this post was submitted on 17 Apr 2024
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submitted 7 months ago by [email protected] to c/[email protected]
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[–] [email protected] 16 points 7 months ago (2 children)

The number of solutions/roots is equal to the highest power x is raised to (there are other forms with different rules and this applies to R and C not higher order systems)

Some roots can be complex and some can be duplicates but when it comes to the real and complex roots, that rule generally holds.

[–] [email protected] 4 points 7 months ago* (last edited 7 months ago) (3 children)

I think you can make arbitrarily complicated roots if you move over to G^n^ which includes the R and C roots...

For example the grade 4 blade (3e1e2e3e4)^2 = 9 in G^4^

Complex roots are covered because the grade 2 blade (e1e2)^2 = -1 making it identical to i so G^n^ (n>=2) includes C.

G^n^ also includes all the scalars (grade 0 blades) so all the real roots are included.

G^n^ also includes all the vectors (grade 1 blades) so any vector with length 3 will square to 9 because u^2 = u dot u = |u|^2 where u is a vector.

All blades will square to a scalar but blades are not the only thing in G^n^ so things get weird with the multivectors(sums of different grades). Any blade with grade n%4 < 2 will square to a positive scalar and the other grades will square to a negative, with the abs of the scalar equal to the norm^2^ of the blade. Can pretty much just make as many roots as you want if you are willing to move into higher dimensional spaces and use a way cooler product.

[–] [email protected] 8 points 7 months ago (1 children)

You lost me at "arbitrarily complicated," sorry.

[–] [email protected] 8 points 7 months ago (1 children)

Lost me at "I think". I don't, apparently.

[–] embed_me 6 points 7 months ago (1 children)

I don't think therefore I never was

[–] [email protected] 7 points 7 months ago

I wish it was that easy! I stopped thinking and I'm still stuck here... Stupid Descartes

[–] [email protected] 3 points 6 months ago (1 children)

I thought this would be related to quaternions, octonions etc. but no, it's multivectors and wedge products. Very neat, I didn't know you could use them like that.

[–] [email protected] 2 points 6 months ago* (last edited 6 months ago)

Oh no, you were right on the money. In G^2^ you have two basis vectors e1 and e2. The geometric product of vectors specifically is equivalent to uv = u dot v + u wedge v.. the dot returns a scalar, the wedge returns a bivector. When you have two vectors be orthonormal like the basis vectors, the dot goes to 0 and you are just left with u wedge v. So e1e2 returns a bivector with norm 1, its the only basis bivector for G^2^.

e1e2^2 = (e1e2)*(e1e2) = e1e2e1e2

A nice thing about the geometric product is its associative so you can rewrite as e1*(e2e1)*e2.. again that middle product is still just a wedge but the wedge product is anti commutative so e2e1 = -e1e2. Meaning you can rewrite the above as e1*(-e1e2)*e2 = -(e1e1)*(e2e2) = -(e1 dot e1)*(e2 dot e2) = -(1)*(1) = -1.. Thus e1e2 squares to -1 and is the same as i. And now you can think of the geometric product of two vectors as uv = u dot v + u wedge v = a + bi which is just a complex number.

In G^3^ you can do the same but now you have 3 basis vectors to work with, e1, e2, e3. Meaning you can construct 3 new basis bivectors e1e2, e2e3, e3e1. You can flip them to be e2e1, e3e2, e1e3 without any issues its just convention and then its the same as quaternions. They all square to -1 and e2e1*e3e2*e1e3 = -e2e1e2e3e1e3 = e2e1e2e1e3e3 = e2e1e2e1 = -1 which is the same as i,j,k of quaternions. So just like in G^2^ the bivectors + scalars form C you get the quaternions in G^3^. Both of them are just bivectors and they work the same way. Octonions and beyond can be made in higher dimensions. Geometric algebra is truly some cool shit.

[–] [email protected] 1 points 6 months ago

Then you can extend to arbitrary algebra