Properties of the logarithm

With a solid in­ter­pre­ta­tion of log­a­r­ithms un­der our belt, we are now in a po­si­tion to look at the ba­sic prop­er­ties of the log­a­r­ithm and un­der­stand what they are say­ing. The defin­ing char­ac­ter­is­tic of log­a­r­ithm func­tions is that they are real val­ued func­tions $$f$$ such that

Prop­erty 1: Mul­ti­ply­ing in­puts adds out­puts.

$$f(x \cdot y) = f(x) + f(y) \tag{1}$$
for all $$x, y \in$$ $$\mathbb R^+$$. This says that when­ever the in­put grows (or shrinks) by a fac­tor of $$y$$, the out­put goes up (or down) by only a fixed amount, which de­pends on $$y$$. In fact, equa­tion (1) alone tells us quite a bit about the be­hav­ior of $$f,$$ and from it, we can al­most guaran­tee that $$f$$ is a log­a­r­ithm func­tion. First, let’s see how far we can get us­ing equa­tion (1) all by it­self:

Prop­erty 2: 1 is mapped to 0.

$$f(1) = 0. \tag{2}$$

This says that the amount the out­put changes if the in­put grows by a fac­tor of 1 is zero — i.e., the out­put does not change if the in­put changes by a fac­tor of 1. This is ob­vi­ous, as “the in­put changed by a fac­tor of 1” means “the in­put did not change.”

Ex­er­cise: Prove (2) from (1).

$$f(x) = f(x \cdot 1) = f(x) + f(1),\text{ so }f(1) = 0.$$

Prop­erty 3: Re­cip­ro­cat­ing the in­put negates the out­put.

$$f(x) = -f\left(\frac{1}{x}\right). \tag{3}$$

This says that the way that grow­ing the in­put by a fac­tor of $$x$$ changes the out­put is ex­actly the op­po­site from the way that shrink­ing the in­put by a fac­tor of $$x$$ changes the out­put. In terms of the “com­mu­ni­ca­tion cost” in­ter­pre­ta­tion, if dou­bling (or tripling, or $$n$$-times-ing) the pos­si­bil­ity space in­creases costs by $$c$$, then halv­ing (or third­ing, or $$n$$-parts-ing) the space de­creases costs by $$c.$$

Ex­er­cise: Prove (3) from (2) and (1).

$$x \cdot \frac{1}{x} = 1,\text{ so }f(1) = f\left(x \cdot \frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right).$$

$$f(1)=0,\text{ so }f(x)\text{ and }f\left(\frac{1}{x}\right)\text{ must be opposites.}$$
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Prop­erty 4: Di­vid­ing in­puts sub­tracts out­puts.

$$f\left(\frac{x}{y}\right) = f(x) - f(y). \tag{4}$$

This fol­lows im­me­di­ately from (1) and (3).

Ex­er­cise: Give an in­ter­pre­ta­tion of (4).

There are at least two good in­ter­pre­ta­tions:

1. $$f\left(x \cdot \frac{1}{y}\right) = f(x) - f(y),$$ i.e., shrink­ing the in­put by a fac­tor of $$y$$ is the op­po­site of grow­ing the in­put by a fac­tor of $$y.$$

2. $$f\left(z \cdot \frac{x}{y}\right) = f(z) + f(x) - f(y),$$ i.e., grow­ing the in­put by a fac­tor of $$\frac{x}{y}$$ af­fects the out­put just like grow­ing the in­put by $$x$$ and then shrink­ing it by $$y.$$

Try trans­lat­ing these into the com­mu­ni­ca­tion cost in­ter­pre­ta­tion if it is not clear why they’re true. <div><div>

Prop­erty 5: Ex­po­nen­ti­at­ing the in­put mul­ti­plies the out­put.

$$f\left(x^n\right) = n \cdot f(x). \tag{5}$$

This says that mul­ti­ply­ing the in­put by $$x$$, $$n$$ times in­curs $$n$$ iden­ti­cal changes to the out­put. In terms of the com­mu­ni­ca­tion cost metaphor, this is say­ing that you can em­u­late an $$x^n$$ digit us­ing $$n$$ differ­ent $$x$$-digits.

Ex­er­cise: Prove (5).

This is easy to prove when $$n \in \mathbb N:$$
$$f\left(x^n\right) = f(\underbrace{x \cdot x \cdot \ldots x}_{n\text{ times}}) = \underbrace{f(x) + f(x) + \ldots f(x)}_{n\text{ times}} = n \cdot f(x).$$

For $$n \in \mathbb Q,$$ this is a bit more difficult; we leave it as an ex­er­cise to the reader. Hint: Use the proof of (6) be­low, for $$n \in \mathbb N,$$ to boot­strap up to the case where $$n \in \mathbb Q.$$

For $$n \in \mathbb R,$$ this is ac­tu­ally not prov­able from (1) alone; we need an ad­di­tional as­sump­tion (such as con­ti­nu­ity) on $$f$$. <div><div>

Prop­erty 5 is ac­tu­ally false, in full gen­er­al­ity — it’s pos­si­ble to cre­ate a func­tion $$f$$ that obeys (1), and obeys (5) for $$n \in \mathbb Q,$$ but which ex­hibits patholog­i­cal be­hav­ior on ir­ra­tional num­bers. For more on this, see patholog­i­cal near-log­a­r­ithms.

This is the first place that prop­erty (1) fails us: 5 is true for $$n \in \mathbb Q,$$ but if we want to guaran­tee that it’s true for $$n \in \mathbb R,$$ we need $$f$$ to be con­tin­u­ous, i.e. we need to en­sure that if $$f$$ fol­lows 5 on the ra­tio­nals it’s not al­lowed to do any­thing in­sane on ir­ra­tional num­bers only.

Prop­erty 6: Root­ing the in­put di­vides the out­put.

$$f(\sqrt[n]{x}) = \frac{f(x)}{n}. \tag{6}$$

This says that, to change the out­put one $$n$$th as much as you would if you mul­ti­plied the in­put by $$x$$, mul­ti­ply the in­put by the $$n$$th root of $$x$$. See Frac­tional digits for a phys­i­cal in­ter­pre­ta­tion of this fact.

Ex­er­cise: Prove (6).

$$(\sqrt[n]{x})^n = x,\text{ so }f\left((\sqrt[n]{x})^n\right)\text{ has to equal }f(x).$$

$$f\left((\sqrt[n]{x})^n\right) = n \cdot f(\sqrt[n]{x}),\text{ so }f(\sqrt[n]{x}) = \frac{f(x)}{n}.$$
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As with (5), (6) is always true if $$n \in \mathbb Q,$$ but not nec­es­sar­ily always true if $$n \in \mathbb R.$$ To prove (6) in full gen­er­al­ity, we ad­di­tion­ally re­quire that $$f$$ be con­tin­u­ous.

Prop­erty 7: The func­tion is ei­ther triv­ial, or sends some in­put to 1.

$$\text{Either f sends all inputs to 0, or there exists a b \neq 1 such that f(b)=1.}\tag{7}$$

This says that ei­ther $$f$$ is very bor­ing (and does noth­ing re­gard­less of its in­puts), or there is some par­tic­u­lar fac­tor $$b$$ such that when the in­put changes by a fac­tor of $$b$$, the out­put changes by ex­actly $$1$$. In the com­mu­ni­ca­tion cost in­ter­pre­ta­tion, this says that if you’re mea­sur­ing com­mu­ni­ca­tion costs, you’ve got to pick some unit (such as $$b$$-digits) with which to mea­sure.

Ex­er­cise: Prove (7).

