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Kamerlingh Onnes the Netherlands Leiden University Leiden, the Netherlands b. 1853 d. 1926 •http://www.nobel.se/physics/laureates 5 Discovery of Superconductivity “As has been said, the experiment left no doubt that, as far as accuracy of measurement went, the resistance disappeared. At the same time, however, s...
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for copyright reasons. Please see: "A current-carrying type II superconductor in the mixed state" from http://phys.kent.edu/pages/cep.htm http://phys.kent.edu/pages cep.htm / ) ) When a current is applied to a type II superconductor (blue rectangular box in the m xed state, the magnetic vortices (blue cylinder...
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interplay of theory and experiment. It would have been very difficult to have arrived at the theory by purely deductive reasoning from the basic equations of quantum mechanics. Even if someone had done so, no one would have believed that such remarkable properties would really occur in nature. But, as you well kn...
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for copyright reasons. Please see: http://nobelprize.org/physics/laureates/1973/index.html Leo Esaki Ivar Giaever Brian David Josephson 1/4 of the prize 1/4 of the prize 1/2 of the prize Japan USA United Kingdom IBM Thomas J. Watson Research Center Yorktown Heights, NY, USA b. 1925 General Electric Co...
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) ⋅ l d dI V = LJ dt , where LJ = Φ 0 2π I cosϕc Φ 0 = flux quantum 483.6 GHz / mV 19 20 •10 SQUID Magnetometers Φ0 I+ V+ 1pF Φ 1.1 µm 1.1 µm I V 20 µm DC SQUID Shunt capacitors ~ 1pF Jct. Size ~ 1.1µm Loop size ~20x20µm2 L SQUID ~ 50pH I ~10 & 20µA c 21 High-Temperature Superconductivity ...
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________ laureates/1987/bednorz-lecture.html http://www.nobel.se/physics/laureates/1987/bednorz-muller-lecture.pdf 23 Perovskite Structure Image removed for copyright reasons. ______________________________________ Please see: Images from http://cst-www.nrl.navy.mil/lattice/struk/perovskite.html •http://cst-www.n...
https://ocw.mit.edu/courses/6-763-applied-superconductivity-fall-2005/41b2882232b1b43823ec488969c21b6d_lecture1.pdf
MRS meetings since their discovery in 1986. Twenty years later, the progress both on the fundamental understanding of these materials and the path towards their industrial applications has been impressive. First-generation wires are now routinely produced in kilometer lengths and used in a variety of large-scale pr...
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Microwave filters in cellular stations Technical Points Low losses, smaller size, sharp filtering Passive microwave devices, Resonators for oscillators Far-infrared bolometers Microwave detectors X-ray detectors SQUID Magnetometers: Magneto-encephalography, NDT Lower surface losses, high quality factors, sma...
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in a superconducting loop • The induced current can be in the opposite direction if we carefully choose a different magnetic field this time Image removed for copyright reasons. • To store and process information as a computer bit, we assign: as state | 0 〉 as state | 1 〉 clockwise Anti-clockwise 35 ...
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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 3: Continuous Dependence On Parameters1 Arguments based on continuity of functions are common in dynamical system analysis. They rarely apply to...
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require existence of a constant M such that 2 |a(¯ x1) − a(¯ x2)| ∃ M |x1 − ¯ x2| ¯ x1, ¯ [t0, tf ] ∈� Rn of (3.1). The proof of both for all ¯ x2 from a neigborhood of a solution x : existence and uniqueness is so simple in this case that we will formulate the statement for a much more general class of integral e...
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(x(� ), �, t)d� � t ≤ [t0, tf ]. (3.4) t0 A proof of the theorem is given in the next section. When a does not depend on the third argument, we have the standard ODE case x˙ (t) = a(x(t), t). In general, Theorem 3.1 covers a variety of nonlinear systems with an infinite dimensional state space, such as feedback i...
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t0 ∃ 1/(2K) we have |xk+1(t) − xk (t)| ∃ t t t0 |a(xk (� ), �, t) − a(xk−1(� ), �, t)|d� ∃ K|xk (� ) − xk−1(� )|d� t0 ∃ 0.5 max {|xk (t) − xk−1(t)|}. t�[t0,tf ] Therefore one can conclude that max {|xk+1(t) − xk (t)|} ∃ 0.5 max {|xk (t) − xk−1(t)|}. t�[t0,tf ] t�[t0,tf ] Hence xk (t) converges exponenti...
