In [1]:
import sympy as sym
import numpy as np
from IPython.display import display,Math
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
In [2]:
print('Critical points: 極値')
display(Math('f(x) = -x^4 + 3x^2'))
display(Math('\\frac{df}{dx} = -4x^3 + 6x'))
display(Math('-4x^3 + 6x = 0'))
display(Math('x = 0, \\pm \\sqrt{\\frac{3}{2}}'))

Critical points: 極値
$\displaystyle f(x) = -x^4 + 3x^2$
$\displaystyle \frac{df}{dx} = -4x^3 + 6x$
$\displaystyle -4x^3 + 6x = 0$
$\displaystyle x = 0, \pm \sqrt{\frac{3}{2}}$

empirical method

In [3]:
x = np.linspace(-5,5,1001)
fx = x**2 * np.exp(-x**2)
dfx = np.diff(fx)/(x[1]-x[0]) # df/dx (differential of x)
# print(np.diff(fx))
print(x[1]-x[0])

localmax = find_peaks(fx)[0]
print(localmax)
localmin = find_peaks(-fx)[0]
print(localmin)

print('The critical points are ' + str(x[localmax]) + ' ' + str(x[localmin]))

plt.plot(x,fx)
plt.plot(x[0:-1],dfx)

plt.plot(x[localmax],fx[localmax],'ro')
plt.plot(x[localmin],fx[localmin],'gs')
plt.show()

0.009999999999999787
[400 600]
[500]
The critical points are [-1.  1.] [0.]

analytic method / symbolic methos

In [4]:
x = sym.symbols('x')
fx = x**2 * sym.exp(-x**2)

dfx = sym.diff(fx)
print(dfx)
critpoints = sym.solve(dfx)
print(critpoints)

-2*x**3*exp(-x**2) + 2*x*exp(-x**2)
[-1, 0, 1]