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K 2 = 1 · so |−K = ∅. Take a hyperplane section H of X. Then there is an n ≥ 0 | � nK| =� ∅ but |H + (n + 1)K| = ∅. Since −K ∼ an effective nonzero s.t. |H + · divisor, H K < 0 and H (H + nK) is eventually negative and H + nK is not effective. Let D ∈ |H + nK|: then |D + K| = ∅ and K · D = K(H + nK) = K H < 0 sinc...
https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01d543f81d0b743f06deed68d9eec77d_lect9.pdf
Again, let H be a hyperplane section: if K H < 0, we can take E = H; if K H = 0, we can take K + nH for n >> 0; so assume K H > 0. Let γ = −K· > 0 so that (H + γK) K = 0. Also, · · · H · K2 (3) (H + γK)2 > H 2 + 2γ(H K) + γ2K = H 2 + · So take β rational and slightly larger than γ to get (K · H)2 (−K 2) > 0...
https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01d543f81d0b743f06deed68d9eec77d_lect9.pdf
as in reduction 2, i.e. K · D ≥ 0 for every effective divisor D s.t. |K + D| = ∅. We will obtain a contradiction. Lemma 1. If X is a minimal surface with p2 = q = 0, K 2 > 0 and K D ≥ 0 for every effective divisor D on X s.t. |K + D| = ∅, then · (1) Pic (X) is generated by ωX = OX (K), and the anticanonical bundle ...
https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01d543f81d0b743f06deed68d9eec77d_lect9.pdf
irreducible, and similarly D is not a multiple. Furthermore, pa(D) = 2 D(D + K) + 1 = 1, showing (2). Next, we claim that the only effective divisor s.t. D + K = ∅ is the zero | divisor. Assume not, i.e. ∃D > 0 s.t. |K + D| = ∅. ∈ D: then since h0(−K) ≥ 1 + K 2 ≥ 2, there is a C ∈ |−K| passing through x. C is an in...
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of positive self-intersection, so by the useful lemma D +nK is not effective for n >> 0. Now, let Δ be an arbitrary effective divisor. Then ∃n ≥ 0 s.t. Δ + nK = 0 | but |Δ + (n + 1)K| = ∅. Take D ∈ |Δ + nK| effective. |D + K| = = 0 from above. Since any divisor is a difference of effective divisors, Pic (X) is generate...
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theorem, (8) 0 → OX (−K) ⊗ Ix ⊗ Iy → OX (−K) → k6 → 0 2 have dimension 3 over k. Taking the long exact sequence, since OX,x/m2 , OX,y/my we find that h0(OX (−K) ⊗ Ix ⊗ Iy) = 0, and get a nonzero section of that sheaf. x � � 4 LECTURES: ABHINAV KUMAR It is a divisor of zero passing through x and y with multiplicit...
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X = X Since q = an X )∗) ∼ = H 1(O∗ X ) = Pic X ∼ (10) This implies that b2 = rank H 2(X, Z) = rank Pic X = 1 contradicting b2 ≥ 5. For positive characteristic, we will sketch a proof: the first proof was given by Zariski, and the second using ´etale cohomology by Artin and by Kurke. Our proof will be by reductio...
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Bluespec Tutorial: Rule Scheduling and Synthesis Michael Pellauer Computer Science & Artificial Intelligence Lab Massachusetts Institute of Technology Based on material prepared by Bluespec Inc, January 2005 March 4, 2005 BST-1 Improving performance via scheduling Latency and bandwidth can be improved by performing...
https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/01e06522cd4ab035dbbbd38279045351_t03_bluespec.pdf
Rule: fetch Predicate: the_bf.i_notFull_ && the_started.get Blocking rules: imem_put, start more urgent rules which can block the execution of this rule (more on urgency later) March 4, 2005 BST-6 3 Static execution order When multiple rules execute in a single clock cycle, they must appear to execute in sequence Th...
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2 decrements register y 0 +1 -1 1 +1 -1 2 x y rule proc0 (cond0); x <= x + 1; endrule rule proc1 (cond1); y <= y + 1; x <= x – 1; endrule rule proc2 (cond2); y <= y – 1; endrule (* descending_urgency = “proc2, proc1, proc0” *) show what happens under different urgency annotations March 4, 2005 BST-11 Example2.bsv Demo ...
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a e r t r n d i t i o e l y t i v March 4, 2005 BST-13 Demo rule splitting: Example 3 (* descending_urgency = "r1, r2" *) // Moving packets from input FIFO i1 rule r1; Tin x = i1.first(); if (dest(x)== 1) o1.enq(x); else o2.enq(x); i1.deq(); if (interesting(x)) c <= c + 1; endrule // Moving packets from input FIFO i2 r...
https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/01e06522cd4ab035dbbbd38279045351_t03_bluespec.pdf
rule urgency Sometimes, an urgency warning or a conflict can be due to a mistake or oversight by the designer (cid:132) A rule may accidentally include an action which shouldn’t be there (cid:132) A rule may accidentally write to the wrong state element (cid:132) A rule predicate might be missing an expression which ...
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synthesized The attributes apply to all methods in the interface March 4, 2005 BST-21 always_ready This attribute has two effects: Asserts that the ready signal for all methods is True (cid:132) It is an error if the tool cannot prove this Removes the associated port in the generated RTL module (cid:132) Any users o...
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1; endrule: cycle method Action start (Tin d_init, Tin r_init) if (r == 0); d <= zeroExtend(d_init); r <= r_init; product <= 0; endmethod method Tout result () if (r == 0); return product; endmethod endmodule: mkMult1 March 4, 2005 BST-26 13 Test bench for Example 1 module mkTest (Empty); // arrays a, b contain the nu...
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Lecture 23: Fault­Tolerant Quantum Computation Scribed by: Jonathan Hodges Department of Nuclear Engineering, MIT December 4, 2003 1 Introduction Before von Neumann proposed classical fault­tolerance in the 1940’s, it was assumed that a compu­ tational device comprised of more than 106 components could not perfor...
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the strings received are 1110101 on a and 0111111 on b. If one performs a bit­wise AND on each successive bit of the bit strings a and b, the result is 0110101. Taking “triples” of bits of this resulting addition, one performs a majority vote. Thus, if one bit has an error probablity of p , two bits in a triple bein...
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a 1 P. Shor – 18.435/2.111 Quantum Computation – Lecture 23 2 specific point, checking the result, then starting the computation again. The probablistic nature of quantum mechanics and the no­cloning theorem make this technique useless for QC. Classically, one might make many copies of the computation to perform a...
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(Kitaev­Solovay Theorem). Given a set of gates on SU(2) ( or SU(k) generally) that generates a dense set in SU(2), then any gate U ∈ SU(2) can be approximated to � using O (logc 1 � ) gates where 1 ≤ c ≤ 2. See Appendix 3 of Nielsen and Chuang for more details. 3.1 Fault Tolerance of σx In order to show that a σx ...
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one of the qubits, where the errors on each qubit are uncorrelated. Then this code will be able to correct for these errors, resulting in a quantum error correcting code and operation that performs σx with fault­tolerance. | | 3.2 Fault Tolerance of σz By using the Steane code above, the equivalent of σz on an unen...
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fault tolerance of the Hadamard gate under this CSS encoding can be seen under the additional ⊥. If the function E(x) represents the act of encoding the bit, the action of a constraint C1 = C2 Hadamard on an encoding qubit must follow the transformations: 1 E(|0�) −→ √ 2 1 E(|1�) −→ √ 2 (E( 0�) + E( 1�)) | | (E( ...
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), thus appropriately adding a phase to the states in the code that are E( 1�). | | | | 3.4 Fault Tolerance of CNOT Gate The σx, σz, and H gates can all be performed on a single encoded qubit with fault­tolerance because these gates are always applied to single qubits. Likewise, given two single­qubit encoded states,...
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added to all states y�, which is just the encoding E( 1�). If an error occurs in any of the two­qubit CNOT operations, this will result in va or vb not being all 0’s or all 1’s, and the CSS code will correct the the appropriate state. | | 4 Error Correction With Fault­Tolerant Precision Both classical and quantum e...
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in time due to the back­action of the CNOT. The stringent requirement of each error not affecting more than a single qubit (or pair in the FT CNOT construction) is not fulfilled. Using the idea of single failure points between qubits, as seen in the FT CNOT construction, we start our parity check register on k qubits ...
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superposition of the states |0� and 1�. The Hadamard transform of this state: | | H ⊗k(α 0� + β 1�) = ( | | α + β √ 2 ) 1 2k−1 � s∈even |s� + ( α − β )√ 2 1 2k−1 � s∈odd |s� (10) Thus if α = β, the state is all zeros and no backaction will occur. The all ones state simply adds the ones vector to the qubits. ...
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18.445 Introduction to Stochastic Processes Lecture 13: Countable state space chains 2 Hao Wu MIT 1 April 2015 Hao Wu (MIT) 18.445 1 April 2015 1 / 5 Recall Suppose that P is irreducible. The Markov chain is recurrent if and only if Px [τ + x < ∞] = 1, for some x. The Markov chain is positive recurrent if and only if ...
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. Theorem Suppose that the Markov chain is irreducible and aperiodic. If the chain is positive recurrent, then lim ||Pn(x, ·) − π||TV = 0. n Hao Wu (MIT) 18.445 1 April 2015 5 / 5 MIT OpenCourseWare http://ocw.mit.edu 18.445 Introduction to Stochastic Processes Spring 2015 For information about citing these materials ...
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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 32: Exam 4 Review 18.01 Fall 2006 Exam 4 Review 1. Trig substitution and trig integrals. 2. Partial fractions. 3. Integrati...
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find C, C = (−2)2 − 2 + 1 (−2 − 1)2 = 1 3 (x → −2) To find A, one method is to plug in the easiest value of x other than the ones we already used (x = 1, −2). Usually, we use x = 0. 1 A (−1)2(2) −1 = + 1 (−1)2 + 1/3 2 and then solve to find A. The Review Sheet handed out during lecture follows on the next p...
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required to cover every case systematically, namely, completing the square1 and long division.2 3. Integration by parts: � b a b � � uv�dx = uv � � � a � b − a u�vdx This is used when u�v is simpler than uv�. (This is often the case if u� is simpler than u.) 4. Arclength: ds = dx2 + dy2. Depending on wheth...
https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/0202fa3893049a6a502c4f7079eca657_exam4_review.pdf
3. 2Long division is used when the degree of P is greater than or equal to the degree of Q. It expresses P (x)/Q(x) = P1(x) + R(x)/Q(x) with P1 a quotient polynomial (easy to integrate) and R a remainder. The key point is that the remainder R has degree less than Q, so R/Q can be split into partial fractions. 3 � ...
https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/0202fa3893049a6a502c4f7079eca657_exam4_review.pdf
x + 1 (To carry out this long division, do not factor the denominator Q(x) = x2 − 2x + 1, just leave it alone.) The quotient x + 2 is a polynomial and is easy to integrate. The remainder term 3x − 2 (x − 1)2 has a numerator 3x − 2 of degree 1 which is less than the degree 2 of the denominator (x − 1)2 . Therefore...
https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/0202fa3893049a6a502c4f7079eca657_exam4_review.pdf
out using the trigonometric substitution u = k tan θ du = k sec2 θdθ. This then leads to sec-tan integrals, and the actual computation for large values of n are long. There are also other cases that we will not cover systematically. Examples are below: 1. If Q(x) = (x − a)m(x − b)n, then the expression is A1 x − a...
https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/0202fa3893049a6a502c4f7079eca657_exam4_review.pdf
2.160 Identification, Estimation, and Learning Lecture Notes No. 3 February 15, 2006 2.3 Physical Meaning of Matrix P The Recursive Least Squares (RLS) algorithm updates the parameter vector t y ) in such a way that the overall squared error may ( ˆ θ(t − 1) based on new data ϕT (t ), be minimal. This is done by...
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has all real eigenvalues. The eigen vectors associated with the individual eigenvalues are also real. Therefore, the matrix ΦΦT can be reduced to a diagonal matrix using a coordinate transformation, i.e. using the eigen vectors as the bases. λ 0 L 0   1   0 M  ∈ R mxm ΦΦT ⇒ D =   M      0 L λm ...
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input data: ϕ(i)L. This direction has been well explored, well excited. Although new data are obtained, the correction to the parameter vector θ(t − 1) is small, if the new input data ϕ(t) is in the same direction as that of λ . See the second figure above. ˆ max The above observations are summarized as follows: 1) ...
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of θ (0) and Po , the (best) estimation thereafter will be different. ˆ Question: How do the initial conditions influence the estimate? The following theorem shows exactly how the RLS algorithm works, given initial conditions. Theorem The Recursive Least Squares (RLS) algorithm minimizes the following cost functio...
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   t − 1  t y )ϕ (t ) + ∑ i y ) ϕ ϕ (i ) T = Pt  (  i = 1 ( )− 1 (i ) + P θ (0) 0    (23) (24) 3 − 1 ˆ − 1) Pt θ(t − 1 − 1 = ϕ ϕ T (t ) − 1 (t ) + P t − 1 Recall Pt ) t [ − 1 ) 1) θ (t ) = P P θ (t + − P ϕ )[ ( t ( t y 1) + − Postmultiplying ϕ (t ) to both sides of (14) ) =θ (t ...
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) −ϕ (t )θ (t − 1)) ( ˆ T (25) (26) (18) Q.E.D. Discussion on the Theorem of RLS ) θ (t ) = arg min θ t     1 ∑ ( i y ) − ϕ (i )θ) (  2 i = 1  1 4 4 42 4 4 43  Squared estimation error   A T 2 + 1 2 1 Weighted squared ) (0)) P 0 ( θ θ ( θ θ ) − − 4 4 4 4 4 4 4 4 from (0)) 3 ) − ...
