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math/0007195 | For MATH, we compute MATH . By the mirror of this calculation and switching MATH and MATH, we obtain: MATH . But by REF , we have MATH . Hence, MATH, so that MATH. |
math/0007195 | REF follows from REF . To get REF , apply REF . Then, REF follows by using REF . |
math/0007195 | Let MATH. Then MATH . |
math/0007195 | For MATH, we compute MATH . |
math/0007195 | By the NAME equation, MATH. Hence, MATH, so that (by REF ) MATH. Now use REF . |
math/0007205 | Its self-adjointness follows from the form REF of MATH. Let us show that this operator is compact. We estimate the NAME norm of MATH: MATH . Thus MATH is a NAME operator. Hence, it is a compact operator CITE. To prove the positivity of MATH, consider the scalar product MATH as MATH. |
math/0007205 | Let us represent REF in the operator form in MATH where MATH has the form REF , and MATH, MATH. Due to the positivity of MATH the homogeneous equation MATH has only the trivial solution. Since MATH is a compact operator, then by the NAME theorem CITE inhomogeneous REF has a unique solution given by MATH with MATH. Due ... |
math/0007206 | Consider a variation MATH of MATH with fixed end-points MATH, MATH. Suppose MATH (yielding the second equation of motion) and MATH. For the variation of the action functional MATH one obtains, MATH . (Note that MATH.) This yields the first equation of motion. The total energy MATH is always a first integral; MATH and M... |
math/0007206 | This follows from the fact that for the dynamical constants MATH, MATH and MATH to be physically feasible, there has to be a value for MATH between MATH and MATH such that REF leads to real MATH. This means that the polynomial on the right hand side must be nonnegative somewhere in the interval MATH. But at MATH it tak... |
math/0007206 | First, the expressions for the logarithmic differentials of MATH are derived in the general case. The reduction to the case of spherical tops is then immediate. Finally, the analytic properties of the differentials are examined. Observe that MATH . This implies MATH such that MATH . By REF , the components of the angul... |
math/0007206 | First, project the vector MATH stereographically from the north pole of the unit sphere into the complex plane. The result is the image of MATH under the NAME transformation REF , that is, MATH. Since MATH, the fromula for MATH follows from REF : MATH . This proves the proposition. |
math/0007206 | The angle MATH is the argument of the curve MATH. It follows that MATH is the imaginary part of the logarithmic differential MATH. REF then follows from REF . |
math/0007206 | Loops occur, if MATH changes sign as MATH moves from MATH to MATH. It follows from REF that this happens if MATH has a solution for MATH in the interval MATH. That equation is equivalent to MATH . Since the left hand side of this equation is smaller than zero for MATH, this equation cannot be fulfilled if MATH and MATH... |
math/0007206 | REF imply that the pullback of the holomorphic differential MATH to the MATH-plane is MATH. The MATH-integral therefore becomes MATH . Since MATH, this implies MATH . The constant has to be MATH for MATH to vanish at MATH. |
math/0007206 | Consider the asymptotic behavior of the logarithmic differential of MATH: MATH . It has only two simple poles on MATH, one at MATH and the other at MATH. For MATH, the asymptotic expansion is MATH . To calculate the asymptotic expansion around the branchpoint MATH, introduce the local parameter MATH. The result is MATH... |
math/0007206 | Put REF in the form MATH and consider separately the integrals MATH and MATH . Substituting the uniformizing variable in the first integral, we get MATH . Using the formula MATH one obtains MATH . The last equality follows from MATH . Now the last integral can easily be solved since MATH. Note that MATH to obtain MATH ... |
math/0007206 | The formula for MATH and MATH follows elementarily from the corresponding integrals REF and the initial conditions. Now consider the logarithmic differential of MATH from REF : MATH . As one can read off from this expression, the differential has three simple poles away from infinity, one at MATH with residue MATH and ... |
math/0007207 | See for example, CITE . |
math/0007207 | We refer to CITE . |
math/0007207 | We refer to REF . |
math/0007207 | The estimate REF is a consequence of REF , if we take REF into account. By using the coercivity of MATH and the NAME inequality we get MATH . By REF we obtain, using the NAME inequality and the boundedness of MATH, MATH . For MATH small enough MATH and REF implies that MATH where MATH is independent of MATH. Therefore ... |
math/0007207 | We refer to REF . |
math/0007207 | By referring to the NAME estimates in REF the proof is analogous as the proof of REF . |
