[math-fun] Iterated square root puzzle
Find a closed form for the the expression sqrt(1/2^(2^0) + sqrt(1/2^(2^1) + sqrt(1/2^(2^2) + ...))). --Dan P.S. No fair using electronic assistance. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ? --Joshua Zucker
How do you suppose problem 8 was intended to read? On 8/14/10, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ?
--Joshua Zucker
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Doesn't it, as shown, turn out to be something like 1 + golden ratio? That seems nice enough for an expression as ugly-looking as that. --Joshua On Fri, Aug 13, 2010 at 9:59 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
How do you suppose problem 8 was intended to read?
On 8/14/10, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ?
This one is very nice --- but unless I've missed something (again!), it does seem an awful lot harder than the others on this page! WFL On 8/14/10, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ?
--Joshua Zucker
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This one was too hard for me, so I dug around a bit and found that it's a special case of a more general formula from Ramanujan (and in fact this specific problem was given by Ramanujan). You can read about it here (skip down to where it says "Ramanujan posed this problem to the 'Journal of Indian Mathematical Society'"): http://en.wikipedia.org/wiki/Nested_radical#Some_identities_of_Ramanujan And also here (look for "Ramanjuan discovered" followed by equation (26) - I used 25 in the URL below to position the page just above it without cutting off the top): http://mathworld.wolfram.com/NestedRadical.html#eqn25 Tom On 8/14/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
This one is very nice --- but unless I've missed something (again!), it does seem an awful lot harder than the others on this page! WFL
On 8/14/10, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ?
--Joshua Zucker
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Two-line solution to Ramanujan's JIMS problem: for n > 0 define h_n == 1 sqrt(1 + 2 sqrt(1 + 3 sqrt(1 + ... + (n-1)sqrt(1 + n(n+2))))); then easily h_n = h_n+1, and by induction 1 sqrt(1 + 2 sqrt(1 + 3 sqrt(1 + ... ))) = h_oo = h_1 = 3 . Begging the question, how might h_n be discovered in the first place? The way the problem is posed, its focus is on the limit as n -> oo of f_n == 1 sqrt(1 + 2 sqrt(1 + 3 sqrt(1 + ... + n))) ; but instead of the far end, it's actually better instead to truncate the front: g_n == n sqrt(1 + (n+1)sqrt(1 + (n+2)sqrt(1 + (n+3)sqrt(1 + ... )))) . Now the desired limit f_oo = g_1 (assuming they exist); and easily g_{n+1} = (g_n/n)^2 - 1 . So consider the iterative process starting from arbitrary real x_1, x_{n+1} = (x_n/n)^2 - 1 . Experiment soon establishes that x_n has in general just two limits: x_n -> -1 for |x_1| < 3 ; x_n -> +oo for |x_1| > 3 ; and furthermore along the critical path, easily x_n = n(n+2) for x_n = 3 . To confirm this bifurcation, write x_n == n(n+2) + y_n ; now by simple algebra, y_{n+1}/y_n = (2 + 4/n + y_n/n^2) > 1 [provided y_n > - 2n (n+2), always true for n > 4]. That is, apart from initial hiccoughs, the deviation of x_n from n(n+2) always increases in the same direction as initially. [A little more work establishes that convergence is linear on the -1 side, divergence doubly exponential on the +oo side.] Finally then, the limit 3 of the "forward" process f_n emerges as the critical point of the "reverse" process g_n . I imagine there must be some well-known analytic technique at work here, but my grasp of the subject is not up to formulating it in any generality. In any case, the sophistication required to discover the result seems to place it an altogether different category from its companions on the webpage where it appeared! http://www.msri.org/specials/festival/activities/InfinityFinite.pdf Fred Lunnon On 8/14/10, Tom Karzes <karzes@sonic.net> wrote:
This one was too hard for me, so I dug around a bit and found that it's a special case of a more general formula from Ramanujan (and in fact this specific problem was given by Ramanujan). You can read about it here (skip down to where it says "Ramanujan posed this problem to the 'Journal of Indian Mathematical Society'"):
http://en.wikipedia.org/wiki/Nested_radical#Some_identities_of_Ramanujan
And also here (look for "Ramanjuan discovered" followed by equation (26) - I used 25 in the URL below to position the page just above it without cutting off the top):
http://mathworld.wolfram.com/NestedRadical.html#eqn25
Tom
On 8/14/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
This one is very nice --- but unless I've missed something (again!), it does seem an awful lot harder than the others on this page! WFL
On 8/14/10, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On somewhat the same topic, how about problem 6 at http://www.msri.org/specials/festival/activities/InfinityFinite.pdf ?
--Joshua Zucker
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participants (4)
-
Dan Asimov -
Fred lunnon -
Joshua Zucker -
Tom Karzes