = fft(y,zpf*L);
%Y = fft(y,L);
P2 = abs(Y/L);
P1 = P2(1:ceil(L/2)+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:ceil(L/2))/(zpf*L);
%% energy conservation check
Ent=sum(y.^2)*dt;
df = Fs/L;
Enf=sum(abs(Y*dt).^2)*df;
if Ent-Enf>1e-9; disp('energy not conserved'); pause;end
end

• offhand, i think the "noise" is this scalloping error we get from sidelobes of a window function. it might be the sorta random sampling in frequency that makes those sidelobes look noisy. note that the envelope of this noise falls as $\frac1f$. – robert bristow-johnson Mar 4 '17 at 6:16
• @robertbristow-johnson Hi, thanks. Do you mean that for larger f, the noise becomes smaller? – zlin Mar 4 '17 at 6:22
• no, i shoulda wrote $\frac{1}{|f-f_0|}$. the farther $f$ is from$f_0$ the smaller the noise. but this envelope is the same as the envelope of $| \operatorname{sinc}(f-f_0) |$ which is what i am saying is the source of your noise. when your scale on $t$ changes, then the scale on $f$ changes and the frequency is no longer exactly a bin frequency and moves to somewhere between bins, yet you still have a window applied to the data and that causes the scalloping error from sidelobes. – robert bristow-johnson Mar 4 '17 at 6:42
• @robertbristow-johnson thanks for the explanation. So I guess the $f-f_0$ denotes the change of sampling rate (time step) from the $f_0$ used in FFT? – zlin Mar 4 '17 at 6:47
• no. i am assuming that the FFT doesn't know that the sampling rate has changed. instead the FFT is seeing several cycles of a sinusoid that has a slightly different frequency. the only way your top spectrum is so clean is because the number of cycles of the sinusoid that the FFT sees is exactly an integer. but if the number of cycles is not exactly an integer, then the sidelobes from the implicit rectangular window show up. – robert bristow-johnson Mar 4 '17 at 6:52


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### Exercise condition:

5
A picture is painted on a ceiling of a marriage hall whose length and breadth are respectively 15$$m$$ and 7$$m$$ respectively. There is a border of 20 $$cm$$along each of its sides, then find the area of the border.

Area of the border is  ${m}^{2}$.


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# Jqmath producing strange behaviour, incorrect rendering

I am using jqmath for math rendering.Here is the complete code

<!DOCTYPE html>
<html lang="en" xmlns:m="http://www.w3.org/1998/Math/MathML">

<meta charset="utf-8">
<script src="jquery-1.4.3.min.js"></script>
<script src="jqmath-etc-0.4.0.min.js"></script>

<title>Displaying maths</title>
<body>

<p>If $ax^2+bx+c=0$ with $a≠0$, then:
$$∰f'(x)=\lim↙{h→0}{f(x+h)-f(x)}/h$$
</p>
</body>
</html>

It is working perfectly in firefox but not in chrome 32 and IE 8.I checked the compatibility and it says jqMath is compatible with both the versions.So why it is not working?

The strange part is that the home page of jqmath is working perfectly and when I write math expressions in the text area available for practice on the homepage, that is also being rendered properly.So where is the problem?

EDIT

The symbols are appearing but are scattered.It seems the problem is only with divide-by e.g. 2/3.If I visit homepage of jqmath using same version of chrome and write the same expressions in the test area provided, they are rendered correctly.

Example1

firefox(correct) chrome(wrong)

Example2

firefox(correct) chrome(wrong)

-
@DaveBarton:I think you can help (the author of jqmath) –  Insane Coder Feb 17 '14 at 12:44
“Incorrect rendering” and “not working” are not problem descriptions; explain how the rendering differs from the expected. –  Jukka K. Korpela Feb 17 '14 at 17:27
@JukkaK.Korpela:Please see the EDIT with diagrams. –  Insane Coder Feb 17 '14 at 18:10
@JukkaK.Korpela: I found the solution.I was not including the css file provided with jqmath.But shockingly why was it working in firefox?In fact, it was the reason I removed that css thinking it is not required. –  Insane Coder Feb 17 '14 at 18:25

:Thanks Dave for the response.Special thanks for giving us the jqmath, an extremely useful library. –  Insane Coder Feb 18 '14 at 10:04


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# What is a simple mathematical model of a star?

I had a discussion at work regarding a recent fusion experiment in China that resulted in temperatures five times hotter than the Sun. Someone mentioned that no one can know the temperature of the Sun. I replied that we have mathematical models of the Sun, but I didn't have any on hand, and I haven't been able to find much on the internet.

So, where can I find a system of equations for modelling a star?

I found this site, which is about the level of complexity I was looking for, but it looks spotty. For instance, $\partial T / \partial t$ and $\partial P / \partial t$ terms seem to come out of nowhere--most everything else is written in terms of $\partial / \partial M$.

I can code numerical solutions to PDEs, but I haven't had much physics or anything, so I didn't know where or what to look for.

EDIT:

I found this really great post by Dr. Brian Koberlein describing a very (very) simple model of a star. He goes on to build upon that simple model here.

