# User Portlet

Gianluca Gorni
Discussions
It is subtle. If you write y' for the velocity, then In[35]:= D[y', y] Out[35]= 0 & because the internal form of y' is Derivative[1][y], which does contain y. My advice is either to introduce the time variable throughout ...
You can always rescale the independent variable, for example this way: independentVariableChange = (a*t + b == x) dependentVariableChange = (g[t] == f[x]) dependentVariableChange /. Solve[independentVariableChange, x][[1]] % /....
It looks like a bug. The correct value for the sum seems to be the absolute value of the formula given by Mathematica: s[n_] = Abs[Sum[(1 - k/(n + 2))^(1/n), {k, 1, n}]]; Table[Abs[s[n] - Sum[(1 - k/(n + 2))^(1/n), {k, 1, n}]], {n, 1, ...
No need for InverseFunction. Simply reverse the pairs of points to interpolate: Plot[x + Sin[x], {x, 0, 10}] ptsReversed = Table[N[{x + Sin[x], x}], {x, 0, 10, 1/10}]; ListPlot@ptsReversed inv = Interpolation[ptsReversed,...
You are not giving a value for L.
You may have a look at my old package [CurvesGraphics6][1] . Much of it has been gradually superseded by built-in functionality, but not everything yet. PhasePlot may help for your problem. [1]: https://www.dimi.uniud.it/gorni/Mma
I am baffled too. Last week I had two polyhedra with rational coordinates, whose RegionIntersection was floating-point. One would be led to think that geometric computing is done with floats. Now we see an example that goes in the opposite...
Floating point numbers are not very suitable for this calculation. I would do it with exact numbers: Reduce[0
Here is a way: DynamicModule[{p1, pwr}, pwr[pt_List, k_Integer] := ReIm[({1, I} . pt)^k]; Manipulate[ Row[{Graphics[{Point[p1]}, PlotRange -> 2, Frame -> True], Graphics[{Point[pwr[p1, 2]]}, PlotRange -> 2, Frame...
You can make a replacement in the output using RealAbs instead of Abs, but then check that the result is correct: f = 1/(x (x - 1)); Integrate[f, x] /. Log[u_] :> Log[RealAbs[u]] FullSimplify[%, Element[x, Reals]] /. Abs ->...