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As we can read in Wolfram Mathworld's article on approximations of $e$, the base of natural logarithm,

An amazing pandigital approximation to e that is correct to $18457734525360901453873570$ decimal digits is given by $$\LARGE \left(1+9^{-4^{6 \cdot 7}}\right)^{3^{2^{85}}}$$ found by R. Sabey in 2004 (Friedman 2004).

The cited paragraph raises two natural questions.

  1. How was it found? I guess that Sabey hasn't used the trial and error method.
  2. Using which calculator can I verify its correctness "to $184\ldots570$ decimal digits"?
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1 Answer 1

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$$\begin{aligned} (1+9^{-4^{42}})^{3^{2^{85}}} &=(1+9^{-4^{42}})^{3^{2*2^{84}}}\\ &=(1+9^{-4^{42}})^{9^{2^{84}}} \\ &=(1+9^{-4^{42}})^{9^{4^{42}}}\\ &=\Bigl(1+\frac1{9^{4^{42}}}\Bigr)^{9^{4^{42}}}\qquad\text{where }=\left(1+\frac1n\right)^n. \end{aligned}$$ This is just the limit definition of $e$ with a large number as an approximation for $\infty$.

Edit: Numberphile just did a video on this, which also gives a pandigital approximation for $\pi$, but it's only accurate up to ten digits.

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  • $\begingroup$ Wow! So it is just a joke..? $\endgroup$
    – Eric
    Sep 28, 2016 at 14:54
  • $\begingroup$ @Eric It’s like a party trick. $\endgroup$ Sep 28, 2016 at 14:56
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    $\begingroup$ @Eric No, it's not easy at all to make it pandigital, all digits from $1$ to $9$ are used exactly once. $\endgroup$ Sep 28, 2016 at 14:57
  • $\begingroup$ @Daniel Fischer Fair enough. Its freaky accuracy isn't a mystery, though. $\endgroup$ Sep 28, 2016 at 14:59
  • $\begingroup$ That's right, all you need is a large $n$ for that. $\endgroup$ Sep 28, 2016 at 14:59

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