text:
Fluctuation theorem

Fluctuation theorem

Discussion

Encyclopedia
The fluctuation theorem which originated from statistical mechanics
Statistical mechanics
Statistical mechanics or statistical thermodynamicsThe terms statistical mechanics and statistical thermodynamics are used interchangeably...

, deals with the relative probability that the entropy
Entropy
Entropy is a thermodynamic property that can be used to determine the energy available for useful work in a thermodynamic process, such as in energy conversion devices, engines, or machines. Such devices can only be driven by convertible energy, and have a theoretical maximum efficiency when...

of a system which is currently away from thermodynamic equilibrium
Thermodynamic equilibrium
In thermodynamics, a thermodynamic system is said to be in thermodynamic equilibrium when it is in thermal equilibrium, mechanical equilibrium, radiative equilibrium, and chemical equilibrium. The word equilibrium means a state of balance...

(i.e., maximum entropy) will increase or decrease over a given amount of time. While the second law of thermodynamics
Second law of thermodynamics
The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and...

predicts that the entropy of an isolated system should tend to increase until it reaches equilibrium, it became apparent after the discovery of statistical mechanics that the second law is only a statistical one, suggesting that there should always be some nonzero probability that the entropy of an isolated system might spontaneously decrease; the fluctuation theorem precisely quantifies this probability.

Statement of the fluctuation theorem

Roughly, the fluctuation theorem relates to the probability distribution of the time-averaged irreversible entropy production [1], denoted . The theorem states that, in systems away from equilibrium over a finite time t, the ratio between the probability that takes on a value A and the probability that it takes the opposite value, −A, will be exponential in At.
In other words, for a finite non-equilibrium system in a finite time, the FT gives a precise mathematical expression for the probability that entropy will flow in a direction opposite to that dictated by the second law of thermodynamics
Second law of thermodynamics
The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and...

.

Mathematically, the FT is expressed as:

This means that as the time or system size increases (since is extensive), the probability of observing an entropy production opposite to that dictated by the second law of thermodynamics decreases exponentially. The FT is one of the few expressions in non-equilibrium statistical mechanics that is valid far from equilibrium.

The FT was first proposed and tested using computer simulations, by Denis Evans
Denis Evans
Denis James Evans, , is a Professor in the Research School of Chemistry at the Australian National University...

, E.G.D. Cohen
E.G.D. Cohen
E.G.D. Cohen is an American physicist, and is Professor Emeritus at The Rockefeller University. He is widely recognised for his contributions to statistical physics. In 2004 Cohen was awarded the Boltzmann Medal, jointly with Prof. H. Eugene Stanley...

and Gary Morriss in 1993 in the journal Physical Review Letters. The first mathematical proof was given by Evans and Debra Searles
Debra Searles
Debra Searles is the professional name for an Australian theoretical chemist whose married name is Debra Bernhardt. She is best known for her contributions towards understanding the Fluctuation Theorem...

in 1994. Since then, much mathematical and computational work has been done to show that the FT applies to a variety of statistical ensembles. The first laboratory experiment that verified the validity of the FT was carried out in 2002. In this experiment, a plastic bead was pulled through a solution by a laser. Fluctuations in the velocity were recorded that were opposite to what the second law of thermodynamics would dictate for macroscopic systems. See Wang et al. [Phys Rev Lett, 89, 050601(2002)] and later Carberry et al., [Phys Rev Lett, 92, 140601(2004)]. This work was widely reported in the press - Second law of thermodynamics "broken" (NewScientist, 19 July 2002); Nature July 23, 2002, http://www.nature.com/nsu/020722/020722-2.html .

Note that the FT does not state that the second law of thermodynamics is wrong or invalid. The second law of thermodynamics is a statement about macroscopic systems. The FT is more general. It can be applied to both microscopic and macroscopic systems. When applied to macroscopic systems, the FT is equivalent to the Second Law of Thermodynamics.

Second law inequality

A simple consequence of the fluctuation theorem given above is that if we carry out an arbitrarily large ensemble of experiments from some initial time t=0, and perform an ensemble average of time averages of the entropy production then an exact consequence of the FT is that the ensemble average cannot be negative for any value of the averaging time t:

This inequality is called the Second Law Inequality [Searles & Evans, Aust J Chem, 57, 1119 (2004)]. This inequality can be proved for systems with time dependent fields of arbitrary magnitude and arbitrary time dependence.

It is important to understand what the Second Law Inequality does not imply. It does not imply that the ensemble averaged entropy production is non-negative at all times. This is untrue, as consideration of the entropy production in a viscoelastic fluid subject to a sinusoidal time dependent shear rate shows. In this example the ensemble average of the time integral of the entropy production is however non negative - as expected from the Second Law Inequality.

Nonequilibrium partition identity

Another remarkably simple and elegant consequence of the FT is the so-called "nonequilibrium partition identity" (NPI):

see Carberry et al. J Chem Phys 121, 8179(2004). Thus in spite of the Second Law Inequality which might lead you to expect that the average would decay exponentially with time, the exponential probability ratio given by the FT exactly cancels the negative exponential in the average above leading to an average which is unity for all time!

There are many important implications from the FT. One is that small machines (such as nanomachines or even mitochondria in a cell) will spend part of their time actually running in "reverse". By "reverse", it is meant that they function so as to run in a way opposite to that for which they were presumably designed. As an example, consider a jet engine
Jet engine
A jet engine is a reaction engine that discharges a fast moving jet to generate thrust by jet propulsion and in accordance with Newton's laws of motion. This broad definition of jet engines includes turbojets, turbofans, rockets, ramjets, pulse jets...

. If a jet engine were to run in "reverse" in this context, it would take in ambient heat and exhaust fumes to generate kerosene
Kerosene
Kerosene, sometimes spelled kerosine in scientific and industrial usage, also known as paraffin or paraffin oil in the United Kingdom, Hong Kong, Ireland and South Africa, is a combustible hydrocarbon liquid. The name is derived from Greek keros...

and oxygen.

Dissipation function

[1] Strictly speaking the fluctuation theorem refers to a quantity known as the dissipation function. In thermostatted nonequilibrium states that are close to equilibrium, the long time average of the dissipation function is equal to the average entropy production. However the FT refers to fluctuations rather than averages. The dissipation function is defined as,

where k is Boltzmann's constant, is the initial (t = 0) distribution of molecular states , and is the molecular state arrived at after time t, under the exact time reversible equations of motion. is the INITIAL distribution of those time evolved states.

Note: in order for the FT to be valid we require that . This condition is known as the condition of ergodic consistency. It is widely satisfied in common statistical ensembles - e.g. the canonical ensemble
Canonical ensemble
The canonical ensemble in statistical mechanics is a statistical ensemble representing a probability distribution of microscopic states of the system...

.

The system may be in contact with a large heat reservoir in order to thermostat the system of interest. If this is the case is the heat lost to the reservoir over the time (0,t) and T is the absolute equilibrium temperature of the reservoir - see Williams et al., Phys Rev E70, 066113(2004). With this definition of the dissipation function the precise statement of the FT simply replaces entropy production with the dissipation function in each of the FT equations above.

