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MathGroup Archive 2006

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Re: Symbolic Derivatives of Unspecified Functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70693] Re: Symbolic Derivatives of Unspecified Functions
  • From: David Bailey <dave at Remove_Thisdbailey.co.uk>
  • Date: Mon, 23 Oct 2006 02:50:52 -0400 (EDT)
  • References: <ehf1mp$5e9$1@smc.vnet.net>

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


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