Functional equation f(f(x)+y) = x+f(y)

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Functional equation f(f(x)+y) = x+f(y)

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Gramos

#1 h Jul 30, 2015, 11:23 AM • 1 Y

Y by Adventure10

Find all functions f : Q → Q such that
$f(f(x)+y)=x+f(y)$ , for x, y ∈ Q.

This post has been edited 1 time. Last edited by math_explorer, Aug 5, 2015, 2:51 AM
Reason: more descriptive title

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pco

#2 h Jul 30, 2015, 11:34 AM • 2 Y

Y by Adventure10, Mango247

Let $P(x,y)$ be the assertion $f(f(x)+y)=x+f(y)$
Let $a=f(0)$

$P(x,0)$ $\implies$ $f(f(x))=x+a$
$P(f(x),y)$ $\implies$ $f(x+y+a)=f(x)+f(y)$
$P(f(x+y),0)$ $\implies$ $f(x+y+a)=f(x+y)+a$
Subtracting, we get $f(x+y)=f(x)+f(y)-a$ and so $f(x)-a$ additive and so, since from $\mathbb Q\to\mathbb Q$ :
$f(x)=cx+a$ and, plugging this back in original equation, $c^2=1$ and $a=0$ and so two solutions :

$\boxed{\text{S1 : }f(x)=x\text{ }\forall x\in\mathbb Q}$ and $\boxed{\text{S2 : }f(x)=-x\text{ }\forall x\in\mathbb Q}$

This post has been edited 1 time. Last edited by Kunihiko_Chikaya, Jul 30, 2015, 10:27 PM

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#3 h Jul 30, 2015, 1:16 PM • 2 Y

Y by Adventure10, Mango247

Gramos wrote:

Find all functions f : Q → Q such that
$f(f(x)+y)=x+f(y)$ , for x, y ∈ Q.

Let $c = f(0)$ so if we put $x = y = 0$ then $f(c) = c$ .
If $y = c$ then $f(f(x) + c) = x + c$ .
If $f(z) = f(t)$ then $f(f(z) + c) = f(f(t) + c) = z + c = t + c$ so $t = 0$ so there exist only one $x$ for every d such $f(x) = d$ .
$f(c+y) = f(y) + c$ and $f(f(y) + c) = y + c$ so $f(f(f(y) + c)) = f(y) + c$ so let $d = f(y) + c$ $\implies$ $f(f(d)) = d$ for every $d$ .
If $y = 0$ then $f(f(x)) = x + c$ but $f(f(x)) = x$ so $c = 0$ .
$f(f(x) + y) = x + f(y)$ ,if $p = f(y)$ $\implies$ $f(p) = y$ and $f(f(x) + f(p)) = x + p$ , $f(x + p) = f(x) + f(p)$ .
So if $f(a) = b$ then $f(b) = a$ $\implies$ only the solutions $f(x) = x$ or $f(x) = -x$ .

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Tan

#4 h Aug 4, 2015, 5:35 PM • 2 Y

Y by Adventure10, Mango247

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