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UNIT NO. 5
โขTOPICS
โขRoots of equation [Bisection method, False position
(Regula Falsi) method and Newton โ Raphson method.]
โขInterpolation [Finite difference, Forward and backward
differences, Interpolation. Newtonโs forward interpolation
formula, Newtonโs backward interpolation formula,
Lagrangeโs interpolation formula and Newtonโs divided
difference formula,,
โขCurve fitting [Least squares curve fitting methods, linear
and nonlinear curve fitting
Roots of equations, Interpolation and Curve fitting
Theory Marks (^) Viva / TW ESE IA CSE Viva Term Work Total Marks 50 30 20 25 25 150 conducted centrally conducted centrally Assignments End semester viva Lab work
Evaluation Process
Application
LINK: https://youtu.be/4fu-DC2jRHA
INTRODUCTION Years ago , we learn to use the quadratic formula to solve F(x). The values calculated with this equation are called the โโrootsโโ. They represents the values of x that make f(x) equal to zero. we can define the roots of an equation as the value of x that makes f(x)=0. Thus , roots are sometimes called the zeros of the equation
Quadratic equation : ๐๐
๐
+bx + c
And its solution is x =
โ๐ยฑ ฮ 2๐
here ฮ = ๐
2
Roots of equation
Types of equation
1..Algebraic Equation
The equation of the form with degree n is called Algebraic equation
Example: ๐
2..Transcendental Equation
A non Algebraic Equation is called Transcendental equation
Intermediate Value theorem If a continuous function f(x) assumes value of opposite sign at the end points of an intervals [a ,b] ie. f(a) f(b) < 0 Then interval [a, b] contains at least one root of the equation f(x) = 0 i.e There exist c โ [a , b] such that f(c) = 0
- In particular Suppose f(๐ฅ
) <0 then root lies in between [๐ฅ
, b] then again bisect this interval to next point ๐ฅ
=
- Repeat this procedure to generate x 1 , x 2 ,,,,, till the root up to desired accuracy is obtained
Solution : Here f(x) = ๐ฅ
โ ๐ฅ โ 11 = 0 NOTE : USE โRADIAN MODEโ IN YOUR CALCULATOR FOR SOLUTION ANY EXAMPLE OF NUMERICAL METHODS First find the interval [a ,b] where f(a) < 0 and f(b) > 0 f(2) = 8 โ 2 - 11 = - 5 < 0 i.e a = 2 f(3) =27 - 3 - 11 = 13 > 0 i.e b = 3 So the root lies in between 2 and 3 Ex - 1 Find the root of equation ๐ฅ 3 โ ๐ฅ โ 11 = 0 using the bisection method up to forth approximation
Solution : Here f(x) =๐๐๐ ๐ฅ โ ๐ฅโ
= 0 First find the interval [a ,b] where f(a) < 0 and f(b) > 0 f(0) = 1 > 0 i.e b = 0 f(1) = - 2.1780 <0 i.e a= 1 So the root lies in between 1 and 0 Ex - 2 Perform the five iteration of the bisection method to obtain a root of the equation f(x) = cos ๐ฅ โ ๐ฅโ
๐ฅ = 0
Solution of equation is 0.
NO of
iterations
a
f(a) <
b
f(b) > 0
F (xn)
1 1 0 ๐ฅ 1 = 0.5 F(๐ฅ 1 ) = 0.0532 > 0
2 1 0.5 ๐ฅ 2 = 0.75 F(๐ฅ 2 ) = - 0.8561 < 0
3 0.75 0.5 ๐ฅ 3 = 0.625 F(๐ฅ 3 ) = - 0.3567 < 0
4 0.625 0.5 ๐ฅ 4 = 0.5625 F(๐ฅ 4 ) = - 0.1413 < 0
NO of iterations a f(a) <0 b f(b) > 0 ๐ฅ๐ = ๐ + ๐ 2 F (xn) 1 0 1 X1= 0.5 F(x1) = - 0.3776 < 0 2 0.5 1 X2= 0.75 F(x2) = 0.0183 > 3 0.5 0.75 X3= 0.625 F(x3) = - 0.186 < 4 0.625 0.75 X4= 0.6875 F(x4) =-0.0853 < 5 0.6875 0.75 X5 =0.71875 F(x5) =-0.0338 < 6 0.71875 0.75 X6 = 0.7344 F(x6)= - 0.0078 < 0 7 0.7344 0.75 X7 = 0.7422 F(x7) = 0.0052 > 8 0.7344 0.7422 X8 = 0.7383 F(x8) = - 0.0013 < 9 0.7383 0.7422 X9 = 0.74025 F(x9)= 0.00195 > 10 0.7383 0.74025 X10 = 0.7393 F(10) =0.0004 > 0 11 0.7383 0.7393 X11= 0. Ans is 0.
Solution : Here f(x) = ๐ฅ
โ 4๐ฅ โ 9 = 0 First find the interval [a ,b] where f(a) < 0 and f(b) > 0 f(2) = - 9 < 0 i.e a = 2 f(3) = 6 > 0 i.e b= 3 So the root lies in between 2 and 3 NO of iterations a f(a) <0 b f(b) > 0 ๐ฅ๐ = ๐ + ๐ 2 F (xn) 1 2 3 4
Ex - 4 Find the root of the equation ๐ฅ
3
โ 4๐ฅ โ 9 = 0 using the bisection method in four stages
Solution : Here f(x) = ๐ฅโ
โ 2 = 0 First find the interval [a ,b] where f(a) < 0 and f(b) > 0 f(0) = - 2 < 0 i.e a = 0 f(1) = 0.7182 > 0 i.e b= 1 So the root lies in between 1 and 0 NO of iterations a f(a) <0 b f(b) > 0 ๐ฅ๐ = ๐๐ ๐ โ ๐๐ ๐ ๐ ๐ โ ๐ ๐ F (xn) 1 0 - 2 1 0.7182 X1= 0.735759 - 0.4644 < 0 2 0.735759 - 0.4644 1 0.7182 X2 = 0.8395 - 0.056293 < 3 0.8395 - 0.056293 1 0.7182 X3 = 0.851184 - 0.006171 < 0 4 0.851184 - 0.006171 1 0.7182 X4 =0. Ex - 1 Find the root of the equation ๐ฅโ
โ 2 = 0 by the Regula false method in four stages
Solution : Here f(x) = ๐๐ โ ๐๐๐๐ โ ๐ = 0 First find the interval [a ,b] where f(a) < 0 and f(b) > 0 f(0) = - 2 < 0 i.e a = 0 f(1) = 1.4597 > 0 i.e b= 1 So the root lies in between 0 and 1 NO of iterations a f(a) <0 b f(b) > 0 ๐ฅ๐ = ๐๐ ๐ โ ๐๐ ๐ ๐ ๐ โ ๐ ๐ F (xn) 1 0 - 2 1 1.4597 X1=0.5781 - 0.1032< 0 2 0.5781 - 0.1032 1 1.4597 X2 =0.6060 - 0.003933 < 3 0.6060 - 0.003933 1 1.4597 X3 = 0.6071 - 0.000005 < 0 4 0.6071 - 0.000005 1 1.4597 X4 =0. Ex โ 2 Find the root of the equation ๐๐ โ ๐๐๐๐ โ ๐ = 0 by the Regula false method in four stages
