In an effort to examine this question on Math.SE, I defined the following functions
c[n_, x_] := (a[n - 1, x] + b[n - 1, x])/4 + x/2;
a[n_, x_] :=
Return[If[n == 0, 0, t = c[n, x]; If[t > x, a[n - 1, x], t]]];
b[n_, x_] :=
Return[If[n == 0, 1, t = c[n, x]; If[t > x, t, b[n - 1, x]]]];
These functions work as expected and return the correct output. However, when I call PiecewiseExpand[a[2, x]] I get the output of
1/4 (1 + 2 x) if x >= 1/22
0 else
This is clearly wrong - a[2, 0.3] = 0.25 while $\frac{1+2\cdot0.3}{4} = 0.4$. For some reason, PiecewiseExpand is not working properly here. Why is this happening, and how can I fix this problem?
Edit: At the suggestion of Nasser, I changed the functions to
c[n_, x_] := (a[n - 1, x] + b[n - 1, x])/4 + x/2;
a[n_, x_] :=
If[n == 0, 0, t = c[n, x]; If[t > x, a[n - 1, x], t, "idk"], "idk1"];
b[n_, x_] :=
If[n == 0, 1, t = c[n, x]; If[t > x, t, b[n - 1, x], "idk"], "idk1"];
Now with PiecewiseExpand[a[1, x], {0 < x < 1}], the output is idk.
If[t > x, a[n - 1, x], t]you need to use theIfwith 3 arguments version.If[t > x, a[n - 1, x], t, "do not know" ]. Since it is not able to decide ift>xwhenxis symbolic. It works withxis a number, since then it is able to decide. And why do you need to use all theseReturnfor? Just get rid of them. $\endgroup$PiecewiseExpandjust returns the fourth argument ("do not know"). $\endgroup$2>xor not ifxis symbolic? Not possible. This is your algorithm. So you need to figure what to do in this case. It works whenxis number. These issues do not show up in say Matlab, because Matlab is all numbers. So no need for the 3 branchIf. But in Mathematica it is needed. $\endgroup$for x∈(0,1)soxin that algorithm is a specific numerical value. Not symbolic. $\endgroup$xhas to be given and it must be a number between 0 and 1. can not be symbolic, Otherwise you can't do it. I said all what I can on this. May be someone else can provide better help. good luck. $\endgroup$