Skip to main content

output formatting - Reduce an equation by putting a new variable


I have the following equation given:


(26−x)⋅√5x−1−(13x+14)⋅√5−2x+12√(5x−1)⋅(5−2x)=18x+32.


In order to solve it, I want to substitute t=√5x−1−√5−2x. This cannot be done directly but with this step the equation could be transformed in the form (t−2)3=0, which can be solved easily.



Is it possible to make this substitution with Mathematica? The only thing I have achieved is to solve it directly with


Solve[(26 - x)*Sqrt[5*x - 1] - (13*x + 14)*Sqrt[5 - 2*x] + 
      12*Sqrt[(5*x - 1)*(5 - 2*x)] == 18*x + 32]

Answer



A standard approach for this kind of task uses Eliminate. It works nicely with polynomial equations, and even though neither this transformation : √5x−1−√5−2x→t nor the original equation : (26−x)√5x−1−(13x+14)√5−2x+12√(5x−1)(5−2x)=18x+32 are of polynomial types, nevertheless we can make some tricks to get exactly (t−2)3=0.


Defining


f[x_] := (26-x) Sqrt[5x-1] - (13x+14) Sqrt[5-2x] + 12 Sqrt[(5x-1)(5-2x)] - 18x - 32

terms like : √(5x−1)(5−2x) should be PowerExpand'ed in order to get the lowest possible order of resulting polynomials (otherwise we would get a ninth order polynomial):


Eliminate[{ f[x] == 0 // PowerExpand, Sqrt[5x-1] - Sqrt[5-2x] == t}, x]



 12 t - 6 t^2 + t^3 == 8

If we didn't play with PowerExpand we would use another definition of function f yielding effectively the same result :


f1[x_] := (26-x) Sqrt[5x-1] - (13x+14) Sqrt[5-2x] + 12 Sqrt[(5x-1)] Sqrt[5-2x] - 18x - 32

Eliminate[{ f1[x] == 0, Sqrt[5x-1] - Sqrt[5-2x] == t}, x]

Now we rewrite the equation we've got -8 + 12 t - 6 t^2 + t^3 == 0 in the following form :



TraditionalForm[-8 + 12 t - 6 t^2 + t^3 == 0 // Factor]

enter image description here


And this is indeed equivalent to our orignial function f[x] :


(t-2)^3 == f[x] /.t-> Sqrt[5x-1] - Sqrt[5-2x] // FullSimplify


 True

Edit



It is remarkable that f being rather a complicated (complex also for real arguments) function has only one simple root :


Reduce[ f[x] == 0, x]


x == 2

which implies t == 2. To make this result more understandable we can plot the contours of f in a complex domain :


GraphicsRow[ Table[ Show[
    ContourPlot @@@ { { g[f[x + I y]], {x, -2, 6}, {y, -4, 4}, Contours -> 37,
                        RegionFunction -> Function[{x, y}, 60 > g[f[x + I y]] > -60],

                        PlotLabel -> Style[ g[HoldForm @ f[x + I y]], Blue, 15],
                        ColorFunction -> "DeepSeaColors"},
                      { Re @ f[x + I y] == 0, {x, -2, 6}, {y, -4, 4},
                                              ContourStyle -> {Red, Thick}},
                      { Im @ f[x + I y] == 0, {x, -2, 6}, {y, -4, 4}, 
                                              ContourStyle -> {Cyan, Thick}} }, 
    Graphics[{ Thickness[0.01], Darker @ Green, Line[{{0.2, 0}, {2.5, 0}}]}]],

                    {g, {Re, Im, Abs}}]]


enter image description here


The color curves denote the following sets of points z in the complex plane



  • red ; where: Re[ f[z]] == 0

  • cyan ; where: Im[ f[z]] == 0

  • green ; the set of z where Im[ f[z]] == 0 restricted to the real domain.


One can easily see that these curves cross only at z == {2,0}.


We complement the plots of the function f restricting to the real domain only (here the green line denotes the region where the imaginary part vanishes):


GraphicsRow[

    Plot[{ Re @ f[x], Im @ f[x]}, {x, ##}, AxesOrigin -> {0, 0}, PlotRange -> Automatic,
         PlotStyle -> { Thick, Thickness[0.008]}, AspectRatio -> 1.1, 
         Epilog -> { { Thickness[0.01], Darker @ Green, Line[{{0.2, 0}, {2.5, 0}}]},
                     { PointSize[0.023], Red, Point[{2, 0}]}}] &
    @@@ {{-3, 3}, {-.1, 3.5}, {0.7, 3}}]

enter image description here


Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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}]
Post a Comment
Read more