Miscellanea Curiosa, Vol. 1: Containing a collection of some of the principal phaenomena in nature, accounted for by the greatest philosophers of this age
History
Miscellanea Curiosa, Vol. 1: Containing a collection of some of the principal phaenomena in nature, accounted for by the greatest philosophers of this age
Natural history; Science -- Early works to 1800; Voyages and travels -- Early works to 1800
which is Expeditiously resolved into Numbers after this manner.
√17 + 4 is equal to 8.1231, whose Logarithm is 0,9097164, and the
fifth part of it is 0,1819433, the Number answering it
1.5203 = [⁵√](√17 + 4).
But the Arithmetical Complement of 0.6577 is 9.8180567, the Number
answering is
0.1819433 = 1/[⁵√](√17 + 4)
and the half difference of these Numbers is 0,4313 = _y_.
Here we may observe, that in the Room of the general Root, we may
advantageously take
_y_ = ½√(_2a_) - ½/[ⁿ√](_2a_)
if the quantity _a_ be pretty large in respect of Unity. As if the
Equation were
5_y_ + 20_y_³ + 16_y_⁵ = 682,
the Logarithm of _2a_ = 3.1348143 whose Fifth part is 0.6269628, the
Number answering is 4.236, and the Number answering the Arithmetical
Complement 9.3730372 is 0.236, the half difference of these Numbers is 2
= _y_.
But if in the aforegoing Equation the Signs are alternately Affirmative
and Negative; or which is the same thing if the Series be after this
manner,
_ny_ + ((1 - _nn_)/(2 × 3))_ny_³ + ((1 - _nn_)/(2 × 3)) ×
((9 - _nn_)/(4 × 5))_ny_⁵ + ((1 - _nn_)/(2 × 3)) ×
((9 - _nn_)/(4 × 5)) × ((25 - _nn_)/(6 × 7))_ny_⁷ + _&c._ = _a_.
The Root of it will be equal to
(1) _y_ = ½[ⁿ√](_a_ + √(_aa_ - 1)) +
½/[ⁿ√](_a_ + √(_aa_ - 1)), or
(2) _y_ = ½[ⁿ√](_a_ + √(_aa_ - 1)) +
½[ⁿ√](_a_ - √(_aa_ - 1)), or
(3) _y_ = ½/[ⁿ√](_a_ - √(_aa_ - 1)) +
½[ⁿ√](_a_ - √(_aa_ - 1)), or
(4) _y_ = ½/[ⁿ√](_a_ - √(_aa_ - 1)) +
½/[ⁿ√](_a_ + √(_aa_ - 1)).
Here it is to be noted, that if (_n_ - 1)/2 be an odd Number, the Sign
of the Root found must be contrary to it.
Let an Equation be propos'd
5_y_ - 20_y_³ + 16_y_⁵ = 6,
whence _n_ = 5, and _a_ = 6, and the Root will be =
½[⁵√](6 + √(35)) + ½/[⁵√](6 + √(35))
or because 6 + √35 = 11.916 whose Logarithm is 1.0761304, and its
Fifth part is 0.2152561, whose Arithmetical Complement is 9.7847439. The
Numbers belonging to these Logarithms are 1.6415 and 0.6091, whose half
Sum is 1.1253 = _y_.
But if it happen that _a_ is less than Unity then the Second Form, as
being more convenient, ought to be pitch'd on. So if the Equation had
been
5_y_ - 20_y_³ + 16_y_⁵ = 61/64,
then _y_ will be =
½[⁵√](61/64 + √(-375/4096)) +
½[⁵√](61/64 - √(-375/4096))
and if the Root of the Fifth Power can by any means be Extracted the
true and possible Root of the Equation, will thence Emerge, tho' the
Expression seems to insinuate an Impossibility. But the Root of the
Fifth Power of the Binomial
61/64 + √(-375/4096) is
¼ + ¼√(-15)
and so the same Root of the Binomial
61/64 + √(-375/4096) is
¼ - ¼√(-15)
the half Sum of which Roots is = ¼ = _y_.
But if that Extraction can not be perform'd, or may seem too difficult,
the thing may be solv'd by the help of a Table of Natural Sines, after
the following manner;
Public-domain text, read in full here on John Shaqi.
Reviews
Reviews
No reviews yet
Be the first to share your thoughts on this work.
Elsewhere in the archive
Join the Discussion
Join the discussion
Sign in to leave a comment or review.
Sign InorCreate an account