Translation:Methodus nova integralium valores per approximationem inveniendi

1.
Among the methods proposed for the numerical approximation of integrals, a prominent place is held by the rules which were developed by and refined by. Specifically, if the value of the integral $\displaystyle \int y \operatorname{d} x$ taken from $x=g$  to $x=h$  is required, then the values of $y$  for these limiting values of $x$  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 $h-g,$ 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.

2.
Let us agree to use a multitude of $n+1$ terms, and let $h-g=\Delta,$  so that the values of $x$  are $g,$  $g+\frac{\Delta}{n},$  $g+\frac{2 \Delta}{n},$  $g+\frac{3 \Delta}{n}$  etc. up to $g+\Delta,$  and correspondingly the values of $y$  are $A,$  $A^{\prime},$  $A^{\prime \prime},$  $A^{\prime \prime \prime}$  etc. up to $A^{(n)}{:}$  finally, let $x=g+\Delta t,$  so that $y$  can also be regarded as a function of $t.$  Let ${Y}$  represent the function

$$\begin{aligned} & A. \frac{(n t-1)(n t-2)(n t-3) \ldots(n t-n)}{(-1)(-2). (-3) \ldots. (-n)} \\ + & A^{\prime}. \frac{n t. (n t-2). (n t-3) \cdots \cdots. (n t-n)}{1. (-1)(-2) \cdots \cdots(1-n)} \\ + & A^{\prime \prime}. \frac{n t. (n t-1). (n t-3) \ldots \ldots(n t-n)}{2. 1 . (-1) \ldots \ldots(2-n)}\\ + & A^{\prime \prime \prime}. \frac{n t. (n t-1). (n t-2) \ldots \ldots(n t-n)}{3. 2 . 1 \ldots \ldots(3-n)}\\ + & \text{ etc. } \\ + &A^{(n)}. \frac{n t(n t-1). (n t-2) \ldots \ldots(n t-n+1)}{n. (n-1)(n-2) \ldots \ldots. 1} \end{aligned}$$

or $\Sigma \frac{A^{(\mu)} T^{(\mu)}}{M^{(\mu)}},$ where $\mu$  represents each of the integers $0$, $1$ , $2$ , $3 \ldots n,$

$T^{(\mu)}=\frac{n t(n t-1)(n t-2)(n t-3) \ldots(n t-n)}{n t-\mu}, $ and $ M^{(\mu)}$ is the value of $T$ for $ n t=\mu.$

It is clear that ${Y}$ represents an integral algebraic function of $t,$  of order $n,$  and its values for each of the $n+1$  values of $t,$  namely $0,$  $\frac{1}{n},$  $\frac{2}{n},$  $\frac{3}{n} \ldots 1$  are equal to the values of $y.$  It is also clear that if $Y^{\prime}$  is another integral function producing the same values of $y$  for the same values of $t,$  then $Y^{\prime}-Y$  will vanish for the same values, and therefore it must be divisible by the factors $t,$  $t-\frac{1}{n},$  $t-\frac{2}{n},$  $t-\frac{3}{n} \ldots t-1$  and therefore also by their product (which is of order $n+1$ ), from which it is clear that $Y^{\prime}$  must, unless it is identical to $Y,$  be of a higher order, meaning that $Y$  is the only integral function among those not exceeding order $n$  which coincides with $y$  for those $n+1$  values. Therefore, if $y$, when expanded into a series of powers of $t$ , breaks off before the term involving $t^{n+1}$ , it will be identical to $Y,$ and if the series converges so quickly as to allow the subsequent terms to be neglected, then the function $Y$  can replace $y$  within the limits $t=0,$  $t=1$  or $x=g,$  $x=h$.

3.
Now our integral $\int y \operatorname{d} x$ is transformed into $\Delta \int y \operatorname{d} t,$  taken from $t=0$  to $t=1,$  and as we have just indicated, we will replace this with $\Delta \int Y \operatorname{d} t.$  Thus by expanding $T^{(\mu)}$  into

$$\alpha t^{n}+\beta t^{n-1}+\gamma t^{n-2}+\delta t^{n-3}+\text{etc. }$$

the integral $\int T^{(\mu)} {d} t$ from $t=0$  to $t=1$  will be

$$=\frac{\alpha}{n+1}+\frac{\beta}{n}+\frac{\gamma}{n-1}+\frac{\delta}{n-2}+\text{etc. }$$

and setting this quantity $=M^{(\mu)} R^{(\mu)},$ the desired integral will be

$$=\Delta (A R+A^{\prime} R^{\prime}+A^{\prime \prime} R^{\prime \prime}+A^{\prime \prime \prime} R^{\prime \prime \prime}+\text{etc.}+A^{(n)} R^{(n)})$$

For example, let us compute the coefficient $R^{\prime \prime}$ for $n=5.$  Here we have

$$\begin{aligned} & T^{\prime \prime}=5^{5} t^{5}-13.5^{4} t^{4}+59.5^{3} t^{3}-107.5^{2} t t+60.5. t \\ & M^{\prime \prime}=2 \times 1 \times(-1) \times(-2) \times(-3)=-12 \end{aligned}$$

Hence $-12 R^{\prime \prime}=\frac{3125}{6}-1625+\frac{7375}{4}-\frac{2675}{3}+150=-\frac{25}{12},$ and therefore $R^{\prime \prime}=\frac{25}{144}.$

The computation can be shortened a bit by setting $2 t-1=u.$ Then we have

$$T^{(\mu)}=\frac{(n u+n)(n u+n-2)(n u+n-4) \ldots(n u-n+4)(n u-n+2)(n u-n)}{2^{n}(n u+n-2 \mu)}$$

Let us set

$$\frac{(n n u u-n n). (n n u u-(n-2)^{2}). (n n u u-(n-4)^{2}). (n n u u-(n-6)^{2}) \ldots}{n n u u-(n-2 \mu)^{2}}=U^{(\mu)},$$

where the numerator should end in $\ldots(nnuu-9)(nnuu-1),$ if $n$  is odd, or in $\ldots (n n u u-4) n u,$  if $n$  is even. Then

$$T^{(\mu)}=\frac{(n u-n+2 \mu) U^{(\mu)}}{2^{n}}.$$

Now the integral $\int T^{(\mu)} \operatorname{d} t$ taken from $t=0$  to $t=1$  is equal to the integral

$$\int \tfrac{1}{2} T^{(\mu)} \operatorname{d} u=\int \frac{n u U^{(\mu)} \operatorname{d} u}{2^{n+1}}+\int \frac{(2 \mu-n) U^{(\mu)} \operatorname{d} u}{2^{n+1}}$$

from $u=-1$ to $u=+1.$

Therefore, by setting

$$U^{(\mu)}=\alpha u^{n-1}+\beta u^{n-3}+\gamma u^{n-5}+\delta u^{n-7}+\text{etc. }$$

(it being evident that the powers $u^{n-2}, u^{n-4}, u^{n-6}$ etc. are absent), the first part $\int \frac{n u U^{(n)} \operatorname{d} u}{2^{n+1}}$  of the integral will vanish for odd values of $n,$  while the other part $\int \frac{(2 \mu-n) U^{(\mu)} \operatorname{d} u}{2^{n+1}}$  will vanish for even values, so that the integral $\int T^{(\mu)} \operatorname{d} t$  becomes

$$=\frac{n}{2^{n}} \left(\frac{\alpha}{n+1}+\frac{\beta}{n-1}+\frac{\gamma}{n-3}+\frac{\delta}{n-5}+\text{etc.}\right)$$

for even values of $n,$ and

$$=\frac{2 \mu-n}{2^{n}} \left(\frac{\alpha}{n}+\frac{\beta}{n-2}+\frac{\gamma}{n-4}+\frac{\delta}{n-6}+\text{etc.}\right)$$

for odd values of $n.$ In our example we have

$ U^{\prime \prime}=(25 u u-25)(25 u u-9)=625 u^{4}-850 u u+225,$ hence $ -12 R^{\prime \prime}=-\tfrac{1}{32} (125-\tfrac{850}{3}+225)=-\tfrac{25}{12} $ as above.

It is worth noting that $U^{(n-\mu)}=U^{(\mu)},$ and therefore $\int T^{(n-\mu)} \operatorname{d} t= \pm \int T^{(\mu)} \operatorname{d} t,$  with the upper sign holding for even $n,$  and the lower sign for odd $n.$  Hence, since it is easy to see that $M^{(n-\mu)}= \pm M^{(\mu)},$  we will always have $R^{(n-\mu)}=R^{(\mu)},$  meaning that the last coefficient is equal to the first, the penultimate to the second, and so on.

4.
We hereby append the numerical values of these coefficients, up to $n=10,$ which were computed by Cotes in Harmonia Mensurarum.

For $n=1,$ or two terms.
 * $R=R^{\prime}=\tfrac{1}{2}$

For $n=2,$ or three terms.
 * $R=R^{\prime \prime}=\tfrac{1}{6},$ $R^{\prime}=\tfrac{2}{3}$

For $n=3,$ or four terms.
 * $R=R^{\prime \prime \prime}=\tfrac{1}{8},$ $R^{\prime}=R^{\prime \prime}=\tfrac{3}{8}$

For $n=4,$ or five terms.
 * ${R}={R}^{\prime \prime \prime}=\tfrac{7}{90},$ $R^{\prime}=R^{\prime \prime \prime}=\tfrac{16}{45},$  $R^{\prime \prime}=\tfrac{2}{15}$

For $n=5,$ or six terms.
 * ${R}={R}^{\scriptscriptstyle V}=\tfrac{19}{288},$ ${R}^{\prime}={R}^{\prime \prime \prime \prime}=\tfrac{25}{9},$  ${R}^{\prime \prime}={R}^{\prime \prime \prime}=\tfrac{25}{1 44}$

For $n=6,$ or seven terms.
 * $R=R^{\scriptscriptstyle VI}=\tfrac{41}{846},$ $R^{\prime}=R^{\scriptscriptstyle V}=\tfrac{9}{35},$  $R^{\prime \prime}=R^{\scriptscriptstyle IV}=\tfrac{9}{280},$  $R^{\prime \prime \prime}=\tfrac{34}{105}$

For $n=7,$ or eight terms.
 * $R=R^{\scriptscriptstyle VII}=\tfrac{751}{17280},$ $R^{\prime}=R^{\scriptscriptstyle VI} = \tfrac{3577}{17280},$  $R^{\prime \prime} = R^{\scriptscriptstyle V} = \tfrac{49}{640},$  $R^{\prime \prime \prime} = R^{\prime \prime \prime \prime} = \tfrac{2989}{17280}$

For $n=8,$ or nine terms.
 * $R=R^{\scriptscriptstyle VIII}=\tfrac{989}{28350},$ $R^{\prime}=R^{\scriptscriptstyle VII}=\tfrac{2944}{14175},$  $R^{\prime \prime}=R^{\scriptscriptstyle VI}=-\tfrac{464}{14175},$  $R^{\prime \prime \prime}=R^{\scriptscriptstyle V}=\tfrac{5248}{14175},$  $R^{\scriptscriptstyle IV}=-\tfrac{454}{2835}$

For $n=9,$ or ten terms.
 * $R = R^{\scriptscriptstyle IX} = \tfrac{2857}{89600},$ $R^{\prime} = R^{\scriptscriptstyle VIII} = \tfrac{15741}{89600},$  $R^{\prime \prime} = R^{\scriptscriptstyle VII} = \tfrac{27}{2240},$  $R^{\prime \prime \prime} = R^{\scriptscriptstyle VI} = \tfrac{1209}{5600},$  $R^{\scriptscriptstyle IV}=R^{\scriptscriptstyle V}=\tfrac{2889}{44800}$

For $n=10,$ or eleven terms.
 * $R=R^{\scriptscriptstyle X}=\tfrac{16067}{598752},$ $R^{\prime}=R^{\scriptscriptstyle IX}=\tfrac{26575}{149688},$  $R^{\prime \prime}=R^{\scriptscriptstyle VIII}=-\tfrac{16175}{199584},$  $R^{\prime \prime \prime}=R^{\scriptscriptstyle VII}=\tfrac{5675}{12474},$  $R^{\scriptscriptstyle IV}=R^{\scriptscriptstyle VI}=-\tfrac{4825}{11088},$  $R^{\scriptscriptstyle V}=\tfrac{17807}{24948}.$

5.
Since the formula $\Delta (A R+A^{\prime} R^{\prime}+A^{\prime} R^{\prime}+A^{\prime \prime} R^{\prime \prime}+\text{etc.}+A^{(n)} R^{(n)})$ exactly represents the integral $\int y \operatorname{d} x$  from $x=g$  to $x=g+\Delta,$  or the integral $\Delta \int y \operatorname{d} t$  from $t=0$  to $t=1$  whenever the expansion of $y$  into a series does not go beyond the power $t^{n}$, 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 $k^{(m)}$ the difference between the true value of the integral $\int t^{m} \mathrm{dt}$  from $t=0$  to $t=1,$  and the value derived from the formula. Then

$$\begin{aligned} & k=1-R-R^{\prime}-R^{\prime \prime}-R^{\prime \prime \prime}-\text{etc.}-R^{(n)} \\ & k^{\prime}=\tfrac{1}{2}-\tfrac{1}{n} (R^{\prime}+2 R^{\prime \prime}+3 R^{\prime \prime \prime}+\text{etc.}+n R^{(n)}) \\ & k^{\prime \prime}=\tfrac{1}{3}-\tfrac{1}{n n} (R^{\prime}+4 R^{\prime \prime}+9 R^{\prime \prime \prime}+\text{etc.}+n n R^{(n)}) \\ & k^{\prime \prime \prime}=\tfrac{1}{4}-\tfrac{1}{n^{2}} (R^{\prime}+8 R^{\prime \prime}+27 R^{\prime \prime \prime}+\text{etc.}+ n^{3} R^{(n)}) \end{aligned}$$

etc. It is evident, therefore, that if $y$ is expanded into a series

$$K+K^{\prime} t+K^{\prime \prime} t t+K^{\prime \prime \prime} t^{3}+\text{etc. }$$

then the difference between the true value of the integral $\int y \operatorname{d} t$ and the approximated value derived from the formula can be expressed as

$$K k+K^{\prime} k^{\prime}+K^{\prime \prime} k^{\prime \prime}+K^{\prime \prime \prime} k^{\prime \prime \prime}+\text{etc.}$$

But evidently, $k,$ $k^{\prime},$  $k^{\prime \prime}$, etc. up to $k^{(n)}$  are all automatically $=0{:}$  thus, the correction of the approximated formula will be

$$K^{(n+1)} {k}^{(n+1)}+{K}^{(n+2)} {k}^{(n+2)}+{K}^{(n+3)} {k}^{(n+3)}+\text{etc.}$$

The nature of quantities $k^{(n+1},$ $k^{(n+2)}$, etc. will be examined more accurately later; here, it suffices to provide the numerical values of the first or second, for each value of $n,$  so that the degree of precision afforded by the approximate formula can be estimated.


