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)=x6−x−1=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)=x3−3x2+4x−5. 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=0x3−9x+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=-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)=2x4−3x2+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) x log 10 ⁡ x = 4.772393 x log 10 ⁡ x = 4.772393 quad xlog_(10)x=4.772393\quad x \log _{10} x=4.772393xlog10⁡x=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)=x−2sin⁡x 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][1−236−13184−1014]
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=[1−111−24122]
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][12−2111221][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=[−4140−5130−102]
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(x0−h)+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}≤5×10−4 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} xlog10⁡x 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}1×10−6.
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′=1−x+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)=ex2−1, 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+c2e−αx 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+1−2uxcosh⁡α+ux−1=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 x∫00.1(1+x)1/2dx.
\(\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}\)

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)=x6−x−1=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=0x6−x−1=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)=x6−x−1
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)=1.56−1.5−1=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)=1.256−1.25−1=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)=1.1256−1.125−1=−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)=x6−x−1
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=x0−f(x0)⋅x1−x0f(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)⋅2−161−(−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=-0.9154∴f(x2)=f(1.0161)=1.01616−1.0161−1=−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=x1−f(x1)⋅x2−x1f(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=2−61⋅1.0161−2−0.9154−61
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=-0.8319∴f(x3)=f(1.0307)=1.03076−1.0307−1=−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=1.0307x2=1.0161 and x3=1.0307
f ( x 2 ) = f ( 1.0161 ) = − 0.9154 and f ( x 3 ) = f ( 1.0307 ) = − 0.8319 f x 2 = f ( 1.0161 ) = − 0.9154 and f x 3 = f ( 1.0307 ) = − 0.8319 f(x_(2))=f(1.0161)=-0.9154″ and “f(x_(3))=f(1.0307)=-0.8319f\left(x_2\right)=f(1.0161)=-0.9154 \text { and } f\left(x_3\right)=f(1.0307)=-0.8319f(x2)=f(1.0161)=−0.9154 and f(x3)=f(1.0307)=−0.8319
∴ x 4 = x 2 − f ( x 2 ) ⋅ x 3 − x 2 f ( x 3 ) − f ( x 2 ) x 4 = 1.0161 − ( − 0.9154 ) ⋅ 1.0307 − 1.0161 − 0.8319 − ( − 0.9154 ) x 4 = 1.1757 ∴ f ( x 4 ) = f ( 1.1757 ) = 1.1757 6 − 1.1757 − 1 = 0.4652 ∴ x 4 = x 2 − f x 2 ⋅ x 3 − x 2 f x 3 − f x 2 x 4 = 1.0161 − ( − 0.9154 ) ⋅ 1.0307 − 1.0161 − 0.8319 − ( − 0.9154 ) x 4 = 1.1757 ∴ f x 4 = f ( 1.1757 ) = 1.1757 6 − 1.1757 − 1 = 0.4652 {:[:.x_(4)=x_(2)-f(x_(2))*(x_(3)-x_(2))/(f(x_(3))-f(x_(2)))],[x_(4)=1.0161-(-0.9154)*(1.0307-1.0161)/(-0.8319-(-0.9154))],[x_(4)=1.1757],[:.f(x_(4))=f(1.1757)=1.1757^(6)-1.1757-1=0.4652]:}\begin{aligned} & \therefore x_4=x_2-f\left(x_2\right) \cdot \frac{x_3-x_2}{f\left(x_3\right)-f\left(x_2\right)} \\ & x_4=1.0161-(-0.9154) \cdot \frac{1.0307-1.0161}{-0.8319-(-0.9154)} \\ & x_4=1.1757 \\ & \therefore f\left(x_4\right)=f(1.1757)=1.1757^6-1.1757-1=0.4652 \end{aligned}∴x4=x2−f(x2)⋅x3−x2f(x3)−f(x2)x4=1.0161−(−0.9154)⋅1.0307−1.0161−0.8319−(−0.9154)x4=1.1757∴f(x4)=f(1.1757)=1.17576−1.1757−1=0.4652
To summarize, after three iterations of the secant method, the approximation of the largest real root α α alpha\alphaα of the equation 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)=x6−x−1=0 is x 4 = 1.1757 x 4 = 1.1757 x_(4)=1.1757x_4=1.1757x4=1.1757. The value of the function at this point is f ( x 4 ) = 0.465365 f x 4 = 0.465365 f(x_(4))=0.465365f\left(x_4\right)=0.465365f(x4)=0.465365, indicating that the root is close to this value.
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)=x3−3x2+4x−5. 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.
Answer:
To 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 – 3x^2 + 4x – 5P(x)=x3−3x2+4x−5, we can use Descartes’ Rule of Signs. This rule states that the number of positive real roots of a polynomial is either equal to the number of sign changes between consecutive coefficients or less than that by an even number. Similarly, the number of negative real roots is determined by the number of sign changes in the coefficients of the polynomial with each x x xxx replaced by − x − x -x-x−x.
Finding the Number of Positive Roots:
The given polynomial is 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 – 3x^2 + 4x – 5P(x)=x3−3x2+4x−5. The coefficients are 1, -3, 4, -5. The sign changes in the coefficients are from 1 to -3, -3 to 4, and 4 to -5. So, there are 3 sign changes.
Number of Positive Roots: 3 or 1 (since it can be less by an even number).
Finding the Number of Negative Roots:
Replace x x xxx with − x − x -x-x−x in P ( x ) P ( x ) P(x)P(x)P(x), we get 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 – 3x^2 – 4x – 5P(−x)=−x3−3x2−4x−5. The coefficients are -1, -3, -4, -5. There are no sign changes.
Number of Negative Roots: 0 (since there are no sign changes).
Calculating P ( 2 ) P ( 2 ) P(2)P(2)P(2) and P ′ ( 2 ) P ′ ( 2 ) P^(‘)(2)P'(2)P′(2) using Synthetic Division:
  1. Calculate P ( 2 ) P ( 2 ) P(2)P(2)P(2):
    We use synthetic division to divide P ( x ) P ( x ) P(x)P(x)P(x) by x − 2 x − 2 x-2x – 2x−2.
    2 1 − 3 4 − 5 2 − 2 4 1 − 1 2 − 1 2 1 − 3 4 − 5 2 − 2 4 1 − 1 2 − 1 {:[2,1,-3,4,-5],[,,2,-2,4],[,1,-1,2,-1]:}\begin{array}{c|ccc} 2 & 1 & -3 & 4 & -5 \\ \hline & & 2 & -2 & 4 \\ \hline & 1 & -1 & 2 & -1 \end{array}21−34−52−241−12−1
    The remainder is -1, so P ( 2 ) = − 1 P ( 2 ) = − 1 P(2)=-1P(2) = -1P(2)=−1.
  2. Calculate P ′ ( 2 ) P ′ ( 2 ) P^(‘)(2)P'(2)P′(2):
    The derivative of a polynomial can be found by multiplying each term by its degree and reducing the degree by one. For 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 – 3x^2 + 4x – 5P(x)=x3−3x2+4x−5, the derivative is P ′ ( x ) = 3 x 2 − 6 x + 4 P ′ ( x ) = 3 x 2 − 6 x + 4 P^(‘)(x)=3x^(2)-6x+4P'(x) = 3x^2 – 6x + 4P′(x)=3x2−6x+4.
    Now, we use synthetic division to find P ′ ( 2 ) P ′ ( 2 ) P^(‘)(2)P'(2)P′(2) by dividing P ′ ( x ) P ′ ( x ) P^(‘)(x)P'(x)P′(x) by x − 2 x − 2 x-2x – 2x−2.
    2 3 − 6 4 6 0 3 0 4 2 3 − 6 4 6 0 3 0 4 {:[2,3,-6,4],[,,6,0],[,3,0,4]:}\begin{array}{c|cc} 2 & 3 & -6 & 4 \\ \hline & & 6 & 0 \\ \hline & 3 & 0 & 4 \end{array}23−6460304
    The remainder is 4, so P ′ ( 2 ) = 4 P ′ ( 2 ) = 4 P^(‘)(2)=4P'(2) = 4P′(2)=4.
To summarize, 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 – 3x^2 + 4x – 5P(x)=x3−3x2+4x−5 has either 3 or 1 positive real roots and 0 negative real roots. Additionally, P ( 2 ) = − 1 P ( 2 ) = − 1 P(2)=-1P(2) = -1P(2)=−1 and P ′ ( 2 ) = 4 P ′ ( 2 ) = 4 P^(‘)(2)=4P'(2) = 4P′(2)=4.
c) Solve x 3 − 9 x + 1 = 0 x 3 − 9 x + 1 = 0 x^(3)-9x+1=0x^3-9 x+1=0x3−9x+1=0 for the root lying between 2 and 4 by the method of false position. Perform two iterations.
Answer:
Let f ( x ) = x 3 − 9 x + 1 f ( x ) = x 3 − 9 x + 1 f(x)=x^(3)-9x+1f(x)=x^3-9 x+1f(x)=x3−9x+1
Here
x x xxx 2 3 4
f ( x ) f ( x ) f(x)f(x)f(x) -9 1 29
x 2 3 4 f(x) -9 1 29| $x$ | 2 | 3 | 4 | | :—: | :—: | :—: | :—: | | $f(x)$ | -9 | 1 | 29 |
Here f ( 2 ) = − 9 < 0 f ( 2 ) = − 9 < 0 f(2)=-9 < 0f(2)=-9<0f(2)=−9<0 and f ( 3 ) = 1 > 0 f ( 3 ) = 1 > 0 f(3)=1 > 0f(3)=1>0f(3)=1>0
∴ ∴ :.\therefore∴ Root lies between 2 and 3
1 s t 1 s t 1^(st)1^{s t}1st iteration :
Here f ( 2 ) = − 9 < 0 f ( 2 ) = − 9 < 0 f(2)=-9 < 0f(2)=-9<0f(2)=−9<0 and f ( 3 ) = 1 > 0 f ( 3 ) = 1 > 0 f(3)=1 > 0f(3)=1>0f(3)=1>0
∴ ∴ :.\therefore∴ Now, Root lies between x 0 = 2 x 0 = 2 x_(0)=2x_0=2x0=2 and x 1 = 3 x 1 = 3 x_(1)=3x_1=3x1=3
x 2 = x 0 − f ( x 0 ) ⋅ x 1 − x 0 f ( x 1 ) − f ( x 0 ) x 2 = 2 − ( − 9 ) ⋅ 3 − 2 1 − ( − 9 ) x 2 = 2.9 x 2 = x 0 − f x 0 ⋅ x 1 − x 0 f x 1 − f x 0 x 2 = 2 − ( − 9 ) ⋅ 3 − 2 1 − ( − 9 ) x 2 = 2.9 {:[x_(2)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))],[x_(2)=2-(-9)*(3-2)/(1-(-9))],[x_(2)=2.9]:}\begin{aligned} & 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=2-(-9) \cdot \frac{3-2}{1-(-9)} \\ & x_2=2.9 \end{aligned}x2=x0−f(x0)⋅x1−x0f(x1)−f(x0)x2=2−(−9)⋅3−21−(−9)x2=2.9
f ( x 2 ) = f ( 2.9 ) = 2.9 3 − 9 ⋅ 2.9 + 1 = − 0.711 < 0 f x 2 = f ( 2.9 ) = 2.9 3 − 9 ⋅ 2.9 + 1 = − 0.711 < 0 f(x_(2))=f(2.9)=2.9^(3)-9*2.9+1=-0.711 < 0f\left(x_2\right)=f(2.9)=2.9^3-9 \cdot 2.9+1=-0.711<0f(x2)=f(2.9)=2.93−9⋅2.9+1=−0.711<0
2 nd 2 nd 2^(“nd “)2^{\text {nd }}2nd iteration :
Here f ( 2.9 ) = − 0.711 < 0 f ( 2.9 ) = − 0.711 < 0 f(2.9)=-0.711 < 0f(2.9)=-0.711<0f(2.9)=−0.711<0 and f ( 3 ) = 1 > 0 f ( 3 ) = 1 > 0 f(3)=1 > 0f(3)=1>0f(3)=1>0
∴ ∴ :.\therefore∴ Now, Root lies between x 0 = 2.9 x 0 = 2.9 x_(0)=2.9x_0=2.9x0=2.9 and x 1 = 3 x 1 = 3 x_(1)=3x_1=3x1=3
x 3 = x 0 − f ( x 0 ) ⋅ x 1 − x 0 f ( x 1 ) − f ( x 0 ) x 3 = 2.9 − ( − 0.711 ) ⋅ 3 − 2.9 1 − ( − 0.711 ) x 3 = 2.9416 x 3 = x 0 − f x 0 ⋅ x 1 − x 0 f x 1 − f x 0 x 3 = 2.9 − ( − 0.711 ) ⋅ 3 − 2.9 1 − ( − 0.711 ) x 3 = 2.9416 {:[x_(3)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))],[x_(3)=2.9-(-0.711)*(3-2.9)/(1-(-0.711))],[x_(3)=2.9416]:}\begin{aligned} & x_3=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)} \\ & x_3=2.9-(-0.711) \cdot \frac{3-2.9}{1-(-0.711)} \\ & x_3=2.9416 \end{aligned}x3=x0−f(x0)⋅x1−x0f(x1)−f(x0)x3=2.9−(−0.711)⋅3−2.91−(−0.711)x3=2.9416
f ( x 3 ) = f ( 2.9416 ) = 2.9416 3 − 9 ⋅ 2.9416 + 1 = − 0.0215 < 0 f x 3 = f ( 2.9416 ) = 2.9416 3 − 9 ⋅ 2.9416 + 1 = − 0.0215 < 0 f(x_(3))=f(2.9416)=2.9416^(3)-9*2.9416+1=-0.0215 < 0f\left(x_3\right)=f(2.9416)=2.9416^3-9 \cdot 2.9416+1=-0.0215<0f(x3)=f(2.9416)=2.94163−9⋅2.9416+1=−0.0215<0
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