Group Abstract Group Abstract

Message Boards Message Boards

0
|
37 Views
|
4 Replies
|
7 Total Likes
View groups...
Share
Share this post:

Why is it impossible to find the zeros of a function in an interval after defining it?

Bill Blair
Posted 2 days ago

As shown in the code below, a function is defined, but why can't we find its zeros over an interval?

f[x_] := x Log[2 - x] /; 0 <= x < 1
f[x_] := (x - 2) Log[x] /; 1 <= x <= 2
f[x_] := -f[-x] /; x < 0
f[x_] := f[x - 2] /; x > 2
Plot[{f[x]}, {x, 0, 10}]

enter image description here

Why is it impossible to find all zeros of a function within a given interval, and how can this problem be resolved?

f[x_] := x Log[2 - x] /; 0 <= x < 1
f[x_] := (x - 2) Log[x] /; 1 <= x <= 2
f[x_] := -f[-x] /; x < 0
f[x_] := f[x - 2] /; x > 2
Plot[{f[x]}, {x, 0, 10}]
Reduce[{f[x] == 0, 0 <= x <= 2026}, x, Reals]

enter image description here
POSTED BY: Bill Blair
4 Replies
Sort By:

I think the following is a cute approach, but it's not very efficient:

Solve[0 == 
  DSolveValue[{y'[x] == 
     D[Piecewise[{{t[x] Log[2 - t[x]], 
         0 <= t[x] < 1}}, (t[x] - 2) Log[t[x]]], t[x]], y[0] == 0,
    t'[x] == 1, t[0] == 0,
    WhenEvent[Mod[x, 2] == 0, t[x] -> t[x] - 2]},
   y[x], {x, 0, 100},
   Method -> {"Events", "MaxEvents" -> 1020} (*not needed unless x goes past 2000*)
   ],
 x
 ]
(*
{{x -> 0}, {x -> 2}, {x -> 4}, {x -> 6}, {x -> 8}, {x -> 10},
 {x -> 12}, {x -> 14}, {x -> 16}, {x -> 18}, {x -> 20},
 {x -> 22}, {x -> 24}, {x -> 26}, {x -> 28}, {x -> 30},
 {x -> 32}, {x -> 34}, {x -> 36}, {x -> 38}, {x -> 40},
 {x -> 42}, {x -> 44}, {x -> 46}, {x -> 48}, {x -> 50},
 {x -> 52}, {x -> 54}, {x -> 56}, {x -> 58}, {x -> 60},
 {x -> 62}, {x -> 64}, {x -> 66}, {x -> 68}, {x -> 70},
 {x -> 72}, {x -> 74}, {x -> 76}, {x -> 78}, {x -> 80},
 {x -> 82}, {x -> 84}, {x -> 86}, {x -> 88}, {x -> 90},
 {x -> 92}, {x -> 94}, {x -> 96}, {x -> 98}, {x -> 100}}
*)

It takes almost 5 seconds. The full 2026 takes 86 seconds (Macbook Pro M4 Max).

For some reason, WRI often plays method option values close to the chest.
POSTED BY: Michael Rogers

Here's a possible workaround:

To echo Gianluca's point from a different perspective, Reduce[] and other symbolic solvers work with the expression and not the definitions of the symbols in the expression — that is, with f[x] as shown in Out[403] in the screenshot, not with the definitions in Downvalues[f]. Of course, they have built-in rules for transforming expressions involving standard built-in functions, but those transformations are not part of the definitions of the functions.
POSTED BY: Michael Rogers
Jim Baldwin
Jim Baldwin
Posted 1 day ago

Writing the function in the following way seems to work:

f[x_] := With[{t = Mod[x, 2]}, If[t < 1, t Log[2 - t], (t - 2) Log[t]]]
Reduce[{f[x] == 0, 0 <= x <= 2026}]
(* (C[1] \[Element] Integers && 0 <= C[1] <= 1013 && 
   x == 2 C[1]) || (C[1] \[Element] Integers && 0 <= C[1] <= 1012 && 
   x == 1 + 2 C[1]) *)
POSTED BY: Jim Baldwin

It seems that your definitions by condition /; supports numerical calculation but not symbolic manipulation. For example,

PiecewiseExpand[f[x], 0 < x < 10]

does nothing. Curiously, f'[1] turns back unevaluated, while f'[1.] gives a numerical value. This is numerical and it works:

FindRoot[f[x] == 0, {x, 3.5}]
POSTED BY: Gianluca Gorni
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard