Mathematics Asked on November 1, 2021
So I conjectured a formula which was proven:
Let $b_r = sum_{d mid r} a_dmu(frac{m}{d})$. 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 $$lim_{k to infty} lim_{n to infty} sum_{r=1}^n a_rfleft(frac{kr}{n}right)frac{k}{n} = left(lim_{s to 1} frac{1}{zeta(s)}sum_{r=1}^infty frac{a_r}{r^s}right)int_0^infty f(x)dx.$$
My question is what is the cardinality of the set of $a_r$?
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 $
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 $sum_{r=1}^n a_rfleft(tfrac{k}{n}rright)tfrac{k}{n}$. But strip $S_{n,k}$ depends on $n$ and $k$, and a value $fleft(tfrac kn rright)$, 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 $(tfrac{k}{n}r, fleft(tfrac kn rright))$ 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 $r$.
Answered by Alex Ravsky on November 1, 2021
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