aa r X i v : . [ m a t h - ph ] M a r Symmetry, Integrability and Geometry: Methods and Applications SIGMA (2011), 022, 12 pages Beyond the Gaussian
Kazuyuki FUJIIDepartment of Mathematical Sciences, Yokohama City University, Yokohama, 236-0027 Japan
E-mail: [email protected]
Received January 12, 2011, in final form February 28, 2011; Published online March 04, 2011doi:10.3842/SIGMA.2011.022
Abstract.
In this paper we present a non-Gaussian integral based on a cubic polynomial,instead of a quadratic, and give a fundamental formula in terms of its discriminant. Itgives a mathematical reinforcement to the recent result by Morozov and Shakirov. We alsopresent some related results. This is simply one modest step to go beyond the Gaussian butit already reveals many obstacles related with the big challenge of going further beyond theGaussian.
Key words: non-Gaussian integral; renormalized integral; discriminant; cubic equation
The
Gaussian is an abbreviation of all subjects related to the Gauss function e − ( px + qx + r ) likethe Gaussian beam, Gaussian process, Gaussian noise, etc. It plays a fundamental role in math-ematics, statistics, physics and related disciplines. It is generally conceived that any attempts togeneralize the Gaussian results would meet formidable difficulties. Hoping to overcome this highwall of difficulties of going beyond the Gaussian in the near future, a first step was introducedin [1]. This paper is its polished version.In the paper [2] the following “formula” is reported: Z Z e − ( ax + bx y + cxy + dy ) dxdy = 1 √− D , (1)where D is the discriminant of the cubic equation ax + bx + cx + d = 0 , and it is given by D = b c + 18 abcd − ac − b d − a d . (2)The formula (1) is of course non-Gaussian. However, if we consider it in the framework of thereal category then (1) is not correct because the left hand side diverges. In this paper we treatonly the real category, and so a , b , c , d , x , y are real numbers.Formally, by performing the change of variable x = tρ , y = ρ for (1) we havel.h.s. of (1) = Z Z e − ρ ( at + bt + ct + d ) | ρ | dtdρ = Z (cid:26)Z e − ( at + bt + ct + d ) ρ | ρ | dρ (cid:27) dt = Z | σ | e − σ dσ Z (cid:12)(cid:12) p ( at + bt + ct + d ) (cid:12)(cid:12) p ( at + bt + ct + d ) dt by the change of variable σ = √ at + bt + ct + dρ . K. FujiiThe divergence comes from Z | σ | e − σ dσ, while the main part is Z (cid:12)(cid:12) p ( ax + bx + cx + d ) (cid:12)(cid:12) p ( ax + bx + cx + d ) dx under the change t → x . As a kind of renormalization the integral may be defined like ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ = Z R (cid:12)(cid:12) p ( ax + bx + cx + d ) (cid:12)(cid:12) p ( ax + bx + cx + d ) dx. However, the right hand side lacks proper symmetry. If we set F ( a, b, c, d ) = Z Z D R e − ( ax + bx y + cxy + dy ) dxdy, where D R = [ − R, R ] × [ − R, R ], then it is easy to see F ( − a, − b, − c, − d ) = F ( a, b, c, d ) . Namely, F is invariant under Z -action. This symmetry is important and must be kept evenin the renormalization process. The right hand side in the “definition” above is clearly notinvariant. Therefore, by modifying it slightly we reach the renormalized integral Definition 1. ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ = Z R p ( ax + bx + cx + d ) dx. (3)We believe that the definition is not so bad (see the Section 4).In the paper we calculate the right hand side of (3) directly , which will give some interestingresults and a new perspective. The result gives a mathematical reinforcement to the result [2]by Morozov and Shakirov. Before stating the result let us make some preparations. The Gamma function Γ( p ) is definedby Γ( p ) = Z ∞ e − x x p − dx ( p >
0) (4)and the Beta function B ( p, q ) is B ( p, q ) = Z x p − (1 − x ) q − dx ( p, q > . Note that the Beta function is rewritten as B ( p, q ) = Z ∞ x p − (1 + x ) p + q dx. See [3] for more detail. Now we are in a position to state the result.eyond the Gaussian 3
Fundamental formula. (I) For
D < Z R p ( ax + bx + cx + d ) dx = C − √− D , (5)where C − = √ B (cid:18) , (cid:19) . (II) For D > Z R p ( ax + bx + cx + d ) dx = C + √ D , (6)where C + = 3 B (cid:18) , (cid:19) . (III) C − and C + are related by C + = √ C − through the identity √ B (cid:18) , (cid:19) = √ B (cid:18) , (cid:19) . (7)Our result shows that the integral depends on the sign of D , and so our question is as follows. Problem.
