Represent non-integer values on the factorial base

Question

I want to compute the representation of the following values using the factorial number system:

• $\pi$
• $e$
• $\phi$

I know how to do it for integer values, but is it feasible for non-integer values?

In particular, is it feasible for irrational values, or even transcendental values?

If yes, can you please suggest an appropriate method?

Thanks

Answer 1

My other answer was answering a very different question.

Looking at Wikipedia's description of fractional values in a factorial number system, it says that $e$ can be represented as $10_F1111111\ldots$, i.e. as $1\times2! +0\times 1! +1\times\frac1{2!}+1\times\frac1{3!}+1\times\frac1{4!}+1\times\frac1{5!} + \cdots$. Note that Wikipedia omits any digits which must be zero so no multiples of $0!$, $\frac1{0!}$ or $\frac1{1!}$ are shown and $71_{10}$ would be represented by $2321_F$.

On a similar basis $\pi$ can be represented by $11_F0031565\ldots$ i.e. as $1\times2! +1\times 1! +0\times\frac1{2!}+0\times\frac1{3!}+3\times\frac1{4!}+1\times\frac1{5!} +5\times\frac1{6!}+6\times\frac1{7!}+5\times\frac1{8!} + \cdots$

The way to calculate this is to have a remainder $r_{n-1}$ after finding the digit which multiplies $\frac{1}{(n-1)!}$, take $d_n=\lfloor nr_{n-1}\rfloor$ and $r_n = nr_{n-1} -d_n$. So with $\pi$, we have $r_1=\pi-3 \approx 0.1415927$ and the calculation of the fractional digits $d_n$ looks like

n   d_n     r_n
1           0.1415927
2   0       0.2831853
3   0       0.8495559
4   3       0.3982237
5   1       0.9911184
6   5       0.9467106
7   6       0.6269741
8   5       0.0157927

On a similar basis $\phi$ can be represented by $1_F1024067\ldots$

Answer 2

You should read the Wikipedia article on the Gamma function defined by $$\Gamma(x) = \int_0^\infty y^{x-1} e^{-x}\,dy$$ which for non-negative integers has the property $$n! = \Gamma(n+1) $$ and in general $$\Gamma(x+1)=x \Gamma(x).$$

You have

• $\Gamma(\pi +1) \approx 7.188$, between $6$ and $24$,
• $\Gamma(e +1) \approx 4.261$, between $2$ and $6$, and
• $\Gamma(\phi +1) \approx 1.449$, between $1$ and $2$,

much as you might expect.

In MSExcel, you can do the first calculation with =EXP(GAMMALN(PI()+1))

Answer 3

This is an old question but I recently came across a nice result by Cantor that fits nicely into posting of the OP.

The first result is probably well known in analytic number theory, and it is based on the humble Eclidean algorithm:

Theorem A: Let $(a_n:n\in\mathbb{N})$ be a sequence of positive integer numbers larger than $1$. Then, any real number $\alpha\in R$ can be uniquely expressed in the form $$\begin{align} \alpha=c_0+\sum^\infty_{n=1}\frac{c_n}{a_1\cdot\ldots\cdot a_n}\tag{0}\label{a-basis} \end{align}$$ where $(c_n:n\in\mathbb{Z}_+)\subset\mathbb{Z}$, satisfy
(i). $c_0\in\mathbb{Z}$, $0\leq c_n<a_n$ for $n\geq1$, and
(ii). $\#\{n\in\mathbb{N}: c_n<a_n-1\}=\infty$.

Comment: the factorial case, which is of interest to the OP, corresponds to the sequence $a_n=n+1$, $n\in\mathbb{N}$.

Proof of Theorem A: In the process of proving the theorem above we also show how the numbers $c_n$ are constructed.
Start with $$ c_0=\lfloor \alpha \rfloor,\qquad\alpha_1=\alpha- c_0=\{\alpha\}$$ For $n\geq1$ define $$ c_n:=\lfloor a_n\alpha_n\rfloor,\qquad \alpha_{n+1}=a_n\alpha_n - c_n=\{a_n\alpha_n\}$$ Recall that the fractional part function $\{x\}:=x-\lfloor x\rfloor$ satisfies $0\leq \{x\}<1$. Thus, $0\leq \alpha_n<1$ for all $n\in\mathbb{N}$. As a consequence, $0\leq a_n\alpha_n<a_n $ and so, $c_n=\lfloor a_n\alpha_n\rfloor$ satisfies $0\leq c_n<a_n$ as stated by the Theorem.
Now we show that \eqref{a-basis} holds. By induction $$\begin{align} \alpha&=c_0+\alpha_1\\ &=c_0+\frac{c_1}{a_1}+\frac{\alpha_2}{a_1}\\ &=c_0+\frac{c_1}{a_1}+\frac{c_2}{a_1a_2}+\frac{\alpha_3}{a_1a_2}\\ &=\ldots\\ &=c_0+\sum^n_{k=1}\frac{c_k}{a_1\cdot\ldots\cdot a_k} + \frac{\alpha_{n+1}}{a_1\cdot\ldots\cdot a_n} \end{align}$$ Consequently, setting $x_n=c_0+\sum^n_{k=1}\frac{c_k}{a_1\cdot\ldots\cdot a_k}$, we get that $$|\alpha-x_n|=\frac{\alpha_{n+1}}{a_1\cdot\ldots\cdot a_n}\leq \frac{1}{2^n}\xrightarrow{n\rightarrow\infty}0$$

