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Sorry, Frank, I'm the real Idiot the original post from Ilian was: area[R_] = Integrate[Boole[x^2 + y^2 R > 0]; FindRoot[area[R] - Pi/2, {R, 1}, WorkingPrecision -> 20] (* Out[4]= {R -> 1.1587284730181215178} *)
Hi, all thanks a lot for your brilliant contributions. But what are the conclusions for the the ' not so deeply involved' users? Do we need further instructions ( more than the documented ones) about using Solve, Roots, Reduce, Setprecision, ... or...
@ Shenghui: I hope, that it's not necessary for an M-newbie to understand, why and how -M- does this, after changing 0.1 to 1/10 @ W. Craig: obviously you've done something completely different ... I wish all of you a happy new year!, Peter
Hi, C ormullion, thanks a lot for your answer. I'll  search and try to find further infos to understand WL!  Peter
Hi, my dear friend! Do you search the following solutions?In[13]:= x = ArcCos[-1.7677669529663687/5]; K = x - Pi/2; Table[(2 i - 1)/2*Pi + (-1)^(i - 1) K, {i, 5}] Out[15]= {1.93216, 4.35102, 8.21535, 10.6342, 14.4985}Peter
Although I'm a Mathematica-newbie, you first should check your code. There are some inconsistencies, for example: VDD vanishes, what about the semicolon ( following ' lamda')? Why are there the many unnecessary parentheses?
Hi Arnoud, I will purchase the program, but before I've some questions. I've got some emails from your company. Should I ask my questions there or at your contact option in my test version. I've found no way to contact you. I think this place here is...
after spending some time with mathematica I've found two other solutions without integration ( circlesegment,-sector)[mcode]In[13]:= Module[{a, b}, a = ArcCos[1 - R R/2]; b = ArcCos[R/2]; FindRoot[2 a - Sin[2 a] + R R*(2 b - Sin[2 b]) == Pi, {R, 1,...