Hey, Daniel; you have described flows:
\[
\frac{d}{dt} y(t) = F(y(t))\\
\]
Where:
\[
F(z) = \frac{d}{dx}\Big{|}_{x=0} f^{\circ x}(z)\\
\]
.....
Now let's describe the path \(y(t)\). Let's call \(y(t) = f^{\circ t}(z)\). This is written as a first order differential equation, which can be translated to a flow equation:
\[
y'(t) = F(y(t))\\
\]
If we trace any path; we hit all the solutions, and hit discontinuities if they are natural....
Any flow equation like this allows us to write \(f^{\circ s}(f^{\circ t}) = f^{\circ s +t}\). But we can start from my perspective, then find this equation naturally.
This is one of the motivations of my "differential bullet product" notation.
....
You can also just write this as matrices multiplying.........
I write this as:
\[
f^{\circ t}(z) = \int_0^t F(z)\,ds\bullet z\\
\]
Which is just an advanced integral of the matrix interpretation....
\[
\frac{d}{dt} y(t) = F(y(t))\\
\]
Where:
\[
F(z) = \frac{d}{dx}\Big{|}_{x=0} f^{\circ x}(z)\\
\]
.....
Now let's describe the path \(y(t)\). Let's call \(y(t) = f^{\circ t}(z)\). This is written as a first order differential equation, which can be translated to a flow equation:
\[
y'(t) = F(y(t))\\
\]
If we trace any path; we hit all the solutions, and hit discontinuities if they are natural....
Any flow equation like this allows us to write \(f^{\circ s}(f^{\circ t}) = f^{\circ s +t}\). But we can start from my perspective, then find this equation naturally.
This is one of the motivations of my "differential bullet product" notation.
....
You can also just write this as matrices multiplying.........
I write this as:
\[
f^{\circ t}(z) = \int_0^t F(z)\,ds\bullet z\\
\]
Which is just an advanced integral of the matrix interpretation....