(b) Which of the following rearrangements (1) x_(n+1)=(x_(n)^(2)-3)/(2),n=0 = sqrt(2x_(n)+3),n=0 (2) ,x_(n+1)= sqrt(2x_(n)+3) is suitable for solving the equation x^(2)-2x=3 using the fixed-point method in 2,4 ? Then use it to find the second approximation, taking x_(0)=2.5. Compute the error bound for your approximations. (c) Successive ap

Question

(b) Which of the following rearrangements (1)

x_(n+1)=(x_(n)^(2)-3)/(2),n>=0
=\sqrt(2x_(n)+3),n>=0

(2)

,x_(n+1)=\sqrt(2x_(n)+3)

is suitable for solving the equation

x^(2)-2x=3

using the fixed-point method in

2,4

? Then use it to find the second approximation, taking

x_(0)=2.5

. Compute the error bound for your approximations. (c) Successive approximations

x_(n)

to the desired root for an equation

f(x)=0

are generated by the iterative scheme

x_(n+1)=(2x_(n)^(3)+4x_(n)^(2)+10)/(3x_(n)^(2)+8x_(n)),n>=0

Use the Newton's method to find the first approximation of the root, starting with

x_(0)=1.5

Please i need solutaion b and c

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Expert Answer to - (b) Which of the following rearrangements (1) x_(n+1)=(x_(n)^(2)-3)/(2),n>=0 = sqrt(2x_(n)+3),n&g

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Solution for - (b) Which of the following rearrangements (1) x_(n+1)=(x_(n)^(2)-3)/(2),n>=0 = sqrt(2x_(n)+3),n&g

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This an additional answer to - (b) Which of the following rearrangements (1) x_(n+1)=(x_(n)^(2)-3)/(2),n>=0 = sqrt(2x_(n)+3),n&g

(Solved): (b) Which of the following rearrangements (1) x_(n+1)=(x_(n)^(2)-3)/(2),n>=0 =\sqrt(2x_(n)+3),n&g ...