MATHEMATICAL PROBLEM. ^ 



gents are negative in even quadrants. To examine this 

 reason on its own principle , let lis suppose the given arc a 

 to begin at B in the figure, not at A, as in article (B) J 

 tlien a — arc B C ::: a quadrant, and nzizl, this makes 

 C ED the second or an even quadrant: through B and C 

 draw the tangents BR, T C R, and the angle T R B is 

 manifestiv right ; that is, C T is perpendicular to B R, but 

 B R is a positive tangent, because B C is an odd quadrant, 

 and T C has been shown to be perpendicular to B R, which 

 invalidates the reason why n should be odd, because the 

 relations of perpendicular lines are not the relations of + 

 and — , or positive and negative quantities. 



(F) Certain mutual relations of a and z may be invest!- Mutual rela- 

 gated in the following manner. Suppose these arcs to vary, ^^°"* ^^'^ ^ '* 

 so as to preserve the equation a-{-zzzt, and we have d-{- zzzt; 



hence d—t — z, but i—^ — it'^'\-r'^) X -r, therefore « — 

 r ' r 



— ^. Now when z is less than — , t^ is less than r% and a 

 r 2 



is less than i; but a and z begin together by (D), or the 

 present article; therefore, when r is a small arc, it is greater 

 than a, and z — a is the greatest, when the angle C O P — 

 45"; on the contrary, when 2;r:Q it is finite, and a maxi- 

 mum, and a is also a maximum, but infinite; hence we 

 have the following equations, a ~ (w— -1) X^r, f:z:«+2~?J2r, 

 where n is any number, whole or fractional, greater than 

 unity. 



(G) These things being premised, we may find z by ap- z found by ap- 

 proximation to any given value of a, thus: put the jjiven arc P''o''""'}'''>"5 ° 

 ^ ^ being given. 

 fl+-Q» or A C D — ^, rn: 1, arc P D — y, its tangent or co- 

 tangent of zzzq; then by the problem §•■ — y=f, but by tri- 



gonometry ty.qzzr'zzl, and q=v+—-]- — -{--^\-^-^- 



1382 f** , 21844 1''3 929569 f'' , „ ^, 



\ \-T. 1-7;: " h &C. Now put 4-, TV, 



155925 6O8IO75 038512875 ^ tj i., 



7Tzr> ^It^j &c. zz b, c, d, €, &c. respectively; and we have 

 t q —gv — u^ + g hv^ — hv'^-\- g cv^ — t'«*+S" dv"^ — d v^ + 

 gev^ — e u'°, &c. — 1; and by reverting the series, as in 

 Emerson's Algebra, page 171, Problem LXII) we get vzz 

 J^ 2 1 