math/0007207 | Put MATH, MATH and MATH. Then we have MATH . For every MATH we denote by MATH the union of all closed cubes MATH such that MATH and MATH. For MATH we define the sets MATH and MATH . Further, we define MATH as the union of all closed cubes MATH with MATH and MATH, and we define MATH as the union of all closed cubes MATH... |
math/0007207 | Let us define MATH and MATH . We apply REF to obtain MATH . Now MATH approaches MATH and MATH tends to zero as MATH. Thus, REF follows by REF is proved. |
math/0007207 | By the strict monotonicity assumption it follows that MATH . Consequently, REF is proved if we can prove that MATH as MATH. The proof of REF will be splitted up into four steps. CASE: We start by showing that MATH . Let us write MATH where the last equality follows from REF . According to REF the map MATH is continuous... |
nlin/0007024 | We construct MATH by taking a quotient of a neighbourhood of the identity section MATH in MATH by a distribution MATH constructed from the linear system. Let MATH be a neighbourhood of a pole MATH not containing any of the other poles, and let MATH be a coordinate on MATH such that MATH at MATH. Then MATH in MATH, wher... |
nlin/0007024 | Under either of REF , the eigenvalues of MATH can be assumed to be distinct, since they are distinct at MATH and since we can, if necessary, replace MATH by a smaller neighbourhood. So we can find a holomorphic gauge transformation MATH such that MATH is the sum of a diagonal polynomial and a term that vanishes to orde... |
nlin/0007024 | Any twistor space is full at a singularity of rank MATH since any MATH satisfying REF is then holomorphic at MATH, and can therefore be generated by a holomorphic vector field in any twistor space. In the irregular case, we construct MATH from the `minimal' twistor space in REF , by cutting out and replacing a neighbou... |
nlin/0007024 | Choose a coordinate MATH on a small disc in MATH such that the pole is at MATH, and extend this to a neighbourhood of MATH in MATH so that MATH is given by MATH. Then we can identify a neighbourhood of MATH with MATH, as before. For each MATH, we have a holomorphic map MATH and hence an element MATH of MATH such that M... |
nlin/0007024 | Let MATH be solution to MATH with constant monodromy, and with constant connection matrices MATH to the special solutions at the poles. Let MATH be nearby points (neither a pole) and let MATH be close to the identity. Then, by integrating the action of MATH on MATH, we have two points MATH, MATH in MATH near MATH. Thes... |
nlin/0007024 | First we note that the Hamiltonians MATH generate the constant gauge transformations. Consider next the flow generated by MATH. We shall find the value of the Hamiltonian vector field at a point of MATH constructed from a global meromorphic REF-form MATH. To do this, we must we must evaluate the gradient of MATH at suc... |
nlin/0007024 | By fixing a base point (disjoint from the poles) and a frame at the base point, we can find a solution MATH for each MATH which depends smoothly on MATH. If the monodromy representation is constant, then we can find a matrix MATH for each MATH such that the monodromy matrices of MATH are constant. If we take MATH close... |
nlin/0007024 | We shall look at the proof in outline. Suppose that the deformation is isomonodromic. Let MATH be a solution to MATH, depending continuously on MATH and with constant monodromy (we have to keep in mind that MATH is multi-valued and singular at the poles). Let MATH be a disc containing a pole (at MATH) and put MATH wher... |
nlin/0007024 | From the definitions, MATH and, in variational notation, MATH . We must show that MATH is skew-symmetric, closed, and non-degenerate. From the first constraint, we have MATH. It follows that MATH . However MATH . The skew-symmetry follows. A similar calculation, starting from MATH shows that MATH is closed. To show tha... |
quant-ph/0007016 | Let MATH denote the worst-case number of comparisons required if MATH and MATH have domain of size MATH. We show that MATH for some (small) constant MATH. Let MATH and consider the subproblem MATH. Using at most MATH comparisons, we can find a claw in MATH with probability at least MATH, provided there is one. We do th... |
quant-ph/0007016 | To apply REF , we will describe a relation MATH for each of our problems. For functions MATH and MATH, we denote by MATH the cardinality of the set MATH. For each problem MATH will be defined by MATH for some appropriate sets MATH and MATH. CASE: Here we suppose that MATH divides MATH. Let MATH be the set of functions ... |
quant-ph/0007021 | We use the notation of REF. For any subset MATH, MATH, let us define MATH are linearly independent. Suppose there is a nontrivial linear combination MATH . Let MATH be a set of largest cardinality such that MATH and let MATH, MATH. We define a vector MATH . Applying MATH to the linear combination above, we have MATH . ... |
