CODEWITHSUNDEEP
symbolic-mathmaple
Rohithsai Sai
Rohithsai Sai

Reputation: 468

Optimize expression symbolically- Find out numerical exception possibilities in equaitons Maple 2018

I am trying to optimize expressions symbolically. I need to find out the maximum value possible for a variable(s), so that the expression still have a valid solution.

For Example:

expr := (a-b)/b-(a^3) # a=(0,10), b= (0,10)

In this eypression b=a^3 is the only case where undefined solution is possible. for a given interval of variables

This looks fine for simple expression. But in reality there are complex equations to solve with more than 2 varibles.

1) First thing is to find out all values of a variable resulting in undefined output

2) assign a symbolic value to the variable so that the undefined result can be eliminated.

I need to optimize the given expression so that it does not have any undefined cases when solving. I understood that when optimizing there always be a condition on variables(in this case variable max is the condition, maximum value the variable can take). output of an expression is always a real value

 OptimizedExpr:=a-b[max]/b[max]-a[3] --> b[max]>a^3 or b[max]<a^3

(it is easy to to say b[max] <> to a^3 and a^3 is the limiting value. in some case it is more reasonable to just ignore other part of limiting value. Hence, I would like to solve for greater than or less than of limiting value)

I would be very glad, how can I find Optimized expressions. I tried using the solve function but observed that expressions are equalled to zero and solving. which is completely opposite to what I was looking. I really do not know is there any way to find out undefined cases in expressions and on what varibale at which value.

I tried to explain the situation at my best and I welcome for any suggested edits or furthur information required.

Upvotes: 2

Views: 90

Answers (1)

acer
acer

Reputation: 7271

You wrote, "b=a^3 is the only case where undefined solution is possible."

Hence it looks like you made a typo with your brackets, and intended the expression to instead be,

expr := (a-b)/(b-a^3);

The discont command handles that example.

discont(expr, b);

                          / 3\ 
                         { a  }
                          \  /

If you have more involved expressions then please show them (and constraints on other parameters such as a would not hurt to know, if you have such).

It's difficult to answer properly if you don't show your actual example, explicitly and in full.

[edit] Ok, you posted a more involved example elsewhere. You can use the discont or the singular commands.

expr2 := tan(a*(b+log(1+(epsilon*a*r)/(c*s)))/2):

ans := [singular(expr2, {c, b})]:
ans := map[2](remove,u->lhs(u)=rhs(u), ans):

lprint(%);
  [{c = 0}, {c = -epsilon*a*r/s},
   {b = - (-2*_Z1*Pi+a*ln((a*epsilon*r+c*s)/(c*s))-Pi)/a}]

eval(expr2, ans[1]);
  Error, numeric exception: division by zero

eval(expr2, ans[2]);
  Error, (in ln) numeric exception: division by zero

eval(expr2, ans[3]):
simplify(%);
  Error, (in tan) numeric exception: division by zero

ans1 := map[2](`=`,c,discont(expr2, c)):
lprint(%);
  {c = 0, c = epsilon*a*r/(s*(exp((2*Pi*_Z2-a*b+Pi)/a)-1)),
   c = -epsilon*a*r/s}

ans2 := map[2](`=`,b,discont(expr2, b)):
lprint(%);
  {b = -(a*ln((a*epsilon*r+c*s)/(c*s))-2*Pi*_Z3-Pi)/a}

eval(expr2, ans1[1]);
  Error, numeric exception: division by zero

eval(expr2, ans1[2]):
  simplify(%) assuming real:
  Error, (in assuming) when calling 'cot'. Received: 'numeric exception: division by zero'

eval(expr2, ans1[3]);
  Error, (in ln) numeric exception: division by zero

eval(expr2, ans2[1]):
simplify(%) assuming real;
  Error, (in assuming) when calling 'cot'. Received: 'numeric exception: division by zero'

Upvotes: 1

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