How to replace log(x) by log(abs(x)) automatically in SymPy - sympy

I am trying to solve differential equation y'=cot(x) using Sympy:
x = symbols("x")
y = Function("y")(x)
dsolve(diff(y, x) - cot(x))
It gives me solution with logarithm: y(x) = C1 + log(sin(x)).
How I can replace in this expression log by combination log + abs: log(sin(x)) -> log(abs(sin(x)))? I can do it manually, but it's not very convenient. Is there any way to do it with SymPy tools (something like subs command)?

Note that the abs isn't actually needed: the integration constant C1 can include I*pi. The form without abs is valid for all complex x whereas the abs form is only valid for real x.
In any case, you can do it like this:
In [6]: sol
Out[6]: y(x) = C₁ + log(sin(x))
In [7]: sol.replace(log, lambda e: log(abs(e)))
Out[7]: y(x) = C₁ + log(│sin(x)│)

Related

sympy substitution of occurences satisfying a given pattern

Is there a way to do a substitution in sympy of all occurences that satisfy a given pattern? For example, if I have an expression that contains multiple occurances of sqrt(anything), is there a way to substitute all those occurences with anything**0.51?
Here is a code example:
import sympy
from sympy import sqrt, exp
from sympy import *
x,y = symbols('x y')
test = sqrt (x+y) + sqrt (exp(y))
in this case I can do substitution manually:
test.subs(sqrt(x+y), (x + y)**0.51).subs(sqrt (exp(y)) , (exp(y))**0.51)
yielding the expected subsitution. But, is there a way to do it one shot so one can easily apply it to long epxressions?
You can use a Wild symbol to do pattern-matched replacements:
In [7]: w = Wild('w')
In [8]: test
Out[8]:
____
_______ ╱ y
╲╱ x + y + ╲╱ ℯ
In [9]: test.replace(sqrt(w), w**0.51)
Out[9]:
0.51
0.51 ⎛ y⎞
(x + y) + ⎝ℯ ⎠
You can use the replace method. Here is the documentation with a few examples.
In your particular case, the square root is a power with exponent 0.5, so we are going to look for those objects:
test.replace(lambda t: t.is_Pow and t.exp is S.Half, lambda t: t.base**0.51)

Sympy - Simplify expression within domain

Can Sympy automatically simplify an expression that includes terms like this one:
cos(x)/(cos(x)**2)**(1/2)
which can be simplified to 1 in the domain that I am interested in 0 <= x <= pi/2 ?
(Examples of other terms that could be simplified in that domain: acos(cos(x)); sqrt(sin(x)**2); sqrt(cos(2*x) + 1); etc.)
If you know the functions that are in your expression (such as sin, cos and tan), you can do the following according to this stack overflow question:
from sympy import *
x = symbols("x", positive=True)
ex = cos(x)/(cos(x)**2)**(S(1)/2)
ex = refine(ex, Q.positive(sin(x)))
ex = refine(ex, Q.positive(cos(x)))
ex = refine(ex, Q.positive(tan(x)))
print(ex)
Note that Q.positive(x*(pi/2-x)) did not help in the process of simplification for trig functions even though this is exactly what you want in general.
But what if you might have crazy functions like polygamma? The following works for some arbitrary choices for ex according to my understanding.
It wouldn't be a problem if the expression was already generated before by SymPy, but if you are inputting the expression manually, I suggest using S(1)/2 or Rational(1, 2) to describe one half.
from sympy import *
# define everything as it would have come from previous code
# also define another variable y to be positive
x, y = symbols("x y", positive=True)
ex = cos(x)/(cos(x)**2)**(S(1)/2)
# If you can, always try to use S(1) or Rational(1, 2)
# if you are defining fractions.
# If it's already a pre-calculated variable in sympy,
# it will already understand it as a half, and you
# wouldn't have any problems.
# ex = cos(x)/(cos(x)**2)**(S(1)/2)
# if x = arctan(y) and both are positive,
# then we have implicitly that 0 < x < pi/2
ex = simplify(ex.replace(x, atan(y)))
# revert back to old variable x if x is still present
ex = simplify(ex.replace(y, tan(x)))
print(ex)
This trick can also be used to define other ranges. For example, if you wanted 1 < x, then you could have x = exp(y) where y = Symbol("y", positive=True).
I think subs() will also work instead of replace() but I just like to be forceful with substitutions, since SymPy can sometimes ignore the subs() command for some variable types like lists and stuff.
You can substitute for a symbol that has the assumptions you want:
In [27]: e = cos(x)/(cos(x)**2)**(S(1)/2) + cos(x)
In [28]: e
Out[28]:
cos(x)
cos(x) + ────────────
_________
╱ 2
╲╱ cos (x)
In [29]: cosx = Dummy('cosx', positive=True)
In [30]: e.subs(cos(x), cosx).subs(cosx, cos(x))
Out[30]: cos(x) + 1

