Among the methods proposed for the numerical approximation of integrals, a prominent place is held by the rules which were developed by Newton and refined by Cotes . Specifically, if the value of the integral
∫
y
d
x
{\textstyle \displaystyle \int y\operatorname {d} x}
taken from
x
=
g
{\textstyle x=g}
to
x
=
h
{\textstyle x=h}
is required, then the values of
y
{\textstyle y}
for these limiting values of
x
{\textstyle x}
and for several other intermediate values progressing by equal increments from first to last, are to be multiplied by certain numerical coefficients. This being done, the sum of the products, multiplied by
h
−
g
,
{\textstyle h-g,}
will supply the desired integral, with greater precision as more terms are used in this operation. Since the principles of this method, which seems to be called into use less frequently by geometers than it should, have nowhere, as far as I know, been fully explained, it will not be out of place to say a few things about them.
Let us agree to use a multitude of
n
+
1
{\textstyle n+1}
terms, and let
h
−
g
=
Δ
,
{\textstyle h-g=\Delta ,}
so that the values of
x
{\textstyle x}
are
g
,
{\textstyle g,}
g
+
Δ
n
,
{\textstyle g+{\frac {\Delta }{n}},}
g
+
2
Δ
n
,
{\textstyle g+{\frac {2\Delta }{n}},}
g
+
3
Δ
n
{\textstyle g+{\frac {3\Delta }{n}}}
etc. up to
g
+
Δ
,
{\textstyle g+\Delta ,}
and correspondingly the values of
y
{\textstyle y}
are
A
,
{\textstyle A,}
A
′
,
{\textstyle A^{\prime },}
A
′
′
,
{\textstyle A^{\prime \prime },}
A
′
′
′
{\textstyle A^{\prime \prime \prime }}
etc. up to
A
(
n
)
:
{\textstyle A^{(n)}{:}}
finally, let
x
=
g
+
Δ
t
,
{\textstyle x=g+\Delta t,}
so that
y
{\textstyle y}
can also be regarded as a function of
t
.
{\textstyle t.}
Let
Y
{\textstyle {Y}}
represent the function
A
.
(
n
t
−
1
)
(
n
t
−
2
)
(
n
t
−
3
)
…
(
n
t
−
n
)
(
−
1
)
(
−
2
)
.
(
−
3
)
…
.
(
−
n
)
+
A
′
.
n
t
.
(
n
t
−
2
)
.
(
n
t
−
3
)
⋯
⋯
.
(
n
t
−
n
)
1.
(
−
1
)
(
−
2
)
⋯
⋯
(
1
−
n
)
+
A
′
′
.
n
t
.
(
n
t
−
1
)
.
(
n
t
−
3
)
…
…
(
n
t
−
n
)
2.1.
(
−
1
)
…
…
(
2
−
n
)
+
A
′
′
′
.
n
t
.
(
n
t
−
1
)
.
(
n
t
−
2
)
…
…
(
n
t
−
n
)
3.2.1
…
…
(
3
−
n
)
+
etc.
+
A
(
n
)
.
n
t
(
n
t
−
1
)
.
(
n
t
−
2
)
…
…
(
n
t
−
n
+
1
)
n
.
(
n
−
1
)
(
n
−
2
)
…
…
.1
{\displaystyle {\begin{aligned}&A.{\frac {(nt-1)(nt-2)(nt-3)\ldots (nt-n)}{(-1)(-2).(-3)\ldots .(-n)}}\\+&A^{\prime }.{\frac {nt.(nt-2).(nt-3)\cdots \cdots .(nt-n)}{1.(-1)(-2)\cdots \cdots (1-n)}}\\+&A^{\prime \prime }.{\frac {nt.(nt-1).(nt-3)\ldots \ldots (nt-n)}{2.1.(-1)\ldots \ldots (2-n)}}\\+&A^{\prime \prime \prime }.{\frac {nt.(nt-1).(nt-2)\ldots \ldots (nt-n)}{3.2.1\ldots \ldots (3-n)}}\\+&{\text{ etc. }}\\+&A^{(n)}.{\frac {nt(nt-1).(nt-2)\ldots \ldots (nt-n+1)}{n.(n-1)(n-2)\ldots \ldots .1}}\end{aligned}}}
or
Σ
A
(
μ
)
T
(
μ
)
M
(
μ
)
,
{\textstyle \Sigma {\frac {A^{(\mu )}T^{(\mu )}}{M^{(\mu )}}},}
where
μ
{\textstyle \mu }
represents each of the integers
0
{\textstyle 0}
,
1
{\textstyle 1}
,
2
{\textstyle 2}
,
3
…
n
,
{\textstyle 3\ldots n,}
T
(
μ
)
=
n
t
(
n
t
−
1
)
(
n
t
−
2
)
(
n
t
−
3
)
…
(
n
t
−
n
)
n
t
−
μ
,
{\textstyle T^{(\mu )}={\frac {nt(nt-1)(nt-2)(nt-3)\ldots (nt-n)}{nt-\mu }},}
and
M
(
μ
)
{\textstyle M^{(\mu )}}
is the value of
T
{\textstyle T}
for
n
t
=
μ
.
{\textstyle nt=\mu .}
It is clear that
Y
{\textstyle {Y}}
represents an integral algebraic function of
t
,
{\textstyle t,}
of order
n
,
{\textstyle n,}
and its values for each of the
n
+
1
{\textstyle n+1}
values of
t
,
{\textstyle t,}
namely
0
,
{\textstyle 0,}
1
n
,
{\textstyle {\frac {1}{n}},}
2
n
,
{\textstyle {\frac {2}{n}},}
3
n
…
1
{\textstyle {\frac {3}{n}}\ldots 1}
are equal to the values of
y
.
{\textstyle y.}
It is also clear that if
Y
′
{\textstyle Y^{\prime }}
is another integral function producing the same values of
y
{\textstyle y}
for the same values of
t
,
{\textstyle t,}
then
Y
′
−
Y
{\textstyle Y^{\prime }-Y}
will vanish for the same values, and therefore it must be divisible by the factors
t
,
{\textstyle t,}
t
−
1
n
,
{\textstyle t-{\frac {1}{n}},}
t
−
2
n
,
{\textstyle t-{\frac {2}{n}},}
t
−
3
n
…
t
−
1
{\textstyle t-{\frac {3}{n}}\ldots t-1}
and therefore also by their product (which is of order
n
+
1
{\textstyle n+1}
), from which it is clear that
Y
′
{\textstyle Y^{\prime }}
must, unless it is identical to
Y
,
{\textstyle Y,}
be of a higher order, meaning that
Y
{\textstyle Y}
is the only integral function among those not exceeding order
n
{\textstyle n}
which coincides with
y
{\textstyle y}
for those
n
+
1
{\textstyle n+1}
values. Therefore, if
y
{\textstyle y}
, when expanded into a series of powers of
t
{\textstyle t}
, breaks off before the term involving
t
n
+
1
{\textstyle t^{n+1}}
, it will be identical to
Y
,
{\textstyle Y,}
and if the series converges so quickly as to allow the subsequent terms to be neglected, then the function
Y
{\textstyle Y}
can replace
y
{\textstyle y}
within the limits
t
=
0
,
{\textstyle t=0,}
t
=
1
{\textstyle t=1}
or
x
=
g
,
{\textstyle x=g,}
x
=
h
{\textstyle x=h}
.
Now our integral
∫
y
d
x
{\textstyle \int y\operatorname {d} x}
is transformed into
Δ
∫
y
d
t
,
{\textstyle \Delta \int y\operatorname {d} t,}
taken from
t
=
0
{\textstyle t=0}
to
t
=
1
,
{\textstyle t=1,}
and as we have just indicated, we will replace this with
Δ
∫
Y
d
t
.
{\textstyle \Delta \int Y\operatorname {d} t.}
Thus by expanding
T
(
μ
)
{\textstyle T^{(\mu )}}
into
α
t
n
+
β
t
n
−
1
+
γ
t
n
−
2
+
δ
t
n
−
3
+
etc.
{\displaystyle \alpha t^{n}+\beta t^{n-1}+\gamma t^{n-2}+\delta t^{n-3}+{\text{etc. }}}
the integral
∫
T
(
μ
)
d
t
{\textstyle \int T^{(\mu )}{d}t}
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
will be
=
α
n
+
1
+
β
n
+
γ
n
−
1
+
δ
n
−
2
+
etc.
{\displaystyle ={\frac {\alpha }{n+1}}+{\frac {\beta }{n}}+{\frac {\gamma }{n-1}}+{\frac {\delta }{n-2}}+{\text{etc. }}}
and setting this quantity
=
M
(
μ
)
R
(
μ
)
,
{\textstyle =M^{(\mu )}R^{(\mu )},}
the desired integral will be
=
Δ
(
A
R
+
A
′
R
′
+
A
′
′
R
′
′
+
A
′
′
′
R
′
′
′
+
etc.
+
A
(
n
)
R
(
n
)
)
{\displaystyle =\Delta (AR+A^{\prime }R^{\prime }+A^{\prime \prime }R^{\prime \prime }+A^{\prime \prime \prime }R^{\prime \prime \prime }+{\text{etc.}}+A^{(n)}R^{(n)})}
For example, let us compute the coefficient
R
′
′
{\textstyle R^{\prime \prime }}
for
n
=
5.
{\textstyle n=5.}
Here we have
T
′
′
=
5
5
t
5
−
13.5
4
t
4
+
59.5
3
t
3
−
107.5
2
t
t
+
60.5.
t
M
′
′
=
2
×
1
×
(
−
1
)
×
(
−
2
)
×
(
−
3
)
=
−
12
{\displaystyle {\begin{aligned}&T^{\prime \prime }=5^{5}t^{5}-13.5^{4}t^{4}+59.5^{3}t^{3}-107.5^{2}tt+60.5.t\\&M^{\prime \prime }=2\times 1\times (-1)\times (-2)\times (-3)=-12\end{aligned}}}
Hence
−
12
R
′
′
=
3125
6
−
1625
+
7375
4
−
2675
3
+
150
=
−
25
12
,
{\textstyle -12R^{\prime \prime }={\frac {3125}{6}}-1625+{\frac {7375}{4}}-{\frac {2675}{3}}+150=-{\frac {25}{12}},}
and therefore
R
′
′
=
25
144
.
{\textstyle R^{\prime \prime }={\frac {25}{144}}.}
The computation can be shortened a bit by setting
2
t
−
1
=
u
.
{\textstyle 2t-1=u.}
Then we have
T
(
μ
)
=
(
n
u
+
n
)
(
n
u
+
n
−
2
)
(
n
u
+
n
−
4
)
…
(
n
u
−
n
+
4
)
(
n
u
−
n
+
2
)
(
n
u
−
n
)
2
n
(
n
u
+
n
−
2
μ
)
{\displaystyle T^{(\mu )}={\frac {(nu+n)(nu+n-2)(nu+n-4)\ldots (nu-n+4)(nu-n+2)(nu-n)}{2^{n}(nu+n-2\mu )}}}
Let us set
(
n
n
u
u
−
n
n
)
.
(
n
n
u
u
−
(
n
−
2
)
2
)
.
(
n
n
u
u
−
(
n
−
4
)
2
)
.
(
n
n
u
u
−
(
n
−
6
)
2
)
…
n
n
u
u
−
(
n
−
2
μ
)
2
=
U
(
μ
)
,
{\displaystyle {\frac {(nnuu-nn).(nnuu-(n-2)^{2}).(nnuu-(n-4)^{2}).(nnuu-(n-6)^{2})\ldots }{nnuu-(n-2\mu )^{2}}}=U^{(\mu )},}
where the numerator should end in
…
(
n
n
u
u
−
9
)
(
n
n
u
u
−
1
)
,
{\textstyle \ldots (nnuu-9)(nnuu-1),}
if
n
{\textstyle n}
is odd, or in
…
(
n
n
u
u
−
4
)
n
u
,
{\textstyle \ldots (nnuu-4)nu,}
if
n
{\textstyle n}
is even. Then
T
(
μ
)
=
(
n
u
−
n
+
2
μ
)
U
(
μ
)
2
n
.
