# 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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Details For BMTE-144 Solved Assignment

## 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 $\alpha$$\alpha$alpha\alpha$\alpha$ of $f\left(x\right)={x}^{6}-x-1=0$$f\left(x\right)={x}^{6}-x-1=0$f(x)=x^(6)-x-1=0f(x)=x^6-x-1=0$f\left(x\right)={x}^{6}-x-1=0$ lying between 1 and 2. Perform three iterations by
i) bisection method
ii) secant method $\left({x}_{0}=2,{x}_{1}=1\right)$$\left({x}_{0}=2,{x}_{1}=1\right)$(x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)$\left({x}_{0}=2,{x}_{1}=1\right)$.
b) Find the number of positive and negative roots of the polynomial $P\left(x\right)={x}^{3}-3{x}^{2}+4x-5$$P\left(x\right)={x}^{3}-3{x}^{2}+4x-5$P(x)=x^(3)-3x^(2)+4x-5P(x)=x^3-3 x^2+4 x-5$P\left(x\right)={x}^{3}-3{x}^{2}+4x-5$. Find $P\left(2\right)$$P\left(2\right)$P(2)P(2)$P\left(2\right)$ and ${P}^{\mathrm{\prime }}\left(2\right)$${P}^{\mathrm{\prime }}\left(2\right)$P^(‘)(2)P^{\prime}(2)${P}^{\mathrm{\prime }}\left(2\right)$ using synthetic division method.
c) Solve ${x}^{3}-9x+1=0$${x}^{3}-9x+1=0$x^(3)-9x+1=0x^3-9 x+1=0${x}^{3}-9x+1=0$ for the root lying between 2 and 4 by the method of false position. Perform two iterations.
1. a) Using ${x}_{0}=-2$${x}_{0}=-2$x_(0)=-2x_0=-2${x}_{0}=-2$ as an initial approximation find an approximation to one of the zeros of
$p\left(x\right)=2{x}^{4}-3{x}^{2}+3x-4$$p\left(x\right)=2{x}^{4}-3{x}^{2}+3x-4$p(x)=2x^(4)-3x^(2)+3x-4p(x)=2 x^4-3 x^2+3 x-4$p\left(x\right)=2{x}^{4}-3{x}^{2}+3x-4$
by using Birge-Vieta method. Perform two iterations.
b) Find by Newton’s method the roots of the following equations correct to three places of decimals
i) $\phantom{\rule{1em}{0ex}}x{\mathrm{log}}_{10}x=4.772393$$\phantom{\rule{1em}{0ex}}x{\mathrm{log}}_{10}x=4.772393$quad xlog_(10)x=4.772393\quad x \log _{10} x=4.772393$\phantom{\rule{1em}{0ex}}x{\mathrm{log}}_{10}x=4.772393$ near $x=6$$x=6$x=6x=6$x=6$
ii) $\phantom{\rule{1em}{0ex}}f\left(x\right)=x-2\mathrm{sin}x$$\phantom{\rule{1em}{0ex}}f\left(x\right)=x-2\mathrm{sin}x$quad f(x)=x-2sin x\quad f(x)=x-2 \sin x$\phantom{\rule{1em}{0ex}}f\left(x\right)=x-2\mathrm{sin}x$ near $x=2$$x=2$x=2x=2$x=2$
1. a) The equation ${x}^{2}+ax+b=0$${x}^{2}+ax+b=0$x^(2)+ax+b=0x^2+a x+b=0${x}^{2}+ax+b=0$ has two real roots $p$$p$pp$p$ and $q$$q$qq$q$ such that $|p|<|q|$$|p|<|q|$|p| < |q||p|<|q|$|p|<|q|$. If we use the fixed point iteration ${x}_{k+1}=\frac{-b}{{x}_{k}+a}$${x}_{k+1}=\frac{-b}{{x}_{k}+a}$x_(k+1)=(-b)/(x_(k)+a)x_{k+1}=\frac{-b}{x_k+a}${x}_{k+1}=\frac{-b}{{x}_{k}+a}$, to find a root then to which root does it converge?
