Inverse Inequality Estimates with Symbolic Computation
IINVERSE INEQUALITY ESTIMATESWITHSYMBOLIC COMPUTATION
CHRISTOPH KOUTSCHAN, MARTIN NEUM ¨ULLER, AND CRISTIAN-SILVIU RADU
Abstract.
In the convergence analysis of numerical methods for solving par-tial differential equations (such as finite element methods) one arrives at certaingeneralized eigenvalue problems, whose maximal eigenvalues need to be esti-mated as accurately as possible. We apply symbolic computation methods tothe situation of square elements and are able to improve the previously knownupper bound, given in “ p - and hp -finite element methods” (Schwab, 1998),by a factor of 8. More precisely, we try to evaluate the corresponding deter-minant using the holonomic ansatz, which is a powerful tool for dealing withdeterminants, proposed by Zeilberger in 2007. However, it turns out that thismethod does not succeed on the problem at hand. As a solution we present avariation of the original holonomic ansatz that is applicable to a larger classof determinants, including the one we are dealing with here. We obtain an ex-plicit closed form for the determinant, whose special form enables us to derivenew and tight upper resp. lower bounds on the maximal eigenvalue, as well asits asymptotic behaviour. Introduction
Interdisciplinary collaborations between different areas of mathematics can be hardwork because of different terminology and the difficulty of recognizing the applica-bility of the methods from one field in the other field. In Linz there is an almost-20-year tradition of bringing together researchers from numerical mathematics andsymbolic computation [20, 21, 15, 1], which at the beginning faced exactly thesekinds of problems. Additionally, there could be the risk that the results are onlyinteresting for one community and not rewarded by the other one. Fortunately,this didn’t happen in our case: in the current work, we use and invent tools at thefrontier of symbolic computation to solve a problem that arose at the frontier ofnumerical analysis research. Hence, this work improves the knowledge and toolsfor both communities.
Mathematics Subject Classification.
Primary 33F10, 65N12; Secondary 65N30, 68W30,65F15, 05A20, 15A15, 15A45.
Key words and phrases.
Zeilberger’s algorithm, inverse inequality, holonomic ansatz, finiteelement method, holonomic function, symbolic determinant evaluation.c (cid:13) a r X i v : . [ c s . S C ] M a y CHRISTOPH KOUTSCHAN, MARTIN NEUM¨ULLER, AND CRISTIAN-SILVIU RADU
Inverse inequalities of the form || v n || X (Ω) ≤ c ( h, n ) || v n || Y (Ω) for all v n ∈ V n , (1) || v n || Z ( ∂ Ω) ≤ c ( h, n ) || v n || Y (Ω) for all v n ∈ V n (2)play an important role in the analysis and design of numerical methods for partialdifferential equations [4, 22, 2, 6] and in the construction of efficient solvers for thearising linear systems of those methods [10, 23]. Bounds for the constants of thetype (1)–(2) have been studied for example in [22, 25, 24, 9, 7], where the asymptoticbehaviour with respect to h and n is covered but usually these constants are overestimated. In many numerical methods a precise knowledge of these constants isrequired, which motivates this work where we use and present tools from symboliccomputation to derive precise estimates.Here Ω ⊂ R d , d ∈ N , is a bounded and open set with sufficiently smooth boundary ∂ Ω, describing a finite element with diameter h > d = 1, often triangles or quadrilaterals for d = 2, usuallytetrahedra or hexahedra for d = 3, . . . ). Let V be some infinite-dimensional spaceof functions defined on Ω such that the solution of the PDE is an element of V .With ( V n ) n ∈ N we denote a family of finite-dimensional (usually closed) subspacesof V whose dimension depends on n ; the desired solution of the PDE is approxi-mated by an element of V n . Moreover we have some given norms ||·|| X (Ω) , ||·|| Y (Ω) and ||·|| Z ( ∂ Ω) which are induced by certain inner products ( · , · ) X (Ω) , ( · , · ) Y (Ω) and( · , · ) Z ( ∂ Ω) , which are used in the analysis of the numerical methods. In general theconstants c and c of (1) and (2) depend on the diameter h and on the param-eter n reflecting the dimension of the space V n . The dependence with respect tothe diameter h is obtained by transforming Equations (1) and (2) to a referencedomain ˆΩ ⊂ R d , i.e. || ˆ v n || X (ˆΩ) ≤ ˆ c ( n ) || ˆ v n || Y (ˆΩ) for all ˆ v n ∈ ˆ V n , (3) || ˆ v n || Z ( ∂ ˆΩ) ≤ ˆ c ( n ) || ˆ v n || Y (ˆΩ) for all ˆ v n ∈ ˆ V n (4)and applying a scaling argument [22, 25, 2]. The more challenging problem is tofind precise estimates for the constants ˆ c and ˆ c with respect to the parameter n .The best possible constants by definition are given byˆ c ( n ) = sup ˆ v n ∈ ˆ V n || ˆ v n || X (ˆΩ) || ˆ v n || Y (ˆΩ) = (cid:118)(cid:117)(cid:117)(cid:116) sup ˆ v n ∈ ˆ V n (ˆ v n , ˆ v n ) X (ˆΩ) (ˆ v n , ˆ v n ) Y (ˆΩ) , (5) ˆ c ( n ) = sup ˆ v n ∈ ˆ V n || ˆ v n || Z ( ∂ ˆΩ) || ˆ v n || Y (ˆΩ) = (cid:118)(cid:117)(cid:117)(cid:116) sup ˆ v n ∈ ˆ V n (ˆ v n , ˆ v n ) Z ( ˆ ∂ Ω) (ˆ v n , ˆ v n ) Y (ˆΩ) . (6)Introducing for ˆ V n the basis functions ( ϕ k ) ≤ k ≤ n , i.e., ˆ V n = span { ϕ , . . . , ϕ n } , wefurther obtain (cid:0) ˆ c ( n ) (cid:1) = sup v n ∈ R n ( K n v n , v n ) (cid:96) ( M n v n , v n ) (cid:96) and (cid:0) ˆ c ( n ) (cid:1) = sup v n ∈ R n ( L n v n , v n ) (cid:96) ( M n v n , v n ) (cid:96) , with the symmetric and positive (semi-) definite matrices K n ( i, j ) := ( ϕ j , ϕ i ) X (ˆΩ) , M n ( i, j ) := ( ϕ j , ϕ i ) Y (ˆΩ) , and L n ( i, j ) := ( ϕ j , ϕ i ) Z ( ˆ ∂ Ω)NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 3 for i, j = 1 , . . . , n . Note that the rows of the matrices are naturally defined via thesecond argument of the inner product and the columns are given naturally via thefirst argument of the inner product. Hence, the constants (ˆ c ) and (ˆ c ) are givenby the largest eigenvalues of the generalized eigenvalue problems K n x n = λ n M n x n and L n x n = µ n M n x n , (7)i.e. (cid:0) ˆ c ( n ) (cid:1) = λ n and (cid:0) ˆ c ( n ) (cid:1) = µ n . In this work we want to solve the problems (3) and (4), i.e., determine estimatesfor ˆ c ( n ) and ˆ c ( n ), for the reference domain ˆΩ = ( − , with( u, v ) X (ˆΩ) = (cid:90) ˆΩ ∂ x u ( x, y ) ∂ x v ( x, y ) d x d y, ( u, v ) Y (ˆΩ) = (cid:90) ˆΩ u ( x, y ) v ( x, y ) d x d y, for u, v ∈ ˆ V n , where ˆ V n is the space of polynomials of degree less than n , i.e.ˆ V n = (cid:8) x i y j : 0 ≤ i, j < n (cid:9) . In Section 2 we state the problem in detail and derive its formulation as a general-ized eigenvalue problem of the form (7). The difficulty is now to find an accurateestimate for the largest eigenvalues λ n and µ n for general parameter n ∈ N . Ofcourse one can compute the eigenvalues exactly for a given fixed parameter n , whichis done for example in [19]. But to derive their exact values or precise estimatesfor a general parameter n one needs techniques from symbolic computation. InSection 3 we use the HolonomicFunctions package [11, 12] to prove a closed-formrepresentation of the characteristic polynomial of our eigenvalue problem, in thespirit of the holonomic ansatz [27] for evaluating determinants. The holonomicansatz is a very powerful method (it was the key to the “holy grail of enumerativecombinatorics” [14]), and a very flexible one, too [16]. For our purposes here wehad to adapt this algorithm, which led to a new variant that is applicable to alarger class of determinants. In Section 4, our representation of the characteristicpolynomial is used to derive and prove the estimates from below and above14 (cid:112) n ( n − n + 1)( n + 2) ≤ ˆ c ( n ) ≤ √ (cid:112) n ( n − n + 1)( n + 2)for the constant ˆ c ( n ) in the inequality || ∂ ˆ x ˆ u n || L (ˆΩ) ≤ ˆ c ( n ) || ˆ u n || L (ˆΩ) for all ˆ u n ∈ ˆ V n . In Lemmas 4.5 and 4.8 we give much sharper estimates for ˆ c ( n ); as an application,they allow to tune the parameters of numerical methods precisely. In Section 5the same representation is used to investigate the asymptotic behaviour of theeigenvalues; on the way we discover some interesting connections to the Taylorexpansions of trigonometric functions. As an encore, we deal with the secondinequality (4) in Section 6. It turns out that it is considerably simpler and we areable to derive the exact value of ˆ c ( n ). CHRISTOPH KOUTSCHAN, MARTIN NEUM¨ULLER, AND CRISTIAN-SILVIU RADU
Throughout the paper, we employ the following notation: ( a ) n denotes the Poch-hammer symbol, also known as rising factorial, defined for all nonnegative integers n by ( a ) n := a · ( a + 1) · · · ( a + n −
1) for n > a ) := 1 . We use (cid:98) x (cid:99) for the floor function, and (cid:100) x (cid:101) for the ceiling function, i.e., the largestinteger below x , resp. the smallest integer above x . For a polynomial p we referto the degree of p with respect to the variable x by deg x ( p ). By δ i,j we denotethe Kronecker delta symbol, i.e., δ i,j = 0 if i (cid:54) = j and δ i,i = 1. If A is the n × n matrix ( a i,j ) ≤ i,j ≤ n , then we use for its determinant the short-hand notationdet( A ) = det ≤ i,j ≤ n ( a i,j ). The determinant of the 0 × The Maximal Eigenvalue Problem
Let the reference domain ˆΩ ⊂ R be defined by ˆΩ := ( − , , the open square ofsize 2 centered around the origin. For n ∈ N and k ∈ { , . . . , n } , we define χ n ( k )and ρ n ( k ) to be the unique integers in { , . . . , n − } satisfying k = χ n ( k ) · n + ρ n ( k ) + 1. In other words, χ n ( k ) := (cid:106) k − n (cid:107) and ρ n ( k ) := k − n. For the rest of this section we fix n ∈ N and write shortly χ ( k ) and ρ ( k ). Byemploying the standard monomial basis ϕ k := x ρ ( k ) t χ ( k ) , we obtain the n × n matrix M n with entries m i,j defined by(8) m i,j := (cid:90) ˆΩ ϕ i ϕ j d x d t (1 ≤ i, j ≤ n )and the n × n matrix K n with entries(9) k i,j := (cid:90) ˆΩ ( ∂ x ϕ i )( ∂ x ϕ j ) d x d t (1 ≤ i, j ≤ n ) . Since ϕ i and ϕ j are just monomials, these integrals can be evaluated in a straight-forward manner: m i,j = (cid:90) − (cid:18)(cid:90) − x ρ ( i ) t χ ( i ) x ρ ( j ) t χ ( j ) d x (cid:19) d t = (cid:90) − − ( − ρ ( i )+ ρ ( j )+1 ρ ( i ) + ρ ( j ) + 1 t χ ( i )+ χ ( j ) d t = 1 − ( − ρ ( i )+ ρ ( j )+1 ρ ( i ) + ρ ( j ) + 1 · − ( − χ ( i )+ χ ( j )+1 χ ( i ) + χ ( j ) + 1 . Similarly k i,j = (cid:90) − (cid:18)(cid:90) − ρ ( i ) ρ ( j ) x ρ ( i ) − t χ ( i ) x ρ ( j ) − t χ ( j ) d x (cid:19) d t = (cid:90) − ρ ( i ) ρ ( j ) 1 − ( − ρ ( i )+ ρ ( j ) − ρ ( i ) + ρ ( j ) − t χ ( i )+ χ ( j ) d t = ρ ( i ) ρ ( j ) 1 − ( − ρ ( i )+ ρ ( j ) − ρ ( i ) + ρ ( j ) − · − ( − χ ( i )+ χ ( j )+1 χ ( i ) + χ ( j ) + 1 , where we assumed that ρ ( i ) + ρ ( j ) >
1; otherwise the integral equals to 0.
NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 5
We are interested in computing the maximal λ n ∈ R such that det( K n − λ n M n ) = 0.In the following we derive an equivalent formulation of this problem that involvessmaller matrices. For this purpose let a i,j := 1 − ( − i + j − i + j − b i,j := ( i − j −
1) 1 − ( − i + j − i + j − , such that the matrix entries m i,j and k i,j can be written as m i,j = a χ ( i )+1 ,χ ( j )+1 · a ρ ( i )+1 ,ρ ( j )+1 k i,j = a χ ( i )+1 ,χ ( j )+1 · b ρ ( i )+1 ,ρ ( j )+1 . This shows that the matrices M n and K n can be written as Kronecker products: M n = A n ⊗ A n and K n = A n ⊗ B n . These representations as Kronecker products are quite natural since the used basisfunctions and the reference domain ˆΩ itself have tensor product structure. Inparticular we then obtain,det( K n − λ n M n ) = det (cid:0) A n ⊗ ( B n − λ n A n ) (cid:1) = det( A n ) n det( B n − λ n A n ) n . So the problem is equivalent to computing the maximal λ n ∈ R such thatdet( B n − λ n A n ) = 0 . Determinant Evaluation
According to the previous discussion, we are now interested in evaluating the de-terminantdet( B n − λA n ) = det ≤ i,j ≤ n (cid:18)(cid:0) − ( − i + j − (cid:1)(cid:16) ( i − j − i + j − − λi + j − (cid:17)(cid:19) for symbolic λ ; the desired maximal eigenvalue λ n is then just the largest rootof the obtained polynomial. We see that the matrix B n − λA n has zeros at allpositions ( i, j ) for which i + j is an odd integer. By applying the permutation(2 , , , . . . , , , , . . . ) to the rows and to the columns of the matrix, we decomposeit into block form and obtaindet( B n − λA n ) = 2 n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A (0) (cid:98) n/ (cid:99) A (1) (cid:100) n/ (cid:101) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 2 n det (cid:16) A (0) (cid:98) n/ (cid:99) (cid:17) · det (cid:16) A (1) (cid:100) n/ (cid:101) (cid:17) where the subscripts indicate the dimensions of the square matrices A (0) and A (1) ,whose entries are independent of the dimension and given by a (0) i,j := (2 i − j − i + 2 j − − λ i + 2 j − , (10) a (1) i,j := 4( i − j − i + 2 j − − λ i + 2 j − . (11) CHRISTOPH KOUTSCHAN, MARTIN NEUM¨ULLER, AND CRISTIAN-SILVIU RADU
Hence the matrices A (0) and A (1) start as follows: A (0) = − λ − λ − λ − λ · · · − λ − λ − λ − λ · · · − λ − λ − λ
11 3511 − λ · · · − λ − λ
11 3511 − λ
13 4913 − λ · · · ... ... ... ... . . . ,A (1) = − λ − λ − λ − λ · · ·− λ − λ − λ − λ · · ·− λ − λ − λ − λ · · ·− λ − λ − λ
11 3611 − λ · · · ... ... ... ... . . . . Theorem 3.1.
Let a (0) i,j and a (1) i,j be defined as in (10) and (11) , then the followingidentities hold for all nonnegative integers n : det A (0) n = det ≤ i,j ≤ n a (0) i,j = ( − n h (0) n · F n ( λ ) , det A (1) n = det ≤ i,j ≤ n a (1) i,j = ( − n h (1) n · λF n − ( λ ) , where F n ( λ ) := ν (cid:88) j =0 ( − j − ν (2 ν − j + 1) n (2 j − ν + n )! λ j with ν = ν ( n ) := (cid:106) n (cid:107) , (12) h ( (cid:96) ) n := 12 n n (cid:89) i =1 (cid:0) ( i − (cid:1) (cid:0) i − (cid:96) + (cid:1) n . (13) Corollary 3.2.
For all nonnegative integers n we have det( B n − λA n ) = ( − n h (0) (cid:98) n/ (cid:99) h (1) (cid:100) n/ (cid:101) λ F n − ( λ ) F n ( λ ) . The key ingredient for the proof of Theorem 3.1 is the following lemma which showsthat the quantities p (0) n,j and p (1) n,j defined there are basically the entries of the lastcolumn of the inverses of A (0) and A (1) , respectively. Lemma 3.3.
With p (0) n,j = 2 n +2 j − (cid:0) (cid:1) n − (cid:0) n + (cid:1) j − ( n − j − n − (cid:88) m =0 2 n − m − (cid:88) k =0 ( − j + m (2 m + 1) k λ m m + k k ! (2 m + k − n − j + 2)! ,p (1) n,j = 4 j − n (4 n − (cid:0) n − (cid:1) j − (2 n − n − j − n − (cid:88) m =0 2 n − m − (cid:88) k =0 ( − j + m (2 m ) k λ m m + k k ! (2 m + k − n − j + 2)! , NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 7 the following identities hold for all nonnegative integers n and for ≤ i ≤ n : n (cid:88) j =1 a (0) i,j p (0) n,j = δ i,n F n ( λ ) , n (cid:88) j =1 a (1) i,j p (1) n,j = δ i,n λF n − ( λ ) . Proof.
These identities can be proven routinely using the holonomic systems ap-proach [26]. We have carried out the necessary calculations using the Holonomic-Functions package [11, 12]. The results are documented in the supplementary elec-tronic material [13].First we derive, using holonomic closure properties and creative telescoping, a (leftGr¨obner) basis for the set of recurrence equations that p (0) n,j satisfies. Again ap-plying closure properties (in this case for multiplication) one obtains recurrencesfor the product a (0) i,j p (0) n,j , and by creative telescoping, for its definite sum, whichwe denote by l i,n (it is the left-hand side of the first identity). Here we face theproblem of poles inside the summation range that are introduced by the certificateof the telescopic relation. We solve this issue by constructing a different certificate,free of the problematic denominators, using an ansatz reminiscent of the polyno-mial ansatz [11, Sec. 3.4]. The recurrences for l i,n have the following form (somepolynomial coefficients are omitted for space reasons):16 n ( n + 1) (2 n + 1) (2 n + 3) (4 n + 1)( i − n + 1) (2 i + 2 n + 3) × (cid:0) i + 2 i + λ − n − n (cid:1) l i,n +2 = ( · · · ) l i +1 ,n + ( · · · ) l i,n +1 + ( · · · ) l i,n , n (2 n + 1)( i − n + 1)(2 i + 2 n + 3) (cid:0) i + 2 i + λ − n − n (cid:1) l i +1 ,n +1 =( · · · ) l i +1 ,n + ( · · · ) l i,n +1 + ( · · · ) l i,n , n − n (2 n − n + 1)(4 n + 1) (4 n + 3)( i − n + 1) ( i − n + 2)(2 i + 2 n + 3) × (cid:0) i + 2 i + λ − n − n (cid:1) l i +2 ,n = ( · · · ) l i +1 ,n + ( · · · ) l i,n +1 + ( · · · ) l i,n . From their support and their leading coefficients it becomes clear that when wewant to use them to compute l i,n for all 1 ≤ i < n , then we have to give the initialconditions l , , l , , l , , and l , . By verifying that they all equal 0 we have shownthat the first identity holds for i < n .For i = n we can construct, by holonomic substitution, a univariate recurrencesatisfied by l n,n . It turns out that the corresponding operator is a left multiple ofthe second-order operator that annihilates F n . Also in this case, the proof can becompleted by checking a few initial conditions. The proof of the second identity isestablished in an analogous way. (cid:3) CHRISTOPH KOUTSCHAN, MARTIN NEUM¨ULLER, AND CRISTIAN-SILVIU RADU
Lemma 3.4.
The following determinant evaluations hold for all nonnegative inte-gers n : det ≤ i,j ≤ n (cid:18) i + 2 j − (cid:19) = 12 n n (cid:89) i =1 (cid:0) ( i − (cid:1) (cid:0) i + (cid:1) n (cid:16) = h (0) n (cid:17) , det ≤ i,j ≤ n (cid:18) i + 2 j − (cid:19) = 12 n n (cid:89) i =1 (cid:0) ( i − (cid:1) (cid:0) i − (cid:1) n (cid:16) = h (1) n (cid:17) . Proof.
These determinants are special cases of Cauchy’s classic double alternant [3]det ≤ i,j ≤ n (cid:18) x i + y j (cid:19) = (cid:89) ≤ i 1) = n (cid:89) i =1 (cid:0) n − i ( n − i )! (cid:1) n (cid:89) i =1 n (cid:0) i + (cid:1) n = 12 n n (cid:89) i =1 (cid:0) ( i − (cid:1) (cid:0) i + (cid:1) n . The second assertion is derived in a completely analogous way. Note also that thesetwo determinants can be proven routinely using the holonomic ansatz [27]. (cid:3) Proof of Theorem 3.1. Lemma 3.3 shows that the vector (cid:0) p ( (cid:96) ) n, , . . . , p ( (cid:96) ) n,n (cid:1) T is, up toa scalar multiple, the n -th column of (cid:0) A ( (cid:96) ) n (cid:1) − for (cid:96) = 0 , 1. Since the entries ofthis vector (and of course those of the matrices A ( (cid:96) ) itself) are polynomials in λ ,this shows that det (cid:0) A (0) n (cid:1) | F n ( λ ) and that det (cid:0) A (1) n (cid:1) | λF n − ( λ ). Note that bothpolynomials F n ( λ ) and λF n − ( λ ) have degree n in λ . Next we argue that also thedeterminants of A (0) n and A (1) n have degree n in λ , which is the maximal possible—taking into account that the matrix entries are linear polynomials in λ . Observe thatthe matrix entries in Lemma 3.4 are precisely lim λ →∞ − a ( (cid:96) ) i,j /λ . Thus Lemma 3.4implies that det (cid:0) A ( (cid:96) ) n /λ (cid:1) = λ − n det (cid:0) A ( (cid:96) ) n (cid:1) converges to a nonzero constant (onlydepending on n ) as λ goes to infinity. Hence deg λ (cid:0) det A ( (cid:96) ) n (cid:1) = n for (cid:96) = 0 , λ . By noting that the polynomials F n ( λ ) are monic andthat the expressions given in Lemma 3.4 are, up to sign, the leading coefficients ofdet (cid:0) A (0) n (cid:1) and det (cid:0) A (1) n (cid:1) , respectively, the assertion of the theorem is proven. (cid:3) Note that our proof of the determinant evaluations in Theorem 3.1 is very rem-iniscent of Zeilberger’s holonomic ansatz [27]. In fact, the only difference is that NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 9 we chose to normalize the vector v n = (cid:0) p (0) n, , . . . , p (0) n,n (cid:1) T in a different way as Zeil-berger would do it: while he suggests the normalization p (0) n,n = 1, we normalize v n such that A (0) n v n = (cid:0) , . . . , , q n ( λ ) (cid:1) T and q n ( λ ) is a monic polynomial withdeg λ ( q n ) = n . (The same discussion applies to A (1) n , of course.)In the original formulation of the holonomic ansatz, i.e., with the normalization p (0) n,n = 1, the final result in the case of success is a holonomic recurrence, i.e., a linearrecurrence with polynomial coefficients, for det (cid:0) A (0) n +1 (cid:1) / det (cid:0) A (0) n (cid:1) . However, thisansatz is not at all guaranteed to succeed: even if the matrix entries are holonomic,this doesn’t mean that the sequence of quotients of consecutive determinants is aholonomic sequence. The determinant of A (0) n is such an example: the polynomials (cid:0) F n ( λ ) (cid:1) n ≥ satisfy the second-order recurrence(4 n + 3) F n +4 ( λ ) + (4 n + 5)(16 n + 40 n − λ + 21) F n +2 ( λ )+ (4 n + 7) λ F n ( λ ) = 0 , which means that (most likely) the quotient F n +2 ( λ ) /F n ( λ ) doesn’t satisfy aholonomic recurrence of any order. (We have strong evidence that this quotient isnon-holonomic, but we haven’t tried to prove this rigorously.) Provided that thisis true, the original holonomic ansatz must fail.Thanks to the additional parameter λ that appears polynomially in the matrixentries, we can identify the determinant of A (0) n in the denominators of the in-verse matrix. Thus a natural normalization of the vector v n would be such that A (0) n v n = (cid:0) , . . . , , det A (0) n (cid:1) T . In that case, the final result would be a holonomicrecurrence for det A (0) n ; hence this variant is applicable when the determinant itselfis a holonomic sequence in n . Unfortunately, that’s not the case for the matrix A (0) n because of the non-holonomic prefactor h (0) n . This explains why we had to choose yetanother normalization, in order to separate the holonomic and the non-holonomicpart of the determinant. For each part then we had to prove a different determinantevaluation: for the holonomic “polynomial part” this was done in Lemma 3.3, forthe non-holonomic “constant part” in Lemma 3.4. It is not unlikely that there aremany more examples of determinants where the original holonomic ansatz fails, butwhere the modifications described here lead to success.At the end of this section we want to briefly discuss an alternative way to derivethe polynomials F n ( λ ). In our above considerations we started with the monomialbasis when formulating the eigenvalue problem. Alternatively, one could employthe Legendre basis leading to the following determinant: D n = det ≤ i,j ≤ n (cid:18)(cid:90) − P (cid:48) i ( x ) P (cid:48) j ( x ) d x − λ (cid:90) − P i ( x ) P j ( x ) d x (cid:19) (note that only the matrix entries on the main diagonal depend on λ ). Indeedany basis ( ϕ k ) ≤ k ≤ n for the space ˆ V n can be used for the computation of theeigenvalues given in (7). So by construction, this determinant leads to the samefamily of polynomials F n ( λ ), and in fact we have that λ det( B n − λA n ) /D n +1 does not depend on λ . Doing the same block decomposition as before, we obtain thetwo families of matrices (cid:18) m (2 m + 1) − δ i,j λ i + 1 (cid:19) ≤ i,j ≤ n and (cid:18) m (2 m − − δ i,j λ i − (cid:19) ≤ i,j ≤ n where m stands for min( i, j ), whose determinants are given by( − n n (cid:0) (cid:1) n F n +1 ( λ ) resp. ( − n n (cid:0) (cid:1) n F n ( λ ) . Note that these determinants are “nicer” than the ones we considered above, be-cause their leading coefficients form holonomic sequences (actually they are hyper-geometric). So it seems that we should have started with this formulation. Butthere is also a drawback: the matrix entries are defined in terms of min( i, j ), whichon the one hand yields nicely structured matrices (constant along “hooks”, with aperturbation on the diagonal) such as − x . . . − x 12 12 12 . . . − x 30 30 . . . − x . . . − x . . . ... ... ... ... ... . . . , but on the other hand requires case distinctions that make the proofs of the relevantidentities (the analog of Lemma 3.3) more complicated.4. Upper and Lower Bounds on the Maximal Root of F n ( λ )In this section we give lower and upper bounds on the maximal root of F n ( λ ).Recall that we are interested in the maximal root ofdet( B n − λA n ) = c n λ F n ( λ ) F n − ( λ ) . We will prove that the maximal root of det( B n − λA n ) is equal to the maximal rootof F n ( λ ). We prove this in Lemma 4.6 which is based on Lemmas 4.2, 4.3, and 4.5,which are technical in nature. A lower and an upper bound on the maximal rootof F n ( λ ) are given in Lemma 4.5. A better upper bound is given in Lemma 4.8.These two lemmas are based on Lemma 4.3. Recall the definition of ν ( n ) = (cid:98) n (cid:99) . Definition 4.1. To simplify notation in this section, we introduce the polynomials (14) f j ( n ) := ( n − j + 1) j j (2 j )! , which correspond (up to sign) to the coefficients of F n ( λ ) : F n ( λ ) = ν ( n ) (cid:88) j =0 ( − j f j ( n ) λ ν ( n ) − j = λ ν ( n ) − f ( n ) λ ν ( n ) − + f ( n ) λ ν ( n ) − − . . . NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 11 In particular, we have f ( n ) = ( n − n ( n − n + 1)( n + 2)8 ,f ( n ) = ( n − ,f ( n ) = ( n − . Lemma 4.2. Let n ∈ N with n ≥ . If λ ∈ R is a root of F n with λ > f ( n ) then F n +1 ( λ ) < .Proof. We distinguish two cases depending on the parity of n . Case n = 2 k + 2 . We have that ν ( n ) = k + 1 and ν ( n + 1) = k + 1. Define G n ( x ) := F n +1 ( x ) − F n ( x ) = k (cid:88) j =0 ( − j − k (2 k − j + 3) k +2 (2 j + 1)! ( j − k − x j . Our goal is to show that G n ( x ) < x > f ( n ); the claim then follows imme-diately by using the assumption that λ is a root of F n . For this purpose we define g j ( n ) to be the absolute value of the coefficient of x j in G n ( x ) so that g j ( n ) = 4 j − k (2 k − j + 3) k +2 (2 j + 1)! ( k − j + 1) . We now want to prove that λ g j ( n ) > g j − ( n ) for 1 ≤ j ≤ k and λ > f ( n ), whichis implied by 12 f ( n ) g j ( n ) > g j − ( n ) , (1 ≤ j ≤ k ) . Substituting for g j ( n ) we obtain12 f ( n )4 j − k ( k − j + 1)(2 k − j + 3) k +2 (2 j + 1)! > j − − k ( k − j + 2)(2 k − j + 5) k +2 (2 j − . Multiplying this inequality by (2 j − j − − k (2 k − j + 5) k , weobtain2 f ( n ) ( k − j + 1)(2 k − j + 3)(2 k − j + 4)2 j (2 j + 1) > ( k − j + 2)(4 k − j + 5)(4 k − j + 6) . Plugging in f ( n ) = (2 k + 1)(2 k + 2)(2 k + 3)(2 k + 4) and substituting j → k − j leads to(16 j + 72 j + 88 j + 16) k + (80 j + 360 j + 424 j + 48) k + (172 j + 774 j + 906 j + 92) k + (196 j + 882 j + 1070 j + 180) k − j − j − j + 16 j + 276 j + 72 > ≤ j ≤ k − 1. Since k > j , the above inequality is true if it is true for k = j .Substituting k = j yields16 j + 152 j + 604 j + 1298 j + 1624 j + 1178 j + 456 j + 72 > , which is obviously true for all j ≥ 0. Now note that G n ( λ ) = F n +1 ( λ ) because F n ( λ ) = 0 by our assumption on λ . Finally note that if k is even then G n ( λ ) = − g (cid:124)(cid:123)(cid:122)(cid:125) < k/ (cid:88) j =1 (cid:0) − g j ( n ) λ + g j − ( n ) (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) < λ j − < , and if k is odd then G n ( λ ) = ( k − / (cid:88) j =0 (cid:0) − g j +1 ( n ) λ + g j ( n ) (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) < λ j < . Case n = 2 k + 1 . We have that ν ( n ) = k and ν ( n + 1) = k + 1. This time let G n ( x ) := F n +1 ( x ) − xF n ( x ) = k (cid:88) j =0 ( − j − k +1 j − k (2 k − j + 3) k +1 (2 j )! ( k − j + 1) (cid:124) (cid:123)(cid:122) (cid:125) =: g j ( n ) x j , and denote by g j ( n ) the absolute value of the coefficient of x j in G n , as before.Again, we want to prove that G n ( x ) < x > f ( n ), but we don’t want torepeat all the arguments from the first case. Instead we discuss how this proofcan be supported by computer algebra techniques. First we want to point out that G n ( x ), being defined as a hypergeometric sum, satisfies a linear recurrence equation,and that inequalities involving such quantities can be proven algorithmically [8].However, our experiments suggest that the present example is computationallytoo expensive, and therefore we let computer algebra enter at a later stage of theproof. In order to prove that λ g j ( n ) > g j − ( n ) for λ > f ( n ) and 1 ≤ j ≤ k , wefocus on the stronger statement f ( n ) g j ( n ) /g j − ( n ) > 1. Elementary calculationsexploiting the hypergeometric nature of g j ( n ) that are analogous to the previouscase lead to the rational function inequality k (2 k + 1)(2 k + 2)(2 k + 3)( k − j + 1)(2 k − j + 3)( k − j + 2)4 j (2 j − k − j + 2)(2 k − j + 2)(4 k − j + 5) > . Now we employ cylindrical algebraic decomposition [5] to establish the correctnessof the previous inequality: naming it ineq , the Mathematica command CylindricalDecomposition[Implies[1 <= j <= k, ineq], {j, k}] yields True in a fraction of a second [13].By the assumption on λ we have that G n ( λ ) = F n +1 ( λ ) because λF n ( λ ) = 0. Theproof is concluded by noting that if k is even then G n ( λ ) = − g (cid:124)(cid:123)(cid:122)(cid:125) < k/ (cid:88) j =1 (cid:0) − g j ( n ) λ + g j − ( n ) (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) < λ j − < k is odd then G n ( λ ) = ( k − / (cid:88) j =0 (cid:0) − g j +1 ( n ) λ + g j ( n ) (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) < λ j < . (cid:3) NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 13 Lemma 4.3. If λ > f ( n ) , then λ f j ( n ) > f j +1 ( n ) for ≤ j ≤ ν ( n ) − .Proof. The problem is equivalent to proving f ( n ) f j ( n ) /f j +1 ( n ) > 1. Substitut-ing (14) for f j ( n ) we obtain( j + 1)(2 j + 1)( n − n ( n + 1)( n + 2)2( n − j − n − j )( n + 2 j + 1)( n + 2 j + 2) > . Again, this can be proven routinely using cylindrical algebraic decomposition [13].Alternatively, we clear denominators and collect terms:(15) (cid:0) j + 3 j − (cid:1) n + 2 (cid:0) j + 3 j − (cid:1) n + (cid:0) j + 13 j + 1 (cid:1) n +2(2 j + 1)(3 j + 1) n − j ( j + 1)(2 j + 1) > . Now observe that 1 ≤ j ≤ ν ( n ) − n > j + 2 and that (15) holds for all n > j + 2 if one can show that it holds for n = 2 j + 2. Substituting n = 2 j + 2into (15) gives 32 j + 208 j + 536 j + 700 j + 424 j + 60 j − > j ≥ (cid:3) Definition 4.4. For n ≥ we define λ n to be the maximal root of F n ( λ ) . We are now ready to give an upper and a lower bound for λ n . Lemma 4.5. For n ≥ the maximal root λ n satisfies m ( n ) ≤ λ n ≤ f ( n ) with m ( n ) := f ( n )2 + (cid:114) f ( n ) − f ( n )= f ( n )2 (cid:32) (cid:115) − 23 ( n − n − n + 3)( n + 4) n ( n − n + 1)( n + 2) (cid:33) . Moreover, λ n < f ( n ) for n ≥ and m ( n ) < λ n for n ≥ .Proof. If n ∈ { , } then obviously m ( n ) = λ n = f ( n ) holds. So let now n ≥ ν := ν ( n ), i.e., ν ≥ 2. In Lemma 4.3 we proved that if λ > f ( n )then λ f j ( n ) > f j +1 ( n ). Consequently, under this assumption on λ , we get: if ν iseven then ν (cid:88) j =2 ( − j f j ( n ) λ ν − j = ν/ − (cid:88) k =1 (cid:0) λ f k ( n ) − f k +1 ( n ) (cid:1) λ ν − k − (cid:124) (cid:123)(cid:122) (cid:125) > f ν ( n ) (cid:124) (cid:123)(cid:122) (cid:125) > > , and if ν is odd then ν (cid:88) j =2 ( − j f j ( n ) λ ν − j = ( ν − / (cid:88) k =1 (cid:0) λ f k ( n ) − f k +1 ( n ) (cid:1) λ ν − k − (cid:124) (cid:123)(cid:122) (cid:125) > > . In particular let now λ ≥ f ( n ). Then F n ( λ ) = λ ν − f ( n ) λ ν − (cid:124) (cid:123)(cid:122) (cid:125) ≥ ν (cid:88) j =2 ( − j f j ( n ) λ ν − j > . Therefore the maximal root of F n ( λ ) cannot exceed f ( n ), which proves the upperbound. Analogously one finds that ν (cid:88) j =3 ( − j f j ( n ) λ ν − j ≤ , which is strict for all n ≥ 6. Then for λ = f ( n ) + (cid:113) f ( n ) − f ( n ) > f ( n )we have F n ( λ ) = λ ν − f ( n ) λ ν − + f ( n ) λ ν − (cid:124) (cid:123)(cid:122) (cid:125) = 0 + ν (cid:88) j =3 ( − j f j ( n ) λ ν − j ≤ . Since F n ( λ ) ≤ x →∞ F n ( x ) = + ∞ , the polynomial F n ( x ) has a root for x ≥ λ . This proves the lower bound. (cid:3) Lemma 4.6. Let n ≥ . Then λ n +1 > λ n .Proof. By Lemma 4.5 we have that λ n > f ( n ). Then by Lemma 4.2 we have that F n +1 ( λ n ) < 0. Since by definition lim x →∞ F n ( x ) = + ∞ , it follows that between λ n and + ∞ the function F n +1 ( x ) takes the value 0 at some point x . In particular λ n < x ≤ λ n +1 . (cid:3) Corollary 4.7. The maximal root of det( B n − λA n ) is equal to the maximal rootof F n ( λ ) . Lemma 4.8. For n ≥ the maximal root λ n satisfies λ n ≤ M ( n ) with M ( n ) := f ( n )3 + (cid:18) f ( n ) (cid:16) p ( n ) + (cid:112) p ( n ) (cid:17)(cid:19) / + (cid:18) f ( n ) (cid:16) p ( n ) − (cid:112) p ( n ) (cid:17)(cid:19) / where the polynomials p and p are given by p ( n ) := 14320 (cid:0) n + 4 n + 8 n + 10 n + 404 n + 796 n − n − n + 16200 (cid:1) ,p ( n ) := 1597196800 ( n − n − n + 3)( n + 4) (cid:0) n + 42 n − n − n − n + 10198 n − n − n + 4705644 n +9619080 n − n − n + 116640000 (cid:1) . Moreover, we have λ n < M ( n ) if n ≥ .Proof. The equality λ n = M ( n ) is easily established for 2 ≤ n ≤ 5, by using thesymbolic simplification capabilities of Mathematica. Next, for n ≥ F n ( λ ) as(16) F n ( λ ) = λ ν ( n ) − (cid:0) λ − f ( n ) λ + f ( n ) λ − f ( n ) (cid:1) + ν ( n ) (cid:88) j =4 ( − j f j ( n ) λ ν ( n ) − j . Obviously, the sum in (16) equals zero if n ∈ { , } since then ν ( n ) = 3. By asimilar argument as in the proof of Lemma 4.5, one sees that this sum is strictly NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 15 positive for n ≥ 8, provided that λ > f ( n ). Note that the maximal root of thepolynomial(17) λ − f ( n ) λ + f ( n ) λ − f ( n )is greater than f ( n ) because the lower bound is the same as the one derived inLemma 4.5 using the same arguments as in its proof. The roots of this third-degreepolynomial can be computed by using Cardano’s formulas, namely we want to solve x + bx + cx + d = 0. The roots of this polynomial are given by y i − b/ i = 1 , , y := α + β and y , := − α + β ± i α − β √ , where α := (cid:18) − Q √ ∆ (cid:19) / and β := (cid:18) − Q − √ ∆ (cid:19) / , where Q := 2 b − bc d and ∆ := (cid:18) (cid:0) c − b (cid:1)(cid:19) + (cid:18) Q (cid:19) . Setting λ = x , b = − f ( n ), c = f ( n ) and d = − f ( n ) we obtain λ i = y i − b/ < > n ≥ 10 and the real root is y − b/ 3. For thecases n = 6 , , , y − b/ 3. By straight-forward calculations one can verify that y − b/ M ( n ) and we get F n ( M ( n )) = 0 for n ∈ { , } . For n ≥ λ ≥ M ( n )we have that (17) is nonnegative which together with the statements about the sumin (16) implies that F n ( λ ) > (cid:3) Since we have proven that for n ≥ m ( n ) < λ n < M ( n ) it follows (fromdividing the inequality by f ( n ) and taking the limit n → ∞ ) that12 + 12 (cid:114) (cid:124) (cid:123)(cid:122) (cid:125) ∼ . ≤ lim n →∞ λ n f ( n ) ≤ 13 + (cid:18) (cid:114) (cid:19) / + (cid:18) − (cid:114) (cid:19) / (cid:124) (cid:123)(cid:122) (cid:125) ∼ . . The previous lemmas indicate how to obtain a sequence of better and better boundsfor λ n : while in Lemma 4.5 the root of the polynomial given by the first three termsof F n ( λ ) yields a lower bound, Lemma 4.8 gives an upper bound by considering thefirst four terms. A more accurate lower bound would follow from taking the firstfive terms, then a better upper bound from the first six terms, etc.5. Asymptotic Behaviour of the Roots Since the matrices M n and K n defined in (8)–(9) are symmetric, it follows thatthe polynomials F n ( λ ) defined in (12) have only real roots, all of which are posi-tive because the coefficients of F n ( λ ) are alternating. When we plot the roots fordifferent n ∈ N we get a very interesting picture, see Figure 1. Moreover, one seesthat the smallest root of F n ( λ ) converges to a specific value as n goes to infinity,and the same is true for the smallest root of F n +1 ( λ ). The situation is similar Figure 1. Distribution of the roots of F n ( λ ) for 2 ≤ n ≤ 50 on alogarithmic scale; for even n the locations of the roots are markedby crosses, for odd n with squares.when considering the second-smallest root, the third-smallest root, and so on. Thefollowing proposition makes this observation precise. Proposition 5.1. Let F n ( λ ) be defined as in (12) and let λ (0) n, < · · · < λ (0) n,n denotethe roots of F n ( λ ) in increasing order, and similarly λ (1) n, < · · · < λ (1) n,n denote theroots of F n +1 ( λ ) . Then for fixed k ∈ N we have lim n →∞ λ (0) n,k = (cid:0) k − (cid:1) π and lim n →∞ λ (1) n,k = k π . Proof. The coefficient of λ j in F n ( λ ) is, according to (12), given by( − j − n (2 n − j + 1) n (2 j )! . We normalize the monic polynomials F n such that their constant coefficient is 1,i.e., we divide F n by ( − − n (2 n + 1) n , and obtain for the coefficient of λ j in thesenormalized polynomials:( − j (2 n − j + 1) n (2 j )!(2 n + 1) n = ( − j (2 j )! · j (2 n − j + 1) j (4 n − j + 1) j . Obviously the second factor is, for fixed j , a rational function in n with numeratorand denominator having the same degree 2 j and the same leading coefficient 16 j ;hence it tends to 1 as n goes to infinity. This means that the power series obtainedas the limit of the normalized polynomials is ∞ (cid:88) j =0 ( − j (2 j )! x j = cos( √ x )whose roots are precisely the limiting values in the assertion. The limit of F n +1 ( λ )can be computed analogously and yields the Taylor expansion of sin( √ x ) / √ x . (cid:3) Recall that we are actually not interested in the smallest root of F n ( λ ) but in thelargest one. Its asymptotic behaviour can be extracted in a similar fashion. NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 17 Proposition 5.2. Let F n ( λ ) be defined as in (12) and let λ n denote the largestroot of F n , as before. Then lim n →∞ λ n n = 1 π . Proof. Let ˆ F n ( λ ) denote the reciprocal polynomial of F n ( λ ), which means thatˆ F n ( λ ) = λ ν ( n ) F n (1 /λ ) where ν ( n ) = (cid:98) n/ (cid:99) is the degree of F n . Then the largestroot of F n equals the reciprocal of the smallest root of ˆ F n . Now consider the familyof polynomials ˆ F n (cid:18) λn (cid:19) = ν ( n ) (cid:88) j =0 (2 j + 1) n ( − n ) j ( n − j )! λ j . The coefficient of λ j in these polynomials tends to ( − − j / (2 j )! as n goes to infinity.Hence in the limit we obtain the power series ∞ (cid:88) j =0 x j ( − j (2 j )! = cos (cid:18) √ x (cid:19) , whose smallest root is π . The claim follows. (cid:3) Note that this result is in accordance with the bounds derived in Section 4, inparticular with the inequality stated at the end of that section: the numerical valueof 8 π − is approximately 0 . F n ( λ ) was taken (in Lemma 4.8), whereasthe lower bound was obtained from a second-degree polynomial (see Lemma 4.5).6. The Boundary Estimate Finally we tackle the second kind of problem, corresponding to Equation (4). Inthis instance it is advantageous to formulate it using the Legendre basis. Thus wehave to solve the eigenvalue problem L n x n = µ n M n x n with the following n × n matrices L n and M n : the ( i, j ) entry of L n is given by P i (1) P j (1) + P i ( − P j ( − M n one has (cid:90) − P i ( x ) P j ( x ) d x. Here the basis functions are the venerable Legendre polynomials P n ( x ). Takinginto account the well-known evaluations P n (1) = 1 and P n ( − 1) = ( − n this isequivalent to finding the roots of the determinant of C n = ( c i,j ) ≤ i,j ≤ n whosematrix entries are given by c i,j := 1 + ( − i + j − δ i,j µ i + 1 . Obviously the matrix C n has zeros at all positions ( i, j ) for which i + j is an oddinteger. As in Section 3 we decompose it into block form and obtaindet( C n ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) C (0) (cid:98) n/ (cid:99) C (1) (cid:100) n/ (cid:101) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = det (cid:16) C (0) (cid:98) n/ (cid:99) (cid:17) · det (cid:16) C (1) (cid:100) n/ (cid:101) (cid:17) where the subscripts indicate the dimension of the square matrices C (0) and C (1) ,whose entries are independent of the dimension and given by c (0) i,j := 2 − δ i,j µ i + 1 and c (1) i,j := 2 − δ i,j µ i − . Theorem 6.1. For all nonnegative integers n we have det (cid:0) C (0) n (cid:1) = ( − n n (cid:0) (cid:1) n µ n − (cid:0) µ − n − n (cid:1) , det (cid:0) C (1) n (cid:1) = ( − n n (cid:0) (cid:1) n µ n − (cid:0) µ − n − n (cid:1) . Proof. By some elementary row operations, the matrix C (0) n is brought to triangularform. First we subtract the first row from rows 2 through n , obtaining the followingmatrix: the (1 , 1) entry is 2 − µ , the remaining entries in the first row are 2, theremaining entries of the first column are µ , and the diagonal entries ( i, i ) are − µ i +1 for i > 1; the rest are zeros. So in order to transform the matrix to lower triangularform, we multiply row i , for 2 ≤ i ≤ n , by i +1 µ and add it to the first row. Thusthe (1 , − µ n (cid:88) i =2 µ i + 1 µ = 25 (cid:0) n + 3 n − µ (cid:1) . It follows that the determinant of C (0) n is25 (cid:0) n + 3 n − µ (cid:1) n (cid:89) i =2 − µ i + 1 = ( − n n (cid:0) (cid:1) n µ n − (cid:0) µ − n − n (cid:1) , as claimed. The evaluation of det (cid:0) C (1) n (cid:1) is obtained in a completely analogousway. (cid:3) Corollary 6.2. For all nonnegative integers n we have det( C n ) = det ≤ i,j ≤ n (cid:16) − i + j − δ i,j µ i + 1 (cid:17) = ( − n (cid:0) (cid:1) n µ n − (cid:16) µ − (cid:106) n (cid:107) − (cid:106) n (cid:107)(cid:17)(cid:16) µ − (cid:108) n (cid:109) − (cid:108) n (cid:109)(cid:17) = ( − n (cid:0) (cid:1) n µ n − (cid:40)(cid:0) µ − n +3 n (cid:1)(cid:0) µ − n + n (cid:1) , if n is even , (cid:0) µ − n +3 n +22 (cid:1)(cid:0) µ − n + n − (cid:1) , if n is odd . The previous corollary now gives an answer to the original eigenvalue problem,namely that the largest eigenvalue µ n of L n x n = µ n M n x n is µ n = (cid:40) n ( n + 3) if n is even , n ( n + 3) + 1 if n is odd . NVERSE INEQUALITY ESTIMATES WITH SYMBOLIC COMPUTATION 19 Outlook and future work In this work we have presented tools from symbolic computation to give preciseestimates for two types of inverse inequalities. 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Christoph Koutschan, Johann Radon Institute for Computational and Applied Math-ematics (RICAM), Austrian Academy of Sciences, Altenberger Straße 69, 4040 Linz,Austria E-mail address : [email protected] Martin Neum¨uller, Institute of Computational Mathematics, Johannes Kepler Univer-sity, Altenberger Straße 69, 4040 Linz, Austria E-mail address : [email protected] Cristian-Silviu Radu, Research Institute for Symbolic Computation (RISC), JohannesKepler University, Altenberger Straße 69, 4040 Linz, Austria E-mail address ::