Comprehensive IGNOU BMTE-144 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTE-144 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTE-144 Assignment Question Paper 2024

bmte-144-question-paper-00121e82-06b0-4090-b5ab-7e9d6642f92d

bmte-144-question-paper-00121e82-06b0-4090-b5ab-7e9d6642f92d

  1. a) Find the largest real root α α alpha\alphaα of f ( x ) = x 6 x 1 = 0 f ( x ) = x 6 x 1 = 0 f(x)=x^(6)-x-1=0f(x)=x^6-x-1=0f(x)=x6x1=0 lying between 1 and 2. Perform three iterations by
    i) bisection method
    ii) secant method ( x 0 = 2 , x 1 = 1 ) x 0 = 2 , x 1 = 1 (x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)(x0=2,x1=1).
b) Find the number of positive and negative roots of the polynomial P ( x ) = x 3 3 x 2 + 4 x 5 P ( x ) = x 3 3 x 2 + 4 x 5 P(x)=x^(3)-3x^(2)+4x-5P(x)=x^3-3 x^2+4 x-5P(x)=x33x2+4x5. Find P ( 2 ) P ( 2 ) P(2)P(2)P(2) and P ( 2 ) P ( 2 ) P^(‘)(2)P^{\prime}(2)P(2) using synthetic division method.
c) Solve x 3 9 x + 1 = 0 x 3 9 x + 1 = 0 x^(3)-9x+1=0x^3-9 x+1=0x39x+1=0 for the root lying between 2 and 4 by the method of false position. Perform two iterations.
  1. a) Using x 0 = 2 x 0 = 2 x_(0)=-2x_0=-2x0=2 as an initial approximation find an approximation to one of the zeros of
p ( x ) = 2 x 4 3 x 2 + 3 x 4 p ( x ) = 2 x 4 3 x 2 + 3 x 4 p(x)=2x^(4)-3x^(2)+3x-4p(x)=2 x^4-3 x^2+3 x-4p(x)=2x43x2+3x4
by using Birge-Vieta method. Perform two iterations.
b) Find by Newton’s method the roots of the following equations correct to three places of decimals
i) x log 10 x = 4.772393 x log 10 x = 4.772393 quad xlog_(10)x=4.772393\quad x \log _{10} x=4.772393xlog10x=4.772393 near x = 6 x = 6 x=6x=6x=6
ii) f ( x ) = x 2 sin x f ( x ) = x 2 sin x quad f(x)=x-2sin x\quad f(x)=x-2 \sin xf(x)=x2sinx near x = 2 x = 2 x=2x=2x=2
  1. a) The equation x 2 + a x + b = 0 x 2 + a x + b = 0 x^(2)+ax+b=0x^2+a x+b=0x2+ax+b=0 has two real roots p p ppp and q q qqq such that | p | < | q | | p | < | q | |p| < |q||p|<|q||p|<|q|. If we use the fixed point iteration x k + 1 = b x k + a x k + 1 = b x k + a x_(k+1)=(-b)/(x_(k)+a)x_{k+1}=\frac{-b}{x_k+a}xk+1=bxk+a, to find a root then to which root does it converge?
b) Estimate the eigenvalues of the matrix
[ 1 2 3 6 13 18 4 10 14 ] 1 2 3 6 13 18 4 10 14 [[1,-2,3],[6,-13,18],[4,-10,14]]\left[\begin{array}{ccc} 1 & -2 & 3 \\ 6 & -13 & 18 \\ 4 & -10 & 14 \end{array}\right][1236131841014]
using the Gershgorin bounds. Draw a rough sketch of the region where the eigenvalues lie.
c) Find the inverse of the matrix
A = [ 1 1 1 1 2 4 1 2 2 ] A = 1 1 1 1 2 4 1 2 2 A=[[1,-1,1],[1,-2,4],[1,2,2]]A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -2 & 4 \\ 1 & 2 & 2 \end{array}\right]A=[111124122]
using Gauss Jordan method.
