In the analytic theory of 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. 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.
Euler derived the formula as connecting a finite sum of products with a finite continued fraction.
\begin{align} a0\left(1+a1\left(1+a2\left( … +an\right) … \right)\right)&=a0+a0a1+a0a1a2+ … +a0a1a2 … an\\ &=\cfrac{a0}{1-\cfrac{a1}{1+a1-\cfrac{a2}{1+a2-\cfrac{\ddots}{\ddots\cfrac{an-1
The identity is easily established by induction on n, and is therefore applicable in the limit: if the expression on the left is extended to represent a 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:
a0+a0a1+a0a1a2+ … +a0a1a2 … an=
a0 | |
1+ |
-a1 | |
1+a1+ |
\cfrac{-a2}{1+a2+} …
-an | |
1+an |
.
If ri are complex numbers and x is defined by
x=1+
infty | |
\sum | |
i=1 |
r1r2 … ri=1+
infty | |
\sum | |
i=1 |
\left(
i | |
\prod | |
j=1 |
rj\right),
x=\cfrac{1}{1-\cfrac{r1}{1+r1-\cfrac{r2}{1+r2-\cfrac{r3}{1+r3-\ddots}}}}
Here equality is to be understood as equivalence, in the sense that the 'th convergent of each continued fraction is equal to the 'th partial sum of the series shown above. So if the series shown is convergent - or uniformly convergent, when the ri's are functions of some complex variable z - then the continued fractions also converge, or converge uniformly.[1]
Theorem: Let
n
n+1
a0,a1,\ldots,an
n | |
\sum | |
k=0 |
k | |
\prod | |
j=0 |
aj=
a0 | |
1+ |
-a1 | |
1+a1+ |
…
-an | |
1+an |
n
b1,\ldots,bn
-b1 | |
1+b1+ |
-b2 | |
1+b2+ |
…
-bn | |
1+bn |
\ne-1.
Proof: We perform a double induction. For
n=1
a0 | |
1+ |
-a1 | |
1+a1 |
=
a0 | |||
|
=
a0(1+a1) | |
1 |
=a0+a0a1 =
1 | |
\sum | |
k=0 |
k | |
\prod | |
j=0 |
aj
-b1 | |
1+b1 |
\ne-1.
n\ge1
We have
-b1 | |
1+b1+ |
-b2 | |
1+b2+ |
…
-bn+1 | |
1+bn+1 |
=
-b1 | |
1+b1+x |
x=
-b2 | |
1+b2+ |
…
-bn+1 | |
1+bn+1 |
\ne-1
by applying the induction hypothesis to
b2,\ldots,bn+1
But if
-b1 | |
1+b1+x |
=-1
b1=1+b1+x
x=-1
-b1 | |
1+b1+ |
-b2 | |
1+b2+ |
…
-bn+1 | |
1+bn+1 |
\ne-1,
Note that for
x\ne-1
1 | |
1+ |
-a | |
1+a+x |
=
1 | |||
|
=
1+a+x | |
1+x |
=1+
a | |
1+x |
;
x=-1-a
Using
a=a1
x=
-a2 | |
1+a2+ |
…
-an+1 | |
1+an+1 |
\ne-1
a1,a2,\ldots,an+1
\begin{align} a0+&a0a1+a0a1a2+ … +a0a1a2a3 … an+1\\ &=a0+a0(a1+a1a2+ … +a1a2a3 … an+1)\\ &=a0+a0(
a1 | |
1+ |
-a2 | |
1+a2+ |
…
-an+1 | |
1+an+1 |
)\\ &=a0(1+
a1 | |
1+ |
-a2 | |
1+a2+ |
…
-an+1 | |
1+an+1 |
)\\ &=a0(
1 | |
1+ |
-a1 | |
1+a1+ |
-a2 | |
1+a2+ |
…
-an+1 | |
1+an+1 |
)\\ &=
a0 | |
1+ |
-a1 | |
1+a1+ |
-a2 | |
1+a2+ |
…
-an+1 | |
1+an+1 |
, \end{align}
As an example, the expression
a0+a0a1+a0a1a2+a0a1a2a3
\begin{align} &a0+a0a1+a0a1a2+a0a1a2a3\\[8pt] ={}&a0(a1(a2(a3+1)+1)+1)\\[8pt] ={}&\cfrac{a0}{\cfrac{1}{a1(a2(a3+1)+1)+1}}\\[8pt] ={}&\cfrac{a0}{\cfrac{a1(a2(a3+1)+1)+1}{a1(a2(a3+1)+1)+1}-\cfrac{a1(a2(a3+1)+1)}{a1(a2(a3+1)+1)+1}}=\cfrac{a0}{1-\cfrac{a1(a2(a3+1)+1)}{a1(a2(a3+1)+1)+1}}\\[8pt] ={}&\cfrac{a0}{1-\cfrac{a1}{\cfrac{a1(a2(a3+1)+1)+1}{a2(a3+1)+1}}}\\[8pt] ={}&\cfrac{a0}{1-\cfrac{a1}{\cfrac{a1(a2(a3+1)+1)}{a2(a3+1)+1}+\cfrac{a2(a3+1)+1}{a2(a3+1)+1}-\cfrac{a2(a3+1)}{a2(a3+1)+1}}}=\cfrac{a0}{1-\cfrac{a1}{1+a1-\cfrac{a2(a3+1)}{a2(a3+1)+1}}}\\[8pt] ={}&\cfrac{a0}{1-\cfrac{a1}{1+a1-\cfrac{a2}{\cfrac{a2(a3+1)+1}{a3+1}}}}\\[8pt] ={}&\cfrac{a0}{1-\cfrac{a1}{1+a1-\cfrac{a2}{\cfrac{a2(a3+1)}{a3+1}+\cfrac{a3+1}{a3+1}-\cfrac{a3}{a3+1}}}}=\cfrac{a0}{1-\cfrac{a1}{1+a1-\cfrac{a2}{1+a2-\cfrac{a3}{1+a3}}}} \end{align}
The exponential function ex is an entire function with a power series expansion that converges uniformly on every bounded domain in the complex plane.
