Mixing
A functional equation of two variables
$begingroup$
Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$
I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.
functional-equations
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
$endgroup$
add a comment |
$begingroup$
Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$
I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.
functional-equations
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
$endgroup$
1
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36
add a comment |
$begingroup$
Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$
I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.
functional-equations
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
$endgroup$
Solve the following functional equation :
$f:Bbb Z rightarrow Bbb Z$, $f(f(x)+y)=x+f(y+2017)$
I have no prior experience with solving functional equation but still tried a bit. I set $x=y=0$ to get $f(f(0))=f(2017)$. Can we apply $f^{-1}$ both sides to get $f(0)=2017$? I am unable to carry this on.
functional-equations
functional-equations
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
edited Jan 3 at 4:26
Cheerful Parsnip
20.9k23397
20.9k23397
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
asked Jan 2 at 8:32
Epsilon zeroEpsilon zero
34118
34118
1
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36
add a comment |
1
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36
1
1
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.
To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$
Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.
Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.
Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$
Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
$endgroup$
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
add a comment |
$begingroup$
You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.
The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?
Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.
answered Jan 2 at 9:19
FerredFerred
763
$endgroup$
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
add a comment |
$begingroup$
Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.
Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$
By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.
Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059243%2fa-functional-equation-of-two-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.
To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$
Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.
Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.
Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$
Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
$endgroup$
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
add a comment |
$begingroup$
Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.
To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$
Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.
Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.
Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$
Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
$endgroup$
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
add a comment |
$begingroup$
Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.
To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$
Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.
Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.
Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$
Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
$endgroup$
Got it! I probably complicated it more than I had to so if anyone sees any way to simply this I would love to hear feedback.
To solve this, let $y=0$ so that
$f(f(x)) = x+f(2017)$.
Let $c=f(2017)$. Then for all $xinmathbb{Z}$,
$$f(f(x)) = x+c$$
Now plugging $x=y=0$ into the original equation we get
$f(f(0)) = f(2017)$. Taking $f$ of both sides yields
$f(f(f(0))) = f(f(2017))$ which is $f(0)+c=2017+c$ so
$f(0) = 2017$.
Now take $f$ of both sides of the original equation to get
$f(f(f(x)+y)) = f(x+f(y+2017))$
which is
$f(x)+y + c = f(x+f(y+2017))$
Setting $y=-2017$ gives
$$f(x)-2017 + c = f(x+f(0)) = f(x+2017)$$.
Now we return again to the original equation with $y=1$. This gives
$f(f(x)+1) = x+f(2018)$ which by the above formula is
$x+f(1)-2017 + c$. So
$$f(f(x)+1)-f(f(x)) = x+f(1)-2017+c - (x+c) = f(1)-2017$$
Now, as $f(f(x)) = x+c$, $f(f(x))$, and thus $f(x)$, can attain all values in $mathbb{Z}$. So for any $kinmathbb{Z}$, there is an $xinmathbb{Z}$ such that $f(x)=k$, and so the above formula becomes
$$f(k+1)-f(k) = f(1)-2017$$
for all $kinmathbb{Z}$. So for all $kinmathbb{Z}$,
$$f(k) = k+c_2$$
for some $c_2$. So the original equation becomes
$$x+y+2c_2=x+y+2017+c_2$$
so $c_2=2017$. Thus the only solution is
$$f(x) = x+2017$$
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 10:28
Erik ParkinsonErik Parkinson
8749
8749
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
add a comment |
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
$begingroup$
(+1) nice! ${}{}$
$endgroup$
– TheSimpliFire
Jan 2 at 10:38
add a comment |
$begingroup$
You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.
The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?
Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.
answered Jan 2 at 9:19
FerredFerred
763
$endgroup$
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
add a comment |
$begingroup$
You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.
The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?
Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.
answered Jan 2 at 9:19
FerredFerred
763
$endgroup$
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
add a comment |
$begingroup$
You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.
The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?
Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.
answered Jan 2 at 9:19
FerredFerred
763
$endgroup$
You could take $y=0$ there and get $f(f(x))=x+f(2017)=x+n$ where $n$ is a constant. Thus $f(f(x))$ is a linear function. You can observe (guess?) what $f(x)$ could be like here. If it is so, then you have no problem taking $f^{-1}$ and you can easily find $n$.
The point is how to prove the nature of $f(x)$ if you know $f(f(x))$ is linear. This is an interesting problem. What is the square root (with integer values) of a linear function?
Now here I'm not sure $f(x)$ will end up being a simple function involving only arithmetic operations, or a function also involving the modulo.
answered Jan 2 at 9:19
FerredFerred
763
edited Jan 2 at 9:20
answered Jan 2 at 9:19
FerredFerred
763
answered Jan 2 at 9:19
FerredFerred
763
answered Jan 2 at 9:19
FerredFerred
763
763
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
add a comment |
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
1
1
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
$begingroup$
This is better posted as a comment. As you don't yet have enough reputation for that, how about answering some other questions and come back? :)
$endgroup$
– TheSimpliFire
Jan 2 at 9:20
add a comment |
$begingroup$
Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.
Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$
By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.
Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
$begingroup$
Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.
Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$
By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.
Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
$begingroup$
Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.
Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$
By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.
Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
Let $n$ be an integer. Then at $f(x)=n$ and $y=0$, the equation $f(f(x)+y)=x+f(y+n)$ becomes $f(n+y)=f(y+n)+xBig|_{f(x)=n}$, so $f(0)=n$.
Now at $f(x)=n+1$, we can derive similar relations, at $y=0$ and $y=-1$ respectively, $$xBig|_{f(x)=n+1}=f(n+1)-f(n)=f(n)-f(n-1)implies 2f(n)=f(n-1)+f(n+1).$$
By induction it can be proven that $2f(n)=f(n-x)+f(n+x)$ or $2f(n)=f(x)+f(2n-x)$ for any integer $x$. Assuming differentiability, $f'(x)-f'(2n-x)=0$ so $f$ is either a linear or periodic function. It cannot be periodic because $f(0)=n$ and $xBig|_{f(x)=0}=f(n)-f(0)$.
Thus $f$ is a linear function and putting it in $f(x)=ax+b$, gives us $b=n$ and $a=1$ since $f(f(x))=x+f(n)$.
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
edited Jan 3 at 7:36
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
answered Jan 2 at 11:08
TheSimpliFireTheSimpliFire
12.6k62360
12.6k62360
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059243%2fa-functional-equation-of-two-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
It's allowed to take $f^{-1}$ on both sides if a function is monotonic. Can you assume that $f$ is monotonic in this case?
$endgroup$
– Matti P.
Jan 2 at 9:01
$begingroup$
Of $f$ is not injective you can't cancel it out
$endgroup$
– Holo
Jan 2 at 9:02
$begingroup$
@MattiP. Not monotonic, to be able to use $f^{-1}$ you need $f$ to be bijective, and to cancel out $f$ you need it to be injective
$endgroup$
– Holo
Jan 2 at 9:04
$begingroup$
It holds, because we have that at $y=-2017$, $f(0)=f(f(x)-2017)-x$ and at $f(x)=2017$, $f(0)=f(0)-x|_{f(x)=2017}$ so that $f(0)=2017$.
$endgroup$
– TheSimpliFire
Jan 2 at 9:19
$begingroup$
$f(x)=x+2017$ is clearly a solution, and probably the only solution, showing nothing else works is proving to be tricky though.
$endgroup$
– Erik Parkinson
Jan 2 at 9:36