matheproject >> Every positive real number x has a g-adic representation

Every positive real number x has a g-adic representation

x∈[0,1[ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=qqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgIGiolaacUfacaaIWaGaaiilaiaaigdacaGGBbaaaa@3BCD@

Gaussian function

[X]

( r n (x)) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=qqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadkhadaWgaaWcbaGaamOBaaqabaGccaGGOaGaamiEaiaacMcacaGGPaaaaa@3B38@

r_{1} (x) ≔ x \cdot g \land r_{n + 1} (x) ≔ (r_{n} (x) - [r_{n} (x)]) \cdot g

and subsequently set

a_{n} (x) ≔ [r_{n} (x)] .

It is easily seen that

0 \leq r_{n} (x) < g

, so that

0 \leq [r_{n} (x)] < g

is valid as well. Therefore we have

a_{n} (x) \in D_{g}

When proving the following please note that the number

x \cdot g - a_{1} (x) = r_{1} (x) - [r_{1} (x)]

is always a member of

[0, 1 [

r_{n} (x \cdot g - a_{1} (x)) = r_{n + 1} (x) for all n \in ℕ^{*}

Proof by induction:

► $r_{1} (x \cdot g - a_{1} (x)) = (x \cdot g - [r_{1} (x)]) \cdot g = r_{2} (x)$

$\begin{array}{l} ▶ r_{n + 1} (x \cdot g - a_{1} (x)) & = (r_{n} (x \cdot g - a_{1} (x)) - [r_{n} (x \cdot g - a_{1} (x))]) \cdot g \\ = (r_{n + 1} (x) - [r_{n + 1} (x)]) \cdot g \\ = r_{n + 2} (x) \end{array}$

a_{n} (x \cdot g - a_{1} (x)) = a_{n + 1} (x) for all n \in ℕ^{*}

The proof is immediate from 1.

For all

n \in ℕ^{*}

the inequality

| x - \sum_{i = 1}^{n} \frac{a_{i} (x)}{g^{i}} | \leq \frac{1}{g^{n}} holds for every x \in [0, 1 [

Proof by induction:

► $| x - \frac{a_{1} (x)}{g} | = | x - \frac{[x \cdot g]}{g} | = \frac{1}{g} | x \cdot g - [x \cdot g] | \leq \frac{1}{g}$

► For the inductive step we employ the induction hypothesis with the number $x \cdot g - a_{1} (x)$ !

$\begin{array}{l} | x - \sum_{i = 1}^{n + 1} \frac{a_{i} (x)}{g^{i}} | & = \frac{1}{g} | x \cdot g - a_{1} (x) - \sum_{i = 1}^{n} \frac{a_{i + 1} (x)}{g^{i}} | \\ = \frac{1}{g} | x \cdot g - a_{1} (x) - \sum_{i = 1}^{n} \frac{a_{i} (x \cdot g - a_{1} (x))}{g^{i}} | \\ \leq \frac{1}{g} \cdot \frac{1}{g^{n}} = \frac{1}{g^{n + 1}} \end{array}$

3. proves the equality

x = \sum_{i = 1}^{\infty} \frac{a_{i} (x)}{g^{i}}

. So we only need to overcome the initial restriction for x.
Now let

x \geq 0

by any positive real. With

n ≔ \min {k \in ℕ | x < g^{k}}

(note:

(g^{n})

is unbounded) we have

\frac{x}{g^{n}} \in [0, 1 [

and from that we can set

x_{i} ≔ a_{i} (\frac{x}{g^{n}})

for

i > 0

and in addition

x_{0} ≔ 0

and get:

\frac{x}{g^{n}} = \sum_{i = 0}^{\infty} \frac{x_{i}}{g^{i}}

and finally have:

x = \sum_{i = 0}^{\infty} \frac{x_{i}}{g^{i - n}} = \sum_{i = - n}^{\infty} \frac{x_{i + n}}{g^{i}} = \sum_{i = - n}^{0} \frac{x_{i + n}}{g^{i}} + \sum_{i = 1}^{\infty} \frac{x_{i + n}}{g^{i}} = x_{0} g^{n} + \dots + x_{n} g^{0} + \sum_{i = 1}^{\infty} \frac{x_{i + n}}{g^{i}}