Talk:Logarithm

Sources for "Feynman's algorithm"
This entire sections seems to be based on a single, vague, two paragraph description in Physics Today (here), with the details being fleshed out through original research. Physics Today may a reliable source, but the article in question is obviously meant for a popular audience, and so is extremely lacking in detail, and does not cite any sources itself from which further detail might be found. If someone can find an academic paper, textbook or reference book that describes this algorithm then I'd be convinced that it belongs in Wikipedia, but as it stands I find its inclusion highly dubious at best. The section title might be taken to imply that the method is known as "Feynman's algorithm" in the literature, but can find no evidence that anyone aside from the editor who added this section called it that. If have no doubt that the method is, in fact valid, but this does not alone merit its inclusion since much of the detail seems to be the result of original research. --RDBury (talk) 03:11, 15 October 2020 (UTC)
 * It seems that this is the first attempt (or one of the first attempts) for computing logarithms on a computer, and that it is its only interest, as it seems to be not specially efficient. So I suggest to reduce this paragraph to a single line, and to move it to the history section. D.Lazard (talk) 03:53, 15 October 2020 (UTC)
 * The algorithm, according to the PT article, was from when Feynman was still at Los Alamos, at which time "computer" meant a person working the computations by hand. The fact that the algorithm itself seems to be optimized for a binary (electronic) computer would seem to contradict this though. The algorithm does seem seem efficient since it eliminates multiplication and division from the computation, but whether more efficient methods are used by modern processors I can't say. There's a trade-off between look-up table size and number of computations, so it's hard to day what 'efficiency' actually means here. For the history section I think we have a similar issue: is there a reference which shows the historical significance of the algorithm? --RDBury (talk) 04:16, 19 October 2020 (UTC)
 * At Los Alamos, Feynman had some IBM machines that weren't quite computers. As well as I know, the fanciest of them could read two numbers off a punched card, multiply them, and then punch the product onto the card. (Especially useful for payroll calculations.) So finding algorithms to do complicated calculations using primitive equipment, might have come up with some interesting algorithms. Gah4 (talk) 16:37, 4 November 2021 (UTC)
 * At Los Alamos, Feynman had some IBM machines that weren't quite computers. As well as I know, the fanciest of them could read two numbers off a punched card, multiply them, and then punch the product onto the card. (Especially useful for payroll calculations.) So finding algorithms to do complicated calculations using primitive equipment, might have come up with some interesting algorithms. Gah4 (talk) 16:37, 4 November 2021 (UTC)

It seems that Feynman rediscovered the "radix method", e.g. as described in "RADIX METHOD OF CALCULATING NATURAL LOGARITHMS IN BINARY NOTATION", which I can't actually find a copy of. But similar things are done with decimal radix as here. I think Danny Hillis's referenced brief description of the method is adequate, and the adaptation of it in the article is even more clear and correct. Dicklyon (talk) 03:11, 17 November 2021 (UTC)

Cent scale
Okay, I'm a "noob" at editing Wikipedia pages (this is my second attempt in 15 years). So here goes:

In the "Particular bases" section of the Logarithm page, it says:

". . . in music theory, where a pitch ratio of two (the octave) is ubiquitous and the cent is the binary logarithm (scaled by 1200) of the ratio between two adjacent equally-tempered pitches in European classical music; . . ."

If you look at the Wikipedia definition of "Cent (music)", and the Wikipedia definition of "Binary logarithm", it appears that there has been a transcription error between these two definitions [Cent (music) & Binary logarithm] and the Logarithm article, which may need to be harmonized.

Based on my music theory classes in college, a cent is 1/100th of the pitch ratio between two adjacent equal-tempered pitches (a interval of a semitone or half-tone) in European music, and 1200 cents make an octave - not a half-tone.

So should the above quoted text in the Logarithm article read: ". . . and the cent is the binary logarithm (scaled by 100) of the ratio between two adjacent equally-tempered-pitches in European classical music; . . ."?

Or would another way of re-writing this be better?

Or am I completely mistaken?

TimeriderTech (talk) 01:10, 4 November 2021 (UTC)


 * I think it's more screwed up than you indicate. It should perhaps say ". . . in music theory, where a pitch ratio of two (the octave) is ubiquitous and the cent is the binary logarithm (scaled by 1200) of the ratio between two pitches; . . .".   Any two pitches, not two notes nominally 100 cents apart.  On the other hand, this is still a bad way to define "the cent" where it means "the interval measured in cents".  Please try to rewrite it to make more cents. Dicklyon (talk) 01:17, 4 November 2021 (UTC)
 * It's just overly specific. The number of cents between any two pitches is 1200x the binary logarithm of their ratio. There is no reason to specify that the two pitches in question be adjacent. —David Eppstein (talk) 01:21, 4 November 2021 (UTC)
 * What you say is correct. But it's hard to incorporate it into a sentence that says "the cent is". Dicklyon (talk) 03:24, 4 November 2021 (UTC)
 * What's wrong with my proposed simpler replacement for the incorrect quoted text above: "the number of cents between any two pitches is the binary logarithm (scaled by 1200) of their ratio" ? —David Eppstein (talk) 07:13, 4 November 2021 (UTC)
 * Nothing wrong with that. I interpreted your "It's just overly specific" as a suggestion to make the smaller change like I first said, with "the cent is".  An alternative using would be "The cent is 1/100 of the difference of logarithms between two pitches a semitone apart" or something like that, where now the log base doesn't even matter.  It's exactly the same problem discussed at length on Talk:Decibel, between trying to say "the decibel is" or trying to say how many decibels a ratio corresponds to.  The standards agencies take the former approach, but I'm not saying I like it.  Dicklyon (talk) 14:36, 4 November 2021 (UTC)

Thank you both for chiming in on this! My attempt to create a new "Cent scale" sub-topic on the talk page for "Logarithm" didn't appear to go as planned, and I'm guessing that somebody silently helped me out. (Thanks! I promise I'll get better at editing.) It also makes me feel better that at least two other editors agree that there's a problem here. Thanks to you both! I'm going to ruminate on a better re-write of the erroneous sentence, incorporating the above input. Hopefully we can all find an alternative phrasing that corrects this mathematical error in a manner most pleasing and clear to read. 64.201.116.75 (talk) 16:33, 4 November 2021 (UTC)

Units and logarithms
Really need a section on the units of a logarithm. For example, what is the unit of Log(10/seconds)? That is a common expression in first order rate equations, like nuclear decay.

From the integral definition of the natural log, "the area under the curve of a plot of 1/x versus x", it is clear that x can have any unit and that the logarithm is always unitless. The "area" under the curve has units of the x-axis times the units of the y-axis which, in all cases, is unity and unitless.

There are some discussion on the web suggestion things like "you can't take the logarithm of a number with units" (which is absurd, scientists and engineers do it all the time), to basically "there is a hidden and highly secret process in which the units disappear". Like "actually log(x*unit) is really log(x*unit/1*unit) so the units cancel", which is wrong.

The fact that the logarithm removes the units means that taking a logarithm is a lossy transform. There is no way, other than external knowledge, that allows the unit to be recovered by taking the exponent of the log. Therefore, e^(Ln(x)) <> x in all cases since the units have been lost. Jsluka (talk) 18:49, 26 October 2022 (UTC)


 * Engineer here - Just because we often do things like take the logarithm of numbers with units does not mean we should. For example, expanding something like log(x*unit) into log(x) + log(unit) implies a solution x to base ^ x = unit (definition of a logarithm), which is really hard to find when the base is in R (like, say, Euler's constant) and you're confined to R or C for the solution x. I don't know how to raise a real number to another real (or even complex) exponent in a way that changes the units from 1 to something like seconds, square meters, or volts. I think we can take logarithms of units when we define them -- Otherwise, I would personally consider something like ln(x) -> R; where x = seconds (in somewhere like R or C) completely nonsense unless one or more of three things is the case:
 * There is a sensible solution to the expression exp(x) = seconds, for x in your given space. Good luck.
 * The argument in question winds up being dimensionless by cancellation.
 * We can "pretend" the units are dimensionless because we plan to just call the units of the solution something like log-seconds anyways.
 * Taking this in a weird direction, it means that we can make equivalence relations between log units in some weird ways. Consider ln(acceleration) == ln(m/s^2) == ln(meters) - 2*ln(seconds), which suggests that ln(seconds) == (ln(meters) - ln(acceleration))/2 == ln( (m / (m/s^2)) ^ (1/2) ) == ln( (seconds^2) ^ (1/2)), which does make sense. I don't know what to do with a result like this, but it does serve your point that clearly one can take the logarithm of a number with units. In fact, one can use a logarithm of pure units not multiplied by anything at all and arrive at sensible answers, just like this. 66.218.139.214 (talk) 19:37, 24 April 2023 (UTC)
 * Reminds me that it seems to me that engineers often write expressions and equations with units factored out, while physicists keep units in. That is, one might say F(Newtons) = m (kg) * a (m/s/s).  The expectation is that one converts to the given units, applies the formula, and then converts the result. Physics will just say F = m a, where the variables keep the units. The former might result in excess conversions, the latter unusual units. Since computer languages normally don't keep units in variables, the engineering way is convenient for programming. And, as with the question here, units go out before the logs. Gah4 (talk) 04:55, 25 April 2023 (UTC)
 * Reminds me that it seems to me that engineers often write expressions and equations with units factored out, while physicists keep units in. That is, one might say F(Newtons) = m (kg) * a (m/s/s).  The expectation is that one converts to the given units, applies the formula, and then converts the result. Physics will just say F = m a, where the variables keep the units. The former might result in excess conversions, the latter unusual units. Since computer languages normally don't keep units in variables, the engineering way is convenient for programming. And, as with the question here, units go out before the logs. Gah4 (talk) 04:55, 25 April 2023 (UTC)

Redundant formula in section on Taylor series
Jacobolus surprisingly reverted an edit that removed a verbatim copy of a line two lines above that line. I'm re-reverting it after [Discussion] with him. MüllerMarcus (talk) 20:21, 20 October 2023 (UTC)


 * No, please do not make excellently accessible featured articles less friendly to novice readers without consensus, and please don't engage in edit warring. The explanation is not "redundant", and the line you call "*identical*" is self evidently different. Please look again more carefully, and try to put yourself in the shoes of a non-technical reader. –jacobolus (t) 20:39, 20 October 2023 (UTC)
 * I made the text a bit more explicit. Does that help? (As always, anyone should feel free to keep working on this who thinks they can improve on it.) –jacobolus (t) 00:58, 21 October 2023 (UTC)
 * that really helps a lot! Thank you for spending time on this, and sorry for the confrontation. MüllerMarcus (talk) 12:21, 21 October 2023 (UTC)