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/sci/ - Science & Math


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File: 114 KB, 371x717, Continuity_Principle.png [View same] [iqdb] [saucenao] [google]
10477587 No.10477587 [Reply] [Original]

Can someone give a simple visual explanation how two circles of radius 2 can still be intersecting when their centers are 5 units apart?

>> No.10477616

>>10477587

How can divergent series have a value? Just fuck those asshole mathematicians, they think they can do anything because Gödel proved that you can make up infinite bullshit in mathematics and still people will be ok without it because axioms lol.

>> No.10477621

if you plot x^2+y^2=2 and (x-5)^2+y^2=2, where, e.g., y is imaginary, you get intersection points like x=5/2 and y=i*sqrt(17)/2. allowing imaginary numbers turns it from intersecting circles to intersecting hyperbolae

>> No.10477628
File: 262 KB, 591x716, Overwhelming_Ideas.png [View same] [iqdb] [saucenao] [google]
10477628

>>10477621
I am not doubting the math. I just am looking for a visual representation (an image) that would make this "obvious".

>> No.10477633

>>10477628
it’s easy; just let z=yi and plot x^2-z^2=2, (x-5)^2-z^2=2. you’ll see two hyperbolas that intersect at two points in the x-z plane

>> No.10477646

here i made the plot for u
https://www.wolframalpha.com/input/?i=plot+x%5E2-y%5E2%3D2,+(x-5)%5E2-y%5E2%3D2

>> No.10477647
File: 219 KB, 899x455, Sorting.gif [View same] [iqdb] [saucenao] [google]
10477647

>>10477633

I am a brainlette. I have no visual plotting devise or the skills to use one if I had it. if it is "easy" could you show an image of it? Please.

>> No.10477648

>>10477647
just click the link in this post >>10477646
you should be able to see how the equations of the lines shown are the same as two circles but y has been swapped with i*y

>> No.10477662
File: 665 KB, 705x865, Special.png [View same] [iqdb] [saucenao] [google]
10477662

>>10477648
I see that, BUT your image should how two circles with their centers 5 units apart of radii 2 that still intersect. In no way looking at your example do I see two circles. YES, I know mathematically this is correct. It is obvious that the solution involves a third "i" dimension and the points should all have 3 coordinates (x, y, i)
and the result is allowed by viewing the transformation from different angles.

Or (most likely) I am wrong and a complete math idiot.

>> No.10477674

>>10477662
I offer a different version: You might be also an idiot, but the one who made the OP pic is an obscuring bastard if he's showing two circles apart and not the actual hyperboloids the equations represent in all the dimensions. That's like taking the two interescting circles, then slicing them into 1D and showing that there are two points which do not touch, but yet if we change the variables they suddenly do in the magical realm of 2D, but we don't see it on the two points 1D plot. Woooooah

>> No.10477676
File: 105 KB, 250x250, BAF69BBE-10DA-4B78-8D66-B100913B928F.gif [View same] [iqdb] [saucenao] [google]
10477676

>>10477662
fine, then maybe think in terms of conic sections. cutting a cone “horizontally” gives you a circle x^2+y^2=const. tilt your cut a little and you get an ellipse, e.g. x^2+(.9y)^2=const. tilt it until you’re parallel with the cone’s side and you get a parabola, x^2+(0y)^2=x^2. tilt it even further and you get a hyperbola x^2-(.1y)^2=const. does that help?

>> No.10477679

>>10477647
>brainlette

How ugly are you if you're interested in complex numbers? Lol, good luck finding a good husband uggo

>> No.10477748

>>10477587
You have to extend your definition of plane curves from 1D subsets of R^2 to 2D subsets of C^2 = R^4

>> No.10477753

>>10477748
reminder C^2 does not equal R^4 and C does not equal R^2, since e.g. there is no multiplication rule for elements of R^2 but there is for C. god i hate topologists

>> No.10477898

>>10477753
OP is looking for a geometric picture of what's happening

>> No.10477911
File: 52 KB, 903x960, 16e.jpg [View same] [iqdb] [saucenao] [google]
10477911

>>10477676
pruty

>> No.10478010
File: 25 KB, 471x484, geek_humor.jpg [View same] [iqdb] [saucenao] [google]
10478010

>>10477587
Can this idea be used to understand quantum entanglement?

If we add a "quantum" dimension to normal 4 dim space-time, then once "Entangled" two particles can be separated in 4 dim space-time but STILL be connected in the quantum dimension.

5 cups of coffee, night be too much.

>> No.10478078

>>10477587
Read up on Bezout's Theorem. A circle is an algebraic plane curve of degree 2, so 2 circles intersect [math]2\cdot 2 = 4[/math] times in the complex projective plane when counted with multiplicity.

It may seem trivial that we sort of define two curves of degree m and n to intersect mn times, but what's remarkable is just extending to the complex plane and allow parallel lines to intersect at infinity is enough to prove the theorem.

t. math undergrad so don't bully me if I'm wrong pls

>> No.10478560
File: 29 KB, 555x485, not_saying_it_was_aliens.jpg [View same] [iqdb] [saucenao] [google]
10478560

>>10478078
>allow parallel lines to intersect at infinity

My mind is blown.
I am not smart enough to grasp impossible ideas,

>> No.10478582

this 13 part series (https://www.youtube.com/watch?v=T647CGsuOVU)) goes in depth about how imaginary number lines can be used to explain how yes they are technically "touching" just in an "imaginary" plane

>> No.10478619

>>10477753
you can sure as shit can define a multiplication in R2
C is just R2 with slightly more algebraic structure, despite having identical geometric structure

>> No.10478705
File: 1.90 MB, 500x500, spinning-optical-illusion.gif [View same] [iqdb] [saucenao] [google]
10478705

>>10478582

Watching the videos now.
So far impressed with visuals and pacing, WELL DONE :-)

>> No.10478783

>>10478619
There are ways to define multiplication on [math]\mathbb{R}^n[/math] for all n, but the most natural one (pointwise) doesn't reduce to multiplication on [math]\mathbb{C}^{n/2}[/math] for even n. [math]\mathbb{C}[/math] is [math]\mathbb{R}[/math] with some added structure, that means you do have to add some structure to get from [math]\mathbb{C}[/math] to [math]\mathbb{R}[/math]

>> No.10478790

>>10477587
https://www.youtube.com/watch?v=T647CGsuOVU
1:51

>> No.10478851
File: 225 KB, 260x308, subjective_color.gif [View same] [iqdb] [saucenao] [google]
10478851

>>10478790

So complex functions need a four dimensional space to view them correctly.

They use computers to represent 3 of the dimensions as "normal 3d space" and the forth dimension as color.

The original image is correct, but just a "shadow" view of the solution in the 2 dimensional world of real numbers.

OK...

Time for some chocolate and a nap.

>> No.10478915

>>10478851
yeah Id recommend watching all 14 videos in that series if you are really interested

>> No.10479512
File: 1.38 MB, 1536x2048, Railroad-Tracks-Perspective[1].jpg [View same] [iqdb] [saucenao] [google]
10479512

>>10478560
It's quite intuitive,your eyes display in a projective plane of sorts. Don't these parallel tracks appear to intersect at infinity?