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the same proof applies when (3.2),(3.3) are replaced by the weaker conditions |a(¯ x1, �, t) − a(¯ x2, �, t)| ∃ K(� )|¯ x1 − ¯ x2| � ¯ x2 ≤ Br (¯ x1, ¯ x0), t0 ∃ � ∃ t ∃ t1, and x, �, t)| ∃ m(t) � ¯ where the functions K(·) and M (·) are integrable over [t0, t1]. x ≤ Br (¯ x0), t0 ∃ � ∃ t ∃ t1, |a(¯ 4 3.2 Continuo...
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R such that |a(¯ x1, �, t, q)−a(¯ x2, �, t, q)| ∃ K|x1−x2| � ¯ x2 ≤ X d, t0 ∃ � ∃ t ∃ tf , q ≤ (q0−d, q0+d); x1, ¯ ¯ ¯ (b) there exists K ≤ R such that |a(¯ x, �, t, q)| ∃ M � ¯ x ≤ X d, t0 ∃ � ∃ t ∃ tf , q ≤ (q0 − d, q0 + d); (c) for every � > 0 there exists � > 0 such that |x0(q1) − ¯ ¯ x0(q2)| ∃ � � q1, q2 ≤ (q...
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x0 . Condition (b) simply bounds a uniformly. Finally, condition (c) means continuous dependence of equations and initial conditions on parameter q. The proof of Theorem 3.2 is similar to that of Theorem 3.1. 3.3 Implications of continuous dependence on parameters This section contains some examples showing how ...
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to x(t) = x(t − t0, ¯ x) means “the value x(t) of the solution of (3.8) with initial conditions x(0) = ¯x”. Remember that this definition makes sense only when uniqueness of solutions is guaranteed, and that x(t, ¯x) may by undefined when |t| is large, in which case we will write x(t, ¯x) = ⊂. x0), where x(t, ¯ Accor...
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, i.e. x(t, ¯ x0) ≥ ¯ x0 is a (a) there exists d > 0 such that x(t, ¯ x) � ¯ x0 as t � ⊂ for all ¯ x satisfying |¯ x0 − ¯ x| < d; (b) for every � > 0 there exists � > 0 such that |x(t, ¯ x) − ¯ x0| < � whenever t → 0 and x − ¯ |¯ x0| < �. In other words, all solutions starting sufficiently close to an asymptotically s...
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converges to infinity, is called the limit set of the “trajectory” t ∈� x(t, ¯x0). x0) � x Theorem 3.4 The limit set of a given trajectory is always closed and invariant under the transformations ¯ x ∈� x(t, ¯ x).
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 4 DENIS AUROUX 1. Pseudoholomorphic Curves For (X 2n, ω) symplectic, J a compatible a.c.s. ∈ J ...
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− 6 + 2k) u is regular if D∂ is onto. Theorem 1. The set J reg(X, β) of J ∈ J (X, ω) s.t. every simple J-holomorphic curve in class β is regular is a Baire subset. For J ∈ J reg(X, β), the subset of simple maps M∗ (X, J, β) ⊂ Mg,k(X, J, β) is smooth and oriented of dimension 2d. g,k Let g(·, ) = ω( , J ) be the a...
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, ∞ to CP1 × CP1 . Away from x = 0, it converges uniformly to x �→ (x, 0). But if you reparameterize to ˜x = nx, 1 ) and away from x = ∞, it converges uniformly to ˜ → (0, 1 x ). x˜ �→ ( n ˜ 1 ˜ x x, ˜ x nx The general idea is: • Identify bubbling regions where sup |dun| → ∞. • Away from those, ∃ convergent s...
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�→ u(zi) for 1 ≤ i ≤ k. The Gromov-Witten invariants are defined as follows: given α1, . . . , αk ∈ H ∗(X), deg (αi) = 2d, � (8) �α1, . . . , αk�g,β = ev∗ 1α1 ∧ · · · ∧ evk ∗αk ∈ Q � [M g,k(X,J,β)] i ev−1(Ci)) (or rather #(ev X (choose Ci trans­ Equivalently, if we represent P D(αi) by a cycle Ci verse to th...
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, M0,k(X, J, dβ) are orbifolds (strata of multiply-covered maps are orbifolded). We can restore transversality by taking domain-dependent Js. More precisely, there d 1 β MIRROR SYMMETRY: LECTURE 4 3 is a universal curve C → M0,k (the fiber over a point is the corresponding curve), and J is now given by a map C ...
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cover. ∞ → For an algebraic geometer, one needs to keep J integrable so X remains an al­ gebraic variety. The moduli space Mg,k is an algebraic stack, as is Mg,k(X, J, β). For an integrable J and fixed j, we have a ∂-operator on sections of u∗T X, and the cokernel of this operator is precisely H 1(Σ, u∗T X). Where du...
