4 ms·

Calculating Logarithms by Hand (2010) [pdf]

108 points by mindcrime 3y ago

082349872349872 3y ago
> all logarithm tables for three hundred years were borrowed from Mr. Briggs’ tables by reducing the number of decimal places. Only in modern times, with the WPA and computing machines, have new tables been independently computed. https://www.feynmanlectures.caltech.edu/I_22.html#:~:text=But%20all,decimal%20places https://www.feynmanlectures.caltech.edu/I_22.html#:~:text=Bu...
eesmith 3y ago
I enjoyed a couple of the footnotes! > “Euler” is pronounced “oiler,” which is why virtually every sports team assembled by the University of Alberta’s math department has been named “the Edmonton Eulers.” "Euler’s mathematical brilliance" is footnoted: "I’m not exaggerating. I’ve taken no less than 13 University level math courses, and Euler’s name showed up in at least 10 of them. The man’s contributions to mathematics are probably more significant than Einstein’s contributions to physics. Find him on Wikipedia when you have a minute or thirty."
- Someone 3y ago
  If you’re a mathematician and you have a conjecture, constant, equation, function or theorem named after you, you can be proud of yourself. If you’re Euler, that’s 3 conjectures, 11 equations, 9 formulas, 4 functions, 2 identities, 10+ constants, 11 theorems, and 2 laws, some of them shared with others (https://en.wikipedia.org/wiki/List_of_things_named_after_Leonhard_Euler https://en.wikipedia.org/wiki/List_of_things_named_after_Leo...) That includes https://en.wikipedia.org/wiki/Euclid–Euler_theorem https://en.wikipedia.org/wiki/Euclid–Euler_theorem, which must have been one of the ‘cooperations’ (Euclid died ±270BC, Euler lived 1707-1783, and his proof was published posthumously) on a theorem that took the longest time in history.
  - eesmith 3y ago
    What are the top 3 mathematicians when ranked by the number of sports teams named after them? ;)
donkeybeer 3y ago
It appears Henry Briggs had used repeated square roots to compute logarithms: https://archived.hpcalc.org/laporte/Briggs%20and%20the%20HP35.htm https://archived.hpcalc.org/laporte/Briggs%20and%20the%20HP3... Mr Briggs had computed the first table of base 10 logarithms for the first few thousand natural numbers: https://en.wikipedia.org/wiki/Henry_Briggs_(mathematician) https://en.wikipedia.org/wiki/Henry_Briggs_(mathematician)
kondor6c 3y ago
My grandfather was a civil engineer for Maricopa County in Arizona (USA, dry desert, last contiguous state to join the union 1912 I believe) from 195[4-9] to like 1998 about. When I started working on advanced algebra while we we driving out to go fishing he told me that they used to have the 2 interns from ASU (AZ state univ.) find a different palo verde tree and go calculate the log that they would use for the bend of a curve for the new road, by hand. If they had different answers they'd both have to go back and redo it. I was astonished because I was just always taught to just push the log button on the calculator. Pretty crazy how things had changed even in curriculum. Thanks for the shared article
credit_guy 3y ago
Here's an even better one, which explains not only how Euler calculated logarithms with the bisection method, but also how he used some smart Taylor series tricks too to get the (natural) logarithms of the integers from 1 to 10, with 25 decimals. http://eulerarchive.maa.org/hedi/HEDI-2005-07.pdf http://eulerarchive.maa.org/hedi/HEDI-2005-07.pdf How good was Euler at that? I checked his results with a multiple precision library (mpmath) and Euler got all the correct digits for 7 out of 10, he rounded off the last decimal in the wrong direction in 2 cases. Just for log(7) he got 21 correct decimals and the last 4 were wrong. Here's the link to Euler's book [1], where these logarithms are on page 118. His values are l1 = 0,00000 00000 00000 00000 00000 l2 = 0,69314 71805 59945 30941 72321 l3 = 1,09861 22886 68109 69139 52452 l4 = 1,38629 43611 19890 61883 44642 l5 = 1,60943 79124 34100 37460 07593 l6 = 1,79175 94692 28055 00081 24773 l7 = 1,94591 01490 55313 30510 54639 l8 = 2.07944 15416 79835 92825 16964 l9 = 2,19722 45773 36219 38279 04905 l10 = 2,30258 50929 94045 68401 79914 The correct values are l1 = 0,00000 00000 00000 00000 00000 l2 = 0,69314 71805 59945 30941 72321 l3 = 1,09861 22886 68109 69139 52452 l4 = 1,38629 43611 19890 61883 44642 l5 = 1,60943 79124 34100 37460 07593 l6 = 1,79175 94692 28055 00081 24774 l7 = 1,94591 01490 55313 30510 53527 l8 = 2.07944 15416 79835 92825 16964 l9 = 2,19722 45773 36219 38279 04905 l10 = 2,30258 50929 94045 68401 79915 [1]https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?article=1100&context=euler-works https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?art...
- thaumasiotes 3y ago
  > and Euler got all the correct digits for 7 out of 10 This isn't the right denominator; you don't get any credit for calculating digits of ln(1).
  - fsckboy 3y ago
    to be fair, you'd get marked off for getting ln(1) wrong, so I think we have to give credit for getting it right. (grading on a scale solves that problem, though i guess the gaussian hadn't been invented yet). the numerator is also "wrong"; of his 3 errors, 2 of them are wrong only "off by one" in the last digit, same direction both times, so without understanding the rounding rules in effect it's hard to know whether what happened is an actual error. ln(7) is a mess though but still at the least significant end.
