Starting from an initial interval (2, 3) containing 5, we will look for consecutive pairs in...

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Starting from an initial interval (2, 3) containing √5, we will look for consecutive pairs in tenths, hundredths, thousandths and so on that can bound √5 (See Strategy 1 in Chapter 3, Day 3 lecture). Then we will choose the lower bound of the interval for the next approximation. For instance, let's call a₁ = 2 be the first approximation and a₂ = 2.2 be the second approximation determined by the lower bound among the two consecutive tenths (2.2, 2.3) that bounds √5 (2.2² <5<2.3²). I i) Provide a3, a4 and the corresponding error bounds €3, 4. ii) Provide the error bounds e, corresponding to the nth approximation an Starting from an initial interval (2, 3) containing √5, we will look for consecutive pairs in tenths, hundredths, thousandths and so on that can bound √5 (See Strategy 1 in Chapter 3, Day 3 lecture). Then we will choose the lower bound of the interval for the next approximation. For instance, let's call a₁ = 2 be the first approximation and a₂ = 2.2 be the second approximation determined by the lower bound among the two consecutive tenths (2.2, 2.3) that bounds √5 (2.2² <5<2.3²). I i) Provide a3, a4 and the corresponding error bounds €3, 4. ii) Provide the error bounds e, corresponding to the nth approximation an

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i The initial approximation a1 is given as 2 and the interval containing 5 is 2 3 The corresponding ... View the full answer