ve the equationx 10 costx)-o, using the initial guesses of xso 15. and o n a. th

Question

ve the equationx 10 costx)-o, using the initial guesses of xso 15. and o n a. the successive approximations should converge to the root near 2 b. the successive approximations should converge to the root near -2 c. the method will fail due to division by zero d. the successive approximations should converge to the root near-3 16 When using the bisection method to solve the equation x+ 10 costn)-O, using the initial guesses of a m 4 and b-2.4 a. the successive approximations should converge to the root near 2 b. the successive approximations should converge to the root near -2 c. the method cannot be used d. the successive approximations should converge to the root near -3 17, The 3-point Gauss quadrature method is: a Slowly convergent b. Generally worse than the 2-point Gauss quadrature method c. Is a good method for interpolation d. Generally better than the 3-point Simpson method Why?

Explanation / Answer

Hi ,

As per the problem statement , Please find the solution for each below:

15) f(x) = x^3+10*cos(x)=0 , initial values 0 , -3

In such case secant method iteration are as follows

So final x =

-1.33213740

from this we can conclude that successive approximations should converge to the root near -2

hence option:b correct

16) Given function f(x) = x^3+10*cos(x) with initial values -4 , 2.4 as follows

As per the bisection method solving ,for converge it took more no of iteration with tolerance level as 0.001

Please find the iterations below:

It converged at point x = -1.3321

From this result we can conclude that "the successive approximations should converge to the root near -2"

correct option : b

17 ) Gauss quadrature method with point level , leads to the better responses .

a) Slowly convergent .

Gauss quadrature method convergence was quick with respect to 3-point rule

b) 3-point Gauss rule was more accurate and faster responsive than 2-point Gauss quadrature method .

Relative Error rate will be less more than 10% lesser in 3-point Gauss rule

c) Yes , it is good for interpolation , within the less approximations end result will be getting having absolute error rating was less or ignorable

d) Comapre to the Simpson rule , 3-point gauss rule was perform well  

Thanks

Step x F(x) |x(i) - x(i-1)| x2 -0.63965987 7.761262071 2.360340128 x3 -1.04984195 3.819979103 0.410182074 x4 -1.44739953 -1.801413198 0.397557588 x5 -1.31999951 0.181793757 0.127400019 x6 -1.33167784 0.006910901 0.011678321 x7 -1.33213933 -0.000029056 0.000461496 x8 -1.33213740 0.000000005 0.00000193

So final x =

-1.33213740

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