\(
\def\ovr{\over\displaystyle}
\def\dst{\displaystyle\strut}
\def\tailfrac{\rlap{\dst \rlap{\phantom{\dst 1\ovr2}}\cdots}}
\def\cofrac#1#2#3{\rlap{\dfrac{#1}{\phantom{#2+}}} \genfrac{}{}{0pt}{0}{}{#2+#3}}
\)
Introduction to Analysis of the Infinite

Leonhard Euler

Book One

Chapter 18

Continued Fractions

Leonhard Euler

Book One

Chapter 18

Continued Fractions

Such are the following expressions: \begin{equation} {a + {\cofrac 1 b {\cofrac 1 c {\cofrac 1 d {\cofrac 1 e {\cofrac 1 f \tailfrac }}}}}} \hbox{or}\qquad\qquad {a + {\cofrac \alpha b {\cofrac \beta c {\cofrac \gamma d {\cofrac \delta e {\cofrac \varepsilon f \tailfrac }}}}}} \end{equation}

In the first, the numerators of all the fractions are unity. Those are the ones we will principally consider. In the second, the numerators are arbitrary numbers.

EXAMPLE 1

*Transform this infinite series*
\begin{equation}
x = 1 - {1\over 2} + {1\over 3} - {1\over 4} + {1\over 5} - \cdots
\end{equation}
*into a continued fraction.*

We accordingly set \(A=1\), \(B=2\), \(C=3\), \(D=4\), etc., and because the given series has a value equal to \(\log 2\), we will have \begin{equation} \log 2 = {\cofrac 1 1 {\cofrac 1 1 {\cofrac 4 1 {\cofrac 9 1 {\cofrac {16} 1 {\cofrac {25} 1 \tailfrac}}}}}} \end{equation}

EXAMPLE 2

*Transform this infinite series*
\begin{equation}
{\pi\over4} = 1 - {1\over 3} + {1\over 5} - {1\over 7} + {1\over 9} - \cdots
\end{equation}
*
where \(\pi\) denotes the circumference of a circle whose diameter equals \(1\),
into a continued fraction.
*

Substituting for \(A\), \(B\), \(C\), \(D\), etc. the numbers \(1\), \(3\), \(5\), \(7\), etc. yields \begin{equation} {\pi\over 4} = {\cofrac 1 1 {\cofrac 1 2 {\cofrac 9 2 {\cofrac {25} 2 {\cofrac {49} 2 \tailfrac}}}}} \end{equation} and so, by inverting the fraction we will have \begin{equation} {4\over\pi} = 1 + {\cofrac 1 2 {\cofrac 9 2 {\cofrac {25} 2 {\cofrac {49} 2 \tailfrac}}}} \end{equation} which is the expression Brouncker first advanced for the quadrature of the circle.

EXAMPLE 3

If given an infinite series such as \begin{equation} x = {1\over m} - {1\over m+n} + {1\over m+2n} - {1\over m+3n} + \cdots \end{equation} on account of \begin{equation} A=m,\quad B=m+n,\quad C=m+2n,\quad \hbox{etc.}, \end{equation} it is transformed into this continued fraction \begin{equation} x = {\cofrac 1 m {\cofrac {mm} {\phantom{n}n} {\cofrac {(m+n)^2} {\phantom{m+n}n} {\cofrac {(m+2n)^2} {\phantom{m+2n}n} {\cofrac {(m+3n)^2} {\phantom{nn}n} \tailfrac}}}}} \end{equation} which by inverting becomes \begin{equation} {1\over x} - m = {\cofrac {mm} {\phantom{n}n} {\cofrac {(m+n)^2} {\phantom{m+n}n} {\cofrac {(m+2n)^2} {\phantom{mm+n}n} {\cofrac {(m+3n)^2} {\phantom{nn}n} \tailfrac } } } } \end{equation}

EXAMPLE 4

Because in §178 above we find that \begin{equation} {\pi\cos{m\pi\over n} \over n\sin{m\pi\over n} } = {1\over m} - {1\over n-m} + {1\over n+m} - {1\over 2n-m} + {1\over 2n+m} - \cdots \end{equation} we will have, for forming the continued fraction \begin{equation} A=m,\quad B=n-m,\quad C=n+m,\quad D=2n-m,\quad \cdots \end{equation} from which \begin{equation} {\pi\cos{m\pi\over n} \over n\sin{m\pi\over n} } = {\cofrac 1 m {\cofrac {mm} {n-2m} {\cofrac {(n-m)^2} {\phantom{2m+}2m} {\cofrac {(n+m)^2} {n-2m} {\cofrac {(2n-m)^2} {\phantom{2m+}2m} {\cofrac {(2n+m)^2} {n-2m} \tailfrac}}}}}} \end{equation}

