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MathGroup Archive 2004

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How to numerically estimate an asymptotic equivalent?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46340] How to numerically estimate an asymptotic equivalent?
  • From: Franck Michel <Franck.Michel at wanadoo.fr>
  • Date: Sat, 14 Feb 2004 22:19:50 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

I have computed a sequence a(n) up to a level N.
I know this sequence as an asymptotic equivalent when n -> +infinity:
     a(n) ~ (n!)^s*A^n*n^beta*M

but I ignore the values of s, A, beta and M, and I would like to perform a
numerical estimation of these 4 values.

Taking ln(a(n)), one gets a linear expression:
s*ln(n!) + ln(A)*n + beta*ln(n) + ln(M)

Therefore, I've used a linear least squares regression on a subsequence of
a(n) for identifying the four parameters s, ln(A), beta and ln(M) with
confidence intervals. I've used Mathematica function Regress.
All seems to work fine but I have a doubt on the confidence intervals.

To check the validity if this method, I used another sequence a(n) which has
a an analogous asymptotic equivalent, and for which the 4 parameters have
been theoretically identified.
The estimation of the parameters s and A is good, but the confidence
intervals are false (below the results of the estimation).

How can I manage to have the good confidence intervals? Or, simply, to have
a proof that some parameters are well estimated?
Are there other possible methods to numerically estimate an asymptotic
equivalent like this one?

Thanks in advance for any suggestion.

Franck Michel

The test in Mathematica:


v=ReadList["vvdp",Record,RecordSeparators->{","} ];

The file vvdp is too large to be included in this post. However, if you are
interested, you can download it at:

http://medicis.polytechnique.fr/~fmichel/vvdp

lv=N[Log[Abs[ToExpression[v]]],50];
<<Statistics`LinearRegression`
do=N[Table[{i,lv[[i]]},{i,400,420}],50];
res=N[Regress[do, {Log[n!],n,Log[n]}, n],50];
Es=Table[res[[1]][[2]][[1]][[i]][[1]],{i,1,4}]

{-1.8710263892729189787784057861852261123361,

  1.000331954678443963842324445264984470289248,

 -0.289755659135106213854544557166413480909547,

 -0.4098253769138103469031112458138752751758}
SE=Table[res[[1]][[2]][[1]][[i]][[2]],{i,1,4}]
{0.000546641311820075664738437775657,
  1.84246075262127962336023535510*10^(-6),
  0.00001049893299985597370639981310624,
  0.000201151099349915791816265931273}

The exact value of s is s=1

The estimation here in Es is very good: 1.000331954... is very close to 1.

But the standard error in SE is false: 1.84246075...*10^(-6) is too small.

The exact value of A is A=3/4
The estimation of ln(A) in Es is good: -0.2897. but the standard error is
false.

The exact value of beta is beta=-1/2
The estimation here -0.289 is not good; the standard error is false.

The exact value of M comes from a small formula, and is approximately 0.23
The estimation of ln(M) is not good and the standard error is false.








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