Page:Newton's Principia (1846).djvu/574

 By arithmetic thus. Let us suppose the thing done, and let TP + e be the correct length of the right line TP as found out by delineation: and thence the correct lengths of the lines OR, BP, and SP, will be OR - $$\scriptstyle \frac{TR}{TP}$$e, BP + e, and $$\scriptstyle \sqrt{SP^{2}+2BPe+ee}=\frac{M^{2}N}{OR^{2}+\frac{2OR\times TR}{TP}e+\frac{TR^{2}}{TP^{2}}ee}$$. Whence, by the method of converging series, we have $$\scriptstyle SP+\frac{BP}{SP}e+\frac{SB^{2}}{2SP^{3}}ee$$, &c., $$\scriptstyle =\frac{M^{2}N}{OR^{2}}+\frac{2TR}{TP}\times\frac{M^{2}N}{OR^{3}}e+\frac{3TR^{2}}{TP^{2}}\times\frac{M^{2}N}{OR^{4}}ee$$, &c. For the given co-efficients $$\scriptstyle \frac{M^{2}N}{OR^{2}}-SP$$, $$\scriptstyle \frac{2TR}{TP}\times\frac{M^{2}N}{OR^{3}}-\frac{BP}{SP}$$, $$\scriptstyle \frac{3TR^{2}}{TP^{2}}\times\frac{M^{2}N}{OR^{4}}-\frac{SB^{2}}{2SP^{3}}$$, putting F, $$\scriptstyle \frac{F}{G}$$, $$\scriptstyle \frac{F}{GH}$$, and carefully observing the signs, we find $$\scriptstyle F+\frac{F}{G}e+\frac{F}{GH}ee=0$$, and $$\scriptstyle e+\frac{ee}{H}=-G$$. Whence, neglecting the very small term $$\scriptstyle \frac{e^2}{H}$$, e comes out equal to - G. If the error $$\scriptstyle \frac{e^2}{H}$$ is not despicable, take $$\scriptstyle -G-\frac{G^2}{H}=e$$.

And it is to be observed that here a general method is hinted at for solving the more intricate sort of problems, as well by trigonometry as by arithmetic, without those perplexed computations and resolutions of affected equations which hitherto have been in use.


 * To cut three right lines given in position, by a fourth right line that shall pass through a point assigned in any of the three, and so as its intercepted parts shall be in a given ratio one to the other.

Let AB, AC, BC, be the right lines given in position, and suppose D to be the given point in the line AC. Parallel to AB draw DG meeting BC



in G; and, taking GF to BG in the given ratio, draw FDE; and FD will be to DE as FG to BG. Q.E.F.