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How do you find square root of 800? - Answers

If you have a non-scientific calculator you can use the Newton-Raphson method. Let f(x) = x2 - 800, and f'(x) = 2x [f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.] Make a guess at the square root of 800, and call is x0. Then calculate xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ... Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good: Suppose you start with x0 = 20 (a pretty poor choice since 202 is 400, which is nowhere near 800). Even so, x3 = 2.843 is less than 1 in 24 thousand from the true value and x4 has an error of less than 1 in 30 billion.



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How do you find square root of 800? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_find_square_root_of_800

If you have a non-scientific calculator you can use the Newton-Raphson method. Let f(x) = x2 - 800, and f'(x) = 2x [f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.] Make a guess at the square root of 800, and call is x0. Then calculate xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ... Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good: Suppose you start with x0 = 20 (a pretty poor choice since 202 is 400, which is nowhere near 800). Even so, x3 = 2.843 is less than 1 in 24 thousand from the true value and x4 has an error of less than 1 in 30 billion.



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https://math.answers.com/math-and-arithmetic/How_do_you_find_square_root_of_800

How do you find square root of 800? - Answers

If you have a non-scientific calculator you can use the Newton-Raphson method. Let f(x) = x2 - 800, and f'(x) = 2x [f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.] Make a guess at the square root of 800, and call is x0. Then calculate xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ... Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good: Suppose you start with x0 = 20 (a pretty poor choice since 202 is 400, which is nowhere near 800). Even so, x3 = 2.843 is less than 1 in 24 thousand from the true value and x4 has an error of less than 1 in 30 billion.

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      If you have a non-scientific calculator you can use the Newton-Raphson method. Let f(x) = x2 - 800, and f'(x) = 2x [f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.] Make a guess at the square root of 800, and call is x0. Then calculate xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ... Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good: Suppose you start with x0 = 20 (a pretty poor choice since 202 is 400, which is nowhere near 800). Even so, x3 = 2.843 is less than 1 in 24 thousand from the true value and x4 has an error of less than 1 in 30 billion.
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