Newton-Raphson Method

Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.
Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.
Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.
Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.
Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.
Newton-Raphson Method for Approximating
We use tangent to approximate the graph of y = f(x), near the point
P (xn, yn), where
yn = f(xn), is small. Let xn+1 be the value of x where that tangent line
crosses the x-axis.
Let tangent = The slope between (x , y) and (xn, yn), is
f `( xn) =
n
n
x x
y y
?
?
………………………………………….… (1)
Since the tangent line crosses the x-axis, y = 0, and yn = f(xn), put in Eq
(1) which becomes
f `( xn) =
n
n
x x
f x
?
? ( )
,
x - xn=
( )
( )
n
n
f x
f x
?
?
,
x = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (2).
Put x= xn+1 in Eq (2) gives
xn+1 = xn -
( )
( )
n
n
f x
f x
?
………………………………………….… (3)
Eq (3) called Newton-Raphson Method, can using this method by the
following
1-Give first approximating to root of equation f(x) = 0. A graph of
y = f(x).
2-Use first approximating to get a second. The second to get a third, and
so on. To go from nth approximation xn to the next approximation xn+1,
by using Eq (3), where f `(x) the derivative of f at xn.
Example 1
65
Solve the following using Newton-Raphson Method
x
1 +1 = 0, start with x0 = -0.5, error % = 0.5 %
Where e % =
1
1
?
? ?
n
n n
x
x x
?%
Sol
f(x) =
x
1 +1, x0 = -0.5,
f `( xn) = -
2
1
x
f(x0) =
0.5
1
?
+1= -1,
f `( x0) = -
( 0.5 )2
1
?
= -4, from Eq (3)
x1 = x0 -
( )
( )
0
0
f x
f x
?
x1 = -0.5 -
4
1
?
?
= -0.75.
By use e % =
1
1
?
? ?
n
n n
x
x x
? % as
e % =
0.75
0.75 ( 0.5)
?
? ? ?
?%
e % = 33%
By use same of new of x1 in Eq (3) as
x2 = x1 -
( )
( )
1
1
f x
f x
?
, ? x2 = -0.937, in same we can find x3 and x4
which use in the following table
n xn f(x) f `( xn) xn+1 e %
0 - 0.5 - 1 - 4 -0.75 33%
1 - 0.75 - 0.333 - 1.77 -0.937 19 %
2 - 0.937 -0.067 - 1.137 -0.997 6 %
3 -0.997 -0.003 - 1.006 - 1.000 0.3 %
To check the answer as:-
1
1
?
+1= -1+1= 0.