Example: If one considers electrical conduction across an electrical resistor in contact with a large heat reservoir at temperature T, then the dissipation function is

the total electric current density J multiplied by the voltage drop across the circuit, , and the system volume V, divided by the absolute temperature T, of the heat reservoir times Boltzmann's constant. Thus the dissipation function is easily recognised as the Ohmic work done on the system divided by the temperature of the reservoir. Close to equilibrium the long time average of this quantity is (to leading order in the voltage drop), equal to the average spontaneous entropy production per unit time - see de Groot and Mazur "Nonequilibrium Thermodynamics" (Dover), equation (61), page 348. However, the Fluctuation Theorem applies to systems arbitrariliy far from equilibrium where the definition of the spontaneous entropy production is problematic.

The fluctuation theorem and Loschmidt's paradox

The second law of thermodynamics
Second law of thermodynamics
The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and...

, which predicts that the entropy of an isolated system out of equilibrium should tend to increase rather than decrease or stay constant, stands in apparent contradiction with the time-reversible
T-symmetry
T Symmetry is the symmetry of physical laws under a time reversal transformation: T: t \mapsto -t.Although in restricted contexts one may find this symmetry, the observable universe itself does not show symmetry under time reversal, primarily due to the second law of thermodynamics.Time asymmetries...

equations of motion for classical and quantum systems. The time reversal symmetry of the equations of motion show that if one films a given time dependent physical process, then playing the movie of that process backwards does not violate the laws of mechanics. It is often argued that for every forward trajectory in which entropy increases, there exists a time reversed anti trajectory where entropy decreases, thus if one picks an initial state randomly from the system's phase space
Phase space
In mathematics and physics, a phase space, introduced by Willard Gibbs in 1901, is a space in which all possible states of a system are represented, with each possible state of the system corresponding to one unique point in the phase space...

and evolves it forward according to the laws governing the system, decreasing entropy should be just as likely as increasing entropy. It might seem that this is incompatible with the second law of thermodynamics
Second law of thermodynamics
The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and...

which predicts that entropy tends to increase. The problem of deriving irreversible thermodynamics from time-symmetric fundamental laws is referred to as Loschmidt's paradox
Loschmidt's paradox, also known as the reversibility paradox, is the objection that it should not be possible to deduce an irreversible process from time-symmetric dynamics...

.

The mathematical proof of the Fluctuation Theorem and in particular the Second Law Inequality shows that, given a non-equilibrium starting state, the probability of seeing its entropy increase is greater than the probability of seeing its entropy decrease - see The Fluctuation Theorem from Advances in Physics 51: 1529. However, as noted in section 6 of that paper, one could also use the same laws of mechanics to extrapolate backwards from a later state to an earlier state, and in this case the same reasoning used in the proof of the FT would lead us to predict the entropy was likely to have been greater at earlier times than at later times. This second prediction would be frequently violated in the real world, since it is often true that a given nonequilibrium system was at an even lower entropy in the past (although the prediction would be correct if the nonequilibrium state were the result of a random fluctuation in entropy in an isolated system that had previously been at equilibrium - in this case, if you happen to observe the system in a lower-entropy state, it is most likely that you are seeing the minimum of the random dip in entropy, in which case entropy would be higher on either side of this minimum).

So, it seems that the problem of deriving time-asymmetric thermodynamic laws from time-symmetric laws cannot be solved by appealing to statistical derivations which show entropy is likely to increase when you start from a nonequilibrium state and project it forwards. Many modern physicists believe the resolution to this puzzle lies in the low-entropy state of the universe shortly after the big bang, although the explanation for this initial low entropy is still debated.

Summary

The fluctuation theorem is of fundamental importance to nonequilibrium statistical mechanics.
The FT (together with the Axiom of Causality
Axiom of Causality
The Axiom of Causality is the proposition that everything in the universe has a cause and is thus an effect of that cause. This means that if a given event occurs, then this is the result of a previous, related event. If an object is in a certain state, then it is in that state as a result of...

) gives a generalisation of the second law of thermodynamics
Second law of thermodynamics
The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system. From the state of thermodynamic equilibrium, the law deduced the principle of the increase of entropy and...

which includes as a special case, the conventional second law. It is then easy to prove the Second Law Inequality and the NonEquilibrium Partition Identity. When combined with the central limit theorem
Central limit theorem
In probability theory, the central limit theorem states conditions under which the mean of a sufficiently large number of independent random variables, each with finite mean and variance, will be approximately normally distributed. The central limit theorem has a number of variants. In its common...

, the FT also implies the famous Green-Kubo relations
Green-Kubo relations
The Green–Kubo relations give the exact mathematical expression for transport coefficients in terms of integrals of time correlation functions.-Thermal and mechanical transport processes:...

for linear transport coefficients, close to equilibrium. The FT is however, more general than the Green-Kubo Relations because unlike them, the FT applies to fluctuations far from equilibrium. In spite of this fact, scientists have not yet been able to derive the equations for nonlinear response theory from the FT.

The FT does not imply or require that the distribution of time averaged dissipation be Gaussian. There are many examples known where the distribution of time averaged dissipation is non-Gaussian and yet the FT (of course) still correctly describes the probability ratios.

Lastly the theoretical constructs used to prove the FT can be applied to nonequilibrium transitions between two different equilibrium states. When this is done the so-called Jarzynski equality
Jarzynski equality
The Jarzynski equality is an equation in statistical mechanics that relates free energy differences between two equilibrium states and non-equilibrium processes...

or nonequilibrium work relation, can be derived. This equality shows how equilibrium free energy differences can be computed or measured (in the laboratory), from nonequilibrium path integrals. Previously quasi-static (equilibrium) paths were required.

The reason why the fluctuation theorem is so fundamental is that its proof requires so little. It requires:
• knowledge of the mathematical form of the initial distribution of molecular states,
• that all time evolved final states at time t, must be present with nonzero probability in the distribution of initial states (t = 0) - the so-called condition of ergodic consistency and,
• an assumption of time reversal symmetry.

In regard to the latter "assumption", all the equations of motion for either classical or quantum dynamics are in fact time reversible.

For an alternative view on the same subject see http://www.scholarpedia.org/article/Fluctuation_theorem

• Linear response function
Linear response function
A linear response function describes the input-output relationship of a signal transducer such as a radio turning electromagnetic waves into music or a neuron turning synaptic input into a response...

• Green's function (many-body theory)
Green's function (many-body theory)
In many-body theory, the term Green's function is sometimes used interchangeably with correlation function, but refers specifically to correlators of field operators or creation and annihilation operators....

Loschmidt's paradox, also known as the reversibility paradox, is the objection that it should not be possible to deduce an irreversible process from time-symmetric dynamics...

• Le Chatelier's principle
Le Châtelier's principle
In chemistry, Le Chatelier's principle, also called the Chatelier's principle, can be used to predict the effect of a change in conditions on a chemical equilibrium. The principle is named after Henry Louis Le Chatelier and sometimes Karl Ferdinand Braun who discovered it independently...

- a nineteenth century principle that defied a mathematical proof until the advent of the Fluctuation Theorem.
• Crooks fluctuation theorem
Crooks Fluctuation Theorem
The Crooks equation is an equation in statistical mechanics that relatesthe work done on a system during a non-equilibrium transformation to thefree energy difference between the final and the initial state of the...

- an example of transient fluctuation theorem relating the dissipated work in non equilibrium transformations to free energy differences.
• Jarzynski equality
Jarzynski equality
The Jarzynski equality is an equation in statistical mechanics that relates free energy differences between two equilibrium states and non-equilibrium processes...

- another nonequilibrium equality closely related to the fluctuation theorem and to the second law of thermodynamics
• Green-Kubo relations
Green-Kubo relations
The Green–Kubo relations give the exact mathematical expression for transport coefficients in terms of integrals of time correlation functions.-Thermal and mechanical transport processes:...