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MIT 6.035 Semantic Analysis Martin Rinard Laboratory for Computer Science Massachusetts Institute of Technology Error Issue • Have assumed no problems in building IR • But are many static checks that need to be done as part of translation • Called Semantic Analysis Goal of Semantic Analysis • Ensure that program obe...
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for v local descriptor for i parameter descriptor for x while (i < v.length) v[i] = v[i]+x; < ldl len ldf sta ldf ldl + lda ldp ldf ldl field descriptor for v local descriptor for i parameter descriptor for x while (i < v.length) v[i] = v[i]+x; while < ldl len ldf sta ldf ldl + lda ldp ldf ldl field descriptor for ...
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names • When to check? – when insert descriptor into local symbol table • Parameter and field symbol tables similar Class Descriptor • When build class descriptor, have – class name and name of superclass – field symbol table – method symbol table • What to check? – Superclass name corresponds to actual class – No na...
https://ocw.mit.edu/courses/6-035-computer-language-engineering-sma-5502-fall-2005/41f7dfc1934579616240a609947159f0_7_semantic_check.pdf
int + double is double, float + double is double • Interesting oddity: C converts float procedure arguments to doubles. Why? Type Inference • Infer types without explicit type declarations • Add is very restricted case of type inference • Big topic in recent programming language research – How many type declaration...
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 2 DENIS AUROUX Reference for today: M. Gross, D. Huybrechts, D. Joyce, “Calabi-Yau Mani­ folds ...
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Ω0,1(X, E) → Ω0,2(X, E) → · · · } ∂ q(X, E) = ker∂/im∂ H ∂ ∂ ∂ Deforming J to a “nearby” J � gives (4) Ω1,0 ⊆ T ∗C = Ω1,0 ⊕ Ω0,1 J J � J is a graph of a linear map (−s) : Ω1 0 (acted , ,1 . J � is determined by Ω1 ,0 → Ω0 � J J J on by i) and Ω0,1 (acted on by i�). s is a section of (Ω1,0)∗ ⊗ Ω0,1 = T1,0 ⊗ Ω...
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(5) [α ⊗ v, α� ⊗ v�] = (α ∧ α�) ⊗ [v, v�] giving it the structure of a differential graded Lie algebra. Proposition 1. J � is integrable ∂s + 2 Proof. We want to check that the bracket of two 0, 1 tangent vectors is still 0, 1, i.e. that 1 [s, s] = 0. ⇔ (6) [ ∂ � + ∂zk � s�k ∂ � + ∂ , ∂z� ∂zk s�k ∂ ∂z...
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integrable complex structures on X}/Diff(X) (or, assuming that Aut(X, J) is discrete, we want that near J, ∃ a universal family X → U ⊂ MCX (complex manifolds, holomorphic fibers ∼= X) s.t. any family of integrable complex structures X � → S induces a map S → U s.t. X pulls back to X �). We have an action of the diffe...
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,1(X, T X 1,0), s(0) = 00. By the above, this should satisfy (12) ∂s(t) + 1 2 [s(t), s(t)] = 0 In particular, s1 = dt t=0 solves ∂s1 = 0. We obtain an infinitesimal action of Diff(X): for (φt), φ0 = id , dt |t=0 = v a vector field, dφ | ds (13) d dt |t=0(−(∂φt)−1 ◦ ∂φt) = − d dt |t=0(∂φt) = −∂v This implies that ...
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phic maps. To first order, this is given by holomorphic vector fields vij on Ui ∩ Uj s.t. vij = −vji and vij + vjk = vik on Uijk. Cech 1-cocycle conditions in the sheaf of holomorphic tangent vector fields. Modding out by holomorphic functions ψi : Ui → Ui (which act by φij �→ ψj φij ψ−1) is precisely modding by the ˇ ...
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that [s1, s1] ∈ Ker (∂). Thus, the primary obstruction to deforming is the class of [s1, s1] in H 2(X, T X 1,0). If it is zero, then there is an s2 s.t. ∂s2 + 1 2 [s1, s1] = 0, and the next obstructure is the class of [s1, s2] ∈ H 2(X, T X 1,0). We are basically attempting to apply by brute force the implicit functi...
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0) = 0. In local coordinates, we have T ∗X 1,0 = Span{dz(t) = dzi − t=0αt ∈ Ωp,q + Ωp+1,q−1 + Ωp−1,q+1 . sij (t)dzj } � X ⇒ Jt (17) αt = i � I,J||I|=p,|J|=q ( αIJ (t)dzi 1 t) ∧ · · · ∧ dzi ) t ) ∧ · · · ∧ dz( t t) ∧ dz( ( j j p q 1 d | Taking dt t=0, the result follows from the product rule. We mostly...
