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randomized test PZ∣X , put g Moreover, ⊃ R( P, Q direction. ) ⊃ R ( det P, Q . By Theorem 10.1, )) (x) = PZ=0∣X=x. Then g is a measurable function. P [Z = 0] = ∑ g(x )P x (x) = EP [g (X )] = ∫ 0 Q [ = ] = ∑ Z 0 x ( ) ( ) = g x Q x [ ( g X EQ )] = ∫ 0 1 1 P [g(X) ≥ t]dt Q [g(X) ≥ t]dt 114 where we applied the formula E...
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� {±∞} . The likelihood R dQ ≤ τ }. Formally, we assume that dP = p(x)dµ and F (x) ≜ ⎧⎪⎪⎪⎪⎪⎪⎪⎪ ⎪⎪⎪⎪⎪⎪⎪⎪⎩ ⎨ log p(x) q(x) +∞ , −∞ , / n a, > , p(x) > 0, x ( ) q x) > p( 0, q(x ( ) > ( ) = 0, q x p x (x) = 0, q(x) = p 0 ) = 0 0 0 Notes: • Q x ( ) exceeds a certain LRT is a deterministic test. The intuition is that upon o...
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[ ( ) { ≥ }] exp τ EQ g X 1 F τ exp{−τ } ⋅ EP [g(X)1{F ≤ τ }] (10.3) (10.4) 115 Below, these and similar inequalities are only checked for the cases of F not taking extended values, but from this remark it should be clear how to treat the general case. • Another useful observation: Q[F = +∞] = P [F = −∞] = 0 . (10.5) ...
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{ F [exp = −∞] , dµ p(x) exp{−F (x)}g(F (x)) + g(−∞)Q F [ = −∞] (10.8) (10.9) (10.10) where we used (10.5) to justify restriction to finite values of F . (1) To show F is a s.s, w e need to show PX∣F = Q PX∣ F (x∣f ) = ( ) ( ∣ ) PX x PF X f x ∣ PF (f ) = X∣F . For the discrete case we have: ef Q(x)1{ P (x) Q(x) PF (f ) ...
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( ∥ d β α D Q ∥ ) Q P ) where d (⋅∥⋅) is the binary divergence. Proof. Use data processing with PZ∣X . Lemma 10.1 (Deterministic tests). ∀E, ∀γ > 0 ∶ P [E] − γQ[E] ≤ P [ log dP dQ > log γ ] Proof. (Discrete version) P [E] − γQ[E] = ∑ x E ∈ p(x) − γq(x ) ≤ ∑ ∈E x ( (x) − γq(x p ))1{p(x)>γq(x)} = [ P log dP dQ > log γ, X...
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value for the Lemma 10.2 (Randomized tests). P [Z = 0] − γQ[Z = 0] ≤ P [ log dP ] d > log γ . Q Proof. Almost identical to the proof of the previous Lemma 10.1: P [Z = 0] − γQ[Z = 0] = ∑ PZ∣X (0 ∣x x )(p(x) − γq(x)) ≤ ∑ PZ∣X (0∣x)(p(x) − γq x (x))1{p(x)> (x)} γq = [ P log ≤ P [ log dP dQ dP dQ > log γ, Z = 0] − Q[ log ...
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of the LLR, whereas Note: To apply the to apply the weak converse Theorem 10.4 we need only to know the expectation of the LLR, i.e., divergence. Simil ( R P, Q is contained in the intersection of a collection say c , at which poin ∗ indexed by γ. of half-spaces > ] [ 10.5 Achievability bounds on R ( P, Q ) R( ) Since ...
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is attained by the following test: ⎧⎪⎪⎪ P log d dQ > τ 1 ⎪⎪ λ log dP ⎨ dQ = τ ⎪⎪⎪⎪ ⎪ log dP dQ < τ 0 ⎩ PZ∣X (0∣x (10.13) ) = where τ ∈ R and λ ∈ [0, 1] are the unique solutions to α = P [log dP dQ > τ ] + λP [log dP dQ = ] τ . Proof of Theorem 10.6. Let t = exp(τ ). Given any test PZ∣X , let g(x) = PZ∣X 0 x to show tha...
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„„„„„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„¶ ] { +EQ[g(X)1 ] dP t dQ > } { ( EP [(1 − g(X))1 dP ] + λP [ d Q „„„„„„„„„„„„„„„„„„„„„„„„„ ·„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„...
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t} { a consequence of the Neyman-Pearson lemma, all the points on the boundary of Remark 10.2. the region R( As ) P, Q are attainable. Therefore − } R(P, Q) = {(α, β) ∶ βα ≤ β ≤ 1 − β1 α . Since α ↦ βα is convex on [0, 1], hence continuous, the region Consequently, the infimum in the definition of βα is in fact a minimum...