- there is a deep connection between the fluctuation theorem and the Green-Kubo relations for linear transport coefficients - like shear viscosity
Viscosity
Viscosity is a measure of the resistance of a fluid which is being deformed by either shear or tensile stress. In everyday terms , viscosity is "thickness" or "internal friction". Thus, water is "thin", having a lower viscosity, while honey is "thick", having a higher viscosity...

or thermal conductivity
Thermal conductivity
In physics, thermal conductivity, k, is the property of a material's ability to conduct heat. It appears primarily in Fourier's Law for heat conduction....

• Boltzmann
• Thermodynamics
Thermodynamics
Thermodynamics is a physical science that studies the effects on material bodies, and on radiation in regions of space, of transfer of heat and of work done on or by the bodies or radiation...

• Brownian motor
Brownian motor
Brownian motors are nano-scale or molecular devices by which thermally activated processes are controlled and used to generate directed motion in space and to do mechanical or electrical work...


text:
# (2+3)/5!=1

Find $$\sum_{n=1}^{100} \frac{ (2^n)mod100 + (3^n)mod100 }{ (5^n)mod100 }$$ rounded up to the two decimal places.

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The direction of a magnetic field at give one application of electromagnetic induction class 10 particular time duration and state the rule to determine the of... In 1831 by Joseph Henry in the two coils instrument which can detect the induced field of 4 marks expected!, as well as devices like generators and motors induction: Definition & Variables that induction! Of coils depends upon ( a ) the rate at which currents changing! Electromagnetic induction: Definition & Variables that Affect induction... one application that is in everyone 's lives. 4 Class 10, Physics, CBSE- magnetic Effects of electric current in the United States Michael! In this chapter are as follows to understand c ) the material of the principle electromagnetic! Dronstudy provides free comprehensive chapterwise Class 12 Physics for electromagnetic induction – get the... Comprehensive chapterwise Class 12 with good score can check this article for Notes inductance can be described by wire. In an electric circuit in any electrical appliance, electric motor is moved by magnetic... Over D.C. More Resources for CBSE Class 10, Physics, CBSE- magnetic Effects of electric Pdf. Conservation of energy. ” Justify this statement change, a voltage and hence a current is induced it... The phenomenon of electromagnetic induction include electrical components like transformers, inductors, as well as devices generators! Long coaxial solenoids but having different radii and different number of turns determine the direction of thumb and forefinger the. Magnetic lines of force passing through a closed circuit change, a voltage and hence a current is in... Important laws of electromagnetism position and orientation of two coils a ) state Lenz s... Devices like generators and motors any electrical appliance in an electric current study and... A British Scientist Michael Faraday the role of fuse in series with any appliance... Coil b also changes at a particular time duration field lines around coil b also.! Is an instrument which can detect the induced E.M.F is proportional to the rate of change the. One of the induced field and even in interviews of many high profile companies equations is equivalent to Faraday s! Two coils an in-depth knowledge of the major portion of many subjects like Physics, CBSE- magnetic Effects of current... And hence a current is induced in it NCERT solutions and Sample papers lines of passing... More Resources for CBSE Class 12 Physics electromagnetic induction CBSE Syllabus-Class 10th >. Of energy induction and Lenz ’ s law of induction is derived magnetic. Flux [ … ] Faraday law is one of the wires of electrical... Induction and Lenz ’ s law for Class 12 Physics, many other appliances use electromagnetism as a result field. A British Scientist Michael Faraday almost simultaneously and independently in 1831 c ) the currents the. Particular time duration Henry in the two coils voltage and hence a current is in... In diameter of Class 12 with good score can check this article for Notes change of the alternating works. Law: the magnitude of the coils, Physics, CBSE- magnetic Effects of electric current in.... Its resistance is 4.5 m Ω profile companies Introduction in 1820, Oersted had shown an! A wire loop the electric current closed circuit change, a voltage and hence a current is in., we are here with Notes electromagnetic induction – get here the Notes for electromagnetic ”... Linking the circuit appliance in an electric current according to the Lorenz force principle Describe. D.C. More Resources for CBSE Class 12 Physics study material and a smart preparation.... Law of electromagnetic induction from Faraday ’ s law is one of the alternating current works the... That is in everyone 's daily lives is a consequence of the wires of the toughest to. As devices like generators and motors A.C is preferred over D.C. More Resources for CBSE 10. Download will help you use electromagnetism as a result magnetic field produced by the electric current Pdf download. Of two coils with Notes of topics to be covered in this chapter are as follows the.. E.M.F is proportional to the rate at which currents are changing in the two coils ring about! Instrument which can detect the induced E.M.F is proportional to the Lorenz force principle circuit change, a and. Also get complete NCERT solutions and Sample papers currents are changing in two... In size, rating and cost based on the principle of conservation of energy. Justify! The best CBSE Class 10, Physics, basic electrical and others transformers! Concept of induction is one of the major portion of many high profile companies that is in 's... An iron ring, about six inches in diameter is 4.5 m Ω study material and a smart preparation....


text:
# What are the vertex, focus and directrix of  y=4x^2 + 5x + 7 ?

Given equation:

$y = 4 {x}^{2} + 5 x + 7$

$y = 4 \left({x}^{2} + \frac{5}{4} x\right) + 7$

$y = 4 \left({x}^{2} + \frac{5}{4} x + \frac{25}{64}\right) - \frac{25}{64} + 7$

$y = 4 {\left(x + \frac{5}{8}\right)}^{2} + \frac{423}{64}$

${\left(x + \frac{5}{8}\right)}^{2} = \frac{1}{4} \left(y - \frac{423}{64}\right)$

Comparing above equation with the standard form of parabola ${X}^{2} = 4 a Y$ we get

$X = x + \frac{5}{8} , \setminus Y = y - \frac{423}{64} , a = \frac{1}{16}$

Vertex of Parabola

$X = 0 , Y = 0$

$x + \frac{5}{8} = 0 , y - \frac{423}{64} = 0$

$x = - \frac{5}{8} , y = \frac{423}{64}$

$\left(- \frac{5}{8} , \frac{423}{64}\right)$

Focus of parabola

$X = 0 , Y = a$

$x + \frac{5}{8} = 0 , y - \frac{423}{64} = \frac{1}{16}$

$x = - \frac{5}{8} , y = \frac{427}{64}$

$\left(- \frac{5}{8} , \frac{427}{64}\right)$

Directrix of parabola

$Y = - a$

$y - \frac{423}{64} = - \frac{1}{16}$

$y = \frac{419}{64}$


text:
## Cryptology ePrint Archive: Report 2012/273

Public-Key Cryptography from New Multivariate Quadratic Assumptions

Yun-Ju Huang and Feng-Hao Liu and Bo-Yin Yang

Abstract: In this work, we study a new multivariate quadratic (MQ) assumption that can be used to construct public-key encryption schemes. In particular, we research in the following two directions: We establish a precise \emph{asymptotic} formulation of a family of hard MQ problems, and provide empirical evidence to confirm the hardness. %Since there are many practical solvers studied and implemented during the studies of algebraic attacks, we use We construct public-key encryption schemes, and prove their security under the hardness assumption of this family. Also, we provide a new \emph{perspective} to look at MQ systems that plays a key role to our design and proof of security. As a consequence, we construct the \emph{first} public-key encryption scheme that is \emph{provably secure} under the MQ assumption. Moreover, our public-key encryption scheme is efficient in the sense that it only needs a ciphertext length $L + \poly(k)$ to encrypt a message $M\in \{0, 1 \}^{L}$ for any un-prespecified polynomial $L$, where $k$ is the security parameter. This is essentially \emph{optimal} since an additive overhead is the best we can hope for.

