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| Science Forum Index » Mathematics Forum » Polynomial Expression of the Diagonals of Pascal's Triangle |
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| Grick |
 Posted: Sat Feb 12, 2005 1:29 am |
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Guest
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Define a general polynomial function Y(n,x) that gives the xth element
of the nth diagonal of Pascal's triangle.
Y(n,x) should be a polynomial of degree n, as I see it. The 1st
diagonal (1,1,1,1...) is obviously of degree 0. Now, the differences of
the 2nd diagonal (1,2,3...) are found in the 1st diagonal, so the 2nd
diagonal must be of degree 1. The differences of the 3rd diagonal are
found in the 2nd diagonal, so the 3rd diagonal must be of degree 3.
I've compiled the coefficients for n=0 to 6. The leading coefficient is
1/(n!),
Details and results found so far are at:
http://img.photobucket.com/albums/v30/Gricksigger/pascal.gif
(picture will be updated) |
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| Robin Chapman |
| Posted: Sat Feb 12, 2005 3:01 am |
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Guest
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Grick wrote:
[quote:c4cc9d1a65]Define a general polynomial function Y(n,x) that gives the xth element
of the nth diagonal of Pascal's triangle.
Y(n,x) should be a polynomial of degree n, as I see it. The 1st
diagonal (1,1,1,1...) is obviously of degree 0. Now, the differences of
the 2nd diagonal (1,2,3...) are found in the 1st diagonal, so the 2nd
diagonal must be of degree 1. The differences of the 3rd diagonal are
found in the 2nd diagonal, so the 3rd diagonal must be of degree 3.
I've compiled the coefficients for n=0 to 6. The leading coefficient is
1/(n!),
Details and results found so far are at:
http://img.photobucket.com/albums/v30/Gricksigger/pascal.gif
(picture will be updated)
[/quote:c4cc9d1a65]
The coefficients of these polynomials are Stirling numbers of the first kind
divided by n!. You may find a web search for "Stirling numbers" informative.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ |
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| A N Niel |
| Posted: Sat Feb 12, 2005 6:49 am |
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[quote:fabda0fe8c]Grick wrote:
Define a general polynomial function Y(n,x) that gives the xth element
of the nth diagonal of Pascal's triangle.
Y(n,x) should be a polynomial of degree n, as I see it. The 1st
diagonal (1,1,1,1...) is obviously of degree 0. Now, the differences of
the 2nd diagonal (1,2,3...) are found in the 1st diagonal, so the 2nd
diagonal must be of degree 1. The differences of the 3rd diagonal are
found in the 2nd diagonal, so the 3rd diagonal must be of degree 3.
I've compiled the coefficients for n=0 to 6. The leading coefficient is
1/(n!),
Details and results found so far are at:
http://img.photobucket.com/albums/v30/Gricksigger/pascal.gif
(picture will be updated)
[/quote:fabda0fe8c]
Of course the nth diagonal is polynomial
x (x-1) (x-2) ... (x-n+1) / n!
which has n factors in the numerator and n in the denominator.
In article <cukd33$pu6$2@newsg3.svr.pol.co.uk>, Robin Chapman
<rjc@ivorynospamtower.freeserve.co.uk> wrote:
[quote:fabda0fe8c]
The coefficients of these polynomials are Stirling numbers of the first kind
divided by n!. You may find a web search for "Stirling numbers" informative.[/quote:fabda0fe8c] |
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