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The function f(x)=x⁻³ −4x has a root at r=2.If the error ei =xi −r after four steps of Newton’s Method is e4 = 10⁻⁶, estimate e5

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Final answer:

The approximate error after the fifth step of Newton's Method, e5, given that the error at the fourth step is 10⁶, can be estimated as (10⁶)², which is 1 × 10⁻¹².

Step-by-step explanation:

You've asked to approximate the error e5 after the fifth step of Newton's Method given that the error e4 is 10⁶ at the fourth step. To solve for e5 using Newton's Method, we would typically use the function f(x) and its derivative f'(x) at the point x4. However, Newton's Method has a general property that the errors ei converge quadratically under certain conditions, which means that if the method is converging, the error is roughly squared in each step. Therefore, if e4 is 10⁶, we can approximate e5 to be (10⁶)².

Following this logic:

e5 ≈ (10⁶)² = (1 × 10⁻⁶) ² = 1 × 10⁻¹²

So the estimated error after the fifth step, e5, is 1 × 10⁻¹².

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