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= 1/vl, at temperatures T < Tc. ≡ (3) Due to the termination of the coexistence line, it is possible to go from the gas phase to the liquid phase continuously (without a phase transition) by going around the critical point. Thus there are no fundamental differences between liquid and gas phases. From a mathematical ...
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phase transition occurs between paramagnetic and ferromagnetic phases of certain substances such as iron or nickel. These materials become spontaneously magnetized below a Curie temperature Tc. There is a discontinuity in magnetization of the substance as the magnetic field h, goes through zero for T < Tc. The phase...
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as the order parameter. In zero field, m vanishes for a paramagnet and is non–zero in a ferromagnetic, i.e. m(T, h = 0) 0 t | | β ∝ � for for T > Tc, T < Tc, (I.20) T )/Tc is the reduced temperature. The singular behavior of the order where t = (Tc − parameter along the coexistence line is therefore indicate...
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. Actually in almost all cases, the same singularity governs both sides and γ+ = γ− = γ. The heat capacity is the thermal response function, and its singularities at zero field are described by the exponent α, i.e. C±(T, h = 0) −α± . t | ∝ | (I.23) Long–range Correlations: Since the response functions are related ...
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system, i.e. (For the time being we treat the magnetization as a scalar quantity.) Substituting the � M = d3 ~r m(~r ). (I.25) above into eq.(I.24) gives kBT χ = � d3~r d3~r ′ ( h m(~r )m(~r ′ ) m(~r ) m(~r ′ ) ) . i i h i − h (I.26) Translational symmetry of a homogeneous system implies that = m is a const...
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Typically such influences occur over a characteristic distance ξ, called the correlation length. (It can be shown rigorously that this function must decay to zero at large separations; in many cases Gc(~r ) decays as exp( ic > ξ.) Let g denote a typical value of the correla­ /ξ) at separations m(~r )m(0) ≡ h ~r | |...
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the nature of the phase transition. For example, the vanishing of the coexistence boundary in the condensation of CO2 has the same singular behavior as that of the phase separation of protein solutions into dilute and dense components. This univer­ sality of behavior needs to be explained. We also noted that the dive...
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of the collection of interacting electrons is excessively complicated. The important degrees of freedom close to the Curie point, whose statistical mechanics is responsible for the phase transition, are long wavelength collective excitations of spins (much like the long wavelength phonons that dominate the heat capa...
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superconductivity, and planar magnets; n = 3 corresponds to classical magnets. While most physical situations occur in three–dimensional space (d = 3), there are also important phenomena on surfaces (d = 2), and in wires (d = 1). Relativistic field theory is described by a similar structure, but in d = 4. As in the case...
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)2 ∇ , � · · · . Including a small magnetic field ~h, that breaks the rotational symmetry, the lowest order terms in the expansion of Φ lead to, = β H � dd x t 2 � m 2(x) + um 4(x) + K 2 ( ∇ m)2 + ~h · · · · − m(x) ~ , (II.1) � which is known as the Landau–Ginzburg Hamiltonian. (The magnetic field also ...
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integrating over (coarse graining) the microscopic degrees of freedom, while constraining their average to ~m(x). It is precisely because of the difficulty of carrying out such a first principles program that we postulate the form of the resulting effective free energy on the basis of symmetries alone. The price paid is...
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N m(x) F � D � m(x), ∂m ∂x , N · · · ≡ � lim →∞ N i=1 � dmiF mi, � mi+1 − a mi , . · · · � (There are some mathematical concerns regarding the existence of functional integrals. The problems are associated with having too many degrees of freedom at short distances, allowing rather badly behaved functio...
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vicinity of the critical point, m is a small quantity, and it is justified to keep only the lowest powers in the expansion of Ψ(m). (We can later check self consistently that the terms left out are indeed small corrections.) The behavior of Ψ(m) depends strongly on the sign of the parameter t. (1) For t > 0, we can...
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· · − Tc) + ) =u + u1(T · · · − (T O Tc) + Tc)2 , (T − − O Tc)2 , (II.5) where a and u are unknown positive constants, dependent upon material properties. The basic idea is that the phenomenological parameters are functions of temperature that can be expanded in a Taylor series in T − Tc. The minimal conditions ...
