paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9901097 | We must show that MATH for some MATH. Set MATH. The sum is then MATH. By REF , this is congruent to MATH regardless of the parity of MATH. Since MATH is odd, MATH, and the result follows. |
math/9901097 | It is clear from REF that MATH, so it will suffice to show that MATH. Choose MATH, and let MATH. We will show by downward induction on MATH that MATH. Using the formula for MATH given in REF , we have MATH . Adding, we get MATH where the last step is REF . Now suppose MATH. By the formula for MATH, MATH. If MATH, then ... |
math/9901097 | Let MATH be the number of size MATH subsets MATH having MATH successions. It is well known (see CITE) that MATH . We see that MATH is the number of subsets of MATH of size MATH which contain the element MATH and have MATH successions. This collection of subsets consists of those size MATH subsets with MATH successions ... |
math/9901097 | We have MATH . The last equality is proven in CITE; it is an application of the NAME convolution formula. The result follows since MATH. Note that the summation from MATH to MATH is correct even when MATH is smaller than MATH because MATH for MATH. |
math/9901097 | Since the MATH's are disjoint, in order to prove the first claim, it suffices to show that MATH. It is obvious from the construction of MATH that MATH has the parity required of an element of MATH for MATH. It only remains to check that MATH is even if and only if MATH. Since MATH is odd, this is a consequence of the f... |
math/9901097 | Suppose that MATH, so that each box contains an even number of balls except for the last one. The MATH-th wall has MATH balls to its left, and since this is an even number, it is in a cell with an even index. Thus, MATH for MATH odd. In this case, MATH and MATH is injective because MATH is uniquely determined by MATH. ... |
math/9901097 | Suppose there is only a boundary ball on the right. Then there is only one outcome of the game; indeed, we have already seen in the proof of REF that each move consists of the ball shifting one box to the left. Now suppose the cell has a ball on each side. The MATH-th move is uniquely determined, and setting MATH and M... |
math/9901097 | Let MATH be the MATH-th column of MATH. Then MATH, and REF completes the proof. |
math/9901097 | We use induction on MATH. The claim is true for MATH, since MATH. Now take MATH and assume that the result holds for MATH. Then MATH where the last step is a version of the NAME convolution formula CITE. |
math/9901097 | We have MATH the last equality is a standard binomial identity CITE. |
math/9901097 | Let MATH be the MATH-th row of MATH. Recall from REF that the map MATH is a bijection of MATH onto MATH. From REF , we have MATH . |
math/9901097 | It only remains to check the final statement. The symplectic lattices of type MATH correspond to MATH, so MATH, MATH, and MATH. Substituting, we obtain another proof of REF . The full affine flag manifold is isomorphic to MATH. In this case, MATH, MATH, and MATH for MATH, and the result follows. |
math/9901097 | The inequality follows immediately from the fact that MATH is a subset of the finite set MATH. These sets are equal when MATH because MATH; thus the formulas for the NAME characteristics of the affine flag varieties of MATH given in REF agree with those given in REF . Also, if MATH, then the NAME characteristics are MA... |
math/9901097 | Let MATH correspond to MATH, so that setting MATH and MATH, we have MATH. We show that the nilpotent map MATH induced on MATH has NAME block structure given by MATH. It is clear from the definition that MATH is the string of MATH balls and MATH walls with MATH walls in the MATH-th cell for MATH and the other cells empt... |
math/9901097 | The proof is similar to the first part of REF . |
math/9901101 | Straightforward calculations show that MATH is a nondegenerate NAME MATH-family, so the universal property gives a nondegenerate homomorphism MATH which satisfies REF implies that MATH is injective: take MATH, where MATH is the gauge action of MATH on MATH. It follows from REF that the coaction identity MATH holds on g... |
math/9901101 | We use the calculus of CITE to handle elements of the crossed product MATH. For each MATH, let MATH denote the corresponding spectral subspace; we write MATH to denote a generic element of MATH. (This subscript convention conflicts with the standard notation for NAME families: each partial isometry MATH is in MATH, and... |
math/9901101 | The corollary follows from REF and NAME 's duality theorem CITE. (NAME though we are using full coactions, NAME 's theorem still applies: the regular representation is an isomorphism of MATH onto the (reduced) crossed product by the reduction of MATH; see CITE). |