{\displaystyle T^{(\mu )}={\frac {(nu-n+2\mu )U^{(\mu )}}{2^{n}}}.}
Now the integral
∫
T
(
μ
)
d
t
{\textstyle \int T^{(\mu )}\operatorname {d} t}
taken from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
is equal to the integral
∫
1
2
T
(
μ
)
d
u
=
∫
n
u
U
(
μ
)
d
u
2
n
+
1
+
∫
(
2
μ
−
n
)
U
(
μ
)
d
u
2
n
+
1
{\displaystyle \int {\tfrac {1}{2}}T^{(\mu )}\operatorname {d} u=\int {\frac {nuU^{(\mu )}\operatorname {d} u}{2^{n+1}}}+\int {\frac {(2\mu -n)U^{(\mu )}\operatorname {d} u}{2^{n+1}}}}
from
u
=
−
1
{\textstyle u=-1}
to
u
=
+
1.
{\textstyle u=+1.}
Therefore, by setting
U
(
μ
)
=
α
u
n
−
1
+
β
u
n
−
3
+
γ
u
n
−
5
+
δ
u
n
−
7
+
etc.
{\displaystyle U^{(\mu )}=\alpha u^{n-1}+\beta u^{n-3}+\gamma u^{n-5}+\delta u^{n-7}+{\text{etc. }}}
(it being evident that the powers
u
n
−
2
,
u
n
−
4
,
u
n
−
6
{\textstyle u^{n-2},u^{n-4},u^{n-6}}
etc. are absent), the first part
∫
n
u
U
(
n
)
d
u
2
n
+
1
{\textstyle \int {\frac {nuU^{(n)}\operatorname {d} u}{2^{n+1}}}}
of the integral will vanish for odd values of
n
,
{\textstyle n,}
while the other part
∫
(
2
μ
−
n
)
U
(
μ
)
d
u
2
n
+
1
{\textstyle \int {\frac {(2\mu -n)U^{(\mu )}\operatorname {d} u}{2^{n+1}}}}
will vanish for even values, so that the integral
∫
T
(
μ
)
d
t
{\textstyle \int T^{(\mu )}\operatorname {d} t}
becomes
=
n
2
n
(
α
n
+
1
+
β
n
−
1
+
γ
n
−
3
+
δ
n
−
5
+
etc.
)
{\displaystyle ={\frac {n}{2^{n}}}\left({\frac {\alpha }{n+1}}+{\frac {\beta }{n-1}}+{\frac {\gamma }{n-3}}+{\frac {\delta }{n-5}}+{\text{etc.}}\right)}
for even values of
n
,
{\textstyle n,}
and
=
2
μ
−
n
2
n
(
α
n
+
β
n
−
2
+
γ
n
−
4
+
δ
n
−
6
+
etc.
)
{\displaystyle ={\frac {2\mu -n}{2^{n}}}\left({\frac {\alpha }{n}}+{\frac {\beta }{n-2}}+{\frac {\gamma }{n-4}}+{\frac {\delta }{n-6}}+{\text{etc.}}\right)}
for odd values of
n
.
{\textstyle n.}
In our example we have
U
′
′
=
(
25
u
u
−
25
)
(
25
u
u
−
9
)
=
625
u
4
−
850
u
u
+
225
,
{\textstyle U^{\prime \prime }=(25uu-25)(25uu-9)=625u^{4}-850uu+225,}
hence
−
12
R
′
′
=
−
1
32
(
125
−
850
3
+
225
)
=
−
25
12
{\textstyle -12R^{\prime \prime }=-{\tfrac {1}{32}}(125-{\tfrac {850}{3}}+225)=-{\tfrac {25}{12}}}
as above.
It is worth noting that
U
(
n
−
μ
)
=
U
(
μ
)
,
{\textstyle U^{(n-\mu )}=U^{(\mu )},}
and therefore
∫
T
(
n
−
μ
)
d
t
=
±
∫
T
(
μ
)
d
t
,
{\textstyle \int T^{(n-\mu )}\operatorname {d} t=\pm \int T^{(\mu )}\operatorname {d} t,}
with the upper sign holding for even
n
,
{\textstyle n,}
and the lower sign for odd
n
.
{\textstyle n.}
Hence, since it is easy to see that
M
(
n
−
μ
)
=
±
M
(
μ
)
,
{\textstyle M^{(n-\mu )}=\pm M^{(\mu )},}
we will always have
R
(
n
−
μ
)
=
R
(
μ
)
,
{\textstyle R^{(n-\mu )}=R^{(\mu )},}
meaning that the last coefficient is equal to the first, the penultimate to the second, and so on.
We hereby append the numerical values of these coefficients, up to
n
=
10
,
{\textstyle n=10,}
which were computed by Cotes in Harmonia Mensurarum .
For
n
=
1
,
{\textstyle n=1,}
or two terms.
R
=
R
′
=
1
2
{\textstyle R=R^{\prime }={\tfrac {1}{2}}}
For
n
=
2
,
{\textstyle n=2,}
or three terms.
R
=
R
′
′
=
1
6
,
{\textstyle R=R^{\prime \prime }={\tfrac {1}{6}},}
R
′
=
2
3
{\textstyle R^{\prime }={\tfrac {2}{3}}}
For
n
=
3
,
{\textstyle n=3,}
or four terms.
R
=
R
′
′
′
=
1
8
,
{\textstyle R=R^{\prime \prime \prime }={\tfrac {1}{8}},}
R
′
=
R
′
′
=
3
8
{\textstyle R^{\prime }=R^{\prime \prime }={\tfrac {3}{8}}}
For
n
=
4
,
{\textstyle n=4,}
or five terms.
R
=
R
′
′
′
=
7
90
,
{\textstyle {R}={R}^{\prime \prime \prime }={\tfrac {7}{90}},}
R
′
=
R
′
′
′
=
16
45
,
{\textstyle R^{\prime }=R^{\prime \prime \prime }={\tfrac {16}{45}},}
R
′
′
=
2
15
{\textstyle R^{\prime \prime }={\tfrac {2}{15}}}
For
n
=
5
,
{\textstyle n=5,}
or six terms.
R
=
R
V
=
19
288
,
{\textstyle {R}={R}^{\scriptscriptstyle V}={\tfrac {19}{288}},}
R
′
=
R
′
′
′
′
=
25
9
,
{\textstyle {R}^{\prime }={R}^{\prime \prime \prime \prime }={\tfrac {25}{9}},}
R
′
′
=
R
′
′
′
=
25
144
{\textstyle {R}^{\prime \prime }={R}^{\prime \prime \prime }={\tfrac {25}{144}}}
For
n
=
6
,
{\textstyle n=6,}
or seven terms.
R
=
R
V
I
=
41
846
,
{\textstyle R=R^{\scriptscriptstyle VI}={\tfrac {41}{846}},}
R
′
=
R
V
=
9
35
,
{\textstyle R^{\prime }=R^{\scriptscriptstyle V}={\tfrac {9}{35}},}
R
′
′
=
R
I
V
=
9
280
,
{\textstyle R^{\prime \prime }=R^{\scriptscriptstyle IV}={\tfrac {9}{280}},}
R
′
′
′
=
34
105
{\textstyle R^{\prime \prime \prime }={\tfrac {34}{105}}}
For
n
=
7
,
{\textstyle n=7,}
or eight terms.
R
=
R
V
I
I
=
751
17280
,
{\textstyle R=R^{\scriptscriptstyle VII}={\tfrac {751}{17280}},}
R
′
=
R
V
I
=
3577
17280
,
{\textstyle R^{\prime }=R^{\scriptscriptstyle VI}={\tfrac {3577}{17280}},}
R
′
′
=
R
V
=
49
640
,
{\textstyle R^{\prime \prime }=R^{\scriptscriptstyle V}={\tfrac {49}{640}},}
R
′
′
′
=
R
′
′
′
′
=
2989
17280
{\textstyle R^{\prime \prime \prime }=R^{\prime \prime \prime \prime }={\tfrac {2989}{17280}}}
For
n
=
8
,
{\textstyle n=8,}
or nine terms.
R
=
R
V
I
I
I
=
989
28350
,
{\textstyle R=R^{\scriptscriptstyle VIII}={\tfrac {989}{28350}},}
R
′
=
R
V
I
I
=
2944
14175
,
{\textstyle R^{\prime }=R^{\scriptscriptstyle VII}={\tfrac {2944}{14175}},}
R
′
′
=
R
V
I
=
−
464
14175
,
{\textstyle R^{\prime \prime }=R^{\scriptscriptstyle VI}=-{\tfrac {464}{14175}},}
R
′
′
′
=
R
V
=
5248
14175
,
{\textstyle R^{\prime \prime \prime }=R^{\scriptscriptstyle V}={\tfrac {5248}{14175}},}
R
I
V
=
−
454
2835
{\textstyle R^{\scriptscriptstyle IV}=-{\tfrac {454}{2835}}}
For
n
=
9
,
{\textstyle n=9,}
or ten terms.
R
=
R
I
X
=
2857
89600
,
{\textstyle R=R^{\scriptscriptstyle IX}={\tfrac {2857}{89600}},}
R
′
=
R
V
I
I
I
=
15741
89600
,
{\textstyle R^{\prime }=R^{\scriptscriptstyle VIII}={\tfrac {15741}{89600}},}
R
′
′
=
R
V
I
I
=
27
2240
,
{\textstyle R^{\prime \prime }=R^{\scriptscriptstyle VII}={\tfrac {27}{2240}},}
R
′
′
′
=
R
V
I
=
1209
5600
,
{\textstyle R^{\prime \prime \prime }=R^{\scriptscriptstyle VI}={\tfrac {1209}{5600}},}
R
I
V
=
R
V
=
2889
44800
{\textstyle R^{\scriptscriptstyle IV}=R^{\scriptscriptstyle V}={\tfrac {2889}{44800}}}
For
n
=
10
,
{\textstyle n=10,}
or eleven terms.
R
=
R
X
=
16067
598752
,
{\textstyle R=R^{\scriptscriptstyle X}={\tfrac {16067}{598752}},}
R
′
=
R
I
X
=
26575
149688
,
{\textstyle R^{\prime }=R^{\scriptscriptstyle IX}={\tfrac {26575}{149688}},}
R
′
′
=
R
V
I
I
I
=
−
16175
199584
,
{\textstyle R^{\prime \prime }=R^{\scriptscriptstyle VIII}=-{\tfrac {16175}{199584}},}
R
′
′
′
=
R
V
I
I
=
5675
12474
,
{\textstyle R^{\prime \prime \prime }=R^{\scriptscriptstyle VII}={\tfrac {5675}{12474}},}
R
I
V
=
R
V
I
=
−
4825
11088
,
{\textstyle R^{\scriptscriptstyle IV}=R^{\scriptscriptstyle VI}=-{\tfrac {4825}{11088}},}
R
V
=
17807
24948
.
{\textstyle R^{\scriptscriptstyle V}={\tfrac {17807}{24948}}.}
Since the formula
Δ
(
A
R
+
A
′
R
′
+
A
′
R
′
+
A
′
′
R
′
′
+
etc.
+
A
(
n
)
R
(
n
)
)
{\textstyle \Delta (AR+A^{\prime }R^{\prime }+A^{\prime }R^{\prime }+A^{\prime \prime }R^{\prime \prime }+{\text{etc.}}+A^{(n)}R^{(n)})}
exactly represents the integral
∫
y
d
x
{\textstyle \int y\operatorname {d} x}
from
x
=
g
{\textstyle x=g}
to
x
=
g
+
Δ
,
{\textstyle x=g+\Delta ,}
or the integral
Δ
∫
y
d
t
{\textstyle \Delta \int y\operatorname {d} t}
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
whenever the expansion of
y
{\textstyle y}
into a series does not go beyond the power
t
n
{\textstyle t^{n}}
, but otherwise only approximates it, it remains to show how to account for the error induced by the immediately following terms. Let us denote generally by
k
(
m
)
{\textstyle k^{(m)}}
the difference between the true value of the integral
∫
t
m
d
t
{\textstyle \int t^{m}\mathrm {dt} }
from
t
=
0
{\textstyle t=0}
to
t
=
1
,
{\textstyle t=1,}
and the value derived from the formula. Then
k
=
1
−
R
−
R
′
−
R
′
′
−
R
′
′
′
−
etc.