ex=1+
infty | |
\sum | |
n=1 |
xn | |
n! |
=1+
infty | |
\sum | |
n=1 |
n | |
\left(\prod | |
i=1 |
x | |
i |
\right)
The application of Euler's continued fraction formula is straightforward:
ex=\cfrac{1}{1-\cfrac{x}{1+x-\cfrac{
1 | |
2 |
x}{1+
1 | |
2 |
x-\cfrac{
1 | |
3 |
x} {1+
1 | |
3 |
x-\cfrac{
1 | |
4 |
x}{1+
1 | |
4 |
x-\ddots}}}}}.
Applying an equivalence transformation that consists of clearing the fractions this example is simplified to
ex=\cfrac{1}{1-\cfrac{x}{1+x-\cfrac{x}{2+x-\cfrac{2x}{3+x-\cfrac{3x}{4+x-\ddots}}}}}
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 ex.
The Taylor series for the principal branch of the natural logarithm in the neighborhood of 1 is well known:
log(1+x)=x-
x2 | |
2 |
+
x3 | |
3 |
-
x4 | |
4 |
+ …
infty | |
= \sum | |
n=1 |
(-1)n+1zn | |
n |
.
This series converges when |x| < 1 and can also be expressed as a sum of products:[2]
log(1+x)=x+(x)\left(
-x | |
2 |
\right)+(x)\left(
-x | \right)\left( | |
2 |
-2x | |
3 |
\right)+(x)\left(
-x | \right)\left( | |
2 |
-2x | \right)\left( | |
3 |
-3x | |
4 |
\right)+ …
Applying Euler's continued fraction formula to this expression shows that
log(1+x)=\cfrac{x}{1-\cfrac{
-x | |
2 |
and using an equivalence transformation to clear all the fractions results in
log(1+x)=\cfrac{x}{1+\cfrac{x}{2-x+\cfrac{22x}{3-2x+\cfrac{32x}{4-3x+\ddots}}}}
This continued fraction converges when |x| < 1 because it is equivalent to the series from which it was derived.[2]
The Taylor series of the sine function converges over the entire complex plane and can be expressed as the sum of products.
\begin{align}\sinx=
infty | |
\sum | |
n=0 |
(-1)n | |
(2n+1)! |
x2n+1&=x-
x3 | |
3! |
+
x5 | |
5! |
-
x7 | |
7! |
+
x9 | |
9! |
- … \\[8pt] &=x+(x)\left(
-x2 | |
2 ⋅ 3 |
\right)+(x)\left(
-x2 | \right)\left( | |
2 ⋅ 3 |
-x2 | |
4 ⋅ 5 |
\right)+(x)\left(
-x2 | \right)\left( | |
2 ⋅ 3 |
-x2 | \right)\left( | |
4 ⋅ 5 |
-x2 | |
6 ⋅ 7 |
\right)+ … \end{align}
\cfrac{x}{1-\cfrac{
-x2 | |
2 ⋅ 3 |
\sinx=\cfrac{x}{1+\cfrac{x2}{2 ⋅ 3-x2+\cfrac{2 ⋅ 3x2}{4 ⋅ 5-x2+\cfrac{4 ⋅ 5x2}{6 ⋅ 7-x2+\ddots}}}}.