• We know the temperature at the core of the sun to considerable precision as attested by precise and correct prediction of the relative flux of various neutrino energies. Jan 25 '14 at 2:08
• Jan 25 '14 at 22:25

The basic problem of modelling a star is covered in a number of textbooks and lecture notes. Try searching for "stellar structure and evolution" or something along those lines. The best readily available lecture notes, IMO, are those of Onno Pols, available here. There was also a similar post on Quora, which you can read too. In the mean time, here's the basic run down.

To construct a reasonable stellar model in a reasonable amount of time, we make several assumptions. We assume that a star is a spherically symmetric, dynamically stable, self-gravitating fluid in local thermodynamic equilibrium. Here's how we unpack all this.

First, spherical symmetry means one spatial co-ordinate. In this case, for a fluid, we can write down the equation of mass conservation: $$\frac{dm}{dr}=4\pi r^2 \rho$$ This just means that an infinitesimal spherical shell of thickness $dr$ at radius $r$ contributes $dm$ to the total mass inside radius $r$. (I'll probably end up calling $m$ the mass co-ordinate.)

Now, if we regard our star as spherically symmetric and dynamically stable, we can kick out the velocity terms and time derivatives in Euler's equation. Supposing gravity is the only external body force, we end up at the equation of hydrostatic equilibrium: $$\frac{dP}{dr}=-\frac{Gm\rho}{r^2}=-\rho g$$

Remember, this follows from the conservation of momentum. To conserve energy, like mass, we say that the contribution to the total luminosity at $r$ is the mass of the shell times the specific energy generation rate $\epsilon$, so we write $$\frac{dL}{dr}=4\pi r^2\rho\epsilon$$

To keep it simple, I've neglected to specify where $\epsilon$ will come from. It generally includes the energy generated by nuclear reactions rates, less the losses due to neutrinos streaming out in some reactions, plus—in some phases—the energy released by contraction. (It can be shown that when a star contracts it heats up but loses energy overall. See the Virial Theorem.) The energy generation depends on the density, temperature and chemical composition of the material. It isn't something we know from first principles. Instead, we use tables of data taken either from detailed calculations or laboratory experiments.

We now have to describe how energy is transported inside the star. The equations are a bit of a mouthful, so I won't write them here, but basically energy can either be transported by radiation or convection, depending on the temperature structure. In either case, you get an equation of the form $dT/dr=$(some right-hand side, see the notes). In the case of radiation, the transport coefficient depends on the opacity of the stellar material, denoted $\kappa$, which itself depends again on the density, temperature and chemical composition. (Strictly speaking, opacity depends on frequency, but we use a specific average opacity: the Rosseland mean opacity.) Like the energy generation rate, this isn't known from first principles: we use tabulated lab data.

Finally, as is usually the case in fluid problems, we have to close the system with an equation of state, which relates the pressure, density, temperature and chemical composition. It's the third equation for which we generally use lab data, although here we do also have some approximate analytic forms.

These four equations (three given + temperature transport) are almost entirely independent of time, so they're sometimes called the structure equations. The three tabulated inputs (energy generation, opacities and equation of state) are sometimes called the matter or microphysics equations.

So, why does a star evolve?

The answer is because the composition changes. Suppose there are $N$ chemical species (${}^1H$, ${}^4He$, etc.), each of whose fractional mass abundance is denoted $X_i$. Then the nuclear reactions convert species $i$ into $j$ at some rate $R_ij$, and we can write a set of equations $$\frac{dX_i}{dt}=\sum_j R_{ij}$$ The rates also depend on the material properties (density, temperature, etc.). Also, in truth, we expect convection to mix material on a dynamical timescale, so we throw in a monstrous diffusion coefficient in those regions.

But that's basically it. Given a composition profile, the structure equations tell you what the star looks like. Then, the reaction rates dictate how the composition changes, and the structure changes accordingly through the matter equations. I haven't gone into details like boundary conditions and whatnot, but if you're still interested, I recommend the notes! They're aimed at a reasonably high level (I'd say late undergrad although there's no reason a second-year couldn't make sense of them) but if you're familiar with other areas of physics it should be a synch.

If you want to build models, you can try using polytropes for very simple (but still useful) models. Or, I'd recommend the Modules for Experiments in Stellar Astrophysics (MESA) package for a fully-fledged, research-grade modelling tool.

The variable $M$ on that page is used instead of the radial $r$ coordinate; $M$ denotes the total mass inside the ball of radius $r$, the cumulative mass. These explanations are clear e.g. from this alternative presentation of the equations of the stellar structure:

https://en.wikipedia.org/wiki/Stellar_structure#Equations_of_stellar_structure

That page also more or less describes where the equations come from and what they mean. On that page you found, $t$ is time and it isn't surprising that the time derivative of the pressure $P$ and absolute temperature $T$ appear in the dynamical equations as well.

There are lots of books on stellar structure

http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=stellar+structure&tag=lubosmotlsref-20

• I read that $M$ is the independent variable, because $r$ changes. I don't know what $a$ and $c$ were in the radiative energy transfer equation. Anyway, I figured out I should be googling "energy transport in stars" instead of "solar model". Thanks. Jan 24 '14 at 18:35
• Sorry, the link to your reference page disappeared so I can't meaningfully tell you what $a,c$ meant on the currently unknown page. Jan 24 '14 at 18:53
• I figured it out. Wikipedia uses the Stefan-Boltzman constant, $\sigma$, whereas the other page uses this $a \cdot c$ term which is related to the Stefan-Boltzman constant by $a=\frac{4\sigma}{c}$, where $c$ is the speed of light, and $a$ is the radiation constant. Jan 24 '14 at 19:51


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Summer Deals - hours only!Up to 80% off on all courses and bundles.-Close
Evaluation Order and Common Problems
Window functions and GROUP BY
Summary and Review

## Instruction

Let's do another exercise.

