$$n^{2}$$ Discritization

Main

We can write, for positive $$n$$,

$$n^{2}=\exp\left(\int_{2-n}^{1}\frac{2}{2-x}\;dx\right)\\$$

Type I - Geometric Integral

Let

$$\prod_{a}^{b}f(x)^{dx}=\lim_{\Delta x\to 0}\prod_{x_{i}}f(x_{i})^{\Delta x}\\$$

then

$$\prod_{a}^{b}f(x)^{dx}=\exp\left(\int_{a}^{b}\log f(x)\;dx\right)\\$$

then we have

$$n^{2}=\lim_{\Delta x\to 0}\prod_{x_{i}}\exp\left(\frac{2}{2-x_{i}}\right)^{\Delta x}\\$$

let the smallest interval between the $$x_{i}$$ instead be $$1$$, an integer discretization ($$\mathrm{ID}$$ of this. Then

$$\mathrm{ID}(n^{2})=\lim_{\Delta x\to 1}\prod_{x_{i}}\exp\left(\frac{2}{2-x_{i}}\right)^{\Delta x}=\exp\left(2\gamma+2\psi(0,n+1)\right)\\$$

where for integer $$n$$

$$\psi(0,n)=\sum_{k=1}^{n-1}\frac{1}{k}-\gamma\\$$

giving for integer $$n$$

$$\mathrm{ID}(n^{2})=\lim_{\Delta x\to 1}\prod_{i=2-n}^{1}\exp\left(\frac{2}{2-i}\right)^{\Delta x}=\exp\left(\sum_{k=1}^{n}\frac{2}{k}\right),\;n\in\mathbb{N^{+}}\\ \mathrm{ID}(n^{2})=\lim_{\Delta x\to 1}\prod_{i=2-n}^{1}\exp\left(\frac{2}{2-i}\right)^{\Delta x}=\prod_{k=1}^{n}\exp\left(\frac{2}{k}\right),\;n\in\mathbb{N^{+}}\\$$

then

$$\lim_{n\to\infty}\frac{\mathrm{ID}(n^{2})}{n^{2}}=e^{2\gamma}\approx 3.17222\\$$

this then tells us that

$$n^{2}\approx\frac{1}{e^{2\gamma}}\prod_{k=1}^{n}\exp\left(\frac{2}{k}\right)\\$$

but we also find a bias

$$n^{2}\sim\frac{1}{e^{2\gamma}}\prod_{k=1}^{n}\exp\left(\frac{2}{k}\right)-n-\frac{1}{3}\\$$

and this relation holds very well for $$n>10$$. This then provides the approximation

$$\sum_{k=1}^{n}\frac{1}{k}\sim\log\left(\sqrt{n^{2}+n+\frac{1}{3}}\right)+\gamma\\$$

Type II - Volterra Product Integral

Let

$$\prod_{a}^{b}(1+f(x)dx)=\lim_{\Delta x\to 0}\prod_{x_{i}}(1+f(x_{i})\Delta x)\\$$

then

$$\prod_{a}^{b}(1+f(x)dx)=\exp\left(\int_{a}^{b}f(x)\;dx\right)\\$$

then we have

$$n^{2}=\lim_{\Delta x\to 0}\prod_{x_{i}}\left(1+\frac{2\Delta x}{2-x_{i}}\right)\\$$

and

$$\mathrm{ID}(n^{2})=\lim_{\Delta x\to 1}\prod_{i=2-n}^{1}\left(1+\frac{2}{2-i}\right)=\frac{(n+1)(n+2)}{2}\\$$

then

$$\lim_{n\to\infty}\frac{n^{2}}{\mathrm{ID}(n^{2})}=2\\$$

Error Function

We can apply the same analysis to the error function integral

$$\mathrm{erf}(x)=\frac{1}{\sqrt{\pi}}\int_{-x}^{x}\exp\left(-t^{2}\right)\;dt\\$$

using the geometric integral we get

$$\mathrm{ID}(\exp\left(\sqrt{\pi}\mathrm{erf}(x)\right))=\lim_{\Delta t\to 0}\prod_{-x}^{x}\left(\exp(\exp(-t^{2})\right)^{\Delta t}\\$$

we find for integer $$n$$, the right hand side is the function

$$f(n)=\exp\left(1+\sum_{k=1}^{n}\frac{2}{\exp(k^{2})}\right)\\$$

giving

$$\sqrt{\pi}\mathrm{erf}(n)\approx 1+\sum_{k=1}^{n}\frac{2}{\exp(k^{2})}\\$$

but we find

$$\lim_{n\to\infty}\left(\frac{1+\sum_{k=1}^{n}\frac{2}{\exp(k^{2})}}{\sqrt{\pi}\mathrm{erf}(n)}\right)\approx 1+2\exp\left(-\pi\right)^{\pi}\\$$

refining to

$$\sqrt{\pi}\mathrm{erf}(n)\approx\frac{1+\sum_{k=1}^{n}\frac{2}{\exp(k^{2})}}{1+2\exp\left(-\pi\right)^{\pi}}\\$$

Then we have the suggestion that

$$\theta_{3}(0,e^{-1})\sim\sqrt{\pi}(1+2e^{-\pi^{2}})\\$$

where $$\theta_{3}$$ is an elliptic theta function of the third kind. But they differ in the $$17^{th}$$ decimal place.

$$n^{3}$$ Discritization

Similar to the first integral we have

$$n^{3}=\exp\left(\int_{3-2n}^{1}\frac{3}{3-x}\;dx\right)\\$$

we can then write

$$n^{3}=\lim_{\Delta x\to 0}\prod_{x_{i}}\exp\left(\frac{3}{3-x_{i}}\right)^{\Delta x}\\ \mathrm{ID}(n^{3})=\prod_{i=3-2n}^{1}\exp\left(\frac{3}{3-i}\right)=e^{3y(2n-1)},\;n\in\mathbb{N^{+}}\\$$

where $$y(n)$$ is a function defined by a recurrence relation

$$y(0)=0,y(1)=\frac{1}{2n},(n-2n)y(n)+(-1-2n+4n)y(n+1)+(1+n-2n)y(2+n)=0\\$$

. but we can also use

$$n^{3}=\exp\left(\int_{2-n}^{1}\frac{3}{2-x}\;dx\right)\\$$

this gives

$$\mathrm{ID}(n^{3})=\exp\left(3\gamma+3\psi(0,n+1)\right)\\$$

which has the same effect as the first $$n^{2}$$ integral. But we find that

$$\sum_{k=1}^{n}\frac{1}{k}\approx\log\left(\sqrt[3]{n^{3}+\frac{3n^{2}}{2}+\frac{7n}{8}+\frac{3}{16}}\right)+\gamma\\$$

although this seems to be a very similar approximation that that of the squared integral.

Polynomials

We can repeat this for many powers of $$n$$ and we find that we obtain a set of polynomials $$P_{m}(n)$$ for the $$m^{th}$$ power of $$n$$ that the discritization applied to. These are

$$P_{0}(n)=1\\ P_{1}(n)=n+\frac{1}{2}\\ P_{2}(n)=n^{2}+n+\frac{1}{3}\\ P_{3}(n)=n^{3}+\frac{3n^{2}}{2}+\frac{7n}{8}+\frac{3}{16}\\ P_{4}(n)=n^{4}+2n^{3}+\frac{5n^{2}}{3}+\frac{2n}{3}+\frac{4}{45}\\ P_{5}(n)=n^{5}+\frac{5n^{4}}{2}+\frac{65n^{3}}{24}+\frac{25n^{2}}{16}+\frac{523n}{1152}+\frac{115}{2304}\\ P_{6}(n)=n^{6}+3n^{5}+4n^{4}+3n^{3}+\frac{13n^{2}}{10}+\frac{3n}{10}+\frac{3}{70}\\$$

these are the asymptotic forms for

$$\exp\left(\left[\sum_{k=1}^{n}\frac{m}{k}\right]-m\gamma\right)\\$$

in the limit of large $$n$$. It would seem that a general expression for $$P_{m}(n)$$ is

$$P_{m}(n)=n^{m}+\frac{m}{2}n^{m-1}+\frac{1}{24}(1+4m+3m^{2})n^{m-2}+\frac{1}{48}(4m+4m^{2}+m^{3})n^{m-3}+\frac{1}{5760}(-81+168m+290m^{2}+120m^{3}+15m^{4})n^{m-4}\cdots\\$$

we see the denominators of each of these terms taking on the form of the Minkowski numbers, $$1,2,24,48,5760,11520,...$$, A053657

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