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José Carlos Santos
science forum Guru
Joined: 25 Mar 2005
Posts: 1111
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 Posted: Mon Jun 26, 2006 3:37 pm Post subject:
A computation
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Hi all,
I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
Best regards,
Jose Carlos Santos |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Mon Jun 26, 2006 4:13 pm Post subject:
Re: A computation
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On Mon, 26 Jun 2006 16:37:44 +0100, José Carlos Santos
<jcsantos@fc.up.pt> wrote:
| Quote: | Hi all,
I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
|
Doing a little work with pencil and paper it seems like the
first 400 or so digits should remain "surprising". Possibly
I dropped a constant somewhere.
| Quote: | Best regards,
Jose Carlos Santos
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************************
David C. Ullrich |
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José Carlos Santos
science forum Guru
Joined: 25 Mar 2005
Posts: 1111
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| Posted: Mon Jun 26, 2006 4:21 pm Post subject:
Re: A computation
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On 26-06-2006 17:13, David C. Ullrich wrote:
| Quote: | I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
Doing a little work with pencil and paper it seems like the
first 400 or so digits should remain "surprising".
|
Indeed!
Care to give an idea of your "little work with pencil and paper"?
Best regards,
Jose Carlos Santos |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Mon Jun 26, 2006 5:12 pm Post subject:
Re: A computation
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In article <4gadaeF1mh4lhU1@individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:
| Quote: | Hi all,
I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
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What's the surprise exactly? |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Mon Jun 26, 2006 5:18 pm Post subject:
Re: A computation
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The World Wide Wade nous a récemment amicalement signifié :
| Quote: | In article <4gadaeF1mh4lhU1@individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:
Hi all,
I suggest that you use some Computer ALgebra System (such as
Mathematica or Maple) in order no compute the first, say, 20 digits
of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
What's the surprise exactly?
|
It looks like pi
--
Patrick |
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M.J.T. Guy
science forum addict
Joined: 03 May 2005
Posts: 62
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| Posted: Mon Jun 26, 2006 5:30 pm Post subject:
Re: A computation
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In article <4gafs4F1m3dkcU1@individual.net>,
=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= <jcsantos@fc.up.pt> wrote:
| Quote: | On 26-06-2006 17:13, David C. Ullrich wrote:
I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
Doing a little work with pencil and paper it seems like the
first 400 or so digits should remain "surprising".
Care to give an idea of your "little work with pencil and paper"?
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I'd guess it involves the Poisson summation formula.
Mike Guy |
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Ignacio Larrosa Cañestro
science forum Guru Wannabe
Joined: 02 May 2005
Posts: 112
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| Posted: Mon Jun 26, 2006 5:52 pm Post subject:
Re: A computation
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En el mensaje:4gadaeF1mh4lhU1@individual.net,
José Carlos Santos <jcsantos@fc.up.pt> escribió:
| Quote: | Hi all,
I suggest that you use some Computer ALgebra System (such as
Mathematica or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
Best regards,
Jose Carlos Santos
|
Try with
(123^{-1}*sum_{-oo < n <+oo} exp(-(n/123)^2))^2
Any relation perhaps with Int(e^(-x^2), x, -inf, inf)?
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Mon Jun 26, 2006 6:18 pm Post subject:
Re: A computation
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In article <44a016f2$0$846$ba4acef3@news.orange.fr>,
"Patrick Coilland" <pcoilland@pcc.fr> wrote:
| Quote: | The World Wide Wade nous a récemment amicalement signifié :
In article <4gadaeF1mh4lhU1@individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:
Hi all,
I suggest that you use some Computer ALgebra System (such as
Mathematica or Maple) in order no compute the first, say, 20 digits
of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
What's the surprise exactly?
It looks like pi
|
Right, it's the square of an infinite Riemann sum for the
integral of exp(-x^2) over R, with interval width 1/10. That
integral equals sqrt(Pi); it's square is Pi. So this is like one
of those calculus exercises on p. 212 in the section on Riemann
integration. I didn't find it surprising. But perhaps the
surprise is how accurate it is, given delta = 1/10. |
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Abstract Dissonance
science forum Guru Wannabe
Joined: 29 Dec 2005
Posts: 201
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| Posted: Mon Jun 26, 2006 6:33 pm Post subject:
Re: A computation
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"José Carlos Santos" <jcsantos@fc.up.pt> wrote in message
news:4gadaeF1mh4lhU1@individual.net...
| Quote: | Hi all,
I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
Best regards,
Jose Carlos Santos
|
Ever heard of the erf function?