Can the result be derived from the method developed in [2] ? A comment is in order. If we treat the Gaussian case ( e − ( ax + bxy + cy ) ) then the integral isreduced to Z R ax + bx + c dx = 2 π √− D (8)if a > D = b − ac <
0. Noting π = √ π √ π )Γ( )Γ(1) = B (cid:18) , (cid:19) (8) should be read as Z R ax + bx + c dx = 2 B (cid:0) , (cid:1) √− D .
The proof is delicate. In order to prevent possible misunderstanding we present a detailed proofin this section.
Proof of (I).
We prove (5) in case of
D < a = 0 in the cubic equation ax + bx + cx + d .Namely, we calculate the integral Z R p ( bx + cx + d ) dx. K. FujiiNoting − D = b (4 bd − c ) > Z R p ( bx + cx + d ) dx = Z R √ b q ( x + cb x + db ) dx = 1 √ b Z R r(cid:16) ( x + c b ) + db − c b (cid:17) dx = 1 √ b Z R r(cid:16) x + bd − c b (cid:17) dx T ≡ bd − c b > = 1 √ b Z R p ( x + T ) dx x = T y = 1 √ b Z R T √ T p ( y + 1) dy = 2 √ b T Z ∞ p ( y + 1) dy y = √ x = 2 √ b T Z ∞ p ( x + 1) dx √ x = 1 √ b T Z ∞ x − ( x + 1) dx = B (cid:0) , (cid:1) √ b T = B (cid:0) , (cid:1) √ b T = √ B (cid:0) , (cid:1) p b (4 bd − c ) = √ B (cid:0) , (cid:1) √− D . (9)Now we consider the general case of a = 0. From the condition D < ax + bx + cx + d = 0. Let us denote it by α . From the equation ax + bx + cx + d = ( x − α )( ax + kx + l ) , aα + bα + cα + d = 0we have easily b = k − aα, c = l − kα, d = − lα. (10)First we assume α = 0. In this case d = 0 and ax + bx + cx + d = x (cid:0) ax + bx + c (cid:1) . Then Z R p x ( ax + bx + c ) dx = Z ∞ p x ( ax + bx + c ) dx + Z −∞ p x ( ax + bx + c ) dx x = y = Z ∞ r y (cid:16) ay + by + c (cid:17) (cid:18) − dyy (cid:19) + Z −∞ r y (cid:16) ay + by + c (cid:17) (cid:18) − dyy (cid:19) = Z ∞ p ( cy + by + a ) dy + Z −∞ p ( cy + by + a ) dy = Z R p ( cy + by + a ) dy (9) ( b → c ; c → b ; d → a ) = √ B (cid:0) , (cid:1) p c (4 ac − b ) = √ B (cid:0) , (cid:1) √− D .
Next, let us calculate the case α = 0: Z R p ( x − α ) ( ax + kx + l ) dx x = y + α = Z R p y { a ( y + α ) + k ( y + α ) + l } dy = Z R p y { ay + (2 aα + k ) y + ( aα + kα + l ) } dy (10) = √ B (cid:0) , (cid:1) p ( aα + kα + l ) { a ( aα + kα + l ) − (2 aα + k ) } = √ B (cid:0) , (cid:1) p ( aα + kα + l ) (4 al − k ) . (11)eyond the Gaussian 5 Key Lemma.
