Tag Info

New answers tagged

1

expr = d + b x + c x + a x^3 - b y - c y - 3 a x^2 y + 3 a x y^2 - a y^3; Collect[(expr /. x -> y + z), z] /. z -> x - y (* d + (b + c) (x - y) + a (x - y)^3 *) Of course, in the real word case, when you do not know apriori, how to simplify the expression, you have no idea of what replacement to apply. But this is exactly the place, where ...


2

I have toyed with code (notice that I use only first solution for Replace). eqs = {S Mex'[ t] == ω e Mey[t] + λ ω e Mny[ t] - Γ R e Mex[t], S Mey'[t] == -ω e Mex[t] - λ ω e Mnx[ t] - Γ R e Mey[t], Mnx'[t] == λ ω n Mey[t] + ω n Mny[ t] - Γ Rnt Mnx[t], Mny'[t] == -λ ω n Mex[t] - ω n Mnx[ t] - Γ Rnt Mny[t], Mex[0] ...


2

My way for this is: eq = (x - a)^2 + (y - b)^2 + (z - c)^2 + d; eq == 0 /. Solve[ForAll[{x, y, z}, -11 - 2 x + x^2 - 4 y + y^2 - 6 z + z^2 == eq]] // TraditionalForm (* {(x-1)^2+(y-2)^2+(z-3)^2-25} *)



Top 50 recent answers are included