# The digit at position $n$ of the number $x$ in base $m$

As a solution to this question, we can define a function $$f_b(x, n)$$ which finds the digit in the $$n$$th position of $$x$$ in base $$b$$.

$$f_b(x, n) = \left\lfloor \frac{x}{b^n} \right\rfloor \bmod b$$

It even works for decimals, for example:

e = 2 . 7  1  8  2  8  1  8  2  8 ...
0  -1 -2 -3 -4 -5 -6 -7 -8 -9 th position


$$f_{10}(e, -8) = 2 \quad \checkmark$$

However, when using a non-integer base, all the digits are fractional.

$$f_\pi(e, [0,-1,-2,\dots]) = [2, 1.717, 0.867, 2.319, \dots]$$

For reference, Wolfram Alpha states that $$e$$ in base $$\pi$$ is actually 2.2021201002111...

Additionally, for negative bases, all the digits of $$f$$ are negative. Wolfram Alpha states that $$e$$ in base $$-10$$ is 3.3223222325590...

Furthermore, the function also screws up if $$x_b$$ is negative. So I am also looking for a function to have a way of differentiating positive and negative numbers (especially important in negative bases, since for example, 21 in decimal would be -39 in negadecimal).

My question is if there is a function which does this but is also valid even for non-integer and negative bases. I'm sure there would be, but I don't know what we need to modify such that this would happen. In other words, is there a function such that

$$f_b\left(\sum_{k=-\infty}^\infty a_k b^k, n \right) = a_n \qquad b\in\mathbb{R}$$

?

I have since then found a recursive method for non-integer but not negative bases :

Start by calculating $$A=\lfloor\log_b n\rfloor$$, then

$$U_1 = f_b(n, A) = \left\lfloor\frac{n}{b^A}\right\rfloor$$ $$U_2 = f_b(n, A-1) = \left\lfloor\frac{n-U_1b^A}{b^{A-1}}\right\rfloor$$ $$U_3 = f_b(n, A-2) = \left\lfloor\frac{n-U_1b^A-U_2b^{A-1}}{b^{A-2}}\right\rfloor$$

etc. The simple problem with this one is when it deals with a negative base, it returns negative digits.

• I think you are mistaken about the "digits" for such irregular bases. The digits are always nonnegative integers. For example, the statement "$e$ in base $\pi$ is $2.20212\ldots$" means $$e = 2 + 2\pi^{-1}+ 0\pi^{-2}+2\pi^{-3}+1\pi^{-4}+2\pi^{-5}+\cdots$$ and the statement "$e$ in base $-10$ is $3.32232\cdots$" means $$e = 3 + 3(-10)^{-1} + 2(-10)^{-2} + 2(-10)^{-3} + 3(-10)^{-4} + 2(-10)^{-5}+\cdots$$ – MPW Feb 27 '19 at 20:48
• @MPW How would one write 15 in base -10? The easiest way is to write -25, right? I know, particularly in irrational bases, that there are loads of different ways to write the same number, but I don't know for a negative base whether for every number there exists a positive number representation... – Infiaria Feb 27 '19 at 20:52

Hint: for starters, try to extract $$a$$, $$b$$, $$c$$ from $$\ a \pi^2 + b\pi +c \$$ or, more generally, the digits from $$a_1/\pi + a_2/\pi^2 + a_3/\pi^3 + a_4/\pi^4 + a_5/\pi^5 + \cdots$$ The traditional mod function is not enough, because $$a\pi^2 + b\pi +c \pmod {\pi} = a \pi^2 +c \neq c$$ since $$a\pi^2$$ isn't an (integer!) multiple of $$\pi$$.
Edit: Instead of $$\!\bmod{\!\! }$$, the function $$\bmod{\!\! _b}$$ that yields $$\bmod_b \,\! : \ (A_k b^k + \cdots + A_2 b^2 + A_1 b + A_0) \mapsto A_0$$ for any nonnegative integers $$A_j \leq |b|$$, seems to do the trick. But I have no definite formula for now. $$% \mod\!\!\!_b: \ (A_k b^k + \cdots + A_2 b^2 + A_1 b + A_0) \mapsto A_0$$
• Yes, that's what the function would intend to do but for any base $b$ ($a_0 + a_1b + a_2b^2 + \cdots + a_{-1}b^{-1} + a_{-2}b^{-2} + \cdots$), not just $\pi$. Of course I can look at something like $20122.1_\pi$ and say, that's $2\pi^4 + 1\pi^2$ etc. I just want a function that does this for any number $x$. – Infiaria Feb 27 '19 at 21:38
• Of course I used $\pi$ as an example only. I think it would be enough to use, (instead of mod function) the function $\!\mod\!\!\!_b$ that yields $$\mod\!\!\!_b: \ (A_k b^k + \cdots + A_2 b^2 + A_1 b + A_0) \mapsto A_0$$ where $A_j$ are integers. However, I don't know for now how to define it as a formula. – eudes Feb 27 '19 at 22:03