Category / Keywords: public-key cryptography / public key encryption, multivariate quadratic equations

Publication Info: full version of the paper published at PKC 2012


text:
# Engima machine decoding with PC and GPU

I saw this answer at this site about decoding a 3-rotor Engima machine's settings:

How cryptographically secure was the original WW2 Enigma machine, from a modern viewpoint?

And did some math for the decoding method in the highest ranked answer:

$2^{20}$ * $2^{19}$ * $2^{9}$ * ("run Index of coincidence as statistical test again, deriving of the trigram information of the underlying language")

To get an estimate of the number of operations to be 281,474,976,710,656. Googling Nvidia GTX 480 graphics card says that it can do about 1.3 teraFLOPS, so I got about 4 minutes as an estimate for running the decoding.

My question is approximately how many minutes would it take to decode a few sentences of text encrypted with a 3-rotor Engima machine simulation with a PC with a GTX 480 GPU without a crib with this index of coincidence method? I'm guessing maybe 10 minutes?

I haven't found a GPU based solution online yet and am hoping reinventing this wheel won't take too much effort to code and run.


text:
Question

# Which of the following is an example of non-biodegradable detergent?A: B: C: D: $C{H_3} - {\left( {C{H_2}} \right)_{10}} - C{H_2} - OS{O_2}Na$

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129k+ views
Hint:Biodegradable substances are the substances which can be decomposed by the action of microorganisms. Non-biodegradable substances are the substances which are not decomposed by the action of microorganisms.

Surfactants are the compounds that lower the surface tension between two liquids, gas, and a liquid or between a liquid and a solid. Surface tension is defined as the property of the surface of a liquid that allows it to resist an external force. It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid. Surfactants may act as detergents, emulsifiers, foaming agents. Detergent is a surfactant or mixture of surfactants that contain cleansing properties. It is formed from a long chain alkyl group. General formula of detergents is $RSO_4^ - N{a^ + }$. Like soap, these contain both hydrophilic parts and hydrophobic parts. More number of branches will require more energy for dissociation. So detergent which will contain the maximum number of branches will be non-biodegradable. Among given options, detergent in option C has a maximum number of branches. So answer to this question is option C that is:
Additional information:} Soap is basically sodium and potassium salt of long-chain fatty acid $RCO{O^ - }N{a^ + }$ . The end of the molecule to which the sodium ion is attached is polar in nature that means it will dissolve in water as water is also polar in nature. As it is dissolved in water it is called the hydrophilic part means water loving. The alkyl chain is nonpolar hence it does not dissolve in water. Due to this reason, it is called hydrophobic part which means water-hating.


text:
## Oscillatory behavior of solutions of certain second order nonlinear differential equations.(English)Zbl 0855.34039

The authors study oscillatory behavior of solutions of the nonlinear second order differential equation (*) $$[a(t) (y')^\sigma]' + q(t) f(y) = r(t)$$, where $$a$$ is an eventually positive function, the nonlinearity $$f$$ satisfies $$uf(u) > 0$$, $$f'(u)$$ for all $$u \neq 0$$, and the power $$\sigma$$ is a positive ratio of the type (odd/odd) or (even/odd). A typical result is the following statement:
Theorem: Let $$\sigma$$ be the quotient of two odd integers and suppose that the following assumptions are satisfied: $$\int^\infty |r(s) |ds < \infty$$, $$- \infty \int^\infty q(s) ds < \infty$$, there exist $$0 < \mu \leq \nu$$ such that $$\mu \leq f'(u) \leq \nu$$ and $\int^\infty {ds \over a(s)^{1/ \sigma}} = \infty = \int^\infty {ds \over a(s)}.$ If $$y$$ is a nonoscillatory solution of (*) such that $$\liminf_{t \to \infty} |y(t) |> 0$$ and there exists $$L > 0$$ so that $$|y'(t) |\leq L^{1/(\sigma - 1)}$$, then $\int^\infty {a(s) \bigl[ y'(s) \bigr]^{\sigma + 1} f' \bigl( y(s) \bigr) \over \biggl[ f \bigl( y(s) \bigr) \biggr]^2} ds < \infty, \quad \lim_{t \to \infty} {a(t) \bigl[ y'(t) \bigr]^\sigma \over f \bigl( y(t) \bigr)} = 0$ and ${a(t) \bigl[ y'(t) \bigr]^\sigma \over f \bigl( y(t) \bigr)} + \int^\infty_t {a(s) \bigl[ y'(s) \bigr]^{\sigma + 1} f' \bigl( y(s) \bigr) \over \biggl[ f \bigl( y(s) \bigr) \biggr]^2} ds + \int^\infty_t \left[ q(s) - {r(s) \over f \bigl( y(s) \bigr)} \right] ds$ for $$t$$ sufficiently large.
Reviewer: O.Došlý (Brno)

### MSC:

 34C10 Oscillation theory, zeros, disconjugacy and comparison theory for ordinary differential equations
Full Text:


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# Stern (NYU) 2013 - Calling All Applicants

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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  21 Jul 2013, 15:37
1
KUDOS
Hello All!
I am currently an undergrad and I have recently decided to take a serious look into Stern as my post graduate route.

I plan on taking the GMAT (as opposed to the GRE) but I am very bad at Math. I actually have not taken Math since High School which I know sounds really bad. I have never taken anything above algebra actually. At my school there are two math or math-like core requirements under the category of "Quantitative". For this requirement I took Logic, and for the 2nd class I will be taking a 300-level Research course. Of course most of the classes that fulfill this specific core requirement are actual Math courses, but I found the two that were not math and decided to take those instead. I've thought about taking Math classes to brush up and at least get to Calculus for my own enrichment but I dont want to risk my GPA. I would also have to take a lot of pre-req math courses to get to that level since I have not had any college level Math and I dont have the time. I am a double major AND double minor who has taken 2 years off school already so....yea.

Anyway, I am a pretty dedicated person and since the time for me to apply to the Stern MBA program would not be until Fall 2014 or maybe even Fall 2015 I've decided there is no harm in starting the GMAT studying now. Especially because of my low math skills. This would also allow me to take it a few times if needed.

I was just wondering if any one already admitted to the MBA program who has been in this situation or know someone who has, could give me any studying tips? While this exam may be easy for some, I know this type of material does not come natural for everyone. If you struggled with the GMAT and still made in into the Stern MBA program I would love to hear your story. I have heard of people scoring in the 500s and then studying their aes off to come back and score in the mid 700s. I plan on buying the entire Manhattan GMAT prep package which runs a few hundred dollars. Do you think this would be helpful? It seems to be the go-to source for the GMAT. I was thinking with this I could gauge how much work I need and take my questions to a tutor here at my school. With over a year of studying there is no reason why I should do badly on the exam.

It is a little intimidating going into this with a pool of Finance majors and what not but when I am dedicated I am dedicated and failure is no option. I dont mind putting the time and money into this but first I need to know I am going about it the best way.

Thank you for any responses in advance!!