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��ned with an ≤ instead of <. {log dP dQ <τ } {log dP τ dQ > τ ] for any τ , then we have λ ∈ (0, 1), and (10.13) is equivalent ¯ with probability λ. probability λ or 1 with 119 t1P[logdPdQ>t]ατt1P[logdPdQ>t]ατ Corollary 10.1. ∀τ ∈ R, there exists (α, β) ∈ R(P, Q) s.t. α = P [ log dP dQ > ] τ β ≤ exp(−τ )P [ log dP dQ ...
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to 0? exponents of the convergence rates of π1 0 ∣ ⎧⎪⎪ π1∣0 → 0 ⎨ ⎪⎪ ⎩π0∣1 → 0 120 MIT OpenCourseWare https://ocw.mit.edu 6.441 Information Theory Spring 2016 For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms.
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MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY...
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(z) = Zn(z) = −Γ(z) = 1 Zn(z) = Yn(z) = Y (z) Y0 = ˆi(z) Y0vˆ(z) D. If line is matched, ZL = Z0, ΓL = 0, Zn(z) = 1 II. Load Impedance Reflected Back to the Source From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. Zn(z = 0) = � Zn z = − = � λ 4 ZL =...
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A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. To match Z1 to RL ⇒ Z1 = RL/Z2 Z2 = Z2 2 RL , Z2 = √ Z1RL. 3 V0coswtYRs IV. Smith Chart Zn(z) = r + jx (z) = 1+Γ(z) Zn 1−Γ2 Γ(z) = Γr + jΓi 1+Γr +jΓi 1−Γ(z) ⇒ r + jx = 1−Γr −jΓi r −Γ2 x = i (1−Γr)2+Γ2 i 2Γi (1−Γr)2+Γ2 i r = � � �...
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50 + Z � � Im [Z(z = −l)] 50 + Re [Z(z = −l)] φ = tan−1 6 V. Standing Wave Parameters From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. vˆ(z) = Vˆ+e−jkz [1 + Γ(z)] ˆi(z) = Y0Vˆ+e−jkz [1 − Γ(z)] ⇒ | ⇒ | vˆ(z) = Vˆ+||1 + Γ(z)| ˆi(z) | | = Y0|Vˆ+||1 − Γ(z)| ...
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Y0Vˆ+ = 1 − |ΓL| = |1 − ΓL| E. If ZL = RL (real), then ΓL real. If ZL > Z0, VSWR = ZL . If ZL < Z0, VSWR = Z0 ZL Z0 7 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. Load Impedance: ZL = Z0 1 + ΓL ejφ | | ejφ 1 ΓL − | | � VSWR + 1 + (VSWR − 1)e jφ � ...
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Introduction to C++ Massachusetts Institute of Technology January 19, 2011 6.096 Lecture 7 Notes: Object-Oriented Programming (OOP) and Inheritance We’ve already seen how to define composite datatypes using classes. Now we’ll take a step back and consider the programming philosophy underlying classes, known as ob...
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pedals, the wheels, etc. OOP allows programmers to pack away details into neat, self-contained boxes (objects) so that they can think of the objects more abstractly and focus on the interactions between them. There are lots of definitions for OOP, but 3 primary features of it are: • Encapsulation: grouping related d...
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you can also think of the interface of a class as the set of buttons each instance of that class makes available. Interfaces abstract away the details of how all the operations are actually performed, allowing the programmer to focus on how objects will use each other’s interfaces – how they interact. This is why C...
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Car and Truck classes share this code. The Vehicle class will be much the same as what we’ve seen before: 1 2 3 4 class Vehicle { protected : string license ; int year ; 2 5 6 public : 7 8 9 10 11 12 13 }; Vehicle ( const string & myLicense , const int myYear ) : license ( myLicense ) , year ( myYea...
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1 class Car : public Vehicle { // Makes Car inherit from Vehicle 2 3 4 public : 5 Car ( const string & myLicense , const int myYear , const string & myStyle ) : Vehicle ( myLicense , myYear ) , style ( myStyle ) {} const string & getStyle () { return style ;} 6 7 8 }; Now class Car has all the data members and ...
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B: 1. Every A object has a B object. For instance, every Vehicle has a string object (called license). 2. Every instance of A is a B instance. For instance, every Car is a Vehicle, as well. Inheritance allows us to define “is-a” relationships, but it should not be used to implement It would be a design error to mak...
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{ return style ;} 6 7 8 9 10 }; 3.2.1 Programming by Difference In defining derived classes, we only need to specify what’s different about them from their base classes. This powerful technique is called programming by difference. Inheritance allows only overriding methods and adding new members and methods. We c...
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If we have a function that expects a Vehicle object, we can safely pass it a Car types. object, because every Car is also a Vehicle. Likewise for references and pointers: anywhere you can use a Vehicle *, you can use a Car *. 5 4.1 virtual Functions There is still a problem. Take the following example: 1 Car c...
