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Reflections and Notes on Euler’s paper E745:
De fractionibus continuis Wallisii
(On the continued fractions of Wallis)
Thomas J Osler, Christopher Tippie and 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 and Euler wishes to provide this. Euler states that Brouncker
discovered the first continued fraction in this sequence and 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 and generalization of this sequence for the first
time.
Section 2.
Euler examines the list of integrals
111 xx
xx
32
22
3
2
1
3
xx
xx
2
5432
4422
53
42
1
5
xx
xx
765432
664422
753
642
1
7
xx
xx
98765432
88664422
9753
8642
1
9
xx
xx, etc…
Using the substitution 22 1 xt we can convert these integrals to the form
1
0
1
0
2
21
1dtt
x
dxx PP
. In Wallis, we find a table of the reciprocal integrals
dxxPQ
1
0
/11/1 . Wallis after much work obtains the following table:
0P 2/1
P
1P 2/3
P
2P
2/5
P
3P
0Q
1
1
11
2
21
3
3
1
11
4
4
2
21
5
5
3
3
1
11
6
6
4
4
2
21
2/1
Q
1 2 2
1
2
31
2 2 4
1 3
4
5
2
31
2 2 4 6
1 3 5
6
7
4
5
2
31
1Q 1
1
3
2
1
2
41
3
5
1
3
2
1
4
6
2
41
5
7
3
5
1
3
2
1
6
8
4
6
2
41
2/3
Q
1
1
4
3
4
2
51
3
6
1
4
3
4
4
7
2
51
5
8
3
6
1
4
3
4
6
9
4
7
2
51
2Q
1
1
5
8
3
2
61
3
7
1
5
8
3
4
8
2
61
5
9
3
7
1
5
8
3
6
10
4
8
2
61
2/5
Q
1
1
6
15
16
2
71
3
8
1
6
15
16
4
9
2
71
5
10
3
8
1
6
15
16
6
11
4
9
2
71
3Q 1
1
7
8
5
2
81
3
9
1
7
8
5
4
10
2
81
5
11
3
9
1
7
8
5
6
12
4
10
2
81
3
The list of integrals that Euler is studying,
1
021 x
dxx P
, are from the row where 2/1Q .
Sections 3 and 4.
Euler now wishes to find expressions, A, B, C, D, E so that
21AB , 22BC , 23CD , 24DE , etc. Then he can write the above list of integrals
as
,1
11
1
A
Axx
xx
,321
1
321
3 ABC
A
BC
xx
xx
,54321
1
54321
5 ABCDE
A
BCDE
xx
xx
,7654321
1
7654321
7 ABCDEFG
A
BCDEFG
xx
xx 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 and “interpolates” to get
,11
1 Axx
x
,21
1
1
AB
Axx
xxx
4
,4321
1
1
4 ABCD
Axx
xx
,654321
1
1
6 ABCDEF
Axx
xxetc…
This is a conjecture totally in the spirit of Wallis.
For convenience, we also use the modern notation AL )0( , BL )1( , CL )2( , etc.
Section 5.
Euler knows that 21
1
02x
dx. He calls this value q. From the first integral
above we see that A
q1
. Since 1AB we now have qA
B1
. From 4BC we get
qBC
44, etc. Euler lists these values and there numerical equivalents in a table.
Difference
636620,01
qA
0,934176
570796,1qB
0,975683
546479,24
qC
0,987813
534292,34
9qD
0,992782
5
527074,49
164
qE
0,995257
522331,5164
259qF
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 2
q . Let us study
this as apparently Euler did. (We use our notation: AL )0( , BL )1( , CL )2( , etc.).
We calculate the following:
)1(
1)2(
2
LL
2
22
1
)1(2
)2(
2)3(
L
LL
)1(2
31
)3(
3)4(
2
222
LLL
22
222
31
)1(42
)4(
4)5(
L
LL
)1(42
531
)5(
5)6(
22
2222
LLL
6
Let us denote the partial product in the Wallis product as
)2)(2(
)12)(12(
66
75
44
53
22
31)(
nn
nnnW .
Then from the above calculations we can write in general
)1(
)12()1(
)1()22(642
)12(531)2(
2222
2222
L
nnW
Ln
nnL
, and
)(
)1()12(
)12(531
)1()2(642)12(
2222
2222
nW
Ln
n
LnnL
.
