Functional equation with $f(2x)$
$begingroup$
Any other solutions(advice) are welcome.
For any $x>0, ;;; 2f(frac{1}{x}+1)+f(2x)=1$
Find all possible f(x).
I wish you luck on a good thing in $2019$.
functional-equations
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
$endgroup$
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
For any $x>0, ;;; 2f(frac{1}{x}+1)+f(2x)=1$
Find all possible f(x).
I wish you luck on a good thing in $2019$.
functional-equations
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
$endgroup$
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
For any $x>0, ;;; 2f(frac{1}{x}+1)+f(2x)=1$
Find all possible f(x).
I wish you luck on a good thing in $2019$.
functional-equations
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
$endgroup$
Any other solutions(advice) are welcome.
For any $x>0, ;;; 2f(frac{1}{x}+1)+f(2x)=1$
Find all possible f(x).
I wish you luck on a good thing in $2019$.
functional-equations
functional-equations
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
edited Jan 23 at 4:57
Jyrki Lahtonen
110k13171378
110k13171378
asked Jan 22 at 19:03
mina_worldmina_world
1749
asked Jan 22 at 19:03
mina_worldmina_world
1749
asked Jan 22 at 19:03
mina_worldmina_world
1749
1749
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Yes, but not quite like that.
Say you'd define the function arbitrarily for $xin(0,2]$. Also, you know how to get $fleft({2over x}+1right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $fleft(5over3right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.
To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $xin(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,infty)$. This in turn gives a piece of $left(1,{5over3}right]$ which sticks nicely to $(0,1]$. Then we get $left[{11over5},3right)$, and so on.
By applying the same equation in reverse, we extend our domain in the different direction, until all $mathbb R$ is covered.
Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, but not quite like that.
Say you'd define the function arbitrarily for $xin(0,2]$. Also, you know how to get $fleft({2over x}+1right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $fleft(5over3right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.
To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $xin(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,infty)$. This in turn gives a piece of $left(1,{5over3}right]$ which sticks nicely to $(0,1]$. Then we get $left[{11over5},3right)$, and so on.
By applying the same equation in reverse, we extend our domain in the different direction, until all $mathbb R$ is covered.
Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
$endgroup$
add a comment |
$begingroup$
Yes, but not quite like that.
Say you'd define the function arbitrarily for $xin(0,2]$. Also, you know how to get $fleft({2over x}+1right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $fleft(5over3right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.
To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $xin(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,infty)$. This in turn gives a piece of $left(1,{5over3}right]$ which sticks nicely to $(0,1]$. Then we get $left[{11over5},3right)$, and so on.
By applying the same equation in reverse, we extend our domain in the different direction, until all $mathbb R$ is covered.
Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
$endgroup$
add a comment |
$begingroup$
Yes, but not quite like that.
Say you'd define the function arbitrarily for $xin(0,2]$. Also, you know how to get $fleft({2over x}+1right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $fleft(5over3right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.
To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $xin(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,infty)$. This in turn gives a piece of $left(1,{5over3}right]$ which sticks nicely to $(0,1]$. Then we get $left[{11over5},3right)$, and so on.
By applying the same equation in reverse, we extend our domain in the different direction, until all $mathbb R$ is covered.
Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
$endgroup$
Yes, but not quite like that.
Say you'd define the function arbitrarily for $xin(0,2]$. Also, you know how to get $fleft({2over x}+1right)$ if you have $f(x)$. Now, $f(1)$ gives you $f(3)$, which in turn gives you $fleft(5over3right)$, which is in $(0,2]$ and hence already defined, maybe differently from what we've got.
To avoid contradiction, let's select a shorter interval (a fundamental domain, as people call it). We define our function arbitrarily for $xin(0,1]$, which we can do in awfully many ways. This gives us its values on $[3,infty)$. This in turn gives a piece of $left(1,{5over3}right]$ which sticks nicely to $(0,1]$. Then we get $left[{11over5},3right)$, and so on.
By applying the same equation in reverse, we extend our domain in the different direction, until all $mathbb R$ is covered.
Apparently, any $f$ which is not a constant has got to be discontinuous at one of the limit points, but that's another story.
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
answered Jan 23 at 10:05
Ivan NeretinIvan Neretin
8,85621535
8,85621535
add a comment |
add a comment |
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