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references: 1 Car c ( " VANITY " , 2003) ; 2 Vehicle & v = c ; 3 cout << v . getDesc () ; This will only call the Car version of getDesc if getDesc is declared as virtual. Once a method is declared virtual in some class C, it is virtual in every derived class of C, even if not explicitly declared as such. However...
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actually implement it, and therefore cannot be instantiated. 5 Multiple Inheritance Unlike many object-oriented languages, C++ allows a class to have multiple base classes: 1 class Car : public Vehicle , public InsuredItem { 2 3 }; ... This specifies that Car should have all the members of both the Vehicle and the ...
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LECTURE 17 LECTURE OUTLINE Review of cutting plane method Simplicial decomposition • • Duality between cutting plane and simplicial • decomposition All figures are courtesy of Athena Scientific, and are used with permission.1 CUTTING PLANE METHOD Start with any x0 X. For k 0, set ≥ ⌘ • xk+1 ⌘ where arg min Fk(x), X x ⌦...
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X is much simpler than minimizing f over X. − 3◆ SIMPLICIAL DECOMPOSITION METHOD f (x0) x0 f (x1) x1 X ˜x2 f (x2) x2 f (x3) x3 ˜x4 x4 = x ˜x1 ˜x3 Level sets of f • (initially x0 Given current iterate xk, and finite set Xk ⌦ x0 { Let x˜k+1 be extreme point of X that solves X, X0 = ⌘ ). } • X minimize ∇ subject to x f (x...
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many ex- treme points), so case (a) must eventually occur. The method will find a minimizer of f over X • in a finite number of iterations. 5 COMMENTS ON SIMPLICIAL DECOMP. Important specialized applications • Variant to enhance e⌅ciency. Discard some of • the extreme points that seem unlikely to “partici- pate” in the o...
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x n ⌘ � ⌅ F (x) = max j=1,...,⌫ y j� x − f (yj) ⇤ [this follows using x�jyj = f (xj) +f (yj), which is ◆f (xj) – the Conjugate Subgra- implied by yj ⌘ dient Theorem] ⌅ 7 ◆ INNER LINEARIZATION OF FNS f (x) F (y) f (y) x0 x1 0 x2 x Slope = y0 O F (x) Slope = y1 f Outer Linearization of f Slope = y2 f y0 y1 f 0 y2 y Inn...
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f is differentiable Ck+1(x) Ck(x) c(x) Slope: f (xk) −⇥ Const. f (x) − xk xk+1 ˜xk+1 x Given Ck: inner linearization of c, obtain xk ⌘ arg min f ⌦� x n ⇤ Obtain x˜k+1 such that (x) +C k(x) ⌅ • � • • • f (xk) ◆c(˜xk+1), −∇ ⌘ x˜k+1} and form Xk+1 = Xk ∪ { 9 NONDIFFERENTIABLE CASE Given Ck: inner linearization of c, obtain...
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2(x), x ⌦� n min f (⌃) +f ⌅ ⌦� 1 n 2 ( ⌃ − ) Primal and dual approximations • • 2,k(x) 1(x) +F min f n x ⌦� F2,k and F • tions of f and f 2 2 f min 1 (⌃) +F 2,k( ⌅ ⌦� n ⌃) − 2,k are inner and outer approxima- x˜i+1 and gi are solutions of the primal or the • dual approximating problem (and corresponding subgradient...
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iβc ωMa − Tïα\iβc ωMa − ï3;3a ûf F Tωρ β [ ψi åb ï3;Xa ï3;ña ï3;ba ï3;ùa ï3;ja PA8w¢† 1ñr T9 síf¢†ÄWo z¢_oo Aí ë o9¢ë9A܆f QwAf © sources unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/hel...
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† AíôAf†í9 ÄëB† ëíf 9v† ¢†Q†ô9†f ÄëB† vëB† fAff†¢†í9 ÄëB†à†í89vo: óm θ Å θi2 9v†¢† Ao í_ ¢†Q†ô9A_í< ¢†m¢ëô9A_í 9ëI†o õàëô† Aío9†ëf; XE t MIT OpenCourseWare https://ocw.mit.edu 2.062J / 1.138J / 18.376J Wave Propagation Spring 2017 For information about citing these materials or our Terms of Use, visit: https://...
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Spring 2010 Semantic Analysis Semantic Analysis Saman Amarasinghe Massachusetts Institute of Technology Massachusetts Institute of Technology Symbol Table Summary • Program Symbol Table (Class Descriptors) • Class Descriptors D Cl – Field Symbol Table (Field Descriptors) i t Pointer to Field Symbol Table for...
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In a stack push a; push b; mul; push c; add; pop x • In single use temporary registers i t t1 = mul a, b x add t1, c x = add t1 c = • Trees – Intermediate values are implicit in the edges St x St x add mul ld c ld a ld b Handling Control Flow Handling Control-Flow (cid:129) Control-Flow Graph • Contr...
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– take small stepsp – don’t try to do too many at once – don t try to do anything too early don’t try to do anything too early – try not to loose any information! Saman Amarasinghe 9 6.035 ©MIT Fall 2006 Outline Outline • Practical Issues in Intermediate t di I t I l ti P i Representation Representation...
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iqueness checks – Uniqueness checks – Type checks Saman Amarasinghe 15 6.035 ©MIT Fall 2006 Flow of control checks Flow of control checks 11 • Flow-control of the program is context • Flow control of the program is context sensitive • Examples: Declaration of a variable should be visible at use – Declarat...
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I l ti P i Representation Representation • What is semantic analysis? • Type systems • What to check? Type Systems Type Systems 17 • A type system is used to for the type • A type system is used to for the type checking • A type system incorporates syntactic constructs of the language – syntactic const...
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• If T is a type expression an array(T, I) is • If T is a type expression an array(T I) is also a type expression – I is a integer constant denoting the number of elements of type T – Example: int foo[128]; ]; [ array(integer, 128) Saman Amarasinghe 26 6.035 ©MIT Fall 2006 Type Expressions: Method Calls T...
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Saman Amarasinghe 29 6.035 ©MIT Fall 2006 A simple typed language A simple typed language • A language that has a sequence of declarations • A language that has a sequence of declarations followed by a single expression P  D; E P  D; E D  D; D | id : T T  char T | cha E  literal | num | id |...
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2006 Parser actions Parser actions 24 E  E1 + E2 E  E1 + E2 integer and { if E1 type == integer and { if E1.type E2 .type == integer then E.type = integer yp g else E.type = type_error yp yp } Saman Amarasinghe 33 6.035 ©MIT Fall 2006 Parser actions Parser actions E  E1 [E2 ] E  E [E ] integ...
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of one type to another type • Example int A; float B; B = B + A • Two types of coercion – widening conversions – narrowing conversions id i i Saman Amarasinghe 37 6.035 ©MIT Fall 2006 Narrowing conversions Narrowing conversions • Conversions that may loose information • Conversions that may loose infor...
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what is its type? Saman Amarasinghe 42 6.035 ©MIT Fall 2006 Overloading Overloading 28 Some operators may have more than one • Some operators may have more than one • type. • Example l E int A, B, C; float X, Y, Z; A = A + B X = X + Y X X + Y • Complicates the type system – Example A = A + X • Wha...
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Local Symbol Table Local Symbol Table • When building the local symbol table, have • When building the local symbol table have a list of local descriptors • What to check for? duplicate variable names – duplicate variable names – shadowed variable names • When to check? k? h h – when insert descriptor into loc...
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else can/should be checked? Saman Amarasinghe 51 6.035 ©MIT Fall 2006 Add Operations Add Operations • What does compiler have? • What does compiler have? – two expressions • What can go wrong? ? Wh t – expressions have wrong type – must both be integers (for example) • So compiler checks type of expressions...
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Expression • What does it do? – Look up variable name. pa • If in local symbol table, reference local descriptor • If in parameter symbol table, error sy bo tab e e o • If in field symbol table, reference field descriptor • If not found, semantic error a ete , – Check that type of array index expression is int...
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• Do semantic checks when build IR • Many correspond to making sure entities are • Many correspond to making sure entities are there to build correct IR • Others correspond to simple sanity checks • Each language has a list that must be checked • Can flag many potential errors at compile time Saman Amarasinghe ...
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18.354J Nonlinear Dynamics II: Continuum Systems Lecture 1 Spring 2015 1 Math basics 1.1 Derivatives and differential equations In this course, we will mostly deal with ordinary differential equations (ODEs) and partial differential equations (PDEs) real-valued scalar or vector fields. Usually, non-bold symbols will be...
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the product: For instance, in the case of a 3D vector field v(t, x), we can obtain a scalar field called divergence another (pseudo-)vector field called curl ∇ · v ≡ ∂ivi, and the gradient matrix ∇ ∧ v ≡ ((cid:15)ijk∂jvk) ∇v ≡ (∂ivj). The (scalar) Laplacian operator (cid:52) in Cartesian coordinates is defined by (cid:52) ...
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by z¯ = x − iy (11) and corresponds to a reflection at the real axis or, equivalently, at the line (cid:61)(z) = 0. Addition of complex numbers is linear z = z1 + z2 = (x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2) = x + iy (12) corresponding to the addition of the two 2D vectors (x1, y1) and (x2, y2). complex multipl...
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exp to the trigonometric sin-and cos-functions. When dealing with axisymmetric problems it is often advantageous to use the polar representation of a complex number √ r = |z| = zz¯ ∈ R+ z = reiφ , 0 , , φ = arctan 2(y, x) ∈ [0, 2π) (18) From the properties of the exp-function, it follows that the multiplication of comp...
