Reflections and Notes on E745 - Euler Archive Home Page

1
<strong>Reflections</strong> <strong>and</strong> <strong>Notes</strong> on Euler’s paper E745:
De fractionibus continuis Wallisii
(On the continued fractions of Wallis)
Thomas J Osler,
Christopher Tippie <strong>and</strong> Kristin Masters
Section 1.
Euler refers to the book by John Wallis “Arithmetica Infinitorum” in which we
find a sequence of continued fractions due to Lord Brouncker. These are given without a
complete derivation <strong>and</strong> Euler wishes to provide this. Euler states that Brouncker
discovered the first continued fraction in this sequence <strong>and</strong> Wallis discovered the
remaining fractions. This is contrary to recent investigations [5] by Jacqueline A. Stedall
which show that Brouncker obtained the entire sequence himself, but never published it.
It was Wallis who made the sequence known to the world. The importance of Euler’s
paper is that he finds a nice derivation <strong>and</strong> generalization of this sequence for the first
time.
Section 2.
Euler examines the list of integrals
x
x
1 xx
1
1
x
1
3
x
xx
2
3
2
2
2
3

2
x
1
5
x
xx
2
3
4
5
2
2
2
3
4
4
4
5
x
1
7
x
xx
2
3
4
5
6
7
2
2
2
3
4
4
4
5
6
6
6
7
x
1
9
x
xx
2
3
4
5
6
7
8
9
2
2
2
3
4
4
4
5
6
6
6
7
8
8
8
, etc…
9
2 2
Using the substitution t 1 x we can convert these integrals to the form
1
0
x
1
P
dx
x
2
1
0
1
t
2
P
dt . In Wallis, we find a table of the reciprocal integrals
1
P
1/ Q
1 / 1 x dx. Wallis after much work obtains the following table:
0
P
0
P P 1
1/ 2
Q 0 1 1 2
1 1
1 2
Q 1 2 2 3
1
1/ 2
1 2
Q 1 1 1 3 4
1
2 1 2
Q 1 4 4 5
1
3/ 2 3 1 2
Q 2 1 3 5 6
1
8 1 2
Q 1 16 6 7
1
5 / 2 15 1 2
P P 2
3/ 2
1 3 2 4
1 1
1 3 2 4
2 2 4 3 5
1
1 3 2 4
1 3 5 4 6
1
2 1 3 2 4
4 4 6 5 7
1
3 1 3 2 4
3 5 7 6 8
1
8 1 3 2 4
16 6 8 7 9
1
15 1 3 2 4
P P 3
5 / 2
1 3 5 2 4 6
1
1
1 3 5 2 4 6
2 2 4 6 3 5 7
1
1 3 5 2 4 6
1 3 5 7 4 6 8
1
2 1 3 5 2 4 6
4 4 6 8 5 7 9
1
3 1 3 5 2 4 6
3 5 7 9 6 8 10
1
8 1 3 5 2 4 6
16 6 8 10 7 9 11
1
15 1 3 5 2 4 6
Q 3 1
5
8
7
1
1
8
2
5
8
7
1
9
3
1
8
2
10
4
5
8
7
1
9
3
11
5
1
8
2
10
4
12
6

3
The list of integrals that Euler is studying,
1
0
x P dx
, are from the row where Q 1/ 2 .
2
1 x
Sections 3 <strong>and</strong> 4.
Euler now wishes to find expressions, A, B, C, D, E so that
2
AB 1 ,
2
BC 2 ,
2
CD 3 ,
DE
2
4
, etc. Then he can write the above list of integrals
as
x
1
x
xx
1
1
A
A
,
1
x
1
3
x
xx
BC
2 3
1
A
ABC
,
1 2 3
x
1
5
x
xx
BCDE
2 3 4 5
1
A
1
ABCDE
,
2 3 4 5
x
1
7
x
xx
BCDEFG
2 3 4 5 6
7
1
A
1
ABCDEFG
2 3 4 5 6
,
7
etc…
The above integrals have odd powers in the numerators. Euler now wishes to find
expressions for integrals with even powers. For this purpose, he studies the third column
of expressions above <strong>and</strong> “interpolates” to get
x 1
1 xx A
1,
xx x
1 xx
1
A
AB
,
1 2

4
x
1
4
x
xx
1
A
ABCD
,
1 2 3 4
x
1
6
x
xx
1
A
1
ABCDEF
2 3 4 5
,
6
etc…
This is a conjecture totally in the spirit of Wallis.
For convenience, we also use the modern notation
L ( 0)
A, L ( 1)
B , L ( 2)
C , etc.
Section 5.
Euler knows that
1
0
1
dx
2
x
. He calls this value q. From the first integral
2
above we see that
q
1 . Since AB 1 we now have B 1 q . From BC 4 we get
A
A
C
4
B
4
, etc. Euler lists these values <strong>and</strong> there numerical equivalents in a table.
q
A
B
C
D
1
q
q
4
q
9q
4
0,636620
1,570796
2,546479
3,534292
Difference
0,934176
0,975683
0,987813
0,992782

