Chapter 3: Applications of Differentiation
Section 3.2: Newton's Method
Example 3.2.3
Apply Newton's method to fx=x2−2 x−1.
Solution
Define f and solve fx=0
Control-drag fx=… Context Panel: Assign Function
fx=x2−2 x−1→assign as functionf
Write fx=0 and press the Enter key.
Context Panel: Solve≻Solve
fx=0
x2−2⁢x−1=0
→solve
x=1+2,x=1−2
Define the Newton iteration function and start an iteration from x=3
Write gx=… Context Panel: Assign Function
gx=x−fx/f′x→assign as functiong
Calculate x1=g3 as the first iterate.
g3. = 2.000000000
Calculate x2=gx1 as the second iterate.
g2. = 3.000000000
The second iterate is the same as the initial iterate, so Newton's method will simply cycle back and forth between these two values. It will never converge to one of the two roots x=1 ±2.
The cycling is examined more closely below.
Start a Newton iteration from a≠1
First iterate: Write ga and press the Enter key.
Context Panel: Simplify≻Simplify
ga
a−2⁢a2−2⁢a−12⁢a−2
= simplify
a+1a−1
Second iterate: Apply g to the simplified ga. Press the Enter key.
Context Panel: Simplify≻Simplify The second iterate is the same as the initial point, so the iteration will cycle between a and a+1/a−1.
g
a+1a−1−2⁢a+12a−12−2⁢a+1a−1−12⁢a+1a−1−2
a
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