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### Young Alan Turing and the Arctangent

With the release of the new film "The Imitation Game", I decided to read the biography this excellent film was based on - Alan Turing: The Enigma. In it, the author Andrew Hodges relates the story that the 15-year-old Alan Turing derived the Maclaurin series for the $$\arctan$$ function, i.e. $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots$ This is trivial using calculus, but it's explicitly stated that young Alan Turing neither knew nor used calculus. How would you derived such a series without calculus?

This is a tricky problem, and I'd suggest first tackling the much easier problem of deriving the Maclaurin series for $$\exp(x)$$ from the relation $$\exp(2x) = \exp(x)\cdot \exp(x)$$. This is an underconstrained relation, so you'll need to assume $$c_0 = 1, c_1 = 1$$.

Getting back to $$\arctan$$, you could start with the half-angle formula for the tangent: $\tan(2x) = \frac{2\tan(x)}{1-{\tan}^2(x)}.$ Now use the Weierstrass-like substitution $$x = \arctan(t)$$ to get $\tan(2\arctan(t)) = \frac{2t}{1-t^2}.$ The right-hand side can be expanded in the usual geometric series fashion to get $\tan(2\arctan(t)) = 2t\cdot (1+t^2+t^4+\ldots).$
Finally, take the $$\arctan$$ of both sides and assume we have the series expansion $$\arctan(x) = c_1 x + c_3 x^3 + c_5 x^5 + \ldots$$. Note that we may ignore the terms with even powers of $$x$$ as $$\arctan(x)$$ is an odd function.

This gives us the setting $2\arctan(t) = \arctan(2t\cdot (1+t^2+t^4+\ldots))$ and expanding as a power series $2(c_1 t + c_3 t^3 + \ldots) = c_1 (2t\cdot (1+t^2+\ldots) + c_3 (2t\cdot (1+t^2+\ldots))^3 + \ldots$
The next step is to line up powers of $$t$$ on both sides and set up a system of simultaneous equations. There's some algebra and combinatorics involved, but we end up with the system of equations $c_{2i+1} = \sum_{j=0}^{i} c_{2j+1}\cdot 2^{2j} \cdot {{i+j} \choose {2j}}.$ Note that this system is underconstrained due to our functional relationship being satisfied by any multiple of the $$\arctan$$ function. We'll assume that $$c_1 = 1$$, but note that this follows from the classical (non-calculus) limit $$\lim_{x\to 0} \frac{\sin(x)}{x} = 1$$.

The first few relations are \begin{align*}
c_1 &= c_1 \\
c_3 &= c_1 + 4\cdot c_3 \\
c_5 &= c_1 + 12\cdot c_3 + 16\cdot c_5 \\
c_7 &= c_1 + 24\cdot c_3 + 80\cdot c_5 + 64\cdot c_7
\end{align*}
Assuming $$c_1 = 1$$ as above we quickly calculate $$c_3 = -\frac{1}{3}, c_5 = \frac{1}{5}, c_7 = -\frac{1}{7}$$, with the pattern being obvious.

That $$c_{2i+1} = \frac{(-1)^i}{2i+1}$$ can be verified by Wolfram Alpha:

An obvious question is whether or not there's a simple demonstration of this; in particular, one that a young Alan Turing may have found. This I don't know (yet).

### Comments

1. Did you mean to link to the book 'Alan Turing: The Enigma' by Andrew Hodges?

http://www.amazon.com/Alan-Turing-Enigma-Inspired-Imitation-ebook/dp/B00M032W92/

1. Yes, thanks! Corrected.

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Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

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Assume we pick Door 1 and then Monty sho…

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