# Tag Info

8

Well, syntax is more demanding than your wish, but come on, even DSolve asks for y[x]. Usage: dChange[expresion, {transformations}, {oldVars}, {newVars}, {functions}] Examples: 1: dChange[ D[η*D[f[η], η], η]/η + (1 - s^2/η^2)*f[η] - f[η]^3 == 0, {η == Sqrt[(1 + z)/(1 - z)]}, {η}, {z}, {f[η]} ] ((z-1)^2 (z+1)((z^2-1) f''[z]+2 z f'[z])-f[z] ...

6

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

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 ...

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 *)

4

The following approach transforms ans1 to ans2 and points the way to more general approaches. As noted in the Question, Simplify returns ans1 unchanged. Introducing a ComplexityFunction that defines fewer symbols as less complex helps some. cf[e_] := LeafCount[e] + 100 Count[e, _Symbol, Infinity] Simplify[ans1, ComplexityFunction -> cf] (* ...

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

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 ...

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, ...

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

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]

2

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

Only top voted, non community-wiki answers of a minimum length are eligible