math/9901101 | Since MATH is discrete, MATH is generated by the set of products MATH, where MATH is a nondegenerate NAME MATH-family and MATH is the canonical homomorphism of MATH into MATH satisfying MATH . Moreover, the crossed product is universal in the sense that for any nondegenerate NAME MATH-family MATH in a MATH-algebra MATH... |
math/9901101 | Since MATH acts freely, there is a labeling MATH and an isomorphism of MATH onto MATH which carries the given action to the action of MATH by right translation CITE. Thus this corollary follows by applying REF to MATH. |
math/9901101 | The above formula extends uniquely to a MATH-homomorphism of MATH into MATH. Since MATH is the enveloping MATH-algebra of MATH, MATH further extends uniquely to a homomorphism of MATH into MATH. The coaction identity obviously holds on the generators (that is, the elements of the bundle MATH), hence on all of MATH. The... |
math/9901101 | Let MATH be the MATH-algebraic bundle over MATH defined by REF , let MATH be the product bundle over MATH whose fiber over MATH is MATH, and give MATH the algebraic operations MATH . Then the space MATH of finitely supported sections becomes a MATH-algebra, which can be identified with a dense MATH-subalgebra of the cr... |
math/9901101 | This follows immediately from REF and NAME 's duality theorem CITE. (See also the parenthetical remark in the proof of REF .) |
math/9901101 | For MATH define MATH by MATH . Then MATH is a MATH-homomorphism, since for MATH we have MATH and MATH . Since MATH is the enveloping MATH-algebra of MATH, MATH extends uniquely to a homomorphism MATH of MATH into MATH. To show MATH is an isomorphism, it suffices to find an inverse for the map MATH, since MATH is the en... |
math/9901101 | It suffices to show that for MATH and MATH, MATH if and only if MATH. Let MATH. Then MATH . Hence MATH for all MATH, which proves the lemma. |
math/9901101 | We need only show that MATH. Take a positive element MATH of MATH. By REF, MATH if and only if MATH (because MATH is positive in MATH), so that MATH if and only if MATH (because MATH is positive in MATH). On the other hand, MATH if and only if MATH. Thus, it suffices to show that for all MATH, MATH and for this it suff... |
math/9901101 | We will show that the space MATH implements a groupoid equivalence (in the sense of CITE) between MATH (acting on the left) and MATH (acting on the right). For the right action we need a continuous open surjection MATH from MATH onto the unit space of MATH. For MATH define MATH. Then MATH is a continuous and open surje... |
math/9901101 | First note that CITE, for example, implies that the full and reduced MATH-algebras of an amenable groupoid coincide. Since MATH is amenable so is the skew product MATH, by CITE; hence MATH and MATH is just the regular representation MATH. The semidirect-product groupoid MATH is also amenable, by REF, since groupoid equ... |
math/9901101 | Since MATH is amenable, so is MATH CITE. Let MATH denote the unique conditional expectation extending the map MATH at the level of MATH-functions. Since MATH is amenable, the regular representation of MATH onto MATH is faithful CITE. By CITE, this implies MATH is faithful in the sense that MATH and MATH imply MATH, and... |
math/9901101 | REF ensures that the hypotheses of REF are satisfied, which gives MATH . The theorem now follows from REF. |
math/9901101 | Let MATH. Since MATH has the largest MATH-norm, it suffices to show MATH is bounded. Choose open bisections (``MATH-sets", in NAME 's terminology) MATH of MATH such that MATH, and a partition of unity MATH subordinate to the open cover MATH of MATH. Then MATH, and MATH. Conclusion: without loss of generality there exis... |
math/9901101 | For notational convenience, throughout this proof we suppress the map MATH, and identify MATH and MATH with their images in MATH. Our strategy is to find a MATH-seminorm on MATH which restricts to the greatest MATH-norm on MATH. This suffices, for then a fortiori the greatest MATH-norm on MATH restricts to the greatest... |
math/9901103 | Let MATH be a faded cosheaf. Put MATH, where MATH is the map defined above by REF for MATH. For each morphism MATH of faded cosheaves MATH and MATH put MATH. Using REF, one can easily check that the functor MATH is well-defined and satisfies the conditions required. |
math/9901103 | MATH follows from the Theorem, MATH is trivial. MATH . Let MATH be the components and MATH inclusion maps of MATH, defined by REF. Consider an arbitrary MATH and let MATH be the point such that MATH (through the isomorphism stated by foreground). Then for MATH we have that MATH iff MATH (since MATH may be identified wi... |
