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I guess that V9 now adds this capability: $Assumptions = { Element[A, Matrices[{m, n}]], Element[B, Matrices[{n, k}]] }; TensorReduce[ Transpose[Transpose[A].Transpose[B]] ] (* Out: B.A *) 43 In the first case PowerExpand comes to the rescue: PowerExpand@Power[p^(n m) q, 1/n] (* Out: p^m q^(1/n) *) Note however that "the transformations made by PowerExpand are correct only if$c$is an integer or$a$and$b$are positive real numbers". Generally speaking, your assumptions can be listed in Reduce, Simplify, or FullSimplify using the ... 37 Solutions to algebraic or transcendental equations are expressed in terms of Root objects whenever it is impossible to find explicit solutions. In general there is no way express roots of 5-th (or higher) order polynomials in terms of radicals. However even higher order algebraic equations can be solved explicitly if an associated Galois group is solvable. ... 30 I've put this code on a GitHub but I don't know what features are needed or what problems it may give. I'm just not using it. But I will incorporate incomming suggestions as soon as I have time. Feedback in form of tests and suggestions very appreciated! (If[DirectoryQ[#], DeleteDirectory[#, DeleteContents -> True]]; CreateDirectory[#]; URLSave[ "... 26 Removing the imaginary portion of an expression is done by doing ComplexExpand[Re[expression]]. Using just Re alone will not work as Re does no evaluation on symbols with unknown complex parts. Now as stated in the problem and the comments above this particular problem requires a fair amount of assumptions. The simplest way to add local assumptions is to ... 26 You can't use replacements that way, because Mathematica does not do replacements on expressions the way they appear to you. To see what I mean, take a look at the FullForm of your expression: x/(y*z) // FullForm Out[1]= Times[x,Power[y,-1],Power[z,-1]] Whereas, the replacement that you're using is Times[y, z]. In general, it is not a good idea to use ... 26 Mathematica does not support this directly. You can do things of this sort using an external package called NCAlgebra. http://math.ucsd.edu/~ncalg/ The relevant documentation may be found at http://math.ucsd.edu/~ncalg/DOWNLOAD2010/DOCUMENTATION/html/NCBIGDOCch4.html#x8-510004.4 In particular have a look at "4.4.8 NCLDUDecomposition[aMatrix, Options]" ... 23 The plan is first get the "external" contour and then use Green's theorem to find its area. r[t_] := {-9 Sin[2 t] - 5 Sin[3 t], 9 Cos[2 t] - 5 Cos[3 t], 0} (*find the intersections*) tr = Quiet@ToRules@Reduce[{r@t1 == r@t2, 0 < t1 < t2 < 2 Pi}, {t1, t2}]; pt = {t1, t2} /. {tr} // Flatten; pts = SortBy[pt, N@# &]; pps = Partition[pts, 2]; Now ... 22 What you have is a MultinormalDistribution. The quadratic and linear forms in the exponential can be rewritten in terms of$-\frac12(\vec{x}-\vec{\mu})^\top\Sigma^{-1}(\vec{x}-\vec{\mu})$where$\vec{\mu}$represents the mean and$\Sigma$the covariance matrix, see the documentation. With this, you can do integrals of the type given in the question by ... 22 A first step would be to implement a convenience function that can automatically apply the method of separation of variables to separable types of equations. To show that the steps could in principle be automated, let me repeat basically the same calculation that I did for cylindrical coordinates with only slight modifications to the heat equation: ClearAll[... 22 I've decided to expand on my comment. Before I delve into the solution, let's all pause for a moment and marvel at the stereographic parametrization of a unit circle: $$\begin{pmatrix}\frac{1-t^2}{1+t^2}\\\frac{2t}{1+t^2}\end{pmatrix}$$ Sometimes also referred to as the Weierstrass substitution, it has often been used as a tool in the solution of algebraic ... 21 The nearest Mathematica has to "types" are Heads of expressions that are Atoms. For example: Through[{AtomQ, Head}[2]] {True, Integer} Through[{AtomQ, Head}[2 + I]] {True, Complex} Through[{AtomQ, Head}["cat"]] {True, String} and so on... There are also somewhat different "types" in the context of Compile. 20 The most direct way to test this is probably the following:$Assumptions = x > 0; Element[x, Reals] // Simplify (* Out[1]= True *) $Assumptions = True; Element[x, Reals] // Simplify (* Out[4]= x ∈ Reals *) So$x>0$seems to imply that$x$is real. 20 After all this time, I came up with a very nice tensor calculus proof of the Hairy Ball Theorem. It only depends on Stokes theorem and standard laws of tensor calculus like the Ricci identity and symmetries of curvature tensors. All the topology is done by Stokes theorem. The remainder of the proof is equational, local and geometrical. It is coordinate/basis ... 19 There is no need to play around with ReplaceAll, Rule, Block, Module or whatever using D, since you have an oparator Derivative really fulfilling your needs while you need not bother if the arguments were defined, so I recommend it to find symbolic derivatives of your function. Remember of shorthands f', f'' to represent first and second derivatives of one-... 19 Disclaimer: This is not a full answer, but perhaps it's a start. From an algebraic stand point this seems like a very hard problem. I attacked it with a more brute force approach. I guess a basis and use LatticeReduce to try to find a Diophantine relation. Note this code only tries to identify roots as the product of integral powers of trig. If it returns ... 19 Here is extensions to @Jens answer (I think) also relying on possible separation of variable. It is not meant as an independent answer, but complements it. First extend his answer to 2D ClearAll[pt, px, x, t, p]; operator = Function[p, D[p, t] - Δ D[p, x, x] - Δ D[p, y, y]]; ansatz = pt[t] px[x] py[y]; pde2 = Expand[Apply[Subtract, operator[ansatz]/... 19 One way is to use an extra argument that acts as a switch. Clear[f]; f[0] = 1; f[1] = 1; f[n_, True] := f[n - 1] + f[n - 2] Example: f7 = f[7, True] (* Out[329]= f[5] + f[6] *) To proceed another step, can do a replacement. f7 /. f[aa_] :> f[aa, True] (* Out[330]= f[3] + 2 f[4] + f[5] *) Can use Nest to repeat this n times. Nest[# /. f[aa_] :&... 18 In this case you can use SeriesCoefficient SeriesCoefficient[Exp[x], {x, 0, n}] 18 I can't help you with functions beyond Reduce, Simplify, ... but I can offer you a tip to help with reducing errors from manual translation. In order to validate that your manual transformation is correct, one can subtract the original expression from the manual transformation and then use Simplify on that expression. Sometimes it will return zero using ... 17 Initially, Mathematica is not designed for such abstract calculations. But, Mathematica is a powerful programming language, so that one can add such functionality easily. See the following examples in related area of differential geometry: calculations in symbolic dimensions Abstract calculations 17 It is assumed that$x$is a real number. Everything else would mathematically not make sense because on complex numbers there does not exist an ordering relation. An example would be to take the expression$\sqrt{x^2}$and to imagine that this is not equal$x$for$x=-\mathbb{i}\$. Therefore the expression is in a general form not simplified In[37]:= Sqrt[x^...