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Massachusetts Institute of Technology 18.413: Error­Correcting Codes Laboratory Professor Daniel A. Spielman Handout 0 February 3, 2004 Signing Up If too many people sign up for the course, I will perform a lottery among those who have signed up, and announce the results by email on Wednesday, Feb 4th. First Re...
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System Identification 6.435 SET 9 – Asymptotic distribution of PEM Munther A. Dahleh Lecture 9 6.435, System Identification 1 Prof. Munther A. Dahleh Central Limit Theorem (Generalization) • Basic Theorem II: Consider are both ARMA processes, possibly correlated, with underlying white noise (bounded 4th moment) Lect...
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are asymptotically efficient if ⇒ is normally distributed. Estimates for accuracy Lecture 9 6.435, System Identification 10 Prof. Munther A. Dahleh Examples ARX: Lecture 9 6.435, System Identification 11 Prof. Munther A. Dahleh MA: Lecture 9 6.435, System Identification 12 Prof. Munther A. Dahleh ARMA: Lecture 9...
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20.110/5.60 Fall 2005 Lecture #1 page 1 Introduction to Thermodynamics Thermodynamics: → Describes macroscopic properties of equilibrium systems → Entirely Empirical → Built on 4 Laws and “simple” mathematics 0th Law ⇒ Defines Temperature (T) 1st Law ⇒ Defines Energy (U) 2nd Law ⇒ Defines Entropy (S) 3rd La...
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20.110/5.60 Fall 2005 Lecture #1 page 3 Two classes of Properties: • Extensive: Depend on the size of the system (n,m,V,…) • Intensive: Independent of the size of the system (T,p, V = V n ,…) The State of a System at Equilibrium: • Defined by the collection of all macroscopic properties that are described by ...
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K) Change of State: (Transformations) • Notation: 3 H2 (g, 5 bar, 100 °C) = 3 H2 (g, 1 bar, 50 °C) initial state final state 2 H2O (ℓ, 1 bar, 50 °C) = 3 H2O (g, 1 bar, 150 °C) initial state final state 20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Ha...
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the warmer to the cooler object. This continues until they are in thermal equilibrium (the heat flow stops). At this point, both bodies are said to have the same “temperature”. This intuitively straightforward idea is formalized in the 0th Law of thermodynamics and is made practical through the development of ther...
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At) Experimental result: A = 0.0036609 = 1/273.15 -273.15 0 100 C Note: t = − 273.15 ° C ( t = − 273.15 ) is called the absolute zero, is special ° C 20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field ...
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8.31451 J K mol − (gas constant) 20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field 20.110/5.60 Fall 2005 Lecture #1 page 9 • Work: “w” = Fw (cid:65)⋅ applie...
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Lecture 7 8.321 Quantum Theory I, Fall 2017 39 Lecture 7 (Sep. 27, 2017) 7.1 Spin Precession In a Magnetic Field Last time, we began discussing the classic example of precession of a spin- 1 particle in a magnetic field. The Hamiltonian of this system is 2 With B = Bzˆ, this becomes The energy levels are giving a level ...
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iωt/2c+|+(cid:105) + eiωt/2c−|−(cid:105) . (7.7) (7.8) (7.9) We now know the state of the system at all times. For example, if we initially have |ψ(cid:105) = |+(cid:105), then which has |ψ(t)(cid:105) = e−iωt/2|+(cid:105) , Prob(cid:0)Sz (cid:126)= (cid:1) =2 1 (7.10) (7.11) for all times. This is why energy eigenstat...
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, Prob(cid:0)Sx = − (cid:126) 2 at time t(cid:1) = sin2 (cid:19) . (cid:18) ωt 2 (7.12) (7.13) (7.14) (7.15) We can check that this is true, but we know it must be true by conserv then have ation of probability. We (cid:104)Sx(cid:105) = cos2 − sin2 (cid:20) (cid:126) 2 (cid:19) (cid:18) ωt 2 (cid:19)(cid:21) (cid:18) ...
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(cid:126)Sy, which gives dSz dt = 0 . dSx dt = iω iω (cid:126) e Szt/ i(cid:126)Sye−iωSzt/(cid:126) = −ωSy(t) . (cid:126) Similarly, using [Sz, Sy] = −i(cid:126)Sx, we find dSy dt = ωSx(t) . (7.17) (7.18) (7.19) (7.20) (7.21) Lecture 7 8.321 Quantum Theory I, Fall 2017 41 We can write these three expression compactly i...
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and p commute, the corresponding operators do not. Suppose we have the Hamiltonian H = p2 2m + V (x) . (7.23) How do we proceed? In the Schr¨odinger picture, we first find the energy eigenkets |j(cid:105) and eigenvalues Ej, which satisfy Once we have found these eigenkets, we can expand an arbitrary state as H|j(cid:105...
