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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.

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    $\begingroup$ 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. $\endgroup$ – Cameron Williams Jul 1 '13 at 4:10
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    $\begingroup$ The fact that $1000$ is roughly $2^{10}$ is very useful, for music, for talking about computer memory, and elsewhere. $\endgroup$ – André Nicolas Jul 1 '13 at 4:13
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ @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. $\endgroup$ – Jeppe Stig Nielsen Jul 1 '13 at 19:08
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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 $1000^n$.

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    $\begingroup$ Got it. So the pattern stems from the fact that $2^{10} = 1024$, which just happens to be close to $1000$. $\endgroup$ – Max Heinritz 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 $2^{30}$ $\endgroup$ – Aaron Meyerowitz Jul 1 '13 at 5:43
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    $\begingroup$ @Max: The "fundamental" coincidence is that $2^7 = 128 \approx 125 = 5^3$. Multiply each side by $2^3$ to get $2^{10} \approx 10^3$. $\endgroup$ – Ilmari Karonen Jul 1 '13 at 7:37
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    $\begingroup$ @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 $2$. $\endgroup$ – Ilmari Karonen Jul 1 '13 at 8:06
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    $\begingroup$ These coincidences are important. Consider $2^{\frac{5}{12}}\approx \frac{4}{3}$. Five semitones approximate a perfect fourth. $\endgroup$ – Eric Jablow Jul 1 '13 at 11:19
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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 $10^3$.

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    $\begingroup$ +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 :-/ $\endgroup$ – Jyrki Lahtonen Jul 1 '13 at 5:32
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    $\begingroup$ 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? $\endgroup$ – Tobias Kienzler 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
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    $\begingroup$ @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

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