使用Python生成具有给定复根的埃尔米特级数

使用Python生成具有给定复根的埃尔米特级数

要生成具有给定复根的埃尔米特级数，请使用Python Numpy中的hermite.hermfromroots()方法。该方法返回一个一维系数数组。如果所有根是实根，则输出数组是实数数组，如果一些根是复数，则输出是复数，即使结果中所有系数都是实数。参数roots是包含根的序列。

步骤

首先，导入所需的库−

from numpy.polynomial import hermite as H

要生成具有给定复根的 Hermite 级数，请使用 hermite.hermfromroots() 方法 −

j = complex(0,1)
print("Result...\n",H.hermfromroots((-j, j)))

获取数据类型-

print("\nType...\n",H.hermfromroots((-j, j)).dtype)

获取形状−

print("\nShape...\n",H.hermfromroots((-j, j)).shape)

示例

from numpy.polynomial import hermite as H

# To generate a Hermite series with given complex roots, use the hermite.hermfromroots() method in Python Numpy.
# The method returns a 1-D array of coefficients. If all roots are real then out is a real array, if some of the roots are complex, then out is complex even if all the coefficients in the result are real.

# The parameter roots are the sequence containing the roots.
j = complex(0,1)
print("Result...\n",H.hermfromroots((-j, j)))

# Get the datatype
print("\nType...\n",H.hermfromroots((-j, j)).dtype)

# Get the shape
print("\nShape...\n",H.hermfromroots((-j, j)).shape)

输出

Result...
   [1.5 +0.j 0. +0.j 0.25+0.j]

Type...
complex128

Shape...
(3,)