# NewtonsIterationMethod ```Newton’s Iteration Method for Finding the Zero of a Function
Introduction: To estimate a zero for a function f  x  we begin by choosing a point x0 ”close” to the


actual zero, calculating the equation of the tangent line thru the point x0 , f  x0  then finding where this
tangent line intersects the x-axis. If we call this point x1 , it should be a better approximation to the zero
we’re looking for. Repeat.
Example: Find a zero for the function f  x   x3  x 2  3
slope = f   x0 
 x , f  x 
0
0
x0
x1
To find the zero of f  x  do the following.
1.
Let x0 be an estimate for the zero.
2.
Then  x0 , f  x0   is a point on the curve. From the point on the curve x0 , f  x0  , if we sight


down the line tangent, where the line crosses the x-axis at x1 is closer to the real zero than x0 .
3.
The equation of the tangent line (slope = mtan  f   x0  ) thru the point  x0 , f  x0   is
y  f   x0  ( x  x0 )  f  x0  . It crosses the x-axis when y  0 . Solving for x  x1 we obtain x1  x0 
4.
Repeat. x2  x1 
f  x1 
f   x1 
, x3  x2 
f  x2 
f   x2 
, … Convergence is often very quick!
f  x0 
f   x0 
Back to our example: From f  x   x3  x 2  3 and f   x   3x 2  2 x we have the formula
xn 1  xn 
xn3  xn2  3
3xn2  2 xn
Set this up on a TI-84
y1  x 3  x 2  3
y2  3 x 2  2 x
2x
x  y1  x  / y2  x   x
We generate the following table of values
n
0
1
2
3
4
xn
2
1.875
1.863793103
1.863706533
1.863706528
Therefore 1.863706528 is the zero of the equation f  x   x3  x 2  3 . This method always works
provided that the initial guess is sufficiently close to the solution and f   xn  is never zero
(why?). Convergence should be fast.
Example: Solve x 3  3 x 2  1  0
Let x0  3 be an initial guess and iterative compute xn1  xn 
n
xn
0
1
2
3
4
3
2.888888888
2.879451567
2.879385242
2.879385242
xn3  3xn2  1
3xn2  6 xn
Exercises: Using Newton’s Formula locate the smallest positive root for each of the following to
within the accuracy of the TI 83/84 calculator.
f  x0 
for that function
f   x0 
a.
Obtain the derivative and write down Newton’s formula x1  x0 
b.
Graph the function to find an integer value close to the root to use as your initial value
c.
Fill in the table below for each function
f  x   x2  5
n
0
1
2
3
4
5
xn
f  x   x2  5
n
0
1
2
3
4
5
xn
f  x   x3  
n
0
1
2
3
4
5
xn
Newton’s Formulas
f  x   x3  3x  17
n
0
1
2
3
4
5
xn
Newton’s Formulas
f  x   x 5  17
f  x   x2  x  1
n
0
1
2
3
4
5
xn
n
0
1
2
3
4
5
xn
```