# Tag Info

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]] ...

3

The direct way works: Times @@ Table[ x - With[{r = Root[#1^3 + a2 #1^2 + a1 #1 + a0 &, i]}, r^3 + r + 1], {i, 3}] // Expand // Simplify // CoefficientList[#, x] & {-1 + 4 a0 - 3 a0^2 + a0^3 - a1 - 2 a0 a1 + 2 a1^2 + a0 a1^2 - a1^3 + a2 + a0 a2 - 2 a0^2 a2 - 3 a1 a2 + 3 a0 a1 a2 + a0 a2^2 - a1 a2^2 + a2^3, ...

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.

2

I upvoted the other responses. That said, there is a better way. CoefficientList[ Resultant[x^3 + a2*x^2 + a1*x + a0, y - (x^3 + x + 1), x], y] (* Out[1179]= {-1 + 4 a0 - 3 a0^2 + a0^3 - a1 - 2 a0 a1 + 2 a1^2 + a0 a1^2 - a1^3 + a2 + a0 a2 - 2 a0^2 a2 - 3 a1 a2 + 3 a0 a1 a2 + a0 a2^2 - a1 a2^2 + a2^3, 3 - 6 a0 + 3 a0^2 + a1 - 2 a1^2 + a1^3 - 2 a2 - ...

2

A slick way is to use the Newton-Girard formulae in conjunction with the handy RootSum[] function: Solve[Table[s[m] == RootSum[Function[x, x^3 + b x^2 + c x + d], Function[r, (r^3 + r + 1)^m]], {m, 3}] ~Join~ Table[-Sum[s[k] e[m - k], {k, m - 1}] - m e[m] == s[m], {m, 3}], Array[e, 3], Array[s, 3]] // Expand ...

1

Another direct approach p1 = #1^3 + a2 #1^2 + a1 #1 + a0 &; p2 = 1 + # + #^3 &; Simplify@CoefficientList[Product[x - p2@Root[p1, i], {i, Exponent[p1@x, x]}], x] Or make it a function that takes two polynomials, p1 and p2, and produces the coefficients of the polynomial p2 evaluated at the roots of p1 ClearAll[f] f = ...

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