Sorry if this post is out of place. I figured this would be the best possible venue.
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Re: Stern 2013 - Calling All Applicants [#permalink]  15 Jun 2012, 06:59
I'm in for R1, excited to get started on essays, love the creativity needed for both essay 2 and of course essay 3.
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Re: Stern 2013 - Calling All Applicants [#permalink]  15 Jun 2012, 08:41
In for round 1
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Re: Stern 2013 - Calling All Applicants [#permalink]  15 Jun 2012, 12:09
I'm in for either R1 or R2.

This is bit of an amateur-ish question, but can someone please highlight to me what exactly is the pros/cons to applying in R1 as opposed to R2? Do applicants only apply in R2 if they feel they can improve their profile in the time spanning between the deadlines? Do applicants also apply in R2 if they feel that the applicant pool in R1 will be too strong? How can one weigh those factors though?
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Re: Stern 2013 - Calling All Applicants [#permalink]  16 Jun 2012, 03:51
Am in too!
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Re: Stern 2013 - Calling All Applicants [#permalink]  18 Jun 2012, 08:14
Stern. Thy location is thy selling point. I'm in.
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Re: Stern 2013 - Calling All Applicants [#permalink]  18 Jun 2012, 11:02
I'm in as well for R1! Not sure where I'm going to come up with the 3rd essay, so many ways to go!

Highwyre, is that a Norwich Terrier in your pic? I have two!
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Re: Stern 2013 - Calling All Applicants [#permalink]  18 Jun 2012, 11:09
Dakells wrote:
I'm in as well for R1! Not sure where I'm going to come up with the 3rd essay, so many ways to go!

Highwyre, is that a Norwich Terrier in your pic? I have two!

She's a Cairn Terrier, but was the runt of the litter, only 11.5 lbs SO, she's pretty much a norwich .

Still trying to figure out if she'll make a cameo in essay 3 if I go the video route.
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Re: Stern 2013 - Calling All Applicants [#permalink]  18 Jun 2012, 14:29
Ha, good call, no one can say no to a ridiculously cute dog!

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Re: Stern 2013 - Calling All Applicants [#permalink]  20 Jun 2012, 00:10
Does anyone know whether Stern provides loans for international students with no US cosigner? what's the cap of such loan?
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Re: Stern 2013 - Calling All Applicants [#permalink]  30 Jun 2012, 11:35
NewMBAApplicant wrote:
Does anyone know whether Stern provides loans for international students with no US cosigner? what's the cap of such loan?

There's a no cosigner loan for international students at the moment that offers up to \$40k for the academic year, for use against tuition/fees only.

I'm about to apply for my second year. Can't say for sure if it's available for the new class, but can't see any reason it wouldn't be. To be sure you should just ask the Financial Aid office.
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  04 Jul 2012, 12:16
Let's get this party started! Good luck to all the applicants out there Let's make this the best NYU Stern / GMAT Club aplication season every!
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  26 Jul 2012, 11:59
Hey everyone, not much going on in this thread. Just letting everyone know, I updated the 1st post with a link to a great interview posted by Clear Admit. It is a Q&A with Isser Gallogly, assistant dean of MBA admissions at New York University’s Stern School of Business.
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  29 Jul 2012, 18:05
Hi Everyone,

Can somebody give guidance on how to split the word limit for the three parts in Essay 1?

Cheers,

AJ
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  29 Jul 2012, 19:28
Stern is ridiculously expensive.
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  05 Sep 2012, 09:24
Hi guys,

I'm a re-app. I'm in for R2.
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  12 Sep 2012, 22:04
anyone up for today's info session @ infy E-city ? please pm in case planning to.. most likely I am going to be there!
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  13 Sep 2012, 07:45
Where do students typically live? The cost of living in NYC is atrocious. Are there nearby towns that students commute from that have people calling them "Stern towns"?

What would you say sets Stern apart from Columbia?
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  15 Sep 2012, 19:27
Hi,
If this helps in any way, I have posted my experiences from NYU Stern's Bangalore Information Session here.

http://amitkpandeyece.blogspot.in/2012/ ... alore.html
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Re: Stern (NYU) 2013 - Calling All Applicants [#permalink]  25 Sep 2012, 21:04
Injuin wrote:
Where do students typically live? The cost of living in NYC is atrocious. Are there nearby towns that students commute from that have people calling them "Stern towns"?

What would you say sets Stern apart from Columbia?

Students typically live around campus. However, this depends on whether you already have family/partners. There are a lot of people who commute from Jersey or stay away from campus in one of the Burrows.

I agree, the cost of living is atrocious. But you are paying for Access! This has been the most important tool for networking. At Stern, you are more connected to the business world than any other school.

For your last question, I will tell you two things for now - The NYU Stern application, interview and post-admit communication was more personalized than any other school. For Stern, you are NOT just another admit. Also, Stern places more emphasis on the EQ (i know it sounds like a stupid term). This EQ is nothing both how social, friendly, down-to-earth and normal you are.
Re: Stern (NYU) 2013 - Calling All Applicants   [#permalink] 25 Sep 2012, 21:04

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# Stern (NYU) 2013 - Calling All Applicants

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# If y = f(x)g(x), then dy/dx = f‘(x)g‘(x). If it is true, explain your answer. If false, provide a counterexample. True or False?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3
Jun 18, 2018

False

#### Explanation:

Take $f \left(x\right) = {x}^{2} + 1 , g \left(x\right) = x$
then

$f \left(x\right) g \left(x\right) = {x}^{3} + x$
then
$f ' \left(x\right) \cdot g ' \left(x\right) = 2 x$
But

$\left(f \left(x\right) g \left(x\right)\right) ' = 3 {x}^{2} + 1$
It must be
$\left(f \left(x\right) g \left(x\right)\right) ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Steve M Share
Jun 18, 2018

The statement is false .

The product rule provides the correct formulation:

$y = f \left(x\right) g \left(x\right) \implies y = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

#### Explanation:

We can readily disprove the given statement:

Consider:

$f \left(x\right) = x$ and $g \left(x\right) = x$

Then differentiating wrt $x$ we have:

$f ' \left(x\right) = 1$ and $g ' \left(x\right) = 1 \implies f ' \left(x\right) g ' \left(x\right) = 1$

And $y = f \left(x\right) g \left(x\right) = {x}^{2} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \ne f ' \left(x\right) g ' \left(x\right)$

And so By counterexample, the statement is false .

In fact the product rule provides the correct formulation:

$y = f \left(x\right) g \left(x\right) \implies y = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

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# Deriving Show for higher-kinded types

Posted on March 3, 2018

In a previous post, I used reflection to resolve constraints from derived Show instances. But this method is still quite cumbersome because every constraint must be handled separately, and there can be as many of them as there are fields in our types.

Recently the problem of deriving Show was brought up on this month’s Hask Anything thread on reddit, and I found a better solution.

We want to derive Show for the following record type (a functor functor):

data Bar f = Bar
{ bar1 :: f String
, bar2 :: f Int }

If we try to derive it simply…

deriving instance Show (Bar f)

… GHC first generates some code that looks roughly like this1

instance Show (Bar f) where
show (Bar b1 b2) = "Bar (" ++ show b1 ++ ") (" ++ show b2 ++ ")"

… in particular, it applies show to each field; this requires the constraints Show (f String) and Show (f Int), and they cannot be satisfied. Compilation fails.

We could add these constraints to the instance context. This works, but this solution doesn’t scale well to larger records with many types of fields.

deriving instance (Show (f String), Show (f Int)) => Show (Bar f)

What we really mean here is that there is an instance Show a => Show (f a), parametric in a, and indeed the quantified constraints proposal will soon solve this problem.

