# User Portlet

John Doty
Discussions
Solve gives you a set of solutions of the form of replacement rules {{x->s1}, {s->s2},...}  where s1 et al. are the solutions. Reduce gives you a reduced form for the input expression. How would you expect Solve to express *this*...
Use Reduce when you want the conditions, Solve when you want the generic solution.
It isn't really the machinery of Solve that's in question. f == 0 /. a -> -5/8 (* False *) -5/8 isn't a root of your equation. Mathematica consistently treats x^(1/3) and CubeRoot[x] as different things, defined differently....
Length is a programming construct, not a part of symbolic math. Solve works on statements of symbolic math. *Mathematica's* symbolic math basically consists of: 1. Problems it can solve. 2. Transformation rules it tries to use to ...
You cannot "declare" that symbol t represents a real number with Element[t, Reals]. You must feed the logical assertion to the function that needs to know it as an assumption. Assuming[Element[t, Reals], LaplaceTransform[UnitStep[t -...
Sign is a programming construct, not an analytic function. However, you can define a generalized function with similar properties and differentiate that: sign[x_] := 2 HeavisideTheta[x] - 1 sign'[x] (* 2 DiracDelta[x] *)
FullSimplify understands how to reduce the root: FullSimplify[Sqrt[3 - 2 Sqrt[2]] (1 + Sqrt[2])] (* 1 *) However, it apparently starts down a different path and never tries this. Reduce the root first, and then simplify: ...
You haven't shown all your code. The usual method for doing this kind of problem is to use Table to evaluate the expression and construct a matrix of the results. Loops are rarely an effective way to get a job done in *Mathematica*.
Thank you Mikayel. It works. The only problem is that extracts exact integers, making calculations slow, but N[] fixes that.
Any expression may be used as the head of an expression. Thus, a = 37[1, 2, 3] is perfectly valid (although strange). Here, a[[0]] is the number 37. You may even mutate it with Set (=): a[[0]]=33 a (* 33[1, 2, 3] *) ...