i need complete answer and clear plz Consider the following

i need complete answer .................and clear plz

Consider the following recursive sequence: x_n = 3x_n-1 n = 1, 2, 3... where x*_0 = 0.995(an estimate of x_0 = 1). Verify that the recursive sequence has the solution x_n = 3^n by showing that is satisfies the recursive equation. Let x_n = x*_n + epsilon_n, where x*_n is the appropriate obtained from the recursive formula and starting from x*_0 = 0.995, and epsilon_n is the error at the nth step. Show that epsilon_n = 3 epsilon_n-1 calculate the error propagated at the 6-th step of the scheme, i.e., epsilon_6, using 3-digit rounding

Solution

It is given in the question that, x_n = 3 * x_{n-1 -------(1)}

Let`s replace \'n\' in the equation with \'n-1\' which gives x_n-1 = 3 * x_{n-2 -------(2)}

combining both (1) and (2), we get x_n = 3 * 3 * x_{n-2 (or)} x_n = 3² * x_{n-2 -------(3)}

This can again be reduced to x_n = 3 * 3 * 3 * x_{n-3 (or)} x_n = 3³ * x_{n-3 -----------(4)}

From (1), (3), (4) we get a generalised expression for x_n = 3ⁱ * x_{n-i ,}   i<=n

let i=n , then the equation becomes x_n = 3ⁿ * x₀ = 3ⁿ

x_n = x\'_n + e_n

==> x₀ = x\'₀ + e₀

==> x₀ - x\'₀ = e₀

Therefore e₀ = 1 - 0.995 = 0.005 --------- (5)

since x\'_n is an approximation obtained from the recursive formula, we can write x\'_n = 3 * x\'_n-1

Thus, e_n = x_n - x\'_n can be written as e_n = (3 * x_n-1 - 3 * x\'_n-1) = 3 * (x_n-1 - x\'_n-1) = 3 * e_n-1

So, e_n = 3 * e_n-1   ---------------------b(i)

e₆ = 3 * e₅ = 3 * 3 * e₄ = 3 * 3 * 3 * e₃ ...................... = 3 * 3 * 3 * 3 * 3 * 3 *e₀

==> e₆ = 3⁶ * e₀ = 729 * 0.005 = 3.65 --------------- b(ii)