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November 14th, 2011, 02:37 PM   #1
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Sequence Trajectory

[color=#000000]In the figure below we see the first 15 terms of the complex sequence ,

[attachment=1:23zjs9hc]n15.gif[/attachment:23zjs9hc]

where it seems that has an asymptotic behavior. Can you prove it?

Plotting for n=100 though, the plot becomes chaotic, something which can be seen in the next figure,

[attachment=0:23zjs9hc]n100.gif[/attachment:23zjs9hc]

can you figure out what goes wrong?



Of course the asymptotic behavior in the first question may vary from computer to computer meaning that the n's for which it holds may be greater than 15.[/color]
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File Type: gif n15.gif (2.6 KB, 423 views)
File Type: gif n100.gif (3.9 KB, 423 views)
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November 14th, 2011, 10:04 PM   #2
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Re: Sequence Trajectory

I don't understand what you're plotting, since it's on a two-dimensional graph (rather than a 3-dimensional graph for N -> C or R -> C or a 4-dimensional graph for C -> C).

Both the real part and the imaginary part of the function you defined oscillate rapidly, at least if you take the usual continuation. There are about e * n! peaks and troughs on the graphs from 1 to n.
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November 15th, 2011, 01:21 AM   #3
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Re: Sequence Trajectory

Quote:
Originally Posted by CRGreathouse
I don't understand what you're plotting, since it's on a two-dimensional graph (rather than a 3-dimensional graph for N -> C or R -> C or a 4-dimensional graph for C -> C).

Both the real part and the imaginary part of the function you defined oscillate rapidly, at least if you take the usual continuation. There are about e * n! peaks and troughs on the graphs from 1 to n.
[color=#000000]
when a complex sequence is plotted on a 2D plane, we wright it in the form . I would like to see another try. [/color]
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November 15th, 2011, 04:55 AM   #4
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Re: Sequence Trajectory

You still haven't described what you actually graphed, since your graph is two-dimensional and the space you describe (N -> C) is three-dimensional.

I described what's going on with the function (real and imaginary parts are oscillating) but I don't know how that corresponds to your graph.
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November 15th, 2011, 05:17 AM   #5
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Re: Sequence Trajectory

I think ZardoZ meant in his previous post



without i in the argument of cos and sin. This sequence can be plotted on a 2D graph.
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November 15th, 2011, 08:38 AM   #6
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Re: Sequence Trajectory

Plotted parametrically?

OK, in that case the problem is just round-off error. I needed 200 digits or so to get it right.

You can see what's going on if you look at the series expansion of exp at 1 -- the first n terms are ignored (as they make integral multiples of 2 pi), the next one is the main term, and the following terms are inconsequential.
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November 15th, 2011, 08:53 AM   #7
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Re: Sequence Trajectory







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November 15th, 2011, 10:16 AM   #8
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Re: Sequence Trajectory

Quote:
Originally Posted by wnvl
No.
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November 15th, 2011, 10:24 AM   #9
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Re: Sequence Trajectory

Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by wnvl
No.
[color=#000000]I am sure he/she meant .

Quote:
OK, in that case the problem is just round-off error. I needed 200 digits or so to get it right.
You have a very strong machine in order to see it after 200 terms, it's a finite memory problem.

[/color]
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November 15th, 2011, 01:58 PM   #10
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Re: Sequence Trajectory

Quote:
Originally Posted by ZardoZ
Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by wnvl
No.
[color=#000000]I am sure he/she meant .
But neither of those right-hand expressions have meaning as limits. One could say

though.

Quote:
Originally Posted by ZardoZ
You have a very strong machine in order to see it after 200 terms, it's a finite memory problem.
I was looking at the first hundred terms only. But it's not hard to look higher as long as you use a slightly more sophisticated method of calculation.
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