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. Since t = a(T Tc) + ; ∂/∂T · · · − ∝ ∂/∂t, and C = ∂2 ∂2f ∂T 2 ∝ − ∂t2 T − βF V � = � � 0 1 8u for t > 0, for t < 0. (II.7) (II.8) We observe a discontinuity, rather than a divergence, in the heat capacity. If we insist upon describing the singularity by a power law, we have to choose the exponent...
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There is also a transverse susceptibility that is always infinite below Tc.) Equation of State: On the critical isotherm t = 0, the magnetization behaves as • m¯ = (h/4u)1/3, i.e. m¯ (t = 0, h) ∼ h1/δ, with δ = 3. (II.10) 18 MIT OpenCourseWare http://ocw.mit.edu 8.334 Statistical Mechanics II: Statistical Physi...
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Wave Energy Generation Jorge Manuel Marques Silva 1 • Bachelor + Master in Electrical Engineering (Energy) in 2015; • Superconductors in Electrical Machines; • Started MIT Portugal PhD Program – Sustainable Energy Systems in 2017: • Renewable Energy (Ocean Waves); • Machine Learning Forecasting; • Predic...
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w.mit.edu/fairuse. 5 • Machine Learning: • Learn and improve from experience without explicit programming; • Environmental variables forecast. © Desert Isle SQL.com. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.ed...
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1.3 Forward Kolmogorov equation Let us again start with the Master equation, for a system where the states can be ordered along a line, such as the previous examples with population size n = 0, 1, 2 · · · , N. We start again with a general Master equation dpn dt = − Rmnpn + Rnmpm . (1.28) m6=n X m6=n X In many relevant...
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x + y, leading to ∂ ∂t ∗ p(x, t) = dy [R(y, x − y)p(x − y) − R(y, x)p(x)] . (1.31) Z We now make a Taylor expansion the first term in the square bracket, but only with respect to the location of the incoming flux, treating the argument pertaining to the separation of the two points as fixed, i.e. R(y, x − y)p(x − y) = R(y...
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∆(x)i ∆t , D(x) ≡ 1 2 dy y2R(y, x) = 1 2 h∆(x)2i ∆t . Z (1.35) (1.36) (1.37) Equation (1.35) is a prototypical description of drift and diffusion which appears in many contexts. The drift term v(x) expresses the rate (velocity) with which the position changes from x due to the transition rates. Given the probabilistic n...
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1.23). Yet the proportions of the two alleles in the population does change from generation to generation. One reason is that some individuals do not reproduce and leave no descendants, while others reproduce many times and have multiple descendants. This is itself a stochastic process and the major source of rapid cha...
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a continuum evolution equation by setting x = n/N ∈ [0, 1], and replacing p(n, t + 1) − p(n, t) ≈ dp(x)/dt, where t is measured in number of generations. Clearly, from Eq. (1.41), there is no drift (cid:10) (cid:11) while the diffusion coefficient is given by v(x) = h(m − n)i = 0 , Dhaploid(x) = 1 2N 2 (m − n)2 = 1 2N x(1...
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chemical reaction that mimicks a mutating population. Consider a system where a reaction between molecules A and B can lead to two outcomes:2 A + B ⇀c A + A or A + B ⇁d B + B , (1.46) at rates c and d. In a “mean-field” approximation the number of A molecules changes as dNA dt = (c − d)NANB = (c − d)NA(N − NA) . (1.47) ...
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by ±1, and hence v(x) = h∆ni N Rn+1,n − Rn−1,n N = N(c − d)x(1 − x) , = = 1 N [cn(N − n) − dn(N − n)] while D(x) = = Rn+1,n + Rn−1,n 2N 2 = 1 2N 2 [cn(N − n) + dn(N − n)] h∆n2i 2N 2 = c + d 2 x(1 − x) . (1.51) (1.52) 2In a sense these reactions mimic the mating process in which the offspring of a heterozygote (a diploid...
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quantified by a parameter s, which is related to c and d by c = 1 4N (1 + s) and d = 1 4N (1 − s) . In the following, we shall employ the nomenclature of population genetics, such that v(x) = s 2 x(1 − x) , and D(x) = 1 4N x(1 − x) . (1.53) (1.54) 1.3.3 Steady states While it is usually hard to solve the Kolmogorov equa...
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(x′) Z + constant, p∗(x) ∝ 1 D(x) exp x v(x′) D(x′) , (cid:21) (cid:20)Z with the proportionality constant set by boundary conditions. Let us examine the case of the dynamics of a fixed population, including mutations, and reproduction with selection. Adding the contributions in Eqs. (1.38), (1.39) and (1.54), we have v...
