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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.
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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 $2.0946$.
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