2
$\begingroup$

Here http://arxiv.org/pdf/0811.4488v1 in Theorem 1 while prooving uniform convergence of

$\sum_{k=0}^{\infty}\lambda^k{\widetilde{X}}^{(2k)}$

it's said that

$|{\widetilde{X}}^{(2k)}| \leq (\max|ru_0^2|)^k \cdot (\max|\frac{1}{pu_0^2}|)^k \cdot \frac{|b-a|^{2k}}{(2k)!}$

Where did they get factor $\frac{1}{(2k)!}$?

1 Answers 1

0

If we take maximums only of $ru_0^2$ and $\frac{1}{pu_0^2}$ and take integrals correctly, we'll get the answer.