## Exercise

Group the auctions based on the day they ended and show the following information: EndDate, the average number of views from auctions on that day, and the ranking based on that average. The highest average should get the rank of 1. Name the last two columns AvgViews and Ranking.

### Stuck? Here's a hint!

Use

RANK() OVER(ORDER BY AVG(Views) DESC


text:
Next: 6.2 Geometrical Representation Up: 6. The Complex Numbers Previous: 6. The Complex Numbers   Index

# 6.1 Absolute Value and Complex Conjugate

6.1   Definition (Complex Numbers, .) We denote the complexification of by , and we call the complex numbers.

6.2   Definition (Absolute value.) In exercise 4.23A we showed that (for any field in which is not a square), if , then

If we are working in , then and hence has a unique square root in , which we denote by and call the absolute value of .

We note that

Also note that for , this definition agrees with our old definition of absolute value in .

6.3   Definition (Real and imaginary parts.) Let and write where . We call the real part of , and we call the imaginary part of (note that the imaginary part of is real), and we write

6.4   Theorem. Let be complex numbers. Then
a)
.
b)
if .
c)
.
d)
.
e)
.
f)
.
g)
.
h)
.
i)
.

Proof: By using properties of the complex conjugate proved in exercise 4.23A, we have

Hence by uniqueness of square roots, . The proofs of b), c), d), e), f), g), h) and i) are left to you.

6.5   Exercise. Prove parts b), c), d), e), f), g), h) and i) of Theorem 6.4. A

6.6   Theorem (Triangle inequality.) Let . Then

Proof: For all ,

 (6.7)

Now since , we have

Hence, from (6.7),

and it follows that

Next: 6.2 Geometrical Representation Up: 6. The Complex Numbers Previous: 6. The Complex Numbers   Index


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# How to derive equivalent martingale measure using Ito's Lemma

Can someone explain how to get equation 27.14 below? I understand the first usage of Ito's Lemma to get $d(\ln f-\ln g)$ but I do not understand how to use Ito's Lemma to go from $d(\ln \frac{f}{g})$ to $d(\frac{f}{g})$. Can someone help to elucidate this process?

## 1 Answer

You have two processes, $X_t:=\log{\frac{f}{g}}$ and $Y_t=\frac{f}{g}$. Note, I use $\log$ for the natural logarithm. Hence we have $Y_t=\exp{(X_t)}$. Therefore, applying Itô:

$$dY_t=\exp{(X_t)}dX_t + \frac{1}{2}\exp{(X_t)}d\langle X,X\rangle_t$$

Using the dynamics of $X_t$, we get

$$dY_t=\frac{f}{g}[-\frac{(\sigma_f-\sigma_g)^2}{2}dt+(\sigma_f-\sigma_g)dz]+\frac{f}{g}\frac{(\sigma_f-\sigma_g)^2}{2}d\langle z, z\rangle_t$$

I think $z$ is a Brownian Motion, such that $d\langle z,z\rangle_t=dt$, which yields the desired result. If $z$ is not a BM, please provide additional information.


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407 493 6601

# Algebra 2 - Determinants & Cramer's Rule

## Introduction

• A determinant is an array of numbers consisting of row elements and column elements that are typically used to solve systems of linear equations.
• In the symbol , the determinant of order $\mathbit{n}$ is defined.
• This algebraic symbol is employed to express the compact form for the sum of all possible products wherein each has $\mathbit{n}$ factors.
• Each element is contained with two subscripts, the first number indicates the row order in which the element is arranged, and the second number denotes the column order in which the element is arranged.

## Order and Evaluating Determinants

• The second order of determinant (also 2 x 2 determinant) is in the form of $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$.
• To illustrate, this is defined as:

$\mathbf{|}{\mathbf{A}}_{\mathbf{2}}\mathbf{|}\mathbf{=}\mathbf{|}\begin{array}{cc}{\mathbf{a}}_{\mathbf{11}}& {\mathbf{a}}_{\mathbf{12}}\\ {\mathbf{a}}_{\mathbf{21}}& {\mathbf{a}}_{\mathbf{22}}\end{array}\mathbf{|}\mathbf{=}{\mathbit{a}}_{\mathbf{11}}{\mathbit{a}}_{\mathbf{22}}\mathbf{-}{\mathbit{a}}_{\mathbf{12}}{\mathbit{a}}_{\mathbf{21}}$

• For instance, if we have  this is evaluated as $\left(8\right)\left(-4\right)-\left(-7\right)\left(-6\right)=-32-42=-74$
• The third order of determinant is described as:

${\mathbit{A}}_{\mathbf{3}}\mathbf{=}|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}|$

• Rewrite this first and second column in order to evaluate the third-order determinant:

${\mathbit{A}}_{\mathbf{3}}\mathbf{=}\mathbf{|}\begin{array}{ccc}{\mathbf{a}}_{\mathbf{11}}& {\mathbf{a}}_{\mathbf{12}}& {\mathbf{a}}_{\mathbf{13}}\\ {\mathbf{a}}_{\mathbf{21}}& {\mathbf{a}}_{\mathbf{22}}& {\mathbf{a}}_{\mathbf{23}}\\ {\mathbf{a}}_{\mathbf{31}}& {\mathbf{a}}_{\mathbf{32}}& {\mathbf{a}}_{\mathbf{33}}\end{array}\mathbf{|}\begin{array}{c}{\mathbf{a}}_{\mathbf{11}}\\ {\mathbf{a}}_{\mathbf{21}}\\ {\mathbf{a}}_{\mathbf{31}}\end{array}\begin{array}{c}{\mathbf{a}}_{\mathbf{12}}\\ {\mathbf{a}}_{\mathbf{22}}\\ {\mathbf{a}}_{\mathbf{32}}\end{array}\mathbf{\to }{\mathbit{A}}_{\mathbf{3}}\mathbf{=}{\mathbit{a}}_{\mathbf{11}}{\mathbit{a}}_{\mathbf{22}}{\mathbit{a}}_{\mathbf{33}}\mathbf{+}{\mathbit{a}}_{\mathbf{12}}{\mathbit{a}}_{\mathbf{23}}{\mathbit{a}}_{\mathbf{31}}\mathbf{+}{\mathbit{a}}_{\mathbf{13}}{\mathbit{a}}_{\mathbf{21}}{\mathbit{a}}_{\mathbf{32}}\mathbf{-}{\mathbit{a}}_{\mathbf{31}}{\mathbit{a}}_{\mathbf{22}}{\mathbit{a}}_{\mathbf{13}}\mathbf{-}{\mathbit{a}}_{\mathbf{32}}{\mathbit{a}}_{\mathbf{23}}{\mathbit{a}}_{\mathbf{11}}\mathbf{-}{\mathbit{a}}_{\mathbf{33}}{\mathbit{a}}_{\mathbf{21}}{\mathbit{a}}_{\mathbf{12}}$

## Basic Properties of Determinants

• When corresponding rows and columns are interchanged, it does not affect the value of the determinant.

Examples:

• Any two rows or columns interchanged alters the sign of the determinant.

Examples:

• Suppose that all elements in a row or column are zero; then the value of this determinant is 0.

Examples:

• If there are any two rows or columns that are identical, the value of this determinant is equal to 0.

Examples:

• If each element in a row or column is multiplied by any number $\mathbit{n}$, the value of the entire determinant is also multiplied by the number $\mathbit{n}\mathbf{.}$

Example:

• If each element in a row or column of a determinant is expressed as a sum of two or more numbers, the determinant can be written as a sum of two or more determinants.

Example:

• Suppose that each element of a row or column is being added $\mathbit{n}$ times the corresponding elements of another row or column; the value of the determinant will not change.

Example:

$|\begin{array}{ccc}5& 9& 3\\ 6& 8& 7\\ 7& 4& 2\end{array}|\mathbf{=}|\begin{array}{ccc}5+4\left(9\right)& 9& 3\\ 6+4\left(8\right)& 8& 7\\ 7+4\left(4\right)& 4& 2\end{array}|\mathbf{=}|\begin{array}{ccc}5& 9& 3\\ 6& 8& 7\\ 7& 4& 2\end{array}|$

## Determinant of a Matrix

• To evaluate the determinant of a matrix, the matrix has to be a square matrix.
• The determinant of a matrix is useful in calculus applications, determining the inverse of a matrix, and solving systems of linear equations.
• To illustrate, we have:

• For a 3 x 3 matrix, we have the following:

$B=\left[\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right]=\left({a}_{1}\right)det\left[\begin{array}{cc}{b}_{2}& {c}_{2}\\ {b}_{3}& {c}_{3}\end{array}\right]-\left({b}_{1}\right)det\left[\begin{array}{cc}{a}_{2}& {c}_{2}\\ {a}_{3}& {c}_{3}\end{array}\right]+\left({c}_{1}\right)det\left[\begin{array}{cc}{a}_{2}& {b}_{2}\\ {a}_{3}& {b}_{3}\end{array}\right]$

## Cramer's Rule

• Cramer's Rule is a procedure that is used to solve for the solutions of a system of linear equations in $\mathbit{n}$ unknowns.
• To illustrate the process of solving a system of linear equations in two unknowns, we have:
• Given the system of linear equations $\left\{\begin{array}{l}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$, then the solution is determined using the following:

• This can also be simplified using the following formulas:

• To illustrate the process of solving a system of linear equations in three unknowns, we have:
• Given the linear system $\left\{\begin{array}{l}{a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}\\ {a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2}\\ {a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3}\end{array}$, then this is expressed as a determinant .

• Hence, we have the following equations to find the solution:

## Berzen's Layer Reduction Method

• Another method of solving systems of linear equations in three unknowns is Berzen's Layer Reduction Method.
• Given a linear system $\left\{\begin{array}{l}{a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}\\ {a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2}\\ {a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3}\end{array}$, then we solve the following determinants: .

