Letter from Newton to John Collins, dated 20 June 1674

37.

SrSir

I thank you for yoryour kind present. Mr Andersons book is very ingenious, & may prove as usefull if his principles be true. But I suspect one of them, namely ytthat yethe bullet moves in a Parabola. This would be so indeed were yethe horizontal celerity of yethe bullet uniform, but I should think its motion decays considerably in yethe flight. Suppose for instance a bullet shot horizontally ~~moves~~ from $A$ moves in yethe line $AE$ , & $AI$ being perpendicular to yethe horizon in it take $AF$ $AG$ , $AH$ , $AI$ , &c in proportion as yethe square numbers $1$ , $4$ , $9$ , $16$ &c: & its certain ytthat if in one moment of time yethe bullet descend as low as $F$ , in yethe next moment it shall descend as low as $G$ , in yethe 3d as low as $H$ &c. And therefore drawing yethe horizontall lines $FB$ , $GC$ , $HD$ , $IE$ ; yethe bullet at yethe end of yethe first moment will be somewhere in yethe line $FB$ suppose at $B$ , & at yethe end of yethe 2d moment it will be somewhere in yethe line $GC$ suppose at $C$ &c. But that $FB$ , $GC$ , $HD$ & $IE$ are in Arithmeticall progression (wchwhich is yethe condition of yethe Parabola) seems not probable; for if it were so, yethe celerity of yethe bullet would increas becaus yethe spaces $AB$ , $BC$ , $CD$ , $DE$ described in equall times are yethe latter bigger ynthan yethe former: whereas I should rather think ytthat yethe celerity decreases very considerably. And perhaps this rule for its decreasing may pretty nearly approach yethe truth, viz: Letting fall yethe perpendiculars $BK$ , $CL$ , $DM$ &c to make $IK$ , $KL$ , $LM$ &c, a decreasing Geometricall progression. If you should have occasion to speak of this to yethe Author, I desire you would not mention me becaus I have no mind to concern my self further about it.

As for yethe method of extracting yethe roots of literal equations, the root of this

y^{3} + a a y - b^{3} = 0

(wchwhich may be a form for all cubic equations) may be thus extracted. Suppose

l

pretty nearly equal to yethe desired root & put

l

in yethe quotient for yethe first term. Then proceed as follows.

\begin{array}{l} \begin{matrix} \begin{matrix} \begin{matrix}  \end{matrix} & (l - \frac{m}{n} - \frac{3 l m m}{n^{3}} + \frac{m^{3}}{n^{4}} \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} l + p = y & y^{3} & l^{3} + 3 l l p + 3 l p p + p^{3} . \\ + a a y & + a^{2} l + a a p . \\ - b^{3} & - b^{3} . \end{matrix} & \begin{array}{l} For brevities sake I have put l^{3} + a a l - b^{3} = m \\ & 3 l l + a a = n . \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} - \frac{m}{n} + q = p & p^{3} & - \frac{m^{3}}{n^{3}} + \frac{3 m m q}{n n} - \frac{3 m q q}{n} + q^{3} . \\ + 3 l p p & + \frac{3 l m m}{n n} + \frac{6 l m}{n} q + 3 l q q . \\ + n p & - m + n q . \\ + m & + m . \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} - \frac{3 l m m}{n^{3}} + r = q & q^{3} & - \frac{27 l^{3} m^{6}}{n^{9}} + \frac{27 l l m^{4}}{n^{6}} r - \frac{9 l m m}{n^{3}} r r + r^{3} . \\ - \frac{3 m}{n} q q & - \frac{27 l l m^{5}}{n^{7}} + \frac{18 l m^{3}}{n^{4}} r - \frac{3 m}{n} r r \\ + 3 l q q & + \frac{27 l^{3} m^{4}}{n^{6}} - \frac{18 l l m m}{n^{3}} r + 3 l r r \\ + \frac{3 m m}{n n} q & - \frac{9 l m^{4}}{n^{5}} + \frac{3 m m}{n n} r \\ - \frac{6 l m}{n} q & + \frac{18 l l m^{3}}{n^{4}} - \frac{6 l m}{n} r \\ + n q & - \frac{3 l m m}{n n} + n r \\ - \frac{m^{3}}{n^{3}} & - \frac{m^{3}}{n^{3}} . \\ + \frac{3 l m m}{n n} & + \frac{3 l m m}{n n} . \end{matrix} \end{matrix} \end{array}

There may be other ways of extracting yethe root of this equation: as for instance if

b

be greater then

a

, the root may be thus exprest

y = b^{3} - \frac{a a}{3 b} + \frac{a^{6}}{81 b^{5}}

+ \frac{a^{8}}{243 b^{7}} &c

. Or thus

y = \frac{b^{3}}{a a} - \frac{b^{9}}{a^{8}} + \frac{3 b^{15}}{a^{14}} - \frac{12 b^{21}}{a^{20}} &c

a

be considerably greater then

b

. The extraction of yethe simple cube root may be done after yethe same manner omitting only yethe 2d terme

a a y

. Or els it may be done as in yethe following example numerall Arithmetic. Suppose

y^{3} = a^{3} + b^{3}

, & that

a

is bigger then

b

, & let yethe divisor for finding yethe terms of yethe Quotient be always tripple yethe square of yethe first term, that is

3 a a

. And yethe form of yethe work will be this.

\begin{array}{l} \begin{array}{l} a^{3} & + & b^{3} & (a + \frac{b^{3}}{3 a a} - \frac{b^{6}}{9 a^{5}} + \frac{5 b^{9}}{81 a^{8}} &c \end{array} \\ \begin{array}{l} a^{3} \\ 0 & + & b^{3} \end{array} \\ \begin{array}{l} \begin{array}{l} b^{3} & + & \frac{b^{6}}{3 a^{3}} & + & \frac{b^{9}}{27 a^{6}} \\ 0 & - & \frac{b^{6}}{3 a^{3}} & - & \frac{b^{9}}{27 a^{6}} \end{array} \end{array} \\ \begin{array}{l} \begin{array}{l} - & \frac{b^{6}}{3 a^{3}} & - & \frac{2 b^{9}}{9 a^{6}} & * & + & \frac{b^{15}}{81 a^{12}} & + & \frac{b^{18}}{729 a^{15}} \\ 0 & - & \frac{5 b^{9}}{27 a^{6}} & * & - & \frac{b^{15}}{81 a^{12}} & - & \frac{b^{18}}{729 a^{15}} \end{array} \end{array} \end{array}

So if

y^{3} = c

, suppose any letter as

a

to be pretty nearly equall to yethe cube root of

c

, & putting

c - a^{3} = b

a^{3} + b = c

extract yethe cube root out of

a^{3} + b

as in yethe former example; wchwhich root

(viz: a + \frac{b^{3}}{3 a a} - \frac{b^{6}}{9 a^{5}}

+ \frac{5 b^{9}}{81 a^{8}} &c)

being once extracted may be kept as a rule for extracting numeral cube roots. But yet (as you suggest) these infinite series are only usefull where yethe roots of equations cannot be attained accurately. For when they can be accurately attaind recours must be had to other methods, this only performing it by approximation. SrSir I am YorYour obliged humble Servant Is. Newton Cambridge. June 20. 1674.

These For Mr John Collins at the Farthing Office in Fenchurch Street in London. 2 Mr Newton about Andersons Parabola and extracting of rootes in Species