From (10) the following equation holds (cid:0) aα + kα + l (cid:1) (cid:0) al − k (cid:1) = 27 a d + 4 ac − abcd − b c + 4 b d = − D. (12)The proof is straightforward but tedious.Therefore, from both (11) and (12) we obtain the formula Z R p ( x − α ) ( ax + kx + l ) dx = √ B (cid:0) , (cid:1) √− D . (cid:4)
Proof of (II).
We prove (6) in case of
D >
0. Let us start with the evaluation of the followingintegral Z R p x ( x − α ) dx for α >
0. Then Z R p x ( x − α ) dx = Z −∞ p x ( x − α ) dx + Z ∞ p x ( x − α ) dx = Z ∞ p x ( x + α ) dx + Z ∞ p x ( x − α ) dx, (13)where the change of variable x → − x for the first term of the right hand side was made.Each term can be evaluated elementarily: Z ∞ p x ( x + α ) dx x = αt = 1 √ α Z ∞ p t ( t + 1) dt = α − Z ∞ t − ( t + 1) dt = α − B (cid:18) , (cid:19) , while Z ∞ p x ( x − α ) dx x = αt = α − Z ∞ p t ( t − dt = α − (Z p t ( t − dt + Z ∞ p t ( t − dt ) = α − (Z p t (1 − t ) dt + Z ∞ p t ( t − dt ) = 2 α − Z p t (1 − t ) dt = 2 α − Z t − (1 − t ) − dt = 2 α − B (cid:18) , (cid:19) , where we have used Z ∞ p t ( t − dt t = s = Z q s (1 − s ) s (cid:18) − dss (cid:19) = Z p s (1 − s ) ds = Z p t (1 − t ) dt. From (13) we have Z R p x ( x − α ) dx = 3 α − B (cid:18) , (cid:19) = 3 B (cid:0) , (cid:1) √ α . K. FujiiNow we consider the special case a = 0 in the cubic equation ax + bx + cx + d . Then by D = b ( c − bd ) > Z R p ( bx + cx + d ) dx = Z ∞−∞ rn b ( x + c b ) − c − bd b o dx = Z ∞−∞ q ( bx − c − bd b ) dx α = c − bd b > = 1 √ b Z ∞−∞ p ( x − α ) dx = 1 √ b Z ∞−∞ p ( x − α ) ( x + α ) dx y = x + α = 1 √ b Z ∞−∞ p y ( y − α ) dy (13) = 1 √ b B (cid:0) , (cid:1) √ α = 3 B (cid:0) , (cid:1) √ αb = 3 B (cid:0) , (cid:1) √ α b = 3 B (cid:0) , (cid:1) p b ( c − bd ) = 3 B (cid:0) , (cid:1) √ D . (14)Next we consider the remaining general case of a = 0. From the condition D > ax + bx + cx + d = 0. We denote one of them by α .Remember the relations b = k − aα , c = l − kα , d = − lα from the equation ax + bx + cx + d = ( x − α ) (cid:0) ax + kx + l (cid:1) , aα + bα + cα + d = 0 . First we assume α = 0. Then ax + bx + cx + d = x ( ax + bx + c )and from D = c ( b − ac ) we have Z R p x ( ax + bx + c ) dx = Z ∞ p x ( ax + bx + c ) dx + Z −∞ p x ( ax + bx + c ) dx x = y = Z ∞ r y (cid:16) ay + by + c (cid:17) (cid:18) − dyy (cid:19) + Z −∞ r y (cid:16) ay + by + c (cid:17) (cid:18) − dyy (cid:19) = Z ∞ p ( cy + by + a ) dy + Z −∞ p ( cy + by + a ) dy = Z R p ( cy + by + a ) dy (14) ( c → b ; b → c ; a → d ) = 3 B (cid:0) , (cid:1) p c ( b − ac ) = 3 B (cid:0) , (cid:1) √ D . (15)For the case α = 0 we obtain the formula Z R p ( x − α ) ( ax + kx + l ) dx x = y + α = Z R p y { a ( y + α ) + k ( y + α ) + l } dy = Z R p y { ay + (2 aα + k ) y + ( aα + kα + l ) } dy (15) = 3 B (cid:0) , (cid:1) p ( aα + kα + l ) { (2 aα + k ) − a ( aα + kα + l ) } = 3 B (cid:0) , (cid:1) p ( aα + kα + l ) ( k − al ) (12) = 3 B (cid:0) , (cid:1) √ D . (cid:4)
Proof of (III).