 * For $n=\phantom{0}1$, we have $k^{\prime \prime}=-\frac{1}{6},$ $k^{\prime \prime \prime}=-\frac{1}{4},$  $k^{\prime \prime \prime \prime}=-\frac{3}{10}$
 * For $n=\phantom{0}2$, we find $k^{\prime \prime \prime}=0,$ $k^{\prime \prime \prime \prime}=-\frac{1}{120},$  $k^{\scriptscriptstyle V}=-\frac{1}{48}$
 * For $n=\phantom{0}3$, it is $k^{\prime \prime \prime \prime}=-\frac{1}{270},$ $\quad k^{\scriptscriptstyle V}=-\frac{1}{108}$
 * For $n=\phantom{0}4 \ldots k^{\scriptscriptstyle V}=0,$ $k^{\scriptscriptstyle VI}=-\frac{1}{2688},$  $k^{\scriptscriptstyle VII}=-\frac{1}{768}$
 * For $n=\phantom{0}5 \ldots k^{\scriptscriptstyle VI}=-\frac{1}{52?00},$ $k^{\scriptscriptstyle VII}=-\frac{11}{15000}$
 * For $n=\phantom{0}6 \ldots k^{\scriptscriptstyle VII}=0,$ $k^{\scriptscriptstyle VIII}=-\frac{1}{38880},$  $k^{\scriptscriptstyle IX}=-\frac{1}{8640}$
 * For $n=\phantom{0}7 \ldots k^{\scriptscriptstyle VIII}=-\frac{167}{10588410},$ $k^{\scriptscriptstyle IX}=-\frac{167}{2352980}$
 * For $n=\phantom{0}8 \ldots k^{\scriptscriptstyle IX}=0,$ $k^{\scriptscriptstyle X}=-\frac{37}{173?1504},$  $k^{\scriptscriptstyle XI}=-\frac{37}{3145728}$
 * For $n=\phantom{0}9 \ldots k^{\scriptscriptstyle X}=-\frac{865}{631351908},$ $k^{\scriptscriptstyle XI}=-\frac{865}{114791256}$
 * For $n=10 \ldots k^{\scriptscriptstyle XI}=0,$ $k^{\scriptscriptstyle XII}=-\frac{26927}{136500000000},$  $k^{\scriptscriptstyle XII}=-\frac{26927}{21000000000}$

For all even values of $n$ here, we observe $k^{(n+1)}=0,$  and furthermore $k^{(n+3)}=\frac{n+3}{2} k^{(n+2)};$  for odd values of $n$, however, we always have $k^{(n+2)}=\frac{n+2}{2} k^{(n+1)}.$  The reason why this occurs can be deduced easily from the following considerations.

In general, let $l^{(m)}$ denote the difference between the true value of the integral $\int (t-\frac{1}{2})^{m} \operatorname{d} t$  from $t=0$  to $t=1,$  and the value derived from the approximate formula, so that we have

$$\begin{array}{c} l^{(m)}=\int (t-\frac{1}{2})^{m} \operatorname{d} t-\left[ (-\frac{1}{2})^{m} R+ (\frac{1}{n}-\frac{1}{2})^{m} R^{\prime}+ (\frac{2}{n}-\frac{1}{2})^{m} R^{\prime \prime}+ (\frac{3}{n}-\frac{1}{2})^{m} R^{\prime \prime \prime}+\right.\text{etc.}\\ \left.+ (\frac{1}{2}-\frac{1}{n})^{m} R^{(n-1)}+ (\frac{1}{2})^{m} R^{(n)}\right] \end{array}$$

with the integral being taken from $t=0$ to $t=1$. Clearly, for odd values of $m$ both the true and approximate integral values vanish: hence $l^{\prime}=0,$  $l^{\prime \prime \prime}=0, l^{\scriptscriptstyle V}=0, l^{\scriptscriptstyle VII}=0$, etc., and generally $l^{(m)}=0$  for all odd values of $m.$  For even values of $m,$  on the other hand, the formula can be written as

$$\begin{array}{c} l^{(m)}=\frac{1}{2^{m}(m+1)}-\frac{2}{n^{m}} ( (\frac{1}{2} n)^{m} R+ (\frac{1}{2} n-1)^{m} R^{\prime}+ (\frac{1}{2} n-2)^{m} R^{\prime \prime}+\text{etc. } \\ +2^{m} R^{ \left(\frac{1}{2} n-2\right)}+R^{ \left(\frac{1}{2} n-1\right)}) \end{array}$$

if $n$ is even; or

$$\begin{aligned} l^{(m)}=\tfrac{1}{2^{n}} (\tfrac{1}{m+1}-\tfrac{2}{n^{n}} (n^{m}&+(n-2)^{m} R^{\prime}+(n-4)^{m} R^{\prime \prime}+\text{etc. } \\ & +3^{m} R^{ \left(\frac{1}{2} n-\frac{3}{2}\right)}+R^{ \left(\frac{1}{2} n-\frac{1}{2}\right)})) \end{aligned}$$

if $n$ is odd.

Therefore, if expanding $y$ in a series according to powers of $t-\tfrac{1}{2}$  yields

$$y=L+L^{\prime} (t-\tfrac{1}{2})+L^{\prime \prime} (t-\tfrac{1}{2})^{2}+L^{\prime \prime \prime} (t-\tfrac{1}{2})^{3}+\text{etc. }$$

then the correction to be applied to the approximate value of the integral $\int y \operatorname{d} t$ from $t=0$  to $t=1$  will be

$$L l+L^{\prime \prime} l^{\prime \prime}+L^{\prime \prime \prime} l^{\prime \prime \prime}+L^{\scriptscriptstyle VI} l^{\scriptscriptstyle VI}+\text{etc. }$$

or rather, since $l^{(m)}$ necessarily vanishes for any integer value of $m$  no greater than $n,$  the correction will be

$$L^{(n+2)} l^{(n+2)}+L^{(n+4)} l^{(n+4)}+L^{(n+6)} l^{(n+6)}+\text{etc. }$$

for even $n,$ or

$$L^{(n+1)} l^{(n+1)}+L^{(n+3)} l^{(n+3)}+L^{(n+5)} l^{(n+5)}+\text{etc. }$$

for odd $n.$

The corrections $l^{(m)}$ can be easily converted to $k^{(m)}$  and vice versa. For if we have

$$(t-\tfrac{1}{2})^{m}=t^{m}-\tfrac{1}{2} m. t^{m-1}+\tfrac{1}{4} \cdot \tfrac{m(m-1)}{1.2} t^{m-2}+\text{etc. }$$

then

$$l^{(m)}=k^{(m)}-\tfrac{1}{2} m k^{(m-1)}+\tfrac{1}{4} \cdot \tfrac{m(m-1)}{1.2} k^{(m-2)}+\text{etc. }$$

And similarly,

$$k^{(m)}=l^{(m)}+\tfrac{1}{2} m l^{(m-1)}+\tfrac{1}{4} \cdot \tfrac{m(m-1)}{1. 2} l^{(m-1)}+\text{etc. }$$

The terms where $l$ is affected by an odd index will be eliminated from the latter formula, and each should only be continued up to the index $n+1$  (inclusive). Therefore, it is clear that we will havefor $n$ even$\phantom{\qquad \qquad k^{(n+3)}=\frac{n+3}{2}. l^{(n+2)}}$ $\begin{aligned} & k^{(n+1)}=0 \\ & k^{(n+2)}=l^{(n+2)} \\ & k^{(n+3)}=\tfrac{n+3}{2}. l^{(n+2)}\end{aligned}$ for $n$ odd$\phantom{\qquad \qquad k^{(n+3)}=\frac{n+3}{2}. l^{(n+2)}}$ $\begin{aligned} & k^{(n+1)}=l^{(n+1)} \\ & k^{(n+2)}=\tfrac{n+2}{2}. l^{(n+1)} \end{aligned}$ |undefined

from which the above observations can be deduced.

6.
Generally speaking, it will therefore be preferable to assign an even value to $n,$ or to employ an odd number of terms, when applying the method of. Indeed, very little precision will be gained by ascending from an even value of $n$ 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 $n$ 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 $n=4,$ the error is approximately expressed by $-\frac{1}{2688} K^{6}$  or $-\frac{1}{2688} L^{6};$  if we set $n=5,$  the error will be approximately $-\frac{11}{52500} K^{6}$  or $-\frac{11}{52500} L^{6},$  thus it will not even be half of the former: on the other hand, for $n=6,$  the error becomes approximately $=-\frac{1}{38880} K^{8}$  or $=-\frac{1}{38880} L^{8},$  and the precision is increased all the more, as the series into which the function has been expanded converges more quickly.

7.
Following these preliminaries regarding the method of, we proceed to a general inquiry, discarding the condition that the values of $x$ progress in an arithmetic progression. We thus address the problem of determining the value of the integral $\int y \operatorname{d} x$ between given limits from some given values of $y,$  either exactly or as closely as possible. Let us assume that the integral is to be taken from $x=g$ to $x=g+\Delta,$  and let us introduce another variable $t=\frac{x-g}{\Delta},$  so that the integral $\Delta \int y \operatorname{d} t$  from $t=0$  to $t=1$  needs to be investigated.

Let $a,$ $a^{\prime},$  $a^{\prime \prime},$  $a^{\prime \prime \prime} \ldots a^{(n)}$  be distinct values of $t,$  let the $n+1$  corresponding values of $y$  be $A,$  $A^{\prime},$  $A^{\prime \prime},$  $A^{\prime \prime \prime} \ldots A^{(n)},$  and let $Y$  denote the following integral algebraic function of order $n{:}$

$$\begin{aligned} & A^{\phantom{\prime \prime}}\frac{ (t-a^{\prime}) (t-a^{\prime \prime}) (t-a^{\prime \prime \prime}) \ldots (t-a^{(n)})}{ (a-a^{\prime}) (a-a^{\prime \prime}) (a-a^{\prime \prime \prime}) \ldots (a-a^{(n)})} \\ +&A^{\prime \phantom{\prime}}\, \frac{(t-a) (t-a^{\prime \prime}) (t-a^{\prime \prime \prime}) \ldots (t-a^{(n)})}{ (a^{\prime}-a) (a^{\prime}-a^{\prime \prime}) (a^{\prime}-a^{\prime \prime \prime}) \ldots (a^{\prime}-a^{(n)})} \\ +&A^{\prime \prime} \frac{(t-a) (t-a^{\prime}) (t-a^{\prime \prime \prime}) \ldots (t-a^{(n)})}{ (a^{\prime \prime}-a) (a^{\prime \prime}-a^{\prime}) (a^{\prime \prime}-a^{\prime \prime}) \ldots (a^{\prime \prime}-a^{(n)})} \\ +& \text{ etc. } \\ +&A^{(n)} \frac{(t-a) (t-a^{\prime})(t-a^{\prime\prime})\ldots(t-a^{(n-1)})}{ (a^{(n)}-a) (a^{(n)}-a^{\prime}) (a^{(n)}-a^{\prime \prime}) \ldots (a^{(n)}-a^{ (n-1)})} \end{aligned}$$

If $t$ is set equal to any of the quantities $a,$  $a^{\prime},$  $a^{\prime \prime},$  $a^{\prime \prime \prime} \ldots a^{(m)}$, it is clear that the values of this function, coincide with the corresponding values of the function $y,$  from which, as concluded in art. 2, we deduce that $Y$ is identical to $y$, provided that $y$  is also an integral algebraic function of order no greater than $n,$  or at least it can take the place of $y,$  if $y$  can be converted into a series of powers of $t$  which exhibits such convergence that it is permissible to neglect the higher order terms.

8.
To evaluate the integral $\int {Y} \operatorname{d} t,$ let us consider each part of ${Y}$  separately. Let $T$ denote the product

$$(t-a) (t-a^{\prime}) (t-a^{\prime \prime}) (t-a^{\prime \prime \prime}) \ldots (t-a^{(n)}),$$

and through the expansion of this product, let

$$T=t^{n+1}+\alpha t^{n}+\alpha^{\prime} t^{n-1}+\alpha^{\prime \prime} t^{n-2}+\text{etc.}+\alpha^{(n)}$$

The numerator of the fraction by which $A$ is multiplied in its respective part of ${Y},$  becomes $=\frac{T}{t-a};$  the numerators in the subsequent parts are likewise $\frac{T}{t-a^{\prime}},$  $\frac{T}{t-a^{\prime \prime}},$  $\frac{T}{t-a^{\prime \prime \prime}}$, etc. The denominators are nothing but the values determined by these numerators if $t$  is set respectively to $a,$  $a^{\prime},$  $a^{\prime \prime},$  $a^{\prime \prime \prime}$ , etc. Let us denote these denominators respectively by $M,$  $M^{\prime},$  $M^{\prime \prime},$  $M^{\prime \prime \prime}$ , etc., so that we have

$$Y=\frac{A T}{M(t-a)}+\frac{A^{\prime} T}{M^{\prime} (t-a^{\prime})}+\frac{A^{\prime \prime} T}{M^{\prime \prime} (t-a^{\prime \prime})}+\text{etc.}+\frac{A^{(n)} T}{M^{(n)} (t-a^{(n)})}$$

When $T=0,$ for $t=a,$  we have the identical equation

$$a^{n+1}+\alpha a^{n}+\alpha^{\prime} a^{n-1}+\alpha^{\prime \prime} a^{n-2}+\text{etc.}+\alpha^{(n)}=0$$

and therefore

$$T=t^{n+1}-a^{n+1}+\alpha (t^{n}-a^{n}) +\alpha^{\prime} (t^{n-1}-a^{n-1})+\alpha^{\prime \prime} (t^{n-2}-a^{n-2})+\text{etc.} +\alpha^{(n-1)}(t-a)$$

Thus, dividing by $t-a$, we get

$$\begin{alignedat}{6} \frac{T}{t-a}=t^{n}&+a t^{n-1}&&+a a t^{n-2}&&+a^{3} t^{n-3}&&+\text{etc.}&&+a^{n} \\ &+\alpha t^{n-1}&&+\alpha a t^{(n-2)}&&+\alpha a a t^{n-3}&&+\text{etc.}&&+\alpha a^{n-1} \\ &&&+\alpha^{\prime} t^{n-2}&&+\alpha^{\prime} a t^{n-3}&&+\text{etc.}&&+\alpha^{\prime} a^{n-2} \\ &&&&&+\alpha^{\prime \prime} t^{n-3^{-}}&&+\text{etc.}&&+\alpha^{\prime \prime} a^{n-3} \\ &&&&&&&+ \text{etc.}&&\text{ etc. } \\ &&&&&&&&&+\alpha^{(n-1)} \end{alignedat}$$

The value of this function for $t={a}$ is obtained as

$$=(n+1) a^{n}+n \alpha a^{n-1}+(n-1) \alpha^{\prime} a^{n-2}+(n-2) \alpha^{\prime \prime} a^{n-3}+\text{etc.}+\alpha^{(n-1)}$$

Hence $M$ is equal to the value of $\frac{\operatorname{d} T}{\operatorname{d} t}$  for $t=a,$  as is evident for other reasons. Similarly, $M^{\prime},$ $M^{\prime \prime},$  $M^{\prime \prime \prime},$  etc. will be the values of $\frac{\operatorname{d} T}{\operatorname{d} t}$  for $t=a^{\prime},$  $t=a^{\prime \prime},$  $t=a^{\prime \prime \prime}$, etc.