erf(x) = 2/sqrt(Pi)*int(exp(-t^2),t=0..x)
==>
erf(x) = 1/sqrt(Pi)*int(exp(-t^2),t=-x..x)
==>
Pi*erf(x)^2 = int(exp(-t^2),t=-x..x)
but
int(exp(-t^2),t=-x..x) = int(exp(-(t/L)^2),t=-L*x..L*x)/L
so
Pi*erf(x)^2 = int(exp(-(t/L)^2),t=-L*x..L*x)/L
taking the limit as x->oo gives
Pi*erf(oo)^2 = int(exp(-(t/L)^2),t=-oo..oo)/L
but erf(oo) = 1
so
Pi = int(exp(-(t/L)^2),t=-oo..oo)/L
(long and winded when one could just integrate directly erf(oo) after a
substitution then squaring)
but
int(exp(-(t/L)^2),t=-oo..oo)/L ~ sum(exp(-(k/L)^2*1/L,k=-oo..oo)
and L determines the rate of convergence in the sum. |
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Han de Bruijn
science forum Guru
Joined: 18 May 2005
Posts: 1285
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| Posted: Tue Jun 27, 2006 7:19 am Post subject:
Re: A computation
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The World Wide Wade wrote:
| Quote: | In article <44a016f2$0$846$ba4acef3@news.orange.fr>,
"Patrick Coilland" <pcoilland@pcc.fr> wrote:
The World Wide Wade nous a récemment amicalement signifié :
In article <4gadaeF1mh4lhU1@individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
It looks like pi
Right, it's the square of an infinite Riemann sum for the
integral of exp(-x^2) over R, with interval width 1/10. That
integral equals sqrt(Pi); it's square is Pi. So this is like one
of those calculus exercises on p. 212 in the section on Riemann
integration. I didn't find it surprising. But perhaps the
surprise is how accurate it is, given delta = 1/10.
|
Due to (a somewhat modified) Shannon's sampling theorem the accuracy is
already quite high for delta < sigma/2 . But in our case 1/10 << 1/2 !!
Some theory in:
http://hdebruijn.soo.dto.tudelft.nl/jaar2005/winter.pdf
Han de Bruijn |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Tue Jun 27, 2006 12:09 pm Post subject:
Re: A computation
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On Mon, 26 Jun 2006 11:18:39 -0700, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
| Quote: | In article <44a016f2$0$846$ba4acef3@news.orange.fr>,
"Patrick Coilland" <pcoilland@pcc.fr> wrote:
The World Wide Wade nous a récemment amicalement signifié :
In article <4gadaeF1mh4lhU1@individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:
Hi all,
I suggest that you use some Computer ALgebra System (such as
Mathematica or Maple) in order no compute the first, say, 20 digits
of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
My guess is that the result will be a surprise to many of you.
What's the surprise exactly?
It looks like pi
Right, it's the square of an infinite Riemann sum for the
integral of exp(-x^2) over R, with interval width 1/10. That
integral equals sqrt(Pi); it's square is Pi. So this is like one
of those calculus exercises on p. 212 in the section on Riemann
integration. I didn't find it surprising. But perhaps the
surprise is how accurate it is, given delta = 1/10.
|
Yes, that's the surprise. If you just plug in some obvious
estimates to compare the integral to the Riemann sum you'd
get a fairly large error, exactly how large depending on
what comparison technique you used, but it's certainly
not going to be accurate to more than a few decimals.
While in fact, if I did the calculations correctly, the
sum is equal to pi up to about 400 or 500 decimals!
You can verify this using Poisson summation (which
here amounts to a standard identity for the theta
function, for example in Dym&McKean "Fourier Series
and Integrals".)
************************
David C. Ullrich |
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Han de Bruijn
science forum Guru
Joined: 18 May 2005
Posts: 1285
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| Posted: Tue Jun 27, 2006 1:27 pm Post subject:
Re: A computation
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David C. Ullrich wrote:
| Quote: | Yes, that's the surprise. If you just plug in some obvious
estimates to compare the integral to the Riemann sum you'd
get a fairly large error, exactly how large depending on
what comparison technique you used, but it's certainly
not going to be accurate to more than a few decimals.
While in fact, if I did the calculations correctly, the
sum is equal to pi up to about 400 or 500 decimals!
You can verify this using Poisson summation (which
here amounts to a standard identity for the theta
function, for example in Dym&McKean "Fourier Series
and Integrals".)
|
Uhm, I've just posted a response in this thread AND a theory why this
must be so. According to my theory, the accuracy is at least:
exp(-(pi.sigma/delta)^2) = exp(-(pi.10)^2) = 2.33629136535946E-0429
Which is according to your findings, but _not_ surprising (anymore):
http://hdebruijn.soo.dto.tudelft.nl/jaar2005/winter.pdf
Han de Bruijn |
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Han de Bruijn
science forum Guru
Joined: 18 May 2005
Posts: 1285
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| Posted: Wed Jun 28, 2006 7:24 am Post subject:
Re: A computation
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Han de Bruijn wrote:
| Quote: | Uhm, I've just posted a response in this thread AND a theory why this
must be so. According to my theory, the accuracy is at least:
exp(-(pi.sigma/delta)^2) = exp(-(pi.10)^2) = 2.33629136535946E-0429
Which is according to your findings, but _not_ surprising (anymore):
http://hdebruijn.soo.dto.tudelft.nl/jaar2005/winter.pdf
|
A safer estimate: # valid decimals >= (pi.sigma/delta)^2/ln(10)
In our case: # valid decimals of Pi >= 428 .
