View source for Euler's continued fraction formula

{{short description|Connects a very general infinite series with an infinite continued fraction.}}
{{for|the continued fraction of Euler's constant ''e''|List of representations of e#As a continued fraction}}
In the [[complex analysis|analytic theory]] of [[generalized continued fraction|continued fractions]], '''Euler's continued fraction formula''' is an identity connecting a certain very general [[infinite series]] with an infinite [[continued fraction]]. First published in 1748, it was at first regarded as a simple identity connecting a finite sum with a finite continued fraction in such a way that the extension to the infinite case was immediately apparent.<ref>{{citation |year=1748 |author=[[Leonhard Euler]]|title=Introductio in analysin infinitorum|volume=I|chapter=18}}</ref> Today it is more fully appreciated as a useful tool in analytic attacks on the general [[convergence problem]] for infinite continued fractions with complex elements.

== The original formula ==
[[Euler]] derived the formula as 
connecting a finite sum of products with a finite [[continued fraction]].

:<math>
\begin{align}
a_0\left(1 + a_1\left(1 + a_2\left(\cdots + a_n\right)\cdots\right)\right) & = a_0 + a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2\cdots a_n \\
& = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \cfrac{\ddots}{\ddots 
\cfrac{a_{n-1}}{1 + a_{n-1} - \cfrac{a_n}{1 + a_n}}}}}}\,
\end{align}
</math>

The identity is easily established by [[mathematical induction|induction]] on ''n'', and is therefore applicable in the limit: if the expression on the left is extended to represent a [[convergent series|convergent infinite series]], the expression on the right can also be extended to represent a convergent infinite [[continued fraction]].

This is written more compactly using [[generalized continued fraction]] notation:
:<math>
a_0 + a_0 a_1 + a_0 a_1 a_2 + \cdots + a_0 a_1 a_2 \cdots a_n =
\frac{a_0}{1 +} \, \frac{-a_1}{1 + a_1 +} \, \cfrac{-a_2}{1 + a_2 +} \cdots \frac{-a_n}{1 + a_n}.
</math>

== Euler's formula ==

If ''r''<sub>''i''</sub> are complex numbers and ''x'' is defined by
:<math>
x = 1 + \sum_{i=1}^\infty r_1r_2\cdots r_i = 1 + \sum_{i=1}^\infty \left( \prod_{j=1}^i r_j \right)\,,
</math>
then this equality can be proved by induction
:<math>
x = \cfrac{1}{1 - \cfrac{r_1}{1 + r_1 - \cfrac{r_2}{1 + r_2 - \cfrac{r_3}{1 + r_3 - \ddots}}}}\,
</math>.

Here equality is to be understood as equivalence, in the sense that the {{mvar|n}}'th [[Convergent (continued fraction)|convergent]] of each continued fraction is equal to the {{mvar|n}}'th partial sum of the series shown above. So if the series shown is convergent &ndash; or ''uniformly'' convergent, when the ''r''<sub>''i''</sub>'s are functions of some complex variable ''z'' &ndash; then the continued fractions also converge, or converge uniformly.<ref>H. S. Wall, ''Analytic Theory of Continued Fractions'', D. Van Nostrand Company, Inc., 1948; reprinted (1973) by Chelsea Publishing Company {{ISBN|0-8284-0207-8}}, p. 17.</ref>

==Proof by induction==
Theorem: Let <math>n</math> be a natural number. For <math>n+1</math> complex values <math>a_0, a_1, \ldots, a_{n}</math>,
:<math>
\sum_{k=0}^n \prod_{j=0}^k a_j = \frac{a_0}{1+} \, \frac{-a_1}{1+a_1+} \cdots \frac{-a_n}{1+a_n} </math>
and for <math>n</math> complex values <math>b_1, \ldots, b_{n}</math>, 
<math>\frac{-b_1}{1+b_1+} \, \frac{-b_2}{1+b_2+} \cdots \frac{-b_n}{1+b_n} \ne -1.</math>

Proof: We perform a double induction. For <math>n=1</math>, we have 
:<math>
\frac{a_0}{1+} \, \frac{-a_1}{1+a_1} = \frac{a_0}{1+\frac{-a_1}{1+a_1}} = \frac{a_0(1+a_1)}{1} = a_0 + a_0 a_1
= \sum_{k=0}^1 \prod_{j=0}^k a_j</math>
and 
:<math>
\frac{-b_1}{1+b_1}\ne -1.</math>
Now suppose both statements are true for some <math>n \ge 1</math>.

We have
<math>\frac{-b_1}{1+b_1+} \, \frac{-b_2}{1+b_2+} \cdots \frac{-b_{n+1}}{1+b_{n+1}}
= \frac{-b_1}{1+b_1+x}</math> where <math>x = \frac{-b_2}{1+b_2+} \cdots \frac{-b_{n+1}}{1+b_{n+1}} \ne -1</math> 

by applying the induction hypothesis to <math>b_2, \ldots, b_{n+1}</math>.

But if <math>\frac{-b_1}{1+b_1+x} = -1</math> implies <math>b_1 = 1+b_1+x</math> implies <math>x = -1</math>, contradiction. Hence 
:<math>\frac{-b_1}{1+b_1+} \, \frac{-b_2}{1+b_2+} \cdots \frac{-b_{n+1}}{1+b_{n+1}} \ne -1,</math>
completing that induction.

Note that for <math>x \ne -1</math>,
:<math>
\frac{1}{1+} \, \frac{-a}{1+a+x} = \frac{1}{1-\frac{a}{1+a+x}} = \frac{1+a+x}{1+x} = 1 + \frac{a}{1+x};</math>
if <math>x=-1-a</math>, then both sides are zero.

Using 
<math>a=a_1</math> and <math>x = \frac{-a_2}{1+a_2+} \, \cdots \frac{-a_{n+1}}{1+a_{n+1}} \ne -1</math>,
and applying the induction hypothesis to the values <math>a_1, a_2, \ldots, a_{n+1}</math>,
:<math>
\begin{align}
a_0 + & a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2a_3 \cdots a_{n+1} \\
&= a_0 + a_0(a_1 + a_1a_2 + \cdots + a_1a_2a_3 \cdots a_{n+1}) \\
&= a_0 + a_0 \big( \frac{a_1}{1+} \, \frac{-a_2}{1+a_2+} \, \cdots \frac{-a_{n+1}}{1+a_{n+1}} \big)\\
&= a_0 \big(1 + \frac{a_1}{1+} \, \frac{-a_2}{1+a_2+} \, \cdots \frac{-a_{n+1}}{1+a_{n+1}} \big)\\
&= a_0 \big(\frac{1}{1+} \, \frac{-a_1}{1+a_1+} \, \frac{-a_2}{1+a_2+} \, \cdots \frac{-a_{n+1}}{1+a_{n+1}} \big)\\
&= \frac{a_0}{1+} \, \frac{-a_1}{1+a_1+} \, \frac{-a_2}{1+a_2+} \, \cdots \frac{-a_{n+1}}{1+a_{n+1}},
\end{align}</math>
completing the other induction.

