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Given $f(x)=A \cdot x^5+B \cdot x^8$ with:

  • $A \in \mathbb{R}^-$
  • $B \in \mathbb{R}^-$
  • $h(z) = w_0 + \sum_{n=1}^\infty h_n \cdot \frac{\left(z-f(w_0)\right)^n}{n!}$
  • $w_0=1$

and: $$h_n=\lim_{w \rightarrow w_0} \frac{d^{n-1}}{dw^{n-1}} \left[\left(\frac{w-w_0}{f(w) - f(w_0)}\right)^n\right]$$ i.e. $$h_n =\lim_{w \rightarrow 1} \frac{d^{n-1}}{dw^{n-1}} \left[\left(\frac{w-1}{A \cdot (w^5-1)+B \cdot (w^8-1)}\right)^n\right]$$

Does $h(z)$ converge? What can be said of $h(-C)^2$ with $C \in \mathbb{R}^+$?

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    $\begingroup$ Your question is answered here mathoverflow.net/questions/418392/… $\endgroup$
    – Nemo
    Commented Apr 11 at 20:58
  • $\begingroup$ $R^-$ is the negative reals? $\endgroup$
    – Gerry Myerson
    Commented Apr 11 at 22:05
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    $\begingroup$ @GerryMyerson yes, updated to use \mathbb{R} instead. $\endgroup$
    – Arthur
    Commented Apr 15 at 8:40
  • $\begingroup$ @MartinNicholson where? I do not see anything relating to the convergence $\endgroup$
    – Arthur
    Commented Apr 16 at 17:11

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