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and we can equate real and imaginary terms. A curious difficulty, however, here presents itself, viz. to decide what value k (which must be integral) has.' In a similar case De Morgan determined the proper value of k in the following manner :Put p=cos 0+i sin õ, then

e-pn cos {cos (x" sin 8) +i sin (x* sin )}d.r

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0

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Now if 0=0, the last integral vanishes; so that we must have k=0, and therefore

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The above investigation is, however, chiefly valuable as suggestive of the result ; it contains no indication of the limits between which n must lie that the last written equations may be true and the integrals not infinite. The integrals have also been obtained by differentiating Son

e-as sin xdx with regard to a and putting a=0 afterwards; but the results obtained are of the form (, za sin æds (n integral), and must therefore be infinite. The following investigation of the values of the integrals seems of interest, as it is rigorous and discriminates between the finite and intinite values. Integrate 1.5*-*"sin y do dy with regard to « first, and we find it

Sodo dx x

co

C

sin ydy;

1

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Уя integrating with regard to x first, we find it

dir
1 + x2n 2n 2n

=r (1+:). S -S.

cosec

(2n> l), (2n<l),

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TT

( 3 *sin ydy.r(1+

r(1+ :)

y

cosec 2n

2n

;

* This method is also given in De Morgan's ' Differential and Integral Calculus,' pp. 630, 576. Some analysts (Oettinger, Bidone, &c.) have not seen any objection to S ton sin xdx being finite for all values of n; but unless we are prepared to write with De Morgan (“Theory of Probabilities, "Encyc. Met.p. 436)

because

P see how this can be admitted.

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m

n

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sin amdx=r(1+.

m

1.

T sin

2m

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1 by taking - =1

1

(in which 2n> 1, so that m may have any value except such as lie between 1 and — 1), and using the relation 1(a)r(1-«)=T COSEC an, we obtain

=r(+2.

1-
Similarly, by integrating So S* e-my cos y dx dy, we find

C =r.
cos xmdx=r(1+

(1+z)cos

. (m between 1 and 00).

The author had calculated a Table of the values of (sin anda, f. cos zude for different values of <; and the curves y=fsin arda, y=S.cos a*da, as obtained from them, were drawn and exhibited to the Section, the discontinuities in each being remarked. [The Tables and curves will be found in the ‘Messenger of Mathematics,' 1871.]

m

2m

0

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On Lambert's Proof of the Irrationality of #, and on the Irrationality of cer

tain other Quantities. By J. W. L. GLAISIER, B.A., F.R.A.Š. The arithmetical quadrature of the circle, that is to say, the expression of the ratio of the circumference to the diameter in the form of a vulgar fraction with both numerator and denominator finite quantities, was shown to be impossible by Lambert in the · Berlin Memoirs ' for 1761; and the proof has since been given in an abridged and modified form by Legendre in the Notes to his · Éléments de Géométrie.'° Although Legendre's method is quite as rigorous as that on which it is founded, still, on the whole, the demonstration of Lambert seems to afford a more striking and convincing proof of the truth of the proposition ; his investigation, however, is given in such detail, and so many properties of continued fractions, now well known, are proved, that it is not very easy to follow his reasoning, which extends over more than thirty pages. The object of the present paper is to exhibit Lambert's demonstration of this important theorem concisely, and in a form free from unnecessary details, and to apply his method to deduce some results with regard to the irrationality of certain circular and other functions. The theorem which Lambert proves, and from which he deduces the irrationality

is that the tangent of a rational arc (i. e. an arc commensurable with the radius) must be irrational; and this he demonstrates by means of an expression for the tangent as a continued fraction, viz.

.22 aca 22 tan

(i) y y- 3y-by- 7y- &c. ' adopting an established notation for continued fractions in which that which follows each minus-sign is written as a factor, to save room. Consider a continued fraction B B BE

(ii) + a,+ a,+&c.' and let Pn be the nth convergent to it; then we know that an

P,=&_PR-,+BPM

9,=4-1+Band

Biß2...Bn
. .

(iii) an

In-19n
* These results can easily be proved by induction.

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nn

-2

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Pr - Pn-1=(-)--1

In-1

and let Ri, R,... Rn...

P Suppose also that the continued fraction (i) is equal to

@ be such that

Ri=«P-BIQ,
R,=c3Ri+B2P,
Rg=a3R2+B3R1,
Rr=nRm-it BnRn-22

n

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Pn

BiBn

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then R,=9P-P, Q, as can be shown by induction ; so that

P Pn
Rn

(iv) Q In

Q9
Now
P Pn Pn+1

PnPn+
Q an In+1

9n 9n+ therefore

P

(-).......Bn+1+(-)++ B1... BA+3+...

in
"

(1
Q
an
9n9n+1

In+19n+2
from (iii). By equating this value of pn to that in (iv), we obtain

Bi... Bn+1 Bi... Bn+2
-Qan

to...
(-)n+iRn=Q

(v) 9n+1

9n+19+2 If P and Q be integers and al... Bi...Br... be also all integers, then from the equations by which Ri... Rn are determined, we see that they also are integers. Now in the case of the continued fraction for tan

y
Ar=(2n-1)y,
Bn= -xa,

9n=(2n-1)99n-1-491-2; and we notice that if x and y be integers, then a ...com...