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Introduction to Simulation - Lecture 6 Krylov-Subspace Matrix Solution Methods Jacob White Thanks to Deepak Ramaswamy, Michal Rewienski, and Karen Veroy Outline • General Subspace Minimization Algorithm – Review orthogonalization and projection formulas • Generalized Conjugate Residual Algorithm – Krylov-subspace – ...
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k r ⇒ − ∑ (cid:71) Mwα i i i 0 = i sα ' Residual Minimizing idea: pick b Mx = − = b k k 1 − to minimize k r ≡ ( k r T ) ( k r ) = 2 2 b − ⎛ ⎜ ⎝ SMA-HPC ©2003 MIT k 1 − ∑ i = 0 α i (cid:71) Mw i T ⎞ ⎛ ⎟ ⎜ ⎠ ⎝ b − k 1 − ∑ i = 0 α i (cid:71) Mw i ⎞ ⎟ ⎠ Arbitrary Subspace methods Residual Minimization Computational Ap...
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M b , and a set of search directions (cid:71) p 1) Generate by orthogonalizing 's j Algorithm Steps (cid:71) w 0 { ,..., (cid:71) w − k 1 } M ' w s j ( Mw ( Mp i j j 1 − − ∑ i = 0 T ) ( ) ( T Mp i Mp i ) ) p i j For = k 0 to − 1 p = w j j r k 2) compute the minimizing solution x T T ) ( ( ) ( ( ) ) T T ) ( ) (...
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j j j ) ) pMp j ) M j Mp j Mp ( + ( r − Normalize Update Solution Update Residual j p j SMA-HPC ©2003 MIT Arbitrary Subspace methods Subspace Selection Criteria ,..., All that matters is the ' k 1 w 0 Criteria for selecting w − { } span w ,..., w − 0 k = − ∑ (cid:71) b Mw i i 0 = } (cid:19) w for N k (cid:71) (ci...
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) { span w 0 x ) 0 ( f x (cid:71) w k ,..., ,..., ∇ } − = 1 x k 1 − ( f x } ) { span r 0 = 0 { span r } k 1 r − ,..., k 1 − ,..., r } If: i Mrα i k 1 − − ∑ i ,..., 0 = r k 1 − } = 0 , } { 0 1 0 k − span r Mr M r (cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10) Krylov Subspace ,......
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+ = r k k x r + − p k 1 + = r k 1 + pα k k Mpα k ( Mr ( Mp 0 k j k − ∑ j = Vector inner products, O(n) Matrix-vector product, O(n) if sparse Vector Adds, O(n) k 1 + T ) ( ) ( T Mp Mp j j ) ) p j O(k) inner products, total cost O(nk) If M is sparse, as k (# of iters) approaches n, 3 O n O n ( ) ) (2 ) total cost = Bet...
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⎡ ⎤ ⎢ ⎥ 1 2 −⎢ ⎥ ⎢ ⎥− 1 ⎥ ⎢ ⎣ ⎦ (cid:8)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:10) M 1 2 (cid:37) − Nodal Equation Form Krylov Methods Nodal Formulation “No-leak Example” Circuit and Matrix 1 2 3 4 1m − m 2 1 − ⎤ ⎡ ⎥ ⎢ 1 2 −⎢ ⎥ ⎥− ⎢ 1 ⎥ ⎢ ⎣ ⎦ (cid:8)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11...
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“leaky” Example Circuit and Matrix 1 2 3 4 1m − m m SMA-HPC ©2003 MIT 1 − 2.01 2.01 ⎤ ⎡ ⎥ ⎢ 1 −⎢ ⎥ ⎥− ⎢ 1 ⎥ ⎢ 1 2.01 ⎣ ⎦ (cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10) M (cid:37) − Nodal Equation Form GCR Performance(Random Rhs) Insulating Leaky 0 10 -1 10 -2 10 -3 10 R E S I...
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Note: for any { 1 r span r , 0 k − i α i ∑ 0 = 0α ≠ 0 0 r − α= 0 SMA-HPC ©2003 MIT kth order polynomial 1 0 i + M r = ( ) ( I M M r ξ k − ) 0 Mr 0 } = span { 0 r Mr , 0 } Krylov Methods Convergence Analysis Basic Properties 0 0 , , x ... } 2) span k M r in GCR, then { 0 r Mr , = th is the k k If for all 0 j ...
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k+1 k+1 ℘(cid:4) th order Therefore Any polynomial which satisfies the zero constraint can be used to get an upper bound on kr + 21 2 SMA-HPC ©2003 MIT Eigenvalues and Vectors Review Basic Definitions Eigenvalues and eigenvectors of a matrix M satisfy eigenvalue (cid:71) (cid:71) u M λ= u i i i eigenvector O...
