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1

I find that getting the answer I want with this type of expression requires the use of a range of techniques. In this case, I would use ComplexExpand[Conjugate[a/b + c/d], {a, d}, TargetFunctions -> Conjugate] giving Conjugate[a]/b + c/Conjugate[d]


3

You could apply PowerExpand to the lefthand side of the 2nd expression. Solve[PowerExpand[z (A z)^k] == m, z]


1

I figured out how to update this function to correctly work on multi-linear expressions. All that's required is one additional call of PolynomialReduce in the third nested If statement. This mirrors the way the second If statement handles the power case. It might be that there should be a similar treatment for the division case (the first If statement), but ...


2

As usual, GroebnerBasis[] is very useful here: det[aa_] := Det[Table[x[li1, li2], {li1, aa /@ {1, 2, 3}}, {li2, aa /@ {4, 5, 6}}]]; origExpr = det[a] det[b]/det[c] + det[d] det[e]/det[f] - det[g] det[h]/det[i]; oeet = origExpr // Expand // Together; varsx = Cases[oeet, x[a_[i_], _], ∞]; vars = Union[Cases[oeet, x[a_[i_], _] :> a, ∞]]; rep = Table[d[ti] ...


1

If you want to substitute 1/x->0 in in expressions then take the limit x->Infinity: y = (10+2*x)/x; Limit[y, x->Infinity] Other ways of doing that is the followings (see bbgodfrey's comment): First@Collect[y, x] Coefficient[y, x, 0]


11

HornerForm[x1 x2 + x1 x3 x4 + x1 x3 x5] (* x1 (x2 + x3 (x4 + x5)) *)


1

I'm sure there are a lot of ways to generate random functions. Here's one: a := RandomChoice[{Times, Plus}] @@ (RandomChoice[{Sqrt, Identity, Power}] /@ (RandomSample[{x, y, z}, 3] RandomInteger[{1, 5}, 3])^RandomInteger[{-3, 3}, 3]); To make a matrix of such functions: Table[a, 3, 3] // MatrixForm Of course the output is different each time. ...


2

The simplest approach is to solve the equations directly with Solve[{Abs[11 + a] == b, Abs[5 + a] == b}] (* {{a -> -8, b -> 3}} *) Note that, Solve[{11 + a == b, 5 + a == -b}] (* {{a -> -8, b -> 3}} *) not "a = 8 and b = 3" as in the question, and Solve[{11 + a == -b, 5 + a == b}] {* {{a -> -8, b -> -3}} *) not "a =8 and b = -3" as ...



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