Will do :) It just finished solving on my system too. Will start extracting expressions for each variable next.

You've been much help.

Yinka

David,

You are amazing. You are going on to the acknowledgement page of my thesis :)

Thanks a lot.

Yinka

Hi David,

Like you said the analytical solution came out quite large and complicated so I went on to solve it numerically, which is shown in the matrix in the attached. However, I'm not sure where the "0." in the first column of the 1st, 4th, and 5th row are coming from.

Can you help look at it please?

Thanks.

Yinka

Attachments:

OK, thank you. The other variables are meant to stay as that such that I can interpret their coefficients as their respective effects on the variable they represent. Yinka

Hi Yinka, I looked at terminal_conditions.nb. What I see there is 28 variables defined. They are eq1 through eq28. Each is set to an expression, not an equation. For example:

eq2 = h - HEH he - HNH hn;

When I look at Example.nb above, it appears you form equations from each of these by equating the expression to 0. But you do not actually do that in terminal_conditions.nb. Notice that in Mathematica the single = is called Set, and assigns a value to a variable. The double == asserts equality.

If this is the case, the first thing to do in terminal_conditions is to form correct equations:

In[1]:= eq2 = h - HEH he - HNH hn == 0

Out[1]= h - he HEH - hn HNH == 0

Then use those with Solve or a chain of Eliminate. I notice several things in the notebook that are of concern. First, you are using variable names beginning with capitals. This can cause problems because Mathematica uses symbol names beginning with capitals for system defined symbols, like E and I. It is better to use lower case and be safe. (However, I don't see any conflicts I recognize.) I also notice "i" is used -- if this is intended to be the imaginary unit, in Mathematica that is I with a capital.

So I think the first thing is to set the eqn to real equations. Then what is needed is a list of the variables, excluding the parameters. I notice that your variable list in the text does not correspond to what is in the Mathematica input. (And the Mathematica variable names should not contain superscripts or subscripts -- that leads to endless trouble.)

Then use Eliminate and or Solve. Here I do see some difficulty. The expressions appear to use variables both in product terms and in exponents. This can be difficult.

Best regards,

David

You're welcome, Yinka. When you have equations and a list of the variables post a notebook again.

--David

Hi Yinka,

I took a look, and I have some questions. In the attached notebook you will see I did two things:

1) In the first step under your recursive solution section I looked at the equations 1, 9-12 you mention. The variable you say you are wanting to solve for do not appear. I suspect you type set them in the text, but use simpler forms for them in the Mathematica input. Could you clarify this?

2) In my second section, I build a rule set that sets all variables in your variable list to 1, and apply it to the equation set. There is a lot left, which I assume is entirely composed of the parameters. (exogenous variables). Does this look correct? Or are some of these typographic errors containing what should be variables.

I am still concerned about the cases in which variables appear both as factors and in exponents. These can be quite difficult. It may be that in that you will need to eliminate as many variables as you can, then solve a remaining system numerically, and then back substitute. This is perhaps good for some uses, but not very satisfying to a theoretician.

In[38]:= Solve[x Exp[x] + x == c, x]

During evaluation of In[38]:= Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

Out[38]= Solve[x + E^x x == c, x]

In[39]:= Solve[Exp[x] + x == c, x]

During evaluation of In[39]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

Out[39]= {{x -> c - ProductLog[E^c]}}

**** Yinka -- I just saw your post as I was writing this. Some is answered, but I thought I wold post anyway since some is not. For the exponent-type terms which do not contain any variable I suggest you assign them a simple symbol name. The less typesetting you do the better off you will be.

-- Best, David

Attachments:

Yes, you are right, I type set them in the text, but use simpler forms for them in the Mathematica input.

Yinka

Please see the attached file for the problem I'm trying to solve and I hope you can suggest how your method could ease my problem. The file contained is terminal_conditions.nb explaining what I'm trying to do and consists of the equation listing. I would prefer to use a more efficient way/ like you proposed above to solve the model here. There is another file not included, which contains my attempt at solving the problem. I've been able to solve for up to 24 of the 28 problems so far, but I am finding it difficult to go past that step because I need the full depiction of each expression to be used in the succeeding step. However, at the current step Mathematica crashes every time I attempt to copy the result for use in the next step. Thanks. Yinka

Attachments:

Maybe this would help too, here is an equation.

eqn = Solve[4 x + 2 y == 28 && x > 0 && y > 0, {x, y}, 
  Integers]; p = {x, y} /. eqn

In[221]:= eqn

Out[221]= {{x -> 1, y -> 12}, {x -> 2, y -> 10}, {x -> 3, 
  y -> 8}, {x -> 4, y -> 6}, {x -> 5, y -> 4}, {x -> 6, y -> 2}}

or

In[220]:= p

Out[220]= {{1, 12}, {2, 10}, {3, 8}, {4, 6}, {5, 4}, {6, 2}}

Paul.

Thanks David, your answer helped me with a dynamic system I was working on.

In addition to the above message, I hope the attached file can help to illustrate how I've been trying to do it and how my very manual approach is creating very large document that at the later stages is then running out of memory to handle.

Attachments: