Euler's Continued Fraction Formula
Theorem
| \(\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j\) | \(=\) | \(\ds a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots + a_0 a_1 a_2 a_3 \cdots a_n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a_0 \paren {1 + a_1 \paren {1 + a_2 \paren { 1 + a_3 \paren {\cdots + a_n } } } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } }\) |
Proof
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
Basis for the Induction
$\map P 1$ is the case:
| \(\ds \sum_{k \mathop = 0}^1 \prod_{j \mathop = 0}^k a_j\) | \(=\) | \(\ds a_0 + a_0 a_1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a_0 \paren {1 + a_1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac {a_0} {\cfrac 1 {1 + a_1 } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac {a_0} {\cfrac {1 + a_1 - a_1} {1 + a_1 } }\) | adding $0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 } }\) |
Since $\dfrac {a_1} {1 + a_1} \ne 1$
Thus $\map P 1$ is seen to hold.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
from which it is to be shown that:
- $\ds \sum_{k \mathop = 0}^{n + 1} \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } } } }$
Induction Step
This is the induction step:
We first note that:
| \(\ds 1 + \frac {a_1} {1 + x}\) | \(=\) | \(\ds \frac {1 + x} {1 + x} + \frac {a_1} {1 + x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 + a_1 + x} {1 + x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac 1 {\cfrac {1 + x} {1 + a_1 + x } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cfrac 1 {\cfrac {1 + a_1 + x - a_1} {1 + a_1 + x } }\) | adding $0$ | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 1 + \frac {a_1} {1 + x}\) | \(=\) | \(\ds \cfrac 1 {1 + \cfrac {-a_1} {1 + a_1 + x } }\) |
Before continuing, we define x as follows:
- $\ds x := \cfrac {-a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } }$
Next, we apply the induction hypothesis to the values $a_1$, $a_2$, $\cdots$ $a_{n + 1}$
| \(\ds \sum_{k \mathop = 0}^{n + 1} \prod_{j \mathop = 0}^k a_j\) | \(=\) | \(\ds a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots + a_0 a_1 a_2 a_3 \cdots a_{n + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a_0 \paren {1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \cdots + a_1 a_2 a_3 \cdots a_{n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a_0 \paren {1 + \cfrac {a_1} {1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {a_4} {1 + a_4 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } } } } }\) | applying the induction hypothesis to the values $a_1$, $a_2$, $\cdots$ $a_{n + 1}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a_0 \paren {\cfrac 1 {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_n } {1 + a_n - \cfrac {a_{n + 1} } {1 + a_{n + 1} } } } } } } } }\) | from $\paren 1$ above |
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>0}:$
- $\ds \sum_{k \mathop = 0}^n \prod_{j \mathop = 0}^k a_j = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots - \cfrac {a_{n - 1} } {1 + a_{n - 1} - \cfrac {a_n} {1 + a_n} } } } } } } $
$\blacksquare$
Examples
Example: $\map {\arcsin} x$
- $\arcsin x = \cfrac x {1 - \cfrac {x^2} {2 \times 3 + x^2 - \cfrac {2 \times 3 \times \paren {3 x}^2} {4 \times 5 + \paren {3 x}^2 - \cfrac {4 \times 5 \times \paren {5 x}^2} {6 \times 7 + \paren {5 x}^2 - \cfrac {6 \times 7 \times \paren {7 x}^2} {\ddots } } } } }$
Example: $\map {\arctan} x$
- $\arctan x = \cfrac x {1 + \cfrac {x^2} {3 - x^2 + \cfrac {\paren {3 x}^2} {5 - 3 x^2 + \cfrac {\paren {5 x}^2} {7 - 5 x^2 + \cfrac {\paren {7 x}^2} {\ddots } } } } }$
Example: $e^x$
- $e^x = \cfrac 1 {1 - \cfrac x {1 + x - \cfrac x {2 + x - \cfrac {2 x } {3 + x - \cfrac {3 x} {4 + x - \cfrac {\ddots} {\ddots } } } } } }$
Example: $\map {\log} {1 + x}$
- $\map \ln {1 + x} = \cfrac x {1 + \cfrac x {2 - x + \cfrac {2^2 x} {3 - 2 x + \cfrac {3^2 x } {4 - 3 x + \cfrac {\ddots} {\ddots } } } } }$
Source of Name
This entry was named for Leonhard Paul Euler.
Sources
- 1748: Leonhard Paul Euler: Introductio in Analysin Infinitorum: Volume $\text {1}$ Chapter $\text {18}$. De Fractionibus Continuis