Find all the bijective functions $f:[0,1]to[0,1]$ such that $x=frac{1}{2}big(f(x)+f^{-1}(x)big)$ for all...
$begingroup$
Find all bijective functions $ f : [0,1] to [0,1]$ that satisfy the equation $$x=frac{1}{2} big(f(x) +f^{-1} (x)big),forall x in[0,1],.$$
I honestly don't know how to approach this. I tried to plug in different values, but this doesn't work.
algebra-precalculus functions inverse functional-equations inverse-function
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
$endgroup$
add a comment |
$begingroup$
Find all bijective functions $ f : [0,1] to [0,1]$ that satisfy the equation $$x=frac{1}{2} big(f(x) +f^{-1} (x)big),forall x in[0,1],.$$
I honestly don't know how to approach this. I tried to plug in different values, but this doesn't work.
algebra-precalculus functions inverse functional-equations inverse-function
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
$endgroup$
$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37
add a comment |
$begingroup$
Find all bijective functions $ f : [0,1] to [0,1]$ that satisfy the equation $$x=frac{1}{2} big(f(x) +f^{-1} (x)big),forall x in[0,1],.$$
I honestly don't know how to approach this. I tried to plug in different values, but this doesn't work.
algebra-precalculus functions inverse functional-equations inverse-function
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
$endgroup$
Find all bijective functions $ f : [0,1] to [0,1]$ that satisfy the equation $$x=frac{1}{2} big(f(x) +f^{-1} (x)big),forall x in[0,1],.$$
I honestly don't know how to approach this. I tried to plug in different values, but this doesn't work.
algebra-precalculus functions inverse functional-equations inverse-function
algebra-precalculus functions inverse functional-equations inverse-function
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
edited Dec 11 '18 at 19:12
Batominovski
33.1k33293
33.1k33293
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
asked Dec 8 '18 at 19:10
JustAnUserJustAnUser
83
83
$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37
add a comment |
$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37
$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Suppose $x in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.
Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n ge 0$.
In particular, $x - na in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.
So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.
(Note that we did not need to assume $f$ is continuous.)
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Suppose $x in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.
Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n ge 0$.
In particular, $x - na in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.
So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.
(Note that we did not need to assume $f$ is continuous.)
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
add a comment |
$begingroup$
Suppose $x in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.
Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n ge 0$.
In particular, $x - na in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.
So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.
(Note that we did not need to assume $f$ is continuous.)
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
add a comment |
$begingroup$
Suppose $x in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.
Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n ge 0$.
In particular, $x - na in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.
So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.
(Note that we did not need to assume $f$ is continuous.)
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
$endgroup$
Suppose $x in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.
Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n ge 0$.
In particular, $x - na in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.
So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.
(Note that we did not need to assume $f$ is continuous.)
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
edited Dec 9 '18 at 3:56
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
answered Dec 8 '18 at 20:49
Hew WolffHew Wolff
2,260716
2,260716
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
add a comment |
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
$begingroup$
Nice solution, thank you! When you choose $a$ at the beginning of your soltuion, is it a constant or can it also be a function?
$endgroup$
– JustAnUser
Dec 8 '18 at 23:45
2
2
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
@JustAnUser. Say you map some value $x$ in the interval to $f(x)$. We let $a=x-f(x)$. A little magic happens and then we conclude that $a=0$. It's just a constant.
$endgroup$
– Mason
Dec 9 '18 at 2:12
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
$begingroup$
By the way, @JustAnUser, I found it very helpful to draw pictures of what the graph of $f$ would look like in different cases. I was too lazy to illustrate the proof, but that's where the proof came from, so you might try that too.
$endgroup$
– Hew Wolff
Dec 9 '18 at 16:20
add a comment |
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$begingroup$
So $f(x)=x$ obviously does the trick. I think it might be the only the continuous function that does this but I don't know how to prove this at first glance. Do we know anything else about the function? A bijection isn't a lot of power over a function. There are many bizarre bijections. Is this question from a text? from a class?
$endgroup$
– Mason
Dec 8 '18 at 19:31
$begingroup$
What if $F(0,x) := x$ and $F(n,x)=f(F(n-1,x))$ for all integer $n$ and real $x$?
$endgroup$
– Somos
Dec 8 '18 at 19:37