Skip to main content

simplifying expressions - How to find the most compact form of an equation


I would like to know whether there is any general approach to obtain the most compact form of an equation with Mathematica.


I used to believe that the FullSimplify command would do that for me, but I recently found out that it will not necessarily give the most compact form. I found out with the following equation, which has been FullSimplified:


$$\left[\sec (\alpha ) \left(\coth \left(L \sqrt{\text{Bo} \cos (\alpha )}\right)+\text{csch}\left(L \sqrt{\text{Bo} \cos (\alpha )}\right) (\text{Bo} L R \sin (\alpha )-1)-R \sin (\alpha ) \sqrt{\text{Bo} \sec (\alpha )}\right)\right]\left(R \sqrt{\text{Bo} \sec (\alpha )}\right)^{-1}$$


Or, in inputform for your convenience:



eq1 = (Sec[α] (Coth[L Sqrt[Bo Cos[α]]] - R Sqrt[Bo Sec[α]] Sin[α] + 
Csch[L Sqrt[Bo Cos[α]]] (-1 + Bo L R Sin[α])))/(R Sqrt[Bo Sec[α]])

However, if I now run the following:


Collect[eq1,Bo] //FullSimplify

I get a much more compact form, namely: $$ -\tan (\alpha )+L \sin (\alpha ) \sqrt{\text{Bo} \sec (\alpha )} \text{csch}\left(L \sqrt{\text{Bo} \cos (\alpha )}\right)+\frac{\tanh \left(\frac{1}{2} L \sqrt{\text{Bo} \cos (\alpha )}\right)}{R \sqrt{\text{Bo} \cos (\alpha )}}$$


Can someone explain why I get a more compact solution when I first collect the equation for a certain parameter and then run FullSimplify and whether there is a general 'recipe' to get the most compact form of an equation?


P.S. sorry for the long equation, but it seems that most of the smaller equations do not have this behaviour which makes sort of intuitive sense



Answer




There are at least two aspects to this. The first is the fact that FullSimplify will not try all possible transformations, even those it is aware of. See: Why does Simplify ignore an assumption? The second is that Mathematica does not see "compact" and "simple" as the same thing; if you give it a different ComplexityFunction it will do much better in this case. Compare:


eq1 = (Sec[α] (Coth[L Sqrt[Bo Cos[α]]] - R Sqrt[Bo Sec[α]] Sin[α] + 
Csch[L Sqrt[Bo Cos[α]]] (-1 + Bo L R Sin[α])))/(R Sqrt[Bo Sec[α]])

short1 = Collect[eq1, Bo] // FullSimplify


-Tan[α] + (
Sec[α] (Bo L R Csch[L Sqrt[Bo Cos[α]]] Sin[α] +
Tanh[1/2 L Sqrt[Bo Cos[α]]]))/(R Sqrt[Bo Sec[α]])


short2 = FullSimplify[eq1, ComplexityFunction -> Composition[StringLength, ToString]]


(Csch[
L Sqrt[Bo Cos[α]]] Sec[α] (-1 + Cosh[L Sqrt[Bo Cos[α]]] +
Bo L R Sin[α]))/(R Sqrt[Bo Sec[α]]) - Tan[α]

The three expressions measured by both LeafCount and my ComplexityFunction:


LeafCount /@ {eq1, short1, short2}

Composition[StringLength, ToString] /@ {eq1, short1, short2}


{59, 51, 49}

{275, 374, 212}

Note that though short1 and short2 appear to be roughly the same length my ComplexityFunction sees short1 as being much longer. That is because it is not a very good metric, and it is using the OutputForm of the expression rendering "2D" text like this:


ToString[short1]



                                                               L Sqrt[Bo Cos[α]]
Sec[α] (Bo L R Csch[L Sqrt[Bo Cos[α]]] Sin[α] + Tanh[-----------------])
2
-Tan[α] + ------------------------------------------------------------------------
R Sqrt[Bo Sec[α]]

The point of all of this is that you need to be quite specific with ComplexityFunction if you want the expression simplified in the way you see as simple or compact.


Comments

Popular posts from this blog

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...