EXAMPLE 1

We previously defined \(e\) to be the number whose logarithm equals \(1\), and found that \begin{equation} {1\over e} = 1 - {1\over 1} + {1\over 1\cdot 2} - {1\over 1\cdot 2\cdot 3} + {1\over 1\cdot 2\cdot 3\cdot 4} - \cdots \end{equation} or alternatively \begin{equation} 1 - {1\over e} = {1\over 1} - {1\over 1\cdot 2} + {1\over 1\cdot 2\cdot 3} - {1\over 1\cdot 2\cdot 3\cdot 4} + \cdots \end{equation} This series will be converted into a continued fraction by setting \begin{equation} A=1,\quad B=2,\quad C=3,\quad D=4,\quad \cdots \end{equation} and when this is done we will have \begin{equation} 1 - {1\over e} = {\cofrac 1 1 {\cofrac 1 1 {\cofrac 2 2 {\cofrac 3 3 {\cofrac 4 4 {\cofrac 5 5 \tailfrac}}}}}} \end{equation} from which, by eliminating the initial asymmetry, we will get \begin{equation} {1\over {e-1}} = {\cofrac 1 1 {\cofrac 2 2 {\cofrac 3 3 {\cofrac 4 4 {\cofrac 4 4 {\cofrac 5 5 \tailfrac}}}}}} \end{equation}

EXAMPLE 2

We previously found that the cosine of any arc which equals its chosen radius is \begin{equation} 1 - {1\over 2} + {1\over 2\cdot 12} - {1\over 2\cdot 12\cdot 30} + {1\over 2\cdot 12\cdot 30\cdot 56} - \cdots \end{equation} Accordingly, let us set \begin{equation} A=1,\quad B=2,\quad C=12,\quad D=30,\quad E=56, \quad \cdots \end{equation} and also set \(x\) to the cosine of an arc which equals its radius. Then we will have \begin{equation} x = {\cofrac 1 1 {\cofrac 1 1 {\cofrac 2 {11} {\cofrac {12} {29} {\cofrac {30} {55} \tailfrac}}}}} \end{equation} or alternatively \begin{equation} { {1\over x} - 1} = {\cofrac 1 1 {\cofrac 2 {11} {\cofrac {12} {29} {\cofrac {30} {55} \tailfrac}}}} \end{equation}

Given the equation \begin{equation} xx=ax+b\,, \end{equation} since \begin{equation} x=a+{b\over x}\,, \end{equation} by substituting into the last term the value of \(x\) already found, we will get \begin{equation} x=a+{\cofrac b a {\dfrac{b}{x}}} \end{equation} and proceding in the same way, we will find an infinite continued fraction \begin{equation} x = a+{\cofrac b a {\cofrac b a {\cofrac b a \ldots}} } \end{equation} which, however, because the numerators \(b\) are not equal to unity, is not so convenient to use.

B) A (a — C) B (b — D) C (c — E) D (d — F etc.

And then we will have, by the nature of division \begin{array}{lll} \displaystyle{A=\alpha B+C}\,, & \quad\hbox{unde}\quad & \displaystyle{ {A\over B}=a+{C\over B }}\,, \phantom{\,,\quad \displaystyle{ {C\over B}= {1\over b+{D\over C}}}}\cr \displaystyle{B=bC+D}\,, & & \displaystyle{ {B\over C}=b+{D\over C}}\,,\quad \displaystyle{ {C\over B}= {1\over b+{D\over C}}} \,, \cr \displaystyle{C=cD+E}\,, & & \displaystyle{ {C\over D}=c+{E\over D}}\,,\quad \displaystyle{ {D\over C}= {1\over c+{E\over D}}} \,, \cr \displaystyle{D=dE+F}\,, & & \displaystyle{ {D\over E}=d+{F\over E}}\,,\quad \displaystyle{ {E\over D}= {1\over d+{F\over E}}} \,, \cr &\quad\hbox{etc.} \cr \end{array} From here, by substituting the latter values into the previous ones, we will have \begin{equation} x = {A\over B} = a + {C\over B} = a + {\cofrac 1 b { {D\over C}}} = a + {\cofrac 1 b {\cofrac 1 c { {E\over D}}}} \end{equation} Finally, \(x\) will be expressed as a quotient purely in terms of the \(a\), \(b\), \(c\), \(d\), etc., we found above, in the following way \begin{equation} x = {a + {\cofrac 1 b {\cofrac 1 c {\cofrac 1 d {\cofrac 1 e {\cofrac 1 f \tailfrac }}}}}} \end{equation}

EXAMPLE 1

Given the fraction \(1461\over 59\), it will be converted into a continued fraction, all of whose numerators are unity, as follows. Let us set up the very same calculation that is normally used to find the greatest common divisor of the numbers \(59\) and \(1461\).

