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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**RE: Symbolic Derivatives of Unspecified Functions**

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