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Topic: Identical limits (Read 737 times) |
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JocK
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Identical limits
« on: Jan 27th, 2005, 9:24am » |
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a0 [in] [bbr], b0 [in] [bbr], n [in] [bbn]0
an+1 = (an + 1)1/3
bn+1 = (bn4 + 1)1/5
prove that a[infty] = b[infty]
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| « Last Edit: Jan 27th, 2005, 9:26am 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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Icarus
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Re: Identical limits
« Reply #1 on: Jan 27th, 2005, 3:29pm » |
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You sure you won't let us take a0 and b0 from [bbc] instead? 
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JocK
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Re: Identical limits
« Reply #2 on: Jan 31st, 2005, 2:21pm » |
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No takers..? Perhaps I should change the problem somewhat:
Why would x3 - x - 1 = 0 and x5 - x4 - 1 = 0 share the same positive root?
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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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towr
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Re: Identical limits
« Reply #3 on: Jan 31st, 2005, 2:53pm » |
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Yeah, I can get that far, and the graph certainly suggest they share the same root. But I haven't a clue how to prove it.
[e] How about something like
::x^3 -x - 1 = 0
x^3 = x + 1
x^5 - x^4 -1 = 0 { repeatedly replace x^3 by x + 1 }
x^2(x + 1) - x(x + 1) - 1 = 0
x^3 - x - 1 = 0
x + 1 - x - 1 = 0
true
So any x for which x^3 = x + 1 must be a solution for x^5 - x^4 -1 = 0 ?!?
::[/e]
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| « Last Edit: Jan 31st, 2005, 3:00pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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JocK
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Re: Identical limits
« Reply #4 on: Jan 31st, 2005, 3:57pm » |
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OK... that's the easy way.
Now the more elegant way: someone dare to cast this observation into graphical form?
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| « Last Edit: Jan 31st, 2005, 3:59pm 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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Aryabhatta
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Re: Identical limits
« Reply #5 on: Jan 31st, 2005, 3:59pm » |
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I think Icarus already knew the solution and was waiting...
Anyway it is easy to prove that the x^3 - x -1 and x^5 - x^4 - 1 each have only 1 real root.
From there it is easy to prove that the roots must be the same:
x^3 = x +1
implies (multiplying by x^2)
x^5 = x^3 + x^2 = x^2 + x + 1
and also
x^4 = x^2 + x (multiply by x)
Thus x^5 = x^4 + 1
Only thing left to prove is that both sequences converge.
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Icarus
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Re: Identical limits
« Reply #6 on: Jan 31st, 2005, 4:23pm » |
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on Jan 31st, 2005, 3:59pm, Aryabhatta wrote:| I think Icarus already knew the solution and was waiting... |
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I did. Polynomial division reveals that x5 - x4 - 1 = (x3 - x - 1)(x2 - x + 1).
The quadratic is easily seen to have no real roots. Showing that the cubic has only one real root is a little harder (unless you remember that signs theorem that I never do). Thus the quintic and cubic share a single real root. And if an and bn converge, they have to converge to it.
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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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Barukh
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Re: Identical limits
« Reply #7 on: Jan 31st, 2005, 11:18pm » |
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on Jan 31st, 2005, 2:21pm, JocK wrote:
Why would x3 - x - 1 = 0 and x5 - x4 - 1 = 0 share the same positive root?
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Why positive?
on Jan 31st, 2005, 4:23pm, Icarus wrote: Polynomial division reveals that x5 - x4 - 1 = (x3 - x - 1)(x2 - x + 1).
…Showing that the cubic has only one real root is a little harder (unless you remember that signs theorem that I never do). |
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The cubic x3 - x – 1 has the same extremum points as x3 – x. The latter has 3 real roots -1, 0, 1, so both cubics have local maximum in the interval I1 = (-1, 0), and local minimum in the interval I2 = (0, 1). Because x3 – x < 1 for x [in] I1, it follows x3 - x – 1 < 0 for x < 1. Therefore, it has only one real root.
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JocK
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Re: Identical limits
« Reply #8 on: Feb 1st, 2005, 1:34pm » |
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on Jan 31st, 2005, 11:18pm, Barukh wrote:
Well.... the geometrical construction I'm interested in demonstrates in one go that both polynomials share a root that has to be positive.
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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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