• To find the value of $\mathbit{z}$ first, we have:

$\mathbit{z}\mathbf{=}\frac{\mathbf{A}\mathbf{F}\mathbf{-}\mathbf{D}\mathbf{C}}{\mathbf{A}\mathbf{E}\mathbf{-}\mathbf{B}\mathbf{D}}$

• After getting the value of $\mathbit{z}$, the next step is to find the value of $\mathbit{x}$ and $\mathbit{y}$ by solving the obtained linear system in two unknowns.

## Cheat Sheet

• In case of $\left\{\begin{array}{l}x+y=1\\ 2x+3y=6\end{array}$, then
• In a system of the linear equation of two variables, we have the following equations: .
• In case of
• In solving a linear system of three unknowns, we have
• In case of

## Blunder Areas

• Do not be confused with the symbol used in determinant of a matrix and absolute value. The determinant of a matrix is not an absolute value symbol.
• When using Cramer's Rule in solving linear systems, the process may fail if the determinant of the coefficient array is zero.
• Cramer's Rule only applies to a square matrix wherein that system has a unique solution and has a non-zero determinant.
• Berzen's Layer Reduction Method only works for a linear system of three variables.


text:
Binomial sum modulo prime power: Part 2

Let $p$ be a prime number and $\omega=\exp(2\pi i/p)$. Recall that we conjectured in this post that

$(*)\qquad\qquad\qquad\displaystyle\sum_{j=0}^{p-1}(1-\omega^j)^n=p\sum_{k=0}^{\lfloor n/p\rfloor}(-1)^{kp}\binom{n}{kp}$

is divisible by $p^{\lceil n/(p-1)\rceil}$. From exercise 2 of this post we know that $(p)$ factors as $(p,\omega-1)^{p-1}$ into prime ideals in the ring $\mathbb Z[\omega]$. So $(\omega-1)^{p-1}\in (p)$. Therefore the right-hand side of $(*)$ is divisible by $p^{\lfloor n/(p-1)\rfloor}$. So we are off by at most one factor of $p$! I believe this approach can be improved upon to account for the extra factor, since we haven’t used any property of the sum.

Furthermore,

$\displaystyle\sum_{j=0}^{p-1}\omega^{-rj}(1-\omega^j)^n=p\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}$

for any $r$. So our best result thus far is the following:

Theorem (weaker version of Fleck’s).

$\displaystyle\sum_{m\equiv r\pmod p}(-1)^m\binom{n}{m}\equiv 0\pmod{p^{\lfloor\frac{n}{p-1}\rfloor-1}}$.

Goal: Improve the floors to ceilings.


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# Expected number of iterations for bozo sort opt algorithm

I'm trying to figure out the upper bound for the number of iterations of the bozo sort opt algorithm, described in this paper on section 3.2: http://www.hermann-gruber.com/pdf/fun07-final.html

I know the maximum number of inversions in this case is nC2, what i'm struguling with is the probability of two numbers in an iteration being an inversion.

Here is the relevant piece of the paper I'm focusing on, I know the answer to the number of iterations is there, I just can't seem to get to it.

## 3.2 Comments on optimized variants of bogo-sort

Though optimizing the running time seems somewhat out of place in the field of pessimal algorithm design, it can be quite revealing for beginners in both fields of optimal and pessimal algorithm design to see how a single optimization step can yield a dramatic speed-up. The very first obvious optimization step in all aforementioned algorithms is to swap two elements only if this makes sense. That is, before swapping a pair, we check if it is an inversion: A pair of positions $$(i, j)$$ in the array $$a[1, \dots, n]$$ is an inversion if $$i < j$$ and $$a[i] > a[j]$$. This leads to optimized variants of bogo-sort and its variations, which we refer to as bogo-sort$$_{opt}$$, bozo-sort$$_{opt}$$, and bozo-sort$$^+_{opt}$$, resp. As there can be at most $$\binom{n}{2}$$ inversions, this number gives an immediate upper bound on the number of swaps for these variants—compare, e.g., to $$\mathcal \Omega(n*n!)$$ swaps carried out by bogo-sort. It is not much harder to give a similar upper bound on the expected number of iterations. As the number of comparisons during a single iteration is in $$\mathcal O(n)$$, we also obtain an upper bound on the expected total number of comparisons:

Lemma 14. The expected number of iterations (resp. comparisons) carried out by the algorithms bogo-sort$$_{opt}$$, bozo-sort$$_{opt}$$, and bozo-sort$$^+_{opt}$$ on a worst-case input $$\overline x$$ of length $$n$$ is at most $$\mathcal O(n^2\log n)$$ (resp. $$\mathcal O(n^3\log n)$$).

• Sure thing, i'll update the question. I'm studying bozo sort opt, an implementation of bozo sort that checks if the randomly selected positions are an inversion before swaping them. – Rodrigo Alvarado Mar 30 at 2:30
• Thank you for update. I have transcribed your image. There are more pages (sorry I saw only one on mobile phone), but it doesn't change the fact that questions should be self-contained and text should be indexable. I see it clearly now, it is Bozo sort, variation of Bogo sort. – Evil Mar 30 at 3:19

The bounds in that paper can be tightened. The tighter bounds in lemma 14 are "..." at most $$O(n\log n)$$ (resp. $$O(n^2\log n))$$.