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Short story $$\vartheta(x) = \arg \left[(\operatorname{Bi}x+i \operatorname{Ai}x)e^{-\frac{2}{3} i (-x)^{3/2}}\right]+\frac{2}{3} \operatorname{Re}\left[(-x)^{3/2}\right]$$ Update: I see that you want use only real functions, so you can expand this as $$\vartheta(x) = \begin{cases} \arctan\frac{\cos \left(\frac{2}{3} (-x)^{3/2}\right) \operatorname{... 17 Looking at the Trace of one which does work: x = Sin[Pi/5] (* Sqrt[5/8 - Sqrt[5]/8] *) Trace[ArcSin[x], TraceInternal -> True] It appears that Mathematica computes the ArcSin numerically and then recognises the result, 0.628319 as possibly equal to Pi/5. To check it computes Sin[Pi/5], and subtracts it from the original argument to see if it gets ... 17 One thing is to make sure you have all the assumptions stated properly. For instance, the first two cases can be handled by passing all the assumptions to FullSimplify FullSimplify[Power[p^(n m) q, 1/n], Assumptions -> {q > 0, p > 0, n ∈ Integers, m ∈ Integers}] p^m q^(1/n) and FullSimplify[(p^n q)/(p^m r), Assumptions -> {q > 0, p &... 17 I use Mathematica in much the same way as you, although in the context of multi-stage physics derivations. Since no one has mentioned it, I'll describe an obvious approach to successive rewrites of an expression. I label each step a calculation with an indexed symbol that I can refer to later. This is preferable to In's and Out's whose numbers can change if ... 17 I believe there are at least three cases treated separately by Derivative. 1) A function defined by a Symbol. This follows the the rule cited in the documentation. g[x___] := f[x]; Derivative[1][g][x] // Trace (* { { g' , { g[#1] <-- Here the rule is being applied , f[#1] } , f'[#1] & } , (f'[#1] &)[x] , f'[x] } ... 16 Use the following representation of the Legendre polynomials:$$ P_n(x) = 2^n \sum_{k=0}^n x^k \binom{n}{k} \binom{\frac{n+k-1}{n}}{n} $$Note that the sum effectively is over k \equiv n \bmod 2. Expand each Legendre polynomial into a sum. Integration with respect to \theta is easy:$$ \int_0^{\pi} \sin^{k_1+k_2+k_3+1} \theta \mathrm{d}\theta ...

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Regarding your general question: from my own experience, a large part of my research findings while still at academia would have been much harder or outright impossible to get without Mathematica, and that applies to both numerical and analytical work. I also know this for many other people. There are a number of areas where it allows you to get through some ...

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If you use the third argument in Solve, i.e. a list of variables to be eliminated (take a look at the Eliminating Variables tutorial in Mathematica) then you'll get the result immediately : Solve[{a b c == -1, a^2/c + b/c^2 == 1, a^2 b + b^2 c + c^2 a == t, a b^5 + b c^5 + c a^5 == res}, {res}, {a, b, c, t}] {{res -> 3}} Edit ...

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