  1. a) Solve the system of equations
0.6 x + 0.8 y + 0.1 z = 1 1.1 x + 0.4 y + 0.3 z = 0.2 x + y + 2 z = 0.5 0.6 x + 0.8 y + 0.1 z = 1 1.1 x + 0.4 y + 0.3 z = 0.2 x + y + 2 z = 0.5 {:[0.6 x+0.8 y+0.1 z=1],[1.1 x+0.4 y+0.3 z=0.2],[x+y+2z=0.5]:}\begin{aligned} 0.6 x+0.8 y+0.1 z & =1 \\ 1.1 x+0.4 y+0.3 z & =0.2 \\ x+y+2 z & =0.5 \end{aligned}0.6x+0.8y+0.1z=11.1x+0.4y+0.3z=0.2x+y+2z=0.5
by LU decomposition method and find the inverse of the coefficient matrix
b) For the linear system of equations
[ 1 2 2 1 1 1 2 2 1 ] [ x 1 x 2 x 3 ] = [ 1 3 5 ] 1 2 2 1 1 1 2 2 1 x 1 x 2 x 3 = 1 3 5 [[1,2,-2],[1,1,1],[2,2,1]][[x_(1)],[x_(2)],[x_(3)]]=[[1],[3],[5]]\left[\begin{array}{ccc} 1 & 2 & -2 \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \\ 5 \end{array}\right][122111221][x1x2x3]=[135]
set up the Gauss-Jacobi and Gauss-Seidal iteration schemes in matrix form. Also check the convergence of the two schemes.
  1. a) Find the dominant eigenvalue and the corresponding eigenvector for the matrix
A = [ 4 14 0 5 13 0 1 0 2 ] A = 4 14 0 5 13 0 1 0 2 A=[[-4,14,0],[-5,13,0],[-1,0,2]]A=\left[\begin{array}{ccc} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{array}\right]A=[41405130102]
using five iterations of the power method and taking y ( 0 ) = [ 111 ] T y ( 0 ) = [ 111 ] T y^((0))=[111]^(T)\boldsymbol{y}^{(0)}=[111]^Ty(0)=[111]T as the initial vector.
b) Solve the system of equations
3 x + 2 y + 4 z = 7 2 x + y + z = 7 x + 3 y + 5 z = 2 3 x + 2 y + 4 z = 7 2 x + y + z = 7 x + 3 y + 5 z = 2 {:[3x+2y+4z=7],[2x+y+z=7],[x+3y+5z=2]:}\begin{aligned} 3 x+2 y+4 z & =7 \\ 2 x+y+z & =7 \\ x+3 y+5 z & =2 \end{aligned}3x+2y+4z=72x+y+z=7x+3y+5z=2
with partial pivoting. Store the multipliers and also write the pivoting vectors.
  1. a) Determine the constants α , β , γ α , β , γ alpha,beta,gamma\alpha, \beta, \gammaα,β,γ in the differentiation formula
y ( x 0 ) = α y ( x 0 h ) + β y ( x 0 ) + γ y ( x 0 + h ) y x 0 = α y x 0 h + β y x 0 + γ y x 0 + h y^(‘)(x_(0))=alphay(x_(0)-h)+betay(x_(0))+gammay(x_(0)+h)\mathrm{y}^{\prime}\left(\mathrm{x}_0\right)=\alpha \mathrm{y}\left(\mathrm{x}_0-\mathrm{h}\right)+\beta \mathrm{y}\left(\mathrm{x}_0\right)+\gamma \mathrm{y}\left(\mathrm{x}_0+\mathrm{h}\right)y(x0)=αy(x0h)+βy(x0)+γy(x0+h)
so that the method is of the highest possible order. Find the order and the error term of the method.
b) The function f ( x ) = ln ( 1 + x ) f ( x ) = ln ( 1 + x ) f(x)=ln(1+x)f(x)=\ln (1+x)f(x)=ln(1+x) is to be tabulated at equispaced points in the interval [ 2 , 3 ] [ 2 , 3 ] [2,3][2,3][2,3] using linear interpolation. Find the largest step size h h hhh that can be used so that the error 5 × 10 4 5 × 10 4 <= 5xx10^(-4)\leq 5 \times 10^{-4}5×104 in magnitude.
c) Using finite differences, show that the data
x x xxx -3 -2 -1 0 1 2 3
f ( x ) f ( x ) f(x)f(x)f(x) 13 7 3 1 1 3 7
x -3 -2 -1 0 1 2 3 f(x) 13 7 3 1 1 3 7| $x$ | -3 | -2 | -1 | 0 | 1 | 2 | 3 | | :— | :— | :— | :— | :— | :— | :— | :— | | $f(x)$ | 13 | 7 | 3 | 1 | 1 | 3 | 7 |
represents a second degree polynomial. Obtain this polynomial using interpolation and find f ( 2.5 ) f ( 2.5 ) f(2.5)f(2.5)f(2.5).
