I have a function:
f[x_] := 1792 - 256 x - 2240 x^2 + 320 x^3 + 448 x^4 - 64 x^5 - 6 x^6 + 6 x^7
interval:
inter = Interval[{-4, 3}]
x:
x = Mean@First@(List @@ inter) // N
and formula:
IntervalIntersection[inter, x - (f[x]/((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)]))]
The formula doesn't calculates correctly because of division f[x]
by interval ((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)])
Though explicitly it can be done by hand, as f[x] == 1349.86
and ((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)]) == Interval[{-52452.4, 45111.9}]
.
So I suppose it f[x]/((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)])
should be eq to Interval[{-0.0257349, 0.0299225}]
but the result is Interval[{-\[Infinity], -0.0257349}, {0.0299225, \[Infinity]}]
1/Interval[{-1, 1}]
? $\endgroup$ – george2079 Apr 26 '17 at 15:03Interval
is evaluated with machine precision arguments, the interval is expanded by roughly machine error. This makes an epsilon difference, either literally$MachineEpsilon
or in the case of0.
,$MinMachineNumber
. Not usually important unless you write code that assumes they are unchanged (or your problem is ill-conditioned). $\endgroup$ – Michael E2 May 26 '17 at 16:19