Notes on Netwon's and Bisection Method

Tarek hajj shehadi
MATH 251 : Chapter II
I) Introduction
‣ Consider an interval ( a, b).
‣ Consider some graph y = f(x), we want to find the value of x such that the graph
y = f(x) intersects the x − axis between the interval ( a, b) at a point we’ve been
looking for which is :
x = r
‣ This interval ( a, b), must satisfy the following properties :
• f(x) is continuous over (a, b).
f(a)
• r ∈ (a, b)
• f(a) × f(b) < 0
‣ Our goal is to find a sequence of roots :
{r 1 , r 2 , … r n }
Such that all these roots lie between interval (a, b).
a
f(b)
<
b
‣ All these roots are found recursively until we have that lim
n→∞
r n = r
‣ r n depends on all of r 1 ⟶ r n−1 .
When n is large, the distance
between r n and r is very close

I) The Bisection Method
‣ The Bisection Method is a method in which the interval ( a, b) gets reduced into half its
size whenever an r i ; 1 ≤ i ≤ n is found and the interval keeps getting reduced in size
until we obtain an accurate approximation of the desired root r.
‣ Step 1 : Let us rename ( a, b) as ( a 0 , b 0 ) .Find r 1 = a 0+b 0
.
‣ Step 2 : Let y = f(r 1 ) . If :
• y = f(r 1 ) = 0 , we are done and r = r 0 .
• f(r 1 ) ∗ f(a 0 ) < 0 , then r ∈ (a 0 , r 1 ) we name it as (a 1 , b 1 ).
• f(r 1 ) ∗ f(a 0 ) > 0 , then r ∈ (r 1 , b 0 ) we name it as (a 1 , b 1 ).
2
f(a 0 )
f(r 1 )
b 0
a 0
_
0
r 1
_
0
r is obviously between r 1 and b 0
f(b 0 )
<
The new interval obtained (a 1 , b 1 ) is half the size of previous (initial) interval ( a 0 , b 0 ) :
b 1 − a 1 = 1 2 (b 0 − a 0 ) and ȁr − r 1 ȁ < (b 1 − a 1 ):
The process is repeated as many times as possible until r n ∈ ( a n , b n ) get a good approximation
b n − a n = 1 (b 2 n 0 − a 0 ) and r n = a n−1+b n−1
2

‣ Theorem : For any recursively defined relation, the order of convergence (how fast we
reach the desired answer) is given by :
a
t n ≤ Ct n−1
• If a = 1, the convergence is said to be linear (very slow)
• If a > 1, the convergence is said to be super linear.
As for the Bisection method, we know that b n − a n ≤ 1 2 (b n−1 − a n−1 ). Let t n = b n − a n , then
we can say that t n−1 = b n−1 − a n−1 .
t n ≤ 1 2 t n−1
C = 1 2 ; a = 1
Therefore, we conclude that the order of convergence of the Bisection algorithm is linear.
very slow convergence
‣ The minimum number of iterations needed to reach the required tolerance is :
Where p is the precision.
k = ⌈(p − 1) log 2 (10) + 1⌉
• Ex : Find the root of f(x) = ln(1 + x) − 1
x+1
; for p = 3
‣ Step 1 : Find two points for which the root is located between them.
We look at domain, x > −1 , so the first integer belong to this domain is x = 0, we compute
f(0), we get f(0) = −1 < 0. We now look at the next integer which is x = 1, we get for
f(1), f(1) = ln(2) − 1 > 0. Notice how f(0) and f(1) are located above and below x axis,
2
respectively. Therefore, the root is located in between (0,1).

‣ Step 2 : Find r 1 = a 0+b 0
.
2
‣ Step 3 : Find the sign of f(r 1 ) ∗ f(a 0 ).
r 1 = 0 + 1
2
= 1 2
f(r 1 ) ∗ f(a 0 ) = f(0) ∗ f ൬ 1 2 ൰ = + ⋯ > 0
‣ Step 4 : Since f(r 1 ) ∗ f(a 0 ) > 0 , then r ∈ (r 1 , b 0 ) we name it as (a 1 , b 1 ). So, the
interval becomes ቀ 1 2 , 1ቁ.
‣ Step 5 : Repeat Step 2
r 2 = 1 + 1 2
2
‣ Step 5 : Repeat Step 3 and so on… it is better to use the formula for number of
iterations for a given precision to know how many times you have to iterate.
= 3 4
II) Newton’s Method
‣ Newton’s Method is a method in which the interval ( a, b) has the desired root in
between and so what we do is take the function y = f(x), draw its tangent at any
point such that this point has an abscissa located between ( a, b) and then intersecting
it with the x − axis.
‣ Again, we seek to find a recursive relation :
r n = r n−1 − f(r n−1)
f ′ (r n−1 )
• Ex : Find the root of f(x) = ln(1 + x) − 1
x+1
; for p = 3
‣ Step 1 : Find two points for which the root is located between them.

We look at domain, x > −1 , so the first integer belong to this domain is x = 0, we compute
f(0), we get f(0) = −1 < 0. We now look at the next integer which is x = 1, we get for f(1),
f(1) = ln(2) − 1 > 0. Notice how f(0) and f(1) are located above and below x axis,
2
respectively. Therefore, the root is located in between (0,1).
‣ Step 2 : Let r 0 = 0.1
r 2 = r 1 − f(r 1)
f ′ (r 1 )
r 3 = r 2 − f(r 2)
f ′ (r 2 )
r 3 = r 2 − f(r 2)
f ′ (r 2 )
r 1 = r 0 − f(r 0)
f ′ (r 0 )
1.1
= 0.1 − [ln(1.1) − 1] = 0.4787
2.1
1.4787
= 0.4787 − [1.4787 ln(1.4787) − 1] = 0.7302
2.4787
1.7302
= 0.7302 − [1.7302 ln(1.7302) − 1] = 0.7628
2.7302
1.7628
= 0.7628 − [1.7628ln(1.7628) − 1] = 0.7632
2.7628
r 4 = r 3 − f(r 3)
f ′ (r 3 )
1.7632
= 0.7632 − [1.7632 ln(1.7632) − 1] = 0.7632
2.7632
‣ Note : a combination of Bisection Method and Newton’s Method can make
approximation of a root much faster
‣ A fast-iterative method to approximate x = √d would be :
r n+1 = (3 − r n 2
d ) r n
2
‣ The root must be located on the open interval ( 0, √3d).
‣ This is useful when asked to compute the square root of an IEEE S.P.R.
√(x) 2 = √m ∗ 2 e = m ′ ∗ 2 e′
‣ Let e = 2k : √(x) 2 = √m ∗ 2 2k = 2 k ∗ √m ; 1 ≤ √m < √2
‣ Let e = 2k + 1 : √(x) 2 = √m ∗ 2 2k+1 = 2 k ∗ √2m ; 1 ≤ √2m ≤ 2
‣ Always start with r 0 = 1, to find the mantissa √m or √2m refer to formula of x = √d