Mixing
A specific question about functions
Clash Royale CLAN TAG#URR8PPP
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Let $Bbb R$ be the set of real numbers. Determine all functions $f : Bbb R toBbb R$ such that, for
all real numbers x and y,$$f(f(x)f(y))+f(x+y)=f(xy)$$My attempt: let's first find some partial answers. If this equation has some polynomial answers of degree $d$ then the coefficient of $x^d$ must be equal at both sides i.e. we must have $$maxd^2,d=d$$which leads to $$d=0,1$$ now let's take the general answer of polynomial kind as $$f(x)=ax+b$$ for some $a$ and $b$. Substituting this in the equation leads to $$f(a^2xy+b^2+abx+aby)+ax+ay+b=axy+b$$which leads to $$a^3xy+ab^2+a^2bx+a^2by+b+ax+ay=axy$$finally we obtain one trivial answer $a=b=0$ and two non-trivial answers $$a=-b=1\a=-b=-1$$therefore the only non-trivial polynomials satisfying the equation above are $$f(x)=x-1\f(x)=-x+1$$I believe these are the only answers of the question. Here turns out two questions:
Are there any other answers except those I found?
If so, what are they?
functions functional-equations
edited 2 days ago
Mason
1,1271223
asked 2 days ago
Mostafa Ayaz
8,5203530
add a comment |Â
up vote
2
down vote
favorite
Let $Bbb R$ be the set of real numbers. Determine all functions $f : Bbb R toBbb R$ such that, for
all real numbers x and y,$$f(f(x)f(y))+f(x+y)=f(xy)$$My attempt: let's first find some partial answers. If this equation has some polynomial answers of degree $d$ then the coefficient of $x^d$ must be equal at both sides i.e. we must have $$maxd^2,d=d$$which leads to $$d=0,1$$ now let's take the general answer of polynomial kind as $$f(x)=ax+b$$ for some $a$ and $b$. Substituting this in the equation leads to $$f(a^2xy+b^2+abx+aby)+ax+ay+b=axy+b$$which leads to $$a^3xy+ab^2+a^2bx+a^2by+b+ax+ay=axy$$finally we obtain one trivial answer $a=b=0$ and two non-trivial answers $$a=-b=1\a=-b=-1$$therefore the only non-trivial polynomials satisfying the equation above are $$f(x)=x-1\f(x)=-x+1$$I believe these are the only answers of the question. Here turns out two questions:
Are there any other answers except those I found?
If so, what are they?
functions functional-equations
edited 2 days ago
Mason
1,1271223
asked 2 days ago
Mostafa Ayaz
8,5203530
Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
1
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Bbb R$ be the set of real numbers. Determine all functions $f : Bbb R toBbb R$ such that, for
all real numbers x and y,$$f(f(x)f(y))+f(x+y)=f(xy)$$My attempt: let's first find some partial answers. If this equation has some polynomial answers of degree $d$ then the coefficient of $x^d$ must be equal at both sides i.e. we must have $$maxd^2,d=d$$which leads to $$d=0,1$$ now let's take the general answer of polynomial kind as $$f(x)=ax+b$$ for some $a$ and $b$. Substituting this in the equation leads to $$f(a^2xy+b^2+abx+aby)+ax+ay+b=axy+b$$which leads to $$a^3xy+ab^2+a^2bx+a^2by+b+ax+ay=axy$$finally we obtain one trivial answer $a=b=0$ and two non-trivial answers $$a=-b=1\a=-b=-1$$therefore the only non-trivial polynomials satisfying the equation above are $$f(x)=x-1\f(x)=-x+1$$I believe these are the only answers of the question. Here turns out two questions:
Are there any other answers except those I found?
If so, what are they?
functions functional-equations
edited 2 days ago
Mason
1,1271223
asked 2 days ago
Mostafa Ayaz
8,5203530
Let $Bbb R$ be the set of real numbers. Determine all functions $f : Bbb R toBbb R$ such that, for
all real numbers x and y,$$f(f(x)f(y))+f(x+y)=f(xy)$$My attempt: let's first find some partial answers. If this equation has some polynomial answers of degree $d$ then the coefficient of $x^d$ must be equal at both sides i.e. we must have $$maxd^2,d=d$$which leads to $$d=0,1$$ now let's take the general answer of polynomial kind as $$f(x)=ax+b$$ for some $a$ and $b$. Substituting this in the equation leads to $$f(a^2xy+b^2+abx+aby)+ax+ay+b=axy+b$$which leads to $$a^3xy+ab^2+a^2bx+a^2by+b+ax+ay=axy$$finally we obtain one trivial answer $a=b=0$ and two non-trivial answers $$a=-b=1\a=-b=-1$$therefore the only non-trivial polynomials satisfying the equation above are $$f(x)=x-1\f(x)=-x+1$$I believe these are the only answers of the question. Here turns out two questions:
Are there any other answers except those I found?