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Simplifying Assumption Almost all NxN matrices have N linearly independent Eigenvectors ↑ (cid:71) u 2 ↓ ↑ (cid:71) u 3 ↓ ↑ (cid:71) u 1 ↓ ↑ (cid:71) u ⎡ ⎢ ⎢ ⎢ ⎣ M N ↓ ⎤ ⎥ ⎥ ⎥ ⎦ (cid:34) (cid:34) (cid:34) ⎡ ↑ (cid:71) ⎢ u = ⎢ λ λ 2 2 ⎢ ↓ ⎣ ↑ (cid:71) u 1 1 ↓ ↑ (cid:71) u λ 3 3 ↓ (cid:34) (cid:34) (cid:34) λ ↑ (cid:71...
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M’s eigenvalues SMA-HPC ©2003 MIT Eigenvalues and Vectors Review Heat Flow Example Incoming Heat Unit Length Rod 1T NT (0)T + - sv T= (0) SMA-HPC ©2003 MIT (1)T + - sv T= (1) Eigenvalues and Vectors Review Heat Flow Example Continued 2 ⎡ ⎢ 1 −⎢ ⎢ 0 ⎢ ⎣ 0 1 0 − (cid:37) 2 (cid:37) (cid:37) 0 − 1 0 ⎤ ⎥ 0 ⎥ ⎥− 1 ⎥ 2 ...
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+ p (cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10) Diagonal a λ 1 +… + a 1 ( f M U U a I = 0 ) ( + a λ 1 + +… a p λ p ) SMA-HPC ©2003 MIT Useful Eigenproperties Spectral Decomposition Decompose arbitrary x in eigencomponents + (cid:71) u + α α 1 1 (cid:71) u N N (cid:71) u 2 2...
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2) If M has only q distinct eigenvalues, the GCR Algorithm converges in at most q steps ( ( ... = Proof: Let ( ) x (cid:4) ℘ q λ 2 λ 1 )( − − − x x x ) λ q ) SMA-HPC ©2003 MIT Summary • Arbitrary Subspace Algorithm – Orthogonalization of Search Directions • Generalized Conjugate Residual Algorithm – Krylov-sub...
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4 Convolution In Lecture 3 we introduced and defined a variety of system properties to which we will make frequent reference throughout the course. Of particular importance are the properties of linearity and time invariance, both because systems with these properties represent a very broad and useful class and be- cau...
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of both continuous- time and discrete-time signals as a linear combination of delayed impulses and the consequences for representing linear, time-invariant systems. The re- sulting representation is referred to as convolution. Later in this series of lec- tures we develop in detail the decomposition of signals as linea...
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al and is similar in its properties to the convolution sum for discrete-time signals and systems. A number of the important properties of convolution that have interpretations and consequences for linear, time-invariant systems are developed in Lecture 5. In the current lecture, we focus on some examples of the evaluat...
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2 0-0 -- *- x[-I]8[n+1] n -1 0 I 2 X[-] x [-2]8[n +2] 0--0-0 -1 0 1 2 n x[o]8[n]+x(I] 8[n -1] + x [-I]8[n+ ]+.-- +X kr =2 x[k]8[n-k] k= -c Signals and Systems TRANSPARENCY 4.2 The convolution sum for linear, time- invariant discrete-time systems expressing the system output as a weighted sum of delayed unit impuls...
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0 LTI: y[n] =E x[k] h[n - k]I Convolution Sum Convolution 4-5 TRANSPARENCY 4.4 Approximation of a continuous-time signal as a linear combination of weighted, delayed, rectangular pulses. [The amplitude of the fourth graph has been corrected to read x(O).] TRANSPARENCY 4.5 As the rectangular pulses in Trans- parency 4....
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) = 0 +o +O k=- o x(kA) hk(t) A +00 =f xT) hT(t) dr If Time-Invariant: LTI: hkj t) = ho(t - kA) h,(t) = he (t - r) +01 f x(r) h(t-7) dr- v(t) 1 -0 Convolution Integral x(t) 0 t ti x (0) h Mt x(kA) oA kA t x (0) x(A) AA y(t) y(t) oA 0 x(t) 0 t t Convolution 4-7 TRANSPARENCY 4.8 Comparison of the convolution su...
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an input that is a unit step and a system impulse response that is a decaying exponential for t > 0. MARKERBOARD 4.2 y(t)f x(r)h(t-r)dr x(t) u (t) h (t )=e~43 u t) x (t) t h (t) x (r) r O 0 h (t-r) t T Convolution MARKERBOARD 4.3 v4egva.z: t). Te Lk (t - -C jT Ct k Lt-T)aT t (0 1 U -e 3 o- eoverkp ~3et ee h*...
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LECTURE 2 Hilbert Symbols Let K be a local field over Qp (though any local field suffices) with char(K) (cid:54)= 2. Note that this includes fields over Q2, since it is the characteristic of the field, and not the residue field, with which we are concerned. Recall from the previous lecture the duality (2.1) Gal2(K) := Galab(K...
