Converting ConditionalExpression to Piecewise

Question

I have Solve[] return a list like {{F->ConditionalExpression[EXPR1, 0<z<a1]},{F->ConditionalExpression[EXPR2, a1<z<a2]},...} and so on -- EXPR* and a1,a2,... are somewhat complicated expressions, containing things like Root[]:

EQ = F^4 (138880 z + 318688 F z + 98415 F^10 z + 6561 F^9 (5 + 4 z) + 
    405 F^5 (625 + 1136 z) + 243 F^6 (-2500 + 5077 z)) == 
    1024 z + 11008 F z + 40192 F^2 z + 33536 F^3 z + 46944 F^6 z + 
    709776 F^7 z + 368388 F^8 z + 4374 F^12 (50 + 43 z) + 
    4374 F^11 (-125 + 202 z);

ftz = Solve[EQ && 0 < F < 1 && 0 < z < 1, F];

e.g. the first conditional:

In[100]:= N[(F /. ftz[[1]])]

Out[100]= ConditionalExpression[
 Root[-1024 z - 11008 z #1 - 40192 z #1^2 - 33536 z #1^3 + 
    138880 z #1^4 + 318688 z #1^5 - 46944 z #1^6 - 709776 z #1^7 - 
    368388 z #1^8 + (253125 + 460080 z) #1^9 + (-607500 + 
       1233711 z) #1^10 + (546750 - 883548 z) #1^11 + (-218700 - 
       188082 z) #1^12 + (32805 + 26244 z) #1^13 + 98415 z #1^14 &, 
  2], 0. < z < 0.674139]

I need to (numerically) integrate this function for given parameter values, which Mathematica refuses to do. However, if I convert the expression into a Piecewise function: for example as

ff = Piecewise[{{ftz[[1]][[1]][[2]][[1]], ftz[[1]][[1]][[2]][[2]]},
    {ftz[[2]][[1]][[2]][[1]], ftz[[2]][[1]][[2]][[2]]},
    {0, z == 0},
    {1, z == 1}}];

it easily computes the integral. Unfortunately, the number of intervals and the breakpoints are parameter dependent, and this must be done automatically as the function is called inside another function.

Does anyone have any suggestions on how to convert such a list of conditionals into a Piecewise function, or how to otherwise compute the integral? Thanks!

Answer 1

That conversion is straightforward, in this case. It is a one-liner

Piecewise[List @@@ Flatten[N@ftz][[All, 2]] ]

Now let me explain. As you have noticed, Solve returns solutions of the form:

{ {_ -> _}, {_ -> _}, ...}

which is what Flatten reduces to

{_ -> _, _ -> _, ...}

Then, I take only the right hand side of each Rule via Flatten[...][[All,2]] leaving us with something of the form

{ConditionalExpression[...], ...}

So, we need to replace ConditionalExpression with List to make it palatable to Piecewise, and the it is done using the shorthand form of Apply at level 1, @@@. Finally, we feed the result into Piecewise and get what you want.

---

As pointed out in the comments, the simpler form

Piecewise[List @@@ Last @@@ N@ftz]

works just as well.

Answer 2

It may not always be appropriate to convert a ConditionalExpression into Piecewise, because the two are not equivalent, and you may be changing the meaning. This is because Piecewise IMPOSES a default value of 0 if the conditions are not satisfied, whereas ConditionalExpression leaves the expression as undefined.

For example, given:

In[1]:=  expr = ConditionalExpression[1, x > 3]

In[2]:=  Simplify[expr, x < -2]
Out[2]=  Undefined

whereas:

In[3]:=  Simplify[Piecewise[{{1, x > 3}}],  x < -2]
Out[3]=  0

But, if you do want to impose the swap, say given:

In[4]:=  sol = Integrate[x^n, {x, 0, 1}]
Out[4]=  ConditionalExpression[1/(1 + n), Re[n] > -1]

... then an easy way to do it is:

In[5]:=  sol /. ConditionalExpression[xx_, yy_] :> Piecewise[{{xx, yy}}]