How to simplify lengthy symbolic expressons in SymPy

I have been working on some integrations and even though the system is working, it takes much more time to work than it should.
The problem is that the expressions are many pages, and even though they are 3 variables only, sy.simplify just crashes the Kernel after 4 hours or so.
Is there a way to make such lengthy expressions more compact?
EDIT:
Trying to recreate a test expression, using cse. I can't really substitute the symbols to make a final expression, equal to the 1st one
sy.var('a:c x')
testexp = sp.log(x)+a*(0.5*x)**2+(b*(0.5*x)**2+b+sp.log(x))/c
r, e = sy.cse(testexp)
FinalFunction = sy.lambdify(r[0:][0]+(a,b,c,x),e[0])
Points = sy.lambdify((a,b,c,x),r[0:][1])
FinalFunction(Points(1,1,1,1),1,1,1,1)
>>>NameError: name 'x1' is not defined
cse(expr) is sometimes a way to get a more compact representation since repeated subexpressions can be replaced with a single symbol. cse returns a list of repeated expressions and a list of expressions (a singleton if you only passed a single expression):
>>> from sympy import solve
>>> var('a:c x');solve(a*x**2+b*x+c, x)
(a, b, c, x)
[(-b + sqrt(-4*a*c + b**2))/(2*a), -(b + sqrt(-4*a*c + b**2))/(2*a)]
>>> r, e = cse(_)
>>> for i in r: pprint(Eq(*i))
...
_____________
╱ 2
x₀ = ╲╱ -4⋅a⋅c + b
1
x₁ = ───
2⋅a
>>> for i in e: pprint(i)
...
x₁⋅(-b + x₀)
-x₁⋅(b + x₀)
You are still going to have long expressions but they will be represented more compactly (and more efficiently for computatation) if cse is able to identify repeated subexpressions.
To use this in SymPy you can create two Lambdas: one to translate the variables into the replacement values and the other to use those values:
>>> v = (a,b,c,x)
>>> Pts = Lambda(v, tuple([i[1] for i in r]+list(v)))
>>> Pts(1,2,3,4)
(2*sqrt(2)*I, 1/2, 1, 2, 3, 4)
>>> Func = Lambda(tuple([i[0] for i in r]+list(v)), tuple(e))
>>> Func(*Pts(1,2,3,4))
(-1 + sqrt(2)*I, -1 - sqrt(2)*I)

Sympy gives unexpected differentiation result when the input is a string

Why is the result of the differentiation not 2*x0 in the following code:
In [54]: import sympy
In [55]: x = [sympy.Symbol('x%d' % i, real=True) for i in range(3)]
In [56]: x
Out[56]: [x0, x1, x2]
In [57]: sympy.diff('x0*x0 + x1*x1 + x2*x2',x[0])
Out[57]: 0
First, the creation of multiple numbered symbols is simpler with
x = sympy.symbols('x0:3', real=True) # returns (x0, x1, x2)
Second, the SymPy function to turn a string into a SymPy expression is sympify. This function is called automatically when you provide input as a string; however, this gives you no control over the interpretation of the string, and "unexpected results" are likely.
In this case, SymPy is not sure that "x0" appearing in the string is the same as x0 you created earlier. After all, your x0 has the additional property of being real, and the symbol from the string has no such assumptions on it. It's Symbol('x0') vs Symbol('x0', real=True); not a match.
This is one of many reasons why throwing a string in a SymPy function is a bad idea. Use sympify, and read about its parameters which control the parsing of input. Specifically, locals parameter is a dictionary mapping pieces of the string to objects you already have in SymPy, precisely what is needed here.
locals = {'x{}'.format(i): x[i] for i in range(3)} # {'x0': x0, 'x1': x1, 'x2': x2}
expr = sympy.sympify('x0*x0 + x1*x1 + x2*x2', locals=locals)
Now you can differentiate expr with respect to any symbols and get expected results
[expr.diff(sym) for sym in x] # [2*x0, 2*x1, 2*x2]
(Another benefit of having an expression before trying diff is that you can invoke diff as a method of the expression, saving the trouble of typing sympy. prefix.)
In your declarations, you should use sympy.symbols that is the reference method (from the documentation and tutorial) to declare variables.
x = [sympy.symbols('x%d' % i, real=True) for i in range(3)]
On top of this, you must pick (from experimentations that I made) either a string in both arguments, as:
sympy.diff('x0*x0 + x1*x1 + x2*x2',str(x[0]))
or symbolic expressions on both sides:
sympy.diff(x[0]*x[0] + x[1]*x[1] + x[2]*x[2], x[0])

How to extract numerator and denominator from polynomial without evaluating?

I have the following expression
A=Symbol('A')
x=Symbol('x')
B=Symbol('B')
C=Symbol('C')
D=Symbol('D')
expression=((A**x-B-C)/(D-1))*(D-1)
n,d=fraction(expression)
I am getting following result:
n=A**x-B-C
d=1
My expected result is
n=(A**x-B-C)*(D-1)
d=(D-1)
Is there way in sympy or need to write customize function to handle that
Use UnevaluatedExpr() to prevent the expression from being evaluated.
from sympy import symbols, fraction, UnevaluatedExpr
A,x,B,C,D = symbols('A x B C D')
expression = (A**x-B-C)/(D-1)*UnevaluatedExpr(D-1)
n,d = fraction(expression)
print(n)
print(d)
This returns
(A**x - B - C)*(D - 1)
D - 1
See the Sympy Advanced Expression Manipulation doc page for more details.