I want to compute the first 3 Taylor series for the function:
f(x) = e^{-x^2} \sin( x - \pi).
I write like this. But got an error message: cannot convert expression to float. How can I fix this?
import sympy
import math
sympy.init_printing(pretty_print=True)
x = sympy.Symbol('x')
f = math.exp(-x**2)*sin(x-(math.pi))
f.series(x0=math.pi, n=3)
You're using math methods instead of sympy methods. Try:
f = sympy.exp(-x**2)*sympy.sin(x-(sympy.pi))
Related
I am trying to apart Exponential function in python.
import sympy as sym
from sympy.abc import t
from sympy import exp
u = (3*(exp(4*t) - 1)*exp(-4*t))/4
apart = sym.apart(u, t)
print(apart)
But i get the error:
exp(4*t) contains an element of the set of generators
it looks like exp() is confusing it. For a workaround
import sympy as sym
from sympy.abc import t,z
from sympy import exp
u = (3*(exp(4*t) - 1)*exp(-4*t))/4
expr = sym.apart(u.subs(exp(t),z), z)
expr = expr.subs(z,exp(t))
Which gives
Out[3]: 3/4 - 3*exp(-4*t)/4
Using 3.7 on conda
Your expression is a univariate, rational function in terms of exp(t):
>>> u.subs(exp(t),y)
3*(y**4 - 1)/(4*y**4)
>>> apart(_)
3/4 - 3/(4*y**4)
>>> _.subs(y, exp(t))
3/4 - 3*exp(-4*t)/4
But SymPy can handle such non-symbol generators so for such an expression sym.apart(u) would have given the same result as shown above. When you said the generator was t it detected the exp(t) and raised the error since an expression like t + exp(t) has two generators that depend on t.
I am using python and html to build a webpage where I can enter a math equation in a textbox and compute the math equation like google calculator.
Are there any libraries or suggestions as to how I may go about doing this?
You're looking for sympy
from sympy import sympify, Symbol
>>> sympify('2 ^ 3').evalf()
8
>>> x = Symbol('x')
>>> sympify('x ^ 2').evalf(subs={x: 2})
4
I recently started learning IPython. I have a problem with Sympy library. It only prints the function instead of exact value.
import matplotlib.pyplot as plt
import numpy as nm
from sympy import *
x,y = symbols("x y")
expr = cos(x)
a = expr.subs(x,1)
print(a)
The output of above program is cos(1). But I want to print the exact value of cos(1).
First of all some nitpicking: “cos(1)” is the most concise way to represent the exact value of cos(1). Numerical representations like “0.54”, “0.540302”, or “0.54030230587” are not exact but are only accurate within the respective number of digits. The fact that cos(1) is not simply translated to one of those numerical approximations is one of the features of SymPy, as it ensures that something like simplify(acos(cos(1))) actually yields 1 and not 0.99999999.
That being said, there are of course purposes where you really need a numerical representation. In this case, you can use the class function n of SymPy expressions or the function N from SymPy, for example:
from sympy import cos, N
expr = cos(1)
print(expr.n()) # 0.540302305868140
print(N(expr)) # 0.540302305868140
I find myself wanting to use approxmations provided as part of the mpmath package, but getting confused on exactly what they are supposed to be doing:
http://docs.sympy.org/dev/modules/mpmath/calculus/approximation.html
What exactly is the difference between a sympy expression and a sympy.mpmath expression ?
If I want a taylor approximation to a symbolic expression without understanding what mpmath package is doing I can do the following:
#Imports
import sympy
import sympy.parsing
import sympy.parsing.sympy_parser
import Library_TaylorApproximation
#Create a sympy expression to approximate
ExampleStringExpression = 'sin(x)'
ExampleSympyExpression = sympy.parsing.sympy_parser.parse_expr(ExampleStringExpression)
#Create a taylor expantion sympy expression around the point x=0
SympyTaylorApproximation = sympy.series(
ExampleSympyExpression,
sympy.Symbol('x'),
1,
4,
).removeO()
#Cast the sympy expressions to python functions which can be evaluated:
VariableNames = [str(var) for var in SympyTaylorApproximation.free_symbols]
PythonFunctionOriginal = sympy.lambdify(VariableNames, ExampleSympyExpression)
PythonFunctionApproximation = sympy.lambdify(VariableNames, SympyTaylorApproximation)
#Evaluate the approximation and the original at a point:
print PythonFunctionOriginal(2)
print PythonFunctionApproximation(2)
#>>> 0.909297426826
#>>> 0.870987413961
However, if I try to do the same thing with mpmath based on the documentation:
TaylorCoefficients = sympy.mpmath.taylor(ExampleSympyExpression, 1, 4 )
print 'TaylorCoefficients', TaylorCoefficients
#>>> TypeError: 'sin' object is not callable
I can try to cram the python function in there (which is callable):
TaylorCoefficients = sympy.mpmath.taylor(PythonFunctionOriginal, 1, 4 )
print 'TaylorCoefficients', TaylorCoefficients
#>>> TaylorCoefficients [mpf('0.8414709848078965'), mpf('0.0'), mpf('0.0'), mpf('0.0'), mpf('-8.3694689805155739e+57')]
But the above does not make any sense, because I know that derivatives cannot be taken of a python function.
I can call the mpmath function sin:
TaylorCoefficients = sympy.mpmath.taylor(sympy.mpmath.sin, 1, 4 )
print 'TaylorCoefficients', TaylorCoefficients
#>>> TaylorCoefficients [mpf('0.8414709848078965'), mpf('0.54030230586813977'), mpf('-0.42073549240394825'), mpf('-0.090050384311356632'), mpf('0.035061291033662352')]
But then I cannot do manipulations on it the way I would want too -> like If I want
SinTimesCos = sympy.mpmath.sin*sympy.mpmath.cos
TaylorCoefficients = sympy.mpmath.taylor(SinTimesCos, 1, 4 )
print 'TaylorCoefficients', TaylorCoefficients
#>>> TypeError: unsupported operand type(s) for *: 'function' and 'function'
Exactly WHAT is an mpmath function ?
It is not a sympy expression, and it is also not a python function. How do I do manipulations on arbitrary expressions?
It would appear that I cannot take approximations of arbitrary sympy expressions in the documentation.
http://docs.sympy.org/dev/modules/mpmath/calculus/approximation.html
How do I take arbitrary approximations ( Pade / Cheby Chev / Fourier )
to arbitrary sympy expressions?
EDIT:
So an example of what I am looking for is the following approximation:
#Start with a sympy expression of (a, b, x)
expressionString = 'cos(a*x)*sin(b*x)*(x**2)'
expressionSympy = sympy.parsing.sympy_parser.parse_expr(expressionString)
#Do not want to decide on value of `a or b` in advance.
#Do want approximation with respect to x:
wantedSympyExpression = SympyChebyChev( expressionSympy, sympy.Symbol('x') )
Result could either be a list of coefficient expressions that are functions of a, and b:
wantedSympyExpressionCoefficients = [ Coef0Expression(a,b), Coef1Expression(a,b), ... , CoefNExpression(a,b)]
OR the result could be the entire sympy expression itself (which is itself a function of a, b):
wantedSympyExpression = Coef0Expression(a,b) + Coef1Expression(a,b) *(x**2) + ... + CoefNExpression(a,b) (x**N)
Note that a and b are not chosen in advance of performing the approximation.
mpmath functions are ordinary Python functions. They simply do their math in arbitrary-precision arithmetic.
But the above does not make any sense, because I know that derivatives cannot be taken of a python function.
You can't take the derivative symbolically, but you can compute an approximation of the derivative by evaluating the function several times and using numerical differentiation techniques. This is what sympy.mpmath.taylor does. Quoting the docs:
The coefficients are computed using high-order numerical differentiation. The function must be possible to evaluate to arbitrary precision.
If you have a SymPy expression and want to evaluate it to arbitrary precision, use evalf, like
sympy.sin(1).evalf(100)
You can use sin(x).evalf(100, subs={x:1}) to replace the x with 1 before evaluating. evalf uses mpmath under the hood, so this will give you the same result that mpmath would, but without having to use mpmath directly.
I apologize if this has been asked already.
I am just learning about SymPy and I'm wondering why it won't spit out a correct answer for what seems to be a simple equation.
from sympy.solvers import solve
from sympy import Symbol, simplify
from sympy.abc import x, alpha, sigma
alpha = Symbol('alpha')
x = Symbol('x')
sigma = Symbol('sigma')
solve((alpha - 0.5*(sigma**2))*((alpha + 0.5*(sigma**2)))**(-1)+ (1/7),sigma**2, simplify = True)
It spits out [2.0* alpha], which I know is incorrect. In fact, the answer should be [2.6666*alpha] or something like that. I'm assuming that SymPy is for some reason converting the number 2.666 to an integer string.
How can I fix this problem? Also, is there any way I could get the fractional form of the solution?
You're probably using Python 2.7, so 1/7 is giving you integer division:
>>> 1/7
0
>>> 1./7
0.14285714285714285
>>> solve((alpha - 0.5*(sigma**2))*((alpha + 0.5*(sigma**2)))**(-1)+ (1/7),sigma**2, simplify = True)
[2.0*alpha]
>>> solve((alpha - 0.5*(sigma**2))*((alpha + 0.5*(sigma**2)))**(-1)+ (1./7),sigma**2, simplify = True)
[2.66666666666667*alpha]
If you want the fractional answer, maybe something like
>>> from sympy import Rational
>>> solve((alpha - (sigma**2)/2)*((alpha + (sigma**2)/2))**(-1)+ Rational(1,7),sigma**2, simplify = True)
[8*alpha/3]
You can also use help(solve) to read the docstring of solve that tells how to use the rational keyword:
>>> solve(x-.3)
[0.300000000000000]
>>> solve(x-.3, rational=True)
[3/10]