Examples of transcendental functions giving almost integers

Informally speaking, an “almost integer” is a real number very close to an integer.

There are some known ways to construct such examples in a systematic way. One is through the use of certain algebraic numbers called Pisot numbers. These numbers α have the property that their powers can get arbitrarly close to integers, that is:

limnα[αn]=0

where [.] is the nearest integer function.

A well-known example is given by the golden ratio φ=1+52, whose powers are increasingly close to integers:

φ19=9349.000107…

φ25=167761.00000596…

Another example comes from numbers of the form eπn.

A well-known example is Ramanujan’s constant:

eπ163=262537412640768743.99999999999925007…

There’s another interesting way to generate almost integers by using the numbers e and π. By using the identity

n=eπn2x=x1/2n=eπn2/x.

we can derive the approximate identity

()n1k=0ek2πn1+n2

which provides a way to construct almost integers with increasing precision:

eπ9+e4π9+e9π9+e16π9+e25π9+e36π9+e49π9+e64π9=1.0000000000010504…

24k=1ek2π25=2.000000000000000000000000000000000310793…

48k=1ek2π49=3.000000000000000000000000000000000000000000000000000000000000000000838654…

So, the question is: is there another way to generate almost integers -with or without increasing precision- by using transcendental functions, as in the previous example?

(Note that there’s a trivial way to do this: By taking a convergent series k=1xk and its limit L, the number 1/Lnk=1xk will be an almost integer, namely close to 1, but I’m looking for a an example like identity (*), or for a different, non trivial one). So, I am looking for an example that may be of the form nk=1f(xk), where f(x) is a transcendental function of x, that is able to generate a set of different almost integers (zero excluded).

Answer

1) If a,bR and a<x<b, then if f(x) is continuous:
baNNn=0f(a+baNn)=baf(t)dt+o(1)N.
2) If [x] denotes largest integer x and a is a non-rational real number, then
limn[na]n=a
i.e. every real number can be approximated by rational numbers arbitrarily well.

3) If β1=1/2,
βr+1=11βr/22,
then
sin(π2(r+1))=βr
4) This one is quite involved.
e25π=13164+75π12log2+6logk6.2619875...×10103,
where

k=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{1-(51841-23184\sqrt{5})Y^{12}}},

where Y is a root of

Y^3+5s^{-1}Y^2-sY-1=0,

where s is such

s=\sqrt[3]{\frac{(t-1)^5}{11+6t+6t^2+t^3+t^4}},

where

t=2\sinh\left(\frac{1}{4}\textrm{arcsinh}\left(\frac{9+\sqrt{5}}{2}\right)\right).

Note that t is an algebraic number.

5) Set

p=\sqrt{2+216\cdot 5^{1/4}-96\cdot 5^{3/4}}

and

k=1-\frac{2}{1+t}\textrm{, }t=\frac{\left(\sqrt{2}+\sqrt{p}\right)^2}{2\cdot 2^{3/4}p^{1/4}\sqrt{2+p}}.

Also

l=\left(1+\frac{2^{3/4}p^{1/4}}{\sqrt{2+p}}\right)^2\frac{4+2\sqrt{5}+\sqrt{2}(3+2\cdot 5^{1/4})}{160}.

Then

\frac{\Gamma\left(\frac{1}{4}\right)^2}{\pi^{3/2}}=\frac{4+k^2-6k^4}{4l}-7.01743379...\times 10^{-107}.

6) (Ramanujan) For |x|<1,

\prod_{k=1}^{\infty}\left(1-x^{p_k}\right)^{-1}=1+\sum^{\infty}_{k=1}\frac{x^{p_1+p_2+\ldots+p_k}}{(1-x)(1-x^2)\ldots(1-x^k)},\tag 1

where p_1,p_2,\ldots, denote the primes in ascending order. The above formula (1) is canceled i.e. the Taylor series on both sides of (1) agree only to the first 22 terms.
(see Bruce C. Berndt. "Ramanujan's Notebooks I." Springer-Verlag, New York Inc. (1985) page 130).

7) This one is inspired from a formula of Ramanujan

Let a,b be positive reals with ab=2\pi and \Psi(x) analytic on \textbf{R}. Let also

M\Psi(s)=\int^{\infty}_{0}\Psi(x)x^{s-1}dx,

be the Mellin transform of \Psi. If also

\phi(x)=Re\left(M\Psi(ix)n^{-ix}\right).

Then

a\sum^{\infty}_{k=0}\Psi\left(ne^{ak}\right)=a\left(1/2-\sum^{\infty}_{k=1}\frac{\Psi^{(k)}(0)}{k!}\frac{n^k}{e^{ak}-1}\right)+c+2\sum^{\infty}_{k=1}\phi(bk),\tag 2

where c=\lim_{h\rightarrow 0}\phi(h)=:\phi(0).

Example

For \Psi(x)=e^{-x}, we get

a\sum^{\infty}_{k=0}e^{-ne^{ak}}=a\left(1/2-\sum^{\infty}_{k=1}\frac{(-1)^kn^k}{k!(e^{ak}-1)}\right)-\gamma-\log n+2\sum^{\infty}_{k=1}\phi(bk),\tag 3

where \phi(x)=Re\left(\Gamma(ix)n^{-ix}\right) and c=\phi(0)=-\gamma-\log n.

For n=1 in (3), we get the formula of Ramanujan and a good approximation of \gamma constant (Euler's constant).

If a=1/N then b=2\pi N and we get as N\rightarrow\infty

\gamma=-\frac{1}{N}\sum^{\infty}_{k=0}\exp\left(-e^{k/N}\right)+\frac{1}{N}\left(\frac{1}{2}-\sum^{\infty}_{k=1}\frac{(-1)^k}{k!\left(e^{k/N}-1\right)}\right)+O\left(e^{-\pi^2N}\right)\tag 4

Also for a=\frac{\log A}{N}, then b=\frac{2\pi N}{\log A} and holds

\gamma=-\frac{\log A}{N}\sum^{\infty}_{k=0}\exp\left(-A^{k/N}\right)+\frac{\log A}{N}\left(\frac{1}{2}-\sum^{\infty}_{k=1}\frac{(-1)^k}{k!(A^{k/N}-1)}\right)+O\left(e^{-\pi^2N/\log A}\right).\tag 5

Set now

k_{10}(N)=\left[\frac{N}{\log A}\log\left(\frac{N\pi^2}{\log A}\right)\right]+1

and

k_{20}(N)=\left[\frac{N\pi^2}{C_N\log A}\right]+1\textrm{, }C_N=P_L\left(\frac{A^{1/N}N\pi^2}{e\log A}\right),

where P_L(x) is the product log function i.e. e^{P_L(x)}P_L(x)=x. Then

\frac{\gamma}{\log A}=-\frac{1}{N}\sum^{k_{10}(N)}_{k=0}\exp\left(-A^{k/N}\right)+\frac{1}{N}\left(\frac{1}{2}-\sum^{k_{20}(N)}_{k=1}\frac{(-1)^k}{k!(A^{k/N}-1)}\right)+O\left(e^{-\pi^2N/\log A}\right).\tag 6

8)
\left(e^{\pi\sqrt{163}}-744\right)^{1/3}=640319.99999999999999999999999939031735...

Attribution
Source : Link , Question Author : Community , Answer Author : Nikos Bagis

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