Let's change the problem a little bit:
20^{-1}*sum_{-oo < n < +oo} exp(-(n/20)^2))^2
Then: # valid decimals of Pi >= (pi.20)^2/ln(10) = 1714 .
The outcome improves quadratically. It is already excellent for
2^{-1}*sum_{-oo < n < +oo} exp(-(n/2)^2))^2
Thus sigma/delta = 1/2 , giving: (pi.2)^2/ln(10) = 17 decimals.
Combs of Gaussian distributions are employed routinely in my software.
Now you all have a clue why that might be a migty good idea.
David Ullrich wrote:
| Quote: | You can verify this using Poisson summation (which
here amounts to a standard identity for the theta
function, for example in Dym&McKean "Fourier Series
and Integrals".)
|
Of course. It's always possible to make it far more difficult than with
the above simple formula, as provided by HdB-Shannon.
Han de Bruijn |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Wed Jun 28, 2006 1:20 pm Post subject:
Re: A computation
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On Wed, 28 Jun 2006 09:24:57 +0200, Han de Bruijn
<Han.deBruijn@DTO.TUDelft.NL> wrote:
| Quote: | Han de Bruijn wrote:
Uhm, I've just posted a response in this thread AND a theory why this
must be so. According to my theory, the accuracy is at least:
exp(-(pi.sigma/delta)^2) = exp(-(pi.10)^2) = 2.33629136535946E-0429
Which is according to your findings, but _not_ surprising (anymore):
http://hdebruijn.soo.dto.tudelft.nl/jaar2005/winter.pdf
A safer estimate: # valid decimals >= (pi.sigma/delta)^2/ln(10)
In our case: # valid decimals of Pi >= 428 .
Let's change the problem a little bit:
20^{-1}*sum_{-oo < n < +oo} exp(-(n/20)^2))^2
Then: # valid decimals of Pi >= (pi.20)^2/ln(10) = 1714 .
The outcome improves quadratically. It is already excellent for
2^{-1}*sum_{-oo < n < +oo} exp(-(n/2)^2))^2
Thus sigma/delta = 1/2 , giving: (pi.2)^2/ln(10) = 17 decimals.
Combs of Gaussian distributions are employed routinely in my software.
Now you all have a clue why that might be a migty good idea.
David Ullrich wrote:
You can verify this using Poisson summation (which
here amounts to a standard identity for the theta
function, for example in Dym&McKean "Fourier Series
and Integrals".)
Of course. It's always possible to make it far more difficult than with
the above simple formula, as provided by HdB-Shannon.
|
Um, the fact that you're evidently not familiar with Poisson
summation does not make it more difficult. In fact there's
Poisson summation going on in that pdf, just not referred to
by that name.
When you post an "estimate" and then post a "safer" estimate
it doesn't inspire confidence. The argument from the
Poisson summation formula is very simple: It shows that
the sum in question is exactly equal to another sum;
in that other sum the first term is much larger than
all the rest...
************************
David C. Ullrich |
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victor_meldrew_666@yahoo.
science forum beginner
Joined: 19 May 2006
Posts: 17
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| Posted: Wed Jun 28, 2006 1:42 pm Post subject:
Re: A computation
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José Carlos Santos wrote:
| Quote: | I suggest that you use some Computer ALgebra System (such as Mathematica
or Maple) in order no compute the first, say, 20 digits of
(10^{-1}*sum_{-oo < n <+oo} exp(-(n/10)^2))^2
|
This relates to a well-known identity involving the theta function
defined by
theta(x) = sum_{n=-infinity}^infinity exp( -pi n^2 x).
This identity states that
theta(x) = x^(-1/2) theta(1/x)
and can be proved by using the Poisson summation formula.
Your number is
[(1/10) theta (1/(100 pi))]^2.
Hence this equals
pi theta (100 pi)^2.
This is
pi[1 + 4 exp(-100 pi^2) + 4 exp(-200 pi^2) + 4 exp(-400 pi^2) + ... ].
(The coefficient of exp(-100 m pi^2) is the number of representations
of m as a sum of two squares of integers).
The term t = exp(-100 pi^2) is tiny: 2.336 x 10^{-429} to four
signficant figures. The higher powers of t are smaller still. Thus
your number exceeds pi by less than 3 x 10^{-428}
(yet exceed pi it does).
The theta function can be extended to complex arguments and
the functional equation was used by Riemann to prove the functional
equation for the zeta function.
For extra credit: prove that
theta(x)^2 = sum_{n=-infinity}^infinity sech(pi n x).
Victor Meldrew |
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