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Cheta with base-change: preliminary results
#14
(08/12/2009, 12:37 PM)Gottfried Wrote: Hi Jay,

there is a simple recursion formula, with which you can generate the inverse matrix.
I used the description at wikipedia for the stencils,
Code:
f(x-2h)= f(x) -2h f'(x)/1! + 4h^2 f''(x)/2! - 8h^3*f'''(x)/3! + 16h^4*f''''(x)/4!
f(x-h) = f(x) -1h f'(x)/1! + 1h^2 f''(x)/2! - 1h^3*f'''(x)/3! +  1h^4*f''''(x)/4!
f(x)   = f(x)  0             0                0                  0
f(x+h) = f(x) +1h f'(x)/1! + 1h^2 f''(x)/2! + 1h^3*f'''(x)/3! +  1h^4*f''''(x)/4!
f(x+2h)= f(x) +2h f'(x)/1! + 4h^2 f''(x)/2! + 8h^3*f'''(x)/3! + 16h^4*f''''(x)/4!


To prepare this for matrix-analysis I rewite this as matrix-equation:

Code:
´
|f(x-2h)|= |1 -2  4  -8 16 |  | f(x)/0!    |
|f(x-h) |= |1 -1  1  -1  1 |  | f'(x)/1!   |
|f(x)   |= |1  0  0   0  0 |* | f''(x)/2!  |
|f(x+h) |= |1  1  1   1  1 |  | f'''(x)/3! |
|f(x+2h)|= |1  2  4   8 16 |  | f''''(x)/4!|
Actually, since I'm trying to find the Taylor series, I don't worry about the factorials. So I work directly with the matrix above, where the first row is [1 -2 4 -8 16].

Inverting this and multiplying by 4!, or (n-1)! in general, I get back a matrix of integer coefficients. For this 5x5 matrix, I get:
Code:
[  0   0  24   0   0]
[  2 -16   0  16  -2]
[ -1  16 -30  16  -1]
[ -2   4   0  -4   2]
[  1  -4   6  -4   1]
The bottom row in general is a simple set of binomial coefficients. Though not easy to see as a pattern from here, the next row up is related to the bottom row, where the middle term is 0, and as we move j terms right or left, we get j times the number below. In general, every odd row can be found by applying this trick to the even row below. But finding the even rows, with the exception of the last row, isn't so simple.

But as I said, I haven't found a general formula. I'll take a look at Henryk's suggestions about the Stirling numbers, to see if I get any help there.

It may turn out that working with the factorials helps simplify the math, so I'll take a look at your matrices later today, when I'm more awake and can focus on them.
~ Jay Daniel Fox
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RE: Cheta with base-change: preliminary results - by jaydfox - 08/12/2009, 02:29 PM



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