Tag Info

Hot answers tagged

6

The False setting can be useful when one wants an integral that is classically divergent. Or when one wants a result without provisos. A downside is greater chance of an incorrect result. A True setting is thus useful for the opposite, e.g. avoidance of finite results for divergent integrals. It can also be useful for more careful checking in multivariate ...


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

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

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 ...


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 ...


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}] ...


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 ...


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 ...



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