Just a quick note, the value \(h'(s)\) satisfies the equation:
\[
h'(s) = \exp(f(s)) f'(s) / f'(h(s))\\
\]
Then this is a first order differential equation which is related to compositional calculus, but I'm not sure how it could help.
Either way, if:
\[
u(s,z) = \int_a^s \dfrac{\exp(f(x)) f'(x)}{f'(z)}\,dx \bullet z\\
\]
Then,
\[
\begin{align}
u(a,z) &= z\\
u'(s,z) &= \dfrac{\exp(f(s))f'(s)}{f'(u(s))}\\
\end{align}
\]
This leaves us the trouble of finding a value \(a\) and \(z\) that matches \(h\).
But I'm not so sure how much the compositional calculus would help, unless you are looking for a way to numerically evaluate this. If that is so, then the formula for this is a little involved (just a slightly more esoteric version than Euler's method). Let \(\{t_j\}_{j=0}^n\) be a partition of \([a,s]\) in descending order: such that \(t_{j} - t_{j+1} = \mathcal{O}(1/n)\); and let \(t_{j} \ge t_j^* \ge t_{j+1}\):
\[
u(s,z) = \lim_{n\to\infty}\Omega_{j=0}^{n-1} z+ \dfrac{\exp(f(t_j^*))f'(t_j^*)}{f'(z)}\left( t_j - t_{j+1}\right)\,\bullet z\\
\]
Which is just saying, if
\[
q_{jn}(z) = z+ \dfrac{\exp(f(t_j^*))f'(t_j^*)}{f'(z)}\left( t_j - t_{j+1}\right)\\
\]
Then,
\[
u(s,z) = \lim_{n\to\infty} q_{0n}(q_{1n}(q_{2n}(...q_{(n-1)n}(z))))\\
\]
This will converge locally for \(|a-s| < \delta\) and \(z\) almost everywhere (sort of). Luckily \(f'\) is nonzero (i believe so). This definition can be extended in a very convoluted manner to a larger domain, it's just fairly difficult. This would be by the defining property of the compositional integral, which is:
\[
\int_{b}^c g(x,z) \,dx \bullet \int_{a}^b g(x,z)\,dx\bullet z = \int_{a}^c g(x,z)\,dx\bullet z\\
\]
This definition works for complex \(s\) as well; but normally I'd write it as a contour integral. Which is:
\[
\int_\gamma g(w,z)\,dw \bullet z = \int_{0}^1 g(\gamma(x),z)\gamma'(x)\,dx\bullet z\\
\]
For \(\gamma\) an arc which satisfies \(\gamma(0)= a\) and \(\gamma(1) = s\). I'm not so sure if this would really help though. I can't see this adding much more to the discussion than just applying Euler's Method on \(h\). The compositional calculus only really develops a use when you start modding out by equivalence classes. But it is definitely helpful at visualizing the interaction between compositions and integrals.
EDIT:
IT seems this is an induced semi-group. Very damned interesting.
The differential equation is separable, so it can be reduced to a semi-group. That is, \(u\) has an alternative representation:
\[
\begin{align}
u(s,z) &= \int_{0}^{A(s)} \frac{dt \bullet z}{f'(z)}\\
A(s) &= \displaystyle \int_a^s \exp(f(x))f'(x)\,dx\\
\end{align}
\]
This is shown by making the substitution \(dt = \exp(f(x))f'(x)\,dx\) in the equation:
\[
u(s,z) = \int_a^s \dfrac{\exp(f(x)) f'(x)}{f'(z)}\,dx \bullet z\\
\]
And if you define:
\[
U(w,z) = \int_{0}^{w} \frac{dt \bullet z}{f'(z)}\\
\]
Then:
\[
\begin{align}
U(0,z) &= z\\
U(w',U(w,z)) &= U(w' + w,z)\\
\end{align}
\]
So, You're differential equation actually reduces into a flow equation, which is very very nice. Essentially then, all we have to worry about to define \(h\) is the pesky semi-group induced by \(1/f'(z)\). Which shouldn't be too too hard.
Just so you understand what a semi group induced by \(1/f'(z)\) means; I mean that every semi group in existence is induced by the identity:
\[
\lim_{\delta\to 0} \frac{U(\delta,z)-z}{\delta} = g(z)\\
\]
Where for \(h\), \(g(z) = 1/f'(z)\); from here, we input \(A(s)\) into the exponent of the semi group--voila, we have \(h\); so long as we choose \(a\) and \(z\) appropriately. I'm happy to explain this more because the compositional calculus includes much of the standard literature, it's just a better way of writing it imo. Where you can apply Leibniz substitutions/Riemann-Stieljtes integration/Flow theory much more compactly. Honestly, it's a well developed shorthand that I developed that isn't really needed. But once you start modding out by equivalence classes, WOAH buckle your seats!
