Re: for all real x and y

alain wrote:

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) =
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) =
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) =
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) =
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 = 0 , we've got :
f1(0)+f1(x) =
(f2(0)+f2(x))*(f3(0)+f3(x))*(f4(0)+f4(x))
+ f1(0)
                    + f1(0) -f2(0)*f3(0)*f4(0) ,
let us put gi(x)=fi(0)+fi(x) and f1(0)
-f2(0)*f3(0)*f4(0) = 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 précédents
-

- Afficher le texte des messages précédents -

Bonsoir,

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

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

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

Alain


i havent got time to check that now , but if you are correct that is intresting.

thanks.

we also have for i mod 4.

fi(2x) = 6*f(i+1)(x)*f(i+2)(x)*f(i+3)(x)

but the big questions are

can fi be expressed in standard functions ?

can it be expressed in an integral ?

regards
tommy1729
.

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