
From:  Martin Kunz 
Subject:  ppval subtracts break from XI  extend documentation? 
Date:  Fri, 12 Aug 2016 10:22:53 +0200 
Useragent:  Mozilla/5.0 (Windows NT 6.1; WOW64; rv:45.0) Gecko/20100101 Thunderbird/45.2.0 
Hi, It took me a while to realise why the evaluation of piecewise polynomials constructed with mkpp does not work as I expected. Consider this example (a test case from ppval): b = 1:3;c = ones (2); pp = mkpp (b, c); ppval (pp, 1.1) I thought we would calculate 1*xi + 1 at xi=1.1 and expected the answer to be 2.1, but in fact this computation yields 1.1. The reason is that ppval subtracts the left break from xi and calculates 1*0.1 + 1 = 1.1. I tried this code in MATLAB and got the same result, so the behaviour is consistent and we should not change it. If the piecewise polynomials are created by spline or interp1, all is fine (probably, I did not test), but when they are generated "by hand" with mkpp great confusion can arise. Therefore I suggest to add a note to the help of ppval and/or mkpp: "Note that ppval evaluates polynomials at XI  BREAKS(i), i.e. subtracts the the lower endpoint of the current interval from XI. This has to be taken into account when creating piecewise polynomial objects with mkpp." By the way, MATLAB includes a remark on this behaviour in the
help on mkpp, but I find it rather cryptic: "Notice
that Martin

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