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Chapter 6: Problem 15

Consider the degree-six polynomial $$ P(x)=924 x^{6}-2772 x^{5}+3150 x^{4}-1680 x^{3}+420 x^{2}-42 x+1 $$ There is no general formula for the roots of a polynomial of degree six, but one can find them easily enough using a computer. a) Make a plot of \(P(x)\) from \(x=0\) to \(x=1\) and by inspecting it find rough values for the six roots of the polynomial-the points at which the function is zero. b) Write a Python program to solve for the positions of all six roots to at least ten decimal places of accuracy, using Newton's method. Note that the polynomial in this example is just the sixth Legendre polynomial (mapped onto the interval from zero to one), so the calculation performed here is the same as finding the integration points for 6-point Gaussian quadrature (see Section 5.6.2), and indeed Newton's method is the method of choice for calculating Gaussian quadrature points.

Short Answer

Expert verified
Plot polynomial and use initial guesses from the plot in Newton's method implemented in Python to find the roots accurately.

Step by step solution

01

Understand the Problem

We need to find the roots of the given degree-six polynomial by first inspecting a plot of the polynomial within the interval from 0 to 1, and then using Newton's method to find the roots accurately using Python.
02

Plot the Polynomial

Use a plotting library such as Matplotlib in Python to graph the polynomial P(x) from x = 0 to x = 1. This visual representation will help identify approximate locations of the roots.
03

Write the Python Program

Write a Python script that utilizes Newton's method to find the roots of the polynomial P(x). Ensure that the roots are calculated to at least ten decimal places of accuracy.
04

Define the Polynomial Function

In Python, define the polynomial function P(x) and its derivative P'(x) for use in Newton's method: P(x) = 924*x**6 - 2772*x**5 + 3150*x**4 - 1680*x**3 + 420*x**2 - 42*x + 1P'(x) = 924*6*x**5 - 2772*5*x**4 + 3150*4*x**3 - 1680*3*x**2 + 420*2*x - 42
05

Implement Newton's Method

Create a function that applies Newton's method to find the roots within an interval. Newton's method formula is: \[ x_{n+1} = x_n - \frac{P(x_n)}{P'(x_n)} \]Use initial guesses based on the rough values observed from the plot.
06

Iterate to Find Roots

Repeat the application of Newton's method iteratively for each root, updating the guess until the difference between consecutive guesses is less than a specified tolerance (e.g., 1e-10).
07

Verify the Results

Check the accuracy of the roots obtained by substituting them back into the polynomial and ensuring that the value is close to zero. This confirms the roots are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's method
Newton's method is a powerful algorithm used to find approximate solutions to real-valued functions. It's especially useful in root-finding problems.

This iterative method starts with an initial guess and refines it to get closer to the actual root. The formula for Newton's method is:


x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}


Here, \(f(x)\) represents our function, and \(f'(x)\) represents its derivative. The algorithm proceeds by:
  • Choosing an initial guess \(x_0\)
  • Using the formula to find a new guess \(x_1\)
  • Repeating the process until the solution converges (the change between consecutive guesses is smaller than a set tolerance).

Newton's method converges quickly if the initial guess is close to the actual root, making it a popular choice in computational mathematics.
Polynomial Roots
Finding the roots of a polynomial means determining the values of \(x\) where the polynomial evaluates to zero. For a polynomial of degree \(n\), there can be up to \(n\) roots.

In this exercise, we work with a degree-six polynomial:


P(x) = 924 x^{6} - 2772 x^{5} + 3150 x^{4} - 1680 x^{3} + 420 x^{2} - 42 x + 1

Plotting is a useful first step in identifying the approximate locations of roots; the graph crosses the x-axis where the polynomial's value is zero.

Once we have rough estimates from the plot, we can refine these values using Newton's method. For accurate results, it's crucial to use an iterative method like Newton's method, which improves our estimate with each iteration, honing in on the true root to a high degree of precision.
Computational Mathematics
Computational mathematics involves using algorithms and numerical methods to solve mathematical problems that might be difficult or impossible to solve analytically.

For polynomial root-finding, analytical solutions are often feasible for polynomials of degree up to four. For higher degrees, like the degree-six polynomial in this exercise, numerical methods become essential.

Python is a popular programming language in this field because of its powerful libraries that facilitate numerical computations. Libraries such as NumPy, SciPy, and Matplotlib help in defining functions, performing complex calculations, and visualizing data respectively.

Mastering computational mathematics includes:
  • Understanding the mathematical theory behind methods like Newton's method
  • Implementing these methods in a programming language
  • Analyzing the results to ensure correctness

Practicing problems like root-finding in polynomials builds a strong foundation in computational approaches, giving insights into both the mathematical and practical implementation sides.
Python Programming in Physics
Python is an invaluable tool in computational physics due to its simplicity and robust library ecosystem. It allows physicists to quickly prototype, test, and solve complex problems.

In this exercise, Python helps us solve for the roots of our polynomial accurately using Newton's method.

