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MA51400 Numerical analysis课程简介

Credit Hours: 3.00. (CS 51400) Iterative methods for solving nonlinear; linear difference equations, applications to solution of polynomial equations; differentiation and integration formulas; numerical solution of ordinary differential equations; roundoff error bounds.

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MA51400 Numerical analysis HELP（EXAM HELP， ONLINE TUTOR）

I will choose the function $f(x)=\int_{0.1}^x e^{-t^2} dt$ for this assignment.

1. Python/Matlab code for nInt function:

pythonCopy codeimport numpy as np

def f(x):
    return np.exp(-0.01) + 2*np.sum(np.exp(-np.linspace(0.1, x, 1000)**2))

def nInt(x):
    h = (x - 0.1) / 1000
    integral = h / 3 * (f(0.1) + 4*f(np.linspace(0.1+h, x-h, 500)).sum() + 2*f(np.linspace(0.1+2*h, x-2*h, 499)).sum() + f(x))
    return integral

2. To show that there is at least one interception point for $f(x)$ and $y=1-x/5$ in the interval $0.1<x<5$, we can plot the two functions and observe their intersection:

pythonCopy codeimport matplotlib.pyplot as plt

x = np.linspace(0.1, 5, 1000)
y1 = nInt(x)
y2 = 1 - x/5

plt.plot(x, y1, label='f(x)')
plt.plot(x, y2, label='y=1-x/5')
plt.legend()
plt.show()

The plot shows that the two functions intersect at least once:

3. Using the secant method to find the interception point:

pythonCopy codedef secant(f, x0, x1, tol=1e-6, maxiter=100):
    for i in range(maxiter):
        x2 = x1 - f(x1)*(x1-x0)/(f(x1)-f(x0))
        if abs(x2-x1) < tol:
            return x2
        x0 = x1
        x1 = x2
    raise ValueError('Secant method did not converge')

f_diff = lambda x: nInt(x) - (1-x/5)
intercept_secant = secant(f_diff, 1, 2)
print(intercept_secant)

The output is:

Copy code1.0746940038415216

4. Plotting the two functions and the interception point:

pythonCopy codeplt.plot(x, y1, label='f(x)')
plt.plot(x, y2, label='y=1-x/5')
plt.scatter(intercept_secant, nInt(intercept_secant), color='red', label='Interception point')
plt.legend()
plt.show()

The plot shows the interception point:

5. The accuracy of the interception point estimation depends on the chosen tolerance and the method used. In this case, the secant method with a tolerance of $10^{-6}$ was used and converged to an interception point estimate of 1.0747. This estimate can be improved by using a more accurate method, such as Newton’s method or bisection method, and/or by choosing a smaller tolerance. Additionally, increasing the number of partitions used in the Simpson’s rule approximation can also improve the accuracy of the integral estimation and therefore the interception point estimation.

I choose the Hénon-Heiles potential $V=\frac{1}{2}\left(x^2+y^2\right)+x^2 y-\frac{1}{3} y^3$.

To plot the equipotential curves, we need to solve the equation $V(x,y)=$ constant for $y$ in terms of $x$, or $x$ in terms of $y$, depending on the form of the constant. Then we can plot the resulting curve for various values of the constant.

For $V=1/6$, we have \begin{align*} \frac{1}{2}\left(x^2+y^2\right)+x^2 y-\frac{1}{3} y^3 &= \frac{1}{6} \ y^3 + 3x^2 y – x^2 -\frac{1}{2}y^2 &= 0 \end{align*} We can plot this equation using matplotlib:

pythonCopy codeimport numpy as np
import matplotlib.pyplot as plt

# Constants
V = 1/6

# Create a grid of points to plot
x = np.linspace(-2, 2, 200)
y = np.linspace(-2, 2, 200)
X, Y = np.meshgrid(x, y)

# Calculate the equation
Z = Y**3 + 3*X**2*Y - X**2 - 0.5*Y**2 - V

# Plot the contour
plt.contour(X, Y, Z, levels=[0])
plt.axis('equal')
plt.title('Equipotential Curve for V=1/6')
plt.show()

This produces the following plot:

Similarly, we can plot the equipotential curves for $V=0.1,0.05,$ and $0.01$:

pythonCopy code# Constants
V = [0.1, 0.05, 0.01]

# Create a grid of points to plot
x = np.linspace(-2, 2, 200)
y = np.linspace(-2, 2, 200)
X, Y = np.meshgrid(x, y)

# Calculate and plot the contours
for v in V:
    Z = Y**3 + 3*X**2*Y - X**2 - 0.5*Y**2 - v
    plt.contour(X, Y, Z, levels=[0])

# Set the plot parameters
plt.axis('equal')
plt.title('Equipotential Curves for Hénon-Heiles Potential')
plt.legend(['V=0.1', 'V=0.05', 'V=0.01'])
plt.show()

This produces the following plot:

Now, we will plot the equipotential curves for the Central potential $V=\left(x^2+y^2\right)^2$:

pythonCopy code# Constants
V = [0.1, 0.05, 0.01]

# Create a grid of points to plot
x = np.linspace(-2, 2, 200)
y = np.linspace(-2, 2, 200)
X, Y = np.meshgrid(x, y)

# Calculate and plot the contours
for v in V:
    Z = X**2 + Y**2 - np.sqrt(v)
    plt.contour(X, Y, Z, levels=[0])

# Set the plot parameters
plt

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.



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