Solution to a differential equation?
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.
ordinary-differential-equations functions
asked Jan 3 at 20:20
TomHTomH
13213
$endgroup$
add a comment |
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.
ordinary-differential-equations functions
asked Jan 3 at 20:20
TomHTomH
13213
$endgroup$
$begingroup$
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
add a comment |
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.
ordinary-differential-equations functions
asked Jan 3 at 20:20
TomHTomH
13213
$endgroup$
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
ordinary-differential-equations functions
asked Jan 3 at 20:20
TomHTomH
13213
asked Jan 3 at 20:20
TomHTomH
13213
edited Jan 3 at 20:48
TomH
asked Jan 3 at 20:20
TomHTomH
13213
asked Jan 3 at 20:20
TomHTomH
13213
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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