Re: Recommendation for numerical differentiation formulas?
- From: spellucci@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx (Peter Spellucci)
- Date: Mon, 17 Jul 2006 13:37:54 +0000 (UTC)
In article <1153137085.886668.152400@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
1940LaSalle@xxxxxxxxx writes:
yes
Peter Spellucci wrote:
In article <1152885890.796128.65800@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
1940LaSalle@xxxxxxxxx writes:
>
>Peter Spellucci wrote:
no, flip in a geometric sense right to left. means
f'(x(n+1))=(1/(12*h))
*(-25*f(x(n+1)+48*f(x(n))-36*f(x(n-1))+16*f(x(n-2))-3*f(x(n-3)))
you need these special one sided formulae only for the first data point
(I named it x(0)) and the last one x(n+1) because these are not covered by
the 3 point compact scheme. all other derivatives come out from the
tridiagonal solve, and you are right: the right hand side is made up from the
table and looking at the indices you see: f(x(0)) up to f(x(n+1)) are needed
there and give you u(1) up to u(n) which correspond to f'(x(1)) up to f'(x(n))
good luck
peter
In thinking about this scheme, it seems I have a few more questions
about the mechanics of it. For the sake of discussion, let's say the
data involved have a total of j points. Then the right side of the
equation would appear to be a matrix (that is, a vector) with one
column and j-1 rows (there are no rows corresponding to x(0) and x(j)
). The left side is the tridiagonal matrix we've mentioned before. I
may be missing something fundamental here, so please bear with me: when
one inverts the matrix on the left and does the appropriate matrix
multiplication, one gets a solution vector (again, one column with j-1
rows)--at least, I believe that's the case. In that solution vector,
the first row would yield the value of u(x(1)), the second, u(x(2)),
etc., up to u(x(j-1)), correct? (What I know of matrix
algebra/manipulation has been acquired purely as needed; I've neverthis is a linear system of equations with a tridiagoanl matrix
taken a formal course.)
and you never invert the matrix, you solve the system by standard
Gaussian elimination with amounts in just some j arithmetic operations
and hence is extremely cheap.
Also: I'd like to verify that all of the coefficients on the left sideyes, if your function data are not subject to noise. noise will
have the same sign. Unless I set up the matrices incorrectly--not out
of the question--I would expect that the results would all have the
same sign if the tabulated function is monotonic with no local maxima
or minima within the range of the independent variable.
blow up by a factor 1/h, hence this may destroy these properties
if it is too large
Last but not least: what's the reference for the one-sided formulas?
I'd like to see/read more.
Thanks again for your patience: this is very valuable.
the linear system is from
Quarteroni, Alfio; Sacco, Riccardo; Saleri, Fausto
Numerical mathematics. (English)
Texts in Applied Mathematics. 37. New York, NY: Springer.
the one sided formula is from
Schmeisser, Gerhard; Schirmeier, Horst
Praktische Mathematik. (German)
de Gruyter Lehrbuch. Berlin-New York: Walter de Gruyter
but there should be other sources containing it it is obtained
simply by Taylor's formula for f(x+k*h) k=0,1,2,3
resp. f(x-k*h) and combining these values linearly such that the f'(x)
obtains with error term h^4
hth
peter
.
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