newton raphson method
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Transcript of newton raphson method
Introduction
As we know from school days , and still we have studied aboutthe solutions of equations like Quadratic equations , cubicalequations and polynomial equations and having roots in the form of x= 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.
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. ++1=0-1 =0-2x -5=0
Transcendental equation :An equation which contains some transcendental functionsSuch as exponential or trigonometric functions.e.g. sin , cos , tan , , , log etc.3x-cosx-1=0logx+2x=0-3x=0Sinx+10x-7=38
• f(x) =0 ,is an given equation
• Starting from an initial point
• Determine the slope of f(x) at x=Call it f’(
Slope =tanѲ=
From here we get
• f’(
Hence ;
• =
• Newton Raphson formula
Algorithm for f(x)=0• Calculate f’(x) symbolically. • Choose an initial guess as given below let [a,b] be any interval such that f(a)<0 and f(b)>0 , then = .• then • Similarly • Then by repetition of this process we can find • At last we reach at a stage where we find .• Then we will stop.• Hence will be the required root of given equation.
NRM by Taylor series :if we have given an equation f(x)=0. be the approximated root of given equation.• Let (h) be the actual root where ‘h’ is very small such that• f()=0From Taylor series expansion on expanding to f() f()=f+ hf’()+ f’’() + f’’’() +…….Now on neglecting higher powers of h• f() + hf’() =0From above h= -
Cont..
Hence first approximation =(;=- Second approximation ;=-
On repeating this process
We get =
This is the required newton Raphson method.
How to solve an example :F(x)= - 2x – 5F’(x) = 3-2Now checking for initial pointF(1) =-6F(2)= -1F(3) =16Hence root lies between (2,3)Initial point () ==2.5From NRM formula= ,putting all these above values in this formula
=
On putting initial value We get first approximate root;=2.164179104Similarly:=2.097135356=2.094555232=2.094551482=2.094551482Hence =Hence 2.094551482 is the required root of given equation.
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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.
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 .