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3

It is definitely a bug. And it can be formuated even more sharply. $Version (* Out[156]= "10.1.0 for Microsoft Windows (64-bit) (March 24, 2015)" *) Define f[n_, a_] := (1 + (-1)^n/n^a)^n If a>0 the limit exists and is equal to unity. If a == 0 we have the problem of the OP, where one might say that "two alternative limits exist". This is ...


1

THIS IS AN EXTENDED COMMENT RATHER THAN AN ANSWER. $Version (* "10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016)" *) fRe[x_, σ_] = 1/(Sqrt[2 π] σ) Exp[-(x^2/(2 σ^2))]; Integrate[fRe[x, σ]/(x - X), {x, -∞, ∞}, PrincipalValue -> True, Assumptions -> σ > 0] (* ConditionalExpression[ -((Sqrt[2]*DawsonF[X/(Sqrt[2]*σ)])/σ), Re[X] > 0 ...


0

I am no expert on Boolean logic, but this may be some start for you: homework = Implies[r, b] && Implies[r, Implies[b, f]] && Implies[b, Implies[f, h]] BooleanConvert[homework] LogicalExpand[homework] BooleanTable[homework, {r, b, f, h}] You could check the documentation and the examples of these functions.


3

For k/z = 1, and integrating by parts: ClearAll["Global`*"] Iv[t_] := Exp[-a*t]*t^2; u[t_] := ArcTanh[Sqrt[(t^2 - 1)/t^2]]; v = Integrate[Iv[t], t]; Du = Simplify@D[u[t], t]; Int == u[t]*v - Integrate[Du*v, t] HoldForm[Integrate[Exp[-a*t]*t^2*ArcTanh[Sqrt[(t^2 - 1)/t^2]], {t, 1, Infinity}] == Limit[-((E^(-a t) (2 + 2 a t + a^2 t^2) ArcTanh[Sqrt[(-1 + ...


4

Preliminary post For k/z = 1 there is a closed form result: I[1,1,a]=((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3) The derivation and the extension to k/z != 1 requires some manual interaction to help Mathematica which we will show in the following. Solution Summary As the integral to be calculated is returned unevaluated ...


5

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 ...


2

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 = ...


4

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, ...



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