I did some study of base 1+i last fall. If you allow for p-adic values, the "infinite repeating ones" (...1111111) is equivalent to "i" and you can start from there and get a second interlocking spiral, rotated by 90 degrees, that completes the plane.
@@TheGrayCuber I worked out some carry rules of addition and subtraction i.e. how to move a '2' or '-1' upwards in the digits. Then you can show that ...222 is equal to 100 which is equal to 2i. It's similar to how ...999+1 equals 0 in normal 10-adics. The same carry rules can also show that 'any' integer can be written in base 1+i, thus proving completeness. Intrigued to learn that i-1 does not have that "bijection" problem though, maybe this will solve something for me
@@TheGrayCuber ...1111 is basically 1b⁰ + 1b¹ + 1b² + ..., which is a well-known geometric series that can be reduced to 1/(1 - b) when |b| < 1. |1+i| is not less than 1, but p-adics use a different notion of distance that essentially flips everything around. While I'm not sure how to properly use it with complex p-adics, I'm going to just ignore that requirement and assume it probably converges in the (1+i)-adics. In that case, we have 1/(1 - (1 + i)) = 1/(1 - 1 - i) = 1/-i = i.
These give the dragon curve so there is likely some deeper connection to that which connects to IFS's and fractals more generally. Essentially sum(d_k*b^k) is a value in the base b where d is some element from the base ring. Writing b = re^(it) gives sum(d_k*r^k*e^(ikt)) One can define F_n(d_n) = d_n*r^n*e^(int) + F_(n-1)(d_(n-1)) The idea is that for whatever base you choose you end up with a pattern plus some extra term. When you let d_n vary over the possible digits you get the pattern but it is both scaled and rotated by the iteration depth which corresponds to the digit place. That is, the "extra term" is a set of points d_n*r^n*e^(int) which once understood then describes the entire process. Also you can then add any set of digits(they don't have to have any special properties. One could also use basis such as matrices or even functor categories.
“We can’t have a complex amount of digits-that doesn’t really make any sense” Someone hasn’t been paying attention in class. The problem you have is not with math, but with its notation
Cool concept! This can be represented as a complex iterated function system with f1(z)=b*z, f2(z)=b*z+1. b is the complex base you're "counting" with, you start with z=0 and you choose either one of the two functions to get the next iterate. What you're looking at is all possible iterates up to a certain cutoff (the max number of digits). It turns out the magnitude and phase angle of b are related to the dynamics of this, and parameterizing the base in polar coordinates would show how the patterns form a bit more clearly. (I haven't watched the original video, so maybe this was already mentioned)
Elementary Theory of Numbers by LeVeque is pretty good! It covers continued fractions, complex integers, and goes into the idea of complex primes which is really cool. Richard Borcherds has a lot of lectures in various topics: www.youtube.com/@richarde.borcherds7998 Michael Penn has videos on some fun topics that walk through calculations ua-cam.com/video/WfKR_MYu_UA/v-deo.htmlsi=znpauTW-0se2uwgr
is there a way to get the sandbox with the sliders being smooth instead of having to load each time? itd be interesting to mentally map the space the two variables sweep out
I did some study of base 1+i last fall. If you allow for p-adic values, the "infinite repeating ones" (...1111111) is equivalent to "i" and you can start from there and get a second interlocking spiral, rotated by 90 degrees, that completes the plane.
That is really cool! How does one demonstrate that …111 = i? Something like -1 = (…111)*(1+i) - (…111) ?
@@TheGrayCuber I worked out some carry rules of addition and subtraction i.e. how to move a '2' or '-1' upwards in the digits. Then you can show that ...222 is equal to 100 which is equal to 2i. It's similar to how ...999+1 equals 0 in normal 10-adics. The same carry rules can also show that 'any' integer can be written in base 1+i, thus proving completeness. Intrigued to learn that i-1 does not have that "bijection" problem though, maybe this will solve something for me
@@TheGrayCuber ...1111 is basically 1b⁰ + 1b¹ + 1b² + ..., which is a well-known geometric series that can be reduced to 1/(1 - b) when |b| < 1. |1+i| is not less than 1, but p-adics use a different notion of distance that essentially flips everything around. While I'm not sure how to properly use it with complex p-adics, I'm going to just ignore that requirement and assume it probably converges in the (1+i)-adics.
