Mapthematics Forums

nonanalytic points are at acos(1/3)/2 and pi/2-acos(1/3)/2. any smooth transition functions could be used, but p1 and p2 need to be adjusted to gain resolution-efficiency (consider off the apothem). h_1=acos(1.0/3)/2 h_2=pi/4-acos(1.0/3)/2 p1=0.6482 (?) p2=6.0 (?) f_2m(x)=p1+f[1]/1!*(x-h_1)+f[2]/2!*...

No. Presumably, the resulting Taylor series would have a finite radius of convergence . Ideally, the radius of convergence would be pi/4 - acos(1/3)/2, so convergence fails at the same point analyticity does. Is it possible to control the radius of convergence in this way? It may be possible, but i...

I'm sorry, I just couldn't find the time. (I was trying to figure out which symbol to use for acos(1/3)/2 and pi/4-acos(1/3)/2 avoiding a, b, c and d. m = acos(-1/3) (Maraldi angle) m_1 = pi - m = acos(1/3) m_2 = pi/2 - m_1 = pi/2 - acos(1/3) (m_3 = pi/4 - m_2 ?) h_1 = m_1 / 2 = acos(1/3)/2 h_2 = m_...

Please wait a while as I would like to derive the C3 class solution on my own. From numerical observation, it seems that the coefficient of the third-order term at the endpoints (actually the fifth-order term due to symmetry) can be left as is or halved to achieve optimal resolution-efficiency. g3 =...

I never imagined that differential coefficients could be so reusable. Great! Class C^1 (First-differentiability): confirmed. f_2(x) = (x-pi/4) * pi/3/acos(1.0/3) + pi/4 Class C^2: confirmed. I now assume f_2(x) = ( x+g*(3*(x-pi/4)-(x-pi/4)**3/(pi/4-acos(1.0/3)/2)**2) - pi/4 )*pi/3/acos(1.0/3)+pi/4 a...

f(a) = constant*a - f(asin(cos(a)/sqrt(2))) I believe you mean plus, not minus? Or f(a) - f(asin(cos(a)/sqrt(2))) = constant*a. My apologies. Corrected. And calm down. It was meant to be a definition, rather than an equation... f(x): tweaking function, f(x) := f_1(x) when x ∈ [0,acos(1/3)/2], f(x) ...

d = f(a) + f(b) + f(c) - pi/2 x = sqrt(1/3) * (f(b) - f(c)) y = f(a) - d/3 = (2*f(a) - f(b) - f(c) + pi/2) / 3 ...But when you do this, it's far too hard to figure out any meaningful characterization of f (even in terms of its inverse function), even in the special case where b = c = asin(cos(a)/sq...

Center defined how? There are many different ways of determining the center of a non-equilateral triangle, further complicated by the fact that we're properly talking about a circular triangle (convex type) rather than the normal kind. Sorry, I don't use arcs in the x-y plane at all. (The story is ...

Looks promising! Good job. ...Should I be worried that the minor axis graph of your C0 projection looks like a radioactive hazard symbol? Thanks! I thought it's Mitsubishi... 8-) k*(a+b+c) = 1 Ah, if we say trilinear coordinates, it will be interpreted as that kind of normalization. Not a good name...

Dihedral projection? As may be well known, it is possible to convert between a hemisphere and a octant without elliptic integrals. hemisphere -> (Lagrange transform) -> 90° gore -> (rotate) -> (Lagrange transform) -> octant Three corners of this octant are from 90°-90°-180° points of the great circl...