We prove the relation (7). Let us make some preparations. For the Gammafunction (4) there are well-known formulas (see for example [3]) B ( x, y ) = Γ( x )Γ( y )Γ( x + y ) ( x, y > , (16)eyond the Gaussian 7Γ( x )Γ(1 − x ) = π sin( πx ) (0 < x < , (17)Γ (cid:16) x (cid:17) Γ (cid:18) x + 12 (cid:19) = √ π x − Γ( x ) = 2 − x Γ (cid:18) (cid:19) Γ( x ) . (18)(18) is called the Legendre’s relation. In the formula we set x = 2 /
3, thenΓ (cid:18) (cid:19) Γ (cid:18) (cid:19) = √ (cid:18) (cid:19) Γ (cid:18) (cid:19) . Multiplying both sides by Γ(1 /
6) givesΓ (cid:18) (cid:19) Γ (cid:18) (cid:19) Γ (cid:18) (cid:19) = √ (cid:18) (cid:19) Γ (cid:18) (cid:19) Γ (cid:18) (cid:19) ⇐⇒ Γ (cid:18) (cid:19) π sin( π ) = √ (cid:18) (cid:19) Γ (cid:18) (cid:19) Γ (cid:18) (cid:19) ⇐⇒ π Γ (cid:18) (cid:19) = √ (cid:18) (cid:19) Γ (cid:18) (cid:19) Γ (cid:18) (cid:19) ⇐⇒ π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) = √ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) ⇐⇒ π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) = √ B (cid:18) , (cid:19) ⇐⇒ π Γ (cid:0) (cid:1) π sin( π ) Γ (cid:0) (cid:1) = √ B (cid:18) , (cid:19) ⇐⇒ √ B (cid:18) , (cid:19) = √ B (cid:18) , (cid:19) , where we have used formulas (16) and (17) several times.The proof of (7) is now complete. (cid:4) In this section let us check whether the renormalized integral (3) is reasonable or not by makinguse of the results in the Section 3.In the introduction we introduced the following integral defined on D R = [ − R, R ] × [ − R, R ] F ( a, b, c, d ) = Z Z D R e − ( ax + bx y + cxy + dy ) dxdy. For this it is easy to see (cid:18) ∂∂a ∂∂d − ∂∂b ∂∂c (cid:19) F ( a, b, c, d )= Z Z D R (cid:0) x · y − x y · xy (cid:1) e − ( ax + bx y + cxy + dy ) dxdy = 0 , (cid:18) ∂∂b ∂∂b − ∂∂a ∂∂c (cid:19) F ( a, b, c, d )= Z Z D R (cid:0) x y · x y − x · xy (cid:1) e − ( ax + bx y + cxy + dy ) dxdy = 0 , K. Fujii (cid:18) ∂∂c ∂∂c − ∂∂b ∂∂d (cid:19) F ( a, b, c, d )= Z Z D R (cid:0) xy · xy − x y · y (cid:1) e − ( ax + bx y + cxy + dy ) dxdy = 0 . (19)On the other hand, if we set F ( a, b, c, d ) = Z R p ( ax + bx + cx + d ) dx = C ± √± D , then we can also verify the same relations: (cid:18) ∂∂a ∂∂d − ∂∂b ∂∂c (cid:19) F ( a, b, c, d ) = 0 , (cid:18) ∂∂b ∂∂b − ∂∂a ∂∂c (cid:19) F ( a, b, c, d ) = 0 , (cid:18) ∂∂c ∂∂c − ∂∂b ∂∂d (cid:19) F ( a, b, c, d ) = 0 . (20)Verification by hand is rather tough, but it can be done easily by use of MATHEMATICA .From (19) and (20) we can conclude that our renormalized integral (3) is reasonable enough. In this section we make some comments on the discriminant (2). See [4] for more details ([4] isstrongly recommended).For the equations f ( x ) = ax + bx + cx + d, f ′ ( x ) = 3 ax + 2 bx + c (21)the resultant R ( f, f ′ ) of f and f ′ is given by R ( f, f ′ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a b c d a b c d a b c a b c
