Talk:Lanczos approximation

TO DO:

• Complete derivation
• Spouge's version
• Section on choice of g and error estimates
• Better ways to calculate A
  • Use of partial fraction decomposition
  • Integer matrices
• Mention corollary at the end of Lanczos paper

Fredrik | tc 10:15, 13 January 2006 (UTC)

arbitrary precision?

I don't believe that lanczos can be used for arbitrary precision because it is a single asymptotic expansion and is thus subject to divergence due to stoke's phenomenon. Does anyone have evidence that it is suitable for arbitrary precision? --njh 04:34, 13 June 2006 (UTC)

@njh
I believe the crux is this: Lanczos' approximation is an interpolation (not extrapolation) formula. In a more familiar context we probably agree that, given f(x) sufficiently nice, there is no necessary obstacle to obtaining arbitrarily precise approximations of f(x0) using an asymptotically convergent series. Say, by passing a polynomial through distinct tabular points in the neighborhood of x0.
For Gamma(z+1) = z!, we can give to arbitrary precision
(1) A close initial guess for z!, z not small
(2) The relationship of z! to (z+k)!, for all integers k.
(3) Definite integral expressions for the true value
These statements are almost exactly the same thing as saying "we can construct an arbitrary-precision interpolating function without the use of tables". Which is not to say it will be easy, only that there is no mystery what is to be done. In Lanczos' formula, a rational interpolation function oscillates on both sides of the true value, the peaks diminishing with increasing g. There are a finite number of oscillations, so after the closest approach, it diverges to infinity. Increasing the expansion order increases the # of oscillations, pushing the best estimate to a larger value of g, and improving that estimate.
After a bit of fiddling, it is easy to show that the oscillations depend only on g (are independent of the input value, z).
Since the limit is approached from both sides, it appears that the series must (HYPOTHESIS) ultimately converge. I still have to check a few things that may take a little time. My questions are:
1. Does the behavior of the interpolating function hold for all n? Or is there some n=n0 beyond which the ripples are nondecreasing and/or diverge? Hypothesis: the properties of the Chebyshev polynomials guarantees convergence.
2. If the series indeed converges, is it to the true value?
This post is updated as I work through the problem. Right now, "update" means "rewrite". Chime in if this is irksome. SlumberlandOSaurus (talk) 15:41, 28 January 2016 (UTC)

Lanczos Coefficients

My question is about correctly interpreting the Lanczos approximation's Coefficients formula.That is the pk(g) formula that begins with a Sum sign in the Coefficients section. Does that formula run only along the bottom line of the triangular coefficient matrix, as suggested by the "a= 0 to k" spec ? If it did, then only one coefficient could be calculated. So it seems to me that each coefficient is obtained by applying that formula to one particular odd row. For example, coefficient P0 would relate to row one only, P1 to row 3 only, .....Pk to row k (last one) only. Can some fan of Lanczos give me some help on this ? Chambala (talk) 00:44, 1 April 2010 (UTC)User Camille Milot March 31, 2010 20:42 Canadian EasternDayligntTimeChambala (talk) 00:44, 1 April 2010 (UTC)

I'm not 100% sure I've got your question right, but you are correct that only one coefficient can be calculated. However this is not the correct result, the statement of the recursion is incomplete. The missing initial value is C(2, 1) = 0. Then the complete list can be constructed. I will fix this shortly. But I think a much better solution is to skip the recursion altogether. The coefficients in question are the nonzero coefficients of the even-order Chebyshev Polynomials of the First Kind: https://en.wikipedia.org/wiki/Chebyshev_polynomials#First_kind. Using the explicit formula on that page, and taking only even values of n, we obtain immediately
$C(2k+1,1)=(-1)^{k}$
$C(2k+1,2a+1)=(-1)^{k-a}k{\frac {(k+a-1)!}{(k-a)!(2a)!}}2^{2a},\;\;a>0$
There then follows mass simplification in the pk(g).
Lastly, if there are no objections, I will convert the whole C(i, j) notation to something readable. C(2k+1, 2a+1) refers to coefficient of the $x^{2a}$ term of Chebyshev polynomial order 2k, so.... if we replace C(2k+1, 2a+1) with $C_{2k,2a}$, the notation will in fact refer to that coefficient.SlumberlandOSaurus (talk) 23:32, 26 January 2016 (UTC)

The notation used for the coefficients formula seems odd. The variable of the sum is an integer L (I know it's a fancy lower-case L in the formula, but I'm not sure how to reproduce that here), yet there's a term in there that has (L - 1/2)! — but the factorial is undefined for non-integers, and Gamma is what we're trying to compute. What is the meaning of this part of the equation? I know Gamma has exact solutions for all the half integers, so my naive thought was that maybe that's what this means. I'm actually trying to implement my own version of Gamma in Java, hence the motivation to understand this. 75.167.103.166 (talk) 02:02, 13 May 2023 (UTC)