−
R
(
n
)
k
′
=
1
2
−
1
n
(
R
′
+
2
R
′
′
+
3
R
′
′
′
+
etc.
+
n
R
(
n
)
)
k
′
′
=
1
3
−
1
n
n
(
R
′
+
4
R
′
′
+
9
R
′
′
′
+
etc.
+
n
n
R
(
n
)
)
k
′
′
′
=
1
4
−
1
n
2
(
R
′
+
8
R
′
′
+
27
R
′
′
′
+
etc.
+
n
3
R
(
n
)
)
{\displaystyle {\begin{aligned}&k=1-R-R^{\prime }-R^{\prime \prime }-R^{\prime \prime \prime }-{\text{etc.}}-R^{(n)}\\&k^{\prime }={\tfrac {1}{2}}-{\tfrac {1}{n}}(R^{\prime }+2R^{\prime \prime }+3R^{\prime \prime \prime }+{\text{etc.}}+nR^{(n)})\\&k^{\prime \prime }={\tfrac {1}{3}}-{\tfrac {1}{nn}}(R^{\prime }+4R^{\prime \prime }+9R^{\prime \prime \prime }+{\text{etc.}}+nnR^{(n)})\\&k^{\prime \prime \prime }={\tfrac {1}{4}}-{\tfrac {1}{n^{2}}}(R^{\prime }+8R^{\prime \prime }+27R^{\prime \prime \prime }+{\text{etc.}}+n^{3}R^{(n)})\end{aligned}}}
etc. It is evident, therefore, that if
y
{\textstyle y}
is expanded into a series
K
+
K
′
t
+
K
′
′
t
t
+
K
′
′
′
t
3
+
etc.
{\displaystyle K+K^{\prime }t+K^{\prime \prime }tt+K^{\prime \prime \prime }t^{3}+{\text{etc. }}}
then the difference between the true value of the integral
∫
y
d
t
{\textstyle \int y\operatorname {d} t}
and the approximated value derived from the formula can be expressed as
K
k
+
K
′
k
′
+
K
′
′
k
′
′
+
K
′
′
′
k
′
′
′
+
etc.
{\displaystyle Kk+K^{\prime }k^{\prime }+K^{\prime \prime }k^{\prime \prime }+K^{\prime \prime \prime }k^{\prime \prime \prime }+{\text{etc.}}}
But evidently,
k
,
{\textstyle k,}
k
′
,
{\textstyle k^{\prime },}
k
′
′
{\textstyle k^{\prime \prime }}
, etc. up to
k
(
n
)
{\textstyle k^{(n)}}
are all automatically
=
0
:
{\textstyle =0{:}}
thus, the correction of the approximated formula will be
K
(
n
+
1
)
k
(
n
+
1
)
+
K
(
n
+
2
)
k
(
n
+
2
)
+
K
(
n
+
3
)
k
(
n
+
3
)
+
etc.
{\displaystyle K^{(n+1)}{k}^{(n+1)}+{K}^{(n+2)}{k}^{(n+2)}+{K}^{(n+3)}{k}^{(n+3)}+{\text{etc.}}}
The nature of quantities
k
(
n
+
1
,
{\textstyle k^{(n+1},}
k
(
n
+
2
)
{\textstyle k^{(n+2)}}
, etc. will be examined more accurately later; here, it suffices to provide the numerical values of the first or second, for each value of
n
,
{\textstyle n,}
so that the degree of precision afforded by the approximate formula can be estimated.
For
n
=
0
1
{\textstyle n={\phantom {0}}1}
, we have
k
′
′
=
−
1
6
,
{\textstyle k^{\prime \prime }=-{\frac {1}{6}},}
k
′
′
′
=
−
1
4
,
{\textstyle k^{\prime \prime \prime }=-{\frac {1}{4}},}
k
′
′
′
′
=
−
3
10
{\textstyle k^{\prime \prime \prime \prime }=-{\frac {3}{10}}}
For
n
=
0
2
{\textstyle n={\phantom {0}}2}
, we find
k
′
′
′
=
0
,
{\textstyle k^{\prime \prime \prime }=0,}
k
′
′
′
′
=
−
1
120
,
{\textstyle k^{\prime \prime \prime \prime }=-{\frac {1}{120}},}
k
V
=
−
1
48
{\textstyle k^{\scriptscriptstyle V}=-{\frac {1}{48}}}
For
n
=
0
3
{\textstyle n={\phantom {0}}3}
, it is
k
′
′
′
′
=
−
1
270
,
{\textstyle k^{\prime \prime \prime \prime }=-{\frac {1}{270}},}
k
V
=
−
1
108
{\textstyle \quad k^{\scriptscriptstyle V}=-{\frac {1}{108}}}
For
n
=
0
4
…
k
V
=
0
,
{\textstyle n={\phantom {0}}4\ldots k^{\scriptscriptstyle V}=0,}
k
V
I
=
−
1
2688
,
{\textstyle k^{\scriptscriptstyle VI}=-{\frac {1}{2688}},}
k
V
I
I
=
−
1
768
{\textstyle k^{\scriptscriptstyle VII}=-{\frac {1}{768}}}
For
n
=
0
5
…
k
V
I
=
−
1
52
?
00
,
{\textstyle n={\phantom {0}}5\ldots k^{\scriptscriptstyle VI}=-{\frac {1}{52?00}},}
k
V
I
I
=
−
11
15000
{\textstyle k^{\scriptscriptstyle VII}=-{\frac {11}{15000}}}
For
n
=
0
6
…
k
V
I
I
=
0
,
{\textstyle n={\phantom {0}}6\ldots k^{\scriptscriptstyle VII}=0,}
k
V
I
I
I
=
−
1
38880
,
{\textstyle k^{\scriptscriptstyle VIII}=-{\frac {1}{38880}},}
k
I
X
=
−
1
8640
{\textstyle k^{\scriptscriptstyle IX}=-{\frac {1}{8640}}}
For
n
=
0
7
…
k
V
I
I
I
=
−
167
10588410
,
{\textstyle n={\phantom {0}}7\ldots k^{\scriptscriptstyle VIII}=-{\frac {167}{10588410}},}
k
I
X
=
−
167
2352980
{\textstyle k^{\scriptscriptstyle IX}=-{\frac {167}{2352980}}}
For
n
=
0
8
…
k
I
X
=
0
,
{\textstyle n={\phantom {0}}8\ldots k^{\scriptscriptstyle IX}=0,}
k
X
=
−
37
173
?
1504
,
{\textstyle k^{\scriptscriptstyle X}=-{\frac {37}{173?1504}},}
k
X
I
=
−
37
3145728
{\textstyle k^{\scriptscriptstyle XI}=-{\frac {37}{3145728}}}
For
n
=
0
9
…
k
X
=
−
865
631351908
,
{\textstyle n={\phantom {0}}9\ldots k^{\scriptscriptstyle X}=-{\frac {865}{631351908}},}
k
X
I
=
−
865
114791256
{\textstyle k^{\scriptscriptstyle XI}=-{\frac {865}{114791256}}}
For
n
=
10
…
k
X
I
=
0
,
{\textstyle n=10\ldots k^{\scriptscriptstyle XI}=0,}
k
X
I
I
=
−
26927
136500000000
,
{\textstyle k^{\scriptscriptstyle XII}=-{\frac {26927}{136500000000}},}
k
X
I
I
=
−
26927
21000000000
{\textstyle k^{\scriptscriptstyle XII}=-{\frac {26927}{21000000000}}}
For all even values of
n
{\textstyle n}
here, we observe
k
(
n
+
1
)
=
0
,
{\textstyle k^{(n+1)}=0,}
and furthermore
k
(
n
+
3
)
=
n
+
3
2
k
(
n
+
2
)
;
{\textstyle k^{(n+3)}={\frac {n+3}{2}}k^{(n+2)};}
for odd values of
n
{\textstyle n}
, however, we always have
k
(
n
+
2
)
=
n
+
2
2
k
(
n
+
1
)
.
{\textstyle k^{(n+2)}={\frac {n+2}{2}}k^{(n+1)}.}
The reason why this occurs can be deduced easily from the following considerations.
In general, let
l
(
m
)
{\textstyle l^{(m)}}
denote the difference between the true value of the integral
∫
(
t
−
1
2
)
m
d
t
{\textstyle \int (t-{\frac {1}{2}})^{m}\operatorname {d} t}
from
t
=
0
{\textstyle t=0}
to
t
=
1
,
{\textstyle t=1,}
and the value derived from the approximate formula, so that we have
l
(
m
)
=
∫
(
t
−
1
2
)
m
d
t
−
[
(
−
1
2
)
m
R
+
(
1
n
−
1
2
)
m
R
′
+
(
2
n
−
1
2
)
m
R
′
′
+
(
3
n
−
1
2
)
m
R
′
′
′
+
etc.
+
(
1
2
−
1
n
)
m
R
(
n
−
1
)
+
(
1
2
)
m
R
(
n
)
]
{\displaystyle {\begin{array}{c}l^{(m)}=\int (t-{\frac {1}{2}})^{m}\operatorname {d} t-\left[(-{\frac {1}{2}})^{m}R+({\frac {1}{n}}-{\frac {1}{2}})^{m}R^{\prime }+({\frac {2}{n}}-{\frac {1}{2}})^{m}R^{\prime \prime }+({\frac {3}{n}}-{\frac {1}{2}})^{m}R^{\prime \prime \prime }+\right.{\text{etc.}}\\\left.+({\frac {1}{2}}-{\frac {1}{n}})^{m}R^{(n-1)}+({\frac {1}{2}})^{m}R^{(n)}\right]\end{array}}}
with the integral being taken from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
. Clearly, for odd values of
m
{\textstyle m}
both the true and approximate integral values vanish: hence
l
′
=
0
,
{\textstyle l^{\prime }=0,}
l
′
′
′
=
0
,
l
V
=
0
,
l
V
I
I
=
0
{\textstyle l^{\prime \prime \prime }=0,l^{\scriptscriptstyle V}=0,l^{\scriptscriptstyle VII}=0}
, etc., and generally
l
(
m
)
=
0
{\textstyle l^{(m)}=0}
for all odd values of
m
.
{\textstyle m.}
For even values of
m
,
{\textstyle m,}
on the other hand, the formula can be written as
l
(
m
)
=
1
2
m
(
m
+
1
)
−
2
n
m
(
(
1
2
n
)
m
R
+
(
1
2
n
−
1
)
m
R
′
+
(
1
2
n
−
2
)
m
R
′
′
+
etc.
+
2
m
R
(
1
2
n
−
2
)
+
R
(
1
2
n
−
1
)
)
{\displaystyle {\begin{array}{c}l^{(m)}={\frac {1}{2^{m}(m+1)}}-{\frac {2}{n^{m}}}(({\frac {1}{2}}n)^{m}R+({\frac {1}{2}}n-1)^{m}R^{\prime }+({\frac {1}{2}}n-2)^{m}R^{\prime \prime }+{\text{etc. }}\\+2^{m}R^{\left({\frac {1}{2}}n-2\right)}+R^{\left({\frac {1}{2}}n-1\right)})\end{array}}}
if
n
{\textstyle n}
is even; or
l
(
m
)
=
1
2
n
(
1
m
+
1
−
2
n
n
(
n
m
+
(
n
−
2
)
m
R
′
+
(
n
−
4
)
m
R
′
′
+
etc.
+
3
m
R
(
1
2
n
−
3
2
)
+
R
(
1
2
n
−
1
2
)
)
)
{\displaystyle {\begin{aligned}l^{(m)}={\tfrac {1}{2^{n}}}({\tfrac {1}{m+1}}-{\tfrac {2}{n^{n}}}(n^{m}&+(n-2)^{m}R^{\prime }+(n-4)^{m}R^{\prime \prime }+{\text{etc. }}\\&+3^{m}R^{\left({\frac {1}{2}}n-{\frac {3}{2}}\right)}+R^{\left({\frac {1}{2}}n-{\frac {1}{2}}\right)}))\end{aligned}}}
if
n
{\textstyle n}
is odd.
