paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0011023 | Let MATH denote the bid on the MATH-th object of the MATH-th disadvantaged bidder. Let MATH be MATH's bid on the MATH-th object. Because the bids of the MATH disadvantaged bidders are within MATH, MATH has no incentive to bid over MATH. Thus, MATH, and MATH. Since bids from different disadvantaged bidders are independe... |
cs/0011023 | Because MATH are symmetric for MATH, we need only show that the probability distribution of MATH is as described in REF . Let MATH . Then MATH . Let MATH . Note that MATH. The probability distribution of MATH equals MATH . |
cs/0011023 | If MATH, the lemma is the same as REF . If MATH, we divide the objects into MATH groups of MATH objects each and employ REF to obtain bids for the first group. We then set the bids for the other MATH groups to the corresponding bids for the first group. We scale every bid by a factor of MATH so that the bids sum to MAT... |
cs/0011023 | From REF and the fact that our game is a zero-sum game, the MATH disadvantaged bidders win MATH objects in total. Since they all use the same bidding algorithm, by symmetry, each of them wins MATH objects. This upper bound of MATH is also a lower bound since the adversary can always win at least MATH objects by employi... |
cs/0011023 | For each MATH, CASE: MATH denotes the expected number of objects MATH wins if MATH uses the uniform probability distribution while the disadvantaged bidders may use any arbitrary probability distribution; CASE: MATH denotes the expected number of objects MATH wins if MATH does not permute his initial bid sequence and t... |
cs/0011023 | Given an initial bid sequence MATH of the MATH disadvantaged bidders, MATH chooses his initial bid sequence to be MATH. Since MATH's bids are different from MATH, in light of REF , we may assume that the disadvantaged bidders permute their bids with the uniform probability distribution. Consequently, the expected numbe... |
cs/0011023 | Given an optimal initial bid sequence MATH of MATH, we show that this sequence can be transformed into a desired sequence MATH without decreasing MATH. Let MATH be the number of MATH's bids that are in MATH. There are three cases. CASE: MATH. For each MATH, let MATH where MATH is the biggest index such that MATH. Then ... |
cs/0011023 | From REF , MATH has an optimal initial bid sequence MATH, such that for all MATH, MATH. If MATH, then it cannot win any object. If MATH, then it can win MATH objects on average. The unit price MATH pays for these objects is strictly greater than MATH . Since the expected number of objects won by such MATH is an integra... |
cs/0011023 | By REF , MATH wins at most MATH objects on average. By REF , this upper bound is also the lower bound of the expected number of objects MATH can win. Then this theorem follows from the fact that our auction is a zero-sum game. |
cs/0011023 | Assume that MATH employs this bidding algorithm. From his budget constraint, he wins at most MATH objects. This upper bound is also a lower bound. To prove this claim by contradiction, assume that MATH wins fewer than MATH objects and thus does not exhaust all his budget. Then, the total number of objects won by the ot... |
cs/0011023 | Let MATH denote the bid on the MATH-th object by the MATH-th disadvantaged bidder. Let MATH be MATH's bid on the MATH-th object. Because MATH, MATH has no incentive to set MATH greater than MATH. Thus, MATH and MATH. Since bids from different disadvantaged bidders are independent, MATH . MATH maximizes MATH as follows:... |
cs/0011025 | Immediately, from the fact that the vector defining MATH is a prefix of the vector defining MATH. |
cs/0011025 | Immediate from the definitions. MATH. |
cs/0011025 | Let MATH be a vector that denotes MATH. Then, by definition of the characteristic function MATH. That is for any MATH holds MATH. Since MATH, MATH. Thus, MATH implies MATH for any sequences MATH and MATH. In particular, this holds for the sequences appearing in MATH. Repetitive application of this argument proves the l... |
cs/0011025 | Immediate from the definitions. MATH. |
cs/0011025 | Let MATH be a vector that denotes MATH. Then, by definition of the characteristic function MATH. That is for any MATH holds MATH. Since MATH, MATH. Thus, MATH for any sequences MATH and MATH. In particular, this holds for the sequences appearing in MATH. Repetitive application of this argument and the transitivity of M... |
cs/0011025 | CASE: Let MATH, and assume that MATH has an infinite NAME. This derivation contains an infinite directed subsequence, that is a subsequence of goals MATH such that the selected atom of MATH, MATH, is a direct descendant of the selected atom of MATH, MATH. There is some MATH, such that for any MATH holds MATH. Let MATH ... |
