Skip to main content

equation solving - Simplify not evaluating completely with Square roots


Here's a simple question. It's no longer high priority that I know this, but it's something that can come in handy later on.


Simplify[a > Sqrt[b]/c + d, a < Sqrt[b]/c + d]


or


Simplify[Sequence @@ ({a > n, a < n} /. {n ->  Sqrt[b]/c + d})]

or


n = Sqrt[b]/c + d;
Simplify[a > n, a < n]

all outputs:


Out: a > Sqrt[b]/c + d


while


Simplify[a > m, a < m]

outputs


Out: False

How come? This isn't an issue of not considering the negative root. The expressions are identical. As demonstrated by Simplify[a > m, a < m], replacing m with more complex expressions aside from having the Sqrt function.


In fact, if I use Surd, no matter the nth root, even or odd, or square root, Simplify will evaluate completely into False.


Why? Is this something that I can fix using Upvalues? -- Finally remember what the "overloading" feature was called.




Answer



The assumptions mechanism used by Simplify will not try to prove or disprove an inequality if the number of variables involved is higher than a built-in limit. To decide polynomial inequalities the assumptions mechanism uses the cylindrical algebraic decomposition algorithm, which has doubly-exponential complexity, hence the limit on the number of variables is low -- by default it is 4. Your inequality contains 5 polynomial variables, since to reduce it to a polynomial system we need to introduce a new variable v to replace Sqrt[b] and add an equation v^2==b.


The limit on the number of variables can be changed using a system option.


In[3]:= SetSystemOptions["SimplificationOptions"->"AssumptionsMaxNonlinearVariables"->5];

In[4]:= Simplify[a > Sqrt[b]/c + d, a < Sqrt[b]/c + d]
Out[4]= False

Comments

Popular posts from this blog

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...

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...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]