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Question

So I conjectured a formula which was proven:

Let $b_r = \sum_{d \mid r} a_d\mu(\frac{m}{d})$. We prove that if the $b_r$'s are small enough, the result is true.

Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r}{r^s}\right)\int_0^\infty f(x)dx.$$

My question is what is the cardinality of the set of $a_r$?

Reason for confusion

Focusing on the L.H.S

This seems to say for every point on the curve can be mapped to $f(x)$ which in turn can be mapped to a coefficient $a_r$ .

$$ x \to f(x) \to a_r $$

Hence, the set has cardinality $ 2^{\aleph_0} $

Focusing on the R.H.S

This seems to say the number of $a_r$ must must be the same as that of the natural numbers.

Hence, the set has cardinality $ \aleph_0 $

Alex Ravsky
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More Anonymous
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  • I don’t undersant the issue. As I understood, in your question you defined ${a_r}$ first as a sequence of non-negative intergers, and later proposed a relation for “arbitrary” $a_r$. I guess, results of mathworker21’s answer a valid even when ${a_r}$ is a sequence of real numbers such that $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$, where $b_r = \sum_{d \mid r} a_d\mu(\frac{m}{d})$ (I guess, there should be $r$ instead of $m$). – Alex Ravsky Jul 29 '20 at 20:42
  • Each sequence ${a_r}$ has countlably many elements, and there are $\frak c$ many disctinct sequenes ${a_r}$ of natural (or real) numbers. Even if we require that $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ then, since there are $\frak c$ such sequences ${b_r}$ and, I guess that Möbius inversion formula injectively maps the set of ${b_r}$’s into the set of ${a_r}$’s, there should be $\frak c$ many disctinct sequenes ${a_r}$. – Alex Ravsky Jul 29 '20 at 20:42
  • So it seems to me each strip has a coefficient $a_r$. In the limit N to infinity each strip can be mapped to a point. Hence there must be as many coefficients as the real number. – More Anonymous Jul 30 '20 at 05:55
  • @AledRavsky I'm sure there is something wrong with the above argument. I'm just not sure what – More Anonymous Jul 30 '20 at 05:56

1 Answers1

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The indices $r$ of the sequence $\{a_r\}$ are natural numbers. Thus $\{a_r\}$ is a sequence of non-negative integers (or, more general, real) numbers. This sequence is predefined by the condition that $a_r$ is the weight (the number of recounts) of $r$-th strip $S_{n,k}$ in the sum $\sum_{r=1}^n a_rf\left(\tfrac{k}{n}r\right)\tfrac{k}{n}$. But strip $S_{n,k}$ depends on $n$ and $k$, and a value $f\left(\tfrac kn r\right)$, corresponding to $S_{n,k}$, depends on $n$ and $k$ too. So to a coefficient $a_r$ corresponds not one strip, and even not a sequence of strips tending to a segment from $(x,0)$ to $(x,f(x))$ for some $x=x(r)$. To $a_r$ corresponds a two-parametric family $\{S_{n,k}\}$ of stips, where $S_{n,k}$ is a strip from $(\tfrac{k}{n}r,0)$ to $(\tfrac{k}{n}r, f\left(\tfrac kn r\right))$ of width $\tfrac{k}{n}$. Thus I don’t see a natural (and, moreover, bijective) map, mapping points $(x,f(x))$ on the graph of $f$ to some $r$.

Alex Ravsky
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