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osition 2.3. The Hilbert symbol satisfies the following properties: (1) Bimultiplicativity. For all a, b, c ∈ K ×, (a, bc) = (a, b) · (a, c). (2) Non-degeneracy. For all a ∈ K ×, if (a, b) = 1 for all b ∈ K ×, then a ∈ (K ×)2. Note that (a, b) = (b, a) trivially. Bimultiplicativity says that we can solve ax2 +by2 = 1 if...
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(x)y, where y ∈ O× K. Then the following are equivalent: (1) x is a square; (2) v(x) is even and y is a square; (3) y mod p is a square in K ×. Note that we may reduce to x ∈ O× K. We offer two proofs: Proof (via Hensel’s Lemma). All explanations aside from that from the final condition are clear. So suppose x mod p is a...
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a + b)π + abπ2, and since abπ2 ∈ p2, we are left with 1 + (a + b)π in the associated graded term, hence multiplication simply corresponds to addition in k. Similarly, for each n ≥ 1, we have (1 + pn)/(1 + pn+1) (cid:39) k by a similar argument, since n + 1 ≤ 2n. Now, σ acts on the filtration as K. Clearly O× O× K ⊇ 1 + ...
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quotients are all isomorphic to finite fields. As a general principle, we can understand many things about A via its associated graded Gr∗A. Definition 2.7. Let A be an abelian group. A filtration on A is a descending sequence of subgroups A =: F0A ⊇ F1A ⊇ F2A ⊇ · · · , and it is said to be complete if A ∼−→ lim ←−n are t...
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= f (y0) + f ((cid:15)1) ≡ x mod F2B, where we have defined y1 := y0 +(cid:15)1. This is an equation of the same form as before, and we may iterate to find a “compatible” system of yn such that f (yn) = x mod Fn+1B for each n ≥ 0, where by “compatible” we mean that for each n we have yn ≡ yn+1 mod Fn+1A. But then there i...
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here we will view it asymmetrically. Suppose a) is a degree 2 extension of K (note that if a is a is not a square, so that K( a nonzero square, then we need only understand K( a) to be the corresponding étale extension of K, isomorphic to K × K). √ √ Claim 2.11. We have (a, b) = 1 if and only if b is a norm for the ext...
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. Theorem 2.12. If L/K is a quadratic extension of local fields, then the norm N : L× → K × is a homomorphism, and N(L×) ⊆ K × is a subgroup of index 2. Example 2.13. Consider C/R. MIT OpenCourseWare https://ocw.mit.edu 18.786 Number Theory II: Class Field Theory Spring 2016 For information about citing these mater...
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Sufficiency Sufficiency MIT 18.443 Dr. Kempthorne Spring 2015 MIT 18.443 Sufficiency 1 Sufficiency Definition Example Theorems Outline 1 Sufficiency Definition Example Theorems MIT 18.443 Sufficiency 2 Sufficiency Definition Example Theorems Sufficient Statistics Definition: Sufficiency X1, X2, . . . , Xn iid with ...
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finition Example Theorems Sufficiency: Example Example 8.8.1A Bernoulli Trials Let X = (X1, . . . , Xn) be the outcome of n i.i.d Bernoulli(θ) random variables The pmf function of X is: p(X | θ) = P(X1 = x1 | θ) × · · · × P(Xn = xn | θ) = θx1 (1 − θ)1−x1 × θx2 (1 − θ)1−x2 × · · · θxn (1 − θ)1−xn = θ xi (1 − θ)(n−...
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we should only need the information of T (X) = t, since the value of X given t reflects only the order information in X which is independent of θ. MIT 18.443 Sufficiency 6 Sufficiency Definition Example Theorems Outline 1 Sufficiency Definition Example Theorems MIT 18.443 Sufficiency 7 Sufficiency Definition Examp...
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ary A If T is sufficient for θ, then the maximum likelihood estimate is a function of T . Rao-Blackwell Theorem Let θˆ be an estimator of θ with E [θˆ2] < ∞ for all θ. Suppose that T is sufficient for θ Define θ˜ = E [θˆ | T ]. Then for all θ, E [(θ˜ − θ)2] ≤ E [(θˆ − θ)2]. The inequality is strict unless θ˜ ≡ θ. ˆ...
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MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Lecture 5 Fall 2013 9/16/2013 Extension of LD to Rd and dependent process. G¨artner-Ellis Theorem Content. 1. Large Deviations in may dimensions 2. G¨artner-Ellis Theorem 3. Large Deviations for Markov chains 1 Large Deviations in Rd Most of the developmen...
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M (θ) < ∞ for all θ is needed. Known counterex­ amples are somewhat involved and can be found in a paper by Dinwoodie [2] which builds on an earlier work of Slaby [5]. The difficulty arises that there is no longer the notion of monotonicity of I(x) as a function of the vector x. This is not the tightest condition and...
