Michael Corral: Vector Calculus

92 CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
To use this program, you should first save the code in Listing 2.1 in a plain text
file called newton.java. You will need the Java Development Kit 9 to compile the
code. In thedirectory wherenewton.java is saved, run thiscommandat acommand
prompt to compile the code: javac newton.java
Then run the program with the initial point (0,0) with this command:
java newton 0 0
Belowistheoutputoftheprogramusing(0,0)astheinitialpoint,truncatedtoshow
the first 10 lines and the last 5 lines:
java newton 0 0
Initial point: (0.0,0.0)
n = 1: (0.0,-1.0)
n = 2: (1.0,-0.5)
n = 3: (0.6065857885615251,-0.44194107452339687)
n = 4: (0.484506572966545,-0.405341511995805)
n = 5: (0.47123972682634485,-0.3966334583092305)
n = 6: (0.47113558510349535,-0.39636450001936047)
n = 7: (0.4711356343449705,-0.3963643379632247)
n = 8: (0.4711356343449874,-0.39636433796318005)
n = 9: (0.4711356343449874,-0.39636433796318005)
n = 10: (0.4711356343449874,-0.39636433796318005)
...
n = 96: (0.4711356343449874,-0.39636433796318005)
n = 97: (0.4711356343449874,-0.39636433796318005)
n = 98: (0.4711356343449874,-0.39636433796318005)
n = 99: (0.4711356343449874,-0.39636433796318005)
n = 100: (0.4711356343449874,-0.39636433796318005)
As you can see, we appear to have converged fairly quickly (after only 8 iterations)
to what appears to be an actual critical point (up to Java’s level of precision), namely
the point (0.4711356343449874,−0.39636433796318005). It is easy to confirm that∇f= 0
atthispoint, eitherbyevaluating ∂f
∂x and∂f ∂y atthepointourselvesorbymodifyingour
program to also print the values of the partial derivatives at the point. It turns out
that both partial derivatives are indeed close enough to zero to be considered zero:
∂f
∂x (0.4711356343449874,−0.39636433796318005)=4.85722573273506×10−17
∂f
∂y (0.4711356343449874,−0.39636433796318005)=−8.326672684688674×10−17
We also have D(0.4711356343449874,−0.39636433796318005)=−8.776075636032301