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RLS theorem from the batch processing case of Pt − 1 = ∑ ϕ ϕ (i ) T t i = 1 − 1 P t = ∑ ϕ ϕ (i ) T − 1 (i ) + P o t i = 1 (i ) to: (28) 4 Other important properties of RLS include: • Convergence of θ (t ) . It can be shown that lim ˆ(t ) θ ˆ θ (t 1) − − t →∞ = 0 See Goodwin and Sin’s book, Ch.3, f...
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) T 1   t ) ( P Pt − 1 ϕ ϕ t − 1 α Pt − 1 − ( ϕ α  t ϕ (t )) + T ) t − 1 ( P  Exercise: Obtain (34) and (35) from (32) and (33). t ϕ (t ))( Pt =  t y ) −ϕ (t )θ (t − 1)) ( ˆ T (32) (33) (34) (35) A drawback of the forgetting factor approach When the system under consideration enters “steady stat...
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ning the whole m -dim space Set P0 =I (the m xm identity matrix) and θ(0) arbitrary Compute ˆ ˆ θ (t ) = θ (t ˆ − 1) + Pt − 1ϕ (t ) t ϕ (t ) ϕ ( P ) t − 1 T t y ) −ϕ (t )θ (t − 1)) ( ( ˆ T (38) (39) where matrix Pt-1 is updated with the same recursive formula as RLS Note that +1 involved in the denomin...
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 e 1    r te ) =  M  = (  e    Consider that each squared error is weighted differently, or Weighted Multi-Output Squared Error: ) r( r T J t (θ ) = ∑ e ( ieWi ˆ θ (t ) = min r ) =∑ ( T ( iy ) Ψ− θ ) W ( r( iy ) Ψ− θ ) ˆ θ (t ) i = 1 T T t t Β Ρ = t t J (θ ) t i = 1 arg θ =Ρ t T t...
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2.160 Identification, Estimation, and Learning Lecture Notes No. 1 February 8, 2006 Mathematical models of real-world systems are often too difficult to build based on first principles alone. Figure by MIT OCW. System Ident cation; “Let the data speak about the system”. ifi Figure by MIT OCW. Image removed fo...
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structure 3. Not available until an actual system has been built 2 Introduction: System Identification in a Nutshell b 3 b 2 b 1 u(t ) y(t ) FIR Finite Impulse Response Model t y ) = ( ( t u b 1 − 1) + ( t u b 2 − 2) ⋅ ⋅ ⋅ + t u − m ) ⋅ ⋅ ⋅ + b m ( Define θ := [b , b , 2 1 ]T m ⋅ ⋅ ⋅ , b ∈ R m ϕ...
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)θ )(−ϕ ) = 0 ( T ∑ t y )ϕ (t ) = ∑ (ϕ (t )θ )ϕ (t ) ( T N t = 1 3 N  ∑ (ϕ(t )ϕ (t )  t =1 T N  θ = ∑ t y )ϕ(t ) (   t =1 = RN ∴ θN = RN ∑ t y )ϕ(t ) −1 ( ˆ N t =1 Question1 What will happen if we repeat the experiment and obtain θˆ again? N Consider the expectation of θN when the experiment is r...
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∑ϕ( −1 ˆ ) ( t e ) t N t =1 Taking expectation E [θ −θ ] = E RN ∑ϕ( −1 0 ˆ N   N t =1  ) t e t ) (  −1 = RN ∑ϕ(t ) ⋅ E [ N t =1 ( t e )] = 0 Question2 Since the true parameter θ is unknown, how do we know how close 0 ˆ N will be toθ ? How many data points, N , do we need to reduce the errorθ −...
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t e ( ) ( s e t )  t = 1 s = 1 T )]ϕ (s )  R − 1  N Assume that {e(t)} is stochastically independent [ ( ) ( t e E s e )] =  )] = 0 s e t e ( ) ( 2 t ≠ s  E [  E [e (t )] = λ t = s Then PN = RN ∑ϕ (t )λϕ (t )RN = λRN − 1 T − 1 N − 1   t = 1   As N increases, RN tends to blow out, but...
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. The convergence of ˆ θN to θ0 may be accelerated if we design inputs such that R is large. IV. The covariance does not depend on the average of the input signal. Only the second moment What will be addressed in 2.160? A) How to best estimate the parameters What type of input is maximally informative? • Inform...
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Introduction to C++ Massachusetts Institute of Technology January 12, 2011 6.096 Lecture 5 Notes: Pointers 1 Background 1.1 Variables and Memory When you declare a variable, the computer associates the variable name with a particular location in memory and stores a value there. When you refer to the variable b...
