3
$\begingroup$

I need to simplify big symbolic solution of equations system by truncating terms that are orders of magnitude less then others in subexpressions like a + b. The solution is big fraction with sum of terms in numerator and denominator like

a b + c ( d - f + g + ...

where every symbol has value (real number). Is it possible to apply rule like this to solution:

replace a + b by a, if Abs[a] >> Abs[b] (say, 1000 times more), by b, if Abs[a] << Abs[b], or leave it as a + b in other cases.

Repeated application of such a rule should result in symbolic expression that is much simpler then exact solution, but is good enough approximation for concrete values of symbols. I am new to Mathematica, searched for similar questions but didn't find any.

Minimal working example: solving eq gives solution that should be simplified by truncating terms basing on parameters alpha and beta below:

v1 = α1 a - β1 b;
v2 = α2 a;
v3 = α3 c;
eq = {-v1 - v2 == 0, v1 + v3 == 0, a + b + c == p};
Solve[eq, {a, b, c}]

α1 = 1000; β1 = 1000; α2 = 1; α3 = 1;
$\endgroup$
4
  • $\begingroup$ Are the single terms available separately as {a, b, c ,d, ...} ? $\endgroup$ Commented Jan 16, 2014 at 14:17
  • 1
    $\begingroup$ If You'd like to apply a rule to the solution, which is a big fraction, it will probably require developing a pattern. If You could post a minimum working example (a simplification of Your big fraction), that would help a lot. $\endgroup$
    – Wojciech
    Commented Jan 16, 2014 at 14:39
  • $\begingroup$ @b.gatessucks: no, terms are created by Solve[], I only have values of parameters. $\endgroup$
    – mvm44
    Commented Jan 16, 2014 at 15:37
  • $\begingroup$ @Wojciech: posted minimum working example $\endgroup$
    – mvm44
    Commented Jan 16, 2014 at 15:39

2 Answers 2

2
$\begingroup$

The following code will do what you ask, given the assumptions that

  • you want to simplify a symbolic sum of terms,
  • your experimental data is encoded in a list of replacements which will cause each of the terms in the sum to give a number upon replacement, and
  • you have a specified tolerance for which terms get neglected - i.e., terms in a sum smaller than tolerance * the biggest term in the sum should get neglected.

Doing this is fairly simple. I ensure the head of the first argument is Plus, since it would be catastrophic if the head were, say, Times. I turn my input into a list of terms. I run the function Thread[Abs[#] > tolerance Max[Abs[#]]]&, which will give True for the terms I want to keep, on the numerical values. I then Pick the terms that gave True.

sumNeglecter[terms, replacements_, tolerance_] := If[MatchQ[terms, _Plus],
    Pick[
      terms, 
      Thread[Abs[#] > tolerance Max[Abs[#]]] &[List @@ terms /. replacements]
    ]
    ,terms]
sumNeglecter[a + b + c + 2 b^c, {a -> 10^6, b -> 10^4, c -> 1}, 0.01]
(*output: a + 2 b^c *)

If what you want to simplify is a fraction, you can easily implement this for the numerator and denominator separately:

termNeglecter[terms_, replacements_, tolerance_] := Piecewise[{
  {sumNeglecter[terms, replacements, tolerance], MatchQ[terms, _Plus]},
  {sumNeglecter[Numerator[terms], replacements, tolerance]/
    sumNeglecter[Denominator[terms], replacements, tolerance], 
    MatchQ[terms, a_/b_]}
  }, terms]

Finally, you can Map this over all levels of your expression if you really require a thorough job:

Map[ termNeglecter[#, replacements, tolerance]&, {terms}, \[Infinity]][[1]]

Beware, though, of

  • Situations where you have many very small terms that together should not be neglected. For example,

    termNeglecter[Sum[u[k]/k, {k, 1, 10000}], {u[_] -> 1}, 0.1]
    

    is about 3.45 times than the original.

  • Situations where you have cancellations. For example, in

    termNeglecter[a - a^b + b, {a -> 10, b -> 1}, 0.1]
    

    the dominant term is b.

  • Situations where nonlinearity is important. 1001.01 is not within 1% of 100, though the exponent is within 1% of 1.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! That was very interesting and helpful. Thanks for mentioning situations where simple truncating doesn't work. $\endgroup$
    – mvm44
    Commented Jan 17, 2014 at 19:28
1
$\begingroup$

Here is a simple general solution I've managed to come up with. Since all the terms are created by Solve, I might as well define them as a list (if You use Flatten with Your Solve result, You should get them in the same form):

data = {a -> 1234, b -> 1.05, c -> 345, d -> -10890, e -> 0.03, f -> 4234, g -> 3.65, 
 h -> 543, i -> -11456, j -> -0.76};

Let's say that the fraction You get as a solution has a following form:

abc = (a + b + c + d + e)/(f + g + h + i + j);

We can now write a function that will do the simplification:

f1[fraction_] := Module[{qwe, qay},
 qwe = (Numerator[fraction] /. {x_ + y_ /; ((Abs[x]/Abs[y] /. data) > 1000) :> x,
    x_ + y_ /; ((Abs[x]/Abs[y] /. data) < 1000) :> y});
 qay = Denominator[fraction] /. {x_ + y_ /; ((Abs[x]/Abs[y] /. data) > 1000) :> x, 
  x_ + y_ /; ((Abs[x]/Abs[y] /. data) < 1000) :> y};
 qwe/qay
];

Now if I call it with abc:

f1[abc]

I get:

(a + c + d)/(f + h + i)

In this particular case abc/.data yields 1.39451 while f1[abc]/.data yields 1.39407.

$\endgroup$
2
  • 1
    $\begingroup$ You're welcome! You can always upvote answers that solve Your problem. $\endgroup$
    – Wojciech
    Commented Jan 17, 2014 at 13:47
  • $\begingroup$ Of course, when I gain enough reputation $\endgroup$
    – mvm44
    Commented Jan 17, 2014 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.