Continued Fraction for Exponential Function

Theorem

$e^x = \cfrac 1 {1 - \cfrac x {1 + x - \cfrac x {2 + x - \cfrac {2 x } {3 + x - \cfrac {3 x} {4 + x - \cfrac {\ddots} {\ddots } } } } } }$

Corollary

$e = \cfrac 1 {1 - \cfrac 1 {2 - \cfrac 1 {3 - \cfrac 2 {4 - \cfrac {\ddots} {\ddots } } } } }$

Proof

$\ds e^x$

$=$

$\ds \sum_{n \mathop = 0}^\infty \frac {x^n } {n!}$

Power Series Expansion for Exponential Function

$\ds $

$=$

$\ds 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots$

$\ds $

$=$

$\ds 1 + \paren 1 \paren x + \paren 1 \paren x \paren {\dfrac x 2} + \paren 1 \paren x \paren {\dfrac x 2} \paren {\dfrac x 3} + \paren 1 \paren x \paren {\dfrac x 2} \paren {\dfrac x 3} \paren {\dfrac x 4} + \cdots$

$\ds $

$=$

$\ds a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + a_0 a_1 a_2 a_3 a_4 + \cdots$

From Euler's Continued Fraction Formula, we have:

$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots } } } } }$

Therefore:

$\ds 1 + \paren 1 \paren x + \paren 1 \paren x \paren {\dfrac x 2} + \paren 1 \paren x \paren {\dfrac x 2} \paren {\dfrac x 3} + \paren 1 \paren x \paren {\dfrac x 2} \paren {\dfrac x 3} \paren {\dfrac x 4} + \cdots$

$=$

$\ds \cfrac 1 {1 - \cfrac {\paren x } {1 + \paren x - \cfrac {\paren {\dfrac x 2} } {1 + \paren {\dfrac x 2} - \cfrac {\paren {\dfrac x 3} } {1 + \paren {\dfrac x 3} - \cfrac {\paren {\dfrac x 4} } {1 + \paren {\dfrac x 4} - \cfrac {\ddots} {\ddots} } } } } }$

$\ds $

$=$

$\ds \cfrac 1 {1 - \cfrac {\paren x } {\paren {1 + \paren x } - \cfrac {\paren {\dfrac x 2} \times 2 } {2 \times \paren {1 + \paren {\dfrac x 2} } - \cfrac {\paren {\dfrac x 3} \times 2 \times 3 } {3 \times \paren {1 + \paren {\dfrac x 3} } - \cfrac {\paren {\dfrac x 4} \times 3 \times 4 } {4 \times \paren {1 + \paren {\dfrac x 4} } - \cfrac {\ddots} {\ddots } } } } } }$

multiplying by $1$

$\ds $

$=$

$\ds \cfrac 1 {1 - \cfrac x {1 + x - \cfrac x {2 + x - \cfrac {2 x } {3 + x - \cfrac {3 x} {4 + x - \cfrac {\ddots} {\ddots } } } } } }$

$\blacksquare$

Also see

Sources

1750: Leonhard Paul Euler: De fractionibus continuis Observationes