{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "\n", "# Fixed-Point Iteration\n", "\n", "### Modules - Root Finding\n", "
\n", "By Eilif Sommer Øyre, Jonas Themsland and Jon Andreas Støvneng \n", "
\n", "\n", "Last edited: March 11th 2018\n", "\n", "---" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## Introduction\n", "\n", "An equation $f(x) = 0$ can always be written as $g(x) = x$. Fixed-point iteration can be applied to approximate the fixed number $r = g(r)$.\n", "\n", "After selecting an initial guess $x_0$, the fixed-point iteration is $x_{n+1} = g(x_n)$. That is,\n", "\n", "\\begin{equation*}\n", "x_0 = \\textrm{initial guess} \\\\\n", "x_1 = g(x_0) \\\\\n", "x_2 = g(x_1) \\\\\n", "x_3 = g(x_2) \\\\\n", "\\vdots\n", "\\end{equation*}\n", "\n", "The algorithm is repeated until either a testing condition $e_{i+1} = \\left|\\:x_{i+1} - x_{i}\\:\\right| < \\alpha$, where $\\alpha$ is some tolerance limit is met, or until a fixed number of iterations $N$ is reached. Note that the method may or may not converge. Note that the choice of $g(x)$ is in general not unique." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 1\n", "\n", "As a simple introductory example we use the fixed-point iteration to solve the equation\n", "\n", "$$\\frac{1}{2}\\sin(x) - x + 1 = 0.$$\n", "\n", "The most natural is to use $g(x)= \\sin(x)/2 + 1$ and then use the fixed-point iteration to solve $g(x)=x$." ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x1:\t1.84147\n", "x2:\t1.96359\n", "x3:\t1.92384\n", "x4:\t1.93832\n", "x5:\t1.93322\n", "x6:\t1.93504\n", "x7:\t1.93439\n", "x8:\t1.93462\n", "x9:\t1.93454\n" ] } ], "source": [ "import math\n", "\n", "x = 1 # initial guess\n", "N = 10 # iterations\n", "\n", "for i in range(1, N):\n", " x = math.sin(x) + 1\n", " print(\"x%i:\\t%.5f\"%(i, x))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Convergence\n", "\n", "For the iteration scheme to return a fixed point $r$, it needs to converge. A criterion for convergence is that the error, $e_i = \\left|\\:r - x_i \\:\\right|$, decreases for each iteration step. This means that the change from $x_i$ to $x_{i+1}$ also has to decrease for each step. The convergence criterion is explained in the following theorem :\n", "\n", "> **Theorem 1** Assume that $g$ is continuously differentiable, that $g(r) = r$, and that $S = \\left|\\:g'(r)\\:\\right| < 1$. Then fixed-point iteration converges linearly with rate $S$ to the fixed point $r$ for initial guesses sufficiently close to $r$. \n", "\n", "Note that in the example above we have $|g'(x)|=\\cos(x)/2 < 1$ which means that the method will converge for all initial guesses.\n", "\n", "The convergence can in fact be of higher order. This is explained in the following theorem :\n", "\n", "> **Theorem 2** Assume that $g$ is $p$ times continuously differentiable and that the fixed-iteration converges to $r$ for some initial guess $x_0$. If $g'(r)=g''(r)=\\cdots=g^{(p-1)}(r)=0$ and $g^{(p-1)}(r)=0$, then the order of convergence is $p$.\n", "\n", "The convergence is said to be of order $p$ when $\\lim_{i\\to\\infty}e_{i+1}/e_{i}^p = \\text{constant}$. \n", "\n", "**Excercise:** Prove theorem 1 using the mean value theorem and the theorem 2 using Taylor's theorem." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 2: Babylonian Method for Finding Square Roots\n", "\n", "As mentioned in the introduction, the choice of $g(x)$ is far from unique. We will now use fixed-point iteration to estimate the square root of a real and positive number $a$. That is, we want to solve $f(x) = x^2-a^2=0$. Two natural choices for $g(x)$ are $g_1(x)=a/x$ and $g_2(x)=x$. However, none of these converges since $|g_1'(\\sqrt a)| = |g_2'(\\sqrt a)| =1$. We can choose the mean as $g(x)$:\n", "\n", "$$g(x) = \\frac{1}{2}\\left(\\frac{a}{x} + x\\right).