Re: Symbolic Derivatives of Unspecified Functions
- To: mathgroup at smc.vnet.net
- Subject: [mg70718] Re: Symbolic Derivatives of Unspecified Functions
- From: misha <iamisha1 at comcast.net>
- Date: Tue, 24 Oct 2006 02:24:30 -0400 (EDT)
- References: <ehf1mp$5e9$1@smc.vnet.net> <ehhqlj$7cm$1@smc.vnet.net>
As suggested, I tried
In[1]:=r[qi_, qj_]=r[qi,qj]
as opposed to
In[1]:=r[qi_, qj_]:=r[qi,qj]
....but I still get an error message,
$IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
Out[1]:=Hold[ri[qi,qj]]
I subsequently take a derivative (ignoring the error message, contrary
to advice sent to me regardin this) after entering a few other functions
and getting similar error messages in the case of unspecified functions
(e.g., ci[Gi_]=ci[gi]).
....
In[n]:= profiti[qi_,qj_,beta_,theta_,gi_,gj_]=r[qi, qj] - ci[Gi]*qi-v*gi^2/2
$IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
$IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
Out[n]:=-(gi^2)*v/2 - qi Hold[ci[Gi]] + Hold[ri[qi,qj]]
In[n+1]:= D[profiti[qi,qj,beta,theta,gi,gj],qi]
Out[n+1]:=del_{qi}Hold[ri[qi,qj]]-Hold[ci[Gi]]
Which is , in terms of correctness, what I want. But I still have many
questions...
1) What does the "$IterationLimit" error message mean and why doesn't
using "=" as opposed to ":=" resolve this (as interpreted the reply
below to suggest)? I read the Help Browser after clicking on "More...",
but that doesn't really clarify things for me in this context. I "just"
want to have ri be a function of qi and qj so that when I take
derivatives of functions that include ri (say w.r.t. qi), I will get
something like D[ri[qi,qj], qi], or, ideally, something that I can
easily render into a TeX file.
Thanks!
Misha
David Bailey wrote:
> misha wrote:
>
>>My apologies for the long post. A brief form of the question is the
>>first sentence of the next paragraph. I'm a very inexperienced
>>Mathematica user and am probably expecting too much and/or being too
>>lazy, but I was wondering whether Mathematica can do something I
>>describe below, and, if so, how to implement it. I also apologize for
>>my likely misuse of notation, terminology, etc.
>>
>>I think my question boils down to whether I can use Mathematica to take
>>derivatives of functions that have no functional form. It seems like
>>this is possible, given an example I found in the help files.
>>D[f[g[x]],x]=f'[g[x]]g'[x]
>>
>>For example, as a first shot, I tried,
>>In[1]:= r[qi_, qj_]=r[qi,qj]
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>Out[1]:= Hold[r[qi,qj]]
>>In[2]:= qi[gi_, gj_]=qi[gi,gj]
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>Out[2]:= Hold[qi[gi,gj]]
>>In[3]:= qj[gi_, gj_]=qj[gi,gj]
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>Out[3]:= Hold[qj[gi,gj]]
>>In[4]:= D[r[qi[gi,gj],qj[gi,gj]],gi]
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>$ IterationLimit::itlim : Iteration limit of 4096 exceeded. More...
>>General::stop:
>> Further output of $IterationLimit::itlim will be suppressed during this
>>calculation. More...
>>Out[4]:= D[Hold[r[Hold[qi[gi,gj]], Hold[qj[gi,gj]]]],gi]
>>
>>I would like something like,
>>r_1*d{qi}/d{gi} + r_2*d{qj}/d{gi}, where r_1 is the partial derivative
>>of r w.r.t. its first argument, r_2 is the partial derivative of r
>>w.r.t. its second argument.
>>
>>So, say I have some functions that assume little or no specific
>>functional form.