If so, what are they?
functions functional-equations
edited 2 days ago
Mason
1,1271223
asked 2 days ago
Mostafa Ayaz
8,5203530
edited 2 days ago
Mason
1,1271223
edited 2 days ago
Mason
1,1271223
edited 2 days ago
Mason
1,1271223
1,1271223
asked 2 days ago
Mostafa Ayaz
8,5203530
asked 2 days ago
Mostafa Ayaz
8,5203530
asked 2 days ago
Mostafa Ayaz
8,5203530
8,5203530
Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
1
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago
add a comment |Â
Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
1
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago
Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
1
1
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago
add a comment |Â
1 Answer
1
active
oldest
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up vote
3
down vote
If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.
Assume $b in mathbbR$ is such that $f(b) = 0$. Then $y=b implies f(0)+f(x+b) = f(xb)$ for all $x in mathbbR$. Now unless $b=1$, we can let $x = fracbb-1$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 implies f(0)+f(x) = f(0)$ for all $x in mathbbR$, i.e. $f equiv 0$. Hence, we may assume $f(b) = 0 implies b=1$.
Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 implies f(x+1) = f(x)-1$ for each $x in mathbbR$. Then $f(0) = 1 implies f(x) = -x+1$ for all $x in mathbbZ$.
Also, for each $x not = 1$, letting $y = fracxx-1$ gives that $f(f(x)f(fracxx-1)) = 0 implies f(x)f(fracxx-1) = 1$. Letting $x = frac12$ shows $f(frac12) = frac12$ and $f(x+1) = f(x)-1$ gives that $f(frack2) = -frack2+1$ for each $k in mathbbZ$. It's then an easy induction (on the size of denominator), using $f(x)f(fracxx-1) = 1$, to conclude that $f(fracpq) = -fracpq+1$ for each $p,q in mathbbZ, q not = 0$. [The point is that $fracxx-1$ at $x = fracpq$ is $fracpp-q$ which has denominator smaller than $q$ in absolute value.]
Hence, $f(x) = -x+1$ for each $x in mathbbQ$, and if $f$ is continuous, this extends to all $x in mathbbR$.
answered 2 days ago
mathworker21
6,4231727
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
@mathworker21: Nice proof!
– quasi
15 hours ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.
Assume $b in mathbbR$ is such that $f(b) = 0$. Then $y=b implies f(0)+f(x+b) = f(xb)$ for all $x in mathbbR$. Now unless $b=1$, we can let $x = fracbb-1$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 implies f(0)+f(x) = f(0)$ for all $x in mathbbR$, i.e. $f equiv 0$. Hence, we may assume $f(b) = 0 implies b=1$.
Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 implies f(x+1) = f(x)-1$ for each $x in mathbbR$. Then $f(0) = 1 implies f(x) = -x+1$ for all $x in mathbbZ$.
Also, for each $x not = 1$, letting $y = fracxx-1$ gives that $f(f(x)f(fracxx-1)) = 0 implies f(x)f(fracxx-1) = 1$. Letting $x = frac12$ shows $f(frac12) = frac12$ and $f(x+1) = f(x)-1$ gives that $f(frack2) = -frack2+1$ for each $k in mathbbZ$. It's then an easy induction (on the size of denominator), using $f(x)f(fracxx-1) = 1$, to conclude that $f(fracpq) = -fracpq+1$ for each $p,q in mathbbZ, q not = 0$. [The point is that $fracxx-1$ at $x = fracpq$ is $fracpp-q$ which has denominator smaller than $q$ in absolute value.]
Hence, $f(x) = -x+1$ for each $x in mathbbQ$, and if $f$ is continuous, this extends to all $x in mathbbR$.
answered 2 days ago
mathworker21
6,4231727
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
@mathworker21: Nice proof!
– quasi
15 hours ago
 |Â
show 1 more comment
up vote
3
down vote
If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.
Assume $b in mathbbR$ is such that $f(b) = 0$. Then $y=b implies f(0)+f(x+b) = f(xb)$ for all $x in mathbbR$. Now unless $b=1$, we can let $x = fracbb-1$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 implies f(0)+f(x) = f(0)$ for all $x in mathbbR$, i.e. $f equiv 0$. Hence, we may assume $f(b) = 0 implies b=1$.
Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 implies f(x+1) = f(x)-1$ for each $x in mathbbR$. Then $f(0) = 1 implies f(x) = -x+1$ for all $x in mathbbZ$.
Also, for each $x not = 1$, letting $y = fracxx-1$ gives that $f(f(x)f(fracxx-1)) = 0 implies f(x)f(fracxx-1) = 1$. Letting $x = frac12$ shows $f(frac12) = frac12$ and $f(x+1) = f(x)-1$ gives that $f(frack2) = -frack2+1$ for each $k in mathbbZ$. It's then an easy induction (on the size of denominator), using $f(x)f(fracxx-1) = 1$, to conclude that $f(fracpq) = -fracpq+1$ for each $p,q in mathbbZ, q not = 0$. [The point is that $fracxx-1$ at $x = fracpq$ is $fracpp-q$ which has denominator smaller than $q$ in absolute value.]
