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How do you find the square root of 350? - Answers

Use the Newton-Raphson method. Let f(x) = x2 - 350 and let f'(x) = 2*x Then start with any guess, x0. The next estimate is x1 = x0 - f(x0)/f'(x0). Continue iterations with xn+1 = xn - f(xn)/f'(xn). If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).



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How do you find the square root of 350? - Answers

https://math.answers.com/other-math/How_do_you_find_the_square_root_of_350

Use the Newton-Raphson method. Let f(x) = x2 - 350 and let f'(x) = 2*x Then start with any guess, x0. The next estimate is x1 = x0 - f(x0)/f'(x0). Continue iterations with xn+1 = xn - f(xn)/f'(xn). If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).



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https://math.answers.com/other-math/How_do_you_find_the_square_root_of_350

How do you find the square root of 350? - Answers

Use the Newton-Raphson method. Let f(x) = x2 - 350 and let f'(x) = 2*x Then start with any guess, x0. The next estimate is x1 = x0 - f(x0)/f'(x0). Continue iterations with xn+1 = xn - f(xn)/f'(xn). If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).

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      Use the Newton-Raphson method. Let f(x) = x2 - 350 and let f'(x) = 2*x Then start with any guess, x0. The next estimate is x1 = x0 - f(x0)/f'(x0). Continue iterations with xn+1 = xn - f(xn)/f'(xn). If you start with x0 = 15 (quite a long way off, given that 152 is only 225), x2 is accurate to 6 decimal places (error = 8 in 10 million) and x3 to 13 dp (2 in 100 trillion).

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