\begin{align} \cosx=
infty | |
\sum | |
n=0 |
(-1)n | |
(2n)! |
x2n&=1-
x2 | |
2! |
+
x4 | |
4! |
-
x6 | |
6! |
+
x8 | |
8! |
- … \\[8pt] &=1+
-x2 | |
2 |
+\left(
-x2 | \right)\left( | |
2 |
-x2 | |
3 ⋅ 4 |
\right)+\left(
-x2 | \right)\left( | |
2 |
-x2 | \right)\left( | |
3 ⋅ 4 |
-x2 | |
5 ⋅ 6 |
\right)+ … \\[8pt]&=\cfrac{1}{1-\cfrac{
-x2 | |
2 |
\therefore\cosx=\cfrac{1}{1+\cfrac{x2}{2-x2+\cfrac{2x2}{3 ⋅ 4-x2+\cfrac{3 ⋅ 4x2}{5 ⋅ 6-x2+\ddots}}}}.
The inverse trigonometric functions can be represented as continued fractions.
\begin{align} \sin-1x=
infty | |
\sum | |
n=0 |
(2n-1)!! | |
(2n)!! |
⋅
x2n+1 | |
2n+1 |
&=x+\left(
1 | |
2 |
\right)
x3 | |
3 |
+\left(
1 ⋅ 3 | |
2 ⋅ 4 |
\right)
x5 | |
5 |
+\left(
1 ⋅ 3 ⋅ 5 | |
2 ⋅ 4 ⋅ 6 |
\right)
x7 | |
7 |
+ … \\[8pt] &=x+x\left(
x2 | |
2 ⋅ 3 |
\right)+x\left(
x2 | \right)\left( | |
2 ⋅ 3 |
(3x)2 | |
4 ⋅ 5 |
\right)+x\left(
x2 | \right)\left( | |
2 ⋅ 3 |
(3x)2 | \right)\left( | |
4 ⋅ 5 |
(5x)2 | |
6 ⋅ 7 |
\right)+ … \\[8pt] &=\cfrac{x}{1-\cfrac{
x2 | |
2 ⋅ 3 |
\sin-1x=\cfrac{x}{1-\cfrac{x2}{2 ⋅ 3+x2-\cfrac{2 ⋅ 3(3x)2}{4 ⋅ 5+(3x)2-\cfrac{4 ⋅ 5(5x2)}{6 ⋅ 7+(5x2)-\ddots}}}}.
\begin{align} \tan-1x=
infty | |
\sum | |
n=0 |
(-1)n
x2n | |
2n+1 |
&=x-
x3 | |
3 |
+
x5 | |
5 |
-
x7 | |
7 |
+ … \\[8pt] &=x+x\left(
-x2 | |
3 |
\right)+x\left(
-x2 | \right)\left( | |
3 |
-3x2 | |
5 |
\right)+x\left(
-x2 | \right)\left( | |
3 |
-3x2 | \right)\left( | |
5 |
-5x2 | |
7 |
\right)+ … \\[8pt] &=\cfrac{x}{1-\cfrac{
-x2 | |
3 |
We can use the previous example involving the inverse tangent to construct a continued fraction representation of π. We note that
\tan-1(1)=
\pi4 | |
, |
And setting x = 1 in the previous result, we obtain immediately
\pi=\cfrac{4}{1+\cfrac{12}{2+\cfrac{32}{2+\cfrac{52}{2+\cfrac{72}{2+\ddots}}}}}.
Recalling the relationship between the hyperbolic functions and the trigonometric functions,
\sinix=i\sinhx
\cosix=\coshx,
i2=-1,
\sinhx=\cfrac{x}{1-\cfrac{x2}{2 ⋅ 3+x2-\cfrac{2 ⋅ 3x2}{4 ⋅ 5+x2-\cfrac{4 ⋅ 5x2}{6 ⋅ 7+x2-\ddots}}}}
\coshx=\cfrac{1}{1-\cfrac{x2}{2+x2-\cfrac{2x2}{3 ⋅ 4+x2-\cfrac{3 ⋅ 4x2}{5 ⋅ 6+x2-\ddots}}}}.
The inverse hyperbolic functions are related to the inverse trigonometric functions similar to how the hyperbolic functions are related to the trigonometric functions,
\sin-1ix=i\sinh-1x
\tan-1ix=i\tanh-1x,
\sinh-1x=\cfrac{x}{1+\cfrac{x2}{2 ⋅ 3-x2+\cfrac{2 ⋅ 3(3x)2}{4 ⋅ 5-(3x)2+\cfrac{4 ⋅ 5(5x2)}{6 ⋅ 7-(5x2)+\ddots}}}}
\tanh-1x=\cfrac{x}{1-\cfrac{x2}{3+x2-\cfrac{(3x)2}{5+3x2-\cfrac{(5x)2}{7+5x2-\ddots}}}}.