Transcript car payment calculator

Section 3.8 – Newton’s
Method
Solving Complicated Equations
There is a well-known formula to solve all quadratic
equations and there exist more complicated
equations to solve third- and fourth-degree
polynomials. Unfortunately, a formula does not exist
for any polynomial with a degree of five or higher.
There are also plenty of other complicated
equations that can not be solved with a formula. For
instance:
cos x  x
If we can not solve these equations algebraically, what
other methods exist to approximate the answer?
A New Method to Solve Equations
Suppose that a car dealer offers to sell you a
car for $18,000 or for payments of $375 per
month for five years. You would like to know
what monthly interest rate the dealer is, in
effect, charging you. To find the answer, you
have to solve the equation:
48 x 1  x   1  x   1  0
60
60
How would you solve such an equation?
A New Method to Solve Equations
Use a graph to help solve:
48 x 1  x   1  x   1  0
60
60
The solution is close to 0.011
The tangent line equation at
x1 = 0.011 is
 0.011, 0.090 
y  0.090  38.534  x  0.011
The x-intercept of the tangent
line is closer to the solution:
0  0.090  38.534  x2  0.011
0.090  38.534 x2  38.534  0.011
1 38.534 x2  38.534  0.011  0.090
x2 x
f  x   48 x 1  x   1  x   1
60
60
0.090
x2  0.011  38.534
x2  0.00866
A New Method to Solve Equations
Use a graph to help solve:
48 x 1  x   1  x   1  0
60
60
The solution is closer to 0.00866
The tangent line equation at
x2 = 0.00866 is
y  0.020  22.231 x  0.00866 
The x-intercept of the tangent
line is closer to the solution:
 0.00866, 0.020 
x3
x2
f  x   48 x 1  x   1  x   1
60
60
0  0.020  22.231 x3  0.00866 
0.020  22.231x3  22.231 0.00866
22.231x3  22.231 0.00866  0.020
0.020
x3  0.00866  22.231
x3  0.00777
A New Method to Solve Equations
Use a graph to help solve:
48 x 1  x   1  x   1  0
60
60
The solution is even closer to
0.00777.
Let’s Generalize the
Algorithm.
f  x   48 x 1  x   1  x   1
60
60
A New Method to Solve Equations
Use a graph to help solve:
f  x  0
The solution is close to xn
x
n
, f  xn 

The tangent line equation at
x = xn is
y  f  xn   f '  xn    x  xn 
The x-intercept of the tangent
line is closer to the solution:
0  f  xn   f '  xn    xn 1  xn 
 f  xn   f '  xn   xn 1  f '  xn   xn
xn 1 xn f '  x   x
n
f  x
n 1
 f '  xn   xn  f  xn 
f  xn 
xn 1  xn 
A Closer Solution:
f '  xn 
Newton’s Method
The Equation
MUST equal 0.
Procedure to solve the equation f (x) = 0:
1. Establish a small positive number Ɛ that determines the
allowable tolerance for estimated solutions (typically
given).
2. Compute f '(x).
3. With a graph or table, choose a number x0 (with f '(x0)≠0)
“close” to a solution of f (x)=0 as an initial estimate.
4. Compute a new approximation with the formula:
xn1  xn 
f  xn 
f ' xn 
f '  xn   0
Each successive approximation is called an iteration.
5. Repeat step 4 until ǀ xn+1 – xn ǀ < Ɛ. The estimate xn+1 then
has the required accuracy.
Example 1
Use Newton’s Method to solve cos x  x . Continue the
iterations until two successive approximations differ by
less than 0.000001. f(x) = 0 : cos x  x  0
0.000001
f '  x    sin x 1
Establish the tolerance:
Find the derivative :
Use a graph to estimate the answer:
Example 1
Use Newton’s Method to solve cos x  x . Continue the
iterations until two successive approximations differ by
less than 0.000001.
Now graph the non-zero side:
Use a graph to estimate the
answer:
y  cos x  x
cos x  x  0
Since the equation equals 0,
the x-intercept(s) are the
solutions
The solution is
approximately 1.
Now continue with Newton’s Method.
Example 1
Use Newton’s Method to solve cos x  x . Continue the
iterations until two successive approximations differ by
less than 0.000001. f(x) = 0 : cos x  x  0
0.000001
Find the derivative : f '  x    sin x  1
Use a graph to estimate the answer: x1  1
Establish the tolerance:
A – B÷C
Compute the iterations:
n
xn
1
1
A
Always Store (->) these
Results in the calculator
xn – f(xn) ÷f ' (xn)
B
f ' (xn) C
-0.459697694
-1.841470985
0.7503638678
2 0.7503638678
-0.018923073 -1.681904953
0.7391128909
3 0.7391128909
-0.000046456
-1.673632544
0.7390851334
4 0.7390851334
-2.8471x10-10
-1.673612029
0.7390851332
5 0.7390851332
f(xn)
The difference is less than 0.000001
The solution of the equation is ~0.739085.
Example 2
Use Newton’s Method to approximate the zeros of
f  x   x3  2x  5 . Continue the iterations until two
successive approximations differ by less than 0.0001.
0.0001
2
Find the derivative : f '  x   3x  2
Use a graph to estimate the answer: x1  2
Establish the tolerance:
A – B÷C
Compute the iterations:
A
Always Store (->) these
Results in the calculator
f(xn)
xn
1
2
-1
10
2.1
2
2.1
0.061
11.23
2.094568121
3
2.094568121
0.0001857232
11.16164684
2.094551482
4
2.094551482
B
f ' (xn) C
xn – f(xn) ÷f ' (xn)
n
The difference is less than 0.0001
The root of the equation is ~2.0946.
Example 3
Use two iterations of Newton’s Method to approximate the
x-coordinates of the intersection of y = x5 and y = -x2 – 2.
x5   x 2  2
x5  x 2  2  0
Equation to solve:
f(x) = 0 :
Establish the tolerance: Two Iterations
f '  x   5x  2 x
Use a graph to estimate the answer: x1  1
4
Find the derivative :
A – B÷C
Compute the iterations:
n
xn
1
-1
2
-5/3
A
Always Store (->) these
Results in the calculator
f(xn)
2
B
f ' (xn) C
3
-8.082304527 35.246913580
xn – f(xn) ÷f ' (xn)
-5/3
-1.4373613543491
End of the second iteration.
The x-coordinate of the intersection is ~ -1.437.
White Board Challenge
Use Newton’s method to find the solution(s)
to the following equation accurate to 3
decimal places:
x  3x  1
2
x  3.303 or
x  0.303
White Board Challenge
Use Newton’s method to approximate the
following expression accurate to 3 decimal
places:
4
13
1.899