Posting again to get better traction.
I am looking for a representation of the expression:
$$ \frac{\partial^n} {\partial\delta^n} e^{Y(\delta,t)}, $$
where $Y(\delta,t) : \mathbb R^2_+ \rightarrow \mathbb R^{n \times n}$ an smooth function matrix-valued function such that $Y(0,t) = 0$, the zero matrix for all $t$. $Y(\delta,t)$ assumes the form:
$$ Y(\delta,t) = \sum_{i=1}^N \sum_{j=1}^M E_{ij} t^i \delta^j $$
where each $E_{i,j}$ is a time-independent matrix. I am aware of the integral representation of the first derivative:
$$ \frac{\partial} {\partial \delta} e^{Y(\delta,t)} = \int_{0}^1 d s_1 \; e^{s_1 Y(\delta,t)} \frac{\partial Y(\delta,t))}{\partial \delta} e^{(1-s_1)Y(\delta,t)} $$
I am interested in a similar integral representation for $\frac{\partial^n} {\partial \delta^n} e^{Y(\delta,t)}$. I tried differentiating under the integral sign and using the integral representation above, but I couldn't see a pattern for what the formula for the $n$-th derivative would be? Can someone help or point to references?
Similar question discussed here.
I differentiated under the integral sign to to obtain the second derivative. Note I am only interested in second (and higher derivatives) evaluated at $\delta = 0$. I've also updated notation above.
Is this correct? I tried differentiating under integral sign again to obtain third derivative evalauted at $\delta = 0$, but I couldn't see an obvious pattern that generalizes to arbitrary $n$-th derivative. I can add my computation for the third derivative later on.
Differentiating under the integral sign we can compute the second, partial derivative with respect to $\delta$ as well. We get:
\begin{align} \frac{\partial^2}{\partial \delta^2} \exp(Y(\delta, t)) & = \int_0^1 ds_1 \frac{\partial}{\partial \delta} \Bigg ( e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t)} \Bigg ), \\ & = \int_0^1 ds_1 \Bigg ( \frac{\partial}{\partial \delta} e^{s_1Y(\delta, t)} \Bigg ) \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t)}, \\ & + \int_0^1 ds_1 e^{s_1Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_1)Y(\delta, t)}, \\ & + \int_0^1 ds_1 e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} \Bigg ( \frac{\partial}{\partial \delta} e^{(1-s_1)Y(\delta, t)} \Bigg ), \\ & := I_1(\delta,t) + I_2(\delta,t) + I_3(\delta,t). \end{align}
Let's analyze each term within the integral one by one:
\begin{align} I_1(\delta,t) & = \int_0^1 \int_0^1 ds_1 ds_2 \; s_1 e^{s_1 s_2Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t)}, \\ I_2(\delta,t) & = \int_0^1 ds_1 e^{s_1Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_1)Y(\delta, t)}, \\ I_3(\delta,t) & = \int_0^1 \int_0^1 ds_1 ds_2 \; (1-s_1) e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1) s_2 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)(1-s_2) Y(\delta, t)}. \end{align}
However, since we are interested in evaluating, \begin{equation} \frac{\partial^2}{\partial \delta^2} \exp(Y(\delta, t)) \Bigg |_{\delta = 0}, \end{equation} using that $Y(0,t) = 0$, the terms $I_1, I_2, I_3$ simply to:
\begin{align} I_1 (0,t) & = \int_0^1 \int_0^1 ds_1 ds_2 s_1 \Bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \Bigg )^2 = \frac{1}{2} \Bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \Bigg )^2 , \\ I_2(0,t) & = \int_0^1 ds_1 \frac{\partial^2 Y( 0, t)}{\partial \delta^2} = \frac{\partial^2 Y( 0, t)}{\partial \delta^2} , \\ I_3 (0,t) & = \int_0^1 \int_0^1 ds_1 ds_2 \; (1-s_1) \Bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \Bigg )^2 = \frac{1}{2} \Bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \Bigg )^2 \end{align}
Therefore, we have:
\begin{align} \frac{\partial^2}{\partial \delta^2} \exp(Y(\delta, t)) \Bigg |_{\delta=0} = \Bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \Bigg )^2 + \frac{\partial^2 Y( 0, t)}{\partial \delta^2}. \end{align}
Some computations for the third derivative. Any chance I'm on the right track? If so, can anyone help me spot a pattern for the $n$-th derivative evaluated at $\delta = 0$.