A typical approach in Python involves using Matplotlib to plot the polynomial and inspect root locations visually. Then, a custom implementation of Newton's method refines these estimates.
\br> Here's a summary of the steps in our Python program:
  • Define the polynomial function \(P(x)\) and its derivative \(P'(x)\).
  • Plot the polynomial using Matplotlib to find rough root approximations.
  • Implement Newton's method to iteratively find root values within a specified tolerance.

Python's readability and ease of use make complex problems approachable, turning challenging textbook exercises into manageable steps.

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Most popular questions from this chapter

Consider a square potential well of width \(w\), with walls of height \(V\) : Using Schrödinger's equation, it can be shown that the allowed energies \(E\) of a single quantum particle of mass \(m\) trapped in the well are solutions of $$ \tan \sqrt{w^{2} m E / 2 \hbar^{2}}= \begin{cases}\sqrt{(V-E) / E} & \text { for the even numbered states, } \\ -\sqrt{E /(V-E)} & \text { for the odd numbered states, }\end{cases} $$ where the states are numbered starting from 0 , with the ground state being state 0 , the first excited state being state 1 , and so forth. a) For an electron (mass \(9.1094 \times 10^{-31} \mathrm{~kg}\) ) in a well with \(V=20 \mathrm{eV}\) and \(w=1 \mathrm{~nm}\), write a Python program to plot the three quantities $$ y_{1}=\tan \sqrt{w^{2} m E / 2 \hbar^{2}}, \quad y_{2}=\sqrt{\frac{V-E}{E}}, \quad y_{3}=-\sqrt{\frac{E}{V-E}}, $$ on the same graph, as a function of \(E\) from \(E=0\) to \(E=20 \mathrm{eV}\). From your plot make approximate estimates of the energies of the first six energy levels of the particle. b) Write a second program to calculate the values of the first six energy levels in electron volts to an accuracy of \(0.001 \mathrm{eV}\) using binary search.

The biochemical process of glycolysis, the breakdown of glucose in the body to release energy, can be modeled by the equations $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-x+n y+x^{2} y_{0} \quad \frac{\mathrm{d} y}{\mathrm{~d} t}=b-a y-x^{2} y $$ Here \(x\) and \(y\) represent concentrations of two chemicals, ADP and F6P, and \(a\) and \(b\) are positive constants. One of the important features of nonlinear equations like these is their stationary points, meaning values of \(x\) and \(y\) at which the derivatives of both variables become zero simultaneously, so that the variables stop changing and become constant in time. Setting the derivatives to zero above, the stationary points of our glycolysis equations are solutions of $$ -x+a y+x^{2} y=0, \quad b-a y-x^{2} y=0 $$ a) Demonstrate analytically that the solution of these equations is $$ x=b, \quad y=\frac{b}{a+b^{2}} $$ b) Show that the equations can be rearranged to read $$ x=y\left(a+x^{2}\right), \quad y=\frac{b}{a+x^{2}} $$ and write a program to solve these for the stationary point using the relaxation method with \(a=1\) and \(b=2\). You should find that the method fails to converge to a solution in this case. c) Find a different way to rearrange the equations such that when you apply the relaxation method again it now converges to a fixed point and gives a solution. Verify that the solution you get agrees with part (a).

A chain of resistors Consider a long chain of resistors wired up like this: All the resistors have the same resistance \(R\). The power rail at the top is at voltage \(V_{+}=\) \(5 V\). The problem is to find the voltages \(V_{1} \ldots V_{N}\) at the internal points in the circuit. a) Using Ohm's law and the Kirchhoff current law, which says that the total net current flow out of (or into) any junction in a circuit must be zero, show that the voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

There is a magical point between the Earth and the Moon, called the \(L_{1}\) Lagrange point, at which a satellite will orbit the Earth in perfect synchrony with the Moon, staying always in between the two. This works because the inward pull of the Earth and the outward pull of the Moon combine to create exactly the needed centripetal force that keeps the satellite in its orbit. Here's the setup: a) Assuming circular orbits, and assuming that the Earth is much more massive than either the Moon or the satellite, show that the distance \(r\) from the center of the Earth to the \(L_{1}\) point satisfies $$ \frac{G M}{r^{2}}-\frac{G m}{(R-r)^{2}}=\omega^{2} r $$ where \(R\) is the distance from the Earth to the Moon, \(M\) and \(m\) are the Earth and Moon masses, \(G\) is Newton's gravitational constant, and \(\omega\) is the angular velocity of both the Moon and the satellite. b) The equation above is a degree-five polynomial equation in \(r\) (also called a quintic equation). Such equations cannot be solved exactly in closed form, but it's straightforward to solve them numerically. Write a program that uses either Newton's method or the secant method to solve for the distance \(r\) from the Earth to the \(L_{1}\) point. Compute a solution accurate to at least four significant figures. The values of the various parameters are: $$ \begin{aligned} G &=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \\\ M &=5.974 \times 10^{24} \mathrm{~kg} \\ m &=7.348 \times 10^{22} \mathrm{~kg} \\ R &=3.844 \times 10^{8} \mathrm{~m} \\ \omega &=2.662 \times 10^{-6} \mathrm{~s}^{-1} \end{aligned} $$ You will also need to choose a suitable starting value for \(r\), or two starting values if you use the secant method.

voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.
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