In that case, we have 1/(1 - (1 + i)) = 1/(1 - 1 - i) = 1/-i = i.
@@haavind man math is so cool
Ratio is always 1+I therefore all 1 according to the formula of geometric series we get 1/(1-(1+i))=1/(-i)=I so that’s proof
These give the dragon curve so there is likely some deeper connection to that which connects to IFS's and fractals more generally.
Essentially sum(d_k*b^k) is a value in the base b where d is some element from the base ring. Writing b = re^(it) gives sum(d_k*r^k*e^(ikt))
One can define F_n(d_n) = d_n*r^n*e^(int) + F_(n-1)(d_(n-1))
The idea is that for whatever base you choose you end up with a pattern plus some extra term. When you let d_n vary over the possible digits you get the pattern but it is both scaled and rotated by the iteration depth which corresponds to the digit place. That is, the "extra term" is a set of points d_n*r^n*e^(int) which once understood then describes the entire process.
Also you can then add any set of digits(they don't have to have any special properties. One could also use basis such as matrices or even functor categories.
“We can’t have a complex amount of digits-that doesn’t really make any sense” Someone hasn’t been paying attention in class. The problem you have is not with math, but with its notation
Fascinating! Cool web tool as well :))
Out of curiosity, what sort of industry/role do you work in? I'd assume a maths-intensive field?
I’m in data analytics, fairly maths-intensive
Cool concept! This can be represented as a complex iterated function system with f1(z)=b*z, f2(z)=b*z+1. b is the complex base you're "counting" with, you start with z=0 and you choose either one of the two functions to get the next iterate. What you're looking at is all possible iterates up to a certain cutoff (the max number of digits). It turns out the magnitude and phase angle of b are related to the dynamics of this, and parameterizing the base in polar coordinates would show how the patterns form a bit more clearly. (I haven't watched the original video, so maybe this was already mentioned)
Where did 0.69 come from 💀
I feel like this exceeds my brain capacity but it's so entertaining and I love your channel
When are we gonna get a math crossover between Combo Class and TheGrayCuber?
I like this topic from Combo Class a lot: ua-cam.com/video/RdnTi-2gahs/v-deo.html
The idea of making i as 1^(1/(1+1)) so that it has cost 4 is fun
This is the most beautiful gift ive ever been given. Are there any resources/literature you'd recommend to read?
Elementary Theory of Numbers by LeVeque is pretty good! It covers continued fractions, complex integers, and goes into the idea of complex primes which is really cool.
Richard Borcherds has a lot of lectures in various topics: www.youtube.com/@richarde.borcherds7998
Michael Penn has videos on some fun topics that walk through calculations ua-cam.com/video/WfKR_MYu_UA/v-deo.htmlsi=znpauTW-0se2uwgr
is there a way to get the sandbox with the sliders being smooth instead of having to load each time? itd be interesting to mentally map the space the two variables sweep out
Unfortunately I don't think that would be easy to do with Tableau, it woudl need to be made using a different software
Do rational numbers in base -1+I map to all complex rationals?
yes!
@@TheGrayCuber Now... Prove it for the reals 🤪
I believe if you dont just use integers but also allow points after a decimal, this base will give you all complex numbers
hey thegraycuber, what did you upload in 2012?
Have you submitted this video to Summer of Math Exposition?
No, I wasn’t really aware of that until submissions ended
I thought you were cuber?
What it this btw
𝕊𝕒𝕟𝕕𝕓𝕠𝕩?
Are you still trying 13bld?
No I will not be attempting 13BLD again
Hi
hi :)
yo
Great video!