quant-ph/0007021 | Essentially, the same proof of REF goes through. Since the query scheme can make an error only if the element is present, we observe that the only vector in the linear combination REF that has a non-zero projection on the space MATH, is the vector MATH. Hence MATH, and the operators MATH continue to be linearly indepen... |
quant-ph/0007021 | Since we are looking at a one probe quantum scheme, MATH. We start by picking a family MATH of sets, MATH, MATH, MATH and MATH for all MATH. By picking the sets greedily CITE, one obtains a family MATH with MATH . Let MATH. Since, MATH, MATH . Hence, MATH are linearly independent. Suppose there is a non-trivial linear ... |
quant-ph/0007021 | The proof of this theorem is similar to the proof of REF . Pick a family MATH of sets, MATH, MATH, MATH, MATH for all MATH, such that MATH. One can prove that MATH, MATH, are linearly independent in exactly the same fashion as REF was proved. The difference is that MATH lie in a vector space of dimension at most MATH i... |
quant-ph/0007021 | For MATH, let MATH denote the function for query MATH, which maps bit strings of length MATH to MATH that is, MATH maps MATH to MATH iff the query scheme given query MATH and bit string MATH evaluates to MATH. Consider a mapping MATH that is, MATH takes a subset of the universe of size at most MATH to a function from b... |
quant-ph/0007021 | A proof very similar to that of REF goes through. We just observe that now the query scheme is a logical disjunction over a family of deterministic query schemes. If the query element is present in the set stored, there is a decision tree in this family that outputs REF. If the query element is not present in the set s... |
quant-ph/0007021 | Suppose there is a classical scheme which stores subsets of size MATH from a universe of size MATH using MATH bits of storage, and answers membership queries using one bit probe with two-sided error at most MATH. Define MATH. Since MATH, MATH. Therefore, MATH . We repeat the query scheme MATH times and accept only if m... |
quant-ph/0007021 | The proof of this theorem is similar to the proof of REF above. We repeat the query scheme MATH times and accept only if more than MATH trials accept. We ``derandomise" the new query scheme in a manner similar to what was done in the proof of REF . We thus get a deterministic query scheme making MATH bit probes and ans... |
quant-ph/0007036 | Since MATH for all MATH we have MATH . Let MATH be the MATH-dimensional vector which has entries indexed by strings MATH and which has MATH as its MATH-th entry. Note that the MATH norm MATH is MATH for all MATH . For any MATH let MATH be defined as MATH . The quantity MATH can be viewed as the total query magnitude wi... |
quant-ph/0007036 | Suppose that MATH is a quantum exact learning algorithm for MATH which makes at most MATH quantum membership queries. If we take MATH then REF implies that there is a set MATH of cardinality at most MATH such that for all MATH we have MATH . Let MATH be any two concepts in MATH . By REF , the probability that MATH outp... |
quant-ph/0007036 | Let MATH be a quantum network which learns MATH and has query complexity MATH . For all MATH we have the following: if MATH's oracle gates are MATH gates, then with probability at least MATH the output of MATH is a representation of a Boolean circuit MATH which computes MATH . Let MATH be all of the concepts in MATH an... |
quant-ph/0007036 | Suppose that MATH is not exact learnable from classical membership queries, that is, for any polynomial MATH there are infinitely many values of MATH such that any learning algorithm for MATH requires more than MATH queries in the worst case. By REF , this means that for any polynomial MATH there are infinitely many va... |
quant-ph/0007036 | If MATH is an eigenvalue of MATH which has corresponding eigenvector MATH then since MATH we have MATH . Without loss of generality we may assume that MATH so MATH for some MATH and MATH for MATH . Thus MATH and hence MATH is in the disk MATH . |
quant-ph/0007045 | We begin by deriving a more usable expression for MATH. MATH where we have used the fact that MATH is periodic of period MATH. Since MATH is one-to-one when restricted to its period MATH, all the kets MATH are mutually orthogonal. Hence, MATH . If MATH, then since MATH is a MATH-th root of unity, we have MATH . On the ... |
quant-ph/0007045 | We begin by noting that MATH where we have made use of the inequalities MATH . It immediately follows that MATH . As a result, we can legitimately use the inequality MATH to simplify the expression for MATH. Thus, MATH . The remaining case, MATH is left to the reader. |
quant-ph/0007045 | Since MATH we know that MATH which can be rewritten as MATH . But, since MATH, it follows that MATH . Finally, since MATH and hence MATH, the above theorem can be applied. Thus, MATH is a convergent of the continued fraction expansion of MATH. |
quant-ph/0007045 | From the above theorem, we know that MATH where MATH is a monotone decreasing sequence of positive reals converging to zero. Thus, MATH . |
quant-ph/0007045 | CASE: The probability of error MATH of finding the hidden state MATH is given by MATH where MATH where MATH is the function that rounds to the nearest integer. Hence, MATH . Thus, MATH CASE: The computational cost of the NAME transform MATH is MATH single qubit operations. The transformations MATH and MATH each carry a... |
quant-ph/0007060 | Since the NAME spaces MATH and MATH have the same (infinite) dimension, it follows from REF of CITE that MATH has a dense set of cyclic vectors in MATH. |
quant-ph/0007060 | Let MATH denote the closed linear span of MATH. Since the infinitesimal generator MATH of the group MATH is positive, NAME 's ``little NAME theorem" REF entails that MATH. However, MATH; that is, MATH is cyclic under operators NAME in MATH over all times (NAME REF). Therefore, MATH is cyclic for MATH. |
quant-ph/0007122 | We first note that the NAME MATH can be obtained as MATH . Therefore MATH. From this, we have MATH for any given MATH. Therefore MATH contains the maximal torus MATH. |
quant-ph/0007122 | For each rotation matrix MATH we easily verify that MATH . |
quant-ph/0007122 | This follows immediately from REF - REF . |
quant-ph/0007122 | We first quote the following fact CITE: For any MATH, there exists a collection of unitary matrices MATH, MATH, and a MATH such that MATH where MATH is a rotation involving MATH and MATH and satisfying REF . For the benefit of the reader and for the sake of self-containedness, we include a direct proof of REF in the Ap... |
quant-ph/0007122 | This is a basic fact which can be found in most basic algebra or group theory books. |
quant-ph/0007123 | Straightforward verification. |
quant-ph/0007123 | The action of the evolution dynamics MATH on the invariant subspace MATH is, by REF , MATH, the identity operator on MATH. Since the component of MATH in MATH is the zero vector, the action of MATH (the MATH identity matrix) on it remains zero for all MATH. |
quant-ph/0007123 | Solve REF for MATH: MATH . Substituting REF into REF , we obtain REF in bra-ket form. |
quant-ph/0007123 | It is obvious that REF holds. To see REF , we have MATH . |
quant-ph/0007123 | The representation REF follows easily from REF . To calculate MATH, write MATH analogous to [REF ], and apply REF to obtain REF ; or apply REF and the properties of the MATH generators commonly used in quantum mechanics [REF , p. REF]. |
quant-ph/0007123 | Use REF . |
quant-ph/0007123 | As in [REF ], we have MATH and, therefore MATH . Let MATH be the orthogonal projection of MATH onto MATH. Then MATH from which, we apply the NAME inequality and obtain MATH . Combining REF , we have established the left half of REF . Next, mimicking [REF - REF ], we have MATH and from MATH, by an application of the NAM... |
quant-ph/0007123 | We have, from REF , MATH . The conclusion follows. |
quant-ph/0007123 | Substituting REF into REF and simplifying, we obtain MATH . The proof follows. |
quant-ph/0007123 | Straightforward verification. |
quant-ph/0007123 | Use the matrix representation REF . |
quant-ph/0007123 | This is the major theorem in REF ; see REF and particularly REF therein. Note also the work by NAME who considered some measurement effects in REF . |
quant-ph/0007123 | By the very definition of MATH in REF , we know that MATH . Therefore REF follows. |
quant-ph/0007123 | By NAME 's formula, MATH we have MATH . |
quant-ph/0007123 | From REF , we have MATH . Now, applying REF , we obtain MATH . Substituting REF into REF , we obtain MATH . |
quant-ph/0007124 | For any MATH, denote MATH . CASE: We have, for MATH, MATH CASE: MATH . |
quant-ph/0007124 | It follows obviously from by REF . |
quant-ph/0007124 | Using REF , we have MATH . Again, from the definition of MATH in REF , we see that REF gives MATH . Therefore REF follows. |
cond-mat/0008422 | Define MATH by MATH . Since they satisfy the following properties: MATH the product MATH is written as MATH where MATH. The square norms of MATH are calculated as MATH . |
cond-mat/0008422 | For a reduced expression MATH, we have MATH . From the NAME REF , we calculate the scalar product as follows: MATH . |
cond-mat/0008422 | It is sufficient to require the conditions MATH and MATH in order to determine the coefficients MATH such that MATH. |
cond-mat/0008422 | The orthogonality for MATH is straightforward from that of the nonsymmetric NAME polynomials REF . We have MATH where the last equality follows from REF . |
cond-mat/0008422 | There exists a MATH-homomorphism MATH defined by MATH . Since MATH does not depend on MATH as REF , we have MATH . Thus we obtain the following relation: MATH . We show that the sum on the left-hand side of the above equation can be replaced by the sum on MATH. Consider the isotropy group MATH for the dominant weight M... |
