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Before we proceed to the application of these transformations, it will be necessary to premise some properties of the functions a and a'.

Let a = cos. - (€) and ' = cos.

e.

COS, e

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1.2.3

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.

+

}-v

• This notation cos. e must not be understood to signify but what is usually written thus, arc (cos. = e). It is true that many authors use cos. m A, Sin.m A, &c. for (cos. A)", Sin. Am ; lest therefore the notation here adopted should appear capricious, it will not be irrelevant to explain its grounds. If o be the characteristic mark of an operation performed on any symbol, x, 0(%) may represent the result of that operation. Now to denote the repetition of the same operation, instead of 010(x)); $($($(x))) ; &c. we may most elegantly write pa(z); ©(x); &c. Thus we use d’x, 43x, <?x, for ddx, AAAX, Ex, &c. By the same analogy, since sin. x, cos. I, tan. x, log. x, &c. are merely characteristic marks to signify certain algebraic operations performed on the symbol x, (such as

VI 1+

+
. &c.

+
1.2.3

2V - 1 &c.) we ought to write sin. ?t for sin. sin. x, log. 3x for log. log. log. x, and so on. To apply this to the inverse functions, we have on CM (x) = pu*m (2). Hence if m = - n p" -"(x) = 0° (x) = x. with the operation (0) performed no times on it, or merely %, that is, &-* (x) must be such a quantity that its nth (0) shall be , or in other words - (x) must represent the nth inverse function. It frequently happens that a peculiar characteristic symbol is appropriated to the inverse function. Let it be t, then o-", X = \", X, and $*x = pan, o-", x = 0% (", 1", *), hence $", 0", * = x, and therefore It" ", = 0+", t. For instance de

dan V=SV, d'V=V.-2-", x =A"x.a" = 1, with the operation of multiplying by a, n times performed on it, and ..-= 1, with the inverse operation so often per formed on it, ami a = 1. Similarly sin.-- = arc (sin.

X = arc (sin. = x). cos. (cos. ? = x) &c.—and ifç=1+ *+ + + &c. x = log.c* and c* = log.-' x.

I= arc

I

1,2

сс

= log -> 1, and CC.....(n)*

=log.--"X, or the nth inverse logarithm of x. It is easy to carry on this idea, and its application to many very difficult operations in the higher branches will evince that it is somewhat more than a mere arbitrary contraction.

Thus, 1 = cvGT

and i'e

wherefore cos. a = :1+and cos. a' = x+x=-.

cant, where c= 1
where c=1+++

+in + 1.2.3 + &c. Hence, a" +-= 2. cos. n«, and 2!" + x

"^* = 2. cos. nx'; 4" – 2** — 2V1.cosin. na, a -aran 2 V-1. sin. na'. Consequently, if k be any arc 1 --- 21". cos. k + x2 = a* {x" ta — 2, cos. k} = 42". sin. (****), sin. (4=**).

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an

In like manner

2(n-1)

n

I I

2 1

3

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1– 2***.cos. k + 2:2* = 44.4. sin. (8473") . sin. (mc. We will now proceed to the application of these equations, and first, in Equation { 1 } for 6 substitute, successively, each of a series of angles

0; 9 +*= ; 9 + = ;.... A + And let the resulting values of w(1) be

q.(t); 2(1);
we have then
gol1). (1)

=a". (1-с2) (1 +12).
1- 21" , cos. 10 + an

{1,2} for the product of the several denominators of the pl!) will be {1—21.

+*}{1-22

1– 21. cos. 9 + *-}.......{1 . cos. 0 +1} = 1- 21". cos, no +224 by Cotes's theorem.

2(1)

I

2

n

2

n

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This equation appears under an imaginary form when a 71, but, since cos. -'e is then a real angle, if we express it in a,

Ca

it will then be free from imaginary symbols; thus
p(!)...
a"(i-e)" (+2-)"

2p. cos. a a}"
2"+2".

n(+a) n(-a)

{ap

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- 2. cos.no

n

4.sin.

· sin.

2

2

:){13}

cosec.

n

cos, no

When e = 1, or the conic section is a parabola, 1=1, and we find 2.(!).....,:(1) (2p)"

= (2p)*-'x 2{. no} A result of such reinarkable simplicity, as deserves a more particular enunciation. Let then, in the diagram, fig. 1, S represent the focus of a parabola A,P,Q, and, having drawn any line SP, make n angles PSP, PSP ....PSP, about S, all equal to each other; draw the axis ASM, and make the angle MSQ = n times MSP; and if L represent the latus rectum, we shall have

SP.SP......SP=L"-. SQ, for, by the polar equation of the curve, SQ= (met

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2

2

3

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Thus, if SP be coincident with SA, and n be odd, cosec. =1, and SP ....SP= = $L".

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but, if SP be perpendicular to SA, and n still odd, cosec. Ź Vē. and SP ....SP={L"

: · {1,5}. If SP be perpendicular to SA, but n of the form, 4m + 2 SP.... SP= P={L.

{1,6}.

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Lastly, if the angle MSP= we shall find, provided n be
of the form 6m + 1,
SP....SP=L".

{1,7}
Let us now resume the general equation {1,2} and first,
let 6 = 0, or, let one of the g(1) terminate in the second vertex,

I

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I

211)

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n

71

na

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n

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na

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2

2

2

we have, for every value of n,
.....
ql") =p". (1+x2)

(2p)." (cos. c) )

4

(sin. {1,8}.

2. Let cos. 120 1, and we obtain qulo..... (1) =p". (1+2)

.

(+)" (2p)" (cos. a)" 1 a)

(cos. {1,9} This embraces all the cases where n is of the form (2m +1) s, and among the rest, when n is any odd number, and one of the glo) terminates in the first vertex ; when n is of the form 4m + 2, and one of the plı) perpendicular to the axis, &c. .

3. Let cos. no = 0; then

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" (

{1,10}. This includes the cases where n is of the form (2m + 1). G

as for instance, where one of the pll) is perpendicular to

the axis and n odd, or, one inclined at an angle, and n of the form 4m + 2, or, lastly, where 6 = or and n = 6m + 3. 4. Let cos. ne=;, then = p* = p". (1+2)*: (1+%*)

{1,11].

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..... (1)

(1+2°) 1-2" +227

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I

n

This takes place whenever n is of the form (6m + 1).

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n

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38

p"

%

{1,12}. Here n must be of the form (3m + 1), and if

§, n is of the form 3m + 1.

We will now proceed to our second equation {2}, and by an operation exactly similar to that from which we obtained the equation {1,1}, we shall find , (2) ..... .(?) = 4" (1 +24)%

1-29". cos. no +220 •{2,1}

{ 1–2e". cos. no ten} a” (a +2-1) 2+

n (0+ c)

R(-a)

I - 26" . cos. no tean

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This transformation in a, however, as it possesses no particular elegance in point of form, and much complexity, we shall henceforward omit, except in a few remarkable instances.

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