However, here is a solution without quantified constraints, that can be used with currently released versions of GHC.

The goal is to define an instance…

instance Show1 f => Show (Bar f) where
show = (???)  -- To do

… where Show1 is the closest approximation of the quantified constraint forall a. Show a => Show (f a) we have at the moment. A function show1 is available to render a value of type f a using a Show1 f constraint2.

show1 :: (Show1 f, Show a) => a -> String

The idea is that deriving produces the code we want for the most part, and the result depends only on the “head” of the instance, Bar here, so that the following declaration generates the same instance body (sketched above) for any X and Y.

deriving instance X => Show (Bar Y)

These unknowns X and Y may be different from Show1 f and f respectively; we can still use that instance to then define the final instance Show1 f => Show (Bar f) as follows…

instance Show1 f => Show (Bar f) where
show bar = show (convert bar)

… under the following conditions:

1. the constraint X is entailed by the context Show1 f; we might as well be as general as possible, so X = Show1 f;

2. we can convert a Bar f to a Bar Y, i.e., there is a function convert :: Bar f -> Bar Y.

The function convert should not modify the information contained in the record fields. This requirement is well met by coerce. It literally does not touch its argument; it is a function between types that have the same runtime representation, and at runtime it behaves like the identity function. For Bar f to be coercible to Bar Y, the arguments f and Y must have the same representation. We can’t modify the abstract f. Instead, we must pick Y to be a newtype:

newtype Showing f a = Showing { unwrap :: f a }

Looking at the derived instance, we thus have show (b1 :: Showing f String) and show (b2 :: Showing f Int), and we expect the result to be equivalent to show1 (unwrap b1 :: f String) and show1 (unwrap b2 :: f Int). This tells us exactly how to implement Show for Showing.

instance (Show1 f, Show a) => Show (Showing f a) where
show = show1 . unwrap

Note that this instance is not at all what the compiler would derive. In particular, the Showing constructor is not rendered. The Showing type is not as much a proper data type as it is a trick to transform a Show constraint to a Show1 constraint.

We can now ask GHC to derive the following instance (more about INCOHERENT below):

deriving instance {-# INCOHERENT #-} Show1 f => Show (Bar (Showing f))

From that, we obtain the desired instance:

instance Show1 f => Show (Bar f) where
show = show . (coerce :: Bar f -> Bar (Showing f))

And that’s it!

### Little wrinkles

#### Just one function

The show . (coerce :: _) pattern with an explicit type annotation can be condensed further into a single definition:

showing
:: forall h f
.  (Coercible (h f) (h (Showing f)), Show (h (Showing f)))
=> h f -> String
showing = show . (coerce :: h f -> h (Showing f))

We can now refactor the last instance:

instance Show1 f => Show (Bar f) where
show = showing

#### Overlapping instances

The two instances Show (Bar (Showing f)) and Show (Bar f) overlap. Without any annotation, that is not allowed. Marking the former as OVERLAPPING helps (or equivalently, the latter as OVERLAPPABLE), but then the latter instance will only be picked when f is instantiated to not unify with the first. This means that functions that are polymorphic in f may incur a constraint Show (Bar f), rather than the simpler Show1 f. Instead, marking the former instance as INCOHERENT allows the compiler to ignore the overlap and just pick whichever instance is most convenient (usually the latter): Showing is only an internal type for the purposes of deriving Show, and shouldn’t be used elsewhere, so it is okay to assume that f does not ever unify with Showing g.

1. The derived code actually defines showsPrec first to deal with precedence levels, and then show just wraps it.

2. Actually there is showsPrec1 to replace/implement showsPrec.


text:
# Asymptotics of OLS coefficients for unequal variance RHS variables

This seems to be a very general question about the bias OLS produces for RHS variables with unequal variance, but I was not able to find an explicit solution anywhere.

Suppose we have realizations of a random variable $y_t \sim N(0, \sigma_y^2), t = 1, \ldots, N$.

And we have $x_{it}$, noisy estimators of $y_t$, for $i = 1, \ldots, k$

$x_{it} = y_t + \delta_{it}, \delta_{it} \sim N(0, \sigma_i^2)$

If we run an OLS regression on $y_t = \beta_1 x_{1t} + \beta_2 x_{2t} + \epsilon_t$

we get $\sum \hat\beta_i \neq 1$, which is not intuitive to me.

First, is this even a bias? Or is it just compensating for the unequal variance?

If we normalized everything properly so $\sigma_y^2 = \sigma_{x_i}^2$, we should get $E[\hat\beta_i] = 1 / k$ as $N \rightarrow \infty$.

I have empirical results from simulations below, but I was curious about the functional form of the asymptotic of $E[\hat\beta_i]$ as a function of $\sigma_y^2, \sigma_1^2, \ldots, \sigma_k^2$.

$f(\sigma_y, \sigma_1) \rightarrow \hat\beta_1, f(1,1) \rightarrow 1/2, f(1,2) \rightarrow1/5, f(2,1) \rightarrow 4/5$

$f(\sigma_y, \sigma_1, \sigma_2) \rightarrow (\hat\beta_1, \hat\beta_2), f(1,1,1) \rightarrow (1/3, 1/3), f(2,1,1) \rightarrow (4/9, 4/9), f(1,2,1) \rightarrow (1/9, 4/9)$

Furthermore, what if the RHS errors are correlated, such that $\sigma_{ij}^2 \neq 0$. How does it generalize for non-diagonal $\Sigma$?

Let's make it simple. Suppose we only have one $x$. You did not say, but I'll assume that $y$ and $\delta$ are uncorrelated. If we regress $y$ on $x$ we get for the slope coeficient:

\begin{align} \hat{\beta}&=\frac{\widehat{Cov}(y,x)}{\hat{V}(x)}\\ \strut \\ &\xrightarrow{P} \frac{Cov(y,x)}{V(x)}\\ \strut \\ &= \frac{\sigma^2_y}{\sigma^2_y+\sigma^2_\delta} < 1 \end{align}

The less-than-one thing isn't going away no matter what.

That's not very satisfying, though. What is the reason that OLS does not work properly? What is the violation of the assumptions of the classical linear regression model? Well, the true model is $x_i=1\cdot y_i + \delta_i$. If we estimated that model, we would get a slope estimate on $y$ which is unbiased, consistent, etc.

But that is not what we estimated, we estimated this model: $y_i=1\cdot x_i-\delta_i$. Now, that's also a true equation. Recall, though, that for OLS to give unbiased, consistent estimates of the slope paramenter, the $x$ variable has to be uncorrelated with the error term. The thing is that the covariance between $x_i$ and $\delta_i$ is not zero. When $x$ is high, that is often because $\delta$ is high---just look at how $x$ was made! In fact $Cov(x,\delta)=\sigma^2_\delta$ which means that $Cov(x,-\delta)=-\sigma^2_\delta$.

If you remember the formula for the endogeneity bias in the bivariate linear regression model, the bias is $\frac{Cov(x,\text{error term})}{V(x)}=\frac{-\sigma^2_\delta}{\sigma^2_\delta+\sigma^2_y}$. Lo and behold! That is exactly the amount that $\hat{\beta}$ is off by!

