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## RE: Table lookup & interpolation [follow-up]

**From**: |
Luke Scharf |

**Subject**: |
RE: Table lookup & interpolation [follow-up] |

**Date**: |
20 Mar 2003 09:15:30 -0500 |

On Wed, 2003-03-19 at 04:34, Ted Harding wrote:
>* On 19-Mar-03 Ted Harding wrote:*
>* > For this case, log(p(h)) is quite close to a linear function of h*
>* > (there is slight curvature). Possibly, linear interpolation of*
>* > log(p) between points may be adequate. If not, quadratic interpolation*
>* > using three consecutive points almost certainly is adequate.*
>* *
>* I just checked: using quadratic interpolation of log(p), the*
>* following compares p(h) with phat(h) where the points used for*
>* interpolation are (h-1000,p(h-1000)), (h+1000,p(h+1000)),*
>* (h+2000,p(h+2000)). The octave commands were:*
>* *
>* y=log(p);c=[];*
>* for i=2:29,*
>* x=h(i);*
>* h1=h(i-1);h2=h(i+1);h3=h(i+2);*
>* y1=y(i-1);y2=y(i+1);y3=y(i+2);*
>* phat=exp((x-h2)*(x-h3)*y1/((h1-h2)*(h1-h3))+ ...*
>* (x-h1)*(x-h3)*y2/((h2-h1)*(h2-h3))+ ...*
>* (x-h1)*(x-h2)*y3/((h3-h1)*(h3-h2)));*
>* c=[c;[p(i) phat]];endfor;c*
I'll give this a try when I get home from work this evening!
Thanks again!
-Luke
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