Mixing
What is wrong with my approach in this problem? Find all functions $f$ such that $f(x + y)f(x − y) = (f(x)...
$begingroup$
Problem:
Find all functions $f : R → R$ such that
$$f(x + y)f(x − y) =
(f(x) + f(y))^2
− 4x^2
f(y)$$
for all $x, y ∈ R$, where R denotes the set of all real numbers.
My approach: I substituted $x=y=0$ which gave $f(0)=0$. I then substituted $x=y=t$ which gave $f(2t)f(0) = 4f(t)(f(t)-t^2) => f(t) = 0, f(t) = t^2 $. Thus the only solutions are $f(x)= 0$ and $f(x) = x^2$.
But this is an olympiad problem (INMO 2011, India). So, there must be something more complicated than this that I am probably missing. My answer matches but the proof given in the official solution is quite long. Here
algebra-precalculus contest-math functional-equations
edited Jan 18 at 17:39
Blue
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
$endgroup$
add a comment |
$begingroup$
Problem:
Find all functions $f : R → R$ such that
$$f(x + y)f(x − y) =
(f(x) + f(y))^2
− 4x^2
f(y)$$
for all $x, y ∈ R$, where R denotes the set of all real numbers.
My approach: I substituted $x=y=0$ which gave $f(0)=0$. I then substituted $x=y=t$ which gave $f(2t)f(0) = 4f(t)(f(t)-t^2) => f(t) = 0, f(t) = t^2 $. Thus the only solutions are $f(x)= 0$ and $f(x) = x^2$.
But this is an olympiad problem (INMO 2011, India). So, there must be something more complicated than this that I am probably missing. My answer matches but the proof given in the official solution is quite long. Here
algebra-precalculus contest-math functional-equations
edited Jan 18 at 17:39
Blue
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
$endgroup$
2
$begingroup$
Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
$endgroup$
– Mindlack
Jan 18 at 17:42
$begingroup$
"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
$endgroup$
– fleablood
Jan 18 at 17:45
2
$begingroup$
You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
$endgroup$
– Aaron
Jan 18 at 17:46
add a comment |
$begingroup$
Problem:
Find all functions $f : R → R$ such that
$$f(x + y)f(x − y) =
(f(x) + f(y))^2
− 4x^2
f(y)$$
for all $x, y ∈ R$, where R denotes the set of all real numbers.
My approach: I substituted $x=y=0$ which gave $f(0)=0$. I then substituted $x=y=t$ which gave $f(2t)f(0) = 4f(t)(f(t)-t^2) => f(t) = 0, f(t) = t^2 $. Thus the only solutions are $f(x)= 0$ and $f(x) = x^2$.
But this is an olympiad problem (INMO 2011, India). So, there must be something more complicated than this that I am probably missing. My answer matches but the proof given in the official solution is quite long. Here
algebra-precalculus contest-math functional-equations
edited Jan 18 at 17:39
Blue
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
$endgroup$
Problem:
Find all functions $f : R → R$ such that
$$f(x + y)f(x − y) =
(f(x) + f(y))^2
− 4x^2
f(y)$$
for all $x, y ∈ R$, where R denotes the set of all real numbers.
My approach: I substituted $x=y=0$ which gave $f(0)=0$. I then substituted $x=y=t$ which gave $f(2t)f(0) = 4f(t)(f(t)-t^2) => f(t) = 0, f(t) = t^2 $. Thus the only solutions are $f(x)= 0$ and $f(x) = x^2$.
