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Let $0<t_0<1$ fixed number , $ n_0$ integer $ \geq 2$ fixed and let $\forall 0<u<1, f(u)= \displaystyle \frac{(1-u)^{n_0} \log(1-u)}{(1-ut_0)^{n_0+1}} $.

Let $0<u_0<1 $ be given. I'm looking a good upper bound (which is an explicit function in $u_0,t_0, n_0$) to $ |f^{(n_0)}(u_0)|$ ($n_0$-th derivative).

I suppose that by Cauchy formula, and choosing the good radius one can have a good upper bound, but I don't know how to do it. Any help?

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    $\begingroup$ $$|f^{(n_0)}(u_0)|=(1-{u_0})^{{n_0}-1} (1-{t_0} {u_0})^{-{n_0}-2}$$ $$\left| \left(\log ^2(1-{u_0}) ({n_0} ({t_0}-1)-{t_0} {u_0}+{t_0})+\log (1-{u_0}) (({n_0} (-{t_0})+{n_0}+{t_0} ({u_0}-1)) \log (1-{t_0} {u_0})+{t_0} {u_0}+{t_0}-2)+(1-{t_0} {u_0}) \log (1-{t_0} {u_0})\right)\right|$$ --- not just an upper bound, but an exact evaluation as an explicit function of $u_0$, $t_0$, $n_0$ (this seems to satisfy what you are asking, correct me if I'm wrong). $\endgroup$
    – Carlo Beenakker
    Commented Jun 6, 2019 at 13:41
  • $\begingroup$ i'll check this formula, thanks for your help $\endgroup$
    – mamiladi
    Commented Jun 9, 2019 at 21:57

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