# The Mathematical Ninja and the Cube Root of 4

The student swam away, thinking almost as hard as he was swimming. The cube root of four? The *square* root was easy enough, he could do that in his sleep. But the cube root?

OK. Breathe. It’s between 1 and 2, obviously. What’s 1.5 cubed? The Mathematical Ninja isn’t going to like that - *three-halves*, that’s what we need. Cube that, it’s $\frac{27}{8}$, or 3.375. Quite a long way short. But - aha! (Breathe) $1.6^3 = 4.096$, should have got that straight away. That’s close enough for two lengths.

The student popped his head out of the water. “About 1.6, sensei.”

The Mathematical Ninja looked into their box of torture devices and emerged with a sinker. “Two lengths of catch-up. You can do better than that.”

The student knew better than to waste energy on grumbling. OK, so a binomial expansion? $\br{4.096-x}^{\frac{1}{3}} \approx 1.6 - \frac{1}{3}\times 4.096^{-\frac{2}{3}} \times x$. Breathe. What’s ugly? $4.096^{-\frac{2}{3}}$ is just $\frac{1}{1.6^2}$. And $x$ is $0.096$ here. Turn.

So, it’s $1.6 - \frac{1}{3}\times \frac{100}{16^2} \times \frac{96}{1000}$. There’s a bit of cancelling can go on there - a factor of 3 and a factor of 16, and a factor of 100 for good measure. $1.6 - \frac{1}{16} \times \frac{2}{10}$ - suddenly it’s looking nicer! Breathe. A discrepancy of $\frac{1}{80}$. That’s 0.0125, which I can subtract. And head up: “1.5875?”

The Mathematical Ninja nodded their head from side to side. “It’s 1.5874,” they muttered, “but I’ll take it.” They handed the student a float. “Two lengths just kick,” they said. “By way of celebration.”

* edited 2018-10-01 to fix a LaTeX typo