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ffer­ ent memory locations • Use polymorphism – calling functions on data without knowing exactly what kind of data it is (more on this in Lectures 7-8) 2 Pointers and their Behavior 2.1 The Nature of Pointers Pointers are just variables storing integers – but those integers happen to be memory ad­ dresses, usuall...
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a variable of type data type * – a “pointer to a data type value.” 2.2.2 Using Pointer Values Once a pointer is declared, we can dereference it with the * operator to access its value: cout << * ptr ; // Prints the value pointed to by ptr , // which in the above example would be x ’s value We can use deferenced poin...
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keyword can be placed within a pointer variable declaration. This is because there are two different variables whose values you might want to forbid changing: the pointer itself and the value it points to. const int * ptr ; declares a changeable pointer to a constant integer. The integer value cannot be changed thro...
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pointer the function returns is invalid. As with any other variable, the value of a pointer is undefined until it is initialized, so it may be invalid. 3 References When we write void f(int &x) {...} and call f(y), the reference variable x becomes another name – an alias – for the value of y in memory. We can decla...
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3.1 The Many Faces of * and & The usage of the * and & operators with pointers/references can be confusing. The * operator is used in two different ways: 1. When declaring a pointer, * is placed before the variable name to indicate that the variable being declared is a pointer – say, a pointer to an int or char, not...
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of using subtraction and addition of pointers to move around between locations in memory, typically between array elements. Adding an integer n to a pointer produces a new pointer pointing to n positions further down in memory. 4.1.1 Pointer Step Size Take the following code snippet: 1 2 3 long arr [] = {6 , l...
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notation, in which you explicitly add your offset to the pointer and dereference the resulting address. For instance, an alternate and functionally identical way to express myArray[3] is *(myArray + 3). 4.2 char * Strings You should now be able to see why the type of a string value is char *: a string is actually a...
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2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 5 - The Finite Element Formulation Prof. K. J. Bathe MIT OpenCourseWare In this system, (X, Y, Z) is the global coordinate system, and (x, y, z) is the local coordinate system for the element i. We want to satisfy the following equations...
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uˆ = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ u1 v1 w1 . . . uN vN wN ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ N is the number of nodes (3N = n) and H is the displacement interpolation matrix. For the moment, let’s assume Su = 0. We use � uˆT = u1 u2 u3 . . . un � Then, we obtain We also assume ε (m) = B (m) uˆ n×1 6×n 6×1 = H (m)u ¯ˆ (m)u¯ ...
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V (m)+ Σ Σ � H Si(m)T f Si(m) f f dSi(m) f � m i Si(m) f uˆ is the unknown to be found. When evaluated on Sf i(m) , i(m) u¯Sf = H Sf i(m) u ¯ˆ With the transformed equation above, we can insert the following identity matrices: H Si(m) f = H (m) � � Sf i(m) � Let u ¯ˆ T = Then u ¯ˆ T = Then u¯ˆ = T ...
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T = [ u1 u2 u3 ]. We want to find: u(1)(x) = H (1) ⎢ ⎡ ⎡ ⎤ ⎤ u1 u1 ⎦ ; u(2)(x) = H (2) ⎢ ⎣ u2 ⎥ ⎣ u2 ⎥ ⎦ u3 u3 3 MIT OpenCourseWare http://ocw.mit.edu 2.092 / 2.093 Finite Element Analysis of Solids and Fluids I Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mi...
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Lecture 1 8.821/8.871 Holographic duality Fall 2014 8.821/8.871 Holographic duality MIT OpenCourseWare Lecture Notes Hong Liu, Fall 2014 Lecture 1 1: HINTS FOR HOLOGRAPHY In this chapter, we will get a favor of the holographic duality. We first study gravity system and derive black hole thermodynamics where holography p...
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in a fixed spacetime (2) The two sides should be be considered as different descriptions of the same quantum system. This duality provides a ”unification”, which has far-reaching implications for both sides of the equation. Maldacena’s original paper has been cited by more than 10,000 times in SLAC database. But the subje...
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itten [1]. 2 Theorem 1 : A theory that allows the construction of a Lorentz-covariant conserved 4-vector current J µ cannot ´ contain massless particles of spin > 1 with non-vanishing values of the conceived charge 2 J 0d3x. Theorem 2 : A theory that allows a conserved Lorentz-covariant stress tensor T µν cannot contai...
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j(helicity) ˆR(θ, k)|k, σ(cid:105) = eiσθ|k, σ(cid:105) ˆ ˆ ˆ where R(θ, k) is the rotational operator by an angle θ around k = k . More about representations of Poincare group can be found in Ref. [2]. The conserved, Lorentz-covariant current is J µ, with the conserved charge ˆQ = J 0d3x; the conserved, Lorentz-covari...
