Shouldn't it be L^2 and L2^2 ?
Furthermore I don't understand why you combine X1 with Y2 and X2 with Y1 in your plot!?
The intersection of the two circles are just (X1/Y1) and (X2/Y2)
Regards
Werner
Werner Exinger wrote:
Shouldn't it be L^2 and L2^2 ?
Yes, you're right. That's was an attemp to see if there are somplification about Discriminant, but aren't, only appear L instead L^2. The measured distances, in that case, are sqrt(L) and sqrt(L2), and must to correct also the constant k. But the general arrange is the same.
Werner Exinger wrote:
Furthermore I don't understand why you combine X1 with Y2 and X2 with Y1 in your plot!?
To see which or if there are strange solutions, introduces by the algebraic manupulations isolating x and y. Without polar representation of the curves, and with that complicated discriminant, I don't see which points could be "strange solutions". Notice that in the interval there are a lot of negative Delta values that carry complex values for x and y.
Best regards.
> I don't see which points could be "strange solutions". Notice that in the interval there are a lot of negative Delta values that carry complex values for x and y.
I still don't see the reason for combining the absicissa from one solution with the ordinate of the second. Points derived that way aren't solutions. And yes, a lot of values for L would produce non-real values for L2 - Thats slowing down calculation and plotting a bit but I found no (quick) way to determine the valid range for the first distance L depending on the constant C (kappa in your sheet) and the distance between the two "focals".
Here is what I get using your sheet without combining coordinates from different points of intersections and its definitely not a quartic but of higher order than four, if its algebraic at all. I scaled to get equal scale on both axis as at first I was confused why the plot is not symmetrical according to the line through the two "focals" ;-)
Regards
Werner
Hi Werner. Thanks for the attached.
"Strange solutions" are those introduced by some algebraic manipulations in the process of isolating variables. For example, from
x-1 = sqrt(3-3x),
you can get
x^2-2x+1 = 3-3x,
then
x^2+x-2=0
which gives x=-2 and x=1. But actually, x=-2 is a "strange solution" introduced by the algebraic manipulation, because don't hold the original equation, and must to be discarded. The only one solution for the original equations is x=1. Maple or mupad solution seems to not introduce those strange solutions. But this must to be, maybe if not full prove, but analyzed at least.
About the complexes values, the "quick" way to prevent them, or, better say, the domain for x, is with the discriminant, i.e. the under radical expression, the Delta(L) in the worksheet. But complex, and function of all other variables: alpha, beta, L, kappa. With a for loop can evaluate it, stored in an intermediate variable for speed the calculations, and continue with x, y1, y2 only when Delta(L) > 0. But that's for speed. Let's the machine do that inutile work.
About x1 vs y1,y2 or x2 vs y1,y2, remember that in geometry, unlike calculus, x and y are equivalents. I don't want to discard x2 prematurely.
Best regards.
Alvaro.
More easy setup, assuming line between focus as y=0, and symetric about the origin. Can be rotated and translated later for general case. For this, Delta have more sense, and can find roots. Unfortunatelly, mupad don't give all roots for the discriminant, only one. So, MC11 better.
Parametrizacion as function of L, canonical parametrization along s = arc length too complicated, can't simplify the integral yet.
Path with holes, aren't analityc. Bye bye complex variable tools. So, imaginary nor algebraic representation could take some time.
For better plots, adaptive steps must neede, like the routines in Amazing Images ebook.
Best regards.
Alvaro.
Attached a mc15 worksheet, saved as mc11, with some "easy" (most precisely, simmetrical) expressions for the discriminant, polar and others properties of the Valery's exponential curve.
Best regards.
Alvaro.