Furthermore, we find the value of the integral $\int \frac{T \operatorname{d} t}{t-a}$ from $t=0$  to $t=1,$  to be:

$$\begin{alignedat}{6} =\tfrac{1}{n+1}&+\tfrac{a}{n}&&+\tfrac{a a}{n-1}&&+\tfrac{a^{3}}{n-2}&&+\text{etc.} &&+a^{n} \\ &+\tfrac{\alpha}{n}&&+\tfrac{\alpha a}{n-1}&&+\tfrac{\alpha a a}{n-2} &&+ \text{etc.}&&+\alpha a^{n-1} \\ &&&+\tfrac{\alpha^{\prime}}{n-1}&&+\tfrac{\alpha^{\prime} a}{n-2} &&+ \text{etc.}&&+\alpha^{\prime} a^{n-2} \\ &&&&&+\tfrac{\alpha^{\prime \prime}}{n-2} &&+ \text{etc.}&&+\alpha^{\prime \prime} a^{n-3} \\ & &&&&&&+ \text{etc.} &&\text{ etc.} \\ & &&&&&&&&+\alpha^{(n-1)} \end{alignedat}$$

Let us arrange these terms in the following order:

$$\begin{aligned} a^{n}&+\alpha a^{n-1}+\alpha^{\prime} a^{n-2}+\alpha^{\prime \prime} a^{n-3}+\text{etc.}+\alpha^{(n-1)} \\ & +\tfrac{1}{2} (a^{n-1}+\alpha a^{n-2}+\alpha^{\prime} a^{n-3}+\text{etc.}+\alpha^{(n-2)}) \\ & +\tfrac{1}{3} (a^{n-2}+\alpha a^{n-3}+\alpha^{\prime} a^{n-4}+\text{etc.}+\alpha^{(n-3)}) \\ & +\tfrac{1}{4} (a^{n-3}+\alpha a^{n-4}+\alpha^{\prime} a^{n-5}+\text{etc.}+\alpha^{(n-4)}) \\ & + \text{etc.} \\ & +\tfrac{1}{n-1} (a a+\alpha a+\alpha^{\prime}) \\ & +\tfrac{1}{n}(a+\alpha) \\ & +\tfrac{1}{n+1} \end{aligned}$$

It is clear that the same quantity arises if, in the product obtained by multiplying the function $T$ by the infinite series

$$t^{-1}+\tfrac{1}{2} t^{-2}+\tfrac{1}{3} t^{-3}+\tfrac{1}{4} t^{-4} \text{ etc., }$$

all terms involving negative powers of $t$ are rejected (or in short, in the integral part of the product, which is an integral function of $t$ ), $t$  is replaced by $a.$  Therefore let us set

$$T (t^{-1}+\tfrac{1}{2} t^{-2}+\tfrac{1}{3} t^{-3}+\tfrac{1}{4} t^{-4}+\text{etc.})=T^{\prime}+T^{\prime \prime}$$

so that $T^{\prime}$ is the integral function of $t$  contained in this product, and $T^{\prime \prime}$  is the other part, namely the series descending to negative infinity. Then the value of the integral $\int \frac{T \operatorname{d} t}{t-a}$ from $t=0$  to $t=1$  will be equal to the value of the function $T^{\prime}$  at $t=a.$  So, if we denote the values of the function

$$\frac{T^{\prime}}{ (\frac{\operatorname{d} T}{\operatorname{d} t})}$$

determined by $t=a,$ $t=a^{\prime},$  $t=a^{\prime \prime},$  $t=a^{\prime \prime \prime}$, etc., up to $t=a^{(n)}$  resp. by $R,$ $R^{\prime},$  $R^{\prime \prime},$  $R^{\prime \prime \prime} \ldots R^{(n)},$  then the integral $\int Y \operatorname{d}$  from $t=0$  to $t=1$  will be

$$=R A+R^{\prime} A^{\prime}+R^{\prime \prime} A^{\prime \prime}+\text{etc.}+R^{(n)} A^{(n)}$$

which, when multiplied by $\Delta$, will give the value, either exact or approximate, of the integral $\int y \operatorname{d} x$ from $x=g$  to $x=g+\Delta$.

9.
These operations are somewhat easier to perform if we introduce another variable $u=2t-1.$ For the sake of brevity, we also write $b=2a-1,$  $b^{\prime}=2a^{\prime}-1,$  $b^{\prime\prime}=2a^{\prime\prime}-1,$  etc. By substituting the value $\frac{1}{2}u+\frac{1}{2}$  for $t,$  let $T$  be transformed into $\frac{U}{2^{n+1}},$  or equivalently let

$$U=(u-b)(u-b^{\prime})(u-b^{\prime\prime})\ldots(u-b^{n})$$

Then $\frac{\operatorname{d} T}{\operatorname{d} t}=\frac{1}{2^{n}} \cdot \frac{\operatorname{d} U}{\operatorname{d} u},$ and hence $M,$  $M^{\prime},$  $M^{\prime\prime}$  etc. are the values of $\frac{1}{2^{n}} \cdot \frac{\operatorname{d} U}{\operatorname{d} u}$  determined by $u=b,$  $u=b^{\prime},$  $u=b^{\prime\prime}$  etc.

Since the series $t^{-1}+\frac{1}{2} t^{-2}+\frac{1}{3} t^{-3}+\frac{1}{4} t^{-4}+$ etc. is nothing but $\log \frac{1}{1-t^{-1}}=\log \frac{1+u^{-1}}{1-u^{-1}}:$  substituting $t=\frac{1}{2}u+\frac{1}{2}$  will transform it into $2u^{-1}+\frac{2}{3}u^{-3}+\frac{2}{5}u^{-5}+\frac{2}{7}u^{-7}+$  etc. Therefore, if we set

$$U(u^{-1}+\tfrac{1}{3}u^{-3}+\tfrac{1}{3}u^{-5}+\tfrac{1}{7}u^{-7}+\text{etc.})=U^{\prime}+U^{\prime\prime},$$

so that $U^{\prime}$ is the integral function of $u$  contained in this product, and $U^{\prime\prime}$  is the other part, which is an infinite descending series, it is clear that

$$T^{\prime}+T^{\prime\prime}=\tfrac{1}{2^{n}}(U^{\prime}+U^{\prime\prime})$$

However, it is clear that $T^{\prime}$, being an integral function of $t,$ will necessarily become an integral function of $u$  as a result of the substitution $t=\frac{1}{2}u+\frac{1}{2}{;}$  on the other hand, $T^{\prime\prime}$ , which contains only negative powers of $t,$  will only generate negative powers of $u$  as a result of the same substitution. Therefore, $U^{\prime}$ will be nothing but $2^{n}T^{\prime}$  transformed by this substitution, and likewise $U^{\prime\prime}$  will be produced from $2^{n}T^{\prime\prime}.$  Consequently, it makes no difference whether we substitute $t=a$  into $\frac{T^{\prime}}{ (\frac{\operatorname{d} T}{\operatorname{d} t})}$  or $u=b$  into $\frac{U^{\prime}}{ (\frac{\operatorname{d} U}{\operatorname{d} u})}.$  From this we conclude that $R,$  $R^{\prime},$  $R^{\prime\prime},$  $R^{\prime\prime\prime}$  etc. are also the values of the function $\frac{U^{\prime}}{ (\frac{\operatorname{d} U}{\operatorname{d} u})}$  determined by $u=b,$  $u=b^{\prime},$  $u=b^{\prime\prime},$  $u=b^{\prime\prime\prime}$  etc.

10.
Before we proceed further, we will illustrate these precepts with an example. Let $n=5,$ and suppose that $a=0,$  $a^{\prime}=\frac{1}{5},$  $a^{\prime \prime}=\frac{2}{5},$  $a^{\prime \prime \prime}=\frac{3}{5},$  $a^{\prime \prime \prime \prime}=\frac{4}{5},$  $a^{\prime \prime \prime \prime \prime}=1.$  Then we have

$$T=t^{6}-3 t^{5}+\tfrac{17}{5} t^{4}-\tfrac{9}{5} t^{3}+\tfrac{274}{625} t t -\tfrac{24}{625} t.$$

Multiplying by $t^{-1}+\frac{1}{2} t^{-2}+\frac{1}{3} t^{-3}+\frac{1}{4} t^{-4}+$ etc., we obtain

$$T^{\prime}=t^{5}-\tfrac{5}{2} t^{4}+\tfrac{67}{30} t^{3}-\tfrac{17}{20} t t+\tfrac{913}{7500} t-\tfrac{19}{7500}$$

Hence the values of the coefficients ${R},$ ${R}^{\prime},$  ${R}^{\prime \prime},$  ${R}^{\prime \prime \prime},$  ${R}^{\prime \prime \prime \prime},$  ${R}^{\prime \prime \prime \prime \prime}$  are expressed by the fractional function

$$\frac{t^5-\frac{2}{5}t^4+\frac{67}{30}t^3-\frac{17}{20}tt+\frac{913}{7500}t-\frac{19}{7500}}{6t^5-15t^4+\frac{68}{5}t^3-\frac{27}{5}tt+\frac{548}{625}t-\frac{24}{625}}$$

wherein the values $0,$ $\frac{1}{5},$  $\frac{2}{5},$  $\frac{3}{5},$  $\frac{4}{5},$  $1$  are subsequently substituted for $t.$  The other method, which is a bit faster, yields $b=-1,$  $b^{\prime}=-\frac{3}{5},$  $b^{\prime \prime}=-\frac{1}{5},$  $b^{\prime \prime \prime}=\frac{1}{5},$  $b^{\prime \prime \prime \prime}=\frac{3}{5},$  $b^{\prime \prime \prime \prime \prime}=1,$

$$\begin{aligned} & U=u^{6}-\tfrac{7}{5} u^{4}+\tfrac{259}{625} u u-\tfrac{9}{625} \\ & U^{\prime}=u^{5}-\tfrac{16}{15} u^{3}+\tfrac{277}{1875} u \end{aligned}$$

from which $R,$ $R^{\prime},$  $R^{\prime \prime}$, etc. will be values of the fractional function

$$\frac{u^{4}-\frac{16}{15} u u+\frac{277}{1875}}{6 u^{4}-\frac{28}{5} u u+\frac{518}{625} }$$

for $u=-1,$ $u=-\frac{3}{5},$  $u=-\frac{1}{5}$, etc. Both methods yield the same numbers given in Art. 4 of Harmonia Mensurarum. However, in such an example like this, where $a,$ $a^{\prime},$  $a^{\prime \prime}$, etc. are all rational quantities, the values of the denominator $\frac{\operatorname{d} T}{\operatorname{d} t}$  are more conveniently computed in the original form, namely $(a-a^{\prime}) (a-a^{\prime \prime}) (a-a^{\prime \prime \prime}) \ldots (a-a^{(n)})$  for $t=a,$  and likewise for the others. The same holds for the denominator $\frac{\operatorname{d} U}{\operatorname{d} u}$, which for $u=b$ becomes $= (b-b^{\prime}) (b-b^{\prime \prime}) (b-b^{\prime \prime \prime}) \ldots (b-b^{(n)}).$

11.
When $a,$ $a^{\prime},$  $a^{\prime \prime}$, etc., are either partially or altogether irrational, it will be useful to transform the fractional function, from which we derive the numbers $R,$  $R^{\prime},$  $R^{\prime \prime}$ , 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 $Z,$ $\zeta,$  $\zeta^{\prime}$  be three indeterminate integral functions of $z,$  and let us seek an integral function which can be substituted for the fraction $\frac{Z}{\zeta}$, as far as $z$  is taken to be any root of the equation $\zeta^{\prime}=0.$   Let us assume that $\zeta$  does not vanish for any of these values of $z$ , or equivalently, that $\zeta$  and $\zeta^{\prime}$  imply no common indeterminate divisor. We will denote the exponents of the highest powers of $z$ in $\zeta$  and $\zeta^{\prime}$  by $k,$  $k^{\prime},$  respectively.

