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Find root ninety eight using Taylor's series? - Answers

The first-order Taylor expansion around x2 is given by sqrt(x2 + a) ~ x + a / 2xSo for sqrt(98), we know the closes perfect square is 81 = 9^2. Therefore: sqrt(98) = sqrt(81 + 17) = sqrt(9^2 + 17) ~ 9 + 17/(2*9) = 9 + 17/18 ~ 9.9444Alternatively, sqrt(98) = sqrt(100 - 2) = sqrt(10^2 - 2) ~ 10 - 2/(2*10) = 10 - 1/10 = 9.9Using a calculator: sqrt(98) ~ 9.899Taylor series will always be an over-approximation.



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Find root ninety eight using Taylor's series? - Answers

https://math.answers.com/other-math/Find_root_ninety_eight_using_Taylor's_series

The first-order Taylor expansion around x2 is given by sqrt(x2 + a) ~ x + a / 2xSo for sqrt(98), we know the closes perfect square is 81 = 9^2. Therefore: sqrt(98) = sqrt(81 + 17) = sqrt(9^2 + 17) ~ 9 + 17/(2*9) = 9 + 17/18 ~ 9.9444Alternatively, sqrt(98) = sqrt(100 - 2) = sqrt(10^2 - 2) ~ 10 - 2/(2*10) = 10 - 1/10 = 9.9Using a calculator: sqrt(98) ~ 9.899Taylor series will always be an over-approximation.



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https://math.answers.com/other-math/Find_root_ninety_eight_using_Taylor's_series

Find root ninety eight using Taylor's series? - Answers

The first-order Taylor expansion around x2 is given by sqrt(x2 + a) ~ x + a / 2xSo for sqrt(98), we know the closes perfect square is 81 = 9^2. Therefore: sqrt(98) = sqrt(81 + 17) = sqrt(9^2 + 17) ~ 9 + 17/(2*9) = 9 + 17/18 ~ 9.9444Alternatively, sqrt(98) = sqrt(100 - 2) = sqrt(10^2 - 2) ~ 10 - 2/(2*10) = 10 - 1/10 = 9.9Using a calculator: sqrt(98) ~ 9.899Taylor series will always be an over-approximation.

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      The first-order Taylor expansion around x2 is given by sqrt(x2 + a) ~ x + a / 2xSo for sqrt(98), we know the closes perfect square is 81 = 9^2. Therefore: sqrt(98) = sqrt(81 + 17) = sqrt(9^2 + 17) ~ 9 + 17/(2*9) = 9 + 17/18 ~ 9.9444Alternatively, sqrt(98) = sqrt(100 - 2) = sqrt(10^2 - 2) ~ 10 - 2/(2*10) = 10 - 1/10 = 9.9Using a calculator: sqrt(98) ~ 9.899Taylor series will always be an over-approximation.

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