Phoebe Floats !. Presented by Bernard Moore Author: Ezra Brown Virginia Polytechnic University The College M athematics J ournal Vol.36 No.2 (March 2005). Inspiration.
Presented by Bernard Moore
Author: Ezra Brown
Virginia Polytechnic University
The College Mathematics Journal Vol.36 No.2 (March 2005)
We place the center of the circle of radius 1 at the origin. We will denote its distance underwater by r. In the case that the density is ½, half the ball is submerged and the depth r=1. Since Phoebe’s density is greater than ½, it’s clear that more than half of Phoebe is submerged, and so we expect its distance underwater to be between 1 and 2.
Do we have a method to solve for r in this cubic polynomial?
Yes. The Cardano-TartagliaFormula.
If Δ>0 the equation has 3 distinct real roots, if Δ=0 the equation has real but multiple roots, and if Δ<0 then we have one real and two complex conjugate pairs.
From seeing the graph below, we can expect that there is a root in the intervals (-1,0), (1,2), and (2,3).
So we compute x1, x2, and x3 as -0.852523, 1.273485, 2.579038. These answers makes sense, we have one for each of the intervals that we listed.
But this r value was with a ball of radius 1, we want to know how deep Phoebe sinks.
be closer to the actual root than is x1.
The following theorem explains the odd behavior that we seen with Newton’s Method. We see that the points x = 1.86693 and .32488 are both repelling points which explains their escaping behavior that we seen in the chart of iterates. We may replace g by N below, where the fixed points of N are the roots of f.
Remember that the starting values for our first table of iterates were -3, 1, and 5 which aren’t near the repelling points at .325 and 1.87. When choosing starting points we will avoid repelling points for better approximations.
Did you know Saturn floats too?!