Finding a Cubic Root: Pencil and Paper Technique

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    Cubic Paper Root
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The discussion focuses on the method for calculating cubic roots using the Newton-Raphson technique. It begins with defining the function f(x) = x^3 - a, where the goal is to find the root X such that f(X) = 0. The process involves selecting an initial guess x₀ and constructing the tangent line L(x) at that point. The next approximation, x₁, is determined by finding the x-intercept of L(x). The iterative formula is provided as xₙ = (2xₙ₋₁³ + a) / (3xₙ₋₁²), which allows for successive approximations to converge on the cubic root. The conversation also touches on the nostalgia for manual calculations and references to cultural works that reflect on the theme of mathematical knowledge.
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I still remember how to extract a square root without a computer but could somebody remind me the technique to find a cubic root just with the pencil and paper?
 
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I would suggest using some Newton-Raphson scheme.

1. Let f(x)=x^{3}-a
You are to find X so that f(X)=0.

2. Pick an initial value x_{0}\to{f}(x_{0})=x_{0}^{3}-a

3. The equation for the tangent line L(x)=at (x_{0},f(x_{0}) is given by:
L(x)=f(x_{0})+f'(x_{0})(x-x_{0})

4- Let the next iteration point be the x-intercept of L(x):
L(x_{1})=0\to{x}_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0}}

5. Or, in this case, the iterative scheme becomes:
x_{n}=x_{n-1}-\frac{x_{n-1}-\frac{a}{x_{n-1}^{2}}}{3}
That is:
x_{n}=\frac{2x_{n-1}^{3}+a}{3x_{n-1}^{2}}, n\geq{1}
 
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I'm actually old enough to remember this. It's been somewhat wisely forgotten. http://www.nist.gov/dads/HTML/cubeRoot.html. You may wish to also check out the Isaac Asimov story, "The Feeling of Power". Kind of haunting, these days.
 
Dick said:
I'm actually old enough to remember this. It's been somewhat wisely forgotten. http://www.nist.gov/dads/HTML/cubeRoot.html. You may wish to also check out the Isaac Asimov story, "The Feeling of Power". Kind of haunting, these days.
Thanks, now it is coming back!
 
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