Skip to main content

simplifying expressions - Simplify equations with pattern assumptions


When simplifying an expression using Simplify[expr, assumptions], the assumptions has to be explicit. Is it possible to simplify an expression with pattern assumptions?


(TransformationFunctions can satisfy some of the needs, but it is not always functional as in the following examples.)


For example


psimp[Sqrt[a[t]^2], a[m_] > 0]


a[t]



psimp[f[x] (1 + a) (1 + b) == 0, f[a_] != 0]



(1 + a) (1 + b) == 0



psimp[(d[t] + b[t] + c[t])^2 + (d[m] + b[m] + c[m])^3 + a[n] + x[n] //
   Expand, {d[f_] + b[f_] + c[f_] == a[f], a[n] + x[n] == s}]


s + a[m]^3 + a[t]^2




psimp[g[x, y]*Derivative[0, 1][f][x, y] + f[x, y]*Derivative[0, 1][g][x, y], 
 (g_)[x_, y_]*Derivative[0, 1][f_][x_, y_] + (f_)[x_, y_]*Derivative[0, 1][g_][x_, y_] == Defer[D[f[x, y]*g[x, y], y]]]


D[f[x, y]*g[x, y], y]




Answer



This can be accomplished by dynamically analyzing the expression and generating the explicit assumptions from the given assumptions pattern.


ClearAll[psimp, psimpStep, ptn2explicit];


ptn2explicit[expr_, pattern_List, RHS_] := 
 With[{cases = Cases[expr, #, {0,Infinity}, Heads -> True] & /@ pattern, 
   ruleLHS = {___, #, ___} & /@ pattern, 
   ruleRHS = pattern /. Verbatim[Pattern][arg_, ___] :> arg},
  MapThread[Verbatim[#1] -> #2 &, {pattern~Append~RHS, #}] & /@ 
   ReplaceList[cases, ruleLHS :> (ruleRHS~Append~RHS)]]

psimp[expr_, cond_] := psimp[expr, {cond}];

psimpStep[expr_, cond_List] := 

 Module[{LHS, RHS, explicitCond, \[UnderBracket], operator, opList, count},
  operator = Equal | Unequal | Greater | Less | GreaterEqual | LessEqual | Element;
  LHS = Cases[cond, (l_~operator~r_) :> l];
  RHS = Cases[cond, (l_~(o : operator)~r_) :> 
     If[o === Equal, \[UnderBracket][][][]@r, r]];
  opList = Cases[cond, (l_~(o : operator)~r_) :> o];
  explicitCond = And @@ Union@
     Flatten[#1~#2~#3 /. ptn2explicit[expr, Variables@#1, #3] & @@@ 
       Transpose[{LHS, opList, RHS}]]; 
  count[e_] := LeafCount[e] - Count[e, \[UnderBracket][][][][_]];

  Simplify[expr, explicitCond, 
    ComplexityFunction -> count] /. \[UnderBracket][][][][x_] :> x]

psimp[expr_, cond_List] := FixedPoint[psimpStep[#, cond] &, expr]

A few explanations are in order here:


(1) Function ptn2explicit generates a list of rules to convert the patterns in expr into explicit expressions. For example,


ptn2explicit[a[t] + b[t] + a[m] + b[m] + c[m], {a[f_], b[f_]}, d[f]]



{{Verbatim[a[f_]] -> a[m], Verbatim[b[f_]] -> b[m], Verbatim[d[f]] -> d[m]}, {Verbatim[a[f_]] -> a[t], Verbatim[b[f_]] -> b[t], Verbatim[d[f]] -> d[t]}}



The ptn2explicit function is a modified version of @halirutan's answer in How to do Cases with multiple related patterns?. (BTW: everybody complains the question in this link was not clear. I guess you know what I want to do in case you also read this post ^_^)


(2) A strange symbol \[UnderBracket] is used here. Because this is the named and visible symbol with greatest lexical order (at least among the first 10^4). As noticed in Why does Simplify ignore an assumption? and references therein, we need to use symbols with greater lexical order for some rules to take effect.


The SubValues \[UnderBracket][][][] is to further reduce the lexical order of this symbol. Otherwise Derivative[__][__][__] has lower order. This particular subvalued one is not the absolute lowest lexical ordered one, but would be enough for most cases.


(3) After simplification, the new expression may contain new symbols that matches the given assumptions pattern. Thus we perform a FixedPoint, until the result stabilizes.


(4) There is a limitation that one has to give names to each patterns in the assumption. For example, psimp[Sqrt[a[t]^2], a[_] > 0] doesn't work but psimp[Sqrt[a[t]^2], a[m_] > 0] works.


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
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

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more