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# newton raphson method

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### newton raphson method

1. 1. Newton raphson method By- Yogesh bhargawa M.Sc. 4th sem. Roll no. 4086
2. 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 2𝑎 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. 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. 4. Transcendental equation : An equation which contains some transcendental functions Such as exponential or trigonometric functions. e.g. sin , cos , tan , 𝑒 𝑥 , 𝑥 𝑒 , log etc. 3x-cosx-1=0 logx+2x=0 𝑒 𝑥 -3x=0 Sinx+10x-7=38
5. 5. Newton Raphson Method : Let us consider an equation f(x)=0 having graphical representation as
6. 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). Slope =tanѲ= 𝑓(𝑥 𝑖)−0 𝑥 𝑖−𝑥 𝑖+1 = 𝑓′ 𝑥𝑖 . From here we get • f’(𝑥𝑖) = 𝑓(𝑥 𝑖) 𝑥 𝑖−𝑥 𝑖+1 . Hence ; • 𝐱 𝐢+𝟏 = 𝐱 𝐢 − 𝐟(𝐱 𝐢) 𝐟′(𝐱 𝐢) • Newton Raphson formula
7. 7. Algorithm for f(x)=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 𝑥0= 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 will stop. • Hence 𝒙𝒊 will be the required root of given equation.
8. 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) f(𝑥0 + ℎ)=f(𝑥0) + hf’(𝑥0)+ ℎ2 2! f’’(𝑥0) + ℎ3 3! f’’’(𝑥0) +……. Now on neglecting higher powers of h • f(𝑥0) + hf’(𝑥0) =0 From above h= - 𝑓(𝑥0) 𝑓′(𝑥0) Cont..
9. 9. Hence first approximation 𝑥1=(𝑥0 + ℎ); 𝑥1=𝑥0- 𝑓(𝑥0) 𝑓′(𝑥0) Second approximation ; 𝑥2=𝑥1- 𝑓(𝑥1) 𝑓′(𝑥1) On repeating this process We get 𝒙 𝒏+𝟏=𝒙 𝒏 − 𝒇(𝒙 𝒏) 𝒇′(𝒙 𝒏) This is the required newton Raphson method.
10. 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 𝑥 𝑛+1=𝑥 𝑛 − 𝑓(𝑥 𝑛) 𝑓′(𝑥 𝑛) ,putting all these above values in this formula
11. 11. 𝑥 𝑛+1 = 𝑥 𝑛 − 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 𝐱 𝟒=𝐱 𝟓 Hence 2.094551482 is the required root of given equation. 23 52 2 3   n nn x xx
12. 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. 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. 14. THANK YOU..!!!!!!