b) Estimate the eigenvalues of the matrix
$\left[\begin{array}{ccc}1& -2& 3\\ 6& -13& 18\\ 4& -10& 14\end{array}\right]$$\left[\begin{array}{ccc}1& -2& 3\\ 6& -13& 18\\ 4& -10& 14\end{array}\right]$[[1,-2,3],[6,-13,18],[4,-10,14]]\left[\begin{array}{ccc} 1 & -2 & 3 \\ 6 & -13 & 18 \\ 4 & -10 & 14 \end{array}\right]$\left[\begin{array}{ccc}1& -2& 3\\ 6& -13& 18\\ 4& -10& 14\end{array}\right]$
using the Gershgorin bounds. Draw a rough sketch of the region where the eigenvalues lie.
c) Find the inverse of the matrix
$A=\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 4\\ 1& 2& 2\end{array}\right]$$A=\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 4\\ 1& 2& 2\end{array}\right]$A=[[1,-1,1],[1,-2,4],[1,2,2]]A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -2 & 4 \\ 1 & 2 & 2 \end{array}\right]$A=\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 4\\ 1& 2& 2\end{array}\right]$
using Gauss Jordan method.
1. a) Solve the system of equations
$\begin{array}{rl}0.6x+0.8y+0.1z& =1\\ 1.1x+0.4y+0.3z& =0.2\\ x+y+2z& =0.5\end{array}$$\begin{array}{r}0.6x+0.8y+0.1z=1\\ 1.1x+0.4y+0.3z=0.2\\ x+y+2z=0.5\end{array}${:[0.6 x+0.8 y+0.1 z=1],[1.1 x+0.4 y+0.3 z=0.2],[x+y+2z=0.5]:}\begin{aligned} 0.6 x+0.8 y+0.1 z & =1 \\ 1.1 x+0.4 y+0.3 z & =0.2 \\ x+y+2 z & =0.5 \end{aligned}$\begin{array}{rl}0.6x+0.8y+0.1z& =1\\ 1.1x+0.4y+0.3z& =0.2\\ x+y+2z& =0.5\end{array}$
by LU decomposition method and find the inverse of the coefficient matrix
b) For the linear system of equations
$\left[\begin{array}{ccc}1& 2& -2\\ 1& 1& 1\\ 2& 2& 1\end{array}\right]\left[\begin{array}{l}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{l}1\\ 3\\ 5\end{array}\right]$$\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]$[[1,2,-2],[1,1,1],[2,2,1]][[x_(1)],[x_(2)],[x_(3)]]=[[1],[3],[5]]\left[\begin{array}{ccc} 1 & 2 & -2 \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \\ 5 \end{array}\right]$\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]$
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=\left[\begin{array}{ccc}-4& 14& 0\\ -5& 13& 0\\ -1& 0& 2\end{array}\right]$$A=\left[\begin{array}{ccc}-4& 14& 0\\ -5& 13& 0\\ -1& 0& 2\end{array}\right]$A=[[-4,14,0],[-5,13,0],[-1,0,2]]A=\left[\begin{array}{ccc} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{array}\right]$A=\left[\begin{array}{ccc}-4& 14& 0\\ -5& 13& 0\\ -1& 0& 2\end{array}\right]$
using five iterations of the power method and taking ${\mathbit{y}}^{\left(0\right)}=\left[111{\right]}^{T}$${\mathbit{y}}^{\left(0\right)}=\left[111{\right]}^{T}$y^((0))=[111]^(T)\boldsymbol{y}^{(0)}=[111]^T${\mathbit{y}}^{\left(0\right)}=\left[111{\right]}^{T}$ as the initial vector.
b) Solve the system of equations
$\begin{array}{rl}3x+2y+4z& =7\\ 2x+y+z& =7\\ x+3y+5z& =2\end{array}$$\begin{array}{r}3x+2y+4z=7\\ 2x+y+z=7\\ x+3y+5z=2\end{array}${:[3x+2y+4z=7],[2x+y+z=7],[x+3y+5z=2]:}\begin{aligned} 3 x+2 y+4 z & =7 \\ 2 x+y+z & =7 \\ x+3 y+5 z & =2 \end{aligned}$\begin{array}{rl}3x+2y+4z& =7\\ 2x+y+z& =7\\ x+3y+5z& =2\end{array}$
with partial pivoting. Store the multipliers and also write the pivoting vectors.
