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Carine science forum beginner
Joined: 05 Apr 2005
Posts: 11

Posted: Mon Jun 19, 2006 5:19 pm Post subject:
Computing a series



Dear all,
I am figthing with a series which doesn't look very complicated but I
can't sort it out.
Here it is:
a[0,0] = 1;
a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]
It turns out that if m+n is even, a[m,n] = 0.
I would like to know its recursive expression.
Any help?
Cheers,
Carine. 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Mon Jun 19, 2006 6:08 pm Post subject:
Re: Computing a series



In article <1150737581.157395.140020@p79g2000cwp.googlegroups.com>,
<csimon@utm.csic.es> wrote:
Quote:  I am figthing with a series which doesn't look very complicated but I
can't sort it out.
Here it is:
a[0,0] = 1;
a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]

It's not a series, or even a sequence, because it has two indices.
And you haven't given us enough information to specify it completely.
Quote:  It turns out that if m+n is even, a[m,n] = 0.

That contradicts your statement a[0,0] = 1.
Robert Israel israel@math.MyUniversity'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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A N Niel science forum Guru
Joined: 28 Apr 2005
Posts: 475

Posted: Mon Jun 19, 2006 6:22 pm Post subject:
Re: Computing a series



In article <1150737581.157395.140020@p79g2000cwp.googlegroups.com>,
<csimon@utm.csic.es> wrote:
Quote:  Dear all,
I am figthing with a series which doesn't look very complicated but I
can't sort it out.
Here it is:
a[0,0] = 1;
a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]

You will need more initial values that a[0,0]...
For example, a[0,1] is not determined from the information you provided.
Quote:  It turns out that if m+n is even, a[m,n] = 0.
I would like to know its recursive expression.

Actually, a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]
is a recursive expression. Perhaps you want a
nonrecursive expression? 

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Gottfried Helms science forum Guru
Joined: 24 Mar 2005
Posts: 301

Posted: Mon Jun 19, 2006 9:15 pm Post subject:
Re: Computing a series



Am 19.06.2006 20:22 schrieb A N Niel:
Quote:  In article <1150737581.157395.140020@p79g2000cwp.googlegroups.com>,
csimon@utm.csic.es> wrote:
Dear all,
I am figthing with a series which doesn't look very complicated but I
can't sort it out.
Here it is:
a[0,0] = 1;
a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]
You will need more initial values that a[0,0]...
For example, a[0,1] is not determined from the information you provided.
It turns out that if m+n is even, a[m,n] = 0.
I would like to know its recursive expression.
Actually, a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]
is a recursive expression. Perhaps you want a
nonrecursive expression?
The definition looks like a (sloppy) definition for a numbertriangle; 
the stirlingnumbertriangle comes into mind, although that
would be (n+1)a[m1,n+1]  a[m1,n] or the like...
Gottfried Helms 

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Ronald Bruck science forum Guru
Joined: 05 Jun 2005
Posts: 356

Posted: Tue Jun 20, 2006 1:08 am Post subject:
Re: Computing a series



In article <1150737581.157395.140020@p79g2000cwp.googlegroups.com>,
<csimon@utm.csic.es> wrote:
Quote:  Dear all,
I am figthing with a series which doesn't look very complicated but I
can't sort it out.
Here it is:
a[0,0] = 1;
a[m,n] = (n+1)a[m1,n+1]  a[m1,n1]
It turns out that if m+n is even, a[m,n] = 0.
I would like to know its recursive expression.
Any help?

Not enough information, as others have pointed out.
Since a[m,*] is given in terms of a[m1,*] (where the *'s can
take different values), you'll need to know a[0,n] not just a[0,0].
The standard trick for finding explicit expressions for a[m,n] is to
consider its formal generating series
f(x,y) = \sum_{m,n} a[m,n] x^m y^n
and to use the recursion information to solve for f[x,y]. Usually you
can identify a[m,n] by some series manipulations. But you don't have
enough information.

Ron Bruck 

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Carine science forum beginner
Joined: 05 Apr 2005
Posts: 11

Posted: Tue Jun 20, 2006 10:25 am Post subject:
Re: Computing a series



Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
This formula is indeed similar to Stirling's but it is not exactly the
same and I can't sort it out.
Cheers,
Carine. 

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Patrick Coilland science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197

Posted: Tue Jun 20, 2006 10:56 am Post subject:
Re: Computing a series



csimon@utm.csic.es nous a récemment amicalement signifié :
Quote:  Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD

These conditions contain a contradiction :
rule 1 : a[0,0] = 1
rule 2 : a[m,n] = 0 if m<n
rule 3 : a[m,n] = 0 if m<0
rule 4 : a[m,n] = 0 if n<0
rule 5 : a[m,n] = (n+1)a[m1,n+1]a[m1,n1 for every (m,n)
a[1,+1] = 0 (rules 2 and 3)
a[1,1] = 0 (rules 3 and 4)
a[0 , 0] = 1 (rule 1)
a[0 , 0] = a[1,+1]  a[1,1] (rule 5)

Patrick 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jun 20, 2006 6:15 pm Post subject:
Re: Computing a series



csimon@utm.csic.es wrote:
Quote:  Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
This formula is indeed similar to Stirling's but it is not exactly the
same and I can't sort it out.
Cheers,
Carine.

For the cases of interest, i.e. m >= 0, n >= 0, m >= n, m + n even, I
make it
a[m,n] = (1)^((m + n)/2) * (m  n  1)!! * C(m, n) 

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Patrick Coilland science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197

Posted: Tue Jun 20, 2006 6:50 pm Post subject:
Re: Computing a series



matt271829news@yahoo.co.uk nous a récemment amicalement signifié :
Quote:  csimon@utm.csic.es wrote:
Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
This formula is indeed similar to Stirling's but it is not exactly
the same and I can't sort it out.
Cheers,
Carine.
For the cases of interest, i.e. m >= 0, n >= 0, m >= n, m + n even, I
make it
a[m,n] = (1)^((m + n)/2) * (m  n  1)!! * C(m, n)

And if I want to compute a[0,0] ?
a[0,0] = (1)^(0) * ( 1)!! * C(0, 0)
= ( 1)!!
what is (1)! ?

Patrick 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jun 20, 2006 7:55 pm Post subject:
Re: Computing a series



Patrick Coilland wrote:
Quote:  matt271829news@yahoo.co.uk nous a récemment amicalement signifié :
csimon@utm.csic.es wrote:
Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
This formula is indeed similar to Stirling's but it is not exactly
the same and I can't sort it out.
Cheers,
Carine.
For the cases of interest, i.e. m >= 0, n >= 0, m >= n, m + n even, I
make it
a[m,n] = (1)^((m + n)/2) * (m  n  1)!! * C(m, n)
And if I want to compute a[0,0] ?
a[0,0] = (1)^(0) * ( 1)!! * C(0, 0)
= ( 1)!!
what is (1)! ?

By convention (1)!! is equal to one (see e.g.
http://mathworld.wolfram.com/DoubleFactorial.html). 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Tue Jun 20, 2006 8:12 pm Post subject:
Re: Computing a series



In article <1150799104.115164.261750@b68g2000cwa.googlegroups.com>,
<csimon@utm.csic.es> wrote:
Quote: 
Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD

As Patrick Coilland pointed out, there's still a problem here, so I'll
assume you meant
a[0,0] = 1
a[m,n] = 0 if m < 0 or n < 0
a[m,n] = (n+1) a[m1,n+1]  a[m1,n1] if m >=0, n>=0 and [m,n] <> [0,0]
The fact that a[m,n] = 0 for m < n does not need to be assumed, it can be
proved by induction on m. Similarly, that a[m,n] = 0 when m+n is odd can
be proved by induction on m. Also that for 0 <= m <= n,
a[m,n] < 0 when m+n == 2 mod 4 and a[m,n] > 0 when m+n == 0 mod 4.
If P_m = sum_{n=0}^m a[m,n] x^n, you have P_0 = 1 and
P_m = P_{m1}'  x P_{m1}. It looks to me like
P_m = (1)^m H_m(x) where H_m is the m'th Hermite polynomial
in the probabilists' convention. Thus
P_m = exp(x^2/2) (d/dx)^n exp(x^2/2)
Robert Israel israel@math.MyUniversity'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Wed Jun 21, 2006 6:17 am Post subject:
Re: Computing a series