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when looking at 0-component of Eq. (7), we have (cid:104)k, σ|J 0|k(cid:48), σ(cid:105) −k−→−k→ q . (cid:48) (2π)3 For massless particles, k2 = k(cid:48)2 = 0. This implies that kµkµ such that k + k(cid:48) = 0 and kµ = (E, 0, 0, E), k(cid:48)µ = (E, 0, 0, −E). In this frame, a rotation by θ around the z-axis has the e...
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= Λµ λ(θ)(cid:104)k(cid:48), j|T ρλ|k, j(cid:105) ρ (θ)Λν ˆ ν only has eigenvalues e±iθ, 1, thus (cid:104)k(cid:48), j|R−1(θ)J µR(θ)|k, j(cid:105) can only be nonzero if j (cid:54) 1 . Otherwise, 2 , j|R−1(θ)T µν ˆR(θ)|k, j(cid:105) can only be nonzero if j (cid:54) 1. Otherwise, Eq. (8) is (13) ˆ Note that Λµ Eq. (7) ...
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Consequence of Electrons as Waves on Free Electron Model • Boundary conditions will produce quantized energies for all free electrons in the material • Two electrons with same spin can not occupy same electron energy (Pauli exclusion principle) Imagine 1-D crystal for now Traveling wave picture Standing wave pic...
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dk dE L π h = 2m = 2k πh dE dk 1 − 2 E 2mE h g(E)=density of states=number of electrons per energy per length n = 2k N = L π F = 2 2mEF hπ or kF = nπ 2 •n=is the number of electrons per unit length, and is determined by the crystal structure and valence •The electron density, n, determines the energy ...
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Rb Cs Cu Ag Au Be Mg Ca Sr Ba Nb Fe Mn Zn Cd Hg 3.25 3.93 4.86 5.20 5.62 2.67 3.02 3.01 1.87 2.66 3.27 3.57 3.71 3.07 2.12 2.14 2.30 2.59 2.65 eF 4.74 eV 3.24 2.12 1.85 1.59 7.00 5.49 5.53 14.3 7.08 4.69 3.93 3.64 5.32 11.1 10.9 9.47 7.47 7.13 TF 5.51 x 104 K 3.77 2.46 2.15 1.84 8.16 6.38 6.42 16.6 8.23 5.44 4.57 4.23 ...
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2.07 2.19 2.41 2.48 2.22 2.30 2.25 2.14 13.6 12.1 10.0 9.46 11.8 11.0 11.5 12.7 Fermi energies, fermi temperatures, fermi waves vectors, and fermi velocities for representative metals* * The table entries are calculated from the values of rs / a0 given in Table 1.1 using m = 9.11 x 10-28 grams. 1.75 1.66 1.51 1.46 1.64...
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e- in a metal): cv = ⎜ ⎛ ∂U ⎞ ⎟ ⎝ ∂T ⎠v U ~ ΔE ⋅ ΔN ~ kbT ⋅[g(EF )⋅ kbT ] ~ g(EF )⋅(kbT )2 U=total energy of electrons in system ⎛ cv = ⎜ ⎝ U ∂ T ∂ ⎞ ⎟ ⎠v ©1999 E.A. Fitzgerald = 2 ⋅ g(EF ) ⋅ kb 2T Right dependence, very close to exact derivation 9 Heat Capacity (cv) of electrons in Metal • Rough deri...
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6.087 Lecture 8 – January 21, 2010 Review Pointers Void pointers Function pointers Hash table 1 Review:Pointers • pointers: int x; int∗ p=&x; • pointers to pointer: int x; int∗ p=&x;int∗∗ pp=&p; • Array of pointers: char∗ names[]={"abba","u2"}; • Multidimensional arrays: int x [20][20]; 1 Review: Stacks • ...
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("%d",∗p); /∗ invalid ∗/ void∗ p; int ∗px=(int∗)p; printf ("%d",∗px); /∗valid ∗/ 5 Function pointers • In some programming languages, functions are first class variables (can be passed to functions, returned from functions etc.). • In C, function itself is not a variable. But it is possible to declare pointer to...
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void ∗ pa , void ∗ pb ) { r e t u r n ( ∗ ( i n t ∗ ) pb − ∗ ( i n t ∗ ) pa ) ; i n ascending o r d e r ∗ / } / ∗ s o r t q s o r t ( a r r , s i z e o f ( a r r ) / s i z e o f ( i n t ) , s i z e o f ( i n t ) , asc ) ; / ∗ s o r t q s o r t ( a r r , s i z e o f ( a r r ) / s i z e o f ( i n t ) , s i z e o ...
