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Isaac Newton’s ‘Of Quadrature by Ordinates’

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Abstract

In Of Quadrature by Ordinates (1695), Isaac Newton tried two methods for obtaining the Newton–Cotes formulae. The first method is extrapolation and the second one is the method of undetermined coefficients using the quadrature of monomials. The first method provides \(n\)-ordinate Newton–Cotes formulae only for cases in which \(n=3,4\) and 5. However this method provides another important formulae if the ratios of errors are corrected. It is proved that the second method is correct and provides the Newton–Cotes formulae. Present significance of each of the methods is given.

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Notes

  1. Newton did not translate the first sentence into Latin.

  2. Cas. 1. Si dentur duæ tantū ordinatæ \(\textit{AK}, \textit{BL}\) fac aream

    $$\begin{aligned} \textit{AKLB}=\frac{\textit{AK}+\textit{BL}}{2}\textit{AB} \end{aligned}$$

       Cas. 2. Si dentur tres \(\textit{AK},\textit{BL}, \textit{CM}\), dic \(\displaystyle \frac{\textit{AK}+\textit{CM}}{2}\textit{AC}=\square \, \textit{AM},\) et rursus \(\displaystyle \frac{\textit{AK}+\textit{BL}}{4}+\frac{\textit{BL}+\textit{CM}}{4}\) in \(\textit{AC}=\textit{AK}+2\textit{BL}+\textit{CM}\) in \(\frac{1}{4}\textit{AC}=\square \,\textit{AM}\) (per Cas. 1) et errorem solutionis prioris esse ad errorem solutionis posterioris ut \(\textit{AC}^{q}\) ad \(\textit{AB}^{q}\) seu 4 ad 1 adeoque solutionum differentiam \(\displaystyle \frac{\textit{AK}-2\textit{BL}+\textit{CM}}{4}\textit{AC}\) esse ad errorem posterioris ut 3 ad 1, et error posterioris erit \(\displaystyle \frac{\textit{AK}-2\textit{BL}+\textit{CM}}{12}\textit{AC}.\) Aufer hunc errorem et solutio posterior evadet

    $$\begin{aligned} \frac{\textit{AK}+4\textit{BL}+\textit{CM}}{6}\textit{AC}=\square \, \textit{AM}.\,\text{ Solutio } \ \mathrm{qu{\ae }sita.} \end{aligned}$$

  3. Cas. 3. Si dentur 4 Ordinatæ AK, BL, CM, DN: dic \( \frac{\textit{AK}+\textit{DN}}{2}\textit{AD}=\square \,\textit{AN}.\) Item \( \frac{\textit{AK}+\textit{BL}}{6}+\frac{\textit{BL}+\textit{CM}}{6}+\frac{\textit{CM}+\textit{DN}}{6}\) in \(\textit{AD}\) (id est \( \frac{\textit{AK}+2\textit{BL}+2\textit{CM}+\textit{DN}}{6}\textit{AD})=\square \,\textit{AN}.\) Et solutionū errores erunt ut \(\textit{AD}^{q}\) ad \(\textit{AB}^{q}\) seu 9 ad 1 adeoque errorum differentia (quæ est solutionū differentia \(\frac{2\textit{AK}-2\textit{BL}-2\textit{CM}+2\textit{DN}}{8}\textit{AD}\)) erit ad errorem posterioris ut 8 ad 1. Aufer hunc errorem et posterior manebit

    $$\begin{aligned} \frac{\textit{AK}+3\textit{BL}+3\textit{CM}+\textit{DN}}{8}AD=\square \, AN. \end{aligned}$$

  4. Cas. 4. Si dentur 5 Ordinatæ, dic (per Cas.2)

    $$\begin{aligned} \frac{\textit{AK}+4\textit{CM}+\textit{EO}}{6}\textit{AE}=\square \,\textit{AO}. \end{aligned}$$

    Item \(\displaystyle \frac{\textit{AK}+4\textit{BL}+\textit{CM}}{12}+\frac{\textit{CM}+4\textit{DN}+\textit{EO}}{12}\) in \(\textit{AE}=\square \,\textit{AO}\) et errores esse ut \(\textit{AE}^{q}\) ad \(\textit{AB}^{q}\) seu 16 ad 1\({}_{[,]}\) et cum errorum differentia sit

    $$\begin{aligned} \frac{\textit{AK}-4\textit{BL}+6\textit{CM}-4\textit{DN}+\textit{EO}}{12}\textit{AE} \end{aligned}$$

    error minoris erit \(\displaystyle \frac{\textit{AK}-4\textit{BL}+6\textit{CM}-4\textit{DN}+\textit{EO}}{180}\textit{AE}\) quem aufer et manebit

    $$\begin{aligned} \frac{7\textit{AK}+32\textit{BL}+12\textit{CM}+32\textit{DN}+7\textit{EO}}{90}\textit{AE}=\square \,\textit{AO}. \end{aligned}$$

  5. Cas. 5. Eodem modo si dentur 7 Ordinatæ, fiet

    $$\begin{aligned} \frac{17\textit{AK}+54\textit{BL}+51\textit{CM}+36\textit{DN}+51\textit{EO}+54\textit{FP}+17\textit{GQ}}{280}\textit{AG}=\square \,\textit{AQ}. \end{aligned}$$

  6. Cas. 6. Et si dentur 9, fiet

    $$\begin{aligned}&217\textit{AK}+1024\textit{BL}+352\textit{CM}+1024\textit{DN}+436\textit{EO}\\&\frac{\qquad \qquad +1024\textit{FP}+352\textit{GQ}+1024\textit{HR}+217\textit{IS}}{5670}\textit{AI}=\square \,\textit{AS}. \end{aligned}$$

  7. Hæ sunt quadraturæ Parabolæ quæ per terminos Ordinatarum omniū transit.

  8. Investigantur etiam hæ quadraturæ in hunc modum. Sit \(a\) summa Ordinatæ primæ et ultimæ, \(b\) summa secundæ et penultimæ\({}_{[,]}\) \(c\) tertiæ & antepenultimæ\({}_{[,]}\) \(d\) summa 4\({}^{\text{ tae }}\) a principio et quartæ a fine\({}_{[,]}\) [\(e\)] summa quintæ a principio et quintæ a fine &c\({}_{[,]}\) \(m\) Ordinata media et Abscissa \(A\) et sit

    $$\begin{aligned} \frac{za+yb+xc+vd+te+sm}{2z+2y+2x+2v+2t+s}A=\square ^{\ae } \mathrm{qu{\ae }sit{\ae },} \end{aligned}$$

    et pro ordinatis primo scribantur termini totidem primi hujus seriei numerorū quadratorum 0.1.4, 9.16.25 &c\({}_{[,]}\) dein termini totidem primi hujus cubicorum 0.1.8.27.64.125 &c\({}_{[,]}\) dein totidem hujus quadrato-quadraticorū 0.1.16.81.256.&c pergendo (si opus est) ad tot series una dempta quot sunt incognitæ quantitates \(z, y, x, v\) &c, et pro quadratura quæsita scribe quadraturam Parabolæ cui hæ ordinatæ congruunt. Et provenient æquationes ex quibus collatis determinabuntur \(z, y, x\) &c.

  9. Ut si Ordinatæ sint quatuor, pono \( a=0+9\ { \& }\ b=1+4. A=3\) et quadraturam\(=9\). Sic fit \(\small \frac{9z+5y}{2z+2y}\times 3=9.\) et inde \(y=3z\). Igitur pro \(y\) scribo \(3z\) et æquatio \(\small \frac{za+yb}{2z+2y}A=\square \) fit \(\displaystyle \tfrac{a+3b}{8}A=\square .\) ut in casu tertio.

  10. Rursus si ordinatæ sint quinque pono \(a=0+16, b=1+9, c=4. A=4, \square =\frac{64}{3}.\) Et sic fit \(\small \frac{16z+10y+4s}{2z+2y+s}\times 4=\frac{64}{3}\) seu \(8z=y+2s.\) Rursus pono

    $$\begin{aligned} a=0+64, b=1+27, c=8,A=4 \text{ et } \square =64. \end{aligned}$$

    et sic fit \(\small \frac{64z+28y+8s}{2z+2y+s}\times 4=64.\) seu \(8z=y+2s\) ut supra. Pono igitur [\(a=0+256, b=1+81, c=16, A=4\) et \(\small \square =\frac{1024}{5}.\) Et sic fit

    $$\begin{aligned} \frac{256z+82y+16s}{2z+2y+s}\times 4=\frac{1024}{5} \end{aligned}$$

    seu \(384z-88s=51y=408z-102s,\) hoc est \(14s=24z\) sive \(7s=12z\) adeoque \(7y=32z.\) Igitur pro \(s\) et \(y\) scribo \(\frac{12}{7}z\) et \(\frac{32}{7}z\) respective et æquatio

    $$\begin{aligned} \frac{za+yb+sm}{2z+2y+s}A=\square \text{ fit } \frac{7a+32b+12m}{90}[A]=\square \end{aligned}$$

    ut in casu quarto.]

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  1. Tokyo Woman’s Christian University, 2-6-1, Zempukuji, Suginami, Tokyo, 167-8585, Japan

    Naoki Osada

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  1. Naoki Osada
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Communicated by : Niccolò Guicciardini.

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Osada, N. Isaac Newton’s ‘Of Quadrature by Ordinates’. Arch. Hist. Exact Sci. 67, 457–476 (2013). https://doi.org/10.1007/s00407-013-0117-1

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