On pages 363 to 365 in “Numerical Analysis” by Gautschi (2012) is a derivation of the Padé approximant of the exponential function.
I am stuck on a step in the beginning of the differentiation below (I typed this verbatim from the book; the relevant statement is at the bottom):
Let $v(t) = t^n (1-t)^m$.
By Leibniz's rule, one finds:
\begin{align} v^{(r)}(t) &= \sum^{r}_{k=0} \binom{r}{k} [t^n]^{(k)} [(1-t)^m]^{(r-k)} \\ &= \sum^r_{k=0} \binom{r}{k}\binom{n}{k} k! t^{n-k} \binom{m}{r-k}(r-k)!(1-t)^{m-r+k}(-1)^{r-k}; \end{align}
hence, in particular,
\begin{align} v^{(r)}(0) &= (-1)^{r-n} \binom{m}{r-n}r! \quad \text{if} \quad r\geq n; \quad v^{(r)}(0)=0 \quad \text{if} \quad r<n;\\ v^{(r)}(1) &= (-1)^m \binom{n}{r-m} r! \quad \text{if} \quad r\geq m; \quad v^{(r)}(1) = 0 \quad \text{if} \quad r < m. \end{align}
Given any integer $q\geq 0$, repeated integration by parts yields
\begin{equation} \int^1_0 \mathrm{e}^{tz} v(t) \mathrm{d}t = \mathrm{e}^z \sum^q_{r=0} (-1)^r \frac{v^{(r)}(1)}{z^{r+1}} - \sum^q_{r=0} (-1)^r \frac{v^{(r)}(0)}{z^{r+1}} + \frac{(-1)^{q+1}}{z^{q+1}} \int^1_0 \mathrm{e}^{tz} v^{(q+1)}(t)\mathrm{d}t. \end{equation}
The following statement I struggle with:
Putting here $q = n+m$, so that $v^{(q+1)}(t) \equiv 0$, ...
So if I understand the thought behind this correctly, then the integral on the RHS is supposed to vanish by choosing $q$ appropriately. But I don't see how this is supposed to happen? The $\equiv$ sign I find extra confusing. Does the author intend to define it to vanish? That doesn't quite make sense to me, so it's probably not it.
What's going on here?