1. a) Determine the constants $\alpha ,\beta ,\gamma$$\alpha ,\beta ,\gamma$alpha,beta,gamma\alpha, \beta, \gamma$\alpha ,\beta ,\gamma$ in the differentiation formula
${\mathrm{y}}^{\mathrm{\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)$${\mathrm{y}}^{\mathrm{\prime }}\left({\mathrm{x}}_{0}\right)=\alpha \mathrm{y}\left({\mathrm{x}}_{0}-\mathrm{h}\right)+\beta \mathrm{y}\left({\mathrm{x}}_{0}\right)+\gamma \mathrm{y}\left({\mathrm{x}}_{0}+\mathrm{h}\right)$y^(‘)(x_(0))=alphay(x_(0)-h)+betay(x_(0))+gammay(x_(0)+h)\mathrm{y}^{\prime}\left(\mathrm{x}_0\right)=\alpha \mathrm{y}\left(\mathrm{x}_0-\mathrm{h}\right)+\beta \mathrm{y}\left(\mathrm{x}_0\right)+\gamma \mathrm{y}\left(\mathrm{x}_0+\mathrm{h}\right)${\mathrm{y}}^{\mathrm{\prime }}\left({\mathrm{x}}_{0}\right)=\alpha \mathrm{y}\left({\mathrm{x}}_{0}-\mathrm{h}\right)+\beta \mathrm{y}\left({\mathrm{x}}_{0}\right)+\gamma \mathrm{y}\left({\mathrm{x}}_{0}+\mathrm{h}\right)$
so that the method is of the highest possible order. Find the order and the error term of the method.
b) The function $f\left(x\right)=\mathrm{ln}\left(1+x\right)$$f\left(x\right)=\mathrm{ln}\left(1+x\right)$f(x)=ln(1+x)f(x)=\ln (1+x)$f\left(x\right)=\mathrm{ln}\left(1+x\right)$ is to be tabulated at equispaced points in the interval $\left[2,3\right]$$\left[2,3\right]$[2,3][2,3]$\left[2,3\right]$ using linear interpolation. Find the largest step size $h$$h$hh$h$ that can be used so that the error $\le 5×{10}^{-4}$$\le 5×{10}^{-4}$<= 5xx10^(-4)\leq 5 \times 10^{-4}$\le 5×{10}^{-4}$ in magnitude.
c) Using finite differences, show that the data
 $x$$x$xx$x$ -3 -2 -1 0 1 2 3 $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ 13 7 3 1 1 3 7
x -3 -2 -1 0 1 2 3 f(x) 13 7 3 1 1 3 7| $x$ | -3 | -2 | -1 | 0 | 1 | 2 | 3 | | :— | :— | :— | :— | :— | :— | :— | :— | | $f(x)$ | 13 | 7 | 3 | 1 | 1 | 3 | 7 |
represents a second degree polynomial. Obtain this polynomial using interpolation and find $f\left(2.5\right)$$f\left(2.5\right)$f(2.5)f(2.5)$f\left(2.5\right)$.
1. a) Derive a suitable numerical differentiation formula of $0\left({h}^{2}\right)$$0\left({h}^{2}\right)$0(h^(2))0\left(h^2\right)$0\left({h}^{2}\right)$ to find ${f}^{\mathrm{\prime }\mathrm{\prime }}\left(2.4\right)$${f}^{\mathrm{\prime }\mathrm{\prime }}\left(2.4\right)$f^(”)(2.4)f^{\prime \prime}(2.4)${f}^{\mathrm{\prime }\mathrm{\prime }}\left(2.4\right)$ with $h=0.1$$h=0.1$h=0.1h=0.1$h=0.1$ given the table
 $x$$x$xx$x$ 0.1 1.2 2.4 3.9 $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ 3.41 2.68 1.37 -1.48
x 0.1 1.2 2.4 3.9 f(x) 3.41 2.68 1.37 -1.48| $x$ | 0.1 | 1.2 | 2.4 | 3.9 | | :— | :— | :— | :— | :— | | $f(x)$ | 3.41 | 2.68 | 1.37 | -1.48 |
b) The position $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ of a particle moving in a line at various times ${x}_{k}$${x}_{k}$x_(k)x_k${x}_{k}$ is given in the following table. Estimate the velocity and acceleration of the particle at $x=1.2$$x=1.2$x=1.2x=1.2$x=1.2$.
