1. Newton's I'v'1ethod
Problem: Given an equation 1(x) = 0, solve for :1; numerically.
• Make an initial gness: ;ro,
xo
1. Newton's Tvlethod
Problem: Given an equation 1(:1;)
1. Newton's rv'Iethocl
= 0,
solve for :1; numerically.
Problem: Given an equation 1(:1;) = 0, solve for:1; numerically.
• Make an initial g'u,ess: Xo.
Now go up to the curve,
xo
Make an in'it'ial r)'l/,eS8: ;1:0.
Now go up to the curve .
• Draw the tangent line.
III
xo
l. Newton's l'viethod
Problem: Given an equation f(x)
1. Newton's l\ilethod
0, solve for ::c numerically.
Problem: Given an equation f(x)
• Make an init·jal g'll.e88: Xo.
Now go up to the curve.
lit Dra.w the tangent line. Its
equation is
:Y
Xu
f(:1;O)
+ f'(::co) (:1:
= 0, solve for x numerically.
• Make an initial g1Le88: Xo.
Now go up to the curve.
• Draw the tangent line. Its
equation is
::co).
:Y
f(:1;O)
+ f'(:ro)(:1; -
:1:0).
• Let .7:1 be in x-intercept of
this tangent line.
1. Newton's rvlethod
Problem: Given an equation f(x)
1. Newton's rvIethocl
= 0,
solve for
:1:
numerically.
Problem: Given an equation f(::c)
• Ma.ke an init'ial gue8S: ::co.
Now go up to the curve.
• Draw the tangent line. Its
equation is
= 0, solve for x numerically.
• Make an in'itial gl1,eS8: :1:0.
Now go up to the curve.
• Draw the tangent line. Its
equation is
:q
y
• Let Xl be in ;r:-intercept of
this tangent line.
C!J
This intercept is given by the formula:
:];1
Xo
f(:ro)
+ f'(:r;o)(:r:
-
:1:0).
• Let :1:1 be in x-intercept of
this ta.ngent line.
• This intercept is given by the formula:
• Now repeat using
:1:1
:];1
as the initial guess.
= :1:0
Newton't) lvIethod
1.
2. Connnentary
Problem: Given an equation f(1:) = 0, solve for x numerically.
• Make an 'initia.l g1J.CSS: :X:o.
Now go up to the curve.
• Draw the tangent line. Its
eqnation is
f(:£o)
y
+ f'(1:0)(:r -
:1:0).
The initial guess, :r:o, was close to the true root. From the picture,
Fr:l,lllC ~" it appears our next estimate :];1,
f(:ro)
:r;l = 1:0 - f'(xo)
is a little closer to the unknown root than :ro was.
The next "iterate", :1:2, calculated from the formula
• Let :X:1 be in 1:-intercept of
this tangent line.
r-1
•
•
I Ius mterccpt
••
IS
glVen by the formula:
• Now repeat using
Xl
:1:1
=
:ro -
f(:ro)
is closer still to the unknown root, see
. ' :1:0
The process continues: Given that an estimate :l: n has already been
calculated, the equation of the tangent line is calculated:
-(f ') .
as the initial guess.
y = f(:r: n )
The intercept 1:2 is given by: :1;2 = :r:1
Fralllc'
7.
+ f'(:rn)(:r
Soction 2: Commentary
;3. Exarnples
The :r:-interccpt is then calculated,
Example 3.1. Find the positive root of the equation :1: 2 = 2.
f (:r) = x 2 - 2.
derivative, f'(x) = 2x.
Solution: The function is
This intercept is la,bcled :r: n +1 and represents our next estimate of the
unknown root.
(1)
The initial guess Xo, and the Newton Iteration formula, equation (1),
together form an a.lgorithm or a procedure of estimating the value of
the root to the equation f(:;;) = O.
Step 1: Compute
Step 2: Construct the iteration formula:
-2
1:11, -
+2
Thus, for this problem, the iteration formula is
This, together with an initial guess of
calculations.
::1:0
l.5 yields the following
Sectioll 3: Examples
Section 3: Examples
Step 3: Construct a table of estimates.
Example 3.2.
near ;DO = 1.5.
Initial guess of
~CO
= 1.5
n
0
1
2
3
4
and iteration formula of
:Cn+l
=
+2
2 = 0 is
or, stated differently, '--_ _ _ _ _--'
Step 1: DiHerentiate f'(:r;) = 3:r;2
:rn+l
Exalllple 3.1.
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Newton's Method
= :r;:3 -;1: 1, xo 1.5
f(:J: n )
:Dn
0.87500000
1.50000000
1.34782608
0.10058217
1.32520039
0.00205836
0.00000092
1.32471817
1.32471795
0.00000000
1.32471795
0.00000000
:D
+ 1 tha.t is
"near" Xo
1.5
Example 3.2.
AcroTgX:
D. P. Story:
dpstOl"y((~\lil.kron.('dll
:r,n, -
The iteration formula is
'--------'
Step 3: Construct the table of estimates.
n
0
1
2
3
4
5
1.
Step 2: Construct the Newton iteration forrnula:
Section 3: Exarnples
f(x)
:r;3
= :r
+ 1 that
is
Solution: The function is f (x)
:1;:3 - 1;
1. The function f is always
defined to make the given equation equivalent to an equatioll of the
form f(;r;) = O. (We just take everything on the right-hand side of
the given equation to the left-hand side. The left-hand side is now an
expression defining f (:r ).
Newton's Method
x 2 2,
f(:r)
:);0 = 1.5
:r:n
f(:r;n)
1.50000000
0.25000000
1.41666667
0.00694445
1.41421568
0.00000600
1.41421356
0.00000000
1.41421356
0.00000000
ecnJatlOn x 2
Find a solution to the equation
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