$$\n", "\n", "In this case we have \n", "\n", "$$\\left|\\:g'(x)\\:\\right|_{x=\\sqrt a}=\\left|\\:\\frac{1}{2}(1-\\frac{a}{x^2})\\:\\right|_{x=\\sqrt a} = 0 < 1$$\n", "\n", "The theorem above thus implies that method converges for some initial guess $x_0$. The method is in fact globally convergent due to the convexity of $f(x)$." ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x1:\t0.53500\n", "x2:\t0.33292\n", "x3:\t0.27159\n", "x4:\t0.26467\n", "x5:\t0.26458\n", "x6:\t0.26458\n", "x7:\t0.26458\n", "x8:\t0.26458\n", "x9:\t0.26458\n" ] } ], "source": [ "a = 0.07 # square of root\n", "x = 1 # initial guess\n", "N = 10 # iterations\n", "\n", "for i in range(1, N):\n", " x = (a/x + x)/2\n", " print(\"x%i:\\t%.5f\"%(i, x))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "This method can also be derived from Newton's method (see the section on Newton's method below). The method was however first used by the Babylonian people long before Newton ." ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## Example 3: Adding Stopping Conditions\n", "\n", "Underneath follows an implementation of the Fixed-Point Iteration with a testing condition $\\left|\\:x_{i+1} - x_i\\:\\right| < \\alpha$. There are other stopping criteria that may be relevant such as the backward error $\\left|\\:f(x_a)- f(0)\\right|$. We need an additional stopping criteria in case convergence fails. We therefore stop the computation if the a given number of iterations $N$ is reached." ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": true }, "outputs": [], "source": [ "def FPI(g, x, maxit=100, alpha=1e-5):\n", " \"\"\" A function using fixed-point iteration to compute \n", " an approximate solution to the fixed point problem g(r) = r.\n", " \n", " Arguments:\n", " g callable function, g(x)\n", " x float. The initial guess\n", " alpha float. Error tolerance\n", " maxit int. Maximum number of iterations\n", " \n", " Returns:\n", " array of each fixed-point approximation\n", " \"\"\"\n", " \n", " result = [x]\n", " x_next = g(x)\n", " i = 0 # Counter\n", " e = abs(x_next - x)\n", " \n", " while e > alpha :\n", " if i > maxit:\n", " print(\"No convergence.\")\n", " break\n", " x = x_next\n", " x_next = g(x_next)\n", " result.append(x)\n", " if x - x_next > e:\n", " print(\"Divergence.\")\n", " break\n", " e = abs(x_next - x)\n", " i += 1\n", " return result" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We want to solve the equation,\n", "\n", "\\begin{equation*}\n", "f(x) = x^4 + x - 1 = 0\n", "%\\label{eq:example}\n", "\\end{equation*}\n", "\n", "using fixed point iteration. First, we rewrite the equation to the form, $x = g(x)$. This particular equation may be rewritten in three ways:\n", "\n", "$$g_1(x) = x = 1 - x^4,$$\n", "\n", "$$g_2(x) = x = \\sqrt{1 - x},$$\n", "\n", "$$g_3(x) = x = \\frac{1 + x^4}{1 + 2x^3}.$$\n", "\n", "Each of these has its own convergence rate. One of the solutions is clearly somewhere between 0 and 1, so we will use $x_0=0.5$ as initial guess." ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": true }, "outputs": [], "source": [ "x_0 = 0.5\n", "N = 100\n", "alpha = 1e-5\n", "\n", "def g1(x): return 1 - x**4\n", "def g2(x): return (1 - x)**(1/4)\n", "def g3(x): return (1 + 3*x**4)/(1 + 4*x**3)\n", "\n", "def print_result(result):\n", " for i in range(len(result)):\n", " print(\"x%i:\\t%.5f\"%(i, result[i]))" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Divergence.