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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 *)


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

I believe the reason for the confusion is the way the Dot product operates. It is a Flat but non-commutative operation when the factors aren't just vectors. In particular, it contracts the last index of the first factor with the first index of the next factor, going from left to right. But in the form $a\cdot\varepsilon\cdot b$, this means that the ...


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


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.


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 ...



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