you can trucate it and it still only a slightly worse approximation:
987/123 = 8.024
98/12 = 8.167
9/1 = 9.000
The last one is pretty bad you should probably not use it
General rules:
Exceptions may be made at the discretion of the mods.
you can trucate it and it still only a slightly worse approximation:
987/123 = 8.024
98/12 = 8.167
9/1 = 9.000
The last one is pretty bad you should probably not use it
9/1 is approximately 8, for extremely large values of 8.
Hello, fellow old nerd.
9/1 ≈ 8
For that last one, how bad are we talking? I need to know soon, I have some important banking software I need to develop.
I wouldn't use it for precise calculations at NASA, but for banking stuff I think it would be fine :)
It depends on the scale of the thing you're using it for.
Have you considered running for Indiana governor? You have the right mindset.
Nah, the engineer probably designed it with a safety factor. You could probably even go 9/0 and be perfectly safe ;)
By the way guys, a very similar approximation for 8, which also starts at 9.000 for n=1, but quickly gets much closer to 8 for increasing n, is:
exp{-2(n-1)} + 8
It approaches 8 about as fast as the above method but this one has a simple formula that is usable in python etc.
16/2 is an almost exact replacement for 8. OP's name includes Fermat, and so he's probably smarter than me though.
Shit like this makes me realise why people become mathematicians. You just play around with numbers and find funny facts about them.
So, years ago in college in Linear Algebra our professor said to us to study about idempotent matrices. So I checked out that wiki page and saw the example for 2x2 matrix, that are composed by the numbers 3, -6, 1 and -2. And I was like wait a second, 3×-2=-6 there's no way they are not relationship there, so I started trying other numbers, and found and proved (using induction) that any n, -n(n-1), 1, -(n-1) is an idempotent matrix. At the test there were no questions about that, and I was short of 0.5 poits to pass the class without having to present a final exam and I told my professor that I spent a lot of time learning that and that even discovered something and proved he pass me the chart and asked me to proved it, after that he gave the missing points. Was really good.
You need to put the name inside the brackets and the link inside the parentheses.
Thanks
I myself once learned 380 digits of π, when I was a crazy high-school kid. My never-attained ambition was to reach the spot, 762 digits out in the decimal expansion, where it goes "999999", so that I could recite it out loud, come to those six 9s, and then impishly say, "and so on!"
—Douglas Hofstadter
That would be an amazing party trick.
Actually come to think of it, even more amazing in the age of smart phones, when it's possible to easily verify to numbers you're reciting.
Then you try to figure out why they do be like that
gonna need this in every base
I'll start with base 2:
1/1 = 1
Base 3:
21 / 12 = 1.1012101210121012
I'm gonna need a formal proof for this.
We should be friends
987654312÷123456789
Change the 21 at the end of the first number to 12 and its perfect. It was only ever 9 away.
WOOOAH🤯
Witch! Begone foul demon, and take your dark sorcery with you!
I just noticed what the numbers are. It really is easy to memorize. So convenient.
Unfortunately, it requires remembering 8, so it kinda defeats the purpose.
You may call it an approxim8ion
gr8 m8, I r8 8/8
The funniest part is that some people will never understand the absolute crusade that some mathematicians might fight over this one day
9876543210987654321 / 1234567890123456789 = 8,0000000729000
I wonder if there’s a related infinite sequence which converges on 8?
This sequence approximates an integer to arbitrary precision, not 8 specifically though, and never perfectly.
I tried it out using other bases, and the rule seems to be that doing this in base n results in n-2 with remainder n-1. So it doesn't ever actually converge, but the remainder becomes small very fast.
never perfectly
eyes you in binary
The sequence in base 2 is only 1/1.
Wonder how close base-16 gets.
FEDCBA987654321 / 123456789ABCDEF
Off by '1.82959E–16' !
(n * 8 + 1) / n
It contains the number 8 though. So how is that useful
Well, simple. Jest substitute that 8 with the above approximation.
strange coincidence
See my other comment, it's no coincide– there's a pattern. I would love to see an actual proof for it though, I don't know enough to say why it behaves that way.