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Topic: FIND A(2005) (Read 464 times) |
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pcbouhid
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FIND A(2005)
« on: Dec 1st, 2005, 9:50am » |
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A sequence of positive real numbers is defined by
a(0) = 1,
a(n+2) = 2a(n) - a(n+1), for all n>0.
Find a(2005).
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JocK
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Re: FIND A(2005)
« Reply #1 on: Dec 1st, 2005, 10:33am » |
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on Dec 1st, 2005, 9:50am, pcbouhid wrote:A sequence of positive real numbers is defined by
a(0) = 1,
a(n+2) = 2a(n) - a(n+1), for all n>0.
Find a(2005).
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This is a second order difference equation. I think you need to prescribe one more a-value (e.g. a(1) = 2 )...?
Or do you want us to specify a(2005) in terms of a(1)...?
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| « Last Edit: Dec 1st, 2005, 10:35am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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pcbouhid
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Re: FIND A(2005)
« Reply #2 on: Dec 1st, 2005, 10:44am » |
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It may, at first, seem that the sequence is not uniquely defined! However, the constraint that the sequence consists of positive numbers allow to deduce the value of a(1).
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JocK
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Re: FIND A(2005)
« Reply #3 on: Dec 1st, 2005, 10:51am » |
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Ahhh... overlooked that particular word.. 
But in that case the solution is trivial, isn't it?
I go for an = 1 for all n.
| hidden: | Reason being that any variation between subsequent terms would 'explode' to positive and negative values, both large in magnitude.
(This is because the characteristic equation x2 = 2 - x has a root -2.)
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| « Last Edit: Dec 1st, 2005, 10:54am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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pcbouhid
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Re: FIND A(2005)
« Reply #4 on: Dec 2nd, 2005, 7:29am » |
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You´re absolutely right, though I have a more deep proof that a(n) = 1, for all n>=0.
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JocK
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Re: FIND A(2005)
« Reply #5 on: Dec 2nd, 2005, 3:13pm » |
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on Dec 2nd, 2005, 7:29am, pcbouhid wrote:
.. educate me!
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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pcbouhid
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Re: FIND A(2005)
« Reply #6 on: Dec 3rd, 2005, 4:42am » |
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Not only you, Jock. Us. The solution isn´t mine.
There it is, in whole:
It may at first seem that the sequence is not uniquely defined! However, the constraint that the sequence consists of positive numbers allows us to deduce the value of a1. We will show that, if a1 != 1, the sequence will eventually contain a negative number.
Letting a(1) = x, we find (===> means "imply"):
a(0) = 1 + 0x,
a(1) = 0 + x ===> x > 0,
a(2) = 2 - x ===> x < 2,
a(3) = -2 + 3x ===> x > 2/3, (1)
a(4) = 6 - 5x ===> x < 6/5,
a(5) = -10 + 11x ===> x > 10/11, ...
It seems clear that, as we calculate more and more terms, x will be "squeezed" between two fractions, both of which are part of a sequence which tends to 1 as n tends to infinity. (It would follow that x = 1.) We verify this intuition below.
Setting x = 0, to isolate the constant terms in (1), we obtain the sequence {b(n)}: 1, 0, 2, -2, 6, -10, ...
We conjecture that, from b(2) = 2, b(3) = -2 onwards, the sequence alternates in sign, with |b(n)+2| > |b(n)|.
We prove this conjecture by mathematical induction.
Consider b(2n) = r, b(2n+1) = -s, where n, r, s are positive integers.
If n = 1, r = s = 2, which alternates in sign, as per the inductive hypothesis.
If n = k, b(2k+2) = b(2(k+1)) = 2r + s > r > 0, and b(2k+3) = b(2(k+1)+1) = - (2r + 3s) < 0, so that |-(2r + 3s)| > s.
That is, b(2k+2) > b(2k) > 0, and b(2k+3) < b(2k+1) < 0.
The result follows by induction; sequence {b(n)} alternates in sign, with |b(n+2)| > |b(n)|.
Setting x = 1, we know from the recurrence relation that a(n) = 1, for all n >= 0.
Therefore, the absolute value of the coefficient of x in sequence (1) must always differ by 1 from the absolute value of the constant term.
More specifically, we have:
a(2n) = b(2n) - (b(2n) - 1)x ===> x < b(2n)/(b(2n) - 1), and
a(2n+1) = b(2n+1) - (b(2n+1) - 1)x ===> x > b(2n+1)/(b(2n+1) - 1). (Note: b(2n+1) < 0.)
Since |b(2n)| and |b(2n+1)| are strictly increasing with n, the limit as n tends to infinity of both {b(2n)/(b(2n) - 1)} and {b(2n+1)/(b(2n+1) - 1)} is 1.
Hence x = 1.
(For any x != 1, there exists n such that x > b(2n)/(b(2n) - 1) or x < b(2n+1)/(b(2n+1) - 1), and hence a(2n) < 0 or a(2n+1) < 0.)
Therefore a(n) = 1, for all n >= 0. Specifically, a(2005) = 1.
Let me know if there are mistakes in the transcription.
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Icarus
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Re: FIND A(2005)
« Reply #7 on: Jan 23rd, 2006, 4:13pm » |
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An alternative way of showing the result: Note that if X and Y are two sequences satisfying an+2 = -an+1 + 2an, then cX + dY also satisfies the condition, for any real numbers c, d. Thus the set of all such sequences form a linear space. Since the sequences are all uniquely determined by their first two elements, the space is 2-dimensional. Hence, if we can identify 2 independent sequences X and Y satisfying the condition, all such sequences are of form cX + dY. Consider the sequence {rn}, for some r. This sequence will satisfy the condition provided r2 = -r + 2, or r2 + r - 2 = 0. The quadratic has 2 solutions: r = 1, r = -2. So every sequence satisfying the formula has form c1n + d(-2)n = c + d(-2)n.
By choosing n > log2 |c/d|, and of the appropriate parity, we get negative terms for any value of c and d except d = 0. Since the sequence in question is positive, d = 0, and the sequence is constant.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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