.Bi...Pr... are so too, and consequently (if P and Q are integers) Ri...Rn... are integers. Bi... Br

Bi... Br+ is The factor by which is multiplied to obtain gr

qr+1

an...

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Br+195 = +

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9r+1

(2r+1)y9,-'98-1

x2
=+
(2r+1)y-

9-1

95 wbich can be made as small as we please by increasing r.

We can therefore from (v), Q being finite, make Rn as small as we please by taking n sufficiently large ; but if P and Q be both integers, Rn must remain an

P

= must be ir

rational ; but tan- =1, so that cannot be rational.

4 The above is in substance Lambert's demonstration ; alterations have been made in points of detail &c., and the notation has been changed.

It may be noticed that the proof does not (as of course it should not) hold good if P and Q be infinite integers; for we cannot make Rn as small as we please in (v) if Q be infinite.

integer whatever value n may have; thus if " be rational

, (= tan")

TT

Legendre proves a theorem which is easily seen to follow directly from Lamberts mode of investigation, viz. that if in the continued fraction

B B B2

(extended to infinity), uit o,+ y +&c.

aby 2, ..., regarded as fractions (@b...B1..., all integers), be all less than unity, e, a then, whether Bu, B. be all positive or all negative, or some positive and some negative, the value of the continued fraction is irrational. He also remarks that go must be irrational; for if it = we should have, from (i), since tan r=0,

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m

> n

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m

and as after some value of r the fractions

&c. must be less than

(2r+l)n' 2r +3' unity, 3 must be irrational if m and n are integers, whence is irrational. The expression of tan v in the form of a continued fraction Lambert obtained by v

va treating sin v or v- too. and cos v or 11.2.3

1.2... in a manner analogous to that in which the greatest common measure of two numbers is found in arithmetic; and Legendre deduced it from a more general theorem he had proved with regard to the conversion of the ratio of two series into a continued fraction. It may be obtained very simply by forming the differential equation corresponding to y= A cos (V 2x+B), viz.

y+y' + 2xy"=0, whence y(«)+(2i+1)y(i+1)+2xy(i+2)=0 by application of Leibnitz's theorem. From this we have

yo

-1
y
1+ 2xy

y'
-1
3+2x2

y"

&c. ;

tan {V(2x) +B} = = b

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therefore

ye
1

-1 2x 23
y V (2x)

1 - 3 - 5-&c. whence, after determining B by putting r=0 and writing (2x) = 0,

vja ps2 tan t=

1- 3- 5- &c. That Lambert's proof is perfectly rigorous and places the fact of the irrationality of a beyond all doubt, is evident to every one who examines it carefully; and considering the small attention that had been paid to continued fractions previously to the time at which it was written, it cannot but be regarded as a very admirable work. From the continued fraction en +1 1 1 1

1 2 1- 2n-1+ on-17 10n-1+&c.' Lambert showed, in the same memoir, that e" is irrational, so that the Napierian logarithm of every rational number is irrational.

We can obtain a little more information about the irrationality of ef, for we have

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1 1 1 2x-1+ 6x + 10x + &c.

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Now any continued fraction in which all the numerators are unity and all the denominators are positive integers must circulate if it be the development of an ex

! pression of the form A+BNC; so that we see that es, when x is integral, cannot be of this form. Taking the expression for the tangent in the form

cot1 =

1 1 1 1 1

2-17 17 3x-2+ 1+ 5.0-2+ &c. we see that when z is an integer, cot? is irrational, but cannot be of the form A+Bc.

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V(1+cot 1)

= (1+cot-2.), and

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sin and

cot

1 COS

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1 cannot either of them be rational unless cot - is of the form BVC, which

ot? is of the form BVC, which not being the case, sin- and cos - are irrational, and cannot be of the form BVC.

1 1 Since cos

2
2 cos

2
1
-1, cos

cannot be rational unless cos ?=BVC, which

B2C would require that cot

a form which we have shown it cannot

- B2C

2 have; so that cos – is irrational. Similar results hold good for the hyperbolic

_! sine and cosine ; that is to say, i (exte *), f(e-e *), and f(exte 5) are irrational.

( It

1 may be remarked that it is easy to show that sin is incommensurable from

1 the series; for if sin 2, then (q eren, as of course we may take it) p 1 1

? 1

1 +...(-)

+(-) 9 1.2.3213

1.2..(9-1):9-1 whence, multiplying both sides by (1.2... 9).29-1,

1

1

2

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1.2..(2+1)x9+1 t..;

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1 And the series on the right-hand side must be intermediate in value to 1

(q+1)** and (9+1)(9+2)(2+3)21)

and is therefore fractional; thus we have

integer = integer + fraction 1 if sin is commensurable. An exactly similar method proves the irrationality of 1 -!

2 16-e *), &c., but gives no result when applied to cosor (este 7). It is

ilent probably true that both the sine and cosine of every rational arc are irrational, though no proof of this bas, I believe, been given; and there is, as Legendre has remarked, very little doubt that a is not only not the square root of a rational quantity, but also not even the root of any algebraical equation with rational coefficients, although the demonstration of this seems difficult. Similar remarks may bo made with respect to e. * This expression can be deduced from (i) by transforming the terms of the latter thus:

1
A- --- -A-1+1

=

B

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