3 ms·

Estimating Logarithms

103 points by surprisetalk 1y ago

xeonmc 1y ago
Protip: since halving and doubling are the same logarithmic distance on either sides of unity, and the logarithmic distance of 2.0 to 5.0 is just a tiny bit larger than that of doubling, this means that you can roughly eyeball the infra-decade fraction by cutting them into thirds
thechao 1y ago
I don't know about powers-of-10; but, you can use something similar to bootstrap logs-in-your-head. So, 2^10=1024. That means log10(2)~3/10=0.3. By log laws: 1 - .3 = 0.7 ~ log10(5). Similarly, log10(3)*9 ~ 4 + log10(2); so, log10(3) ~ .477. Other prime numbers use similar "easy power rules". Now, what's log10(80)? It's .3*3 + 1 ~ 1.9. (The real value is 1.903...). The log10(75) ~ .7*2+.477 = 1.877 (the real answer is 1.875...). Just knowing some basic "small prime" logs lets you rapidly calculate logs in your head.
- madcaptenor 1y ago
  For log(3) I prefer the "musical" approximation 2^19 ~ 3^12. This is a "musical" fact because it translates into 2^(7/12) ~ 3/2 - that is, seven semitones make a perfect fifth). Together with log(2) ~ 3/10 that gives log(3) ~ 19/40. Also easy to remember: 7^4 = 2401 ~ 2400. log(2400) = log(3) + 3 log(2) + 2 ~ 19/40 + 3 * 12/40 + 2 = 135/40, so you get log(7) ~ 135/160 = 27/32 = 0.84375.
  - thechao 1y ago
    These are both great! I learned most of these old tricks from my dad & grandfather.
  - thechao 1y ago
    I'll double-reply, since I think people might also appreciate this... there's a pretty easy way to find the square of 2-leading-digit numbers (and square-roots, too.) Direct square: 54^2 = (50 + 4)^2 = 2500 + 2 * 50 * 4 + 4^2. Direct square-root: 3800 = (61^2=3721) + r ~ (61 + (3800-3721)/(2*61) = 61 + 79/122. To check: (61 + 79/122)^2= 3,800.419. You can estimate the "overhead" by noting that 79/122 ~ 2/3 => (2/3)^2 = 0.444. Of course, my grandfather would've just used a slide-rule directly.
briian 1y ago
So much of economics maths/stats is built on this one little trick. It's still pretty cool to me that A this works and B it can be used to do so much.
fdspopofe 1y ago
[dead]
nicoburns 1y ago
[flagged]
saulpw 1y ago
Here's all you really need to know about logs when estimating in your head: The number of digits minus one is the magnitude (integer). Then add the leading digit like so: 1x = ^0.0 2x = ^0.3 (actually ^0.301...) pi = ^0.5 (actually ^0.497...) 5x = ^0.7 (actually ^0.699...) Between these, you can interpolate linearly and it's fine for estimating. Also 3x is close enough to pi to also be considered ^0.5. In fact, if all you're doing is estimating, you don't even really need to know the above log table. Just use the first digit of the original number as the first digit past the decimal. So like 6000 would be ^3.6 (whereas it's actually ^3.78). It's "wrong" but not that far off if you're using logarithmetic for napkin math.
- xeonmc 1y ago
  And this is also the basis of the fast inverse square root algorithm. Floating point numbers are just linear interpolations between octaves.
- thomasahle 1y ago
  What is this ^notation? Looks like 5x=^0.699 means log_10(5)=0.699.
  - xeonmc 1y ago
    5 = 10^0.699
  - saulpw 1y ago
    It's magnitude notation. ^X is short for 10^X.
- dhosek 1y ago
  The pi = ^0.5 bit reminds me of a college physics professor who was fond of using the shortcut π²=10.
  - madcaptenor 1y ago
    This is wrong, π² = g, the acceleration due to gravity.
gus_massa 1y ago
Typo near the top, in case someone knows the author: > log(100)≤log(N)<log(100) There is a missing 0 in the last log. It should be > log(100)≤log(N)<log(1000)
- deleted 1y ago
  [deleted]
- obrhubr 1y ago
  Thanks for pointing this out :), I fixed it!
stevefan1999 1y ago
Related Wikipedia entry: https://en.wikipedia.org/wiki/Logarithmic_number_system https://en.wikipedia.org/wiki/Logarithmic_number_system Also related: https://blog.timhutt.co.uk/fast-inverse-square-root/ https://blog.timhutt.co.uk/fast-inverse-square-root/ (I see that someone already mentioned fast inverse square root algorithm is related to this, which is famously used by John Carmack which is one of my hero who led me into tech industry, despite I didn't end up in gaming industry)
teo_zero 1y ago
The artcle makes a weird use of the notation for successive exponentiations. It's difficult to represent the concept here as only text is allowed, but shortly the author uses a^b^c to mean (a^b)^c. This is counter intuitive as in most mathematical contexts exponentiation is right-associative, that is a^b^c means a^(b^c). A pair of parentheses would remove any doubts!