5
E
F
4 16
4,527074
9q
0,995257
9 25
q 5,522331
4 16
Euler now makes the following troubling statement:
“Here I attached a third column, which exhibits the numeric values of these letters to
show more clearly the extent that these numbers increase according to the law of
uniformity. This does not happen if I take a false value in the place of q.”
Just what is this law of uniformity? Euler hints that the value
2
q has a special value.
Somehow it creates some uniform flow to the numbers A, B, C as they increase.
We can start with any value for A, then all the remaining letters are determined by
this choice. However, there is something special about the choice
q
2
. Let us study
this as apparently Euler did. (We use our notation:
L ( 0)
A, L ( 1)
B , L ( 2)
C , etc.).
We calculate the following:
L (2)
2
1
L(1)
L (3)
2
2
L(2)
2
2 L(1)
2
1
2 2 2
3 1 3
L ( 4)
2
L(3)
2 L(1)
L (5)
2
4
L(4)
2 2
2 4 L(1)
2 2
1 3
2 2 2 2
5 1 3 5
L ( 6)
2 2
L(5)
2 4 L(1)

6
Let us denote the partial product in the Wallis product as
1 3 3 5 5 7 (2n
1)(2n
1)
W(
n)
.
2 2 4 4 6 6 (2n)(2n)
Then from the above calculations we can write in general
2 2 2
2
( 1 3 5 (2n
1)
L 2n)
2 2 2
2 4 6 (2n
2)
2 L(1)
W ( n
(2n
1)
1) , <strong>and</strong>
L(1)
2 2 2
2
( 2 4 6 (2n)
L(1)
L 2n
1)
2 2 2
1 3 5 (2n
1)
2
(2n
1) L(1)
.
W ( n)
Since
2
limW ( n)
we get the two limits
(5.1)
L(2n)
lim
2n
1
L(2n)
lim
2n
2
L(1)
<strong>and</strong>
(5.2)
L(2n
lim
2n
1)
1
L(1)
.
2
From these last two results we see that
L ( n)
lim exists <strong>and</strong> equals 1 if <strong>and</strong> only if
n
L (1)
2
. Otherwise, if
L (1)
2
, the limit does not exist, but the separate limits through
even <strong>and</strong> odd values on n do exist. Thus it is clear that the choice
2
q A L(1) is
very special.
Euler will work intuitively in the next sections to move “uniformly” from A to B
to C <strong>and</strong> so on in the hope of getting the right result. Later in the paper, he will
demonstrate that his intuitive guess is correct.
Section 6.

7
Euler now hunts for A, B, C, D, E. He states his problem in a more general form:
Problem:
To find a series of letters A, B, C, D, etc…progressing by the law of uniformity in such a
way that AB =ff; BC=
f
a
2
; CD=
f
2a
2
; etc.
Given f <strong>and</strong> a we wish to have
2
AB f ,
2
BC ( f a) ,
CD
2
( f 2a)
, etc.
Section 7.
Euler begins with the problem of finding expressions A, B, C, D, etc., such that
2
AB f ;
2
BC ( f a) ;
CD
2
( f 2a)
; etc.
Noticing that
f
a
2
f
a
2
f
2
2 a
4
, Euler asks how he might increase
these two factors on the LHS so that the product is exactly
2
f .
We start with
(7.1)
2
AB f ,
<strong>and</strong> assume that the desired increase is given by
a s / 2
A f ,
2 A'
a s / 2
B f ,
2 B'
where s , A’ <strong>and</strong> B’ are to be determined. Multiplying by 2 we get
(7.2)
s
2A 2 f a ,
A'
s
2B 2 f a .
B'
Multiplying the above together we get using (7.1)
(7.3)
s
s
2 f
A'
B'
2
f a 2 f a 4 ,

8
which can also be written as
2
2 2 s s
s
4AB 4 f a (2 f a)
(2 f a)
.
B'
A'
A'
B'
Since
4 4 f
2
AB we have
2
2 s s
s
a (2 f a)
(2 f a)
,
B'
A'
A'
B'
<strong>and</strong> multiplying by
A ' B'
we get
(7.4)
2
2
a A'
B'
A'
s(2
f a)
B'
s(2
f a)
s .
We are free to choose s as we like, <strong>and</strong> a reasonable choice to simplify (7.4) is
(7.5)
2
s a .
After we divide (7.4) by
2
a we get
2
A ' B'
A'(2
f a)
B'(2
f a)
a .
Notice that this last expression can be written as
( A ' (2 f a))(
B'
(2 f a))
4 f
2
or
(7.6)
( f
2
A ' 2 f a)(
B'
2 f a)
4 .
We also notice that because of (7.5), (7.2) has become the important equations
(7.7)
2A
2 f
a
2
a
, <strong>and</strong>
A'
2
a
2B 2 f a .
B'
Section 8.
We now wish to determine appropriate expressions for A’ <strong>and</strong> B’. We see from
(7.6) that these expressions should start as A' 4 f <strong>and</strong> B'
4 f . How does a fit
with
4 f ? Since we have been examining the three equally spaced numbers
a
f , f ,
2

9
<strong>and</strong>
a
f , it seems reasonable to think of 4 f 2a
, 4 f , <strong>and</strong> 4 f 2a
. Thus we now
2
try the equations
(8.1)
A '
4 f
2a
s'
, <strong>and</strong>
A''
s'
B ' 4 f 2a
,
B''
where s’ , A’’, <strong>and</strong> B’’ are to be found. Using (8.1) to remove A’ <strong>and</strong> B’ from (7.6) we get
(8.2)
s'
s'
2 f
A''
B''
2
f 3a
2 f 3a
4 .
Now compare (8.2) with (7.3). If in (7.3) we replace
s s'
, A'
A'',
B'
B'
' ,
<strong>and</strong>
a
3a
we get (8.2). Therefore (7.4) becomes after these substitutions
(8.3)
3 s
2 2
2
a A''
B''
A''
s'(2
f 3a)
B''
s'(2
f 3a)
' .
We are free to select a value for s’ <strong>and</strong> a nice choice to simplify (8.3) is
(8.4)
2 2
s ' 3 a .
Dividing (8.3) by by s’ get
2 2
A ''
B''
A''(2
f 3a)
B''(2
f 3a)
3 a .
Corresponding to (7.6), this last expression can be written as
(8.5)
( f
2
A '' 2 f 3a)(
B''
2 f 3a)
4 .
Notice that equations (8.1) are now the important relations
(8.6)
A '
4 f
2a
2
3 a
A''
2
, <strong>and</strong>
2 2
3 a
B ' 4 f 2a
.
B''
Section 9 <strong>and</strong> 10.
Next we seek appropriate expressions for A’’ <strong>and</strong> B’’. As we reasoned above to
get (8.1) we now try
(9.1)
s''
A '' 4 f 2a
<strong>and</strong>
A'''
s''
B '' 4 f 2a
.
B'''

10
Using (9.1) to remove A’’ <strong>and</strong> B’’ from (8.5) we now have (corresponding to (8.2))
(9.2)
s''
s''
2 f
A'''
B'''
2
f 5a
2 f 5a
4 ,
<strong>and</strong> by a similar reasoning we will arrive at
(9.3)
2 2
5 a
A '' 4 f 2a
<strong>and</strong>
A'''
2 2
5 a
B '' 4 f 2a
.
B'''
Summarizing our results for (7.7), (8.6) <strong>and</strong> (9.3) we see
2A
2 f
a
2
a
, <strong>and</strong>
A'
2
a
2B 2 f a ,
B'
A '
4 f
2a
2
3 a
A''
2
, <strong>and</strong>
2 2
3 a
B ' 4 f 2a
,
B''
2 2
5 a
A '' 4 f 2a
<strong>and</strong>
A'''
2 2
5 a
B '' 4 f 2a
.
B'''
Section 11.
The general pattern is now emerging <strong>and</strong> we can write our continued fractions as
(11.1) 2A
2 f a
2
2 2
2 2
a 3 a 5 a
4 f 2a
4 f 2a
4 f 2a
,
<strong>and</strong>
(11.2) 2B
2 f a
2
2 2
2 2
a 3 a 5 a
4 f 2a
4 f 2a
4 f 2a
.
Notice that (11.2) is obtained from (11.1) by replacing f by
f
a . In a similar way,
replacing f by
f
2 a in (11.1) gives us
2C
2 f
3a
a
4 f
2
6a
3
4 f
2
2
a
6a
5
4 f
2
2
a
6a
,
replacing f by
f
3 a in (11.1) yields

11
2D
2 f
5a
2
a
4 f 10a
3
4 f
2
2
a
10a
5
4 f
2
2
a
10a
,
<strong>and</strong> replacing f by
f
4 a in (11.1) yields
2E
2 f
7a
2
a
4 f 14a
3
4 f
2
2
a
14a
5
4 f
2
2
a
14a
<strong>and</strong> so forth.
Section 12.
f
a
The continued fractions of Brouncker are obtained from the above by writing
1. We get
2 2 2
1 3 5 4
2A 1
,
2 2 2
2 2 2
1 3 5
2B 3
,
6 6 6
2 2 2
1 3 5 16
2C 5
,
10 10 10
2 2 2
1 3 5 9
2E 7
,
14 14 14 4
2 2 2
1 3 5 256
2F 9
.
18 18 18 9
Section 13.
Euler remind us that he previously derived
2 2 2
1 3 5 4
2A 1
from
2 2 2
the Gregory Leibniz series
1 1 1
(13.1) 1
.
4 3 5 7
Euler has a neat way of doing this. From (13.1) we have

12
(13.2) 1
4
.
So
(13.3)
4
1
1
1
1
1
1
1
1
1
1
.
From (13.1) <strong>and</strong> (13.2) we have
(13.4)
1
3
.
So proceeding as before
(13.5)
1
1
1
3
1
3
3
3
9
1
3
9
3
9
1 3
3
3
9
1
.
Combining (13.3) with (13.5) we get
(13.6)
4
1
1
1
1
1
1
3
1
3
9
1
1
2
1
9
3
1
.
From 13.1) <strong>and</strong> (13.4) we have
(13.7)
1
5
.
So proceeding as before
(13.8)
1
1
1
5
1
5
5
5
25
1
5
25
5
25
1 5
5
25
.
1
5

13
Combining (13.6) with (13.8) we get
4
1
2
1
9
3
1
1
2
3
1
9
5
25
1
5
1
2
2
1
9
25
1
5
,
etc.
Section 14.
Euler now poses the problem to find representations for A, B, C, …, as infinite
products <strong>and</strong> ratios of integrals.
Problem.
To investigate the values of the individual letters, expressed first (1) through continual
products, then (2) however expressed through integral formulas, for the series A, B, C, D,
etc…that continues according to the law of uniformity in such a way that
AB
ff ; BC
( f
2
a)
; CD
( f
2
2a)
; etc...
Section 15.
Since
A
2
f
B
,
B
( f
a)
C
2
,
C
( f
2a)
D
2
, etc., we have
A
2
f
B
( f
2
f
a)
C
2
f
( f
2
C
a)
2
( f
2
f
a)
2
( f
2a)
D
2
.

14
(Notice the morphing of the continued fraction A into the Wallis like product!)
Continuing in this way we get the infinite product
A
ff ( f
( f
2
2a)
( f
2
a)
( f
2
4a)
( f
2
3a)
( f
2
6a)
( etc...
2
5a)
( etc...
which can be written for the purpose of convergence as
(15.1)
A
f
f ( f aa)
( f a)(
f a)
( f
( f
2a)(
f
3a)(
f
4a)
3a)
( f
( f
4a)(
f
5a)(
f
6a)
5a)
etc...
The above product <strong>and</strong> the products for B, C, D, have been checked with Mathematica
( a 1, <strong>and</strong> f 1,
2, 3, 4.).
Section 16.
Euler has previously derived the following lemma:
Integrating from 0 to 1 we have
n
x
1
m 1
x
n
x
n
k
m k
m
m k n
m n
m k 2n
m 2n
m k 3n
m 3n
m k 4n
...
m 4n
n
x
1
x
n
x
n
k
.

15
(Here we see eighteenth century mathematics. In the above expression, the
infinite product on the RHS l diverges to infinity while the integral tends to zero. No
doubt Euler is aware of this, <strong>and</strong> the results he will obtain are valid.)
We will be more careful than Euler <strong>and</strong> avoid x . To derive an improvement on the
above expression for the integral
1
m 1
x dx
I we make the substitution
n ( n k )/ n
1 x
0
(16.1)
t
n
x
<strong>and</strong> get
I
1 m
k
1
n
n
1
n
t
(1
t)
dt
1
n
m
n
m k
n
k
n
.
Using z ( z)
( z 1)
we get
I
1
n
m
n
m k
n
k
n
m k
n
m
n
m
n
m k
n
1
1
k
n
n

16
( m k)
m
m k
n
m n
n
n
m
m
n
k
n
n
n
1
1
k
n
n
.
Continuing in this way we get
I
( m k)(
m k
m(
m n)(
m
m ( r 1) n k
n)(
m k 2n)
( m k rn)
n
n
,
2n)(
m 3n)
( m rn)
m k ( r 1) n n
n
Using (16.1) <strong>and</strong> the beta integral we can finally write our improved form of Euler’s
lemma
(16.2)
1
0
1
x
m 1
x
dx
n ( n
k ) / n
( m k)(
m k
m(
m n)(
m
m ( r 1) n 1
n)(
m k 2n)
( m k rn)
x dx
.
( n k ) / n
2n)(
m 3n)
( m rn)
n
1 x
1
0
Next Euler lets
n 2 a , m f <strong>and</strong> k a in (16.2) to get
x
1
f
1
x
x
2a
f
f
a
f
f
3a
2a
f
f
5a
...
3a
x
1
x
x
2a
(Note the typo in the third factor in the denominator of the RHS, 3a should be 4a.)
1
0
1
x
f
x
1
2a
dx
1/ 2
( f
f ( f
a)(
f
2a)(
f
3a)(
f
4a)(
f
f 2( r 1) a 1
5a)
( f (2r
1) a)
x dx
.
2 1/ 2
6a)
( f 2ra)
a
1 x
1
0

17
a
r
a
f
r
a
f
ra
f
a
r
f
a
f
a
f
a
f
f
a
f
a
f
a
f
2
2
1
2
3/
2
1
2
)
2
(
)
1)
(2
(
)
6
)(
4
)(
2
(
)
5
)(
3
)(
(
Using
2)
1/
(
)
lim(
1)
(
2)
1/
(
n
x
n
x
n
x
n
(see (16.5) below) we get
(16.3)
2
1/
2
2
)
2
(
)
1)
(2
(
)
6
)(
4
)(
2
(
)
5
)(
3
)(
(
lim
1
1
0
2
1/
2
1
r
a
f
a
ra
f
a
r
f
a
f
a
f
a
f
f
a
f
a
f
a
f
x
dx
x
r
a
f
Next Euler replaces f by
a
f in the last expression to get
(16.4)
1
2
2
)
1)
(2
(
)
2)
(2
(
)
5
)(
3
)(
(
)
6
)(
4
)(
2
(
lim
1
1
0
2
1/
2
1
r
a
f
a
a
r
f
a
r
f
a
f
a
f
a
f
a
f
a
f
a
f
x
dx
x
r
a
a
f
Lemma on limits
(16.5)
2)
1/
(
)
(
1)
(
2)
1/
(
lim
n
x
n
x
n
x
n
Proof:

18
Using Stirlings formula
( m 1) lim 2
m m
m m e for large m we get
m
( x
( x
n 1/ 2)
lim
n 1) n
( x
n
1/ 2)(
( x
x
n
n)(
x
1/ 2)
n)
( x
( x n
n)
e
1/ 2)
( x
n)
e
( x
n
1/ 2)
.
Now the numerator can be written as
( x
n
1/ 2)(
x
n
1/ 2)
( x
n
1/ 2)
e
( x
n
1/ 2)
( x
n)
1
1
2( x
( x
n)
n)
( x
n
1/ 2)
(1
1
2( x
n)
( x
n
1/ 2)
e
( x
n
1/ 2)
<strong>and</strong> so we have for large n
( x
( x
n 1/ 2)
n 1)
lim
n
( x
n)
1
1
2( x
( x
n)
( x
n)
( x n
n)(
x
1/ 2)
n)
( x
(1
n)
e
1
2( x
( x
n)
n)
( x
n
1/ 2)
e
( x
n
1/ 2)
lim
n
1
1
2( x
( x
n)
n)
(
1/ 2)
(1
1
2( x
n)
( x
n
1/ 2)
e
1/ 2
lim( x
n
n)
(
1/ 2)
This checks with Mathematica <strong>and</strong> completes our lemma.
Section 17.
Euler gets the ratios of integrals
A
x
f
1
a 1
x
x
a
2
:
x
1
f
1
x
x
a
2
.

19
B
x
f 2a
1
1
x
x
:
a
2
x
f
1
a 1
x
x
a
2
C
x
f 3a
1
1
x
x
:
a
2
x
f 2a
1
1
x
x
a
2
etc…
We will show that Euler has made a rare error here. The right h<strong>and</strong> sides should be
multiplied by f,
f a , <strong>and</strong> f 2 a . This error has been checked with Mathematica.
Euler now divides (16.4) by (16.3)
1
0
1
0
1
1
x
f a 1
x
x
f
x
2a
1
2a
dx
dx
1/ 2
(1/ 2
f ( f 2a)
( f
lim
r ( f a)(
f a)
( f
2a)(
f
3a)(
f
4a)
( f
3a)
( f
(2r
(2r
2) a)(
f
1) a)(
f
(2r
(2r
2) a)
1) a)
f
2a
f
2a
r
r
1/ 2
1
.
f
r 1/ 2
2a
Since 1 we have
f
r 1
2a

20
(17.1)
1
0
1
0
1
1
x
f a 1
x
x
f
x
2a
1
2a
dx
dx
1/ 2
1/ 2
f ( f 2a)
( f
lim
r ( f a)(
f a)
( f
2a)(
f
3a)(
f
4a)
( f
3a)
( f
(2r)
a)(
f
(2r
1) a)(
f
(2r
2) a)
(2r
1) a)
Comparing this with Euler’s (15.1)
A
f
f ( f aa)
( f a)(
f a)
( f
( f
2a)(
f
3a)(
f
4a)
3a)
( f
( f
4a)(
f
5a)(
f
6a)
5a)
etc...
we see that they differ by the factor f.
This shows that
1
x
f a 1
2a
1 x
0
A L(0)
1
(17.2)
f 1
f f
x dx
0
1
x
2a
dx
1/ 2
1/ 2
.
(For convenience, we use the modern notation
L ( 0)
A, L ( 1)
B , L (2) 3, etc.)
Replacing f by
f ma in (17.2) we get
(17.3)

21
1
0
1
0
x
1
x
1
f ma a 1
x
x
2a
f ma 1
2a
dx
1/ 2
dx
1/ 2
( f ma)(
f ma 2a)
( f ma 2a)(
f
lim
r ( f ma a)(
f ma a)
( f ma 3a)(
f
( f ma
( f ma
(2r)
a)(
f
(2r
1) a)(
f
ma (2r
2) a)
ma (2r
1) a)
ma
ma
4a)
3a)
<strong>and</strong>
(17.4)
1
0
1
0
x
1
f ma a 1
x
1
x
f ma 1
x
2a
2a
dx
1/ 2
dx
1/ 2
L(
m)
.
f ma
(This completes our demonstration of Euler’ omission of the factor
f ma from (17.4) .
In section 27, equation (27.3) as well as in section 30 Euler does not omit this factor in
essentially the same equations.)
Euler now poses a generalization of his previous problem.
A MORE GENERAL PROBLEM
To find a series A, B, C, D, etc. that proceeds uniformly in such a way that
AB
ff
c;
BC
( f
a)
2
c;
CD
( f
2a)
2
c;
DE
( f
3a)
2
c;
where, in the case of single products, the letter f is increased by a quantity a.

22
PREVIOUS SOLUTION FOR CONTINUAL FRACTIONS
Section 18.
Euler begins with the problem of finding expressions A, B, C, D, etc., such that
AB
f
2
2
2
c ; BC ( f a)
c ; CD ( f 2a)
c ; etc. The solution is an alteration
on the previous work. The following sections 19 to 23 are a very minor variation on
sections 7 to 11.
Section 19.
We start with
2
(19.1) AB f c ,
<strong>and</strong> assume that the desired increase is given by
a s / 2
A f ,
2 A'
a s / 2
B f ,
2 B'
where s , A’ <strong>and</strong> B’ are to be determined. Multiplying by 2 we get
(19.2)
s
2A 2 f a ,
A'
s
2B 2 f a .
B'
Multiplying the above together we get using (19.1)
s s
2
(19.3) 2 f a 2 f a 4 f 4c
,
A B
which can also be written as
2
2 2 s s
s
4AB 4 f a (2 f a)
(2 f a)
.
B'
A'
A'
B'
Since
2
4AB 4 f 4c
we have

23
2
2 s s
s
a 4c
(2 f a)
(2 f a)
,
B'
A'
A'
B'
<strong>and</strong> multiplying by
A ' B'
we get
(19.4)
( s
2
2
a 4c)
A'
B'
A'
s(2
f a)
B'
s(2
f a)
.
We are free to choose s as we like, <strong>and</strong> a reasonable choice to simplify (19.4) is
2
(19.5 divide (19.4) by a 4c
we get
2
A'
B'
A'(2
f a)
B'(2
f a)
a 4c
.
Notice that this last expression can be written as
or
2
( A'
(2 f a))(
B'
(2 f a))
4 f 4c
2
(19.6) ( A'
2 f a)(
B'
2 f a)
4 f 4c
.
We also notice that because of (19.5), (19.2) has become the important equations
(19.7)
2A
2 f
a
a
2
4c
, <strong>and</strong>
A'
2
a 4c
2B 2 f a .
B'
Section 20.
We now wish to determine appropriate expressions for A’ <strong>and</strong> B’. We see from
(19.6) that these expressions should start as A' 4 f <strong>and</strong> B'
4 f . How does a
fit with
4 f ? Since we have been examining the three equally spaced numbers
f
a
2
, f
, <strong>and</strong>
a
f , it seems reasonable to think of 4 f 2a
, 4 f , <strong>and</strong> 4 f 2a
. Thus we now
2
try the equations
(20.1)
A '
4 f
2a
s'
, <strong>and</strong>
A''
s'
B ' 4 f 2a
,
B''

24
where s’ , A’’, <strong>and</strong> B’’ are to be found. Using (20.1) to remove A’ <strong>and</strong> B’ from (19.6) we
get
s'
s'
2
(20.2) 2 f 3a
2 f 3a
4 f 4c
.
A''
B''
Now compare (20.2) with (19.3). If in (19.3) we replace
s s'
, A'
A'',
B'
B''
, <strong>and</strong> a 3a
we get (20.2). Therefore (19.4) becomes after
these substitutions
(20.3)
( 3
s
2 2
2
a 4c)
A''
B''
A''
s'(2
f 3a)
B''
s'(2
f 3a)
' .
We are free to select a value for s’ <strong>and</strong> a nice choice to simplify (20.3) is
2 2
(20.4) s'
3 a 4c
.
Dividing (20.3) by by s’ get
2 2
A''
B''
A''(2
f 3a)
B''(2
f 3a)
3 a 4c
.
Corresponding to (19.6), this last expression can be written as
2
(20.5) ( A''
2 f 3a)(
B''
2 f 3a)
4 f 4c
.
Notice that equations (20.1) are now the important relations
(20.6)
A '
4 f
2a
2 2
3 a
A''
4c
, <strong>and</strong>
2 2
3 a 4c
B ' 4 f 2a
.
B''
Section 21 <strong>and</strong> 22.
Next we seek appropriate expressions for A’’ <strong>and</strong> B’’. As we reasoned above to
get (20.1) we now try
(21.1)
s''
A '' 4 f 2a
<strong>and</strong>
A'''
s''
B '' 4 f 2a
.
B'''
Using (9.1) to remove A’’ <strong>and</strong> B’’ from (20.5) we now have (corresponding to (20.2))

25
s''
s''
2
(21.2) 2 f 5a
2 f 5a
4 f 4c
,
A'''
B'''
<strong>and</strong> by a similar reasoning we will arrive at
(21.3)
2 2
5 a 4c
A '' 4 f 2a
<strong>and</strong>
A'''
2 2
5 a 4c
B '' 4 f 2a
.
B'''
Summarizing our results for (19.7), (20.6) <strong>and</strong> (21.3) we see
2A
2 f
a
a
2
4c
, <strong>and</strong>
A'
2
a 4c
2B 2 f a ,
B'
A '
4 f
2a
2 2
3 a
A''
4c
, <strong>and</strong>
2 2
3 a 4c
B ' 4 f 2a
,
B''
2 2
5 a 4c
A '' 4 f 2a
<strong>and</strong>
A'''
2 2
5 a 4c
B '' 4 f 2a
.
B'''
Section 23.
The general pattern is now emerging <strong>and</strong> we can write our continued fractions as
(23.1) 2A
2 f a
2
a 4c
2 2
3 a 4c
2 2
5 a 4c
4 f 2a
4 f 2a
4 f 2a
,
<strong>and</strong>
(23.2) 2B
2 f a
2
a 4c
2 2
3 a 4c
2 2
5 a 4c
4 f 2a
4 f 2a
4 f 2a
.
Notice that (23.2) is obtained from (23.1) by replacing f by
f
a . In a similar way,
replacing f by
f
2 a in (23.1) gives us
2C
2 f
3a
2
a
4 f
4c
6a
2
3 a
4 f
2
4c
6a
2
5 a
4 f
2
4c
6a
,
replacing f by
f
3 a in (23.1) yields
2D
2 f
5a
a
4 f
2
4c
10a
2
3 a
4 f
2
4c
10a
2
5 a
4 f
2
4c
10a
,

26
<strong>and</strong> replacing f by
f
4 a in (23.1) yields
2E
2 f
7a
a
4 f
2
4c
14a
2
3 a
4 f
2
4c
14a
2
5 a
4 f
2
4c
14a
<strong>and</strong> so forth.
Section 24.
With
AB
f
2
2
2
2
c , BC ( f a)
c CD ( f 2a)
c , DE ( f 3a)
c
we have
A
f
2
B
c
,
B
( f
a)
C
2
c
,
C
( f
2a)
D
2
c
, etc., so
A
Section 25.
f
2
B
c
( f
2
f c
2
a)
C
c
( f
( f
2
c)
C
2
a)
c
2
( f c)
2
( f a)
c
( f
2a)
D
2
c
.
The product emerging is
(25.1)
2
( f c)
2
( f a)
c
( f
( f
2a)
3a)
2
c
c
( f
( f
4a)
5a)
c
c
( f
( f
6a)
7a)
A 2
2
2
2
2
c
.
c
This product will not converge as written. Squaring we can write
2
2
2
2
( ff c)((
f 2a)
c)
(( f 2a)
c)((
f 4a)
c)
(25.2) A ( ff c)
etc...
2
2
2
2
(( f a)
c)((
f a)
c)
(( f 3a)
c)((
f 3a)
c)
which does converge.
Case 1 in which c = - bb
Section 26.
Replacing c by
2
b in (23.1) we get

27
2 2 2 2 2 2 2 2
a 4b
3 a 4b
5 a 4b
(26.1) 2A
2 f a
4 f 2a
4 f 2a
4 f 2a
which can also be written as
( a 2b)(
a 2b)
2A 2 f a
.
(3a
2b)(3a
2b)
4 f 2a
(5a
2b)(5a
2b)
4 f 2a
(7a
2b)(7a
2b)
4 f 2a
4 f 2a
etc...
Returning to (25.1) we have
A
2 2
( f b )
2
( f a)
b
2
( f
( f
2a)
3a)
2
2
b
b
2
2
( f
( f
4a)
5a)
2
2
b
b
2
2
( f
( f
6a)
7a)
2
2
b
b
2
2
which we rewrite in the convergent form
( f b)(
f 2a
b)
( f 2a
b)(
f 4a
b)
(26.2) A ( f b)
etc...
( f a b)(
f a b)
( f 3a
b)(
f 3a
b)
There is a misprint in the last factor. It should be
( f
( f
2a
3a
b)(
f
b)(
f
4a
3a
b)
.
b)
Section 27.
This section is a slight variation on section 16. Euler begins again with his lemms:
Integrating from 0 to 1 we have
n
x
1
m 1
x
x
n
n
k
m k
m
m k n
m n
m k 2n
m 2n
m k 3n
m 3n
m k 4n
...
m 4n
n
x
1
x
n
x
n
k
.
Again he lets n 2 a <strong>and</strong> k a , but he changes m from f to m f b <strong>and</strong> gets
f b 1
f a b f 3a
b f 5a
b x x x x
(27.1) .
f b f 2a
b f 4a
b
2a
2a
(1 x ) 1 x

28
Next he lets
n 2 a <strong>and</strong> k a , but now he changes m from f to m f a b
f a b 1
f 2a
b f 4a
b f 6a
b x x x x
(27.2) .
f a b f 3a
b f 5a
b
2a
2a
1 x 1 x
<strong>and</strong> gets
Dividing (27.2) by (27.1) he gets the product in (26.2). Therefore
f a b 1
f b 1
x x x x
(27.3) A ( f b)
: .
2a
2a
1 x 1 x
Note that this time he has the correct factor ( f b)
which he neglected in (17.2)
when b 0.
Section 28.
Euler exhibits the example of (26.1)
2A
2 f
a
2
a
4 f
2
4b
2a
2
3 a
4 f
2
4b
2a
2
2
5 a
4 f
2
4b
2a
2
With f 2 , a b 1 we have AB
2
f
2
b 3 , BC ( f
2
a)
2
b 8 ,
CD ( f
2
2a)
2
b 15, DE ( f
2
3a)
2
b 24, etc.
(28.1) 2A 3
3 5 21 45 77
6 6 6 6 6
.
Euler writes:
“But it will be this in continual products: A
3 3 5 5 7 7 9 9
2 4 4 6 6 8 8 10
etc...
Then it will be considered this way in integral formulas: A
x x xx x
: . ”
1 xx 1 xx

29
We can also write the infinite product using (26.2)
A
( f b)(
f 2a
b)
( f 2a
b)(
f 4a
b)
( f b)
etc...
( f a b)(
f a b)
( f 3a
b)(
f 3a
b)
3 3 5 5 7 7 9 9
(28.2) A 1
.
2 4 4 6 6 8 8 10
Using (27.3) we get the ratio of integrals
f a b 1
f b 1
x x x x
(28.3) A ( f b)
: .
2a
2a
1 x 1 x
A
x x xx x
: .
1 xx 1 xx
Since the above two integrals are 1 <strong>and</strong> / 4 we get
(28.4) A 4/
.
Section 29.
Now we let
2
c b in (23.1)
Case 2 in which c = + bb
2A
2 f
a
2
a
4 f
4c
2a
2
3 a
4 f
2
4c
2a
2
5 a
4 f
2
4c
2a
to get
(29.1) 2A
2 f a
2
a
2
4b
2 2
3 a
2
4b
2 2
5 a
2
4b
4 f 2a
4 f 2a
4 f 2a
.
In the infinite product (26.2)
A
( f b)(
f 2a
b)
( f 2a
b)(
f 4a
b)
( f b)
etc...
( f a b)(
f a b)
( f 3a
b)(
f 3a
b)
we can replace b by either bi or bi to get
(29.2)

30
A
( f b 1)( f 2a
b 1) ( f 2a
b 1)( f 4a
b 1)
( f b 1)
etc...
( f a b 1)( f a b 1) ( f 3a
b 1)( f 3a
b 1)
<strong>and</strong>
(29.3)
A
( f b 1)( f 2a
b 1) ( f 2a
b 1)( f 4a
b 1)
( f b 1)
etc...
( f a b 1)( f a b 1) ( f 3a
b 1)( f 3a
b 1)
Multiplying (29.2) times (29.3) we get
A
2
( ff
bb)
(( f
2
bb)((
f 2a)
bb)
2
2
a)
bb)((
f a)
bb
(( f
(( f
2a)
3a)
bb)((
f
bb)((
f
4a)
3a)
2
2
( ff
2
2
bb)
bb)
etc...
which is the same as our (25.1) with
Section 30.
If we replace b by bi in (27.3)
2
c b .
A
( f
b)
x
f
1
a
b 1
x
2a
x
:
x
f
1
x
.
2a
x
b 1
we get
(30.1) A ( f b 1)
f a 1 b 1
f 1 b 1
x x x
:
x
.
1
2a
x 1
2a
x
<strong>and</strong> if we replace b by
bi we get
f a 1 b 1
f 1 b 1
x x x x
(30.2) A ( f b 1)
:
.
2a
2a
1 x 1 x
Section 31.
If we multiply (30.1) times (30.2) we get,

31
x
f
a
1 b
1
x
x
f
a
1 b
1
x
(31.1)
A
2
( ff
bb)
1
x
x
2a
f 1 b 1
x
x
f
1
1 b
x
1
2a
x
Section 32.
1
x
2a
1
x
2a
In the denominator of (31.1) we set
x
1
f
1
x
x
2a
V
<strong>and</strong> get
x
b
1
V
x
b
1
V.
Now make the substitutions
The sum:
( x
x
b 1 b 1
V
p
Difference:
( x
x
b 1 b 1
V
q
<strong>and</strong> get
(32.1)
( x
b
1 b 1 pp qq
V x V
4
Section 33.
If we replace x by exp(logx ) in p <strong>and</strong> q of the previous section we get
p
( e
b log x
1
e
b log x
1
)
V ,
q
( e
b log x
1
e
b log x
1
)
V ,
(Here Euler writes lx for our
log x .) Using the identities
1
1
1
1
e e 2cos <strong>and</strong> e e 2 1sin
with
blog x we have
p 2 V cos <strong>and</strong> q 2 1 V sin .

32
Using (32.1) we see
(33.1)
pp qq
4
(
V cos
)
2
(
V sin
)
2
which is the denominator of (31.1).
Section 34.
The same argument with numerator of (31.1) would have given Euler
a
2 a
( x V cos ) ( x V
sin
)
2
<strong>and</strong> thus (31.1) can be expressed as
(34.1)
A
2
( ff
(
bb)
x
(
a
V cos
V cos
)
)
2
2
(
(
x
a
V sin
V sin
)
2
)
2
where we recall
V
x
1
f
1
x
x
2a
<strong>and</strong>
blog x.
Section 35.
Euler remarks briefly on the integrals (I need help with the Latin here.)
x
f
1
x cos( blog
x)
1
x
2a
<strong>and</strong>
x
f
1
xsin(
blog
x)
.
2a
1 x
Section 36.
Euler begins by writing the formula for integration by parts
P Q PQ Q P .
f 1
He then takes P cos( blogx)
<strong>and</strong> Q x x
<strong>and</strong> gets
f
f 1
x
(36.1) x x cos( blog
x)
cos( blog
x)
( blog
x)
f

33
f 1
A second integration by parts uses P sin ( blogx)
<strong>and</strong> Q x x
<strong>and</strong> we get
f
f 1
x
b f 1
(36.2) x xsin(
blog
x)
sin( blog
x)
x x cos( blog
x)
f
f
Now, starting with (36.1) we can replace the integral on the RHS with (36.2) to obtain
1
0
x
f
1
cos( blog
x)
dx
x
f
f
cos( blog
x)
b
f
x
f
f
sin( blog
x)
b
f
1
0
x
f
1
cos( blog
x)
dx
which simplifies to
1
0
x
f
1
cos( blogx)
dx
x
f
f
cos( blogx)
bx
f
f
2
sin( blogx)
b
f
2
2
1
0
x
f
1
cos( blogx)
dx .
Solving for the integral we get
f
f 1
x
(36.3) x x cosblx
( f cos( blog
x)
bsin(
blog
x));
.
ff bb
In a similar way, starting with (36.2) we get
f
f 1
x
(36.4) x xsin(
blog
x)
( f sin( blog
x)
bcos(
blog
x))
ff bb
(Note that neither of the above integrals connect with the integrals in previous sections.
All of those had
1 x 2a in the denominator.) Perhaps Euler was showing his skill in the
use of complex numbers.
References

34
[1] Euler, L., De Fractionibus Continuis Wallisii, (E745), on the web at the Euler
Archive, http://www.math.dartmouth.edu/~euler/.
[2] Khrushchev, S., A recovery of Brouncker’s proof for the quadrature continued
fractions, Publicacions matermatiques, 50(2006), pp. 2-42.
[3] Nunn, T. Percy, The Arithmetic of Infinites.: A School Introduction to the Integral
Calculus, (Part 1), The Mathematical Gazette, 5(1910), pp. 345-356.
[4] Nunn, T. Percy, The Arithmetic of Infinites.: A School Introduction to the Integral
Calculus, (Part 2), The Mathematical Gazette, 5(1911), pp. 377-386.
[5] Stedall, Jacqueline A., Catching Proteus: The Collaborations of Wallis <strong>and</strong>
Brouncker. I. Squaring the Circle, <strong>Notes</strong> <strong>and</strong> Records of the Royal Society of London,
Vol. 54, No. 3, (Sep., 2000), pp. 293 -316
[6] Wallis, John, The Arithmetic of Infinitesimals, (Translated from Latin by Jacqueline
A. Stedall), Springer Verlag, New York, 2004.