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I'm interested in this function

$$ h(m,n) = \operatorname{sgn}\Bigl(\lim_{w \to \infty}{\sum_{x_1=1}^{w}\dotsi\sum_{x_u=1}^{w}\dfrac{1}{1 + wp^2(m,n,x_1,\dotsc,x_u)}}\Bigr) $$

where $p$ is a polynomial in $u+2$ variables and integer coefficients. Can we simplify it to $$ h(m,n) = \operatorname{sgn}\Bigl(\dfrac{f(m,n)}{g(m,n)}\Bigr) $$

without the limit?

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  • $\begingroup$ What’s $p$? Is $u$ given? Why the parenthesis closes there? $\endgroup$
    – Alessandro Della Corte
    Commented Sep 10, 2022 at 6:28
  • $\begingroup$ @AlessandroDellaCorte $p$ is a polynomial with $u$ unknown and a specific degree $\endgroup$
    – raoof
    Commented Sep 10, 2022 at 6:33
  • $\begingroup$ So $m,n,w,u,x_1,\dots,x_u$ are all non-negative integers and $\epsilon$ a real number, aren't they? And is $p$ a polynomial with real coefficients? $\endgroup$
    – Pietro Majer
    Commented Sep 10, 2022 at 7:41
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    $\begingroup$ You are asking for the sign of a sum of positive values: and you want to know if at the limit it is zero or 1, is that correct? Basically you are asking if dominated conevergence applies? Well, if the sums are finite for $\epsilon=1$, since the term being summed is monotonous in $\epsilon$, the answer is yes, it is zero, thank you Lebesgue. $\endgroup$
    – username
    Commented Sep 10, 2022 at 8:03
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    $\begingroup$ By editing the question to remove the mention of $\epsilon$, you have made the existing answer invalid. $\endgroup$
    – JRN
    Commented Oct 30, 2022 at 8:10

1 Answer 1

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Assuming you mean $\epsilon$ is a real positive number and $p$ is a real-valued function in $2+u$ variables, you want a limit of a sum of non-negative numbers, $$\lim_{\epsilon\to0}\sum_{x\in\mathbb N_+^u}\frac{\epsilon^2}{\epsilon^2+p^2(n,m,x)}.$$ For $\epsilon\searrow 0$, each coefficient converges decreasing either to $1$ or $0$, according whether $p(n,m,x)=0$ or not. Therefore, if the sum is finite for at least one $\epsilon$, by Beppo Levi the limit is just the number of non-negative integer zeros $x\in\mathbb N_+^u$ of $p(n,m,\cdot\}$; and if $\text{sgn}$ is the usual signum function, the final expression is $1$ or $0$ according whether $p(n,m,\cdot)$ has a zero $x\in\mathbb N_+^u$ or not.

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  • $\begingroup$ So everything reduces to checking if the sum is finite for at least one $\epsilon>0$ (hence for all, since each term is concave w.r.to $\epsilon$). If e.g. $p$ is a constant, the sum is infinite for all $\epsilon>0$. $\endgroup$
    – Pietro Majer
    Commented Sep 10, 2022 at 8:37
  • $\begingroup$ $p$ is a integer valued function. what I have in mind is whether it can be simplified to $f(m,n)/g(m,n)$ without the limit so my function could have some undefined values $\endgroup$
    – raoof
    Commented Sep 10, 2022 at 8:53
  • $\begingroup$ @PietroMajer You say Beppo-Levi, I say Lebesgue : Potayto/Potahto... : ) $\endgroup$
    – username
    Commented Sep 10, 2022 at 11:31
  • $\begingroup$ I really mean beppi levi (monotone convergence; no need of finiteness if it was increasing); though lebesgue would also do (dominated convergence) $\endgroup$
    – Pietro Majer
    Commented Sep 10, 2022 at 12:00
  • $\begingroup$ The question has been changed; $\epsilon$ is now no longer mentioned. Your answer has become invalid. $\endgroup$
    – JRN
    Commented Oct 30, 2022 at 8:07

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