But this is an olympiad problem (INMO 2011, India). So, there must be something more complicated than this that I am probably missing. My answer matches but the proof given in the official solution is quite long. Here
algebra-precalculus contest-math functional-equations
algebra-precalculus contest-math functional-equations
edited Jan 18 at 17:39
Blue
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
edited Jan 18 at 17:39
Blue
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
edited Jan 18 at 17:39
Blue
48.6k870156
edited Jan 18 at 17:39
Blue
48.6k870156
edited Jan 18 at 17:39
Blue
48.6k870156
48.6k870156
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
asked Jan 18 at 17:37
Shubhraneel PalShubhraneel Pal
45939
45939
2
$begingroup$
Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
$endgroup$
– Mindlack
Jan 18 at 17:42
$begingroup$
"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
$endgroup$
– fleablood
Jan 18 at 17:45
2
$begingroup$
You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
$endgroup$
– Aaron
Jan 18 at 17:46
add a comment |
2
$begingroup$
Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
$endgroup$
– Mindlack
Jan 18 at 17:42
$begingroup$
"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
$endgroup$
– fleablood
Jan 18 at 17:45
2
$begingroup$
You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
$endgroup$
– Aaron
Jan 18 at 17:46
2
2
$begingroup$
Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
$endgroup$
– Mindlack
Jan 18 at 17:42
$begingroup$
Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
$endgroup$
– Mindlack
Jan 18 at 17:42
$begingroup$
"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
$endgroup$
– fleablood
Jan 18 at 17:45
$begingroup$
"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
$endgroup$
– fleablood
Jan 18 at 17:45
2
2
$begingroup$
You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
$endgroup$
– Aaron
Jan 18 at 17:46
$begingroup$
You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
$endgroup$
– Aaron
Jan 18 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.
$endgroup$
add a comment |
$begingroup$
As @Jyrki Lahtonen and others pointed out, you cannot conclude $f(x)equiv 0$ or $f(x)equiv x^2$ in the current step. But a few more observations can lead you to the desired conclusion. Assume $tne 0$ and $f(t)=0$. Then we have
$$
f(x+t)f(x-t) = f(x)^2,quadforall xinBbb R.
$$ Suppose $f(-t) = (-t)^2=t^2$. Then we also have
$$
f(x-t)f(x+t)=f(x)^2+2t^2 f(x) +t^4-4x^2t^2.
$$ Thus it holds
$$
f(x) = 2x^2-frac{t^2}{2},quadforall xinBbb R.
$$ But we can choose $x$ such that $2x^2-frac{t^2}{2}ne 0$ and $2x^2-frac{t^2}{2}ne x^2$. This leads to a contradiction.
From this, we know that $f(t)=0$ implies $f(-t)=0$. This means either $f(t)=f(-t)=0$ or $f(t)=f(-t)=t^2$ for all $tne 0$. So $f$ is an even function. Now we interchange $x$ and $y$ in the original functional equation. Then we get
$$
f(x+y)f(y-x)=(f(x) + f(y))^2
− 4y^2
f(x).
$$ Since $f(y-x)=f(x-y)$, we have
$$
4x^2f(y)=4y^2f(x),
$$ and for $xne 0$, $yne 0$,
$$
frac{f(x)}{x^2}=frac{f(y)}{y^2}=text{constant}.
$$ Thus, we conclude that if there exists $t_0ne 0$ such that $f(t_0)=0$, then $f(t)equiv 0$, and otherwise $f(t)equiv t^2$.
answered Jan 18 at 18:29
SongSong
15.8k1739
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.
$endgroup$
add a comment |
$begingroup$
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.
$endgroup$
add a comment |
$begingroup$
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.
$endgroup$
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.
answered Jan 18 at 17:46
community wiki
Jyrki Lahtonen
add a comment |
add a comment |
$begingroup$
As @Jyrki Lahtonen and others pointed out, you cannot conclude $f(x)equiv 0$ or $f(x)equiv x^2$ in the current step. But a few more observations can lead you to the desired conclusion. Assume $tne 0$ and $f(t)=0$. Then we have
$$
f(x+t)f(x-t) = f(x)^2,quadforall xinBbb R.
$$ Suppose $f(-t) = (-t)^2=t^2$. Then we also have
$$
f(x-t)f(x+t)=f(x)^2+2t^2 f(x) +t^4-4x^2t^2.
$$ Thus it holds
$$
f(x) = 2x^2-frac{t^2}{2},quadforall xinBbb R.
$$ But we can choose $x$ such that $2x^2-frac{t^2}{2}ne 0$ and $2x^2-frac{t^2}{2}ne x^2$. This leads to a contradiction.
