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How do you find the cube root of a decimal fraction? - Answers
It depends on the fraction. Sometimes it helps to convert the decimal into a rational fraction. For example, 0.296296... recurring = 296/999 = 8/27. Now so cuberoot(8/27) = cuberoot(8)/cuberoot(27) = 2/3 = 0.66... recurring. This method only works if the cuberoot is a simple rational fraction. In general, however, the best option is numerical iteration, using the Newton Raphson method. To find the cuberoot of k, let f(x) = x^3 - k. then finding the cuberoot of k is equivalent to finding the 0 of f(x). Let f'(x) = 3*x^2 [the derivative of f(x)] Start with an approximate answer, x(0). Then for n = 0, 1, 2, ... Let x(n+1) = x(n) - f(x(n))/f'(x(n)) The sequence x(0), x(1), x(2), ... will converge to the cuberoot of k. The iteration equation, given above, is much easier to read if the n and n+1 are read as suffices, but the new and "improved" browser cannot handle suffices!
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How do you find the cube root of a decimal fraction? - Answers
It depends on the fraction. Sometimes it helps to convert the decimal into a rational fraction. For example, 0.296296... recurring = 296/999 = 8/27. Now so cuberoot(8/27) = cuberoot(8)/cuberoot(27) = 2/3 = 0.66... recurring. This method only works if the cuberoot is a simple rational fraction. In general, however, the best option is numerical iteration, using the Newton Raphson method. To find the cuberoot of k, let f(x) = x^3 - k. then finding the cuberoot of k is equivalent to finding the 0 of f(x). Let f'(x) = 3*x^2 [the derivative of f(x)] Start with an approximate answer, x(0). Then for n = 0, 1, 2, ... Let x(n+1) = x(n) - f(x(n))/f'(x(n)) The sequence x(0), x(1), x(2), ... will converge to the cuberoot of k. The iteration equation, given above, is much easier to read if the n and n+1 are read as suffices, but the new and "improved" browser cannot handle suffices!
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How do you find the cube root of a decimal fraction? - Answers
It depends on the fraction. Sometimes it helps to convert the decimal into a rational fraction. For example, 0.296296... recurring = 296/999 = 8/27. Now so cuberoot(8/27) = cuberoot(8)/cuberoot(27) = 2/3 = 0.66... recurring. This method only works if the cuberoot is a simple rational fraction. In general, however, the best option is numerical iteration, using the Newton Raphson method. To find the cuberoot of k, let f(x) = x^3 - k. then finding the cuberoot of k is equivalent to finding the 0 of f(x). Let f'(x) = 3*x^2 [the derivative of f(x)] Start with an approximate answer, x(0). Then for n = 0, 1, 2, ... Let x(n+1) = x(n) - f(x(n))/f'(x(n)) The sequence x(0), x(1), x(2), ... will converge to the cuberoot of k. The iteration equation, given above, is much easier to read if the n and n+1 are read as suffices, but the new and "improved" browser cannot handle suffices!
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- og:descriptionIt depends on the fraction. Sometimes it helps to convert the decimal into a rational fraction. For example, 0.296296... recurring = 296/999 = 8/27. Now so cuberoot(8/27) = cuberoot(8)/cuberoot(27) = 2/3 = 0.66... recurring. This method only works if the cuberoot is a simple rational fraction. In general, however, the best option is numerical iteration, using the Newton Raphson method. To find the cuberoot of k, let f(x) = x^3 - k. then finding the cuberoot of k is equivalent to finding the 0 of f(x). Let f'(x) = 3*x^2 [the derivative of f(x)] Start with an approximate answer, x(0). Then for n = 0, 1, 2, ... Let x(n+1) = x(n) - f(x(n))/f'(x(n)) The sequence x(0), x(1), x(2), ... will converge to the cuberoot of k. The iteration equation, given above, is much easier to read if the n and n+1 are read as suffices, but the new and "improved" browser cannot handle suffices!
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