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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[m-1,n+1] - a[m-1,n-1]
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.
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

<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[m-1,n+1] - a[m-1,n-1]

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.

Robert Israel israel@math.MyUniversity'sInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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

<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[m-1,n+1] - a[m-1,n-1]

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[m-1,n+1] - a[m-1,n-1]
is a recursive expression. Perhaps you want a
non-recursive expression?
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[m-1,n+1] - a[m-1,n-1] 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[m-1,n+1] - a[m-1,n-1] is a recursive expression. Perhaps you want a non-recursive expression? The definition looks like a (sloppy) definition for a number-triangle;

the stirling-number-triangle comes into mind, although that
would be (n+1)a[m-1,n+1] - a[m-1,n] or the like...

Gottfried Helms
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

<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[m-1,n+1] - a[m-1,n-1] 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[m-1,*] (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
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
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

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[m-1,n+1]-a[m-1,n-1 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
matt271829-news@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

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)
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

matt271829-news@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= 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
matt271829-news@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: matt271829-news@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= 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).
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

<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

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[m-1,n+1] - a[m-1,n-1] 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_{m-1}' - x P_{m-1}. 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
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=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_{m-1}' - x P_{m-1}. 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!/((m-2k)! k! 2^k) (-x)^(m-2k) (-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
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=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_{m-1}' - x P_{m-1}. 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!/((m-2k)! k! 2^k) (-x)^(m-2k) (-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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