# Division by an interval

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]}]

Can you help me to resolve it ?

• It seems the issue is that your definition of a reciprocal of an interval is different from the one mathematica uses. What would you expect for 1/Interval[{-1, 1}] ? – george2079 Apr 26 '17 at 15:03
• One should keep in mind that every time Interval 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 of 0., $MinMachineNumber. Not usually important unless you write code that assumes they are unchanged (or your problem is ill-conditioned). – Michael E2 May 26 '17 at 16:19

f[x]/((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)]) equals to Interval[{-[Infinity], -0.0257349}, {0.0299225, [Infinity]}]. Please check the geometric representation of it.

In:

Clear[f, x, a, b, inter]
f[x_] := 1792 - 256 x - 2240 x^2 + 320 x^3 + 448 x^4 - 64 x^5 -
6 x^6 + 6 x^7;
inter = Interval[{-4, 3}];
x = Mean@First@(List @@ inter) // N;
IntervalIntersection[inter,
x - (f[x]/((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)]))];
f[x]/((1/4) f'[x] + (3/4) f'[x + (2/3) (inter - x)]);

a = f[x] (*1349.859375*)
b = ((1/4) f'[x] + (3/4) f'[
x + (2/3) (inter -
x)]) (*Interval[{-52452.38917824089,45111.86612654332}]*)

a/b (*Interval[{-\[Infinity],-0.025734945464790555},{0.\
029922490264834276,\[Infinity]}]*)

(*geometric representation of a/b*)
Plot[a/x1, {x1, -52452.38917824089, 45111.86612654332},
Filling -> Axis]


Out:

1349.86
Interval[{-52452.4, 45111.9}]
Interval[{-\[Infinity], -0.0257349}, {0.0299225, \[Infinity]}]


• ok, how can I divide Interval[{-\[Infinity], -0.0257349}, {0.0299225, \[Infinity]}] into two Interval[{-\[Infinity], -0.0257349}] and Interval[{0.0299225, \[Infinity]}] ? – Johanna Apr 26 '17 at 15:29
• Interval[{-[Infinity], -0.0257349}, {0.0299225, [Infinity]}] is a 2D interval. Interval[{-[Infinity], -0.0257349}] is 1D interval. I am not sure it's feasible. – UnchartedWorks Apr 26 '17 at 15:52
• you can do this: Interval /@ List @@ Interval[{-\[Infinity], -0.0257349}, {0.0299225, \[Infinity]}]` – george2079 Apr 26 '17 at 18:45