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New answers tagged symbolic

1

If you look at the output of Solve[Z + Y #1 + X #1^2 - Conjugate[Y] #1^3 + Conjugate[Z] #1^4 &[x] == 0, x, Quartics -> True] you'll see that the solutions in radicals are far more complicated. How do I treat solutions in terms of Root? I leave them that way. Root is nice. It's handier to have solutions ordered by magnitude than by principal values ...

2

The first result can be obtained using Dirichlet regularization: Sum[n, {n, 1, Infinity}, Regularization -> "Dirichlet"] -(1/12) The second can not be obtained, though. I don't have enough smarts to know if this is because it would actually require different regularization, or that Mma just doesn't know how to handle this case.

18

Usage: dChange[expresion, {transformations}, {oldVars}, {newVars}, {functions}] You can also skip {} if a list has only one element. (Update) Examples: 0. wave equation in retarded/advanced coordinates dChange[ D[u[x, t], {t, 2}] == c^2 D[u[x, t], {x, 2}], {a == x + c t, r == x - c t}, {x, t}, {a, r}, {u[x, t]} ] c Derivative[1, 1][u][a, r] == ...

6

UnitConvert[Quantity[3, "PlanckConstant"], "ReducedPlanckConstant"] /. x_?NumericQ :> RootApproximant[x/Pi]*Pi Quantity[6*Pi, "ReducedPlanckConstant"]

4

One way is shown below. Alternatives include using Simplify with Assumptions, again on testing that the eigenvalues are nonnegative. Resolve[ ForAll[p, 0 <= p <= 2, And @@ Thread[ Eigenvalues[{{1, 0, 0, Sqrt[1 - p]}, {0, 0, 0, 0}, {0, 0, p, 0}, {Sqrt[1 - p], 0, 0, 1 - p}}] >= 0]]] (* Out[9]= True *)

2

Your transformation is not generally true; if you provide FullSimplify with your assumptions you get a better result: expr = (Sqrt[2] Sqrt[Ea - g L m])/Sqrt[m]; FullSimplify[expr, m > 0] Sqrt[-2 g L + (2 Ea)/m]

3

One can certainly compute a[n] for arbitrary positive integer $n$. a[n_] := 1 - Sum[Binomial[n, k] 2^(n - k - 1) a[k], {k, 0, n - 2}] - 2 n a[n - 1]; a[0] := 1; a[1] := -1; a[2] := 3; a[15] (* $-694475294514315$ *) ListLogPlot[Table[a[i], {i, 1, 20}]] Just note that many values of a[i] are negative and won't show up on the ListLogPlot. However, ...

4

You need to tell M that all symbols are real: expr = (1 + (x + I*y)/2 + (x + I*y)^2/12)/(1 - (x + I*y)/ 2 + (x + I*y)^2/12); ComplexExpand@ Abs @expr

3

Update Since you gave a good proof, that $2/\pi$ is the correct solution, Mathematica is obviously failing at the task. The question is, why. Analysis of Mathematica's behavior First, the sum you gave is no Riemann sum: You defined the intervals as $$\Delta x_k=\frac{1}{n+1/k}$$ so $k/n$ does not lie within the appropriate subinterval, e.g. for $n=10$, ...

2

fixed in 10.1 (windows): code: Clear[x] Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi

4

Yesterday I found the approach below with Hold/ReleaseHold on v10.0.0 on Win8.1 achieves the same result as v8.0.4, namely, it gives a limit of $\frac{2}{\pi}$. ReleaseHold@Limit[ Hold[ Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}] ], n -> Infinity] (* 2/Pi *) However, on v10.0.2 on Linux, this approach gives the result shown below...as does ...

5

You can gain insight by evaluating and then simplifying the indefinite integrals. Integrate[(1 + b x + c y)/(1 + e x + f y + I η), y, x, Assumptions -> {x ∈ Reals, y ∈ Reals, b ∈ Reals, c ∈ Reals, e ∈ Reals,f ∈ Reals, η ∈ Reals, η != 0}]; ans = Collect[%, {ArcTan[η/(1 + e x + f y)], Log[1 + e^2 x^2 + 2 f y + f^2 y^2 + 2 e (x + f x y) + η^2]}, ...

5

One way to approach this is to look for the source of the problem. In this case, the outer integral (on y) is irrelevant to the problem, because even the simpler integral int = Integrate[(1 + b x)/(1 + e x + I η), {x, -1/2, 1/2}, Assumptions -> {b ∈ Reals, c ∈ Reals, e ∈ Reals, η ∈ Reals, η != 0}] gives a conditional expression. But now the answer ...

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