Therefore, if expanding
y
{\textstyle y}
in a series according to powers of
t
−
1
2
{\textstyle t-{\tfrac {1}{2}}}
yields
y
=
L
+
L
′
(
t
−
1
2
)
+
L
′
′
(
t
−
1
2
)
2
+
L
′
′
′
(
t
−
1
2
)
3
+
etc.
{\displaystyle y=L+L^{\prime }(t-{\tfrac {1}{2}})+L^{\prime \prime }(t-{\tfrac {1}{2}})^{2}+L^{\prime \prime \prime }(t-{\tfrac {1}{2}})^{3}+{\text{etc. }}}
then the correction to be applied to the approximate value of the integral
∫
y
d
t
{\textstyle \int y\operatorname {d} t}
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
will be
L
l
+
L
′
′
l
′
′
+
L
′
′
′
l
′
′
′
+
L
V
I
l
V
I
+
etc.
{\displaystyle Ll+L^{\prime \prime }l^{\prime \prime }+L^{\prime \prime \prime }l^{\prime \prime \prime }+L^{\scriptscriptstyle VI}l^{\scriptscriptstyle VI}+{\text{etc. }}}
or rather, since
l
(
m
)
{\textstyle l^{(m)}}
necessarily vanishes for any integer value of
m
{\textstyle m}
no greater than
n
,
{\textstyle n,}
the correction will be
L
(
n
+
2
)
l
(
n
+
2
)
+
L
(
n
+
4
)
l
(
n
+
4
)
+
L
(
n
+
6
)
l
(
n
+
6
)
+
etc.
{\displaystyle L^{(n+2)}l^{(n+2)}+L^{(n+4)}l^{(n+4)}+L^{(n+6)}l^{(n+6)}+{\text{etc. }}}
for even
n
,
{\textstyle n,}
or
L
(
n
+
1
)
l
(
n
+
1
)
+
L
(
n
+
3
)
l
(
n
+
3
)
+
L
(
n
+
5
)
l
(
n
+
5
)
+
etc.
{\displaystyle L^{(n+1)}l^{(n+1)}+L^{(n+3)}l^{(n+3)}+L^{(n+5)}l^{(n+5)}+{\text{etc. }}}
for odd
n
.
{\textstyle n.}
The corrections
l
(
m
)
{\textstyle l^{(m)}}
can be easily converted to
k
(
m
)
{\textstyle k^{(m)}}
and vice versa. For if we have
(
t
−
1
2
)
m
=
t
m
−
1
2
m
.
t
m
−
1
+
1
4
⋅
m
(
m
−
1
)
1.2
t
m
−
2
+
etc.
{\displaystyle (t-{\tfrac {1}{2}})^{m}=t^{m}-{\tfrac {1}{2}}m.t^{m-1}+{\tfrac {1}{4}}\cdot {\tfrac {m(m-1)}{1.2}}t^{m-2}+{\text{etc. }}}
then
l
(
m
)
=
k
(
m
)
−
1
2
m
k
(
m
−
1
)
+
1
4
⋅
m
(
m
−
1
)
1.2
k
(
m
−
2
)
+
etc.
{\displaystyle l^{(m)}=k^{(m)}-{\tfrac {1}{2}}mk^{(m-1)}+{\tfrac {1}{4}}\cdot {\tfrac {m(m-1)}{1.2}}k^{(m-2)}+{\text{etc. }}}
And similarly,
k
(
m
)
=
l
(
m
)
+
1
2
m
l
(
m
−
1
)
+
1
4
⋅
m
(
m
−
1
)
1.2
l
(
m
−
1
)
+
etc.
{\displaystyle k^{(m)}=l^{(m)}+{\tfrac {1}{2}}ml^{(m-1)}+{\tfrac {1}{4}}\cdot {\tfrac {m(m-1)}{1.2}}l^{(m-1)}+{\text{etc. }}}
The terms where
l
{\textstyle l}
is affected by an odd index will be eliminated from the latter formula, and each should only be continued up to the index
n
+
1
{\textstyle n+1}
(inclusive). Therefore, it is clear that we will have
from which the above observations can be deduced.
Generally speaking, it will therefore be preferable to assign an even value to
n
,
{\textstyle n,}
or to employ an odd number of terms, when applying the method of Cotes . Indeed, very little precision will be gained by ascending from an even value of
n
{\textstyle n}
to the next highest odd one, as the error remains of the same order, although affected by a slightly smaller coefficient. Conversely, ascending from an odd value of
n
{\textstyle n}
to the next highest even one will increase the order of the error by two, and the coefficient being significantly reduced, so the precision will increase. So if five terms are used, that is, for
n
=
4
,
{\textstyle n=4,}
the error is approximately expressed by
−
1
2688
K
6
{\textstyle -{\frac {1}{2688}}K^{6}}
or
−
1
2688
L
6
;
{\textstyle -{\frac {1}{2688}}L^{6};}
if we set
n
=
5
,
{\textstyle n=5,}
the error will be approximately
−
11
52500
K
6
{\textstyle -{\frac {11}{52500}}K^{6}}
or
−
11
52500
L
6
,
{\textstyle -{\frac {11}{52500}}L^{6},}
thus it will not even be half of the former: on the other hand, for
n
=
6
,
{\textstyle n=6,}
the error becomes approximately
=
−
1
38880
K
8
{\textstyle =-{\frac {1}{38880}}K^{8}}
or
=
−
1
38880
L
8
,
{\textstyle =-{\frac {1}{38880}}L^{8},}
and the precision is increased all the more, as the series into which the function has been expanded converges more quickly.
Following these preliminaries regarding the method of Cotes , we proceed to a general inquiry, discarding the condition that the values of
x
{\textstyle x}
progress in an arithmetic progression. We thus address the problem of determining the value of the integral
∫
y
d
x
{\textstyle \int y\operatorname {d} x}
between given limits from some given values of
y
,
{\textstyle y,}
either exactly or as closely as possible. Let us assume that the integral is to be taken from
x
=
g
{\textstyle x=g}
to
x
=
g
+
Δ
,
{\textstyle x=g+\Delta ,}
and let us introduce another variable
t
=
x
−
g
Δ
,
{\textstyle t={\frac {x-g}{\Delta }},}
so that the integral
Δ
∫
y
d
t
{\textstyle \Delta \int y\operatorname {d} t}
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
needs to be investigated.
Let
a
,
{\textstyle a,}
a
′
,
{\textstyle a^{\prime },}
a
′
′
,
{\textstyle a^{\prime \prime },}
a
′
′
′
…
a
(
n
)
{\textstyle a^{\prime \prime \prime }\ldots a^{(n)}}
be distinct values of
t
,
{\textstyle t,}
let the
n
+
1
{\textstyle n+1}
corresponding values of
y
{\textstyle y}
be
A
,
{\textstyle A,}
A
′
,
{\textstyle A^{\prime },}
A
′
′
,
{\textstyle A^{\prime \prime },}
A
′
′
′
…
A
(
n
)
,
{\textstyle A^{\prime \prime \prime }\ldots A^{(n)},}
and let
Y
{\textstyle Y}
denote the following integral algebraic function of order
n
:
{\textstyle n{:}}
A
′
′
(
t
−
a
′
)
(
t
−
a
′
′
)
(
t
−
a
′
′
′
)
…
(
t
−
a
(
n
)
)
(
a
−
a
′
)
(
a
−
a
′
′
)
(
a
−
a
′
′
′
)
…
(
a
−
a
(
n
)
)
+
A
′
′
(
t
−
a
)
(
t
−
a
′
′
)
(
t
−
a
′
′
′
)
…
(
t
−
a
(
n
)
)
(
a
′
−
a
)
(
a
′
−
a
′
′
)
(
a
′
−
a
′
′
′
)
…
(
a
′
−
a
(
n
)
)
+
A
′
′
(
t
−
a
)
(
t
−
a
′
)
(
t
−
a
′
′
′
)
…
(
t
−
a
(
n
)
)
(
a
′
′
−
a
)
(
a
′
′
−
a
′
)
(
a
′
′
−
a
′
′
)
…
(
a
′
′
−
a
(
n
)
)
+
etc.
+
A
(
n
)
(
t
−
a
)
(
t
−
a
′
)
(
t
−
a
′
′
)
…
(
t
−
a
(
n
−
1
)
)
(
a
(
n
)
−
a
)
(
a
(
n
)
−
a
′
)
(
a
(
n
)
−
a
′
′
)
…
(
a
(
n
)
−
a
(
n
−
1
)
)
{\displaystyle {\begin{aligned}&A^{\phantom {\prime \prime }}{\frac {(t-a^{\prime })(t-a^{\prime \prime })(t-a^{\prime \prime \prime })\ldots (t-a^{(n)})}{(a-a^{\prime })(a-a^{\prime \prime })(a-a^{\prime \prime \prime })\ldots (a-a^{(n)})}}\\+&A^{\prime {\phantom {\prime }}}\,{\frac {(t-a)(t-a^{\prime \prime })(t-a^{\prime \prime \prime })\ldots (t-a^{(n)})}{(a^{\prime }-a)(a^{\prime }-a^{\prime \prime })(a^{\prime }-a^{\prime \prime \prime })\ldots (a^{\prime }-a^{(n)})}}\\+&A^{\prime \prime }{\frac {(t-a)(t-a^{\prime })(t-a^{\prime \prime \prime })\ldots (t-a^{(n)})}{(a^{\prime \prime }-a)(a^{\prime \prime }-a^{\prime })(a^{\prime \prime }-a^{\prime \prime })\ldots (a^{\prime \prime }-a^{(n)})}}\\+&{\text{ etc. }}\\+&A^{(n)}{\frac {(t-a)(t-a^{\prime })(t-a^{\prime \prime })\ldots (t-a^{(n-1)})}{(a^{(n)}-a)(a^{(n)}-a^{\prime })(a^{(n)}-a^{\prime \prime })\ldots (a^{(n)}-a^{(n-1)})}}\end{aligned}}}
If
t
{\textstyle t}
is set equal to any of the quantities
a
,
{\textstyle a,}
a
′
,
{\textstyle a^{\prime },}
a
′
′
,
{\textstyle a^{\prime \prime },}
a
′
′
′
…
a
(
m
)
{\textstyle a^{\prime \prime \prime }\ldots a^{(m)}}
, it is clear that the values of this function, coincide with the corresponding values of the function
y
,
{\textstyle y,}
from which, as concluded in art. 2, we deduce that
Y
{\textstyle Y}
is identical to
y
{\textstyle y}
, provided that
y
{\textstyle y}
is also an integral algebraic function of order no greater than
n
,
{\textstyle n,}
or at least it can take the place of
y
,
{\textstyle y,}
if
y
{\textstyle y}
can be converted into a series of powers of
t
{\textstyle t}
which exhibits such convergence that it is permissible to neglect the higher order terms.
To evaluate the integral
∫
Y
d
t
,
{\textstyle \int {Y}\operatorname {d} t,}
let us consider each part of
Y
{\textstyle {Y}}
separately. Let
T
{\textstyle T}
denote the product
(
t
−
a
)
(
t
−
a
′
)
(
t
−
a
′
′
)
(
t
−
a
′
′
′
)
…
(
t
−
a
(
n
)
)
,
{\displaystyle (t-a)(t-a^{\prime })(t-a^{\prime \prime })(t-a^{\prime \prime \prime })\ldots (t-a^{(n)}),}
and through the expansion of this product, let
T
=
t
n
+
1
+
α
t
n
+
α
′
t
n
−
1
+
α
′
′
t
n
−
2
+
etc.
