A functional equation (another)
$begingroup$
I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$
I am actually not sure about existence of a solution.
convex-analysis functional-equations
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
$endgroup$
add a comment |
$begingroup$
I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$
I am actually not sure about existence of a solution.
convex-analysis functional-equations
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
$endgroup$
add a comment |
$begingroup$
I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$
I am actually not sure about existence of a solution.
convex-analysis functional-equations
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
$endgroup$
I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$
I am actually not sure about existence of a solution.
convex-analysis functional-equations
convex-analysis functional-equations
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
edited Dec 13 '18 at 12:51
Ethan Bolker
42.6k549113
42.6k549113
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
asked Dec 13 '18 at 12:50
S. CurryS. Curry
92
92
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all if we plug in $x=1$ we get
$$f(1) = frac 12 + frac 14 f(1). $$
This implies
$$f(1) = frac 23$$
so it appears that you cannot have $f(1)=1$.
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
$endgroup$
add a comment |
$begingroup$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
$c=$any constant.
$$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
After simplification :
$$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
We put $(3)$ in $(1)$ :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
After simplification :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
Comparing to Eq.$(2)$ we see that :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
$endgroup$
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%2f3037964%2fa-functional-equation-another%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all if we plug in $x=1$ we get
$$f(1) = frac 12 + frac 14 f(1). $$
This implies
$$f(1) = frac 23$$
so it appears that you cannot have $f(1)=1$.
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
$endgroup$
add a comment |
$begingroup$
First of all if we plug in $x=1$ we get
$$f(1) = frac 12 + frac 14 f(1). $$
This implies
$$f(1) = frac 23$$
so it appears that you cannot have $f(1)=1$.
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
$endgroup$
add a comment |
$begingroup$
First of all if we plug in $x=1$ we get
$$f(1) = frac 12 + frac 14 f(1). $$
This implies
$$f(1) = frac 23$$
so it appears that you cannot have $f(1)=1$.
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
$endgroup$
First of all if we plug in $x=1$ we get
$$f(1) = frac 12 + frac 14 f(1). $$
This implies
$$f(1) = frac 23$$
so it appears that you cannot have $f(1)=1$.
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
answered Dec 13 '18 at 12:56
flawrflawr
11.5k32446
11.5k32446
add a comment |
add a comment |
$begingroup$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
$c=$any constant.
$$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
After simplification :
$$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
We put $(3)$ in $(1)$ :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
After simplification :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
Comparing to Eq.$(2)$ we see that :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
$endgroup$
add a comment |
$begingroup$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
$c=$any constant.
$$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
After simplification :
$$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
We put $(3)$ in $(1)$ :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
After simplification :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
Comparing to Eq.$(2)$ we see that :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
$endgroup$
add a comment |
$begingroup$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
$c=$any constant.
$$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
After simplification :
$$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
We put $(3)$ in $(1)$ :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
After simplification :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
Comparing to Eq.$(2)$ we see that :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
$endgroup$
$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
$c=$any constant.
$$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
After simplification :
$$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
We put $(3)$ in $(1)$ :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
After simplification :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
Comparing to Eq.$(2)$ we see that :
$$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
answered Dec 14 '18 at 11:55
JJacquelinJJacquelin
43.7k21853
43.7k21853
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%2f3037964%2fa-functional-equation-another%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
