mr_e_man wrote:(Notice, we couldn't have done this symmetrization so easily before I had expanded it. Or could we?)
Yes, we could!
![Cool 8)](./images/smilies/cool.gif)
No need to take a month to expand and simplify the 4-fold product for a spider; just leave the product as is. I'll explain.
ICN5D wrote:This plots a triangular array of 3 toruses standing on their rim. My next, more recent idea (experiment) was to define a mirror-image (upside-down) triangle array and call the 3rd variable "w" (when it's really z for the plotter) , and then compare it to this one above. Then, try to take all the terms containing w and insert them in to the equation above. It will produce a 4-variable equation that should morph between the triangular arrays by rotation on plane zw. Or, so it seems. It's an experiment I've never tried before, and it's a total shot in the dark, but could totally work, too.
You have the right idea here. It's basically what I did with the symmetry that swaps a and b (the two major radii), and swaps z and w, and rotates in the x,y-plane by 360°/(2n) (in this case 60°, thus replacing the triangle by its mirror image). But I didn't know how to enforce this symmetry and add the appropriate w terms without expanding the product.
My new approach doesn't rely on symmetry; it can take any two 3D objects as cross-sections, defined by polynomial equations F(x,y,z,0)=0 and F(x,y,0,w)=0, provided that they have the same 2D cross-section F(x,y,0,0)=0, and construct a 4D object F(x,y,z,w)=0. Of course it's not unique; any terms can be added with zw as a factor, to get a different shape.
Clearly, from the properties of polynomials, F(x,y,z,0) is exactly all of those monomials in F(x,y,z,w) that don't have w as a factor. So F(x,y,z,w) - F(x,y,z,0) is exactly all of the terms that do have w as a factor. Similarly, F(x,y,0,0) is all of the terms that have neither z nor w as a factor, and F(x,y,z,0) - F(x,y,0,0) is all of the terms that have z but not w as a factor. So we can decompose F into four components as follows:
F(x,y,z,w) = [F(x,y,0,0)] + [F(x,y,z,0) - F(x,y,0,0)] + [F(x,y,0,w) - F(x,y,0,0)] + [zw P(x,y,z,w)],
where the first component has neither z nor w, the second component has z but not w, the third component has w but not z, and the fourth component has both z and w. This can be rearranged to
F(x,y,z,w) = F(x,y,z,0) + F(x,y,0,w) - F(x,y,0,0) + zw P(x,y,z,w).
For our 2n-legged tiger, we have an array of n tori in the x,y,z-space, a different array of n tori in the x,y,w-space, and a common array of 2n circles in the x,y-plane:
F(x,y,z,0) = prod
k=0n-1( (x²+y²+z²+d²-c² - 2
A2k•
r)² - 4b²z² - 4(
b2k•
r)² )
= prod
k=0n-1( (x²+y²+z²+d²-c² - 2A(x cos(2πk/n) + y sin(2πk/n)))² - 4b²z² - 4b²(y cos(2πk/n) - x sin(2πk/n))² ),
F(x,y,0,w) = prod
k=0n-1( (x²+y²+w²+d²-c² - 2
B2k•
r)² - 4a²w² - 4(
a2k•
r)² )
= prod
k=0n-1( (x²+y²+w²+d²-c² - 2B(x cos(π(2k+1)/n) + y sin(π(2k+1)/n)))² - 4a²w² - 4a²(x sin(π(2k+1)/n) - y cos(π(2k+1)/n))² ),
F(x,y,0,0) = prod
k=02n-1( x²+y²+d²-c² - 2
dk•
r )
= prod
k=0n-1 (x²+y²+d²-c² - 2
d2k•
r) (x²+y²+d²-c² - 2
d2k-1•
r).
(Now how can I rewrite that dot product...
d2k•
r =
A2k•
r +
b2k•
r= (a + b cos(π/n))/sin(π/n) (x cos(2πk/n) + y sin(2πk/n)) + b (y cos(2πk/n) - x sin(2πk/n))
= (ax cos(2πk/n) + bx cos(2πk/n + π/n) + ay sin(2πk/n) + by sin(2πk/n + π/n)) / sin(π/n)
= (a (x cos(2kπ/n) + y sin(2kπ/n)) + b (x cos((2k+1)π/n) + y sin((2k+1)π/n))) / sin(π/n),
d2k-1•
r =
A2k•
r -
b2k•
r= (a + b cos(π/n))/sin(π/n) (x cos(2πk/n) + y sin(2πk/n)) - b (y cos(2πk/n) - x sin(2πk/n))
= (ax cos(2πk/n) + bx cos(2πk/n - π/n) + ay sin(2πk/n) + by sin(2πk/n - π/n)) / sin(π/n)
= (a (x cos(2kπ/n) + y sin(2kπ/n)) + b (x cos((2k-1)π/n) + y sin((2k-1)π/n))) / sin(π/n). )
So we add the two n-fold products, and subtract the 2n-fold product, and finally add (or subtract) some z²w² terms, to get an equation for a 2n-legged tiger. But this doesn't work well for the ordinary 4-legged tiger; the "remainder" is fully 8th degree, with lots of terms that would be impossible to guess, if we want the correct exact shape:
F(x,y,z,w) = (...) + z²w² (12(x²+y²-c²)² - 8(a²-b²)(x²-y²) - 8(a²+b²)c² + 4(a²+b²)² - 8(a²-b²)² - 64a²b² + 12(x²+y²-c²)(z²+w²) + 4(a²-b²)(z²-w²) + 4z⁴ + 6z²w² + 4w⁴).
In contrast, my previous approach, of partially expanding the product and adding w² terms where they look like they should fit to match z², only requires further subtracting 64a²b²z²w² at the end to get an exact tiger. But my new approach is much easier for larger n. And we have no way to tell what a "correct" 2n-legged tiger should look like (beyond its general topology); no way to tell what terms should be added at the end, with either approach.
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