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One way to do it is with pattern replacement: factorFromSum = Sum[a_*b_, {var_, lmts__}] /; FreeQ[b, var] :> b*Sum[a, {var, lmts}]; Sum[2*b0*b1*Subscript[x, i], {i, 1, n}] + Sum[b1^2*Subscript[x, i]^4, {i, 1, n}] /....
It seems that `Region` makes a rough plot to display quickly.
You can see from the plot that the primitive is discontinuous at the point where you evaluate it: Block[{\[Beta] = 1, \[ScriptCapitalD]LS = 1, Rs = 2}, ReImPlot[S, {R, Rs - \[Beta] \[ScriptCapitalD]LS - 1, Rs + \[Beta]...
At the first attempt the plot range was random, but I was lucky. Then I collected the coordinates of the intersection with the mouse, using the "get coordinates" tool. I used that value as seed for `FindRoot`.
Are you sure it is `F(lambda)` instead of `F[lambda]`? If you define `c1 = RandomReal[3,1]` you get a list instead of a number: is that really what you want? I would simply write `c1 = RandomReal[3]`.
As for the primitive of `Abs[ff[x,m,s]]`, I get back the input, not 0. The square of `ff` is a simple sum of Gaussians, there is no problem in finding a primitive in terms of the error function `Erf`, and in calculating the limits at infinity.
Just avoid the construct `f[var]` in the definition of `myPlot`: myPlot[f_, var_, a_, b_] := Manipulate[Plot[f, {var, a, b}, Epilog -> {PointSize[.02], Red, Point[{pnt, f /. var -> pnt}]}], {pnt, a, b}] This...
I don't know, but maybe because with this definition the four vectors are positively oriented: In[8]:= v2 = {0, 1, 0, 0}; v3 = {0, 0, 1, 0}; v4 = {0, 0, 0, 1}; cross = Cross[v2, v3, v4]; Det[{v2, v3, v4, cross}] ...
Sorry, try this: With[{u = 1.63}, Plot[{u*M0*Theta[QoverM*M0, M0]/10^4, u*M1*Theta[QoverM*M1, M1]/10^4, u*M2*Theta[QoverM*M2, M2]/10^4}, {QoverM, 0, 1}, PlotRange -> All, PlotStyle -> {{Blue, Dashed}, {Green}, {Red,...
With `DateObject` it may be easier to manipulate further: Map[DateObject[#, DateFormat -> {"Day", "/", "Month", "/", "YearShort"}] &, ita[[1]]]