Numerical Methods Assignment

Added on - 29 Apr 2020

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NUMERICAL METHODS
ContentsGraphical Method..................................................................................................................................2Bisection Method..................................................................................................................................3False Position Method...........................................................................................................................4Increment Search Method.....................................................................................................................5Simple Fixed point Iteration..................................................................................................................6Newton Raphson Method.....................................................................................................................6Secant Method......................................................................................................................................7Brents Method......................................................................................................................................7Multiple roots........................................................................................................................................8System of Non-linear Equation..............................................................................................................8
Graphical MethodConsider the examplef(x)=x37x+3SolutionThis equation can also be rearranged asx= (x3+3¿/7The roots of the original equationx37x+3=0is represented by the intersection of the graphsy= (x3+3¿/7andy=xThe iteration formulae for the above equation will be:xn+1=(x¿¿n¿¿3+3)/7¿¿Where, n = 0, 1, 2, 3 and so onThe first iteration:x1=(x¿¿0¿¿3+3)7¿¿=(2¿¿3+3)7¿= 1.571428The initial approximationx0is always an assumed value and in this case it’s considered asx0= 22ndIterationx2=(x¿¿1¿¿3+3)7¿¿=(1.571428¿¿3+3)7¿= 0.9829173rdIteration
x3=(x¿¿2¿¿3+3)7¿¿=(0.982917¿¿3+3)7¿= 0.5642294thIterationx4=(x¿¿3¿¿3+3)7¿¿=(0.564229¿¿3+3)7¿= 0.4542285thIterationx5=(x¿¿4¿¿3+3)7¿¿=(0.454228¿¿3+3)7¿= 0.44195966thIterationx6=(x¿¿5¿¿3+3)7¿¿=(0.4419596¿¿3+3)7¿= 0.440903Upon further iteration the root can be found out to be 0.441Bisection MethodConsider the equationf(x)=x23starting on [1, 2]Solutionf(1)=123= -2 (negative value)f(2)=223= 1 (positive value)In this case, the starting point is already provided. But if it is not provided, we need to find 2different approximation of x which yields a negative and positive value.1stIterationx1=(a+b)2=(1+2)2= 1.5(Here, a and b are the starting points)Substituting,x1in the equationf(1.5)=1.523= -0.75This gives a negative value, which means we can neglect the previous negative value interval. Thenew interval becomes [1.5, 2]2ndIterationx2=(a+b)2=(1.5+2)2= 1.75(Here, a and b are the starting points)Substituting,x1in the equationf(1.75)=1.7523= 0.0625
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