Full marks! Extra credit: Product[(4*Cos[(2/3)^n*x] - Sec[(2/3)^n*x])/3, {n, ∞}] == Sinc[y] What is y? --rwg
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?
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.
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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.
Julian sez it's good out to |argz| ≤ 2π/3, past which the limit is (2 Re(sqrt(z0))-|sqrt(z0|)^2/3 . [...]
MR> (I am having trouble identifying some of your characters, but I think I
know what it says ... )
Of course, it depends on the branch of sqrt(z) you are using. Let me take Arg(z) to be in the half-open interval (-pi, pi] , and w = sqrt(z) to be in the sector -pi/2 < Arg(w) <= pi/2 .
Write z = r exp(i theta) , where theta = Arg(z) . From the rhombus with vertices at 0 , z , |z| and z + |z| , it is clear that the argument of z + |z| is half the argument of z . Moreover, z + |z| = 2 r cos (theta / 2) exp(i theta / 2) . The branch of sqrt(z) implies that sqrt(z |z|) also has argument theta / 2 . Thus, z + |z| - sqrt(z |z|) also has argument theta / 2 ... well, maybe . We have z + |z| - sqrt(z |z|) = r (2 cos (theta / 2) - 1) exp(i theta / 2) , so as long as the factor 2 cos (theta / 2) - 1 is non-negative, the argument is theta / 2 ; otherwise add or subtract pi .
This shows that one step of the recurrence multiplies z by (2 cos (theta / 2) - 1) exp(-i theta / 2) , where theta = Arg(z) , and if theta is in the interval [-2 pi / 3 , 2 pi / 3] , the argument gets halved (and thus stays in that interval). Thus, if the initial z_0 is in the sector [-2 pi / 3 , 2 pi / 3] , we multiply by the infinite product prod_{k = 1, infinity} (2 cos (theta / 2^k) - 1) exp(-i theta / 2^k) , which is a telescoping product. Indeed, write the factor as (exp(3 i theta / 2^(k+1)) + exp(-3 i theta / 2^(k+1))) / ((exp(i theta / 2^(k+1)) + exp(-i theta / 2^(k+1))) exp(i theta / 2^k)) so that z_n = z_0 prod_{k = 1, n} ( expression above ) = z_0 (exp(3i theta / 2) - exp(-3i theta / 2)) (exp(i theta / 2^(n+1)) - exp(-i theta / 2^(n+1))) / ((exp(i theta / 2) - exp(-i theta / 2)) (exp(3i theta / 2^(n+1)) - exp(-3i theta / 2^(n+1))) exp(i theta (1 - 2^(-n)))) = r exp(i theta / 2^n) sin(3 theta / 2) sin(theta / 2^(n+1)) / (sin (theta / 2) sin(3 theta / 2^(n+1)) (hopefully no typos there!) Now the limit is easy to evaluate as r sin(3 theta / 2) / (3 sin (theta / 2)) = r (2 cos theta + 1) / 3 = (2 Re(z_0) + |z_0|) / 3 .
If z_0 is outside the sector [-2 pi / 3 , 2 pi / 3] , then z_1 is within the sector. For example, if 2 pi / 3 < Arg(z_0) <= pi , then z_0 + |z_0| - sqrt(z_0 |z_0|) = r (2 cos (theta / 2) - 1) exp(i theta / 2) , as before. But now the factor 2 cos (theta / 2) - 1 is negative, so the argument is (theta / 2) - pi . Thus z_1 is in the sector -2 pi / 3 < Arg(z_1) <= -pi / 2 . In this case, the limit is (2 Re(z_1) + |z_1|) / 3 . I leave it to those interested to express it in terms of z_0 .
By the way, I haven't seen this before, on the Putnam, or elsewhere, but that doesn't mean it couldn't have been a Putnam problem.
Michael Reid