Divide $\zeta$ by $\zeta^{\prime},$  as is usual, until the order of the remainder is less than $k^{\prime};$  let the remainder be $=\frac{\zeta^{\prime \prime}}{\lambda},$  and let its order be $=k^{\prime \prime},$  so that $\frac{1}{\lambda} z^{k^{\prime \prime}}$  is the highest order term of the remainder; we will denote the quotient of this division by $\frac{p}{\lambda}.$  Similarly, divide the function $\zeta^{\prime}$  by $\zeta^{\prime \prime},$  let the residue $\frac{\zeta^{\prime \prime \prime}}{\lambda^{\prime}}$  of order $k^{\prime \prime \prime}$  be obtained as $\frac{p^{\prime}}{\lambda^{\prime}};$  then again from the division of the function $\zeta^{\prime \prime}$  by $\zeta^{\prime \prime \prime},$  let the residue $\frac{\zeta^{\prime \prime \prime \prime}}{\lambda^{\prime \prime}}$  of order $k^{\prime \prime \prime \prime}$  be obtained as $\frac{p^{\prime \prime}}{\lambda^{\prime \prime}}$, and so on, until in the series of functions $\zeta^{\prime \prime},$  $\zeta^{\prime \prime \prime},$  $\zeta^{\prime \prime \prime \prime},$  etc., each having its highest term affected by a coefficient of $1,$  we arrive at $\zeta^{(m)}=1.$   It is easy to see that this must eventually happen, since none of the functions $\zeta,$  $\zeta^{\prime},$  $\zeta^{\prime \prime},$  $\zeta^{\prime \prime \prime},$  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 $0.$  Thus, we will have a series of equations:

$\begin{aligned} & \zeta^{\prime \prime \phantom{\prime \prime \prime}}=\lambda \zeta-p \zeta^{\prime} \\ & \zeta^{\prime \prime \prime \phantom{\prime \prime}}=\lambda^{\prime} \zeta^{\prime}-p^{\prime} \zeta^{\prime \prime} \\ & \zeta^{\prime \prime \prime \prime \phantom{\prime}}=\lambda^{\prime \prime} \zeta^{\prime \prime}-p^{\prime \prime} \zeta^{\prime \prime \prime} \\ & \zeta^{\prime \prime \prime \prime \prime}=\lambda^{\prime \prime \prime} \zeta^{\prime \prime \prime}-p^{\prime \prime \prime} \zeta^{\prime \prime \prime \prime} \end{aligned}$ etc., up to $ \zeta^{(m)}=\lambda^{(m-2)} \zeta^{(m-2)}-p^{(m-2)} \zeta^{(m-1)}$ |undefined

where $\zeta^{\prime \prime},$ $\zeta^{\prime \prime \prime},$  $\zeta^{\prime \prime \prime \prime},$  etc., $\zeta^{(m)}$  are integral functions of $z,$  of orders $k^{\prime \prime},$  $k^{\prime \prime \prime},$  $k^{\prime \prime \prime \prime}\dots k^{(m)};$  where the numbers $k^{\prime},$  $k^{\prime \prime},$  $k^{\prime \prime \prime},$  etc., continuously decrease until the last one $k^{(m)}=0;$  and $p,$  $p^{\prime},$  $p^{\prime \prime},$  $p^{\prime \prime \prime}$, etc., are integral functions of $z$  of orders $k-k^{\prime},$  $k^{\prime}-k^{\prime \prime},$  $k^{\prime \prime}-k^{\prime \prime \prime},$  $k^{\prime \prime \prime}-k^{\prime \prime \prime \prime}$ , etc. (except in the case where $k<k^{\prime},$  where it is clear that we must set $p=0$ ).

Having prepared in this manner, we form a second series of integral functions of $z,$ which we call $\eta,$  $\eta^{\prime},$  $\eta^{\prime \prime},$  $\eta^{\prime \prime \prime},$  etc., up to $\eta^{(m)}.$  Indeed, let us set $\eta=1,$  $\eta^{\prime}=0,$  and for the remaining functions we derive each from the preceding two according to the same rule by which the functions $\zeta,$  $\zeta^{\prime},$  $\zeta^{\prime \prime},$  $\zeta^{\prime \prime \prime}$, etc., are related to each other, namely through the following equations:

$\begin{aligned} &\eta^{\prime \prime \phantom{\prime \prime \prime}} =\lambda \eta-p \eta^{\prime} \\ &\eta^{\prime \prime \prime \phantom{\prime \prime}} =\lambda^{\prime} \eta^{\prime}-p^{\prime} \eta^{\prime \prime} \\ &\eta^{\prime \prime \prime \prime \phantom{\prime}} =\lambda^{\prime \prime} \eta^{\prime \prime}-p^{\prime \prime} \eta^{\prime \prime \prime} \\ &\eta^{\prime \prime \prime \prime \prime} =\lambda^{\prime \prime \prime} \eta^{\prime \prime \prime}-p^{\prime \prime \prime} \eta^{\prime \prime \prime \prime} \end{aligned}$ etc., up to $ \eta^{(m)} =\lambda^{(m-2)} \eta^{(m-2)}-p^{(m-2)} \eta^{(m-1)}$ |undefined Clearly $\eta^{\prime \prime}=\lambda$ is of order $0$  here; $\eta^{\prime \prime \prime}=-\lambda p^{\prime}$  is of order $k^{\prime}-k^{\prime \prime},$  and likewise the subsequent functions $\eta^{\prime \prime \prime \prime},$  $\eta^{\prime \prime \prime \prime \prime}$, etc., are of orders $k^{\prime}-k^{\prime \prime \prime},$  $k^{\prime}-k^{\prime \prime \prime \prime}$ , etc., so that the last one $\eta^{(m)}$  is of order $k^{\prime}-k^{(m-1)}.$

Next we consider a "third" series of functions, $\zeta-\zeta \eta,$ $\zeta^{\prime}-\zeta \eta^{\prime},$  $\zeta^{\prime \prime}-\zeta \eta^{\prime \prime},$  $\zeta^{\prime \prime \prime}-\zeta \eta^{\prime \prime \prime},$  etc., among which any three consecutive terms will manifestly have a similar relation, namely,

$$\begin{aligned} \zeta^{\prime \prime}-\zeta \eta^{\prime \prime}&=\lambda(\zeta-\zeta \eta)-p (\zeta^{\prime}-\zeta \eta^{\prime}) \\ \zeta^{\prime \prime \prime}-\zeta \eta^{\prime \prime \prime}&=\lambda^{\prime} (\zeta^{\prime}-\zeta \eta^{\prime})-p^{\prime} (\zeta^{\prime \prime}-\zeta \eta^{\prime \prime}) \\ \zeta^{\prime \prime \prime \prime}-\zeta \eta^{\prime \prime \prime \prime}&=\lambda^{\prime \prime} (\zeta^{\prime \prime}-\zeta \eta^{\prime \prime})-p^{\prime \prime} (\zeta^{\prime \prime \prime}-\zeta \eta^{\prime \prime \prime}) \end{aligned}$$

Now, the first of these functions is $=0,$ the second is $=\zeta^{\prime},$  hence it is easily inferred that each is divisible by $\zeta^{\prime}.$

Moreover, it follows without difficulty that we can replace the fraction $\frac{Z}{\zeta},$ with the integral function $Z \eta^{(m)},$  provided that no values are assigned to ${z}$  other than those which are roots of the equation $\zeta^{\prime}=0;$  for it is clear that the difference $\frac{Z (1-\zeta \eta^{(m)})}{\zeta}$  must vanish for such a value of $z,$  since $1-\zeta \eta^{(m)}=\zeta^{(m)}-\zeta \eta^{(m)}$  is divisible by $\zeta^{\prime}.$

Instead of the function $Z \eta^{(m)},$ we can also take the remainder which arises upon dividing it by $\zeta^{\prime},$  whose order will be lower than the order of the function $\zeta^{\prime}.$

Indeed, this remainder can be immediately and more conveniently extracted using the following algorithm. We form the following equations:

$\begin{aligned} Z^{\phantom{\prime \prime \prime}} &=q^{\prime} \zeta^{\prime}+Z^{\prime} \\ Z^{\prime \phantom{\prime \prime}} &=q^{\prime \prime} \zeta^{\prime \prime}+Z^{\prime \prime} \\ Z^{\prime \prime \phantom{\prime}} &=q^{\prime \prime \prime} \zeta^{\prime \prime \prime}+Z^{\prime \prime \prime} \\ Z^{\prime \prime \prime} &=q^{\prime \prime \prime \prime} \zeta^{\prime \prime \prime \prime}+Z^{\prime \prime \prime \prime} \end{aligned}$ etc., up to $Z^{(m-1)} =q^{(m)} \zeta^{(m)}+Z^{(m)}$ |undefined by dividing $Z$ by $\zeta^{\prime},$  then the remainder of the first division $Z^{\prime}$  by $\zeta^{\prime \prime},$  then the remainder of the second division by $\zeta^{\prime \prime \prime},$  and so forth. Since the remainder always belongs to an order lower than the divisor, the order of the functions $Z^{\prime},$ $Z^{\prime \prime},$  $Z^{\prime \prime \prime},$  $Z^{\prime \prime \prime},$  etc. will be respectively lower than $k^{\prime},$  $k^{\prime \prime},$  $k^{\prime \prime \prime},$  $k^{\prime \prime \prime},$  etc.; while the last $Z^{(m)}$  necessarily becomes $=0,$  since the divisor $\zeta^{(m)}$  is $=1.$  Therefore, we have

$$Z=q^{\prime} \zeta^{\prime}+q^{\prime \prime} \zeta^{\prime \prime}+q^{\prime \prime \prime} \zeta^{\prime \prime \prime}+q^{\prime \prime \prime \prime} \zeta^{\prime \prime \prime \prime}+\text{etc.}+q^{(m)} \zeta^{(m)}$$

Moreover, since only the roots of the equation $\zeta^{\prime}=0$ are taken for $z,$  it follows that $\zeta^{\prime}=0,$  $\zeta^{\prime \prime}=\zeta \eta^{\prime \prime},$  $\zeta^{\prime \prime \prime}=\zeta \eta^{\prime \prime \prime},$  $\zeta^{\prime \prime \prime \prime}=\zeta \eta^{\prime \prime \prime \prime}$  etc., and under the same restriction, it follows that

$$\frac{Z}{\zeta}=q^{\prime \prime} \eta^{\prime \prime}+q^{\prime \prime \prime} \eta^{\prime \prime \prime}+q^{\prime \prime \prime \prime} \eta^{\prime \prime \prime \prime}+\text{etc.}+q^{(m)} \eta^{(m)}$$

However, the order of this expression will necessarily be less than than $k^{\prime}{:}$ since the order of the quotients $q^{\prime \prime},$  $q^{\prime \prime \prime},$  $q^{\prime \prime \prime},$  etc. must be less than $k^{\prime}-k^{\prime \prime},$  $k^{\prime \prime}-k^{\prime \prime \prime},$  $k^{\prime \prime \prime}-k^{\prime \prime \prime},$  etc., the order of each part $q^{\prime \prime} \eta^{\prime \prime},$  $q^{\prime \prime \prime} \eta^{\prime \prime \prime},$  $q^{\prime \prime \prime \prime} \eta^{\prime \prime \prime \prime},$  etc. will be less than $k^{\prime}-k^{\prime \prime},$  $k^{\prime}-k^{\prime \prime \prime},$  $k^{\prime}-k^{\prime \prime \prime},$  etc.

Finally we observe that if it so happens that among the values of the indeterminate $z,$ those that need to be substituted in the fraction $\frac{Z}{\zeta}$  are a mixture of rationals and irrationals, it will be more practical to separate them and only include the latter in the equation $\zeta^{\prime}=0.$  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.

12.
Here is an example of the transformation explained in the preceding article. Let the given fractional function be

$$\frac{z^{6}-\frac{50}{39} z^{4}+\frac{283}{715} z z-\frac{256}{15015}}{7 z^{6}-\frac{105}{13} z^{4}+\frac{315}{143} z z-\frac{35}{429}}$$

where $z$ indefinitely represents the roots of the equation

$$z^{7}-\tfrac{21}{13} z^{5}+\tfrac{105}{143} z^{3}-\tfrac{35}{429} z=0.$$

If we wanted to include all seven roots here, we would descend to a sixth-order integral function. However, for the rational value $z=0,$ the calculation of the fraction is straightforward, giving the value $\frac{256}{1225},$  so excluding this root from the equation of sixth degree, we have:

$$z^{6}-\tfrac{21}{13} z^{4}+\tfrac{105}{143} z z-\tfrac{35}{429}=0$$

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:

$$\begin{aligned} & \zeta^{\phantom{\prime \prime \prime \prime}}=7 z^{6}-\tfrac{105}{13} z^{4}+\tfrac{315}{143} z z-\tfrac{35}{429} \\ & \zeta^{\prime \phantom{\prime \prime \prime}}=z^{6}-\tfrac{21}{13} z^{4}+\tfrac{105}{143} z z-\tfrac{35}{429} \\ & \zeta^{\prime \prime \phantom{\prime \prime}}=z^{4}-\tfrac{10}{11} z z+\tfrac{5}{33} \\ & \zeta^{\prime \prime \prime \phantom{\prime}}=z z-\tfrac{3}{7} \\ & \zeta^{\prime \prime \prime \prime}=1 \\ & \begin{alignedat}{2} & \lambda^{\phantom{\prime \prime \prime \prime}}=\tfrac{13}{42} && \qquad p=\tfrac{13}{6} \\ & \lambda^{\prime \phantom{\prime \prime \prime}}=-\tfrac{4719}{280} &&\qquad p^{\prime}=-\tfrac{4719}{280} z z+\tfrac{3333}{280} \\ & \lambda^{\prime \prime \phantom{\prime \prime}}=-\tfrac{147}{8} && \qquad p^{\prime \prime}=-\tfrac{147}{8} z z+\tfrac{777}{88} \end{alignedat}\\ & \eta^{\phantom{\prime \prime \prime \prime}}=1 \\ & \eta^{\prime \phantom{\prime \prime \prime}}=0 \\ & \eta^{\prime \prime \phantom{\prime \prime }}=\tfrac{13}{42} \\ & \eta^{\prime \prime \prime \phantom{\prime}}=\tfrac{20449}{3920} z z-\tfrac{14443}{3920} \\ & \eta^{\prime \prime \prime \prime}= \tfrac{61347}{640} z^{4} - \tfrac{127413}{1120} z z+ \tfrac{120263}{4480} \\ & \begin{alignedat}{2} & Z^{\phantom{\prime \prime \prime \prime} }=z^{6}-\tfrac{50}{39} z^{4}+\tfrac{283}{715} z z-\tfrac{256}{15015} ; &&\qquad q^{\prime}=1 \\ & Z^{\prime \phantom{\prime \prime \prime}}=\tfrac{1}{3} z^{4}-\tfrac{22}{65} z z+\tfrac{323}{5005} &&\qquad q^{\prime \prime}=\tfrac{1}{3} \\ & Z^{\prime \prime \phantom{\prime \prime}}=-\tfrac{76}{2145} z z+\tfrac{632}{45045} &&\qquad q^{\prime \prime \prime}=-\tfrac{76}{2145} \\ & Z^{\prime \prime \prime \phantom{\prime}}=-\tfrac{4}{3465} &&\qquad q^{\prime \prime \prime \prime}=-\tfrac{4}{3465}\end{alignedat} \end{aligned}$$