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void ∗ arg ) { s t r u c t node∗ np =( s t r u c t node ∗ ) p ; p r i n t f ( "%d " , np−>data ) ; } a p p l y ( phead , p r i n t , NULL ) ; 10 Callback (cont.) Counting nodes: void d o t o t a l ( void ∗ p , void ∗ arg ) { s t r u c t node∗ np =( s t r u c t node ∗ ) p ; i n t ∗ p t o t a l ∗ p t o t a ...
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drawfn f p [ 4 ] = {& draw_square ,& draw_rec ,& d r a w _ c i r c l e ,& draw_poly } ; void draw ( s t r u c t shape∗ ps ) { ( ∗ f p [ ps−>t y p e ] ) ( ps ) ; / ∗ c a l l t h e c o r r e c t f u n c t i o n ∗ / } 13 6.087 Lecture 8 – January 21, 2010 Review Pointers Void pointers Function pointers Hash tab...
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MULTIPLIER 31 s t r u c t wordrec { char ∗ word ; unsigned long count ; s t r u c t wordrec ∗ n e x t ; } ; / ∗ hash bucket ∗ / s t r u c t wordrec ∗ t a b l e [ MAX_LEN ] ; 17 Hash table: example unsigned long h a s h s t r i n g ( const char ∗ s t r ) { unsigned long hash =0; while ( ∗ s t r ) { } hash=...
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7. Rational Cherednik algebras and Hecke algebras for varieties with group actions 7.1. Twisted differential operators. Let us recall the theory of twisted differential oper­ ators (see [BB], section 2). Let X be a smooth affine algebraic variety over C. Given a closed 2-form ω on X, the algebra Dω(X) of differential op...
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2,cl, given by the De Rham differential acting from 1-forms to closed plex of sheaves Ω1 X 2-forms (sitting in degrees 1 and 2, respectively). If X is projective then this space is isomorphic to H2,0(X, C) ⊕ H1,1(X, C). We refer the reader to [BB], Section 2, for details. ΩX → X X Remark 7.1. One can show that Dω...
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OX (Z) ⊗OX → We have an exact sequence of OX -modules: 0 → OX → OX (Z) → i∗N φ − → 0 Thus we have a natural surjective map of OX -modules ξZ : T X → OX (Z)/OX . 7.3. The Cherednik algebra of a variety with a finite group action. We will now generalize the definition of Ht,c(G, h) to the global case. Let X be an affi...
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fficients. Definition 7.2. A Dunkl-Opdam operator for (X, G) is an element of Dω/t(X)r[c] given by the formula (7.1) D := tLv − · fY (x) · (1 − g), � 2c(Y, g) 1 − λY,g (Y,g)∈S where λY,g is the eigenvalue of g on the conormal bundle to Y , v ∈ Γ(X, T X) is a vector field on X, and fY ∈ OX (Z) is an element of the ...
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and G is a subgroup in GL(h). Let v be a constant vector field, and let fY (x) = (αY , v)/αY (x), where αY ∈ h∗ is a nonzero functional vanishing on Y . Then the operator D is just the usual Dunkl-Opdam operator Dv in the complex reflection case (see Section 2.5). This implies that all the Dunkl-Opdam operators in the...
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(the principal symbol homomorphism). The homomorphism ψ is clearly surjective, and our job is to show that it is injective (this is the nontrivial part of the proof). In each degree, ψ is a morphism of finitely generated OG -modules. Therefore, to check its injectivity, it suffices to check the injectivity on the form...
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� has been accomplished already. We are done. Remark 7.9. The following remark is meant to clarify the proof of Theorem 7.7. In the case X = h, the proof of Theorem 7.7 is based, essentially, on the (fairly nontrivial) fact that the usual Dunkl-Opdam operators Dv commute with each other. It is therefore very import...
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by affine G-invariant open sets. Then the quotient variety X/G exists. For any affine open set U in X/G, let U � be the preimage of U in X. Then we can define the algebra Ht,c,0(G, U �) as above. If U ⊂ V , we have an obvious restriction map Ht,c,0(G, V �) Ht,c,0(G, U �). The gluing axiom is clearly satisfied. Thus the c...
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relatively prime to the characteristic. 51 � (2) The construction and main properties of the (sheaves of) Cherednik algebras of alge­ braic varieties can be extended without significant changes to the case when X is a complex analytic manifold, and G is not necessarily finite but acts properly discon­ tinuously. In t...
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1 = z, z2, . . . , zd near y. A Dunkl-Opdam operator near y for the vector field ∂z can be written in the form � 1 n−1 2c(Y, gm) m ( z m=1 Conjugating this operator by the formal expression zη(Y ) := (zm)η(Y )/m, we get 1 − λm (g − 1) + η(Y )). ∂ ∂z D = + Y,g ∂ z η(Y ) ◦ D ◦ z−η(Y ) = ∂ ∂z + n−1 1 � 2c(Y,...
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such Y will be called a reflection hypersurface). For (Y, g) ∈ S, let GY be the subgroup of G whose elements act trivially on Y . This group is obviously cyclic; let nY = |GY |. Let CY be the conjugacy class in BG corresponding to a small circle going counterclockwise around the image of Y in X/G, and TY be a represe...
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we will sometimes drop x0 form the notation. The main result of this section is the following theorem. Theorem 7.15. Assume that H2(X, C) = 0. Then A = Hτ (G, X) is a flat formal defor­ mation of A0, which means A = A0[[τ ]] as a module over C[[τ ]]. Example 7.16. Let h be a finite dimensional vector space, and G be a...
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C+ be a complex number with a positive imaginary part, and G�� = G � (Q∨ ⊕ ηQ∨) be the double affine Weyl group. Then Hτ (h, G��) is (one of the versions of) the double affine Hecke algebra of Cherednik ([Ch]), and it is flat by Theorem 7.15. The fact that this algebra is flat was proved by Cherednik, Sahi, Noumi, Stokma...
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, k = 1, . . . , nj . Define the Hecke algebra Hτ (Σ) of Σ to be generated over C[[τ ]] by the same generators al, bl, cj with defining relations l nj � (cj − e 2πji/nj e τkj ) = 0, c1c2 · · · cm = � albla−1b−1 . l l k=1 l Thus Hτ (Σ) is a deformation of C[Γ]. This deformation is flat if H is a Euclidean plane ...
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see that Hτ (Σ) fails to be flat in the following “forbidden” cases: g = 0, m = 2, (n1, n2) = (n, n); m = 3, (n1, n2, n3) = (2, 2, n), (2, 3, 3), (2, 3, 4), (2, 3, 5). Indeed, the orbifold Euler characteristic of a closed surface Σ of genus g with m special points x1, . . . , xm whose orders are n1, . . . , nm is χ...
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1 → π1(X, x) → orb (X/G, x) G → → π1 1. 54 � � 7.8. Hecke algebras of wallpaper groups and del Pezzo surfaces. The case when H is the Euclidean plane (i.e., Γ is a wallpaper group) deserves special attention. If there are elliptic elements, this reduces to the following configurations: g = 0 and m = 3, (n1, n2...
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Γ, H) be the projector to an eigenspace of c. Consider the “spherical” subalgebra Bτ (Γ, H) := eHτ (Γ, H)e. � · Theorem 7.19 (Etingof, Oblomkov, Rains, [EOR]). (i) If � = 0 then the algebra Bτ (Γ, H) is commutative, and its spectrum is an affine del Pezzo surface. More precisely, in the case (2, 2, 2, 2), it is a ...
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H N with the action of ΓN = SN �ΓN . If H is a Euclidean or Lobachevsky plane, then by Theorem 7.15 Hτ (ΓN , X N ) is a flat deformation of the group algebra C[ΓN ]. If N > 1, this algebra has one more essential parameter than for N = 1 (corresponding to reflections in SN ). In the Euclidean case, one expects that an...
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let M be a module over H1,c,η,0,G,X which is a locally free coherent sheaf when restricted to X �/G. Then the restriction of M to X �/G is a G-equivariant D-module on X � which is coherent and locally free as an O-module. Thus, M corresponds to a locally constant sheaf (local system) on X �/G, which gives rise to a ...
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Y − η(Y ))/nY . Proposition 7.22. The functor KZ maps the category Cc,η to the category of representations of the algebra Hτ (c,η)(G, X). Proof. The result follows from the corresponding result in the linear case (which we have already proved) by restricting M to the union of G-translates of a neighborhood of a ge...
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o­ sition 7.22 and it deforms flatly the module regG. This implies Hτ (c,η)(G, X) is flat over C[[τ ]]. Remark 7.23. When X is not simply connected, the theorem is still true under the as­ sumption π2(X) ⊗ C = 0 (i.e. H2( � X� is the universal cover of X), and the proof is contained in [E1]. X, C) = 0, where 56 7....
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1 double affine Hecke algebra. If we set three of the four Ti’s 2 = 1, we get the double affine Hecke algebra of type satisfying the undeformed relation Ti A1. More precisely, this algebra is generated by T1, . . . , T4 with relations T2 2 = T3 2 = T4 2 = 1, (T1 − t)(T1 + t−1) = 0, T1T2T3T4 = q. Another presentation ...
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2 = 1. The Hecke algebra of the partial braid group is then defined to be the group algebra of B plus an extra relation: (T − q1)(T + q2) = 0. A common way to present this Hecke algebra is to renormalize the generators so that one has the following relations: T XT = X −1, T −1Y T −1 = Y −1, Y −1X −1Y XT 2 = q, (T −...
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