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Everything is logarithms

305 points by E-Reverance

by xelxebar

3 subcomments

The baseless log here is just a torsor [0]!
Lots of things are torsors: position, currency values, calendar dates etc. the vales themselves are arbitrary, and translating/scaling them by some value doesn't make a functional difference. Torsors let us talk about these things without needing to make such an arbitrary choice a priori.
In the case of baseless logs, the underlying set is "information units", i.e. log 2 is bits, log e is nats, log 10 is digits, etc. The conversion factors give us the torsor's group, and picking a privileged unit is just a trivialization of the torsor.
The vector division notation is, similarly, encoding a g-torsor in precisely the same way as length units are.
The examples so far are all torsors with abelian groups, but specifying position both requires choosing an origin and a length unit. The group of this torsor is a suitable semidirect product between translation and scaling, which gives a non-abelian group.
Most of the time we just implicitly choose a trivialization, which often causes confusion because it identifies objects with operations on them, e.g. conflating vectors as positions with vectors as translations. The author's treatise on problems with geometric algebra [1] even brings up this point!
[0]:https://math.ucr.edu/home/baez/torsors.html
[1]:https://alexkritchevsky.com/2024/02/28/geometric-algebra.htm...

by helterskelter

4 subcomments

Logs are awesome. I started a math textbook from the 1920's a while ago, and all the calculations relied on tabulated logs, where you would convert the number to a log in a table to reduce the operation's degree, then convert back to the ordinary representation. This would reduce operations like finding cubed roots to division, would could be converted to log-log to be further reduced to subtraction before you would restore to ordinary notation. It feels like you're using a magic wormhole or something when you're doing this stuff by hand, it's really neat.

by badlibrarian

3 subcomments

This essay needs a type system. Every time it says “log” it should say: log of what, into what?
It’s like audio where people say "dB" as if it answers the next question. Relative to what, measured how, and weighted for whom?
Author should brush up on https://en.wikipedia.org/wiki/Lie_theory

by orc00

1 subcomments

Charles Petzold's The Lost Art of Logarithms is a great read (still a work in progress).
https://www.lostartoflogarithms.com/

by aesthesia

1 subcomments

I think what's going on with the complex logarithm is basically the same as the logarithm that outputs the set of all possible bases for a vector space. The complex logarithm produces a Z-torsor, and the basis logarithm produces a GL(V)-torsor. There's probably some way to represent a choice of branch cut as a part of the choice of the base of the complex logarithm, and similarly the choice of a specific basis as part of the choice of base of the vector space base logarithm.

by amavect

2 subcomments

>You might ask: if we have a baseless logarithm log(N), do we also have a “baseless exponential”?

Sure we can, with some naive algebra. If we can take log(x,base) and drop the base, then we can also take pow(base,x) and drop the base. Since bits=log(2), then pow(bits)=2. You can probably connect it to the reverse of things, like integrals.

Also, for fun, I'll play with some notation tricks.

  log(freq) = pitch
  freq = pow(pitch)
  octave = log(2)

  400*Hz = 100*Hz*4  // the frequency 400 Hz equals 4 times 100 Hz
  log(400*Hz) = log(100*Hz) + log(4)
  log(400*Hz) = log(100*Hz) + 2*log(2)
  log(400*Hz) = log(100*Hz) + 2*octave
  log(400*Hz) = log(100*Hz) + 2*octave  // the pitch of 400 Hz equals 2 octaves above the pitch of 100 Hz

  cent = log(2)/1200
  A4 = log(440*Hz)
  B4 = A4 + 200*cent  // the pitch B4 equals 200 cents above A4
  B4 = log(440*Hz) + 200*log(2)/1200
  B4 = log(440*Hz) + log(2^(2/12))
  B4 = log(440*Hz * 2^(2/12))
  pow(B4) = 493.883 Hz  // the frequency of B4 equals 493.883 Hz

I like the intuition that baseless logarithm notation gives, and it also avoids needing to choose a specific reference point. I can also directly calculate by choosing an arbitrary base:

  pow(log(440*Hz) + 200*log(2)/1200)
  exp(ln(440) + 200*ln(2)/1200)

by GL26

0 subcomment

The same idea comes up in physics. In quantum physics, the action S appears as the logarithm-like quantity behind the amplitude e^iS/(h^bar). In statistical mechanics, entropy is the logarithm of the number of possible microstates Omega : S = log(Omega). Although the concepts come from different parts of physics, they both reflect the same principle: using a log as a way to turn multiplicative relationships into additive ones.

by adrian_b

1 subcomments

The term "baseless logarithm" is really nonsensical and using it would be a great mistake.
Nonetheless, where the author of TFA is correct is that logarithms are a single physical quantity, like length, area or volume, and that choosing the so called "base" is choosing the unit of measurement for logarithms.
Logarithms are included in the dimensional formulae of many derived physical quantities, e.g. for describing the attenuation or amplification of waves during their propagation, where one uses quantities like logarithm per length and logarithm per time.
Changing the "base" of logarithms modifies the numeric values of all derived physical quantities exactly in the same manner as changing any other fundamental unit of measurement, like the unit of length or the unit of time.
Like for any physical quantity, the complete value of a logarithm is independent of the unit of measurement, because it is the product between the numeric value and the unit of measurement. When the unit of measurement is changed, both the numeric value and the unit are changed and the product stays the same (i.e. the logarithm corresponds to the same ratio, regardless what base is used to compute a numeric value for the logarithm).
Nowadays, the unit of logarithms is normally chosen between the octave (binary logarithms), neper (hyperbolic logarithms) or bel (decimal logarithms).
The units of measurement for logarithms are not the bases, but the logarithms of the bases, which is why e.g. the value of the number "e", the base of the hyperbolic logarithms, is never needed in any computation. The only values that are needed are "ln 2" or its inverse "log2 e", which are used to convert the numeric values of logarithms when the unit of measurement is changed between those corresponding to binary logarithms and to hyperbolic logarithms (a.k.a. natural logarithms, but there is nothing more "natural" about hyperbolic logarithms than about any other kind of logarithms).

by anArbitraryOne

0 subcomment

I can't believe he called normal logarithms 'based'

by kfse

2 subcomments

All this would be way more interesting if it actually helped to demonstrate a novel mathematical fact. Right now it's more like notational play.

by amelius

1 subcomments

Does this answer the question of why we see hyperoperations until exponentiation in physics, but not higher?

by myzek

1 subcomments

Wasn't there some scientific paper recently that proved that every operation can be represented as a logarithm? Like, the same as every logic gate can be derived from NAND gates

by math_dandy

0 subcomment

tl;dr: Being a homomorphism from a multiplicative structure into an additive structure isn't enough to grant it the logarithm title.
Although logarithms are certainly ubiquitous in mathematics, I don't think that the mappings that the article's author identifies as logarithms are appropriately viewed as such.
I can't endorse viewing dimension as a logarithm. It appears superficially logarithm-like because we typically (and somewhat unfortunately) write the direct sum of n copies of a vector space V as V^n rather than nV. Writing nV, we simply get the dimension identity dim(nV) = n dim(V). Writing nV instead of V^n also conveniently frees up V^n for the tensor product of n copies of V, with corresponding dimension identity dim(V^n) = dim(V)^n. So I don't think there's any "multiplicative-to-additive" business going on here at all.
Also, I don't think it's advisable to view the p-adic valuation ord_p as a logarithm, even though it's a homomorphisms from the multiplicative group of the rational or p-adic field into the additive group of the rational field. In fact, in many number theoretic contexts, the ratio log_p/ord_p is of particular interest.
I think a good rule of thumb for viewing a mapping as some kind of logarithm is that it has to have some relation with the Taylor expansion of log(1 + x) around x=0. Being a homomorphism from a multiplicative structure into an additive structure isn't enough to get the logarithm title.

by jongjong

2 subcomments

That's a lot of ways to think about logarithms.
Logarithms are laughably simple once you've fully internalized the meaning of the log function; it simply answers the question:
"To what power must I raise the base to get the argument?"
This is why the output tapers out as you increase the argument; because even if you increase the argument exponentially, you only need a fixed increment in the power to reach that number... So if you increase the argument only by a fixed amount (linearly) instead of exponentially, then it makes sense that the output will grow sub-linearly.
I remember when I was doing algebra with logs many years ago at school, I was applying rules to remove the log from one side of the equation.
Then when I got to uni, I had to revise the rules but it was kind of silly of me because those rules can be trivially derived if you just think about what the log function means. Turns out I had been solving equations with logs throughout school without understanding what they even meant... It's only at university that I actually bothered to learn them.
Actually TBH. I didn't even fully understand powers for some time even though I was doing calculus with them at school. I only fully understood powers once I properly internalized the concept of k-ary trees as a proxy.
It's one thing to be able to apply something, another to understand it. And I think to innovate with something, as a tool, it's not enough to be able to apply it. You must understand it.

by saulpw

0 subcomment

This sentiment (not necessarily the content) is what I'm striving to communicate with Mag World[0] (website and podcast so far).
[0] magworld.pw

by psychoslave

0 subcomment

IIRC, Knuth use lg for logarithm base 2.

by monkamonme

0 subcomment

[flagged]

by yaccb3

1 subcomments

Look, the whole thing actually makes sense and the core idea is pretty cool because it's true that a lot of stuff in math looks identical. But in my opinion this is way too much of a macro-level overgeneralization and you risk throwing everything into the same pot, which ends up diluting the actual point of things.I mean, if you take a hammer and a meat mallet, at the end of the day they're both chunks of metal used to hit stuff, but if you bunch them together without making any distinction, you lose track of why you use one to drive nails into a wall and the other to prep cutlets.Saying everything is just one big logarithm is a nice mental exercise, but I feel like it flattens out the differences too much and makes you lose the practical utility of the individual math tools, which are meant to solve completely different problems.

by SadErn

0 subcomment

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