Hence, the integral function equivalent to the given fraction is derived to be:

$$-\tfrac{1859}{16800} z^{4}-\tfrac{1573}{29400} z z+\tfrac{7947}{39200}$$

13.
To determine the degree of precision of our integral formula ${R} A+{R}^{\prime} A^{\prime}+{R}^{\prime \prime} A^{\prime \prime}+\text{etc.}+{R}^{(n)} A^{(n)},$ let us generally set

$${R} {a}^{m}+{R}^{\prime} {a}^{\prime m}+{R}^{\prime \prime} {a}^{\prime \prime m}+\text{etc.}+{R}^{(n)} a^{(n) m}=\tfrac{1}{m+1} - k^{(m)}$$

so that $k^{(m)}$ is the difference between the true and approximate values of the integral $\int t^{m} \operatorname{d} t$  taken from $t=0$  to $t=1$. Thus, expanding each fraction into a series gives us

$$\begin{aligned} & \frac{R}{t-a}+\frac{R^{\prime}}{t-a^{\prime}}+\frac{R^{\prime \prime}}{t-a^{\prime \prime}}+\text{etc.}+\frac{R^{(n)}}{t-a^{(n)}} \\ & \quad=(1-k) t^{-1}+ (\tfrac{1}{2}-k^{\prime}) t^{-2}+ (\tfrac{1}{3}-k^{\prime \prime}) t^{-3}+ (\tfrac{1}{4}-k^{\prime \prime \prime}) t^{-4}+\text{etc. } \\ & \quad=t^{-1}+\tfrac{1}{2} t^{-2}+\tfrac{1}{3} t^{-3}+\tfrac{1}{4} t^{-4}+\text{etc.}-\Theta \end{aligned}$$

where

$$\Theta=k t^{-1}+k^{\prime} t^{-2}+k^{\prime \prime} t^{-3}+k^{\prime \prime \prime} t^{-4}+\text{etc.,}$$

or rather (since we already know that $k,$ $k^{\prime},$  $k^{\prime \prime},$  $k^{\prime \prime \prime}$  etc. up to $k^{(n)}$  vanish automatically)

$$\Theta=k^{(n+1)} t^{-(n+2)}+k^{(n+2)} t^{-(n+3)}+k^{(n+3)} t^{-(n+4)}+\text{etc.}$$

Multiplying this by $T$ gives

$$T \left(\frac{R}{t-a}+\frac{R^{\prime}}{t-a^{\prime}}+\frac{R^{\prime \prime}}{t-a^{\prime \prime}}+\text{etc.}+\frac{R^{(n)}}{t-a^{(n)}}\right)=T^{\prime}+T^{\prime \prime}-T \Theta.$$

The first part of this equation is an integral function of $t,$ of order $n,$  and its values for $t=a,$  $t=a^{\prime},$  $t=a^{\prime \prime}$  etc. are $M R,$  $M^{\prime} R^{\prime},$  $M^{\prime \prime} R^{\prime \prime}$  etc.: therefore, since the same holds for the function $T^{\prime},$  it is clear from the method by which the numbers $R,$  $R^{\prime},$  $R^{\prime \prime}$  etc. are determined that this first part of the equation must be identical to $T^{\prime},$  and thus $T^{\prime \prime}=T \Theta.$  Therefore, $\Theta$  can be found by expanding the fraction $\frac{T^{\prime \prime}}{T},$  and accordingly the coefficients $k^{(n+1)},$  $k^{(n+2)}$  etc. can be determined. These being found, the correction to the approximate value of our integral $\int y \operatorname{d} t$ will be

$$=k^{(n+1)} K^{(n+1)}+k^{(n+2)} K^{(n+2)}+\text{etc.,}$$

if the series into which $y$ has been expanded is

$$y=K+K^{\prime} t+K^{\prime \prime} t t+K^{\prime \prime \prime} t^{3}+\text{etc. }$$

14.
If desired, the correction can be expressed in terms of the coefficients of powers of $t-\frac{1}{2}$  in the series

$$y=L+L^{\prime} (t-\tfrac{1}{2})+L^{\prime \prime} (t-\tfrac{1}{2})^{2}+L^{\prime \prime \prime} (t-\tfrac{1}{2})^{3}+\text{etc. }$$

which is

$$=l^{(n+1)} L^{(n+1)}+l^{(n+2)} L^{(n+2)}+l^{(n+3)} L^{(n+3)}+\text{etc. }$$

if we generally write $l^{(m)}$ for the correction to the approximate value of the integral $\int (t-\frac{1}{2})^{m} \operatorname{d} t.$  These corrections $l^{(m)}$  are related to the corrections $k^{(m)}$  via the equation

$$l^{(m)}=k^{(m)}-\tfrac{1}{2} m k^{(m-1)}+\tfrac{1}{4} \cdot \tfrac{m. m-1}{1. 2} k^{(m-2)}-\tfrac{1}{8} \cdot \tfrac{m. m-1. m-2}{1. 2 . 3} k^{(m-3)}+\text{etc. }$$

Thus we can independently determine them, by considering that the substitution $t=\frac{1}{2} u+\frac{1}{2}$ transforms the function $\Theta$  into

$$\begin{aligned} & 2 k (u^{-1}-u^{-2}+u^{-3}-u^{-4}+\text{etc.}) \\ + & 4 k^{\prime} (u^{-2}-2 u^{-3}+3 u^{-4}-4 u^{-5}+\text{etc.}) \\ + & 8 k^{\prime \prime} (u^{-3}-3 u^{-4}+6 u^{-5}-10 u^{-6}+\text{etc.}) \\ + & 16 k^{\prime \prime \prime} (u^{-4}-4 u^{-5}+10 u^{-6}-20 u^{-7}+\text{etc.}) \\ + & \text{ etc. } \end{aligned}$$

or into

$$\begin{alignedat}{1} 2 k u^{-1}&+4 (k^{\prime}-\tfrac{1}{2}) u^{-2}+8 (k^{\prime \prime}-\tfrac{1}{2} \cdot 2 k^{\prime}+\tfrac{1}{4} k) u^{-3} \\ &+16 (k^{\prime \prime \prime}-\tfrac{1}{2} \cdot 3 k^{\prime \prime}+\tfrac{1}{4} \cdot 3 k^{\prime}-\tfrac{1}{8} k) u^{-4}+\text{etc. } \end{alignedat}$$

or into

$$2 l u^{-1}+4 l^{\prime} u^{-2}+8 l^{\prime \prime} u^{-3}+16 l^{\prime \prime \prime} u^{-4}+\text{etc. }$$

or finally, since we know a priori that $l,$ $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc. up to $l^{(n)}$  automatically vanish, into

$$2^{n+2} l^{(n+1)} u^{-(n+2)}+2^{n+3} l^{(n+2)} u^{-(n+3)}+2^{n+4} l^{(n+4)} u^{-(n+4)}+\text{etc. }$$

But $\Theta=\frac{T^{\prime \prime}}{T};$ therefore, since the substitution $t=\frac{1}{2} u+\frac{1}{2}$  transforms $T,$  $T^{\prime \prime}$  into $\frac{U}{2^{n+1}},$  $\frac{U^{\prime \prime}}{2^{n}},$  (art. 9), the same substitution will transform the function $\Theta$  into $\frac{2 U^{\prime \prime}}{U}.$  Therefore, if we let $\Omega$  denote the series  expansion of the fraction $\frac{U^{\prime \prime}}{U},$  we will have

$$\Omega=2^{n+1} l^{(n+1)} u^{-(n+2)}+2^{n+2} l^{(n+2)} u^{-(n+3)}+2^{n+3} l^{(n+3)} u^{-(n+4)}+\text{etc. }$$

in this manner, as many of the coefficients $l^{(n+1)},$ $l^{(n+2)}$  etc. as are desired may be found.

Thus in the example of art. 10, we find

$$\begin{aligned} & U^{\prime \prime}=-\tfrac{176}{13125} u^{-1}-\tfrac{304}{28125} u^{-3}-\tfrac{2576}{309375} u^{-5}-\text{etc. } \\ & \Omega=-\tfrac{176}{13125} u^{-7}-\tfrac{832}{28125} u^{-9}-\tfrac{189856}{4296875} u^{-11}-\text{etc. } \end{aligned}$$

and so the correction to the approximate value of the integral is

$$=-\tfrac{11}{52500} L^{\scriptscriptstyle VI}-\tfrac{13}{112500} L^{\scriptscriptstyle VIII}-\tfrac{5933}{137500000} L^{\scriptscriptstyle X}-\text{etc. }$$

15.
The coefficient $K^{(m)}$ in the series expansion of the function $y$  is, by the theorem of, equal to the value of the expression

$$\frac{1}{1.2 \cdot 3 \ldots m} \cdot \frac{\operatorname{d}^{m} y}{\operatorname{d} t^{m}} $$

or

$$\frac{\Delta^{m}}{1.2 \cdot 3 \ldots m} \cdot \frac{\operatorname{d}^{m} y}{\operatorname{d} x^{m}}$$

for $t=0$ or $x=g;$  similarly, the coefficient $L^{(m)}$  is the value of the same expression at $t=\frac{1}{2}$  or $u=0$  or $x=g+\frac{1}{2} \Delta{:}$  We assign the order $m$  to both coefficients. Therefore, generally speaking, our integration will be exact for orders up to and including $n$, no matter which values are chosen for $a,$ $a^{\prime},$  $a^{\prime \prime} \ldots a^{(n)}$. However, there is nothing to prevent the precision from being raised to a higher degree by judiciously selecting the values of these quantities. Thus, as we have already seen, in Cotes's method, i.e., for $a=0,$ $a^{\prime}=\frac{1}{n},$  $a^{\prime \prime}=\frac{2}{n},$  $a^{\prime \prime \prime}=\frac{3}{n}$, etc., the precision is automatically extended to order $n+1$  whenever $n$  is even. Generally, it is evident that if the values of $a,$ $a^{\prime},$  $a^{\prime \prime},$  $a^{\prime \prime \prime}$, etc. are chosen so that in the function $T^{\prime \prime}$  or $U^{\prime \prime}$ , one or more terms vanish from the beginning, the precision will be increased by as many degrees beyond order $n$  as there are terms vanishing. Hence, it is easily inferred that when the number of quantities allowed to be chosen is $n+1,$ by properly determining them, the precision can always be raised to order $2 n+1$  inclusive. Thus, with only $n+1$ terms, the same order of precision can be achieved, which would require $2 n+1$  or $2 n+2$  terms if we were following the method of.

16.
The whole matter revolves around determining, for any given value of $n$, a function $T$ of the form $t^{n+1}+\alpha t^{n}+\alpha^{\prime} t^{n-1}+\alpha^{\prime \prime} t^{n-2}$  etc., such that when the product

$$T (t^{-1}+\tfrac{1}{2} t^{-2}+\tfrac{1}{3} t^{-3}+\tfrac{1}{4} t^{-4}+\text{etc.})$$

is expanded into powers $t^{-1},$ $t^{-2},$  $t^{-3} \ldots t^{-(n+1)}$, all of the coefficients are found to be $0;$  or if one prefers, a function $U$  of the form $u^{n+1}+\beta u^{n}+\beta^{\prime} u^{n-1}+\beta^{\prime \prime} u^{n-2}+$  etc., whose product with $u^{-1}+\frac{1}{3} u^{-3}+\frac{1}{3} u^{-5}+\frac{1}{7} u^{-7}+$  etc. is free of the powers $u^{-1},$ $u^{-2},$  $u^{-3},$  $u^{-4} \ldots u^{-(n+1)}.$  The latter method will be somewhat simpler, since it is easily seen that the coefficients of $U$, to satisfy the prescribed condition, must alternately vanish, i.e., $\beta=0,$  $\beta^{\prime \prime}=0,$  $\beta^{\prime \prime \prime \prime}=0$ , etc., so around half of the work can already be considered complete. Let us consider some of the simpler cases.

I. For $n=0,$ the sole coefficient of $t^{-1}$  in the product

$$(t+\alpha) (t^{-1}+\frac{1}{2} t^{-2}+\frac{1}{3} t^{-3}+\text{etc.})$$

must vanish. This yields $\alpha=\frac{1}{2},$ hence $T=t-\frac{1}{2}.$  Similarly, $U=u.$

II. For $n=1,$ the determination of $T$  depends on two equations:

$$\begin{aligned} & 0=\frac{1}{3}+\frac{1}{2} \alpha+\alpha^{\prime} \\ & 0=\frac{1}{4}+\frac{1}{3} \alpha+\frac{1}{2} \alpha^{\prime} \end{aligned}$$

From which we deduce $\alpha=-1,$ $\alpha^{\prime}=\frac{1}{6},$  hence $T=t t-t+\frac{1}{6}.$  The determination of $U$  yields a single equation:

$$0=\frac{1}{3}+\beta^{\prime}$$

Thus, $\beta^{\prime}=-\frac{1}{3},$ leading to $U=u u-\frac{1}{3}.$

III. For $n=2,$ the function $T$  is determined by three equations:

$$\begin{aligned} & 0=\frac{1}{4}+\frac{1}{3} \alpha+\frac{1}{2} \alpha^{\prime}+\alpha^{\prime \prime} \\ & 0=\frac{1}{5}+\frac{1}{4} \alpha+\frac{1}{3} \alpha^{\prime}+\frac{1}{2} \alpha^{\prime \prime} \\ & 0=\frac{1}{6}+\frac{1}{5} \alpha+\frac{1}{4} \alpha^{\prime}+\frac{1}{3} \alpha^{\prime \prime} \end{aligned}$$

This yields $\alpha=-\frac{3}{2}, \alpha^{\prime}=\frac{3}{5}, \alpha^{\prime \prime}=-\frac{1}{20},$ hence $T=t^{3}-\frac{3}{2} t t+\frac{3}{5} t-\frac{1}{20}.$  To determine $U,$  a single equation suffices:

$$0=\frac{1}{5}+\frac{1}{3} \beta^{\prime}$$

Thus, $\beta^{\prime}=-\frac{3}{5},$ leading to $U=u^{3}-\frac{3}{5} u.$

However, we will not further pursue this method, which leads to more and more complicated calculations. Instead, we will proceed to the genuine source of the general solution.

