# Infinite solutions to a function.

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.

• 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. – Peter Foreman Apr 21 at 22:09
• That's why I said to ignore constants, which is what I think you're referring to but I'm not sure. – John Miller Apr 21 at 22:11
• 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 – Henry Apr 21 at 22:11
• @Henry I meant to put $\infty$ not $k$ sorry, and the function $f(x)$ should be continuous everywhere. – John Miller Apr 21 at 22:12
• 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. – 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.

• 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. – John Miller Apr 21 at 22:32
• Oh sorry I forgot to add the square root. – John Miller Apr 21 at 22:33
• 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. – John Miller Apr 21 at 22:35
• But you added the solution $i$ to the function am I not incorrect? – John Miller Apr 21 at 22:45
• Oh I should probably change the question again then. – John Miller Apr 21 at 22:48