59) 1461 (24 118 ———— 281 236 ——— 45) 59 (1 45 —— 14) 45 (3 42 —— 3) 14 (4 12 —— 2) 3 (1 2 — 1) 2 (2 2 — 0

From these quotients we will get \begin{equation} {1461\over 59} = {24 + {\cofrac 1 1 {\cofrac 1 3 {\cofrac 1 4 {\cofrac 1 1 {\displaystyle {1\over 2}} }}}}} \end{equation}

EXAMPLE 2

Even decimal fractions can be converted in the very same way. Let us consider \begin{equation} \sqrt 2 = 1.41421356 = {141\,421\,356\over 100\,000\,000}\,, \end{equation} from which we set up this calculation \begin{array}{r|r|r} 100\,000\,000 & 141\,421\,356 & 1 \cr 82\,842\,712 & 100\,000\,000 & 2 \cr \hline 17\,157\,288 & 41\,421\,356 & 2 \cr 14\,213\,560 & 34\,314\,576 & 2 \cr \hline 2\,943\,728 & 7\,106\,780 & 2 \cr 2\,438\,648 & 5\,887\,456 & 2 \cr \hline 505\,080 & 1\,219\,324 & 2 \cr 418\,728 & 1\,010\,160 & 2\cr \hline \hbox{etc.}\;\; & 209\,364 \end{array} From this calculation all the denominators are now seen to be \(2\), and moreover \begin{equation} \sqrt 2 = {1 + {\cofrac 1 2 {\cofrac 1 2 {\cofrac 1 2 {\cofrac 1 2 \tailfrac }}}}} \end{equation} the reason for which has already been explained above.

EXAMPLE 3

But the number \(e\), whose logarithm equals \(1\), is at this point worth particular attention. We have \begin{equation} e=2.718281828459 \,, \end{equation} and so \begin{equation} {e-1\over2} = 0.8591409142295 \,, \end{equation} whose decimal fraction, if handled in the above manner, will yield the following quotients \begin{array}{r|r|r} 8\,591\,409\,142\,295 & 10\,000\,000\,000\,000 & 1 \cr 8\,451\,545\,146\,224 & 8\,591\,409\,142\,295 & 6 \cr \hline 139\,863\,996\,071 & 1\,408\,590\,857\,704 & 10 \cr 139\,312\,557\,916 & 1\,398\,639\,960\,710 & 14 \cr \hline 551\,438\,155 & 9\,950\,896\,994 & 18 \cr 550\,224\,488 & 9\,925\,886\,790 & 22 \cr \hline 1\,213\,667 & 25\,010\,204 & \quad\hbox{etc.}\cr \end{array} If that calculation were painstakingly continued further toward the value of \(e\) itself, then these quotients would be obtained \begin{equation} 1,\quad 6,\quad 10,\quad 14,\quad 18,\quad 22,\quad 26,\quad 30,\quad 34,\quad \cdots \end{equation} which, except for the first, sets out an arithmetic progression, from which it is clear that \begin{equation} {e-1\over 2} = {\cofrac 1 1 {\cofrac 1 6 {\cofrac 1 {10} {\cofrac 1 {14} {\cofrac 1 {18} {\cofrac 1 {22} \tailfrac }}}}}} \end{equation} and the rationale of the fraction can be obtained from infinitesimal calculus.

EXAMPLE 1

Let us express the ratio of the diameter to the circumference using numbers which are as economical as possible, such that the accuracy cannot be increased unless larger numbers are introduced. If the known decimal fraction \begin{equation} 3.1415926535\cdots \end{equation} is evolved by continued division in the manner set out previously, we will obtain the following quotients \begin{equation} 3,\quad 7,\quad 15,\quad 1,\quad 292,\quad 1,\quad 1,\quad \cdots \,, \end{equation} from which the following fractions will be formed \begin{equation} {1\over 0},\quad {3\over 1},\quad {22\over 7},\quad {333\over 106},\quad {355\over 113},\quad {103993\over 33102},\quad \cdots \,. \end{equation} The second fraction now shows that the diameter being to the circumference as \(1:3\) can certainly not be given more accurately without larger numbers. The third fraction gives the Archimedean ratio \(7:22\), and the fifth that of Metius, which comes so close to the true value that the error is less than \({1\over 113\cdot 33102}\). The rest of these fractions alternate between being larger and smaller than the true value.

EXAMPLE 2

Let us express, using smallest numbers, the approximate ratio of the day to the average solar year. Since such a year is \(\,365^d\,5^h\,48'\,55''\), the year will contain, as an ordinary fraction \begin{equation} 365 {\textstyle{20935\over 86400}} \end{equation} days. So the only thing needed is to evolve this fraction, which will give the following quotients \begin{equation} 4,\quad 7,\quad 1,\quad 6,\quad 1,\quad 2,\quad 2,\quad 4 \end{equation} yielding the fractions \begin{equation} {0\over 1},\quad {1\over 4},\quad {7\over 29},\quad {8\over 33},\quad {55\over 227},\quad {63\over 260},\quad {181\over 747},\quad \cdots \end{equation} Therefore the hours, along with the minutes and seconds which surpass \(365\) days make about one day in four years, which is the origin of the Julian calendar. But more precisely, \(33\) years fill up \(8\) days, or \(747\) years \(181\) days, from which it follows that in \(400\) years there are \(97\) extra days. Hence, whereas the Julian calendar inserts \(100\) days during this interval, the Gregorian calendar converts three of the leap years into regular years.

Translated from Latin by Todd Doucet in 2019.

First Draft

First Draft