of Edinburgh, Session 1883-84. 301 
is satisfied. We have thus the problem “ to find an arc equal to its 
own tangent.” 
Having applied, at the end of the radius OA, a tangent to the 
circle, let us suppose two points thence to set out with equal 
velocities, the one q to travel along the tangent, the others to move 
along the circumference of the circle, carrying with it an indefinitely 
extended radius Op. At first this secant will precede the point q, 
and by the time q has reached the distance Ah, equal to the quad- 
rant AB, the intersection has moved off to an infinite distance along 
to tangent. After this the point of intersection reappears on the 
other side and comes up to A just when q reaches the distance Ac 
equal to the half -circumference ABC. The intersection now chases 
the travelling point q and comes up to it at Q, somewhat before p 
has reached the third quadrant. The first root of our equation, 
that is AQ or ABCP, is below the value x = In the same way 
we readily perceive that the next root is when p> has made more 
than an entire revolution by nearly a quadrant, that is when it has 
come to P' a little before B ; the root, then, is rather less than § 77 . 
The numerical values may be found by a very rapid approxima- 
tion. Assuming x = \tt = An, we compute the arc of which this is 
the tangent ; we regard the length of this arc as a new tangent, 
compute the length of the arc thereto belonging, and so continue 
until there be no change within the limits of the precision at which 
we aim. Three or four operations exhaust the precision of seven- 
place tables. The successive approximations are thus represented 
by the symbols 
tan"^(|7r), 
tan-2(|7r), 
and so on, so that we have ic = tan~®(§7r) ; where 00 stands for an 
indefinitely large integer number. 