As an example, the expression <math>a_0 + a_0a_1 + a_0a_1a_2 + a_0a_1a_2a_3</math> can be rearranged into a continued fraction.
:<math> \begin{align}
& a_0 + a_0a_1 + a_0a_1a_2 + a_0a_1a_2a_3 \\[8pt]
= {} & a_0(a_1(a_2(a_3 + 1) + 1) + 1) \\[8pt]
= {} & \cfrac{a_0}{\cfrac{1}{a_1(a_2(a_3 + 1) + 1) + 1}} \\[8pt]
= {} & \cfrac{a_0}{\cfrac{a_1(a_2(a_3 + 1) + 1) + 1}{a_1(a_2(a_3 + 1) + 1) + 1} - \cfrac{a_1(a_2(a_3 + 1) + 1)}{a_1(a_2(a_3 + 1) + 1) + 1}} = \cfrac{a_0}{1 - \cfrac{a_1(a_2(a_3 + 1) + 1)}{a_1(a_2(a_3 + 1) + 1) + 1}} \\[8pt]
= {} & \cfrac{a_0}{1 - \cfrac{a_1}{\cfrac{a_1(a_2(a_3 + 1) + 1) + 1}{a_2(a_3 + 1) + 1}}} \\[8pt]
= {} & \cfrac{a_0}{1 - \cfrac{a_1}{\cfrac{a_1(a_2(a_3 + 1) + 1)}{a_2(a_3 + 1) + 1} + \cfrac{a_2(a_3 + 1) + 1}{a_2(a_3 + 1) + 1} - \cfrac{a_2(a_3 + 1)}{a_2(a_3 + 1) + 1}}} = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2(a_3 + 1)}{a_2(a_3 + 1) + 1}}} \\[8pt]
= {} & \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{\cfrac{a_2(a_3 + 1) + 1}{a_3 + 1}}}} \\[8pt]
= {} & \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{\cfrac{a_2(a_3 + 1)}{a_3 + 1} + \cfrac{a_3 + 1}{a_3 + 1} - \cfrac{a_3}{a_3 + 1}}}} = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \cfrac{a_3}{1 + a_3}}}}
\end{align}</math>
This can be applied to a sequence of any length, and will therefore also apply in the infinite case.

== Examples==
=== The exponential function ===
The exponential function ''e''<sup>''x''</sup> is an [[entire function]] with a power series expansion that converges uniformly on every bounded domain in the complex plane.

:<math>
e^x = 1 + \sum_{n=1}^\infty \frac{x^n}{n!} = 1 + \sum_{n=1}^\infty \left(\prod_{i=1}^n \frac{x}{i}\right)\,
</math>

The application of Euler's continued fraction formula is straightforward:

:<math>
e^x = \cfrac{1}{1 - \cfrac{x}{1 + x - \cfrac{\frac{1}{2}x}{1 + \frac{1}{2}x - \cfrac{\frac{1}{3}x}
{1 + \frac{1}{3}x - \cfrac{\frac{1}{4}x}{1 + \frac{1}{4}x - \ddots}}}}}.\,
</math>

Applying an [[generalized continued fraction#The equivalence transformation|equivalence transformation]] that consists of clearing the fractions this example is simplified to

:<math>
e^x = \cfrac{1}{1 - \cfrac{x}{1 + x - \cfrac{x}{2 + x - \cfrac{2x}{3 + x - \cfrac{3x}{4 + x - \ddots}}}}}\,
</math>

and we can be certain that this continued fraction converges uniformly on every bounded domain in the complex plane because it is equivalent to the power series for ''e''<sup>''x''</sup>.

=== The natural logarithm ===
The [[mercator series|Taylor series]] for the [[principal branch]] of the natural logarithm in the neighborhood of 1 is well known:

:<math>
\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots =
\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n}}{n}.\,
</math>

This series converges when {{Math|{{!}}x{{!}} < 1}} and can also be expressed as a sum of products:<ref name="Abel">This series converges for |''x''|&nbsp;<&nbsp;1, by [[Abel's test]] (applied to the series for log(1&nbsp;&minus;&nbsp;''x'')).</ref>

:<math>
\log (1+x) = x + (x)\left(\frac{-x}{2}\right) + (x)\left(\frac{-x}{2}\right)\left(\frac{-2x}{3}\right) + (x)\left(\frac{-x}{2}\right)\left(\frac{-2x}{3}\right)\left(\frac{-3x}{4}\right) + \cdots
</math>

Applying Euler's continued fraction formula to this expression shows that

:<math>
\log (1+x) = \cfrac{x}{1 - \cfrac{\frac{-x}{2}}{1+\frac{-x}{2}-\cfrac{\frac{-2x}{3}}{1+\frac{-2x}{3}-\cfrac{\frac{-3x}{4}}{1+\frac{-3x}{4}-\ddots}}}}
</math>

and using an equivalence transformation to clear all the fractions results in

:<math>
\log (1+x) = \cfrac{x}{1+\cfrac{x}{2-x+\cfrac{2^2x}{3-2x+\cfrac{3^2x}{4-3x+\ddots}}}}
</math>


