导数应用（一）：差分计算（导数）

1.数学背景

导数：$\frac{dy}{dx}=\frac{y(x_i)-y(x_{i-1})}{x_i-x_{x-i}}$

差分：$\frac{\Delta Y}{\Delta X}=\frac{Y_i-Y_{i-1}}{X_i-X_{i-1}}$

然而由于数据特性，大多数情况$\Delta X$ 远远无法逼近 $dx$
由此衍生出各种差分形式，但主要思想万变不离其宗

2.代码

按照惯例，先放图：

• 蓝色曲线为 $y_1=sin(x)$
• 黄色曲线为 $y_2=(sin(x){)}^{\prime }=cos(x)$
• 红色曲线为 $y_3=(sin(x){)}^{\prime \prime }=-sin(x)$

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np

# 一阶中心差分
def diff_1st_mid(series,dx):
	arr = []
	for i in range(0,len(series),1):
		if i ==0:
			arr.append((series[1] - series[0])/dx)
		elif i== len(series)-1:
			arr.append((series[len(series)-1] - series[len(series)-2])/dx)
		else:
			arr.append((series[i+1] - series[i-1])/ (2 * dx) )

	return arr

# 二阶中心差分
def diff_2nd_mid(series,dx):
	arr = []
	dx2 = dx * dx
	for i in range(0,len(series),1):
		if i ==0:
			arr.append((series[2] -2*series[1]  + series[0])/dx2)
		elif i== len(series)-1:
			arr.append((series[len(series)-1] - 2*series[len(series)-2] +series[len(series)-3])/dx2)
		else:
			arr.append((series[i+1] - 2 *series[i] +  series[i-1])/ (dx2) )

	return arr

X=np.linspace(-np.pi,np.pi,1000,endpoint=True)

Y = np.sin(X)
Diff1 = np.cos(X)

DD1 = diff_1st_mid(Y,np.pi * 2 /(len(X)))
DD2 = diff_2nd_mid(Y,np.pi * 2 /(len(X)))

plt.plot(X,Y)

plt.plot(X,DD1, color = 'yellow', linewidth=1, linestyle="--")
plt.plot(X,DD2, color = 'red', linewidth=1, linestyle="--")

print(DD1)

plt.show()