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Is that will work as analytic continuation for mandelbrot function, if not why?

Mandelbrot function can be defined with Catalan numbers as $\sum_{n=1}^{\infty}\frac {(2n-2)!}{(n-1)!(n)!}x^n$

From (Infinite series as integral representation) we know that $\sum_{k=x}^{\infty} f(k)=-2\pi \int_{-\infty}^{\infty}\frac {F (x-1/2+it)}{(e^{\pi t}+e^{-\pi t})^2}dt$, which works for divergent series (and this is antiderivative without +C). So it should technically work.

I'm not saying that this is possible to give exact value of this integral, but at least we could count it by numerical methods

PS: If that will work, pls post image of this function in complex plane

Wreior
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  • What does it mean to analytically continue a set? – FShrike Feb 23 '23 at 18:20
  • You right, terminologii of mandelbrot set is actualy use to call the limits of convergence. I will change it to mandelbrot function – Wreior Feb 23 '23 at 18:22
  • It shouldn't "technically" work until we have a technical proof of your conjectural equation – FShrike Feb 23 '23 at 18:45
  • It is proven in link. I mean, I hope it is sufficient proof. But I tried this in some divergent series and that works (but you have to be of choosing x which do not cross singularity) – Wreior Feb 23 '23 at 18:55

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The following will work as an analytical continuation. Take $$\sum_{n=1}^\infty (-1)^n \frac{x^n(2n-2)!}{(n-1)!n!} $$ Then, rewriting as a contour integral it becomes $$-\int_C \frac{\csc(\pi n)}{2i} x^n \frac{(2n-2)!}{(n-1)!n!} dn$$ Where $n!$ is of course the gamma function, but I've written it as the factorial to emphasize the correspondence between this integral and the original series. A linear contour from $1/2 < c < 1$ will work, in particular, we can compute the series as $$-\int_{3/4 - i \infty}^{3/4 + i \infty} \frac{\csc(\pi n)}{2i} x^n \frac{(2n-2)!}{(n-1)!n!} dn$$ For instance, here is what the function looks like along the real line, with the original function in blue and the continuation in orange. I suspect this integral doesn't converge for $x$ negative, so some more work might be required to get a continuation there (this may suggest there is a branch cut there by the way).
enter image description here

Looks like I was correct about that branch cut, the series evaluates to simply $-\frac{1}{2}\sqrt{1-4x}+\frac{1}{2}$. Which is of course the solution to $f(x) = f(x)^2 + x$ (i.e. the mandelbrot iteration).

Caleb Briggs
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  • Actualy my theorem is stronger than yours. If you take my integral and use integration by parts with changeing limits of integration, it will be the same. But I will get extra limits which correct my form to do not get divergent. Actualy you proved that I am right, but in different way so I will acept it anyway – Wreior Feb 24 '23 at 07:52
  • @Wreior Would you mind posting your approach to this problem? I think there is a branch cut on this function, so I'm interested in how your theorem deals with that – Caleb Briggs Feb 24 '23 at 08:26
  • Oh. By theorem my integral whould be convergent, but you would have to count it analyticly. Damn. I will think about it maybe there is a way – Wreior Feb 24 '23 at 10:52
  • Wait just a minute. Mandelbrot function is just $-\frac{1}{2}\sqrt{1-4x}+\frac{1}{2}$? But it's set of convergence is fractal? – Wreior Feb 24 '23 at 15:24
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    @Wreior The convergence set of the Mandelbrot function is not fractal. It looks like this question has been asked before, see here; https://math.stackexchange.com/questions/3925497/proving-that-the-limit-of-the-mandelbrot-polynomials-co-efficients-are-the-cata – Caleb Briggs Feb 24 '23 at 18:30