On some functional equations with several functions of two variables

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F. Radó,
Institutul de Calcul

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F. Radó, Sur quelques équations fonctionnelles avec plusieures fonctions à deux variables. (French) Mathematica (Cluj) 1 (24) 1959 321–339.

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Mathematica Cluj

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Published by the Romanian Academy  Publishing House

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1222-9016

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1959-Rado-On-some-equations
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ON SOME FUNCTIONAL EQUATIONS WITH SEVERAL FUNCTIONS OF TWO VARIABLES*)

byF. RADÓin Cluj

Functions representable by nomograms with aligned points of genus 0 have the form
(1)
z = H − 1 [ F ( x ) + G ( y ) ] , z = H − 1 [ F ( x ) + G ( y ) ] , z=H^(-1)[F(x)+G(y)],z=H−1[F(x)+G(y)],
Or F , G , H F , G , H F,G,HF,G,Hare continuous and monotonic functions in the narrow sense. Each of the following conditions is necessary and sufficient for a function f ( x , y ) f ( x , y ) f(x,y)f(x,y), continuous and strictly monotonic (with respect to each variable), i.e. of the form (1):
Condition T : f ( x 1 , y 2 ) = f ( x 2 , y 1 ) , f ( x 1 , y 3 ) = f ( x 3 , y 1 ) T : f ( x 1 , y 2 ) = f ( x 2 , y 1 ) , f ( x 1 , y 3 ) = f ( x 3 , y 1 ) T:f(x_(1),y_(2))=f(x_(2),y_(1)),f(x_(1),y_(3))=f(x_(3),y_(1))T:f(x1,y2)=f(x2,y1),f(x1,y3)=f(x3,y1)
→ f ( x 2 , y 3 ) = f ( x 3 . y 2 ) → f ( x 2 , y 3 ) = f ( x 3 . y 2 ) rarr f(x_(2),y_(3))=f(x_(3).y_(2))→f(x2,y3)=f(x3.y2)
Condition B : f ( x 1 , y 2 ) = f ( x 2 , y 1 ) , f ( x 1 , y 3 ) = f ( x 2 , y 2 ) = B : f ( x 1 , y 2 ) = f ( x 2 , y 1 ) , f ( x 1 , y 3 ) = f ( x 2 , y 2 ) = B:f(x_(1),y_(2))=f(x_(2),y_(1)),f(x_(1),y_(3))=f(x_(2),y_(2))=B:f(x1,y2)=f(x2,y1),f(x1,y3)=f(x2,y2)=
= f ( x 3 , y 1 ) → f ( x 2 , y 3 ) = f ( x 3 , y 2 ) = f ( x 3 , y 1 ) → f ( x 2 , y 3 ) = f ( x 3 , y 2 ) =f(x_(3),y_(1))rarr f(x_(2),y_(3))=f(x_(3),y_(2))=f(x3,y1)→f(x2,y3)=f(x3,y2)
Condition R : f ( x 1 , y 3 ) = f ( x 2 , y 1 ) , f ( x 1 , y 4 ) = f ( x 2 , y 2 ) , f ( x 3 , y 3 ) = R : f ( x 1 , y 3 ) = f ( x 2 , y 1 ) , f ( x 1 , y 4 ) = f ( x 2 , y 2 ) , f ( x 3 , y 3 ) = R:f(x_(1),y_(3))=f(x_(2),y_(1)),f(x_(1),y_(4))=f(x_(2),y_(2)),f(x_(3),y_(3))=R:f(x1,y3)=f(x2,y1),f(x1,y4)=f(x2,y2),f(x3,y3)=
= f ( x 4 , y 1 ) → f ( x 3 , y 4 ) = f ( x 4 , y 2 ) = f ( x 4 , y 1 ) → f ( x 3 , y 4 ) = f ( x 4 , y 2 ) =f(x_(4),y_(1))rarr f(x_(3),y_(4))=f(x_(4),y_(2))=f(x4,y1)→f(x3,y4)=f(x4,y2)
Using these conditions, I solved the functional equation of associativity in [12].
(2) f [ f ( x , y ) , z ] = f [ x , f ( y , z ) ] (2) f [ f ( x , y ) , z ] = f [ x , f ( y , z ) ] {:(2)f[f(x","y)","z]=f[x","f(y","z)]:}(2)f[f(x,y),z]=f[x,f(y,z)]
and of bisymmetry
(3)
f [ f ( u , x ) , f ( y , v ) ] = f [ f ( u , y ) , f ( x , v ) ] f [ f ( u , x ) , f ( y , v ) ] = f [ f ( u , y ) , f ( x , v ) ] f[f(u,x),f(y,v)]=f[f(u,y),f(x,v)]f[f(u,x),f(y,v)]=f[f(u,y),f(x,v)]
for the class of continuous and strictly monotonic functions. For the same class of functions, equations (2) and (3) were solved for the first time by J. ACZÉL, [1], [2], [3].
In this note I will use the condition B B BBto solve the generalization of the equation of associativity for 4 unknown functions from which I will also obtain the solutions of the generalization of the functional equation of bisymmetry and transitivity.
These equations are closely related to the problem of decomposing a function of three variables by superimposing functions of two variables. This decomposition allows the construction of compound nomograms.
  1. Conditions for a function of three variables to be of the form F [ φ ( x , y ) , z ] F [ φ ( x , y ) , z ] F[varphi(x,y),z]F[φ(x,y),z]If the functions F F FFAnd φ φ varphiφare not subject to any restrictions, any function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)can be written in the form
(1) f ( x , y , z ) = F [ φ ( x , y ) , z ] . (1) f ( x , y , z ) = F [ φ ( x , y ) , z ] . {:(1)f(x","y","z)=F[varphi(x","y)","z].:}(1)f(x,y,z)=F[φ(x,y),z].
However, if we ask that F F FFAnd φ φ varphiφare differentiable, then the following condition is necessary and sufficient for that f f ffeither of the form (1)
f x z ′ ′ f y ′ − f y z ′ ′ f x ′ = 0 f x z ′ ′ f y ′ − f y z ′ ′ f x ′ = 0 f_(xz)^('')f_(y)^(')-f_(yz)^('')f_(x)^(')=0fxz′′fy′−fyz′′fx′=0
(GOURSAT condition [7]).
We can easily see:
THEOREM 1. The necessary and sufficient condition for the function to continue f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)either of the form (1), with F F FFAnd φ φ varphiφcontinuous and strictly monotonous, is the implication
(2) f ( x 1 , y 1 , z 1 ) = f ( x 2 , y 2 , z 1 ) → f ( x 1 , y 1 , z 2 ) = f ( x 2 , y 2 , z 2 ) (2) f ( x 1 , y 1 , z 1 ) = f ( x 2 , y 2 , z 1 ) → f ( x 1 , y 1 , z 2 ) = f ( x 2 , y 2 , z 2 ) {:(2)f{:(x_(1)","y_(1)","z_(1)):}=f{:(x_(2)","y_(2)","z_(1)):}rarr f{:(x_(1)","y_(1)","z_(2)):}=f{:(x_(2)","y_(2)","z_(2)):}:}(2)f(x1,y1,z1)=f(x2,y2,z1)→f(x1,y1,z2)=f(x2,y2,z2)
  1. The equation of generalized associativity. The functional equation
(3) g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] (3) g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] {:(3)g[varphi(x","y)","z]=h[x","psi(y","z)]:}(3)g[φ(x,y),z]=h[x,ψ(y,z)]
Given 4 unknown functions, generalize the equation of associativity and its various modifications, for example:
(4) h [ x , h ( y , z ) ] = h [ z , h ( y , x ) ] (4) h [ x , h ( y , z ) ] = h [ z , h ( y , x ) ] {:(4)h[x","h(y","z)]=h[z","h(y","x)]:}(4)h[x,h(y,z)]=h[z,h(y,x)]
(Grassmann associativity),
(Tarki associativity),
(5) g [ g ( x , y ) , z ] = g [ x , g ( z , y ) ] (5) g [ g ( x , y ) , z ] = g [ x , g ( z , y ) ] {:(5)g[g(x","y)","z]=g[x","g(z","y)]:}(5)g[g(x,y),z]=g[x,g(z,y)]
(6) g [ g ( x , y ) , z ] = g [ y , g ( z , x ) ] (6) g [ g ( x , y ) , z ] = g [ y , g ( z , x ) ] {:(6)g[g(x","y)","z]=g[y","g(z","x)]:}(6)g[g(x,y),z]=g[y,g(z,x)]
(cyclic associativity), the half-symmetry equation [12], etc. M. HOSSZÚ solved equations (4), (5), (6) under the assumption that the solutions are continuous and strictly monotonic functions, and equation (4) under the more restrictive assumptions that the solutions admit first-order partial derivatives and are strictly monotonic [9].
We will also give the solution of equation (3) under the condition of continuity and strict monotonicity. We will further assume that the functions φ ( x , y ) , ψ ( x , y ) , g ( x , y ) , h ( x , y ) φ ( x , y ) , ψ ( x , y ) , g ( x , y ) , h ( x , y ) varphi(x,y),psi(x,y),g(x,y),h(x,y)φ(x,y),ψ(x,y),g(x,y),h(x,y)pulse be defined x ∈ ⟨ a , b ⟩ , y ∈ ⟨ a , b ⟩ x ∈ ⟨ a , b ⟩ , y ∈ ⟨ a , b ⟩ x in(:a,b:),y in(:a,b:)x∈⟨has,b⟩,y∈⟨has,b⟩They take their values ​​in the same
interval (closed, open, or half-open), the equation φ ( x , y 0 ) = z 0 φ ( x , y 0 ) = z 0 varphi(x,y_(0))=z_(0)φ(x,y0)=z0has one (only) solution x x xxFor y 0 , z 0 ∈< a , b > y 0 , z 0 ∈< a , b > y_(0),z_(0)∈<a,b >y0,z0∈<has,b>and this condition is satisfied by the other three functions (the interval ⟨ a , b ⟩ ⟨ a , b ⟩ (:a,b:)⟨has,b⟩form with each of the operations φ , ψ , g , h φ , ψ , g , h varphi,psi,g,hφ,ψ,g,ha quasigroup).
Theorem 2. Under these hypotheses, the general solution of the functional equation (3) is the following system of functions;
ρ ( x , y ) = H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] (7) g ( x , y ) = H 2 − 1 [ G 1 ( x ) + G 2 ( y ) ] g ( x , y ) = H 3 − 1 [ H 1 ( x ) + G 2 ( y ) ] h ( x , y ) = H 3 − 1 [ F 1 ( x ) + H 2 ( y ) ] , ρ ( x , y ) = H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] (7) g ( x , y ) = H 2 − 1 [ G 1 ( x ) + G 2 ( y ) ] g ( x , y ) = H 3 − 1 [ H 1 ( x ) + G 2 ( y ) ] h ( x , y ) = H 3 − 1 [ F 1 ( x ) + H 2 ( y ) ] , {:[rho(x","y)=H_(1)^(-1){:[F_(1)(x)+G_(1)(y)]:}],[(7)g(x","y)=H_(2)^(-1){:[G_(1)(x)+G_(2)(y)]:}],[g(x","y)=H_(3)^(-1){:[H_(1)(x)+G_(2)(y)]:}],[h(x","y)=H_(3)^(-1){:[F_(1)(x)+H_(2)(y)]:}","]:}ρ(x,y)=H1−1[F1(x)+G1(y)](7)g(x,y)=H2−1[G1(x)+G2(y)]g(x,y)=H3−1[H1(x)+G2(y)]h(x,y)=H3−1[F1(x)+H2(y)],
oit F 1 , G 1 , G 2 , H 1 , H 2 , H 3 F 1 , G 1 , G 2 , H 1 , H 2 , H 3 F_(1),G_(1),G_(2),H_(1),H_(2),H_(3)F1,G1,G2,H1,H2,H3are continuous and strictly monotonic functions. Substituting (7) into (3), we have
H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] = H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] , H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] = H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] , H_(3)^(-1)[F_(1)(x)+G_(1)(y)+G_(2)(z)]=H_(3)^(-1)[F_(1)(x)+G_(1)(y)+G_(2)(z)],H3−1[F1(x)+G1(y)+G2(z)]=H3−1[F1(x)+G1(y)+G2(z)],
Therefore, functions (7) satisfy equation (3).
Now suppose that the functions are continuous and strictly monotonic φ , ψ , g , h φ , ψ , g , h varphi,psi,g,hφ,ψ,g,hsatisfy (3) and denote the two members of equation (3) by f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)
(8) f ( x , y , z ) = g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] . (8) f ( x , y , z ) = g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] . {:(8)f(x","y","z)=g[varphi(x","y)","z]=h[x","psi(y","z)].:}(8)f(x,y,z)=g[φ(x,y),z]=h[x,ψ(y,z)].
By applying theorem 1
(9) { f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) → f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) . (9) { f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) → f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) . {:(9){{:[f{:(x_(2)","y_(1)","z_(1)):}=f{:(x_(1)","y_(2)","z_(1)):}=f{:(x_(1)","y_(1)","z_(2)):}rarr],[f{:(x_(1)","y_(2)","z_(2)):}=f{:(x_(2)","y_(1)","z_(2)):}=f{:(x_(2)","y_(2)","z_(1)):}.]:}:}(9){f(x2,y1,z1)=f(x1,y2,z1)=f(x1,y1,z2)→f(x1,y2,z2)=f(x2,y1,z2)=f(x2,y2,z1).
Let's demonstrate that the function of two variables f ( x , y , z 1 ) f ( x , y , z 1 ) f(x,y,z_(1))f(x,y,z1)checks the condition B B BB. Either
(10) { f ( x 1 , y 2 , z 1 ) = f ( x 2 , y 1 , z 1 ) f ( x 1 , y 3 , z 1 ) = f ( x 2 , y 2 , z 1 ) = f ( x 2 , y 1 , z 1 ) (10) { f ( x 1 , y 2 , z 1 ) = f ( x 2 , y 1 , z 1 ) f ( x 1 , y 3 , z 1 ) = f ( x 2 , y 2 , z 1 ) = f ( x 2 , y 1 , z 1 ) {:(10){{:[f{:(x_(1)","y_(2)","z_(1)):}=f{:(x_(2)","y_(1)","z_(1)):}],[f{:(x_(1)","y_(3)","z_(1)):}=f{:(x_(2)","y_(2)","z_(1)):}=f{:(x_(2)","y_(1)","z_(1)):}]:}:}(10){f(x1,y2,z1)=f(x2,y1,z1)f(x1,y3,z1)=f(x2,y2,z1)=f(x2,y1,z1)
and note by z 2 z 2 z_(2)z2the value z z zzfor which
we have
f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 1 , z 2 ) f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 1 , z 2 ) f(x_(2),y_(1),z_(1))=f(x_(1),y_(1),z_(2))f(x2,y1,z1)=f(x1,y1,z2)
f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) f(x_(2),y_(1),z_(1))=f(x_(1),y_(2),z_(1))=f(x_(1),y_(1),z_(2))f(x2,y1,z1)=f(x1,y2,z1)=f(x1,y1,z2)
and using (9)
(11)
f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) f(x_(1),y_(2),z_(2))=f(x_(2),y_(1),z_(2))=f(x_(2),y_(2),z_(1))f(x1,y2,z2)=f(x2,y1,z2)=f(x2,y2,z1)
We deduce from (10) and (11)
f ( x 2 , y 2 , z 1 ) = f ( x 1 , y 3 , z 1 ) = f ( x 1 , y 2 , z 2 ) f ( x 2 , y 2 , z 1 ) = f ( x 1 , y 3 , z 1 ) = f ( x 1 , y 2 , z 2 ) f(x_(2),y_(2),z_(1))=f(x_(1),y_(3),z_(1))=f(x_(1),y_(2),z_(2))f(x2,y2,z1)=f(x1,y3,z1)=f(x1,y2,z2)
and applying (9)
(12) f ( x 1 , y 3 , z 2 ) = f ( x 2 , y 2 , z 2 ) = f ( x 2 , y 3 , z 1 ) (12) f ( x 1 , y 3 , z 2 ) = f ( x 2 , y 2 , z 2 ) = f ( x 2 , y 3 , z 1 ) {:(12)f{:(x_(1)","y_(3)","z_(2)):}=f{:(x_(2)","y_(2)","z_(2)):}=f{:(x_(2)","y_(3)","z_(1)):}:}(12)f(x1,y3,z2)=f(x2,y2,z2)=f(x2,y3,z1)
Taking into account again (10) and (11)
f ( x 3 , y 1 , z 1 ) = f ( x 2 , y 2 , z 1 ) = f ( x 2 , y 1 , z 2 ) , f ( x 3 , y 1 , z 1 ) = f ( x 2 , y 2 , z 1 ) = f ( x 2 , y 1 , z 2 ) , f(x_(3),y_(1),z_(1))=f(x_(2),y_(2),z_(1))=f(x_(2),y_(1),z_(2)),f(x3,y1,z1)=f(x2,y2,z1)=f(x2,y1,z2),
and applying (9) we have
(13) f ( x 2 , y 2 , z 2 ) = f ( x 3 , y 1 , z 2 ) = f ( x 3 , y 2 , z 1 ) . (13) f ( x 2 , y 2 , z 2 ) = f ( x 3 , y 1 , z 2 ) = f ( x 3 , y 2 , z 1 ) . {:(13)f{:(x_(2)","y_(2)","z_(2)):}=f{:(x_(3)","y_(1)","z_(2)):}=f{:(x_(3)","y_(2)","z_(1)):}.:}(13)f(x2,y2,z2)=f(x3,y1,z2)=f(x3,y2,z1).
Relations (12) and (13) give
(14) f ( x 2 , y 3 , z 1 ) = f ( x 3 , y 2 , z 1 ) (14) f ( x 2 , y 3 , z 1 ) = f ( x 3 , y 2 , z 1 ) {:(14)f{:(x_(2)","y_(3)","z_(1)):}=f{:(x_(3)","y_(2)","z_(1)):}:}(14)f(x2,y3,z1)=f(x3,y2,z1)
SO ( 10 ) → ( 14 ) ( 10 ) → ( 14 ) (10)rarr(14)(10)→(14)which shows that the function f ( x , y , z 1 ) f ( x , y , z 1 ) f(x,y,z_(1))f(x,y,z1)check the condition B B BBIt follows that
φ ( x , y ) = Φ [ f ( x , y , z 1 ) ] φ ( x , y ) = Φ [ f ( x , y , z 1 ) ] varphi(x,y)=Phi[f(x,y,z_(1))]φ(x,y)=Φ[f(x,y,z1)]
similarly checks the condition B B BB, and consequently
(15) φ ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] . (15) φ ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] . {:(15)varphi(x","y)=H^(-1){:[F_(1)(x)+G_(1)(y)]:}.:}(15)φ(x,y)=H−1[F1(x)+G1(y)].
We substitute this expression into (3) and set y = y ′ , z = z 0 y = y ′ , z = z 0 y=y^('),z=z_(0)y=y′,z=z0:
h [ x , ψ ( y ′ , z 0 ) ] = g { H 1 − 1 [ F 1 ( x ) + G 1 ( y ′ ) ] , z 0 } . h [ x , ψ ( y ′ , z 0 ) ] = g { H 1 − 1 [ F 1 ( x ) + G 1 ( y ′ ) ] , z 0 } . h[x,psi(y^('),z_(0))]=g{H_(1)^(-1)[F_(1)(x)+G_(1)(y^('))],z_(0)}.h[x,ψ(y′,z0)]=g{H1−1[F1(x)+G1(y′)],z0}.
By noting ψ ( y ′ , z 0 ) = y ψ ( y ′ , z 0 ) = y psi(y^('),z_(0))=yψ(y′,z0)=ywe have
(16) h ( x , y ) = H 3 − 1 [ F 1 ( x ) + H 2 ( y ) ] (16) h ( x , y ) = H 3 − 1 [ F 1 ( x ) + H 2 ( y ) ] {:(16)h(x","y)=H_(3)^(-1){:[F_(1)(x)+H_(2)(y)]:}:}(16)h(x,y)=H3−1[F1(x)+H2(y)]
By a further substitution in (3) we find
(17) g { H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] , z } = H 3 − 1 { F 1 ( x ) + H 2 [ ψ ( y , z ) ] } (17) g { H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] , z } = H 3 − 1 { F 1 ( x ) + H 2 [ ψ ( y , z ) ] } {:(17)g{H_(1)^(-1)[F_(1)(x)+G_(1)(y)]","z}=H_(3)^(-1){:{F_(1)(x)+H_(2)[psi(y","z)]}:}:}(17)g{H1−1[F1(x)+G1(y)],z}=H3−1{F1(x)+H2[ψ(y,z)]}
By asking y = y 0 y = y 0 y=y_(0)y=y0
(18)
g ( x , y ) − H 3 − 1 [ Φ ( x ) + X ( y ) ] g ( x , y ) − H 3 − 1 [ Φ ( x ) + X ( y ) ] g(x,y)-H_(3)^(-1)[Phi(x)+X(y)]g(x,y)−H3−1[Φ(x)+X(y)]
and (17) becomes
(19) Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] + X ( z ) = F 1 ( x ) + H 2 [ ψ ( y , z ) ] . (19) Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] + X ( z ) = F 1 ( x ) + H 2 [ ψ ( y , z ) ] . {:(19)PhiH_(1)^(-1){:[F_(1)(x)+G_(1)(y)]:}+X(z)=F_(1)(x)+H_(2)[psi(y","z)].:}(19)ΦH1−1[F1(x)+G1(y)]+X(z)=F1(x)+H2[ψ(y,z)].
By putting here x = x 0 x = x 0 x=x_(0)x=x0
(20) ψ ( x , y ) = H 2 − 1 [ Ψ ( x ) + X ( y ) ] . (20) ψ ( x , y ) = H 2 − 1 [ Ψ ( x ) + X ( y ) ] . {:(20)psi(x","y)=H_(2)^(-1)[Psi(x)+X(y)].:}(20)ψ(x,y)=H2−1[Ψ(x)+X(y)].
We substitute (20) into equation (19):
or
Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] + X ( z ) = F 1 ( x ) + Ψ ( y ) + X ( z ) Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] + X ( z ) = F 1 ( x ) + Ψ ( y ) + X ( z ) PhiH_(1)^(-1)[F_(1)(x)+G_(1)(y)]+X(z)=F_(1)(x)+Psi(y)+X(z)ΦH1−1[F1(x)+G1(y)]+X(z)=F1(x)+Ψ(y)+X(z)
Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] = [ F 1 ( x ) + G 1 ( y ) ] + [ Ψ ( y ) − G 1 ( y ) ] . Φ H 1 − 1 [ F 1 ( x ) + G 1 ( y ) ] = [ F 1 ( x ) + G 1 ( y ) ] + [ Ψ ( y ) − G 1 ( y ) ] . PhiH_(1)^(-1)[F_(1)(x)+G_(1)(y)]=[F_(1)(x)+G_(1)(y)]+[Psi(y)-G_(1)(y)].ΦH1−1[F1(x)+G1(y)]=[F1(x)+G1(y)]+[Ψ(y)−G1(y)].
By doing y = y = y=y=const, we see that
Φ H 1 − 1 ( ξ ) = ξ + a , Φ H 1 − 1 ( ξ ) = ξ + a , PhiH_(1)^(-1)(xi)=xi+a,ΦH1−1(ξ)=ξ+has,
SO
Φ ( x ) = H 1 ( x ) + a Ψ ( x ) = G 1 ( x ) + a . Φ ( x ) = H 1 ( x ) + a Ψ ( x ) = G 1 ( x ) + a . {:[Phi(x)=H_(1)(x)+a],[Psi(x)=G_(1)(x)+a.]:}Φ(x)=H1(x)+hasΨ(x)=G1(x)+has.
By noting X ( x ) + a = G 2 ( x ) X ( x ) + a = G 2 ( x ) X(x)+a=G_(2)(x)X(x)+has=G2(x)we finally have
g ( x , y ) = H 3 − 1 [ H 1 ( x ) + G 2 ( y ) ] ψ ( x , y ) = H 2 − 1 [ G 1 ( x ) + G 2 ( y ) ] . g ( x , y ) = H 3 − 1 [ H 1 ( x ) + G 2 ( y ) ] ψ ( x , y ) = H 2 − 1 [ G 1 ( x ) + G 2 ( y ) ] . {:[g(x","y)=H_(3)^(-1){:[H_(1)(x)+G_(2)(y)]:}],[psi(x","y)=H_(2)^(-1){:[G_(1)(x)+G_(2)(y)]:}.]:}g(x,y)=H3−1[H1(x)+G2(y)]ψ(x,y)=H2−1[G1(x)+G2(y)].
By combining formulas (15) and (16) with these last two, we obtain solution (7) of the functional equation (3).
3. Special cases. We can now easily obtain the solutions of the various special cases of equation (3).
a) The equation of associativity. We have
g ( x , y ) = φ ( x , y ) = h ( x , y ) = ψ ( x , y ) . g ( x , y ) = φ ( x , y ) = h ( x , y ) = ψ ( x , y ) . g(x,y)=varphi(x,y)=h(x,y)=psi(x,y).g(x,y)=φ(x,y)=h(x,y)=ψ(x,y).
In this case, formulas (7) represent the same function. In [12] we demonstrated that relations
f ( x , y ) = H − 1 [ F ( x ) + G ( y ) ] f ( x , y ) = H ∗ − 1 [ F ∗ ( x ) + G ∗ ( y ) ] f ( x , y ) = H − 1 [ F ( x ) + G ( y ) ] f ( x , y ) = H ∗ − 1 [ F ∗ ( x ) + G ∗ ( y ) ] {:[f(x","y)=H^(-1)[F(x)+G(y)]],[f(x","y)=H^(**-1)[F^(**)(x)+G^(**)(y)]]:}f(x,y)=H−1[F(x)+G(y)]f(x,y)=H*−1[F*(x)+G*(y)]
it results
F ∗ ( x ) = a F ( x ) + b , G ∗ ( x ) = a G ( x ) + c , H ∗ ( x ) = a H ( x ) + b + c , F ∗ ( x ) = a F ( x ) + b , G ∗ ( x ) = a G ( x ) + c , H ∗ ( x ) = a H ( x ) + b + c , F^(**)(x)=aF(x)+b,G^(**)(x)=aG(x)+c,H^(**)(x)=aH(x)+b+c,F*(x)=hasF(x)+b,G*(x)=hasG(x)+c,H*(x)=hasH(x)+b+c,
Or a , b , c a , b , c a,b,chas,b,care constants. Therefore, the first and fourth formulas (7) imply
H 2 = G 1 + a , H 3 = H 1 + a , H 2 = G 1 + a , H 3 = H 1 + a , H_(2)=G_(1)+a,quadH_(3)=H_(1)+a,H2=G1+has,H3=H1+has,
the second and the third
H 1 = G 1 + b , H 3 = H 2 + b H 1 = G 1 + b , H 3 = H 2 + b H_(1)=G_(1)+b,quadH_(3)=H_(2)+bH1=G1+b,H3=H2+b
and the third and the fourth
H 1 = F 1 + c , H 2 = G 2 + c . H 1 = F 1 + c , H 2 = G 2 + c . H_(1)=F_(1)+c,quadH_(2)=G_(2)+c.H1=F1+c,H2=G2+c.
Therefore
G 1 = F 1 + c − b , G 2 = F 1 + a − b H 1 = F 1 + c , H 2 = F 1 + a − b + c , H 3 = F 1 + a + c G 1 = F 1 + c − b , G 2 = F 1 + a − b H 1 = F 1 + c , H 2 = F 1 + a − b + c , H 3 = F 1 + a + c {:[G_(1)=F_(1)+c-b","quadG_(2)=F_(1)+a-b],[H_(1)=F_(1)+c","quadH_(2)=F_(1)+a-b+c","quadH_(3)=F_(1)+a+c]:}G1=F1+c−b,G2=F1+has−bH1=F1+c,H2=F1+has−b+c,H3=F1+has+c
it results
F 1 [ φ ( x , y ) ] + c = F 1 ( x ) + F 1 ( y ) + c 1 − b F 1 [ φ ( x , y ) ] + c = F 1 ( x ) + F 1 ( y ) + c 1 − b F_(1)[varphi(x,y)]+c=F_(1)(x)+F_(1)(y)+c_(1)-bF1[φ(x,y)]+c=F1(x)+F1(y)+c1−b
By noting H ( x ) = F 1 ( x ) − b H ( x ) = F 1 ( x ) − b H(x)=F_(1)(x)-bH(x)=F1(x)−b, We have
φ ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] φ ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] varphi(x,y)=H^(-1)[H(x)+H(y)]φ(x,y)=H−1[H(x)+H(y)]
b) The Grassmann equation. By the notation h ( x , y ) = − h ′ ( y , x ) h ( x , y ) = − h ′ ( y , x ) h(x,y)=-h^(')(y,x)h(x,y)=−h′(y,x), equation (4) becomes
h ′ [ h ′ ( x , y ) , z ] = h [ x , h ( y , z ) ] h ′ [ h ′ ( x , y ) , z ] = h [ x , h ( y , z ) ] h^(')[h^(')(x,y),z]=h[x,h(y,z)]h′[h′(x,y),z]=h[x,h(y,z)]
therefore we obtain (4) by setting in (3)
h = ψ = g ′ = φ ′ . h = ψ = g ′ = φ ′ . h=psi=g^(')=varphi^(').h=ψ=g′=φ′.
As mentioned above, we arrive at the relationships
H 1 = H 2 + a , F 1 = G 2 + a , F 1 = b H 1 + c , G 1 = b G 2 + d , H 1 = b H 3 + c 1 + d , H 1 = H 2 + a , F 1 = G 2 + a , F 1 = b H 1 + c , G 1 = b G 2 + d , H 1 = b H 3 + c 1 + d , {:[H_(1)=H_(2)+a","F_(1)=G_(2)+a","quadF_(1)=bH_(1)+c","quadG_(1)=bG_(2)+d","],[H_(1)=bH_(3)+c_(1)+d","]:}H1=H2+has,F1=G2+has,F1=bH1+c,G1=bG2+d,H1=bH3+c1+d,
hence
G 1 = b 2 H 2 + a b 2 + b c − a b + d , G 2 = b H 2 + a b + c − a , G 1 = b 2 H 2 + a b 2 + b c − a b + d , G 2 = b H 2 + a b + c − a , G_(1)=b^(2)H_(2)+ab^(2)+bc-ab+d,quadG_(2)=bH_(2)+ab+c-a,G1=b2H2+hasb2+bc−hasb+d,G2=bH2+hasb+c−has,
SO
(21) h ( x , y ) = ψ ( x , y ) = H 2 − 1 [ b 2 H 2 ( x ) + b H 2 ( y ) + α ] , (21) h ( x , y ) = ψ ( x , y ) = H 2 − 1 [ b 2 H 2 ( x ) + b H 2 ( y ) + α ] , {:(21)h(x","y)=psi(x","y)=H_(2)^(-1){:[b^(2)H_(2)(x)+bH_(2)(y)+alpha]:}",":}(21)h(x,y)=ψ(x,y)=H2−1[b2H2(x)+bH2(y)+α],
which coincides with the result of Mr. Hosszú
Si b 2 + b − 1 ≠ 0 b 2 + b − 1 ≠ 0 b^(2)+b-1!=0b2+b−1≠0noting H ( x ) = H 2 ( x ) + α b 2 + b − 1 H ( x ) = H 2 ( x ) + α b 2 + b − 1 H(x)=H_(2)(x)+(alpha)/(b^(2)+b-1)H(x)=H2(x)+αb2+b−1we have
(22) h ( x , y ) = H 2 − 1 [ b 2 H ( x ) + b H ( x ) ] ; (22) h ( x , y ) = H 2 − 1 [ b 2 H ( x ) + b H ( x ) ] ; {:(22)h(x","y)=H_(2)^(-1)[b^(2)H(x)+bH(x)];:}(22)h(x,y)=H2−1[b2H(x)+bH(x)];
if b 2 + b − 1 = 0 b 2 + b − 1 = 0 b^(2)+b-1=0b2+b−1=0noting H = H 2 ( x ) H = H 2 ( x ) H=H_(2)(x)H=H2(x)
(22)
h ( x , y ) = H − 1 [ 1 − ε 5 2 H ( x ) + 1 + ε 5 2 H ( y ) + α ] , ε = ± 1 h ( x , y ) = H − 1 [ 1 − ε 5 2 H ( x ) + 1 + ε 5 2 H ( y ) + α ] , ε = ± 1 h(x,y)=H^(-1)[(1-epsisqrt5)/(2)H(x)+(1+epsisqrt5)/(2)H(y)+alpha],epsi=+-1h(x,y)=H−1[1−ε52H(x)+1+ε52H(y)+α],ε=±1
It can be seen by direct verification that the functions (22) and (22') satisfy equation (4), therefore all continuous and strictly monotonic solutions of this equation are given by formulas (22) and ( 22 ′ ) ( 22 ′ ) (22^('))(22′), Or H ( x ) H ( x ) H(x)H(x)is continuous and strictly monotonic.
c) Tarki's equation. The functional equation (3) becomes Tarki's equation (5) by particularization
g = φ = h = ψ ′ g = φ = h = ψ ′ g=varphi=h=psi^(')g=φ=h=ψ′
Using formulas (7) we obtain as above
g ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] . g ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] . g(x,y)=H^(-1)[H(x)+H(y)].g(x,y)=H−1[H(x)+H(y)].
Therefore, equation (5) is equivalent to the equation of associativity.
(6) becomes
d) The equation of cyclic associativity. By permuting x x xxAnd y y yy, the equation therefore by setting in (3).
g [ g ( y , x ) , z ] = g [ x , g ( z , y ) ] g [ g ( y , x ) , z ] = g [ x , g ( z , y ) ] g[g(y,x),z]=g[x,g(z,y)]g[g(y,x),z]=g[x,g(z,y)]
g = h = φ ′ = ψ ′ g = h = φ ′ = ψ ′ g=h=varphi^(')=psi^(')g=h=φ′=ψ′
We obtain equation (6). We find that this equation is also equivalent to the equation of associativity.
e) The equation of half-symmetry.
f [ j ( y , x ) , z ] = ∫ [ / ( y , z ) , x ] f [ j ( y , x ) , z ] = ∫ [ / ( y , z ) , x ] f[j(y,x),z]=int[//(y,z),x]f[j(y,x),z]=∫[/(y,z),x]
is the special case of (3) for
g = ψ = h ′ = φ ′ . g = ψ = h ′ = φ ′ . g=psi=h^(')=varphi^(').g=ψ=h′=φ′.
We find
f ( x , y ) = H − 1 [ H ( x ) + G ( y ) ] . f ( x , y ) = H − 1 [ H ( x ) + G ( y ) ] . f(x,y)=H^(-1)[H(x)+G(y)].f(x,y)=H−1[H(x)+G(y)].
Let us consider two further special cases of equation (3), in which the unknown function appears g ( x , y ) g ( x , y ) g(x,y)g(x,y)and its inverses. The equation z = g ( x , y ) z = g ( x , y ) z=g(x,y)z=g(x,y)resolved in relation to x x xxis written x = g ( y , z ) x = g ( y , z ) x=g(y,z)x=g(y,z), and resolved with respect to y y yyis written y = g ( z , x ) y = g ( z , x ) y=g(z,x)y=g(z,x)f
) The functional equation
(23)
g [ g ¯ ( x , y ) , z ] = g [ x , g ~ ( y , z ) ] . g [ g ¯ ( x , y ) , z ] = g [ x , g ~ ( y , z ) ] . g[ bar(g)(x,y),z]=g[x, tilde(g)(y,z)].g[g¯(x,y),z]=g[x,g~(y,z)].
The theorem ≥ ≥ >=≥dome
g ¯ ( x , y ) = H 1 − 1 [ − G 2 ( x ) + H 3 ( y ) ] g ~ ( x , y ) = G 2 − 1 [ H 3 ( x ) − H 1 ( y ) ] h = g , φ = g ¯ , ψ = g ~ H 1 = F 1 + a , H 2 = G 2 + a , − G 2 = F 1 + b , H 3 = G 1 − b , H 2 = c G 1 + d , − H 1 = c G 2 + e , G 2 = c H 2 + d + e . g ¯ ( x , y ) = H 1 − 1 [ − G 2 ( x ) + H 3 ( y ) ] g ~ ( x , y ) = G 2 − 1 [ H 3 ( x ) − H 1 ( y ) ] h = g , φ = g ¯ , ψ = g ~ H 1 = F 1 + a , H 2 = G 2 + a , − G 2 = F 1 + b , H 3 = G 1 − b , H 2 = c G 1 + d , − H 1 = c G 2 + e , G 2 = c H 2 + d + e . {:[ bar(g)(x","y)=H_(1)^(-1){:[-G_(2)(x)+H_(3)(y)]:}],[ tilde(g)(x","y)=G_(2)^(-1){:[H_(3)(x)-H_(1)(y)]:}],[h=g","quad varphi= bar(g)","quad psi= tilde(g)],[H_(1)=F_(1)+a","quadH_(2)=G_(2)+a","-G_(2)=F_(1)+b","quadH_(3)=G_(1)-b","],[H_(2)=cG_(1)+d","-H_(1)=cG_(2)+e","quadG_(2)=cH_(2)+d+e.]:}g¯(x,y)=H1−1[−G2(x)+H3(y)]g~(x,y)=G2−1[H3(x)−H1(y)]h=g,φ=g¯,ψ=g~H1=F1+has,H2=G2+has,−G2=F1+b,H3=G1−b,H2=cG1+d,−H1=cG2+e,G2=cH2+d+e.
It follows that c = 1 , d = − b , c = b − a c = 1 , d = − b , c = b − a c=1,d=-b,c=b-ac=1,d=−b,c=b−hasAnd
(24) g ( x , y ) = H − 1 [ G ( x ) − G ( y ) ] . (24) g ( x , y ) = H − 1 [ G ( x ) − G ( y ) ] . {:(24)g(x","y)=H^(-1)[G(x)-G(y)].:}(24)g(x,y)=H−1[G(x)−G(y)].
On the other hand, function (24) satisfies equation (23).
g) The functional equation
(25) g [ g ~ ( x , y ) , z ] = g [ x , g ¯ ( y , z ) ] (25) g [ g ~ ( x , y ) , z ] = g [ x , g ¯ ( y , z ) ] {:(25)g[ tilde(g)(x","y)","z]=g[x"," bar(g)(y","z)]:}(25)g[g~(x,y),z]=g[x,g¯(y,z)]
to the general solution
g ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] g ( x , y ) = H − 1 [ H ( x ) + H ( y ) ] g(x,y)=H^(-1)[H(x)+H(y)]g(x,y)=H−1[H(x)+H(y)]
Therefore, equation (25) is equivalent to the equation of associativity.
4. The equation of generalized bisymmetry. The functional equation
(26) f [ g ( u , x ) , h ( y , v ) ] = φ [ ψ ( u , y ) , χ ( x , v ) ] (26) f [ g ( u , x ) , h ( y , v ) ] = φ [ ψ ( u , y ) , χ ( x , v ) ] {:(26)f[g(u","x)","h(y","v)]=varphi[psi(u","y)","chi(x","v)]:}(26)f[g(u,x),h(y,v)]=φ[ψ(u,y),χ(x,v)]
with 6 unknown functions was solved by Mr. Hosszú under the assumptions of differentiability and strict monotonicity [8]. Let us take up this equation again and look for its continuous and strictly monotonic solutions.
Theorem 3. The continuous, strictly monotonic, and invertible solutions with respect to x x xxel y y yy(quasigroup property) of equation (26) are
(27) { f ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] φ ( x , y ) = H − 1 [ F 4 ( x ) + G 4 ( y ) ] g ( x , y ) = F 1 − 1 [ F 2 ( x ) + G 2 ( y ) ] ψ ( x , y ) = F 4 − 1 [ F 2 ( x ) + F 3 ( y ) ] h ( x , y ) = G 1 − 1 [ F 3 ( x ) + G 3 ( y ) ] χ 1 ( x , y ) = G 4 − 1 [ G 2 ( x ) + G 3 ( y ) ] (27) { f ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] φ ( x , y ) = H − 1 [ F 4 ( x ) + G 4 ( y ) ] g ( x , y ) = F 1 − 1 [ F 2 ( x ) + G 2 ( y ) ] ψ ( x , y ) = F 4 − 1 [ F 2 ( x ) + F 3 ( y ) ] h ( x , y ) = G 1 − 1 [ F 3 ( x ) + G 3 ( y ) ] χ 1 ( x , y ) = G 4 − 1 [ G 2 ( x ) + G 3 ( y ) ] {:(27){{:[f(x","y)=H^(-1){:[F_(1)(x)+G_(1)(y)]:},varphi(x","y)=H^(-1){:[F_(4)(x)+G_(4)(y)]:}],[g(x","y)=F_(1)^(-1){:[F_(2)(x)+G_(2)(y)]:},psi(x","y)=F_(4)^(-1){:[F_(2)(x)+F_(3)(y)]:}],[h(x","y)=G_(1)^(-1){:[F_(3)(x)+G_(3)(y)]:},chi_(1)(x","y)=G_(4)^(-1){:[G_(2)(x)+G_(3)(y)]:}]:}:}(27){f(x,y)=H−1[F1(x)+G1(y)]φ(x,y)=H−1[F4(x)+G4(y)]g(x,y)=F1−1[F2(x)+G2(y)]ψ(x,y)=F4−1[F2(x)+F3(y)]h(x,y)=G1−1[F3(x)+G3(y)]χ1(x,y)=G4−1[G2(x)+G3(y)]
where the nine functions of one variable are continuous and strictly monotonic.
For the proof, let
g ( ξ , η ) = g ′ ( η , ξ ) et φ ( ξ , η ) = φ ′ ( η , ξ ) g ( ξ , η ) = g ′ ( η , ξ ) et φ ( ξ , η ) = φ ′ ( η , ξ ) g(xi,eta)=g^(')(eta,xi)"et"varphi(xi,eta)=varphi^(')(eta,xi)g(ξ,η)=g′(η,ξ)Andφ(ξ,η)=φ′(η,ξ)
and let's put in ( 26 ) v = v 0 ( 26 ) v = v 0 (26)v=v_(0)(26)v=v0
f [ g ′ ( x , u ) , h ( y , v 0 ) ] = φ ′ [ χ ( x , v 0 ) , ψ ( u , y ′ ) ] . f [ g ′ ( x , u ) , h ( y , v 0 ) ] = φ ′ [ χ ( x , v 0 ) , ψ ( u , y ′ ) ] . f[g^(')(x,u),h(y,v_(0))]=varphi^(')[chi(x,v_(0)),psi(u,y^('))].f[g′(x,u),h(y,v0)]=φ′[χ(x,v0),ψ(u,y′)].
Through the change in notation
k ( ξ , η ) = f [ ξ , h ( η , v 0 ) ] (28) l ( ξ , η ) = φ ′ [ χ ( ξ , v 0 ) , η ] k ( ξ , η ) = f [ ξ , h ( η , v 0 ) ] (28) l ( ξ , η ) = φ ′ [ χ ( ξ , v 0 ) , η ] {:[k(xi","eta)=f[xi","h(eta,v_(0))]],[(28)l(xi","eta)=varphi^(')[chi(xi,v_(0))","eta]]:}k(ξ,η)=f[ξ,h(η,v0)](28)L(ξ,η)=φ′[χ(ξ,v0),η]
equation (26) becomes
k [ g ′ ( x , u ) , y ] = l [ x , ψ ( u , y ) ] k [ g ′ ( x , u ) , y ] = l [ x , ψ ( u , y ) ] k[g^(')(x,u),y]=l[x,psi(u,y)]k[g′(x,u),y]=L[x,ψ(u,y)]
which is of the form (3). Theorem 2 gives
g ( x , y ) = g ′ ( y , x ) = F 1 − 1 [ F 2 ( x ) + G 3 ( y ) ] ψ ( x , y ) = F 1 − 1 [ F 2 ( x ) + F 3 ( y ) ] k ( x , y ) = H − 1 [ F 1 ( x ) + F 3 ( y ) ] l ( x , y ) = H − 1 [ G 2 ( x ) + F 4 ( y ) ] . g ( x , y ) = g ′ ( y , x ) = F 1 − 1 [ F 2 ( x ) + G 3 ( y ) ] ψ ( x , y ) = F 1 − 1 [ F 2 ( x ) + F 3 ( y ) ] k ( x , y ) = H − 1 [ F 1 ( x ) + F 3 ( y ) ] l ( x , y ) = H − 1 [ G 2 ( x ) + F 4 ( y ) ] . {:[g(x","y)=g^(')(y","x)=F_(1)^(-1){:[F_(2)(x)+G_(3)(y)]:}],[psi(x","y)=F_(1)^(-1){:[F_(2)(x)+F_(3)(y)]:}],[k(x","y)=H^(-1){:[F_(1)(x)+F_(3)(y)]:}],[l(x","y)=H^(-1){:[G_(2)(x)+F_(4)(y)]:}.]:}g(x,y)=g′(y,x)=F1−1[F2(x)+G3(y)]ψ(x,y)=F1−1[F2(x)+F3(y)]k(x,y)=H−1[F1(x)+F3(y)]L(x,y)=H−1[G2(x)+F4(y)].
Let's substitute the expression found for the first formula (28) k ( ξ , η ) k ( ξ , η ) k(xi,eta)k(ξ,η)and let's ask ξ = x , h ( η , v 0 ) = y ξ = x , h ( η , v 0 ) = y xi=x,h(eta,v_(0))=yξ=x,h(η,v0)=y; noting G 1 ( y ) = F 3 [ h ¯ ( v 0 , y ) ] G 1 ( y ) = F 3 [ h ¯ ( v 0 , y ) ] G_(1)(y)=F_(3)[ bar(h)(v_(0),y)]G1(y)=F3[h¯(v0,y)], we obtain
f ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] f ( x , y ) = H − 1 [ F 1 ( x ) + G 1 ( y ) ] f(x,y)=H^(-1)[F_(1)(x)+G_(1)(y)]f(x,y)=H−1[F1(x)+G1(y)]
Let's substitute the expression for the second formula (28) l ( ξ , η ) l ( ξ , η ) l(xi,eta)L(ξ,η), let's ask x = ψ ( ξ , v 0 ) , y = τ 1 x = ψ ( ξ , v 0 ) , y = τ 1 x=psi(xi,v_(0)),y=tau_(1)x=ψ(ξ,v0),y=τ1and note G 4 ( x ) = G 2 [ χ ¯ ( v 0 , x ) ] G 4 ( x ) = G 2 [ χ ¯ ( v 0 , x ) ] G_(4)(x)=G_(2)[ bar(chi)(v_(0),x)]G4(x)=G2[χ¯(v0,x)]; we obtain
or
φ ′ ( x , y ) = H − 1 [ G 4 ( x ) + F 4 ( y ) ] φ ( x , y ) = H − 1 [ F 4 ( x ) + G 4 ( y ) ] φ ′ ( x , y ) = H − 1 [ G 4 ( x ) + F 4 ( y ) ] φ ( x , y ) = H − 1 [ F 4 ( x ) + G 4 ( y ) ] {:[varphi^(')(x","y)=H^(-1){:[G_(4)(x)+F_(4)(y)]:}],[varphi(x","y)=H^(-1){:[F_(4)(x)+G_(4)(y)]:}]:}φ′(x,y)=H−1[G4(x)+F4(y)]φ(x,y)=H−1[F4(x)+G4(y)]
Substitute the expressions found into equation (26) for g ( x , y ) g ( x , y ) g(x,y)g(x,y), ψ ( x , y ) , f ( x , y ) ψ ( x , y ) , f ( x , y ) psi(x,y),f(x,y)ψ(x,y),f(x,y)And φ ( x , y ) φ ( x , y ) varphi(x,y)φ(x,y):
F 2 ( u ) + G 2 ( x ) + G 1 [ h ( y , v ) ] = F 2 ( u ) + F 3 ( y ) + G 2 [ χ ( x , v ) ] . F 2 ( u ) + G 2 ( x ) + G 1 [ h ( y , v ) ] = F 2 ( u ) + F 3 ( y ) + G 2 [ χ ( x , v ) ] . F_(2)(u)+G_(2)(x)+G_(1)[h(y,v)]=F_(2)(u)+F_(3)(y)+G_(2)[chi(x,v)].F2(u)+G2(x)+G1[h(y,v)]=F2(u)+F3(y)+G2[χ(x,v)].
By grouping the terms
G 1 [ h ( y , v ) ] − F 3 ( y ) = G 4 [ χ ( x , v ) ] − G 2 ( x ) G 1 [ h ( y , v ) ] − F 3 ( y ) = G 4 [ χ ( x , v ) ] − G 2 ( x ) G_(1)[h(y,v)]-F_(3)(y)=G_(4)[chi(x,v)]-G_(2)(x)G1[h(y,v)]−F3(y)=G4[χ(x,v)]−G2(x)
the first member does not depend on x x xxthe second one does not depend on y y yy. therefore both members are functions of v v vvonly, either G ( v ) G ( v ) G(v)G(v); SO
h ( y , v ) = G 1 − 1 [ F 3 ( y ) + G 3 ( v ) ] , χ ( x , v ) = G 4 − 1 [ G 2 ( x ) + G 3 ( v ) ] h ( y , v ) = G 1 − 1 [ F 3 ( y ) + G 3 ( v ) ] , χ ( x , v ) = G 4 − 1 [ G 2 ( x ) + G 3 ( v ) ] h(y,v)=G_(1)^(-1)[F_(3)(y)+G_(3)(v)],quad chi(x,v)=G_(4)^(-1)[G_(2)(x)+G_(3)(v)]h(y,v)=G1−1[F3(y)+G3(v)],χ(x,v)=G4−1[G2(x)+G3(v)]
Or
h ( x , y ) = G 1 − 1 [ F 3 ( x ) + G 3 ( y ) ] (29) x ( x , y ) = G 4 − 1 [ G 2 ( x ) + G 3 ( y ) ] . h ( x , y ) = G 1 − 1 [ F 3 ( x ) + G 3 ( y ) ] (29) x ( x , y ) = G 4 − 1 [ G 2 ( x ) + G 3 ( y ) ] . {:[h(x","y)=G_(1)^(-1){:[F_(3)(x)+G_(3)(y)]:}],[(29)x(x","y)=G_(4)^(-1){:[G_(2)(x)+G_(3)(y)]:}.]:}h(x,y)=G1−1[F3(x)+G3(y)](29)x(x,y)=G4−1[G2(x)+G3(y)].
We have just obtained all the formulas (27).
On the other hand, the functions (27) form a system of solutions for equation (26), which can be seen by direct verification.
5. Geometric application. Equation (26) leads to the following generalization of Thomsen's theorem [6]:
Theorem, 1. Let F 1 , F 2 , F 3 F 1 , F 2 , F 3 F_(1),F_(2),F_(3)F1,F2,F3three families of curves in the xy plane that enjoy the following property: if the points M , S M , S M,SM,Sof figure 1 are on the same curve of the family F 2 F 2 F_(2)F2, and the points N , R N , R N,RN,Ron the same curve of f 3 f 3 f_(3)f3, then the points P , Q P , Q P,QP,Qare located on the same curve of F 1 F 1 F_(1)F1Under these assumptions, the three families coincide and their equation is
F ( x ) + G ( y ) = const F ( x ) + G ( y ) = const F(x)+G(y)="const"F(x)+G(y)=const
Let the equations of the families of curves be
f 1 ) f ( x , y ) = f 1 ) f ( x , y ) = f_(1))f(x,y)=f1)f(x,y)=const
( f 2 ) g ( x , y ) = ( f 2 ) g ( x , y ) = (f_(2))g(x,y)=(f2)g(x,y)=const
f 3 ) h ( x , y ) = f 3 ) h ( x , y ) = f_(3))h(x,y)=f3)h(x,y)=const.
We have
(30) g ( x 1 , y 2 ) = g ( x 2 , y 1 ) , h ( x 1 , y 3 ) = h ( x 3 , y 1 ) → h ( x 2 , y 3 ) = f ( x 3 , y 2 ) (30) g ( x 1 , y 2 ) = g ( x 2 , y 1 ) , h ( x 1 , y 3 ) = h ( x 3 , y 1 ) → h ( x 2 , y 3 ) = f ( x 3 , y 2 ) {:(30)g{:(x_(1)","y_(2)):}=g{:(x_(2)","y_(1)):}","h{:(x_(1)","y_(3)):}=h{:(x_(3)","y_(1)):}rarr h{:(x_(2)","y_(3)):}=f{:(x_(3)","y_(2)):}:}(30)g(x1,y2)=g(x2,y1),h(x1,y3)=h(x3,y1)→h(x2,y3)=f(x3,y2)
Note
s = g ( x 1 , y 2 ) = g ( x 2 , y 1 ) , s = g ( x 1 , y 2 ) = g ( x 2 , y 1 ) , s=g(x_(1),y_(2))=g(x_(2),y_(1)),s=g(x1,y2)=g(x2,y1),
We can write using the inverse function notation already employed.
x 2 = g ¯ ( y 1 , s ) y 2 = g ~ ( s , x 1 ) x 3 = h ¯ ( y 1 , t ) y 3 = h ~ ( t , x 1 ) x 2 = g ¯ ( y 1 , s ) y 2 = g ~ ( s , x 1 ) x 3 = h ¯ ( y 1 , t ) y 3 = h ~ ( t , x 1 ) {:[x_(2)= bar(g){:(y_(1)","s):},y_(2)= tilde(g){:(s","x_(1)):}],[x_(3)= bar(h){:(y_(1)","t):},y_(3)= tilde(h){:(t","x_(1)):}]:}x2=g¯(y1,s)y2=g~(s,x1)x3=h¯(y1,t)y3=h~(t,x1)
Condition (30) becomes the functional equation
(31) f [ g ¯ ( y 1 , s ) , h ~ ( t , x 1 ) ] = f [ h ¯ ( y 1 , t ) , g ~ ( s , x 1 ) ] (31) f [ g ¯ ( y 1 , s ) , h ~ ( t , x 1 ) ] = f [ h ¯ ( y 1 , t ) , g ~ ( s , x 1 ) ] {:(31)f[ bar(g)(y_(1),s)"," tilde(h)(t,x_(1))]=f[ bar(h)(y_(1),t)"," tilde(g)(s,x_(1))]:}(31)f[g¯(y1,s),h~(t,x1)]=f[h¯(y1,t),g~(s,x1)]
which is a special case of equation (26). Applying Theorem 3 and taking into account χ = g , h = ψ ¯ χ = g , h = ψ ¯ chi=g,h= bar(psi)χ=g,h=ψ¯, we obtain by a simple calculation
f ( x , y ) = H 1 − 1 [ F ( x ) + G ( y ) ] g ( x , y ) = H 2 − 1 [ F ( x ) + G ( y ) ] h ( x , y ) = H 3 − 1 [ F ( x ) + G ( y ) ] f ( x , y ) = H 1 − 1 [ F ( x ) + G ( y ) ] g ( x , y ) = H 2 − 1 [ F ( x ) + G ( y ) ] h ( x , y ) = H 3 − 1 [ F ( x ) + G ( y ) ] {:[f(x","y)=H_(1)^(-1)[F(x)+G(y)]],[g(x","y)=H_(2)^(-1)[F(x)+G(y)]],[h(x","y)=H_(3)^(-1)[F(x)+G(y)]]:}f(x,y)=H1−1[F(x)+G(y)]g(x,y)=H2−1[F(x)+G(y)]h(x,y)=H3−1[F(x)+G(y)]
  1. The generalized transitivity equation. The transitivity equation
(32) f [ f ( x , t ) , f ( y , t ) ] = f ( x , y ) (32) f [ f ( x , t ) , f ( y , t ) ] = f ( x , y ) {:(32)f[f(x","t)","f(y","t)]=f(x","y):}(32)f[f(x,t),f(y,t)]=f(x,y)
was studied by A.R. Schweitzer by transforming it into a partial differential equation [13], [14]. Under the assumptions of continuity and strict monotonicity, it was solved by M. Hosszú [10]. The solution of the more general equation is also found in the same note.
(33) f [ φ ( x , t ) , ψ ( y , t ) ] = g ( x , y ) (33) f [ φ ( x , t ) , ψ ( y , t ) ] = g ( x , y ) {:(33)f[varphi(x","t)","psi(y","t)]=g(x","y):}(33)f[φ(x,t),ψ(y,t)]=g(x,y)
but only for monotonic functions, which admit first-order partial derivatives. We give its continuous and strictly monotonic solution, again by reduction to the equation of generalized associativity.
Theorem: 5. The solutions of equation (33), which are continuous, strictly monotonic, and invertible with respect to x x xxAnd y y yysoul given by the formulas
(34) { f ( x , y ) = H − 1 [ F 1 ( x ) − G 1 ( y ) ] g ( x , y ) = H − 1 [ F 2 ( x ) − G 2 ( y ) ] φ ( x , y ) = F 1 − 1 [ F 2 ( x ) − G 3 ( y ) ] ψ ( x , y ) = G 1 − 1 [ G 2 ( x ) − G 3 ( y ) ] (34) { f ( x , y ) = H − 1 [ F 1 ( x ) − G 1 ( y ) ] g ( x , y ) = H − 1 [ F 2 ( x ) − G 2 ( y ) ] φ ( x , y ) = F 1 − 1 [ F 2 ( x ) − G 3 ( y ) ] ψ ( x , y ) = G 1 − 1 [ G 2 ( x ) − G 3 ( y ) ] {:(34){{:[f(x","y)=H^(-1){:[F_(1)(x)-G_(1)(y)]:}],[g(x","y)=H^(-1){:[F_(2)(x)-G_(2)(y)]:}],[varphi(x","y)=F_(1)^(-1){:[F_(2)(x)-G_(3)(y)]:}],[psi(x","y)=G_(1)^(-1){:[G_(2)(x)-G_(3)(y)]:}]:}:}(34){f(x,y)=H−1[F1(x)−G1(y)]g(x,y)=H−1[F2(x)−G2(y)]φ(x,y)=F1−1[F2(x)−G3(y)]ψ(x,y)=G1−1[G2(x)−G3(y)]
Or F 1 , F 2 , G 1 , G 2 , G 3 , H F 1 , F 2 , G 1 , G 2 , G 3 , H F_(1),F_(2),G_(1),G_(2),G_(3),HF1,F2,G1,G2,G3,Hare continuous and strictly monotonic functions.
