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Hans Dolhaine
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What do you mean by "cannot be used for further calculation"? If I copy your result du[x] to to a new variable dux then I get a number with dux /. {C1 -> .2, C2 -> 1.2, C3 -> .66, C4 -> .9, Ks -> 1, Subscript[q, 1] -> .01, x -> 5, l ->...
If your sum is finite, (and if it it this what you mean ) sum = a*1 + a*2 +a*3 +.....+ a*k sum = a* ( 1 + 2 + ..... + k ) sum = a* k *(k + 1 ) / 2
I don't know exactly what you mean. Something like this perhaps? pc[n_] := Module[{}, Table[ {Cos[2 Pi j /n], Sin[2 Pi j/n], 0}, {j, 0, n}] ] e1 = Join[{{0, 0, 1}}, #] & /@ Partition[pc[8], 2, 1]; e2 =...
I think you have already what you want Solution to your 1st equation fd = Exp[-c t] (c1 (-1 + Exp[(c - g) t] + D0 (c - g)))/(c - g) ODE D[fd, t] - c1 Exp[-g t] - c fd Rewrite the solution fdx = fd /. t -> x Then ...
Hello Stefano, long time ago I learnt that it is better not to use variables with a "complicated" structure like Superscript and so on. So it seems to be better to use "simple" expressions. You could change that like \[CurlyPhi] = ...
Is there a function....?, may be, but certainly there is a general formula (even valid in R^n). I really appreciate that "RegionDistance"-stuff, but I prefer an old fashioned understandable way. The line is given by two points v[x] and v[y] and can...
Or maybe this is a bit better p1 = {.5, 2}; p2 = {3.7, 4.5}; g1[t_] := p1 (1 - t) + p2 t pl1 = Graphics[{Red, Arrow[{g1[.1], p1}], Blue, Arrow[{g1[.9], p2}]}]; pl2 = Graphics[{Thick, Red, Table[Line[{g1[2 j/10],...
What do you think of Interpolation[data] ?
Hello Pierre, with my stone-age-old Version 7 both integrals give zero. Perhaps you should check this numerically. Your integrand is, let's say, somewhat special (for a > b I wrote a = b + x and x >0 ). But I am quite sure that there are...
There is something wrong with your code, which I have modified to get a solution. I have no idea about your initial/boundary conditions. At least the code below produces plots and shows what you could do sol = NDSolve[ { ...