| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I'm looking for a general expression for the a- and b-coefficients in:
n k k+1 j k n k-1 j
SUM(i ) = SUM(a n ) and: n = SUM( SUM( b i ) )
i=1 j=1 j i=1 j=0 j
Examples:
n 2 n
k=1 SUM(i ) = 1/2 * ( n + n ) n = SUM( 1 )
i=1 i=1
n 2 3 2 2 n
k=2 SUM(i ) = 1/6 * ( 2n + 3n + n ) n = SUM( 2i - 1 )
i=1 i=1
n 3 4 3 2 3 n 2
k=3 SUM(i ) = 1/4 * ( n + 2n + n ) n = SUM( 3i - 3i + 1 )
i=1 i=1
n 4 5 4 3 4 3 2
k=4 SUM(i ) = 1/30 * ( 6n + 15n + 10n ) n = SUM( b i + b i + b i + b )
i=1 3 2 1 0
...
I wrote a small program to work out the first column (for k=1...8) and I
calculated the the b's for k=1...3 in the second column, but what I really
need is a general formula or some sort of recursive scheme...
Ideas, solutions, pointers anybody?
Thanks in advance...
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1678.1 | hope this is useful | HERON::BUCHANAN | The was not found. | Wed Oct 21 1992 14:23 | 36 |
>I'm looking for a general expression for the a- and b-coefficients in: > > n k k+1 j k n k-1 j > SUM(i ) = SUM(a n ) and: n = SUM( SUM( b i ) ) > i=1 j=1 j i=1 j=0 j > b = -(-1)^(k-j) C(k,j) j a = 1/(k+1) k+1 (l<k) a = (1/l+1) sum(j=l+2,...,k+1) (-1)^(j-l) a j!/(j-l)! l+1 j Of the top of my head, I don't see a non-recursive expression for a . j Note: a = 1/2 k Note: a = 0 for t > 0. k-2t > n 4 5 4 3 >k=4 SUM(i ) = 1/30 * ( 6n + 15n + 10n ) > i=1 Small typo: n 4 5 4 3 k=4 SUM(i ) = 1/30 * ( 6n + 15n + 10n -n) i=1 Andrew. | |||||
| 1678.2 | binomial of course -- thanks! | BRSTR2::SYSMAN | Dirk Van de moortel | Thu Oct 22 1992 06:51 | 41 |
re .-1
this is very helpfull: thanks to the b's you specified, it's all quite
simple. I hadn't noticed that the b's are binomial coefficients...
k n k k
n = SUM[i - (i-1) ] why didn't I see that???
i=1
now, starting from this it's very easy to find the a's iteratively:
k+1 k+1 k+1
n = SUM(i - (i-1) )
k+1 k+1
= SUM(i ) - SUM(i-1)
k+1 k+1 k
= SUM(i ) - SUM(i + (k+1)i - ... )
k+1 k+1 k k-1
= SUM(i ) - SUM(i ) + (k+1)SUM(i ) - C(k+1,2)SUM(i ) - ...
\___________________/
cancel out
k k-1
= (k+1)SUM(i ) - C(k+1,2)SUM(i ) + ...
so
k 1 k+1 k-1
SUM(i ) = ----- ( n + C(k+1,2)SUM(i ) - ... )
k+1
k
so we clearly have SUM(i ) expressed in terms of sums of lower powers
from which your a's can easily be derived...
Thanks for clearing it up...
Dirk
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| 1678.3 | more | HERON::BUCHANAN | The was not found. | Thu Oct 22 1992 10:23 | 28 |
Let me make a small correction to the recursive formula for a:
(l<k) a = (1/l+1) sum(j=l+2,...,k+1) (-1)^(j-l) a C(j,j-l)
l+1 j
Each term a is actually a function of k as well as j. We can write
j
(k-j-1)
a = (-1) * (k!/j!) * f
j k-j
where the subscript of f will run from -1 to k.
Then the equation above becomes:
(t>=0) f = - sum(j=-1,...,t-1) f /(t+1-j)!
t j
Values of f for small t starting with -1 are:
1, -1/2, 1/12, 0, -1/6!, 0, 1/(6.7!), 0, -3/10!, 0, 10/12!, 0, ...
Anyone spot the pattern? I haven't nailed it down yet.
If you are exploring this problem, you may find the MAPLE command:
rsolve({y(n)=y(n-1)+n^12,y(0)=0},y);
saves you time, for various values of k. (Here k=12.)
Cheers,
Andrew.
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| 1678.4 | 3D::ROTH | Geometry is the real life! | Thu Oct 22 1992 13:30 | 11 | |
I believe these numbers are coefficients of Bernoulli polynomials
and elegant algorithms for getting them are in books on combinatorics
(like Knuth, the art of computer programming, etc.) so you may want
to look there for more info.
I remember playing with those sums when I was a kid and I made a
variety of ways to calculate them, filling many pieces of paper with
my blundering attempts. It's the kind of thing I thought "math" was
at the time.
- Jim
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