In doc of sympy http://docs.sympy.org/latest/modules/integrals/integrals.html we can read:
The manualintegrate module has functions that return the steps used (see the module docstring for more information).
but calling help(sympy.integrals.manualintegrate) we get:
Help on function manualintegrate in module sympy.integrals.manualintegrate:
manualintegrate(f, var)
manualintegrate(f, var)
Compute indefinite integral of a single variable using an algorithm that
resembles what a student would do by hand.
Unlike ``integrate``, var can only be a single symbol.
Examples
========
>>> from sympy import sin, cos, tan, exp, log, integrate
>>> from sympy.integrals.manualintegrate import manualintegrate
>>> from sympy.abc import x
>>> manualintegrate(1 / x, x)
log(x)
>>> integrate(1/x)
log(x)
>>> manualintegrate(log(x), x)
x*log(x) - x
>>> integrate(log(x))
x*log(x) - x
>>> manualintegrate(exp(x) / (1 + exp(2 * x)), x)
atan(exp(x))
>>> integrate(exp(x) / (1 + exp(2 * x)))
RootSum(4*_z**2 + 1, Lambda(_i, _i*log(2*_i + exp(x))))
>>> manualintegrate(cos(x)**4 * sin(x), x)
-cos(x)**5/5
>>> integrate(cos(x)**4 * sin(x), x)
-cos(x)**5/5
>>> manualintegrate(cos(x)**4 * sin(x)**3, x)
cos(x)**7/7 - cos(x)**5/5
>>> integrate(cos(x)**4 * sin(x)**3, x)
cos(x)**7/7 - cos(x)**5/5
>>> manualintegrate(tan(x), x)
-log(cos(x))
>>> integrate(tan(x), x)
-log(sin(x)**2 - 1)/2
See Also
========
sympy.integrals.integrals.integrate
sympy.integrals.integrals.Integral.doit
sympy.integrals.integrals.Integral
I don't see step by step solution.
You are looking at the docstring of the function manualintegrate, not of the module manualintegrate. The module is here and it says
This module also provides functionality to get the steps used to evaluate a
particular integral, in the integral_steps function. This will return
nested namedtuples representing the integration rules used.
The integral_steps function is documented thus:
Returns the steps needed to compute an integral. This function attempts to mirror what a student would do by hand as closely as possible. SymPy Gamma uses this to provide a step-by-step explanation of an integral. The code it uses to format the results of this function can be found at https://github.com/sympy/sympy_gamma/blob/master/app/logic/intsteps.py.
Unless you are using SymPy Gamma, the output of integral_steps will be hard to read. Example:
from sympy.integrals.manualintegrate import integral_steps
integral_steps(x*sin(3*x), x)
returns
PartsRule(u=x, dv=sin(3*x), v_step=URule(u_var=_u, u_func=3*x, constant=1/3, substep=ConstantTimesRule(constant=1/3, other=sin(_u), substep=TrigRule(func='sin', arg=_u, context=sin(_u), symbol=_u), context=sin(_u), symbol=_u), context=sin(3*x), symbol=x), second_step=ConstantTimesRule(constant=-1/3, other=cos(3*x), substep=URule(u_var=_u, u_func=3*x, constant=1/3, substep=ConstantTimesRule(constant=1/3, other=cos(_u), substep=TrigRule(func='cos', arg=_u, context=cos(_u), symbol=_u), context=cos(_u), symbol=_u), context=cos(3*x), symbol=x), context=-cos(3*x)/3, symbol=x), context=x*sin(3*x), symbol=x)
It's much more readable on SymPy Gamma site.
Related
I am new to sympy, and I cannot understand why the result of the following piece of code does not results in f(x)=0
from sympy import *
f = Function('f')
x = Symbol('x')
simplify(Eq(f(x)+1,1))
When SymPy rewrites x + x as 2*x that is automatic rewriting. Not everything is automatic, however, as you have seen. If you want to know what value of f(x) makes that Equality true, you can solve for it:
>>> solve(Eq(f(x) + 1, 1), f(x))
[0]
I find it strange that sympy cannot evaluate integrate(sec(x+1)**2, x) when it can evaluate integrate(sec(x)**2, x). I've restricted the domain of x just in case and I still can't evaluate the integral of `sec(x+1)**2.
x, y, z = symbols('x, y, z', real=True, positive=True)
Why does sympy struggle with this?
This is arguably a bug in SymPy but you can work around by rewriting sec:
>>> integrate(sec(x+1)**2, x)
Integral(sec(x + 1)**2, x)
>>> _.rewrite(cos) # sin or tan works too
Integral(cos(x + 1)**(-2), x)
>>> _.doit()
-2*tan(x/2 + 1/2)/(tan(x/2 + 1/2)**2 - 1)
>>> simplify(_)
tan(x + 1)
I want to solve the following equation for x with SymPy:
(Note that the equation can be simplified as mentioned in the comments, I copied it verbatim from an example in a legal document.)
According to my understanding, this translates to the following SymPy expression:
from sympy import Sum, solve
from sympy.abc import k, x
solve(350 - 18500 + Sum(182.94 * (1/(1+x)**(k/12)), (k, 1, 120)), x)
However, when I run this, the result is empty:
[]
What am I doing wrong?
solve probably shouldn't give [] but you will get better results from nsolve for this expression using a guess for x near 0:
>>> from sympy.abc import k, x
>>> from sympy import nsolve
eq = 350 - 18500 + Sum(182.94 * (1/(1+x)**(k/12)), (k, 1, 120))
>>> nsolve(eq, 0)
0.0397546543274819
>>> eq.subs(x,_).round(2)
0
I'm trying to setup sympy to calculate derivatives. When I test it with simple equation, I'm finding the same answer (equality is true between sympy calculation and my own calculation). However when I try with more complicated ones, when it doesnt work (I checked answers with wolfram alpha too).
Here is my code:
from __future__ import division
from sympy import simplify, cos, sin, expand
from sympy import *
x, y, z, t = symbols('x y z t')
k, m, n = symbols('k m n', integer=True)
f, g, h = symbols('f g h', cls=Function)
equation = (x**3*y-x*y**3)/(x**2+y**2)
equation2 = (x**4*y+4*x**2*y**3-y**5)/((x**2+y**2)**2)
pprint(equation)
print ""
pprint(equation2)
print diff(equation,x) == equation2
This is a common "gotcha" in Sympy. For creating symbolic equalities, you should use sympy.Eq and not = or == (see the tutorial). For your example,
Eq(equation.diff(x), equation2).simplify()
True
Note, as above, that you may have to call simplify() in order to see wheather the Eq object corresponds to True or False
I want to take the derivative of the following function
y=(np.log(x))/(1+x),
if I am using sympy, it gives me the following error
from sympy import *
y1=Derivative((np.log(x))/(1+x), x)
print y1
sequence too large; cannot be greater than 32
Do it like this:
>>> from sympy import *
>>> var('x')
x
>>> y1 = diff(log(x)/(1+x))
>>> y1
-log(x)/(x + 1)**2 + 1/(x*(x + 1))
As Sanjeev mentioned in a comment you need to define variables in one way or another.
np.log in your code would be a function that accepts a numerical value and returns a numerical value; sympy needs to see a function names that it knows in formal terms, such as log
In this context, you need to use sympy's diff function, rather than Derivative.