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Page:The Painted Veil - Maugham - 1925.djvu/131

 “I have a suspicion that you don’t very much like my husband,” she said with cool irony.

“I respect him. He has brains and character; and that, I may tell you, is a very unusual combination. I don’t suppose you know what he is doing here, because I don’t think he’s very expansive with you. If any man single-handed can put a stop to this frightful epidemic he’s going to do it. He’s doctoring the sick, cleaning the city up, trying to get the drinking water pure. He doesn’t mind where he goes nor what he does. He’s risking his life twenty times a day. He’s got Colonel Yü in his pocket and he’s induced him to put the troops at his disposal. He’s even put a little pluck into the magistrate and the old man is really trying to do something. And the nuns at the convent swear by him. They think he’s a hero.”

“Don’t you?”

“After all this isn’t his job, is it? He’s a bacteriologist. There was no call for him to come here. He doesn’t give me the impression that he’s moved by compassion for all these dying Chinamen. Watson was different. He loved the human race. Though he was a missionary it didn’t make any difference to him if they were Christian, Buddhist or Confucian; they were just human beings. Your husband isn’t here because he cares a damn if a hundred thousand Chinese die of cholera; he isn’t here either in the interests of science. Why is he here?”

“You’d better ask him.”

“It interests me to see you together. I sometimes


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 cially all inventions tending to improve mechanical appliances—and to these you are referring, I think—are already made to order, so to speak at the present time. In the future the Public Department of Inventions will undoubtedly organize a branch of service whose exclusive object it will be to invent mechanical improvements of all kinds. It is evident that these professional inventors will receive the highest wages, like the best workers in any other branch. And special rewards will be paid for every new invention of great importance. An invention, for instance, by which the hours of labor of all workers would be reduced considerably might be rewarded by paying to the inventor a pension for the rest of his life; and all other inventions might be rewarded in proportion. You will admit that to arrange such matters would meet with little difficulty under a system of social co-operation. And, before all other things, the future system will have the advantage that the individual discoverer and inventor will get the reward, and not some idle, good-for-nothing parasite.

I am satisfied that such an arrangement would solve this problem. And yet, the question of improving machinery reminds me of another objection. You have already stated that in consequence of the introduction of machinery a large number of workers becomes superfluous, and that under the new system more "hands" would be made idle than even to-day. How would you overcome that obstacle?

That difficulty exists within our present social state of affairs only; it could not exist under the system proposed by the socialists. To-day, of course, every new and improved machine becomes a curse to so many workers whom it throws out of work. In the new society, the people who become superfluous in one branch are employed to produce other commodities and luxuries, or their becoming superfluous causes a reduction of the hours of labor for all other workers.

I don't quite understand what you mean.

Let us suppose the United States and Canada to be united into one communistic commonwealth. It would form a world by itself and its inhabitants would be able to raise and produce whatever human beings may need and desire for their

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Page:The life of Charlotte Brontë (IA lifeofcharlotteb01gaskrich).pdf/142

 man's daughter," "taking interest in politics ever since she was five years old," and holding frequent discussions with such of the girls as were Dissenters and Radicals, was sure to have made herself as much acquainted as she could with the condition of those to whom she was opposed in opinion.

The bulk of the population were Dissenters, principally Independents. In the village of Heckmondwike, at one end of which Roe Head is situated, there were two large chapels, belonging to that denomination, and one to the Methodists, all of which were well filled two or three times on a Sunday, besides having various prayer-meetings, fully attended, on week-days. The inhabitants were a chapel-going people, very critical about the doctrine of their sermons, tyrannical to their ministers, and violent Radicals in politics. A friend, well acquainted with the place when Charlotte Brontë was at school, has described some events which occurred then among them:—

"A scene, which took place at the Lower Chapel at Heckmondwike, will give you some idea of the people at that time. When a newly-married couple made their appearance at chapel, it was the custom to sing the Wedding Anthem, just after the last prayer, and as the congregation was quitting the chapel. The band of singers who performed this ceremony expected to have money given them, and often passed the following night in drinking; at least, so said the minister of the place; and he determined to put an end to this custom. In this he was supported by many members of the chapel


text:
Translation:Summatio quarandum serierum singularium

1.
Among the remarkable truths to which the theory of the division of the circle has opened the way, the summation proposed in Disquisitiones Arithmeticae art. 356 claims not the last place for itself, not only because of its particular elegance and wonderful fecundity, which will be explained more fully on another occasion, but also because its rigorous demonstration is not burdened by uncommon difficulties. Of course, this should have been expected, since the difficulties do not fall so much into the theorem itself, but rather into a limitation of the theorem, which was then ignored, but whose demonstration is immediately available and easily derived from the theory explained in the present work. The theorem is presented there in the following form. Supposing $n$ to be a prime number, denoting all of the quadratic residues modulo $n$  between the limits 1 and $n-1$  (incl.) indefinitely by $a,$  denoting all the non-residues between the same limits by $b,$  denoting by $\omega$  the arc $\frac{360^{\circ}}{n},$  and denoting by $k$  any fixed integer not divisible by $n,$  we have

 * I. for values of $n$ which are of the form $4m+1,$

$$\begin{aligned} \Sigma \cos a k \omega & =-\frac{1}{2} \pm \frac{1}{2} \surd{n} \\ \Sigma \cos b k \omega & =-\frac{1}{2} \mp \frac{1}{2} \surd{n}, \text{ and therefore } \\ \Sigma \cos a k \omega -\Sigma \cos b k \omega & = \pm \surd{n} \\ \Sigma \sin a k \omega & =0 \\ \Sigma \sin b k \omega & =0 \end{aligned}$$

 * II. for values of $n$ which are of the form $4m+3,$

$$\begin{aligned} \Sigma \cos a k \omega & =-\frac{1}{2} \\ \Sigma \cos b k \omega & =-\frac{1}{2} \\ \Sigma \sin a k \omega & = \pm \frac{1}{2} \surd{n} \\ \Sigma \sin b k \omega & =\mp \frac{1}{2} \surd{n} \\ \Sigma \sin a k \omega -\Sigma \sin b k \omega & = \pm \surd{n} \end{aligned}$$

These sums have been demonstrated with all rigor in loc. cit. , and the only remaining difficulty is in determining the sign to be assigned to the radical quantity. It can easily be shown that this sign depends only on the number $k,$ that the same sign must hold for all values of $k$  that are quadratic residues modulo $n,$  and that the opposite sign must hold for all values of $k$  that are quadratic non-residues modulo $n.$  Therefore, the whole matter depends upon the case $k=1,$  and it is evident that as soon as the sign for this value is known, the signs for all other values of $k$  will immediately follow. But in this very question, which at first glance seems to be among the easier ones, we encounter unforeseen difficulties, and the method with which we have made progress so far completely denies us further help.

2.
It would not be out of place, before proceeding further, to work out some examples of our summation by numerical calculation. However, it will be convenient to preface this with some general observations.

 * I. If in the case where $n$ is a prime number of the form $4m+1,$  all quadratic residues of $n$  lying between 1 and $\frac{1}{2}(n-1)$  (inclusive) are denoted indefinitely by $a^{\prime},$  and all non-residues between the same limits are denoted by $b^{\prime},$  then certainly all $n-a^{\prime}$  are included in $a,$  and all $n-b^{\prime}$  are included in $b.$  Therefore, since $a^{\prime},$  $b^{\prime},$  $n-a^{\prime},$  $n-b^{\prime}$  together clearly exhaust the entire set of numbers $1,$  $2,$  $3 \ldots n-1,$  all $a^{\prime}$  together with all $n-a^{\prime}$  include all $a,$  and likewise all $b^{\prime}$  together with all $n-b^{\prime}$  include all $b.$  Hence we have

$$\begin{aligned} \Sigma \cos ak\omega &= \Sigma \cos a^{\prime}k\omega + \Sigma \cos (n-a^{\prime})k\omega \\ \Sigma \cos bk\omega &= \Sigma \cos b^{\prime}k\omega + \Sigma \cos (n-b^{\prime})k\omega \\ \Sigma \sin ak\omega &= \Sigma \sin a^{\prime}k\omega + \Sigma \sin (n-a^{\prime})k\omega \\ \Sigma \sin bk\omega &= \Sigma \sin b^{\prime}k\omega + \Sigma \sin (n-b^{\prime})k\omega \end{aligned}$$

 * Now, considering that $\cos (n-a^{\prime})k\omega = \cos a^{\prime}k\omega,$ $\cos (n-b^{\prime})k\omega = \cos b^{\prime}k\omega,$  $\sin (n-a^{\prime})k\omega = -\sin a^{\prime}k\omega,$  $\sin (n-b^{\prime})k\omega = -\sin b^{\prime}k\omega,$  it is clear that

$$\begin{aligned} \Sigma \sin ak\omega &= \Sigma \sin a^{\prime}k\omega - \Sigma \sin a^{\prime}k\omega = 0 \\ \Sigma \sin bk\omega &= \Sigma \sin b^{\prime}k\omega - \Sigma \sin b^{\prime}k\omega = 0 \end{aligned}$$

 * The summation of cosines, on the other hand, takes on the form

$$\begin{aligned} \Sigma \cos ak\omega &= 2 \Sigma \cos a^{\prime}k\omega \\ \Sigma \cos bk\omega &= 2 \Sigma \cos b^{\prime}k\omega \end{aligned}$$

 * from which it follows that

$$\begin{aligned} 1 + 4\Sigma \cos a^{\prime}k\omega &= \pm \surd{n} \\ 1 + 4\Sigma \cos b^{\prime}k\omega &= \mp \surd{n} \\ 2 \Sigma \cos a^{\prime}k\omega - 2 \Sigma \cos b^{\prime}k\omega &= \pm \surd{n} \end{aligned}$$

 * II. In the case where $n$ is of the form $4m+3,$  the complement of any quadratic residue $a$  modulo $n$  will be a quadratic non-residue, and the complement of any quadratic non-residue $b$  will be a quadratic residue; therefore, all $n-a$  will coincide with all $b,$  and all $n-b$  will coincide with all $a.$  Hence we conclude

$$\Sigma \cos a k \omega=\Sigma \cos (n-b) k \omega=\Sigma \cos b k \omega$$

 * and so, since $a$ and $b$  together fill out all the numbers $1,$  $2,$  $3 \ldots n-1,$  and therefore

$$\Sigma \cos a k \omega+\Sigma \cos b k \omega=\cos k \omega+\cos 2 k \omega+\cos 3 k \omega+\text{etc.}+\cos (n-1) k \omega=-1,$$

 * the summations

$$\begin{aligned} & \Sigma \cos a k \omega=-\frac{1}{2} \\ & \Sigma \cos b k \omega=-\frac{1}{2} \end{aligned}$$

 * are automatically evident. Similarly,

$$\Sigma \sin a k \omega=\Sigma \sin (n-b) k \omega=-\Sigma \sin b k \omega$$

 * and from this it is clear how the summations

$$\begin{aligned} & 2\Sigma \sin a k \omega= \pm \surd{n} \\ & 2\Sigma \sin b k \omega= \mp \surd{n} \end{aligned}$$

 * depend upon each other.

3.
Now here are some examples of numerical computations:

 * I. For $n=5,$ there is one value of $a^{\prime},$  namely $a^{\prime}=1,$  and one value of $b^{\prime},$  namely $b^{\prime}=2;$  and these are

$$\cos \omega=+0{,}3090169944 \qquad \qquad \cos 2 \omega=-0{,}8090169944$$

 * Hence $1+4 \cos \omega=+\surd{5},$ $1+4 \cos 2 \omega=-\surd{5}.$

 * II. For $n=13,$ there are three values of $a^{\prime},$  namely $1,$  $3,$  $4,$  and an equal number of values of $b^{\prime},$  namely $2,$  $5,$  $6,$  from which we compute

$$\begin{array}{r} \cos \omega\phantom{1} = + 0{,}8854560257 \\ \cos 3\omega = + 0{,}1205366803 \\ \cos 4 \omega = - 0{,}3546048870 \\ \hline \text{Sum} = + 0{,}6513878190 \end{array} \qquad \qquad \begin{array}{r} \cos 2\omega = + 0{,}5680647467 \\ \cos 5\omega = - 0{,}7485107482 \\ \cos 6 \omega = - 0{,}9709418174 \\ \hline \text{Sum} = - 1{,}1513878189 \end{array}$$

 * Hence $1+4 \Sigma \cos a^{\prime} \omega=+\surd{13},$ $1+4 \Sigma \cos b^{\prime} \omega=-\surd{13}.$

 * III. For $n=17,$ we have four values of $a^\prime,$  namely $1,$  $2,$  $4,$  $8,$  and an equal number of values of $b^\prime,$  namely $3,$  $5,$  $6,$  $7.$  From this, we compute the cosines

$$\begin{array}{r} \cos \omega\phantom{1} =+0{,}9324722294 \\ \cos 2 \omega =+0{,}7390089172 \\ \cos 4 \omega =+0{,}0922683595 \\ \cos 8 \omega =-0{,}9829730997 \\ \hline \text{Sum} =+0{,}7807764064 \end{array} \qquad \qquad \begin{array}{r} \cos 3 \omega =+0{,}4457383558 \\ \cos 5 \omega =-0{,}2736629901 \\ \cos 6 \omega =-0{,}6026346364 \\ \cos 7 \omega =-0{,}8502171357 \\ \hline \text{Sum} =-1{,}2807764065 \end{array}$$

 * Hence $1+4 \Sigma \cos a^{\prime} \omega=+\surd{17},1+4 \Sigma \cos b^{\prime} \omega=-\surd{17}.$

