Recursive definition of a Gevrey-class function Given the following Gevrey-class function $\Phi:\mathbb{R} \rightarrow \mathbb{R}$
$$\Phi_{s,T}(t) = 
\begin{cases}
\begin{align}
0 \quad & t \le 0 \\
1 \quad & t \ge T \\
\frac{\int\limits_{0}^t\Lambda_{s,T}(\tau) \text{d}\tau}{\int\limits_{0}^T\Lambda_{s,T}(\tau) \text{d}\tau} \quad & t \in (0,T),
\end{align}
\end{cases}
$$
where 
$$\Lambda_{s,T}(t) = e^{-\frac{1}{[(1-t/T)t/T]^s}}, \quad s \in \mathbb{N^+}, \: T\in\mathbb{R}. $$
I know there must be a recurrence for the $(k+1)^{\text{st}}$-derivative of $\Phi_{s,T}$:
$$\Phi_{s,T}^{(k+1)}(t) = f\left(\Phi_{s,T}^{(k)}(t), \Phi_{s,T}^{(k-1)}(t)\right), \quad t \in (0,T).$$
Does anybody know what it is and how it can be proven inductively?
 A: Lemma. The $(k+1)^{\text{st}}$ derivative of $\Phi_{s,T}$ is recursivly defined as follows:
$$
\begin{align}
\Phi_{s,T}^{(k+1)}(t) & = \frac{\Lambda_{s,T}^{(k)}(t)}{\int\limits_0^T\Lambda_{s,T}(\tau)\text{d}\tau}, \quad k=0,...,n_t\\
\Lambda_{s,T}^{(k+1)}(t) & = \sum\limits_{j=0}^{k}
\left(
\begin{array}{c}
k \\ j
\end{array}
\right)\Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) \\
\Lambda_{s,T}^{(0)}(t) = \Lambda_{s,T}(t) & = e^{-p(t)^{-s}}\\
\lambda^{(j+1)}(t) & = p^{-1}(t)\left(\lambda^{(j)}(t)p'(t)(-s-j)-\lambda^{(j-1)}(t)p''(t)j\left(s+\frac{j-1}{2}\right)\right), \quad j=1,...,k \\
\lambda^{(0)}(t) = \lambda(t) & = -p(t)^{-s} \\
\lambda^{(1)}(t) = \lambda'(t) & = sp(t)^{-s-1}p'(t) \\
p(t) &= \left(1-\frac{t}{T}\right)\frac{t}{T}.
\end{align}
$$
Proof. By induction.
$$
\Lambda_{s,T}
\left\{
\begin{align}
(k=0)\qquad & \Lambda_{s,T}^{(1)}(t) 
= \left(
\begin{array}{c}
0 \\ 0
\end{array}
\right)
\Lambda_{s,T}^{(0)}(t)\lambda^{(1)}(t) = e^{-p(t)^{-s}}sp(t)^{-s-1}p'(t) \\
(k \implies k+1) \quad & \left(\Lambda_{s,T}^{(k)}(t)\right)' 
= \left(
\sum\limits_{j=0}^{k-1}
\left(
\begin{array}{c}
k-1 \\ j
\end{array}
\right)\Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+1)}(t)
\right)' \\
& = \sum\limits_{j=0}^{k-1}
\left(
\begin{array}{c}
k-1 \\ j
\end{array}
\right)
\left(
\Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+1)}(t)
\right)' \\
& = \sum\limits_{j=0}^{k-1}
\left(
\begin{array}{c}
k-1 \\ j
\end{array}
\right)
\left(
\underbrace{
\Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t)
}_{(i)}
+ 
\underbrace{
\Lambda_{s,T}^{(k-1-j)}(t)\lambda^{(j+2)}(t)
}_{(ii)}
\right)
\end{align}
\right.
$$
Note that (ii) of the actual and (i) of the successive sum index match such that the overall sum is given as
$$ 
\begin{align}
\left(\Lambda_{s,T}^{(k)}(t)\right)' &= \Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) + \left(1 + \left(\begin{array}{c}k-1\\1\end{array}\right)\right)\Lambda_{s,T}^{(k-j-1)}(t)\lambda^{(j+2)}(t) \\
& + 
\left(
\left(
\begin{array}{c}
k-1 \\ 1
\end{array}
\right)
+
\left(
\begin{array}{c}
k-1 \\ 2
\end{array}
\right)
\right)
\Lambda_{s,T}^{(k-j-2)}(t)\lambda^{(j+3)}(t) + ... + \Lambda_{s,T}^{(0)}(t)\lambda^{(k+1)}(t),
\end{align}
$$
where the coefficients evaluate to
$$ 
\begin{align}
\left(
\begin{array}{c}
k-1 \\ j
\end{array}
\right)
+
\left(
\begin{array}{c}
k-1 \\ j+1
\end{array}
\right)
& = \frac{(k-1)!}{j!(k-1-j)!} + \frac{(k-1)!}{(j+1)!(k-2-j)!} \\
& = \frac{(k-1)!((j+1)+(k-1-j))}{j!(k-2-j)!(j+1)(k-1-j)} \\
& = \frac{k!}{(j+1)!(k-1-j)!} \\
& = 
\left(
\begin{array}{c}
k \\ j+1
\end{array}
\right).
\end{align}
$$
This proves the hypothesis
$$ \left(\Lambda_{s,T}^{(k)}(t)\right)' = \sum\limits_{j=0}^{k}
\left(
\begin{array}{c}
k \\ j
\end{array}
\right)\Lambda_{s,T}^{(k-j)}(t)\lambda^{(j+1)}(t) 
= \Lambda_{s,T}^{(k+1)}(t).
$$
Similarly, we have for $\lambda$ that (arguments omitted)
$$
\lambda
\left\{
\begin{align}
(j=1) \quad & \lambda^{(2)} = p^{-1}
\left(
\lambda^{(1)}p'(-s-1)-s\lambda^{(0)}p''
\right) \\
(j\implies j+1) \quad & \left(\lambda^{(j)}\right)'
= \left( p^{-1} \left( \lambda^{(j-1)}p'(-s-j+1) - \lambda^{(j-2)}p''\sum\limits_{i=0}^{j-2}(s+i) \right)\right)' \\
& = -p^{-2}p' \left( \lambda^{(j-1)}p'(-s-j+1) - \lambda^{(j-2)}p''\sum\limits_{i=0}^{j-2}(s+i) \right) \\
& + p^{-1} \left( (-s-j+1)\left(\lambda^{(j-1)}p'\right)' - \sum\limits_{i=0}^{j-2}(s+i) \left(\lambda^{(j-2)}p''\right)' \right) \\
& = p^{-1}\left(p'\lambda^{(j)}(-1+(-s-j+1)) \\
- \lambda^{(j-1)}p''\left( \sum\limits_{i=0}^{j-2}(s+i)+(s+j-1) \right)
\right) \\
& = p^{-1} \left( \lambda^{(j)}p'(-s-j) - \lambda^{(j-1)}p''\sum\limits_{i=0}^{j-1}(s+i) \right) \\
& = \lambda^{(j+1)}.
\end{align}
\right.
$$
$\Box$
