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I'm looking for $d^*$, the smallest value of $d$ such that the following holds: $$\frac{\left(\sum_{i=1}^d i^{-p}\right)^2}{\sum_{i=1}^d i^{-2p}}\ge \frac{1}{2}\frac{\zeta(p)^2}{\zeta(2p)}$$

with $\zeta$ referring to Riemann Zeta function.

Empirically this seems to scale as the function below, why?

$$d^*\approx c_1 \exp\left(\frac{c_2}{p-1}\right)$$

enter image description here Notebook

Edit

Mathematica user found a way to express $c_1$ and $c_2$ in terms of Euler gamma constant and $\sqrt{2}$

enter image description here

Motivation: $\frac{\zeta(p)^2}{\zeta(2p)}$ gives "effective rank" of infinite-dimensional linear regression problem with eigenvalues $1^{-p},2^{-p},\ldots$ (ie, how many observations are needed for prediction). This question gives $d$ such that finite-dimensional truncation after $d$ terms has similar effective rank.

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    $\begingroup$ You may try to use $\sum_{i = 1}^d {i^{ - s} } \sim \zeta (s) - \frac{{d{}^{1 - s}}}{{s - 1}}$ for large $d$. $\endgroup$
    – Gary
    May 24, 2023 at 1:13
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    $\begingroup$ I spent hours with this interesting problem without any viable and simple answer. Basically, I repeated your calculations (just over a wider range). $\endgroup$ May 24, 2023 at 8:24
  • $\begingroup$ @Gary thanks, I see same behavior using this approximation. The problem becomes to determine why the following is linear, and what are values of slope and intercept $$\log \left(2 \zeta \left(2 \left(1+\frac{1}{r}\right)\right) \left(\zeta \left(1+\frac{1}{r}\right)-r \left(e^x\right)^{-1/r}\right)^2\right)-\log \left(\zeta \left(1+\frac{1}{r}\right)^2 \left(\zeta \left(2 \left(1+\frac{1}{r}\right)\right)-\frac{\left(e^x\right)^{1-2 \left(\frac{1}{r}+1\right)}}{2 \left(\frac{1}{r}+1\right)-1}\right)\right)$$ $\endgroup$ May 24, 2023 at 18:27

1 Answer 1

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ùToo long for comments.*

The problem is to find, for a given value of $p$, the zero of function $$f(d)=2 \zeta (2 p) \left(H_d^{(p)}\right){}^2-\zeta (p)^2 H_d^{(2 p)}$$

Making the problem continuous, let $d=e^x$ and the function is quite nice (except that it shows a minimum value for small values of $x$.

Consider instead $$g(x)=\log \left(2 \zeta (2 p) \left(H_{e^x}^{(p)}\right){}^2\right)-\log \left(\zeta (p)^2 H_{e^x}^{(2 p)}\right)$$ which is very smooth and Newton method converges quitae fast.

Trying for $p=1.1$ and the ridculous $x_0=1$, the iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.00000 \\ 1 & 4.90872 \\ 2 & 8.71146 \\ 3 & 11.1061 \\ 4 & 11.6858 \\ 5 & 11.7114 \\ 6 & 11.7115 \\ \end{array} \right)$$

Repeating you calculations for $1.001 \leq p \leq 1.200$, the fit you proposed is almost perfect; what I obtained for the empirical model $$x=-a + \frac b {p-1}$$ is $$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & 0.56830 & 5.95\times 10^{-4} & \{0.56712,0.56947\} \\ b & 1.22797 & 6.57\times 10^{-6} & \{1.22796,1.22799\} \\ \end{array}$$

I have not been able to find any way to conclude using asymptotic analysis.

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  • $\begingroup$ Interesting! BTW, using Gary's approximation for harmonic number the results are identical in float64 precision for $p<1.05$, and it gives the same constants. Follow-up question here $\endgroup$ May 24, 2023 at 18:25
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    $\begingroup$ BTW, the following seems to give a perfect fit for $p<1.1$ -- $$x=\frac{\log \left(\sqrt{2}+2\right)}{p-1}-\gamma$$ $\endgroup$ May 24, 2023 at 20:11

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