definite_integral of max function [closed]
 
  
 
  
 
 
 
 
 
 
    
        
        asked 2 years ago  
 
 Max Alekseyev gravatar image  
 
 
 
 
 
The following code returns 1/2, however it can be easily verified that 
∫10max
What's wrong?
 
from sage.symbolic.integration.integral import definite_integral
var('x')
definite_integral(max(x,1-x),x,0,1)
 
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Closed for the following reason
      the question is answered, right answer was accepted by
      Max Alekseyev 

      close date 2024-05-29 02:21:27.569198
 
 
Comments
1
use max_symbolic
FrédéricC gravatar imageFrédéricC (
    
        2 years ago
    )
Thanks! max_symbolic does lead to the correct answer, but I worry that using max silently produces an incorrect answer - should Sage give an error or a warning about the issue at least?
Max Alekseyev gravatar imageMax Alekseyev (
    
        2 years ago
    )
This is a common pitfall. You could have tried to see what max(x,1-x) answers.
FrédéricC gravatar imageFrédéricC (
    
        2 years ago
    )
This is weird - both max(x,1-x) and min(x,1-x) return x. At very least this breaks the identity \max(x,1-x) + \min(x,1-x) = 1.
Max Alekseyev gravatar imageMax Alekseyev (
    
        2 years ago
    )
With symbolic arguments, max and min return the first argument. Hence:
sage: max(x,1-x) + min(x,1-x)                                                   
2*x
However, 
sage: max(x,1-x) + min(1-x,x)                                                   
1
Juanjo gravatar imageJuanjo (
    
        2 years ago
    )

            
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