Difference Equations

Our local choice was Matthews & Walker Mathematical Methods of Physics.

Always used to be popular with the turbulent flow brigade - often gave divergent power series that could only be tamed with Shank's. The pure mathematicians used to cringe at the abuse of method but they could not deny that the results it predicted matched experimental data well!

Works best on alternating series with poor convergence or divergence!

I was once very interested in extended convergence tricks and have used the rational approximation for Log(1+x) in anger several times.

Log(1+x) = x(6+x)/(6+4x) where -1/2 < x < 1

The other common one sometimes useful is

Sqrt(x) = (1+3x)/(3+x) where -1/2 < x < 2

Only really any good if you have fast hardware divide.

These tricks sometimes allow a rough polynomial approximation soluble analytically to provide a much better initial input guess for more rapid convergence of an iterative method like N-R.

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Hi Martin, (I also hated Green's functions as a student) Well I'm sure you know this, but it came as a revelation to me many years after graduating. The Green's function is just the pulse response of the system. If someone had told me that in school things would have made a lot more sense! I sometimes wish I could take some of those courses over again. It seems like I might understand them better now.

George H.

I recall the materials course where the prof explained everything using tensor calculus. Unfortunately, nobody had previously bothered to teach us tensor calculus. All the lectures were pure gibberish. I'm not at all sure how I passed that course.

Good stuff. Did you ever read Forman Acton's "Numerical Methods That Work"?

Cheers

Phil Hobbs

Bell curve.

Best regards, Spehro Pefhany

Yes - but along time ago. I don't possess a copy any more.

One memorable fun version of 3 term Shanks with terrible numerical stability (but you can divide it out to get something much better) is difference of geometric mean over arithmetic mean

given the partial sums a, b, c

x' = (a*c-b^2)/(a+c-2*b)

And if you feed it 1,3,7 it gives

x' = (1*7-3^2)/(1+7-2*3) = (7-9)/(8-6) = -1

Users of twos compliment arithmetic take note!

A numerically stable version that is formally equivalent is

x' = (a+2b+c)/4 - (a-c)^2/(a+c-2b)/4

Subject to typos and lapses of memory...