 * IV. For $n=3,$ there is one value of $a,$  namely $a=1,$  which corresponds to

$$\sin \omega=+0.8660254038 \quad\quad \phantom{\sin \omega=+0.8660254038}$$

 * Hence, $2 \sin \omega=+\surd{3}.$

 * V. For $n=7,$ there are three values of $a,$  namely $1,$  $2,$  $4\text{:}$  hence we have the sines

$$\begin{array}{rl} \sin \omega\phantom{1} =+0{,}7818314825 \phantom{,}&\\ \sin 2 \omega =+0{,}9749279122 \phantom{,}&\\ \sin 4 \omega =-0{,}4338837391 \phantom{,}&\\ \begin{array}{c}\hline \phantom{a}\text{Sum} =+1{,}3228756556,\end{array}& \text{hence } 2 \Sigma \sin a \omega=+\surd{7}. \end{array}$$

 * VI. For $n=11,$ the values of $a$  are $1,$  $3,$  $4,$  $5,$  $9,$  which correspond to sines

$$\begin{array}{rl} \sin \omega\phantom{1} =+0{,}5406408175 \phantom{,}&\\ \sin 3 \omega =+0{,}9898214419 \phantom{,}&\\ \sin 4 \omega =+0{,}7557495744 \phantom{,}&\\ \sin 5 \omega =+0{,}2817325568 \phantom{,}&\\ \sin 9 \omega =-0{,}9096319954 \phantom{,}&\\ \begin{array}{r}\hline \phantom{a}\text{Sum} =+1{,}6583123952,\end{array}& \text{hence } 2 \Sigma \sin a \omega=+\surd{11}. \end{array}$$

 * VII. For $n=19,$ the values of $a$  are $1,$  $4,$  $5,$  $6,$  $7,$  $9,$  $11,$  $16,$  $17,$  which correspond to sines

$$\begin{array}{rl} \sin \omega\phantom{11}=+0{,}3246994692 \phantom{,}&\\ \sin 4 \omega\phantom{1}=+0{,}9694002659 \phantom{,}&\\ \sin 5 \omega\phantom{1}=+0{,}9965844930 \phantom{,}&\\ \sin 6 \omega\phantom{1}=+0{,}9157733267 \phantom{,}&\\ \sin 7 \omega\phantom{1}=+0{,}7357239107 \phantom{,}&\\ \sin 9 \omega\phantom{1}=+0{,}1645945903 \phantom{,}&\\ \sin 11 \omega=-0{,}4759473930 \phantom{,}&\\ \sin 16 \omega=-0{,}8371664783 \phantom{,}&\\ \sin 17 \omega=-0{,}6142127127 \phantom{,}&\\ \begin{array}{r}\hline \phantom{a}\text{Sum}=+2{,}1794494718,\end{array}&\text{hence } 2 \Sigma \sin a \omega=+\surd{19}. \end{array}$$

4.
In all these examples the radical quantity has a positive sign, and the same is easily confirmed for larger values $n=23,$ $n=29,$  etc. , from which a strong likelihood emerges that this holds generally. However, the proof of this phenomenon cannot be sought from the principles set forth in loc. cit. , and must be regarded as deserving of a thorough investigation. Therefore, the purpose of this commentary is to present a rigorous proof of this most elegant theorem, which has been attempted in vain in various ways for many years, and was finally achieved successfully through careful and subtle considerations. At the same time, we will bring the theorem itself, with its elegance preserved or rather enhanced, to a much greater generality. Finally, in the conclusion, we will reveal a remarkable and close connection between this summation and another very important arithmetic theorem. We hope that not only will geometers be gratified by the results of these investigations, but also that the methods, which may well be useful on other occasions, will be deemed worth of their attention.

5.
Our proof relies on the consideration of a specific type of series, whose terms depend on expressions of the form

$$\frac{(1-x^{m})(1-x^{m-1})(1-x^{m-2}) \ldots(1-x^{m-\mu+1})}{(1-x)(1-x x)(1-x^{3}) \ldots (1-x^{\mu})}$$

For the sake of brevity, we will denote such a fraction by $(m, \mu),$ and we will first present some general observations about such functions.

 * I. Whenever $m$ is a positive integer smaller than $\mu,$  the function $(m, \mu)$  clearly vanishes, since the numerator involves the factor $1-x^{0}.$  For $m = \mu,$  the factors in the numerator will be identical, but in the reverse order compared to the factors in the denominator, so that $(\mu, \mu) = 1.$  Finally, in the case where $m$  is a positive integer greater than $\mu,$  we have the formulas

$$\begin{aligned} & (\mu+1, \mu) = \frac{1-x^{\mu+1}}{1-x} = (\mu+1,1) \\ & (\mu+2, \mu) = \frac{(1-x^{\mu+2})(1-x^{\mu+1})}{(1-x)(1-x x)} = (\mu+2,2) \\ & (\mu+3, \mu) = \frac{(1-x^{\mu+3})(1-x^{\mu+2})(1-x^{\mu+1})}{(1-x)(1-xx)(1-x^{3})} = (\mu+3,3) \text{ etc.} \end{aligned}$$

 * or more generally,

$$(m, \mu) = (m, m-\mu)$$

 * II. Furthermore, it is easily confirmed that, in general,

$$(m, \mu+1)=(m-1, \mu+1)+x^{m-\mu-1}(m-1, \mu)$$

 * and likewise

$$\begin{aligned} (m-1, \mu+1)&=(m-2, \mu+1)+x^{m-\mu-2}(m-2, \mu) \\ (m-2, \mu+1)&=(m-3, \mu+1)+x^{m-\mu-3}(m-3, \mu) \\ (m-3, \mu+1)&=(m-4, \mu+1)+x^{m-\mu-4}(m-4, \mu) \text{ etc.} \end{aligned}$$

 * which continues until

$$\begin{aligned} (\mu+2, \mu+1) & =(\mu+1, \mu+1)+x(\mu+1, \mu) \\ & =(\mu, \mu)+x(\mu+1, \mu) \end{aligned}$$

 * and therefore, as long as $m$ is a positive integer greater than $\mu+1,$

$$(m, \mu+1)=(\mu, \mu) +x(\mu+1, \mu)+x x(\mu+2, \mu)+x^{3}(\mu+3, \mu)+\text{etc.} +x^{m-\mu-1}(m-1, \mu)$$

 * Hence it is clear that if, for any given value of $\mu,$ the function $(m, \mu)$  is integral for all positive integer values of $m,$  then the function $(m, \mu+1)$  must also be integral. Therefore, since this assumption holds for $\mu=1,$  the same will hold for $\mu=2,$  and thus for $\mu=3$  etc ., i . e . in general for any positive integer value of $m$  the function $(m, \mu)$  will be integral, or in other words the product

$$(1-x^{m})(1-x^{m-1})(1-x^{m-2}) \ldots(1-x^{m-\mu+1})$$

 * will be divisible by

$$(1-x)(1-x^{2})(1-x^{3}) \ldots(1-x^{\mu}).$$

6.
We will now consider two series, each of which can be used to achieve our goal. The first series is

$$1-\frac{1-x^{m}}{1-x}+\frac{(1-x^{m})(1-x^{m-1})}{(1-x)(1-xx)}-\frac{(1-x^{m})(1-x^{m-1})(1-x^{m-2})}{(1-x)(1-xx)(1-x^{3})}+\ldots$$

or

$$1-(m, 1)+(m, 2)-(m, 3)+(m, 4)-\ldots$$

which for the sake of brevity we will denote by $f(x, m).$ It is immediately clear that when $m$  is a positive integer, this series terminates after its ${m+1}^{\text{st}}$  term (which is $= \pm 1$ ), and therefore in this case, the sum must be a finite integral function of $x.$  Furthermore, according to observation II of article 5, it is clear that in general, for any value of $m,$  we have

$$\begin{aligned} 1 & =1 \\ -(m, 1) & =-(m-1,1)-x^{m-1} \\ +(m, 2) & =+(m-1,2)+x^{m-2}(m-1,1) \\ -(m, 3) & =-(m-1,3)-x^{m-3}(m-1,2) \ldots \end{aligned}$$

and therefore

$$f(x, m)=1-x^{m-1}-(1-x^{m-2})(m-1,1)+(1-x^{m-3})(m-1,2)-(1-x^{m-4})(m-1.3)+\ldots$$

But it is also clear that

$$\begin{aligned} & (1-x^{m-2})(m-1,1)=(1-x^{m-1})(m-2,1) \\ & (1-x^{m-3})(m-1,2)=(1-x^{m-1})(m-2,2) \\ & (1-x^{m-4})(m-1,3)=(1-x^{m-1})(m-2,3) \ldots \end{aligned}$$

from which we deduce the equation

$$

7.
Since for $m=0$ we have $f(x, m)=1,$  we obtain, from the formula we have just found,

$$\begin{aligned} & f(x, 2)=1-x \\ & f(x, 4)=(1-x)(1-x^{3}) \\ & f(x, 6)=(1-x)(1-x^{3})(1-x^{5}) \\ & f(x, 8)=(1-x)(1-x^{3})(1-x^{5})(1-x^{7}) \text{ etc.} \end{aligned}$$

or more generally, for any even value of $m,$

$$

On the other hand, since for $m=1$ we have $f(x, m) = 0,$  we have

$$\begin{aligned} & f(x, 3) = 0 \\ & f(x, 5) = 0 \\ & f(x, 7) = 0 \text{ etc.} \end{aligned}$$

or, in general, for any odd value of $m,$

$$f(x, m) = 0$$

Indeed, the latter sum could have already been derived from the fact that in the series

$$1-(m, 1)+(m, 2)-(m, 3)+\text{etc.}+(m, m-1)-(m, m),$$

the last term destroys the first, the penultimate destroys the second, etc.

8.
For our purpose it suffices to consider the case where $m$ is a positive odd integer. However, it will not hurt to add a few remarks about the cases where $m$ is fractional or negative, due to the singularity of the matter. Clearly in these cases our series will not be interrupted, but will diverge to infinity. Moreover, it is easily seen that the series diverges whenever the value of $x$ is less than 1, so its summation should be restricted to values of $x$  which are greater than 1.

According to formula [1] in article 6, we have

$$\begin{aligned} & f(x,-2)=\frac{1}{1-\frac{1}{x}} \\ & f(x,-4)=\frac{1}{1-\frac{1}{x}} \cdot \frac{1}{1-\frac{1}{x^{3}}} \\ & f(x,-6)=\frac{1}{1-\frac{1}{x}} \cdot \frac{1}{1-\frac{1}{x^{3}}} \cdot \frac{1}{1-\frac{1}{x^{5}}} \text{ etc.} \end{aligned}$$

so that for negative, integral, even values of $m,$ the function $f(x, m)$  can also be assigned a value with finitely many terms. For the remaining values of $m,$ we will convert the function $f(x, m)$  into an infinite product using the following method.

As $m$ approaches negative infinity, the function $f(x, m)$  converges to

$$1+\frac{1}{x-1}+\frac{1}{x-1} \cdot \frac{1}{x x-1}+\frac{1}{x-1} \cdot \frac{1}{x x-1} \cdot \frac{1}{x^{3}-1}+\text{etc.}$$

Therefore, this series is equal to the infinite product

$$\frac{1}{1-\frac{1}{x}} \cdot \frac{1}{1-\frac{1}{x^{3}}} \cdot \frac{1}{1-\frac{1}{x^{5}}} \cdot \frac{1}{1-\frac{1}{x^{7}}} \text{ etc. to infinity}$$

Moreover, since it is generally true that

$$f(x, m)=f(x, m-2 \lambda). (1-x^{m-1})(1-x^{m-3})(1-x^{m-5}) \ldots (1-x^{m-2 \lambda+1})$$

we have

$$\begin{aligned} f(x, m) & =f(x,-\infty) \cdot(1-x^{m-1})(1-x^{m-3})(1-x^{m-5}) \text{ etc. to infinity} \\ & =\frac{1-x^{m-1}}{1-x^{-1}} \cdot \frac{1-x^{m-3}}{1-x^{-3}} \cdot \frac{1-x^{m-5}}{1-x^{-5}} \cdot \frac{1-x^{m-7}}{1-x^{-7}} \text{ etc. to infinity} \end{aligned}$$

whose factors clearly converge to unity.

The case $m=-1$ deserves special attention. Here we have

$$f(x,-1)=1+x^{-1}+x^{-3}+x^{-6}+x^{-10}+\text{etc.}$$

It follows that this series can be expressed as an infinite product

$$\frac{1-x^{-2}}{1-x^{-1}} \cdot \frac{1-x^{-2}}{1-x^{-3}} \cdot \frac{1-x^{-6}}{1-x^{-5}} \text{ etc.}$$

or, by replacing $x$ with $x^{-1},$

$$1+x+x^{3}+x^{6}+\text{etc.}=\frac{1-x x}{1-x} \cdot \frac{1-x^{4}}{1-x^{3}} \cdot \frac{1-x^{6}}{1-x^{5}} \cdot \frac{1-x^{8}}{1-x^{7}} \text{ etc.}$$

This equality between two somewhat complicated expressions, to which we will return on another occasion, is indeed very remarkable.

9.
Secondly, we consider the series

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

or

$$1+x^{\frac{1}{2}}(m, 1)+x(m, 2)+x^{\frac{3}{2}}(m, 3)+x x(m, 4)+\text{etc.}$$

which we will denote by $F(x, m).$ We will restrict this discussion to the case where $m$  is a positive integer, so that the series always terminates at the $m+1^{\text{st}}$  term, which is $=x^{\frac{1}{2} m}(m, m).$  Since

$$(m, m)=1, \quad(m, m-1)=(m, 1), \quad(m, m-2)=(m, 2) \text{ etc.}$$

the above series can also be expressed as:

$$F(x, m)=x^{\frac{1}{2} m}+x^{\frac{1}{2}(m-1)}(m, 1)+x^{\frac{1}{2}(m-2)}(m, 2)+x^{\frac{1}{2}(m-3)}(m, 3)+\text{etc.}$$

Hence we have

$$\begin{aligned} (1+x^{\frac{1}{2} m+\frac{1}{2}}) F(x, m) =1&+x^{\frac{1}{2}}(m, 1)+x(m, 2)+x^{\frac{3}{2}}(m, 3)+\text{etc.} \\ & +x^{\frac{1}{2}}. x^{m}+x. x^{m-1}(m, 1)+x^{\frac{3}{2}}. x^{m-2}(m, 2)+\text{etc.} \end{aligned}$$

Therefore, since we have (art. 5, II)

$$\begin{aligned} (m, 1)+x^{m} & =(m+1,1) \\ (m, 2)+x^{m-1}(m, 1) & =(m+1,2) \\ (m, 3)+x^{m-2}(m, 2) & =(m+1,3) \text{ etc.,} \end{aligned}$$

we obtain the result

$$

But $F(x, 0)=1;$ therefore we have

$$\begin{aligned} & F(x, 1)=1+x^{\frac{1}{2}} \\ & F(x, 2)=(1+x^{\frac{1}{2}})(1+x) \\ & F(x, 3)=(1+x^{\frac{1}{2}})(1+x)(1+x^{3}) \text{ etc.,} \end{aligned}$$

or in general

$$

10.
Having made these preliminary observations, let us now proceed towards our objective. Since the squares $1,$ $4,$  $9 \ldots (\frac{1}{2}(n-1))^2$  are all incongruent to each other modulo $n,$  it is clear that their minimal residues modulo $n$  must be identical to the numbers $a,$  and therefore

$$\begin{aligned} &\Sigma \cos a k \omega=\cos k \omega+\cos 4 k \omega+\cos 9 k \omega+\text{etc.}+\cos (\frac{1}{2}(n-1))^2 k \omega \\ &\Sigma \sin a k \omega=\sin k \omega+\sin 4 k \omega+\sin 9 k \omega+\text{etc.}+\sin (\frac{1}{2}(n-1))^2 k \omega \end{aligned}$$

Similarly, since the same squares $1,$ $4,$  $ 9 \ldots (\frac{1}{2}(n-1))^2$  are congruent to $(\frac{1}{2}(n+1))^2,$  $(\frac{1}{2}(n+3))^2,$  $(\frac{1}{2}(n+5))^2 \ldots (n-1)^2$  in reverse order, we have

$$\begin{aligned} &\Sigma \cos a k \omega=\cos (\frac{1}{2}(n+1))^2 k \omega+\cos (\frac{1}{2}(n+3))^2 k \omega+\text{etc.}+\cos (n-1)^2 k \omega \\ &\Sigma \sin a k \omega=\sin (\frac{1}{2}(n+1))^2 k \omega+\sin (\frac{1}{2}(n+3))^2 k \omega+\text{etc.}+\sin (n-1)^2 k \omega \end{aligned}$$

Therefore, assuming

$$\begin{alignedat}{2} &T=1+\cos k \omega+\cos 4 k \omega+\cos 9 k \omega+\text{etc.}+\cos (n-1)^2 k \omega \\ & U= \sin k \omega+\sin 4 k \omega+\sin 9 k \omega+\text{etc.}+\sin (n-1)^2 k \omega \end{alignedat}$$

we will have

$$\begin{aligned} 1+2 \Sigma \cos a k \omega & =T \\ 2 \Sigma \sin a k \omega & =U \end{aligned}$$

Hence it is clear that the summations proposed in article 1 depend on the summations of the series $T$ and $U.$  We will therefore direct our discussion to these, and complete it in a general way that it includes not only prime values of $n$  but composite values as well. Let us also suppose that the number $k$ is relatively prime to $n;$  for the case where $k$  and $n$  have a common divisor can be reduced to this one without any difficulty.

11.
Let us denote the imaginary quantity $\surd{-1}$ by $i,$  and let

$$\cos k \omega+i \sin k \omega=r$$

so that $r^{n}=1,$ or equivalently $r$  is a root of the equation $r^{n}-1=0.$  It is easy to see that the numbers $k,$  $2 k,$  $3 k \ldots (n-1) k$  are all indivisible by $n$  and are incongruent to each other modulo $n\text{:}$  therefore, the powers of $r$

$$1, r, r^{2}, r^{3} \ldots r^{n-1}$$

will all be distinct, and each of them will satisfy the equation $x^{n}-1=0.$ Because of this, these powers represent all of the roots of the equation $x^{n}-1=0.$

These conclusions would be invalid if $k$ had a common divisor with $n.$  For if $\nu$  were such a common divisor, then $k\mathrm{.}\frac{n}{\nu}$  would be divisible by $n,$  and hence a power less than $r^{n},$  say $r^{\frac{n}{\nu}},$  would be equal to unity. In this case, therefore, the powers of $r$ up to the $\frac{n}{\nu}^{th}$  would all be roots of the equation $x^{n}-1=0,$  and indeed they would be distinct roots, if $\nu$  were the greatest common divisor of $k$  and $n.$  In our case, where $k$  and $n$  are assumed to be prime to each other, $r$  can conveniently be called a proper root of the equation $x^{n}-1=0.$  In the other case, where $k$  and $n$  have a (greatest) common divisor $\nu,$  we will say that $r$  is an improper root of that equation. Clearly in the latter case, $r$ would be a proper root of the equation $x^{\frac{n}{\nu}}-1=0.$  The simplest improper root is unity, and in the case where $n$  is a prime number, there are no other improper roots whatsoever.

12.
If we now set

$$W=1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}$$

it is clear that $W=T+i U,$ so that $T$  is the real part of $W,$  and $U$  is obtained from the imaginary part of $W$  by suppressing the factor $i.$  The whole matter is therefore reduced to finding the sum $W\text{:}$  for this purpose, either the series considered in article 6 or the one we have shown how to sum in article 9 can be used, although the former is less suitable in the case where $n$  is an even number. Nonetheless, we hope that it will be agreeable to the reader if we treat the case where $n$ is odd according to both methods.

Let us first suppose that $n$ is an odd number, that $r$  is an arbitrary proper root of the equation $x^{n}-1=0,$  and that in the function $f(x, m)$  we set $x=r$  and $m=n-1.$  Then clearly

$$\begin{alignedat}{3} & \frac{1-x^{m}}{1-x}&=\frac{1-r^{-1}}{1-r}&=-r^{-1} \\ & \frac{1-x^{m-1}}{1-x x}&=\frac{1-r^{-2}}{1-r r}&=-r^{-2} \\ & \frac{1-x^{m-2}}{1-x^{3}}&=\frac{1-r^{-3}}{1-r^{3}}&=-r^{-3} \text{ etc.} \end{alignedat}$$

up to

$$\begin{alignedat}{3} &\frac{1-x}{1-x^{m}}&=\frac{1-r^{-m}}{1-r^{m}}&=-r^{-m}\phantom{\text{etc.}} \end{alignedat}$$

(It will not be superfluous to mention that these equations are valid only to the extent that $r$ is assumed to be a proper root: for if $r$  were an improper root, the numerator and denominator of some of these fractions would simultaneously vanish, and thus the fractions would become indeterminate).

From this, we derive the following equation:

$$\begin{aligned} f(r, n-1) & =1+r^{-1}+r^{-3}+r^{-6}+\text{etc.}+r^{-\frac{1}{2}(n-1) n} \\ & =(1-r)(1-r^{3})(1-r^{5}) \ldots(1-r^{n-2}) \end{aligned}$$

The same equation will still hold if we substitute $r^{\lambda}$ for $r,$  where $\lambda$  is any arbitrary integer relatively prime to $n,$  for then $r^{\lambda}$  will also be a proper root of the equation $x^{n}-1=0.$  Let us write $r^{n-2}$  instead of $r,$  or equivalently $r^{-2}.$  Then

$$1+r^{2}+r^{6}+r^{12}+\text{etc.}+r^{(n-1) n}=(1-r^{-2})(1-r^{-6})(1-r^{-10}) \ldots(1-r^{-2(n-2)})$$

Next, let us multiply both sides of this equation by

$$r \cdot r^{3} \cdot r^{5} \ldots \cdot r^{(n-2)}=r^{\frac{1}{4}(n-1)^{2}}$$

Since

$$\begin{alignedat}{4} r^{2+\frac{1}{4}(n-1)^{2}} & =r^{\frac{1}{4}(n-3)^{2}},\quad & r^{(n-1) n+\frac{1}{4}(n-1)^{2}}&=r^{\frac{1}{4}(n+1)^{2}} \\ r^{6+\frac{1}{4}(n-1)^{2}} & =r^{\frac{1}{4}(n-5)^{2}},\quad & r^{(n-2)(n-1)+\frac{1}{4}(n-1)^{2}}&=r^{\frac{1}{4}(n+3)^{2}} \\ r^{12+\frac{1}{4}(n-1)^{2}} & =r^{\frac{1}{4}(n-7)^{2}},\quad & r^{(n-3)(n-2)+\frac{1}{4}(n-1)^{2}}&=r^{\frac{1}{4}(n+5)^{2}} \text{ etc.} \end{alignedat}$$

we get the following equation

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

or, by rearranging the terms of the first member,

$$

13.
The factors of the second member of the equation [5] can also be written as

$$\begin{aligned} r-r^{-1} & =-(r^{n-1}-r^{-n+1}) \\ r^{3}-r^{-3} & =-(r^{n-3}-r^{-n+3}) \\ r^{5}-r^{-5} & =-(r^{n-5}-r^{-n+5}) \text{ etc.} \end{aligned}$$

up to

$$r^{n-2}-r^{-n+2}=-(r^{2}-r^{-2})\phantom{\text{ etc.}}$$

in which case the equation takes the following form:

$$W=(-1)^{\frac{1}{2}(n-1)}(r^{2}-r^{-2})(r^{4}-r^{-4})(r^{6}-r^{-6}) \ldots (r^{n-1}-r^{-n+1})$$

Multiplying this equation by [5] in its original form, we obtain

$$W^{2}=(-1)^{\frac{1}{2}(n-1)}(r-r^{-1})(r^{2}-r^{-2})(r^{3}-r^{-3}) \ldots (r^{n-1}-r^{-n+1})$$

where $(-1)^{\frac{1}{2}(n-1)}$ is either $+1$  or $-1,$  depending on whether $n$  is of the form $4 \mu+1$  or $4 \mu+3.$  Therefore,

$$W^{2}= \pm r^{\frac{1}{2} n(n-1)}(1-r^{-2})(1-r^{-4})(1-r^{-6}) \ldots(1-r^{-2(n-1)})$$

But it is clear that $r^{-2},$ $r^{-4},$  $r^{-6} \ldots r^{-2 n+2}$  are precisely the roots of the equation $x^{n}-1=0,$  except for the root $x=1.$  Hence the following equation must hold

$$(x-r^{-2})(x-r^{-4})(x-r^{-6}) \ldots(x-r^{-2 n+2})=x^{n-1}+x^{n-2}+x^{n-3}+\text{etc.}+x+1$$

and setting $x=1,$ we find that

$$(1-r^{-2})(1-r^{-4})(1-r^{-6}) \ldots(1-r^{-2 n+2})=n$$

Since it is evident that $r^{\frac{1}{2} n(n-1)}=1,$ our equation becomes:

$$

In the case where $n$ is of the form $4 \mu+1,$  we have:

$W= \pm \surd{n},$ and therefore $T= \pm \surd{n},$ $U=0$

On the other hand, in the case where $n$ is of the form $4 \mu+3,$  we have:

$W= \pm i \surd{n},$ and therefore $T=0,$ $U= \pm \surd{n}$

14.
The method of the previous article determines only the absolute values of $T$ and $U,$  and leaves their signs ambiguous, so it is necessary to determine whether $T$  (in the first case) and $U$  (in the second case) are equal to $+\surd{n}$  or $-\surd{n}$. However, at least when $k=1,$ this can be deduced from equation [5] in the following way. Since, for $k=1,$

$$\begin{aligned} r-r^{-1} & =2 i \sin \omega \\ r^{3}-r^{-3} & =2 i \sin 3 \omega \\ r^{5}-r^{-5} & =2 i \sin 5 \omega \text{ etc., } \end{aligned}$$

this equation is transformed into

$$W=(2 i)^{\frac{1}{2}(n-1)} \sin \omega \sin 3 \omega \sin 5 \omega \ldots \sin (n-2) \omega$$

Now, in the case where $n$ is of the form $4 \mu+1,$  in the series of odd numbers

$$1,3,5,7 \ldots \frac{1}{2}(n-3), \frac{1}{2}(n+1) \ldots(n-2)$$

there can be found $\frac{1}{4}(n-1)$ which are less than $\frac{1}{2} n,$  and these clearly correspond to positive sines. On the other hand, the remaining $\frac{1}{4}(n-1)$ will be larger than $\frac{1}{2} n,$  and these correspond to negative sines. Therefore, the product of all the sines must be equal to a product of a positive quantities, multiplied by the factor $(-1)^{\frac{1}{4}(n-1)},$ and thus $W$  will be equal to the product of a positive real quantity with $i^{n-1},$  or $1,$  since $i^{4}=1$  and $n-1$  is divisible by 4. That is, the quantity $W$ will be a positive real quantity, and hence we must have

$$W=+\surd{n}, \quad T=+\surd{n}$$

In the second case, where $n$ is of the form $4 \mu+3,$  in the series of odd numbers

$$1,3,5,7 \ldots \frac{1}{2}(n-1), \frac{1}{2}(n+3) \ldots(n-2)$$

the first $\frac{1}{4}(n+1)$ will be smaller than $\frac{1}{2} n,$  and the remaining $\frac{1}{4}(n-3)$  will be larger. Among the sines of the arcs $\omega,$ $3 \omega,$  $5 \omega \ldots (n-2) \omega,$  therefore, $\frac{1}{4}(n-3)$  will be negative, and thus $W$  will be the product of $i^{\frac{1}{2}(n-1)}$  with a positive real quantity and $(-1)^{\frac{1}{4}(n-3)};$  the third factor is $=i^{\frac{1}{2}(n-3)},$  which when combined with the first, gives $i^{n-2}=i,$  since $i^{n-3}=1.$  Therefore we must have

