0
$\begingroup$

I have an expression that involves many Sign functions. I want to expand this in the piecewise format.

My function is:

exp = 2 Sign[t] (1/2 (-1 + e) - 1/2 (1 + e) Sign[t]) + (1/2 (-1 + e1) + 
    1/2 (1 + e1) Sign[d - t1]) (-1 + Sign[t1]) - (1/2 (-e + e1) + 
    1/2 (e + e1) Sign[d + t - t1]) (-1 + 
    Sign[-t + t1]) - (1/2 (-1 + e2) + 1/2 (1 + e2) Sign[d - t2]) (-1 +
     Sign[t2]) + (1/2 (-e + e2) + 1/2 (e + e2) Sign[d + t - t2]) (-1 +
     Sign[-t + t2]) + 
 2 (1/2 (-e1 + e2) + 1/2 (e1 + e2) Sign[t1 - t2]) Sign[-t1 + t2]

Here, all variables can be regarded as real.

Using the answer provided in Converting HeavisideTheta[]s and Sign[]s functions to a single Piecewise[], I also borrow a user-defined function:

ToPiecewise[f_] := 
 PiecewiseExpand[
  f /. {Sign[x_] :> Piecewise[{{1, x >= 0}, {-1, x < 0}}]}]

What I want to eavluate is

ToPiecewise[exp]

but this is extremely slow! (Furthermore, aborting (Alt + .) does not work and I had to quit the kernel.) Surprisingly, if I consider a single term, for example the first one and evaluate

ToPiecewise[2 Sign[t] (1/2 (-1 + e) - 1/2 (1 + e) Sign[t])]

then the answer comes immediately. This happens for all other single terms. Why this strange behavior happen?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Consider:

exp = 2 Sign[t] (1/2 (-1 + e) - 1/2 (1 + e) Sign[t]) + (1/2 (-1 + e1) + 
    1/2 (1 + e1) Sign[d - t1]) (-1 + Sign[t1]) - (1/2 (-e + e1) + 
    1/2 (e + e1) Sign[d + t - t1]) (-1 + 
    Sign[-t + t1]) - (1/2 (-1 + e2) + 1/2 (1 + e2) Sign[d - t2]) (-1 +
     Sign[t2]) + (1/2 (-e + e2) + 1/2 (e + e2) Sign[d + t - t2]) (-1 +
     Sign[-t + t2]) + 
 2 (1/2 (-e1 + e2) + 1/2 (e1 + e2) Sign[t1 - t2]) Sign[-t1 + t2]

ToPiecewise[exp[[1]]]
ToPiecewise[exp[[2]]]
ToPiecewise[exp[[3]]] ...

![enter image description here

All the terms have different conditions. This makes the overall condition extremely complicated.

A way around this is to give every term its own conditions like:

ToPiecewise /@ exp

enter image description here

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. However, it is sad since what I want is to simplify the whole expression into "single" piecewise command. I think this is not too hard, because the total case is $2^8=256$ which is not too large, where 8 comes from the different expressions inside Sign. $\endgroup$ Commented Oct 9, 2021 at 15:54
  • $\begingroup$ Counting and multiplying the different conditions gives 7840 cases. $\endgroup$ Commented Oct 9, 2021 at 18:35
  • $\begingroup$ I think your 7840 is obtained by $7840=2\times 4\times 4\times 5\times 7\times 7$. However, there is redundancy in some of the 6 terms. For example, for the 4th term, the condition t1-t2 >=0 and True is redundant. $\endgroup$ Commented Oct 10, 2021 at 2:15
  • $\begingroup$ Oh, I also forgot to mention that I am not interested when Sign evaluates to Sign[0]. In other words, I am only interested when t != 0, t1 - t2 != 0, t2 != 0, t1 != 0, d - t1 != 0, d - t2 != 0, -t + t1 != 0, d + t - t1 != 0, d + t - t2 != 0, -t + t2 != 0. Then, the total case is 256. $\endgroup$ Commented Oct 10, 2021 at 2:36

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.