Hence, $f(x) = -x+1$ for each $x in mathbbQ$, and if $f$ is continuous, this extends to all $x in mathbbR$.
answered 2 days ago
mathworker21
6,4231727
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
@mathworker21: Nice proof!
– quasi
15 hours ago
 |Â
show 1 more comment
up vote
3
down vote
up vote
3
down vote
If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.
Assume $b in mathbbR$ is such that $f(b) = 0$. Then $y=b implies f(0)+f(x+b) = f(xb)$ for all $x in mathbbR$. Now unless $b=1$, we can let $x = fracbb-1$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 implies f(0)+f(x) = f(0)$ for all $x in mathbbR$, i.e. $f equiv 0$. Hence, we may assume $f(b) = 0 implies b=1$.
Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 implies f(x+1) = f(x)-1$ for each $x in mathbbR$. Then $f(0) = 1 implies f(x) = -x+1$ for all $x in mathbbZ$.
Also, for each $x not = 1$, letting $y = fracxx-1$ gives that $f(f(x)f(fracxx-1)) = 0 implies f(x)f(fracxx-1) = 1$. Letting $x = frac12$ shows $f(frac12) = frac12$ and $f(x+1) = f(x)-1$ gives that $f(frack2) = -frack2+1$ for each $k in mathbbZ$. It's then an easy induction (on the size of denominator), using $f(x)f(fracxx-1) = 1$, to conclude that $f(fracpq) = -fracpq+1$ for each $p,q in mathbbZ, q not = 0$. [The point is that $fracxx-1$ at $x = fracpq$ is $fracpp-q$ which has denominator smaller than $q$ in absolute value.]
Hence, $f(x) = -x+1$ for each $x in mathbbQ$, and if $f$ is continuous, this extends to all $x in mathbbR$.
answered 2 days ago
mathworker21
6,4231727
If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.
Assume $b in mathbbR$ is such that $f(b) = 0$. Then $y=b implies f(0)+f(x+b) = f(xb)$ for all $x in mathbbR$. Now unless $b=1$, we can let $x = fracbb-1$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 implies f(0)+f(x) = f(0)$ for all $x in mathbbR$, i.e. $f equiv 0$. Hence, we may assume $f(b) = 0 implies b=1$.
Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 implies f(x+1) = f(x)-1$ for each $x in mathbbR$. Then $f(0) = 1 implies f(x) = -x+1$ for all $x in mathbbZ$.
Also, for each $x not = 1$, letting $y = fracxx-1$ gives that $f(f(x)f(fracxx-1)) = 0 implies f(x)f(fracxx-1) = 1$. Letting $x = frac12$ shows $f(frac12) = frac12$ and $f(x+1) = f(x)-1$ gives that $f(frack2) = -frack2+1$ for each $k in mathbbZ$. It's then an easy induction (on the size of denominator), using $f(x)f(fracxx-1) = 1$, to conclude that $f(fracpq) = -fracpq+1$ for each $p,q in mathbbZ, q not = 0$. [The point is that $fracxx-1$ at $x = fracpq$ is $fracpp-q$ which has denominator smaller than $q$ in absolute value.]
Hence, $f(x) = -x+1$ for each $x in mathbbQ$, and if $f$ is continuous, this extends to all $x in mathbbR$.
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
6,4231727
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
@mathworker21: Nice proof!
– quasi
15 hours ago
 |Â
show 1 more comment
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
@mathworker21: Nice proof!
– quasi
15 hours ago
1
1
You haven't prove the existence of $b$.
– greedoid
2 days ago
You haven't prove the existence of $b$.
– greedoid
2 days ago
1
1
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
Yes I have. $x=y=0$ implies $f(f(0)^2) = 0$. This is how I got $f(0)^2 = 1$.
– mathworker21
2 days ago
1
1
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
Also, I note that for each $x not = 1$, $f(f(x)f(fracxx-1)) = 0$. Am I missing something?
– mathworker21
2 days ago
2
2
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
@greedoid I don't see an issue. Right away you can take $x=y=0$ to see that $f(f(0)^2) = 0$. So that gives you the existence of some $b$ for which $f(b) = 0$. But anyways the paragraph starting with "Assume" is just the implication $f(b) = 0 implies b=1$.
– mathworker21
2 days ago
1
1
@mathworker21: Nice proof!
– quasi
15 hours ago
@mathworker21: Nice proof!
– quasi
15 hours ago
 |Â
show 1 more comment
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Can we manipulate this to say that $f(0)in 0,pm 1$?
– Mason
2 days ago
I don't think so, or if is it this is not a strong background implication. How can you prove that for example $f(0)ne 2$?
– Mostafa Ayaz
2 days ago
1
imo-official.org/problems/IMO2017SL.pdf
– greedoid
2 days ago