# Ask Uncle Colin: fourth roots

Dear Uncle Colin,

How would you find $\sqrt[4]{923521}$ without a calculator?

-- Some Quite Recherché Technique

Hi, SQRT!

I have a few possible techniques here. The first is “do some clever stuff with logarithms”, the second is “do some clever stuff with known squares” and the last is “do some clever stuff with calculus.”

Without a calculator, I can easily see that 923,521 is a shade below $10^6$, which makes its base-10 logarithm somewhere a shade below 6. Knowing that $\log_{10}(3) \approx 0.477$ means $\log_{10}(9) \approx 0.954$, so $\log_{10}(923,521) \approx 5.96$ is a fair estimate.

The logarithm of its fourth root is a quarter of that, 1.49. That’s a little more than $\log(30) \approx 1.477$ and a little less than $\log(\sqrt{1000}) = 1.5$ - which means the fourth root of 923,521 is somewhere in the region of 31. In fact, 31 is a plausible answer, since the last digit is 1; multiplying 31 by itself repeatedly would show it to be the correct answer.

Option 2 is “doing clever stuff with known squares” - in particular, finding the square root and the square root of the result. I know that $\sqrt{923,521}$ is in the neighbourhood of 1000 - in fact, 960 looks like an even better bet, as $960^2 = 921,600$, which is pretty close to the given number.

Difference of two squares is a handy thing here: 923,521 - 921,600 = 1,921, which is $n^2 - 960^2$, but also $(n-960)(n+960)$. A moment’s thought shows that $n=961$ makes that work. And, as any fule kno, $961 = 31^2$.

The final approach is the most reliable, generally; the other two rely on saying “this is probably a whole number”, but the Newton-Raphson approach doesn’t.

The Newton-Raphson method takes a function - for example, $f(x) = x^4 - 923,521$ and an initial guess (let’s say $x_0=40$, assuming we’re not thinking very hard) and iteratively improves the guess using the equation:

$f’(x_n)\left(x_{n+1} - x_n\right) = 0 - f(x_n)$

You don’t normally see it written this way (instead, it’s usually given explicitly as $x_{n+1} = x_n - \frac{f(x_n)}{f’(x_n)}$) - but I like to affirm that “this is just following the tangent line until it hits the $x$-axis”.

But let’s freeze-frame here a second: if we do this as stated, we’re going to end up taking cubes and fourth powers of some pretty awful numbers as soon as we leave the safe haven of our initial, well-chosen guesses. Instead, it’s going to be much nicer to use $f(x)= x^2 - 923,521$ to find the square root of our number, and then square root the result.

So! $f’(x) = 2x$, and let’s pick $x_0 = 1000$, a reasonable guess. Then, $x_1 = 1000 - \frac{76,479}{2,000}$, which works out to be $1000 - 38.2395$, or $961.7605$ (which is already pretty good.)

To get the next iteration, we’ll need to do some work. Squaring 961.7605 without a calculator would be unenviable, but fortunately we have some short cuts! We know it’s $1000 - \frac{76,479}{2000}$, so squaring it gives $1,000,000 - 76,479 + 38.2395^2$. The first two terms sum to 923,521!

That’s good, because we want $x_2 = x_1 - \frac{961.705^2 - 923,521}{1923.521}$, and the top is just $38.2395^2$. I would treat that as $(50-11.7605)^2 = 2500 - 1176 + 140$-odd, or in the region of 1464. In ballpark terms, $\frac{1464}{1923}$ is certainly around three-quarters, and it’s plausible that it’d be bang on the 0.7605 we need to bring it down to 961. Someone with more patience than I have might like to check by hand!

Then, once we have 961, we could reasonably apply the same technique again; however, we know that $961 = 31^2$ because we’ve paid attention, so we get the same answer as before.

Hope that helps!

- Uncle Colin