Since 2
)(lim nW we get the two limits
(5.1) )1(
2
2
)2(lim
12
)2(lim
Ln
nL
n
nL and
(5.2) 2
)1(
12
)12(lim
L
n
nL.
From these last two results we see that n
nL )(lim exists and equals 1 if and only if
2)1(L . Otherwise, if
2)1(L , the limit does not exist, but the separate limits through
even and odd values on n do exist. Thus it is clear that the choice 2
)1(LAq is
very special.
Euler will work intuitively in the next sections to move “uniformly” from A to B
to C and 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=2
af ; CD=2
2af ; etc.
Given f and a we wish to have 2fAB , 2)( afBC , 2)2( afCD , etc.
Section 7.
Euler begins with the problem of finding expressions A, B, C, D, etc., such that
2fAB ; 2)( afBC ; 2)2( afCD ; etc.
Noticing that 422
22 a
fa
fa
f , Euler asks how he might increase
these two factors on the LHS so that the product is exactly 2f .
We start with
(7.1) 2fAB ,
and assume that the desired increase is given by
'
2/
2 A
safA ,
'
2/
2 B
safB ,
where s , A’ and B’ are to be determined. Multiplying by 2 we get
(7.2) '
22A
safA ,
'22
B
safB .
Multiplying the above together we get using (7.1)
(7.3) 24'
2'
2 fB
saf
A
saf ,
8
which can also be written as
'')2(
')2(
'44
222
BA
saf
A
saf
B
safAB .
Since 244 fAB we have
'')2(
')2(
'
22
BA
saf
A
saf
B
sa ,
and multiplying by '' BA we get
(7.4) 22 )2(')2(''' safsBafsABAa .
We are free to choose s as we like, and a reasonable choice to simplify (7.4) is
(7.5) 2as .
After we divide (7.4) by 2a we get
2)2(')2(''' aafBafABA .
Notice that this last expression can be written as
24))2('))(2('( fafBafA
or
(7.6) 24)2')(2'( fafBafA .
We also notice that because of (7.5), (7.2) has become the important equations
(7.7) '
222
A
aafA , and
'22
2
B
aafB .
Section 8.
We now wish to determine appropriate expressions for A’ and B’. We see from
(7.6) that these expressions should start as fA 4' and fB 4' . How does a fit
with f4 ? Since we have been examining the three equally spaced numbers 2
af , f ,
9
and 2
af , it seems reasonable to think of af 24 , f4 , and af 24 . Thus we now
try the equations
(8.1) ''
'24'
A
safA , and
''
'24'
B
safB ,
where s’ , A’’, and B’’ are to be found. Using (8.1) to remove A’ and B’ from (7.6) we get
(8.2) 24''
'32
''
'32 f
B
saf
A
saf .
Now compare (8.2) with (7.3). If in (7.3) we replace ''',''',' BBAAss ,
and aa 3 we get (8.2). Therefore (7.4) becomes after these substitutions
(8.3) 222 ')32(''')32('''''''3 safsBafsABAa .
We are free to select a value for s’ and a nice choice to simplify (8.3) is
(8.4) 223' as .
Dividing (8.3) by by s’ get
223)32('')32('''''' aafBafABA .
Corresponding to (7.6), this last expression can be written as
(8.5) 24)32'')(32''( fafBafA .
Notice that equations (8.1) are now the important relations
(8.6) ''
324'
22
A
aafA , and
''
324'
22
B
aafB .
Section 9 and 10.
Next we seek appropriate expressions for A’’ and B’’. As we reasoned above to
get (8.1) we now try
(9.1) '''
''24''
A
safA and
'''
''24''
B
safB .
10
Using (9.1) to remove A’’ and B’’ from (8.5) we now have (corresponding to (8.2))
(9.2) 24'''
''52
'''
''52 f
B
saf
A
saf ,
and by a similar reasoning we will arrive at
(9.3) '''
524''
22
A
aafA and
'''
524''
22
B
aafB .
Summarizing our results for (7.7), (8.6) and (9.3) we see
'
222
A
aafA , and
'22
2
B
aafB ,
''
324'
22
A
aafA , and
''
324'
22
B
aafB ,
'''
524''
22
A
aafA and
'''
524''
22
B
aafB .
