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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?
>
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