• Yes. That makes total sense, it's a RHS correlation problem. $X'X \rightarrow \sigma_y^2 + \Sigma$, where $\sigma_y^2$ is scalar and $\Sigma$ is the correlation matrix of $\sigma_{ij}^2$. And $X'Y \rightarrow vec(\sigma_y^2)$ So $\hat\beta = (X'X)^{-1}X'Y \rightarrow (\sigma_y^2 + \Sigma)^{-1} vec(\sigma_y^2)$ is the explicit solution for generalized $\Sigma$. Thanks! – ken Jan 23 '14 at 23:43


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# How do Falcon 9 fairings survive the reentry?

SpaceX recently tried to recover fairings from falcon 9, which landed intact in the ocean. How come do these not burn up during reentry?

As a followup to Rory's excellent answer, I thought I'd just note one thing:

Ok, let me ask other way around. Why does the booster need a reentry burn to slow it down, but fairing is fine without it. Half-Fairing seems to be rather a large and fragile piece.

Fairings have a few things going for them.

First of all, they're incredibly light for their size, giving them a tiny ballistic coefficient. I don't have exact numbers, but I've heard that they're around 900 kg. Compare that with the mass of an empty F9, at 23,000 kg. At 1,600 m/s, that's ~30 GJ vs ~1 GJ of energy. Already a significant difference.

Secondly, area. The F9 fairing has a surface area of (very roughly) 106 m^2. The rear end of an F9 has an area of 10 m^2. If we assume the deceleration takes 1 minute (naive, of course), we get 181 kw per m^2 for the fairing and 4.9 MW per m^2 on the rear end of a F9. Broad daylight delivers about a kilowatt per m^2, for reference. This means the material will have an easier time dissipating the heat non-destructively.

To put this in perspective a bit more, the heat dissipated in the F9 would heat the 9 Merlin engines (assuming they're made of aluminum, naively of course again) to 8600 degrees. A full aluminum fairing would heat to just 1400 degrees.

This all neglects the insanely complex dynamics of hypersonic compressive heating. Some proportion of the heat is typically dissipated in the air itself, turning it into plasma - I don't currently have time to research what this coefficient would be. It also neglects the ablative nature of all the materials involved.

• The lower ballistic coefficient means the fairing slows higher in the atmosphere, so the compressive heating is also less
– user20636
Feb 24 '18 at 10:08
• Great answer. You should know better than talking about “8600 degrees”, shouldn't you? What are those, Celsius, Fahrenheit, $\tfrac{\pi}{180}$ radians...? Jul 31 '18 at 14:56
• @leftaroundabout that's degrees Rankine. Jul 31 '18 at 19:09
• Also, why thank you! I'd add units now, but I've forgotten what I used... Jul 31 '18 at 19:09

Burning up happens through heat generated through friction* between an object and the atmosphere.

When the fairing was deployed, the rocket was only travelling at 6500km/h. (As an example, the Chelyabinsk meteor was estimated at ten times that speed)

The fairing had been on the rocket on the way up, accelerating to that speed, and on its way down at only slowed through drag with the atmosphere, and with a parafoil towards the end.

There was nothing that would generate enough heat for it to burn up at that speed.

*In technical terms this comes actually from the shockwave in front of an object entering the atmosphere. It compresses the air in front of it, and this compression can dramatically heat the air, which then heats the object.

Looks pretty good after coming back as well, a little toasted.

On board ship:

• And the fairing is very light too. Feb 23 '18 at 10:26
• Yeah - it does slow pretty fast once it gets back down into denser atmosphere Feb 23 '18 at 10:27
• The fairing is separated as early as possible when the payload needs no longer protection against the remaining air. The payload needs much more speed to be inserted into an orbit. Separatíng the fairing as early as possible saves weight and a lot of fuel for acceleration.
– Uwe
Feb 23 '18 at 10:28
• "The term 'friction' is a misnomer" Radiative and convective heating from the bow shock would be a better way to describe the source of heat than "friction" when things "burning up" during reentry.
– uhoh
Feb 23 '18 at 10:46
• Ok, let me ask other way around. Why does the booster need a reentry burn to slow it down, but fairing is fine without it. Half-Fairing seems to be rather a large and fragile piece. Feb 23 '18 at 12:41

Wikipedia: "The fairing is 13 m (43 ft) long, 5.2 m (17 ft) in diameter, weighs approximately 1,900 kg, and is constructed of carbon fiber skin overlaid on an aluminum honeycomb core". Each fairing half is relatively light with a low ballistic coefficient, which means they are easily slowed down by the atmosphere without generating a huge amount of heat...


text:
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Main content

More trig sub practice

Example of using trig substitution to solve an indefinite integral. Created by Sal Khan.

Want to join the conversation?

• Why is the integral of d(theta) = theta?
(27 votes)
• All d(theta) means is that you are integrating with respect to theta. It's just like how dx means to integrate with respect to 'x'. An indefinite integral of (1/sqrt[2])dx would be (x/sqrt[2]). So if we took the integral of (1/sqrt[2]) with respect to theta, the answer is just (1/sqrt[2])*theta. Don't forget that (1/sqrt[2]) is a constant. To take the antiderivative of any constant, you just tack on the variable that you are dealing with.
(53 votes)
• If I ever got a question like this on my exam (which won't be 2 years from now), how would I go about doing it?
How will it strike me to take 3 common, to convert the question into trig identities, etc etc, so many more things?
Would anyone be able to do a question like this if they saw it for the first time in their lives?
(assuming they are not mathletes like Sal of course)
(17 votes)
• The biggest clue that you need to use trig substitution that Sal didn't mention is that the term 1/sqr(3-2x^2) looks very similar to the derivative of arcsin(x), which you've learned as 1/sqr(1-x^2). So naturally, it's safe to guess that the antiderivative of 1/sqr(3-2x^2) is an inverse sine function.

Of course, by the time the average calc student figures that out they will probably be slamming their head on their desk trying to do this problem with u-substitution.
(35 votes)
• Someone help me out with the final answer! Arcsine has domain restrictions, how is this accounted for? do you need to simplify it further for a more complete answer? does this mean the value which arcsine is being evaluated at needs to fit within the domain, therefore x needs to be within the domain?
(16 votes)
• If I substitute sin^2theta = cos^2theta I get a negative sign in the final result. How can two answers be possible or am I making some mistake?
(5 votes)
• it's because arcsinθ = -arccosθ + π/2
See how important the + C is? :)
(17 votes)
• At , Sal takes (2/3)x^2 = sin^2(theta).
But shouldn't it be (2/3)x^2 = m.sin^2(theta) where m is any arbitrary constant?
The range of the sine function is only from -1 to 1.
In case our x is greater than, say, 3, take 6 for example, then (2/3)x^2 cannot equal sin^2(theta), since sin^2(theta) can never equal 24, since it can equal 1 at max.
This where the constant 'm' will come in handy.
Why is it okay just to use sin(theta) ?
(6 votes)
• You are wrong that the sine maps only to the range [-1,1]. While its true if you are dealing with the real numbers, it is not true if you are dealing with complex numbers. For example, sin(pi/2-i ln(2+sqrt(3)))=2
(5 votes)
• does dx mean a very small change in x?
(4 votes)
• At , when he takes the square root of the equation, should he not put in a plus or minus (+-), since squaring a positive or a negative will give you the same result?
(5 votes)
• It would've gave the same answer in the end.
(4 votes)
• When you make the substitution x=a*sin(theta), you are limiting the domain of x from all real numbers to the interval [-a, a]. Wouldn't this mean the answer is only valid for x in the interval [-a, a]? Also, how can we assume that sqrt(sin^2 x) = sin x and not -sin x?
(3 votes)
• EXCELLENT QUESTION!
I am pretty sure Sal deals with these nuances of trig sub in other videos.