+
α
(
n
)
{\displaystyle T=t^{n+1}+\alpha t^{n}+\alpha ^{\prime }t^{n-1}+\alpha ^{\prime \prime }t^{n-2}+{\text{etc.}}+\alpha ^{(n)}}
The numerator of the fraction by which
A
{\textstyle A}
is multiplied in its respective part of
Y
,
{\textstyle {Y},}
becomes
=
T
t
−
a
;
{\textstyle ={\frac {T}{t-a}};}
the numerators in the subsequent parts are likewise
T
t
−
a
′
,
{\textstyle {\frac {T}{t-a^{\prime }}},}
T
t
−
a
′
′
,
{\textstyle {\frac {T}{t-a^{\prime \prime }}},}
T
t
−
a
′
′
′
{\textstyle {\frac {T}{t-a^{\prime \prime \prime }}}}
, etc. The denominators are nothing but the values determined by these numerators if
t
{\textstyle t}
is set respectively to
a
,
{\textstyle a,}
a
′
,
{\textstyle a^{\prime },}
a
′
′
,
{\textstyle a^{\prime \prime },}
a
′
′
′
{\textstyle a^{\prime \prime \prime }}
, etc. Let us denote these denominators respectively by
M
,
{\textstyle M,}
M
′
,
{\textstyle M^{\prime },}
M
′
′
,
{\textstyle M^{\prime \prime },}
M
′
′
′
{\textstyle M^{\prime \prime \prime }}
, etc., so that we have
Y
=
A
T
M
(
t
−
a
)
+
A
′
T
M
′
(
t
−
a
′
)
+
A
′
′
T
M
′
′
(
t
−
a
′
′
)
+
etc.
+
A
(
n
)
T
M
(
n
)
(
t
−
a
(
n
)
)
{\displaystyle Y={\frac {AT}{M(t-a)}}+{\frac {A^{\prime }T}{M^{\prime }(t-a^{\prime })}}+{\frac {A^{\prime \prime }T}{M^{\prime \prime }(t-a^{\prime \prime })}}+{\text{etc.}}+{\frac {A^{(n)}T}{M^{(n)}(t-a^{(n)})}}}
When
T
=
0
,
{\textstyle T=0,}
for
t
=
a
,
{\textstyle t=a,}
we have the identical equation
a
n
+
1
+
α
a
n
+
α
′
a
n
−
1
+
α
′
′
a
n
−
2
+
etc.
+
α
(
n
)
=
0
{\displaystyle a^{n+1}+\alpha a^{n}+\alpha ^{\prime }a^{n-1}+\alpha ^{\prime \prime }a^{n-2}+{\text{etc.}}+\alpha ^{(n)}=0}
and therefore
T
=
t
n
+
1
−
a
n
+
1
+
α
(
t
n
−
a
n
)
+
α
′
(
t
n
−
1
−
a
n
−
1
)
+
α
′
′
(
t
n
−
2
−
a
n
−
2
)
+
etc.
+
α
(
n
−
1
)
(
t
−
a
)
{\displaystyle T=t^{n+1}-a^{n+1}+\alpha (t^{n}-a^{n})+\alpha ^{\prime }(t^{n-1}-a^{n-1})+\alpha ^{\prime \prime }(t^{n-2}-a^{n-2})+{\text{etc.}}+\alpha ^{(n-1)}(t-a)}
Thus, dividing by
t
−
a
{\textstyle t-a}
, we get
T
t
−
a
=
t
n
+
a
t
n
−
1
+
a
a
t
n
−
2
+
a
3
t
n
−
3
+
etc.
+
a
n
+
α
t
n
−
1
+
α
a
t
(
n
−
2
)
+
α
a
a
t
n
−
3
+
etc.
+
α
a
n
−
1
+
α
′
t
n
−
2
+
α
′
a
t
n
−
3
+
etc.
+
α
′
a
n
−
2
+
α
′
′
t
n
−
3
−
+
etc.
+
α
′
′
a
n
−
3
+
etc.
etc.
+
α
(
n
−
1
)
{\displaystyle {\begin{alignedat}{6}{\frac {T}{t-a}}=t^{n}&+at^{n-1}&&+aat^{n-2}&&+a^{3}t^{n-3}&&+{\text{etc.}}&&+a^{n}\\&+\alpha t^{n-1}&&+\alpha at^{(n-2)}&&+\alpha aat^{n-3}&&+{\text{etc.}}&&+\alpha a^{n-1}\\&&&+\alpha ^{\prime }t^{n-2}&&+\alpha ^{\prime }at^{n-3}&&+{\text{etc.}}&&+\alpha ^{\prime }a^{n-2}\\&&&&&+\alpha ^{\prime \prime }t^{n-3^{-}}&&+{\text{etc.}}&&+\alpha ^{\prime \prime }a^{n-3}\\&&&&&&&+{\text{etc.}}&&{\text{ etc. }}\\&&&&&&&&&+\alpha ^{(n-1)}\end{alignedat}}}
The value of this function for
t
=
a
{\textstyle t={a}}
is obtained as
=
(
n
+
1
)
a
n
+
n
α
a
n
−
1
+
(
n
−
1
)
α
′
a
n
−
2
+
(
n
−
2
)
α
′
′
a
n
−
3
+
etc.
+
α
(
n
−
1
)
{\displaystyle =(n+1)a^{n}+n\alpha a^{n-1}+(n-1)\alpha ^{\prime }a^{n-2}+(n-2)\alpha ^{\prime \prime }a^{n-3}+{\text{etc.}}+\alpha ^{(n-1)}}
Hence
M
{\textstyle M}
is equal to the value of
d
T
d
t
{\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}}
for
t
=
a
,
{\textstyle t=a,}
as is evident for other reasons. Similarly,
M
′
,
{\textstyle M^{\prime },}
M
′
′
,
{\textstyle M^{\prime \prime },}
M
′
′
′
,
{\textstyle M^{\prime \prime \prime },}
etc. will be the values of
d
T
d
t
{\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}}
for
t
=
a
′
,
{\textstyle t=a^{\prime },}
t
=
a
′
′
,
{\textstyle t=a^{\prime \prime },}
t
=
a
′
′
′
{\textstyle t=a^{\prime \prime \prime }}
, etc.
Furthermore, we find the value of the integral
∫
T
d
t
t
−
a
{\textstyle \int {\frac {T\operatorname {d} t}{t-a}}}
from
t
=
0
{\textstyle t=0}
to
t
=
1
,
{\textstyle t=1,}
to be:
=
1
n
+
1
+
a
n
+
a
a
n
−
1
+
a
3
n
−
2
+
etc.
+
a
n
+
α
n
+
α
a
n
−
1
+
α
a
a
n
−
2
+
etc.
+
α
a
n
−
1
+
α
′
n
−
1
+
α
′
a
n
−
2
+
etc.
+
α
′
a
n
−
2
+
α
′
′
n
−
2
+
etc.
+
α
′
′
a
n
−
3
+
etc.
etc.
+
α
(
n
−
1
)
{\displaystyle {\begin{alignedat}{6}={\tfrac {1}{n+1}}&+{\tfrac {a}{n}}&&+{\tfrac {aa}{n-1}}&&+{\tfrac {a^{3}}{n-2}}&&+{\text{etc.}}&&+a^{n}\\&+{\tfrac {\alpha }{n}}&&+{\tfrac {\alpha a}{n-1}}&&+{\tfrac {\alpha aa}{n-2}}&&+{\text{etc.}}&&+\alpha a^{n-1}\\&&&+{\tfrac {\alpha ^{\prime }}{n-1}}&&+{\tfrac {\alpha ^{\prime }a}{n-2}}&&+{\text{etc.}}&&+\alpha ^{\prime }a^{n-2}\\&&&&&+{\tfrac {\alpha ^{\prime \prime }}{n-2}}&&+{\text{etc.}}&&+\alpha ^{\prime \prime }a^{n-3}\\&&&&&&&+{\text{etc.}}&&{\text{ etc.}}\\&&&&&&&&&+\alpha ^{(n-1)}\end{alignedat}}}
Let us arrange these terms in the following order:
a
n
+
α
a
n
−
1
+
α
′
a
n
−
2
+
α
′
′
a
n
−
3
+
etc.
+
α
(
n
−
1
)
+
1
2
(
a
n
−
1
+
α
a
n
−
2
+
α
′
a
n
−
3
+
etc.
+
α
(
n
−
2
)
)
+
1
3
(
a
n
−
2
+
α
a
n
−
3
+
α
′
a
n
−
4
+
etc.
+
α
(
n
−
3
)
)
+
1
4
(
a
n
−
3
+
α
a
n
−
4
+
α
′
a
n
−
5
+
etc.
+
α
(
n
−
4
)
)
+
etc.
+
1
n
−
1
(
a
a
+
α
a
+
α
′
)
+
1
n
(
a
+
α
)
+
1
n
+
1
{\displaystyle {\begin{aligned}a^{n}&+\alpha a^{n-1}+\alpha ^{\prime }a^{n-2}+\alpha ^{\prime \prime }a^{n-3}+{\text{etc.}}+\alpha ^{(n-1)}\\&+{\tfrac {1}{2}}(a^{n-1}+\alpha a^{n-2}+\alpha ^{\prime }a^{n-3}+{\text{etc.}}+\alpha ^{(n-2)})\\&+{\tfrac {1}{3}}(a^{n-2}+\alpha a^{n-3}+\alpha ^{\prime }a^{n-4}+{\text{etc.}}+\alpha ^{(n-3)})\\&+{\tfrac {1}{4}}(a^{n-3}+\alpha a^{n-4}+\alpha ^{\prime }a^{n-5}+{\text{etc.}}+\alpha ^{(n-4)})\\&+{\text{etc.}}\\&+{\tfrac {1}{n-1}}(aa+\alpha a+\alpha ^{\prime })\\&+{\tfrac {1}{n}}(a+\alpha )\\&+{\tfrac {1}{n+1}}\end{aligned}}}
It is clear that the same quantity arises if, in the product obtained by multiplying the function
T
{\textstyle T}
by the infinite series
t
−
1
+
1
2
t
−
2
+
1
3
t
−
3
+
1
4
t
−
4
etc.,
{\displaystyle t^{-1}+{\tfrac {1}{2}}t^{-2}+{\tfrac {1}{3}}t^{-3}+{\tfrac {1}{4}}t^{-4}{\text{ etc., }}}
all terms involving negative powers of
t
{\textstyle t}
are rejected (or in short, in the integral part of the product, which is an integral function of
t
{\textstyle t}
),
t
{\textstyle t}
is replaced by
a
.
{\textstyle a.}
Therefore let us set [ 1]
T
(
t
−
1
+
1
2
t
−
2
+
1
3
t
−
3
+
1
4
t
−
4
+
etc.
)
=
T
′
+
T
′
′
{\displaystyle T(t^{-1}+{\tfrac {1}{2}}t^{-2}+{\tfrac {1}{3}}t^{-3}+{\tfrac {1}{4}}t^{-4}+{\text{etc.}})=T^{\prime }+T^{\prime \prime }}
so that
T
′
{\textstyle T^{\prime }}
is the integral function of
t
{\textstyle t}
contained in this product, and
T
′
′
{\textstyle T^{\prime \prime }}
is the other part, namely the series descending to negative infinity. Then the value of the integral
∫
T
d
t
t
−
a
{\textstyle \int {\frac {T\operatorname {d} t}{t-a}}}
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
will be equal to the value of the function
T
′
{\textstyle T^{\prime }}
at
t
=
a
.
{\textstyle t=a.}
So, if we denote the values of the function
T
′
(
d
T
d
t
)
{\displaystyle {\frac {T^{\prime }}{({\frac {\operatorname {d} T}{\operatorname {d} t}})}}}
determined by
t
=
a
,
{\textstyle t=a,}
t
=
a
′
,
{\textstyle t=a^{\prime },}
t
=
a
′
′
,
{\textstyle t=a^{\prime \prime },}
t
=
a
′
′
′
{\textstyle t=a^{\prime \prime \prime }}
, etc., up to
t
=
a
(
n
)
{\textstyle t=a^{(n)}}
resp. by
R
,
{\textstyle R,}
R
′
,
{\textstyle R^{\prime },}
R
′
′
,
{\textstyle R^{\prime \prime },}
R
′
′
′
…
R
(
n
)
,
{\textstyle R^{\prime \prime \prime }\ldots R^{(n)},}
then the integral
∫
Y
d
{\textstyle \int Y\operatorname {d} }
from
t
=
0
{\textstyle t=0}
to
t
=
1
{\textstyle t=1}
will be
=
R
A
+
R
′
A
′
+
R
′
′
A
′
′
+
etc.