This continued fraction converges when |''x''| < 1 because it is equivalent to the series from which it was derived.<ref name="Abel"/>

=== The trigonometric functions===

The [[Taylor series]] of the [[sine]] function converges over the entire complex plane and can be expressed as the sum of products.
:<math> \begin{align} 
\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} & = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots \\[8pt]
& = x + (x)\left(\frac{-x^2}{2 \cdot 3}\right) + (x)\left(\frac{-x^2}{2 \cdot 3}\right)\left(\frac{-x^2}{4 \cdot 5}\right) + (x)\left(\frac{-x^2}{2 \cdot 3}\right)\left(\frac{-x^2}{4 \cdot 5}\right)\left(\frac{-x^2}{6 \cdot 7}\right) + \cdots 
\end{align}</math>
Euler's continued fraction formula can then be applied
:<math>\cfrac{x}{1 - \cfrac{\frac{-x^2}{2 \cdot 3}}{1 + \frac{-x^2}{2 \cdot 3} - \cfrac{\frac{-x^2}{4 \cdot 5}}{1 + \frac{-x^2}{4 \cdot 5} - \cfrac{\frac{-x^2}{6 \cdot 7}}{1 + \frac{-x^2}{6 \cdot 7} - \ddots}}}}</math>
An equivalence transformation is used to clear the denominators: 
:<math> \sin x = \cfrac{x}{1 + \cfrac{x^2}{2 \cdot 3 - x^2 + \cfrac{2 \cdot 3x^2}{4 \cdot 5 - x^2 + \cfrac{4 \cdot 5x^2}{6 \cdot 7 - x^2 + \ddots}}}}.</math>
The same [[argument]] can be applied to the [[cosine]] function:
:<math> \begin{align}
\cos x =  \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} & = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \\[8pt]
& = 1 + \frac{-x^2}{2} + \left(\frac{-x^2}{2}\right)\left(\frac{-x^2}{ 3 \cdot 4}\right) + \left(\frac{-x^2}{2}\right)\left(\frac{-x^2}{ 3 \cdot 4}\right)\left(\frac{-x^2}{ 5 \cdot 6}\right) + \cdots \\[8pt] & = \cfrac{1}{1 - \cfrac{\frac{-x^2}{2}}{1 + \frac{-x^2}{2} - \cfrac{\frac{-x^2}{3 \cdot 4}}{1 + \frac{-x^2}{3 \cdot 4} - \cfrac{\frac{-x^2}{5 \cdot 6}}{1 + \frac{-x^2}{5 \cdot 6} - \ddots}}}}
\end{align}</math>
:<math> \therefore \cos x = \cfrac{1}{1 + \cfrac{x^2}{2 - x^2 + \cfrac{2x^2}{3 \cdot 4 - x^2 + \cfrac{3 \cdot 4x^2}{5 \cdot 6 - x^2 + \ddots}}}}.</math>
==== The inverse trigonometric functions ====
The [[inverse trigonometric functions]] can be represented as continued fractions.
:<math>
\begin{align}
\sin^{-1} x = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!} \cdot \frac{x^{2n+1}}{2n+1} & = x + \left( \frac{1}{2} \right) \frac{x^3}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \frac{x^5}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \frac{x^7}{7} + \cdots \\[8pt]
& = x + x \left(\frac{x^2}{2 \cdot 3}\right) + x \left(\frac{x^2}{2 \cdot 3}\right)\left(\frac{(3x)^2}{4 \cdot 5}\right) + x \left(\frac{x^2}{2 \cdot 3}\right)\left(\frac{(3x)^2}{4 \cdot 5}\right)\left(\frac{(5x)^2}{6 \cdot 7}\right) + \cdots \\[8pt]
& = \cfrac{x}{1 - \cfrac{\frac{x^2}{2 \cdot 3}}{1 + \frac{x^2}{2 \cdot 3} - \cfrac{\frac{(3x)^2}{4 \cdot 5}}{1 + \frac{(3x)^2}{4 \cdot 5} - \cfrac{\frac{(5x)^2}{6 \cdot 7}}{ 1 + \frac{(5x)^2}{6 \cdot 7} - \ddots}}}}
\end{align}
</math>
An equivalence transformation yields
:<math> \sin^{-1} x = \cfrac{x}{1 - \cfrac{x^2}{2 \cdot 3 + x^2 - \cfrac{2 \cdot 3 (3x)^2}{4 \cdot 5 +(3x)^2 - \cfrac{4 \cdot 5 (5x)^2}{6 \cdot 7 + (5x)^2 - \ddots}}}}.</math>
The continued fraction for the [[Gregory's series|inverse tangent]] is straightforward:
:<math>
\begin{align}
\tan^{-1} x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n + 1}}{2n + 1}  & = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \\[8pt]
& = 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 \\[8pt]
& = \cfrac{x}{1 - \cfrac{\frac{-x^2}{3}}{1 + \frac{-x^2}{3} - \cfrac{\frac{-3x^2}{5}}{1 + \frac{-3x^2}{5} - \cfrac{\frac{-5x^2}{7}}{1 + \frac{-5x^2}{7} - \ddots}}}} \\[8pt]
& = \cfrac{x}{1 + \cfrac{x^2}{3 - x^2 + \cfrac{(3x)^2}{5 - 3x^2 + \cfrac{(5x)^2}{7 - 5x^2 + \ddots}}}}.
\end{align}
</math>

