1. IF P, be the numerator of the 7th convergent to the continued fraction whose quotients are

I, and if P be the numerator of the 7th convergent to the fraction whose quotients are In In-1

prove that

and that n is the number of the recurring quotients 9, P Pan be the nth and 27th convergents to √N, 12...24, if Qn' Q2

prove that Qan
that Q2 = 2PQ, and P2 = 2P,2 + (− 1)”+1.

2n

For the rider, it is convenient to assume the following notation, which is employed in a tract on The Expression of a Quadratic Surd as a Continued Fraction, by Thomas Muir, M.A., Glasgow, 1874.

Let K (abc...) denote the numerator of the last convergent to the continued fraction a+

1 1 1

b+c+7 regarded as the result of an operation on the quotients a, b, c...l.

P

Then if be the 7th convergent to the continued fraction which expresses (N),

=

=

=

by (a).

59

{2A.K (Aq,..q1) +K (Aq,··q2)} K (2,··q1) + K (Aq,..q,). K (9,··9.) K (9,..q1) {2A.K (91··q1) +2K (¶ ̧··Z2)}

{A.K(Aq,..q,)+K(Aq,··q2)}K(q,··q,)+K(Aq,··q,){A.K(q,··q,)+K(q,··92)}

2K (q,..q1). K (Aq ̧··q1)

K(Aq,q,A). K (q1--q1) + K (Aq,q,)*

2K (q,··q1). K (Aq ̧·q1)

K (Aq ̧...q,)2 — K (Aq ̧...q ̧A).K (q ̧···q.)

(2).

Now, if we consider the continued fraction whose quotients

are A, 1992

29 1, A, in number n + 1, and take the

difference of the two last convergents,

K(Aq,...q) K(Aq,.....q,A)

(− 1)"

=

2. Prove that one of the values of

log {1+ cos 20+ √(−1) sin 20} is log (2 cos✪) + 0 √(− 1),

is

Prove also that one of the values of

sin (cos +√(− 1) sin0}

cos1√(sin ) + √(− 1) log {√√(sin @) + √(1 + sin 0)}

when ✪ is between 0 and 7.

and

If

2

i=-1,

log (1+ cos 20+ i sin 20) = † log (4 cos3 0) + (nπ + 0) i,

log (1+ cos 20+ i sin 20)

= log(2 cos20) + log(1+i tan 0),

expanding log(1+i tan0), and equating the coefficients of i, we have Gregory's Series

sina=√(1 − sinė), cosa=√(sin0), eß = √√/ (sin @) + √√(1 + sin @).

3. Prove that the equation to the circle which cuts at right angles three circles whose equations are given in the form (x − a)2 + (y − b2) = c2 is

Prove that the diameter of the circle which cuts at right angles the three escribed circles of the triangle ABC is

α

sin A

(1+ cos B cos C+ cos C cos A + cos A cos B)3.

The condition that the circle

x2 + y2 — 2a,x− 2b,y + a‚2 + b2 — c‚2 = 0,

should cut at right angles the circle

x2 + y2 — 2ax - 2by + a2 + b2 − c2 = 0

may be written

a+b,*-c-2aa, -2bb,+ a2 + b2 — c2 = 0,

eliminating a, b, and a2+b2 - c2, we have the required result. In the rider take as axes of x and y the exterior and interior bisectors of the angle A; let a, ẞ, y be the centres of the escribed circles, and ", " ", their radii, and D the diameter of the circumscribing circle

3

Aa Aß cot

= 2D cos

COS

if then the equation to the orthotomic circle be

K(ix+y)=Px+Qy+R,

2 P

K

B

(左)

A

C

K=4D2 cos

4. Prove the relation (x) - $ (0) = xp' (0x), where e is a proper fraction, stating the conditions subject to which it is true.

Hence deduce Maclaurin's Theorem for the expansion of f(x) in ascending powers of x.