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How can i use expressions from previous results?

Posted 10 years ago

Hi Everyone,

I am finding it difficult to use expressions generated from previous results when I applied "%". Basically I am trying to solve a system of large simultaneous equations by gradual combination. The error message is due to the fact that the output are expressed with "?", while in the new input argument, they and the newly added equations needed to be written with "==".

So, my question would be how can I map these previous expressions/results into the new input with the instruction "==" for each element therein.

Thanks.

Yinka

POSTED BY: olayinka oyekola
24 Replies

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

You've been much help.

Yinka

POSTED BY: olayinka oyekola
Posted 10 years ago

Thanks, Yinka, but not necessary. It was one line in a 10th generation language. But you could cite Mathematica. ;-)

POSTED BY: David Keith

David,

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

Thanks a lot.

Yinka

POSTED BY: olayinka oyekola
Posted 10 years ago

Oh -- and to answer your last question, I was setting your variables to 1 and evaluating the equations just for the purpose of seeing the parameters by themselves.

POSTED BY: David Keith

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:
POSTED BY: olayinka oyekola
Posted 10 years ago

I don't know -- it may be just a very small number. It could go away if Chop is used. You could set the remaining undefined variables to zero and evaluate it to get its magnitude. I suspect it is so small that it will not affect numerical results once values are substituted for the remaining variables.

POSTED BY: David Keith

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

POSTED BY: olayinka oyekola
Posted 10 years ago

Hi Yinka,

Attached is a notebook. I did not do anything special. I just used

Timing[sol = Solve[eqns, varlist]]

And Mathematica solved the system in 27 minutes on my 4-core Z800. I notice that one variable - ue - solved to 0. The rest solved to very messy expressions.

So you wind up with a nested list of rules as the solution. The expression for any of the varialbes can be had as variableName/.First[sol] The use of First clears the outer bracket of the solution set. You might get simpler expressions using sol2 = Simplify[sol], but be prepared to wait for a long time.

These expressions for the variables could be evaluated to a numerical result by using a list of rules to define the values of the parameters.

Best,

David

Attachments:
POSTED BY: David Keith
Posted 10 years ago

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

POSTED BY: David Keith

Thank you very much David. I will go through the code to clean it up and hopefully it starts to work.

Thanks again.

Yinka

POSTED BY: olayinka oyekola
Posted 10 years ago

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

--David

POSTED BY: David Keith

Hi David,

I have attached a notebook with just the definitions of model expressions and the list of model variables. The task is to solve for those 28 listed variables in terms of the remaining variables (these are exogenous) and the model parameters.

I hope I have done these listings correctly.

Thanks.

Yinka

Attachments:
POSTED BY: olayinka oyekola
Posted 10 years ago

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

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POSTED BY: David Keith

Hi David, Thanks a lot.

  1. In response to question 1, the type set appearance and the appearance in the input of Mathematica are kind of different. In the Mathematica input, the variables to solve for, which is relevant to equations 1, 9-12 are rr, ue, un, ke, kn. That is, while I put in the text section subscript t for time in the model variable, in the input this is not the case. Also, when a model variable has a superscript e or n, this is put in front of the variable - hence, e.g., ue in the input is u superscript e subscript t in the text section.

  2. What is left is entirely correct. All of them are the exogenous variables and the model parameters. Yes, I have solved this numerically in MATLAB, but wasn't quite satisfactory to my boss because the numerical value didn't achieve what I needed it to. So, he said I must first derive them analytically - he had like to see that before - he allows parameter value specification.

So, are those two steps that set variables to 1 and eliminated them in the next section part of what I need to be doing or you just used that to ascertain which are the endogenous variables?

Best wishes,

Yinka

POSTED BY: olayinka oyekola

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

Yinka

POSTED BY: olayinka oyekola

Further, I am attaching the pdf of the model expressions that I hoped to solve using Mathematica.

The expressions that appear in the Mathematica inputs with exponents are parameters. E.g. steady state ratio (ee/ke) raised to power nue is written as (ee/ke)^nue. No variables that I intended to solve for has exponents with it.

I think by creating the model list, what you referring to is for me to do what "syms" does in MATLAB, I have attempted that by varlist={.} in the terminal_conditions.nb but I am skeptical about if that's the way to do it in Mathematica.

So, given the list in the pdf file, I am hoping to have a solution for each of the 28 variables in terms of only the remaining variables and model parameters.

Please let me know if I am making any progress. Thanks.

Yinka

Attachments:
POSTED BY: olayinka oyekola
Attachments:
POSTED BY: olayinka oyekola

Thanks Paul.

POSTED BY: olayinka oyekola
Posted 10 years ago

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.

POSTED BY: Paul Cleary

Thank you David.

POSTED BY: olayinka oyekola

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

POSTED BY: Jake Trexel
Posted 10 years ago

You can use Eliminate:

In[1]:= eqs = {x + y + z == 3, 2 x + 3 y - z == 5, x + z == 4};

In[2]:= Solve[eqs]

Out[2]= {{x -> 4, y -> -1, z -> 0}}

In[3]:= eq4 = Eliminate[eqs[[{1, 2}]], x]

Out[3]= -1 + 3 z == y

In[4]:= eq5 = Eliminate[eqs[[{2, 3}]], x]

Out[4]= -1 + z == y

In[5]:= sol1 = Solve[{eq4, eq5}]

Out[5]= {{y -> -1, z -> 0}}

In[6]:= sol2 = Solve[eqs[[1]] /. sol1]

Out[6]= {{x -> 4}}
POSTED BY: David Keith

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:
POSTED BY: olayinka oyekola

Hi David,

That message meant to be left on the discussion for this problem (still finding my way around both the use of mathematica and its messaging board). Anyway, it is meant to be an addendum to my post containing the attachment, "terminal_conditions.nb".

Thanks.

Yinka

POSTED BY: olayinka oyekola
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