By setting
ψ ( y , t ) = z , ψ ( y , t ) = z , psi(y,t)=z,ψ(y,t)=z,
hence y = Ψ ¯ ( t , z ) y = Ψ ¯ ( t , z ) y= bar(Psi)(t,z)y=Ψ¯(t,z), equation (33) becomes
(35) f [ φ ( x , t ) , z ] = g [ x , ψ ¯ ( l , z ) ] (35) f [ φ ( x , t ) , z ] = g [ x , ψ ¯ ( l , z ) ] {:(35)f[varphi(x","t)","z]=g[x"," bar(psi)(l","z)]:}(35)f[φ(x,t),z]=g[x,ψ¯(L,z)]
It suffices to apply Theorem 2 and perform a simple calculation.
The solution to equation (32) is obtained from formulas (34) by setting f = g = φ = ψ f = g = φ = ψ f=g=varphi=psif=g=φ=ψ
(36) f ( x , y ) = F − 1 [ F ( x ) − F ( y ) ] (36) f ( x , y ) = F − 1 [ F ( x ) − F ( y ) ] {:(36)f(x","y)=F^(-1)[F(x)-F(y)]:}(36)f(x,y)=F−1[F(x)−F(y)]
  1. Pseudo-sums with three terms. We saw in no. 2 that if the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)admits both decompositions of the form (8)
    or
    the functions φ , ψ , g φ , ψ , g varphi,psi,gφ,ψ,gAnd h h hhare expressed by formulas (7), therefore
(8) f ( x , y , z ) = g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] (8) f ( x , y , z ) = g [ φ ( x , y ) , z ] = h [ x , ψ ( y , z ) ] {:(8)f(x","y","z)=g[varphi(x","y)","z]=h[x","psi(y","z)]:}(8)f(x,y,z)=g[φ(x,y),z]=h[x,ψ(y,z)]
f ( x , y , z ) = H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] f ( x , y , z ) = H 3 − 1 [ F 1 ( x ) + G 1 ( y ) + G 2 ( z ) ] f(x,y,z)=H_(3)^(-1)[F_(1)(x)+G_(1)(y)+G_(2)(z)]f(x,y,z)=H3−1[F1(x)+G1(y)+G2(z)]
(37) f ( x , y , z ) = K − 1 [ F ( x ) + G ( y ) + H ( z ) ] (37) f ( x , y , z ) = K − 1 [ F ( x ) + G ( y ) + H ( z ) ] {:(37)f(x","y","z)=K^(-1)[F(x)+G(y)+H(z)]:}(37)f(x,y,z)=K−1[F(x)+G(y)+H(z)]
Conversely, given (37), if we set φ ( x , y ) = F ( x ) + G ( y ) φ ( x , y ) = F ( x ) + G ( y ) varphi(x,y)=F(x)+G(y)φ(x,y)=F(x)+G(y), g ( x , y ) = K − 1 [ x + H ( y ) ] , ψ ( x , y ) = G ( x ) + H ( y ) , h ( x , y ) = K − 1 [ F ( x ) + y ] g ( x , y ) = K − 1 [ x + H ( y ) ] , ψ ( x , y ) = G ( x ) + H ( y ) , h ( x , y ) = K − 1 [ F ( x ) + y ] g(x,y)=K^(-1)[x+H(y)],psi(x,y)=G(x)+H(y),h(x,y)=K^(-1)[F(x)+y]g(x,y)=K−1[x+H(y)],ψ(x,y)=G(x)+H(y),h(x,y)=K−1[F(x)+y], SO f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)admits the decompositions (8). We will say that the function
(37) is a pscudo-sum with three terms, if the functions F , G , H , K F , G , H , K F,G,H,KF,G,H,Kcontinuous and strictly monotonic subs. It follows from what we have just said:
theorem: 6. The necessary and sufficient condition for the function to be continuous and strictly monotonic f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)either a pseudo-sum with three terms is the existence of decompositions (8).
Consequently, a third decomposition (8) results from the two decompositions.
f ( x , y , z ) = l [ k ( x , z ) , y ] f ( x , y , z ) = l [ k ( x , z ) , y ] f(x,y,z)=l[k(x,z),y]f(x,y,z)=L[k(x,z),y]
The decompositions (8) are equivalent to the two implications
(38)
{ f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) → f ( x 2 , y 1 , z 2 ) = f ( x 1 , y 2 , z 2 ) f ( x 1 , y 1 , z 2 ) = f ( x 1 , y 2 , z 1 ) → f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) { f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) → f ( x 2 , y 1 , z 2 ) = f ( x 1 , y 2 , z 2 ) f ( x 1 , y 1 , z 2 ) = f ( x 1 , y 2 , z 1 ) → f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) {{:[f{:(x_(2)","y_(1)","z_(1)):}=f{:(x_(1)","y_(2)","z_(1)):}rarr f{:(x_(2)","y_(1)","z_(2)):}=f{:(x_(1)","y_(2)","z_(2)):}],[f{:(x_(1)","y_(1)","z_(2)):}=f{:(x_(1)","y_(2)","z_(1)):}rarr f{:(x_(2)","y_(1)","z_(2)):}=f{:(x_(2)","y_(2)","z_(1)):}]:}{f(x2,y1,z1)=f(x1,y2,z1)→f(x2,y1,z2)=f(x1,y2,z2)f(x1,y1,z2)=f(x1,y2,z1)→f(x2,y1,z2)=f(x2,y2,z1)
The implications (38) entail
(39) { f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) → f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) . (39) { f ( x 2 , y 1 , z 1 ) = f ( x 1 , y 2 , z 1 ) = f ( x 1 , y 1 , z 2 ) → f ( x 1 , y 2 , z 2 ) = f ( x 2 , y 1 , z 2 ) = f ( x 2 , y 2 , z 1 ) . {:(39){{:[f{:(x_(2)","y_(1)","z_(1)):}=f{:(x_(1)","y_(2)","z_(1)):}=f{:(x_(1)","y_(1)","z_(2)):}rarr],[f{:(x_(1)","y_(2)","z_(2)):}=f{:(x_(2)","y_(1)","z_(2)):}=f{:(x_(2)","y_(2)","z_(1)):}.]:}:}(39){f(x2,y1,z1)=f(x1,y2,z1)=f(x1,y1,z2)→f(x1,y2,z2)=f(x2,y1,z2)=f(x2,y2,z1).
We will demonstrate that, conversely, (39) implies (38), that is, Theorem 7 holds.
The implication (39) is necessary and sufficient for the function to be continuous and strictly monotonic. f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)that is, a pseudo-sum with three terms.
Implication (39) is identical to (9), encountered when solving the functional equation of generalized associativity. We have seen that (39) implies that f ( x , y , z 0 ) f ( x , y , z 0 ) f(x,y,z_(0))f(x,y,z0)is of the form (1), Taking into account the symmetrical form of (39) with respect to x , y , z x , y , z x,y,zx,y,zwe only have f ( x , y 0 , z ) f ( x , y 0 , z ) f(x,y_(0),z)f(x,y0,z)And f ( x 0 , y , z ) f ( x 0 , y , z ) f(x_(0),y,z)f(x0,y,z)have similar expressions.
For y 1 y 1 y_(1)y1And z 1 z 1 z_(1)z1fixed, we can note
f ( x , y , z 1 ) = K − 1 [ F ( x ) + G ( y ) ] (40) f ( x , y 1 , z ) = Ψ − 1 [ Φ ( x ) + H ( z ) ] f ( x , y , z 1 ) = K − 1 [ F ( x ) + G ( y ) ] (40) f ( x , y 1 , z ) = Ψ − 1 [ Φ ( x ) + H ( z ) ] {:[f{:(x","y","z_(1)):}=K^(-1)[F(x)+G(y)]],[(40)f{:(x","y_(1)","z):}=Psi^(-1)[Phi(x)+H(z)]]:}f(x,y,z1)=K−1[F(x)+G(y)](40)f(x,y1,z)=Ψ−1[Φ(x)+H(z)]
We substitute expressions (40) into (39)
(41) F ( x 2 ) + G ( y 1 ) = F ( x 1 ) + G ( y 2 ) , Φ ( x 2 ) + H ( z 1 ) = Φ ( x 1 ) + H ( z 2 ) → Ψ − 1 [ Φ ( x 2 ) + H ( z 2 ) ] = K − 1 [ F ( x 2 ) + G ( y 2 ) ] . (41) F ( x 2 ) + G ( y 1 ) = F ( x 1 ) + G ( y 2 ) , Φ ( x 2 ) + H ( z 1 ) = Φ ( x 1 ) + H ( z 2 ) → Ψ − 1 [ Φ ( x 2 ) + H ( z 2 ) ] = K − 1 [ F ( x 2 ) + G ( y 2 ) ] . {:[(41)F{:(x_(2)):}+G{:(y_(1)):}=F{:(x_(1)):}+G{:(y_(2)):}","quad Phi{:(x_(2)):}+H{:(z_(1)):}=Phi{:(x_(1)):}+H{:(z_(2)):}rarr],[Psi^(-1)[Phi(x_(2))+H(z_(2))]=K^(-1)[F(x_(2))+G(y_(2))].]:}(41)F(x2)+G(y1)=F(x1)+G(y2),Φ(x2)+H(z1)=Φ(x1)+H(z2)→Ψ−1[Φ(x2)+H(z2)]=K−1[F(x2)+G(y2)].
We are watching x 1 x 1 x_(1)x1And x 2 x 2 x_(2)x2as independent variables, the assumptions of implication (41) allow us to express G ( y 2 ) G ( y 2 ) G(y_(2))G(y2)And H ( z 2 ) H ( z 2 ) H(z_(2))H(z2)depending on x 1 x 1 x_(1)x1And x 2 x 2 x_(2)x2( y 1 y 1 y_(1)y1And z 1 z 1 z_(1)z1are fixed); we substitute them in the conclusion of the same implication and obtain the functional equation
Ψ − 1 [ 2 Φ ( x 2 ) − Φ ( x 1 ) + H ( z 1 ) ] = K − 1 [ 2 F ( x 2 ) − F ( x 1 ) + G ( y 1 ) ] . Ψ − 1 [ 2 Φ ( x 2 ) − Φ ( x 1 ) + H ( z 1 ) ] = K − 1 [ 2 F ( x 2 ) − F ( x 1 ) + G ( y 1 ) ] . Psi^(-1)[2Phi(x_(2))-Phi(x_(1))+H(z_(1))]=K^(-1)[2F(x_(2))-F(x_(1))+G(y_(1))].Ψ−1[2Φ(x2)−Φ(x1)+H(z1)]=K−1[2F(x2)−F(x1)+G(y1)].
Note
Φ ( x 2 ) = x , Φ ( x 1 ) = y Φ ( x 2 ) = x , Φ ( x 1 ) = y Phi(x_(2))=x,quad Phi(x_(1))=yΦ(x2)=x,Φ(x1)=y
And
(42) K Ψ − 1 [ x + H ( z 1 ) ] − G ( y 1 ) = φ ( x ) , F Φ − 1 ( x ) = ψ ( x ) ; (42) K Ψ − 1 [ x + H ( z 1 ) ] − G ( y 1 ) = φ ( x ) , F Φ − 1 ( x ) = ψ ( x ) ; {:(42)KPsi^(-1)[x+H(z_(1))]-G{:(y_(1)):}=varphi(x)","FPhi^(-1)(x)=psi(x);:}(42)KΨ−1[x+H(z1)]−G(y1)=φ(x),FΦ−1(x)=ψ(x);
we have
φ ( 2 x − y ) = 2 ψ ( x ) − ψ ( y ) , φ ( 2 x − y ) = 2 ψ ( x ) − ψ ( y ) , varphi(2x-y)=2psi(x)-psi(y),φ(2x−y)=2ψ(x)−ψ(y),
what becomes, through the change of function
(43) 2 ψ ( x 2 ) = χ ( x ) , (43) 2 ψ ( x 2 ) = χ ( x ) , {:(43)2psi((x)/(2))=chi(x)",":}(43)2ψ(x2)=χ(x),
the functional equation
χ ( x + y ) = φ ( x ) + ψ ( y ) . χ ( x + y ) = φ ( x ) + ψ ( y ) . chi(x+y)=varphi(x)+psi(y).χ(x+y)=φ(x)+ψ(y).
This equation has the solution [12]
φ ( x ) = a x + b , ψ ( x ) = a x + c , χ ( x ) = a x + b + c ; φ ( x ) = a x + b , ψ ( x ) = a x + c , χ ( x ) = a x + b + c ; varphi(x)=ax+b,quad psi(x)=ax+c,quad chi(x)=ax+b+c;φ(x)=hasx+b,ψ(x)=hasx+c,χ(x)=hasx+b+c;
taking into account (43), 2 c = b + c 2 c = b + c 2c=b+c2c=b+c, SO c = b c = b c=bc=bLet's return to (42)
(44) F ( ξ ) = a Φ ( ξ ) + b K ( ξ ) = a Ψ ( ξ ) + b ′ (44) F ( ξ ) = a Φ ( ξ ) + b K ( ξ ) = a Ψ ( ξ ) + b ′ {:[(44)F(xi)=a Phi(xi)+b],[K(xi)=a Psi(xi)+b^(')]:}(44)F(ξ)=hasΦ(ξ)+bK(ξ)=hasΨ(ξ)+b′
Or b ′ = b + G ( y 1 ) − a H ( z 1 ) b ′ = b + G ( y 1 ) − a H ( z 1 ) b^(')=b+G(y_(1))-aH(z_(1))b′=b+G(y1)−hasH(z1)The
second formula (40) becomes
a Ψ [ f ( x , y 1 , z ) ] + b ′ = a Φ ( x ) + b + a H ( z ) + b ′ − b , a Ψ [ f ( x , y 1 , z ) ] + b ′ = a Φ ( x ) + b + a H ( z ) + b ′ − b , a Psi[f(x,y_(1),z)]+b^(')=a Phi(x)+b+aH(z)+b^(')-b,hasΨ[f(x,y1,z)]+b′=hasΦ(x)+b+hasH(z)+b′−b,
or taking into account (44)
K [ f ( x , y 1 , z ) ] = F ( x ) + a H ( z ) + b ′ − b , K [ f ( x , y 1 , z ) ] = F ( x ) + a H ( z ) + b ′ − b , K[f(x,y_(1),z)]=F(x)+aH(z)+b^(')-b,K[f(x,y1,z)]=F(x)+hasH(z)+b′−b,
by writing H 1 ( z ) H 1 ( z ) H_(1)(z)H1(z)For a H ( z ) + b ′ − b a H ( z ) + b ′ − b aH(z)+b^(')-bhasH(z)+b′−b
f ( x , y 1 , z ) = K − 1 [ F ( x ) + H 1 ( z ) ] . f ( x , y 1 , z ) = K − 1 [ F ( x ) + H 1 ( z ) ] . f(x,y_(1),z)=K^(-1)[F(x)+H_(1)(z)].f(x,y1,z)=K−1[F(x)+H1(z)].