Section 11.
The general pattern is now emerging and we can write our continued fractions as
(11.1) af
a
af
a
af
aafA
24
5
24
3
2422
22222
,
and
(11.2) af
a
af
a
af
aafB
24
5
24
3
2422
22222
.
Notice that (11.2) is obtained from (11.1) by replacing f by af . In a similar way,
replacing f by af 2 in (11.1) gives us
af
a
af
a
af
aafC
64
5
64
3
64322
22222
,
replacing f by af 3 in (11.1) yields
11
af
a
af
a
af
aafD
104
5
104
3
104522
22222
,
and replacing f by af 4 in (11.1) yields
af
a
af
a
af
aafE
144
5
144
3
144722
22222
and so forth.
Section 12.
The continued fractions of Brouncker are obtained from the above by writing
1af . We get
4
2
5
2
3
2
112
222
A ,
6
5
6
3
6
132
222
B ,
16
10
5
10
3
10
152
222
C ,
4
9
14
5
14
3
14
172
222
E ,
9
256
18
5
18
3
18
192
222
F .
Section 13.
Euler remind us that he previously derived 4
2
5
2
3
2
112
222
A from
the Gregory Leibniz series
(13.1) 7
1
5
1
3
11
4.
Euler has a neat way of doing this. From (13.1) we have
12
(13.2) 14
.
So
(13.3) 1
1
11
11
1
1
1
14.
From (13.1) and (13.2) we have
(13.4) 3
1.
So proceeding as before
(13.5) 1
3
93
31
93
31
993
31
3
3
1
11.
Combining (13.3) with (13.5) we get
(13.6)
13
92
11
13
931
11
11
11
4.
From 13.1) and (13.4) we have
(13.7) 5
1.
So proceeding as before
(13.8) 1
5
255
51
255
51
25255
51
5
5
1
11.
13
Combining (13.6) with (13.8) we get
15
252
92
11
15
2553
92
11
13
92
11
4,
etc.
Section 14.
Euler now poses the problem to find representations for A, B, C, …, as infinite
products and 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
...;)2(;)(; 22 etcafCDafBCffAB
Section 15.
Since B
fA
2
, C
afB
2)(,
D
afC
2)2(, etc., we have
D
af
af
f
af
Cf
C
af
f
B
fA
2
2
2
2
2
2
22 )2(
)()()(.
14
(Notice the morphing of the continued fraction A into the Wallis like product!)
Continuing in this way we get the infinite product
...()5()3()(
...()6()4()2(222
222
etcafafaf
etcafafafffA
which can be written for the purpose of convergence as
(15.1)
...)5)(5(
)6)(4(
)3)(3(
)4)(2(
))((
)(etc
afaf
afaf
afaf
afaf
afaf
aafffA
The above product and the products for B, C, D, have been checked with Mathematica
( 1a , and ,1f 2, 3, 4.).
Section 16.
Euler has previously derived the following lemma:
Integrating from 0 to 1 we have
.
1
...4
4
3
3
2
2
1
1
n knn
n knn
m
x
xx
nm
nkm
nm
nkm
nm
nkm
nm
nkm
m
km
x
xx
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, and the results he will obtain are valid.)
We will be more careful than Euler and avoid x . To derive an improvement on the
above expression for the integral
1
0
/)(
1
1nknn
m
x
dxxI we make the substitution
(16.1) nxt
and get
n
km
n
k
n
m
ndttt
nI n
k
n
m1
)1(1 11
.
Using )1()( zzz we get
n
n
k
n
km
n
m
n
m
n
km
n
km
n
k
n
m
nI
1
11
16
n
n
k
n
nkm
n
nm
n
nm
n
nkm
m
km
1
1)(
.
Continuing in this way we get
n
n
k
n
nrkm
n
nrm
rnm
rnkm
nmnmnmm
nkmnkmkmI
)1(
)1(
)(
)(
)3)(2)((
)2)()(( ,
Using (16.1) and the beta integral we can finally write our improved form of Euler’s
lemma
(16.2)
1
0
/)(
1)1(
1
0
/)(
1
1)(
)(
)3)(2)((
)2)()((
1nknn
nrm
nknn
m
x
dxx
rnm
rnkm
nmnmnmm
nkmnkmkm
x
dxx .