Sup­pose $$f$$ does not send all in­puts to $$0$$, and let $$x$$ be an in­put that $$f$$ sends to some $$y \neq 0.$$ Then $$f(\sqrt[y]{x}) = \frac{f(x)}{y} = 1.$$%%note: You may be won­der­ing, “what if $$y$$ is nega­tive, or a frac­tion?” If so, see Strange roots. Short ver­sion: $$\sqrt[-3/4]{x}$$ is perfectly well-defined.%%

$$b$$ is $$\sqrt[y]{x}.$$ We know that $$b \neq 1$$ be­cause $$f(b) = 1$$ whereas, by (2), $$f(1) = 0$$. <div><div>

Prop­erty 8: If the func­tion is con­tin­u­ous, it is ei­ther triv­ial or a log­a­r­ithm.

$$\text{If f(b)=1 then } f(b^x) = x. \tag{8}$$

This prop­erty fol­lows im­me­di­ately from (5). Thus, (8) is always true if $$x$$ is a ra­tio­nal, and if $$f$$ is con­tin­u­ous then it’s also true when $$x$$ is ir­ra­tional.

Prop­erty (8) states that if $$f$$ is non-triv­ial, then it in­verts ex­po­nen­tials with base $$b.$$ In other words, $$f$$ counts the num­ber of $$b$$-fac­tors in $$x$$. In other words, $$f$$ counts how many times you need to mul­ti­ply $$1$$ by $$b$$ to get $$x$$. In other words, $$f = \log_b$$!

Many texts take (8) to be the defin­ing char­ac­ter­is­tic of the log­a­r­ithm. As we just demon­strated, one can also define log­a­r­ithms by (1) as con­tin­u­ous non-triv­ial func­tions whose out­puts grow by a con­stant (that de­pends on $$y$$) when­ever their in­puts grow by a fac­tor of $$y$$. All other prop­er­ties of the log­a­r­ithm fol­low from that.

If you want to re­move the “con­tin­u­ous” qual­ifier, you’re still fine as long as you stick to ra­tio­nal in­puts. If you want to re­move the “non-triv­ial” qual­ifier, you can in­ter­pret the func­tion $$z$$ that sends ev­ery­thing to zero as $$\log_\infty$$. Allow­ing $$\log_\infty$$ and re­strict­ing our­selves to ra­tio­nal in­puts, ev­ery func­tion $$f$$ that satis­fies equa­tion (1) is iso­mor­phic to a log­a­r­ithm func­tion.

In other words, if you find a func­tion whose out­put changes by a con­stant (that de­pends on $$y$$) when­ever its in­put grows by a fac­tor of $$y$$, there is ba­si­cally only one way it can be­have. Fur­ther­more, that func­tion only has one de­gree of free­dom — the choice of $$b$$ such that $$f(b)=1.$$ As we will see next, even that de­gree of free­dom is rather paltry: All log­a­r­ithm func­tions be­have in es­sen­tially the same way. As such, if we find any $$f$$ such that $$f(x \cdot y) = f(x) + f(y)$$ (or any phys­i­cal pro­cess well-mod­eled by such an $$f$$), then we im­me­di­ately know quite a bit about how $$f$$ be­haves.

Parents:

• I’m hav­ing trou­ble pars­ing in­ter­pre­ta­tion #1 -- which part is sup­posed to map onto the right hand side of equa­tion (4)?

• May need to build the in­tu­ition that know­ing how f(x) be­haves tells me how f(c*x) is differ­ent from f(c).

(You’re us­ing the lan­guage of “grow­ing the in­put,” but I just see a static in­put called x.)

• (8) doesn’t fol­low from (5). The as­sump­tion in (5) was than $$n$$ ranged over nat­u­rals, not re­als. In fact, (1) only im­plies (8) if you also re­quire the func­tion to be con­tin­u­ous.

(1) es­sen­tially says $$f$$ is a ho­mo­mor­phism from $$(\mathbb{R}^{>0},\cdot)$$ to $$(\mathbb{R},+)$$. To gen­er­ate a func­tion satis­fy­ing (1) but not (8), we need only com­pose $$log$$ (choose a base) with an au­to­mor­phism in the ad­di­tive group and show that the com­po­si­tion is not a mul­ti­ple of a log­a­r­ithm. We can get such an au­to­mor­phism by con­sid­er­ing $$\mathbb{R}$$ as an in­finite di­men­sional vec­tor space over the ra­tio­nals and, for ex­am­ple, swap­ping two di­men­sions.

• (5) was in­tended to as­sume that $$n \in \mathbb R^{\ge 1},$$ or pos­si­bly $$\in \mathbb R^{\ge 0}$$ if you want an easy way to prove (6). In that case, how does (8) not fol­low from (5)? (If $$f(x^y)=yf(x)$$ in gen­eral, then $$f(b^n)=nf(b)$$ and $$f(b)=1 \implies f(b^n)=n,$$ un­less I’m miss­ing some­thing.)

• The proof of (5) only goes through for $$n\in\mathbb{N}$$.

You can prove a ver­sion of (8) from (5), namely, $$f(b)=1\Rightarrow f(b^q)=q$$ for $$q\in\mathbb{Q}$$, but this doesn’t pin down $$f$$ com­pletely, un­less you in­clude a con­ti­nu­ity con­di­tion.

• tl;dr: I did some read­ing on re­lated top­ics, and it turns out that (1) may be suffi­cient to define log­a­r­ithms if we take as an ax­iom that ev­ery set is Lebesgue mea­surable (which is in­com­pat­i­ble with the ax­iom of choice). Other­wise, we need to add an ad­di­tional con­di­tion to (1).

(1) states that $$f(x\cdot y)=f(x)+f(y)$$. Given a func­tion $$g$$ satis­fy­ing this con­di­tion, we can gen­er­ate an ad­di­tional func­tion satis­fy­ing this con­di­tion by com­pos­ing $$g$$ with a func­tion $$h$$, where $$h(x+y)=h(x)+h(y)$$:

$$h(g(x\cdot y))=h(g(x))+h(g(y))$$

$$h$$, as defined, is a solu­tion to Cauchy’s func­tional equa­tion. The fam­ily of func­tions given by $$h(x)=ch(x)$$ for some con­stant $$c$$ is always a solu­tion, giv­ing the usual log­a­r­ithm fam­ily. The ex­is­tence of other solu­tions is in­de­pen­dent of ZF. When they do ex­ist they are always patholog­i­cal and gen­er­ate non-Lebesgue mea­surable sets (for more, see this stack­ex­change link).

We can prove the ex­is­tence of such solu­tions in ZFC by not­ing that the solu­tions of the Cauchy func­tional equa­tion are ex­actly the ho­mo­mor­phisms from the ad­di­tive group of $$\mathbb{R}$$ to it­self. The real num­bers form an in­finite di­men­sional vec­tor space over the field $$\mathbb{Q}$$. Lin­ear trans­for­ma­tions from the vec­tor space to it­self trans­late into ho­mo­mor­phisms from the group to it­self. Since the ax­iom of choice im­plies that any vec­tor space has a ba­sis, we can, for ex­am­ple, find a non-triv­ial lin­ear trans­for­ma­tion by swap­ping two ba­sis vec­tors. This in turn in­duces a ho­mo­mor­phism from the group to it­self. (The Wikipe­dia page gives the gen­eral form of a solu­tion to this func­tional, which turn out to be all the lin­ear trans­for­ma­tions on the vec­tor space.)

(I’m not say­ing that this ar­ti­cle should dis­cuss ax­iom­a­ti­za­tions of set the­ory, but it doesn’t seem good to make state­ments that are only true if you as­sume, e.g., an un­usual al­ter­na­tive to the ax­iom of choice.)

Wikipe­dia proves that the patholog­i­cal solu­tions must all be dense in $$\mathbb{R}$$, so to ex­clude them, we can adopt any num­ber of con­di­tions. Wikipe­dia points at ”$$f$$ is con­tin­u­ous”, ”$$f$$ is mono­tonic on any in­ter­val”, and ”$$f$$ is bounded on any in­ter­val”. Con­ti­nu­ity seems to be most in­tu­itive; once we have defined the value of the func­tion on the ra­tio­nals (which we can do with ba­si­cally the ar­gu­ments already on this page), the rest of its val­ues are de­ter­mined.

• How are these changes? (start­ing at prop 5, through the end)