We are now varying y 1 y 1 y_(1)y1holding z 1 z 1 z_(1)z1fixed. In the last formula only the function H 1 H 1 H_(1)H1varies with y 1 y 1 y_(1)y1, SO
f ( x , y , z ) = K − 1 [ F ( x ) + ψ ( y , z ) ] = h [ x , ψ ( y , z ) ] . f ( x , y , z ) = K − 1 [ F ( x ) + ψ ( y , z ) ] = h [ x , ψ ( y , z ) ] . f(x,y,z)=K^(-1)[F(x)+psi(y,z)]=h[x,psi(y,z)].f(x,y,z)=K−1[F(x)+ψ(y,z)]=h[x,ψ(y,z)].
Similarly, f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)admits two similar decompositions: once with x , y x , y x,yx,ygrouped and a second time with x , z x , z x,zx,zgrouped. Theorem 7 follows from Theorem 6.
Observations. 1) If f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)is a two-term pseudo-sum for arbitrary fixed values ​​of any variable, it does not follow that f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)is a pseudo-sum with three terms, as the example shows us
f ( x , y , z ) = F ( x ) G ( y ) + H ( z ) . f ( x , y , z ) = F ( x ) G ( y ) + H ( z ) . f(x,y,z)=F(x)G(y)+H(z).f(x,y,z)=F(x)G(y)+H(z).
  1. If the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)is a pseudo-sum in relation to x , z x , z x,zx,zFor y y yy, arbitrarily fixed and at the same time in relation to y , z y , z y,zy,zFor x x xxfixed arbitrarily, it does not follow that it is a pseudo-sum with respect to x , y x , y x,yx,yFor z z zzfixed, as can be seen from the example
f ( x , y , z ) = [ F ( x ) + G ( y ) ] M ( x ) H ( z ) . f ( x , y , z ) = [ F ( x ) + G ( y ) ] M ( x ) H ( z ) . f(x,y,z)=[F(x)+G(y)]M(x)H(z).f(x,y,z)=[F(x)+G(y)]M(x)H(z).
  1. The geometric interpretation of Theorem 7. We consider the points A ( x 2 , y 1 , z 1 ) , B ( x 1 , y 2 , z 1 ) , C ( x 1 , y 1 , z 2 ) , A ′ ( x 1 , y 2 , z 2 ) , B ′ ( x 2 , y 1 , z 2 ) A ( x 2 , y 1 , z 1 ) , B ( x 1 , y 2 , z 1 ) , C ( x 1 , y 1 , z 2 ) , A ′ ( x 1 , y 2 , z 2 ) , B ′ ( x 2 , y 1 , z 2 ) A(x_(2),y_(1),z_(1)),B(x_(1),y_(2),z_(1)),C(x_(1),y_(1),z_(2)),A^(')(x_(1),y_(2),z_(2)),B^(')(x_(2),y_(1),z_(2))HAS(x2,y1,z1),B(x1,y2,z1),C(x1,y1,z2),HAS′(x1,y2,z2),B′(x2,y1,z2), C ′ ( x 2 , y 2 , z 1 ) C ′ ( x 2 , y 2 , z 1 ) C^(')(x_(2),y_(2),z_(1))C′(x2,y2,z1)(fig. 2). Implication (39) expresses that if the points A , B , C A , B , C A,B,CHAS,B,Care on the same level surface of the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z), then the points A ′ , B ′ , C ′ A ′ , B ′ , C ′ A^('),B^('),C^(')HAS′,B′,C′are also on the same level surface. In other words: if we try to construct octahedra with two faces parallel to the coordinate plane x y x y xyxytwo faces parallel to the plane y z y z yzyz, two parallels to z x z x zxzxand with two curved faces formed by level surfaces of the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)Then these octahedra can be constructed; they close. The planes parallel to the coordinate planes and the level surfaces of the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x,y,z)forming a spatial fabric, the octahedra considered above are the tissue octahedra. Condition (39) expresses that the tissue octahedra close.
On the other hand, if the curved surfaces of the
Fig. 2.
fabrics have the equations
K − 1 [ F ( x ) + G ( y ) + H ( z ) ] = const , K − 1 [ F ( x ) + G ( y ) + H ( z ) ] = const , K^(-1)[F(x)+G(y)+H(z)]="const",K−1[F(x)+G(y)+H(z)]=const,
Then the fabric is topologically equivalent to a fabric formed by four families of parallel planes, which is called a regular fabric. Here is the interpretation of Theorem 7: The necessary and sufficient condition for a spatial fabric to be topologically equivalent to a regular fabric is that all the fabric octahedra close. We have recovered a well-known result from the geometry of fabrics [6].
9. Separation of variables. To represent the function nomographically
(45) y = f ( x 1 , x 2 , … , x n ) (45) y = f ( x 1 , x 2 , … , x n ) {:(45)y=f{:(x_(1)","x_(2)","dots","x_(n)):}:}(45)y=f(x1,x2,…,xn)
we are looking for the separation of variables in the form
f ( x 1 , … , x n ) = F [ φ 1 ( x 1 , … , x p 1 ) , φ 2 ( x p 1 + 1 , … , x p 2 ) (46) … , φ r ( x p r − 1 + 1 , … , x p ) , x p r + 1 , … , x n ] f ( x 1 , … , x n ) = F [ φ 1 ( x 1 , … , x p 1 ) , φ 2 ( x p 1 + 1 , … , x p 2 ) (46) … , φ r ( x p r − 1 + 1 , … , x p ) , x p r + 1 , … , x n ] {:[f{:(x_(1)","dots","x_(n)):}=F{:[varphi_(1)(x_(1),dots,x_(p_(1)))","varphi_(2)(x_(p_(1))+1,dots,x_(p_(2))):}],[(46){: dots","varphi_(r)(x_(p_(r-1)+1),dots,x_(p))","x_(p_(r)+1)","dots","x_(n)]:}]:}f(x1,…,xn)=F[φ1(x1,…,xp1),φ2(xp1+1,…,xp2)(46)…,φr(xpr−1+1,…,xp),xpr+1,…,xn]
If each of the equations
(-47) { ξ 1 = φ 1 ( x 1 , … , x p 1 ) ξ 2 = φ 2 ( x p 1 + 1 , … , x p 2 ) ξ r = φ r ( x p r − 1 , … , x p r ) y = F ( ξ 1 , ξ 2 , … , ξ r , x p r + 1 , … , x n ) (-47) { ξ 1 = φ 1 ( x 1 , … , x p 1 ) ξ 2 = φ 2 ( x p 1 + 1 , … , x p 2 ) ξ r = φ r ( x p r − 1 , … , x p r ) y = F ( ξ 1 , ξ 2 , … , ξ r , x p r + 1 , … , x n ) {:(-47){{:[xi_(1)=varphi_(1){:(x_(1)","dots","x_(p_(1))):}],[xi_(2)=varphi_(2){:(x_(p_(1)+1)","dots","x_(p_(2))):}],[xi_(r)=varphi_(r){:(x_(p_(r-1))","dots","x_(p_(r))):}],[y=F{:(xi_(1)","xi_(2)","dots","xi_(r)","x_(p_(r)+1)","dots","x_(n)):}]:}:}(-47){ξ1=φ1(x1,…,xp1)ξ2=φ2(xp1+1,…,xp2)ξr=φr(xpr−1,…,xpr)y=F(ξ1,ξ2,…,ξr,xpr+1,…,xn)
is representable nomographically, so we can construct a compound nomogram for equation (45).
L. BAL and I established the necessary and sufficient conditions for the existence of the decomposition (46), in the case of functions F F FFAnd φ i φ i varphi_(i)φidifferentiable [4], [5], in the form
∂ ∂ x j ( j x i ′ j x i + 1 ′ ) = 0 (48) ( k = 1 , 2 , … , r ) ( i = p k − 1 + 1 , … , p k − 1 ) ( p 0 = 0 ) ( j = 1 , … , p k − 1 , p k + 1 , … , n ) ∂ ∂ x j ( j x i ′ j x i + 1 ′ ) = 0 (48) ( k = 1 , 2 , … , r ) ( i = p k − 1 + 1 , … , p k − 1 ) ( p 0 = 0 ) ( j = 1 , … , p k − 1 , p k + 1 , … , n ) (del)/(delx_(j))((j_(x_(i))^('))/(j_(x_(i+1))^(')))=0quad{:[(48)(k=1","2","dots","r)],[{:(i=p_(k-1)+1","dots","p_(k)-1):}{:(p_(0)=0):}],[{:(j=1","dots","p_(k-1)","p_(k)+1","dots","n):}]:}∂∂xj(jxi′jxi+1′)=0(48)(k=1,2,…,r)(i=pk−1+1,…,pk−1)(p0=0)(j=1,…,pk−1,pk+1,…,n)
In the demonstration we used
Lemma 1. If the function f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)allows decompositions
(49) f ( x 1 , … , x n ) = F k [ x 1 , … , x p l − 1 , φ k ( x p k − 1 + 1 , … , x p k ) , x p k + 1 , … , x n ] (49) f ( x 1 , … , x n ) = F k [ x 1 , … , x p l − 1 , φ k ( x p k − 1 + 1 , … , x p k ) , x p k + 1 , … , x n ] {:(49)f{:(x_(1)","dots","x_(n)):}=F_(k){:[x_(1)","dots","x_(p_(l-1))","varphi_(k)(x_(p_(k-1)+1),dots,x_(p_(k)))","x_(p_(k)+1)","dots","x_(n)]:}:}(49)f(x1,…,xn)=Fk[x1,…,xpL−1,φk(xpk−1+1,…,xpk),xpk+1,…,xn]
( k = 1 , 2 , … , r ) ( p 0 = 0 ) ( k = 1 , 2 , … , r ) ( p 0 = 0 ) (k=1,2,dots,r)(p_(0)=0)(k=1,2,…,r)(p0=0), then the decomposition (46) is valid. Here the functions F k , φ k F k , φ k F_(k),varphi_(k)Fk,φkare not subject to any restrictive conditions.
To establish the conditions under which the function (45) admits the decomposition (46) in the case of functions F F FFAnd φ j φ j varphi_(j)φjcontinuous and strictly monotonic, we state the following lemma, the proof of which presents no difficulty.
Lemma 2. For the function f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn), continuous and strictly monotonic with respect to each variable, i.e., of the form
(50) f ( x 1 , … , x n ) = F [ φ ( x 1 , … , x p ) , x p + i , … , x n ] (50) f ( x 1 , … , x n ) = F [ φ ( x 1 , … , x p ) , x p + i , … , x n ] {:(50)f{:(x_(1)","dots","x_(n)):}=F{:[varphi(x_(1),dots,x_(p))","x_(p+i)","dots","x_(n)]:}:}(50)f(x1,…,xn)=F[φ(x1,…,xp),xp+i,…,xn]
Involvement is both necessary and sufficient.
(51) { f ( x 1 , x 2 , … x p . x p + 1 , … , x n ) = f ( y 1 , y 2 , … , y p , x p + 1 , … , x n ) → f ( x 1 , x 2 , … , x p , y p + 1 , … , y n ) = f ( y 1 , y 2 , … , y p , y p + 1 , … , y n ) (51) { f ( x 1 , x 2 , … x p . x p + 1 , … , x n ) = f ( y 1 , y 2 , … , y p , x p + 1 , … , x n ) → f ( x 1 , x 2 , … , x p , y p + 1 , … , y n ) = f ( y 1 , y 2 , … , y p , y p + 1 , … , y n ) {:(51){{:[f{:(x_(1)","x_(2)","dotsx_(p).x_(p+1)","dots","x_(n)):}=f{:(y_(1)","y_(2)","dots","y_(p)","x_(p+1)","dots","x_(n)):}rarr],[f{:(x_(1)","x_(2)","dots","x_(p)","y_(p+1)","dots","y_(n)):}=f{:(y_(1)","y_(2)","dots","y_(p)","y_(p+1)","dots","y_(n)):}]:}:}(51){f(x1,x2,…xp.xp+1,…,xn)=f(y1,y2,…,yp,xp+1,…,xn)→f(x1,x2,…,xp,yp+1,…,yn)=f(y1,y2,…,yp,yp+1,…,yn)
Theorem 2 generalizes Theorem 1.
From Lemmas 1 and 2, it immediately follows:
THEOREM 8. For the function to be continuous and strictly monotonic f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)either of the form (46) with F F FFAnd φ i φ i varphi_(i)φicontinuous and strictly monotonic, it is necessary and sufficient that the relations
( i k = 0 , 1 ; j k = 0 , 1 i k = 0 , 1 ; j k = 0 , 1 i_(k)=0,1;j_(k)=0,1ik=0,1;jk=0,1valid for
i s = 0 , j s = 1 , i σ = j σ = 0 , ( s = p k − 1 + 1 , … , p k ) ( σ = 1 , … , p k − 1 , p k + 1 , … , n ) ( k = 1 , 2 , … , r ) , ( p 0 = 0 ) i s = 0 , j s = 1 , i σ = j σ = 0 , ( s = p k − 1 + 1 , … , p k ) ( σ = 1 , … , p k − 1 , p k + 1 , … , n ) ( k = 1 , 2 , … , r ) , ( p 0 = 0 ) {:[i_(s)=0","j_(s)=1","i_(sigma)=j_(sigma)=0",",{:(s=p_(k-1)+1","dots","p_(k)):}],[,{:(sigma=1","dots","p_(k-1)","p_(k)+1","dots","n):}],[,(k=1","2","dots","r)","{:(p_(0)=0):}]:}is=0,js=1,iσ=jσ=0,(s=pk−1+1,…,pk)(σ=1,…,pk−1,pk+1,…,n)(k=1,2,…,r),(p0=0)
i s = 0 , j s = 1 , i σ = j σ = 1 i s = 0 , j s = 1 , i σ = j σ = 1 i_(s)=0,quadj_(s)=1,quadi_(sigma)=j_(sigma)=1is=0,js=1,iσ=jσ=1
s s ssAnd σ σ sigmaσhaving the same meanings as above.
10. Generalization of pseudo-sums. In the decompositions (49) es r r rrfunctions φ k φ k varphi_(k)φkdo not contain common variables. If we
assume that the function f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)admits decompositions, in which a variable appears under several internal functions, then we will see that a more particular form will result for f f ff.
Let us first consider the example
(53) f ( x , y , z , u ) = g [ φ ( x , y , z ) , u ] = h [ x , ψ ( y , z , u ) ] (53) f ( x , y , z , u ) = g [ φ ( x , y , z ) , u ] = h [ x , ψ ( y , z , u ) ] {:(53)f(x","y","z","u)=g[varphi(x","y","z)","u]=h[x","psi(y","z","u)]:}(53)f(x,y,z,u)=g[φ(x,y,z),u]=h[x,ψ(y,z,u)]
where the functions φ , ψ , g , h φ , ψ , g , h varphi,psi,g,hφ,ψ,g,hare continuous and monotonic.
