Re: Mixed Numerical Derivatives

*To*: mathgroup at smc.vnet.net*Subject*: [mg87164] Re: Mixed Numerical Derivatives*From*: Szabolcs Horvát <szhorvat at gmail.com>*Date*: Thu, 3 Apr 2008 05:13:08 -0500 (EST)*Organization*: University of Bergen*References*: <fsveco$sp7$1@smc.vnet.net> <fsvosj$5at$1@smc.vnet.net>

[note: message sent to comp.soft-sys.math.mathematica] dh wrote: > Hi, > > I have version 6 and can not find ND. It is in the NumericalCalculus package. > > Nevertheless, you could try an iterative approach. E.g.: > > ND[ ND[fun[x,y],{x,2},1] ,{y,2}, 1] This will appear to work for simple functions, but I would not recommend it because the inner ND is evaluated /before/ the outer ND, and produces a "semi-symbolic" result, i.e. something containing both 'y' and floating point numbers. This might not work at all for complicated functions that /must/ take numerical arguments, and even for simple functions ND cannot check for convergence, etc., so it is problematic. A similar solution is to define one's own "double-derivative" function which avoids evaluation of the "inner" ND explicitly, e.g. ND2[expr_, x_, y_, x0_, y0_] := Module[{fun, a}, fun[a_?NumericQ] := ND[expr /. y -> a, x, x0]; ND[fun[a], a, y0] ] ND2[Sin[x y], x, y, 1, 1] ... but I am still not comfortable with this approach ... > > hope this helps, Daniel > > > > jwmerrill at gmail.com wrote: > >> How can I come up with the Hessian of a function, at a particular > >> point, which can only be evaluated numerically? If I had a symbolic > >> function, I could do something like > > >> In[123]:= > >> D[x^3 + z*y^-1 + z^(1/2), {{x, y, z}, 2}] /. {x -> 3, y -> 5, > >> z -> 12} > > >> Out[123]= {{18, 0, 0}, {0, 24/ > >> 125, -1/25}, {0, -1/25, -1/(96 Sqrt[3])}} > > >> The function I'm interested in, though, can only be calculated > >> numerically. Using ND, I can find the diagonal elements of the > >> Hessian: > > >> In[92]:= rules = Last[ > >> FindMaximum[{Total[logPr[vdt, ddt, var, #] & /@ testData], ddt > 0, > >> var > 0}, {{vdt, .9}, {ddt, 120}, {var, 90}}]] > > >> Out[92]= {vdt -> 0.95945, ddt -> 151.097, var -> 103.255} > > >> In[111]:= Needs["NumericalCalculus`"] > > >> In[124]:= ND[ > >> Total[logPr[vdtp, ddt, var, #] & /@ testData] /. rules, {vdtp, 2}, > >> Evaluate[vdt /. rules]] > > >> Out[124]= -64.4011 > > >> But what about the off diagonal elements?