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#1 h Apr 17, 2025, 3:01 PM

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find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

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#2 h Apr 18, 2025, 1:16 AM • 1 Y

Y by luci1337

luci1337 wrote:

find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

Suppose first that $f(1)=0$ .
Here $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$ .
$P(1,x)\Rightarrow f(f(x))=-f(2)$
So $P(x,y)$ becomes $4f(x)f(x+y)-2f(x^2)=xf(2x)-yf(2)$ , and fixing $x=j$ we get that $f$ is linear. Testing, we find no solutions.

Otherwise, $f(1)\ne0$ ; we can define $g:\mathbb R\to\mathbb R$ such that $g(x)=\frac{f(x)}{f(1)}$ .
$P(0,x)\Rightarrow2f(0)f(x)=f(0)$
If $f(0)\ne0$ then $f(x)=\frac12$ for all $x$ which isn't a valid solution, so $f(0)=0$ .
$P(x,0)\Rightarrow 2f(x)^2-f\left(x^2\right)=\frac{xf(2x)}2$
Now $P(x,y)$ rewrites as:
$4f(x)f(x+y)=4f(x)^2+xf(f(y)).$ $P(1,x)\Rightarrow f(f(x))=4f(1)f(x+1)-4f(1)^2$
Now $P(x,y)$ rewrites as:
$4f(x)f(x+y)=4f(x)^2+4xf(1)f(y+1)-4xf(1)^2,$ or, taking $y\mapsto y-1$ :
$g(x)g(x+y-1)=g(x)^2+xg(y)-x.$
Call this assertion $Q(x,y)$ , and recall that $g(0)=0$ and $g(1)=1$ .

$Q(x+1,0)\Rightarrow g(x)g(x+1)=g(x+1)^2-x-1$
$Q(x,2)\Rightarrow g(x)g(x+1)=g(x)^2+xg(2)-x$
Comparing these, we get $g(x+1)^2=g(x)^2+xg(2)+1$ . Squaring the equation we got from $Q(x,2)$ , we get:
$g(x)^2\left(g(x)^2+xg(2)+1\right)=\left(g(x)^2+xg(2)-x\right)^2$ which rearranges as:
$g(x)^2(x(2-g(2))+1)=x^2(g(2)-1)^2.\qquad(*)$
If $2-g(2)\ne0$ we will be able to find some $x\ne0$ such that $x(2-g(2))+1<0$ , then for size reasons $x^2(g(2)-1)^2=0$ so $g(2)=1$ . But then, $(*)$ reduces to $g(x)^2(x+1)=0$ , so $g(x)=0$ for all $x\ne-1$ . This contradicts $g(1)=1$ , so this case is impossible. Thus $g(2)=2$ and $(*)$ becomes $g(x)^2=x^2$ .

Suppose there are some $a,b\ne0$ with $g(a)=a$ and $g(b)=-b$ .
$Q(a,b)\Rightarrow g(a+b-1)=a-b-1$
and since $g(a+b-1)\in\{a+b-1,1-a-b\}$ testing both cases gives $a=1$ .
$Q(b,b)\Rightarrow g(2b-1)=1$
and since $g(2b-1)\in\{2b-1,1-2b\}$ , testing both cases gives $b=1$ .
This is a contradiction, so either $g(x)=x$ for all $x$ or $g(x)=-x$ for all $x$ . Since $g(1)=1$ , we have $g(x)=x$ for al $x$ , so $f(x)=xf(1)$ . Testing in the original equation, we get
dang
no solution.

This post has been edited 4 times. Last edited by jasperE3, Apr 18, 2025, 6:36 AM

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pco

#3 h Apr 18, 2025, 6:11 AM • 1 Y

Y by jasperE3

jasperE3 wrote:

luci1337 wrote:

find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

... we get $\boxed{f(x)=x}$ .

Which, unfortunately, is not a solution.

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#4 h Apr 18, 2025, 6:38 AM

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pco wrote:

jasperE3 wrote:

luci1337 wrote:

find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

... we get $\boxed{f(x)=x}$ .

Which, unfortunately, is not a solution.

WOWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW
thanks

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#5 h Apr 18, 2025, 4:31 PM

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jasperE3 wrote:

luci1337 wrote:

find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

Suppose first that $f(1)=0$ .
Here $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$ .
$P(1,x)\Rightarrow f(f(x))=-f(2)$
So $P(x,y)$ becomes $4f(x)f(x+y)-2f(x^2)=xf(2x)-yf(2)$ , and fixing $x=j$ we get that $f$ is linear. Testing, we find no solutions.

Otherwise, $f(1)\ne0$ ; we can define $g:\mathbb R\to\mathbb R$ such that $g(x)=\frac{f(x)}{f(1)}$ .
$P(0,x)\Rightarrow2f(0)f(x)=f(0)$
If $f(0)\ne0$ then $f(x)=\frac12$ for all $x$ which isn't a valid solution, so $f(0)=0$ .
$P(x,0)\Rightarrow 2f(x)^2-f\left(x^2\right)=\frac{xf(2x)}2$
Now $P(x,y)$ rewrites as:
$4f(x)f(x+y)=4f(x)^2+xf(f(y)).$ $P(1,x)\Rightarrow f(f(x))=4f(1)f(x+1)-4f(1)^2$
Now $P(x,y)$ rewrites as:
$4f(x)f(x+y)=4f(x)^2+4xf(1)f(y+1)-4xf(1)^2,$ or, taking $y\mapsto y-1$ :
$g(x)g(x+y-1)=g(x)^2+xg(y)-x.$
Call this assertion $Q(x,y)$ , and recall that $g(0)=0$ and $g(1)=1$ .

$Q(x+1,0)\Rightarrow g(x)g(x+1)=g(x+1)^2-x-1$
$Q(x,2)\Rightarrow g(x)g(x+1)=g(x)^2+xg(2)-x$
Comparing these, we get $g(x+1)^2=g(x)^2+xg(2)+1$ . Squaring the equation we got from $Q(x,2)$ , we get:
$g(x)^2\left(g(x)^2+xg(2)+1\right)=\left(g(x)^2+xg(2)-x\right)^2$ which rearranges as:
$g(x)^2(x(2-g(2))+1)=x^2(g(2)-1)^2.\qquad(*)$
If $2-g(2)\ne0$ we will be able to find some $x\ne0$ such that $x(2-g(2))+1<0$ , then for size reasons $x^2(g(2)-1)^2=0$ so $g(2)=1$ . But then, $(*)$ reduces to $g(x)^2(x+1)=0$ , so $g(x)=0$ for all $x\ne-1$ . This contradicts $g(1)=1$ , so this case is impossible. Thus $g(2)=2$ and $(*)$ becomes $g(x)^2=x^2$ .

Suppose there are some $a,b\ne0$ with $g(a)=a$ and $g(b)=-b$ .
$Q(a,b)\Rightarrow g(a+b-1)=a-b-1$
and since $g(a+b-1)\in\{a+b-1,1-a-b\}$ testing both cases gives $a=1$ .
$Q(b,b)\Rightarrow g(2b-1)=1$
and since $g(2b-1)\in\{2b-1,1-2b\}$ , testing both cases gives $b=1$ .
This is a contradiction, so either $g(x)=x$ for all $x$ or $g(x)=-x$ for all $x$ . Since $g(1)=1$ , we have $g(x)=x$ for al $x$ , so $f(x)=xf(1)$ . Testing in the original equation, we get
dang
no solution.

thanks u buddy

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