00 0 3 a b c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (22)It is easy to calculate (22) and the result becomes1 a R ( f, f ′ ) = 27 a d + 4 ac − abcd − b c + 4 b d = − D. On the other hand, if α , β , γ are three solutions of f ( x ) = 0 in (21), then the followingrelations are well–known α + β + γ = − ba , αβ + αγ + βγ = ca , αβγ = − da . From these it is easy to see α + β + γ = − ba , α + β + γ = b − aca , α + β + γ = − b + 3 a d − abca ,α + β + γ = b + 4 a bd + 2 a c − ab ca . The author owes the calculation to Hiroshi Oike. eyond the Gaussian 9If we set∆ = ( α − β )( α − γ )( β − γ )the discriminant D is given by D = a ∆ . Let us calculate ∆ directly. For the Vandermonde matrix V = α β γα β γ = ⇒ | V | = − ∆we obtain by some manipulations of determinant∆ = ( −| V | ) = | V || V T | = | V V T | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) α + β + γ α + β + γ α + β + γ α + β + γ α + β + γ α + β + γ α + β + γ α + β + γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ba b − aca − ba b − aca − b + 3 a d − abca b − aca − b + 3 a d − abca b + 4 a bd + 2 a c − ab ca (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ba b − aca ba − ca − ad − bca − b + 2 aca − ad − bca abd + 2 ac − b ca (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ba b − aca ba − ca − ad − bca − ca − ad − bca abd + 2 ac − b ca (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ba − ca ba − ca − ad + bca − ca − ad − bca − bd + 2 c a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ba − ca b − ac a bc − ad a bc − ad a c − bd a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 3 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b − ac a bc − ad a bc − ad a c − bd a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 a − (cid:8) ( bc − ad ) − b − ac )( c − bd ) (cid:9) . This result is very suggestive. In fact, from the cubic equation ax + bx + cx + d = 0we have three data A = b − ac, B = bc − ad, C = c − bd, AX + BX + C = 0then the discriminant is just B − AC . This is very interesting. Problem.
Clarify the above connection.
As a result we have D = − (cid:8) ( bc − ad ) − b − ac )( c − bd ) (cid:9) = b c + 18 abcd − ac − b d − a d . In this section we calculate some quantities coming from the integral.The expectation value h x i is formally given by h x i = Z Z x e − ( ax + bx y + cxy + dy ) dxdy Z Z e − ( ax + bx y + cxy + dy ) dxdy = − ∂∂a log (cid:26)Z Z e − ( ax + bx y + cxy + dy ) dxdy (cid:27) , so renormalized expectation values h x i RN , h x y i RN , h xy i RN , h y i RN are defined as Definition 2. h x i RN = − ∂∂a log (cid:26) ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ (cid:27) , h x y i RN = − ∂∂b log (cid:26) ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ (cid:27) , h xy i RN = − ∂∂c log (cid:26) ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ (cid:27) , h y i RN = − ∂∂d log (cid:26) ‡ Z Z R e − ( ax + bx y + cxy + dy ) dxdy ‡ (cid:27) . From the integral forms (5) and (6) it is easy to calculate the above. Namely, we have h x i RN = 18 bcd − c − ad D , h x y i RN = 2 bc + 18 acd − b d D , h xy i RN = 2 b c + 18 abd − ac D , h y i RN = 18 abc − b − a d D , where D = b c + 18 abcd − ac − b d − a d .We can calculate other quantities like h x y i RN or h x y i RN by use of these ones, which willbe left to readers. In this paper we calculated the non-Gaussian integral (1) in a direct manner and, moreover,calculated some renormalized expectation values. It is not clear at the present time whetherthese results are useful enough or not. It would be desirable to accumulate many supportingevidences. Some