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| @Michael > As you also may know, one can derive the singular solution without integration. > Blockquote No, I didn't know that. |
| I am afraid I do not understand your problem, but what do you think about this? Define a tangent-vector tvec[u_, f_] := Module[{}, abl = D[f, x]; v1 = {u, f /. x -> u}; v2 = {u + .1, (f /. x -> u) + .1 (abl /. x -> u)};... |
| Ok. I have a stone-age-old version of Mathematica, and I am afraid the 3D procedure won't run on my system. What is returned by the 3D function? If these were 3D vectors (look at the output ), you might get 2 D vectors by vecs3D = Flatten[... |
| Perhaps change Times to another function res = CForm[mult[(N[5/3]), (s/N[1/9])^(N[1/13])]] later then res /. mult -> Times |
| @Henrik: Coool. Greetings, Hans |
| Here is an more elemantary way: 3.48382834 e - 09 /. Plus[x_, Times[y_, e]] -> y 10^x |
| Dear Sukanta, ok. theta is unknown. I am afraid that your problem is next to impossiible, this can be due to numeric problems and / or the existence of multiple solutions. At least I got several times error messages (fail to converge...). Look... |
| The differences? Gianlucas method use 2 Jordandecomposition, my method translates your problem to a sort of Gauss-elimination, which after all could be described as similar. Pros and cons? I don't know, I just wanted to see if this method works. ... |
| Exactly! Given a and m a = {a1, a2, a3} m = {m1, m2, m3} Define with an arbitrary parameter x f = x a - m Then Cross[a, m] Cross[f, a] |
| ERROR of mine! If you want the area of the triangle you have to divide the foregoing results by 2. |