Re: Mixed Numerical Derivatives

*To*: mathgroup at smc.vnet.net*Subject*: [mg87159] Re: Mixed Numerical Derivatives*From*: dh <dh at metrohm.ch>*Date*: Wed, 2 Apr 2008 05:58:00 -0500 (EST)*References*: <fsveco$sp7$1@smc.vnet.net>

Hi, I have version 6 and can not find ND. Nevertheless, you could try an iterative approach. E.g.: ND[ ND[fun[x,y],{x,2},1] ,{y,2}, 1] 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? >