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g(θ1, θ2) = 0, we have that (θ1 2 + θ1θ2 + θ2 2). x1 − θ1 − θ2 = 0, 1 2 x2 − θ2 − θ1 = 0, 1 2 2 from which we have θ1 = x1 − x2, θ2 = x2 − x1 4 3 2 3 4 3 2 3 Then So we need to find 2 I(x1, x2) = (x1 + x2 − x1x2). 2 2 3 2 (x1 + x 2 − x1x2) 2 inf x1,x2 3 s.t. 2x1 + x2 ≥ 5 (x ∈ F ) This becomes a non-li...
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1) which gives x1 = 10 11 , x2 = 35 11 and I(x1, x2) = 5.37. Thus lim sup n 1 n log P( Sn n ∈ F ) ≤ −5.37 Applying the lower bound part of the Cram´er’s Theorem we obtain lim inf n 1 n log P( Sn n ∈ F ) ≥ lim inf n 1 n log P( Sn n ∈ F o) =≥ = − 5.37 (by continuity of I). 3 (3) (4) − inf 2...
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the i.i.d. case n φn(θ) = = 1 n 1 n log E[exp(n(θ, n−1Sn))] log M n(θ) = log M (θ) = log E[exp((θ, X1))]. Loosely speaking G¨artner-Ellis Theorem says that when convergence φn(θ) → φ(θ) (5) takes place for some limiting function φ, then under certain additional technical assumptions, the large deviation...
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supθ(θa − M (θ)) when a > µ = E[X]. But now we are dealing with the multidimensional case where such an identity does not make sense. 3 Large Deviations for finite state Markov chains Let Xn be a finite state Markov chain with states Σ = {1, 2, . . . , N }. The transition matrix of this Markov chain is P = (Pi,j , ...
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have strictly positive compo­ nents. This theorem can be found in many books on linear algebra, for example [4]. The following corollary for the Perron-Frobenious Theorem shows that the essentially the rate of growth of the sequence of matrices Bn is ρn. Specifically, Corollary 1. For every vector φ = (φj , 1 ≤ j ≤ ...
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Let ρ(Pθ) denote its Perron-Frobenious eigenvalue. Theorem 4. The sequence 1 Sn = tions bounds with rate function I(x) = n p1 n p f (Xn) satisfies the large devia­ 1≤i≤k θ∈Rd ((θ, x)−log ρ(Pθ)). Specifically, 6 for every state i0 ∈ Σ, closed set F ⊂ Rd and every open set U ⊂ Rd, the fol­ lowing holds: l...
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Perron-Frobenious theory in fact can be used to show that such a differentiability indeed takes place. Details can be found in the book by Lancaster [3], Theorem 7.7.1. References [1] A. Dembo and O. Zeitouni, Large deviations techniques and applications, Springer, 1998. [2] IH Dinwoodie, A note on the upper boun...
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3.032 Mechanical Behavior of Materials Fall 2007 Using U(r): Measure parameters for U(r) in physical model to predict stresses that are high enough for elastic instabilities to occur (e.g., nucleation of defects in crystals). Images removed due to copyright restrictions. Please see: Fig. 1c in Gouldstone, Andrew, et a...
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Properties of Polymer Nanofibers." Macromolecules 40 (2007): 8483-8489. Electrospun PEO could be used for filters, composites, fuel cells, drug delivery, cell scaffolds, etc. [Rutledge Group, MIT] Internal energy inside polymer nanofiber increases for fibers of R < 5 nm. Curgul, Rutledge and Van Vliet, Macromolecu...
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18. Div grad curl and all that Theorem 18.1. Let A ⊂ Rn be open and let f : A −→ R be a differ­ entiable function. If �r : I −→ A is a flow line for �f : A −→ Rn, then the function f ◦ �r : I −→ R is increasing. Proof. By the chain rule, d(f �r) ◦ dt (t) = �f (�r(t)) · �r�(t) = �r�(t) �r�(t) ≥ 0. · � Corollary...
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jˆ+ ˆı + on vector fields: Definition 18.5. Let A ⊂ R3 be an open subset and let F� : A −→ R3 be a vector field. The divergence of F� is the scalar function, which is defined by the rule div F� : A −→ R, div F� (x, y, z) = � · F� (x, y, z) = ∂f ∂x + ∂f ∂y + ∂f . ∂z 1 The curl of F� is the vector field c...
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F� = �0, and it is called incompressible if the divergence is zero, div F� = 0. : −→ � 3R Proposition 18.7. Let f be a scalar field and F� a vector field. (1) If f is C2, then curl(grad f ) = �0. Every conservative vector field is rotation free. (2) If F� is C2, then div(curl F� ) = 0. The curl of a vector field is ...
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