Theorem 6 shows us that the functions f ( x , y , z 0 , u ) f ( x , y , z 0 , u ) f(x,y,z_(0),u)f(x,y,z0,u)And f ( x , y 0 , z , u ) f ( x , y 0 , z , u ) f(x,y_(0),z,u)f(x,y0,z,u)are pseudo-sums with three terms
(54) { f ( x , y , z 0 , u ) = K − 1 [ F ( x ) + G ( y ) + H ( u ) ] f ( x , y 0 , z , u ) = K 1 − 1 [ F 1 ( x ) + G 1 ( z ) + H 1 ( u ) ] . (54) { f ( x , y , z 0 , u ) = K − 1 [ F ( x ) + G ( y ) + H ( u ) ] f ( x , y 0 , z , u ) = K 1 − 1 [ F 1 ( x ) + G 1 ( z ) + H 1 ( u ) ] . {:(54){{:[f{:(x","y","z_(0)","u):}=K^(-1)[F(x)+G(y)+H(u)]],[f{:(x","y_(0)","z","u):}=K_(1)^(-1){:[F_(1)(x)+G_(1)(z)+H_(1)(u)]:}.]:}:}(54){f(x,y,z0,u)=K−1[F(x)+G(y)+H(u)]f(x,y0,z,u)=K1−1[F1(x)+G1(z)+H1(u)].
By setting in the first equality (54) y = y 0 y = y 0 y=y_(0)y=y0and in the second z = z 0 z = z 0 z=z_(0)z=z0, we obtain two representations for the same two-term pseudo-sum. We have [12]
F 1 ( x ) = a F ( x ) + b H 1 ( x ) = a H ( x ) + a G ( y 0 ) − G 1 ( z 0 ) + c K 1 ( x ) = a K ( x ) + b + c F 1 ( x ) = a F ( x ) + b H 1 ( x ) = a H ( x ) + a G ( y 0 ) − G 1 ( z 0 ) + c K 1 ( x ) = a K ( x ) + b + c {:[F_(1)(x)=aF(x)+b],[H_(1)(x)=aH(x)+aG{:(y_(0)):}-G_(1){:(z_(0)):}+c],[K_(1)(x)=aK(x)+b+c]:}F1(x)=hasF(x)+bH1(x)=hasH(x)+hasG(y0)−G1(z0)+cK1(x)=hasK(x)+b+c
therefore the second equality (54) is written
f ( x , y 0 , z , u ) = K − 1 [ F ( x ) + G 1 ( z ) + G ( y 0 ) − 1 a G 1 ( z 0 ) + H ( u ) ] f ( x , y 0 , z , u ) = K − 1 [ F ( x ) + G 1 ( z ) + G ( y 0 ) − 1 a G 1 ( z 0 ) + H ( u ) ] f(x,y_(0),z,u)=K^(-1)[F(x)+G_(1)(z)+G(y_(0))-(1)/(a)G_(1)(z_(0))+H(u)]f(x,y0,z,u)=K−1[F(x)+G1(z)+G(y0)−1hasG1(z0)+H(u)]
We vary y 0 y 0 y_(0)y0leaving z 0 z 0 z_(0)z0fixed; then F , H , G F , H , G F,H,GF,H,GAnd K K KKdoes not change, and
G 1 ( z ) + G ( y 0 ) − 1 a G 1 ( z 0 ) = G ( y 0 , z ) G 1 ( z ) + G ( y 0 ) − 1 a G 1 ( z 0 ) = G ( y 0 , z ) G_(1)(z)+G(y_(0))-(1)/(a)G_(1)(z_(0))=G(y_(0),z)G1(z)+G(y0)−1hasG1(z0)=G(y0,z)
SO
(55) f ( x , y ′ , z , u ) = K − 1 [ F ( x ) + G ( y , z ) + H ( u ) ] (55) f ( x , y ′ , z , u ) = K − 1 [ F ( x ) + G ( y , z ) + H ( u ) ] {:(55)f(x","y^(')","z","u)=K^(-1)[F(x)+G(y","z)+H(u)]:}(55)f(x,y′,z,u)=K−1[F(x)+G(y,z)+H(u)]
Therefore, (53) implies (55), and obviously (55) implies (53). This result can be generalized:
Theorem 9. Decompositions
(56) f ( x 1 , … , x n ) = g [ φ ( x 1 , … , x q ) , x q + 1 , … , x n ] = = h [ x 1 , … , x p , Ψ ( x p + 1 , … , x n ) ] (56) f ( x 1 , … , x n ) = g [ φ ( x 1 , … , x q ) , x q + 1 , … , x n ] = = h [ x 1 , … , x p , Ψ ( x p + 1 , … , x n ) ] {:[(56)f{:(x_(1)","dots","x_(n)):}=g{:[varphi(x_(1),dots,x_(q))","x_(q+1)","dots","x_(n)]:}=],[=h{:[x_(1)","dots","x_(p)","Psi(x_(p+1),dots,x_(n))]:}]:}(56)f(x1,…,xn)=g[φ(x1,…,xq),xq+1,…,xn]==h[x1,…,xp,Ψ(xp+1,…,xn)]
( p < q ) ( p < q ) (p < q)(p<q), Or φ , ψ , g , h φ , ψ , g , h varphi,psi,g,hφ,ψ,g,hare continuous and strictly monotonic functions, are necessary and sufficient for the function f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)either in the form
(57) f ( x 1 , … , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H ( x q + 1 , … , x n ) ] (57) f ( x 1 , … , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H ( x q + 1 , … , x n ) ] {:[(57)f{:(x_(1)","dots","x_(n)):}=K^(-1)[F(x_(1),dots,x_(p))+G(x_(p+1),dots,x_(q))+],[+H(x_(q+1),dots,x_(n))]]:}(57)f(x1,…,xn)=K−1[F(x1,…,xp)+G(xp+1,…,xq)++H(xq+1,…,xn)]
F , G , H , K F , G , H , K F,G,H,KF,G,H,Kbeing continuous and strictly monotonic.
Formula (57) can be put in the form (56), so it remains to establish
( 56 ) → ( 57 ) ( 56 ) → ( 57 ) (56)rarr(57)(56)→(57)We have seen that this implication is valid for n = 4 n = 4 n=4n=4.
We assume it is valid for n − 1 ≧ 4 n − 1 ≧ 4 n-1 >= 4n−1≧4and we will demonstrate it for n n nn. Because n ≧ 5 n ≧ 5 n >= 5n≧5, there are at least two elements in one of the variable groups x 1 , … , x p ; x p + 1 , … , x q ; x q + 1 , … , x n x 1 , … , x p ; x p + 1 , … , x q ; x q + 1 , … , x n x_(1),dots,x_(p);x_(p+1),dots,x_(q);x_(q+1),dots,x_(n)x1,…,xp;xp+1,…,xq;xq+1,…,xnLet's assume, for the sake of clarity, that it's the third one. It follows from the induction hypothesis
(58) { f ( x 1 , … , x n − 2 , x n − 1 , x n 0 ) = K − 1 [ F ( x 1 , … , x p ) + + G ( x p + 1 , … , x q ) + H ′ ( x q + 1 , … , x n − 1 , x n − 2 ) ] f ( x 1 , … , x n − 2 , x n − 1 0 , x n ) = K 1 − 1 [ F 1 ( x 1 , … , x p ) + + G 1 ( x p + 1 , … , x q ) + H 1 ( x q + 1 , … , x n − 2 , x n ) ] (58) { f ( x 1 , … , x n − 2 , x n − 1 , x n 0 ) = K − 1 [ F ( x 1 , … , x p ) + + G ( x p + 1 , … , x q ) + H ′ ( x q + 1 , … , x n − 1 , x n − 2 ) ] f ( x 1 , … , x n − 2 , x n − 1 0 , x n ) = K 1 − 1 [ F 1 ( x 1 , … , x p ) + + G 1 ( x p + 1 , … , x q ) + H 1 ( x q + 1 , … , x n − 2 , x n ) ] {:(58){{:[f{:(x_(1)","dots","x_(n-2)","x_(n-1)","x_(n)^(0)):}=K^(-1)[F(x_(1),dots,x_(p))+],[+G(x_(p+1),dots,x_(q))+H^(')(x_(q+1),dots,x_(n-1),x_(n-2))]],[f{:(x_(1)","dots","x_(n-2)","x_(n-1)^(0)","x_(n)):}=K_(1)^(-1){:[F_(1)(x_(1),dots,x_(p))+:}],[{:+G_(1)(x_(p+1),dots,x_(q))+H_(1)(x_(q+1),dots,x_(n-2),x_(n))]:}]:}:}(58){f(x1,…,xn−2,xn−1,xn0)=K−1[F(x1,…,xp)++G(xp+1,…,xq)+H′(xq+1,…,xn−1,xn−2)]f(x1,…,xn−2,xn−10,xn)=K1−1[F1(x1,…,xp)++G1(xp+1,…,xq)+H1(xq+1,…,xn−2,xn)]
We have
K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + (59) + H ′ ( x q + 1 , … , x n − 2 , x n − 1 0 ) ] = K 1 − 1 [ F 1 ( x 1 , … , x p ) + + G 1 ( x p + 1 , … , x q ) + H 1 ( x q − 1 , … , x n − 2 , x n 0 ) ] K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + (59) + H ′ ( x q + 1 , … , x n − 2 , x n − 1 0 ) ] = K 1 − 1 [ F 1 ( x 1 , … , x p ) + + G 1 ( x p + 1 , … , x q ) + H 1 ( x q − 1 , … , x n − 2 , x n 0 ) ] {:[K^(-1)[F(x_(1),dots,x_(p))+G(x_(p+1),dots,x_(q))+],[(59)+H^(')(x_(q+1),dots,x_(n-2),x_(n-1)^(0))]=K_(1)^(-1){:[F_(1)(x_(1),dots,x_(p))+:}],[{:+G_(1)(x_(p+1),dots,x_(q))+H_(1)(x_(q-1),dots,x_(n-2),x_(n)^(0))]:}]:}K−1[F(x1,…,xp)+G(xp+1,…,xq)+(59)+H′(xq+1,…,xn−2,xn−10)]=K1−1[F1(x1,…,xp)++G1(xp+1,…,xq)+H1(xq−1,…,xn−2,xn0)]
By fixing all variables except x 1 x 1 x_(1)x1And x p + 1 x p + 1 x_(p+1)xp+1, we obtain two representations of the same two-term pseudo-sum, therefore K 1 = a K + a ′ [ 12 ] K 1 = a K + a ′ [ 12 ] K_(1)=aK+a^(')[12]K1=hasK+has′[12]By substituting this expression of K 1 K 1 K_(1)K1in the second equation (58) and writing for 1 a F 1 , 1 a G 1 , 1 a H 1 − a ′ 1 a F 1 , 1 a G 1 , 1 a H 1 − a ′ (1)/(a)F_(1),(1)/(a)G_(1),(1)/(a)H_(1)-a^(')1hasF1,1hasG1,1hasH1−has′simply F 1 , G 1 , H 1 F 1 , G 1 , H 1 F_(1),G_(1),H_(1)F1,G1,H1, we obtain the same form for the second equation (58) with the only modification that instead of K 1 K 1 K_(1)K1we will have K K KKIn (59) the arguments of the function K − 1 K − 1 K^(-1)K−1must be equal, therefore
F 1 ( x 1 , … , x p ) = F ( x 1 , … , x p ) , G 1 ( x p + 1 , … , x q ) = G ( x p + 1 , … , x q ) F 1 ( x 1 , … , x p ) = F ( x 1 , … , x p ) , G 1 ( x p + 1 , … , x q ) = G ( x p + 1 , … , x q ) F_(1)(x_(1),dots,x_(p))=F(x_(1),dots,x_(p)),G_(1)(x_(p+1),dots,x_(q))=G(x_(p+1),dots,x_(q))F1(x1,…,xp)=F(x1,…,xp),G1(xp+1,…,xq)=G(xp+1,…,xq)
The second equation (58) can be written
f ( x 1 , … , x n − 2 , x n − 1 o , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H 1 ( x q + 1 , … , x n − 2 , x n ) ] f ( x 1 , … , x n − 2 , x n − 1 o , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H 1 ( x q + 1 , … , x n − 2 , x n ) ] {:[f{:(x_(1)","dots","x_(n-2)","x_(n-1)^(o)","x_(n)):}=K^(-1)[F(x_(1),dots,x_(p))+G(x_(p+1),dots,x_(q))+],[{:+H_(1)(x_(q+1),dots,x_(n-2),x_(n))]:}]:}f(x1,…,xn−2,xn−1o,xn)=K−1[F(x1,…,xp)+G(xp+1,…,xq)++H1(xq+1,…,xn−2,xn)]
We vary x n − 1 o x n − 1 o x_(n-1)^(o)xn−1oleaving x n o x n o x_(n)^(o)xnofixed; then the functions F , G , K − 1 F , G , K − 1 F,G,K^(-1)F,G,K−1do not change and we get
f ( x 1 , x 2 , … , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H ( x q + 1 , … , x n − 1 , x n ) ] f ( x 1 , x 2 , … , x n ) = K − 1 [ F ( x 1 , … , x p ) + G ( x p + 1 , … , x q ) + + H ( x q + 1 , … , x n − 1 , x n ) ] {:[f{:(x_(1)","x_(2)","dots","x_(n)):}=K^(-1)[F(x_(1),dots,x_(p))+G(x_(p+1),dots,x_(q))+],[+H(x_(q+1),dots,x_(n-1),x_(n))]]:}f(x1,x2,…,xn)=K−1[F(x1,…,xp)+G(xp+1,…,xq)++H(xq+1,…,xn−1,xn)]
Theorem 9 has been proven.
Now suppose that f ( x , y , z , u ) f ( x , y , z , u ) f(x,y,z,u)f(x,y,z,u)admits the following decompositions:
It follows from what we have just shown
f ( x , y , z , u ) = L − 1 [ F ( x , y ) + H ( z ) + K ( u ) ] = = L 1 − 1 [ F 1 ( x , y ) + G 1 ( y ) + K 1 ( u ) ] f ( x , y , z , u ) = L − 1 [ F ( x , y ) + H ( z ) + K ( u ) ] = = L 1 − 1 [ F 1 ( x , y ) + G 1 ( y ) + K 1 ( u ) ] {:[f(x","y","z","u)=L^(-1)[F(x","y)+H(z)+K(u)]=],[=L_(1)^(-1){:[F_(1)(x","y)+G_(1)(y)+K_(1)(u)]:}]:}f(x,y,z,u)=L−1[F(x,y)+H(z)+K(u)]==L1−1[F1(x,y)+G1(y)+K1(u)]
By assigning constant values ​​to y y yyAnd z z zzAs we can see from above, we can assume L 1 = L L 1 = L L_(1)=LL1=LThen we ask z = z = z=z=const, and we find F ( x , y ) = F 1 ( x , z 0 ) + G 1 ( y ) = F ( x ) + G ( y ) F ( x , y ) = F 1 ( x , z 0 ) + G 1 ( y ) = F ( x ) + G ( y ) F(x,y)=F_(1)(x,z_(0))+G_(1)(y)=F(x)+G(y)F(x,y)=F1(x,z0)+G1(y)=F(x)+G(y), SO
(61) f ( x , y , z , u ) = L − 1 [ F ( x ) + G ( y ) + H ( z ) + K ( u ) ] (61) f ( x , y , z , u ) = L − 1 [ F ( x ) + G ( y ) + H ( z ) + K ( u ) ] {:(61)f(x","y","z","u)=L^(-1)[F(x)+G(y)+H(z)+K(u)]:}(61)f(x,y,z,u)=L−1[F(x)+G(y)+H(z)+K(u)]
Therefore, the three decompositions (60) are necessary and sufficient for the function f ( x , y , z , u ) f ( x , y , z , u ) f(x,y,z,u)f(x,y,z,u)that is, a pseudo-sum with four terms.
