Solution to a differential equation?












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$begingroup$


Suppose $f'(x)=frac{f(2x)-f(x)}{x}$ for all $x in (0,1]$, what are the solutions?



In my problem, $f'(x)$ is also continuous negative.










share|cite|improve this question











$endgroup$












  • $begingroup$
    it works for $f(x)=c$ where $c < 0$
    $endgroup$
    – Vasya
    Jan 3 at 20:51










  • $begingroup$
    Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
    $endgroup$
    – TomH
    Jan 3 at 21:04










  • $begingroup$
    Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
    $endgroup$
    – LutzL
    Jan 3 at 21:06












  • $begingroup$
    Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
    $endgroup$
    – TomH
    Jan 4 at 8:06










  • $begingroup$
    I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
    $endgroup$
    – TomH
    Jan 4 at 11:14
















0












$begingroup$


Suppose $f'(x)=frac{f(2x)-f(x)}{x}$ for all $x in (0,1]$, what are the solutions?



In my problem, $f'(x)$ is also continuous negative.










share|cite|improve this question











$endgroup$












  • $begingroup$
    it works for $f(x)=c$ where $c < 0$
    $endgroup$
    – Vasya
    Jan 3 at 20:51










  • $begingroup$
    Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
    $endgroup$
    – TomH
    Jan 3 at 21:04










  • $begingroup$
    Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
    $endgroup$
    – LutzL
    Jan 3 at 21:06












  • $begingroup$
    Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
    $endgroup$
    – TomH
    Jan 4 at 8:06










  • $begingroup$
    I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
    $endgroup$
    – TomH
    Jan 4 at 11:14














0












0








0





$begingroup$


Suppose $f'(x)=frac{f(2x)-f(x)}{x}$ for all $x in (0,1]$, what are the solutions?



In my problem, $f'(x)$ is also continuous negative.










share|cite|improve this question











$endgroup$




Suppose $f'(x)=frac{f(2x)-f(x)}{x}$ for all $x in (0,1]$, what are the solutions?



In my problem, $f'(x)$ is also continuous negative.







ordinary-differential-equations functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 20:48







TomH

















asked Jan 3 at 20:20









TomHTomH

13213




13213












  • $begingroup$
    it works for $f(x)=c$ where $c < 0$
    $endgroup$
    – Vasya
    Jan 3 at 20:51










  • $begingroup$
    Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
    $endgroup$
    – TomH
    Jan 3 at 21:04










  • $begingroup$
    Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
    $endgroup$
    – LutzL
    Jan 3 at 21:06












  • $begingroup$
    Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
    $endgroup$
    – TomH
    Jan 4 at 8:06










  • $begingroup$
    I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
    $endgroup$
    – TomH
    Jan 4 at 11:14


















  • $begingroup$
    it works for $f(x)=c$ where $c < 0$
    $endgroup$
    – Vasya
    Jan 3 at 20:51










  • $begingroup$
    Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
    $endgroup$
    – TomH
    Jan 3 at 21:04










  • $begingroup$
    Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
    $endgroup$
    – LutzL
    Jan 3 at 21:06












  • $begingroup$
    Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
    $endgroup$
    – TomH
    Jan 4 at 8:06










  • $begingroup$
    I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
    $endgroup$
    – TomH
    Jan 4 at 11:14
















$begingroup$
it works for $f(x)=c$ where $c < 0$
$endgroup$
– Vasya
Jan 3 at 20:51




$begingroup$
it works for $f(x)=c$ where $c < 0$
$endgroup$
– Vasya
Jan 3 at 20:51












$begingroup$
Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
$endgroup$
– TomH
Jan 3 at 21:04




$begingroup$
Right, although I would also like $f'(x)<0$, which excludes constant function. More generally, it works for all linear functions $a-bx$ with $b>0$. Also, if $f$ is a polynomial, then it must be linear. Are there any other functions than linear (and constant)?
$endgroup$
– TomH
Jan 3 at 21:04












$begingroup$
Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
$endgroup$
– LutzL
Jan 3 at 21:06






$begingroup$
Trying $f(x)=ax^b$ gives as condition $b=2^b-1$ which has solutions $b=0$ and $b=1$, so all functions $f(x)=ax+c$.
$endgroup$
– LutzL
Jan 3 at 21:06














$begingroup$
Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
$endgroup$
– TomH
Jan 4 at 8:06




$begingroup$
Thanks! Any ideas how to prove that there are no continuously differentiable non-linear functions that satisfy the equation?
$endgroup$
– TomH
Jan 4 at 8:06












$begingroup$
I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
$endgroup$
– TomH
Jan 4 at 11:14




$begingroup$
I'm wondering whether Weierstrass Theorem could be used here? Intuitively, since $f(x)$ is continuous, it can be closely approximated by polynomials. The only polynomials that (exactly) satisfy the equation are of degree 1, i.e. linear. This seems to give a strong hint that $f(x)$ itself must be linear. But I'm not famililar with these techniques, so I would appreciate some pointers.
$endgroup$
– TomH
Jan 4 at 11:14










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