Learning about functions containing derivatives?
$begingroup$
Ideally, I'm looking to "solve" $$f(x)=x-f'(x-s),quad where,, x ,>=0,quad s=1, quad and quad f(-s) = 0$$
and the moving on to the same format but this equation instead:
$$f(x)=sign(x)-sign(f'(x-s))$$
But I ended up with a ton of questions while attempting to solve it. I'd break it into smaller questions but they seem to my untrained eye to be simple answers about properties on functions of this type.
- As far as I can tell this is a chaotic function? So there is no solution correct?
- Is this generally true of any function containing the derivative of
a translated copy? I know it's not generally true if we remove the
translation. - Is there a name for this "recursive derivative" so I
can research more about it, or is it just a blanket term of
differential equation? - I noticed that no matter how I calculated the
sequence, as long as the spacing was even the generated sequence was
identical for each term (scaled based on $s$), so f(1) depends on the
spacing used to get there, not sensitivity to initial conditions, or even accuracy, is there a name for this property, and/or is the $s$ spacing considered
part of initial conditions?
Assuming I didn't define the function correctly, here's the first few terms:
$$0,1,1,3,2,6,2,11,-1,21,-12,44,-44,101,-131,247$$
derivatives
asked Dec 4 '18 at 11:24
BlackBlack
1046
$endgroup$
add a comment |
$begingroup$
Ideally, I'm looking to "solve" $$f(x)=x-f'(x-s),quad where,, x ,>=0,quad s=1, quad and quad f(-s) = 0$$
and the moving on to the same format but this equation instead:
$$f(x)=sign(x)-sign(f'(x-s))$$
But I ended up with a ton of questions while attempting to solve it. I'd break it into smaller questions but they seem to my untrained eye to be simple answers about properties on functions of this type.
- As far as I can tell this is a chaotic function? So there is no solution correct?
- Is this generally true of any function containing the derivative of
a translated copy? I know it's not generally true if we remove the
translation. - Is there a name for this "recursive derivative" so I
can research more about it, or is it just a blanket term of
differential equation? - I noticed that no matter how I calculated the
sequence, as long as the spacing was even the generated sequence was
identical for each term (scaled based on $s$), so f(1) depends on the
spacing used to get there, not sensitivity to initial conditions, or even accuracy, is there a name for this property, and/or is the $s$ spacing considered
part of initial conditions?
Assuming I didn't define the function correctly, here's the first few terms:
$$0,1,1,3,2,6,2,11,-1,21,-12,44,-44,101,-131,247$$
derivatives
asked Dec 4 '18 at 11:24
BlackBlack
1046
$endgroup$
add a comment |
$begingroup$
Ideally, I'm looking to "solve" $$f(x)=x-f'(x-s),quad where,, x ,>=0,quad s=1, quad and quad f(-s) = 0$$
and the moving on to the same format but this equation instead:
$$f(x)=sign(x)-sign(f'(x-s))$$
But I ended up with a ton of questions while attempting to solve it. I'd break it into smaller questions but they seem to my untrained eye to be simple answers about properties on functions of this type.
- As far as I can tell this is a chaotic function? So there is no solution correct?
- Is this generally true of any function containing the derivative of
a translated copy? I know it's not generally true if we remove the
translation. - Is there a name for this "recursive derivative" so I
can research more about it, or is it just a blanket term of
differential equation? - I noticed that no matter how I calculated the
sequence, as long as the spacing was even the generated sequence was
identical for each term (scaled based on $s$), so f(1) depends on the
spacing used to get there, not sensitivity to initial conditions, or even accuracy, is there a name for this property, and/or is the $s$ spacing considered
part of initial conditions?
Assuming I didn't define the function correctly, here's the first few terms:
$$0,1,1,3,2,6,2,11,-1,21,-12,44,-44,101,-131,247$$
derivatives
asked Dec 4 '18 at 11:24
BlackBlack
1046
$endgroup$
Ideally, I'm looking to "solve" $$f(x)=x-f'(x-s),quad where,, x ,>=0,quad s=1, quad and quad f(-s) = 0$$
and the moving on to the same format but this equation instead:
$$f(x)=sign(x)-sign(f'(x-s))$$
But I ended up with a ton of questions while attempting to solve it. I'd break it into smaller questions but they seem to my untrained eye to be simple answers about properties on functions of this type.
- As far as I can tell this is a chaotic function? So there is no solution correct?
- Is this generally true of any function containing the derivative of
a translated copy? I know it's not generally true if we remove the
translation. - Is there a name for this "recursive derivative" so I
can research more about it, or is it just a blanket term of
differential equation? - I noticed that no matter how I calculated the
sequence, as long as the spacing was even the generated sequence was
identical for each term (scaled based on $s$), so f(1) depends on the
spacing used to get there, not sensitivity to initial conditions, or even accuracy, is there a name for this property, and/or is the $s$ spacing considered
part of initial conditions?
Assuming I didn't define the function correctly, here's the first few terms:
$$0,1,1,3,2,6,2,11,-1,21,-12,44,-44,101,-131,247$$
derivatives
derivatives
asked Dec 4 '18 at 11:24
BlackBlack
1046
asked Dec 4 '18 at 11:24
BlackBlack
1046
edited Dec 4 '18 at 12:45
Black
asked Dec 4 '18 at 11:24
BlackBlack
1046
asked Dec 4 '18 at 11:24
BlackBlack
1046
asked Dec 4 '18 at 11:24
BlackBlack
1046
1046
add a comment |
add a comment |
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