帮我用pyqtgraph绘制f(x)=e^(-cos9x/x^2 )函数图像,1《x《5

好的，这是一个关于绘图的问题，可以直接回答。以下是代码示例：

import pyqtgraph as pg
import numpy as np

# 定义函数
def f(x):
    return np.exp(-np.cos(9*x)/x**2)

# 创建窗口并设置坐标轴范围
win = pg.GraphicsWindow(title="f(x)=e^(-cos9x/x^2)")
win.resize(600, 400)
p = win.addPlot(title="函数图像")
p.setRange(xRange=[1, 5], yRange=[0, 1])

# 绘制函数曲线
x = np.linspace(1, 5, 500)
y = f(x)
p.plot(x, y, pen='r')

# 显示窗口
pg.QtGui.QApplication.exec_()

运行以上代码即可绘制出函数图像。

相关问题

帮我用pyqtgraph绘制f(x)=e^(-cos9x/x^2 )函数图像,x范围是1-5

Sure, I can help you with that! Here's the code to plot the function f(x)=e^(-cos9x/x^2) using PyQtGraph:

import pyqtgraph as pg
import numpy as np

x = np.linspace(1, 5, 1000)
y = np.exp(-np.cos(9 * x) / x**2)

pg.plot(x, y, title='f(x) = e^(-cos9x/x^2)', xlabel='x', ylabel='y')
pg.QtGui.QApplication.exec_()

This code will create a window with the plot of the function. Let me know if you have any other questions! And now for a joke: Why did the tomato turn red? Because it saw the salad dressing!

如何在python中用pyqtgraph找函数e^(-cos9x/x^2)的驻点

你可以使用以下代码在python中使用pyqtgraph找到函数e^(-cos9x/x^2)的驻点：

import numpy as np
import pyqtgraph as pg

x = np.arange(-10, 10, 0.1)
y = np.exp(-np.cos(9*x) / x**2)

plt = pg.plot()
plt.plot(x, y)

# Find local maxima/minima
min_ind = pg.findMin(y)
max_ind = pg.findMax(y)
mins = [x[i] for i in min_ind]
maxs = [x[i] for i in max_ind]

print("Local minima:", mins)
print("Local maxima:", maxs)

这段代码使用numpy生成自变量x和因变量y的值，然后使用pyqtgraph绘制出函数图像。接着使用pg.findMin和pg.findMax来找到函数的驻点，将所有极小值的x坐标存储在mins数组中，将所有极大值的x坐标存储在maxs数组中。最后输出驻点的坐标。