  1. a) Derive a suitable numerical differentiation formula of 0 ( h 2 ) 0 h 2 0(h^(2))0\left(h^2\right)0(h2) to find f ( 2.4 ) f ( 2.4 ) f^(”)(2.4)f^{\prime \prime}(2.4)f(2.4) with h = 0.1 h = 0.1 h=0.1h=0.1h=0.1 given the table
x x xxx 0.1 1.2 2.4 3.9
f ( x ) f ( x ) f(x)f(x)f(x) 3.41 2.68 1.37 -1.48
x 0.1 1.2 2.4 3.9 f(x) 3.41 2.68 1.37 -1.48| $x$ | 0.1 | 1.2 | 2.4 | 3.9 | | :— | :— | :— | :— | :— | | $f(x)$ | 3.41 | 2.68 | 1.37 | -1.48 |
b) The position f ( x ) f ( x ) f(x)f(x)f(x) of a particle moving in a line at various times x k x k x_(k)x_kxk is given in the following table. Estimate the velocity and acceleration of the particle at x = 1.2 x = 1.2 x=1.2x=1.2x=1.2.
x x xxx 1.0 1.2 1.4 1.6 1.8 2.0 2.2
f ( x ) f ( x ) f(x)f(x)f(x) 2.72 3.32 4.06 4.96 6.05 7.39 9.02
x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 f(x) 2.72 3.32 4.06 4.96 6.05 7.39 9.02| $x$ | 1.0 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 | 2.2 | | :— | :— | :— | :— | :— | :— | :— | :— | | $f(x)$ | 2.72 | 3.32 | 4.06 | 4.96 | 6.05 | 7.39 | 9.02 |
c) Take 10 figure logarithm to bases 10 from x = 300 x = 300 x=300x=300x=300 to x = 310 x = 310 x=310x=310x=310 by unit increment. Calculate the first derivative of log 10 x log 10 x log_(10)x\log _{10} xlog10x when x = 310 x = 310 x=310x=310x=310.
  1. a) Show that 1 + μ 2 δ 2 = 1 + δ 2 2 1 + μ 2 δ 2 = 1 + δ 2 2 sqrt(1+mu^(2)delta^(2))=1+(delta^(2))/(2)\sqrt{1+\mu^2 \delta^2}=1+\frac{\delta^2}{2}1+μ2δ2=1+δ22 where μ μ mu\muμ and δ δ delta\deltaδ are the average and central differences operators, respectively.
b) A table of values is to be constructed for the function f ( x ) f ( x ) f(x)f(x)f(x) given by f ( x ) = 1 1 + x f ( x ) = 1 1 + x f(x)=(1)/(1+x)f(x)=\frac{1}{1+x}f(x)=11+x in the interval [ 1 , 4 ] [ 1 , 4 ] [1,4][1,4][1,4] with equal step length. Determine the spacing h h hhh such that quadratic interpolation gives result with accuracy 1 × 10 6 1 × 10 6 1xx10^(-6)1 \times 10^{-6}1×106.
c) Using the classical R-K method of O ( h 4 ) O h 4 O(h^(4))O\left(h^4\right)O(h4) calculate approximate solution of the IVP, y = 1 x + 4 y , y ( 0 ) = 1 y = 1 x + 4 y , y ( 0 ) = 1 y^(‘)=1-x+4y,y(0)=1y^{\prime}=1-x+4 y, y(0)=1y=1x+4y,y(0)=1 at x = 0.6 x = 0.6 x=0.6x=0.6x=0.6, taking h = 0.1 h = 0.1 h=0.1h=0.1h=0.1 and 0.2 . Use extrapolation technique to improve the accuracy.
