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anonymous
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 inequality   « on: Jun 16th, 2003, 5:50pm » Quote Modify Remove

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

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

hint: geometric progression
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william wu

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 Re: inequality   « Reply #1 on: Jun 17th, 2003, 9:43pm » Quote Modify

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

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James Fingas
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 Re: inequality   « Reply #2 on: Jun 19th, 2003, 9:51am » Quote Modify

The exact upper bound is 0.263036536, based on studying the sumi=-inf..inf(ai5-ai6) with a1 on the interval [0.52/3..0.54/3]. 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,a16/5,a2,a26/5,a 3,...
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.
 « Last Edit: Jun 19th, 2003, 10:19am by James Fingas » IP Logged

towr
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 Re: inequality   « Reply #3 on: Jun 19th, 2003, 12:51pm » Quote Modify

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

which makes sense to me, since every term a2^(n-1)*5 - a2^(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..
 « Last Edit: Jun 19th, 2003, 1:03pm by towr » IP Logged

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James Fingas
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 Re: inequality   « Reply #4 on: Jun 20th, 2003, 9:34am » Quote Modify

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 0.54/3 to 0.52/3.

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.
 « Last Edit: Jun 20th, 2003, 9:36am by James Fingas » IP Logged

anonymous
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 Re: inequality   « Reply #5 on: Jun 20th, 2003, 2:41pm » Quote Modify Remove

Does anyone wants me to post the proof?

here's another hint: telescopic sum
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James Fingas
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 Re: inequality   « Reply #6 on: Jun 23rd, 2003, 10:36am » Quote Modify

The only bounding geometric sum that I've tried has been sumi=1..infa5*i-a6*i, but this doesn't do well as a gets near 1. I proved that if you define bi = a1i, then sumi=1..inf(bi - bi+1)*bi+2 was bounded less than 1/2, but this doesn't necessarily mean anything because the relationship between bi and bi+1 isn't the same as the relationship between ai and ai+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).
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