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17 42 From this point forward, we will drop the subscript H on the Heisenberg picture operators. Using a commutator identity, we then have dx dt = = = 1 i(cid:126) 1 i(cid:126) p m (cid:21) p2 2m p 2m x, (cid:20) x, (cid:16)(cid:104) . (cid:105) p + p (cid:104) x, (cid:105)(cid:17) p 2m This is the expected result, but...
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.2.1 Example: Charged Particle in a Uniform Electric Field Consider a charged particle in a uniform electric field, which has the Hamiltonian H = p2 m 2 − qE(t) . x (7.32) In the Schr¨odinger picture, this is a messy problem to solve. In the Heisenberg picture, however, the problem is not difficult at all. Using Eq. (7.30...
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dx dt = p m , dp dt = −mω2x . Solving these equations gives x(t) = x(0) cos(ωt) + p(0) mω sin(ωt) , p(t) = −mωx(0) sin(ωt) + p(0) cos(ωt) . (7.39) (7.40) Finding these equations in the Schr¨odinger picture is messy, though it can be done; in the Heisenberg picture, the result was immediate. Keep in mind that these are ...
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 10 DENIS AUROUX 1. The Quintic (contd.) Recall that we had a quintic mirror family Xˇ ψ with L...
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rewrite the Picard-Fuchs equation in the form (4) 3 d4 � ˇ [Ω] + dz4 k=0 ck(z) [ ˇ Ω] = 0 dk dzk 1 2 DENIS AUROUX �3 ckWk = 0. By Griffiths transversality ( dk Ω has no (0, 3)­ ˇ dzk k=0 Then W4 + component unless k ≥ 3), W0 = W1 = W2 = 0. Moreover, � d2Ωˇ Xˇ dz2 � � dΩˇ Xˇ dz d2Ωˇ dz2 d3Ωˇ dz3 d...
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2 c 1950750 c 2 1 q 2 − 2 2 c 10277490000 c 6 1 q 3 + · · · 3 2 MIRROR SYMMETRY: LECTURE 10 3 Now we can describe the mirror symmetry: there exists a basis of H 2(X, Z) ∼= Z (where X is the original quintic) given by the Poincar´e dual {e} of a hyperplane s.t., writing [B + iω] = te, q = exp(2πit) = exp(2...
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� q d3 n3 1 − � qd n1 q 2 + 27 n3 + 8 d>0 = 5 + n1q + 8 n2 + � � n1 q 3 + 64 n4 + 27 n2 + 8 � n1 q 4 + 64 · · · Matching these gives c1 = −5. • • n1 = 575 5 c2 · = c2 2875 : classical algebraic geometry tells us that 2875 is the number of lines on a quintic, c2 = 1. • n2 = 609250 (had been cal...
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um” intersection theory involving J-holomorphic disks. On the complex side, we look at intersections of subvarieties and holomorphic maps/extensions of bundles/sheaves. Thus, the complex side is governed by “classical” algebraic geometry, and all the “quantum” information is on the symplectic side. For this, we wil...
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λi | λi → ∞}. The Floer complex CF (L0, L1) is the free Λ-module Λ|L0∩L1| generated by L0 ∩ L1. Our goal is to define a differential δ s.t. HF (L0, L1) = H ∗(CF, δ) is invariant under Hamiltonian isotopies. The motivation for this was to understand Arnold’s conjecture on Lagrangian intersections. From that point of v...
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Lecture 7 Burke’s Theorem and Networks of Queues Eytan Modiano Massachusetts Institute of Technology Eytan Modiano Slide 1 �Burke’s Theorem • An interesting property of an M/M/1 queue, which greatly simplifies combining these queues into a network, is the surprising fact that the output of an M/M/1 queue with arriva...
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, the (forward) departure process is Poisson • By the same type of argument, the state (packets in system) left by a (forward) departure is independent of the past departures – In backward process the state is independent of future arrivals Eytan Modiano Slide 4 NETWORKS OF QUEUES Exponential Exponential Poiss...
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Slow truck effect Short packets Long packet queue queue queue • Example of bunching from slow truck effect long packets require long service at each node – – Shorter packets catch up with the long packets • Similar to phenomenon that we experience on the roads – Slow car is followed by many faster cars becaus...
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P(n n i i = k i 1 i 1 where ρi = λi µi • That is, in steady state the state of node i (ni) is independent of the states of all other nodes (at a given time) Independent M/M/1 queues – – Surprising result given that arrivals to each queue are neither Poisson nor independent – Similar to Kleinrock’s independenc...