application(s) will be reported elsewhere [5].eyond the Gaussian 11At this stage we can consider a further generalization. Namely, for the general degree n polynomial f ( x ) = a x n + a x n − + · · · + a n − x + a n the (non-Gaussian) integral becomes Z R n p f ( x ) dx. (23)The discriminant D of the equation f ( x ) = 0 is given by the resultant R ( f, f ′ ) of f and f ′ like 1 a R ( f, f ′ ) = ( − n ( n − D ⇐⇒ D = ( − n ( n − R ( f, f ′ ) /a where R ( f, f ′ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a · · · a n − a n a a · · · a n − a n . . . . . . a a · · · a n − a n na ( n − a · · · a n − na ( n − a · · · a n − . . . . . . na ( n − a · · · a n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , see (21) and (22).For example, for n = 4 R ( f, f ′ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a a a a a a a a a
00 0 a a a a a a a a a a a a a a a a a
00 0 0 4 a a a a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and D = 256 a a − a a − a a − a a − a a a + a a a + 16 a a a − a a a − a a a + 144 a a a a − a a a a + 144 a a a a − a a a a + 18 a a a a + 18 a a a a − a a a a a , and for n = 5 R ( f, f ′ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a a a a a a a a a a a a a a a a a
00 0 0 a a a a a a a a a a a a a a a a a a a a a a a a a a
00 0 0 0 5 a a a a a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) D = 3125 a a − a a a a − a a a a + 2000 a a a a + 2250 a a a a − a a a a + 256 a a + 2000 a a a a − a a a a + 2250 a a a a − a a a a a a + 160 a a a a a − a a a a + 1020 a a a a a − a a a a − a a a a + 825 a a a a + 560 a a a a a − a a a − a a a a a + 144 a a a a + 108 a a a − a a a − a a a a + 160 a a a a a − a a a a + 1020 a a a a a + 560 a a a a a − a a a a a a + 144 a a a a + 24 a a a a a − a a a a − a a a a a + 24 a a a a a + 356 a a a a a a − a a a a a − a a a a a + 18 a a a a a + 108 a a a − a a a a a + 16 a a a + 16 a a a a − a a a a + 256 a a − a a a a − a a a + 144 a a a a − a a + 144 a a a a − a a a a − a a a a a + 18 a a a a + 16 a a a − a a a − a a a + 18 a a a a a − a a a − a a a a + a a a a . However, to write down the general case explicitly is not easy (almost impossible).
Problem.
Calculate (23) for n = 4 ( and n = 5) directly. The wall called Gaussian is very high and not easy to overcome, and therefore hard work willbe needed.Recently the subsequent paper [6] by Morozov and Shakirov appeared. Our works are deeplyrelated to so-called non-linear algebras, so we will make some comments on this point in a nearfuture. As a general introduction to them see for example [7].The author thanks referees and Hiroshi Oike, Ryu Sasaki for many useful suggestions andcomments.
References [1] Fujii K., Beyond Gaussian: a comment, arXiv:0905.1363.[2] Morozov A., Shakirov Sh., Introduction to integral discriminants,
J. High Energy Phys. (2009), no. 12,002, 39 pages, arXiv:0903.2595.[3] Whittaker E.T., Watson G.N., A course of modern analysis, Cambridge University Press, Cambridge, 1996.[4] Satake I., Linear algebra, Shokabo, Tokyo, 1989 (in Japanese).[5] Fujii K., Beyond the Gaussian. II. Some applications, in progress.[6] Morozov A., Shakirov Sh., New and old results in resultant theory,
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