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I am trying to understand the proof in [1] for Theorem 2.3. In particular, there is a line with the following inequality

$\sum_{j=0}^p \frac{|t|^j}{j!} ||N||^j \le C_1(1 + |t|)^p$

with $C_1$ and $p$ somehow a function of $N$ has 1's on its superdiagonal, i.e.,

$$N = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 0 & 0\\\ 0 & 0 & 1 & \cdots & 0 & 0 & 0\\\\ \vdots & \vdots & \ddots & \cdots & \vdots & \vdots & \vdots \\\\ 0 & 0 & 0 & \cdots & 0 & 1 & 0 \\\\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\\\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{pmatrix} $$

I have no idea where this inequality comes from. I found this discussion but it seems to have the same logical jump from a partial sum of an exponential to an inequality given some constant. Is there some general bound result that I am missing? Thanks!

[1] Sideris, Thomas C. Ordinary differential equations and dynamical systems. Vol. 2. Paris: Atlantis Press, 2013., link

esquire70
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    If you have two expressions $a_0+a_1|t|+...+a_p|t|^p$ and $b_0+b_1|t|+...+b_p|t|^p$, both with positive coefficients, then their quotient is bounded in both directions. – Lutz Lehmann Sep 24 '22 at 12:15
  • Thanks! Just to clarify, in the terminology @LutzLehmann gave, $a_j = ||N||^j/j!$ while $b_j=0$ for $j=0,\ldots,p-1$ while $b_p = C_1$ – esquire70 Sep 24 '22 at 12:45
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    No, the $b_j$ are the coefficient of the binomial power, $b_j=\binom{p}{j}$. $C_1$ is an upper bound of the quotient. – Lutz Lehmann Sep 24 '22 at 13:21

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