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Let

Vn+1 <vnk where k is a positive quantity <l. Then raising the product V,V,V3... Ur... to the power 1 – k, we have V, v, -* . Vz. V, -* . Vz. v in which the factors are alternately greater and less than 1; and also the pairs of factors V1.0,-%; 0,-*. v, ; v, -*.v3, &c., are alternately greater and less than 1. (Cf. Art. 450.)

519. If v, always decreases as n increases and is always positive, then according as the series whose general term is vn is convergent or divergent, so also is the product whose general factor is

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For let log (1 + vn) = Un, so that 1 + Vn =

= e un em – 1

Un

Un
Then

1 +

+
| 2 3 4

Un

2

3

Un

+

+

Un

n

nor

can

This series cannot be zero or infinity for any finite value of Un

log (1 + on)

Vn (including zero). That is, neither

log (1 + vn)

Vn be infinite for

any

value of n. .. (by Art. 453) according as the series whose general term is Vm is convergent or divergent, so is the series whose general term is log (1 + vn), and so also is the product whose general factor is

1 + Uno

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The same argument applies to the product whose general factor is 1 – Vn, where vn <l; which of course cannot be infinite but may be zero.

Example. The limit of the coefficient of an in the expansion of (1+x)m is zero or infinity according as m is greater or less than – 1. For the coefficient of x is

m
m+1

+1

m+1 2

3 This product is always divergent, because the sum

1+3+3+... is divergent. Hence the product is zero or infinity according as m +1 is positive or negative. (See Art. 462.)

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520. It was shown in Arts. 416, 418 that } (ac" + x-n) is the same function of } (x + 2c-?) as cos na is of cos a: and that xn - x-n is the same function of 1 (2+) as

is of cos a.

sin na sin a

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Put x = ea; then } (x+2-1) = 1 (ea +e-a)= cosh a; } (x – 2-4) sinh a; 3 (+c") = cosh a; } (e” – ch”)=sinh a.

Thus cosh na is the same function of cosh a as cos na is of

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Hence the second formulæ of Arts. 419, 420, 421 become cosh no

3п

2n-1
= 2n-1 (cosh A - cos
cosh 0 - cos

-COS
2n
2n

2n

1 sinh 2n-1 sinh 0 ( cosh 0 – cos cosh 0

T

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(cosh

T

n

- COS

n

n

) 3.) ... (cosh
2.)

-),
2n-1 (cosh 0 – cos a) {cosh (a + 2)}
... fooab 6 - coe (2+->

a =)}

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cosh no - cos na=

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COS

n

cosh 0 –

2n-2 a +

n

521. T'o resolve cos 0 and cosh O into factors.
We have cos na and cosh na equal respectively to

37

2n-1 2n-1 ( cos a – COS

2n

2n

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-),

(
m) (c

T 2n-1 ( cosh a - COS

2n)

(cosh

cosh a

3T COS

2n)

cosh a

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2n 1 COS

T 212

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Now cos (2n – 1) 7/2n = cos λπ/2n.
λπ)

2η –λ
..
2n

2n
λπ

2na
COS
cosh a-

COS
2n)

2n

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: a – )
(co

) (cosh

); 20-^ = sin

cosh a

λπ ?

+ sinh? a. 2n

T

sino a

sino a

2

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sino a

a)... (sin"

a) (sino a) (sino

a) a).

T

.. if n is even, multiplying first and last factors &c., we have

37T

n-1 2n-1 (sin 2n 2n

2n 37

n-1 2n-1 (sin + sinh? a.

+ sinh a

+ sinh? a. 2n 2n

2n Here put na=0. Thus cos Q and cosh 0 are equal respectively to 8: 37

n-1 1

?

?
2n
2n

2n

2

a) ... (sir

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Here put 0= 0. Thus

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Divide each of the above two equations by this last. Thus
sino 0/n
sin? 0/n

sinon
1
sin 2n sino 37/2n sino (n − 1) 7/2n
sinh? d?n

sinh? 6 m cosh 0 =(1+

7

sino 37/2n/ sin" (n-1) 7/2n) In these last equations, make n infinite, while 6 remains constant and finite. Then 0/n and XT/2n become zero: and, for the ratio of the functions sin and sinh we may substitute that of the quantities O'n and do/2n themselves. Thus

1

to infinity
3п
20

20
0

57

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008 0=1-1-(3){1-
coah o= {1+ (*)}{1+(?)}{1+2)... to infinity.

1+( }

T

1

These products are convergent for all finite values of 0, because the sum of the series 2012 1 1 1

+

52 72 is convergent: the nth term being of the form (2n + 1)“, where -C=-2 <-1,

G+ 5+ 6+ ...

) )

(it

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The first formulæ may easily be remembered, by giving to 6

37 57 the values

&c. which make each side vanish. 2' 2 2

TT

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:. if n is odd, multiplying second and last factors &c.,

1 sin na = 2n-1 sin a ( sin

sin? 2n

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Here put 0=0. Thus

2. n=2n-1 sin?“ sin?

n

n

T

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Thus Divide each of the above two equations by this last.

0
sin? 0/n

sinon
sin 0= n sin

1
sinon

sin(n − 1) 7/2n
0
0/2

sinh? 6 m
sinh 8

1+

1+ ? |

sino (n 1) 7/2n

2n)

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n

T

(1
(

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m sinh

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n

Here make n infinite. Thus

sin A=0

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9=0{1-(93{1-()}(1-(39) ...to infinity.

)} sinh 6 – 1+%(%)(4

{1+0)}{1+(29)} {1+ ( 3 ) } ... to infinity.

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2

= +

37

-1

COS na

COS a

COS

+

λ0

n

(PAT + 8)},

B
B)};

cosh na

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ز

λ=0

n

These products are also convergent.

The first formula may easily be remembered by giving to e the values O, T, 21 &c., which make each side vanish.

523. To resolve cos 0 cos and cosh 0 – cos o into factors. We have

2λπ cos = 21-1 P = 0? {c

2λπ cos = 2n-1 plan-1 cosh a - cos

{ where P denotes the product of the factors obtained by giving to À the integral values between the assigned limits.

Now cos (287/12 + B) = cos {2 (n − 1) 7/n-B}, .. if n is even, 1 = {n gives cos (287/n + B) =- cos B, and cos

1= -1 = 2n-1 (cos’ a - cosB) P. {co cosh na cos

2-1 =2n-1 (cosh2 a - cosa B) P.

2λπ {cosh a-cos

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-- {cos a = con (- )} food a–006 (32-->)} B{

= ,

(2T+B){cosh a – cos (P = B)} {cosh a B)}

pin (CNT - 8)

B

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2λπ
= sin?

n
In - 22
COS

TT FR
n

2λπ

a

n

+ sinh? a.

n

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