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Tutorial #2 Verilog Simulation Toolflow Tutorial Notes Courtesy of Christopher Batten % v c s m i p s . v s i m v / % . - R P P & % v c s 6.884 – Spring 2005 02/16/05 T02 – Verilog 1 Figure by MIT OCW. A Hodgepodge Of Information − CVS source management system − Browsing a CVS repository with viewcvs − Makef...
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T02 – Verilog 4 CVS Basics Common CVS commands – cvs checkout pname – cvs update pname – cvs commit [filelist] Commit your changes – cvs add [filelist] – cvs diff Checkout a working copy Update working dir vs. repos Add new files/dirs to repos See how working copy differs Set the $CVSEDITOR environment vari...
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05 02/16/05 T02 – Verilog 7 CVS – Multiple Users F.1 checkout checkout User A User B F.1 F.1 6.884 – Spring 2005 02/16/05 T02 – Verilog 8 CVS – Multiple Users F.2 commit User A User B F.1 F.2 6.884 – Spring 2005 02/16/05 T02 – Verilog 9 CVS – Multiple Users F.2 commit Conflict! User A ...
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Æ Makefiles Why not use Makefiles to start out with? – Dependency tracking is less necessary – Difficult to implement some operations Why are we changing now? – Makefiles are more familiar to many of you – Dependency tracking will become more useful with the addition of test binary generation and Bluespec compila...
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batten/mips2stage % mkdir build % cd build % ../configure.pl ../config/mips2stage.mk % make simv % make self_test.bin % make self_test.vmh % ./simv +exe=self_test.vmh % make run-tests You can just use make run-tests and the dependency tracking will cause the simulator and the tests to be built before running the...
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05 T02 – Verilog 19 Writing SMIPS Assembly You can find the assembly format for each instruction in the SMIPS processor spec next to the instruction tables Use self_test.S as an example 6.884 – Spring 2005 02/16/05 T02 – Verilog 20 Writing SMIPS Assembly Our assembler accepts three types of register sp...
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1 mtc0 r2, r21 loop: beq zero, zero, loop nop .set reorder TEST_CODEEND Assembler directive which tells the assembler not to reorder instructions – programmer is responsible for filling in the delay slot 6.884 – Spring 2005 02/16/05 T02 – Verilog 23 Use smips-objdump for Disassembly Eventually the disassem...
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a0000 lui $k0,0x0 101c: 275a1400 addiu $k0,$k0,5120 1020: 03400008 jr $k0 1024: 42000010 rfe 00001100 <__testexcep>: 1100: 401a6800 mfc0 $k0,$13 1104: 00000000 nop <snip> 00001400 <__testcode>: 1400: 24010001 li $at,1 1404: 4081a800 mtc0 $at,$21 0001408 <loop>: 1408: 1000ffff b 1408 <loop> ... 141c: 3c080000 lu...
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408 <loop> ... 141c: 3c080000 lui $t0,0x0 1420: 8d081530 lw $t0,5424($t0) 1424: 3c01dead lui $at,0xdead 1428: 3421beef ori $at,$at,0xbeef 142c: 11010003 beq $t0,$at,143c <loop+34> ... 1438: 0000000d break 143c: 24080001 li $t0,1 1440: 4088a800 mtc0 $t0,$21 1444: 1000ffff b 1444 <loop+3c> 6.884 – Spring 2005 0...
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: 1 [pc=00001004] [ireg=08000500] [rd1=00000000] [rd2=00000000] [wd=00001008] tohost= 0 CYC: 2 [pc=00001400] [ireg=00000000] [rd1=00000000] [rd2=00000000] [wd=00000000] tohost= 0 CYC: 3 [pc=00001404] [ireg=24010001] [rd1=00000000] [rd2=xxxxxxxx] [wd=00000001] tohost= 0 CYC: 4 [pc=00001408] [ireg=4081a800] [rd1=xxxxx...
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ilog 27 Trace Output Instead of Waveforms It is sometimes very useful to use $display calls from the test harness to create cycle-by-cycle trace output instead of pouring through waveforms #include <smipstest.h> TEST_SMIPS TEST_CODEBEGIN .set noat addiu r1, zero, 1 mtc0 r1, r21 loop: beq zero, zero, loop .set at ...
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[ireg=1000ffff] [rd1=00000000] [rd2=00000000] [wd=00001410] tohost= 1 CYC: 6 [pc=00001408] [ireg=00000000] [rd1=00000000] [rd2=00000000] [wd=00000000] tohost= 1 6.884 – Spring 2005 02/16/05 T02 – Verilog 28 Final Notes Lab Assignment 1 – Don’t worry about cvs/make for now since I will be finishing setting this u...
