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There's a calculator with 10 functions exp, square, sin, cos, tan and their inverses. We need to convert 0 to 1 to 2 to 3 to -3. Note: Addition, subtraction, multiplication and division are not there

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    $\begingroup$ Is the calculator in degrees or radians? $\endgroup$ – Rand al'Thor Oct 11 '16 at 23:10
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    $\begingroup$ Does the calculator allow complex intermediate values? $\endgroup$ – 2012rcampion Oct 12 '16 at 0:38
  • $\begingroup$ This is a fascinating broken-calculator variation and makes me wonder about the smallest subset of push-button functions (perhaps standard calculator keys, perhaps not) that could produce any rational number $\endgroup$ – humn Oct 12 '16 at 4:39
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(This began as combined answers of Gareth McCaughan and humn, and has progressed nicely.)

As spoilered as it gets:

0$\cos $1$\exp $ $e$ $ x^2 $ $e^2$ $\log $2$\exp $ $e^2$ $\surd $ $e$ $\surd $ $e^{1/2}$ $\log $ $\frac{1}{2}$ $\cos^{-1}$ $\frac{\pi}{3}$ $\tan $ $\sqrt{3}$ $x^2 $3$ e^x $ $e^3$ $\tan^{-1}$ $\alpha ~ \small \big( \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{e^3}1 \big)$ $\sin $ $\small \dfrac{e^3}{\sqrt{1+e^6}}$ $\cos^{-1}$ $\beta ~ \small \big( \beta = \frac\pi2{-}\alpha \, , ~ \sin\beta = \dfrac1{\sqrt{1+e^6}} \big)$ $\tan $ $\small \dfrac1{e^3}$ $\log $−3

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  • $\begingroup$ Do we even need the description of the intermediate functions anymore, seeing as the short form doesn't use $x+1$ anymore, and only uses $-x$ in one place? $\endgroup$ – 2012rcampion Oct 12 '16 at 3:17
  • $\begingroup$ Cleaned up, @2012rcampion. Thanks to your improvement, the buttons might well fit on a single line by themselves now. $\endgroup$ – humn Oct 12 '16 at 4:14
  • $\begingroup$ Spoilertagging doesn't really work with that keyboard-style formatting: the only thing that's hidden is the result of each operation, which is easy to deduce anyway since we know the starting point. $\endgroup$ – Rand al'Thor Oct 12 '16 at 13:52
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Here's the missing piece in humn's otherwise excellent answer: how to get from 3 to -3.

First of all, it's enough to get from exp(3) to exp(-3); in other words, we need to take a reciprocal. It's then enough to get from atan(exp(3)) to atan(exp(-3)), which means getting from $x$ to $\pi/2-x$ for some $x$. But we can do that by taking sin and then acos.

So the specific sequence of operations is:

exp atan sin acos tan log.

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Here are the given functions:

$ e^x $ $\rm ln $ $ x^2 $ $ \surd$ $\rm sin$ $\rm asin$ $\rm cos$ $\rm acos$ $\rm tan$ $\rm atan$

These functions can be contructed from those:

$ |x| $ $ \, : ~~~ x $ $ x^2 $ $x^2$ $ \surd$ $|x|$ $\raise-1ex\strut$

$ 1{+}x $ $ : ~~~ x $ $ \surd$ $\surd x$ $\rm atan$ $ \theta $ ($\tan \theta = \surd x$) $\rm cos$ $ \frac1{\sqrt{1+x}} $ $\rm ln $ $ -\ln \sqrt{1+x} $
      $ |x| $ (as constructed) $ \ln \sqrt{1+x} $ $ e^x $ $ \sqrt{1+x} $ $ x^2 $ $1+x$

This gets to 3:

0 $ 1{+}x $ 1 $ 1{+}x $ 2 $ 1{+}x $ 3

To convert 3 to -3, see Gareth McCaughan’s answer or the combined wiki post.

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  • 0 to 1:

    $\exp(0)=1$.

  • 1 to 2:

    $\cos^2(\tan^{-1}(1))=\frac{1}{1^2+1}=\frac{1}{2}$; take the inverse to get $2$.

  • 2 to 3:

    $\tan^2(\cos^{-1}(2^{-1}))=\Big(\frac{1}{2^{-1}}\Big)^2-1=3$.

  • 3 to -3:

    $\log((\exp(3))^{-1})=-3$.

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  • $\begingroup$ Just to clarify: $\cos^2(x)=\left(\cos(x)\right)^2$, not $\cos(\cos(x))$, right? $\endgroup$ – 2012rcampion Oct 12 '16 at 0:06
  • $\begingroup$ How do you perform 1/x? $\endgroup$ – humn Oct 12 '16 at 0:08
  • $\begingroup$ @2012rcampion Yep. Squaring is allowed, according to the question, even though multiplication isn't. $\endgroup$ – Rand al'Thor Oct 12 '16 at 0:14
  • $\begingroup$ @humn Taking inverses is allowed according to the question. $\endgroup$ – Rand al'Thor Oct 12 '16 at 0:14
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    $\begingroup$ Inverses of the functions given $\endgroup$ – humn Oct 12 '16 at 0:15

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