 $x$$x$xx$x$ 1 1.2 1.4 1.6 1.8 2 2.2 $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ 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$$x=300$x=300x=300$x=300$ to $x=310$$x=310$x=310x=310$x=310$ by unit increment. Calculate the first derivative of ${\mathrm{log}}_{10}x$${\mathrm{log}}_{10}x$log_(10)x\log _{10} x${\mathrm{log}}_{10}x$ when $x=310$$x=310$x=310x=310$x=310$.
1. a) Show that $\sqrt{1+{\mu }^{2}{\delta }^{2}}=1+\frac{{\delta }^{2}}{2}$$\sqrt{1+{\mu }^{2}{\delta }^{2}}=1+\frac{{\delta }^{2}}{2}$sqrt(1+mu^(2)delta^(2))=1+(delta^(2))/(2)\sqrt{1+\mu^2 \delta^2}=1+\frac{\delta^2}{2}$\sqrt{1+{\mu }^{2}{\delta }^{2}}=1+\frac{{\delta }^{2}}{2}$ where $\mu$$\mu$mu\mu$\mu$ and $\delta$$\delta$delta\delta$\delta$ are the average and central differences operators, respectively.
b) A table of values is to be constructed for the function $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ given by $f\left(x\right)=\frac{1}{1+x}$$f\left(x\right)=\frac{1}{1+x}$f(x)=(1)/(1+x)f(x)=\frac{1}{1+x}$f\left(x\right)=\frac{1}{1+x}$ in the interval $\left[1,4\right]$$\left[1,4\right]$[1,4][1,4]$\left[1,4\right]$ with equal step length. Determine the spacing $h$$h$hh$h$ such that quadratic interpolation gives result with accuracy $1×{10}^{-6}$$1×{10}^{-6}$1xx10^(-6)1 \times 10^{-6}$1×{10}^{-6}$.
c) Using the classical R-K method of $O\left({h}^{4}\right)$$O\left({h}^{4}\right)$O(h^(4))O\left(h^4\right)$O\left({h}^{4}\right)$ calculate approximate solution of the IVP, ${y}^{\mathrm{\prime }}=1-x+4y,y\left(0\right)=1$${y}^{\mathrm{\prime }}=1-x+4y,y\left(0\right)=1$y^(‘)=1-x+4y,y(0)=1y^{\prime}=1-x+4 y, y(0)=1${y}^{\mathrm{\prime }}=1-x+4y,y\left(0\right)=1$ at $x=0.6$$x=0.6$x=0.6x=0.6$x=0.6$, taking $h=0.1$$h=0.1$h=0.1h=0.1$h=0.1$ and 0.2 . Use extrapolation technique to improve the accuracy.
1. a) Compute the values of
$1={\int }_{0}^{1}\frac{dx}{1+{x}^{2}}$$1={\int }_{0}^{1} \frac{dx}{1+{x}^{2}}$1=int_(0)^(1)(dx)/(1+x^(2))1=\int_0^1 \frac{d x}{1+x^2}$1={\int }_{0}^{1}\frac{dx}{1+{x}^{2}}$
by using the trapezoidal rule with $h=0.5,0.25,0.125$$h=0.5,0.25,0.125$h=0.5,0.25,0.125h=0.5,0.25,0.125$h=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}^{\mathrm{\prime }}=2xy,y\left(1\right)=1\text{at}x=1.5\text{with}h=0.1$${y}^{\mathrm{\prime }}=2xy,y\left(1\right)=1\text{at}x=1.5\text{with}h=0.1$y^(‘)=2xy,y(1)=1″ at “x=1.5” with “h=0.1y^{\prime}=2 x y, y(1)=1 \text { at } x=1.5 \text { with } h=0.1${y}^{\mathrm{\prime }}=2xy,y\left(1\right)=1\text{at}x=1.5\text{with}h=0.1$
If the exact solution is $y\left(x\right)={e}^{{x}^{2}-1}$$y\left(x\right)={e}^{{x}^{2}-1}$y(x)=e^(x^(2)-1)y(x)=e^{x^2-1}$y\left(x\right)={e}^{{x}^{2}-1}$, find the error.
c) Show that ${u}_{x}={c}_{1}{e}^{{\alpha }_{x}}+{c}_{2}{e}^{-{\alpha }_{x}}$${u}_{x}={c}_{1}{e}^{{\alpha }_{x}}+{c}_{2}{e}^{-{\alpha }_{x}}$u_(x)=c_(1)e^(alpha _(x))+c_(2)e^(-alpha _(x))u_x=c_1 e^{\alpha_x}+c_2 e^{-\alpha_x}${u}_{x}={c}_{1}{e}^{{\alpha }_{x}}+{c}_{2}{e}^{-{\alpha }_{x}}$ is a solution of the difference equation
${u}_{x+1}-2{u}_{x}\mathrm{cosh}\alpha +{u}_{x-1}=0\text{.}$${u}_{x+1}-2{u}_{x}\mathrm{cosh}\alpha +{u}_{x-1}=0\text{.}$u_(x+1)-2u_(x)cosh alpha+u_(x-1)=0″. “u_{x+1}-2 u_x \cosh \alpha+u_{x-1}=0 \text {. }${u}_{x+1}-2{u}_{x}\mathrm{cosh}\alpha +{u}_{x-1}=0\text{.}$
1. a) Using the following table of values, find approximately by Simpson’s rule, the arc length of the graph $y=\frac{1}{x}$$y=\frac{1}{x}$y=(1)/(x)y=\frac{1}{x}$y=\frac{1}{x}$ between the points $\left(1,1\right)$$\left(1,1\right)$(1,1)(1,1)$\left(1,1\right)$ and $\left(5,\frac{1}{5}\right)$$\left(5,\frac{1}{5}\right)$(5,(1)/(5))\left(5, \frac{1}{5}\right)$\left(5,\frac{1}{5}\right)$
 $x$$x$xx$x$ 1 2 3 4 5
x 1 2 3 4 5| $x$ | 1 | 2 | 3 | 4 | 5 | | :— | :— | :— | :— | :— | :— |
 $\sqrt{\frac{1+{x}^{4}}{{x}^{4}}}$$\sqrt{\frac{1+{x}^{4}}{{x}^{4}}}$sqrt((1+x^(4))/(x^(4)))\sqrt{\frac{1+x^4}{x^4}}$\sqrt{\frac{1+{x}^{4}}{{x}^{4}}}$ 1.414 1.031 1.007 1.002 1.001
sqrt((1+x^(4))/(x^(4))) 1.414 1.031 1.007 1.002 1.001| $\sqrt{\frac{1+x^4}{x^4}}$ | 1.414 | 1.031 | 1.007 | 1.002 | 1.001 | | :—: | :— | :— | :— | :— | :— |
b) i) Calculate the third-degree Taylor polynomial about ${x}_{0}=0$${x}_{0}=0$x_(0)=0x_0=0${x}_{0}=0$ for $f\left(x\right)=\left(1+x{\right)}^{1/2}$$f\left(x\right)=\left(1+x{\right)}^{1/2}$f(x)=(1+x)^(1//2)f(x)=(1+x)^{1 / 2}$f\left(x\right)=\left(1+x{\right)}^{1/2}$.
ii) Use the polynomial in part (i) to approximate $\sqrt{1.1}$$\sqrt{1.1}$sqrt1.1\sqrt{1.1}$\sqrt{1.1}$ and find a bound for the error involved.
iii) Use the polynomial in part (i) to approximate ${\int }_{0}^{0.1}\left(1+x{\right)}^{1/2}dx$${\int }_{0}^{0.1} \left(1+x{\right)}^{1/2}dx$int_(0)^(0.1)(1+x)^(1//2)dx\int_0^{0.1}(1+x)^{1 / 2} d x${\int }_{0}^{0.1}\left(1+x{\right)}^{1/2}dx$.
$$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 $\alpha$$\alpha$alpha\alpha$\alpha$ of $f\left(x\right)={x}^{6}-x-1=0$$f\left(x\right)={x}^{6}-x-1=0$f(x)=x^(6)-x-1=0f(x)=x^6-x-1=0$f\left(x\right)={x}^{6}-x-1=0$ lying between 1 and 2. Perform three iterations by
i) bisection method
ii) secant method $\left({x}_{0}=2,{x}_{1}=1\right)$$\left({x}_{0}=2,{x}_{1}=1\right)$(x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)$\left({x}_{0}=2,{x}_{1}=1\right)$.
i) Bisection method
Here ${x}^{6}-x-1=0$${x}^{6}-x-1=0$x^(6)-x-1=0x^6-x-1=0${x}^{6}-x-1=0$
Let $f\left(x\right)={x}^{6}-x-1$$f\left(x\right)={x}^{6}-x-1$f(x)=x^(6)-x-1f(x)=x^6-x-1$f\left(x\right)={x}^{6}-x-1$
${1}^{\text{st}}$${1}^{\text{st}}$1^(“st “)1^{\text {st }}${1}^{\text{st}}$ iteration :
Here $f\left(1\right)=-1<0$$f\left(1\right)=-1<0$f(1)=-1 < 0f(1)=-1<0$f\left(1\right)=-1<0$ and $f\left(2\right)=61>0$$f\left(2\right)=61>0$f(2)=61 > 0f(2)=61>0$f\left(2\right)=61>0$
$\therefore$$\therefore$:.\therefore$\therefore$ Now, Root lies between 1 and 2
${x}_{0}=\frac{1+2}{2}=1.5$${x}_{0}=\frac{1+2}{2}=1.5$x_(0)=(1+2)/(2)=1.5x_0=\frac{1+2}{2}=1.5${x}_{0}=\frac{1+2}{2}=1.5$
$f\left({x}_{0}\right)=f\left(1.5\right)={1.5}^{6}-1.5-1=8.8906>0$$f\left({x}_{0}\right)=f\left(1.5\right)={1.5}^{6}-1.5-1=8.8906>0$f(x_(0))=f(1.5)=1.5^(6)-1.5-1=8.8906 > 0f\left(x_0\right)=f(1.5)=1.5^6-1.5-1=8.8906>0$f\left({x}_{0}\right)=f\left(1.5\right)={1.5}^{6}-1.5-1=8.8906>0$
${2}^{\text{nd}}$${2}^{\text{nd}}$2^(“nd “)2^{\text {nd }}${2}^{\text{nd}}$ iteration :
Here $f\left(1\right)=-1<0$$f\left(1\right)=-1<0$f(1)=-1 < 0f(1)=-1<0$f\left(1\right)=-1<0$ and $f\left(1.5\right)=8.8906>0$$f\left(1.5\right)=8.8906>0$f(1.5)=8.8906 > 0f(1.5)=8.8906>0$f\left(1.5\right)=8.8906>0$
$\therefore$$\therefore$:.\therefore$\therefore$ Now, Root lies between 1 and 1.5
${x}_{1}=\frac{1+1.5}{2}=1.25$${x}_{1}=\frac{1+1.5}{2}=1.25$x_(1)=(1+1.5)/(2)=1.25x_1=\frac{1+1.5}{2}=1.25${x}_{1}=\frac{1+1.5}{2}=1.25$
$f\left({x}_{1}\right)=f\left(1.25\right)={1.25}^{6}-1.25-1=1.5647>0$$f\left({x}_{1}\right)=f\left(1.25\right)={1.25}^{6}-1.25-1=1.5647>0$f(x_(1))=f(1.25)=1.25^(6)-1.25-1=1.5647 > 0f\left(x_1\right)=f(1.25)=1.25^6-1.25-1=1.5647>0$f\left({x}_{1}\right)=f\left(1.25\right)={1.25}^{6}-1.25-1=1.5647>0$
${3}^{rd}$${3}^{rd}$3^(rd)3^{r d}${3}^{rd}$ iteration :
Here $f\left(1\right)=-1<0$$f\left(1\right)=-1<0$f(1)=-1 < 0f(1)=-1<0$f\left(1\right)=-1<0$ and $f\left(1.25\right)=1.5647>0$$f\left(1.25\right)=1.5647>0$f(1.25)=1.5647 > 0f(1.25)=1.5647>0$f\left(1.25\right)=1.5647>0$
$\therefore$$\therefore$:.\therefore$\therefore$ Now, Root lies between 1 and 1.25
${x}_{2}=\frac{1+1.25}{2}=1.125$${x}_{2}=\frac{1+1.25}{2}=1.125$x_(2)=(1+1.25)/(2)=1.125x_2=\frac{1+1.25}{2}=1.125${x}_{2}=\frac{1+1.25}{2}=1.125$
$f\left({x}_{2}\right)=f\left(1.125\right)={1.125}^{6}-1.125-1=-0.0977<0$$f\left({x}_{2}\right)=f\left(1.125\right)={1.125}^{6}-1.125-1=-0.0977<0$f(x_(2))=f(1.125)=1.125^(6)-1.125-1=-0.0977 < 0f\left(x_2\right)=f(1.125)=1.125^6-1.125-1=-0.0977<0$f\left({x}_{2}\right)=f\left(1.125\right)={1.125}^{6}-1.125-1=-0.0977<0$
After three iterations, the interval containing the largest real root $\alpha$$\alpha$alpha\alpha$\alpha$ is narrowed down to $\left[1.125,1.25\right]$$\left[1.125,1.25\right]$[1.125,1.25][1.125,1.25]$\left[1.125,1.25\right]$. The midpoint of this interval, ${x}_{3}=1.1875$${x}_{3}=1.1875$x_(3)=1.1875x_3=1.1875${x}_{3}=1.1875$, is the best approximation of the root after three iterations using the bisection method.
ii) Secant method $\left({x}_{0}=2,{x}_{1}=1\right)$$\left({x}_{0}=2,{x}_{1}=1\right)$(x_(0)=2,x_(1)=1)\left(x_0=2, x_1=1\right)$\left({x}_{0}=2,{x}_{1}=1\right)$.
Let $f\left(x\right)={x}^{6}-x-1$$f\left(x\right)={x}^{6}-x-1$f(x)=x^(6)-x-1f(x)=x^6-x-1$f\left(x\right)={x}^{6}-x-1$
${1}^{\text{st}}$${1}^{\text{st}}$1^(“st “)1^{\text {st }}${1}^{\text{st}}$ iteration :
${x}_{0}=1$${x}_{0}=1$x_(0)=1x_0=1${x}_{0}=1$ and ${x}_{1}=2$${x}_{1}=2$x_(1)=2x_1=2${x}_{1}=2$
$f\left({x}_{0}\right)=f\left(1\right)=-1$$f\left({x}_{0}\right)=f\left(1\right)=-1$f(x_(0))=f(1)=-1f\left(x_0\right)=f(1)=-1$f\left({x}_{0}\right)=f\left(1\right)=-1$ and $f\left({x}_{1}\right)=f\left(2\right)=61$$f\left({x}_{1}\right)=f\left(2\right)=61$f(x_(1))=f(2)=61f\left(x_1\right)=f(2)=61$f\left({x}_{1}\right)=f\left(2\right)=61$
$\therefore {x}_{2}={x}_{0}-f\left({x}_{0}\right)\cdot \frac{{x}_{1}-{x}_{0}}{f\left({x}_{1}\right)-f\left({x}_{0}\right)}$$\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)}$:.x_(2)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))\therefore x_2=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}$\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)}$
${x}_{2}=1-\left(-1\right)\cdot \frac{2-1}{61-\left(-1\right)}$${x}_{2}=1-\left(-1\right)\cdot \frac{2-1}{61-\left(-1\right)}$x_(2)=1-(-1)*(2-1)/(61-(-1))x_2=1-(-1) \cdot \frac{2-1}{61-(-1)}${x}_{2}=1-\left(-1\right)\cdot \frac{2-1}{61-\left(-1\right)}$
${x}_{2}=1.0161$${x}_{2}=1.0161$x_(2)=1.0161x_2=1.0161${x}_{2}=1.0161$
$\therefore f\left({x}_{2}\right)=f\left(1.0161\right)={1.0161}^{6}-1.0161-1=-0.9154$$\therefore f\left({x}_{2}\right)=f\left(1.0161\right)={1.0161}^{6}-1.0161-1=-0.9154$:.f(x_(2))=f(1.0161)=1.0161^(6)-1.0161-1=-0.9154\therefore f\left(x_2\right)=f(1.0161)=1.0161^6-1.0161-1=-0.9154$\therefore f\left({x}_{2}\right)=f\left(1.0161\right)={1.0161}^{6}-1.0161-1=-0.9154$
${2}^{\text{nd}}$${2}^{\text{nd}}$2^(“nd “)2^{\text {nd }}${2}^{\text{nd}}$ iteration :
${x}_{1}=2$${x}_{1}=2$x_(1)=2x_1=2${x}_{1}=2$ and ${x}_{2}=1.0161$${x}_{2}=1.0161$x_(2)=1.0161x_2=1.0161${x}_{2}=1.0161$
$f\left({x}_{1}\right)=f\left(2\right)=61\text{and}f\left({x}_{2}\right)=f\left(1.0161\right)=-0.9154$$f\left({x}_{1}\right)=f\left(2\right)=61\text{and}f\left({x}_{2}\right)=f\left(1.0161\right)=-0.9154$f(x_(1))=f(2)=61″ and “f(x_(2))=f(1.0161)=-0.9154f\left(x_1\right)=f(2)=61 \text { and } f\left(x_2\right)=f(1.0161)=-0.9154$f\left({x}_{1}\right)=f\left(2\right)=61\text{and}f\left({x}_{2}\right)=f\left(1.0161\right)=-0.9154$
$\therefore {x}_{3}={x}_{1}-f\left({x}_{1}\right)\cdot \frac{{x}_{2}-{x}_{1}}{f\left({x}_{2}\right)-f\left({x}_{1}\right)}$$\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)}$:.x_(3)=x_(1)-f(x_(1))*(x_(2)-x_(1))/(f(x_(2))-f(x_(1)))\therefore x_3=x_1-f\left(x_1\right) \cdot \frac{x_2-x_1}{f\left(x_2\right)-f\left(x_1\right)}$\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)}$
${x}_{3}=2-61\cdot \frac{1.0161-2}{-0.9154-61}$${x}_{3}=2-61\cdot \frac{1.0161-2}{-0.9154-61}$x_(3)=2-61*(1.0161-2)/(-0.9154-61)x_3=2-61 \cdot \frac{1.0161-2}{-0.9154-61}${x}_{3}=2-61\cdot \frac{1.0161-2}{-0.9154-61}$
${x}_{3}=1.0307$${x}_{3}=1.0307$x_(3)=1.0307x_3=1.0307${x}_{3}=1.0307$
$\therefore f\left({x}_{3}\right)=f\left(1.0307\right)={1.0307}^{6}-1.0307-1=-0.8319$$\therefore f\left({x}_{3}\right)=f\left(1.0307\right)={1.0307}^{6}-1.0307-1=-0.8319$:.f(x_(3))=f(1.0307)=1.0307^(6)-1.0307-1=-0.8319\therefore f\left(x_3\right)=f(1.0307)=1.0307^6-1.0307-1=-0.8319$\therefore f\left({x}_{3}\right)=f\left(1.0307\right)={1.0307}^{6}-1.0307-1=-0.8319$
${3}^{rd}$${3}^{rd}$3^(rd)3^{r d}${3}^{rd}$ iteration :
${x}_{2}=1.0161\text{and}{x}_{3}=1.0307$${x}_{2}=1.0161\text{and}{x}_{3}=1.0307$x_(2)=1.0161″ and “x_(3)=1.0307x_2=1.0161 \text { and } x_3=1.0307