103 points by surprisetalk 1y ago

Protip: since halving and doubling are the same logarithmic distance on either sides of unity, and the logarithmic distance of 2.0 to 5.0 is just a tiny bit larger than that of doubling, this means that you can roughly eyeball the infra-decade fraction by cutting them into thirds

thechao 1y ago

I don't know about powers-of-10; but, you can use something similar to bootstrap logs-in-your-head. So, 2^10=1024. That means log10(2)~3/10=0.3. By log laws: 1 - .3 = 0.7 ~ log10(5). Similarly, log10(3)*9 ~ 4 + log10(2); so, log10(3) ~ .477. Other prime numbers use similar "easy power rules". Now, what's log10(80)? It's .3*3 + 1 ~ 1.9. (The real value is 1.903...). The log10(75) ~ .7*2+.477 = 1.877 (the real answer is 1.875...). Just knowing some basic "small prime" logs lets you rapidly calculate logs in your head.

madcaptenor 1y ago
For log(3) I prefer the "musical" approximation 2^19 ~ 3^12. This is a "musical" fact because it translates into 2^(7/12) ~ 3/2 - that is, seven semitones make a perfect fifth). Together with log(2) ~ 3/10 that gives log(3) ~ 19/40. Also easy to remember: 7^4 = 2401 ~ 2400. log(2400) = log(3) + 3 log(2) + 2 ~ 19/40 + 3 * 12/40 + 2 = 135/40, so you get log(7) ~ 135/160 = 27/32 = 0.84375.
- thechao 1y ago
  These are both great! I learned most of these old tricks from my dad & grandfather.
- thechao 1y ago
  I'll double-reply, since I think people might also appreciate this... there's a pretty easy way to find the square of 2-leading-digit numbers (and square-roots, too.) Direct square: 54^2 = (50 + 4)^2 = 2500 + 2 * 50 * 4 + 4^2. Direct square-root: 3800 = (61^2=3721) + r ~ (61 + (3800-3721)/(2*61) = 61 + 79/122. To check: (61 + 79/122)^2= 3,800.419. You can estimate the "overhead" by noting that 79/122 ~ 2/3 => (2/3)^2 = 0.444. Of course, my grandfather would've just used a slide-rule directly.

briian 1y ago

So much of economics maths/stats is built on this one little trick. It's still pretty cool to me that A this works and B it can be used to do so much.

fdspopofe 1y ago

[dead]

nicoburns 1y ago

[flagged]

saulpw 1y ago

Here's all you really need to know about logs when estimating in your head: The number of digits minus one is the magnitude (integer). Then add the leading digit like so: 1x = ^0.0 2x = ^0.3 (actually ^0.301...) pi = ^0.5 (actually ^0.497...) 5x = ^0.7 (actually ^0.699...) Between these, you can interpolate linearly and it's fine for estimating. Also 3x is close enough to pi to also be considered ^0.5. In fact, if all you're doing is estimating, you don't even really need to know the above log table. Just use the first digit of the original number as the first digit past the decimal. So like 6000 would be ^3.6 (whereas it's actually ^3.78). It's "wrong" but not that far off if you're using logarithmetic for napkin math.

xeonmc 1y ago
And this is also the basis of the fast inverse square root algorithm. Floating point numbers are just linear interpolations between octaves.
thomasahle 1y ago
What is this ^notation? Looks like 5x=^0.699 means log_10(5)=0.699.
- xeonmc 1y ago
  5 = 10^0.699
- saulpw 1y ago
  It's magnitude notation. ^X is short for 10^X.
dhosek 1y ago
The pi = ^0.5 bit reminds me of a college physics professor who was fond of using the shortcut π²=10.
- madcaptenor 1y ago
  This is wrong, π² = g, the acceleration due to gravity.

gus_massa 1y ago

Typo near the top, in case someone knows the author: > log(100)≤log(N)<log(100) There is a missing 0 in the last log. It should be > log(100)≤log(N)<log(1000)

deleted 1y ago
[deleted]
obrhubr 1y ago
Thanks for pointing this out :), I fixed it!

stevefan1999 1y ago

Related Wikipedia entry: https://en.wikipedia.org/wiki/Logarithmic_number_system https://en.wikipedia.org/wiki/Logarithmic_number_system Also related: https://blog.timhutt.co.uk/fast-inverse-square-root/ https://blog.timhutt.co.uk/fast-inverse-square-root/ (I see that someone already mentioned fast inverse square root algorithm is related to this, which is famously used by John Carmack which is one of my hero who led me into tech industry, despite I didn't end up in gaming industry)

teo_zero 1y ago

The artcle makes a weird use of the notation for successive exponentiations. It's difficult to represent the concept here as only text is allowed, but shortly the author uses a^b^c to mean (a^b)^c. This is counter intuitive as in most mathematical contexts exponentiation is right-associative, that is a^b^c means a^(b^c). A pair of parentheses would remove any doubts!