=== A continued fraction for {{pi}} ===
We can use the previous example involving the inverse tangent to construct a [[William_Brouncker,_2nd_Viscount_Brouncker#Brouncker's_formula|continued fraction]] representation of [[Pi|π]]. We note that

:<math>
\tan^{-1} (1) = \frac\pi4 ,
</math>

And setting ''x'' = 1 in the previous result, we obtain immediately

:<math>
\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \ddots}}}}}.\,
</math>

=== The hyperbolic functions===
Recalling the relationship between the [[hyperbolic function]]s and the trigonometric functions,
:<math> \sin ix = i \sinh x </math>
:<math> \cos ix = \cosh x ,</math>
And that  <math> i^2 = -1,</math> the following continued fractions are easily derived from the ones above:
:<math> \sinh x = \cfrac{x}{1 - \cfrac{x^2}{2 \cdot 3 + x^2 - \cfrac{2 \cdot 3x^2}{4 \cdot 5 + x^2 - \cfrac{4 \cdot 5x^2}{6 \cdot 7 + x^2 - \ddots}}}}</math>
:<math> \cosh x = \cfrac{1}{1 - \cfrac{x^2}{2 + x^2 - \cfrac{2x^2}{3 \cdot 4 + x^2 - \cfrac{3 \cdot 4x^2}{5 \cdot 6 + x^2 - \ddots}}}}.</math>

==== The inverse hyperbolic functions====
The [[inverse hyperbolic functions]] are related to the inverse trigonometric functions similar to how the hyperbolic functions are related to the trigonometric functions,
:<math> \sin^{-1} ix = i \sinh^{-1} x </math>
:<math> \tan^{-1} ix = i \tanh^{-1} x ,</math>
And these continued fractions are easily derived:
:<math> \sinh^{-1} x = \cfrac{x}{1 + \cfrac{x^2}{2 \cdot 3 - x^2 + \cfrac{2 \cdot 3 (3x)^2}{4 \cdot 5 - (3x)^2 + \cfrac{4 \cdot 5 (5x^2)}{6 \cdot 7 - (5x^2) + \ddots}}}}</math>
:<math> \tanh^{-1} x = \cfrac{x}{1 - \cfrac{x^2}{3 + x^2 - \cfrac{(3x)^2}{5 + 3x^2 - \cfrac{(5x)^2}{7 + 5x^2 - \ddots}}}}.</math>

== See also ==
* [[Gauss's continued fraction]]
* [[Engel expansion]]
* [[List of topics named after Leonhard Euler]]

== References ==
{{reflist}}

{{Leonhard Euler}}

[[Category:Continued fractions]]
[[Category:Leonhard Euler]]