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15.093 Optimization Methods Lecture 8: Robust Optimization 1 Papers • B. and Sim, The Price of Robustness, Operations Research, 2003. • B. and Sim, Robust Discrete optimization, Mathematical Programming, 2003. 2 Structure Motivation Data Uncertainty Robust Mixed Integer Optimization Robust 0-1 Optimization • ...
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• It allows to control the degree of conservatism of the solution; • It is computationally tractable both practically and theoretically. 5 Data Uncertainty minimize c x subject to Ax ≤ b ′ l ≤ x ≤ u xi ∈ Z, i = 1, . . . , k, Slide 6 Slide 7 WLOG data uncertainty affects only A and c, but not the vector b. Slid...
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. • We will guarantee that if nature behaves like this then the robust solution will be feasible deterministically. Even if more than Γi change, then the robust solution will be feasible with very high probability. Slide 9 Slide 10 2 6.1 Problem minimize ′ c x + max {S0| S0⊆J0,|S0|≤Γ0} dj xj | | ) ( j∈S0 X ...
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. . , k. ∈ 6.3 Proof Given a vector x ∗ , we define: ∗ βi(x ) = max {Si| Si⊆Ji,|Si|=Γi} ∗ aˆij xj | . | ) ( j∈Si X Slide 11 Slide 12 Slide 13 This equals to: ∗ βi(x ) = max s.t. j∈Ji X j∈Ji X 0 ≤ ∗ aˆij xj zij | | zij zij Γi 1 ≤ ≤ i, j ∀ ∈ Ji. Slide 14 Dual: βi(x ∗ ) = min pij + Γizi s.t. j∈...
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is the 6.5 Probabilistic Guarantee 6.5.1 Theorem 2 ∗ Let x be an optimal solution of robust MIP. (a) If A is subject to the model of data uncertainty U: Pr ∗ a˜ijxj > bi j X 1 ≤ 2n ! µ) (1 −   n n l l=⌊ν⌋ X (cid:18) (cid:19) n = Ji | (b) As n , ν = Γi+n and µ = ν | 2 ν − ⌊  ; bound is tight. ⌋ → ...
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Approx bound Bound 2 0 10 −1 10 −2 10 −3 10 −4 10 0 1 2 3 4 5 Γ i 6 7 8 9 10 Γ 0 2.8 36.8 82.0 200 Violation Probability 0.5 4.49 × 10−1 5.71 × 10−3 5.04 × 10−9 0 Optimal Value 5592 5585 5506 5408 5283 Reduction 0% 0.13% 1.54% 3.29% 5.50% • w˜i are independently distributed and...
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• • • 8.1 Remarks Slide 23 • Examples: the shortest path, the minimum spanning tree, the minimum assignment, the traveling salesman, the vehicle routing and matroid inter­ section problems. • Other approaches to robustness are hard. Scenario based uncertainty: minimize max(c1x, c2x) subject to x ∈ X. ′ ′ is NP...
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(cj + max(dj − θ, 0)) xj j X Slide 26 dn+1 = 0. dn ≥ dl+1, . . . ≥ ≥ min x∈X,dl≥θ≥dl+1 θΓ + n l cj xj + (dj j=1 X j=1 X θ)xj = − dlΓ + min x∈X n l cj xj + (dj j=1 X j=1 X n dl)xj = Zl − l ∗ Z = min l=1,...,n+1 dlΓ + min x∈X cjxj + (dj j=1 X j=1 X dl)xj. − 8.4 Theorem 3 Slide 27 • Algori...
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7 8923 9059 9627 10049 10146 10355 10619 10619 ¯ % change in Z(Γ) 0 % 0.056 % 1.145 % 2.686 % 9.125 % 13.91 % 15.00 % 17.38 % 20.37 % 20.37 % σ(Γ) 501.0 493.1 471.9 454.3 396.3 371.6 365.7 352.9 342.5 340.1 % change in σ(Γ) 0.0 % -1.6 % -5.8 % -9.3 % -20.9 % -25.8 % -27.0 % -29.6 % -31.6 % ...
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Recursion and Intro to Coq Armando Solar Lezama Computer Science and Artificial Intelligence Laboratory M.I.T. With content from Arvind and Adam Chlipala. Used with permission. September 21, 2015 September 21, 2015 L02-1 Recursion and Fixed Point Equations Recursive functions can be thought of as solutions o...