Next Euler lets an 2 , fm and ak in (16.2) to get
aa
f
x
xx
af
af
af
af
f
af
x
xx
22
1
1...
3
5
2
3
1
(Note the typo in the third factor in the denominator of the RHS, 3a should be 4a.)
1
0
2/12
1)1(2
1
0
2/12
1
1)2(
))12((
)6)(4)(2(
)5)(3)((
1 a
arf
a
f
x
dxx
raf
arf
afafaff
afafaf
x
dxx .
17
a
ra
f
ra
f
raf
arf
afafaff
afafaf
2
2
1
2/32
12
)2(
))12((
)6)(4)(2(
)5)(3)((
Using )2/1()(lim
)1(
)2/1(nx
nx
nx
n (see (16.5) below) we get
(16.3)
2/12
2)2(
))12((
)6)(4)(2(
)5)(3)((lim
1
1
0
2/12
1
ra
fa
raf
arf
afafaff
afafaf
x
dxx
ra
f
Next Euler replaces f by af in the last expression to get
(16.4)
12
2))12((
))22((
)5)(3)((
)6)(4)(2(lim
1
1
0
2/12
1
ra
fa
arf
arf
afafaf
afafaf
x
dxx
r
a
af
Lemma on limits
(16.5) )2/1()()1(
)2/1(lim nx
nx
nx
n
Proof:
18
Using Stirlings formula mm
memmm 2lim)1( for large m we get
)()(
)2/1()2/1(
)()(
)2/1()2/1(lim
)1(
)2/1(nxnx
nxnx
n enxnx
enxnx
nx
nx.
Now the numerator can be written as
)2/1(
)2/1(
)2/1(
)2/1()2/1(
)(2
11()(
)(2
11)(
)2/1()2/1(
nx
nx
nx
nxnx
enx
nxnx
nx
enxnx
and so we have for large n
)2/1(
2/1
)2/1(
)2/1(
)()(
)2/1(
)2/1(
)2/1(
)(lim
)(2
11()(
)(2
11lim
)()(
)(2
11()(
)(2
11)(
lim)1(
)2/1(
nx
enx
nxnx
enxnx
enx
nxnx
nx
nx
nx
n
nx
n
nxnx
nx
nx
nx
n
This checks with Mathematica and completes our lemma.
Section 17.
Euler gets the ratios of integrals
a
f
a
af
x
xx
x
xxA
2
1
2
1
1
:
1
.
19
a
af
a
af
x
xx
x
xxB
2
1
2
12
1
:
1
a
af
a
af
x
xx
x
xxC
2
12
2
13
1
:
1
etc…
We will show that Euler has made a rare error here. The right hand sides should be
multiplied by f, af , and af 2 . This error has been checked with Mathematica.
Euler now divides (16.4) by (16.3)
.
12
2/12
))12()()12((
))22()()22((
)3)(3(
)4)(2(
))((
)2(lim
1
1
1
0
2/1(2
1
1
0
2/12
1
ra
f
ra
f
arfarf
arfarf
afaf
afaf
afaf
aff
x
dxx
x
dxx
r
a
f
a
af
Since 1
12
2/12
ra
f
ra
f
we have
20
(17.1)
))12()()12((
))22()()2((
)3)(3(
)4)(2(
))((
)2(lim
1
1
1
0
2/12
1
1
0
2/12
1
arfarf
arfarf
afaf
afaf
afaf
aff
x
dxx
x
dxx
r
a
f
a
af
Comparing this with Euler’s (15.1)
...)5)(5(
)6)(4(
)3)(3(
)4)(2(
))((
)(etc
afaf
afaf
afaf
afaf
afaf
aafffA
we see that they differ by the factor f.
This shows that
(17.2) f
L
f
A
x
dxx
x
dxx
a
f
a
af
)0(
1
1
1
0
2/12
1
1
0
2/12
1
.
(For convenience, we use the modern notation AL )0( , BL )1( , 3)2(L , etc.)
