## August 30, 2020

### Euler’s Continued Fraction Formula

#### Posted by John Baez I’ve been reading about Ramanujan. His mastery of continued fractions made me realize how bad I am at manipulating them. Here’s something much more basic: a proof that

$\frac{4}{\pi} = 1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \frac{7^2}{2 + \ddots}}}}$

It illustrates a method called Euler’s continued fraction formula.

There’s nothing new about this — it goes back to around 1748. But it might be fun if you haven’t seen it already.

Here is Euler’s continued fraction formula:

$a_0 + a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2\cdots a_n =$ $\frac{a_0}{1 - \frac{a_1}{1 + a_1 - \frac{a_2}{1 + a_2 - \frac{\ddots}{\ddots \frac{a_{n-1}}{1 + a_{n-1} - \frac{a_n}{1 + a_n}}}}}}\,$

The easiest way to convince yourself of it is to take the right-hand side for some smallish value of $n$ and keep simplifying it in obvious ways — putting sums of fractions on common denominators, etc. — until you get the left-hand side.

It’s a nice way to turn finite sums into fractions… and then, by taking a limit, infinite series into continued fractions.

For example, starting with

$\begin{array}{ccl} \arctan x &=& \displaystyle{ \int_0^x \frac{d t}{1 + t^2} } \\ \\ & = & \displaystyle{ \int_0^x \left(1 - t^2 + t^4 - t^6 \cdots \right) dt } \\ \\ &=& \displaystyle{ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots } \\ \\ &=& x + x \left(\frac{-x^2}{3}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right)\left(\frac{-5x^2}{7}\right) + \cdots \end{array}$

and then applying Euler’s continued fraction formula, we get

$\begin{array}{ccl} \arctan x &=& x + x \left(\frac{-x^2}{3}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right)\left(\frac{-5x^2}{7}\right) + \cdots \\ \\ &=& \frac{x}{1 - \frac{\frac{-x^2}{3}}{1 + \frac{-x^2}{3} - \frac{\frac{-3x^2}{5}}{1 + \frac{-3x^2}{5} - \frac{\frac{-5x^2}{7}}{1 + \frac{-5x^2}{7} - \ddots}}}} \\ \\ &=& \frac{x}{1 + \frac{x^2}{3 - x^2 + \frac{(3x)^2}{5 - 3x^2 + \frac{(5x)^2}{7 - 5x^2 + \ddots}}}} \end{array}$

where in the last step we did a lot of multiplying numerators and denominators by the same odd number.

All this works within the radius of convergence of the geometric series, namely for $|x| < 1$. But with some more care it works at $x = 1$, which is nice because $\arctan 1 = \frac{\pi}{4}$. So we get

$\frac{\pi}{4} = \frac{1}{1 + \frac{1^2}{3 - 1 + \frac{3^2}{5 - 3 + \frac{5^2}{7 - 5 + \cdots}}}}$

or in other words

$\frac{4}{\pi} = 1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \frac{7^2}{2 + \ddots}}}}$

You can see other interesting examples of this method on Wikipedia.

Posted at August 30, 2020 6:49 AM UTC

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## 1 Comment & 0 Trackbacks

### Re: Euler’s Continued Fraction Formula

I wonder if anything came of the ‘continued fraction coalgebra’ mentioned here.

Hmm, quite an interesting connection being mooted in the post there, from Conway’s rational tangles to their completion in some kind of reals meets infinite tangles fusion.

Posted by: David Corfield on August 30, 2020 10:41 AM | Permalink | Reply to this

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