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2.997 Decision­Making in Large­Scale Systems MIT, Spring 2004 February 17 Handout #6 Lecture Note 4 1 Average­cost Problems In the average cost problems, we aim at finding a policy u which minimizes Ju(x) = lim sup 1 T →∞ T E � T −1 � � � � x0 = 0 . gu(xt) � t=0 (1) Since the state space is finite, it c...
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., x∗, . . . . . ., x∗, . . . . . ., x∗, . . . . . . , � �� � � �� � � �� � λ1 u h(x) λ2 u Let Ti(x), i = 1, 2, . . . be the stages corresponding to the ith visit to state x∗, starting at state x. Let ⎡ u(x) = E ⎣ λi t=Ti (x) �Ti+1 (x)−1 gu(xt) Ti+1(x) − Ti(x) ⎤ ⎦ Intuitively, we must have λi u(x), since ...
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=0 1 We can now speculate about a version of Bellman’s equation for computing λ∗ and h∗. Approximating J ∗(x, T ) as in (3, we have J ∗(x, T + 1) = min ga(x) + Pa(x, y)J ∗(y, T ) � a λ∗(T + 1) + h∗(x) + o(T ) = min ga(x) + Pa(x, y) [λ∗T + h∗(y) + o(T )] � � � y � y � a � λ∗ + h∗(x) = mina ga(x) + �...
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T h∗ ≡ Tu∗ h∗. Then, and Ju∗ (x) = λ∗, ∀x, J ∗ u∗ (x) ≤ Ju(x), ∀u. Proof: Let u = (u1, u2, . . . ). Let N be arbitrary. Then TuN −1 h∗ ≥ T h∗ = λ∗e + h∗ TuN −2 (h∗ + λ∗e) TuN −2 TuN −1 h∗ ≥ = TuN −2 h∗ + λ∗e ≥ T h∗ + λ∗e = h∗ + 2λ∗e 2 Then Thus,we have T1T2 · · · TN −1h∗ ≥ N λ∗e + h∗ � E N −1 � � gu...
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exists, then the average cost Ju∗ (x) is the same for all initial states. It is easy to come up with examples where this is not the case. For instance, consider the case when the transition probability is an identity matrix, i.e., the state visits itself every time, and each state incurs different transition costs g(...
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5 Properties of Linear, Time-Invariant Systems In this lecture we continue the discussion of convolution and in particular ex- plore some of its algebraic properties and their implications in terms of linear, time-invariant (LTI) systems. The three basic properties of convolution as an algebraic operation are that it i...
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the result. The consequence of this for the intercon- nection of LTI systems is that a parallel combination can be collapsed into a single system whose impulse response is the sum of the two individual ones. In looking at and understanding the algebraic properties of convolution, it is worthwhile to recognize that conv...
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arity-the fact that for a linear system (whether or not it is time-invariant), if the input is zero for all t (or n), then the output is zero also. In this lecture we illustrate the properties discussed above with some systems. The problems associated with this lecture provide the opportunity to explore these propertie...
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[k] h[n-k] =x[n] * h[n] k=-o Convolution Integral: +xt x1(t) =f x (r) 8 (t -T) dT -00 +00 y(t) =f X(T) h(t-T) dT = x(t) * h(t) -00 Commutative: x[n] * h [n] = h[n] *x[n] x(t) * h(t) = h(t) *x(t) TRANSPARENCY 5.2 The commutativity property of convolution. u[n] -> au[n] =anu[n] * u[n] u(t) * eatu(t) = e-atu(t) *u(t)...
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parallel can be collapsed into a single LTI system. Signals and Systems MARKERBOARD 5.1 XNVEMTlI L M p4mnpd eSPMC I - ) It 0 ov ,vro o. 4 (4 ->4-k (V~ -T ~400 V,00 K0 04 oJt 1 U '.~O -~Or U at -46~ MARKERBOARD 5.2 -ov- LII ~ievs COIIASOL I + tV(*)O t<0o L\T ea' '-t4ety , vx ,c \ 0 0 -T 0 0 Q74' 0 +9A 6 S0 Pcc...
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8tt) -A) kvtYO.Yrnf _ d LA. +) =8C4) u. C-0= u, U t)Ut + MIT OpenCourseWare http://ocw.mit.edu Resource: Signals and Systems Professor Alan V. Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For inf...
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6.581J / 20.482J Foundations of Algorithms and Computational Techniques in Systems Biology Professor Bruce Tidor Professor Jacob K. White 6.581 / 20.482 6.581J / 20.482J Foundations of Algorithms and Computational Techniques in Systems Biology Professor Bruce Tidor Professor Jacob K. White
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MATH 18.152 COURSE NOTES - CLASS MEETING # 6 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 6: Laplace’s and Poisson’s Equations We will now study the Laplace and Poisson equations on a domain (i.e. open connected subset) Ω ⊂ Rn. Recall that (0.0.1) The Laplace equation is n def∆ = ∑ 1 i=...
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B 1 ( ) ( J1 t, x, y, z , J2 t, x, y, z , J3 t, x, y, z ( ) ( )) ) ) )) is is the electric field is the magnetic induction the current density Maxwell’s equations are (1.1.1) (1.1.2) ∂tE − ∇ × ∇ × + tB ∂ B − = J, = , E 0 ∇ ⋅ E = ρ, = ∇ ⋅ B 0. − Recall that ∇× is the curl operator, = Let’s look for steady-state solutions...
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) u z and let f z to e differentiable at z0 if ( ) = ( ) + b iv z be a complex-valued function analysis. Let z = + x iy (where x, y R) b ∈ (where u, v R a complex number, ). We recall that f is said e ∈ exists. If the limit exists, we denote it by f A fundamen x0 tal result of complex analysis is the following: f is diff...
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iii)the solution varies continuously with the data. Let Ω ⊂ Rn be a domain with a Lipschitz boundary, and let N denote the unit outward normal ˆ vector to ∂Ω. We consider the PDE (2.0.6) ∆u(x ) = ( ) f x , ∈ x Ω, supplemented by some boundary conditions. The following boundary conditions are known to lead to well-posed...
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Uniqueness via the Energy Method In this section, we address the question of uniqueness for solutions to the equation (0.0.3), sup- plemented by suitable boundary conditions. As in the case of the heat equation, we are able to provide a simple proof based on the energy method. Theorem 3.1. Let Ω ⊂ Rn be a smooth, bound...
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0 on ∂Ω, we have that w 0 in Ω, which ≡ (3.0.10) which implies that (3.0.11) and we can argue as b ∫ w∇ ˆN w dσ = −α ∫ w2 dσ ≤ 0, ∂Ω ∂ Ω efore conclude that w 0 in Ω. = 0, ∫ ∣∇w∣2 Ω ≡ 4 MATH 18.152 COURSE NOTES - CLASS MEETING # 6 ∣ ∇ ˆN w ∂Ω Now in the Neumann case, we have that ything w is constant in Ω. But now we ...
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case only; the proof is similar for other values of n. Let’s also assume that x is the origin; as we will see, we will be able to treat the case of general x by reducing it to on R2. For a ball of radius r, we have that the origin. We will work with polar coordinates (r, θ) r dθ. Note also that along ∂Br 0 , we have th...
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ˆ ) = 1 2π ∫ ∂B1 0 ( ) ∇ ( ) ( ) u σ σ dσ . ˆ N 1 2π ∫ B1(0) ∆ u(y ) d2y. But ∆u 0 since u is harmonic, so we have shown that = (4.0.17) ′ g (r) = 0, and we have shown (4.0.12b) for x 0. = T o prove (4.0.12a), we use polar coordinate integration and (4.0.12b) (in the case x 0) to obtain = MATH 18.152 COURSE NOTES - CL...
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+ ) ( = (4.0.12b) 2 ∫ ω 2 2R BR x ) ( x. for general u y d2y, ( ) (cid:3) 5. Maximum Principle Let’s now discuss another amazing property verified by harmonic functions. The property, known as the strong maximum principle, says that most harmonic functions achieve their maximums and minimums only on the interior of Ω. T...
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) ( u y d2 ) ( y = 1 ∣ ( )∣ B p {∫ Br z ( ) u y d2y ( ) + ∫ B Br z ( ) / u y d2y ( ) } {∣B r )∣ ( )∣ ( ) + ∫ z z u B Br z ( ) / 2 u(y) d y} ≥ 1 ∣ ( )∣ B p {∣Br (z)∣u(z) + m(∣B(p)∣ − ∣Br(z)∣)} . Rearranging inequality (5.0.22), we conclude that (5.0.23) u (z) ≤ m. Com bining (5.0.21) and (5.0.23), we conclude that (5.0....
https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf
to the data f, g C Ω , then inciple ∈ (1) (Comparison Pr 2 ) If f ≥ g on ∂Ω and f ≠ g, then uf > ug in Ω. (2) (Stability Estimate) For any x ∈ Ω, we have that ∣ uf x ( ) − u ( )∣ ≤ g x ( ) ∣ max f y y ∂Ω ∈ − )∣ ( g y . Proof. We first prove the Comparison we see that w solves Principle. Let w = u f − ug . Then by subtra...
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the case f = t of the corollary now follows directly from applying the Stability (cid:3) MIT OpenCourseWare http://ocw.mit.edu 18.152 Introduction to Partial Differential Equations. Fall 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf