I have rewritten this question.

Take a series, $$ S_k = \sum_{n=1}^k u_n $$ where $S_k \in \mathbb{C}$ forms a divergent series as $k \to \infty$. Now take a function $f_n(x)$ such that there are infinite roots of this function wherein each successive root (these are the only roots to the equation) is the next term in the series $S_k$, $$ f_n(x) = (x-S_1)(x-S_2)(x-S_3)\dots . $$ Apart from the trivial cases whereby we just increase the power or change the coefficient such that, $$ f_n(x) = i_0 \cdot (x-S_1)^{i_1}(x-S_2)^{i_2}(x-S_3)^{i_3}\dots $$ are there any other functions that cross these points exclusively.

  • $\begingroup$ There are obviously an infinite number of curves that fit such a description because the points that are not equal to $0$ can take any value. $\endgroup$ – Peter Foreman Apr 21 at 22:09
  • $\begingroup$ That's why I said to ignore constants, which is what I think you're referring to but I'm not sure. $\endgroup$ – John Miller Apr 21 at 22:11
  • $\begingroup$ What you are saying is indeed unclear. Your $S_n$ only seems to be defined for positive integers and takes positive values unbounded above (though perhaps $S_0=0$). It is also unclear whether the second part of the question is related to the first, or whether your functions are supposed to be continuous or series $\endgroup$ – Henry Apr 21 at 22:11
  • $\begingroup$ @Henry I meant to put $\infty$ not $k$ sorry, and the function $f(x)$ should be continuous everywhere. $\endgroup$ – John Miller Apr 21 at 22:12
  • $\begingroup$ I think you mean $S_n = \sum_\limits{k=1}^n \frac{1}{k}$ and then $S_n$ are the partial sums of the Harmonic series. $\endgroup$ – Doug M Apr 21 at 22:17

From your description I think you want to define some sequence $$a_n=\sum_{k=1}^n b_k$$ and then have a function $f(z)$ with roots at every $z=a_n$. We can take $$f_n(z)=(z-a_1)(z-a_2)\dots(z-a_{n-1})(z-a_n)$$ Then the function $f(z)$ can defined by $$f(z)=\lim_{n\to\infty}f_n(z)$$ which gives a continuous definition of $f(z)$ for any sequence $a_n$ which diverges to $\pm\infty$. But if $a_n$ converges to some value $L\in\mathbb{R}$, then such a function $f(z)$ will be discontinuous at $z=L$ as $f(z)$ will have infinite roots in the neighborhood of $z=L$.

Assuming that $a_n$ diverges, since the function $g(z)\cdot f(z)$ where $g(z)\gt0$ for all $z\in\mathbb{C}$ has the same roots as $f(z)$ we can say that $g(z)\cdot f(z)$ is also a valid function that fits your description. For example, if we have $g(z)=e^{kz}$ for some $k\in\mathbb{R}$ then we have an infinite number of valid functions that fit such a description.

  • $\begingroup$ Thank you for clearing up my explanation of the problem. But for the second part, by including $g(x)$ you have added the solution $x = \sqrt{-1}$ which is not part of the series. I was talking about functions whose only solutions are those from the series. I'm going to make another edit. $\endgroup$ – John Miller Apr 21 at 22:32
  • $\begingroup$ Oh sorry I forgot to add the square root. $\endgroup$ – John Miller Apr 21 at 22:33
  • $\begingroup$ I think there are infinite solutions but only if you keep adding equivalent terms, as in you can only change the function $f_n(x)$ by increasing the power of any element, not by adding new terms. $\endgroup$ – John Miller Apr 21 at 22:35
  • $\begingroup$ But you added the solution $i$ to the function am I not incorrect? $\endgroup$ – John Miller Apr 21 at 22:45
  • $\begingroup$ Oh I should probably change the question again then. $\endgroup$ – John Miller Apr 21 at 22:48

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