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        riddles >> putnam exam (pure math) >> inequality
        
(Message started by: anonymous on Jun 16^th, 2003, 5:50pm)

Title: inequality
Post by anonymous on Jun 16^th, 2003, 5:50pm

Let a₁ be a real number such that 0 < a₁ < 1
Define a_n+1=(a_n)² for n=1,2,3,...

Prove that
sum_{n=1}^{infinity} (a_n-a_n+1)a_n+2 < 1/3

hint: geometric progression

Title: Re: inequality
Post by william wu on Jun 17^th, 2003, 9:43pm

Didn't really get anywhere with this one, aside from the following perhaps trivial observations:

http://www.ocf.berkeley.edu/~wwu/images/riddles/inequality_geoProgression_attempt.png

Title: Re: inequality
Post by James Fingas on Jun 19^th, 2003, 9:51am

The exact upper bound is 0.263036536, based on studying the sum_i=-inf..inf(a_i⁵-a_i⁶) with a₁ on the interval [0.5^2/3..0.5^4/3]. The maximum happens exactly when a₁=0.5.

UPDATE: whoops! no it's not. The sum is a little larger for a₁=0.5134 (not the exact maximal value). The actual sum is roughly the same.

ADDED: consider the sequence a₁,a₁^6/5,a₂,a₂^6/5,a₃,...
You can see that this is a monotone decreasing function, and therefore splits the interval [0,1] into alternating bands. Half are included in the sum, and half aren't.

Title: Re: inequality
Post by towr on Jun 19^th, 2003, 12:51pm

I get the maximum very close to 1
(at least higher than 5/6)

which makes sense to me, since every term a^{2^(n-1)*5} - a^{2^(n-1)*6} has a maximum for a = (5/6)^{(1/2)^(n-1)}
which gets ever closer to 1 for increasing n.

the maximum is the same for all terms though, so summing those won't give an upper limit..

Title: Re: inequality
Post by James Fingas on Jun 20^th, 2003, 9:34am

towr,

I think you're saying that the sum generally gets larger as a₁ goes closer to 1. That is true, but I am considering the two-sided infinite sum (from -inf to inf), which effectively takes the limit as you go nearer and nearer to 1. For the two-sided infinite sum, choosing a₁=0.25 is equivalent to choosing a₁=0.0625 or a₁=0.5, so I restricted my considerations to a band around 0.5, from 0.5^4/3 to 0.5^2/3.

The value for the whole sum ranges from 0.263032 to 0.263037 as far as my analysis can tell, depending on what you pick for a₁. The maximum value for the two-sided infinite sum seems to occur for a₁ around 0.515, but I haven't found a reason why.

Title: Re: inequality
Post by anonymous on Jun 20^th, 2003, 2:41pm

Does anyone wants me to post the proof?

here's another hint: telescopic sum

Title: Re: inequality
Post by James Fingas on Jun 23^rd, 2003, 10:36am

The only bounding geometric sum that I've tried has been sum_i=1..infa^5*i-a^6*i, but this doesn't do well as a gets near 1. I proved that if you define b_i = a₁ⁱ, then sum_i=1..inf(b_i - b_i+1)*b_i+2 was bounded less than 1/2, but this doesn't necessarily mean anything because the relationship between b_i and b_i+1 isn't the same as the relationship between a_i and a_i+1.

I don't have a good idea for a telescoping sum, although that would be nice.

Well I just figured out a way to brute-force it. We know that the two-sided infinite sums are the same for all a₁ that are the same up to squaring a number of times. We also know that the sum is less than a₁⁵ plus (1-a₁). So I calculated the first three terms of the sum, then replaced the rest of the sum with a₁⁴⁰, added (1-a₁), and found that the result was less than 1/3 for a₁ in the range 0.8 to 0.9. Since you can map all two-sided infinite sums into this range, all such sums add to less than 1/3. The one-sided sums are always smaller than the two-sided sums. QED (but not very satisfying).