As in the question and the papers, we will fix a set of $$n$$ elements that are totally ordered. The array $$a=(a_1, a_2, \cdots, a_n)$$ will be a permutation of the $$n$$ elements. A pair of element $$\{a_i, a_j\}$$ is a good pair if $$i\lt j$$ and $$a[i]\lt a[j]$$. It is a bad pair if $$i\lt j$$ and $$a[i]\gt a[j]$$. A bad pair is also call an inversion.

Fact on inversion. Swapping a bad pair strictly decreases the number of all inversions. An array is sorted if and only if the number of all inversions is 0. The maximum number of all inversions is the same as the numbers of all pairs, $$\binom n2$$.

Now we will use the fact above to show the tightened version of lemma 14 for one case.

### The expected number of comparisons carried out by bozo-sort$$_{\text{opt}}$$ on a worst-case input $$x$$ of length $$n$$ is at most $$4n^2\log n + (n-1)$$, where $$\log$$ is the natural logarithm.

Proof: Suppose there are $$m$$ bad pairs in the array. A run of procedure rand. transpose hit one with probability $$p = m/n^2$$, so the expected number of runs of that procedure until we hit a bad pair is $$1/p=n^2/m$$. Since the number of bad pairs decreases each time we swap a bad pair, the expected number of runs of that procedure until the array is sorted is bounded by $$\sum_{m=1}^{\binom n2}\frac{n^2}m=n^2\sum_{m=1}^{\binom n2}\frac1m\le 2n^2\log n$$

Note that each procedure rand. transpose makes one comparison. The effect of running procedure sorted before every $$n-1$$ procedure rand. transpose will at most double number of comparisons besides the initial $$n-1$$ comparisons. Q.E.D.

The above proof is adapted from Fun sort or the chaos of unordered binary search by T. Biedl, T. Chan, E. D. Demaine, R. Fleischer, M. Golin, J.A. King, and J. I. Munro. Discrete Applied Mathematics, 144(3):231–236, 2004.

### Exercises

Exercise 1. Show that $$\sum_{m=1}^{\binom n2}\frac1m\le 2\log n.$$ (Hint, here is the rate of divergence of harmonic series).

Exercise 2. (One minute or less.) Show that $${m}/\binom n2$$ is the probability of procedure rand. transpose in bozo-sort$$^+_{\text{opt}}$$ hitting one bad pair when there are $$m$$ bad pairs.

Exercise 3. Show that the expected number of bad pairs found (and, swapped) by a call to procedure randomly permute in bogo-sort$$_{\text{opt}}$$ when there are $$m$$ bad pairs is $$\Theta(m(n-1)/n^2)$$. Show the expected number of total comparisons until the array is sorted is at most by $$\Theta(n^2\log n)$$.


text:
# Find the derivative of the following function using logarithmic differentiation?

Mar 30, 2017

see below

#### Explanation:

Steps in Logarithmic Differentiation
1. Take natural logarithms of both sides of an equation y=f(x) and use the Laws of Logarithms to simplify.
2. Differentiate implicitly with respect to x
3. Solve the resulting equation for y’
4. Replace y with original equation

Let $y = \frac{5 \sqrt{x} {\left({x}^{2} + 1\right)}^{4}}{{\left(2 x + 1\right)}^{3} {\tan}^{5} \left({x}^{2}\right)}$ and lets rewrite the equation as $y = \frac{5 {x}^{\frac{1}{2}} {\left({x}^{2} + 1\right)}^{4}}{{\left(2 x + 1\right)}^{3} {\tan}^{5} \left({x}^{2}\right)}$

Now use the following Properties of Logarithms to expand the original problem

color(red)(log_b(xy)=log_b x+log_b y
color(red)(log_b(x/y)=log_b x-log_b y
color(red)(log_bx^n=n log_b x
That is,

color(blue)(ln y=ln((5x^(1/2)(x^2+1)^4)/((2x+1)^3tan^5(x^2)))

color(blue)(ln y =5ln x^(1/2)+ln(x^2+1)^4-ln(2x+1)^3-ln (tan x^2)^5

color(blue)(ln y =5/2 ln x+4 ln(x^2+1)-3ln(2x+1)-5 ln(tan x^2)

color(blue)(d/dx (ln y =5/2 ln x+4 ln(x^2+1)-3ln(2x+1)-5 ln(tan x^2))

color(blue)(1/y dy/dx = 5/2 1/x + 4 ((2x)/(x^2+1))-3(2/(2x+1))-5((2xsec^2(x^2))/tan x^2)

$\textcolor{b l u e}{\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{10 x} + \frac{8 x}{{x}^{2} + 1} - \frac{6}{2 x + 1} - \frac{10 x {\sec}^{2} \left({x}^{2}\right)}{\tan} {x}^{2}}$

color(blue)( dy/dx = [5/(10x)+(8x)/(x^2+1)-6/(2x+1)-(10xsec^2(x^2))/tan x^2]*y

color(blue)( dy/dx = [5/(10x)+(8x)/(x^2+1)-6/(2x+1)-(10xsec^2(x^2))/tan x^2]*[(5sqrtx (x^2+1)^4)/((2x+1)^3tan^5(x^2))]


