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greater than unity. Hence the fraction is at greatest

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

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Here C the numerator of p, is a finite positive integer.

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Now, by hypothesis, each of the remainders P1, P2... 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.

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 nth convergent is incommensurable, the whole fraction is incommensurable.

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If r 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 cannot be commensurable.

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As before this is impossible. Hence 2 cannot be commensurable.

Observe. This proposition includes the last.

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3.

Given tan > 0 and cos 0 > 1 - 102, show that

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7. (2 cos 0-1) (2 cos 20 − 1) (2 cos 22 0 − 1)

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9.tan 10 tan 10 + 1 tan 10 + ...ad inf. = 1/0 – cot 0.

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11. Find the limits, when x is zero, of

(cos x) cotx, (1-sin 2x)cosecx, (cos ax)cosec2 ßx, (sec ax) cot3 fx.

12.

2+ √3 2+√(2+√3) 2+√{2+√(2+ √3)}

π

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13. If measures an acute angle, 0>

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Ө

3 sin 0

2 + cos 0

+ has its least positive value, 0> √3-1.

cos

15. The chord of a circle subtending an angle 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

2-1

-2

(4x-1) (4x-1)... (4" x − 1) = x2 - μxn−1 + vxn−2 — ... + (− 1)"

=

then (4-1) (421)... (4" - 1) aλ. 2n cn -μ. 2n-1 Cn-1 + ... + (− 1)" co where a = the arc, and cm

the chord of a/2m.

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Find the following limits, when x=0; or 0 = 0; (18—33).

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40. sinh 3x = 4 sinh® x + 3 sinh .

41. The terms in the expansion of cosh nx and of sinh nx are taken alternately from the expansion of (cosh x + sinh x)”.

43. cosech2 x + 1 coth2 x.

-

42. sech*x+tanh x= 1. 44. Cosh and Coth are always greater than 1; Sech and Tanh are always less than 1; Sinh and Cosech may

value.

have any

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ad inf.

49. 0 = tanh 0 + } tanh* 0 + } tanh 0 + ...

Sum the following series to infinity: (50-55).

50. x sinh a + x° sinh 2a + * sinh 3a + ...

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