+
R
(
n
)
A
(
n
)
{\displaystyle =RA+R^{\prime }A^{\prime }+R^{\prime \prime }A^{\prime \prime }+{\text{etc.}}+R^{(n)}A^{(n)}}
which, when multiplied by
Δ
{\textstyle \Delta }
, will give the value, either exact or approximate, of the integral
∫
y
d
x
{\textstyle \int y\operatorname {d} x}
from
x
=
g
{\textstyle x=g}
to
x
=
g
+
Δ
{\textstyle x=g+\Delta }
.
These operations are somewhat easier to perform if we introduce another variable
u
=
2
t
−
1.
{\textstyle u=2t-1.}
For the sake of brevity, we also write
b
=
2
a
−
1
,
{\textstyle b=2a-1,}
b
′
=
2
a
′
−
1
,
{\textstyle b^{\prime }=2a^{\prime }-1,}
b
′
′
=
2
a
′
′
−
1
,
{\textstyle b^{\prime \prime }=2a^{\prime \prime }-1,}
etc. By substituting the value
1
2
u
+
1
2
{\textstyle {\frac {1}{2}}u+{\frac {1}{2}}}
for
t
,
{\textstyle t,}
let
T
{\textstyle T}
be transformed into
U
2
n
+
1
,
{\textstyle {\frac {U}{2^{n+1}}},}
or equivalently let
U
=
(
u
−
b
)
(
u
−
b
′
)
(
u
−
b
′
′
)
…
(
u
−
b
n
)
{\displaystyle U=(u-b)(u-b^{\prime })(u-b^{\prime \prime })\ldots (u-b^{n})}
Then
d
T
d
t
=
1
2
n
⋅
d
U
d
u
,
{\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}={\frac {1}{2^{n}}}\cdot {\frac {\operatorname {d} U}{\operatorname {d} u}},}
and hence
M
,
{\textstyle M,}
M
′
,
{\textstyle M^{\prime },}
M
′
′
{\textstyle M^{\prime \prime }}
etc. are the values of
1
2
n
⋅
d
U
d
u
{\textstyle {\frac {1}{2^{n}}}\cdot {\frac {\operatorname {d} U}{\operatorname {d} u}}}
determined by
u
=
b
,
{\textstyle u=b,}
u
=
b
′
,
{\textstyle u=b^{\prime },}
u
=
b
′
′
{\textstyle u=b^{\prime \prime }}
etc.
Since the series
t
−
1
+
1
2
t
−
2
+
1
3
t
−
3
+
1
4
t
−
4
+
{\textstyle t^{-1}+{\frac {1}{2}}t^{-2}+{\frac {1}{3}}t^{-3}+{\frac {1}{4}}t^{-4}+}
etc. is nothing but
log
1
1
−
t
−
1
=
log
1
+
u
−
1
1
−
u
−
1
:
{\textstyle \log {\frac {1}{1-t^{-1}}}=\log {\frac {1+u^{-1}}{1-u^{-1}}}:}
substituting
t
=
1
2
u
+
1
2
{\textstyle t={\frac {1}{2}}u+{\frac {1}{2}}}
will transform it into
2
u
−
1
+
2
3
u
−
3
+
2
5
u
−
5
+
2
7
u
−
7
+
{\textstyle 2u^{-1}+{\frac {2}{3}}u^{-3}+{\frac {2}{5}}u^{-5}+{\frac {2}{7}}u^{-7}+}
etc. Therefore, if we set
U
(
u
−
1
+
1
3
u
−
3
+
1
3
u
−
5
+
1
7
u
−
7
+
etc.
)
=
U
′
+
U
′
′
,
{\displaystyle U(u^{-1}+{\tfrac {1}{3}}u^{-3}+{\tfrac {1}{3}}u^{-5}+{\tfrac {1}{7}}u^{-7}+{\text{etc.}})=U^{\prime }+U^{\prime \prime },}
so that
U
′
{\textstyle U^{\prime }}
is the integral function of
u
{\textstyle u}
contained in this product, and
U
′
′
{\textstyle U^{\prime \prime }}
is the other part, which is an infinite descending series, it is clear that
T
′
+
T
′
′
=
1
2
n
(
U
′
+
U
′
′
)
{\displaystyle T^{\prime }+T^{\prime \prime }={\tfrac {1}{2^{n}}}(U^{\prime }+U^{\prime \prime })}
However, it is clear that
T
′
{\textstyle T^{\prime }}
, being an integral function of
t
,
{\textstyle t,}
will necessarily become an integral function of
u
{\textstyle u}
as a result of the substitution
t
=
1
2
u
+
1
2
;
{\textstyle t={\frac {1}{2}}u+{\frac {1}{2}}{;}}
on the other hand,
T
′
′
{\textstyle T^{\prime \prime }}
, which contains only negative powers of
t
,
{\textstyle t,}
will only generate negative powers of
u
{\textstyle u}
as a result of the same substitution. Therefore,
U
′
{\textstyle U^{\prime }}
will be nothing but
2
n
T
′
{\textstyle 2^{n}T^{\prime }}
transformed by this substitution, and likewise
U
′
′
{\textstyle U^{\prime \prime }}
will be produced from
2
n
T
′
′
.
{\textstyle 2^{n}T^{\prime \prime }.}
Consequently, it makes no difference whether we substitute
t
=
a
{\textstyle t=a}
into
T
′
(
d
T
d
t
)
{\textstyle {\frac {T^{\prime }}{({\frac {\operatorname {d} T}{\operatorname {d} t}})}}}
or
u
=
b
{\textstyle u=b}
into
U
′
(
d
U
d
u
)
.
{\textstyle {\frac {U^{\prime }}{({\frac {\operatorname {d} U}{\operatorname {d} u}})}}.}
From this we conclude that
R
,
{\textstyle R,}
R
′
,
{\textstyle R^{\prime },}
R
′
′
,
{\textstyle R^{\prime \prime },}
R
′
′
′
{\textstyle R^{\prime \prime \prime }}
etc. are also the values of the function
U
′
(
d
U
d
u
)
{\textstyle {\frac {U^{\prime }}{({\frac {\operatorname {d} U}{\operatorname {d} u}})}}}
determined by
u
=
b
,
{\textstyle u=b,}
u
=
b
′
,
{\textstyle u=b^{\prime },}
u
=
b
′
′
,
{\textstyle u=b^{\prime \prime },}
u
=
b
′
′
′
{\textstyle u=b^{\prime \prime \prime }}
etc.
Before we proceed further, we will illustrate these precepts with an example. Let
n
=
5
,
{\textstyle n=5,}
and suppose that
a
=
0
,
{\textstyle a=0,}
a
′
=
1
5
,
{\textstyle a^{\prime }={\frac {1}{5}},}
a
′
′
=
2
5
,
{\textstyle a^{\prime \prime }={\frac {2}{5}},}
a
′
′
′
=
3
5
,
{\textstyle a^{\prime \prime \prime }={\frac {3}{5}},}
a
′
′
′
′
=
4
5
,
{\textstyle a^{\prime \prime \prime \prime }={\frac {4}{5}},}
a
′
′
′
′
′
=
1.
{\textstyle a^{\prime \prime \prime \prime \prime }=1.}
Then we have
T
=
t
6
−
3
t
5
+
17
5
t
4
−
9
5
t
3
+
274
625
t
t
−
24
625
t
.
{\displaystyle T=t^{6}-3t^{5}+{\tfrac {17}{5}}t^{4}-{\tfrac {9}{5}}t^{3}+{\tfrac {274}{625}}tt-{\tfrac {24}{625}}t.}
Multiplying by
t
−
1
+
1
2
t
−
2
+
1
3
t
−
3
+
1
4
t
−
4
+
{\textstyle t^{-1}+{\frac {1}{2}}t^{-2}+{\frac {1}{3}}t^{-3}+{\frac {1}{4}}t^{-4}+}
etc., we obtain
T
′
=
t
5
−
5
2
t
4
+
67
30
t
3
−
17
20
t
t
+
913
7500
t
−
19
7500
{\displaystyle T^{\prime }=t^{5}-{\tfrac {5}{2}}t^{4}+{\tfrac {67}{30}}t^{3}-{\tfrac {17}{20}}tt+{\tfrac {913}{7500}}t-{\tfrac {19}{7500}}}
Hence the values of the coefficients
R
,
{\textstyle {R},}
R
′
,
{\textstyle {R}^{\prime },}
R
′
′
,
{\textstyle {R}^{\prime \prime },}
R
′
′
′
,
{\textstyle {R}^{\prime \prime \prime },}
R
′
′
′
′
,
{\textstyle {R}^{\prime \prime \prime \prime },}
R
′
′
′
′
′
{\textstyle {R}^{\prime \prime \prime \prime \prime }}
are expressed by the fractional function
t
5
−
2
5
t
4
+
67
30
t
3
−
17
20
t
t
+
913
7500
t
−
19
7500
6
t
5
−
15
t
4
+
68
5
t
3
−
27
5
t
t
+
548
625
t
−
24
625
{\displaystyle {\frac {t^{5}-{\frac {2}{5}}t^{4}+{\frac {67}{30}}t^{3}-{\frac {17}{20}}tt+{\frac {913}{7500}}t-{\frac {19}{7500}}}{6t^{5}-15t^{4}+{\frac {68}{5}}t^{3}-{\frac {27}{5}}tt+{\frac {548}{625}}t-{\frac {24}{625}}}}}
wherein the values
0
,
{\textstyle 0,}
1
5
,
{\textstyle {\frac {1}{5}},}
2
5
,
{\textstyle {\frac {2}{5}},}
3
5
,
{\textstyle {\frac {3}{5}},}
4
5
,
{\textstyle {\frac {4}{5}},}
1
{\textstyle 1}
are subsequently substituted for
t
.
{\textstyle t.}
The other method, which is a bit faster, yields
b
=
−
1
,
{\textstyle b=-1,}
b
′
=
−
3
5
,
{\textstyle b^{\prime }=-{\frac {3}{5}},}
b
′
′
=
−
1
5
,
{\textstyle b^{\prime \prime }=-{\frac {1}{5}},}
b
′
′
′
=
1
5
,
{\textstyle b^{\prime \prime \prime }={\frac {1}{5}},}
b
′
′
′
′
=
3
5
,
{\textstyle b^{\prime \prime \prime \prime }={\frac {3}{5}},}
b
′
′
′
′
′
=
1
,
{\textstyle b^{\prime \prime \prime \prime \prime }=1,}
U
=
u
6
−
7
5
u
4
+
259
625
u
u
−
9
625
U
′
=
u
5
−
16
15
u
3
+
277
1875
u
{\displaystyle {\begin{aligned}&U=u^{6}-{\tfrac {7}{5}}u^{4}+{\tfrac {259}{625}}uu-{\tfrac {9}{625}}\\&U^{\prime }=u^{5}-{\tfrac {16}{15}}u^{3}+{\tfrac {277}{1875}}u\end{aligned}}}
from which
R
,
{\textstyle R,}
R
′
,
{\textstyle R^{\prime },}
R
′
′
{\textstyle R^{\prime \prime }}
, etc. will be values of the fractional function
u
4
−
16
15
u
u
+
277
1875
6
u
4
−
28
5
u
u
+
518
625
{\displaystyle {\frac {u^{4}-{\frac {16}{15}}uu+{\frac {277}{1875}}}{6u^{4}-{\frac {28}{5}}uu+{\frac {518}{625}}}}}
for
u
=
−
1
,
{\textstyle u=-1,}
u
=
−
3
5
,
{\textstyle u=-{\frac {3}{5}},}
u
=
−
1
5
{\textstyle u=-{\frac {1}{5}}}
, etc. Both methods yield the same numbers given in Art. 4 of Harmonia Mensurarum . However, in such an example like this, where
a
,
{\textstyle a,}
a
′
,
{\textstyle a^{\prime },}
a
′
′
{\textstyle a^{\prime \prime }}
, etc. are all rational quantities, the values of the denominator
d
T
d
t
{\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}}
are more conveniently computed in the original form, namely
(
a
−
a
′
)
(
a
−
a
′
′
)
(
a
−
a
′
′
′
)
…
(
a
−
a
(
n
)
)
{\textstyle (a-a^{\prime })(a-a^{\prime \prime })(a-a^{\prime \prime \prime })\ldots (a-a^{(n)})}
for
t
=
a
,
{\textstyle t=a,}
and likewise for the others. The same holds for the denominator
d
U
d
u
{\textstyle {\frac {\operatorname {d} U}{\operatorname {d} u}}}
, which for
u
=
b
{\textstyle u=b}
becomes
=
(
b
−
b
′
)
(
b
−
b
′
′
)
(
b
−
b
′
′
′
)
…
(
b
−
b
(
n
)
)
.