text:
# WMAP5 and the Cluster Mass Function - Astrophysics

WMAP5 and the Cluster Mass Function - Astrophysics - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: The recently revised cosmological constraints from the Five-Year WMAP dataameliorate previous tension between cosmological constraints from the microwavebackground and from cluster abundances. We demonstrate that the revisedestimates of cosmological parameters are in excellent agreement with the massfunction of X-ray clusters in the Sloan Digital Sky Survey. Velocitysegregation between galaxies and the underlying dark matter could cause virialmass estimates to be biased, causing the mass scale of the mass function to beoffset from the true value. Modest velocity segregation$\sigma {gxy}-\sigma {DM}$=1.13$^{+0.06} {-0.05}$ is sufficient to match themass function to the Five-Year WMAP results. When the new WMAP results arecombined with constraints from supernovae and baryon acoustic oscillations,there is no need for velocity segregation$\sigma {gxy}-\sigma {DM}$=1.05$\pm$0.05. This result agrees withexpectations for velocity segregation from state-of-the-art numericalsimulations of clusters. Together with the improved agreement between the newWMAP results and recent cosmic shear measurements, this result demonstratesthat the amplitude of large-scale structure in the nearby universe matches thatpredicted from the structure seen in the microwave background. The newconstraint we place on velocity segregation in clusters indicates that virialmass estimates for clusters are reasonably accurate. This result suggests thatfuture cluster surveys will be able to probe both cosmological parameters andfundamental cluster physics.

Autor: Kenneth Rines, Antonaldo Diaferio, Priyamvada Natarajan

Fuente: https://arxiv.org/


text:
# how to show equivalence relation and its classes

i am stumbling again in proving things in maths. the task is to prove that this statement

$A \sim B : \Longleftrightarrow \sum_{a\in A} a = \sum_{b\in B} b$

is an Equivalence Relation on Power Set of {0,1,2} with total order $\leq$.

my start is this:

a) Reflexiv: $A \sim A \Longrightarrow \sum_{a\in A} a = \sum_{a\in A} a$ this is OK
b) Symmetric:.. how do i do this, how can i use the notion of total order here?
i am really stuck, :( i also need to show the equivalence classes here

thanks a lot for help in advance

-

To show that $\sim$ is symmetric you must show that if $A\sim B$, then $B\sim A$. But this is easy, because equality is symmetric: if $A\sim B$, then by definition $\sum_{a\in A}a=\sum_{b\in B}b$, so $\sum_{b\in B}b=\sum_{a\in A}a$, and therefore $B\sim A$, again by the definition of $\sim$.

Transitivity of $\sim$ follows similarly from transitivity of equality. Assume that $A\sim B$ and $B\sim C$. Then by the definition of $\sim$ we know that $\sum_{a\in A}a=\sum_{b\in B}b$ and $\sum_{b\in B}b=\sum_{c\in C}c$; can you now put the pieces together to conclude that $A\sim C$?

You’ll have one equivalence class for each possible sum of a subset of $\{0,1,2\}$. There are only eight subsets, so you can simply make a list:

$$\begin{array}{rcc} \text{Subset}:&\varnothing&\{0\}&\{1\}&\{2\}&\{0,1\}&\{0,2\}&\{1,2\}&\{0,1,2\}\\ \text{Sum}:&0&0&1&2&1&2&3&3 \end{array}$$

Now which subsets have the same sums and therefore belong to the same equivalence class?

-
thanks but how do i include the total order in the proof, –  doniyor Nov 16 '12 at 21:40
@doniyor: It doesn’t actually enter into the argument in any way: the equivalence relation $\sim$ is completely independent of the order $\le$ on $\{0,1,2\}$. –  Brian M. Scott Nov 16 '12 at 21:43
oh okay, thanks. but can you please explain me again how you came to know the equivalence classes? i cannot find the way to come to this intuitivly :( –  doniyor Nov 16 '12 at 21:46
@doniyor: $A\sim B$ means that $\sum_{a\in A}a=\sum_{b\in B}b$: the subsets $A$ and $B$ have the same sum. Two subsets of $\{0,1,2\}$ are $\sim$-related if and only if they have the same sum. By definition of equivalence class two subsets of $\{0,1,2\}$ are in the same equivalence class if and only if they’re $\sim$-related. Put the two together: $A$ and $B$ are in the same equiv. class if and only if they have the same sum. Thus, there must be one equiv. class for each of the four possible sums, and as you can see from the table, each equiv. class has exactly two members. –  Brian M. Scott Nov 16 '12 at 21:52
perfect! Great Help! thanks a lot –  doniyor Nov 16 '12 at 21:57

Note that mapping a set $S\subset \{0,1,2\}$ to its element sum $\sum_{x\in S} x$ is a function $f$ (to be presice: a function from the power set of $\{0,1,2\}$ to e.g. the reals, but these details don't evenmatter). Note that any relation of the from $$A\sim B:\iff f(A)=f(B)$$ is an equivalence relation, by the very properties of the role-model of equivalence: equality

• $A\sim A$ because $f(A)=f(A)$.
• $A\sim B$ implies $B\sim A$ because $f(A)=f(B)$ is just the same as $f(B)=f(A)$.
• If $A\sim B$ and $B\sim C$ then $A\sim C$ because if $f(A)=f(B)=f(C)$, then $f(A)=f(C)$.
-
great! Thank you –  doniyor Nov 17 '12 at 4:56


