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ali666

#1 h Aug 19, 2006, 8:31 PM • 2 Y

Y by Adventure10, Mango247

find all valued functions $f$ such that for all real $x,y$ :
$f(x-y)=f(x)f(y)$

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Rust

#2 h Aug 19, 2006, 9:02 PM • 2 Y

Y by Adventure10, Mango247

One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$ , then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

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ali666

#3 h Aug 19, 2006, 9:57 PM • 2 Y

Y by Adventure10, Mango247

Rust wrote:

One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$ , then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

thank Rust.
a similar problem : http://www.mathlinks.ro/Forum/viewtopic.php?t=97604

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#4 h Apr 30, 2025, 6:03 AM

Y by

Rust wrote:

One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$ , then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?

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#5 h Apr 30, 2025, 6:21 AM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

Rust wrote:

One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$ , then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?

Because if $f(a)=-1$ for some $a$ then from $f(x)(1-f(2x))=0$ we have:
$f\left(\frac a2\right)(1-f(a))=0$ and so $f\left(\frac a2\right)=0$ , but this contradicts $f\left(\frac a2\right)\in\{-1,1\}$ .

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#6 h Apr 30, 2025, 6:25 AM

Y by

jasperE3 wrote:

Blackbeam999 wrote:

Rust wrote:

One solution is f(x)=0.
y=0 give f(x)(1-f(0))=0. If exist x, suth that $f(x)\not =0$ , then f(0)=1.
x=y give $f(y)^{2}=1$ and x=2y give f(y)(1-f(2y))=0.
Therefore only 2 solutions:
1. $f(x)=0 \ \forall x\in R,$
2. $f(x)=1 \ \forall x\in R.$

How can I prove that there is no a,b exist such that f(a)=1 , f(b)= -1?

Because if $f(a)=-1$ for some $a$ then from $f(x)(1-f(2x))=0$ we have:
$f\left(\frac a2\right)(1-f(a))=0$ and so $f\left(\frac a2\right)=0$ , but this contradicts $f\left(\frac a2\right)\in\{-1,1\}$ .

Oh thanks

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