From this, we know that $f(t)=0$ implies $f(-t)=0$. This means either $f(t)=f(-t)=0$ or $f(t)=f(-t)=t^2$ for all $tne 0$. So $f$ is an even function. Now we interchange $x$ and $y$ in the original functional equation. Then we get
$$
f(x+y)f(y-x)=(f(x) + f(y))^2
− 4y^2
f(x).
$$ Since $f(y-x)=f(x-y)$, we have
$$
4x^2f(y)=4y^2f(x),
$$ and for $xne 0$, $yne 0$,
$$
frac{f(x)}{x^2}=frac{f(y)}{y^2}=text{constant}.
$$ Thus, we conclude that if there exists $t_0ne 0$ such that $f(t_0)=0$, then $f(t)equiv 0$, and otherwise $f(t)equiv t^2$.
answered Jan 18 at 18:29
SongSong
15.8k1739
$endgroup$
add a comment |
$begingroup$
As @Jyrki Lahtonen and others pointed out, you cannot conclude $f(x)equiv 0$ or $f(x)equiv x^2$ in the current step. But a few more observations can lead you to the desired conclusion. Assume $tne 0$ and $f(t)=0$. Then we have
$$
f(x+t)f(x-t) = f(x)^2,quadforall xinBbb R.
$$ Suppose $f(-t) = (-t)^2=t^2$. Then we also have
$$
f(x-t)f(x+t)=f(x)^2+2t^2 f(x) +t^4-4x^2t^2.
$$ Thus it holds
$$
f(x) = 2x^2-frac{t^2}{2},quadforall xinBbb R.
$$ But we can choose $x$ such that $2x^2-frac{t^2}{2}ne 0$ and $2x^2-frac{t^2}{2}ne x^2$. This leads to a contradiction.
From this, we know that $f(t)=0$ implies $f(-t)=0$. This means either $f(t)=f(-t)=0$ or $f(t)=f(-t)=t^2$ for all $tne 0$. So $f$ is an even function. Now we interchange $x$ and $y$ in the original functional equation. Then we get
$$
f(x+y)f(y-x)=(f(x) + f(y))^2
− 4y^2
f(x).
$$ Since $f(y-x)=f(x-y)$, we have
$$
4x^2f(y)=4y^2f(x),
$$ and for $xne 0$, $yne 0$,
$$
frac{f(x)}{x^2}=frac{f(y)}{y^2}=text{constant}.
$$ Thus, we conclude that if there exists $t_0ne 0$ such that $f(t_0)=0$, then $f(t)equiv 0$, and otherwise $f(t)equiv t^2$.
answered Jan 18 at 18:29
SongSong
15.8k1739
$endgroup$
add a comment |
$begingroup$
As @Jyrki Lahtonen and others pointed out, you cannot conclude $f(x)equiv 0$ or $f(x)equiv x^2$ in the current step. But a few more observations can lead you to the desired conclusion. Assume $tne 0$ and $f(t)=0$. Then we have
$$
f(x+t)f(x-t) = f(x)^2,quadforall xinBbb R.
$$ Suppose $f(-t) = (-t)^2=t^2$. Then we also have
$$
f(x-t)f(x+t)=f(x)^2+2t^2 f(x) +t^4-4x^2t^2.
$$ Thus it holds
$$
f(x) = 2x^2-frac{t^2}{2},quadforall xinBbb R.
$$ But we can choose $x$ such that $2x^2-frac{t^2}{2}ne 0$ and $2x^2-frac{t^2}{2}ne x^2$. This leads to a contradiction.
