Wild power pie

Many people are aware of {\pi} being a number between 3 and 4, and some also know that {e} is between 2 and 3. Although the difference {\pi-e} is less than 1/2, it’s enough to place the two constants in separate buckets on the number line, separated by an integer.

When dealing with powers of {e}, using {e>2} is frequently wasteful, so it helps to know that {e^2>7}. Similarly, {\pi^2<10} is way more precise than {\pi<4}. To summarize: {e^2} is between 7 and 8, while {\pi^2} is between 9 and 10.

Do any two powers of {\pi} and {e} have the same integer part? That is, does the equation {\lfloor \pi^n \rfloor = \lfloor e^m \rfloor} have a solution in positive integers {m,n}?

Probably not. Chances are that the only pairs {(m,n)} for which {|\pi^n - e^m|<10} are {m,n\in \{1,2\}}, the smallest difference attained by {m=n=1}.

Indeed, having {|\pi^n - e^m|<1} implies that {|n\log \pi - m|<e^{-m}}, or put differently, {\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n \,\pi^n}}. This would be an extraordinary rational approximation… for example, with {n=100} it would mean that {\log \pi = 1.14\ldots} with the following {50} digits all being {0}. This isn’t happening.

Looking at the continued fraction expansion of {\log \pi} shows the denominators of modest size {[1; 6, 1, 10, 24, \dots]}, indicating the lack of extraordinarily nice rational approximations. Of course, can use them to get good approximations, {\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n^2}}, which leads to {\pi^n\approx e^m} with small relative error. For example, dropping {24} and subsequent terms yields the convergent {87/76}, and one can check that {\pi^{76} = 6.0728... \cdot 10^{37}} while {e^{87} = 6.0760...\cdot 10^{37}}.

Trying a few not-too-obscure constants with the help of mpmath library, the best coincidence of integer parts that I found is the following: the 13th power of the golden ratio {\varphi = (\sqrt{5}+1)/2} and the 34th power of Apèry’s constant {\zeta(3) = 1^{-3}+2^{-3}+3^{-3}+4^{-4}+\dots} both have integer part 521.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s