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 2010-11-14, 04:00 #67 3.14159     May 2010 Prime hunting commission. 24×3×5×7 Posts Now, back to the main focus of the thread.. Some obscure theory on primes of the form 2p-1. What about them? Want to find one? P95 is your friend. Why don't you try 264354649 - 1? By the way; What is the status of 264354649 - 1? (A number of about 19.37 million digits.) If it's uncharted territory, perhaps sm88 can pursue actually looking for one, as opposed to the usual business of seeing patterns where none exist, and babbling nonsense no one understands. Try these values for p, for 2p-1, if that one is composite. These are all probably uncharted territory anyway, and I'm a tad too lazy to go looking around for each exponent's status, although I'm sure that at least one has already been meddled with at least once. Code: 60687901 38938813 45615587 60254287 73008427 80300791 62029711 50742259 78463417 86100557 70978823 30219743 42805117 97802563 54000169 45921541 67627421 50714737 39618451 49718023 80976547 71843129 82795997 37144193 32807759 52988981 85507519 81177307 29024783 94450157 49717169 84537977 52294097 53169451 70975439 62525147 Last fiddled with by 3.14159 on 2010-11-14 at 04:16
2010-11-14, 13:31   #68
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by 3.14159 Now, back to the main focus of the thread.. Some obscure theory on primes of the form 2p-1. What about them? Want to find one? P95 is your friend. Why don't you try 264354649 - 1? By the way; What is the status of 264354649 - 1? (A number of about 19.37 million digits.) If it's uncharted territory, perhaps sm88 can pursue actually looking for one, as opposed to the usual business of seeing patterns where none exist, and babbling nonsense no one understands. Try these values for p, for 2p-1, if that one is composite. These are all probably uncharted territory anyway, and I'm a tad too lazy to go looking around for each exponent's status, although I'm sure that at least one has already been meddled with at least once. Code: 60687901 38938813 45615587 60254287 73008427 80300791 62029711 50742259 78463417 86100557 70978823 30219743 42805117 97802563 54000169 45921541 67627421 50714737 39618451 49718023 80976547 71843129 82795997 37144193 32807759 52988981 85507519 81177307 29024783 94450157 49717169 84537977 52294097 53169451 70975439 62525147
get me the worlds fastest computer and I'll care if i can't get pari to print the biggest prime known in under 30 minutes then I'm not worth it to help anyone find one. so unless I retest and get faster than I did before by 25% I'm not worth it to be searching for them. apparently that was checking if it was prime because it took 31 ms to calculate this time.

Last fiddled with by science_man_88 on 2010-11-14 at 13:39

 2010-11-14, 15:11 #69 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts my computer isn't worth breaking to be in the record books, plain and simple
2010-11-14, 15:20   #70
3.14159

May 2010
Prime hunting commission.

24·3·5·7 Posts

Quote:
 Originally Posted by science_man_88 my computer isn't worth breaking to be in the record books, plain and simple

A random c160, completely factored: 1613 * 16685818316410231 * 15694342916407420598282061127 * 21443356185490234674701067126317 * 11753380988799343801639257420578787695318692321305544085583187497380249619645889.

Last fiddled with by 3.14159 on 2010-11-14 at 15:25

2010-11-14, 15:20   #71
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

203008 Posts

Quote:
 Originally Posted by 3.14159 .. P95 is your friend..
all that means to me is a prime of 95 digits.

2010-11-14, 15:37   #72
3.14159

May 2010
Prime hunting commission.

110100100002 Posts

Quote:
 Originally Posted by science_man_88 all that means to me is a prime of 95 digits.
Hey, look! P95!

And, an actual p95;

82449289611552559574990925283007152491333479550118068579078519555198400142339056437465979206437.

Last fiddled with by 3.14159 on 2010-11-14 at 15:45

2010-11-14, 15:44   #73
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

203008 Posts

Quote:
 Originally Posted by 3.14159 Hey, look! P95!
that helps me none

2010-11-14, 16:19   #74
CRGreathouse

Aug 2006

592510 Posts

Quote:
 Originally Posted by science_man_88 that helps me none
Prime95 is (Mersenne prime)-testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it.

Last fiddled with by CRGreathouse on 2010-11-14 at 16:19

2010-11-14, 16:21   #75
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by CRGreathouse Prime95 is (Mersenne prime)-testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it.
I have before and it would take me over 50 days to test one exponent pointless to me.

that and it teaches me nothing about them.

Last fiddled with by science_man_88 on 2010-11-14 at 16:22

2010-11-14, 16:42   #76
CRGreathouse

Aug 2006

3·52·79 Posts

Quote:
 Originally Posted by science_man_88 I have before and it would take me over 50 days to test one exponent pointless to me. that and it teaches me nothing about them.
I don't see why 50 days/test makes it bad. But I agree that it doesn't teach you anything.

2010-11-14, 17:12   #77
3.14159

May 2010
Prime hunting commission.

24×3×5×7 Posts

Quote:
 Originally Posted by CRGreathouse Prime95 is (Mersenne prime)-testing software. Pi is apparently suggesting that, rather than think about the problem, you apply some cycles to it.

No one's even made it clear what the supposed "problem" is, besides looking for some Mersenne number.

It's really simple; P95 is your friend.

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