Function equation in Taiwan TST

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Function equation in Taiwan TST

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Source: 2023 Taiwan TST Round 2 Independent Study 1-A

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6 posts

#1 h Apr 5, 2023, 5:54 AM

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Find all functions $f : \mathbb{R} \to \mathbb{R}$ , such that
$f\left(xy+f(y)\right)f(x)=x^2f(y)+f(xy)$ for all $x,y \in \mathbb{R}$

Proposed by chengbilly

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429 posts

hakN

#2 h Apr 5, 2023, 10:04 AM • 2 Y

Y by PRMOisTheHardestExam, bin_sherlo

Solution

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1301 posts

#3 h Apr 13, 2023, 8:07 PM • 1 Y

Y by PRMOisTheHardestExam

Weird Solution

This post has been edited 1 time. Last edited by CANBANKAN, May 4, 2023, 4:06 AM

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108 posts

#4 h May 3, 2023, 12:48 PM • 1 Y

Y by PRMOisTheHardestExam

this was hard

took nearly an hour

Solution

This post has been edited 4 times. Last edited by NoctNight, May 4, 2023, 12:11 AM

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108 posts

#5 h May 3, 2023, 12:51 PM

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what are d-ratings of taiwan tst round 2 quizzes generally?

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1069 posts

gghx

#6 h May 4, 2023, 1:36 AM • 3 Y

Y by PRMOisTheHardestExam, IAmTheHazard, VicKmath7

why so complicated

Solution

motivation

This post has been edited 3 times. Last edited by gghx, May 4, 2023, 4:07 AM

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46 posts

#7 h May 4, 2023, 1:58 AM • 1 Y

Y by PRMOisTheHardestExam

bruh beaten by gghx by literal 16 minutes
i guess i did $P(x, 0) \implies f^2(0)f(x) = (x^2 + 1)f(0)$ ; if $f^2(0)\neq 0$ the.. wait nvm gghx just made a typo its the same thing sad ge
i guess i also did $P(x, 1)$ implies $f(1) = 0 \implies f(x)^2 = f(x) \implies f(x) \in \{0, 1\}$ for all $x$ , then if $f(y) = 1$ then $P(1, y)$ is $1 + f(xy) = 0 \implies f(xy) = -1 \notin \{0, 1\}$ contradiction
noctnight ur funny

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#8 h May 4, 2023, 2:30 AM • 1 Y

Y by PRMOisTheHardestExam

Meanwhile I am the biggest loser here. I spent more then 3 hours.

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1069 posts

gghx

#9 h May 4, 2023, 3:43 AM • 11 Y

Y by cosmicgenius, CANBANKAN, PRMOisTheHardestExam, squarc_rs3v2m, hakN, IMUKAT, Aryan-23, IAmTheHazard, egxa, Scilyse, navier3072

CANBANKAN wrote:

Meanwhile I am the biggest loser here. I spent more then 3 hours.

skill issue bro do better

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5000 posts

#10 h Nov 15, 2023, 11:52 PM

Y by

The answer is $f \equiv 0$ and $f(x)=x$ only, which clearly work. Let $P(x,y)$ denote the assertion.
From $P(x,0)$ we find that $f(f(0))f(x)=x^2f(0)+f(0)$ , so either $f(0)=0$ or $f$ is (nonlinear) quadratic. Quadratic functions can easily be shown to not work, so it follows that $f(0)=0$ .

If $f(a)=f(b)$ , comparing $P(a,b)$ with $P(b,a)$ implies $a^2=b^2$ or $f(a)=f(b)=0$ . On the other hand, if some $c$ has $f(c)=0$ then $P(x,c)$ yields $f(cx)f(x)=f(cx)$ , so for any $x$ we either have $f(cx)=0$ or $f(x)=1$ . But the latter only occurs at at most two values of $x$ , so $f \equiv 0$ on all but (at most) two reals. Repeating this with a different choice of $c$ yields $f \equiv 0$ , which is one solution. Otherwise, suppose that $f$ is injective at $0$ . I then claim that it's actually injective in general. Indeed, if $f(a)=f(-a) \neq 0$ for some $a \neq 0$ , then for any $x \neq 0$ either comparing $P(x,a)$ and $P(-x,-a)$ or $P(x,-a)$ and $P(-x,a)$ yields $f(x)=f(-x)$ , since $ax+f(a)=-ax+f(a)=0$ can never hold. But then comparing $P(x,y)$ with $P(-x,y)$ for some $x,y \neq 0$ implies $f(xy+f(y))=f(-xy+f(y))$ : a clear contradiction if we take $x=f(y)/y \neq 0$ , since now $f(2f(y))=0$ contradicting injectivity at $0$ .

