Re: for all real x and y

"alainverghote@xxxxxxxx" <alainverghote@xxxxxxxx> writes:

On 27 jan, 18:37, tommy1729 <tommy1...@xxxxxxxxx> wrote:
alain wrote:
On 27 jan, 00:24, tommy1729 <tommy1...@xxxxxxxxx>
wrote:
for all real x and y ( and preferably complex too )

f1(x+y) =3D
f2(y)f3(x)f4(x)+f2(x)f3(y)f4(x)+f2(x)f3(x)f4(y)
+ f2(x)f3(y)f4(y)+f2(y)f3(x)f4(y)+f2(y)f3(y)f4(x)

f2(x+y) =3D
f1(y)f3(x)f4(x)+f1(x)f3(y)f4(x)+f1(x)f3(x)f4(y)
+ f1(x)f3(y)f4(y)+f1(y)f3(x)f4(y)+f1(y)f3(y)f4(x)

f3(x+y) =3D
f1(y)f2(x)f4(x)+f1(x)f2(y)f4(x)+f1(x)f2(x)f4(y)
+ f1(x)f2(y)f4(y)+f1(y)f2(x)f4(y)+f1(y)f2(y)f4(x)

f4(x+y) =3D
f1(y)f2(x)f3(x)+f1(x)f2(y)f3(x)+f1(x)f2(x)f3(y)
+ f1(x)f2(y)f3(y)+f1(y)f2(x)f3(y)+f1(y)f2(y)f3(x)

with f1 f2 f3 and f4 distinct.

regards
tommy1729

Bonjour,

for y =3D 0 , we've got :
f1(0)+f1(x) =3D
(f2(0)+f2(x))*(f3(0)+f3(x))*(f4(0)+f4(x))
+ f1(0)
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 + f1(0) -f2(0)*f3(0)*f4(0) ,
let us put gi(x)=3Dfi(0)+fi(x) and f1(0)
-f2(0)*f3(0)*f4(0) =3D c1
...ci,

Alain

???

im not sure that is correct , since i see " minus f2 " for example.

even if correct , what are you trying to tell me ?

thanks for your reply though.

regards
tommy1729- Masquer le texte des messages pr=E9c=E9dents -

- Afficher le texte des messages pr=E9c=E9dents -

Bonsoir,

for y =3D 0 , put gi(x)=3Dfi(0)+fi(x)
c1 =3D f1(0)-f2(0)*f3(0)*f4(0) , ci =3D fi(0) -...

we'll have :
g1(x) =3D g2(x)*g3(x)*g4(x)+c1
g2(x) =3D g1(x)*g3(x)*g4(x)+c2
g3(x) =3D g1(x)*g2(x)*g4(x)+c3
g4(x) =3D g1(x)*g2(x)*g3(x)+c4

So g1(x)^2 - c1*g1(x) =3D g2(x)^2 - c2*g2(x) =3D...
=3D g1*g2(x)*g3(x)*g4(x) , a common value

What I get (with some help from Maple) is this.

First of all, with x=y=0, either all f_i(0) = 0 or all are square roots
of 1/6, with any signs allowed as long as the product of all of them is
positive. Note that there's a symmetry: you can multiply any two of the f_i
by -1 and preserve the equations. So we may assume wlog either all
f_i(0) = 0 or all f_i(0) = +1/sqrt(6).

If all f_i(0)=0, put y=0 to get all f_i(x) = 0.

If all f_i(0) = 1/sqrt(6), put y=0 and solve the resulting equations.
Using Groebner basis methods, I find the following solutions:

1) all f_i(x) = 0
2) all f_i(x) are sqrt(1/6)
3) one f_i(x) = 35/48 sqrt(6) and the others are -7/12 sqrt(6)
4) two f_i(x) = -7/12 sqrt(6) + 3/4 sqrt(14) and the other two are
-7/12 sqrt(6) - 3/4 sqrt(14).

(1) clearly can't happen if f_i(0)=1/sqrt(6): if all f_i(x) = 0 then all
f_i(x+y) = 0.
(2) is possible.
(3) and (4) are impossible, because they would produce wrong values for
f_i(2x)

Conclusion:
This system has only eight solutions, all constant:
all f_i(x) = 0
all f_i(x) = sqrt(1/6)
two f_i(x) = sqrt(1/6), two -sqrt(1/6)
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

Relevant Pages

  • Re: Proofs of theorems and Symmetries
    ... >> symmetry property such that both the hypothesis and the conclusion are ... >> T is explicitely symmetric with regards to S at any stage. ... One example that comes to mind is the theory of finite-dimensional ...
    (sci.math.research)
  • doubt with equations
    ... because i am trying to prove some symmetry in my proof. ... Regards, ... J.Anil Kumar. ... Prev by Date: ...
    (comp.text.tex)
  • Re: fov study
    ... symmetry is bad in games with ranged attacks? ...
    (rec.games.roguelike.development)