Robert Israel wrote:
Quote:  In article <1150799104.115164.261750@b68g2000cwa.googlegroups.com>,
csimon@utm.csic.es> wrote:
Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
As Patrick Coilland pointed out, there's still a problem here, so I'll
assume you meant
a[0,0] = 1
a[m,n] = 0 if m < 0 or n < 0
a[m,n] = (n+1) a[m1,n+1]  a[m1,n1] if m >=0, n>=0 and [m,n] <> [0,0]
The fact that a[m,n] = 0 for m < n does not need to be assumed, it can be
proved by induction on m. Similarly, that a[m,n] = 0 when m+n is odd can
be proved by induction on m. Also that for 0 <= m <= n,
a[m,n] < 0 when m+n == 2 mod 4 and a[m,n] > 0 when m+n == 0 mod 4.
If P_m = sum_{n=0}^m a[m,n] x^n, you have P_0 = 1 and
P_m = P_{m1}'  x P_{m1}. It looks to me like
P_m = (1)^m H_m(x) where H_m is the m'th Hermite polynomial
in the probabilists' convention. Thus
P_m = exp(x^2/2) (d/dx)^n exp(x^2/2)
.... of course, I meant 
P_m = exp(x^2/2) (d/dx)^m exp(x^2/2)
Moreover, the coefficients a[m,n] can be computed from this using Faa
di Bruno's
formula for derivatives of f(g(x)) (in this case take f = exp and g(x)
= x^2/2).
Note that g'(x) = x, g''(x) = 1 and g'''(x) = 0. Thus Faa di Bruno's
formula says
P_m = sum_{k=0}^{floor(m/2)} m!/((m2k)! k! 2^k) (x)^(m2k) (1)^k
i.e. if n = m  2 k, a[m,n] = (1)^(m+k) m!/(n! k! 2^k)
Robert Israel israel@math.MyUniversity'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Carine science forum beginner
Joined: 05 Apr 2005
Posts: 11

Posted: Fri Jun 23, 2006 8:20 am Post subject:
Re: Computing a series



Thanks Robert! That was exactly what we needed. We were intuiting it
would converge to sin type function but couldn't formalize it.
Cheers,
Carine.
Robert Israel a écrit :
Quote:  Robert Israel wrote:
In article <1150799104.115164.261750@b68g2000cwa.googlegroups.com>,
csimon@utm.csic.es> wrote:
Thanks for your answers.
Sorry, I made typo mistakes and I forgot to precise something
important: a[m,n] = 0 if m<n and if m or n<0.
So if
a[0,0] = 1
a[m,n] = (n+1)a[m1,n+1]a[m1,n1]
Therefore
a[m,n] = 0 when m+n is ODD
As Patrick Coilland pointed out, there's still a problem here, so I'll
assume you meant
a[0,0] = 1
a[m,n] = 0 if m < 0 or n < 0
a[m,n] = (n+1) a[m1,n+1]  a[m1,n1] if m >=0, n>=0 and [m,n] <> [0,0]
The fact that a[m,n] = 0 for m < n does not need to be assumed, it can be
proved by induction on m. Similarly, that a[m,n] = 0 when m+n is odd can
be proved by induction on m. Also that for 0 <= m <= n,
a[m,n] < 0 when m+n == 2 mod 4 and a[m,n] > 0 when m+n == 0 mod 4.
If P_m = sum_{n=0}^m a[m,n] x^n, you have P_0 = 1 and
P_m = P_{m1}'  x P_{m1}. It looks to me like
P_m = (1)^m H_m(x) where H_m is the m'th Hermite polynomial
in the probabilists' convention. Thus
P_m = exp(x^2/2) (d/dx)^n exp(x^2/2)
... of course, I meant
P_m = exp(x^2/2) (d/dx)^m exp(x^2/2)
Moreover, the coefficients a[m,n] can be computed from this using Faa
di Bruno's
formula for derivatives of f(g(x)) (in this case take f = exp and g(x)
= x^2/2).
Note that g'(x) = x, g''(x) = 1 and g'''(x) = 0. Thus Faa di Bruno's
formula says
P_m = sum_{k=0}^{floor(m/2)} m!/((m2k)! k! 2^k) (x)^(m2k) (1)^k
i.e. if n = m  2 k, a[m,n] = (1)^(m+k) m!/(n! k! 2^k)
Robert Israel israel@math.MyUniversity'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 


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