cs/0008001 | Suppose there is such a cycle. Letting MATH be the vertex at one end of REF-edge, we can trace around the cycle, giving a sequence of vertices MATH, where MATH is the vertex at the other end of REF-edge. The assignment has MATH for MATH, and MATH, and hence it violates REF . Only If. Suppose the assignment violates a t... |
cs/0008001 | The ``if" portion of this proof is covered by REF . The ``only if" portion is proved by induction on the number of variables in the antecedent of the transitivity constraint (REF .) That is, assume a transitivity constraint containing MATH variables in the antecedent is violated and that all other violated constraints ... |
cs/0008001 | The ``if" portion of this proof is covered by REF . The ``only if" portion is proved by induction on the number of variables in the antecedent of the transitivity constraint (REF .) Assume a transitivity constraint with MATH variables is violated, and that no transitivity constraint with fewer variables in the antecede... |
cs/0008001 | We consider the case where MATH. The general statement of the proposition then holds by induction on MATH. Define assignment MATH to be: MATH . We consider two cases: CASE: If MATH, then any cycle in MATH through MATH must contain a REF-edge other than MATH. Hence adding this edge does not introduce any transitivity vi... |
cs/0008001 | We note that any cycle in MATH must be present in MATH and have the same edge labeling. Thus, if MATH has no cycle with a single REF-edge, then neither does MATH. |
cs/0008001 | Suppose that MATH is satisfiable, that is, there is some assignment MATH such that MATH. Then by REF we can find an assignment MATH such that MATH. Furthermore, since the construction of MATH by REF preserves the values assigned to all variables in MATH, and these are the only relational variables occurring in MATH, we... |
cs/0008001 | Our proof is an adaptation of a proof by CITE that MATH has a bisection bandwidth of at least MATH. That is, one would have to remove at least MATH edges to split the graph into two parts of equal size. Observe that MATH has MATH vertices and MATH edges. These edges are split so that MATH are in MATH and MATH are in MA... |
cs/0008001 | Classify the parity of face MATH as ``even" when MATH is even, and as ``odd" otherwise. Observe that no two faces of the same parity can have a common edge. Divide the set of split faces into two subsets: those with even parity and those with odd. Both of these subsets are edge independent, and one of them must have at... |
cs/0008001 | Suppose there is an edge-independent set of MATH split faces. For each split face, choose one edge in MATH and one edge in MATH bordering that face. For each value MATH, define assignment MATH (respectively, MATH), to the variables representing edges in MATH (respectively, MATH) as follows. For an edge MATH that is not... |
cs/0008001 | Partition the set of split faces into four sets: EE, EO, OE, and OO, where face MATH is assigned to a set according to the values of MATH and MATH: CASE: Both MATH and MATH are even. CASE: MATH is even and MATH is odd. CASE: MATH is odd and MATH is even. CASE: Both MATH and MATH are odd. Each of these sets is vertex in... |
cs/0008001 | For any ordering of the variables in MATH, partition them into two sets MATH and MATH such that those in MATH come before those in MATH, and such the number of variables that are in MATH are equally split between MATH and MATH. Suppose there is a vertex-independent set of MATH split faces. For each value MATH, we defin... |
cs/0008010 | REF illustrates the recursive construction of such a polygon, for all MATH of the form MATH. The shortest flipturn sequence for the polygon includes only diagonal flipturns and therefore has length MATH. Another sequence, which we believe to be the longest, requires twelve flipturns to remove every MATH vertices. REF i... |
cs/0008010 | When MATH is a multiple of MATH, the polygon consists of a horizontally symmetric rectangular `comb' with MATH `teeth'; if MATH is not a multiple of MATH, we add a small rectangular notch in a bottom corner of the polygon. See REF . (We consider a rectangle to be a comb with one tooth.) Both the teeth and the gaps betw... |
cs/0008010 | CASE: Suppose some corner of MATH is not a vertex of MATH. Some edge of MATH lies on a line separating the missing corner from the interior of MATH. This edge contains a diagonal lid. CASE: Without loss of generality, suppose MATH does not contain the top left and top right vertices of MATH. REF implies that MATH has a... |
cs/0008010 | We achieve the stated upper bound by performing an orthogonal extended flipturn only when no diagonal pockets are available. By REF , we are forced to perform an orthogonal flipturn on a polygon MATH if and only if all four corners of MATH are also vertices of MATH. Let MATH be a nonconvex orthogonal MATH-gon with boun... |
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