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bi For the fraction = where, by the last theorem, p, is not ay - Pi

b greater than unity. Hence the fraction is at greatest which

a-1 is less than unity.

546. Theorem IV. An endless continued fraction of the second class, in which the numerator and denominator of every component are finite integers, and in which every remainder is positive and less than unity, must be incommensurable.

For, if possible, let the fraction be commensurable and equal

B
to where A and B are finite positive integers.
A
B b

a, B-5,4 С
Then
A

say.
, - P1

B B
Here C the numerator of pı is a finite positive integer.

D

E
Similarly P2 =

&c., where D, E are positive

D' integers.

Now, by hypothesis, each of the remainders Pi, Pg ... is less than unity. Hence B, C, D, E... form a series of positive integers which are in descending order of magnitude and yet infinite in number: this is absurd. Hence the endless continued fraction cannot be commensurable.

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547. Since the convergents of a continued fraction of the second class (if positive) are in ascending order, such a fraction resembles a series all of whose terms are positive. Hence moreover such a fraction cannot have more than one finite limit.

It should be observed that the assigned characteristics may be supposed to begin to hold after any finite number of components. For example, the nth convergent of any continued fraction must be commensurable if its components are commensurable. Hence, if the remainder after the nt convergent is incommensurable, the whole fraction is incommensurable.

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If is taken large enough, in every component from m2

onwards, the denominator will exceed the 'numerator (2r + 1) n by more than unity. Hence, by Theorem III., every remainder from this point onwards will be positive and less than unity. Hence, by Theorem IV., the fraction must be incommensurable. But it is not Hence t cannot be commensurable.

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As before this is impossible. Hence 72 cannot be commensurable.

Observe. This proposition includes the last.

EXAMPLES XIX.

2. cot 0 <3 and >>

20

1

6 1. tan 0 > and <

2 A < 2 - A2

0

2
3. Given tan 6 > 0 and cos 0 > 1 - 40, show that
sin 0 >0 - 17 63.

1 0
4. Given tan > 0, show that cot 0 < ;

and further

A 4 1 that cot 0 <

o 3.

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5. If x > 1, cot >1- x + x?.

1 + aca
6. sin 0 > tan 0 } tan3 0.
7. (2 cos 6 - 1) (2 cos 20 - 1) (2 cos 22 0 - 1)

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9. } tan 10 + 1 tan į0 + štan 0 + ... ad inf. = 1/0 - cot 6.

1 10. tan-11 + tan-?? +

ad inf. na + n + 1

4.

T

-1 1

+ tan-1

+...

4

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11. Find the limits, when x is zero, of
(cos x) cotx, (1 - sin 2x)cosec <, (cos ax coseca Bx, (sec ax)cotFx.

2 + 73 2+ (2+ /3) 2 + 1/{2+(2+3)}
12.
4
4

4

9 ad inf.

3 sin 0 13. If 0 measures an acute angle, 0 >

2 + cos 0

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15. The chord of a circle subtending an angle 0 at the centre = 2 sin 10.

16. Without using any trigonometrical formula, but assuming only that the chord of a circle changes its sign without changing its magnitude when the arc changes its sign without changing its magnitude, and that the chord is nearly equal to the arc when both are small, prove the following three propositions for approximating to the value of the arc :

(1) Subtract the chord of the arc from 8 times the chord of the half-arc and divide the result by 3.

(2) Add the chord of the arc to 256 times the chord of the quarter-arc, subtract 40 times the chord of the half-arc, and divide the result by 45.

(3) If (4x - 1) (42 x - 1)...(4" x - 1) = 1a" - xn-1 + vx-2 –... + (-1)" then (4-1) (4 – 1)...(4" - 1)a= 1.2Cm - M. 2n-1cm-1 + ... + (-1)* Co –

Cnwhere a =

the arc,
and cm
the chord of

loge x
17. Find the limit, when x =

l, of

-1"
Find the following limits, when x=0; or 6 = 0; (18—33).
a 600

X – sin x
18.
19.

20.
203

a/2m.

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sino poc

1 - cos qoc

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24.

26.

1 - cos pa

8- (sin 0 + sinh 6 20). . 25. 6-6 (cosh 0 – cos 0 + 0%). m sin X – sin ma

sino nx - sino ma

27.
X (cos » – cosma)
2 sin 3. - 3 sin 2c

sin 2x + 2 sino x - 2 sin x

29. X - sin x

cos X - cosx (sin 2+ sin 2.c – sin 3x)2 (COS X — 2 cos 2x + cos 3x)3 •

28.

30.

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Prove the following statements: (34–49). 34. cosh 2x = coshx + sinh” x = 2 cosh2 x – 1=1 + 2 sinhạ x.

+

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41. The terms in the expansion of cosh nx and of sinh nx are taken alternately from the expansion of (cosh ac + sinh x)". 42. sech? 2 + tanho x = = 1. 43. cosecho x + 1

cotho x. 44. Cosh and Coth are always greater than 1; Sech and Tanh are always less than 1; Sinh and Cosech may have any

1 value.

45. cosh y cosec X + sinh y cot x > 1. 46. If cos x = sech y; then sin x = = tanh y and tan x == sinh y. 47. If sin x = sech y; then cos x = = tanh y and cot x

sinh g. 22 22 32 48. cosh a =

... ad inf.
12 32
0 = tanh 0 + 3 tanho 6 + 3 tanho 6 +...

ad inf.
Sum the following series to infinity: (50—55).
50. a sinh a + ao sinh 2a + ° sinh 3a + ...

X2

23 51. a cosh at cosh 2a + cosh 3a + 12

3

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49.

1

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