4. 4
As we know from school days , and still we have studied
about the solutions of equations like Quadratic equations
, Cubical equations & polynomial equations and having
roots in the form of
𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
where a, b , c are the coefficient of equation.
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 the easiest of them is Newton Raphson
Method Which has to be discussed in this power point
presentation.
Newton Raphson method is a numerical technique
which is used to find the roots of Algebraic &
transcendental Equations .
5. 6
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
6. 7
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
Newton Raphson Method :
Let us consider an equation 𝑓(𝑥) = 0
having graphical representation as
7. 8
.
• f(x) =0 , is a given equation
• Starting from an initial point 𝑥0
• Determine the slope of f(x) at x=𝑥0 .Call 𝑓’(𝑥0).
Slope ;
= tanѲ
𝑓′(
Hence ;
• Newton Raphson formula
Algorithm for f(x)=0
• Calculate f’(x) symbolically.
8. Choose an initial guess 𝑥0 as given below let [a,b] be any interval such that 𝑓 𝑎 < 0 𝑎𝑛𝑑 𝑓 𝑏 > 0
.
•
. Similarly
.
• 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 will stop.
• Hence 𝒙𝒊 will be the required root of given equation.
NRM by Taylor series :-
if we have given an equation 𝑓(𝑥) = 0.
𝑥0 be the approximated root of given equation.
9. 10
• 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)
f(f’’’(𝑥0) +…….
Now on neglecting higher powers of h
f(𝑥0) + hf’(𝑥0) =0
From above h=
Hence first approximation 𝑥1=(𝑥0 + ℎ);
10. Second approximation ;
On repeating this process
We get
This is the required newton Raphson method.
How to solve an example :
F(x)= 𝑥3- 2x – 5
F’(x) = 3𝑥2-2
Now checking for initial point ;
F(3) =16
F(2)= -1
F(3) =16
Hence root lies between (2,3)
Initial point
(
From NRM formula
Putting all these above values in this
formula
11. .
12
𝑛+1 𝑛
𝑥 = 𝑥 −
n n
3 x 2
2
2 x 5
𝑛+1 𝑛
𝑥 = 𝑥 −
n
n n
3 x 2
2
2 x 5
x 3
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
12. 13
• To find the square root of any no.
• To find the inverse
• To find inverse square root.
• Root of any given equation.
Application of NRM :
13. 16
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