bumbledraven 3y ago
Related: Hartzler, "A two-and-one-half-place logarithm table", The Mathematics Teacher, Vol. 53, No. 3 (Mar 1960) https://www.jstor.org/stable/27956101 https://www.jstor.org/stable/27956101 Bayer, "Setting up an approximate antilog table", The Mathematics Teacher, Vol. 55, No. 3 (Mar 1962) https://www.jstor.org/stable/27956560 https://www.jstor.org/stable/27956560 Doerfler, "Dead Reckoning: Calculating Without Instruments", Taylor Trade Publishing (Sep 1993) https://www.amazon.com/Dead-Reckoning-Calculating-Without-Instruments/dp/0884150879 https://www.amazon.com/Dead-Reckoning-Calculating-Without-In...
mindcrime 3y ago
I was randomly re-watching some Youtube video on solving exponential equations and was poking around solving some, using the log() function in R to do the logs, and got to thinking about the "old days" when you had to look these things up in log tables. And that led me to think "Wait, how did the first log tables get written in the first place? And for that matter, how exactly do you work out a log value the manual way anyway???" A few moments of Googling turned this up. Thought some other folks might find this topic interesting as well. One fun aspect of all this, is that it forces you to consider that thinking of exponentiation as "repeated multiplication" is a bit dodgy when you allow for decimal exponents. I mean, given 3^2.5, what does it mean to say "multiply 3 by itself 2.5 times?" Actually, thinking about that was the thing that really set me off on an exploration of how the heck they worked out these log values in the beginning.
- arbitrandomuser 3y ago
  A very good exposition of this topic can be found in the Feynman lectures https://www.feynmanlectures.caltech.edu/I_22.html https://www.feynmanlectures.caltech.edu/I_22.html
- movpasd 3y ago
  The basic idea of how to generalise "repeated multiplication" to the whole real continuum is to think about multiplication as a kind of scaling. Going from 2^n to 2^(n+1) is ultimately about scaling the result up by a factor of x2. So, 2^(n+0.5) ought to be about scaling up halfway to x2, so that if you were to repeat that operation again, you got x2; and that's precisely sqrt(2). That reasoning gets you all the rationals. Since rationals are dense in the reals, impose continuity and you get all reals. This is the root of the algebraic definition of exponentials. But where there are reals there is calculus, and we also get a lot of insight from a differential definition. The key insight that leads to this definition is that if we could break down the exponential to 2^(n+epsilon) for a very small epsilon, we could stack up O(1/epsilon) of them to get wherever we like on the real continuum. So it makes sense that the definition dy/dx = a.y should produce the same function. This notion of "break down an operation into infinitesimal bits that stack up" can be taken seriously and formalised, and if you do that you end up with the theory of Lie groups and Lie algebras.
sumnole 3y ago
What, did you guys forget your slide rules at home?
- mindcrime 3y ago
  I've always kinda wanted to buy an actual slide rule, just for the lulz of it. Haven't gotten around to it yet. But one of these days...
mikewarot 3y ago
I can't really follow the algorithm Euler used... I've come up with one that is likely equivalent, and easy to do with a 4 function calculator. For log10.... Scale X to be between 1 and 10 For each time you divide by 10, add 1.0 to the log each time you have to multiply by 10, subtract 1.0 from the log Add the decimal point to the log Loop: Take X' (the scaled value) to the 10th power square it (store to memory, if you have it, this is x'^2) square that (which is x'^4) square that (now x'^8) multiply that by x'^2 you now have X'^10 the number now has to be divided N times to be between 1.0 and 10, and N to the log if you need more digits, goto Loop Here it is, taking log10(3) 3 0. 59,049 0.4 51,537,752.07320113 0.47 13,220,708.19480807 0.477 16.31350185342626 0.4771 133.4971414230401 0.47712 17.97710116675744 0.477121 352.530441082974 0.4771212 296,460.0951963823 0.47712125 52,439.97032955288 0.477121254 15,726,220.94397862 0.4771212547
- nickcw 3y ago
  Very nice algorithm - no square roots and 1 digit per iteration guaranteed. I converted this into python to understand it better def log10(x): """ Find 10 decimal places of log10(x) using calculator functions only """ assert x > 0, "can't log <= 0" log = 0 while x >= 10: log += 1 x /= 10 while x < 1: log -= 1 x *= 10 # x is now [1, 10) decimal = 0.1 for i in range(10): x10 = x**10 # or calculate as above so as not to use power function while x10 >= 10: log += decimal x10 /= 10 x = x10 decimal /= 10 print(f"log10(x) = {log:.10f}, x = {x:.10f}") log10(3) Which gives log10(x) = 0.4000000000, x = 5.9049000000 log10(x) = 0.4700000000, x = 5.1537752073 log10(x) = 0.4770000000, x = 1.3220708195 log10(x) = 0.4771000000, x = 1.6313501853 log10(x) = 0.4771200000, x = 1.3349714142 log10(x) = 0.4771210000, x = 1.7977101167 log10(x) = 0.4771212000, x = 3.5253044107 log10(x) = 0.4771212500, x = 2.9646009506 log10(x) = 0.4771212540, x = 5.2439970092 log10(x) = 0.4771212547, x = 1.5726220230
  - physPop 3y ago

