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.. log e2 = log(1 + i tan 6) - log(1 — i tan 6).

.. 2i02 i(tan ✪ — } tan3 0 + tan3 0 — etc.) (Art. 130) .. 0 = tan 0 – 1 tan3 0 + } tan30 — etc.

which is Gregory's Series.

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(1)

This series is convergent if tan 0 or <1, i.e., if lies

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Sch. This series may also be obtained by reverting (5) in Cor. 2, Art. 156.

Cor. 1. If tan = x, we have from (1)

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a series which is very slowly convergent, so that a large number of terms would have to be taken to calculate to a close approximation. We shall therefore show how series, which are more rapidly convergent, may be obtained from Gregory's series.

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a series which converges much more rapidly than (3) of

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In this way it is found that = 3.141592653589793 ........

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70

1 -1 99

are much more convenient for pur

poses of numerical calculation than the series for tan-1 1.

239

Example. Find the numerical value of π to 6 figures by

Machin's series,

165. Given sin = x sin (0 + «); expand 0 in a Series of Ascending Powers of x.

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0 sin 20 sin 40+ sin 60 sin 80+....

=

166. Given tan x = n tan 0; expand x in Powers of n.

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.. 2 ix2i0+ log(1+me-2i0)-log (1+ me2i0)

= 2i0 — m (e2io − e−2i0) +

m2

(etio — e-4io) —

2

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(1) When n is even. If r = 0, we obtain from (1) a real

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responding factors are x —- 1 and x + 1.

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1,

If we put

- 2 additional roots, since

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1 roots are found by

(2) When n is odd. The only real root is 1, found by putting r = 0 in (1); the other n putting r=1, 2, 3,...

n- - 1

in (1) or (2) in succession.

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168. Resolve " + 1 into Factors.

Since cos (2r+1)π ±√ −1 sin (2r+1)π = − 1, where r is any integer, and

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.. ï”= cos (2r+1)π ± √−1 sin (2r+1)π.

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(1) When n is even. There is no real root; the n roots are all imaginary, and are found by putting

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