Summation of General Logarithms

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Theorem

Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.

Let $\ds \prod_{\map R i} a_i$ denote a product over $R$.


Let the fiber of truth of $R$ be finite.

Then:

$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$


Proof

The proof proceeds by induction.

First let:

$S := \set {a_i: \map R i}$

We have that $S$ is finite.

Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered.

Let $S$ have $m$ elements, identified as:

$S = \set {s_1, s_2, \ldots, s_m}$

For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $\map P n$ be the proposition:

$\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$


$\map P 0$ is the case:

\(\ds \map {\log_b} {\prod_{i \mathop = 1}^0 s_i}\) \(=\) \(\ds \log_b 1\) Definition of Vacuous Product
\(\ds \) \(=\) \(\ds 0\) Logarithm of 1 is 0
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \log_b s_i\) Definition of Vacuous Summation


Basis for the Induction

$\map P 1$ is the case:

\(\ds \map {\log_b} {\prod_{i \mathop = 1}^1 s_i}\) \(=\) \(\ds \log_b s_1\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^1 \log_b s_i\) Definition of Summation

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \map {\log_b} {\prod_{i \mathop = 1}^k s_i} = \sum_{i \mathop = 1}^k \log_b s_i$


from which it is to be shown that:

$\ds \map {\log_b} {\prod_{i \mathop = 1}^{k + 1} s_i} = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$


Induction Step

This is the induction step:


\(\ds \map {\log_b} {\prod_{i \mathop = 1}^{k + 1} s_i}\) \(=\) \(\ds \map {\log_b} {\paren {\prod_{i \mathop = 1}^k s_i} \times s_{k + 1} }\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \map {\log_b} {\prod_{i \mathop = 1}^k s_i} + \log_b s_{k + 1}\) Sum of General Logarithms
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k \log_b s_i + \log_b s_{k + 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{k + 1} \log_b s_i\) Definition of Summation

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\ds \map {\log_b} {\prod_{i \mathop = 1}^n s_i} = \sum_{i \mathop = 1}^n \log_b s_i$


Hence, by definition of $S$:

$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$

$\blacksquare$


Sources

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