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Old 2020-08-01, 03:23   #12
LaurV
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Quote:
Originally Posted by R. Gerbicz View Post
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...).

Last fiddled with by LaurV on 2020-08-01 at 03:50
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Old 2020-08-01, 10:03   #13
R. Gerbicz
 
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"Robert Gerbicz"
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Quote:
Originally Posted by LaurV View Post
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:
Code:
cnt=0;solve(x=-1,2,cnt+=1;print(cnt" "x);x^3)
So it is doing at most 259 iterations.
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Old 2020-08-16, 11:02   #14
JeppeSN
 
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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, (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
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