Now we turn to the issue of uniqueness. A simple inspection of a telescopic series yields $$\begin{align} \sum^{\infty}_{k=1}\frac{a_{n+k}-1}{a_{n+1}\cdot\ldots\cdot a_{n+k}} = \sum^{\infty}_{k=1}\Big(\frac{1}{a_{n+1}\cdot\ldots\cdot a_{n+k-1}}-\frac{1}{a_{n+1}\cdot\ldots\cdot a_{n+k}}\Big)=1\tag{1}\label{one} \end{align}$$ for any $n\in\mathbb{Z}_+$. Suppose $$\alpha=b_0 + \sum^\infty_{n=0}\frac{b_n}{a_1\cdot\ldots\cdot a_n}$$ where $(b_n:n\in\mathbb{Z}_+)$ satisfy (i) and (ii) of the Theorem. Identity \eqref{one} implies that $$0\leq \alpha-b_0=\sum^\infty_{n=0}\frac{b_n}{a_1\cdot\ldots\cdot a_n}<\sum^\infty_{n=0}\frac{a_n-1}{a_1\cdot\ldots\cdot a_n}=1$$ Hence $b_0=\lfloor \alpha\rfloor = c_0$. If the sequence $b\neq c$, let $n$ be the smallest integer (which must be at least $1$) for which $c_n\neq b_n$. Without loss of generality, we may assume that $c_n>b_n$. It follows that $$\sum^\infty_{k=n}\frac{c_k}{a_1\cdot\ldots\cdot a_k}=\sum^\infty_{k=n}\frac{b_k}{a_1\cdot\ldots\cdot a_k}<\infty$$ Since $0\leq b_n<a_n-1$ infinitely often, and $0\leq c_n\leq a_n-1$ for all $n\in\mathbb{N}$, we have that $$\begin{align} \frac{1}{a_1\cdot\ldots\cdot a_n}&\leq \frac{c_n-b_n}{a_1\cdot\ldots\cdot a_n}= \sum^\infty_{k=n+1}\frac{b_k-c_k}{a_1\cdot\ldots\cdot a_k}\\ &< \sum^\infty_{k=n+1}\frac{a_k-1}{a_1\cdot\ldots\cdot a_k}\\ &=\frac{1}{a_1\cdot\ldots a_n}\sum^\infty_{k=1}\frac{a_{n+k}-1}{a_{n+1}\cdot\ldots\cdot a_{n+k}}=\frac{1}{a_1\cdot\ldots a_n} \end{align}$$ which is a contradiction. Hence $c_n=b_n$ for all $n\in\mathbb{Z}_+$.

Comment: a similar procedure can be use to show that any $c_0\in\mathbb{Z}_+$ can be written uniquely as $$c_0=d_n(a_n\cdot\ldots\cdot a_1)+\ldots + d_1a_1+d_0$$ where $0<d_n\leq a_{n+1}-1$, and $0\leq d_k\leq a_{k+1}-1$ for $0\leq k<n$.

The next result by G. Cantor gives a sufficient condition for irrationality of $\alpha$ in the $a$-base.

Theorem (Cantor): Let $a$ be a sequence as in the previous theorem. If $\alpha\in\mathbb{R}$ is such that the corresponding $c_n$ in the conclusion of the previous theorem satisfy $c_n>0$ infinitely often, and each prime number $p$ divides infinitely many of the numbers in $a_n$, then $\alpha$ is irrational.

Example: Perhaps the simplest example for Cantor's theorem is $e=2+\sum^\infty_{n=1}\frac{1}{(n+1)!}$.

Proof of Cantor's theorem: Suppose $\alpha=\frac{p}{q}$ where $p,q$ are integers, $q>0$ and $g.c.d(p, q)=1$. There is $N$ large enough such that $q$ divides $a_1\cdot\ldots\cdot a_N$. From $$\begin{align} \alpha -c_0&=\sum^N_{n=1}\frac{c_n}{a_1\cdot\ldots\cdot a_n}+\sum^\infty_{n=N+1}\frac{c_n}{a_1\cdot\ldots\cdot a_n}\\ &= \sum^N_{n=1}\frac{c_n}{a_1\cdot\ldots\cdot a_n} +\frac{1}{a_1\cdot\ldots a_N}\sum^\infty_{n=1}\frac{c_{n+N}}{a_{N+1}\cdot\ldots\cdot a_{n+N}}, \end{align} $$ it follows that $$\begin{align} 0<y=a_1\cdot\ldots\cdot a_N\Big(\frac{p-c_0q}{q} - \sum^N_{n=1}\frac{c_n}{a_1\cdot\ldots\cdot a_n} \Big)&=\sum^\infty_{n=1}\frac{c_{N+n}}{a_{N+1}\cdot\ldots\cdot a_{N+n}}\\ &<\sum^\infty_{n=1}\frac{a_{N+n}-1}{a_{N+1}\cdot\ldots\cdot a_{N+n}}=1 \end{align}$$ On the other hand, $y\in\mathbb{N}$ which is a contradiction. Therefore $\alpha\in\mathbb{R}\setminus\mathbb{Q}$.

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Reference:

1. Cantor G., Gesammelte Abhandlungen Mathematischen und Philosophischen, Zermelo, E. editor, Springer-Verlag, Berlin 1932, pp. 35-42