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Computing a Fixed Point • Recursion requires repeated application of a function • Self application allows us to recreate the original term • Consider: W = (x. x x) (x. x x) • Notice b-reduction of W leaves W : W  W • Now to get F (F (F (F ...))) we insert F in W: WF = (x.F (x x)) (x.F (x x)) which b-reduce...
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(n=0, False, f(n-1)) H2 = f.n.Cond(n=0, True, f(n-1)) Can we express odd using Y ? substituting “H2 odd” for even odd  odd = H1 (H2 odd) = H odd where H = = Y H f. H1 (H2 f) September 21, 2015 L02-7 Self-application and Paradoxes Self application, i.e., (x x) is dangerous. Suppose: u ...
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endent Types – A draft is available online (http://adam.chlipala.net/cpdt/) – most of what it covers goes beyond the scope of 6.820. • Another popular book: Bertot & Casteran, Interactive Theorem Proving and Program Development (Coq'Art) – https://www.labri.fr/perso/casteran/CoqArt/ • A popular online book that us...
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� reduction • rewrite H – use (potentially quantified) equality [H] to rewrite in the conclusion. • intros – move quantified variables and/or hypotheses "above the double line. • apply thm – apply a named theorem, reducing the goal into one new subgoal for each of the theorem's hypotheses, if any. September ...
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6.895 Essential Coding Theory September 8, 2004 Lecturer: Madhu Sudan Scribe: Piotr Mitros Lecture 1 1 Administrative Madhu Sudan To do: • Sign up for scribing – everyone must scribe, even listeners. • Get added to mailing list • Look at problem set 1. Part 1 due in 1 week. 2 Overview of Class Historical ov...
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1 1 0 1 1 ⎞ ⎟ ⎟ ⎠ 1 1 0 1 (b1, b2, b3, b4) −→ (b1, b2, b3, b4) G· Here, the multiplication is over F2. Claim: If a, b ∈ { 0, 1} , a = b then a · G and b · G differ in ≥ 3 coordinates. This implies that we can correct any one bit error, since with a one bit error, we will be one bit away from the correct stri...
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0. We can weaken it by asking how few coordinates can x = y − z be non­zero on, given that xH = 0? 0. We’ll make subclaim 1: If x has only 1 non­zero entry, then x H =� We’ll make subclaim 2: If x has only 2 non­zero entries, then x H =� Subclaim 1 is easy to verify. If we have exactly one non­zero value, x H is just...
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) = 26 rows, and: � xG5| x ∈ {0, 1} 26 � = {y| yH5 = 0} G5 also has full column rank, so it maps bit strings uniquely. Note that our efficiency is now 31 , so we’re asymptotically approaching 1. In general, we can encode 26 n − log2(n + 1) bits to n bits, correcting 1 bit errors. 2.1 Decoding We can figure out if ...
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2 Theoretical Bounds 2.2.1 Definitions Define the Hamming Distance as Δ(x, y) = i xi � yi, or the number of bits by which x and y differ. Define a ball around string x of radius (integer) t as B(x, t) = {y ∈ Σn Δ(x, y) ≤ t}, or the set of strings that differ by at most t bits. We can describe a code in terms of the max...
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) ≤ 2n Taking log of both sides, k ≤ n − log2(n + 1) Notice that the 26 bit Hamming code is as good as possible: 31 − 5 = 26 3 Themes Taking strings and writing them so they differ in many coordinates. This is called an error correcting code. In general, you have a finite alphabet Σ. Popular examples: {0, 1}, ASCI...
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algorithms (computationally efficient — polynomial, linear, or sublinear time) • Decoding algorithms • Theoretical bounds/limitations • Applications Hamming’s paper did all of the above — it described a code, an efficient encoding algorithm (poly­ time. Challenge: Construct linear­time encoding algorithm for Hamming co...
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1 Sequence 1.1 Probability & Information We are used to dealing with information presented as a sequence of letters. For example, each word in English languate is composed of m = 26 letters, the text itself includes also spaces and punctuation marks. Similarly in biology the blueprint for any organism is the string of ...
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. . + pm)N = (cid:48) (cid:88) {Nα} pN1 1 pN2 2 · · · pNm m × N ! α=1 Nα! (cid:81)m , (1.3) (1.4) where the sum is restricted so that (cid:80)m α=1 Nα = N . Note that because of normalization, both sides of the above equation are equal 1. The terms within the sum on the right-hand side are known the multinomial probabi...
