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I was fiddling with numbers when I noticed that $$50 \times 1.05^{168} \times \frac{12600}{727767941} \approx \pi$$ I understand it's an approximation. Does anyone know why?

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    $\begingroup$ That's probably just a coincidence. $\endgroup$ – Arthur Nov 10 '15 at 15:52
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    $\begingroup$ Ho many digits accuracy does it give you? That's a fairly random expression - given any huge numerator, we can likely find some denominator that gets somewhat close. $\endgroup$ – Thomas Andrews Nov 10 '15 at 15:53
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    $\begingroup$ If you gave us a reason or something interesting about that denominator, then it might be worth exploring, or if these values came out of another question, but just blindly, a number close to $\pi$ is a dime a dozen - there are lots of them. There isn't a "reason" for it - there are lots of expressions equally close to every real number. $\endgroup$ – Thomas Andrews Nov 10 '15 at 16:01
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    $\begingroup$ There are occasionally really incredible approximations of $\pi$. For example: $$\frac{\ln 262 537 412 640 768 744}{\sqrt{163}}$$ is unreasonably close, because there is a deep reason for $e^{\pi\sqrt{163}}$ to be remarkably close to an integer. $\endgroup$ – Thomas Andrews Nov 10 '15 at 16:24
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    $\begingroup$ $262537412640768744=640320^3+744.$ $\endgroup$ – Lucian Nov 10 '15 at 16:44
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By my calculator this expression only agrees with $\pi$ to about 13 decimal digits. Since there are 22 separate digits in the expression, we should expect that very many different expressions of that shape approximate $\pi$ with a similar precision -- there's nothing particular remarkable about one of them, nor any "explanation" other than it just happens to be close to $\pi$.

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  • $\begingroup$ Thanks. My calculator and my low coding abilities did not enable me to see to what extent it agrees with pi. $\endgroup$ – TabulaSmaragdina Nov 10 '15 at 15:57
  • $\begingroup$ In fact, $1.05^{168}\times\frac{80034}{92454253}$ is a better approximation with fewer digits, but $\frac{80143857}{25510582}$ without the power of $1.05$ is even better. $\endgroup$ – Henning Makholm Nov 10 '15 at 16:41

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