In Lucky Numbers, Feynman described how he beat a man with the abacus to find the cube root of the number $1729.03$.
The number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.
Some part of his explanation is (surprisingly) a bit confusing and implicit.
Why is $1.03$ "only one part in nearly $2000$"? He later used $\frac{1}{1728}$. But $1.03 \ne \frac{1}{2000}$ and $1.03 \ne \frac{1}{1728}$. Also it is unclear how he came up with $\frac{1}{1728}$ from the outset.
Can you elaborate on his claim that "the cube root's excess is one-third of the number's excess" based on calculus? Was he using $\mathrm{d}y = f'(x) \cdot \mathrm{d}x$ at $x = 1728$, where $y = f(x) = x^{\frac{1}{3}}$? But $f'(x) = \frac{1}{3}x^{-\frac{2}{3}}$, so $\mathrm{d}y \ne \frac{1}{3}\mathrm{d}x$ as he claimed, missing $x^{-\frac{2}{3}} = 1728^{-\frac{2}{3}} = \frac{1}{144}$. I do see $\frac{1}{3} \cdot \frac{1}{144} = \frac{1}{3} \cdot \frac{1}{1728} \cdot 12 = \frac{1}{1728} \cdot 4$, that is, "find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12)". But his explanation does not follow that logic. He did not mention $x^{-\frac{2}{3}}$ at all and seemed to pull 1/1728 out from thin air, so as to serve his "illusion" that "the cube root's excess is one-third of the number's excess". I would like some clarification here.
He was referring to the relative difference: $\frac{1.03}{1728} \approx 0.000596$, which is slightly more than $\frac{1}{2000} = 0.0005$.
Let $f(x) = \sqrt[3]{x}$. Suppose that you want to approximate $f(t + \epsilon)$, given that you know $f(t)$ (in the specific example, $f(1728) = 12$.
Let's use the tangent line approximation at the point $(t, \sqrt[3]{t})$. The slope of the tangent line is $f'(t) = \frac{1}{3}t^{-2/3}$. So the equation of the tangent line is:
$$\tilde{f}(x) = \frac{1}{3}t^{-2/3}(x - t) + \sqrt[3]{t} $$ $$\tilde{f}(t + \epsilon) = \frac{1}{3}t^{-2/3}\epsilon + \sqrt[3]{t} $$
So, “the cube root's excess” would seem to be $\frac{1}{3}t^{-2/3}\epsilon$. At first, I was confused like you, wondering how the $t^{-2/3}$ went away. But then it dawned on me that Feynman was speaking of the “excess” in relative rather than absolute terms. In which case:
Which is consistent with his statement. So, he was using the tangent-line approximation, just in confusingly-worded terms. Here's an equivalent formulation that makes the relative excess more apparent.
$$\tilde{f}(t + \epsilon) = \sqrt[3]{t}(1 + \frac{\epsilon}{3t}) $$
Plugging $t = 1728$ and $\epsilon = 1.03$ gives the approximation:
$$\sqrt[3]{1729.03} \approx 12(1 + \frac{1.03}{3 \times 1728}) = 12 + \frac{1.03}{3 \times 144} \approx 12.002384259259259$$
For comparison, the actual value (to double-precision) is $12.002383785691718$, so the error is only 39 parts per billion.
Of course, if you're not the kind of person who can multiply or divide by 1728 in your head, you can round off the numbers in the “excess” term.
$$\sqrt[3]{1729.03} \approx 12(1 + \frac{1}{3 \times 2000}) = 12 + \frac{1}{500} = 12.002$$
Which isn't as good of an approximation, but it's still a relative error of only 32 parts per million, so not bad.