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6

It is not quite clear, what do you you want to get out of the answer. Would you like to compare Maple and Mma and understand, which one is better ? Or would you like to understand the alternative forms of taking this integral? Or the reason, why the results of Marple and Mma are different? Or transform the Mma result in terms of xand y? Or, ...


6

I think this does what you want: Less @@ SortBy[Defer[Subscript[ω, #]] & /@ {1, 2, 11, 22}, N @@ # &]


4

With $Assumptions = {0 <= pdw <= 1, 0 <= pwd <= 1}; trM = {{1 - pdw, pdw}, {pwd, 1 - pwd}}; dmP = DiscreteMarkovProcess[{1, 0}, trM]; gD = GammaDistribution[a, b]; the mean and variance of the described process for a time slice at time t are mean = Expectation[(m[t] - 1)*g, {m \[Distributed] dmP, g \[Distributed] gD}] variance = ...


4

There may be a better way to do this, but perhaps 4 Fm Derivative[1][Fm] /. Function[b_] :> b // Evaluate // Function 4 Cos[#1] Sin[#1]& 1/4 Fp (2 Fp - Fp^3/Fm^2 + 4 Fm Derivative[1][Fm] Derivative[1][Fp]) /. Function[b_] :> b // Evaluate // Function 1/4 Cos[#1] Sin[#1] (2 Cos[#1] Sin[#1]-Cos[#1]^3 Sin[#1]+4 Cos[#1] Sin[#1] ...


4

Put assumptions in: Clear[a, c] Integrate[ q^2 ((4 (c π))/((a q^2 - c) (c + a q^2))), {q, 0, ∞}, Assumptions -> {a, c} ∈ Reals] (* ConditionalExpression[(Sqrt[c] π^2)/a^(3/2), (a > 0 && c > 0) || (a < 0 && c < 0)] *)


4

This will produce a simplified form: Simplify[ D[D[E^(I (-k r + t \[Omega]))/r, x], y] /. Power[x^2 + y^2 + z^2, n_] :> Power[HoldForm[r], 2 n]]


3

I may have misunderstood the aims, If so, I apologize. For the first question: f[n_] := GraphDistance[CompleteGraph[5, EdgeWeight -> #], 1, 2] & /@ RandomVariate[ExponentialDistribution[1], {n, 10}]; : This generates a sample of size n of graph distances between vertex 1 and 2. You can visualize: Histogram[f[10000]] Estimating ...


2

One could go very low-level and try to define an operator just in terms of basic properties such as linearity, but I chose to go for a middle ground and assume that we're dealing with a linear differential operator. The coefficients in the operator are called a[i] where i is a symbolic summation index corresponding to the order of the derivative in each term ...


2

Depending on your specific needs the straightforward way to do this is Series[p[x+e,y-e,t],{e,0,1}] or Series[p[x+ex,y-ey,t],{ex,0,1},{ey,0,1}] If you want to extract the terms proportional to $\epsilon$ you can get the correct coefficients with SeriesCoefficient[p[x + e, y - e, t], {e, 0, 1}] or if that suits you better with ...


2

Another way using the built-in coordinate transforms : define a rule to perform the transformation : rule = Rule[x^2 + y^2 + z^2, Simplify[TransformedField["Cartesian" -> "Spherical", x^2 + y^2 + z^2, {x, y, z} -> {\[Rho], \[Theta], \[Phi]}], Assumptions -> {\[Rho] > 0}]] expr = D[D[E^(I (-k r ...


2

Use rules to do both the forwards and the backwards substitutions. Step 1: deriv = D[D[E^(I*((-k)*r + t \[Omega]))/r /. r -> Sqrt[x^2 + y^2 + z^2], x], y] Step 2: PowerExpand[deriv /. {x^2 + y^2 + z^2 -> r^2}]


2

Merely my own variation of the existing answer by mfvonh: expr = 1/4 Fp (2 Fp - Fp^3/Fm^2 + 4 Fm Derivative[1][Fm] Derivative[1][Fp]); Fm = Sin[#] &; Fp = Sin[#] Cos[#] &; FullSimplify[expr /. (x_ &) :> x] Function @@ {%} 1/32 (3 + 7 Cos[2 #1]) Sin[2 #1]^2 1/32 (3 + 7 Cos[2 #1]) Sin[2 #1]^2 &


1

Following closely the documentation for DiscreteMarkovProcess (I am no expert though…) M = {{1 - w, w}, {d, 1 - d}} P = DiscreteMarkovProcess[{1, 0}, M] Mean[P[n]] // FullSimplify Variance[P[n]] // FullSimplify Graph[P] data = RandomFunction[P /. w -> 1/2 /. d -> 1/4, {0, 10}] ListPlot[data, Filling -> Axis, Ticks -> ...


1

Another replacement version: expr /. (x_ &) :> x /. x_ :> (x &)


1

Postfix-Definition Clear@"`*" Sin*Cos // f[r_] := Sin[r]*Cos[r] Sin*Cos // f[Pi/7] (* out *) Cos[Pi/7] Sin[Pi/7] Prefix-Definition Clear@"`*" g[r_][Sin*Cos] := Sin[r]*Cos[r] g[Pi/11][Sin*Cos] (*out*) Cos[Pi/11] Sin[Pi/11] Just 1 letter more to type



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