Limited functions calculator

Question

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

Answer 1

^{(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

Answer 2

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.

Answer 3

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.

Answer 4

• 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:

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