equation solving - Is there a better way to get the eight points of intersection?

I want to get the eight points of intersection from the equations 2 Abs[x] + Abs[y] == 1 and Abs[x] + 2 Abs[y] == 1. To solve these equations, I tried

Solve[{2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}, {x, y}]

but could only get the four points. I then tried

y /. Quiet@Solve[#, y] /. 
   Abs[x_] -> {x, -x} & /@ {2 Abs[x] + Abs[y] == 1,  Abs[x] + 2 Abs[y] == 1}

p = ({x, y} /. Solve[y == #] & /@ {{1 - 2 x, (1 - x)/2}, {(1 - x)/2, (

      1 + x)/2}, {(1 + x)/2, 1 + 2 x}, {1 + 2 x, -1 - 2 x}, {-1 - 2 x,
       1/2 (-1 - x)}, {1/2 (-1 - x), 
      1/2 (-1 + x)}, {1/2 (-1 + x), -1 + 2 x}, {-1 + 2 x, 1 - 2 x}})~
  Flatten~1

{{1/3, 1/3}, {0, 1/2}, {-(1/3), 1/3}, {-(1/2), 0}, {-(1/3), -(1/3)}, {0, -(1/2)}, {1/3, -(1/3)}, {1/2, 0}}

It works, but I don't like it. Could you recommend a better method?

enter image description here

Answer

Actually, the way I interpret your ContourPlot, there are only 4 points of intersection (red/blue curves). So I do interpret your question such that you want to include intersection with the coord axes as well. I hope this helps:

eq1 = 2 Abs[x] + Abs[y] == 1;
eq2 = Abs[x] + 2 Abs[y] == 1;

p = {x, y} /.
(Solve[#, {x, y}] & /@
 {{eq1, eq2},
  {x == 0, eq2},
  {eq1, y == 0}})~Flatten~1

Comments

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...

Blog

Search This Blog