  1. a) Compute the values of
1 = 0 1 d x 1 + x 2 1 = 0 1 d x 1 + x 2 1=int_(0)^(1)(dx)/(1+x^(2))1=\int_0^1 \frac{d x}{1+x^2}1=01dx1+x2
by using the trapezoidal rule with h = 0.5 , 0.25 , 0.125 h = 0.5 , 0.25 , 0.125 h=0.5,0.25,0.125h=0.5,0.25,0.125h=0.5,0.25,0.125. Improve this value by using the Romberg’s method. Compare your result with the true value.
b) Use modified Euler’s method to find the approximate solution of IVP
y = 2 x y , y ( 1 ) = 1 at x = 1.5 with h = 0.1 y = 2 x y , y ( 1 ) = 1 at x = 1.5 with h = 0.1 y^(‘)=2xy,y(1)=1″ at “x=1.5” with “h=0.1y^{\prime}=2 x y, y(1)=1 \text { at } x=1.5 \text { with } h=0.1y=2xy,y(1)=1 at x=1.5 with h=0.1
If the exact solution is y ( x ) = e x 2 1 y ( x ) = e x 2 1 y(x)=e^(x^(2)-1)y(x)=e^{x^2-1}y(x)=ex21, find the error.
c) Show that u x = c 1 e α x + c 2 e α x u x = c 1 e α x + c 2 e α x u_(x)=c_(1)e^(alpha _(x))+c_(2)e^(-alpha _(x))u_x=c_1 e^{\alpha_x}+c_2 e^{-\alpha_x}ux=c1eαx+c2eαx is a solution of the difference equation
u x + 1 2 u x cosh α + u x 1 = 0 . u x + 1 2 u x cosh α + u x 1 = 0 . u_(x+1)-2u_(x)cosh alpha+u_(x-1)=0″. “u_{x+1}-2 u_x \cosh \alpha+u_{x-1}=0 \text {. }ux+12uxcoshα+ux1=0.
  1. a) Using the following table of values, find approximately by Simpson’s rule, the arc length of the graph y = 1 x y = 1 x y=(1)/(x)y=\frac{1}{x}y=1x between the points ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) and ( 5 , 1 5 ) 5 , 1 5 (5,(1)/(5))\left(5, \frac{1}{5}\right)(5,15)
x x xxx 1 2 3 4 5
x 1 2 3 4 5| $x$ | 1 | 2 | 3 | 4 | 5 | | :— | :— | :— | :— | :— | :— |
1 + x 4 x 4 1 + x 4 x 4 sqrt((1+x^(4))/(x^(4)))\sqrt{\frac{1+x^4}{x^4}}1+x4x4 1.414 1.031 1.007 1.002 1.001
sqrt((1+x^(4))/(x^(4))) 1.414 1.031 1.007 1.002 1.001| $\sqrt{\frac{1+x^4}{x^4}}$ | 1.414 | 1.031 | 1.007 | 1.002 | 1.001 | | :—: | :— | :— | :— | :— | :— |
b) i) Calculate the third-degree Taylor polynomial about x 0 = 0 x 0 = 0 x_(0)=0x_0=0x0=0 for f ( x ) = ( 1 + x ) 1 / 2 f ( x ) = ( 1 + x ) 1 / 2 f(x)=(1+x)^(1//2)f(x)=(1+x)^{1 / 2}f(x)=(1+x)1/2.
ii) Use the polynomial in part (i) to approximate 1.1 1.1 sqrt1.1\sqrt{1.1}1.1 and find a bound for the error involved.
iii) Use the polynomial in part (i) to approximate 0 0.1 ( 1 + x ) 1 / 2 d x 0 0.1 ( 1 + x ) 1 / 2 d x int_(0)^(0.1)(1+x)^(1//2)dx\int_0^{0.1}(1+x)^{1 / 2} d x00.1(1+x)1/2dx.
\(b=c\:cos\:A+a\:cos\:C\)

BMTE-144 Sample Solution 2024

bmte-144-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

bmte-144-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

  1. a) Find the largest real root α α alpha\alphaα of f ( x ) = x 6 x 1 = 0 f ( x ) = x 6 x 1 = 0 f(x)=x^(6)-x-1=0f(x)=x^6-x-1=0f(x)=x6x1=0 lying between 1 and 2. Perform three iterations by
    i) bisection method
    ii) secant method ( x 0 = 2 , x 1 = 1 ) x 0 = 2 , x 1 = 1 (x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)(x0=2,x1=1).
Answer:
i) Bisection method
Here x 6 x 1 = 0 x 6 x 1 = 0 x^(6)-x-1=0x^6-x-1=0x6x1=0
Let f ( x ) = x 6 x 1 f ( x ) = x 6 x 1 f(x)=x^(6)-x-1f(x)=x^6-x-1f(x)=x6x1
1 st 1 st 1^(“st “)1^{\text {st }}1st iteration :
Here f ( 1 ) = 1 < 0 f ( 1 ) = 1 < 0 f(1)=-1 < 0f(1)=-1<0f(1)=1<0 and f ( 2 ) = 61 > 0 f ( 2 ) = 61 > 0 f(2)=61 > 0f(2)=61>0f(2)=61>0
:.\therefore Now, Root lies between 1 and 2
x 0 = 1 + 2 2 = 1.5 x 0 = 1 + 2 2 = 1.5 x_(0)=(1+2)/(2)=1.5x_0=\frac{1+2}{2}=1.5x0=1+22=1.5
f ( x 0 ) = f ( 1.5 ) = 1.5 6 1.5 1 = 8.8906 > 0 f x 0 = f ( 1.5 ) = 1.5 6 1.5 1 = 8.8906 > 0 f(x_(0))=f(1.5)=1.5^(6)-1.5-1=8.8906 > 0f\left(x_0\right)=f(1.5)=1.5^6-1.5-1=8.8906>0f(x0)=f(1.5)=1.561.51=8.8906>0
2 nd 2 nd 2^(“nd “)2^{\text {nd }}2nd iteration :
Here f ( 1 ) = 1 < 0 f ( 1 ) = 1 < 0 f(1)=-1 < 0f(1)=-1<0f(1)=1<0 and f ( 1.5 ) = 8.8906 > 0 f ( 1.5 ) = 8.8906 > 0 f(1.5)=8.8906 > 0f(1.5)=8.8906>0f(1.5)=8.8906>0
:.\therefore Now, Root lies between 1 and 1.5
x 1 = 1 + 1.5 2 = 1.25 x 1 = 1 + 1.5 2 = 1.25 x_(1)=(1+1.5)/(2)=1.25x_1=\frac{1+1.5}{2}=1.25x1=1+1.52=1.25
f ( x 1 ) = f ( 1.25 ) = 1.25 6 1.25 1 = 1.5647 > 0 f x 1 = f ( 1.25 ) = 1.25 6 1.25 1 = 1.5647 > 0 f(x_(1))=f(1.25)=1.25^(6)-1.25-1=1.5647 > 0f\left(x_1\right)=f(1.25)=1.25^6-1.25-1=1.5647>0f(x1)=f(1.25)=1.2561.251=1.5647>0
3 r d 3 r d 3^(rd)3^{r d}3rd iteration :
Here f ( 1 ) = 1 < 0 f ( 1 ) = 1 < 0 f(1)=-1 < 0f(1)=-1<0f(1)=1<0 and f ( 1.25 ) = 1.5647 > 0 f ( 1.25 ) = 1.5647 > 0 f(1.25)=1.5647 > 0f(1.25)=1.5647>0f(1.25)=1.5647>0
:.\therefore Now, Root lies between 1 and 1.25
x 2 = 1 + 1.25 2 = 1.125 x 2 = 1 + 1.25 2 = 1.125 x_(2)=(1+1.25)/(2)=1.125x_2=\frac{1+1.25}{2}=1.125x2=1+1.252=1.125
f ( x 2 ) = f ( 1.125 ) = 1.125 6 1.125 1 = 0.0977 < 0 f x 2 = f ( 1.125 ) = 1.125 6 1.125 1 = 0.0977 < 0 f(x_(2))=f(1.125)=1.125^(6)-1.125-1=-0.0977 < 0f\left(x_2\right)=f(1.125)=1.125^6-1.125-1=-0.0977<0f(x2)=f(1.125)=1.12561.1251=0.0977<0
After three iterations, the interval containing the largest real root α α alpha\alphaα is narrowed down to [ 1.125 , 1.25 ] [ 1.125 , 1.25 ] [1.125,1.25][1.125,1.25][1.125,1.25]. The midpoint of this interval, x 3 = 1.1875 x 3 = 1.1875 x_(3)=1.1875x_3=1.1875x3=1.1875, is the best approximation of the root after three iterations using the bisection method.
ii) Secant method ( x 0 = 2 , x 1 = 1 ) x 0 = 2 , x 1 = 1 (x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)(x0=2,x1=1).
Let f ( x ) = x 6 x 1 f ( x ) = x 6 x 1 f(x)=x^(6)-x-1f(x)=x^6-x-1f(x)=x6x1
1 st 1 st 1^(“st “)1^{\text {st }}1st iteration :
x 0 = 1 x 0 = 1 x_(0)=1x_0=1x0=1 and x 1 = 2 x 1 = 2 x_(1)=2x_1=2x1=2
f ( x 0 ) = f ( 1 ) = 1 f x 0 = f ( 1 ) = 1 f(x_(0))=f(1)=-1f\left(x_0\right)=f(1)=-1f(x0)=f(1)=1 and f ( x 1 ) = f ( 2 ) = 61 f x 1 = f ( 2 ) = 61 f(x_(1))=f(2)=61f\left(x_1\right)=f(2)=61f(x1)=f(2)=61
x 2 = x 0 f ( x 0 ) x 1 x 0 f ( x 1 ) f ( x 0 ) x 2 = x 0 f x 0 x 1 x 0 f x 1 f x 0 :.x_(2)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))\therefore x_2=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}x2=x0f(x0)x1x0f(x1)f(x0)
x 2 = 1 ( 1 ) 2 1 61 ( 1 ) x 2 = 1 ( 1 ) 2 1 61 ( 1 ) x_(2)=1-(-1)*(2-1)/(61-(-1))x_2=1-(-1) \cdot \frac{2-1}{61-(-1)}x2=1(1)2161(1)
x 2 = 1.0161 x 2 = 1.0161 x_(2)=1.0161x_2=1.0161x2=1.0161
f ( x 2 ) = f ( 1.0161 ) = 1.0161 6 1.0161 1 = 0.9154 f x 2 = f ( 1.0161 ) = 1.0161 6 1.0161 1 = 0.9154 :.f(x_(2))=f(1.0161)=1.0161^(6)-1.0161-1=-0.9154\therefore f\left(x_2\right)=f(1.0161)=1.0161^6-1.0161-1=-0.9154f(x2)=f(1.0161)=1.016161.01611=0.9154
2 nd 2 nd 2^(“nd “)2^{\text {nd }}2nd iteration :
x 1 = 2 x 1 = 2 x_(1)=2x_1=2x1=2 and x 2 = 1.0161 x 2 = 1.0161 x_(2)=1.0161x_2=1.0161x2=1.0161
f ( x 1 ) = f ( 2 ) = 61 and f ( x 2 ) = f ( 1.0161 ) = 0.9154 f x 1 = f ( 2 ) = 61 and f x 2 = f ( 1.0161 ) = 0.9154 f(x_(1))=f(2)=61″ and “f(x_(2))=f(1.0161)=-0.9154f\left(x_1\right)=f(2)=61 \text { and } f\left(x_2\right)=f(1.0161)=-0.9154f(x1)=f(2)=61 and f(x2)=f(1.0161)=0.9154
x 3 = x 1 f ( x 1 ) x 2 x 1 f ( x 2 ) f ( x 1 ) x 3 = x 1 f x 1 x 2 x 1 f x 2 f x 1 :.x_(3)=x_(1)-f(x_(1))*(x_(2)-x_(1))/(f(x_(2))-f(x_(1)))\therefore x_3=x_1-f\left(x_1\right) \cdot \frac{x_2-x_1}{f\left(x_2\right)-f\left(x_1\right)}x3=x1f(x1)x2x1f(x2)f(x1)
x 3 = 2 61 1.0161 2 0.9154 61 x 3 = 2 61 1.0161 2 0.9154 61 x_(3)=2-61*(1.0161-2)/(-0.9154-61)x_3=2-61 \cdot \frac{1.0161-2}{-0.9154-61}x3=2611.016120.915461
x 3 = 1.0307 x 3 = 1.0307 x_(3)=1.0307x_3=1.0307x3=1.0307
f ( x 3 ) = f ( 1.0307 ) = 1.0307 6 1.0307 1 = 0.8319 f x 3 = f ( 1.0307 ) = 1.0307 6 1.0307 1 = 0.8319 :.f(x_(3))=f(1.0307)=1.0307^(6)-1.0307-1=-0.8319\therefore f\left(x_3\right)=f(1.0307)=1.0307^6-1.0307-1=-0.8319f(x3)=f(1.0307)=1.030761.03071=0.8319
3 r d 3 r d 3^(rd)3^{r d}3rd iteration :
x 2 = 1.0161 and x 3 = 1.0307 x 2 = 1.0161 and x 3 = 1.0307 x_(2)=1.0161″ and “x_(3)=1.0307x_2=1.0161 \text { and } x_3=1.0307