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Can you make $e$ with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9?

You can only use the digits once.

The allowed operators are listed beloẉ.

  • Addition
  • Subtraction
  • Multiplication
  • Division
  • Square root
  • Exponents
  • Log, Cos, Sin

You shall not use a operation not listed on the list.

You can only use up to 20 operators.

That is all - Have fun!

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    $\begingroup$ I presume arranging the symbols to form the shape of the letter is not acceptable, despite the phrasing of the question. $\endgroup$
    – keshlam
    Commented Oct 16, 2022 at 14:50
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    $\begingroup$ Spoiler: Assuming “Log” is natural log, you can just do $2^{3/Log(8)}$. Of course, this does not work in base 10 log. $\endgroup$
    – shalop
    Commented Oct 16, 2022 at 19:50
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    $\begingroup$ Are we allowed to concatenate digits into larger numbers (e.g. '1' + '2' = '12')? $\endgroup$
    – Galen
    Commented Oct 16, 2022 at 22:35
  • $\begingroup$ @shalop Use ln instead of Log. $\endgroup$
    – David Hammen
    Commented Oct 17, 2022 at 4:59
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    $\begingroup$ @DavidHammen Log was specifically used in the list of permitted operations. I think shalop has made the right notational choice here. $\endgroup$
    – Arthur
    Commented Oct 18, 2022 at 12:42

2 Answers 2

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Since we are allowed to use $\sin$ and $\cos$, we can exploit the identity $e^{ix}=\cos x+i\sin x$ to hit $e$ on the nose: $$e=e^{i(-i)}=\cos i-i\sin i=\cos\sqrt{1+2-3+4-5}-\sqrt{6-7}\sin\sqrt{8-9}$$

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  • $\begingroup$ Ah, super clever! You can avoid the unary minus (negation is not listed as an allowed operation) while keeping the stylish numerical order of the digits by using, for example, $1+2-3+4-5$ combined with $6-7$ and $8-9$. $\endgroup$
    – Bass
    Commented Oct 16, 2022 at 8:35
  • $\begingroup$ @Bass Right, I thought subtraction included negation... $\endgroup$
    – Parcly Taxel
    Commented Oct 16, 2022 at 9:41
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    $\begingroup$ Yeah, that's clearly OP's view too, and since they both basically do the same thing and use the same symbol, the difference is indeed pure semantics. On this site of course, my favourite antics are of the "ped-" variety :-) $\endgroup$
    – Bass
    Commented Oct 16, 2022 at 10:21
  • $\begingroup$ (+1) Mathematically speaking, I think $e^{ix} = \cos x + i\sin x$ only holds for real values of $x$. Otherwise one could do stuff like this: $$1 = 1^i = (e^{2\pi i})^i = e^{-2\pi}.$$ $\endgroup$
    – Yanko
    Commented Oct 18, 2022 at 20:05
  • $\begingroup$ @Yanko $(a^b)^c\neq a^{bc}$ in general. $$1^i = e^{\log(1^i)} = e^{i \log(1)} = e^{i (\ln|1| + i \arg(1))} = e^{-\arg(1)} = e^0 = 1$$ $\endgroup$
    – user170231
    Commented Oct 18, 2022 at 20:18
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I'm not beating Parcly, but here is a classic champion that I looked up at WolframMathWorld, using only algebraic operations (and thus not hitting exactly):$$\left(1 + 9^{-4^{6\cdot 7}}\right)^{3^{2^{85}}}$$It uses $e\approx (1 + 1/n)^n$, and tries to make $n$ as large as possible. It is apparently correct to $18\,457\,734\,525\,360\,901\,453\,873\,570$ digits.

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    $\begingroup$ Really as stated, amazing. Using WolframAlpha 3^2^85 and 9^4^42 both has the same number of digits and finishes with ......3648803841. Just do not see how to prove those are the same values. $\endgroup$
    – z100
    Commented Oct 17, 2022 at 20:58
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    $\begingroup$ @z100 It's not a very difficult argument, using only standard power rules and the fact that $9$ and $4$ are squares:$$9^{4^{42}}=(3^2)^{(2^2)^{42}}=3^{2\cdot 2^{2\cdot 42}}=3^{2^{1+84}}$$Alternately, using your WolframAlpha result, you can note that the two expressions are both clearly powers of 3. And powers of 3 with the same final digit and the same number of digits must be the same number (two powers of 3 with the same number of digits must differ by a factor of 1, 3 or 9, and only one of those alternatives preserves the final digit). $\endgroup$
    – Arthur
    Commented Oct 17, 2022 at 21:24
  • $\begingroup$ Really cool! How did you compute the number of correct digits? $\endgroup$
    – doetoe
    Commented Nov 2, 2022 at 16:59
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    $\begingroup$ @doetoe It was a very hard calculation. The main crux of the proof was looking at the page I linked to, and copy-pasting that number from there. $\endgroup$
    – Arthur
    Commented Nov 2, 2022 at 18:00

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