The document discusses the Newton-Raphson method, a numerical technique for finding the roots of algebraic and transcendental equations. It outlines the algorithm, starting with an initial guess and iteratively refining the approximation until a root is found. Limitations of the method are also mentioned, such as potential issues when the derivative is zero or the method getting stuck at local extrema.

1. NEWTON
RAPHSON
METHOD

2. Introduction
As we know from school days , and still we have studied
about the solutions of equations like Quadratic
equations , cubical equations and polynomial equations
and having roots in the
form of
x
=
−𝑏± 𝑏2−4𝑎𝑐
2
𝑎
where a, b , c are the
coefficient of equ.
But nowadays it is very difficult to
remember formulas for higher degree polynomial equations
. Hence to remove these difficulties there are few
numerical methods, one of them is Newton Raphson
Method.
Which has to be discussed in this power point presentation.

3. Newton Raphson
Method :
Newton Raphson method is a numerical technique
which is used to find the roots of
Algebraic & transcendental Equations .
Algebraic Equations :
An equation of the form of quadratic or
polynomial. e.g. 𝑥4+𝑥2+1=0
𝑥8-1 =0
𝑥3-2x -5=0

4. Transcendental
equation :
An equation which contains some
transcendental functions Such as
exponential or trigonometric
functions.
e.g. sin , cos , tan , 𝑒𝑥 , 𝑥𝑒 , log etc.
3x-cosx-1=0
logx+2x=0
𝑒𝑥-3x=0 Sinx+10x-7=38

5. Newton Raphson
Method :
Let us consider an equation f(x)=0 having graphical representation as

6. • f(x) =0 ,is an given equation
• Starting from an initial point 𝑥0
• Determine the slope of f(x) at x=𝑥0 .Call it f’(
𝑥0).
𝑖
𝑓(𝑥𝑖)−0
Slope =tanѲ= = 𝑓′ 𝑥 .
𝑥𝑖−𝑥𝑖+1
From here we get
𝑖
• f’(𝑥 ) =
𝑓(𝑥𝑖)
𝑥𝑖−𝑥𝑖+1
.
Hence
;
• 𝐱𝐢+𝟏 = 𝐱𝐢− 𝐢
𝐟(𝐱 )
𝐟′(𝐱𝐢)
• Newton Raphson
formula

7. Algorithm for f(x)=0
0
• Calculate f’(x) symbolically.
• Choose an initial guess 𝑥0as given below
let [a,b] be any interval such that f(a)<0 and f(b)>0 ,
then
𝑥 =𝑎+𝑏
.
2
• then 𝑥1 = 𝑥0
− 𝑓 𝑥0
𝑓′ 𝑥0
.
• Similarly 𝑥2 =𝑥1
− 𝑓 𝑥1
𝑓′(𝑥1)
.
• Then by repetition of this process we can find 𝑥3 ,𝑥4 ,𝑥5
……
• At last we reach at a stage where we find 𝒙𝒊+𝟏 = 𝒙𝒊.
• Then we willstop.
• Hence 𝒙𝒊 will be the required root of given equation.

8. NRM by Taylor
series :
if we have given an equation f(x)=0.
𝑥0 be the approximated root of given equation.
• Let (𝑥0 +h) be the actual root where ‘h’ is very small such
that
• f(𝑥0 + ℎ)=0
From Taylor series expansion on expanding to f(𝑥0 + ℎ)
ℎ2 ℎ3
f(𝑥0 + ℎ)=f(𝑥0) + hf’(𝑥0)+ 2!
f’’(𝑥0) + 3!
f’’’(𝑥0) +…….
Now on neglecting higher powers of h
• f(𝑥0) + hf’(𝑥0) =0
𝑓(𝑥0)
From above h= -
𝑓′(𝑥0)
Cont.
.

9. Hence first approximation 𝑥1=(𝑥0 +
ℎ);
𝑥1=𝑥0- 0
𝑓(𝑥 )
𝑓′(𝑥0)
Second approximation ;
2
𝑥 =𝑥
-
1
𝑓(𝑥 )
1 𝑓′(𝑥1)
On repeating this
process We get
𝒙𝒏+𝟏=𝒙𝒏 − 𝒇(𝒙𝒏)
𝒇′(𝒙𝒏)
This is the required newton Raphson
method.

10. How to solve an example :
F(x)= 𝑥3- 2x – 5
F’(x) = 3𝑥2-2
Now checking for initial point
F(1) =-6
F(2)= -1
F(3) =16
Hence root lies between (2,3)
Initial point (𝑥0) =
2+3
2
=2.5
From NRM formula
𝑛 +1
𝑛
𝑓′(𝑥𝑛)
𝑥 =𝑥 − 𝑓(𝑥𝑛)
putting all these above values in this
formula

11. 𝑛 +1 𝑛
𝑥 = 𝑥 −
On putting initial value 𝑥0 = 2.5
We get first approximate root;
𝑥1=2.164179104
Similarly:
𝑥2=2.097135356
𝑥3=2.094555232
𝑥4=2.094551482
𝑥5=2.094551482
Hence 𝐱𝟒=𝐱𝟓
Hence 2.094551482 is the required root of given equation.
n
n n
3 x 2
 2
 2 x  5
x 3

12. Application of NRM
To find the square root of any no.
To find the inverse
To find inverse square root.
Root of any given equation.

13. Limitations
of NRM
• F’(x)=0 is real disaster for this method
• F’’(x)=0 causes the solution to diverse
• Sometimes get trapped in local maxima
and minima .

14. THANK YOU..!!!!!!