17.
Given a continued fraction

$$\varphi = \cfrac{v}{ w+\cfrac{v^{\prime}}{ w^{\prime}+\cfrac{ v^{\prime \prime}}{ w^{\prime \prime}+\cfrac{ v^{\prime \prime \prime}}{ w^{\prime \prime \prime}+\text{ etc.}}}}}$$

a series of fractions that are succesively closer to it can be found using the following algorithm. Two series of quantities, $V,$ $V^{\prime},$  $V^{\prime \prime},$  $V^{\prime \prime \prime}$  etc., and $W,$  $W^{\prime},$  $W^{\prime \prime},$  $W^{\prime \prime \prime}$  etc., are formed using the formulas

$$\begin{array}{ll} V=0 & \quad W=1 \\ V^{\prime}=v & \quad W^{\prime}=w W \\ V^{\prime \prime}=w^{\prime} V^{\prime}+v^{\prime} V &\quad W^{\prime \prime}=w^{\prime} W^{\prime}+v^{\prime} W \\ V^{\prime \prime \prime}=w^{\prime \prime} V^{\prime \prime}+v^{\prime \prime} V^{\prime} &\quad W^{\prime \prime \prime}=w^{\prime \prime} W^{\prime \prime}+v^{\prime \prime} W^{\prime} \\ V^{\prime \prime \prime}=w^{\prime \prime \prime} V^{\prime \prime \prime}+v^{\prime \prime \prime} V^{\prime \prime} &\quad W^{\prime \prime \prime \prime}=w^{\prime \prime \prime} W^{\prime \prime \prime}+v^{\prime \prime \prime} W^{\prime \prime} \end{array}$$

and so on, where:

$$\begin{aligned} & \frac{V}{W}=0 \\ & \frac{V^{\prime}}{W^{\prime}}=\frac{v}{w} \\ &\frac{V^{\prime \prime}}{W^{\prime \prime}}=\cfrac{v}{w+\cfrac{v^{\prime}}{w^{\prime}}} \\ & \frac{V^{\prime \prime \prime}}{W^{\prime \prime \prime}}=\cfrac{v}{w+\cfrac{v^{\prime}}{w^{\prime}+\cfrac{v^{\prime \prime}}{w^{\prime \prime}}}} \end{aligned}$$

and so forth. Furthermore, it is evident, or easily confirmed from the preceding equations, that:

$$\begin{alignedat}{3} & V W^{\prime}&&-V^{\prime} W&&=-v \\ & V^{\prime} W^{\prime \prime}&&-V^{\prime \prime} W^{\prime}&&=+v v^{\prime} \\ & V^{\prime \prime} W^{\prime \prime \prime}&&-V^{\prime \prime \prime} W^{\prime \prime}&&=-v v^{\prime} v^{\prime \prime} \\ & V^{\prime \prime \prime} W^{\prime \prime \prime \prime}&&-V^{\prime \prime \prime \prime} W^{\prime \prime \prime}&&=+v v^{\prime} v^{\prime \prime} v^{\prime \prime \prime} \end{alignedat}$$

and so on. Hence, it is clear that in the series:

$$\frac{v}{W W^{\prime}}-\frac{v v^{\prime}}{W^{\prime} W^{\prime \prime}}+\frac{v v^{\prime} v^{\prime \prime}}{W^{\prime \prime} W^{\prime \prime \prime}}-\frac{v v^{\prime} v^{\prime \prime} v^{\prime \prime \prime}}{W^{\prime \prime \prime} W^{\prime \prime \prime \prime}}+\text{etc. }$$

the first term is $=\frac{V^{\prime}}{W^{\prime}}$

the sum of the first two terms is $=\frac{V^{\prime \prime}}{W^{\prime \prime}}$

the sum of the first three terms is $=\frac{V^{\prime \prime \prime}}{W^{\prime \prime \prime}}$

the sum of the first four terms is $=\frac{V^{\prime \prime \prime \prime}}{W^{\prime \prime \prime \prime}}$

and so on; therefore, the series itself, either extended infinitely or until it is terminated, represents the continued fraction $\varphi$. Additionally, from this, the difference between $\varphi$ and each of the approximating fractions $\frac{V^{\prime}}{W^{\prime}},$  $\frac{V^{\prime \prime}}{W^{\prime \prime}},$  $\frac{V^{\prime \prime \prime}}{W^{\prime \prime \prime}}$, etc., is obtained.

By changing $t$ to $\frac{1}{u},$  in formula 33 in art. 14 of Disquisitionum generalium circa seriem infinitam, we easily obtain the transformation of the series

$$\varphi=u^{-1}+\tfrac{1}{3} u^{-3}+\tfrac{1}{5} u^{-5}+\tfrac{1}{7} u^{-7}+\text{etc. }$$

into the following continued fraction:

$$\cfrac{1}{u-\cfrac{\tfrac{1}{3}}{u-\cfrac{\tfrac{2.2}{3.5}}{u-\cfrac{\tfrac{3.3}{5.7}}{u-\cfrac{\tfrac{4.4}{7.9}}{u-\text{ etc.}}}}}}$$

so that we have

$$\begin{aligned} & v=1, v^{\prime}=-\frac{1}{3}, v^{\prime \prime}=-\frac{4}{15}, v^{\prime \prime \prime}=-\frac{9}{35}, v^{\prime \prime \prime \prime}=-\frac{16}{63} \text{ etc. } \\ & w=w^{\prime}=w^{\prime \prime}=w^{\prime \prime \prime}=w^{\prime \prime \prime \prime} \text{ etc.}=u. \end{aligned}$$

Hence for $V,$ $V^{\prime},$  $V^{\prime \prime},$  $V^{\prime \prime \prime}$, etc., $W,$  $W^{\prime},$  $W^{\prime \prime},$  $W^{\prime \prime \prime}$ , etc., we obtain the following values.

$$\begin{aligned} & V=0 && W=1 \\ & V^{\prime}=1 && W^{\prime}=u \\ & V^{\prime \prime}=u && W^{\prime \prime}=u u-\frac{1}{3} \\ & V^{\prime \prime \prime}=u u-\frac{4}{15}, && W^{\prime \prime \prime}=u^{3}-\frac{3}{5} u \\ & V^{\prime \prime \prime \prime}=u^{3}-\frac{11}{21} u, && W^{\prime \prime \prime \prime}=u^{4}-\frac{6}{7} u u+\frac{3}{35} \\ & V^{\scriptscriptstyle V}=u^{4}-\frac{7}{9} u u+\frac{64}{945} && W^{\scriptscriptstyle V}=u^{5}-\frac{10}{9} u^{3}+\frac{5}{21} u \\ & V^{\scriptscriptstyle VI}=u^{5}-\frac{34}{3} \frac{4}{3} u^{3}+\frac{1}{5} u, && W^{\scriptscriptstyle VI}=u^{6}-\frac{15}{11} u^{4}+\frac{5}{11} u u-\frac{5}{231} \\ & V^{\scriptscriptstyle VII}=u^{6}-\frac{50}{39} u^{4}+\frac{283}{715} u u-\frac{256}{15015}, && W^{\scriptscriptstyle VII}=u^{7}-\frac{21}{13} u^{5}+\frac{105}{143} u^{3}-\frac{35}{429} u \text{ etc } \end{aligned}$$

With careful attention, it becomes clear that all of the functions $V,$ ${V}^{\prime},$  $V^{\prime \prime},$  $V^{\prime \prime \prime}$, etc., ${W},$  ${W}^{\prime},$  $W^{\prime \prime},$  ${W}^{\prime \prime \prime}$ , etc., will be integral functions of the indeterminate $u;$  the highest term in $V^{(m)}$  will be $u^{m-1},$  and the powers $u^{m-2},$  $u^{m-4},$  $u^{m-6}$ , etc., are absent; while the highest term in $W^{(m)}$  will be $u^{m}$ , and the powers $u^{m-1},$  $u^{m-3},$  $u^{m-5}$ , etc., are absent. By what has been shown above, we have

$$\varphi=\frac{1}{W W^{\prime}}+\frac{1}{3 W^{\prime} W^{\prime \prime}}+\frac{2. 2}{3 . 3 . 5 W^{\prime \prime} W^{\prime \prime \prime}}+\frac{2. 2 . 3 . 3}{3 . 3 . 5 . 5 . 7 W^{\prime \prime \prime} W^{\prime \prime \prime \prime}}+\frac{2. 2 . 3 . 3 . 4 . 4}{3 . 3 . 5 . 5 . 7 . 7 . 9 W^{\prime \prime \prime \prime} W^{\prime \prime \prime \prime \prime}}+\text{etc.}$$

and thus generally

$$\begin{aligned} \varphi-\frac{V^{(m)}}{W^{(m)}}= & \frac{2. 2 . 3 . 3 \ldots \ldots m. m}{3. 3 . 5 . 5 \ldots(2 m-1)(2 m+1) W^{(m)} W^{(m+1)}} \\ & +\frac{2. 2 . 3 . 3 \ldots \ldots(m+1)(m+1)}{3. 3 . 5 . 5 \ldots \ldots(2 m+1)(2 m+3) W^{(m+1)} W^{(m+2)}} \\ & + \text{etc. } \end{aligned}$$

Therefore, if $\varphi-\frac{V^{(m)}}{W^{(m)}}$ is converted into a descending series, its first term will be

$$=\frac{2. 2 . 3 . 3 \ldots m m\, u^{-(2 m+1)}}{3. 3 . 5 . 5 \ldots (2 m-1)(2 m+1)}.$$

The product $\varphi W^{(m)}$ will indeed be composed of the integral function $V^{(m)}$  and an infinite series whose first term is

$$=\frac{2. 2 . 3 . 3 \ldots m m\, u^{-(m+1)}}{3. 3 . 5 . 5 \ldots(2 m-1)(2 m+1)}.$$

Therefore, a function $U$ of order $n+1$  is automatically obtained, which satisfies the condition established in the previous article, namely, that the product $\varphi U$  be free from powers $u^{-1},$  $u^{-2},$  $u^{-3} \ldots u^{-(n+1)}.$  Namely, it is none other than $W^{(n+1)},$  and it is evident that $U^{\prime}$  will be equal to $V^{(m+1)},$  and that the first term of $U^{\prime \prime}$  is

$$=\frac{2. 2 . 3 . 3 \ldots \ldots(n+1)(n+1)}{3. 3 . 5 . 5 \ldots \ldots(2 n+1)(2 n+3)}. u^{-(n+2)}$$

Therefore, if for $b,$ $b^{\prime},$  $b^{\prime \prime} \ldots b^{(n)}$  are taken to be the roots of the equation $W^{(n+1)}=0$, and the values of the coefficients $R,$  $R^{\prime},$  $R^{\prime \prime} \ldots R^{(n)}$  are obtained through the precepts stated above, our integral formula will enjoy precision of order $2 n+1$ , and its correction will be expressed approximately by

$$\frac{1}{2^{2 n+2}} \cdot \frac{2. 2 . 3 . 3 \ldots \ldots(n+1)(n+1)}{3. 3 . 5 . 5 \ldots \ldots(2 n+1)(2 n+3)} L^{(2 n+2)}=\frac{1. 1 . 2 . 2 . 3 . 3 \ldots \ldots(n+1)(n+1)}{2. 6 . 6 . 10 . 10 . 14 \ldots \ldots(4 n+2)(4 n+6)} L^{(2 n+2)}$$

18.
The previous articles describe how to find suitable functions $U$ for each value of the number $n$, but only successively, as one moves from smaller to larger values. However, we easily observe that these functions can generally be expressed as:

$$\begin{aligned} u^{n+1}-\frac{(n+1) n}{2. (2 n+1)} u^{n-1} & +\frac{(n+1) n(n-1)(n-2)}{2. 4(2 n+1)(2 n-1)} u^{n-3}-\frac{(n+1) n(n-1)(n-2)(n-3)(n-4)}{2. 4 . 6 . (2 n+1)(2 n-1)(2 n-3)} u^{n-5} + \text{ etc. } \end{aligned}$$

or, if we use the symbol $F$ as in the commentary cited above, as:

$$u^{n+1} F (-\tfrac{1}{2} n,-\tfrac{1}{2}(n+1),- (n+\tfrac{1}{2}), u^{-2})$$

This induction can easily be converted into a rigorous demonstration by the well-known method, or, if one prefers, with the assistance of formula 19 in the aforementioned discussion. The function $U$, if desired, can also be expressed in the reverse order of terms by:

$$\pm \frac{3. 5 . 7 \ldots(n+1)}{(n+3)(n+5) \ldots(2 n+1)} \cdot u F (-\tfrac{1}{2} n, \tfrac{1}{2}(n+3), \tfrac{3}{2}, u u)$$

for even $n$, with the upper or lower sign depending on whether $\frac{1}{2} n$ is even or odd, or by:

$$\pm \frac{1.3. 5 \ldots n}{(n+2)(n+4) \ldots(2 n+1)} F (-\tfrac{1}{2}(n+1), \tfrac{1}{2} n+1, \tfrac{1}{2}, u u)$$

for odd $n$, with the upper or lower sign depending on whether $\frac{1}{2}(n+1)$ is even or odd.

The function $U^{\prime}$ does not admit an equally simple general expression; however, from the origin of quantities $V,$  $V^{\prime},$  $V^{\prime \prime}$, etc., it is clear that the last term of $U^{\prime}$  for even $n$  becomes:

$$= \pm \frac{2. 2 . 4 . 4 . 6 . 6 \ldots \ldots n. n}{3. 5 . 7 . 9 . 11 .13 \ldots \ldots(2 n-1)(2 n+1)}$$

with the upper or lower sign depending on whether $\frac{1}{2} n$ is even or odd.

