LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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The Mercator series, its value at $1$ and the product law determine $\log$ on all positive reals, while the series alone is only local

Statement

There is exactly one function $f : (0, \infty) \to R$ such that

$f (x y) = f (x) + f (y) (x, y > 0)$

and

$f (1 + u) = \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{u^{n}}{n} (- 1 < u \leq 1) .$

That function is the natural logarithm. The series condition itself is local; the product law is the continuation rule.

Facts & Assumptions

Given: A function $f$ satisfying the two displayed conditions.

[L1]

For $- 1 < u \leq 1$ , $\log (1 + u) = \sum_{n = 1}^{\infty} (- 1)^{n + 1} u^{n} / n$ ([The power series for log(1+x) on (-1,1], including the Abel endpoint](/item/thm-log-one-plus-x-power-series)).

[L2]

The natural logarithm satisfies $\log (x y) = \log x + \log y$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L3]

For every real $r$ , there is a natural $n \geq 1$ with $r < n$ (Every complete ordered field is Archimedean).

[F1]

Natural powers satisfy $x^{0} = 1$ and $x^{n + 1} = x^{n} x$ (Integer powers $a^{m}$ ).

[L4]

The induction principle proves a property for every natural once the base and successor steps are established (The principle of mathematical induction).

Proof

technique · direct

1.1

The published natural logarithm satisfies the local series condition by [L1] and the product law by [L2], so an extension exists.

L1L2

1.2

From [F1], induction gives $2^{n} \geq n + 1$ for every natural $n$ : equality holds at $n = 0$ , and $2^{n + 1} = 2 \cdot 2^{n} \geq 2 (n + 1) \geq n + 2$ .

F1L4algebra

1.3

Repeated use of the product law, justified by induction, gives $f (2^{k}) = k f (2)$ , and the series condition at $u = 1$ determines $f (2)$ .

F1L4given

2.1

Given $x > 0$ , use [L3] to choose $k \geq 1$ with $x < k$ . Then $2^{k} \geq k + 1 > k > x$ by step 1.2. Put $y = x / 2^{k}$ and $u = y - 1$ ; thus $0 < y < 1$ and $- 1 < u < 0$ .

L3step 1.2algebra

3.1

Since $x = 2^{k} y = 2^{k} (1 + u)$ , the product law and the local series condition force $f (x) = k f (2) + f (1 + u) = k \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} + \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{u^{n}}{n} .$

step 2.1step 1.3given

4.1

Formula 3.1 forces the value of any extension at every $x > 0$ , so at most one extension exists. Together with step 1.1, that unique extension is $\log$ .

step 3.1step 1.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 111 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Henry Ricardo, The Equivalence of Definitions of the Natural Logarithm Function (standard reference, not scraped)

The Mercator series, its value at 11 and the product law determine log⁡log on all positive reals, while the series alone is only local