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/sci/ - Science & Math

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15342472 No.15342472 [Reply] [Original]

was messing around with desmos and found these functions

f(x) = (x^n)/a

g(x) = f(x)^2 - f(x)

the roots of g(x) = the nth root of a

I found this interesting so I went to wolfram alpha and plugged in a test

https://www.wolframalpha.com/input?i=%28%28x%5E3%29%2F27%29%5E2+-+%28%28x%5E3%29%2F27%29

if you scroll down you'll see three points on the circumference of a circle with one of them being 3 - the third root of 27

I found this even more interesting so I plugged in a really big number for n and got the following

https://www.wolframalpha.com/input?i=%28%28x%5E100%29%2F100000%29%5E2+-+%28%28x%5E100%29%2F100000%29

if you wait for Wolfram to compute all the roots you'll find a 100 vertex NGon with a radius of the hundredth root of 100000

that got me thinking if I could do some kind of limit and substitute x for re^i*theta and to my surprise, it actually worked. and by adjusting theta you can rotate the points of the NGon around the origin

https://www.wolframalpha.com/input?i=%28x%5E4%29-%28%289%5E2%29e%5E%28i+*+pie%2F2%29%29%3D0

can someone who actually went to a university explain what's happening here? I learned calculus from youtube videos so I really have no idea, I'm just plugging in equations.

>> No.15342480

>>15342472
You're actually retarded, if x is a root, then f(x) = 0 or 1, for the latter case this means (x^n)/a=1.

>> No.15342496

>>15342480

? so aggressive, maybe we're just having a misunderstanding?

square root of two written with those equations ((x^2)/2)^2 - ((x^2)/2) where x = root 2

https://www.wolframalpha.com/input?i=%28%28x%5E2%29%2F2%29%5E2+-+%28%28x%5E2%29%2F2%29

>> No.15342518

>>15342480
wait no I see what you're saying now, but I'm referring to the imaginary roots as well. not just the real solutions. You're right, I am pretty stupid, but what's new?

>> No.15343465

make pics and post them here from your wolfram alpha calculations, aint nobody clicking that shit and running those calculations

>> No.15344101

>>15342472
The roots of g(x) are x^(2n) - ax^n = 0 which are exactly x^n - a = 0 (and x=0).
The roots of x^n -a = 0 are (a)^(1/n) (The positive nth root is prefered but actually any can be taken) times those of x^n - 1 = 0. Hope this helps :)

>> No.15344384

>>15344101
Oh, so this is just the nth roots of unity with extra steps?

>> No.15344443

>>15344384
Yeah, exactly. Usually things that have some rotational symmetry along a point, are related to the roots of unit, special points if it's along the point 0+0i. More interestingly, if you have a set of finite points with rotational symmetry they correspond to scaled + translated + initial rotations of roots of unity.