Maybe these functions are wrongly set. I deduced them by writing: x ^ n + y ^ n = 1 whose derived equation is: x ' * x ^ (n-1) + y ' * y ^ (n-1) = 0 and the solution : x ' = - y ^ (n-1) and y ' = x ^ (n-1) Is it possible that there is a transform function f such that (d / dx) qin (f (x)) = qos (f (x))? I'm looking for something simpler but I do not know if it's possible. Le dimanche 27 mai 2018 à 21:43:56 UTC+1, Allan Wechsler <acwacw@gmail.com> a écrit : I have followed this exploration of "squircular trigonometry" with some interest. It feels to me like an important piece of the "soul" of circular trigonometry is missing from the proposed generalization, and I wonder if the gap can be filled. Namely, sin and cos are part of a system of functions that are intimately related by a cycle of differential equations. D sin = cos, D cos = -sin, and so on. The derivative formulas for qin and qos are not nearly as pretty. Perhaps this means we have mis-parametrized these functions? Is it possible that there is a transformation function f such that (d/dx) qin(f(x)) = qos(f(x))? And, if so, maybe the whole qin-qos universe might look prettier viewed through the lens of this transformation? To put it another way: if we permit a point to travel around the squircle at non-constant speed, so its angle is theta(t), then can we find a function theta so that D y = x? On Sun, May 27, 2018 at 3:53 PM, françois mendzina essomba2 via math-fun <math-fun@mailman.xmission.com> wrote: I keep in memory (qos(x))^4+(qin(x))^4=1 and : qin(sigma/2)=1, qin(sigma/4)=2^(-1/4) . I looked for the equivalent of the formula Gamma(1-z)*Gamma(z)=Pi/(sin(z* Pi)) with the function qin(x). I imagined that the following equalities are correct, I could verify numerically. integrate((1-x^4)^(-3/4), x, 0, 1)=(sigma/2)/qin(sigma/2); integrate(x^4*(1-x^8)^(-7/8), x, 0, 1)=(sigma/4)/qin(sigma/4); but I was wrong about this equality, it is not correct. integrate(x^12*(1-x^16)^(-15/ 16), x, 0, 1) = (sigma/8)/qin(sigma/8); Why ??? The first answer may be related to the nature of the squircle itself, on an axis, its area can not be divided into several equal parts. We can only get 2 or 4 equal parts. but I remain convinced that there must be this type of formula ______________________________ _________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/ cgi-bin/mailman/listinfo/math- fun