{\textstyle =(b-b^{\prime })(b-b^{\prime \prime })(b-b^{\prime \prime \prime })\ldots (b-b^{(n)}).}
When
a
,
{\textstyle a,}
a
′
,
{\textstyle a^{\prime },}
a
′
′
{\textstyle a^{\prime \prime }}
, etc., are either partially or altogether irrational, it will be useful to transform the fractional function, from which we derive the numbers
R
,
{\textstyle R,}
R
′
,
{\textstyle R^{\prime },}
R
′
′
{\textstyle R^{\prime \prime }}
, etc., into an integral function. Since an elementary explanation of this transformation cannot be found in algebraic books, we will provide one here. Specifically, let
Z
,
{\textstyle Z,}
ζ
,
{\textstyle \zeta ,}
ζ
′
{\textstyle \zeta ^{\prime }}
be three indeterminate integral functions of
z
,
{\textstyle z,}
and let us seek an integral function which can be substituted for the fraction
Z
ζ
{\textstyle {\frac {Z}{\zeta }}}
, as far as
z
{\textstyle z}
is taken to be any root of the equation
ζ
′
=
0.
{\textstyle \zeta ^{\prime }=0.}
Let us assume that
ζ
{\textstyle \zeta }
does not vanish for any of these values of
z
{\textstyle z}
, or equivalently, that
ζ
{\textstyle \zeta }
and
ζ
′
{\textstyle \zeta ^{\prime }}
imply no common indeterminate divisor. We will denote the exponents of the highest powers of
z
{\textstyle z}
in
ζ
{\textstyle \zeta }
and
ζ
′
{\textstyle \zeta ^{\prime }}
by
k
,
{\textstyle k,}
k
′
,
{\textstyle k^{\prime },}
respectively.
Divide
ζ
{\textstyle \zeta }
by
ζ
′
,
{\textstyle \zeta ^{\prime },}
as is usual, until the order of the remainder is less than
k
′
;
{\textstyle k^{\prime };}
let the remainder be
=
ζ
′
′
λ
,
{\textstyle ={\frac {\zeta ^{\prime \prime }}{\lambda }},}
and let its order be
=
k
′
′
,
{\textstyle =k^{\prime \prime },}
so that
1
λ
z
k
′
′
{\textstyle {\frac {1}{\lambda }}z^{k^{\prime \prime }}}
is the highest order term of the remainder; we will denote the quotient of this division by
p
λ
.
{\textstyle {\frac {p}{\lambda }}.}
Similarly, divide the function
ζ
′
{\textstyle \zeta ^{\prime }}
by
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
let the residue
ζ
′
′
′
λ
′
{\textstyle {\frac {\zeta ^{\prime \prime \prime }}{\lambda ^{\prime }}}}
of order
k
′
′
′
{\textstyle k^{\prime \prime \prime }}
be obtained as
p
′
λ
′
;
{\textstyle {\frac {p^{\prime }}{\lambda ^{\prime }}};}
then again from the division of the function
ζ
′
′
{\textstyle \zeta ^{\prime \prime }}
by
ζ
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime },}
let the residue
ζ
′
′
′
′
λ
′
′
{\textstyle {\frac {\zeta ^{\prime \prime \prime \prime }}{\lambda ^{\prime \prime }}}}
of order
k
′
′
′
′
{\textstyle k^{\prime \prime \prime \prime }}
be obtained as
p
′
′
λ
′
′
{\textstyle {\frac {p^{\prime \prime }}{\lambda ^{\prime \prime }}}}
, and so on, until in the series of functions
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
ζ
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime },}
ζ
′
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime \prime },}
etc., each having its highest term affected by a coefficient of
1
,
{\textstyle 1,}
we arrive at
ζ
(
m
)
=
1.
{\textstyle \zeta ^{(m)}=1.}
It is easy to see that this must eventually happen, since none of the functions
ζ
,
{\textstyle \zeta ,}
ζ
′
,
{\textstyle \zeta ^{\prime },}
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
ζ
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime },}
etc., cannot have a common indeterminate divisor with the preceding one, and therefore, a division without remainder cannot happen as long as the divisor is of order greater than
0.
{\textstyle 0.}
Thus, we will have a series of equations:
ζ
′
′
′
′
′
=
λ
ζ
−
p
ζ
′
ζ
′
′
′
′
′
=
λ
′
ζ
′
−
p
′
ζ
′
′
ζ
′
′
′
′
′
=
λ
′
′
ζ
′
′
−
p
′
′
ζ
′
′
′
ζ
′
′
′
′
′
=
λ
′
′
′
ζ
′
′
′
−
p
′
′
′
ζ
′
′
′
′
{\textstyle {\begin{aligned}&\zeta ^{\prime \prime {\phantom {\prime \prime \prime }}}=\lambda \zeta -p\zeta ^{\prime }\\&\zeta ^{\prime \prime \prime {\phantom {\prime \prime }}}=\lambda ^{\prime }\zeta ^{\prime }-p^{\prime }\zeta ^{\prime \prime }\\&\zeta ^{\prime \prime \prime \prime {\phantom {\prime }}}=\lambda ^{\prime \prime }\zeta ^{\prime \prime }-p^{\prime \prime }\zeta ^{\prime \prime \prime }\\&\zeta ^{\prime \prime \prime \prime \prime }=\lambda ^{\prime \prime \prime }\zeta ^{\prime \prime \prime }-p^{\prime \prime \prime }\zeta ^{\prime \prime \prime \prime }\end{aligned}}}
etc., up to
ζ
(
m
)
=
λ
(
m
−
2
)
ζ
(
m
−
2
)
−
p
(
m
−
2
)
ζ
(
m
−
1
)
{\textstyle \zeta ^{(m)}=\lambda ^{(m-2)}\zeta ^{(m-2)}-p^{(m-2)}\zeta ^{(m-1)}}
where
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
ζ
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime },}
ζ
′
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime \prime },}
etc.,
ζ
(
m
)
{\textstyle \zeta ^{(m)}}
are integral functions of
z
,
{\textstyle z,}
of orders
k
′
′
,
{\textstyle k^{\prime \prime },}
k
′
′
′
,
{\textstyle k^{\prime \prime \prime },}
k
′
′
′
′
…
k
(
m
)
;
{\textstyle k^{\prime \prime \prime \prime }\dots k^{(m)};}
where the numbers
k
′
,
{\textstyle k^{\prime },}
k
′
′
,
{\textstyle k^{\prime \prime },}
k
′
′
′
,
{\textstyle k^{\prime \prime \prime },}
etc., continuously decrease until the last one
k
(
m
)
=
0
;
{\textstyle k^{(m)}=0;}
and
p
,
{\textstyle p,}
p
′
,
{\textstyle p^{\prime },}
p
′
′
,
{\textstyle p^{\prime \prime },}
p
′
′
′
{\textstyle p^{\prime \prime \prime }}
, etc., are integral functions of
z
{\textstyle z}
of orders
k
−
k
′
,
{\textstyle k-k^{\prime },}
k
′
−
k
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime },}
k
′
′
−
k
′
′
′
,
{\textstyle k^{\prime \prime }-k^{\prime \prime \prime },}
k
′
′
′
−
k
′
′
′
′
{\textstyle k^{\prime \prime \prime }-k^{\prime \prime \prime \prime }}
, etc. (except in the case where
k
<
k
′
,
{\textstyle k<k^{\prime },}
where it is clear that we must set
p
=
0
{\textstyle p=0}
).
Having prepared in this manner, we form a second series of integral functions of
z
,
{\textstyle z,}
which we call
η
,
{\textstyle \eta ,}
η
′
,
{\textstyle \eta ^{\prime },}
η
′
′
,
{\textstyle \eta ^{\prime \prime },}
η
′
′
′
,
{\textstyle \eta ^{\prime \prime \prime },}
etc., up to
η
(
m
)
.
{\textstyle \eta ^{(m)}.}
Indeed, let us set
η
=
1
,
{\textstyle \eta =1,}
η
′
=
0
,
{\textstyle \eta ^{\prime }=0,}
and for the remaining functions we derive each from the preceding two according to the same rule by which the functions
ζ
,
{\textstyle \zeta ,}
ζ
′
,
{\textstyle \zeta ^{\prime },}
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
ζ
′
′
′
{\textstyle \zeta ^{\prime \prime \prime }}
, etc., are related to each other, namely through the following equations:
η
′
′
′
′
′
=
λ
η
−
p
η
′
η
′
′
′
′
′
=
λ
′
η
′
−
p
′
η
′
′
η
′
′
′
′
′
=
λ
′
′
η
′
′
−
p
′
′
η
′
′
′
η
′
′
′
′
′
=
λ
′
′
′
η
′
′
′
−
p
′
′
′
η
′
′
′
′
{\textstyle {\begin{aligned}&\eta ^{\prime \prime {\phantom {\prime \prime \prime }}}=\lambda \eta -p\eta ^{\prime }\\&\eta ^{\prime \prime \prime {\phantom {\prime \prime }}}=\lambda ^{\prime }\eta ^{\prime }-p^{\prime }\eta ^{\prime \prime }\\&\eta ^{\prime \prime \prime \prime {\phantom {\prime }}}=\lambda ^{\prime \prime }\eta ^{\prime \prime }-p^{\prime \prime }\eta ^{\prime \prime \prime }\\&\eta ^{\prime \prime \prime \prime \prime }=\lambda ^{\prime \prime \prime }\eta ^{\prime \prime \prime }-p^{\prime \prime \prime }\eta ^{\prime \prime \prime \prime }\end{aligned}}}
etc., up to
η
(
m
)
=
λ
(
m
−
2
)
η
(
m
−
2
)
−
p
(
m
−
2
)
η
(
m
−
1
)
{\textstyle \eta ^{(m)}=\lambda ^{(m-2)}\eta ^{(m-2)}-p^{(m-2)}\eta ^{(m-1)}}
Clearly
η
′
′
=
λ
{\textstyle \eta ^{\prime \prime }=\lambda }
is of order
0
{\textstyle 0}
here;
η
′
′
′
=
−
λ
p
′
{\textstyle \eta ^{\prime \prime \prime }=-\lambda p^{\prime }}
is of order
k
′
−
k
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime },}
and likewise the subsequent functions
η
′
′
′
′
,
{\textstyle \eta ^{\prime \prime \prime \prime },}
η
′
′
′
′
′
{\textstyle \eta ^{\prime \prime \prime \prime \prime }}
, etc., are of orders
k
′
−
k
′
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime \prime },}
k
′
−
k
′
′
′
′
{\textstyle k^{\prime }-k^{\prime \prime \prime \prime }}
, etc., so that the last one
η
(
m
)
{\textstyle \eta ^{(m)}}
is of order
k
′
−
k
(
m
−
1
)
.