The function $U^{\prime \prime}=\varphi W^{(n+1)}-V^{(n+1)}$, whose first term we have already given in the previous article, can also be computed by a recursive algorithm, since we generally have:

$$\begin{aligned} & \varphi W^{\prime \prime}-V^{\prime \prime}=w^{\prime} (\varphi W^{\prime}-V^{\prime})+v^{\prime}(\varphi W-V) \\ & \varphi W^{\prime \prime \prime}-V^{\prime \prime \prime}=w^{\prime \prime} (\varphi W^{\prime \prime}-V^{\prime \prime})+v^{\prime \prime} (\varphi W^{\prime}-V^{\prime}) \\ & \varphi W^{\prime \prime \prime \prime}-V^{\prime \prime \prime \prime}=w^{\prime \prime \prime} (\varphi W^{\prime \prime \prime}-V^{\prime \prime \prime})+v^{\prime \prime \prime} (\varphi W^{\prime \prime}-V^{\prime \prime}) \end{aligned}$$

and so on, and therefore in the case at hand,

$$\varphi W^{(m+2)}-V^{(m+2)}=u (\varphi W^{(m+1)}-V^{(m+1)})-\frac{(m+1)^{2}}{(2 m-1)(2 m+1)} (\varphi W^{(m)}-V^{(m)})$$

Thus, we find

$$\begin{alignedat}{6} \varphi W-V&=&u^{-1}&+&\frac{1}{3} u^{-3}&+&\frac{1}{5} u^{-5}&+&\frac{1}{7} u^{-7}&+&\text{ etc. } \\ \varphi W^{\prime}-V^{\prime}&=&\frac{1}{3} u^{-2}&+&\frac{1}{5} u^{-4}&+&\frac{1}{7} u^{-6}&+&\frac{1}{9} u^{-8}&+&\text{ etc. } \\ \varphi W^{\prime \prime}-V^{\prime \prime}&=&\frac{4}{45} u^{-3}&+&\frac{8}{105} u^{-5}&+&\frac{4}{63} u^{-7}&+&\frac{112}{2079} u^{-9}&+&\text{ etc. } \\ \varphi W^{\prime \prime \prime}-V^{\prime \prime \prime}&=&\frac{4}{175} u^{-4}&+&\frac{8}{315} u^{-6}&+&\frac{4}{165} u^{-8}&+&\frac{16}{715} u^{-10}&+&\text{ etc. } \end{alignedat}$$

etc., which can also be represented by series as follows:

$$\begin{alignedat}{7} \varphi W-V&=&u^{-1} &(1+&\frac{1.2}{2.3} u^{-2}&+&\frac{1.2 . 3 . 4}{2 . 4 .3 .5} u^{-4}&+&\frac{1 . 2 . 3 . 4 . 5 . 6}{2 . 4 .6 . 3 . 5 . 7} u^{-6}&+&\text{ etc.}) \\ \varphi W^{\prime}-V^{\prime}&=&\frac{1}{3} u^{-2} &(1+&\frac{2.3}{2.5} u^{-4}&+&\frac{2 . 3 . 4 . 5}{2.4 . 5.7} u^{-4}&+&\frac{2.3 . 4 . 5 . 6 . 7}{2 . 4 . 6 . 5 . 7 . 9} u^{-6}&+&\text{ etc.}) \\ \varphi W^{\prime \prime}-V^{\prime \prime}&=&\frac{4}{45} u^{-3} &(1+&\frac{3 . 4}{2 . 7} u^{-2}&+&\frac{ 3 . 4 . 5 . 6}{2 . 4 . 7 . 9} u^{-4}&+&\frac{3.4 . 5 . 6 . 7 . 8}{2 . 4 . 6 . 7 . 9 . 11} u^{-6}&+&\text{ etc.}) \\ \varphi W^{\prime \prime \prime}-V^{\prime \prime \prime}&=&\frac{4}{175} u^{-4} &(1+&\frac{4.5}{2.9} u^{-2}&+&\frac{4.5 .6 .7}{2.4 .9 .11} u^{-4}&+&\frac{4.5 . 6 . 7.8 .9}{2.4 . 6.9 .11 .13} u^{-6}&+&\text{ etc.}) \end{alignedat}$$

etc. Following this induction, we will generally have

$$\begin{aligned} U^{\prime \prime}= & \varphi W^{(n+1)}-V^{(n+1)} \text{ equal to the product of } \\ & \frac{2. 2 . 3 . 3 . 4 . 4 \ldots \ldots(n+1). (n+1)}{3. 3 . 5 . 5 . 7 . 7 . 9 \ldots(2 n+1)(2 n+3)} u^{-(n+2)} \end{aligned}$$

with the infinite series

$$1+\frac{(n+2)(n+3)}{2(2 n+5)} u^{-2}+\frac{(n+2)(n+3)(n+4)(n+5)}{2. 4 . (2 n+5)(2 n+7)} u^{-4}+\text{etc.}$$

or, if one prefers, with $F (\frac{1}{2} n+1, \frac{1}{2} n+\frac{3}{2}, n+\frac{5}{2}, u^{-2}).$ This induction can also be easily elevated to full certainty either by the well-known method or with the help of formula 19 in the aforementioned discussion.

19.
Since it suffices to know either of the functions $T$ or $U$, we have preferred the determination of the latter as simpler. This determination, relying on the expansion of the series $ u^{-1}+\frac{1}{3} u^{-3}+\frac{1}{5} u^{-5}+$ etc. into a continued fraction, could have been derived by similar reasoning from the expansion of the series $ t^{-1}+\frac{1}{2} t^{-2}+\frac{1}{3} t^{-3}+\frac{1}{4} t^{-4}+$  etc. into a continued fraction:

$$ \cfrac{1}{t-\cfrac{\frac{1}{2}}{1-\cfrac{\frac{1}{6}}{t-\cfrac{\frac{2}{6}}{1-\cfrac{\frac{2}{10}}{t-\cfrac{\frac{3}{10}}{1-\text{ etc.}}}}}}} $$

However, we arrive at the same conclusion by considering that $T$ is none other than $\frac{U}{2^{n+1}}$  or $\frac{W^{(n+1)}}{2^{n+1}},$  with $u$  replaced by $2t-1.$  In this way, the functions successively adopted for $T$  can be determined by the following algorithm:

$$ \begin{alignedat}{2} W^{\phantom{\prime \prime \prime \prime}} & =1 \\ \tfrac{1}{2} W^{\prime \phantom{\prime \prime \prime}} & =t-\tfrac{1}{2} \\ \tfrac{1}{4} W^{\prime \prime \phantom{\prime \prime}} & = (t-\tfrac{1}{2}) \cdot \tfrac{1}{2} W^{\prime}-\tfrac{1. 1}{2 . 6} W&&=t t-t+\tfrac{1}{6} \\ \tfrac{1}{8} W^{\prime \prime \prime \phantom{\prime}} & = (t-\tfrac{1}{2}) \cdot \tfrac{1}{4} W^{\prime \prime}-\tfrac{2. 2}{6 . 10} \cdot \tfrac{1}{2} W^{\prime}&&=t^{3}-\tfrac{3}{2} t t+\tfrac{3}{5} t-\tfrac{1}{20} \\ \tfrac{1}{10} W^{\prime \prime \prime \prime} & = (t-\tfrac{1}{2}) \cdot \tfrac{1}{8} W^{\prime \prime \prime}-\tfrac{3.3}{10. 14} \cdot \tfrac{1}{4} W^{\prime \prime}&&=t^{4}-2 t^{3}+\tfrac{9}{7} t t-\tfrac{2}{7} t+\tfrac{1}{70} \end{alignedat} $$

etc. By induction, it follows that generally

$$ T=t^{n+1}-\frac{(n+1)^{2}}{1. (2 n+2)} t^{n}+\frac{(n+1)^{2}. n n}{1. 2 . (2 n+2)(2 n+1)} t^{n-1}-\frac{(n+1)^{2}. n n. (n-1)^{2}}{1. 2 . 3 . (2 n+2)(2 n+1). 2 n} t^{n-2}+\text{etc.} $$

or $T=t^{n+1} F (-(n+1),-(n+1),-2(n+1), t^{-1})$, which can be easily demonstrated. If one prefers, $T$ can also be expressed with the inverse order of terms, as $$ \pm \frac{1.2. 3 . 4 \ldots(n+1)}{2. 6 . 10 . 14 \ldots. (4 n+2)} F(n+2,-(n+1), 1, t) $$

where the upper sign holds for odd $n,$, and the lower for even $n$. Finally, in a similar manner, one finds that $T^{\prime \prime}$ is equal to the product of

$$ \frac{1. 1 . 2 . 2 . 3 . 3 \ldots \ldots(n+1). (n+1)}{2. 6 . 6 . 10 . 10 . 14 \ldots \ldots (4 n+2). (4 n+6)} t^{-(n+2)} $$

with the infinite series

$$ 1+\frac{(n+2)^{2}}{1. (2 n+4)} t^{-1}+\frac{(n+2)^{2}(n+3)^{2}}{1. 2 . (2 n+4)(2 n+5)} t^{-2}+\frac{(n+2)^{2}. (n+3)^{2}(n+4)^{2}}{1. 2 . 3 . (2 n+4)(2 n+5)(2 n+6)} t^{-3}+\text{etc.} $$

or with $F (n+2, n+2,2 n+4, t^{-1})$.

20.
Since the powers $u^n,$ $u^{n-2},$  $u^{n-4},$  etc. are absent from the function $U$, the roots of the equation $U=0$  will always occur in pairs of equal magnitude but opposite signs, except that for even values of $n$ , it is necessary to include a singular root $0$. Once the roots are found, the values of coefficients $R,$ $R^{\prime},$  $R^{\prime \prime}$  etc. will be obtained according to the method in art. 11, through an integral function of $u$. For odd values of $n$, this function will be of the form:

$$\gamma u^{n-1}+\gamma^{\prime} u^{n-3}+\gamma^{\prime \prime} u^{n-5}+\text{etc. }$$

For even values, if the coefficient corresponding to the root $u=0$ is excluded, it will be of the form:

$$\gamma u^{n-2}+\gamma^{\prime} u^{n-4}+\gamma^{\prime \prime} u^{n-6}+\text{etc. }$$

The example in art. 12 illustrates this reduction for $n=6.$ It is clear that opposite values of $u$  always correspond to equal coefficients. Of course, in the case where n is even, the coefficient corresponding to the root $u=0$ can be easily assigned a priori. This coefficient will be obtained if $u=0$ is substituted into $\frac{U^{\prime}}{ (\frac{d U}{d u})}.$  We have already provided the value of the numerator $U'$  for $u=0$  in art. 18, and the value of the denominator is

$$= \pm \frac{3. 5 . 7 \ldots. . (n+1)}{(n+3)(n+5) \ldots. (2 n+1)}= \pm \frac{3. 3 . 5 . 5 . 7 . 7 \ldots. (n+1)(n+1)}{3. 5 . 7 . 9 . 11 \ldots. (2 n+1)}.$$

Thus the sought coefficient is:

$$= \left(\frac{2 . 4 . 6 . 8 \ldots \ldots n}{3 . 5 . 7 . 9 \ldots \ldots(n+1)}\right)^{2}$$

21.
The integral function of $u$ which produces the coefficients ${R},$  ${R}^{\prime},$  ${R}^{\prime \prime}$  etc. in the case we are treating here can also be derived independently of the general method of art. 11, in the following manner. By differentiating the equation

$$\varphi-\frac{U^{\prime}}{U}=\frac{U^{\prime \prime}}{U},$$

substituting $\frac{\operatorname{d} \varphi}{\operatorname{d} u}=\frac{1}{1-u u},$ and multiplying by $U U(u u-1),$  we obtain

$$(u u-1) U^{\prime} \frac{\operatorname{d} U}{\operatorname{d} u}-U \left(\frac{\operatorname{d} U^{\prime}}{\operatorname{d} u} . (u u-1)+U\right)=(u u-1) U U \frac{\operatorname{d} (\frac{U^{\prime \prime}}{U})}{\operatorname{d} u}$$

The terms on the left-hand side of this equation clearly constitute an integral function of $u{:}$ thus it follows that on the right-hand side, the coefficients of powers of $u$  with negative exponents must destroy each other.

But $\frac{\operatorname{d} \frac{U^{\prime \prime}}{U}}{\operatorname{d} u}$ produces an infinite series starting from the term

$$- \left(\frac{1.2 . 3 . 4 \ldots \ldots(n+1)}{1 . 3 . 5 . 7 \ldots . (2 n+1)}\right)^{2} u^{-(2 n+4)},$$

which when multiplied by $(u u-1) U U$ can only result in a constant quantity

$$- \left(\frac{1 . 2 . 3 . 4 \ldots(n+1)}{1.3 . 5 . 7 \ldots(2 n+1)}\right)^{2}$$

From this, we deduce that

$$(u u-1) U^{\prime} \frac{\operatorname{d} U}{\operatorname{d} u}+ \left(\frac{1.2 . 3 . 4 \ldots . (n+1)}{1 . 3 . 5 . 7 \ldots(2 n+1)}\right)^{2}$$

must be divisible by $U,$ and therefore the fractional function $\frac{U^{\prime}}{ (\frac{\operatorname{d} U}{\operatorname{d} u})}$  which produces the coefficients ${R}, {R}^{\prime}, {R}^{\prime \prime}$  etc., will be equivalent to the integral function

$$- \left(\frac{1 . 3 . 5 . 7 \cdots . (2 n+1)}{1 . 2 . 3 . 4 . \cdots(n+1)} U^{\prime}\right)^{2}. (u u-1)$$

This function, which is of order $2 n+2,$ clearly implying only even powers of $u$, can be replaced with the remainder arising from its division by $U$ , which will be of order $n$  or $n-1,$  depending on whether $n$  is even or odd. However, if in the former case we prefer to exclude the coefficient corresponding to the root $u=0,$ we can instead replace it with its remainder upon division by $\frac{U}{u}$, which will only be of order $n-2.$

22.
In order to make available that which is required for the application of the method thus far presented, it seemed appropriate to append, for successive values of the number $n,$ numerical values of the quantities $a,$  $a^{\prime},$  $a^{\prime \prime}$  etc., and also of the coefficients ${R},$  ${R}^{\prime},$  ${R}^{\prime \prime}$  etc., computed to sixteen figures, together with their logarithms to ten figures.

I. First term, $n=0.$ $\begin{aligned} U&=u, \;\; U^{\prime}=1, \;\; T=t-\frac{1}{2}, \;\; T^{\prime}=1 \\ a&=0{.}5 \\ R&=1 \end{aligned}$

Correction to the integral formula approximately $=\frac{1}{12} L^{\prime \prime}.$

II. Second term, $n=1.$ $\begin{aligned} U\phantom{{}^{\prime}}&=u u-\frac{1}{2}, \quad U^{\prime}=u \\ T\phantom{{}^{\prime}}&=t t-t+\frac{1}{6}, \quad T^{\prime}=t-\frac{1}{2} \\ a\phantom{{}^{\prime}}&=0{.}2113248654\;\;051871 \\ a^{\prime}&=0{.}7886751345\;\;948129 \\ R\phantom{{}^{\prime}}&=R^{\prime}=\frac{1}{2} \end{aligned}$

Correction approximately $=\frac{1}{180} L^{\prime \prime \prime \prime}$

III. Third term, $n=2.$ $\begin{alignedat}{2} &U&&=u^{3}-\frac{3}{5} u, \;\;U^{\prime}=u u-\frac{4}{15}\\ &T&&=t^{3}-\frac{3}{2} t t+\frac{3}{5} t-\frac{1}{20}, \;\; T^{\prime}=t t-t+\frac{11}{60}\\ &a&&=0{.}1127016653\;\;792583\\ &a^{\prime}&&=0{.}5\\ &a^{\prime \prime}&&=?{.}8872983346 \;\; 207417\\ &R&&=R^{\prime \prime}=\frac{5}{18}\\ &R^{\prime}&&=\frac{4}{9}\\ \end{alignedat}$

Correction approximately $=\frac{1}{2800} L^{\scriptscriptstyle VI}.$

IV. Fourth term, $n=3.$ $\begin{array}{rl} U^{\phantom{\prime}}&=u^{4}-\frac{6}{7} u u+\frac{3}{35} \\ U^{\prime}&=u^{3}-\frac{11}{21} u \\ T^{\phantom{\prime}}&=t^{4}-2 t^{3}+\frac{9}{7} t t-\frac{2}{7} t+\frac{1}{70} \\ T^{\prime}&=t^{3}-\frac{3}{2} t t+\frac{13}{2} t-\frac{5}{84} \\ & \;\;\quad a^{\phantom{\prime \prime \prime}}=0{.}0694318442 \;\; 029754 \\ & \;\;\quad a^{\prime \phantom{\prime \prime}}=0{.}3300094782 \;\; 075677 \\ & \;\;\quad a^{\prime \prime\phantom{\prime}}=0{.}6699905217\;\;924323 \\ & \;\;\quad a^{\prime \prime \prime}=0{.}9305681557 \;\; 970246\\ R^{\phantom{\prime}}&=R^{\prime \prime \prime}=0{.}1739274225\;\;687284 \;\;\;\log. =9{.}2403680612 \\ R^{\prime}&=R^{\prime \prime\phantom{\prime}}=0{.}3260725774\;\;312716 \;\;\;\log. =9{.}5133142764 \end{array}$ |undefined

General expression for the coefficients

$$-\tfrac{35}{144} u u+\tfrac{17}{48}$$

Correction approximately $=\frac{1}{44100} L^{\scriptscriptstyle VIII}$

V. Fifth term, $n=4.$

$\begin{array}{rlrl} & U^{\phantom{\prime}}=u^{5}-\frac{10}{9} u^{3}+\frac{5}{21} u \\ & U^{\prime}=u^{4}-\frac{7}{9} u u+\frac{64}{945} \\ & T^{\phantom{\prime}}=t^{5}-\frac{5}{2} t^{4}+\frac{20}{9} t^{3}-\frac{5}{6} t t+\frac{5}{42} t-\frac{1}{252} &&\\ & T^{\prime}=t^{4}-2 t^{3}+\frac{47}{36} t t-\frac{11}{36} t+\frac{137}{75600} &&\\ & \;\;\quad a^{\phantom{\prime \prime \prime \prime}}=0{,}0469100770\;\;306680 &&\\ & \;\;\quad a^{\prime\phantom{\prime \prime \prime}}=0{,}2307653449 \;\; 471585 &&\\ & \;\;\quad a^{\prime \prime\phantom{\prime \prime}}=0{,}5 &&\\ & \;\;\quad a^{\prime \prime \prime\phantom{\prime}}=0{,}7692346550 \;\; 528415&& \\ & \;\;\quad a^{\prime \prime \prime \prime}=0{,}9530899229\;\;693320 && \\ & R^{\phantom{\prime \prime}}={R}^{\prime \prime \prime \prime}=0{,}1184634425\;\;280945 &\log .=&9{,}0735843490 \\ & {R}^{\prime\phantom{\prime}}={R}^{\prime \prime \prime \phantom{\prime}}=0{,}2393143352 \;\; 496832 & =& 9{,}3789687142 \\ & R^{\prime \prime}=\frac{64}{225}^{\phantom{\prime}}=0{,}2844444444 \;\; 444444 &=& 9{,}4539974559 \end{array}$ |undefined

General expression for the coefficients, excluding ${R}^{\prime \prime},$

$$-\tfrac{91}{400} u u+\tfrac{1099}{3600}$$

Correction approximately $=\frac{1}{698544} L^{\scriptscriptstyle X}$

VI. Sixth term, $n=5.$

$\begin{array}{rlrl} U^{\phantom{\prime}}= & u^{6}-\frac{15}{11} u^{4}+\frac{5}{11} u u-\frac{5}{231} \\ U^{\prime}= & u^{5}-\frac{34}{33} u^{3}+\frac{1}{5} u \\ T^{\phantom{\prime}}= & t^{6}-3 t^{5}+\frac{75}{22} t^{4}-\frac{20}{11} t^{3}+\frac{5}{11} t t-\frac{1}{22} t+\frac{1}{924} \\ T^{\prime}= & t^{5}-\frac{5}{2} t^{4}+\frac{74}{33} t^{3}-\frac{19}{22} t t+\frac{29}{220} t-\frac{7}{1320} \\ & a^{\phantom{\prime \prime \prime \prime \prime}}=0{,}0337652428 \;\; 984240 \\ & a^{\prime \phantom{\prime \prime \prime \prime}}=0{,}1693953067 \;\; 668678 \\ & a^{\prime \prime \phantom{\prime \prime \prime}}=0{,}3806904069 \;\; 584015 \\ & a^{\prime \prime \prime \phantom{\prime \prime}}=0{,}6193095930 \;\; 415985 \\ & a^{\prime \prime \prime \prime \phantom{\prime}}=0{,}8306046932 \;\; 331322 \\ & a^{\prime \prime \prime \prime \prime}=0{,}9662347571 \;\; 015760 \\ R^{\phantom{\prime \prime}}= & R^{\prime \prime \prime \prime \prime}=0{,}0856622461 \;\; 895852 \quad \log. = 8{,}9327894580 \\ R^{\prime\phantom{\prime}}= & R^{\prime \prime \prime \prime \phantom{\prime}}=0{,}1803807865 \;\; 240693 \quad \phantom{\log\;\, = .} 9{,}2561902763 \\ R^{\prime \prime}= & R^{\prime \prime \prime \phantom{\prime \prime}}=0{,}2339569672\;\;863455 \quad \phantom{\log\;\, = .} 9{,}3691359831 \end{array}$ |undefined

General expression for the coefficients

$$-\tfrac{77}{800} u^{4}-\tfrac{7}{75} u u+\tfrac{23}{96}$$

Correction approximately $=\frac{1}{11099088} L^{\scriptscriptstyle XII}$

VII. Seventh term, $n=6.$ $\begin{array}{rl} U^{\phantom{\prime}}=&u^{7}-\frac{21}{13} u^{5}+\frac{105}{143} u^{3}-\frac{35}{429} u \\ U^{\prime}=&u^{6}-\frac{50}{39} u^{4}+\frac{283}{715} u u-\frac{256}{15015} \\ T^{\phantom{\prime}}=&t^{7}-\frac{7}{2} t^{6}+\frac{63}{13} t^{5}-\frac{175}{52} t^{4}+\frac{175}{143} t^{3}-\frac{63}{286} t t+\frac{7}{429} t-\frac{1}{3432} \\ T^{\prime}=&t^{6}-3 t^{5}+ \frac{535}{156} t^{4}-\frac{145}{78} t^{3}+\frac{1377}{2860} t t-\frac{223}{4290} t+\frac{323}{240240} \\ & a^{\phantom{\prime \prime \prime \prime \prime \prime}}=0{.}0254460438 \;\; 286202 \\ & a^{\prime \phantom{\prime \prime \prime \prime \prime}}=0{.}1292344072 \;\; 003028 \\ & a^{\prime \prime \phantom{\prime \prime \prime \prime}}=0{.}2970774243 \;\; 113015 \\ & a^{\prime \prime \prime \phantom{\prime \prime \prime}}=0{.}5 \\ & a^{\prime \prime \prime \prime \phantom{\prime \prime}}=0{.}7029225756 \;\; 886985 \\ & a^{\prime \prime \prime \prime \prime \phantom{\prime}}=0{.}8707655927 \;\; 996972 \\ & a^{\prime \prime \prime \prime \prime \prime}=0{.}9745539561 \;\; 713798 \\ R^{\phantom{\prime \prime \prime}}=&R^{\prime \prime \prime \prime \prime \prime}=0{.}0647424830\;\;844348 \quad \log. =8{.}8111893529 \\ R^{\prime \phantom{\prime \prime}}=&R^{\prime \prime \prime \prime \prime \phantom{\prime}}=0{.}1398526957\;\;446384 \quad \phantom{\log. =.} 9{.}1456708421 \\ R^{\prime \prime \phantom{\prime}}=&R^{\prime \prime \prime \prime \phantom{\prime \prime}}=0{.}1909150252\;\;525595 \quad \phantom{\log. =.} 9{.}2808401093 \\ R^{\prime \prime \prime}=&\frac{256}{1225}\,\;=0{.}2089795918 \;\; 367347 \quad \phantom{\log. = .} 9{.}3201038766 \end{array}$ |undefined

General expression for the coefficients, excluding $R^{\prime \prime \prime}$

$$-\tfrac{1859}{16800} u^4 - \tfrac{1573}{29400} uu + \tfrac{7947}{39200}$$

Correction approximately $=\tfrac{1}{176679360} L^{\scriptscriptstyle XIV}$

23.
Coronidis loco, we will illustrate the effectiveness of our method by computing the value of the integral

$$\int \frac{d x}{\log x}$$

from $x=100000$ to $x=200000.$


 * I. From the first term, we have $\Delta R A=8390{,}394608$
 * II. From the second terms we get $\ldots \left\{\begin{array}{ll}\Delta R A&=4271{,}810097 \\ \Delta R^{\prime} A^{\prime}&=4134{,}144502 \\ \hline \text{Summa}&=8405{,}954599\end{array}\right..$
 * III. From the third terms $\ldots \ldots \left\{\begin{array}{ll}\Delta R A&=2390{,}572772 \\ \Delta R^{\prime} A^{\prime}&=3729{,}064270 \\ \Delta R^{\prime \prime} A^{\prime \prime}&=2286{,}599733 \\ \hline \text{Summa}&=8406{,}236775\end{array}\right.$
 * IV. From the fourth terms $\ldots \ldots\left\{\begin{array}{ll}\Delta R A&=1501{,}957053 \\ \Delta R^{\prime} A^{\prime}&=2763{,}769240 \\ \Delta R^{\prime \prime} A^{\prime \prime}&=2711{,}454637 \\ \Delta R^{\prime \prime \prime} A^{\prime \prime \prime}&=1429{,}062040 \\ \hline \text{Summa}&=8406{,}242970\end{array}\right.$
 * V. From the fifth terms $\ldots \ldots \left\{\begin{array}{ll}\Delta R A&=1024{,}879445 \\ \Delta R^{\prime} A^{\prime}&=2041{,}833335 \\ \Delta R^{\prime \prime} A^{\prime \prime}&=2386{,}601133 \\ \Delta R^{\prime \prime \prime} A^{\prime \prime \prime}&=1980{,}509616 \\ \Delta R^{\prime \prime \prime \prime} A^{\prime \prime \prime \prime}&=\phantom{0}972{,}419588 \\ \hline \text{Summa}&=8406{,}243117\end{array}\right.$
 * VI. From the sixth terms $\ldots \ldots \ldots \left\{\begin{array}{ll}\Delta R A&=\phantom{0}741{,}912854 \\ \Delta R^{\prime} A^{\prime}&=1545{,}757256 \\ \Delta R^{\prime \prime} A^{\prime \prime}&=1976{,}737668 \\ \Delta R^{\prime \prime \prime} A^{\prime \prime \prime}&=1950{,}466223 \\ \Delta R^{\prime \prime \prime \prime} A^{\prime \prime \prime \prime}&=1488{,}588550 \\ \Delta R^{\scriptscriptstyle V} A^{\scriptscriptstyle V}&=\phantom{0}702{,}780570 \\ \hline \text{Summa}&=8406{,}243121\end{array}\right.$
 * VII. From the seventh terms $\ldots \ldots \left\{\begin{array}{ll}\Delta R A&=\phantom{0}561{,}1213804 \\ \Delta R^{\prime} A^{\prime}&=1202{,}0551998 \\ \Delta R^{\prime \prime} A^{\prime \prime}&=1621{,}6290819 \\ \Delta R^{\prime \prime \prime} A^{\prime \prime \prime}&=1753{,}4212406 \\ \Delta R^{\prime \prime \prime \prime} A^{\prime \prime \prime \prime}&=1584{,}9790252 \\ \Delta R^{\scriptscriptstyle V} A^{\scriptscriptstyle V}&=1152{,}0681116 \\ \Delta R^{\scriptscriptstyle VI} A^{\scriptscriptstyle VI}&=\phantom{0}530{,}9690816 \\ \hline \text{Summa}&=8406{,}2431211\end{array}\right.$

In the calculations of, the value of the integral was found to be $=8406{,}24312.$