Replacing f by maf in (17.2) we get
(17.3)
21
))12()()12((
))22()()2((
)3)(3(
)4)(2(
))((
)2)((lim
1
1
1
0
2/12
1
1
0
2/12
1
armafarmaf
armafarmaf
amafamaf
amafamaf
amafamaf
amafmaf
x
dxx
x
dxx
r
a
maf
a
amaf
and
(17.4) maf
mL
x
dxx
x
dxx
a
maf
a
amaf
)(
1
1
1
0
2/12
1
1
0
2/12
1
.
(This completes our demonstration of Euler’ omission of the factor maf 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
;)3(;)2(;)(; 222 cafDEcafCDcafBCcffAB
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
cfAB 2 ; cafBC 2)( ; cafCD 2)2( ; 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
(19.1) cfAB 2 ,
and assume that the desired increase is given by
'
2/
2 A
safA ,
'
2/
2 B
safB ,
where s , A’ and B’ are to be determined. Multiplying by 2 we get
(19.2) '
22A
safA ,
'22
B
safB .
Multiplying the above together we get using (19.1)
(19.3) cfB
saf
A
saf 4422 2 ,
which can also be written as
'')2(
')2(
'44
222
BA
saf
A
saf
B
safAB .
Since cfAB 444 2 we have
23
'')2(
')2(
'4
22
BA
saf
A
saf
B
sca ,
and multiplying by '' BA we get
(19.4) 22 )2(')2(''')4( safsBafsABAca .
We are free to choose s as we like, and a reasonable choice to simplify (19.4) is
(19.5 divide (19.4) by ca 42 we get
caafBafABA 4)2(')2(''' 2 .
Notice that this last expression can be written as
cfafBafA 44))2('))(2('( 2
or
(19.6) cfafBafA 44)2')(2'( 2 .
We also notice that because of (19.5), (19.2) has become the important equations
(19.7) '
422
2
A
caafA , and
'
422
2
B
caafB .
Section 20.
We now wish to determine appropriate expressions for A’ and B’. We see from
(19.6) that these expressions should start as fA 4' and fB 4' . How does a
fit with f4 ? Since we have been examining the three equally spaced numbers 2
af , f
, and 2
af , it seems reasonable to think of af 24 , f4 , and af 24 . Thus we now
try the equations
(20.1) ''
'24'
A
safA , and
''
'24'
B
safB ,
24
where s’ , A’’, and B’’ are to be found. Using (20.1) to remove A’ and B’ from (19.6) we
get
(20.2) cfB
saf
A
saf 44
''
'32
''
'32 2 .
Now compare (20.2) with (19.3). If in (19.3) we replace
''',''',' BBAAss , and aa 3 we get (20.2). Therefore (19.4) becomes after
these substitutions
(20.3) 222 ')32(''')32(''''''')43( safsBafsABAca .
We are free to select a value for s’ and a nice choice to simplify (20.3) is
(20.4) cas 43' 22 .
Dividing (20.3) by by s’ get
caafBafABA 43)32('')32('''''' 22.
Corresponding to (19.6), this last expression can be written as
(20.5) cfafBafA 44)32'')(32''( 2.
Notice that equations (20.1) are now the important relations
(20.6) ''
4324'
22
A
caafA , and
''
4324'
22
B
caafB .
Section 21 and 22.
Next we seek appropriate expressions for A’’ and B’’. As we reasoned above to
get (20.1) we now try
(21.1) '''
''24''
A
safA and
'''
''24''
B
safB .
Using (9.1) to remove A’’ and B’’ from (20.5) we now have (corresponding to (20.2))
25
(21.2) cfB
saf
A
saf 44
'''
''52
'''
''52 2 ,
and by a similar reasoning we will arrive at
(21.3) '''
4524''
22
A
caafA and
'''
4524''
22
B
caafB .
Summarizing our results for (19.7), (20.6) and (21.3) we see
'
422
2
A
caafA , and
'
422
2
B
caafB ,
''
4324'
22
A
caafA , and
''
4324'
22
B
caafB ,
'''
4524''
22
A
caafA and
'''
4524''
22
B
caafB .
Section 23.
The general pattern is now emerging and we can write our continued fractions as
(23.1) af
ca
af
ca
af
caafA
24
45
24
43
24
422
22222
,
and
(23.2) af
ca
af
ca
af
caafB
24
45
24
43
24
422
22222
.
Notice that (23.2) is obtained from (23.1) by replacing f by af . In a similar way,
replacing f by af 2 in (23.1) gives us
af
ca
af
ca
af
caafC
64
45
64
43
64
4322
22222
,
replacing f by af 3 in (23.1) yields
af
ca
af
ca
af
caafD
104
45
104
43
104
4522
22222
,
26
and replacing f by af 4 in (23.1) yields
af
ca
af
ca
af
caafE
144
45
144
43
144
4722
22222
and so forth.
Section 24.
With cfAB 2 , cafBC 2)( cafCD 2)2( , cafDE 2)3(
we have
B
cfA
2
, C
cafB
2)(,
D
cafC
2)2(, etc., so
D
caf
caf
cf
caf
Ccf
C
caf
cf
B
cfA
2
2
2
2
2
2
22 )2(
)(
)(
)(
)(
)(.
Section 25.
The product emerging is
(25.1)
caf
caf
caf
caf
caf
caf
caf
cfA
2
2
2
2
2
2
2
2
)7(
)6(
)5(
)4(
)3(
)2(
)(
)(.
This product will not converge as written. Squaring we can write
(25.2) ...))3)(()3((
))4)(()2((
)))(()((
))2)((()(
22
22
22
22 etc
cafcaf
cafcaf
cafcaf
cafcffcffA
which does converge.
Case 1 in which c = - bb
Section 26.
Replacing c by 2b in (23.1) we get
27
(26.1) af
ba
af
ba
af
baafA
24
45
24
43
24
422
22222222
which can also be written as
...24
)27)(27(24
)25)(25(24
)23)(23(24
)2)(2(22
etcaf
babaaf
babaaf
babaaf
babaafA .
Returning to (25.1) we have
22
22
22
22
22
22
22
22
)7(
)6(
)5(
)4(
)3(
)2(
)(
)(
baf
baf
baf
baf
baf
baf
baf
bfA
which we rewrite in the convergent form
(26.2) ...)3)(3(
)4)(2(
))((
)2)(()( etc
bafbaf
bafbaf
bafbaf
bafbfbfA
There is a misprint in the last factor. It should be )3)(3(
)4)(2(
bafbaf
bafbaf.
Section 27.
This section is a slight variation on section 16. Euler begins again with his lemms:
Integrating from 0 to 1 we have
.
1
...4
4
3
3
2
2
1
1
n knn
knnn
m
x
xx
nm
nkm
nm
nkm
nm
nkm
nm
nkm
m
km
x
xx
Again he lets an 2 and ak , but he changes m from f to bfm and gets
(27.1) .1)1(4
5
2
3
2
1
2 a
bf
a x
xx
x
xx
baf
baf
baf
baf
bf
baf
28
Next he lets an 2 and ak , but now he changes m from f to bafm and gets
(27.2) .115
6
3
42
2
1
2 a
baf
a x
xx
x
xx
baf
baf
baf
baf
baf
baf
Dividing (27.2) by (27.1) he gets the product in (26.2). Therefore
(27.3) .1
:1
)(2
1
2
1
a
bf
a
baf
x
xx
x
xxbfA
Note that this time he has the correct factor )( bf which he neglected in (17.2)
when .0b
Section 28.
Euler exhibits the example of (26.1)
af
ba
af
ba
af
baafA
24
45
24
43
24
422
22222222
With 2f , 1ba we have 322 bfAB , 8)( 22 bafBC ,
15)2( 22 bafCD , 24)3( 22 bafDE , etc.
(28.1) 6
77
6
45
6
21
6
5
6
332A .
Euler writes:
“But it will be this in continual products: ...108
99
86
77
64
55
42
33etcA
Then it will be considered this way in integral formulas: .1
:1 xx
xxx
xx
xxA ”
29
We can also write the infinite product using (26.2)
...)3)(3(
)4)(2(
))((
)2)(()( etc
bafbaf
bafbaf
bafbaf
bafbfbfA
(28.2) 108
99
86
77
64
55
42
331A .
Using (27.3) we get the ratio of integrals
(28.3) .1
:1
)(2
1
2
1
a
bf
a
baf
x
xx
x
xxbfA
.1
:1 xx
xxx
xx
xxA
Since the above two integrals are 1 and 4/ we get
(28.4) /4A .
Case 2 in which c = + bb
Section 29.
Now we let 2bc in (23.1)
af
ca
af
ca
af
caafA
24
45
24
43
24
422
22222
to get
(29.1) af
ba
af
ba
af
baafA
24
45
24
43
24
422
22222222
.
In the infinite product (26.2)
...)3)(3(
)4)(2(
))((
)2)(()( etc
bafbaf
bafbaf
bafbaf
bafbfbfA
we can replace b by either bi or bi to get
(29.2)
30
...)13)(13(
)14)(12(
)1)(1(
)12)(1()1( etc
bafbaf
bafbaf
bafbaf
bafbfbfA
and
(29.3)
...)13)(13(
)14)(12(
)1)(1(
)12)(1()1( etc
bafbaf
bafbaf
bafbaf
bafbfbfA
Multiplying (29.2) times (29.3) we get
...))3)(()3((
))4)(()2((
))(()((
))2)((()(
22
22
22
22 etc
bbafbbaf
bbafbbaf
bbafbbaf
bbafbbffbbffA
which is the same as our (25.1) with 2bc .
Section 30.
If we replace b by bi in (27.3)
.1
:1
)(2
1
2
1
a
bf
a
baf
x
xx
x
xxbfA
we get
(30.1) .1
:1
)1(2
11
2
11
a
bf
a
baf
x
xx
x
xxbfA
and if we replace b by bi we get
(30.2) .1
:1
)1(2
11
2
11
a
bf
a
baf
x
xx
x
xxbfA
Section 31.
If we multiply (30.1) times (30.2) we get,
31
(31.1)
a
bf
a
bf
a
baf
a
baf
x
xx
x
xx
x
xx
x
xx
bbffA
2
11
2
11
2
11
2
11
2
11
11)(
Section 32.
In the denominator of (31.1) we set Vx
xx
a
f
2
1
1 and get
.11 VxVx bb
Now make the substitutions
The sum: pVxx bb 11(
Difference: qVxx bb 11(
and get
(32.1) 4
( 11 qqppVxVx bb
Section 33.
If we replace x by )exp(logx in p and q of the previous section we get
,)( 1log1log Veep xbxb
,)( 1log1log Veeq xbxb
(Here Euler writes lx for our xlog .) Using the identities
cos211 ee and sin1211 ee
with xb log we have
cos2 Vp and sin12 Vq .
32
Using (32.1) we see
(33.1) 22 )sin()cos(4
VVqqpp
which is the denominator of (31.1).
Section 34.
The same argument with numerator of (31.1) would have given Euler
22 )sin()cos( VxVx aa
and thus (31.1) can be expressed as
(34.1) 22
22
2
)sin()cos(
)sin()cos()(
VV
VxVxbbffA
aa
where we recall a
f
x
xxV
2
1
1and xb log .
Section 35.
Euler remarks briefly on the integrals (I need help with the Latin here.)
a
f
a
f
x
xbxxand
x
xbxx
2
1
2
1
1
)logsin(
1
)logcos(.
Section 36.
Euler begins by writing the formula for integration by parts
PQPQQP .
He then takes )logcos( xbP and xxQ f 1 and gets
(36.1) )log()logcos()logcos(1 xbxbf
xxbxx
ff
33
A second integration by parts uses )log( sin xbP and xxQ f 1 and we get
(36.2) )logcos()logsin()logsin( 11 xbxxf
bxb
f
xxbxx f
ff
Now, starting with (36.1) we can replace the integral on the RHS with (36.2) to obtain
dxxbxf
bxb
f
x
f
bxb
f
xdxxbx f
fff )logcos()logsin()logcos()logcos(
1
0
11
0
1
which simplifies to
dxxbxf
bxb
f
bxxb
f
xdxxbx f
fff )logcos()logsin()logcos()logcos(
1
0
1
2
2
2
1
0
1 .
Solving for the integral we get
(36.3) ));logsin()logcos((cos1 xbbxbfbbff
xblxxx
ff
.
In a similar way, starting with (36.2) we get
(36.4) ))logcos()logsin(()logsin(1 xbbxbfbbff
xxbxx
ff
(Note that neither of the above integrals connect with the integrals in previous sections.
All of those had ax 21 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 and
Brouncker. I. Squaring the Circle, Notes and 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.
