/sci/ - Science & Math » Thread #10477587

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Simple visual explantioin needed

Anonymous Mon 18 Mar 14:25:19 2019 No.10477587 View Reply Original Report

Quoted By: >>10477616 >>10477748 >>10478010 >>10478078 >>10478790

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

Anonymous

Anonymous Mon 18 Mar 2019 14:41:49 No.10477616 Report

Quoted By:

>>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.

Anonymous

Anonymous Mon 18 Mar 2019 14:45:53 No.10477621 Report

Quoted By: >>10477628

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

Anonymous

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Anonymous Mon 18 Mar 2019 14:49:24 No.10477628 Report

Quoted By: >>10477633

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

Anonymous

Anonymous Mon 18 Mar 2019 14:52:42 No.10477633 Report

Quoted By: >>10477647

>>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

Anonymous

Anonymous Mon 18 Mar 2019 14:59:24 No.10477646 Report

Quoted By: >>10477648

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

Anonymous

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Anonymous Mon 18 Mar 2019 15:01:57 No.10477647 Report

Quoted By: >>10477648 >>10477679

>>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.

Anonymous

Anonymous Mon 18 Mar 2019 15:03:14 No.10477648 Report

Quoted By: >>10477662

>>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

Anonymous

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Anonymous Mon 18 Mar 2019 15:11:24 No.10477662 Report

Quoted By: >>10477674 >>10477676

>>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.

Anonymous

Anonymous Mon 18 Mar 2019 15:20:35 No.10477674 Report

Quoted By:

>>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

Anonymous

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Anonymous Mon 18 Mar 2019 15:22:02 No.10477676 Report

Quoted By: >>10477911

>>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?

Anonymous

Anonymous Mon 18 Mar 2019 15:22:22 No.10477679 Report

Quoted By:

>>10477647
>brainlette

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

Anonymous

Anonymous Mon 18 Mar 2019 15:52:29 No.10477748 Report

Quoted By: >>10477753

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

Anonymous

Anonymous Mon 18 Mar 2019 15:55:26 No.10477753 Report

Quoted By: >>10477898 >>10478619

>>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

Anonymous

Anonymous Mon 18 Mar 2019 17:06:00 No.10477898 Report

Quoted By:

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

Anonymous

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Anonymous Mon 18 Mar 2019 17:13:50 No.10477911 Report

Quoted By:

>>10477676
pruty

Anonymous

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Anonymous Mon 18 Mar 2019 18:04:24 No.10478010 Report

Quoted By:

>>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.

Anonymous

Anonymous Mon 18 Mar 2019 18:36:43 No.10478078 Report

Quoted By: >>10478560

>>10477587
Read up on Bezout's Theorem. A circle is an algebraic plane curve of degree 2, so 2 circles intersect $2\cdot 2 = 4$ 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

Anonymous

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Anonymous Mon 18 Mar 2019 22:31:12 No.10478560 Report

Quoted By: >>10479512

>>10478078
>allow parallel lines to intersect at infinity

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

Anonymous

Anonymous Mon 18 Mar 2019 22:39:21 No.10478582 Report

Quoted By: >>10478705

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

Anonymous

Anonymous Mon 18 Mar 2019 22:55:51 No.10478619 Report

Quoted By: >>10478783

>>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

Anonymous

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Anonymous Mon 18 Mar 2019 23:34:27 No.10478705 Report

Quoted By:

>>10478582

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

Anonymous

Anonymous Tue 19 Mar 2019 00:19:51 No.10478783 Report

Quoted By:

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

Anonymous

Anonymous Tue 19 Mar 2019 00:26:00 No.10478790 Report

Quoted By: >>10478851

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

Anonymous

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Anonymous Tue 19 Mar 2019 00:55:12 No.10478851 Report

Quoted By: >>10478915

>>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.

Anonymous

Anonymous Tue 19 Mar 2019 01:31:14 No.10478915 Report

Quoted By:

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

Anonymous

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Anonymous Tue 19 Mar 2019 05:50:11 No.10479512 Report

Quoted By:

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

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