math/9901106 | Suppose MATH, where MATH and MATH are non-trivial. Then no non-trivial element of MATH is conjugate to an element of MATH. This can be seen as follows. Let MATH be the normal closure of MATH in MATH. Let MATH be the natural projection. Then there is an isomorphism MATH such that the restriction of MATH to MATH is the i... |
math/9901106 | Standard arguments show that MATH is irreducible and MATH-irreducible. It follows that MATH is non-trivial, non-infinite-cyclic, and indecomposable CITE. The incompressibility of each MATH shows that MATH injects into MATH. We now apply REF . MATH . |
math/9901106 | Let MATH be a tapered regular neighborhood of MATH. Thus MATH is a REF-ball containing MATH such that MATH, MATH is isotopic in MATH rel MATH to a diameter of MATH, and MATH is tamely embedded in MATH except at MATH. Let MATH. (We call MATH the exterior of MATH. We also use this term for the closure of the complement o... |
math/9901106 | Let MATH denote the center of the group MATH. Recall that MATH is an infinite cyclic group generated by MATH and that MATH. Recall also that a free product of two non-trivial groups has trivial center and that any element of finite order in a free product is conjugate to an element of one of the factors. (See CITE.) We... |
math/9901106 | We first outline the proof and then fill in the details with a sequence of lemmas. The construction of the MATH will have a pattern similar to that of the NAME arc. MATH will be parametrized as MATH, and for each integer MATH we will have that MATH meets MATH in three properly embedded arcs MATH, MATH, and MATH, where ... |
math/9901106 | REF shows a three component tangle in a REF-ball. REF shows a two component tangle in a REF-ball. By CITE and CITE these two tangles are excellent. Let MATH and MATH be their respective exteriors. We glue MATH and MATH together as indicated in REF to obtain the exterior MATH of the union of the arcs MATH, MATH, and MAT... |
math/9901110 | It's clear that (since trivalent graphs are connected) the MATH relation defines inductively the value of MATH on any trivalent graph, as long as the definition does not introduce inconsistency. The MATH relation is a difference of two MATH relations involving graphs with zero or one internal vertices. Because of this,... |
math/9901110 | It's enough to proceed locally, in the neighborhood of a point MATH. For ease of notation, assume that all the MATH intersect at MATH. Choose a neighborhood MATH of MATH that can be written as in REF as MATH, with MATH. Then the blowup of MATH along MATH is MATH, the blowup of this space along MATH is MATH, and so fort... |
math/9901110 | If MATH, then their intersection at this stage will be empty. In this case, MATH, which, by hypothesis, we've already blown up; and in the coordinates over MATH, there exist points with either all the points in MATH identical, or all the points in MATH identical, but not both. On the other hand, if MATH, MATH and MATH ... |
math/9901110 | For any MATH-chain MATH in MATH, we have MATH using NAME 's theorem and MATH. |
math/9901110 | Let MATH be the set of vertices not in MATH that are joined to a vertex in MATH. By hypothesis, MATH. Now, MATH (the configuration space appropriate to this graph) is the pullback in MATH and each summand in the fibered chain MATH will be the restriction of a product of two maps, from MATH and MATH, to MATH. The two ma... |
math/9901110 | In this case, we can independently translate the vertices of two disjoint subsets of MATH within MATH without changing the map to MATH. The translation is in any direction in MATH if MATH or along the direction MATH (otherwise), but in either case the map to MATH factors through a map to a bundle of strictly lower dime... |
math/9901110 | The map on MATH that takes MATH to MATH and leaves all other coordinates unchanged is an orientation-reversing automorphism on the stratum that takes the fibered chain to itself (swapping and negating the spheres corresponding to the edges MATH - MATH and MATH - MATH). |
math/9901110 | The map on MATH that takes MATH to MATH, MATH to MATH, and leaves all other coordinates fixed is an orientation-preserving autmorphism on the stratum that takes the fibered chain to its negative (swapping and negating the spheres corresponding to the edges MATH - MATH and MATH - MATH and negating the sphere correspondi... |
math/9901110 | Suppose the face corresponding to the subset MATH has external edges. If some vertex MATH is connected to exactly one external vertex, either MATH is on the knot, in which case MATH (as in the previous section) cannot be connected and the face vanishes by REF ; or MATH is not on the not, in which case MATH has a subgra... |
math/9901110 | See CITE, and especially REF . |
