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"## Newton's Method ##\n",
"\n",
"Newton's Method is a technique for approximating a root to an equation of the form $f(x)=0$. What is required is an initial estimate for the root, called $x_1$, and an iterative formula:\n",
"\n",
"$$x_{n+1} = x_n - \\frac{f(x_n)}{f^{\\prime}(x_n)}$$\n",
"\n",
"This produces a sequence $x_1, x_2, x_3, \\ldots $ which converges to the root (hopefully).\n",
"\n",
"\n",
"\n",
"An implementation of Newton's Method is shown in the code block below. You can change the function $f$, and the initial\n",
"guess $x_1$. Clicking evaluate will run two iterations of Newton's method and return the next two approximations. Feel free to add some more lines\n",
"of code to find more iterations - just duplicate the last two lines of code and update the index:\n",
"\n",
"```python\n",
"x4=NewtonIt(x3);\n",
"print(x4);\n",
"```\n",
"\n",
"### Run the calculations yourself...###"
]
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"text": [
"1\n",
"0.870073695796\n",
"0.871221414263\n"
]
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"source": [
"import math\n",
"import numpy as np\n",
"from sympy import *\n",
"\n",
"x = Symbol('x'); # declare the variable for function\n",
"y = cos(x)+5*x-5; # change this to whatever function you like\n",
"dy = y.diff(x); # python will compute the derivative of y\n",
"f = lambdify(x, y, 'numpy') # convert y to a lambda function f so we can evaluate it as 'f(4)', etc.\n",
"df = lambdify(x, dy, 'numpy') # convert dy to a lambda function df so we can evaluate it as 'df(4)', etc.\n",
"def NewtonIt(x):\n",
" return x-(f(x)/df(x)); # Newtons Iterative Formula which we are calling \"NewtonIt\"\n",
"\n",
"x1=1; # initial guess\n",
"print(x1); # prints x1 in output window below\n",
"x2=NewtonIt(x1); # one iteration of Newtons Method - output below\n",
"print(x2);\n",
"x3=NewtonIt(x2); # a second iteration of Newtons Method\n",
"print(x3);"
]
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" "
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"
Using a loop to compute more iterations...
\n",
"\n",
"If you have Python programming experience you could write a `for` loop to automate the iterations. Replace everything from the initial guess down (line 6 and below) with something like:\n",
"\n",
"```python\n",
"xn=1; # initial guess\n",
"print(xn);\n",
"for i in range(10):\n",
" xn=NewtonIt(xn)\n",
" print('{:.20f}'.format(xn));\n",
"```\n",
"\n",
"This will produce 10 iterations, each value shown to 20 decimal places.
\n",
"\n",
"You could get even fancier and run a `while` loop where the stopping condition is that two iterations agree to some specified number of decimal places.\n"
]
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"1\n",
"0.87007369579620918998\n",
"0.87122141426291899169\n",
"0.87122151449508777876\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n",
"0.87122151449508855592\n"
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"source": [
"import math\n",
"import numpy as np\n",
"from sympy import *\n",
"\n",
"x = Symbol('x'); # declare the variable for function\n",
"y = cos(x)+5*x-5; # change this to whatever function you like\n",
"dy = y.diff(x); # python will compute the derivative of y\n",
"f = lambdify(x, y, 'numpy') # convert y to a lambda function f so we can evaluate it as 'f(4)', etc.\n",
"df = lambdify(x, dy, 'numpy') # convert dy to a lambda function df so we can evaluate it as 'df(4)', etc.\n",
"def NewtonIt(x):\n",
" return x-(f(x)/df(x)); # Newtons Iterative Formula which we are calling \"NewtonIt\"\n",
"\n",
"xn = 1; # initial guess\n",
"print(xn);\n",
"for i in range(10):\n",
" xn=NewtonIt(xn)\n",
" print('{:.20f}'.format(xn));"
]
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"---\n",
"\n",
"Sample code to see how to differentiate using sympy\n",
"\n",
"https://stackoverflow.com/questions/9876290/how-do-i-compute-derivative-using-numpy"
]
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"cell_type": "code",
"execution_count": 19,
"metadata": {},
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"name": "stdout",
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"text": [
"2*x + 1/x\n"
]
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"data": {
"text/plain": [
"array([ 3., 3., 3., 3., 3.])"
]
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"source": [
"from sympy import *\n",
"import numpy as np\n",
"x = Symbol('x')\n",
"y = x**2 + ln(x)\n",
"yprime = y.diff(x)\n",
"print(yprime)\n",
"\n",
"f = lambdify(x, yprime, 'numpy')\n",
"f(np.ones(5))"
]
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