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Topic: Double Inequality (Read 639 times) |
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Barukh
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Double Inequality
« on: Jan 17th, 2006, 8:25am » |
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Find at least one natural number n for which the following double inequality holds:
1/2 * 6/7 * … * (5n+1)/(5n+2) < 1/2006 < 5/6 * 10/11 * … * (5n+5)/(5n+6)
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Barukh
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Re: Double Inequality
« Reply #1 on: Jan 21st, 2006, 9:06am » |
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Sorry, but these days problems posted (especially in Easy section) leave the first page too fast... 
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Eigenray
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Re: Double Inequality
« Reply #2 on: Jan 22nd, 2006, 1:50am » |
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Well, it's easy to show such an n exists. Let Pn be the LHS, and Qn the RHS. Note Qn > 5/3 Pn, so if Pn > 1/2006, then Qn+1 > 5/3 (5n+10)/(5n+11) /2006 > 1/2006.
Finding n is a big trickier.
| hidden: | First note -log(1-x) > x for x>0, so
-log Pn > log 2 + 1/7 + ... + 1/(5n+2)
> log 2 + 1/5 [int]75n+7 dx/x
= 1/5 log((5n+7)*32/7) > 1/5 log(4(5n+7))
> log 2006 when (5n+7) > 1/4 * 20065.
On the other hand,
[int] -log(1-1/x)dx = log(x-1) + log (1-1/x)-x,
and since the second term is decreasing in x,
[int]ab -log(1-1/x)dx < log((b-1)/(a-1)).
Since the integrand is also decreasing, it follows that
-log Qn < log(6/5) + 1/5 [int]65n+6 -log(1-1/x)
< log(6/5) + 1/5 log((5n+5)/5)
< log 2006 when (5n+5) < 5(5/6*2006)5.
Thus both inequalities are satisfied when
(1/4) 20065 < 5(n+1) < (56/65) 20065.
As x1/x is decreasing, 56/65 > 1 (in fact, > 2). So the difference between the two sides above is huge, and we may safely take, say,
n = [(1/10)*20065]. |
I spent quite a while trying to get nice bounds I could easily compute by hand. At one point I had it down to noting e121/120/6 < 1/2 (using, for example, -log(1-x) < x + 11/20 x2 for x<1/11), but it still didn't feel quite right to me.
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| « Last Edit: Jan 22nd, 2006, 2:06am by Eigenray » |
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SWF
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Re: Double Inequality
« Reply #3 on: Jan 23rd, 2006, 5:25pm » |
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Rewriting trying to get 5*n in denominators gives: (5n+1)/(5n+2) = (1+1/(5n)) / (1+2/(5n)). After shifting index and being careful with n=0, (5n+5)/(5n+6) becomes 5m/(5m+1)= 1/(1+1/(5m)).
I used the following inequalities with 0<x<1:
x - x2/2 < ln(1+x) < x
[sum] for n= 1 to N of (1/n^2) < [pi]^2/6
gamma + ln(N) < [sum] for n= 1 to N of (1/n) < gamma + ln(N+1)
gamma is Euler's constant= 0.5772156649...
Taking ln() and using the inequalities gives
10035*exp( [pi]2/15 - gamma ) < N < 20065*exp(-gamma) - 2
or 1.1e15 < N < 1.82e16
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Barukh
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Re: Double Inequality
« Reply #4 on: Jan 24th, 2006, 2:07am » |
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Hmm... The solution I am aware of does give an exact value...
Hint: Could you use even more inequalities?
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Aryabhatta
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Re: Double Inequality
« Reply #5 on: Jan 24th, 2006, 10:21am » |
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I think this should be moved to the medium section!
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SWF
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Re: Double Inequality
« Reply #6 on: Jan 24th, 2006, 9:08pm » |
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Barukh, what do you mean by a solution with an exact value? Do you mean it give the exact value for the highest and lowest possible values of n, or do you mean there is a value that happens to be the same for both the upper and lower limit? I was trying to find a wide range of n that work. One way to improve on the range is to use x/(1+x/2) < ln(1+x). The following gives a fairly simple (although I had trouble discovering it) way to bring the upper and lower bounds together:
If you note that for n>0 the 1/2*...*(5n+1)/(5n+2) term is always less than 1/(10n)^.2, and the 5/6*...*(5n+5)/(5n+6) is always greater than 1/(10n)^.2 (show by induction), upper and lower bounds on n are exactly what Eigenray gave: 2006^5/10. One inequalty says n must be greater than 2006^5/10 and the other says n must be less, but 2006^5/10 is not an integer, so this is not quite right, but should not be hard to fix for sticklers.
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Eigenray
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Re: Double Inequality
« Reply #7 on: Jan 24th, 2006, 9:42pm » |
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on Jan 24th, 2006, 2:07am, Barukh wrote:| Hmm... The solution I am aware of does give an exact value... |
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In fact, for any 0 < r < 1/2, we have
1/2 * 6/7 * … * (5n+1)/(5n+2) < r < 5/6 * 10/11 * … * (5n+5)/(5n+6),
where n=floor[ 1/(5r5) ].
Taking r=1/2006, this is exactly (20065-1)/5.
Is that better?
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Barukh
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Re: Double Inequality
« Reply #8 on: Jan 24th, 2006, 11:19pm » |
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on Jan 24th, 2006, 9:42pm, Eigenray wrote:
In fact, for any 0 < r < 1/2, we have
1/2 * 6/7 * … * (5n+1)/(5n+2) < r < 5/6 * 10/11 * … * (5n+5)/(5n+6),
where n=floor[ 1/(5r5) ].
Taking r=1/2006, this is exactly (20065-1)/5.
Is that better? |
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Yes, it is. 
But probably I ought to analyze your original post better.
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