How to solve this equation with pari gp ...

[LaurV]

Originally Posted by R. Gerbicz

Good question, it is not binary search,

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...).

[R. Gerbicz]

Originally Posted by LaurV

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,

Ok, but for f(x)=x^3 it is weaker than the binary search. Just try this:

cnt=0;solve(x=-1,2,cnt+=1;print(cnt" "x);x^3)

So it is doing at most 259 iterations.

[JeppeSN]

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:

solve(x=-1, 2, x^3)
solve(x=-1+0.1, 2+0.1, (x-0.1)^3) - 0.1
solve(x=-1+0.1, 2+0.1, print(x);(x-0.1)^3) - 0.1

with usual realprecision default. /JeppeSN