math/9901110 | It's natural to proceed by cases, depending on the parities of the weight diagram MATH. MATH parity even, MATH parity even or MATH parity odd, MATH parity odd In either of these two cases, the fibered chain MATH must be both odd and even under the antipode map on MATH and hence vanishes. MATH parity odd, MATH parity ev... |
math/9901110 | These are both REF-dimensional vector spaces, so a canonical non-zero map between them will suffice. In general, a linear map from MATH to MATH is a bilinear map from MATH and MATH to MATH. So suppose MATH and MATH. Since MATH and MATH are both subspaces of MATH, MATH is a subspace of MATH and MATH is a subspace of MAT... |
math/9901110 | This follows immediately from the familiar equation in anti-commuting algebras (such as MATH), MATH which in turn follows from the equation for elements of degree REF MATH by induction. |
math/9901110 | This is just an iterated version of REF above, with the further note that the isomorphism is independent of the order of the MATH by REF . |
math/9901110 | Given an element MATH of the second vector space, the element MATH (using the inclusions of MATH in MATH) is a non-zero element of MATH which pdoesn't depend on the particular ordering. (By REF , both MATH and MATH change sign under a simple transposition.) |
math/9901110 | By REF , and REF , we have MATH . |
math/9901110 | As noted above, the natural injection MATH induces an injection MATH. Similarly, the natural surjective map MATH induces a surjective map MATH, whose kernel is generated by elements of the form MATH, MATH. For any two elements MATH, MATH, pick an arbitrary preimage MATH of MATH. Then MATH is independent of the choice o... |
math/9901110 | For each MATH, we have, by REF , an isomorphism MATH so, combining these (and using MATH), MATH . |
math/9901111 | According to REF, MATH where the summation is over all pairs MATH of disjoint subsets of MATH such that MATH has MATH elements, MATH. It is convenient to set apart the term of the sum in REF corresponding to the distinguished partition MATH, MATH, MATH . This term will be called distinguished. Let us show that MATH for... |
math/9901111 | The values of all of the not distinguished summands in REF equal zero for the same reason as in REF . Namely, if MATH, then the factor MATH in REF is zero. Now let MATH. If there is MATH such that MATH, then there is a pair of numbers MATH such that MATH and MATH. In that case the factor MATH of the first product in RE... |
math/9901111 | According to REF, MATH is the sum over all pairs MATH of disjoint subsets of MATH such that MATH has MATH elements, and MATH has MATH elements. By REF , all terms in this sum but the distinguished one are equal to zero. The distinguished summand is zero since the factor MATH is zero. |
math/9901111 | By REF , it is enough to show that the distinguished summands MATH and MATH of respectively MATH and MATH are equal. We have MATH . MATH and MATH contain factors of five types: MATH , MATH , MATH , MATH , MATH. Comparing the factors of each type in MATH and MATH we conclude that MATH. REF are proved. |
math/9901111 | According to REF , the poles of the NAME are of the form MATH (modulo MATH). Consider REF . Modulo MATH, the poles of the NAME on the left are at MATH, whereas the poles of the NAME on the right are at MATH, that is, elsewhere. Thus the poles of the NAME on the right must come from the poles of the MATH's on the left. ... |
math/9901111 | For a natural number MATH, define MATH . Let MATH, MATH, be the weight functions associated in REF with the space MATH for MATH. Define a MATH-valued function MATH by MATH . We have MATH that is, MATH . The coordinates of the functions MATH and MATH satisfy the resonance relations of REF . Let us write one of the relat... |
math/9901111 | To prove the formula it is enough to compare the residues of all functions at the point MATH. |
math/9901111 | Recall that if we set MATH, where we add MATH only if MATH, and MATH, then MATH . We have moved the MATH to the middle using the rule MATH. Using the same rule to move MATH to the right, we obtain MATH . This expression is therefore independent of MATH, and can thus be replaced by its value at MATH, which is, by defini... |
math/9901111 | For generic MATH, the NAME MATH has only simple MATH-poles. The kernel of the residue of this matrix at a MATH-pole is described in REF . It follows from REF that a function MATH obeys the condition MATH with respect to the vector MATH at a point MATH if and only if for any MATH-pole MATH of MATH the vector MATH belong... |
math/9901111 | The only non-trivial cases are when MATH. Let MATH. By REF , MATH is holomorphic and MATH is holomorphic. Hence by REF , MATH obeys MATH. Let MATH. The fact, that MATH satisfies the condition MATH if MATH satisfies the condition MATH, follows from identities REF and is similar to the proof of the identities REF. The ca... |
math/9901111 | The statement is trivial if MATH. Suppose that MATH, and let MATH with MATH. Let MATH where MATH . Our goal is to prove the identity MATH for any MATH. Using REF, we can rewrite REF as MATH . Now REF follows from identities REF and the condition MATH for the function MATH. A similar consideration shows that the Lemma h... |
math/9901111 | MATH . It is easy to see that the expression in the parentheses can be written as MATH for suitable coefficients MATH. Thus the expression is zero. The other equalities to zero are proved similarly. |
math/9901111 | We prove the first relation. MATH . At this point it is easy to move the MATH-matrices to the left, with the desired result. |
math/9901111 | The Lemma follows from REF and the normalization property MATH. |
math/9901111 | For a given integer MATH, a relation involving MATH exists, if for some MATH, there is MATH such that MATH . Let MATH. It means that for all MATH we have MATH . Here MATH. This is equivalent to MATH . This inequality contradicts REF. Let MATH. It means that for all MATH we have MATH . This is equivalent to MATH . This ... |
math/9901111 | There are only four cases. We prove the Lemma for each of them. CASE: For all MATH, MATH . CASE: For all MATH, MATH . CASE: There is MATH, such that MATH . CASE: There is MATH, such that MATH . In REF , we have MATH. This means that there is MATH such that MATH which can be rewritten as MATH . Combining with REF we get... |
math/9901111 | By REF , there are MATH and MATH such that MATH . Here MATH. This means that the transformation MATH can be applied to MATH, see Sec. CASE: We prove that the product of transformations MATH can be applied to MATH and gives MATH. We assume that MATH, the proof for MATH is similar. For all MATH, define MATH by MATH. REF ... |
math/9901111 | The proof of this Lemma is similar to the proof of REF , but now we move over the total circle to the left. Namely, by REF , there are MATH and MATH such that REF is satisfied. This means that the transformation MATH can be applied to MATH. We prove that the product of transformations MATH can be applied to MATH and gi... |
math/9901111 | First we prove that REF implies the MATH fusion rules. Indeed, the inequality MATH implies MATH for all MATH. Similarly, the inequality MATH implies MATH for all MATH. Hence, MATH . Notice that the number MATH has the same parity as MATH. Therefore, MATH and MATH . The vector MATH obeys the fusion rules for MATH. Hence... |
math/9901111 | The NAME anti-symmetry of MATH means MATH for all MATH. The vanishing conditions of REF imply MATH for all neutral delta functions MATH. For generic MATH, let MATH be the restriction of MATH to MATH. Here the summation is over integers MATH such that MATH. Let MATH be its image under the action of the transfer matrix. ... |
math/9901111 | The NAME anti-symmetry of MATH gives the relation REF. The condition MATH implies MATH for all MATH. The vanishing conditions of REF imply MATH for all bad MATH. Now the Theorem is proved similarly to the proof of REF considering the function MATH and taking the limit of the equation MATH as MATH tends to zero. |
math/9901112 | Suppose there is a MATH and a sequence MATH such that MATH but MATH . Then MATH and hence MATH . By REF, MATH where MATH is invertible in MATH, MATH, MATH. Thus, REF implies MATH and hence MATH, and MATH . Consequently, MATH with MATH yields MATH . Applying REF again we infer MATH contradicting our hypothesis MATH. Thi... |
math/9901112 | From the proof of REF one infers MATH and hence MATH can be decomposed as MATH where MATH . From MATH one concludes MATH and hence MATH . Denote by MATH the orthogonal projection onto MATH then MATH . In order to prove that MATH we argue as follows: Suppose there is a sequence MATH such that MATH . Hence MATH . Next, l... |
math/9901112 | CASE: Clearly MATH since MATH using MATH. Moreover, by MATH for MATH choosing MATH is sufficient to bound the first integral in REF. REF is obvious from REF. CASE: For any MATH . Taking MATH sufficiently small and MATH sufficiently large such that the first and third integrals in REF are sufficiently small uniformly wi... |
math/9901112 | By NAME.-Nagy's dilation theorem (see the corresponding result in REF for contractions), one infers MATH where MATH is the minimal self-adjoint dilation of MATH in MATH. Then the existence of MATH and REF yield MATH . In order to prove REF one can argue as follows. Consider the contraction MATH . According to REF, ever... |