$W=+i \surd{n},$ and $U=+\surd{n}$

15.
We will now show how the same conclusions can be deduced from the series considered in article 9. Let us write $-y^{-1}$ in place of $x^{\frac{1}{2}}$  in equation [4], so that

$$1-y^{-1} \frac{1-y^{-2 m}}{1-y^{-2}}+y^{-2} \frac{(1-y^{-2 m})(1-y^{-2 m+2})}{(1-y^{-2})(1-y^{-4})}-y^{-3} \frac{(1-y^{-2 m})(1-y^{-2 m+2})(1-y^{-2 m+2})}{(1-y^{-2})(1-y^{-4})(1-y^{-6})}+\text{etc.}$$

up to the $m+1^{\text{st}}$ term will be

$$

If we take $y$ to be a proper root of the equation $y^{n}-1=0,$  say $r,$  and at the same time we set $m=n-1,$  then we have

$$\begin{aligned} & \frac{1-y^{-2 m}}{1-y^{-2}}=\frac{1-r^{2}}{1-r^{-2}}=-r^{2} \\ & \frac{1-y^{-2 m+2}}{1-y^{-4}}=\frac{1-r^{4}}{1-r^{-4}}=-r^{4} \\ & \frac{1-y^{-2 m+4}}{1-y^{-6}}=\frac{1-r^{6}}{1-r^{-6}}=-r^{6} \text{ etc.} \end{aligned}$$

up to

$$\frac{1-y^{-2}}{1-y^{-2 m}}=\frac{1-r^{2 n-2}}{1-r^{-2 n+2}}=-r^{2 n-2}$$

where it should be noted that none of the denominators $1-r^{-2},$ $1-r^{-4}$  etc. will be $=0.$ Hence equation [7] takes the form

$$1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}=(1-r^{-1})(1+r^{-2})(1-r^{-3}) \ldots(1+r^{-n+1})$$

in the second member of this equation, if we multiply the first term by the last, the second term by the penultimate, etc. , then we obtain

$$\begin{aligned} & (1-r^{-1})(1+r^{-n+1})=r-r^{-1} \\ & (1+r^{-2})(1-r^{-n+2})=r^{n-2}-r^{-n+2} \\ & (1-r^{-3})(1+r^{-n+3})=r^{3}-r^{-3} \\ & (1+r^{-4})(1-r^{-n+4})=r^{n-4}-r^{-n+4} \text{ etc.} \end{aligned}$$

From these products, it is easy to see that the product

$$(r-r^{-1})(r^{3}-r^{-3})(r^{5}-r^{-5}) \ldots(r^{n-4}-r^{-n+4})(r^{n-2}-r^{-n+2})$$

will be

$$=1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}=W$$

This equation is identical to equation [5] in article 12, which was derived from the first series, so the rest of the argument can be carried out in the same way as in articles 13 and 14.

16.
We now move on to the other case, where $n$ is an even number. First let $n$ be of the form $4 \mu+2,$  or equivalently an oddly even number. It is clear that the numbers $\frac{1}{4} n n,$ $(\frac{1}{2} n+1)^{2}-1,$  $(\frac{1}{2} n+2)^{2}-4,$  etc. , or in general $(\frac{1}{2} n+\lambda)^{2}-\lambda \lambda,$ can be divided by $\frac{1}{2} n$  to produce odd quotients, and thus they are congruent to $\frac{1}{2} n$  modulo $n.$  Hence, if $r$  is a proper root of the equation $x^{n}-1=0,$  and thus $r^{\frac{1}{2} n}=-1,$  it follows that

$$\begin{aligned} r^{(\frac{1}{2} n)^{2}} & =-1 \\ r^{(\frac{1}{2} n+1)^{2}} & =-r \\ r^{(\frac{1}{2} n+2)^2} & =-r^{4} \\ r^{(\frac{1}{2} n+3)^{2}} & =-r^{9} \text{ etc.} \end{aligned}$$

Hence, in the series

$$1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}$$

the term $r^{(\frac{1}{2} n)^{2}}$ destroys the first term, the following term destroys the second term, etc. , and therefore

$$W=0, \quad T=0, \quad U=0$$

17.
There remains the case where $n$ is of the form $4 \mu,$  or evenly even. Here, in general, $(\frac{1}{2} n+\lambda)^{2}-\lambda \lambda$ will be divisible by $n,$  and therefore

$$r^{(\frac{1}{2} n+\lambda)^{2}}=r^{\lambda \lambda}$$

Hence, in the series

$$1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}$$

the term $r^{(\frac{1}{2} n)^{2}}$ will be equal the first term, the following term will be equal to the second term, etc. , so that

$$W=2(1+r+r^{4}+r^{9}+\text{etc.}+r^{(\frac{1}{2} n-1)^{2}})$$

Let us now suppose that in equation [7] of article 15, we set $m=\frac{1}{2} n-1,$ and for $y$  we substitute a proper root $r$  of the equation $y^{n}-1=0.$  Then just as in article 15, the equation takes the form

$$1+r+r^{4}+\text{etc.}+r^{(\frac{1}{2} n-1)^{2}}=(1-r^{-1})(1+r^{-2})(1-r^{-3}) \ldots(1-r^{-\frac{1}{2} n+1})$$

or

$$

Furthermore, since $r^{\frac{1}{2} n}=-1,$ and thus

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

and since the product of the factors $-r^{\frac{1}{2} n-2},$ $-r^{\frac{1}{2} n-4},$  $-r^{\frac{1}{2} n-6}$  etc. up to $-r^{2}$ is $=(-1)^{\frac{1}{4} n-1} r^{\frac{1}{16} n n-\frac{1}{4} n},$  the previous equation can also be expressed as

$$W=2(-1)^{\frac{1}{4} n-1} r^{\frac{1}{16} n n-\frac{1}{4} n}(1-r^{-1})(1-r^{-2})(1-r^{-3})(1-r^{-4}) \ldots(1-r^{-\frac{1}{2} n+1})$$

Since

$$\begin{aligned} & 1-r^{-1}=-r^{-1}(1-r^{-n+1}) \\ & 1-r^{-2}=-r^{-2}(1-r^{-n+2}) \\ & 1-r^{-3}=-r^{-3}(1-r^{-n+3}) \text{ etc.} \end{aligned}$$

we have

$$\begin{aligned} (1-r^{-1})&(1-r^{-2})(1-r^{-3}) \ldots(1-r^{-\frac{1}{2} n+1}) \\ =&(-1)^{\frac{1}{2} n-1} r^{-\frac{1}{8} n n+\frac{1}{4} n}(1-r^{-\frac{1}{2} n-1})(1-r^{-\frac{1}{2} n-2})(1-r^{-\frac{1}{2} n-3}) \ldots(1-r^{-n+1}) \end{aligned}$$

and therefore

$$W=2(-1)^{\frac{3}{4} n-2} r^{-\frac{1}{16} n n}(1-r^{-\frac{1}{2} n-1})(1-r^{-\frac{1}{2} n-2})(1-r^{-\frac{1}{2} n-3}) \ldots(1-r^{-n+1})$$

Multiplying this value of $W$ by the one we previously found, and adjoining the factor $1-r^{-\frac{1}{2} n}$  to both sides, we get

$$(1-r^{-\frac{1}{2} n}) W^{2}=4(-1)^{n-3} r^{-\frac{1}{4} n}(1-r^{-1})(1-r^{-2})(1-r^{-3}) \ldots(1-r^{-n+1})$$

But we have

$$\begin{aligned} 1-r^{-\frac{1}{2} n}&=2 \\ (-1)^{n-3}&=-1 \\ r^{-\frac{1}{4} n}&=-r^{\frac{1}{4} n} \\ (1-r^{-1})(1-r^{-2})(1-r^{-3}) &\ldots(1-r^{-n+1})=n \end{aligned}$$

From which it finally follows that

$$

Now it can be easily seen that $r^{\frac{1}{4} n}$ is either $=+i$  or $=-i,$  depending on whether $k$  is of the form $4 \mu+1$  or $4 \mu+3.$  And since

$$2 i=(1+i)^{2}, \quad-2 i=(1-i)^{2}$$

we will have, in the case where $k$ is of the form $4 \mu+1,$

$W= \pm(1+i) \surd{n},$ and thus $T=U=\pm \surd{n}$

and in the other case, where $k$ is of the form $4 \mu+3,$

$W= \pm(1-i) \surd{n},$ and thus $T=-U= \pm \surd{n}$

18.
The method of the previous article provided the absolute values of the functions $T$ and $U,$  and determined the conditions under which equal or opposite signs should be given to them. But the signs themselves are not yet determined at this point. We will supply this for the case $k=1,$ as follows.

Let $\rho=\cos \frac{1}{2} \omega+i \sin \frac{1}{2} \omega,$ so that $r=\rho \rho$  and $\rho^{n}=-1.$  It is clear that equation [8] can be expressed as

$$W=2(1+\rho^{n-2})(1+\rho^{-4})(1+\rho^{n-6})(1+\rho^{-8}) \ldots(1+\rho^{-n+4})(1+\rho^{2})$$

or, by arranging the factors in a different order,

$$W=2(1+\rho^{2})(1+\rho^{-4})(1+\rho^{6})(1+\rho^{-8}) \ldots(1+\rho^{-n+4})(1+\rho^{n-2})$$

Now we have

$$\begin{aligned} & 1+\rho^{2}=2 \rho \cos \frac{1}{2} \omega \\ & 1+\rho^{-4}=2 \rho^{-2} \cos \omega \\ & 1+\rho^{+6}=2 \rho^{3} \cos \frac{3}{2} \omega \\ & 1+\rho^{-8}=2 \rho^{-4} \cos 2 \omega \text{ etc.} \end{aligned}$$

up to

$$\begin{aligned} & 1+\rho^{-n+4}=2 \rho^{-\frac{1}{2} n+2} \cos (\frac{1}{4} n-1) \omega \\ & 1+\rho^{n-2}=2 \rho^{\frac{1}{2} n-1} \cos (\frac{1}{4} n-\frac{1}{2}) \omega \end{aligned}$$

Therefore, we have:

$$W=2^{\frac{1}{2} n} \rho^{\frac{1}{4} n} \cos \frac{1}{2} \omega \cos \omega \cos \frac{3}{2} \omega \ldots \cos (\frac{1}{4} n-\frac{1}{2}) \omega$$

The cosines in this product are clearly positive, but the factor $\rho^{\frac{1}{4}n}$ becomes $=\cos 45^{\circ}+i \sin 45^{\circ}=(1+i) \surd{\frac{1}{2}}.$  Hence we conclude that $W$  is the product of $1+i$  and a positive real quantity, so we must have

$$W=(1+i) \cdot \surd{n}, \quad T=+\surd{n}, \quad U=+\surd{n}$$

19.
It will be worthwhile to gather together here all of the summations we have evaluated so far. In general, we have

$$ \begin{array}{|c|c|c|} \hline T=&U=&\text{when }n\text{ is of the form}\\ \hline \pm\surd{n}&\pm\surd{n}&4\mu\\ \pm\surd{n}&0&4\mu+1\\ 0&0&4\mu+2\\ 0&\pm\surd{n}&4\mu+3\\ \hline \end{array} $$

and in the case where $k$ is assumed to be $=1,$  the positive sign must be assigned to the radical quantity. All of the things which had been observed by induction in article 3, for the first few values of $n$, have now been demonstrated with all rigor, and nothing remains but to determine the signs for other values of $k$ in all cases. But before this task can be undertaken in all generality, it will be necessary to first consider more closely the cases in which $n$ is either a prime number or a power of a prime number.

20.
Let $n$ be a prime odd number. Then it is clear from what was explained in article 10 that $W=1+2 \Sigma r^{a}=1+2 \Sigma R^{a k},$ where we set $R=\cos \omega+i \sin \omega,$  and $a$  denotes all of the quadratic residues of $n$  between $1$  and $n-1$  indefinitely. But if we also denote indefinitely by $b$ all the quadratic non-residues between the same limits, it is seen without any difficulty that all of the numbers $a k$  will be congruent modulo $n$  to either all of $a$  or all of $b,$  without respect to order, depending on whether $k$  is a quadratic residue or non-residue modulo $n$. Therefore, in the former case, we have

$$W=1+2 \Sigma R^{a}=1+R+R^{4}+R^{9}+\text{etc.}+R^{(n-1)^{2}}$$

and thus $W=+\surd{n},$ if $n$  is of the form $4 \mu+1,$  and $W=+i \surd{n},$  if $n$  is of the form $4 \mu+3.$

On the other hand, in the case where $k$ is a quadratic non-residue modulo $n,$  we have

$$W=1+2 \Sigma R^{b}$$

Hence, since it is clear that all integers $a$ and $b$  together complete the complex integer numbers $1,$  $2,$  $3 \ldots,$  and thus

$$\Sigma R^{a}+\Sigma R^{b}=R+R^{2}+R^{3}+\text{etc.}+R^{n-1}=-1$$

we have

$$W=-1-2 \Sigma R^{a}=-(1+R+R^{4}+R^{9}+\text{etc.}+R^{(n-1)^{2}})$$

and thus $W=-\surd{n}$ if $n$  is of the form $4\mu+1,$  and $W=-i\surd{n}$  if $n$  is of the form $4\mu+3.$

Hence we conclude:

 * first, if $n$ is of the form $4\mu+1,$  and $k$  is a quadratic residue modulo $n,$

$$T=+\surd{n}, \quad U=0$$

 * second, if $n$ is of the form $4\mu+1,$  and $k$  is a quadratic non-residue modulo $n,$

$$T=-\surd{n}, \quad U=0$$

 * third, if $n$ is of the form $4\mu+3,$  and $k$  is a quadratic residue modulo $n,$

$$T=0, \quad U=+\surd{n}$$

 * fourth, if $n$ is of the form $4\mu+3,$  and $k$  is a quadratic non-residue modulo $n,$

$$T=0, \quad U=-\surd{n}$$

21.
Let $n$ be a square or higher power of an odd prime $p$, and let $n = p^{2 \chi} q,$  where $q$  is either $1$  or $p.$  It is first of all important to observe here that if $\lambda$  is any integer not divisible by $p^{\chi},$  then we have

$$\begin{aligned} & r^{\lambda \lambda}+r^{(\lambda+p^{\chi} q)^{2}}+r^{(\lambda+2 p^{\chi} q)^{2}}+r^{(\lambda+3 p^{\chi} q)^{2}}+\text{etc.}+r^{(\lambda+n-p^{\chi} q)^{2}} \\& \quad=r^{\lambda \lambda}\left\{1+r^{2 \lambda p^{\chi} q}+r^{4 \lambda p^{\chi} q}+r^{6 \lambda p^{\chi} q}+\text{etc.}+r^{2 \lambda(n-p^{\chi} q)}\right\}=\frac{r^{\lambda \lambda}(1-r^{2 \lambda n})}{1-r^{2 \lambda p^{\chi} q}}=0 \end{aligned}$$

From this it is easy to see that

$$W=1+r^{p^{2 \chi}}+r^{4 p^{2 \chi}}+r^{9 p^{2 \chi}}+\text{etc.}+r^{(n-p^{\chi})^{2}}$$

Indeed, the remaining terms of the series

$$1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}}$$

can be distributed into $(p^{\chi}-1) q$ partial sums, each of which has $p^{\chi}$  terms, and is seen to vanish by applying the transformation given above.

Hence it follows, in the case where $q=1,$ or where $n$  is a power of a prime number with an even exponent, that

$W=p^{\chi}=+\surd{n},$ and therefore $T=+\surd{n},$ $U=0$

On the other hand, in the case where $q=p,$ or where $n$  is a power of a prime number with an odd exponent, let us set $r^{p^{2\chi}}=\rho,$  where $\rho$  is a proper root of the equation $x^{p}-1=0,$  specifically $\rho=\cos \frac{k}{p} 360^{\circ}+i \sin \frac{k}{p} 360^{\circ}.$  Then

$$W=1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(p^{\chi+1}-1)^{2}}=p^{\chi}(1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(p-1)^{2}})$$

But the sum of the series $1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(p-1)^{2}}$ has been determined in the preceding article, and from this we conclude that

$W= \pm \surd{n}=T,$ if $p$ is of the form $4 \mu+1 $ $W= \pm i \surd{n}=i U,$ if $p$  is of the form $4 \mu+3$

with a positive or negative sign depending on whether $k$ is a quadratic residue or a non-residue modulo $p.$

22.
The following proposition, which is easily derived from that which has been set forth in articles 20 and 21, will be of considerable use to us below. Let

$$W^{\prime}=1+r^{h}+r^{4 h}+r^{9 h}+\text{etc.}+r^{h(n-1)^{2}}$$

where $h$ is any integer not divisible by $p.$  Then in the case where $n=p,$  or where $n$  is a power of $p$  with an odd exponent, we have

$W^{\prime}=W$, if $h$ is a quadratic residue modulo $p$

$ W^{\prime}=-W$, if $h$ is a quadratic non-residue modulo $p$

For it is clear that $W^{\prime}$ arises from $W$  if $kh$  is substituted for $k.$  In the former case, $k$  and $kh$  will be the same, and in the latter different, insofar as they are quadratic residues or non-residues modulo $p.$

However, in the case where $n$ is a power of $p$  with an even exponent, it is clear that $W^{\prime}=+\surd{n},$  and therefore always $W^{\prime}=W.$

23.
In articles 20, 21, 22 we considered odd prime numbers and their powers. It remains, therefore, to consider the case where $n$ is a power of two.

For $n=2,$ it is clear that $W=1+r=0.$

For $n=4,$ we obtain $W=1+r+r^{4}+r^{9}=2+2 r.$  Hence $W=2+2 i,$  whenever $k$  is of the form $4 \mu+1,$  and $W=2-2 i,$  whenever $k$  is of the form $4 \mu+3.$

For $n=8,$ we have $W=1+r+r^{4}+r^{9}+r^{16}+r^{25}+r^{36}+r^{49}=2+4 r+2 r^{4}$  $=4 r.$  Hence

$W=(1+i) \surd{8}$ whenever $k$ is of the form $8 \mu+1 $

$W=(-1+i) \surd{8},$ whenever $k$  is of the form $8 \mu+3$

$W=(-1-i) \surd{8},$ whenever $k$  is of the form $8 \mu+5$

$W=(1-i) \surd{8},$ whenever $k$  is of the form $8 \mu+7$

If $n$ is a higher power of two, let $n=2^{2 \chi} q,$  so that $q$  is either equal to 1 or 2, and $\chi$  is greater than 1. It must first of all be observed here that if $\lambda$ is an integer not divisible by $2^{\chi-1},$  then we have

$$\begin{aligned} & r^{\lambda \lambda}+r^{(\lambda+2^{\chi} q)^{2}}+r^{(\lambda+2.2^{\chi} q)^{2}}+r^{(\lambda+3.2^{\chi} q)^{2}}+\text{etc.}+r^{(\lambda+n-2^{\chi} q)^{2}} \\ & \quad=r^{\lambda \lambda}\left\{1+r^{2^{\chi+1} \lambda q}+r^{2.2^{\chi+1} \lambda q}+r^{3.2^{\chi+1} \lambda q}+\text{etc.}+r^{(2 n-2^{\chi+1} q) \lambda}\right\}=\frac{r^{\lambda \lambda}(1-r^{2 \lambda n})}{1-r^{2^{\chi+1} \lambda q}}=0 \end{aligned}$$

Hence it is easy to see that

$$W=1+r^{2^{2 \chi-2}}+r^{4.2^{2 \chi-2}}+r^{9.2^{2 \chi-2}}+\text{etc.}+r^{(n-2^{\chi-1})^{2}}$$

Let us set $r^{2^{2 \chi-2}}=\rho.$ Then $\rho$  will be a root of the equation $x^{4 q}-1=0,$  and in fact $\rho=\cos \frac{k}{4 q} 360^{\circ}+i \sin \frac{k}{4 q} 360^{\circ}.$  Thus we have

$$\begin{aligned} W & =1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(2^{\chi+1} q-1)^{2}} \\ & =2^{\chi-1}(1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(4 q-1)^{2}}) \end{aligned}$$

But the sum of the series $1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(4 q-1)^{2}}$ is determined by what we have already explained in the cases $n=4,$  $n=8.$  Hence we conclude that

 * in the case where $q=1,$ or where $n$  is a power of 4,

$ W=(1+i) 2^{\chi}=(1+i) \surd{n},$ if $k$ is of the form $4 \mu+1$ $W=(1-i) 2^{\chi}=(1-i) \surd{n},$ if $k$  is of the form $4 \mu+3$

 * which are the exact formulas already given for $n=4;$

 * in the case where $q=2,$ or where $n$  is a power of two with an odd exponent greater than 3,

$W=(1+i) 2^{\chi} \surd{2}\phantom{-}=(1+i) \surd{n},$ if $k$ is of the form $8 \mu+1$

$W=(-1+i) 2^{\chi} \surd{2}=(-1+i) \surd{n},$ if $k$  is of the form $8 \mu+3$

$W=(-1-i) 2^{\chi} \surd{2}=(-1-i) \surd{n},$ if $k$  is of the form $8 \mu+5$

$W=(1-i) 2^{\chi} \surd{2}\phantom{-}=(1-i) \surd{n},$ if $k$  is of the form $8 \mu+7$

 * which also precisely match the formulas we provided for $n=8.$

24.
It will also be worthwhile to determine the ratio of the sum of the series

$$W' = 1 + r^h + r^{4h} + r^{9h} + \text{etc.} + r^{h(n-1)^2}$$

to $W,$ where $h$  is an arbitrary odd integer. Since $W'$ arises from $W$  by replacing $k$  with $kh,$  the value of $W'$  will depend on the form of the number $kh$  in the same way as $W$  depends on the form of the number $k.$  Let us set $\frac{W'}{W} = l.$  Then it is clear that

I. In the case where $n = 4,$ or any higher power of two with an even exponent,

$l = 1,$ if $h$ is of the form $4\mu+1$

$l = -i,$ if $h$  is of the form $4\mu+3,$  and $k$  is of the form $4\mu+1$

$l = +i,$ if $h$  is of the form $4\mu+3,$  and $k$  is of the same form

II. In the case where $n = 8,$ or any higher power of two with an odd exponent,

$l = 1,$ if $h$ is of the form $8\mu+1,$

$l = -1,$ if $h$  is of the form $8\mu+5,$

$l = +i,$ if either $h$  is of the form $8\mu+3,$  and $k$  is of the form $4\mu+1,$

$\phantom{l = +i} \quad$ or $h$  is of the form $8\mu+7,$  and $k$  is of the form $4\mu+3,$

$l = -i,$ if either $h$  is of the form $8\mu+3,$  and $k$  is of the form $4\mu+3,$

$\phantom{l = -i}\quad$ or $h$  is of the form $8\mu+7,$  and $k$  is of the form $4\mu+1.$

With this, the determination of $W$ in those cases where $n$  is a prime number or a power of a prime number is complete. It remains, therefore, for us to finish those cases where $n$ is composed of several prime factors, to which end the following theorem paves the way.

25.
Let $n$ be the product of two relatively prime positive integers $a$  and $b,$  and set

$$\begin{aligned} & P=1+r^{a a}+r^{4 a a}+r^{9 a a}+\text{etc.}+r^{(b-1)^{2} a a} \\ & Q=1+r^{b b}+r^{4 b b}+r^{9 b b}+\text{etc.}+r^{(a-1)^{2} b b} \end{aligned}$$

Then I claim that $W=PQ.$

Proof. Let $\alpha$ indefinitely denote the numbers $0,$  $1,$  $2,$  $3 \ldots a-1,$  let $\beta$  indefinitely denote the numbers $0,$  $1,$  $2,$  $3 \ldots b-1,$  and let $\nu$  indefinitely denote the numbers $0,$  $1,$  $2,$  $3 \ldots n-1.$  Then it is clear that

$$P=\Sigma r^{a a \beta \beta}, \quad Q=\Sigma r^{b b \alpha \alpha}, \quad W=\Sigma r^{\nu \nu}$$

Thus, we have $PQ=\Sigma r^{a a \beta \beta +b b \alpha \alpha},$ where all possible values of $\alpha$  and $\beta$  are to be substituted. Furthermore, because $2ab\alpha\beta=2\alpha\beta n,$ we have $PQ=\Sigma r^{(a \beta+b \alpha)^{2}}.$  But it is clearly seen, without difficulty, that the individual values of $a \beta+b \alpha$  are distinct from each other, and each is equal to some value of $\nu.$  Thus, we have $PQ=\Sigma r^{\nu \nu}=W.$

It should also be noted that $r^{a a}$ is a proper root of the equation $x^{b}-1=0,$  and $r^{b b}$  is a proper root of the equation $x^{a}-1=0.$

26.
Now let $n$ be the product of three mutually prime numbers $a,$  $b,$  $c.$  Then clearly if we set $bc = b',$  then $a$  and $b'$  will be relatively prime. Therefore, $W$ is a product of two factors:

$$\begin{aligned} & 1+r^{aa}+r^{4aa}+r^{9aa}+\text{etc.}+r^{(b'-1)^2aa} \\ & 1+r^{b'b'}+r^{4b'b'}+r^{9b'b'}+\text{etc.}+r^{(a-1)^2b'b'} \end{aligned}$$

However, since $r^{aa}$ is a proper root of the equation $x^{bc} - 1 = 0,$  the first factor will be the product of two factors

$$\begin{aligned} & 1+\rho^{bb}+\rho^{4bb}+\rho^{9bb}+\text{etc.}+\rho^{(c-1)^2bb} \\ & 1+\rho^{cc}+\rho^{4cc}+\rho^{9cc}+\text{etc.}+\rho^{(b-1)^2cc} \end{aligned}$$

if we set $r^{aa} = \rho.$ Hence it is clear that $W$  is the product of three factors:

$$\begin{aligned} & 1+r^{bbcc}+r^{4bbcc}+r^{9bbcc}+\text{etc.}+r^{(a-1)^2bbcc} \\ & 1+r^{aacc}+r^{4aacc}+r^{9aacc}+\text{etc.}+r^{(b-1)^2aacc} \\ & 1+r^{aabb}+r^{4aabb}+r^{9aabb}+\text{etc.}+r^{(c-1)^2aabb} \end{aligned}$$

where $r^{bbcc},$ $r^{aacc},$  and $r^{aabb}$  are proper roots of the equations $x^a - 1 = 0,$  $x^b - 1 = 0,$  $x^c - 1 = 0,$  respectively.

27.
From this it is easily concluded that in general, if $n$ is the product of any prime factors $a,$  $b,$  $c,$  etc. , then $W$ will be a product of as many factors

$$\begin{aligned} & 1+r^{\frac{n n}{a a}}+r^{\frac{4 n n}{a a}}+r^{\frac{9 n n}{a a}}+\text{etc.}+r^{\frac{(a-1)^{2} n n}{a a}} \\ & 1+r^{\frac{n n}{b b}}+r^{\frac{4 n n}{b b}}+r^{\frac{9 n n}{b b}}+\text{etc.}+r^{\frac{(b-1)^{2} n n}{b b}} \\ & 1+r^{\frac{n n}{c c}}+r^{\frac{4 n n}{c c}}+r^{\frac{9 n n}{c c}}+\text{etc.}+r^{\frac{(c-1)^{2} n n}{c c}} \text{ etc.} \end{aligned}$$

where $r^{\frac{n n}{a b}},$ $r^{\frac{n n}{ b b}},$  $r^{\frac{n n}{c c}}$  etc. are proper roots of the equations $x^{a}-1=0,$ $x^{b}-1=0,$  $x^{c}-1=0$  etc.

28.
Out of these principles, a passage to the complete determination of $W$ for any given value of $n$  has appeared before us. Let $n$ be decomposed into factors $a,$  $b,$  $c,$  etc. , which are either distinct prime numbers or powers of distinct prime numbers. Let $r^{\frac{n n}{a a}}=A,$ $r^{\frac{n n}{b b}}=B,$  $r^{\frac{n n}{c c}}=C,$  etc. , and let $A,$ $B,$  $C,$  etc. be the respective roots of the equations $x^{a}-1=0,$ $x^{b}-1=0,$  $x^{c}-1=0,$  etc. Then $W$ is the product of the factors

$$\begin{aligned} & 1+A+A^{4}+A^{9}+\text{etc.}+A^{(u-1)^{2}} \\ & 1+B+B^{4}+B^{9}+\text{etc.}+B^{(b-1)^{2}} \\ & 1+C+C^{4}+C^{9}+\text{etc.}+C^{(c-1)^{2}} \text{ etc.} \end{aligned}$$

But each of these factors can be determined by the methods explained in articles 20, 21, 23. Hence, the value of the product can also be known. It will be useful to collect the rules for determining these factors here. When the root $A$ is $=\frac{k n}{a} \cdot \frac{360^{0}}{a},$  the sum $1+A+A^{4}+A^{9}+\text{etc.}+A^{(a-1)^{2}},$  which we shall denote by $L,$  will be determined by the number $\frac{k n}{a},$  in the same way that $W$  was determined by $k$  in our general discussion. We have already distinguished twelve cases:

 * I. If $a$ is a prime number of the form $4 \mu+1,$  say $=p,$  or a power of such a prime number with an odd exponent, and at the same time $\frac{k n}{a}$  is a quadratic residue modulo $p,$  then $L=+\surd{a}.$

 * II. If $\frac{k n}{a}$ is a quadratic non-residue modulo $p,$  then $L=-\surd{a}.$

 * III. If $a$ is a prime number of the form $4 \mu+3,$  say $=p,$  or a power of such a prime number with an odd exponent, and at the same time $\frac{k n}{a}$  is a quadratic residue modulo $p,$  then $L=+i \surd{a}.$

 * IV. If, with the rest of the assumptions as in III, $\frac{k n}{a}$ is a quadratic non-residue modulo $p,$  then $L=-i \surd{a}.$

 * V. If $a$ is a square number, or a higher power of a prime number (with an even exponent), then $L=+\surd{a}.$

 * VI. If $a=2,$ then $L=0.$

 * VII. If $a=4$ or a higher power of two with an even exponent, and also $\frac{k n}{a}$  is of the form $4 \mu+1,$  then $L=(1+i) \surd{a}.$

 * VIII. If, with the rest of the assumptions as in VII, $\frac{k n}{a}$ is of the form $4 \mu+3,$  then $L=(1-i) \surd{a}.$

 * IX. If $a=8,$ or a higher power of two with an odd exponent, and at the same time $\frac{k n}{a}$  is of the form $8 \mu+1,$  then $L=(1+i) \surd{a}.$

 * X. If, with the rest of the assumptions as in IX, $\frac{k n}{a}$ is of the form $8 \mu+3,$  then $L=(-1+i) \surd{a}.$

 * XI. If, with the rest of the assumptions as in IX, $\frac{k n}{a}$ is of the form $8 \mu+5,$  then $L=(-1-i) \surd{a}.$

 * XII. If, with the rest of the assumptions as in IX, $\frac{k n}{a}$ is of the form $8 \mu+7,$  then $L=(1-i) \surd{a}.$

29.
For example, let $n=2520=8. 9 . 5 . 7$ and $k=13.$  In this case, we have

for $a=8,$ by case XII, $L=(1-i) \surd{8}$

for the factor 9, by case V, the corresponding sum will be $=\surd{9}$

for the factor 5, by case II, the corresponding sum will be $=-\surd{5}$

for the factor 7, by case III, the corresponding sum will be $=+i \surd{7}$ Hence, we get $W=(1-i) \cdot(-i) \cdot \surd{2520}=(-1-i) \surd{2520}.$

If, for the same value of $n,$ we set $k = 1,$  then

for the factor 8, the sum is $(-1+i) \surd{8}$

for the factor 9, the sum is $\surd{9}$

for the factor 5, the sum is $\surd{5}$

for the factor 7, the sum is $-i \surd{7}$ Hence, the product is $W = (1+i) \surd{2520}.$

30.
Another method of finding the sum $W$ in a general manner is suggested by that which was set forth in articles 22 and 24. Set $\cos \omega+i \sin \omega=\rho,$ and

$$\rho^{\frac{n n}{a a}}=\alpha, \quad \rho^{\frac{n n}{b b}}=\beta, \quad \rho^{\frac{n n}{c c}}=\gamma \text{ etc.}$$

so that we have $r=\rho^{k},$ $A=\alpha^{k},$  $B=\beta^{k},$  $C=\gamma^{k}$  etc. Then

$$1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(n-1)^{2}}$$

will be a product of factors

$$\begin{aligned} & 1+\alpha+\alpha^{4}+\alpha^{9}+\text{etc.}+\alpha^{(a-1)^{2}} \\ & 1+\beta+\beta^{4}+\beta^{9}+\text{etc.}+\beta^{(b-1)^{2}} \\ & 1+\gamma+\gamma^{4}+\gamma^{9}+\text{etc.}+\gamma^{(c-1)^{2}} \text{ etc.} \end{aligned}$$

and therefore $W$ will be a product of factors

$$\begin{aligned} & w=1+\rho+\rho^{4}+\rho^{9}+\text{etc.}+\rho^{(n-1)^{2}} \\ & \mathfrak{A}=\frac{1+A+A^{4}+A^{9}+\text{etc.}+A^{(a-1)^{2}}}{1+\alpha+\alpha^{4}+\alpha^{9}+\text{etc.}+a^{(a-1)^{2}}} \\ & \mathfrak{B}=\frac{1+B+B^{4}+B^{9}+\text{etc.}+B^{(b-1)^{2}}}{1+\beta+\beta^{4}+\beta^{9}+\text{etc.}+\beta^{(b-1)^2}} \\ & \mathfrak{C}=\frac{1+C+C^{4}+C^{9}+\text{etc.}+C^{(c-1)^{2}}}{1+\gamma+\gamma^{4}+\gamma^{9}+\text{etc.}+\gamma^{(c-1)^{2}}} \text{ etc.} \end{aligned}$$

Now, the first factor $w$ is determined by the discussion in article 19 above; the remaining factors $\mathfrak{A},$  $\mathfrak{B},$  $\mathfrak{C}$  etc. come from the formulas of articles 22 and 24, which are collected here again so that they can all be considered together. Twelve cases must be distinguished here, namely

 * I. If $a$ is a prime number (odd) $=p,$  or a power of such a number with an odd exponent, and $k$  is a quadratic residue modulo $p,$  then the corresponding factor will be $\mathfrak{A}=+1.$

 * II. If, with the rest of the assumptions as in I, $k$ is a quadratic non-residue modulo $p,$  then $\mathfrak{A}=-1.$

 * III. If $a$ is the square of an odd prime number, or a higher power with an even exponent, then $\mathfrak{A}=+1.$

 * IV. If $a$ is $=4,$  or a higher power of two with an even exponent, and $k$  is of the form $4 \mu+1,$  then $\mathfrak{A}=+1.$

 * V. If, with the rest of the assumptions as in IV, $k$ is of the form $4 \mu+3,$  and $\frac{n}{a}$  is of the form $4 \mu+1,$  then $\mathfrak{A}=-i.$

 * VI. If, with the rest of the assumptions as in IV, $k$ is of the form $4 \mu+3,$  and $\frac{n}{a}$  is of the form $4 \mu+3,$  then $\mathfrak{A}=+i.$

 * VII. If $a$ is $=8,$  or a higher power of two with an odd exponent, and $k$  is of the form $8 \mu+1,$  then $\mathfrak{A}=+1.$

 * VIII. If, with the rest of the assumptions as in VII, $k$ is of the form $8 \mu+5,$  then $\mathfrak{A}=-1.$

 * IX. If, with the rest of the assumptions as in VII, $k$ is of the form $8 \mu+3,$  and $\frac{n}{a}$  is of the form $4 \mu+1,$  then $\mathfrak{A}=+i.$

 * X. If, with the rest of the assumptions as in VII, $k$ is of the form $8 \mu+3,$  and $\frac{n}{a}$  is of the form $4 \mu+3,$  then $\mathfrak{A}=-i.$

 * XI. If, with the rest of the assumptions as in VII, $k$ is of the form $8 \mu+7,$  and $\frac{n}{a}$  is of the form $4 \mu+1,$  then $\mathfrak{A}=-i.$

 * XII. If, with the rest of the assumptions as in VII, $k$ is of the form $8 \mu+7,$  and $\frac{n}{a}$  is of the form $4 \mu+3,$  then $\mathfrak{A}=+i.$

We omit the case where $a=2;$ indeed, in this case $\mathfrak{A}$  would be $\frac{0}{0},$  or indeterminate, but then anyway $W=0$.

The remaining factors $\mathfrak{B},$ $\mathfrak{C},$  etc. depend in the same way on $b,$ $c,$  etc. , as $\mathfrak{A}$ depends on $a.$

31.
According to this second method, the first example in article 29 is as follows:

 * The factor $w$ is $=(1+i) \surd{2520}$

 * For $a=8,$ the corresponding factor $\mathfrak{A}$  is, by case VIII, $=-1$

 * The second factor $9$ corresponds to a factor $+1$  (by case III)

 * The factor $5$ corresponds to a factor $-1$  (by case II)

 * The factor $7$ corresponds to a factor $-1$  (by case II)

Hence, the product $W=(-1-i) \surd{2520}$ is obtained, as in article 29.

32.
Since the value of $W$ can be determined using two methods, one of which is based on the relations of the numbers $\frac{n k}{a},$  $\frac{n k}{b},$  $\frac{n k}{c},$  etc. with the numbers $a,$ $b,$  $c,$  etc. , and the other depending on the relations of $k$ with the numbers $a,$  $b,$  $c,$  etc. , there must be a certain conditional connection between all these relations, so that each of them must be determinable from the others. Let us suppose that all the numbers $a,$ $b,$  $c,$  etc. are odd prime numbers, and take $k=1.$ Let the factors $a,$  $b,$  $c,$  etc. be distributed into two classes, one of which contains those that are of the form $4\mu+1,$ and which are denoted by $p,$  $p^{\prime},$  $p^{\prime\prime},$  etc. , and the other consisting of those that are of the form $4\mu+3,$ and which are denoted by $q,$  $q^{\prime},$  $q^{\prime\prime},$  etc. We will designate the multitude of the latter by $m.$ Having done this, we observe first that $n$  will be of the form $4\mu+1,$  when $m$  is even (which also applies to the case where the factors of the other class are completely absent, or where $m=0$ ), whereas $n$  will be of the form $4\mu+3,$  when $m$  is odd. Now the determination of $W$ is achieved by the first method as follows. Let numbers $P,$ $P^{\prime},$  $P^{\prime\prime},$  etc. , $Q,$ $Q^{\prime},$  $Q^{\prime\prime},$  etc. be determined from the relations between the numbers $\frac{n}{p},$ $\frac{n}{p^{\prime}},$  $\frac{n}{p^{\prime \prime}}$  etc., $\frac{n}{q},$  $\frac{n}{q^{\prime}},$  $\frac{n}{q^{\prime \prime}}$  and the numbers $p,$  $p^{\prime},$  $p^{\prime\prime},$  etc. , $q,$ $q^{\prime},$  $q^{\prime\prime},$  etc. , respectively, by setting

$P=+1,$ if $\frac{n}{p}$ is a quadratic residue modulo $p$

$P=-1,$ if $\frac{n}{p}$  is a quadratic non-residue modulo $p$

and likewise for the rest. Then $W$ will be the product of the factors $P \surd{p},$  $P^{\prime} \surd{p^{\prime}},$  $P^{\prime\prime} \surd{p^{\prime\prime}},$  etc. , $iQ \surd{q},$ $iQ^{\prime} \surd{q^{\prime}},$  $iQ^{\prime\prime} \surd{q^{\prime\prime}},$  etc. , and hence

$$W=P P^{\prime} P^{\prime\prime} \ldots Q Q^{\prime} Q^{\prime\prime} \ldots i^{m} \surd{n}$$

By the second method, or rather directly by the rules from article 19,

$W=+\surd{n},$ if $n$ is of the form $4\mu+1,$ or equivalently, if $m$  is even

$W=+i \surd{n},$ if $n$  is of the form $4\mu+3,$  or if $m$  is odd Both cases may be included together in the following formula:

$$W=i^{m m} \surd{n}$$

Hence it follows that

$$P P^{\prime} P^{\prime\prime} \ldots Q Q^{\prime} Q^{\prime\prime} \ldots=i^{m m-m}$$

But $i^{m m-m}$ is $=1$  whenever $m$  is of the form $4\mu$  or $4\mu+1,$  and $=-1$  whenever $m$  is of the form $4\mu+2$  or $4\mu+3,$  and from this we deduce the following very elegant

''Let $a,$ $b,$  $c,$  etc. denote positive odd prime numbers that are not equal to each other, and let their product be $=n.$ Let $m$  be the number of the form $4\mu + 3$  among them, so that the other numbers are of the form $4\mu + 1.$  Then the multitude of those numbers among $a,$  $b,$  $c,$  etc. such that $\frac{n}{a},$ $\frac{n}{b},$  $\frac{n}{c},$  etc. are quadratic non-residues, will be even whenever $m$ is of the form $4\mu$  or $4\mu + 1,$  but odd whenever $m$  is of the form $4\mu + 2$  or $4\mu + 3.$ ''

By setting e. g. $a=3,$ $b=5,$  $c=7,$  $d=11,$  we have three numbers of the form $4\mu + 3,$  namely $3,$  $7,$  and $11;$  and we have $5. 7 . 11 R 3;$ $3. 7 . 11 R 5;$ $3. 5 . 11 R 7;$ $3. 5 . 7 N 11,$ so there is a unique $\frac{n}{d}$  which is a quadratic non-residue modulo $d.$

33.
The celebrated fundamental theorem concerning quadratic residues is nothing but a special case of the theorem just developed. By limiting the multitude of the numbers $a,$ $b,$  $c,$  etc. to two, it is evident that if only one of them, or neither, is of the form $4 \mu+3,$ then we must have simultaneously $a R b,$  $b R a,$  or simultaneously $a N b,$  $b N a.$  On the other hand, if both are of the form $4 \mu+3,$  then one of them must be a quadratic non-residue modulo the other, and the other a quadratic residue modulo the one. And so a fourth demonstration has been given for this most important theorem, the first and second demonstration having been given in Disquisitiones Arithmeticae, and the third recently in a special commentary (Commentt. T. XVI). We will present two other proofs in the future, based again on completely different principles. It is exceedingly surprising that this most beautiful theorem, which at first so obstinately eluded all attempts, could be approached later by methods so very distant from one other.

34.
In fact, the remaining theorems, which act as a supplement to the fundamental theorem, that is, by which the prime numbers for which $-1,$ $2,$  and $-2$  are quadratic residues or non-residues may be identified, can also be derived from the same principles. Let us start with the residue $+2.$

Set $n=8a,$ where $a$  is a prime number, and let $k=1.$  Then by the method of article 28, $W$  will be the product of two factors, of which one will be $+\surd{a}$  or $+i\surd{a},$  if $8,$  or equivalently $2,$  is a quadratic residue modulo $a;$  or else $-\surd{a}$  or $-i\surd{a},$  if 2 is a quadratic non-residue modulo $a.$  The second factor is

$(1+i)\surd{8},$ if $a$ is of the form $8\mu+1$

$(-1+i)\surd{8},$ if $a$  is of the form $8\mu+3$

$(-1-i)\surd{8},$ if $a$  is of the form $8\mu+5$

$(1-i)\surd{8},$ if $a$  is of the form $8\mu+7$

But by article 18, we will always have $W=(1+i)\surd{n}.$ Dividing this value by the four values of the second factor, it is clear that the first factor must be

$ +\surd{a},$ if $a$ is of the form $8\mu+1$

$-i\surd{a},$ if $a$  is of the form $8\mu+3$

$-\surd{a},$ if $a$  is of the form $8\mu+5$

$+i\surd{a},$ if $a$  is of the form $8\mu+7$

From this it follows automatically that $2$ must be a quadratic residue modulo $a$  in the first and fourth cases, and in the second and third cases it must be a quadratic non-residue.

35.
Prime numbers for which $-1$ is a quadratic residue or non-residue are easily recognized with the help of the following theorem, which is also quite memorable by itself.

The product of the two factors

$$\begin{aligned} & W^{\prime}=1+r^{-1}+r^{-4}+\text{etc.}+r^{-(n-1)^{2}} \\ & W=1+r+r^{4}+\text{etc.}+r^{(n-1)^{2}} \end{aligned}$$

is $=n,$ if $n$  is odd, or $=0,$  if $n$  is odd even, or $=2 n,$  if $n$  is evenly even.

Proof. Since it is clear that

$$\begin{aligned} W & =r+r^{4}+r^{9}+\text{etc.}+r^{n n} \\ & =r^{4}+r^{9}+\text{etc.}+r^{(n+1)^{2}} \\ & =r^{9}+\text{etc.}+r^{(n+2)^{2}} \text{ etc.} \end{aligned}$$

the product $W {W}^{\prime}$ can also be presented as

$$\begin{aligned} &\phantom{ +, } 1+r+r^{4}+r^{9}+\text{etc.}+r^{(n-1)^{2}} \\ &+ r^{-1}(r+r^{4}+r^{9}+r^{16}+\text{etc.}+r^{n n}) \\ &+ r^{-4}(r^{4}+r^{9}+r^{16}+r^{25}+\text{etc.}+r^{(n+1)^{2}}) \\ &+ r^{-9}(r^{9}+r^{16}+r^{25}+r^{36}+\text{etc.}+r^{(n+2)^{2}}) \\ &\text{etc.} \\ &+ r^{-(n-1)^{2}}(r^{(n-1)^{2}}+r^{n n}+r^{(n+1)^{2}}+r^{(n+2)^{2}}+\text{etc.}+r^{(2 n-2)^{2}}) \end{aligned}$$

which, when summed vertically, produces

$$\begin{aligned} &\phantom{+}n \\ &+r(1+r r+r^{4}+r^{6}+\text{etc.}+r^{2 n-2}) \\ &+r^{4}(1+r^{4}+r^{8}+r^{12}+\text{etc.}+r^{4 n-4}) \\ &+r^{9}(1+r^{6}+r^{12}+r^{18}+\text{etc.}+r^{6 n-6}) \\ &+\text{etc.} \\ &+r^{(n-1)^{2}}(1+r^{2 n-2}+r^{4 n-4}+r^{6 n-6}+\text{etc.}+r^{2(n-1)^{2}}) \end{aligned}$$

Now if $n$ is odd, each part of this sum, except the first $n,$  will be $=0.$  For the second part is clear $\frac{r(1-r^{2 n})}{1-r r},$  the third $\frac{r^{4}(1-r^{4 n})}{1-r^{4}},$  etc. When $n$ is even, it is also be necessary to study the part

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

which is $=n r^{\frac{1}{4} n n}.$ In the former case, we therefore obtain $W W^{\prime}=n,$  but in the latter, $=n+n r^{\frac{1}{4} n n}.$  But $r^{\frac{1}{4} n n}$  will be $=+1,$  if $n$  is evenly even, and thus $W W^{\prime}=2 n.$  On the other hand, if $n$  is oddly even, then $r^{\frac{1}{4} n n}=-1,$  and thus $W W^{\prime}=0.$  Q. E. D.

36.
Already from article 22, it is clear that if $n$ is an odd prime number, then $\frac{W^{\prime}}{W}$  will be equal to $+1$  or $-1,$  depending on whether $-1$  is a quadratic residue or a non-residue modulo $n.$  Hence in the former case, we must have $W^{2}=+n,$  in the latter $W^{2}=-n;$  wherefore, by article 13, we conclude that the former case can only occur when $n$  is of the form $4 \mu+1,$  and the latter case when $n$  is of the form $4 \mu+3.$

Finally, from the combination of conditions for the residues $+2$ and $-1,$  it naturally follows that $-2$  is a quadratic residue modulo any prime number of the form $8\mu+1$  or $8\mu+3,$  and it is a quadratic non-residue modulo any prime number of the form $8\mu+5$  or $8\mu+7.$


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Page:The Painted Veil - Maugham - 1925.djvu/132

 wonder how you behave when you’re alone. When I’m there you’re acting, both of you, and acting damned badly, by George. You’d neither of you get thirty bob a week in a touring company if that’s the best you can do.”

“I don’t know what you mean,” smiled Kitty, keeping up a pretence of frivolity which she knew did not deceive.

“You’re a very pretty woman. It’s funny that your husband should never look at you. When he speaks to you it sounds as though it were not his voice but somebody else’s.”

“Do you think he doesn’t love me?” asked Kitty in a low voice, hoarsely, putting aside suddenly her lightness.

“I don’t know. I don’t know if you fill him with such a repulsion that it gives him goose-flesh to be near you or if he’s burning with a love that for some reason he will not allow himself to show. I’ve asked myself if you’re both here to commit suicide.”

Kitty had seen the startled glance and then the scrutinising look Waddington gave them when the incident of the salad took place.

“I think you’re attaching too much importance to a few lettuce leaves,” she said flippantly. She rose. “Shall we go home? I’m sure you want a whisky and soda.”

“You’re not a heroine at all events. You’re frightened to death. Are you sure you don’t want to go away?”

“What has it got to do with you?”


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 comfort and happiness. And then let us suppose all the different branches of production to be organized into trades unions, or whatever you may call it, for the purpose of producing all the necessaries and luxuries of life according to the best, most rational process conducted by the directions of the scientists of the age. Let us furthermore suppose the people who are the mediaries between the producer and the consumer to be organized like the producers, also the people who carry on the business of transportation and communication, and supposed their income to be regulated according to what the producers receive for their labor; and let it be supposed that seven hours be sufficient for everybody to work in order to produce whatever is needed, and to do all other business for the whole commonwealth. Let us finally suppose that by some great invention the entire department of machinery and mechanism would have to be revolutionized, as it may be any day at the present period, by the steam engine being supplanted by the electro-dynamic motor. By such a sudden change one-fourth of all the workers hitherto employed in that department might become superfluous, as the quantity of products to be brought forth by that department could be created by applying the new invention in three quarters of the time formerly necessary for the process of production. Under the new social system the introduction of such an invention would be a blessing at once for all mankind; for, the one-fourth of workers having become superfluous could be employed to produce luxuries, that formerly could not be produced for the lack of laboring force and time. So, you see, that every new invention, even if it makes ever so many workers superfluous in one branch, will result in an advantageous improvement of the condition of the whole. But, the time will certainly come when all desires of mankind for comfort and enjoyment will be fully satisfied, and if then a new invention should be made, it will be applied for the purpose of reducing the hours of labor of all workers; and so, I think, I have proved to you that our society with its easily manageable organization has in its hands the means of using all new improvements, discoveries or inventions, yea, even every unexpected favor from nature's bountiful treasury for the benefit of every individual, consequently for the benefit of the whole.


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 “I’ll help you.”

“Are you going to fall to my look of secret sorrow? Look at my profile and tell me if my nose isn’t a trifle too long.”

He gazed at her reflectively, that malicious, ironical look in his bright eyes, but mingled with it, a shadow, like a tree standing at a river’s edge and its reflection in the water, was an expression of singular kindliness. It brought sudden tears to Kitty’s eyes.

“Must you stay?”

“Yes.”

They passed under the flamboyant archway and walked down the hill. When they came to the compound they saw the body of the dead beggar. He took her arm, but she released herself. She stood still.

“It’s dreadful, isn’t it?”

“What? Death.”

“Yes. It makes everything else seem so horribly trivial. He doesn’t look human. When you look at him you can hardly persuade yourself that he’s ever been alive. It’s hard to think that not so very many years ago he was just a little boy tearing down the hill and flying a kite.”

She could not hold back the sob that choked her.


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 But, right here, I would like to make the very same objection you made only a short while ago. You said—and I think you were right—that it would be nonsensical to ask an unemployed cigarmaker to cultivate land out west. Even if you gave him all the implements necessary for successful cultivation, he would starve on the most fertile ground donated to him. Therefore, if we suppose that under the new social system an invention had been made by which 10,000 cigarmakers would become superfluous, what would you do with them? Would you have them till the soil, or what would they do?

You forget two important points. While to-day these 10,000 workers would be left to shift for themselves, that's to say: while they would be altogether helpless, the new society, in which all inventions and improvements will immediately benefit everyone, will have the means to gradually employ these 10,000 workers in other branches, where they will be occupied with some labor similar to their former trade, and they will soon learn it; the cost of such changes will be easily paid out of the gain accruing from the introduction of the new invention for the whole society. The second point of importance you forget, is that the division of labor will continually go on, and all work will finally be reduced to a few simple movements of the workers' hands in turning on some piece of machinery, or changing its speed, or stopping it, etc., a task that will be easy to learn within a very short time; consequently the change from one occupation to another will not cause serious trouble neither to the whole people, nor to the individuals undergoing such change; and, what to-day causes disaster to hundreds and thousands, will, under the new system benefit every single individual of the whole commonwealth.

I suppose, of course, that the organization as you pictured it to me as existing in the United States and Canada, should be extended all over the world?

Communism and Socialism will only then be a blessing to all mankind when it shall extend over every land of the globe; and when this should be the case, it will result in a glorious condition of those of whose future happiness and wealth


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FEW days later Waddington, sitting with Kitty, a long glass of whisky and soda in his hand, began to speak to her of the convent.

“The Mother Superior is a very remarkable woman,” he said. “The Sisters tell me that she belongs to one of the greatest families in France, but they won’t tell me which; the Mother Superior, they say, doesn’t wish it to be talked of.”

“Why don’t you ask her if it interests you?” smiled Kitty.

“If you knew her you’d known to ask her an indiscreet question.”

“She must certainly very remarkable if she can impress you with awe.’

“I am the bearer of a message from her to you. She has asked me to say that, though of course you may not wish to adventure into the very centre of the epidemic, if you do not mind that it will give her great pleasure to show you the convent.”

“It’s very kind of her. I shouldn’t have thought she was aware of my existence.”

“I’ve spoken about you; I go there two or three times a week just now to see if there’s anything I can do; and I daresay your husband has told them about you. You must be prepared to find that they have an unbounded admiration for him.”

“Are you a Catholic?”

His malicious eyes twinkled and his funny little face was puckered with laughter.


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 “Why are you grinning at me?” asked Kitty.

“Can any good come out of Galilee? No, I’m not a Catholic. I describe myself as a member of the Church of England, which I suppose is an inoffensive way of saying that you don’t believe in anything very much. When the Mother Superior came here ten years ago she brought seven nuns with her and of those all but three are dead. You see, at the best of times, Mei-tan-fu is not a health resort. They live in the very middle of the city, in the poorest district, they work very hard and they never have a holiday.”

“But are there only three and the Mother Superior now?”

“Oh, no, more have taken their places. There are six of them now. When one of them died of cholera at the beginning of the epidemic two others came up from Canton.”

Kitty shivered a little.

“Are you cold?”

“No, it was only some one walking over my grave.”

“When they leave France they leave it for ever. They’re not like the Protestant missionaries who have a year’s leave every now and then. I always think that must be the hardest thing of all. We English have no very strong attachment to the soil, we can make ourselves at home in any part of the world, but the French, I think, have an attachment to their country which is almost a physical bond. They’re never really at ease when they’re out of it. It always seems to me very moving that these


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or the absolute decree and predestination of both good and evil, is the sixth article of the Muslim's creed. The orthodox belief is that whatever hath or shall come to pass in this world, whether it be good or bad, proceedeth entirely from the divine will, and is irrevocably fixed and recorded in the preserved tablet (Lahw-ul-Mahfúz).

