Newton's Method for the Square Root of 37
Using Newton's method (and no calculator square-root button), find $\sqrt{37}$ correct to three decimal places.
State the function you apply Newton's method to, the iteration formula, your starting point, and the iterates until three decimals stabilize.
Hints
- Newton's method for $f(x) = 0$ iterates $x_{k+1} = x_k - f(x_k)/f'(x_k)$. Choose $f$ so that its root is $\sqrt{37}$.
- With $f(x) = x^2 - 37$ the update simplifies to $x_{k+1} = \tfrac{1}{2}\left(x_k + 37/x_k\right)$. Start from the nearest perfect square: $x_0 = 6$.
- One step from $6$ gives $6 + 1/12 = 6.08\overline{3}$; a second step barely moves it. Compare with the Taylor expansion $\sqrt{36 + 1} \approx 6 + \frac{1}{2\cdot 6}$.
Worked Solution
How to Think About It: Newton's method finds a root of $f$ by repeatedly replacing $f$ with its tangent line. For square roots, choose $f(x) = x^2 - 37$; the tangent-line root has a very simple form and a good starting point is the nearest perfect square.
Quick Estimate: $36 < 37 < 49$, so $\sqrt{37}$ is just above $6$. The first-order Taylor expansion $\sqrt{36 + 1} \approx 6 + \frac{1}{2 \cdot 6} = 6.0833$ already lands on the three-decimal answer $6.083$; the true value is slightly less because $\sqrt{\cdot}$ is concave.
Formal Solution:
*Step 1 -- Set up.* Take $f(x) = x^2 - 37$, so $f'(x) = 2x$. Newton's update is $$x_{k+1} = x_k - \frac{x_k^2 - 37}{2x_k} = \frac{1}{2}\left(x_k + \frac{37}{x_k}\right).$$
*Step 2 -- Iterate from $x_0 = 6$.* $$x_1 = \frac{1}{2}\left(6 + \frac{37}{6}\right) = \frac{1}{2}\left(6 + 6.1\overline{6}\right) = 6.08\overline{3},$$ $$x_2 = \frac{1}{2}\left(6.08333 + \frac{37}{6.08333}\right) = \frac{1}{2}\left(6.08333 + 6.08219\right) = 6.08276.$$ The third decimal has stabilized: $x_1 = 6.0833$ and $x_2 = 6.0828$ both round to $6.083$.
*Step 3 -- Check.* $6.083^2 = 37.0029$ and $6.082^2 = 36.9907$, so $\sqrt{37}$ lies between $6.082$ and $6.083$ and rounds to $6.083$. (The exact value is $6.0827625\ldots$.)
*Step 4 -- Why it converges so fast.* Near a simple root the Newton error satisfies $e_{k+1} \approx \frac{f''}{2f'} e_k^2 = \frac{e_k^2}{2\sqrt{37}}$. Starting with $e_0 \approx 0.083$, one step gives $e_1 \approx 5.7 \times 10^{-4}$ and two steps give $e_2 \approx 2.7 \times 10^{-8}$, matching the iterates above.
Answer: $\sqrt{37} \approx 6.083$ (Newton iterates $6 \to 6.0833 \to 6.08276$; exact value $6.08276\ldots$).
Intuition
Newton's method converges quadratically near a simple root: the number of correct digits roughly doubles each step, so starting from the nearby perfect square $36$ gives three decimals after one iteration and about seven after two. This update, the Babylonian method, is the same first-order Taylor idea behind the classic $\sqrt{a^2 + b} \approx a + \frac{b}{2a}$ mental-math shortcut, and it is the workhorse for implied volatility solvers and yield-to-maturity calculations.