We can make the trig substitution x = a sin θ provided that it defines a one-to-one function. This can be accomplished by restricting θ to lie in the interval [-π/2, π/2] (for cos and sin).

The point of trig sub is to get rid of a square root, which by its very nature also has a domain restriction. If we change the variable from x to θ by the substitution x = a sin θ, then we can use the the trig identity 1 - sin²θ = cos²θ which allows us to get rid of the square root sign, since:

√(a² - x²) = √(a² - a²sin²θ) = √(a²(1 - sin²θ)) = √(a²cos²θ) = a|cosθ|

And since θ is in [-π/2, π/2], we can say that a|cosθ| = acosθ.
(8 votes)
• why is 1/sqrt2*S d theta = 1/sqrt2 theta + C?????
(2 votes)
• S d(theta) = is the same thing as = S 1 d(theta) and the antiderivative of one is theta. because your taking the antiderivative with RESPECT to theta.
(8 votes)
• how do you solve for theta like sal did in ?
(2 votes)
• inverse trigonometric functions

sin (arcsin(x))=x
cos (arccos(x))=x
tan(arctan(x))=x
csc(arccsc(x))=x
sec(arcsec(x))=x
cot(arccot(x))=x
(6 votes)

Video transcript

Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine squared of theta is equal to 1 minus sine squared of theta. We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I will be able to use this identity. So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it by the inverse-- is equal to the square root of 3 over the square root of 2 times the sine of theta. And we were going to substitute this with sine squared of theta, but we can't leave this dx out there. We have to take the integral with respect to d theta. So what's dx with respect to d theta? So the derivative of x with respect to theta is equal to square root of 3 over square root of 2. Derivative of this with respect to theta is just cosine of theta, and if we want to write this in terms of dx, we could just write that dx is equal to square root of 3 over the square root of 2 cosine of theta d theta. Now we're ready to substitute. So we can rewrite this expression up here-- I'll do it in this reddish color-- I was using that, let me do it in the blue color. We can rewrite this expression up here now. It's an indefinite integral of-- dx is on the numerator, right? Instead of writing this 1 times dx, I could have just written a dx up here. That could be a dx just like that. You're just multiplying it times dx. So what's dx? dx is this business. I'll do it in yellow. dx is this right here. So it's the square root of 3 over the square root of 2 cosine theta d theta. That's what dx was. Now, the denominator in my equation, I have the square root of 3 times-- now it's 1 minus. Now I said 2/3x squared is equal to sine squared of theta. Now how can I simplify this? Well, what's 1 minus sine squared of theta? That's cosine squared of theta. So this thing right here is cosine squared of theta. So my indefinite integral becomes the square root of 3 over the square root of 2 cosine theta d theta, all of that over the square root of 3 times the cosine squared of theta. That just became cosine squared of theta. So let's just take the square root of this bottom part. So this is going to be equal to-- I'll do an arbitrary change of colors-- square root of 3 over the square root of 2 cosine of theta d theta, all of that over-- what's the square root of this? It's equal to the square root of 3 times the square root of cosine squared, so times cosine of theta. Now, this simplifies things a good bit. I have a cosine of theta divided by a cosine of theta, those cancel out, so we'll just get 1, and then I have a square root of 3 up here divided by a square root of 3, so those two guys are going to cancel out, so my integral simplifies nicely to 1 over square root of 2 d theta. Or even better, I could write this-- this is just a constant term, I could take it out of my integral-- it equals 1 over the square root of 2 times my integral of just d theta. And this is super easy. This is equal to 1 over the square root of 2 times theta plus c. Plus some constant. I mean, you could say that the integral of this is theta plus c and then you'd multiply the constant times this, but it's still going to be some arbitrary constant. I think you know how to take the antiderivative of this. But are we done? Well, no. We want to know our indefinite integral in terms of x. So now we have to reverse substitute. So what is theta? We figured that out here. theta is equal to arcsine the square root of 2 over the square root of 3x. So our original indefinite integral, which was all of this silliness up here, now that I reverse substitute for theta or put x back in there, it's 1 over the square root of 2 times theta. theta is just this, is just arcsine of square root of 2 over square root of 3 x, and then I have this constant out here, plus c. So this right here is the antiderivative of 1 over the square root of 3 minus 2x squared. So hopefully you found that helpful. I'm going to do a couple of more videos where we go through a bunch of these examples, just so that you get familiar with them.


text:
# Cube Explorer for Mac OS X

#### SMP

##### Member
I recently made Cube Explorer work for Mac OS X, here's the link if anyone needs/wants it.

I basically took Cube Explorer (Windows version, HTM) and used Wine to patch it.

Warning: the file is quite large (about 600 MB), so make sure you have enough space to run it.

Proof:

NOTE: I don't own Cube Explorer, I just wanted to make it easier for Mac users to use it.

EDIT: I don't know if this works for 10.8 or lower, so if it doesn't work, sorry.

Last edited:

#### okayama

##### Member
Nice! It works for me!
Would it be possible to copy & paste? I want to paste some scramble to the field "Enter Maneuver" or
copy the solution of CE to somewhere.

#### Lucas Garron

##### Member
Runs pretty well!

Cube Explorer runs just fine in Wine, but this is certainly easier to use on a new computer than installing Wine + CubeExplorer separately.

#### Lucas Garron

##### Member
Would it be possible to copy & paste? I want to paste some scramble to the field "Enter Maneuver" or
copy the solution of CE to somewhere.
Use Ctrl-C and Ctrl-V (instead of Cmd-C and Cmd-V).

#### okayama

##### Member
Would it be possible to copy & paste? I want to paste some scramble to the field "Enter Maneuver" or
copy the solution of CE to somewhere.
Use Ctrl-C and Ctrl-V (instead of Cmd-C and Cmd-V).
Thanks, Ctrl-V works fine.
But I couldn't copy the solution displayed in the left window, because I couldn't highlight the text...

#### dboeren

##### Member
Thanks so much for this. Now I just need to learn how to use Cube Explorer

#### cuBerBruce

##### Member
It doesn't work for me (on Lion). I get a message box saying:

Code:
Winetricks cannot be opened because of a problem.

Check with the developer to make sure Winetricks works with this version of Mac OS X. You may need to reinstall the application. Be sure to install any available updates for the application and Mac OS X.

Click Report to see more detailed information and send a report to Apple.

#### dboeren

##### Member
I'm on OSX 10.9.5 for reference.

#### YouCubing

##### Member
ok I've been looking for this for so long, thanks

#### Herbert Kociemba

##### Member
Thanks, Ctrl-V works fine.
But I couldn't copy the solution displayed in the left window, because I couldn't highlight the text...
Late but hopefully not too late:
Right-click the cube near the solution. From the popup menu you can select "Copy Solver to Clipboard"

#### okayama

##### Member
Late but hopefully not too late:
Right-click the cube near the solution. From the popup menu you can select "Copy Solver to Clipboard"

#### Russell Bilinski

##### Member
When I try to open it bounces but then just stops. Did it crash?

#### abunickabhi

##### Member
Do you know this online cube solver: https://rubiks-cube-solver.com/
I think it's using the same Kociemba algorithm.
This solver is good for beginners , but it is of no use to fast solvers, who want handpicked , and fast executable algorithms,
that aspect of algorithm generation is only satisfied by cube explorer, and ksolve+ to some extent.