My Math Forum Sequence Trajectory

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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 $\color{red}\bf a_n=ne^{2\pi i e n!}$,

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

where it seems that $a_n$ 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?

$\color{red}\rule{350pt}{1.5pt}$

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]
Attached Images
 n15.gif (2.6 KB, 423 views) n100.gif (3.9 KB, 423 views)

 November 14th, 2011, 10:04 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
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]$a_n=ne^{2\pi i e n!}=n\left(\cos\left(2\pi i e n!\right)+i\sin\left(2\pi i e n!\right)\right)$
when a complex sequence is plotted on a 2D plane, we wright it in the form $a_n=\left<n\cos\left(2\pi i e n!\right),n\sin\left(2\pi i e n!\right)\right=>=$. I would like to see another try. [/color]

 November 15th, 2011, 04:55 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
 November 15th, 2011, 05:17 AM #5 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Sequence Trajectory I think ZardoZ meant in his previous post $a_n=\left=$ without i in the argument of cos and sin. This sequence can be plotted on a 2D graph.
 November 15th, 2011, 08:38 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
 November 15th, 2011, 08:53 AM #7 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Sequence Trajectory $e\cdot n!=\left \{ 1+1+\frac{1}{2!}+...+\frac{1}{n!}+\frac{1}{n+1!}+\ frac{1}{n+2!} +... \right \}n!$ $=integer+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)} +\frac{1}{(n+1)(n+2)(n+3)} +...$ $\lim_{n \to \infty }e\cdot n!=integer$ $\lim_{n \to \infty }a_n=\lim_{n \to \infty }\left=$ $=\lim_{n \to \infty }\left$ $=\left$
November 15th, 2011, 10:16 AM   #8
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Re: Sequence Trajectory

Quote:
 Originally Posted by wnvl $\lim_{n \to \infty }e\cdot n!=integer$
No.

November 15th, 2011, 10:24 AM   #9
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Re: Sequence Trajectory

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by wnvl $\lim_{n \to \infty }e\cdot n!=integer$
No.
[color=#000000]I am sure he/she meant $\lim_{n\to \infty}e\cdot n!=integer+\lim_{n\to\infty}\sum_{k=n+1}^{\infty}\ frac{1}{k!}$ .

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]

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 $\lim_{n \to \infty }e\cdot n!=integer$
No.
[color=#000000]I am sure he/she meant $\lim_{n\to \infty}e\cdot n!=integer+\lim_{n\to\infty}\sum_{k=n+1}^{\infty}\ frac{1}{k!}$ .
But neither of those right-hand expressions have meaning as limits. One could say
$e\cdot n!\bmod{2\pi}=\frac1n+O\left(\frac{1}{n^2}\right)$
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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