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) (Nα log Nα − Nα) = −N · (cid:18) Nα N (cid:88) α α (cid:19) log (cid:18) Nα N (cid:19) . (Stirling’s approximation for N ! is used for all Nα (cid:29) 1.) The above formula is closely related to the entropy of mixing in thermodynamics, and quite generally for any set of probabilities {pα}, we can define a mixing entro...
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4, then Eq. (1.8) reduces to 0, which is consistent—we gain no information. On the other hand, if pA = pT = 0 and pC = pG = 1 2 , then I = 2 − (cid:88) G,C 1 2 log2 2 = 1 bit per base. 1.2 Evolving Probabilities As organisms reproduce the underlying genetic information is passed on to subsequent gen- eration. The copyi...
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shed it by comparing such single nucleotide polymorphisms (SNPs). Non-synonymous mutations are not necessarily deleterious and may lead to viable off-spring. 1.2.2 Classical Genetics The study of heredity began long before the molecular structure of DNA was understood. Several thousand years of experience breeding anima...
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+ 1) = m (cid:88) β=1 παβpβ(τ ), or in matrix form (cid:126)p(τ + 1) = ←→π (cid:126)p(τ ) = ←→π τ (cid:126)p(1), (1.9) where the last identity is obtained by recursion, assuming that the transition probability matrix remains the same. Probabilities must be normalized to unity, and thus the transition probabilities are ...
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(cid:88) α Rαβ = 0, or Rββ = − (cid:88) α(cid:54)=β Rαβ. dpα(t) dt = (cid:88) β(cid:54)=α (Rαβpβ(t) − Rβαpα(t)) , (1.14) (1.15) which is known as the Master equation. 1.2.4 Steady state Because of the conservation of probability in Eqs. (1.10) and (1.14), the transition probability ←− matrix ←→π , and by extension the ...
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µ1 µ2 −µ1 (cid:19) (cid:18) p1 p2 (cid:19) . (1.18) The above 2 × 2 transition rate matrix has the following two eigenvectors (cid:18) −µ2 µ1 µ2 −µ1 (cid:19) (cid:18) µ1 µ1+µ2 µ2 µ1+µ2 (cid:19) = 0, and (cid:18) −µ2 µ1 µ2 −µ1 (cid:19) (cid:18) 1 −1 (cid:19) = −(µ1 + µ2) (cid:18) 1 −1 (cid:19) . (1.19) −→ p∗ with eigenv...
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µ1+µ2 (cid:33) µ1 µ2 −µ1 + e−(µ1+µ2)t µ2 − e−(µ1+µ2)t µ2 µ1+µ2 µ1+µ2 = (cid:32) µ1 µ1+µ2 µ2 µ1+µ2 (cid:19) + µ2 µ1 + µ2 (cid:18) 1 −1 (cid:19)(cid:21) . (1.21) At long times the probabilities to find state A1 or A2 are in the ratios µ1 to µ2 as dictated by the steady state eigenvector. The rate at which the probabilitie...
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stays the same, or changes by unity. Thus the transition rate matrix only has non-zero terms along or adjoining to the diagonal. For example Rn,n+1 = µ2(n + 1), and Rn,n−1 = µ1(N − n + 1), (1.22) where the former indicates that a population of n + 1 A1s can decrease by one if any one of them mutates to A2, while the po...
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mean numbers of constituents evolve to the steady state with N ∗ A/N ∗ B = b/a. However, in a system where the number of particles is small, for example for a variety of proteins within a cell, the mean number may not be representative, and the entire dis- tribution is relevant. The probability to find a state with NA =...
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2.160 System Identification, Estimation, and Learning Lecture Notes No. 5 February 22, 2006 4. Kalman Filtering 4.1 State Estimation Using Observers In discrete-time form a linear time-varying, deterministic, dynamical system is represented by xt +1 = t x A t + u B t t (1 ) nx1 is a n-dimensional state vect...
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yı = H t xıt t t + u B t t L ( y + t t − yı ) t (3) To differentiate the estimated state from the actual state of the physical system, the estimated state residing in the real-time simulator is denoted xıt . With this feedbackthe state of the simulator will follow the actual state of the real system, and ther...