cs/0011025 | Assume that MATH. This means that there exists an occurrence MATH of a variable MATH and a replacement MATH, such that MATH. On the other hand, MATH is rigid on MATH. Thus, the replacement cannot be extended to a substitution MATH, such that MATH. This means, that MATH is non-linear in it variables, that is, MATH appea... |
cs/0011025 | Assume for the sake of contradiction, that MATH is not ground. Thus, MATH has at least one variable occurrence say MATH. Then, MATH, contradicting that MATH. |
cs/0011025 | CASE: Let MATH. This means that for any term MATH, MATH. In particular, MATH. Clearly, MATH is a subterm of MATH. Thus, by REF MATH, that is, MATH. Thus, MATH. CASE: Let MATH. If MATH, then MATH holds. Thus, at least one of those terms is not equal to MATH and MATH. Hence, MATH, proving the second statement of the lemm... |
cs/0011025 | Let MATH be non-empty and let MATH. Then, REF allow to conclude monotonicity and subterm properties for the argument positions occupied by the instances of MATH and to mimic the proof of REF . |
cs/0011025 | Let MATH be a substitution. If MATH, MATH and the proof is done. Otherwise, exists some MATH, such that MATH. By REF MATH. Thus, MATH. This means, that for any MATH, for all occurrences MATH holds that MATH. But now, the pseudo-rigidity condition is applicable, and MATH. |
cs/0011025 | Suppose the above condition is satisfied for MATH. Take any MATH and any clause MATH such that MATH exists. Suppose, that MATH is a body atom, such that MATH and that MATH is a computed answer substitution for MATH. Then MATH is identical to MATH, and thus, MATH is identical to MATH. Since MATH is rigid on MATH and MAT... |
cs/0011025 | We base our proof on the notion of directed sequence. Let MATH be non-terminating, that is, MATH has an infinite derivation. By REF it has an infinite directed derivation as well. Let MATH be this infinite directed derivation. We denote MATH and MATH. There is a clause MATH, and substitutions MATH and MATH such that MA... |
cs/0011025 | We define MATH as following: MATH where MATH is the MATH-th element in the sequence, generated by REF for the directed sequence MATH with MATH. The sequence MATH is well-defined: if MATH, on one hand we get that MATH, and on the other hand, MATH, that is MATH. For MATH only one of those definitions is applicable. The r... |
cs/0011025 | Since MATH and MATH are queries in some of the NAME of the simply moded queries and of the simply moded program, they are simply moded CITE. Thus, the output positions, both of MATH and of MATH, are occupied by distinct variables. Since MATH we can claim that MATH, up to variables renaming. Thus, REF becomes applicable... |
cs/0011025 | We base the choice of MATH on the NAME. More precisely, we define MATH if there is a well-moded and simply moded goal MATH and there is a directed sequence MATH in the NAME for MATH, such that the selected atom of MATH is MATH and the selected atom of MATH is MATH and MATH, MATH. Let MATH be a lest NAME model of MATH. ... |
cs/0011025 | If MATH and MATH are well-moded then MATH is well-moded as well CITE. Analogously, MATH is simply moded CITE. Thus, the input positions of MATH are ground and the output positions of MATH are occupied by distinct variables. Therefore, MATH cannot affect the input positions of MATH, and MATH, that is, MATH. Since MATH i... |
cs/0011025 | We define MATH in the following way: MATH . The properties of MATH follow from the corresponding properties of MATH and MATH. Let MATH be a clause. We have to prove that for any substitution MATH, such that MATH and MATH are ground and MATH holds that MATH. Then, by the property of MATH stated above there is some MATH,... |
cs/0011026 | Consider a flat folding of a mountain-valley pattern, and let MATH be consecutive creases with the same direction. The portion MATH of the segment can be in one of three configurations (see REF ): CASE: The portion forms a ``spiral" with MATH being the outermost edge of the spiral, and MATH being the innermost; or CASE... |
cs/0011026 | Let MATH be maximum such that MATH all have the same direction. By the mingling property, either MATH or MATH. In the former case, MATH is a foldable end, so we have the desired result. A generalization of the latter case is that we have MATH all with the same orientation, and MATH. If MATH, then MATH is a foldable end... |
cs/0011026 | This is obvious because folding a foldable end is equivalent to chopping off a portion of the segment. Thus, if we take a flat folding of the original pattern, remove that portion of the segment, and double up the number of layers for the adjacent portion of the segment, we have a flat folding of the new object. |
cs/0011026 | Let MATH be a crimpable pair, and assume by symmetry that MATH is a mountain and MATH is a valley. Consider a flat folding MATH of the original segment, such as the one in REF (left). We orient our view to regard the segment MATH as flipping over during the folding, so that the remainder of the (unfolded) segment keeps... |
