Chapter XIV: Appendix

_An Analytical Resolution of certain Equations of the Third, Fifth,
Seventh, Ninth Powers, and so on _ad Infinitum_, in finite Terms,
after the manner of _Cardan_'s Rules for Cubicks. By Mr. _A. Moivre_,
Transact. _Nᵒ 309_._

Let (_n_) be any Number, (_y_) an unknown Quantity, or Root of the Equation, (_a_) a Quantity altogether known, or what they call _Homogeneum Comparationis_: And let the Relation of these Quantities to each other be exprest by the Equation.

_ny_ + ((_nn_ - 1)/(2 × 3))_ny_³ + ((_nn_ - 1)/(2 × 3)) × ((_nn_ - 9)/(4 × 5))_ny_⁵ + ((_nn_ - 1)/(2 × 3)) × ((_nn_ - 9)/(4 × 5)) × ((_nn_ - 25)/(6 × 7))_ny_⁷, _&c._ = _a_.

Its plain from the Nature of this Series, that if _n_ be any odd Number (that is an Integer, it matters not whether Affirmative or Negative) then the Series will Terminate, and the Equation arising will be one of the above defin'd, whose Root is

(1) _y_ = ½[ⁿ√](√(1 + _aa_) + _a_) -
½/[ⁿ√](√(1 + _aa_) + _a_) or,

(2) _y_ = ½[ⁿ√](√(1 + _aa_) + _a_) -
½[ⁿ√](√(1 + _aa_) - _a_) or,

(3) _y_ = ½/[ⁿ√](√(1 + _aa_) - _a_) -
½[ⁿ√](√(1 + _aa_) - _a_) or,

(4) _y_ = ½/[ⁿ√](√(1 + _aa_) - _a_) -
½/[ⁿ√](√(1 + _aa_) + _a_)

For Example, Let the Root of this Equation of the Fifth Power be required

5_y_ + 20_y_³ + 16_y_⁵ = 4

in which case _n_ is = 5, and _a_ = 4, and the Root, according to the first Form, is

_y_ = ½[⁵√](√(17 + 4)) - ½/[⁵√](√(17 + 4))

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;

To the Radius 1 let _a_ = 61/64 = 0,95112 the Sine of some Arch which is therefore 72° 23', whose Fifth part (because _n_ is equal to 5) is 14° 28' the Sine of it is 0.24981 = ¼ nearly.

The same is the Method of proceeding in Equations of higher Dimensions.

_A Discourse concerning the Action of the Sun and Moon on Animal
Bodies; and the Influence which This may have in many Diseases. By
_Richard Mead_, M. D. F. R. S._