Problem in solving a functional equation
I have the following functional equation with me:
$f(x-y)=f(x)f(y)+g(x)g(y)$
$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions
Firstly interchanging $x$ and $y$ tells me that the function is an even function.
Furthermore $x=y$ gives me
$f(0) = f(x)^2 + g(x)^2$.
Setting $y=0$ in the original equation gives me
$f(x) = f(x)f(0) + g(x)g(0)$.
Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?
functional-equations
edited Dec 1 '18 at 21:36
the_fox
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
add a comment |
I have the following functional equation with me:
$f(x-y)=f(x)f(y)+g(x)g(y)$
$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions
Firstly interchanging $x$ and $y$ tells me that the function is an even function.
Furthermore $x=y$ gives me
$f(0) = f(x)^2 + g(x)^2$.
Setting $y=0$ in the original equation gives me
$f(x) = f(x)f(0) + g(x)g(0)$.
Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?
functional-equations
edited Dec 1 '18 at 21:36
the_fox
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16
add a comment |
I have the following functional equation with me:
$f(x-y)=f(x)f(y)+g(x)g(y)$
$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions
Firstly interchanging $x$ and $y$ tells me that the function is an even function.
Furthermore $x=y$ gives me
$f(0) = f(x)^2 + g(x)^2$.
Setting $y=0$ in the original equation gives me
$f(x) = f(x)f(0) + g(x)g(0)$.
Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?
functional-equations
edited Dec 1 '18 at 21:36
the_fox
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
I have the following functional equation with me:
$f(x-y)=f(x)f(y)+g(x)g(y)$
$f(x)=cos x $ and $g(x)=sin x$ is an obvious solution but I have some trouble proving it. I have obtained the following simple results by simple substitutions
Firstly interchanging $x$ and $y$ tells me that the function is an even function.
Furthermore $x=y$ gives me
$f(0) = f(x)^2 + g(x)^2$.
Setting $y=0$ in the original equation gives me
$f(x) = f(x)f(0) + g(x)g(0)$.
Now from here I can clearly note that if $f(0)=0$ then both $f(x)$ and $g(x)$ are equal to zero. However I am unable to proceed from here. How do I proceed from here? My book just mentions that since $f(x)[1-f(0)]=g(x)g(0)$, $f(0) = 1$ and $g(0)=0$. How do I conclude this?
functional-equations
functional-equations
edited Dec 1 '18 at 21:36
the_fox
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
edited Dec 1 '18 at 21:36
the_fox
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
edited Dec 1 '18 at 21:36
the_fox
2,53711431
edited Dec 1 '18 at 21:36
the_fox
2,53711431
edited Dec 1 '18 at 21:36
the_fox
2,53711431
2,53711431
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 6:30
saisanjeevsaisanjeev
947212
947212
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16
add a comment |
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16
add a comment |
1 Answer
1
active
oldest
votes
Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.
Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).
So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
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%2f3021063%2fproblem-in-solving-a-functional-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.
Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).
So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
add a comment |
Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.
Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).
So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
add a comment |
Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.
Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).
So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
Assume that $g(0)$ is not zero. Then $g(x) = dfrac{1 - f(0)}{g(0)}f(x) = Cf(x)$, so $f(x-y) = (1 + C) f(x)f(y)$.
Setting $x = 2y$ gives $$f(y) = (1 + C) f(2y) f(y)$$ that is $f(2y)$ would be a constant (as long as $f(y)$ is not $0$).
So to get an interesting solution we must assume that $g(0) = 0$. $f(0) = 1$ follows.
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
answered Dec 1 '18 at 20:34
user58697user58697
1,829512
1,829512
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
add a comment |
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
Thanks for the help. Could you please help me in solving the equation further to prove that cosx and sinx are the only possibilities for f(x) and g(x)?
– saisanjeev
Dec 2 '18 at 7:04
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%2f3021063%2fproblem-in-solving-a-functional-equation%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
The equation also admits the constant solutions $f(x) = c$ and $g(x) = sqrt{c-c^2}$ with $0le cle 1$.
– Joey Zou
Dec 1 '18 at 19:16