12
$\begingroup$

See the complete list here: http://en.wikipedia.org/wiki/Power_of_two#Powers_of_1024.

I'm wondering if there's a mathematical explanation for the relationship or if it's just coincidence.

$\endgroup$
9
  • 7
    $\begingroup$ It's because $2^{10}$ is close to $1000$. $2^{20} = 2^{10+10} = 2^{10}\times2^{10} \approx 1000\times 1000$. The pattern carries on from here. As you can see, the error grows and that's because $2^x$ is a convex function (opens upward) and so the errors get compounded. $\endgroup$ Jul 1 '13 at 4:10
  • 1
    $\begingroup$ The fact that $1000$ is roughly $2^{10}$ is very useful, for music, for talking about computer memory, and elsewhere. $\endgroup$ Jul 1 '13 at 4:13
  • 1
    $\begingroup$ @AndréNicolas I'm afraid I cannot think of where $2^{10}$ applies in music, let alone the approximation $1000 \approx 1024$. Nice riddle. $\endgroup$
    – Kaz
    Jul 1 '13 at 14:09
  • 1
    $\begingroup$ @Kaz: The audible spectrum is 20Hz-20.000 Hz. In music, an octave is a frequency band from f to 2× f Hz. Hence, the audible spectrum is approximately 10 octaves $\endgroup$
    – MSalters
    Jul 1 '13 at 15:13
  • 2
    $\begingroup$ @Kaz One approximation that is often seen in music is $(3:2)^{12} \approx (2:1)^7$, or twelve perfect fifths equal seven octaves. (The error is called the Pythagorean comma.) As powers of primes that is $3^{12} \approx 2^{19}$. The approximation from this question, when "reduced", is $5^3 \approx 2^8$ which musically could be seen as $(5:4)^3 \approx (2:1)$, the interpretation being "three 'perfect' major thirds equal one octave". For example C-E, E-G#, G#-C. Of course such approximations can never be exact because of unique factorization of integers. $\endgroup$ Jul 1 '13 at 19:08
26
$\begingroup$

Since $2^{10}=1024$: $$2^{10n}=(1000+24)^n=1000^n+24\cdot 1000^{n-1}n+...$$ Thus, as long as $24n$ remains a lot smaller than $1000$, then $2^{10n}$ will be near $1000^n$.

$\endgroup$
6
  • 3
    $\begingroup$ Got it. So the pattern stems from the fact that $2^{10} = 1024$, which just happens to be close to $1000$. $\endgroup$ Jul 1 '13 at 5:01
  • $\begingroup$ I'd say so. Cool that one can do $2^{20}$ mentally given $24^2=576$ and with a certain amount of effort $2^{30}$ $\endgroup$ Jul 1 '13 at 5:43
  • 9
    $\begingroup$ @Max: The "fundamental" coincidence is that $2^7 = 128 \approx 125 = 5^3$. Multiply each side by $2^3$ to get $2^{10} \approx 10^3$. $\endgroup$ Jul 1 '13 at 7:37
  • 1
    $\begingroup$ @Fabio: While that's also a handy coincidence (e.g. it's why $1,2,5,10,20,50,100,\dotsc$ is such a nice approximation to a geometric series), it not nearly as close as $2^7\approx5^3$. In fact, $8^3=2^9=512$, which is off from $10^3=1000$ by almost exactly a factor of $2$. $\endgroup$ Jul 1 '13 at 8:06
  • 6
    $\begingroup$ These coincidences are important. Consider $2^{\frac{5}{12}}\approx \frac{4}{3}$. Five semitones approximate a perfect fourth. $\endgroup$ Jul 1 '13 at 11:19
13
$\begingroup$

A good "explanation" is that $\log_{10} 2 = 0.3010$.

Hence, $\log_{10} 2^{10} = 10 \log_{10} 2 = 3.01$, hence $2^{10}$ is very close to $10^3$.

$\endgroup$
6
  • 5
    $\begingroup$ +1: But my thoughts run the opposite direction. Because $2^{10}$ is approximately $10^3$ we have $\log_{10}2\approx0.3$, which is a useful bit for a class of mental arithmetic estimations. Undoubtedly you too played with a calculator enough in your youth to have $\log_{10}2$ memorized up to a not so useful number of decimal places, but I think this is still good for teaching :-/ $\endgroup$ Jul 1 '13 at 5:32
  • 3
    $\begingroup$ Agreed. That is why the word explanation is in quotes. I memorized that $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$, and the rest of the logs of single digits can be derived from there. $\endgroup$
    – Calvin Lin
    Jul 1 '13 at 5:40
  • $\begingroup$ @CalvinLin You can use $\log(a^nb^m)=n\log a + m\log b$ to obtain the logs of $4=2^2,6=2\cdot3,8=2^3,9=3^3$; but how to you obtain those of 5 and 7? Interpolation? $\endgroup$ Jul 1 '13 at 6:32
  • $\begingroup$ @TobiasKienzler $\log 5 = 1 - \log 2$. For $\log 7$, I interpolate. (Or use $\log 7 = \frac 12 \log 49 \approx \frac12 \log 50 = 1-\frac 12 \log 2$.) $\endgroup$
    – mrf
    Jul 1 '13 at 8:00
  • 1
    $\begingroup$ @mrf You can also interpolate with $\log 48$ (since we have $\log 3$). Essentially, it becomes the bisection of roots game, E.g. $\log 48\times 50 < \log 49*49 < \log 50*50$. $\endgroup$
    – Calvin Lin
    Jul 1 '13 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.