sympy中的多元泰勒近似

Question

我的目标是使用sympy 编写多维泰勒近似，其中

• 使用尽可能多的内置代码，
• 计算两个变量的给定函数的截断泰勒近似
• 返回结果没有 Big-O-remainder term，例如在sin(x)=x - x**3/6 + O(x**4)。

到目前为止，这是我尝试过的：

方法 1

天真地，可以为每个变量组合两次 series 命令，但不幸的是，这不起作用，正如此示例中函数 sin(x*cos(y)) 所示：

sp.sin(x*sp.cos(y)).series(x,x0=0,n=3).series(y,x0=0,n=3)
>>> NotImplementedError: not sure of order of O(y**3) + O(x**3)

方法 2

基于this post我先写了一个一维泰勒近似：

def taylor_approximation(expr, x, max_order):
    taylor_series = expr.series(x=x, n=None)
    return sum([next(taylor_series) for i in range(max_order)])

使用 1D 示例检查它可以正常工作

mport sympy as sp
x=sp.Symbol('x')
y=sp.Symbol('y')
taylor_approximation(sp.sin(x*sp.cos(y)),x,3)

>>> x**5*cos(y)**5/120 - x**3*cos(y)**3/6 + x*cos(y)

但是，如果我知道在 x 和 y 中进行两个扩展的链式调用，sympy 就会挂断

# this does not work
taylor_approximation(taylor_approximation(sp.sin(x*sp.cos(y)),x,3),y,3)

有人知道如何解决这个问题或以其他方式实现它吗？

Answer 1

您可以使用 expr.removeO() 从表达式中删除大 O。

---

Oneliner：expr.series(x, 0, 3).removeO().series(y, 0, 3).removeO()

Answer 2

这是一个用于 Sympy 的多元泰勒级数展开：

def Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree):
    """
    Mathematical formulation reference:
    https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/3%3A_Topics_in_Partial_Derivatives/Taylor__Polynomials_of_Functions_of_Two_Variables
    :param function_expression: Sympy expression of the function
    :param variable_list: list. All variables to be approximated (to be "Taylorized")
    :param evaluation_point: list. Coordinates, where the function will be expressed
    :param degree: int. Total degree of the Taylor polynomial
    :return: Returns a Sympy expression of the Taylor series up to a given degree, of a given multivariate expression, approximated as a multivariate polynomial evaluated at the evaluation_point
    """
    from sympy import factorial, Matrix, prod
    import itertools

    n_var = len(variable_list)
    point_coordinates = [(i, j) for i, j in (zip(variable_list, evaluation_point))]  # list of tuples with variables and their evaluation_point coordinates, to later perform substitution

    deriv_orders = list(itertools.product(range(degree + 1), repeat=n_var))  # list with exponentials of the partial derivatives
    deriv_orders = [deriv_orders[i] for i in range(len(deriv_orders)) if sum(deriv_orders[i]) <= degree]  # Discarding some higher-order terms
    n_terms = len(deriv_orders)
    deriv_orders_as_input = [list(sum(list(zip(variable_list, deriv_orders[i])), ())) for i in range(n_terms)]  # Individual degree of each partial derivative, of each term

    polynomial = 0
    for i in range(n_terms):
        partial_derivatives_at_point = function_expression.diff(*deriv_orders_as_input[i]).subs(point_coordinates)  # e.g. df/(dx*dy**2)
        denominator = prod([factorial(j) for j in deriv_orders[i]])  # e.g. (1! * 2!)
        distances_powered = prod([(Matrix(variable_list) - Matrix(evaluation_point))[j] ** deriv_orders[i][j] for j in range(n_var)])  # e.g. (x-x0)*(y-y0)**2
        polynomial += partial_derivatives_at_point / denominator * distances_powered
    return polynomial

这是一个二变量问题的验证，遵循以下练习和答案：https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/3%3A_Topics_in_Partial_Derivatives/Taylor__Polynomials_of_Functions_of_Two_Variables

# Solving the exercises in section 13.7 of https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/3%3A_Topics_in_Partial_Derivatives/Taylor__Polynomials_of_Functions_of_Two_Variables
from sympy import symbols, sqrt, atan, ln

# Exercise 1
x = symbols('x')
y = symbols('y')
function_expression = x*sqrt(y)
variable_list = [x,y]
evaluation_point = [1,4]
degree=1
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))
degree=2
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))

# Exercise 3
x = symbols('x')
y = symbols('y')
function_expression = atan(x+2*y)
variable_list = [x,y]
evaluation_point = [1,0]
degree=1
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))
degree=2
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))

# Exercise 5
x = symbols('x')
y = symbols('y')
function_expression = x**2*y + y**2
variable_list = [x,y]
evaluation_point = [1,3]
degree=1
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))
degree=2
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))

# Exercise 7
x = symbols('x')
y = symbols('y')
function_expression = ln(x**2+y**2+1)
variable_list = [x,y]
evaluation_point = [0,0]
degree=1
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))
degree=2
print(Taylor_polynomial_sympy(function_expression, variable_list, evaluation_point, degree))

对结果执行simplify() 会很有用。

Answer 3

多元泰勒展开式

def mtaylor(funexpr,x,mu,order=1):

    nvars = len(x)
    hlist = ['__h' + str(i+1) for i in range(nvars)]
    command=''
    command="symbols('"+'  '.join(hlist) +"')"
    hvar = eval(command)
    #mtaylor is utaylor for specificly defined function
    t = symbols('t')
    #substitution
    loc_funexpr = funexpr
    for i in range(nvars):
        locvar = x[i]
        locsubs = mu[i]+t*hvar[i]
        loc_funexpr = loc_funexpr.subs(locvar,locsubs)
    #calculate taylorseries
    g = 0
    for i in range(order+1):
        g+=loc_funexpr.diff(t,i).subs(t,0)*t**i/math.factorial(i)

    #resubstitute
    for i in range(nvars):
        g = g.subs(hlist[i],x[i]-mu[i])

    g = g.subs(t,1)    
    return g

测试一些功能

x1,x2,x3,x4,x5 = symbols('x1 x2 x3 x4 x5')
funexpr=1+x1+x2+x1*x2+x1**3
funexpr=cos(funexpr)
x=[x1,x2,x3,x4,x5]
mu=[1,1,1,1,1]
mygee = mtaylor(funexpr,x,mu,order=4)
print(mygee)