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Newton's equation \(y^{3}-2 y-5=0\) has a root near \(y=2\). Starting with \(y_{0}=2\), compute \(y_{1}, y_{2}\), and \(y_{3}\), the next three Newton-Raphson estimates for the root.
in Mathematics by Platinum (93,184 points) | 430 views

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Let \(f(y)=y^{3}-2 y-5\). Then \(f^{\prime}(y)=3 y^{2}-2\), and the Newton Method produces the recurrence
y_{n+1}=y_{n}-\frac{y_{n}^{3}-2 y_{n}-5}{3 y_{n}^{2}-2}=\frac{2 y_{n}^{3}+5}{3 y_{n}^{2}-2}
(there was no good case for simplification here). Start with the estimate \(y_{0}=2\). Then \(y_{1}=21 / 10=2.1\). It follows that (to calculator accuracy) \(y_{2}=2.094568121\) and \(y_{3}=2.094551482\). These are almost the numbers that Newton obtained (see the notes). But Newton in effect used a rounded version of \(y_{2}\), namely \(2.0946\).
by Platinum (93,184 points)

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