[math-fun] Aaronson's puzzles
JA> For anyone who wants a little puzzle to start the week: if I set z_0 = i, and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to infinity? -- James Tougher: Suppose z_(n+1) = z_n + |z_n| - sqrt(z_n |z_n|). Why is z_∞ = 1/3? What is it for general z_0? Spoiler below. JA> Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp). Has anyone pointed out that this is the ground-level orbital period, so a zero-altitude satellite in polar orbit will synchronize latitudes with the point mass? --rwg Spoiler: (2Re(z)+|z|)/3. This HAD to be on a Putnam.
On Fri, May 18, 2012 at 7:40 AM, Bill Gosper <billgosper@gmail.com> wrote:
JA>
For anyone who wants a little puzzle to start the week: if I set z_0 = i, and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to infinity?
-- James
Tougher: Suppose z_(n+1) = z_n + |z_n| - sqrt(z_n |z_n|). Why is z_∞ = 1/3? What is it for general z_0? Spoiler below.
JA>
Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp).
Has anyone pointed out that this is the ground-level orbital period, so a
zero-altitude satellite in polar orbit will synchronize latitudes with the point mass? --rwg Spoiler: (2Re(z0)+|z0|)/3. This HAD to be on a Putnam.
Oops, this only holds for Re(z0)≥0 ! Re(z0)<0 is more interesting. Help, Julian!-) --rwg
(Repair at bottom) On Sat, May 19, 2012 at 12:23 AM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, May 18, 2012 at 7:40 AM, Bill Gosper <billgosper@gmail.com> wrote:
JA>
For anyone who wants a little puzzle to start the week: if I set z_0 = i, and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to infinity?
-- James
Tougher: Suppose z_(n+1) = z_n + |z_n| - sqrt(z_n |z_n|). Why is z_∞ = 1/3? What is it for general z_0? Spoiler below.
JA>
Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp).
Has anyone pointed out that this is the ground-level orbital period, so a
zero-altitude satellite in polar orbit will synchronize latitudes with the point mass? --rwg Spoiler: (2Re(z0)+|z0|)/3. This HAD to be on a Putnam.
Oops, this only holds for Re(z0)≥0 ! Re(z0)<0 is more interesting. Help, Julian!-) --rwg
Julian sez it's good out to |argz| ≤ 2π/3, past which the limit is (2 Re(sqrt(z0))-|sqrt(z0|)^2/3 . Check: In[808]:= tst[z_, n_Integer: 9] := Nest[#1 + Abs[#1] - Sqrt[#1 Abs[#1]] &, z, n] -> If[Abs[Arg[z]] > 2*π/3, (2*Re[Sqrt[z]] - Abs[Sqrt[z]])^2, Abs[z] + 2 Re[z]]/3 In[809]:= tst[I - 9., 19] Out[809]= 2.3877 - 0.0000148113 I -> 2.3877 --rwg If this was on a Putnam, they're even worse than I thought.
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Bill Gosper