cs/0011026 | By REF , the pattern is mingling, and hence by REF we can find a crimpable pair or a foldable end. Making this fold preserves flat foldability by REF , so by induction the result holds. |
cs/0011026 | First note that it is trivial to check in constant time whether a pair of consecutive folds form a crimp or whether an end is foldable. We begin by testing all such folds, and hence in linear time have a linked list of all possible folds at this time. We also maintain reverse pointers from each symbol in the string to ... |
cs/0011026 | Performing an all-layers simple fold that is not allowable forbids us from all-layers simple folding certain creases, and hence the resulting segment cannot be completely folded after that point. Therefore, only allowable folds can be in the sequence. It remains to show that performing an allowable fold preserves folda... |
cs/0011026 | Let MATH be the complement string of MATH (that is, the complement of each letter of MATH), and let MATH be the reverse string of MATH. The fold at position MATH of MATH is allowable precisely if the first MATH characters of the suffix of MATH starting in the MATH-nd position are identical to the suffix of MATH startin... |
cs/0011026 | For MATH, valley fold MATH if exactly one of MATH and MATH is in MATH. After these folds, as we travel along the steps corresponding to MATH, we travel in the MATH direction for elements that belong to MATH and in the MATH direction for elements that belong to MATH. Because the sums of elements of both MATH and MATH ar... |
cs/0011026 | If either MATH or MATH is folded without having the staircase confined between the MATH-coordinates of MATH and MATH, the rectangular frame MATH would intersect with the staircase and would make the other of MATH and MATH impossible to fold. Hence the staircase must be brought between the MATH-coordinates of MATH and M... |
cs/0011037 | The last rule of a derivation of MATH must be one of MATH, MATH, MATH, MATH, MATH or MATH. In each of these cases the claim is trivial. For example, in the case MATH we conclude by two applications of MATH that MATH. |
cs/0011037 | Induction on MATH , using REF : If MATH is a variable or a constant the claim is obvious. If MATH is of the form MATH then by REF either MATH or MATH. We apply the induction hypothesis to the corresponding subterm and can type MATH by the same rule used to type MATH. If MATH is of the form MATH, then MATH must be typed... |
cs/0011037 | Induction on MATH shows that only conversions have to be considered. The only non-trivial case is handled in REF . |
cs/0011037 | Induction on MATH and case distinction according to REF : If MATH is of the form MATH then MATH has to be of type MATH and hence MATH has to be empty (since there is no elimination rule for the type MATH). If MATH is of the form MATH or MATH, then MATH has to be empty as well, for otherwise the term would not be normal... |
cs/0011037 | Induction on MATH , using the fact that MATH is typed and therefore in the case of an application only one of the terms can contain the variable MATH free (compare REF ). For instance, if MATH is MATH, then the last rule of a derivation of MATH must be one of MATH, MATH, MATH, MATH, MATH or MATH. In each of these cases... |
cs/0011037 | Induction on MATH and case distinction according to REF . The case MATH is trivial. In the case MATH the MATH cannot be empty (for otherwise the term would have an arrow type), so we can apply the induction hypothesis to the (by REF ) shorter term MATH. Similarly for MATH, where we apply the induction hypothesis to MAT... |
cs/0011037 | REF shows that for closed terms of type MATH the usual length and the number of free variables coincide (due to the typing MATH of the MATH function). The number of free variables trivially does not increase when reducing the term to normal form. |
cs/0011037 | We prove this by induction on MATH. The only non-trivial case is when MATH is an application (note that the case MATH does not occur since MATH is typed). So let MATH be of the form MATH. We distinguish cases whether MATH is a list or not. If MATH is a list with MATH entries then, as MATH is typed, we know that MATH mu... |
cs/0011037 | We prove this by induction on the definition of the relation MATH. Case MATH via MATH. We distinguish whether MATH is a list or not. Subcase MATH is a list with MATH entries. Then by typing restrictions and from REF we know that MATH. Also MATH has to be of the form MATH. MATH is again a list with MATH entries, so MATH... |
cs/0011037 | We only treat unary functions. Let MATH be an almost closed normal list. Then MATH, as MATH. Therefore an upper bound for the number of steps is MATH, which is polynomial in the length of the input MATH. |
cs/0011038 | By REF , MATH . Since MATH and MATH, we use NAME 's inequality CITE on sums of independent bounded random variables to have REF . |
cs/0011038 | REF follow from REF becomes straightfoward when restated in terms of distance as follows. For any MATH with MATH, there is a node MATH on the path between MATH and MATH in MATH such that MATH . Furthermore, if MATH then MATH is distinct from MATH and MATH. |
cs/0011038 | This corollary follows from REF and simple algebra. |
cs/0011038 | See REF. |
cs/0011038 | The proof is straightforward. Note that the more unbalanced MATH is, the smaller its g-depth is. |
cs/0011038 | See REF. |
cs/0011038 | See REF. |
cs/0011038 | The inequalities follow from REF , respectively. |
cs/0011038 | The proof is by induction on MATH. The base case follows from the fact that the statement holds for MATH at line REF. The induction step follows from the use of a relevant triplet at line REF. |
cs/0011038 | We analyze the time and space complexities separately as follows. Time complexity. Line REF takes MATH time. Line REF takes MATH total time to examine MATH triplets for each MATH. As for the repeat at line REF, lines REF, and REF take MATH time to search through MATH. For the MATH-th iteration of the repeat where MATH,... |
cs/0011038 | There are two directions, both using the following equation. By line REF, MATH . MATH . To prove by contradiction, assume MATH or MATH in MATH. If MATH, then MATH, and by MATH, MATH is an internal node in MATH. By MATH, the triplet formed by MATH is not small. Thus, by MATH and REF , MATH. By symmetry, if MATH, then MA... |
cs/0011038 | There are two directions. MATH . From lines REF , MATH . Thus, whether MATH and MATH are leaves or internal nodes in MATH, by MATH, MATH, and MATH, MATH . By line REF , MATH . Then, since MATH and thus MATH, by MATH, we have MATH . By symmetry, MATH. Thus, the test of line REF fails. MATH . To prove by contradiction, a... |
cs/0011038 | By REF , for every node MATH strictly between MATH and MATH in MATH, there exists a leaf MATH with MATH. To choose MATH, there are two cases: REF both MATH and MATH are internal nodes in MATH, and REF MATH or MATH is a leaf in MATH. CASE: By REF , let MATH and MATH. By MATH, neither MATH nor MATH is small. To fix the n... |
cs/0011038 | The two statements are proved as follows. CASE: This statement follows directly from the initialization of MATH at line REF, the deletions from MATH at line REF, and the insertions into MATH at lines REF . The proof is by induction on MATH. CASE: MATH. By MATH, MATH, MATH, MATH, and REF , MATH is a splitting triplet fo... |
cs/0011038 | The proof is by induction on MATH. CASE: MATH. By REF , MATH, and the greedy selection of line REF, line REF constructs MATH without edge lengths. Then, MATH holds trivially. MATH follows from MATH, MATH, and line REF. MATH follows from MATH, MATH and the use of REF at line REF. Induction hypothesis: MATH, MATH, and MA... |
cs/0011038 | By REF , MATH if MATH . Similarly, by REF , MATH if MATH . We choose MATH. Consequently, MATH. By REF , with probability at least MATH, NAME outputs MATH, and MATH and MATH hold, which correspond to the two statements of the theorem. |
cs/0011042 | Assume MATH is order-consistent and MATH. Let MATH be an answer set for MATH. Since MATH is order-consistent, so is MATH. By REF , there is a splitting sequence MATH for MATH such that all MATH-components of MATH are signed. Notice that MATH is also a splitting sequence for MATH, and that all MATH-components of MATH ar... |
cs/0011045 | We first show that given any arrangement completely filling the matrix, rearranging the numbers to become monotonic in one coordinate does not increase the overall spread. It is obvious that rearranging the numbers in any way within the same line does not change the spread in that line, thus the rearrangement for full ... |
cs/0011045 | The proof is a generalization of NAME 's proof of the main theorem in CITE. We use induction on MATH and the largest dimension size, MATH. The base cases of MATH and MATH are trivial. Suppose we have a matrix with the largest dimension size MATH (largest coordinate value MATH). By the induction hypothesis, the herringb... |
cs/0011045 | By REF , herringbone arrangement of a completely filled cube maximizes the smalls sequence. A process similar to the derivation of the bounding smalls sequence creates a bounding bigs sequence. For the bigs sequence, however, we start instead with the largest element in the cell with the largest coordinate value and wo... |
cs/0011045 | We shall assume for simplicity that MATH is odd. For even MATH the argument works in a similar way. We can divide the matrix into MATH pieces by cutting through the middle of each face, including the middle line into both sides that are separated by it. For MATH see REF . The initial and the last pieces, coordinate-wis... |
