New answers tagged

2

You can "help" FindGeneratingFunction by specifying the function space you want it to explore: FindGeneratingFunction[{1, 4, 6, 4, 1}, x, FunctionSpace -> "Polynomial"] This returns the $1 + 4 x + 6 x^2 + 4 x^3 + x^4$ polynomial you expected.


0

With q = {x, Φ} V = (x^2*m^2 + Φ^2*I)/2 fv = Function[{x, Φ}, Evaluate[Total[Flatten[D[V, {q}]]]]][1, pi/8] (* m^2 + (I pi)/8 *) gives the desired result, if I understand the question correctly. Addendum If, as suggested in a comment, it is necessary that x and Φ have explicit t dependence, the following can be used instead. q = {x[t], Φ[t]} V = ...


5

Direct Integration Possible issues in performing the integrations include choice of Assumptions, branch cuts in the integrands, and how limits are taken. Addressing the first of these gives five solutions. il[f_, s_, t_] := Module[{r}, 1/(2 π I) Integrate[f Exp[s t], {s, r - I ∞, r + I ∞}, Assumptions -> r > 2 && t > 0]] ...


4

This is a method that is similar to Oleksandr R.'s answer: The following tells Mathematica to render Primed symbols in superscripted box form: MakeBoxes[Primed[x_], StandardForm] := SuperscriptBox[ToBoxes[x], "\[Prime]"] And this tells Mathematica to parse a superscripted box structure internally as Primed, and not as Derivative: ...


6

It is because there is no general solution for your problem, even for b being a positive integer: When b is odd, the maximum is positive infinity, while for even b it is not. You may want to use things like Maximize[-x^# + a*x, x] & /@ Table[i, {i, 5}] to observe.



Top 50 recent answers are included