20200801, 03:23  #12 
Romulan Interpreter
"name field"
Jun 2011
Thailand
5^{3}×79 Posts 
Methink is some Brent/Newton variation (it computes the tangents, and their Ox intersection to get next point). You may force the "abort" by playing with real precision defaults.
Edit: yep, the manual says it uses Brent (draw a secant from a to b, it intersects Ox, that is the new point, it matched with your output, well, Brent Method is a bit more complex, but that is the idea, and it should be very easy to implement a Brent(function, a, b, epsilon) to work as intended, give me some minutes for a recursive version...). Last fiddled with by LaurV on 20200801 at 03:50 
20200801, 10:03  #13  
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Quote:
Code:
cnt=0;solve(x=1,2,cnt+=1;print(cnt" "x);x^3) 

20200816, 11:02  #14 
"Jeppe"
Jan 2016
Denmark
2^{4}·11 Posts 
I thought the problem would go away if you shifted away from zero (where the floating point number representation can shift almost arbitrarily), but it still cannot do any of:
Code:
solve(x=1, 2, x^3) solve(x=1+0.1, 2+0.1, (x0.1)^3)  0.1 solve(x=1+0.1, 2+0.1, print(x);(x0.1)^3)  0.1 
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