Re: for all real x and y

On 27 Jan., 21:50, hagman <goo...@xxxxxxxxxxxxx> 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

Esp. f1(0+0) = 6 f2(0) f3(0) f4(0), hence
f1(0)^2 = 6 f1(0) f2(0) f3(0) f4(0).
By symmetry,
f1(0)^2 = f2(0)^2 = f3(0)^2 = f4(0)^2 = 6 f1(0) f2(0) f3(0) f4(0)
With c := f1(0) all values fk(0) are just +- c.
Thus either f1(0)=f2(0)=f3(0)=f4(0)=c or exactly two
of f2(0),f3(0),f4(0) are -c and the other is +c.
We may restrict to the first case because the eqautions
remain correct if we replace exactly two of the functions
with their negative.
Then we conclude that c^2 = 6 c^4, i.e. c = +- 1/sqrt(6).
Again, wlog c=+1/sqrt(6).

Assume that the fk are sufficiently smooth, i.e
fk(x) = c + ak x^n +O(x^{n+1}) for some n>=1.
From
f1(x+0) = f2(0)f3(x)f4(x) + f2(x)f3(0)f4(x) + f2(x)f3(x)f4(0)
+ f2(0)f3(0)f4(x) + f2(x)f3(0)f4(0) + f2(0)f3(x)f4(0)
we conclude thus
c + a1 x^n = c (c + a3 x^n) (c + a4 x^n)
+ c (c + a2 x^n) (c + a3 x^n)
+ c (c + a2 x^n) (c + a4 x^n)
+ c^2 (c + a2 x^n)
+ c^2 (c + a3 x^n)
+ c^2 (c + a4 x^n) + O(x^{n+1})
Hence
a1 = 3*c^2 (a2 + a3 + a4) = (a2+a3+a4)/2
or 3 a1 = a1+a2+a3+a4.
By symmetry
3 a1 = 3 a2 = 3 a3 = 3 a4 = a1+a2+a3+a4
and thus a1 = a2 = a3 = a4 = 0.

Hence (up to a little playing around with signs), the only
smooth solutions are
f1(x) = ... = f4(x) = 1/sqrt(6).

hagman

In the last sentence, I should have written holomorphic instead of
smooth
.

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