Regards, James. Hope I can help Tommy.
\[
h'(s) = \exp(f(s)) f'(s) / f'(h(s))\\
\]
Then this is a first order differential equation which is related to compositional calculus, but I'm not sure how it could help.
Either way, if:
\[
u(s,z) = \int_a^s \dfrac{\exp(f(x)) f'(x)}{f'(z)}\,dx \bullet z\\
\]
Then,
\[
\begin{align}
u(a,z) &= z\\
u'(s,z) &= \dfrac{\exp(f(s))f'(s)}{f'(u(s))}\\
\end{align}
\]
This leaves us the trouble of finding a value \(a\) and \(z\) that matches \(h\).
But I'm not so sure how much the compositional calculus would help, unless you are looking for a way to numerically evaluate this. If that is so, then the formula for this is a little involved (just a slightly more esoteric version than Euler's method). Let \(\{t_j\}_{j=0}^n\) be a partition of \([a,s]\) in descending order: such that \(t_{j} - t_{j+1} = \mathcal{O}(1/n)\); and let \(t_{j} \ge t_j^* \ge t_{j+1}\):
\[
u(s,z) = \lim_{n\to\infty}\Omega_{j=0}^{n-1} z+ \dfrac{\exp(f(t_j^*))f'(t_j^*)}{f'(z)}\left( t_j - t_{j+1}\right)\,\bullet z\\
\]
Which is just saying, if
\[
q_{jn}(z) = z+ \dfrac{\exp(f(t_j^*))f'(t_j^*)}{f'(z)}\left( t_j - t_{j+1}\right)\\
\]
Then,
\[
u(s,z) = \lim_{n\to\infty} q_{0n}(q_{1n}(q_{2n}(...q_{(n-1)n}(z))))\\
\]
This will converge locally for \(|a-s| < \delta\) and \(z\) almost everywhere (sort of). Luckily \(f'\) is nonzero (i believe so). This definition can be extended in a very convoluted manner to a larger domain, it's just fairly difficult. This would be by the defining property of the compositional integral, which is:
\[
\int_{b}^c g(x,z) \,dx \bullet \int_{a}^b g(x,z)\,dx\bullet z = \int_{a}^c g(x,z)\,dx\bullet z\\
\]
This definition works for complex \(s\) as well; but normally I'd write it as a contour integral. Which is:
\[
\int_\gamma g(w,z)\,dw \bullet z = \int_{0}^1 g(\gamma(x),z)\gamma'(x)\,dx\bullet z\\
\]
For \(\gamma\) an arc which satisfies \(\gamma(0)= a\) and \(\gamma(1) = s\). I'm not so sure if this would really help though. I can't see this adding much more to the discussion than just applying Euler's Method on \(h\). The compositional calculus only really develops a use when you start modding out by equivalence classes. But it is definitely helpful at visualizing the interaction between compositions and integrals.
EDIT:
IT seems this is an induced semi-group. Very damned interesting.
The differential equation is separable, so it can be reduced to a semi-group. That is, \(u\) has an alternative representation:
\[
\begin{align}
u(s,z) &= \int_{0}^{A(s)} \frac{dt \bullet z}{f'(z)}\\
A(s) &= \displaystyle \int_a^s \exp(f(x))f'(x)\,dx\\
\end{align}
\]
This is shown by making the substitution \(dt = \exp(f(x))f'(x)\,dx\) in the equation:
\[
u(s,z) = \int_a^s \dfrac{\exp(f(x)) f'(x)}{f'(z)}\,dx \bullet z\\
\]
And if you define:
\[
U(w,z) = \int_{0}^{w} \frac{dt \bullet z}{f'(z)}\\
\]
Then:
\[
\begin{align}
U(0,z) &= z\\
U(w',U(w,z)) &= U(w' + w,z)\\
\end{align}
\]
So, You're differential equation actually reduces into a flow equation, which is very very nice. Essentially then, all we have to worry about to define \(h\) is the pesky semi-group induced by \(1/f'(z)\). Which shouldn't be too too hard.
Just so you understand what a semi group induced by \(1/f'(z)\) means; I mean that every semi group in existence is induced by the identity:
\[
\lim_{\delta\to 0} \frac{U(\delta,z)-z}{\delta} = g(z)\\
\]
Where for \(h\), \(g(z) = 1/f'(z)\); from here, we input \(A(s)\) into the exponent of the semi group--voila, we have \(h\); so long as we choose \(a\) and \(z\) appropriately. I'm happy to explain this more because the compositional calculus includes much of the standard literature, it's just a better way of writing it imo. Where you can apply Leibniz substitutions/Riemann-Stieljtes integration/Flow theory much more compactly. Honestly, it's a well developed shorthand that I developed that isn't really needed. But once you start modding out by equivalence classes, WOAH buckle your seats!
Regards, James. Hope I can help Tommy.