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How do we find value of square roots? - Answers

There is no simple way. One way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.If you want to find the square root of k, define f(x) = x^2 – k.Then finding the square root of k is equivalent to solving f(x) = 0.Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:Suppose you want to find the square root of 5 and you start with x0 = 5 (a pretty poor choice since you want the square root of 5, not 5).Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.There is another method which resembles long division. However, the divisor - which is the estimate - is a sequence that converges to the square root from below. It is not easy to explain it using this browser.Finally, there is the method of bracketing hundreds.Suppose you want the square root of 7.4 = 2^2 < 7 < 3^2 = 9 so 2 < sqrt(7) < 3Next 676 = 26^2 < 700 < 27^2 = 729 so 2.6 < sqrt(7) < 2.7Next go for 70000, and so on.



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How do we find value of square roots? - Answers

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

There is no simple way. One way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.If you want to find the square root of k, define f(x) = x^2 – k.Then finding the square root of k is equivalent to solving f(x) = 0.Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:Suppose you want to find the square root of 5 and you start with x0 = 5 (a pretty poor choice since you want the square root of 5, not 5).Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.There is another method which resembles long division. However, the divisor - which is the estimate - is a sequence that converges to the square root from below. It is not easy to explain it using this browser.Finally, there is the method of bracketing hundreds.Suppose you want the square root of 7.4 = 2^2 < 7 < 3^2 = 9 so 2 < sqrt(7) < 3Next 676 = 26^2 < 700 < 27^2 = 729 so 2.6 < sqrt(7) < 2.7Next go for 70000, and so on.



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

How do we find value of square roots? - Answers

There is no simple way. One way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.If you want to find the square root of k, define f(x) = x^2 – k.Then finding the square root of k is equivalent to solving f(x) = 0.Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:Suppose you want to find the square root of 5 and you start with x0 = 5 (a pretty poor choice since you want the square root of 5, not 5).Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.There is another method which resembles long division. However, the divisor - which is the estimate - is a sequence that converges to the square root from below. It is not easy to explain it using this browser.Finally, there is the method of bracketing hundreds.Suppose you want the square root of 7.4 = 2^2 < 7 < 3^2 = 9 so 2 < sqrt(7) < 3Next 676 = 26^2 < 700 < 27^2 = 729 so 2.6 < sqrt(7) < 2.7Next go for 70000, and so on.

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      There is no simple way. One way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.If you want to find the square root of k, define f(x) = x^2 – k.Then finding the square root of k is equivalent to solving f(x) = 0.Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:Suppose you want to find the square root of 5 and you start with x0 = 5 (a pretty poor choice since you want the square root of 5, not 5).Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.There is another method which resembles long division. However, the divisor - which is the estimate - is a sequence that converges to the square root from below. It is not easy to explain it using this browser.Finally, there is the method of bracketing hundreds.Suppose you want the square root of 7.4 = 2^2 < 7 < 3^2 = 9 so 2 < sqrt(7) < 3Next 676 = 26^2 < 700 < 27^2 = 729 so 2.6 < sqrt(7) < 2.7Next go for 70000, and so on.
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