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I'm writing a program to compute an exact differential for my physics laboratory. I know that I can set real domain or positive (from sympy import *):
x, y, z = symbol('x y z', positive = True)
My problem is to specify domain for example n>1. Is it possible?
In my output I'm getting an expresion like |n^2-1| and with setting this domain n>1 I would accept output like n^2-1 (without absolute value "||")
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SymPy variables are objects of Symbols class. Symbol() function's argument is a string containing symbol which can be assigned to a variable. A symbol may be of more than one alphabets. SymPy also has a Symbols() function that can define multiple symbols at once.
Note that by default in SymPy the base of the natural logarithm is E (capital E ). That is, exp(x) is the same as E**x .
The assignment operator, denoted by the “=” symbol, is the operator that is used to assign values to variables in Python. The line x=1 takes the known value, 1, and assigns that value to the variable with name “x”. After executing this line, this number will be stored into this variable.
For assumptions on symbols, you can use positive or negative:
p = Symbol('p', positive=True)
But this can only define p>0 (or p<0 if you use negative=True).
For more complex expression refinement, try refine(expression, assumption):
In [1]: n = Symbol('n')
In [2]: refine(Abs(n-1), Q.positive(n-1))
Out[2]: n - 1
In [3]: refine(Abs(n-1))
Out[3]: │n - 1│
That is, you create the assumption Q.positive(n-1), that is n > 1, and pass it to refine.
There is currently work in progress to port this assumption style to other algorithms, but support is still incomplete (simplify appears not to recognize this kind of assumption).
It is expected that support of Q.statement( ... ) will be extended in future versions of SymPy, as there is currently a lot of work in progress on this.
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