Now, from $P(-1,y)$ and $P(-1,-y)$ we obtain $f(f(-y)+y)=f(f(y)-y)=f(y)+f(-y)$ , so $f(-y)+y=f(y)-y:=k \implies f(y)-f(-y)=2y$ for some $k \in \mathbb{R}$ (dependent on $y$ ). Solving then yields $f(k)=2k$ , but then we also have $f(k)-f(-k)=2k \implies f(-k)=0$ , so $k=0$ . Thus $f(y)=y$ for all $y$ , as desired. $\blacksquare$

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#11 h Mar 24, 2024, 2:02 PM

Y by

chengbilly wrote:

Find all functions $f : \mathbb{R} \to \mathbb{R}$ , such that
$f\left(xy+f(y)\right)f(x)=x^2f(y)+f(xy)$ for all $x,y \in \mathbb{R}$

Proposed by chengbilly

Let $P(x,y)$ be the assertion of the problem. One answer is $\boxed{f(x)=0\; \forall x\in \mathbb R}$ . So Let's assume that $f$ is not zero constant.

Step 1: $f(t)=0 \leftrightarrow t=0$ .
Define $c:= f(0)$ . Then
$P(x,0)\rightarrow cx^2 + c = f(x)f(c) \rightarrow \text{if}\; c\neq 0 \;\text{then}\; f(c)\neq 0 \;\text{and}\; f(x) = \frac{c}{f(c)}(x^2+1)\; \forall x\in \mathbb R,$ which does not work in $P(x,y)$ . So $c=0$ . Assume that $f(t)=0$ for some $t\neq 0$ .
$P(t,y), P(y,t) \rightarrow t^2f(y)+f(ty)=0,\; f(ty) = f(ty)f(y)\rightarrow f(y)\in \{0,1\}$ If $f$ is not constant so $f(y)=1$ for some $y$ and then $f(ty)=-t^2 \notin \{0,1\}$ , a contradiction.

Step 2: $f$ is injective.
Assume that $f(a)=f(b)=s$ for some $a>b$ . From last step we infer that $s\neq 0$ .
$P(a,b) , P(b,a) \rightarrow sf(ab+s) = a^2s+f(ab) = b^2s+f(ab) \rightarrow a^2=b^2 \rightarrow a=-b.$ So we have that if $f(x)=f(y)\neq 0$ then $x+y=0$ .
$P(a,1) , P(-a,1) \rightarrow a^2f(1) + s = sf(a+f(1)) = sf(-a+f(1)) \rightarrow f(a+f(1)) = f(-a+f(1)).$ So $a+f(1) = -a+f(1)=0$ or $a+f(1)-a+f(1)=0$ which imply $f(1)=0\; \text{or}\; a=0$ , contradiction.

Step 3: $f(-y) = f(y)-2y$ for all $y\in \mathbb R$ .
$P(-1,y), P(-1,-y) \rightarrow f(y)+f(-y) = f(-1)f(f(y)-y) = f(-1)f(f(-y)+y) \rightarrow f(y)-y = f(-y)+y \rightarrow f(-y) = f(y)-2y.$
Step 4: $f(1)=1$ and $f(-1)=-1$ .
Define $a: = f(1)$ and $b:= f(-1)$ . from previous step we know $a=b+2$ .
$P(1,1) \rightarrow f(a+1) = 2 \rightarrow f(-a-1) = -2a.$ Then we have
$P(-a-1,1)\rightarrow -2ab= -2a+a(a+1)^2 \rightarrow (a-1)(a+4)=0.$ In addition from $P(1,-1-a)$ we know $f(-3a-1)=-4$ . So if $f(1)=a=-4$ then $f(-3a-1) = f(11)=f(1)=-4$ which is contradiction because $f$ is injective. So $a=1$ .

Step 5: $f(x) =x$ for all real $x\neq \frac{1}{2}$ .

$P(x,1) \rightarrow f(x+1) = 1+\dfrac{x^2}{f(x)} \rightarrow f(x)-1 = \dfrac{(x-1)^2}{f(x-1)} = \dfrac{(x-1)^2}{2x-2+f(1-x)}$ On the other hand
$P(-x,1) \rightarrow f(1-x) = 1+\dfrac{x^2}{f(-x)} = 1+\dfrac{x^2}{f(x)-2x}.$ So
$\dfrac{(x-1)^2}{f(x)-1} - 2(x-1) = 1 + \dfrac{x^2}{f(x)-2x}\rightarrow f(x)^2(2x-1) - f(x)\left(4x^2-2x\right)+(2x^3-x^2) =0$ $\rightarrow (2x-1)(f(x)-x)^2 =0. \rightarrow f(x) =x \; \forall x\neq \frac{1}{2}.$
Step 6: $\boxed{f(x) =x \; \forall x \in \mathbb R}$ .
It is just enough to prove $f(\frac{1}{2})=\frac{1}{2}$ . We have
$P(-\frac{1}{2},1) \rightarrow f(\frac{1}{2}) = 1+ \dfrac{(-\frac{1}{2})^2}{f(-\frac{1}{2})} = \frac{1}{2}.$

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