\n", "x0:\t0.50000\n", "x1:\t0.93750\n" ] } ], "source": [ "result_g1 = FPI(g1, x_0, N, alpha)\n", "print_result(result_g1)" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x0:\t0.50000\n", "x1:\t0.84090\n", "x2:\t0.63157\n", "x3:\t0.77909\n", "x4:\t0.68557\n", "x5:\t0.74883\n", "x6:\t0.70794\n", "x7:\t0.73514\n", "x8:\t0.71739\n", "x9:\t0.72912\n", "x10:\t0.72143\n", "x11:\t0.72650\n", "x12:\t0.72317\n", "x13:\t0.72536\n", "x14:\t0.72392\n", "x15:\t0.72487\n", "x16:\t0.72425\n", "x17:\t0.72465\n", "x18:\t0.72439\n", "x19:\t0.72456\n", "x20:\t0.72445\n", "x21:\t0.72452\n", "x22:\t0.72447\n", "x23:\t0.72451\n", "x24:\t0.72448\n", "x25:\t0.72450\n" ] } ], "source": [ "result_g2 = FPI(g2, x_0, N, alpha)\n", "print_result(result_g2)" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "x0:\t0.50000\n", "x1:\t0.79167\n", "x2:\t0.72986\n", "x3:\t0.72453\n", "x4:\t0.72449\n" ] } ], "source": [ "result_g3 = FPI(g3, x_0, N, alpha)\n", "print_result(result_g3)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The method does not converge for $g_1$, but both $g_2$ and $g_3$ converges towards the fixed number $0.72449$, which is a solution of the equation $f(x) = 0$. The method converges much faster for $g_3$ than for $g_2$.\n", "\n", "This result can be verified by calculating the convergence rates $S = \\left|\\: g'(r) \\:\\right|$:\n", "\n", "\\begin{equation*}\n", "\\left|\\:g_1'(0.72449)\\:\\right| \\approx 1.521 > 1, \\\\\n", "\\left|\\:g_2'(0.72449)\\:\\right| \\approx 0.657 < 1, \\\\\n", "\\left|\\:g_3'(0.72449)\\:\\right| \\approx 0.000 < 1. \\\\\n", "\\end{equation*}\n", "\n", "Theorem 1 tells us that the method will converge for $g_2$ and $g_3$ for some inital guess. Moreover, from theorem 2 we know that the method converges linearly for $g_2$ and the order of convergence for $g_3$ is two.\n", "\n", "**Exercise:** Verify that the convergence for $g_2$ is linear and that the convergence of $g_3$ is of second order by using the numerical results above." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## In comparison with the Newton-Rhapson method\n", "\n", "The Newton-Rhapson method is a special case of fixed-point iteration where the convergence rate is zero - the fastest possible. The method estimates the root of a differentiable function $f(x)$ by iteratively calculating the expression\n", "\n", "\\begin{equation*}\n", "x_{n+1} = x_n -\\frac{f(x_n)}{f'(x_n)}.\n", "\\end{equation*}\n", "\n", "By using theorem 2, we can show that Newton's method is of second order (if $f'(r)\\neq 0$). If $f''(r)=0$, then the method has a third order convergence.\n", "\n", "This Newton iteration may be rewritten as a fixed point iteration\n", "\n", "\\begin{equation*}\n", "x_{n+1} = g(x_n), \\: n = 1, 2 ,3, ...\n", "\\end{equation*}\n", "\n", "where\n", "\n", "\\begin{equation*}\n", "r = g(r) = r - \\frac{f(r)}{f'(r)}\n", "%\\label{eq:newton}\n", "\\end{equation*}\n", "\n", "Provided that the iteration converges to a fixed point $r$ of $g$. From this we obtain\n", "\n", "$$\\frac{f(r)}{f'(r)} = 0 \\: \\Rightarrow \\: f(r) = 0$$\n", "\n", "and thus the fixed point $r$ is a root of $f$.\n", "\n", "The Newton's method is further discussed in the module [Root Finding - Newton-Rhapson Method](https://nbviewer.jupyter.org/urls/www.numfys.net/media/notebooks/newton_raphson_method.ipynb)." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Final comments\n", "\n", "We have now used the fixed-point iteration method to solve the equation $f(x) = 0$ by rewriting it as a fixed-point problem. This is a simple method and which is easy to implement, but unlike the [Bisection Method](https://nbviewer.jupyter.org/urls/www.numfys.net/media/notebooks/bisection_method.ipynb) it only converges if the initial guess is sufficiently close to the root $r$. It is in general only locally convergent. However, the convergence rate may, or may not, be faster than that of the Bisection Method, which is 1/2.\n" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## References\n", "\n", " Sauer, T.: Numerical Analysis international edition, second edition, Pearson 2014 \n", " Gautschi, W.: Numerical Analysis, second edition, Birkhäuser 2012 (https://doi.org/10.1007/978-0-8176-8259-0)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.4" } }, "nbformat": 4, "nbformat_minor": 2 }