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Use the Newton-Raphson method, with 3 as starting point, to find a fraction that is within $10^{-8}$ of $\sqrt{10}$. Show (without using the square root button) that your answer is indeed within $10^{-8}$ of the truth.
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Solution: The number $\sqrt{10}$ is the unique positive solution of the equation $f(x)=0$ where $f(x)=x^{2}-10$. We use the Newton Method to approximate a solution of this equation.

Let $x_{0}$ be our initial estimate of the root, and let $x_{n}$ be the $n$-th improved estimate. Note that $f^{\prime}(x)=2 x$. The Newton Method recurrence is therefore
$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}=x_{n}-\frac{x_{n}^{2}-10}{2 x_{n}}$
To make the expression on the right more beautiful, and calculations easier, it is useful to manipulate it a bit. We get
$x_{n+1}=x_{n}-\frac{x_{n}}{2}+\frac{10}{2 x_{n}}=\frac{1}{2}\left(x_{n}+\frac{10}{x_{n}}\right) .$

\begin{aligned}
&\text { Compute, starting with } x_{0}=3 \text {. Then } x_{1}=(1 / 2)\left(x_{0}+10 / x_{0}\right)= \\
&(1 / 2)(3+10 / 3)=19 / 6 \text {. And } x_{2}=(1 / 2)(19 / 6+60 / 19)=721 / 228 \text {. } \\
&\text { We could go on calculating with fractions-and there is interesting } \\
&\text { mathematics involved-but from here on we switch to the calculator. } \\
&\text { If we allow the }=\text { sign to be used sloppily, we get } x_{1}=3.166666667 \text {. } \\
&\text { Then } x_{2}=(1 / 2)\left(x_{1}+10 / x_{1}\right)=3.162280702, \text { and } x_{3}=3.16227766, \\
&\text { and } x_{4}=3.16227766 \text {. } \\
&\text { The calculator says that } x_{3}=x_{4} \text { to } 8 \text { decimal places. We can therefore } \\
&\text { dare hope that } 3.16227766 \text { is close enough. One way of checking is to } \\
&\text { let } a=3.16227765 \text { and } b=3.16227767 \text {. A quick calculation shows-if } \\
&\text { the squaring button can be trusted, and it is one of the ones that can } \\
&\text { be that } f(a)<0 \text { while } f(b)>0 \text {. } \\
&\text { Thus the function } f(x) \text { changes sign as } x \text { goes from } a \text { to } b \text {. It follows by } \\
&\text { the Intermediate Value Theorem that } f(x)=0 \text { has a solution (namely } \\
&\sqrt{10} \text { ) between } a \text { and } b \text {. Since } \sqrt{10} \text { lies in the interval }(a, b), \text { and the } \\
&\text { distance from } 3.16227766 \text { to either } a \text { or } b \text { is } 10^{-8}, \text { it follows that the } \\
&\text { distance from } 3.16227766 \text { to } \sqrt{10} \text { is less than } 10^{-8} \text {. }
\end{aligned}
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