Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
Reduce[{(4 I1^2 (-2 + b (-2 + I1)^2 + 2 I1 - I1^2) + 
  c^2 (-3 - I1^4 + b (-3 + I1^2)^2) - 
  4 c I1 (-2 + I1^2 - I1^3 + b (6 - 3 I1 - 2 I1^2 + I1^3))) ((
 c (-1 + I1) (c - 2 I1 + c I1))/(b I1^2 (a - r)))^b < 0, c > 0, 
 c < 1, b > 1, r > a, r > 0, a > 0}, I1, Reals]

The code above cannot be "reduced" in Mathematica. Could someone tell me what I am doing wrong?

share|improve this question

closed as off-topic by MarcoB, Karsten 7., Bob Hanlon, ilian, C. E. Sep 29 '15 at 19:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Karsten 7., Bob Hanlon, ilian, C. E.
If this question can be reworded to fit the rules in the help center, please edit the question.

    
it seems that your first term is a product $ A B^b<0 $ with $ A=c^2 \left(b \left(\text{I1}^2-3\right)^2-\text{I1}^4-3\right)-4 c \text{I1} \left(b \left(\text{I1}^3-2 \text{I1}^2-3 \text{I1}+6\right)-\text{I1}^3+\text{I1}^2-2\right)+4 \text{I1}^2 \left(b (\text{I1}-2)^2-\text{I1}^2+2 \text{I1}-2\right), $ $ B=\left(\frac{c (\text{I1}-1) (c \text{I1}+c-2 \text{I1})}{b \text{I1}^2 (a-r)}\right)^b $, $ A,B \neq 0, $ so why not formulate the problem into two: $ A<0, B>0 $ and $ A>0, B<0, n\; \text{odd}. $ – gpap Sep 20 '13 at 11:09
    
even if I do that I still have the same reduction problem. – bubbly Sep 20 '13 at 11:24
1  
I don't think you do. Try With[{A = (4 I1^2 (-2 + b (-2 + I1)^2 + 2 I1 - I1^2) + c^2 (-3 - I1^4 + b (-3 + I1^2)^2) - 4 c I1 (-2 + I1^2 - I1^3 + b (6 - 3 I1 - 2 I1^2 + I1^3))), B = ((c (-1 + I1) (c - 2 I1 + c I1))/(b I1^2 (a - r)))}, Reduce[{A < 0, B > 0, 0 < c < 1, b > 1, r > a, r > 0, a > 0}, I1, Reals]] (valid for any $ b>1$), and replace A > 0, B < 0 which will be valid for only odd bs. – gpap Sep 20 '13 at 11:30
    
Yes, that works ! Thanks! – bubbly Sep 20 '13 at 16:17