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1

The matrix Clear[a,B,d,j,M]; m = {{B/2 - d - j/2 + a, (Sqrt[2]*j)/2, M}, {(Sqrt[2]*j)/2, B/2 + a, 0}, {M, 0, (-3*B)/2 - d + j/2 + a}}; has eigenvalues determined by the characteristic polynomial of (maximal) degree 3. In the absence of any other information, we get three Root objects as the eigenvalues, which is Mathematica's way of preserving all ...


2

You could also do something like the following (taking advantage of TransformedField: solidHarmonicS[l_?IntegerQ, m_?IntegerQ, x_, y_, z_] := Module[{r, θ, ϕ, xx, yy, zz}, FullSimplify@ Evaluate[ TransformedField["Spherical" -> "Cartesian", r^l SphericalHarmonicY[l, m, θ, ϕ], {r, θ, ϕ} -> {xx, yy, zz}]] /. {xx -> x, yy ...


0

This can be implemented by appropriate replacement rules, though they are a bit fiddly to get just right (note in particular the exact form of the first one). The implementation below uses partial memoization as in this question, and it will only match the case when $l$ and $m$ are integers; otherwise, it will go through a direct SphericalHarmonicY ...


2

Your expression simplifies to this $$\vec X \vec X^T A + A^T \vec X \vec X^T$$ using just these rules Unprotect[D, Transpose, Dot]; (*Derivative rules*) D[Tr[A_], X_] := Tr[D[A, X]] D[Transpose[A_], X_] := D[A, X]\[Transpose] D[A_ .B_, X_] := D[A, X].B + A.D[B, X] (*Tranpose rules*) 0\[Transpose] := 0 1\[Transpose] := 1 (A_\[Transpose])\[Transpose] := A ...


2

A workaround might be to express Abs in terms of Conjugate, so: a = ComplexExpand[Abs[Zeta[x + I y]], TargetFunctions -> Conjugate] $\sqrt{\zeta (x-i y) \zeta (x+i y)}$ FullSimplify[D[a, {x, 2}] + D[a, {y, 2}]] $\frac{\zeta '(x-i y) \zeta '(x+i y)}{\sqrt{\zeta (x-i y) \zeta (x+i y)}}$ I don't know if this is the correct result, but it seems ...


1

If you want the term $\frac{\partial f}{\partial r}$ to disappear you need to introduce new function which would be: w2 = f[r, θ] r which means that you have to make a substitution f -> w2/r, this way: lapla1 /. f -> (w2[#, #2]/# &) // Simplify // ExpandAll If you once used f or w, don't change theirs definitions, use a new one, you ...


3

Mathematica treats Abs as if it were (complex) differentiable, which, unfortunately, is not the case. It returns Abs' when it is differentiated. That indicates that Mathematica does not know what to do with the input. One problem with the OP's problem is that the Chain Rule does not apply to expressions of the form D[Abs[u[x,y]], x]. Another problem is ...


2

Please see answer by @michael-e2 as it is much better. I'm using the chain rule, here, for arbitrary functions (which appears to be what Mathematica is doing), but this is not correct for |z| which is not complex differentiable. Are you sure this is a bug? Consider replacing Abs and Zeta with arbitrary functions foo and bar: D[ foo[ bar[x + I y]], {x, 2}] ...


4

Basing on the following thread: Change variables in differential expressions and using great code by Jens for visualisation purposes (I have replaced part [vars__Symbol] with [vars__] because you are using Substripted names which are not Symbols, but that's only about visualisation). You can do the following in the first step: M2 = M /. f -> (f[#, #2, ...


3

Maybe you would like to automate the process even more, by defining a function that does all the steps in your calculation at once: definiteIntegral[integrand_, {x_, xMin_, xMax_}] := Module[{antiDerivative = Integrate[integrand, x]}, Simplify[Subtract @@ (antiDerivative /. x -> {xMax, xMin})]] definiteIntegral[ 1/Sqrt[(ϵ + u)^2 + Δ^2], {x, m, ξ - ...


3

You can do this: In[20]:= Integrate[1/Sqrt[(ϵ + u)^2 + Δ^2], ϵ] // FullSimplify Out[20]= Log[u + ϵ + Sqrt[Δ^2 + (u + ϵ)^2]] In[21]:= F2[ϵ_] = % Out[21]= Log[u + ϵ + Sqrt[Δ^2 + (u + ϵ)^2]] Note that I used =, not := to make sure % gets expanded before the definition is recorded.



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