From this, we know that $f(t)=0$ implies $f(-t)=0$. This means either $f(t)=f(-t)=0$ or $f(t)=f(-t)=t^2$ for all $tne 0$. So $f$ is an even function. Now we interchange $x$ and $y$ in the original functional equation. Then we get
$$
f(x+y)f(y-x)=(f(x) + f(y))^2
− 4y^2
f(x).
$$ Since $f(y-x)=f(x-y)$, we have
$$
4x^2f(y)=4y^2f(x),
$$ and for $xne 0$, $yne 0$,
$$
frac{f(x)}{x^2}=frac{f(y)}{y^2}=text{constant}.
$$ Thus, we conclude that if there exists $t_0ne 0$ such that $f(t_0)=0$, then $f(t)equiv 0$, and otherwise $f(t)equiv t^2$.
answered Jan 18 at 18:29
SongSong
15.8k1739
$endgroup$
As @Jyrki Lahtonen and others pointed out, you cannot conclude $f(x)equiv 0$ or $f(x)equiv x^2$ in the current step. But a few more observations can lead you to the desired conclusion. Assume $tne 0$ and $f(t)=0$. Then we have
$$
f(x+t)f(x-t) = f(x)^2,quadforall xinBbb R.
$$ Suppose $f(-t) = (-t)^2=t^2$. Then we also have
$$
f(x-t)f(x+t)=f(x)^2+2t^2 f(x) +t^4-4x^2t^2.
$$ Thus it holds
$$
f(x) = 2x^2-frac{t^2}{2},quadforall xinBbb R.
$$ But we can choose $x$ such that $2x^2-frac{t^2}{2}ne 0$ and $2x^2-frac{t^2}{2}ne x^2$. This leads to a contradiction.
From this, we know that $f(t)=0$ implies $f(-t)=0$. This means either $f(t)=f(-t)=0$ or $f(t)=f(-t)=t^2$ for all $tne 0$. So $f$ is an even function. Now we interchange $x$ and $y$ in the original functional equation. Then we get
$$
f(x+y)f(y-x)=(f(x) + f(y))^2
− 4y^2
f(x).
$$ Since $f(y-x)=f(x-y)$, we have
$$
4x^2f(y)=4y^2f(x),
$$ and for $xne 0$, $yne 0$,
$$
frac{f(x)}{x^2}=frac{f(y)}{y^2}=text{constant}.
$$ Thus, we conclude that if there exists $t_0ne 0$ such that $f(t_0)=0$, then $f(t)equiv 0$, and otherwise $f(t)equiv t^2$.
answered Jan 18 at 18:29
SongSong
15.8k1739
edited Jan 18 at 19:49
answered Jan 18 at 18:29
SongSong
15.8k1739
answered Jan 18 at 18:29
SongSong
15.8k1739
answered Jan 18 at 18:29
SongSong
15.8k1739
15.8k1739
add a comment |
add a comment |
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Well, you wrote that for each $t$, $f(t)=0$ or $f(t)=t^2$. That does not mean that either $f=0$ or $f(x)=x^2$ for all $x$. (an example: $f(x)=x^2$ if $x >0$, $f(x)=0$ else).
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– Mindlack
Jan 18 at 17:42
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"(an example: f(x)=x2 if x>0, f(x)=0 else)" That's a poor example as $0^2 =0$. But you are coorect. $f(t)$ may equal $0$ for some $t$ and equal $t^2$ for others.
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– fleablood
Jan 18 at 17:45
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You are trapped at the so-called point wise trap. If $(f(x)-a)(f(x)-b)=0$, it is the case that for every $x$, either $f(x)=a$ or $f(x)=b$, not that, $f(x)=a$ for every $x$ or $f(x)=b$ for every $x$. Also, I think you should post olympiad-related stuff to AoPS, people there are much more comfortable with olympiad-level problems/ideas than those in here, who are usually not as good, in solving olympiad-level problems.
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– Aaron
Jan 18 at 17:46