>>Di=[(beta*(gi+gj))/(1-theta)][gi/(gi+gj) + theta*gj/(gi+gj)]
>>alpha=beta/(1-theta) (obviously this is specific)
>>Ci(Gi)=ci(Gi)*qi, (ci(Gi) has no specific form)
>>Gi(gi, gj, beta, theta) = (1 - alpha)*gi + alpha*theta*gj
>>pi_i(gi,gj,qi,qj,beta,theta)=ri(qi,qj)-ci(Gi)*qi-(v*gi^2)/2, where ri
>>and ci do not have functional forms. (I'd also like to work with a less
>>specific form of (v*gi^2)/2, but I'll set that aside for now...).
>>
>>I want to take derivatives, such as
>>
>>derivative of pi_i with respect to qi, a first order condition, or FOC
>>
>>(1) d{pi_i}/d{qi} = 0
>>
>>then the derivative of (1) w.r.t. qi and qj, and the total derivative of
>>(1) w.r.t gi, assuming qi and qj are functions of gi (and gj).
>>
>>I get some results that include expressions with "Hold" and problems
>>with iterations.
>>
>>Here are more details about the above problems and what I want
>>Mathematica to do...
>>
>>This may be obvious, but this is a two-stage oligopoly model with
>>investment. With specific functional forms one can do something like
>>the following:
>>
>>Solving the second (i.e., last) stage of the game in Cournot competition
>>(i.e., firms i and j maximize profit, pi_i, pi_j, choosing qi and qj,
>>respectively) will yield expressions for qi and qj in terms of the other
>>variables, gi, gj, beta, and theta. This is done by setting the first
>>derivative of pi_i w.r.t. qi equal to zero, then solving for qi. Then,
>>using these expressions, you move to the first stage and solve for the
>>optimal investments, gi and gj, by substituting the above qi and qj into
>>pi_i, setting the first derivative of pi_i w.r.t. gi to zero, and
>>solving for gi, yielding expression in terms of beta and theta.
>>
>>However, without assuming functional forms, it gets a little
>>hairier...making a number of common assumptions, such as d{ri}/d{qj}<0,
>>d{ri}/(d{qi}d{qj})<0, d{ci}/d{G}<0, d^2{ci}/d{G^2} >0, symmetry
>>(d{ri}/(d{qi}d{qj}) = d{rj}/(d{qj}d{qi}) = r_ij, d^2{ri}/d{qi^2} =
>>d^2{rj}/{qj^2} = r_ii, and some others, gets a FOC that looks something like
>>
>>(2) d{pi_i}/d{qi}=d{ri}/d{qi}-d{ci}/d{qi} = 0
>>
>>Then, since there's no functional form, one obviously cannot solve for
>>qi, so the goal is to find the slope of the "reaction function",
>>d{qj}/d{qi}, which turns out to be (in this model)
>>
>>(3) d{qj}/d{qi}=-(d^2{ri}/d{qi^2})/(d^2{ri}/d{qi}d{qj}).
>>
>>This comes from totally differentiating (2) w.r.t. qi and qj.
>>
>>Then, one can solve the the changes in qi and qj w.r.t. changes in gi,
>>d{qi}/d{gi} and d{qj}/d{gi}, by totally differentiating (2) and its
>>equivalent for pi_j w.r.t. gi.
>>
>>There is far more than this, but if I can get this done in Mathematica,
>>my life will be far easier. Ideally, I can get "nice" expressions that
>>can be exported into a TeX file via Mathematica.
>>
>>Many thanks in advance for reading and responding.
>>
>>Misha
>>
>>
>
> Hello,
>
> As you point out at the start, it is possible to take derivatives of
> expressions that contain undefined functions - but in that case, don't
> try to define them!
>
> r[qi_, qj_]=r[qi,qj]
>
> You do not want a definition for r, and this definition is infinitely
> recursive (which is why you get an error message). As a general point,
> there is no point in ignoring an error like that - you have to discover
> why it is happening!
>
> If later, you want the function r to have a specific form, don't use a
> function definition, just use a replacement rule, for example:
>
> r[a,b]^2 /. r[x_,y_]->Exp[(x+y)^2]
>
> Manipulating expressions with replacement rules is often much more
> convenient than doing everything with definitions, because you can use a
> replacement rule when and where you want it.
>
> I hope this gives you a start at least.
>
> David Bailey
> http://www.dbaileyconsultancy.co.uk
>