Accuracy of approximating $\pi$ as n-th root of an integer

I was watching a YouTube short the other day, and he talked about approximating $$\pi$$ as n-th root of an integer. So, I was like, 'surely I can just calculate $$\pi^n$$ and round it'.

After that, I realize that I can actually count how many digits does the approximation got right. So, I write simple code in python to just get a feel of how good the approximation could get, and this is what I found

1. The plot is surprisingly looked linear, at least up to 27 (which is the highest I can go with my limited coding ability
2. The equation for that linear extrapolation is $$k = Mn$$, where $$k$$ is the number of correct decimal place, and $$n$$ is the power.

I have 2 question regarding this plot.

1. Is it really close to linear, or is it just a coincidence because I work with small number.
2. What is the value of $$M$$? So far, I only get like 0.53... And, is it rational?
• drive.google.com/file/d/1z3CNM8fKIiSL6EOOlXO1D7YL7dfqWIEs is the link to the figure if you can to have a quick look Oct 29, 2022 at 15:02
• Using 10M binary digits (about 3M decimal digits) for calculations with MPFR via Haskell library Rounded, for $n=1,10,100,...$ I get $k=0,6,51,501,4975,49720,497155$. It looks like it might be very slightly sublinear? code: let pi_ = pi :: Rounded TowardNearest 10000000 in [- ceiling (logBase 10 (abs (pi_ - (rint_round_ (pi_ ** n))**(1/n) :: Rounded TowardNearest 10000000))) | m <- [0..6], let n = 10^m ] Oct 29, 2022 at 15:55
• @Claude I think the slight sublinear trend you're noticing is just that the number of correct digits has both a linear term and some sublinear terms, and as $n$ increases, the sublinear terms contribute proportionately less. Oct 29, 2022 at 16:15

We can get a reasonable guess at what's going to happen just by the philosophy that probably writing $$\pi$$ as the $$n^{\text{th}}$$ root of an integer shouldn't be any more or less efficient at conveying the value of $$\pi$$ than just writing down the decimal digits. In other words, we can expect that writing down $$\pi \approx \sqrt[27]{26\,487\,841\,119\,103}$$ and $$\pi \approx 3.141\,592\,653\,589\,793$$ should be approximately equally accurate, because both of these have put in $$16$$ digits of effort into approximating $$\pi$$. (In this particular example, the $$27^{\text{th}}$$ root is only a tiny bit worse: it gets the $$3$$ at the end wrong.)

I'm actually not sure whether this intuition suggests counting the digits in "$$27$$" or not - if we don't count them, the $$27^{\text{th}}$$ root performs slightly better than the decimal expansion. Either way, that won't make a big difference, since $$n$$ is tiny compared to $$\pi^n$$.

If we believe this intuitive argument, then since the integer part of $$\pi^n$$ has about $$\log_{10}(\pi^n) = n \log_{10}\pi \approx 0.497 n$$ digits, it should correctly predict about $$0.497n$$ digits of $$\pi$$.

That's just intuition, not a proof. Here's an algebraic argument for the same estimate.

Suppose that we approximate $$\pi$$ as $$\sqrt[n]{m}$$, where $$m$$ is the closest integer to $$\pi^n$$. This guarantees to us that $$|m - \pi^n| < \frac12$$, and usually the error will be around $$\frac14$$. Let's write this as $$\left|\frac{m}{\pi^n} - 1\right| < \frac1{2\pi^n}.$$

To estimate the error, write $$\pi = \sqrt[n]{m} + \delta$$. Then $$\frac{\sqrt[n]m}{\pi} = 1 - \frac{\delta}{\pi}$$, or $$\frac{m}{\pi^n} = (1 - \frac{\delta}{\pi})^n$$. Substituting this into the inequality above, we get $$\left|\left(1 - \frac{\delta}{\pi}\right)^n - 1\right| < \frac1{2\pi^n}.$$ The linear approximation of $$(1+x)^n$$ is $$1 + nx$$, which has an $$O(x^2)$$ error as $$x \to 0$$; it is reasonable to use it here, since in fact $$\frac{\delta}{\pi} \to 0$$ as $$n\to \infty$$. This tells us that $$\left|1 - \frac{n\delta}{\pi} - 1 \right| \lessapprox \frac1{2\pi^n}$$ or $$|\delta| \lessapprox \frac1{2n \pi^{n-1}}$$.

We have $$k$$ correct digits when $$|\delta| \approx 10^{-k}$$, so in general the number of correct digits is about $$-\log_{10} |\delta| \approx \log_{10} (2n\pi^{n-1}) = \log_{10}(\tfrac{2}{\pi} n) + n \log_{10}\pi.$$ This gives us a result very similar to the intuition we had: dropping the $$\log_{10}(\frac2\pi n)$$, which is a sublinear term, we predict about $$0.497 n$$ correct digits.

This assumes that we don't get much luckier in approximation of $$\pi^n$$ by an integer, which sometimes happens: for example, $$\pi^3 \approx 31.006$$, and as a result, $$\sqrt[3]{31} \approx 3.14138$$ gets us more correct digits than we expected. It could even be true (but I don't think there's any reason to suspect it) that $$\pi^n$$ regularly gets much closer than expected to an integer, which could upset the trend line. Both the intuition and the algebraic estimate above assume $$\pi$$ behaves like any other real number of about its size, which is probably true but hard to prove.