cs/0011046 | Although findSlot(root) does not guarantee to find an available position for a MATH operation for illegitimate MATH, if findSlot(root) does return MATH, then from the logic of findSlot, MATH is a child of some node of MATH, and thus MATH will contain MATH as a result of MATH. For a deleteMin operation, MATH returns som... |
cs/0011046 | Let MATH denote the path of nodes examined via recursion for a given invocation of MATH. Observe that MATH is called within this invocation of verify iff MATH is the rightmost path within MATH (otherwise swAncestor would return a non-MATH value). By construction, MATH for MATH sets up the leftmost path in MATH to the r... |
cs/0011046 | Each insert and deleteMin operation invokes verify(root), however such operations also change the active tree. With respect to the sequence of verify(root) calls starting from the initial state, each change to the active tree either occurs in a subtree previously visited by verify or occurs in a subtree not yet visited... |
cs/0011046 | REF establishes that REF hold after MATH operations. In the rest of this proof, we assume that REF hold. For the sake of generality we suppose that MATH is only loosely balanced: assume there exist constants MATH and MATH so that for any MATH, MATH, the minimum height MATH taken over all subtrees of MATH with MATH node... |
gr-qc/0011067 | We use the definitions, constructions and notations of the proof of CITE. Thus, let MATH be a coordinate neighborhood of the form MATH, with MATH and MATH, in which MATH is the graph of a MATH function MATH, and in which MATH is the graph of a semi - convex function MATH. Here MATH is a locally MATH hypersurface in MAT... |
gr-qc/0011067 | It is shown in CITE that MATH is locally the graph of a semi-convex function. That is, there is a coordinate system MATH so that MATH is given by a graph MATH where MATH is MATH and MATH is convex. Define new coordinates by MATH for MATH and MATH. In these coordinates MATH is given by MATH . |
gr-qc/0011067 | We choose a coordinate system MATH on a open set MATH containing MATH as in REF so that MATH is given by MATH where MATH is convex. We may assume that the point MATH has coordinates MATH. We also assume that MATH is of the form MATH for MATH an open convex set in MATH and that MATH takes values in the interval MATH. Th... |
gr-qc/0011067 | Let MATH be a convex normal neighborhood of MATH, having closure disjoint from MATH. For MATH sufficiently small the distance sphere MATH is contained in MATH, is compact, and agrees with the geodesic sphere of radius MATH centered at MATH. Then, MATH restricted to MATH achieves a minimum at some point MATH, say. Let M... |
gr-qc/0011069 | One computes MATH where in the second line we have performed a partial integration and used the equations of motion. On the other hand MATH where in the second line we have used that MATH, which holds because MATH is a Killing field. This gives MATH . So, in order to prove the lemma, we must show that MATH on MATH. The... |
gr-qc/0011069 | The proof of the above result is divided into two parts. In the first part, we show that the time ordered products can be normalized in such a way that the interacting current density MATH is covariantly conserved. A proof of this requires the demonstration of a corresponding set of NAME identities, see REF below. A pr... |
gr-qc/0011069 | We want to prove the NAME identities by an induction on MATH and the total degree of the NAME monomials defined by MATH. Let us write MATH for the left hand side minus the right hand side of REF , that is, the anomaly and let us inductively assume that the NAME identities can be satisfied for some MATH and some order M... |
hep-th/0011015 | By REF and CITE, the NAME - NAME equations in the NAME gauge have unique global solutions with finite energy for all smooth initial data MATH with support in the base of MATH. Moreover, these solutions are smooth in all variables. We shall show that their support properties are as stated above. To see this, we note tha... |
hep-th/0011015 | REF (see also CITE) establishes the existence and uniqueness of finite energy solutions for sufficiently small smooth initial data. In particular, if the initial data for MATH and MATH are smooth, have compact support and are bounded by a sufficiently small constant and the initial data for MATH are computed from the g... |
hep-th/0011015 | The leading term MATH in the asymptotic expansion of MATH corresponds to the MATH contribution to the series REF . The next term is given by the vacuum expectation value of the MATH contribution which has the form MATH . Plugging into the integral the expression given in REF and taking locality into account, one obtain... |
hep-th/0011015 | As MATH and MATH can be replaced in any given order of perturbation theory by MATH for sufficiently large MATH, the statement follows from the preceding lemma by extracting the asymptotically leading contribution in MATH from the function MATH appearing there. |
hep-th/0011237 | Since the metrics MATH and MATH are conformally equivalent as expressed in REF , the corresponding connections are related by MATH (see for example, REF). Now let MATH be vector fields tangent to MATH, and denote the normal projections corresponding to MATH with respect to MATH and MATH by MATH and MATH, respectively. ... |
hep-th/0011237 | There is a one - to - one correspondence between wedges MATH in dS and pairs of lightlike rays in MATH of the form MATH, where MATH are unit vectors in Euclidean MATH satisfying MATH . Namely, MATH and MATH are the unique unit vectors such that MATH . Conversely, MATH is the set of all MATH satisfying MATH . In particu... |
hep-th/0011237 | MATH contains MATH if and only if the apices of MATH are contained in the closure of MATH compare REF. The MATH- and MATH-components in ambient MATH of these two apices are given by MATH respectively, where MATH and MATH . Hence, taking into account the symmetry MATH, the two apices are contained in the closure of MATH... |
hep-th/0011237 | As mentioned in the proof of REF , there are unique unit vectors MATH corresponding to a wedge MATH such that MATH if and only if both inequalities MATH hold. The unit vectors MATH corresponding to MATH and hence to MATH are determined by the equation MATH to be MATH . Let now MATH. Then MATH must satisfy MATH and MATH... |
hep-th/0011237 | The first statement has been established already in REF , while the latter one is a consequence of the subsequent REF . |
hep-th/0011237 | Let MATH and MATH satisfy REF. Then REF entails MATH. Moreover, the intersection of the wedges MATH is non - empty, because by REF it contains a double cone MATH where MATH and MATH . For the proof of the second part of the statement we proceed to ambient NAME space and denote by MATH the NAME transformation correspond... |
hep-th/0011237 | The claimed properties follow from the corresponding properties of the underlying NAME net, taking into account the specific features of MATH established in REF. |
hep-th/0011237 | The first step is to show that MATH arises from the linear map MATH in MATH defined by MATH . Since MATH leaves the set of spacelike vectors invariant. Denoting MATH for spacelike MATH the claim is that MATH . To see this, let MATH be the map defined by the right hand side of the above equation. Recall CITE that the MA... |
math-ph/0011001 | The proof uses standard compact operator results, see for example, CITE. First note that the operator MATH is compact. This is straightforward: since MATH as MATH, it follows that MATH is the norm limit as MATH of the finite rank operators defined by MATH for MATH and MATH otherwise, and thus is compact. Since MATH, MA... |
math-ph/0011001 | To get a contradiction, assume MATH, MATH, satisfying REF , is a solution of REF . Multiplying REF by MATH, and summing with respect to MATH from MATH to MATH we get MATH . If MATH the imaginary part of MATH is negative (see REF and the discussion following it) and thus, if some MATH is nonzero then the left side of RE... |
math-ph/0011001 | REF is of the form MATH in MATH, where MATH and MATH . We show first that MATH. Indeed, let MATH and set MATH. Then we see that MATH and, by multiplying with MATH and summing over MATH we get MATH . Note that, because MATH, the following quantity is real: MATH implying that MATH with (compare REF ) MATH . Let MATH. Obv... |
math-ph/0011001 | The substitution MATH, and MATH leads to the following system of equations for MATH and MATH . We now let MATH and, for MATH, MATH. We use again the NAME alternative and, as in the previous proofs, we need only to show the absence of a solution of the homogeneous equation at MATH. We thus multiply the homogeneous equat... |
math-ph/0011001 | Special care is only needed near MATH. The system REF - REF now reads MATH . We take MATH and MATH. The system becomes MATH . The system REF is of the form MATH where MATH are in MATH. We prove that the homogeneous equation has no nontrivial solutions: MATH implies MATH. Let MATH, MATH, MATH for MATH and MATH, MATH and... |
math-ph/0011001 | All but the last claim has already been shown. The last statement is a standard NAME theorem (note that MATH is the NAME transform along the imaginary line). |
math-ph/0011001 | We can write REF (with MATH) as MATH . It is easy to check, in view of the fact that MATH and MATH are MATH, that MATH. Furthermore MATH is convergent as MATH. Thus MATH as MATH. Since now the l.h.s. of REF converges to zero and MATH does not, the equality REF is only consistent if MATH. |
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