Similarly,
\begin{equation} \frac{\partial^3}{\partial \delta^3} \exp(Y(\delta, t)) = \frac{\partial}{\partial \delta} I_1(\delta,t) + \frac{\partial}{\partial \delta} I_2(\delta,t) + \frac{\partial}{\partial \delta} I_3(\delta,t) := F_1(\delta,t) + F_2(\delta,t) + F_3(\delta,t). \end{equation}
\vspace{0.1in}
Writing the double and triple integrals as,
\begin{equation} \int_0^1 \int_0^1 ds_1 ds_2 = \int_{[0,1]^2} ds \quad \quad \int_0^1 \int_0^1 \int_0^1 ds_1 ds_2 ds_3 = \int_{[0,1]^3} ds \end{equation}
we compute each of the terms above one by one:
\begin{align} F_1(\delta,t) & = \int_{[0,1]^3} ds \; s^2_1s_2 e^{s_1 s_2 s_3 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1 s_2 (1-s_3) Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^3} ds \; s^2_1(1-s_2) e^{s_1 s_2Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2) s_3Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2) (1-s_3) Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^3} ds \; s_1(1-s_1) e^{s_1 s_2Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1) s_3 Y(\delta, t))} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1) (1-s_3) Y(\delta, t))}, \\ & + \int_{[0,1]^2} ds \; s_1 e^{s_1 s_2Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{s_1(1-s_2)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^2} ds \; s_1 e^{s_1 s_2Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1(1-s_2)Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_1)Y(\delta, t))}. \end{align}
Evaluating at $\delta = 0$ yields,
\begin{align} F_1(0,t) & = \int_{[0,1]^3} ds \; s^2_1s_2 \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3 + \int_{[0,1]^3} ds \; s^2_1(1-s_2) \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3 + \int_{[0,1]^3} ds \; s_1(1-s_1) \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3, \\ & + \int_{[0,1]^2} ds \; s_1 \frac{\partial^2 Y( 0, t)}{\partial \delta^2} \frac{\partial Y( 0, t)}{\partial \delta} + \int_{[0,1]^2} ds \; s_1 \frac{\partial^2 Y( 0, t)}{\partial \delta^2} \frac{\partial Y( 0, t)}{\partial \delta}. \end{align} \begin{align} F_2(\delta,t) & = \int_{[0,1]^2} ds \; s_1 e^{s_1 s_2 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1 (1-s_2) Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_1)Y(\delta, t)}, \\ & + \int_{[0,1]^2} ds (1-s_1) e^{s_1Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_1) s_2 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_1) (1-s_2) Y(\delta, t)}, \\ & + \int_{[0,1]} ds e^{s_1Y(\delta, t)} \frac{\partial^3 Y( \delta, t)}{\partial \delta^3} e^{(1-s_1)Y(\delta, t)} \end{align}
Evaluating at $\delta = 0$ yields,
\begin{align} F_2(0,t) & = \int_{[0,1]^2} ds \; s_1 \frac{\partial Y( 0, t)}{\partial \delta} \frac{\partial^2 Y( 0, t)}{\partial \delta^2} + \int_{[0,1]^2} ds (1-s_1) \frac{\partial^2 Y( 0, t)}{\partial \delta^2} \frac{\partial Y( 0, t)}{\partial \delta} + \int_{[0,1]} ds \frac{\partial^3 Y( 0, t)}{\partial \delta^3} \end{align}
\begin{align} F_3(\delta,t) & = \int_{[0,1]^3} ds \; (1-s_1) s_2 e^{s_1 s_3 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_1 (1-s_3) Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_2 (1-s_1)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_2)(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^3} ds \; s_2(1-s_1) e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_2 (1-s_1) s_3 Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_2 (1-s_1) (1-s_3) Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_2)(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^3} ds \; (1-s_2)(1-s_1)^2 e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_2 (1-s_1)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_2)(1-s_1)Y(\delta, t))} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_2)(1-s_1)s_3Y(\delta, t))}, \\ & + \int_{[0,1]^2} ds \; (1-s_1) e^{s_1Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{s_2 (1-s_1)Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{(1-s_2)(1-s_1)Y(\delta, t))}, \\ & + \int_{[0,1]^2} ds \; (1-s_1) e^{s_1Y(\delta, t)} \frac{\partial Y( \delta, t)}{\partial \delta} e^{s_2 (1-s_1)Y(\delta, t)} \frac{\partial^2 Y( \delta, t)}{\partial \delta^2} e^{(1-s_2)(1-s_1)Y(\delta, t))} \end{align}
Evaluating at $\delta = 0$ yields,
\begin{align} F_3(0,t) & = \int_{[0,1]^3} ds \; (1-s_1) s_2 \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3 + \int_{[0,1]^3} ds \; s_2(1-s_1) \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3 + \int_{[0,1]^3} ds \; (1-s_2)(1-s_1)^2 \bigg ( \frac{\partial Y( 0, t)}{\partial \delta} \bigg )^3, \\ & + \int_{[0,1]^2} ds \; (1-s_1) \frac{\partial^2 Y( 0, t)}{\partial \delta^2} \frac{\partial Y( 0, t)}{\partial \delta} + \int_{[0,1]^2} ds \; (1-s_1) \frac{\partial Y( 0, t)}{\partial \delta} \frac{\partial^2 Y( 0, t)}{\partial \delta^2}. \end{align}