math/9901112 | Clearly MATH is analytic for MATH since MATH for all MATH by REF . An application of REF then yields REF. |
math/9901112 | Consider MATH . Then MATH and by REF, MATH and hence MATH since MATH can be chosen arbitrarily small. |
math/9901112 | Let MATH and denote MATH . Then MATH proves that MATH is purely absolutely continuous, MATH by standard arguments (see, for example, CITE). Thus, MATH . By REF, MATH defines a quadratic form MATH proving REF. The representation REF (with MATH) implies MATH and hence MATH for some MATH implies MATH for all MATH . In par... |
math/9901112 | It suffices to consider MATH . Since MATH and MATH are self-adjoint, MATH is clearly analytic in MATH and satisfies MATH for MATH . Relation REF is then an elementary consequence of the second resolvent equation, MATH the fact MATH, and simply follows by multiplying the right-hand sides of REF in either order. |
math/9901112 | It suffices to prove REF. Applying REF repeatedly one infers MATH where we used MATH, MATH for MATH with MATH (compare REF), and REF. |
math/9901112 | It suffices to consider MATH and MATH . Since MATH by the NAME hypothesis on MATH and the fact MATH as MATH, the scalar NAME function MATH satisfies MATH . By standard results (see, for example, CITE, CITE), REF yields MATH where MATH is a finite measure, MATH . Moreover, the fact that MATH is uniformly bounded with re... |
math/9901112 | Let MATH. By REF we infer MATH . Adding REF, differentiating REF with respect to MATH proves REF for MATH. The result extends to all MATH by continuity of MATH in MATH-norm. |
math/9901112 | By REF, and REF one infers for MATH sufficiently large, MATH by repeated use of REF. Analytic continuation of REF with respect to MATH then proves REF. |
math/9901112 | First we prove REF in the case MATH. The monotone convergence theorem, REF, and REF then yield MATH . By the spectral theorem applied to MATH one obtains MATH and hence MATH . Decomposing the self-adjoint trace class operator MATH into its positive and negative parts, MATH the monotone convergence theorem yields MATH u... |
math/9901112 | An explicit computation shows MATH for all MATH . Since MATH for MATH as a result of analytic continuation, one obtains MATH proving REF. (Here the interchange of the MATH and MATH integrals follows from NAME 's theorem considering REF in the weak sense.) |
math/9901118 | The operator MATH has norm less than or equal to MATH and MATH is not an eigenvalue of MATH (see Appendix). So MATH never vanishes for any MATH, MATH. First note that MATH . Then we have MATH . Since MATH is a rank REF operator, MATH where MATH as in REF below. On the other hand, we define MATH as follows, MATH . Then ... |
math/9901118 | First note that MATH . From REF, the right hand side of REF is given by MATH where the prime MATH indicates differentiation with respect to MATH. From REF, in order to compute the asymptotics of MATH, we need the asymptotics of MATH as MATH uniformly in MATH. Note that the jump matrix MATH has the following factorizati... |
math/9901118 | REF follows from REF . From REF, we have MATH where MATH solves the RHP REF. From REF, MATH. Hence we have MATH . REF is obtained by taking derivative of REF and using REF. (The differentiability of the infinite product in REF follows from the uniform error estimate in the proof of REF .) |
math/9901118 | First note that the statistics of the second row of MATH is the same as the second column of MATH. This follows immediately by considering the fact that if MATH has a MATH-increasing subsequence, then MATH has a MATH-decreasing subsequence, together with the combinatorial results of NAME referred to in the Introduction... |
math/9901118 | From REF, we have MATH for any MATH. From the definition of MATH with MATH and MATH in REF, the second integral is MATH. We use the relation between MATH and MATH in REF, and the relation between MATH and MATH in REF, to express the first integral in terms of MATH. Then we have MATH . When MATH, a simple computation sh... |
math/9901118 | From the definition of MATH, the left hand side of REF is MATH . Setting MATH, MATH and MATH, the above expression becomes MATH . On the other hand, the right hand side of REF is MATH . For the third integral in REF, NAME expansions of MATH and MATH give us MATH since the integration when MATH vanishes. Evaluating the ... |
math/9901118 | Set MATH . Define MATH . Using the definition of the NAME coefficient MATH, MATH . But MATH for MATH and also MATH for MATH. Therefore REF is zero for MATH. This shows that for fixed MATH, MATH is a polynomial in MATH of degree MATH which is orthogonal to MATH. Thus, for some MATH, MATH . Now the evenness of MATH, MATH... |
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