{\textstyle k^{\prime }-k^{(m-1)}.}
Next we consider a "third" series of functions,
ζ
−
ζ
η
,
{\textstyle \zeta -\zeta \eta ,}
ζ
′
−
ζ
η
′
,
{\textstyle \zeta ^{\prime }-\zeta \eta ^{\prime },}
ζ
′
′
−
ζ
η
′
′
,
{\textstyle \zeta ^{\prime \prime }-\zeta \eta ^{\prime \prime },}
ζ
′
′
′
−
ζ
η
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime }-\zeta \eta ^{\prime \prime \prime },}
etc., among which any three consecutive terms will manifestly have a similar relation, namely,
ζ
′
′
−
ζ
η
′
′
=
λ
(
ζ
−
ζ
η
)
−
p
(
ζ
′
−
ζ
η
′
)
ζ
′
′
′
−
ζ
η
′
′
′
=
λ
′
(
ζ
′
−
ζ
η
′
)
−
p
′
(
ζ
′
′
−
ζ
η
′
′
)
ζ
′
′
′
′
−
ζ
η
′
′
′
′
=
λ
′
′
(
ζ
′
′
−
ζ
η
′
′
)
−
p
′
′
(
ζ
′
′
′
−
ζ
η
′
′
′
)
{\displaystyle {\begin{aligned}\zeta ^{\prime \prime }-\zeta \eta ^{\prime \prime }&=\lambda (\zeta -\zeta \eta )-p(\zeta ^{\prime }-\zeta \eta ^{\prime })\\\zeta ^{\prime \prime \prime }-\zeta \eta ^{\prime \prime \prime }&=\lambda ^{\prime }(\zeta ^{\prime }-\zeta \eta ^{\prime })-p^{\prime }(\zeta ^{\prime \prime }-\zeta \eta ^{\prime \prime })\\\zeta ^{\prime \prime \prime \prime }-\zeta \eta ^{\prime \prime \prime \prime }&=\lambda ^{\prime \prime }(\zeta ^{\prime \prime }-\zeta \eta ^{\prime \prime })-p^{\prime \prime }(\zeta ^{\prime \prime \prime }-\zeta \eta ^{\prime \prime \prime })\end{aligned}}}
Now, the first of these functions is
=
0
,
{\textstyle =0,}
the second is
=
ζ
′
,
{\textstyle =\zeta ^{\prime },}
hence it is easily inferred that each is divisible by
ζ
′
.
{\textstyle \zeta ^{\prime }.}
Moreover, it follows without difficulty that we can replace the fraction
Z
ζ
,
{\textstyle {\frac {Z}{\zeta }},}
with the integral function
Z
η
(
m
)
,
{\textstyle Z\eta ^{(m)},}
provided that no values are assigned to
z
{\textstyle {z}}
other than those which are roots of the equation
ζ
′
=
0
;
{\textstyle \zeta ^{\prime }=0;}
for it is clear that the difference
Z
(
1
−
ζ
η
(
m
)
)
ζ
{\textstyle {\frac {Z(1-\zeta \eta ^{(m)})}{\zeta }}}
must vanish for such a value of
z
,
{\textstyle z,}
since
1
−
ζ
η
(
m
)
=
ζ
(
m
)
−
ζ
η
(
m
)
{\textstyle 1-\zeta \eta ^{(m)}=\zeta ^{(m)}-\zeta \eta ^{(m)}}
is divisible by
ζ
′
.
{\textstyle \zeta ^{\prime }.}
Instead of the function
Z
η
(
m
)
,
{\textstyle Z\eta ^{(m)},}
we can also take the remainder which arises upon dividing it by
ζ
′
,
{\textstyle \zeta ^{\prime },}
whose order will be lower than the order of the function
ζ
′
.
{\textstyle \zeta ^{\prime }.}
Indeed, this remainder can be immediately and more conveniently extracted using the following algorithm. We form the following equations:
Z
′
′
′
=
q
′
ζ
′
+
Z
′
Z
′
′
′
=
q
′
′
ζ
′
′
+
Z
′
′
Z
′
′
′
=
q
′
′
′
ζ
′
′
′
+
Z
′
′
′
Z
′
′
′
=
q
′
′
′
′
ζ
′
′
′
′
+
Z
′
′
′
′
{\textstyle {\begin{aligned}Z^{\phantom {\prime \prime \prime }}&=q^{\prime }\zeta ^{\prime }+Z^{\prime }\\Z^{\prime {\phantom {\prime \prime }}}&=q^{\prime \prime }\zeta ^{\prime \prime }+Z^{\prime \prime }\\Z^{\prime \prime {\phantom {\prime }}}&=q^{\prime \prime \prime }\zeta ^{\prime \prime \prime }+Z^{\prime \prime \prime }\\Z^{\prime \prime \prime }&=q^{\prime \prime \prime \prime }\zeta ^{\prime \prime \prime \prime }+Z^{\prime \prime \prime \prime }\end{aligned}}}
etc., up to
Z
(
m
−
1
)
=
q
(
m
)
ζ
(
m
)
+
Z
(
m
)
{\textstyle Z^{(m-1)}=q^{(m)}\zeta ^{(m)}+Z^{(m)}}
by dividing
Z
{\textstyle Z}
by
ζ
′
,
{\textstyle \zeta ^{\prime },}
then the remainder of the first division
Z
′
{\textstyle Z^{\prime }}
by
ζ
′
′
,
{\textstyle \zeta ^{\prime \prime },}
then the remainder of the second division by
ζ
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime },}
and so forth. Since the remainder always belongs to an order lower than the divisor, the order of the functions
Z
′
,
{\textstyle Z^{\prime },}
Z
′
′
,
{\textstyle Z^{\prime \prime },}
Z
′
′
′
,
{\textstyle Z^{\prime \prime \prime },}
Z
′
′
′
,
{\textstyle Z^{\prime \prime \prime },}
etc. will be respectively lower than
k
′
,
{\textstyle k^{\prime },}
k
′
′
,
{\textstyle k^{\prime \prime },}
k
′
′
′
,
{\textstyle k^{\prime \prime \prime },}
k
′
′
′
,
{\textstyle k^{\prime \prime \prime },}
etc.; while the last
Z
(
m
)
{\textstyle Z^{(m)}}
necessarily becomes
=
0
,
{\textstyle =0,}
since the divisor
ζ
(
m
)
{\textstyle \zeta ^{(m)}}
is
=
1.
{\textstyle =1.}
Therefore, we have
Z
=
q
′
ζ
′
+
q
′
′
ζ
′
′
+
q
′
′
′
ζ
′
′
′
+
q
′
′
′
′
ζ
′
′
′
′
+
etc.
+
q
(
m
)
ζ
(
m
)
{\displaystyle Z=q^{\prime }\zeta ^{\prime }+q^{\prime \prime }\zeta ^{\prime \prime }+q^{\prime \prime \prime }\zeta ^{\prime \prime \prime }+q^{\prime \prime \prime \prime }\zeta ^{\prime \prime \prime \prime }+{\text{etc.}}+q^{(m)}\zeta ^{(m)}}
Moreover, since only the roots of the equation
ζ
′
=
0
{\textstyle \zeta ^{\prime }=0}
are taken for
z
,
{\textstyle z,}
it follows that
ζ
′
=
0
,
{\textstyle \zeta ^{\prime }=0,}
ζ
′
′
=
ζ
η
′
′
,
{\textstyle \zeta ^{\prime \prime }=\zeta \eta ^{\prime \prime },}
ζ
′
′
′
=
ζ
η
′
′
′
,
{\textstyle \zeta ^{\prime \prime \prime }=\zeta \eta ^{\prime \prime \prime },}
ζ
′
′
′
′
=
ζ
η
′
′
′
′
{\textstyle \zeta ^{\prime \prime \prime \prime }=\zeta \eta ^{\prime \prime \prime \prime }}
etc., and under the same restriction, it follows that
Z
ζ
=
q
′
′
η
′
′
+
q
′
′
′
η
′
′
′
+
q
′
′
′
′
η
′
′
′
′
+
etc.
+
q
(
m
)
η
(
m
)
{\displaystyle {\frac {Z}{\zeta }}=q^{\prime \prime }\eta ^{\prime \prime }+q^{\prime \prime \prime }\eta ^{\prime \prime \prime }+q^{\prime \prime \prime \prime }\eta ^{\prime \prime \prime \prime }+{\text{etc.}}+q^{(m)}\eta ^{(m)}}
However, the order of this expression will necessarily be less than than
k
′
:
{\textstyle k^{\prime }{:}}
since the order of the quotients
q
′
′
,
{\textstyle q^{\prime \prime },}
q
′
′
′
,
{\textstyle q^{\prime \prime \prime },}
q
′
′
′
,
{\textstyle q^{\prime \prime \prime },}
etc. must be less than
k
′
−
k
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime },}
k
′
′
−
k
′
′
′
,
{\textstyle k^{\prime \prime }-k^{\prime \prime \prime },}
k
′
′
′
−
k
′
′
′
,
{\textstyle k^{\prime \prime \prime }-k^{\prime \prime \prime },}
etc., the order of each part
q
′
′
η
′
′
,
{\textstyle q^{\prime \prime }\eta ^{\prime \prime },}
q
′
′
′
η
′
′
′
,
{\textstyle q^{\prime \prime \prime }\eta ^{\prime \prime \prime },}
q
′
′
′
′
η
′
′
′
′
,
{\textstyle q^{\prime \prime \prime \prime }\eta ^{\prime \prime \prime \prime },}
etc. will be less than
k
′
−
k
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime },}
k
′
−
k
′
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime \prime },}
k
′
−
k
′
′
′
,
{\textstyle k^{\prime }-k^{\prime \prime \prime },}
etc.
Finally we observe that if it so happens that among the values of the indeterminate
z
,
{\textstyle z,}
those that need to be substituted in the fraction
Z
ζ
{\textstyle {\frac {Z}{\zeta }}}
are a mixture of rationals and irrationals, it will be more practical to separate them and only include the latter in the equation
ζ
′
=
0.
{\textstyle \zeta ^{\prime }=0.}
For rational values, there will be no need for calculation; for irrational values, however, the calculation will be simpler the lower the degree of the integral function to which the fraction can be reduced.
Here is an example of the transformation explained in the preceding article. Let the given fractional function be
z
6
−
50
39
z
4
+
283
715
z
z
−
256
15015
7
z
6
−
105
13
z
4
+
315
143
z
z
−
35
429
{\displaystyle {\frac {z^{6}-{\frac {50}{39}}z^{4}+{\frac {283}{715}}zz-{\frac {256}{15015}}}{7z^{6}-{\frac {105}{13}}z^{4}+{\frac {315}{143}}zz-{\frac {35}{429}}}}}
where
z
{\textstyle z}
indefinitely represents the roots of the equation
z
7
−
21
13
z
5
+
105
143
z
3
−
35
429
z
=
0.
{\displaystyle z^{7}-{\tfrac {21}{13}}z^{5}+{\tfrac {105}{143}}z^{3}-{\tfrac {35}{429}}z=0.}
If we wanted to include all seven roots here, we would descend to a sixth-order integral function. However, for the rational value
z
=
0
,
{\textstyle z=0,}
the calculation of the fraction is straightforward, giving the value
256
1225
,
{\textstyle {\frac {256}{1225}},}
so excluding this root from the equation of sixth degree, we have:
z
6
−
21
13
z
4
+
105
143
z
z
−
35
429
=
0
{\displaystyle z^{6}-{\tfrac {21}{13}}z^{4}+{\tfrac {105}{143}}zz-{\tfrac {35}{429}}=0}
from which it is easily foreseen that there will arise a fourth-order integral function. Now, from the application of the preceding rules, the following sequences emerge:
ζ
′
′
′
′
=
7
z
6
−
105
13
z
4
+
315
143
z
z
−
35
429
ζ
′
′
′
′
=
z
6
−
21
13
z
4
+
105
143
z
z
−
35
429
ζ
′
′
′
′
=
z
4
−
10
11
z
z
+
5
33
ζ
′
′
′
′
=
z
z
−
3
7
ζ
′
′
′
′
=
1
λ
′
′
′
′
=
13
42
p
=
13
6
λ
′
′
′
′
=
−
4719
280
p
′
=
−
4719
280
z
z
+
3333
280
λ
′
′
′
′
=
−
147
8
p
′
′
=
−
147
8
z
z
+
777
88
η
′
′
′
′
=
1
η
′
′
′
′
=
0
η
′
′
′
′
=
13
42
η
′
′
′
′
=
20449
3920
z
z
−
14443
3920
η
′
′
′
′
=
61347
640
z
4
−
127413
1120
z
z
+
120263
4480
Z
′
′
′
′
=
z
6
−
50
39
z
4
+
283
715
z
z
−
256
15015
;
q
′
=
1
Z
′
′
′
′
=
1
3
z
4
−
22
65
z
z
+
323
5005
q
′
′
=
1
3
Z
′
′
′
′
=
−
76
2145
z
z
+
632
45045
q
′
′
′
=
−
76
2145
Z
′
′
′
′
=
−
4
3465
q
′
′
′
′
=
−
4
3465