108 points by mindcrime 3y ago

> all logarithm tables for three hundred years were borrowed from Mr. Briggs’ tables by reducing the number of decimal places. Only in modern times, with the WPA and computing machines, have new tables been independently computed. https://www.feynmanlectures.caltech.edu/I_22.html#:~:text=But%20all,decimal%20places https://www.feynmanlectures.caltech.edu/I_22.html#:~:text=Bu...

eesmith 3y ago

I enjoyed a couple of the footnotes! > “Euler” is pronounced “oiler,” which is why virtually every sports team assembled by the University of Alberta’s math department has been named “the Edmonton Eulers.” "Euler’s mathematical brilliance" is footnoted: "I’m not exaggerating. I’ve taken no less than 13 University level math courses, and Euler’s name showed up in at least 10 of them. The man’s contributions to mathematics are probably more significant than Einstein’s contributions to physics. Find him on Wikipedia when you have a minute or thirty."

Someone 3y ago
If you’re a mathematician and you have a conjecture, constant, equation, function or theorem named after you, you can be proud of yourself. If you’re Euler, that’s 3 conjectures, 11 equations, 9 formulas, 4 functions, 2 identities, 10+ constants, 11 theorems, and 2 laws, some of them shared with others (https://en.wikipedia.org/wiki/List_of_things_named_after_Leonhard_Euler https://en.wikipedia.org/wiki/List_of_things_named_after_Leo...) That includes https://en.wikipedia.org/wiki/Euclid–Euler_theorem https://en.wikipedia.org/wiki/Euclid–Euler_theorem, which must have been one of the ‘cooperations’ (Euclid died ±270BC, Euler lived 1707-1783, and his proof was published posthumously) on a theorem that took the longest time in history.
- eesmith 3y ago
  What are the top 3 mathematicians when ranked by the number of sports teams named after them? ;)

donkeybeer 3y ago

It appears Henry Briggs had used repeated square roots to compute logarithms: https://archived.hpcalc.org/laporte/Briggs%20and%20the%20HP35.htm https://archived.hpcalc.org/laporte/Briggs%20and%20the%20HP3... Mr Briggs had computed the first table of base 10 logarithms for the first few thousand natural numbers: https://en.wikipedia.org/wiki/Henry_Briggs_(mathematician) https://en.wikipedia.org/wiki/Henry_Briggs_(mathematician)

kondor6c 3y ago

My grandfather was a civil engineer for Maricopa County in Arizona (USA, dry desert, last contiguous state to join the union 1912 I believe) from 195[4-9] to like 1998 about. When I started working on advanced algebra while we we driving out to go fishing he told me that they used to have the 2 interns from ASU (AZ state univ.) find a different palo verde tree and go calculate the log that they would use for the bend of a curve for the new road, by hand. If they had different answers they'd both have to go back and redo it. I was astonished because I was just always taught to just push the log button on the calculator. Pretty crazy how things had changed even in curriculum. Thanks for the shared article

credit_guy 3y ago

Here's an even better one, which explains not only how Euler calculated logarithms with the bisection method, but also how he used some smart Taylor series tricks too to get the (natural) logarithms of the integers from 1 to 10, with 25 decimals. http://eulerarchive.maa.org/hedi/HEDI-2005-07.pdf http://eulerarchive.maa.org/hedi/HEDI-2005-07.pdf How good was Euler at that? I checked his results with a multiple precision library (mpmath) and Euler got all the correct digits for 7 out of 10, he rounded off the last decimal in the wrong direction in 2 cases. Just for log(7) he got 21 correct decimals and the last 4 were wrong. Here's the link to Euler's book [1], where these logarithms are on page 118. His values are l1 = 0,00000 00000 00000 00000 00000 l2 = 0,69314 71805 59945 30941 72321 l3 = 1,09861 22886 68109 69139 52452 l4 = 1,38629 43611 19890 61883 44642 l5 = 1,60943 79124 34100 37460 07593 l6 = 1,79175 94692 28055 00081 24773 l7 = 1,94591 01490 55313 30510 54639 l8 = 2.07944 15416 79835 92825 16964 l9 = 2,19722 45773 36219 38279 04905 l10 = 2,30258 50929 94045 68401 79914 The correct values are l1 = 0,00000 00000 00000 00000 00000 l2 = 0,69314 71805 59945 30941 72321 l3 = 1,09861 22886 68109 69139 52452 l4 = 1,38629 43611 19890 61883 44642 l5 = 1,60943 79124 34100 37460 07593 l6 = 1,79175 94692 28055 00081 24774 l7 = 1,94591 01490 55313 30510 53527 l8 = 2.07944 15416 79835 92825 16964 l9 = 2,19722 45773 36219 38279 04905 l10 = 2,30258 50929 94045 68401 79915 [1]https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?article=1100&context=euler-works https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?art...

thaumasiotes 3y ago
> and Euler got all the correct digits for 7 out of 10 This isn't the right denominator; you don't get any credit for calculating digits of ln(1).
- fsckboy 3y ago
  to be fair, you'd get marked off for getting ln(1) wrong, so I think we have to give credit for getting it right. (grading on a scale solves that problem, though i guess the gaussian hadn't been invented yet). the numerator is also "wrong"; of his 3 errors, 2 of them are wrong only "off by one" in the last digit, same direction both times, so without understanding the rounding rules in effect it's hard to know whether what happened is an actual error. ln(7) is a mess though but still at the least significant end.

bumbledraven 3y ago

Related: Hartzler, "A two-and-one-half-place logarithm table", The Mathematics Teacher, Vol. 53, No. 3 (Mar 1960) https://www.jstor.org/stable/27956101 https://www.jstor.org/stable/27956101 Bayer, "Setting up an approximate antilog table", The Mathematics Teacher, Vol. 55, No. 3 (Mar 1962) https://www.jstor.org/stable/27956560 https://www.jstor.org/stable/27956560 Doerfler, "Dead Reckoning: Calculating Without Instruments", Taylor Trade Publishing (Sep 1993) https://www.amazon.com/Dead-Reckoning-Calculating-Without-Instruments/dp/0884150879 https://www.amazon.com/Dead-Reckoning-Calculating-Without-In...

mindcrime 3y ago

I was randomly re-watching some Youtube video on solving exponential equations and was poking around solving some, using the log() function in R to do the logs, and got to thinking about the "old days" when you had to look these things up in log tables. And that led me to think "Wait, how did the first log tables get written in the first place? And for that matter, how exactly do you work out a log value the manual way anyway???" A few moments of Googling turned this up. Thought some other folks might find this topic interesting as well. One fun aspect of all this, is that it forces you to consider that thinking of exponentiation as "repeated multiplication" is a bit dodgy when you allow for decimal exponents. I mean, given 3^2.5, what does it mean to say "multiply 3 by itself 2.5 times?" Actually, thinking about that was the thing that really set me off on an exploration of how the heck they worked out these log values in the beginning.

arbitrandomuser 3y ago
A very good exposition of this topic can be found in the Feynman lectures https://www.feynmanlectures.caltech.edu/I_22.html https://www.feynmanlectures.caltech.edu/I_22.html
movpasd 3y ago
The basic idea of how to generalise "repeated multiplication" to the whole real continuum is to think about multiplication as a kind of scaling. Going from 2^n to 2^(n+1) is ultimately about scaling the result up by a factor of x2. So, 2^(n+0.5) ought to be about scaling up halfway to x2, so that if you were to repeat that operation again, you got x2; and that's precisely sqrt(2). That reasoning gets you all the rationals. Since rationals are dense in the reals, impose continuity and you get all reals. This is the root of the algebraic definition of exponentials. But where there are reals there is calculus, and we also get a lot of insight from a differential definition. The key insight that leads to this definition is that if we could break down the exponential to 2^(n+epsilon) for a very small epsilon, we could stack up O(1/epsilon) of them to get wherever we like on the real continuum. So it makes sense that the definition dy/dx = a.y should produce the same function. This notion of "break down an operation into infinitesimal bits that stack up" can be taken seriously and formalised, and if you do that you end up with the theory of Lie groups and Lie algebras.

sumnole 3y ago

What, did you guys forget your slide rules at home?

mindcrime 3y ago
I've always kinda wanted to buy an actual slide rule, just for the lulz of it. Haven't gotten around to it yet. But one of these days...

mikewarot 3y ago

I can't really follow the algorithm Euler used... I've come up with one that is likely equivalent, and easy to do with a 4 function calculator. For log10.... Scale X to be between 1 and 10 For each time you divide by 10, add 1.0 to the log each time you have to multiply by 10, subtract 1.0 from the log Add the decimal point to the log Loop: Take X' (the scaled value) to the 10th power square it (store to memory, if you have it, this is x'^2) square that (which is x'^4) square that (now x'^8) multiply that by x'^2 you now have X'^10 the number now has to be divided N times to be between 1.0 and 10, and N to the log if you need more digits, goto Loop Here it is, taking log10(3) 3 0. 59,049 0.4 51,537,752.07320113 0.47 13,220,708.19480807 0.477 16.31350185342626 0.4771 133.4971414230401 0.47712 17.97710116675744 0.477121 352.530441082974 0.4771212 296,460.0951963823 0.47712125 52,439.97032955288 0.477121254 15,726,220.94397862 0.4771212547

nickcw 3y ago
Very nice algorithm - no square roots and 1 digit per iteration guaranteed. I converted this into python to understand it better def log10(x): """ Find 10 decimal places of log10(x) using calculator functions only """ assert x > 0, "can't log <= 0" log = 0 while x >= 10: log += 1 x /= 10 while x < 1: log -= 1 x *= 10 # x is now [1, 10) decimal = 0.1 for i in range(10): x10 = x**10 # or calculate as above so as not to use power function while x10 >= 10: log += decimal x10 /= 10 x = x10 decimal /= 10 print(f"log10(x) = {log:.10f}, x = {x:.10f}") log10(3) Which gives log10(x) = 0.4000000000, x = 5.9049000000 log10(x) = 0.4700000000, x = 5.1537752073 log10(x) = 0.4770000000, x = 1.3220708195 log10(x) = 0.4771000000, x = 1.6313501853 log10(x) = 0.4771200000, x = 1.3349714142 log10(x) = 0.4771210000, x = 1.7977101167 log10(x) = 0.4771212000, x = 3.5253044107 log10(x) = 0.4771212500, x = 2.9646009506 log10(x) = 0.4771212540, x = 5.2439970092 log10(x) = 0.4771212547, x = 1.5726220230
- physPop 3y ago