好的，所以我一直在尝试编码一种“天真的”方法来计算复杂形式的标准傅里叶级数的系数。我想我离得很近，但是有一些奇怪的行为。这可能比编程问题更像是一个数学问题，但是我在math.stackexchange上already asked并得到了零答案。这是我的工作代码：
import matplotlib.pyplot as plt
import numpy as np
def coefficients(fn, dx, m, L):
"""
Calculate the complex form fourier series coefficients for the first M
waves.
:param fn: function to sample
:param dx: sampling frequency
:param m: number of waves to compute
:param L: We are solving on the interval [-L, L]
:return: an array containing M Fourier coefficients c_m
"""
N = 2*L / dx
coeffs = np.zeros(m, dtype=np.complex_)
xk = np.arange(-L, L + dx, dx)
# Calculate the coefficients for each wave
for mi in range(m):
coeffs[mi] = 1/N * sum(fn(xk)*np.exp(-1j * mi * np.pi * xk / L))
return coeffs
def fourier_graph(range, L, c_coef, function=None, plot=True, err_plot=False):
"""
Given a range to plot and an array of complex fourier series coefficients,
this function plots the representation.
:param range: the x-axis values to plot
:param c_coef: the complex fourier coefficients, calculated by coefficients()
:param plot: Default True. Plot the fourier representation
:param function: For calculating relative error, provide function definition
:param err_plot: relative error plotted. requires a function to compare solution to
:return: the fourier series values for the given range
"""
# Number of coefficients to sum over
w = len(c_coef)
# Initialize solution array
s = np.zeros(len(range))
for i, ix in enumerate(range):
for iw in np.arange(w):
s[i] += c_coef[iw] * np.exp(1j * iw * np.pi * ix / L)
# If a plot is desired:
if plot:
plt.suptitle("Fourier Series Plot")
plt.xlabel(r"$t$")
plt.ylabel(r"$f(x)$")
plt.plot(range, s, label="Fourier Series")
if err_plot:
plt.plot(range, function(range), label="Actual Solution")
plt.legend()
plt.show()
# If error plot is desired:
if err_plot:
err = abs(function(range) - s) / function(range)
plt.suptitle("Plot of Relative Error")
plt.xlabel("Steps")
plt.ylabel("Relative Error")
plt.plot(range, err)
plt.show()
return s
if __name__ == '__main__':
# Assuming the interval [-l, l] apply discrete fourier transform:
# number of waves to sum
wvs = 50
# step size for calculating c_m coefficients (trap rule)
deltax = .025 * np.pi
# length of interval for Fourier Series is 2*l
l = 2 * np.pi
c_m = coefficients(np.exp, deltax, wvs, l)
# The x range we would like to interpolate function values
x = np.arange(-l, l, .01)
sol = fourier_graph(x, l, c_m, np.exp, err_plot=True)
现在，每个系数乘以2 / N的系数。但是，我在我教授的打字笔记中得到了这一和的推导，其中不包括2 / N的因数。当我导出the form myself时，我得出的公式的系数是1 / N，无论我尝试了什么技巧，都不能取消。我在math.stackexchange上问过发生了什么，但没有得到答案。
我确实注意到的是，添加1 / N会实际减少实际解和傅立叶级数之间的差异，但这仍然不正确。所以我尝试了2 / N并获得更好的结果。我真的想弄清楚这一点，所以在尝试学习快速傅立叶变换之前，我可以为基本的傅立叶级数写一个漂亮，干净的算法。
那我在做什么错呢？
最佳答案
假设c_n由A_n给出，如mathworld
同上c_n = 1/T \int_{-T/2}^{T/2}f(x)e^{-2ipinx/T}dx
我们可以解析地计算coeffs c_n（这是与梯形积分进行比较的好方法）
k = (1-2in)/2
c_n = 1/(4*pi*k)*(e^{2pik} - e^{-2pik})
因此，您的系数可能正确计算了（两条错误曲线看起来都一样）
现在请注意，重构
f时，您将coeff c_0最多添加到c_m但是重构应该在
c_{-m}到c_m时进行因此，您缺少一半的系数。
修正后的系数函数和理论系数
import matplotlib.pyplot as plt
import numpy as np
def coefficients(fn, dx, m, L):
"""
Calculate the complex form fourier series coefficients for the first M
waves.
:param fn: function to sample
:param dx: sampling frequency
:param m: number of waves to compute
:param L: We are solving on the interval [-L, L]
:return: an array containing M Fourier coefficients c_m
"""
N = 2*L / dx
coeffs = np.zeros(m, dtype=np.complex_)
xk = np.arange(-L, L + dx, dx)
# Calculate the coefficients for each wave
for mi in range(m):
n = mi - m/2
coeffs[mi] = 1/N * sum(fn(xk)*np.exp(-1j * n * np.pi * xk / L))
return coeffs
def fourier_graph(range, L, c_coef, ref, function=None, plot=True, err_plot=False):
"""
Given a range to plot and an array of complex fourier series coefficients,
this function plots the representation.
:param range: the x-axis values to plot
:param c_coef: the complex fourier coefficients, calculated by coefficients()
:param plot: Default True. Plot the fourier representation
:param function: For calculating relative error, provide function definition
:param err_plot: relative error plotted. requires a function to compare solution to
:return: the fourier series values for the given range
"""
# Number of coefficients to sum over
w = len(c_coef)
# Initialize solution array
s = np.zeros(len(range), dtype=complex)
t = np.zeros(len(range), dtype=complex)
for i, ix in enumerate(range):
for iw in np.arange(w):
n = iw - w/2
s[i] += c_coef[iw] * (np.exp(1j * n * ix * 2 * np.pi / L))
t[i] += ref[iw] * (np.exp(1j * n * ix * 2 * np.pi / L))
# If a plot is desired:
if plot:
plt.suptitle("Fourier Series Plot")
plt.xlabel(r"$t$")
plt.ylabel(r"$f(x)$")
plt.plot(range, s, label="Fourier Series")
plt.plot(range, t, label="expected Solution")
plt.legend()
if err_plot:
plt.plot(range, function(range), label="Actual Solution")
plt.legend()
plt.show()
return s
def ref_coefficients(m):
"""
Calculate the complex form fourier series coefficients for the first M
waves.
:param fn: function to sample
:param dx: sampling frequency
:param m: number of waves to compute
:param L: We are solving on the interval [-L, L]
:return: an array containing M Fourier coefficients c_m
"""
coeffs = np.zeros(m, dtype=np.complex_)
# Calculate the coefficients for each wave
for iw in range(m):
n = iw - m/2
k = (1-(1j*n)/2)
coeffs[iw] = 1/(4*np.pi*k)* (np.exp(2*np.pi*k) - np.exp(-2*np.pi*k))
return coeffs
if __name__ == '__main__':
# Assuming the interval [-l, l] apply discrete fourier transform:
# number of waves to sum
wvs = 50
# step size for calculating c_m coefficients (trap rule)
deltax = .025 * np.pi
# length of interval for Fourier Series is 2*l
l = 2 * np.pi
c_m = coefficients(np.exp, deltax, wvs, l)
# The x range we would like to interpolate function values
x = np.arange(-l, l, .01)
ref = ref_coefficients(wvs)
sol = fourier_graph(x, 2*l, c_m, ref, np.exp, err_plot=True)
关于python - 寻找傅立叶系数算法，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/58808303/