Source file: math1110-summer2014-newtons_method.mw 

Choose "EditExecuteWorksheet" to see results of computations. 

[Ostebee & Zorn's Calculus 4.6 #6] Approximate the solution of  "x tan(x) = 1" on the interval [0,pi/2] using Newton's method. 

>	f := x -> x*tan(x)-1: 'f(x)' = f(x);

 

(1)

 

>	diff(f(x),x);

 

(2)

 

>	fp := x->tan(x)+x*(1+tan(x)^2): 'diff(f(x),x)' = fp(x);

 

(3)

 

>	# We'll iterate n times. n := 4: # Our initial guess is 1. x[0] := 1; for i from 1 to n do   'f' = evalf(f(x[i-1]));   'df/dx' = evalf(fp(x[i-1]));    x[i] := evalf(x[i-1] - f(x[i-1])/fp(x[i-1])): end do;

 

 

 

 

 

 

 

 

 

 

 

 

 

	(4)

 

>	# Check our answer... x[n]*tan(x[n])=1;

 

(5)

 

>	# Our solutions... print("Newton's Method:"); for i from 0 to n do   x[i]; end do; # Maple's solution... print("Maple's solution:"); fsolve(x*tan(x)=1,x=1);

 

 

 

 

 

 

 

 

	(6)

 

Using Maple's "NewtonsMethod" command... 

>	with(Student[Calculus1]):

 

>	NewtonsMethod(x*tan(x)-1, x = 1);

 

(7)

 

>	NewtonsMethod(x*tan(x)-1, x = 1, output = sequence);

 

(8)

 

>	NewtonsMethod(x*tan(x)-1, x = 1, view = [0.8..1.1, DEFAULT], output = plot);

 

>	NewtonsMethod(x*tan(x)-1, x = 1, view = [0.8..1.1, DEFAULT], output = animation);

 

>	NewtonsMethod(x^3 - x, x = 2, view = [0..3, DEFAULT], output = plot);

 

>	NewtonsMethod(x^3 - x, x = 2, output = animation);

 

>	NewtonsMethod(x^2 + x + 1, x=2, output=animation,iterations = 15);

 

>	? NewtonsMethod

 

>