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with noise, and the state transition of the actual process is to some extent disturbed by noise. If stochastic properties of these noise sources are available, state estimation may be performed more effectively than simply using sensor signals as noise-free signals and estimating the sate based on noise-free state ...
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m t ( ) t X ( ) − ⎟ ⎜ ⎢ Y ⎟ ⎜ ⎢ ⎣ (t Z ) − m (t) ⎠ ⎝ ( ) t C XYZ = Z ) − ( ) m t x If mx=my=mz=0 t Y ( ) − ( ) mY t t Z ( ) − )(tm Z ⎤ ⎥ ) ⎥ ⎥ ⎦ (4) 3 [ 2 ⎡ X E ⎢ [ C XYZ (t ) = ⎢ t X E ⎣ [ ⎢ t X E (t )] )] ( ) ( t Y )] ( ) ( t Z [ )] ( ) ( t X E t Y [ 2 (t )] Y E [ ( ) ( t Y E t Z ...
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xt , y , z , x , y , z ) t 2 t 1 Z Y X Z Y X 1 2 2 2 t 2 t 1 t 2 1 1 1 C XYZ ( t t 1, 2 ⎡ ) = ⎢ ⎢ ⎢ ⎣ [ t X E ( [ ( t Y E [ ( t Z E 1) 1) 1) t X ( 2 t X ( 2 t X ( 2 )] )] )] [ t X E ( ) 1 [ ( ) t Y E 1 [ ( ) t Z E 1 2 )] t Y ( 2 )] 2 )] t Y ( t Y ( [ t X E ( ) 1 [ ( ) t Y E 1 [ ( ) t ...
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process, the state xt driven by wt is a S igure 4 . S , is a deterministic term. In random process. The second term on the right hand side, u B t the following stochastic state estimation, this deterministic part of inputs is not important, since its influence upon the state xt is completely predictable and hence ...
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If the noise signals at any two time slices are uncorrelated, , t ) = E [v ⋅ CV ( s t T ] v s = ,0 t∀ ≠ s (10) (11) (12) the noise is called “White”. (We will discuss why this is called white later in the following chapter.) Note that, if t= s , the above covariance is that of the first order density, i.e. a...
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measurement noise , v ∈ R t nx1 � x1 � x1 n n ∈ R × , and H ∈ R tv have zero mean values, � xn t . Assume E[wt]=0, E[vt]=0. and that they have the following covariance matrices: t CV ),( s = [ vv E ⋅ t s T ] = t CW ),( s t[ = w E ⋅ w T s ] = CWV ( s , t ) = w E [ vt ⋅ T ] s 0 ⎧ ⎨ R...
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uation (8) and the output eq Rudolf E. Kalman solved this problem around 1960. Kalman Filter: two major points of his seminal workin 1960. I) II) If we assume that the optimal filter is linear, then the Kalman filter is the state estimator having the smallest unconditioned error covariance among all linear filter...
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t t − 1 t Note E[vt]=0 Correction of the state estimate Assimilating the new measurement yt, we can update the state estimate in proportion to the output estimation error. xıt = xı t − 1 t + Kt ( y − t x H ı t ) t − 1 t (21) (22) (23) Equation (23) provides a structure of linear filter in recursive form...
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Topic 0 Notes Jeremy Orloff 0 18.04 course introduction This class is an adaptation of a class originally taught by Andre Nachbin. He deserves most of the credit for the course design. The topic notes were written by me with many corrections and improvements contributed by Jörn Dunkel. Of course, any responsibility f...
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ed from R. Rosales 18.04 OCW 1999) Do not be fooled by the fact things start slow. This is the kind of course where things keep on building up continuously, with new things appearing rather often. Nothing is really very hard, but the total integration can be staggering - and it will sneak up on you if you do not watc...
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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 10: Solutions to Laplace’s Equation In C...
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V = Φ ∇ Φ d d ⎣ d d ⎦ (cid:118)∫ S i da = ∇ Φ d dV = 0 2 ∫ V on S, Φ d = 0 or ∇Φ d i da = 0 Φ d = 0 ⇒ Φ a = Φ b on S ∇Φ d i da = 0 ⇒ ∂Φ a = ∂n ∂Φ b on S ⇒ E ∂n na = E nb on S 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 10 Page 1 of 8 A problem is uniquely posed when ...
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= 0 ) 1. Try product solution: Φ (x, y ) = Χ x Y ( (y ) ) ( Y y ) ( 2 d Χ x Y dx 2 + ( X )x 2 d ) ( y dy 2 = 0 Multiply through by 1 : XY 2 1 d Χ X dx 2 = − 2 1 d Y Y dy 2 = − k 2 k=separation constant only a function of x only a function of y 2 d Χ dx 2 = − k Χ 2 ; 2 d Y dy 2 2 = k Y 2. Zer...
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cos kxe + D cos kxe -ky 3 4 = E 1 sin kx sinh ky + E2 sin kx cosh ky + E3 cos kx sinh ky + E 4 cos kx cosh ky 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 10 Page 4 of 8 4. Parallel Plate Electrodes Neglecting end effects, ( Φ x . ) Boundary conditions are: Φ (x = 0) = Φ 0 , Φ...
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Lecture 10 Page 5 of 8 Electric field lines: dy dx = E y = E x x y ydy = xdx 2 y 2 = 2 x 2 + C 2 y = x + y 0 − x 0 (field line passes through (x0, y0)) 2 2 2 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 10 Page 6 of 8 6. Spatially Periodic Potential Sheet Φ (x, y )...
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a sin ay Electric Field Lines: dy dx = E y E x ⎧−cot ay ⎪ = ⎨ ⎪ ⎩+cot ay x > 0 x < 0 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 10 Page 7 of 8 x > 0 cos ay e -ax = constant x < 0 cos ay e +ax = constant 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Z...
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6.895 Theory of Parallel Systems Lecture 9 Analysis of Cilk Scheduler Lecturer: Michael A. Bender Scribe: Alexandru Caraca¸s, C. Scott Ananian Lecture Summary 1. The Cilk Scheduler We review the Cilk scheduler. 2. Location of Shallowest Thread We define the depth of a thread and the shallowest thread. Next, We pr...
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(which is not in a deque). (c) If the deque is empty and the processor is unable to execute α’s parent (because the parent is busy), then the processor work steals. 3. Procedure α Syncs. If there are no outstanding children, then continue: we’re properly synced. Oth- erwise (when there are no outstanding children), ...
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Structural lemma) Consider any processor p at time t. Let u0 be the current root thread (of procedure α0) executing on processor p. Let u1, u2, . . . , uk be the threads (of procedures α1, α2, . . . , αk ) in p’s deque ordered from bottom to top. Let d(ui) be the depth of thread ui in DAG G. Then, d(u0) ≥ d(u1) > · ...
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Change in the deque of a processor when the processor returns from a procedure. • Case 1: Steal. A steal removes the top entry from the deque (see Figure 2). The processor performing the steal begins executing uk with an empty deque, trivially satisfying the inequality as in the base case. The processor from whom th...
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to ua through u0 which has length d(u0) + 1. Therefore, d(ua) ≥ d(u0) + 1, 9-3 uk uk−1 . . . u2 u1 u0 =⇒ uk uk−1 . . . u2 u1 ua Figure 4: Change in the deque of a processor when the processor reaches a sync point or continues a procedure. · · · u0 ua · · · α0 Figure 5: A piece of a computatio...
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a shallowest thread may be the currently-executing thread instead of the thread at the top of the deque. Since the base case satisfies the property, and every action maintains the property, then at any time t in the execution for all deques d(u0) ≥ d(u1) > · · · > d(uk−1) > d(uk ). =⇒ uk uk−1 . . . u2 u1 u0...
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2) by making a (back) edge to every spawn thread from the parent’s continuation thread. Note that if a sync is immediately followed the spawn, we need to insert an extra continuation thread before the sync to prevent cycles in the DAG. Figure 8 shows the execution graph of Figure 9 augmented with these edges. We sh...
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ors left in the DAG, but it is neither being executed nor at the top of the deque. Observation 8 (Depth of Augmented DAG G(cid:2)) The critical path of the augmented DAS G(cid:2) is 2T∞. We can now get to a spawned thread via our back edge from the continuation edge, adding a distance of one to the longest path to ...
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argument based on two buckets to help in proving the bound. 9-6 Figure 10: The work and steal buckets used in the accounting argument. Theorem 10 (Cilk Scheduler execution time bound) Consider the execution of any fully strict mul­ tithreaded computation with work T1 and critical path T∞ by Cilk’s work-stealing alg...
