Re: [R] matrix of higher order differences

From: Ravi Varadhan <rvaradhan_at_jhmi.edu>
Date: Wed, 27 Apr 2011 17:41:22 -0400

Peter, I have indeed worked with Gregory-Newton and divided differences in my very first numerical analysis course a couple of decades ago! However, I am perplexed by the particular form of this matrix where the differences are stored along the diagonals. I know that this is not the *same* as the Wronskian, but was just wondering whether it is an established matrix that is some kind of an *ian* like Hermitian, Jacobian, Hessian, Wronskian, Laplacian, ...

Best,
Ravi.



Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University

Ph. (410) 502-2619
email: rvaradhan_at_jhmi.edu

-----Original Message-----
From: peter dalgaard [mailto:pdalgd_at_gmail.com] Sent: Wednesday, April 27, 2011 4:59 PM
To: Ravi Varadhan
Cc: R Help
Subject: Re: [R] matrix of higher order differences

On Apr 27, 2011, at 21:34 , Ravi Varadhan wrote:

> My apologies in advance for being a bit off-topic, but I could not quell my curiosity.
>
> What might one do with a matrix of all order finite differences? It seems that such a matrix might be related to the Wronskian (its discrete analogue, perhaps).
>
> http://en.wikipedia.org/wiki/Wronskian

Not quite, I think. This is one function at different values of x, the Wronskian is about n different functions.

Tables of higher-order differences were used fundamentally for interpolation and error detection in tables of function values (remember those?), but rarely computed to the full extent - usually only until the effects of truncation set in and the differences start alternating in sign.

>
> Ravi.
> -------------------------------------------------------
> Ravi Varadhan, Ph.D.
> Assistant Professor,
> Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University
>
> Ph. (410) 502-2619
> email: rvaradhan_at_jhmi.edu
>
>
> -----Original Message-----
> From: r-help-bounces_at_r-project.org [mailto:r-help-bounces_at_r-project.org] On Behalf Of Petr Savicky
> Sent: Wednesday, April 27, 2011 11:01 AM
> To: r-help_at_r-project.org
> Subject: Re: [R] matrix of higher order differences
>
> On Wed, Apr 27, 2011 at 11:25:42AM +0000, Hans W Borchers wrote:

>> Jeroen Ooms <jeroenooms <at> gmail.com> writes:
>> 
>>> 
>>> Is there an easy way to turn a vector of length n into an n by n matrix, in
>>> which the diagonal equals the vector, the first off diagonal equals the
>>> first order differences, the second... etc. I.e. to do this more
>>> efficiently:
>>> 
>>> diffmatrix <- function(x){
>>> 	n <- length(x);
>>> 	M <- diag(x);
>>> 	for(i in 1:(n-1)){
>>> 		differences <- diff(x, dif=i);
>>> 		for(j in 1:length(differences)){
>>> 			M[j,i+j] <- differences[j]
>>> 		}
>>> 	}
>>> 	M[lower.tri(M)] <- t(M)[lower.tri(M)];
>>> 	return(M);
>>> }
>>> 
>>> x <- c(1,2,3,5,7,11,13,17,19);
>>> diffmatrix(x);
>>> 
>> 
>> I do not know whether you will call the appended version more elegant,
>> but at least it is much faster -- up to ten times for length(x) = 1000,
>> i.e. less than 2 secs for generating and filling a 1000-by-1000 matrix.
>> I also considered col(), row() indexing:
>> 
>>    M[col(M) == row(M) + k] <- x
>> 
>> Surprisingly (for me), this makes it even slower than your version with
>> a double 'for' loop.
>> 
>> -- Hans Werner
>> 
>> # ----
>> diffmatrix <- function(x){
>> 	n <- length(x)
>> 	if (n == 1) return(x)
>> 
>> 	M <- diag(x)
>> 	for(i in 1:(n-1)){
>> 		x <- diff(x)           # use 'diff' in a loop
>> 		for(j in 1:(n-i)){     # length is known
>> 			M[j, i+j] <- x[j]  # and reuse x
>> 		}
>> 	}
>> 	M[lower.tri(M)] <- t(M)[lower.tri(M)]
>> 	return(M)
>> }
>> # ----

>
> Hi.
>
> The following avoids the inner loop and it was faster
> for x of length 100 and 1000.
>
> diffmatrix2 <- function(x){
> n <- length(x)
> if (n == 1) return(x)
> A <- matrix(nrow=n+1, ncol=n)
> for(i in 1:n){
> A[i, seq.int(along=x)] <- x
> x <- diff(x)
> }
> M <- matrix(A, nrow=n, ncol=n)
> M[upper.tri(M)] <- t(M)[upper.tri(M)]
> return(M)
> }
>
> Reorganizing an (n+1) x n matrix into an n x n matrix
> shifts i-th column by (i-1) downwards. In particular,
> the first row becomes the main diagonal. The initial
> part of each of the remaining rows becomes a diagonal
> starting at the first component of the original row.
>
> Petr Savicky.
>
> ______________________________________________
> R-help_at_r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help_at_r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes_at_cbs.dk  Priv: PDalgd_at_gmail.com

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Received on Wed 27 Apr 2011 - 21:55:27 GMT

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