Of this doctrine Muhammad makes great use in his Qurán, and all those who have had any practical acquaintance with the lives of Muammadans, know well to what extent it influences the daily life of every Muslim. It


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 Rh is not only urged as a source of consolation in every trial, but as a palliation of every crime. "It was written in my taqdír" (fate), is an excuse familiar to every European who has had much intercourse with Muslim servants or soldiers.

The following is a translation of an Arabic treatise on the subject: "Faith in the decrees of God, is that we believe in our heart and confess with our tongue that the most High God hath decreed all things so that nothing can happen in the world, whether it respects the conditions and operations of things, or good and evil, or obedience and disobedience, or faith and infidelity, or sickness and health, or riches and poverty, or life and death, that is not contained in the written tablet of the decrees of God. But God hath so decreed good works, obedience, and faith, that He ordains and wills them, and that they may be under His decree, His salutary direction, His good pleasure and command. On the contrary, God hath decreed, and does ordain and determine evil, disobedience, and infidelity; yet without His salutary direction, good pleasure and command, but being only by way


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100 of seduction, indignation, and prohibition. But whosoever shall say that God is not delighted with good faith, or that God hath not an indignation against evil and unbelief, he is certainly an infidel."


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Page talk:Nature - Volume 1.pdf/13

Personal equation of astronomical observers
I do not know what is meant to be referred to by "Letter on, 3" in the first item listed on this page. Page 3 is an advertising page, and in any case printed well before the letter on page 85 which it is supposedly in response to. I have skimmed the letters section of every single issue (before, after, and including the letters section on page 85), reading in detail every letter relating to astronomy, and the only reference to that first one is the response by van Stadt on page 337.

A later letter by George C. Thompson, on page 507, does use the phrase "personal equations" with reference to astronomical observation, but in a somewhat different sense and in response to the article "How large seems the moon?"

I considered whether the glyphs that appear to be "3;" might actually be "By" with some ink faded, so that the entry would actually read "Letter on, By Dr. Van Stadt", but it really does look like a "3" when I look at it closely. My best guess at the moment is that that portion of the entry is simply spurious in its entirety, rather than just having a mis-numbered page (as some other index entries do). --Superiority (talk) 21:11, 22 January 2024 (UTC)

title: Page talk:Nature - Volume 1.pdf/13


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 women should make just that sacrifice. I suppose if I were a Catholic it would seem very natural to me.”

Kitty looked at him coolly. She could not quite understand the emotion with which the little man spoke and she asked herself whether it was a pose. He had drunk a good deal of whisky and perhaps he was not quite sober.

“Come and see for yourself,” he said, with his bantering smile, quickly reading her thought. “It’s not nearly so risky as eating a tomato.”

“If you’re not frightened there’s no reason why I should be.”

“I think it’ll amuse you. It’s like a little bit of France.”

HEY crossed the river in a sampan. A chair was waiting for Kitty at the landing-stage and she was carried up the hill to the water-gate. It was through this that the coolies came to fetch water from the river and they hurried to and fro with huge buckets hanging from the yoke on their shoulder, splashing the causeway so that it was as wet as though it had heavily rained. Kitty’s bearers gave short, sharp cries to urge them to make way.

“Of course all business is at a standstill,” said Waddington, walking by her side. “Under normal circumstances you have to fight your way through


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 the coolies carrying loads up and down to the junks.”

The street was narrow and winding so that Kitty lost all sense of the direction in which she was going. Many of the shops were closed. She had grown used on the journey up to the untidiness of a Chinese street, but here was the litter of weeks, garbage and refuse; and the stench was so horrible that she had to put her handkerchief to her face. Passing through Chinese cities she had been incommoded by the staring of the crowd, but now she noticed that no more than an indifferent glance was thrown at her. The passers-by, scattered rather than as usual thronging, seemed intent on their own affairs. They were cowed and listless. Now and then as they went by a house they heard the beating of gongs and the shrill, sustained lament of unknown instruments. Behind those closed doors one was lying dead.

“Here we are,” said Waddington at last.

The chair was set down at a small doorway, surmounted by a cross, in a long white wall, and Kitty stepped out. He rang the bell.

“You mustn’t expect anything very grand, you know. They’re miserably poor.”

The door was opened by a Chinese girl, and after a word or two from Waddington she led them into a little room on the side of the corridor. It contained a large table covered with a chequered oil-cloth and round the walls was a set of stiff chairs. At one end of the room was a statue, in plaster, of the Blessed Virgin. In a moment a nun came in,


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408

title: Page:Nature - Volume 1.pdf/422


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Page:The life of Charlotte Brontë (IA lifeofcharlotteb01gaskrich).pdf/143

 and congregation; but so strong was the democratic element, that he met with the most violent opposition, and was often insulted when he went into the street. A bride was expected to make her first appearance, and the minister told the singers not to perform the anthem. On their declaring they would, he had the large pew which they usually occupied locked; they broke it open; from the pulpit he told the congregation that, instead of their singing a hymn, he would read a chapter. Hardly had he uttered the first word, before up rose the singers, headed by a tall, fierce-looking weaver, who gave out a hymn, and all sang it at the very top of their voices, aided by those of their friends who were in the chapel. Those who disapproved of the conduct of the singers, and sided with the minister, remained seated till the hymn was finished. Then he gave out the chapter again, read it, and preached. He was just about to conclude with prayer, when up started the singers and screamed forth another hymn. These disgraceful scenes were continued for many weeks, and so violent was the feeling, that the different parties could hardly keep from blows as they came through the chapel-yard. The minister, at last, left the place, and along with him went many of the most temperate and respectable part of the congregation, and the singers remained triumphant.

"I believe that there was such a violent contest respecting the choice of a pastor, about this time, in the upper chapel at Heckmondwike, that the Riot Act had to be read at a church-meeting."

Certainly, the soi-disant Christians who forcibly


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 short and plump, with a homely face, red cheeks and merry eyes. Waddington, introducing Kitty to her, called her Seur St. Joseph.

“C’est la dame du docteur?” she asked, beaming, and then added that the Mother Superior would join them directly.

Sister St. Joseph could speak no English and Kitty’s French was halting; but Waddington, fluent, voluble and inaccurate, maintained a stream of facetious comment which convulsed the good-humoured nun. Her cheerful, easy laughter not a little astonished Kitty. She had an idea that the religious were always grave and this sweet and childlike merriment touched her.

HE door opened, to Kitty’s fancy not quite naturally but as though it swung back of itself on its hinges, and the Mother Superior entered the little room. She stood for an instant on the threshold and a grave smile hovered upon her lips as she looked at the laughing Sister and Waddington’s puckered, clownish face. Then she came forward and held out her hand to Kitty.

“Mrs. Fane?” She spoke in English with a good deal of accent, but with a correct pronunciation, and she gave the shadow of a bow. “It is a great pleasure to me to make the acquaintance of the wife of our good and brave doctor.”

Kitty felt that the Superior’s eyes held her in a


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Page:1955 North Dakota Session Laws.pdf/28

20

Subdivision 8.

Subdivision 8a.

Subdivision 9.


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 long and unembarrassed look of appraisal. It was so frank that it was not uncivil; you felt that here was a woman whose business it was to form an opinion of others and to whom it never occurred that subterfuge was necessary. With a dignified affability she motioned to her visitors to take chairs and herself sat down. Sister St. Joseph, smiling still but silent, stood at the side but a little behind the Superior.

“I know you English like tea,” said Mother Superior, “and I have ordered some. But I must make my excuses if it is served in the Chinese fashion. I know that Mr. Waddington prefers whisky, but that I am afraid I cannot offer him.”

She smiled and there was a hint of malice in her grave eyes.

“Oh, come, ma mère, you speak as if I were a confirmed drunkard.”

“I wish you could say that you never drink, Mr. Waddington.”

“I can at all events say that I never drink except to excess.”

The Mother Superior laughed and translated into French for Sister St. Joseph the flippant remark. She looked at him with lingering, friendly eyes.

“We must make allowances for Mr. Waddington because two or three times when we had no money at all and did not know how we were to feed our orphans Mr. Waddington came to our rescue.”

The convert who had opened the door for them now came in with a tray on which were Chinese


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Page:The life of Charlotte Brontë (IA lifeofcharlotteb01gaskrich).pdf/144

 ejected Mr. Redhead at Haworth, ten or twelve years before, held a very heathen brotherhood with the soi-disant Christians of Heckmondwike; though the one set might be called members of the Church of England, and the other Dissenters.

The letter from which I have taken the above extract relates throughout to the immediate neighbourhood of the place where Charlotte Brontë spent her school days, and describes things as they existed at that very time. The writer says,—"Having been accustomed to the respectful manners of the lower orders in the agricultural districts, I was, at first, much disgusted and somewhat alarmed at the great freedom displayed by the working classes of Heckmondwike and Gomersall to those in a station above them. The term 'lass' was as freely applied to any young lady, as the word 'wench' is in Lancashire. The extremely untidy appearance of the villages shocked me not a little, though I must do the housewives the justice to say that the cottages themselves were not dirty, and had an air of rough plenty about them (except when trade was bad), that I had not been accustomed to see in the farming districts. The heap of coals on one side of the house-door, and the brewing tubs on the other, and the frequent perfume of malt and hops as you walked along, proved that fire and "home-brewed" were to be found at almost every man's hearth. Nor was hospitality, one of the main virtues of Yorkshire, wanting. Oat-cake, cheese, and beer, were freely pressed upon the visitor.

"There used to be a yearly festival, half religious,


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Page:พรบ วิ ปกครอง ๒๕๓๙ (๐๑).pdf/12

 When paragraph 1(1), (2), (3), or (4) has been observed and the authority making the administrative order intends to retain the effects of the original order, he shall note the facts and his intention down in the original order or annex the note thereto and must also notify the participants of such intention in writing.

As regards (2), (3), and (4), they must be observed before the conclusion of the appellate proceedings according to Part 5 of this Chapter or the specific law thereon, or if no appeal is required, before the administrative order is brought into the consideration of the person having the power to consider and rule upon its correctness.

An administrative order shall become effective against a person from the time he is notified thereof.

An administrative order shall remain effective for as long as it is not revoked or does not conclude for reasons of time or for other reasons.

When an administrative order concludes, the authority shall have the power to summon the possessor of any document or other object created in connection with the making of the said administrative order and bearing a text or symbol showing the existence of such administrative order to return it or, when it is subject to his ownership, bring it before the authority to further be marked with a symbol showing its conclusion.

An administrative order containing insignificant errors or mistakes may be amended by an authority at any time.

When an administrative order is amended in accordance with paragraph 1, the relevant persons must be notified as appropriate for the case. In this event, the authority may demand a relevant person to send the administrative order or any document or other object created in connection with the making of the said administrative order for the amendment.

Subject to section 48, in the case that any administrative order is not issued by a Minister and there is no law designating a specific appellate process within the administrative sector, a participant may lodge an appeal against the administrative order by filing his appeal with the authority making the administrative order within fifteen days from the day he is notified of the said order.

The appeal must be made in writing and specify the points of the protest and the facts or laws cited also.

title: Page:พรบ วิ ปกครอง ๒๕๓๙ (๐๑).pdf/12


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Page:The life of Charlotte Brontë (IA lifeofcharlotteb01gaskrich).pdf/145

 half social, held at Heckmondwike, called 'The Lecture.' I fancy it had come down from the times of the Nonconformists. A sermon was preached by some stranger at the Lower Chapel, on a week day evening, and the next day two sermons in succession were delivered at the Upper Chapel. Of course, the service was a very long one, and as the time was June, and the weather often hot, it used to be regarded by myself and my companions as no pleasurable way of passing the morning. The rest of the day was spent in social enjoyment; great numbers of strangers flocked to the place; booths were erected for the sale of toys and gingerbread (a sort of 'Holy Fair'); and the cottages having had a little extra paint and white-washing, assumed quite a holiday look.

"The village of Gomersall" (where Charlotte Brontë's friend 'Mary' lived with her family), "which was a much prettier place than Heckmondwike, contained a strange-looking cottage, built of rough unhewn stones, many of them projecting considerably, with uncouth heads and grinning faces carved upon them; and upon a stone above the door was cut, in large letters, '.' It was erected by a man in the village, opposite to the house of his enemy, who had just finished for himself a good house, commanding a beautiful view down the valley, which this hideous building quite shut out."

Fearless—because this people were quite familiar to all of them—amidst such a population, lived and walked the gentle Miss Wooler's eight or nine pupils. She


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 cups, a tea-pot, and a little plate of the French cakes called Madeleines.

“You must eat the Madeleines,” said the Mother Superior, “because Sister St. Joseph made them for you herself this morning.”

They talked of commonplace things. The Mother Superior asked Kitty how long she had been in China and if the journey from Hong-Kong had greatly tired her. She asked her if she had been in France and if she did not find the climate of Hong-Kong trying. It was a conversation, trivial but friendly, which gained a peculiar savour from the circumstances. The parlour was very quiet, so that you could hardly believe that you were in the midst of a populous city. Peace dwelt there. And yet all round about the epidemic was raging and the people, terrified and restless, were kept in check but by the strong will of a soldier who was more than half a brigand. Within the convent walls the infirmary was crowded with sick and dying soldiers, and of the orphans in the nuns’ charge a quarter were dead.

Kitty, impressed she hardly knew why, observed the grave lady who asked her these amiable questions. She was dressed in white and the only colour on her habit was the red heart that burned on her breast. She was a woman of middle age, she might have been forty or fifty, it was impossible to say, for there were few wrinkles on her smooth, pale face, and you received the impression that she was far from young chiefly from the dignity of her bearing, her assurance, and the emaciation of her