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2.4.

532. Now we have sin {r + (-1)" 0} = sin 0. If then is any angle lying between ro தா T and r + ŽI, where r has some assigned integral value,

1
1 sin 1.3 sin 1.3.5 sin?

$
sin
2

+... =(-1)"($-17). 3 2.4 5 .

6 7 Writing here = 1+ +4, we have if y is any angle lying between (r - 1) 7 and ra, where r has some assigned integral value,

1 cos 1.3 coso y 1.3.5 cos? cos y + 2 3 2.4 5 2.4.6 7

= (-1)" (y + + - ro). Hence any angle y may be expanded in powers of sin y or

+

+

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T

of cos y.

The calculation of 1.

+

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+

533. The two series of Art. 529 enable us to calculate the value of an arc in terms of its sine or tangent. In particular we may find from them the ratio of the circumference to the diameter of a circle ; i.e. the value of 7.

Using the inverse symbol to represent the circular measure of an acute angle, we have

1 203 1. 3 co 1.3.5 27 sin-1 3 = 2 +

2 · 3 2.45 2.4.67
tan-?x= – 1 203 + 1 205 – 7 x + ...
534. In the first of the above equations, leta = 2. Then

= ft. Thus
1 1 3 1 3.5 1

3.5.7 1
+
6 2 3. 24 2 5. 27 13 7. 210

14 9. 213
1 9 3.5 3.5.7
3 +

g 10.26 7.210 9.215 Taking 3 terms of this series, we have = 3.14 approximately.

After the first few terms, this series does not converge rapidly enough for convenience of calculation.

sin-1

=

T

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535. In the second of the equations of Art. 533, put a = 1. Then tan-lx = 1.

Thus

1 1 1 1 1
1
4 3 5 7 9 11
1 1 1

+
8 1.3 5.7 9.11
This series is very slowly convergent.

We may however adapt the inverse-tangent series for the purposes of calculation by expressing 4. as the sum or difference of suitable angles.

The excellence of a series for purposes of calculation depends on (1) the rapidity of its convergence, and (2) the simplicity of the operations to be performed. 536. We shall make use of the equation

T-t tan-1 T=tan-it + tan-1

1+ T't Here T and t may be chosen at will, and we shall attempt to find simple expressions for the remaining term.

537. First, put t = 1.
Then if T=1, we have
41 = tan-? 1 + tan-1

.(1). And if T

we have

tan- f = tan-* } - tan-7. Substituting in (1) for tan-1} or tan-} from this last equation we have 41 = 2 tan-1} - tan-17

(2), 1 = 2 tan-1} + tan-17....

(3) 538. Next, put t = . If T=1;

17 = tan-+ 1 + tan- s. If

tan- s = tan- [ + tanIf

T=; tan-3 = tan-'4 + tan- 35. .. (adding) 1. =3 tan-1 + tan-..

(4). This is convenient, since 99 has the simple divisors 9 and 11.

.

1

1 3

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-1

-11

T=;

3

-1

23•

7

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T=;

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7

it; tan-1

-1 7

17

tan-1}

+ tan

: -1 9 46

1 239

9

46

-1

239.

539. Next, put t=}.
If

T=1; =tan-} + tan- 3.
If

tan- ; = tan-*} + tan-? 17
If

T =
If
T= 4; tan-18

tan-?} - tan-1
.. (adding) 41 = 4 tan-} - tan-

..(5). This series is quickly converging, but the divisor 239 is inconveniently large.

540. We have thus found five substitutions for 7 by means of which its value may be calculated more or less conveniently. Thus

1 1 1

3.4 5. 42 7.43

1 1 1
+}x 1
3.9

(1). 5.92 7.93

1 1 1
17= 1

3. 4 5.42 7.43
1 1

1
-7*(1
3.49 5. 492 7.493

:(2).

+

+==+*(1-54**---)
(1

..........

+

+

+

+

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1 1 1

+
3.9 5.92 7.93
1
1

1
+

+
3.49 5.492

7.493
1
1

1
+
3.16 162 7.163
52 54 56

+
3.992 5.994 7.996
1 1 1

+
3.52 5.54 7.56
1
1

1
1

3.2392 5.2394 7.2396

) +...) ......... (3).

..)

.......... (4) .)

-

) (5).

(1

(1 (1

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5 99

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Х 239

+

Series (1) is known as Euler's; series (5) as Machin's; the other series were suggested by Hutton, whose method has been followed in the above derivations.

tan-1

239 =

-1

999

The last series may be simplified by noting that

= tan-1

- 7o-tanwhich the student should prove.

541. We will now prove the important proposition that and also tra are incommensurable. For this purpose it will be necessary to give a short account of continued fractions.

Continued Fractions.

542. A continued fraction is a fraction of the form bi

bi b2 b3
and is usually written shortly

а1
+
A2 = Az =

+

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b
A2+ &c.

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We will give a few theorems on fractions of the form bi bz

where an, 61, A2, b, ... ai · Cl2 03 are all positive. Such continued fractions are said to be of the second class.

bi bj bg The simple fractions

are called the components: the

α, α, ας fractions obtained by stopping at these are called the convergents: and the fractions obtained by omitting them are called the remainders.

bn+1 bx+2 Thus being the nth component,

is the nth re

an+1 – An+2 mainder.

b,

b, The first convergent is

a,b, +; the second

and а.

6aa, b,
ba'

n

bin an

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Cl2

So on.

Also, if F is the fraction, and on the nth remainder,
bi
b

b,
F

and generally Pn-1
; P1
A1 P1
Ala - P2

an - Pn

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Lastly, if cn is the nth convergent of F, and

Yn
the nth

converg

ent of pii

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and Cn+1

543. Theorem I. The convergents are in ascending order of magnitude, provided that they are all positive.

bi

b,
For we have

an
dj - Yn-1

Az - Vn Assuming, then, that a convergent yn of the nth order is greater than

Vn-1 of the (n − 1)th order, we see that Cn+1 of the (n + 1)th order is greater than on of the nth order.

b Now clearly the 2nd convergent

is greater than the

a, -b/a bi 1st convergent

Hence, universally, a convergent is increased

а1 by taking an additional component.

544. Theorem II. If the denominator of every component exceeds the numerator by unity at least, the continued fraction is itself positive and not greater than unity. For assuming that yn of the nth order is not greater than unity,

b Cn+1 of the (n + 1)th order is at greatest

which is not greater than unity; and is positive, since ay is not less than Yn:

b, Now clearly the 1st convergent is not greater than unity.

ai Hence no convergent can be greater than unity; and all are positive.

And, therefore, the fraction itself is positive but not greater than unity.

545. Theorem III. If the denominator of every component exceeds the numerator by unity at least, while that of the first exceeds its numerator by more than unity, the continued fraction is itself less than unity. J..T.

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