Coherent Sequence is Partial Sum of P-adic Expansion/Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $p$ be a prime number.

Let $\sequence {\alpha_n}$ be a coherent sequence.

For all $n \in \N$, let $\sequence {b_{j, n} }_{0 \le j \le n}$ be a sequence such that:

$(1) \quad \displaystyle \alpha_n = \sum_{j \mathop = 0}^n b_{j,n} p^j$
$(2) \quad \forall j: 0 \le j \le n: 0 \le b_{j, n} < p$


Then:

$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n b_{i, i} p^i$


Proof

The theorem is proved by induction:


Basis for the Induction

$n = 0$

By definition: $\alpha_0 = b_{0, 0}$

This is the basis for the induction.


Induction Hypothesis

This is our induction hypothesis:

$\alpha_k = \displaystyle \sum_{i \mathop = 0}^k b_{i, i} p^i$

Now we need to show true for $n = k + 1$:

$\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$


Induction Step

This is our induction step.

From Difference of Consecutive terms of Coherent Sequence there exists $c_{k + 1} \in \Z$:

$(a) \quad 0 \le c_{k + 1} < p$
$(b) \quad \alpha_{k + 1} = c_{k + 1} p^{k + 1} + \alpha_k$

From the induction hypothesis:

$\alpha_{k + 1} = c_{k + 1} p^{k + 1} + \sum_{j \mathop = 0}^k b_{j, j} p^j$

From the Zero Padded Basis Representation it follows that the two sequences:

$\sequence {b_{j, k + 1} }_{0 \le j \le k + 1}$

and

$\sequence {b_{0, 0}, b_{1, 1}, \dots, b_{k - 1, k - 1}, b_{k, k}, c_{k + 1}}$

are equal.

Then:

$\forall j \in \closedint 0 {k + 1} : b_{j, k + 1} = b_{j, j}$

It follows that:

$\alpha_{k + 1} = \displaystyle \sum_{i \mathop = 0}^{k + 1} b_{i, i} p^i$


By induction:

$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n b_{i, i} p^i$

$\blacksquare$


Sources