We find by complete induction:
Theorem 10. Decompositions
(62)
f ( x 1 , … , x n ) = g i [ φ i ( x 1 , … , x i − 1 , x i + 1 , … , x n ) , x i ] f ( x 1 , … , x n ) = g i [ φ i ( x 1 , … , x i − 1 , x i + 1 , … , x n ) , x i ] f(x_(1),dots,x_(n))=g_(i)[varphi_(i)(x_(1),dots,x_(i-1),x_(i+1),dots,x_(n)),x_(i)]f(x1,…,xn)=gi[φi(x1,…,xi−1,xi+1,…,xn),xi]
( i = 2 , … , n ) ( i = 2 , … , n ) (i=2,dots,n)(i=2,…,n), Or φ i φ i varphi_(i)φiAnd g i g i g_(i)giare continuous and strictly monotonic functions, are necessary and sufficient for the function f f ffeither a pseudosum at n n nnterms
(63) f ( x 1 , … , x n ) = F − 1 [ F 1 ( x 1 ) + F 2 ( x 2 ) + … + F n ( x n ) ] (63) f ( x 1 , … , x n ) = F − 1 [ F 1 ( x 1 ) + F 2 ( x 2 ) + … + F n ( x n ) ] {:(63)f{:(x_(1)","dots","x_(n)):}=F^(-1){:[F_(1)(x_(1))+F_(2)(x_(2))+dots+F_(n)(x_(n))]:}:}(63)f(x1,…,xn)=F−1[F1(x1)+F2(x2)+…+Fn(xn)]
F i F i F_(i)FiAnd F F FFbeing continuous and strictly monotonic functions.
Theorem 10 generalizes Theorem 6. Theorem 7 generalizes as follows:
Theorem 11. The necessary and sufficient condition for the function to be continuous and strictly monotonic f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)either a pseudo-sum to n ⩾ 3 n ⩾ 3 n >= 3n⩾3The term is that the relationships
(64) f ( x 1 ( i 1 ) , x 2 ( i 2 ) , … , x n ( i n ) ) = f ( x 1 ( j 1 ) , x 2 ( j 2 ) , … , x n ( j n ) ) (64) f ( x 1 ( i 1 ) , x 2 ( i 2 ) , … , x n ( i n ) ) = f ( x 1 ( j 1 ) , x 2 ( j 2 ) , … , x n ( j n ) ) {:(64)f(x_(1)^((i_(1)))","x_(2)^((i_(2)))","dots","x_(n)^((i_(n))))=f(x_(1)^((j_(1)))","x_(2)^((j_(2)))","dots","x_(n)^((j_(n)))):}(64)f(x1(i1),x2(i2),…,xn(in))=f(x1(j1),x2(j2),…,xn(jn))
valid for indices i k = 0 , 1 , j k = 0 , 1 i k = 0 , 1 , j k = 0 , 1 i_(k)=0,1,j_(k)=0,1ik=0,1,jk=0,1who check ∑ k = 1 n i k = ∑ k = 1 n j k = 1 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 1 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=1∑k=1nik=∑k=1njk=1, lead to the same relations (64) for ∑ k = 1 n i k = ∑ k = 1 n j k = 2 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 2 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=2∑k=1nik=∑k=1njk=2.
For n = 3 n = 3 n=3n=3This theorem coincides with Theorem 7. We will demonstrate the sufficiency of the condition by induction, assuming it for n − 1 n − 1 n-1n−1We admit that relations (64) with ∑ k = 1 n i k = ∑ k = 1 n j k = 1 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 1 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=1∑k=1nik=∑k=1njk=1lead to the same relationships with ∑ k = 1 n i k = ∑ k = 1 n j k = 2 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 2 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=2∑k=1nik=∑k=1njk=2and let's demonstrate that f f ffis a pseudo-sum.
Let us consider the function
φ ( x 1 , … , x n − 1 ) = f ( x 1 , … , x n − 1 , x n ( 0 ) ) φ ( x 1 , … , x n − 1 ) = f ( x 1 , … , x n − 1 , x n ( 0 ) ) varphi(x_(1),dots,x_(n-1))=f(x_(1),dots,x_(n-1),x_(n)^((0)))φ(x1,…,xn−1)=f(x1,…,xn−1,xn(0))
and suppose that
(65) φ ( x 1 ( i 1 ) , … , x n − 1 ( i n − 1 ) ) = ( x 1 ( i 1 ) , … , x n − 1 ( i n − 1 ) ) , (65) φ ( x 1 ( i 1 ) , … , x n − 1 ( i n − 1 ) ) = ( x 1 ( i 1 ) , … , x n − 1 ( i n − 1 ) ) , {:(65)varphi(x_(1)^((i_(1)))","dots","x_(n-1)^((i_(n-1))))=(x_(1)^((i_(1)))","dots","x_(n-1)^((i_(n-1))))",":}(65)φ(x1(i1),…,xn−1(in−1))=(x1(i1),…,xn−1(in−1)),
if i k = 0 , 1 , j k = 0 , 1 , ∑ k = 1 n − 1 i k = ∑ k = 1 n − 1 j k = 1 i k = 0 , 1 , j k = 0 , 1 , ∑ k = 1 n − 1   i k = ∑ k = 1 n − 1   j k = 1 i_(k)=0,1,j_(k)=0,1,sum_(k=1)^(n-1)i_(k)=sum_(k=1)^(n-1)j_(k)=1ik=0,1,jk=0,1,∑k=1n−1ik=∑k=1n−1jk=1Let's determine x n ( 1 ) x n ( 1 ) x_(n)^((1))xn(1)in the
following way:
φ ( x 1 ( 1 ) , x 2 ( 0 ) , … , x n − 1 ( 0 ) ) = f ( x 1 ( 0 ) , … , x n − 1 ( 0 ) , x n ( 1 ) ) . φ ( x 1 ( 1 ) , x 2 ( 0 ) , … , x n − 1 ( 0 ) ) = f ( x 1 ( 0 ) , … , x n − 1 ( 0 ) , x n ( 1 ) ) . varphi(x_(1)^((1)),x_(2)^((0)),dots,x_(n-1)^((0)))=f(x_(1)^((0)),dots,x_(n-1)^((0)),x_(n)^((1))).φ(x1(1),x2(0),…,xn−1(0))=f(x1(0),…,xn−1(0),xn(1)).
So the function f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)checks (64) for ∑ k = 1 n i k = ∑ k = 1 n j k = 1 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 1 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=1∑k=1nik=∑k=1njk=1, therefore also for ∑ k = 1 n i k = ∑ k = 1 n j k = 2 ∑ k = 1 n   i k = ∑ k = 1 n   j k = 2 sum_(k=1)^(n)i_(k)=sum_(k=1)^(n)j_(k)=2∑k=1nik=∑k=1njk=2Taking into account the definition of the function φ ( x 1 , … , x n − 1 ) φ ( x 1 , … , x n − 1 ) varphi(x_(1),dots,x_(n-1))φ(x1,…,xn−1)We see that conditions (65) are satisfied for ∑ k = 1 n − 1 i k = ∑ k = 1 n − 1 j k = 2 ∑ k = 1 n − 1   i k = ∑ k = 1 n − 1   j k = 2 sum_(k=1)^(n-1)i_(k)=sum_(k=1)^(n-1)j_(k)=2∑k=1n−1ik=∑k=1n−1jk=2Therefore, relations (65) for ∑ k = 1 n − 1 i k = ∑ k = 1 n − 1 j k == 1 ∑ k = 1 n − 1   i k = ∑ k = 1 n − 1   j k == 1 sum_(k=1)^(n-1)i_(k)=sum_(k=1)^(n-1)j_(k)==1∑k=1n−1ik=∑k=1n−1jk==1lead to relations (65) for
∑ k = 1 n − 1 i k = ∑ k = 1 n − 1 j k = 2 , ∑ k = 1 n − 1   i k = ∑ k = 1 n − 1   j k = 2 , sum_(k=1)^(n-1)i_(k)=sum_(k=1)^(n-1)j_(k)=2,∑k=1n−1ik=∑k=1n−1jk=2,
Therefore, we obtain, by applying the induction hypothesis
Done
φ ( x 1 , … , x n − 1 ) = F − 1 [ F 1 ( x 1 ) + … + F n − 1 ( x n − 1 ) ] . φ ( x 1 , … , x n − 1 ) = F − 1 [ F 1 ( x 1 ) + … + F n − 1 ( x n − 1 ) ] . varphi(x_(1),dots,x_(n-1))=F^(-1)[F_(1)(x_(1))+dots+F_(n-1)(x_(n-1))].φ(x1,…,xn−1)=F−1[F1(x1)+…+Fn−1(xn−1)].
f ( x 1 , … , x n − 1 , x n ( 0 ) ) = F − 1 [ F 1 ( x 1 ) + … + F n − 1 ( x n − 1 ) ] f ( x 1 , … , x n − 1 , x n ( 0 ) ) = F − 1 [ F 1 ( x 1 ) + … + F n − 1 ( x n − 1 ) ] f(x_(1),dots,x_(n-1),x_(n)^((0)))=F^(-1)[F_(1)(x_(1))+dots+F_(n-1)(x_(n-1))]f(x1,…,xn−1,xn(0))=F−1[F1(x1)+…+Fn−1(xn−1)]
and also
f ( x 1 , … , x n − 2 , x n − 1 ( 0 ) , x n ) = Φ − 1 [ Φ 1 ( x 1 ) + … + Φ n − 2 ( x n − 2 ) + Φ n ( x n ) ] . f ( x 1 , … , x n − 2 , x n − 1 ( 0 ) , x n ) = Φ − 1 [ Φ 1 ( x 1 ) + … + Φ n − 2 ( x n − 2 ) + Φ n ( x n ) ] . f(x_(1),dots,x_(n-2),x_(n-1)^((0)),x_(n))=Phi^(-1)[Phi_(1)(x_(1))+dots+Phi_(n-2)(x_(n-2))+Phi_(n)(x_(n))].f(x1,…,xn−2,xn−1(0),xn)=Φ−1[Φ1(x1)+…+Φn−2(xn−2)+Φn(xn)].
By posing in the last two relationships x 3 = x 3 ( 0 ) , … , x n = x n ( 0 ) x 3 = x 3 ( 0 ) , … , x n = x n ( 0 ) x_(3)=x_(3)^((0)),dots,x_(n)=x_(n)^((0))x3=x3(0),…,xn=xn(0)we have two forms of writing for f ( x 1 , x 2 , x 3 ( 0 ) , … , x n ( 0 ) ) f ( x 1 , x 2 , x 3 ( 0 ) , … , x n ( 0 ) ) f(x_(1),x_(2),x_(3)^((0)),dots,x_(n)^((0)))f(x1,x2,x3(0),…,xn(0))from which we obtain that Φ Φ PhiΦcan be chosen equal to F F FFWe immediately
F i = Φ i ( i = 1 , 2 , … , n − 2 ) F i = Φ i ( i = 1 , 2 , … , n − 2 ) F_(i)=Phi_(i)quad(i=1,2,dots,n-2)Fi=Φi(i=1,2,…,n−2)
SO
f ( x 1 , … , x n − 2 , x n − 1 ( 0 ) , x n ) = F − 1 [ F 1 ( x 1 ) + … + F n − 2 ( x n − 2 ) + Φ n ( x n ) ] f ( x 1 , … , x n − 2 , x n − 1 ( 0 ) , x n ) = F − 1 [ F 1 ( x 1 ) + … + F n − 2 ( x n − 2 ) + Φ n ( x n ) ] f(x_(1),dots,x_(n-2),x_(n-1)^((0)),x_(n))=F^(-1)[F_(1)(x_(1))+dots+F_(n-2)(x_(n-2))+Phi_(n)(x_(n))]f(x1,…,xn−2,xn−1(0),xn)=F−1[F1(x1)+…+Fn−2(xn−2)+Φn(xn)]
We vary x n − 1 ( 0 ) x n − 1 ( 0 ) x_(n-1)^((0))xn−1(0)leaving x n ( 0 ) x n ( 0 ) x_(n)^((0))xn(0)fixed
Likewise
f ( x 1 , … , x n ) = F − 1 [ F 1 ( x 1 ) + … + F n − 2 ( x n − 2 ) + Φ ( x n − 1 , x n ) ] f ( x 1 , … , x n ) = F − 1 [ F 1 ( x 1 ) + … + F n − 2 ( x n − 2 ) + Φ ( x n − 1 , x n ) ] f(x_(1),dots,x_(n))=F^(-1)[F_(1)(x_(1))+dots+F_(n-2)(x_(n-2))+Phi(x_(n-1),x_(n))]f(x1,…,xn)=F−1[F1(x1)+…+Fn−2(xn−2)+Φ(xn−1,xn)]
f ( x 1 , … , x n ) = G − 1 [ G 1 ( x ) + … + G n − 4 ( x n − 4 ) + ψ ( x n − 3 , x n − 2 ) + + G n − 1 ( x n − 1 ) + G n ( x n ) ] f ( x 1 , … , x n ) = G − 1 [ G 1 ( x ) + … + G n − 4 ( x n − 4 ) + ψ ( x n − 3 , x n − 2 ) + + G n − 1 ( x n − 1 ) + G n ( x n ) ] {:[f{:(x_(1)","dots","x_(n)):}=G^(-1){:[G_(1)(x)+dots+G_(n-4)(x_(n-4))+psi(x_(n-3),x_(n-2))+:}],[{:+G_(n-1)(x_(n-1))+G_(n)(x_(n))]:}]:}f(x1,…,xn)=G−1[G1(x)+…+Gn−4(xn−4)+ψ(xn−3,xn−2)++Gn−1(xn−1)+Gn(xn)]
By comparing the last two relationships, we obtain that f ( x 1 , … , x n ) f ( x 1 , … , x n ) f(x_(1),dots,x_(n))f(x1,…,xn)is a pseudo-sum to n n nnterms.
Thus we have demonstrated the sufficiency of the stated condition. Direct verification shows that it is also necessary. The theo-

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