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 Cheta with base-change: preliminary results bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/12/2009, 09:03 AM (This post was last modified: 08/12/2009, 09:06 AM by bo198214.) (08/12/2009, 02:15 AM)jaydfox Wrote: So I calculate sexp(m*2^p), for m an integer between -64 and 64, for example, and p equal to -108. Because I'm only using reals, convergence is assured. I'm using sexp(0)=0, because Andrew's slog is centered at 0, so I can have slog(sexp(0)) and expect an exact answer. Anyway, I can then use those 129 points to calculate the various derivatives (actually, I use a matrix inversion that directly gives me the coefficients of the approximated Taylor series). Once I have a Taylor series, I can then move away from the real line. See the following wikipedia article for the general idea of how the grid of points can be used to get derivatives: http://en.wikipedia.orirg/wiki/Five-point_stencil But thats what I asked you. You make an interpolating polynomial through the points sexp(m*2^p) and then you can get the derivatives as the derivatives of the polynomial. Its described on that wiki page. Quote:Instead of a 5-point stencil, I use a 129-point stencil. I don't yet have a formula for the coefficients for the stencil, so I have to use a matrix inversion to derive them "the hard way". Well the interpolating polynomial can be computed as the Lagrange or the Newton Polynomial (with which I was incidentally concerned in the last time considering Ansus formulas) Lets consider the equidistant Newton interpolation polynomial for $N+1$ points $x_0,x_0+h,x_0+2h,\dots,x_0+Nh$: $N_N(x)=\sum_{n=0}^N \left(t\\n\right) \sum_{k=0}^n (-1)^{n-k} \left(n\\k\right) f(x_0+hk)$ where $t=(x-x_0)/h$. To compute the coefficients we need to compute the coefficients of $\left(t\\n\right)$ or more specifically the coefficients of $t(t-1)\dots (t-n+1)$, but these are the well know Stirling numbers of the first kind: $t(t-1)\dots (t-n+1) = \sum_{j=0}^n s(n,j) x^j$ Plug this into the Newton formula: $N_N(x)=\sum_{j=0}^N t^j \sum_{n=j}^N \frac{s(n,j)}{n!} \sum_{k=0}^n (-1)^{n-k} \left(n\\k\right) f(x_0+hk)$ Then we want to have the coefficients at $a:=x_0+hN/2$, let $d=hN/2$ $t^j = \frac{(x-x_0)^j}{h^j} = h^{-j} (x-a+d)=h^{-j}\sum_{i=0}^j \left(j\\i\right) d^{j-i} (x-a)^i$ Into the previous formula: $N_N(x)=\sum_{i=0}^N (x-a)^i \sum_{j=i}^N \left(j\\i\right) \frac{d^{j-i}}{h^j} \sum_{n=j}^N \frac{s(n,j)}{n!} \sum_{k=0}^n (-1)^{n-k} \left(n\\k\right) f(x_0+hk)$ So these are coefficients of the stencil without matrix inversion. In your described case $a=0$, $h=2^p$, $x_0=-Mh$, $d=Mh$, $M=64$. Maybe there is a more symmetric formula but this should at least work. And you still didnt tell me what FMA means! « Next Oldest | Next Newest »

 Messages In This Thread Cheta with base-change: preliminary results - by jaydfox - 08/11/2009, 05:16 PM RE: Cheta with base-change: preliminary results - by bo198214 - 08/11/2009, 06:45 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/11/2009, 06:52 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/11/2009, 08:03 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/11/2009, 09:12 PM RE: Cheta with base-change: preliminary results - by sheldonison - 08/12/2009, 01:31 AM RE: Cheta with base-change: preliminary results - by jaydfox - 08/12/2009, 02:15 AM RE: Cheta with base-change: preliminary results - by Gottfried - 08/12/2009, 12:37 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/12/2009, 02:29 PM RE: Cheta with base-change: preliminary results - by Gottfried - 08/12/2009, 03:47 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/12/2009, 04:49 PM RE: Cheta with base-change: preliminary results - by bo198214 - 08/11/2009, 08:11 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/11/2009, 08:45 PM RE: Cheta with base-change: preliminary results - by bo198214 - 08/11/2009, 08:54 PM RE: Cheta with base-change: preliminary results - by jaydfox - 08/12/2009, 12:24 AM

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