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the number of dollars in the steal bucket is O(P T∞), with high probability. We give an exact definition of high probability during the course of the proof. In the rest of the lecture, we prove the previous lemma. We introduce several concepts to prove the bound. Observation 13 (Dollars) Per time-step P dollars ente...
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1 Rounds As long as no processor is stealing, the computation is progressing efficiently. What divide the computation into rounds, which (more or less) contain the same number of steals (see Figure 11). Definition 16 (Round) A round of work stealing attempts is a set of at least P and at most 2P − 1 steal attempts. We...
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done at random and there are Proof P places that a processor could steal from. Hence, the probability that a critical thread would be stolen during a round is 1/P , and the probability that the critical thread will not be stolen is 1 − 1/P . Hence, for C rounds we obtain the probability that a critical thread is no...
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Base case: During last time-step t, last thread in G(cid:1) is uL and it is critical and executed (see Figure 12). Induction step: Suppose during time-step t, thread uj is critical. During time-step t − 1, either uj is also critical, or if not it is because it has un-executed predecessors in G(cid:2). Some threads ui...
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A delay sequence occurs if ∀i at least πi rounds occur while ui is critical and un-executed. A delay sequence is a combinatorial object. A delay sequence occurs if there is a directed path in G(cid:2) in which one of the threads in the path is critical at each time-step. Lemma 21 If at least 2P (2T∞ + R) steal attem...
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∞ + R) steal attempts occur (cid:5) (cid:4) ≤ Pr (cid:5) exists delay sequence (U, R, Π) that occurs (cid:6) ≤ ≤ Pr [(U, R, Π) occurs] delay sequences (U, R, Π) (cid:1) Number of Delay Sequences 9-10 ⎛ ⎝ (cid:2) Maximum Probability that any delay sequence occurs ⎞ ⎠ . Lemma 23 (Probability of Del...
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than partitioning R into 2T∞ pieces. We simply count up all the possible ways of partitioning the path. Thus, by multiplying the number of paths with the number of ways in which a path can be partitioned we obtain the number of delay sequences, which is at most (cid:1) 22T∞ 2T∞ + R 2T∞ (cid:2) . Our goal is to...
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considerably large, while the second term is exponentially small in R. We want to choose a value for R such that the product of the two terms is small. We let R = 2CT∞ where C is a constant. Replacing in the previous equation we have: Pr [any (U, R, Π) occurs] ≤ (cid:1) [2e(C + 1)]1/C e (cid:2)R . (4) Definiti...
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the possible events we have shown that with high probability there are few steal attempts. With probability at least 1 − ε the number of steal attempts is O(P T∞ + lg ε We have started our analysis using an accounting argument, based on a work bucket and a steal bucket.We have shown that at the end of the computati...
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y x (cid:12) (cid:13) ey x x . ≤ Figure 13: Death-bed formulae. 13
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16.920J/SMA 5212 Numerical Methods for PDEs Lecture 5 Finite Differences: Parabolic Problems B. C. Khoo Thanks to Franklin Tan SMA-HPC ©2002 NUS Outline • Governing Equation • Stability Analysis • 3 Examples • Relationship between σand λh • Implicit Time-Marching Scheme • Summary SMA-HPC ©2002 NUS 2 G...
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u u 2 + − j 1 2 x ∆ u −+ j 1 + ( O x ∆ 2 ) which is second-order accurate. • Schemes of other orders of accuracy may be constructed. SMA-HPC ©2002 NUS 5 Stability Analysis Discretization We obtain at x 1 x 2 : : du 1 dt du 2 dt = = υ 2 x ∆ υ 2 x ∆ ( u − o 2 u u + 1 2 ) (...
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��     u υ   N   ∆ 2 x  0 1 − 0 A SMA-HPC ©2002 NUS 7 Stability Analysis PDE to Coupled ODEs Or in compact form (cid:71) du dt (cid:71) (cid:71) Au b + = where (cid:71) u (cid:71) b [ u = 1 o u υ  =  2 x ∆  u 2 0 0 0 T ] u − 1 N Nu υ 2 x ∆ T    We hav...
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− diagonalizes the matrix by E A N 1) matrix formed by the ( N − 1) columns 1E AE− = Λ where Λ = λ  1        λ 2 0 0         1 λ − N SMA-HPC ©2002 NUS 10 Stability Analysis Coupled ODEs to Uncoupled ODEs Starting from (cid:71) (cid:71) Au b + = Premultiplicati...
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