The respective eccentricities
of the hyperbola x2
= 1 and its conjugate y2
= 1 are denoted by e and f(e).
(f(e) + f(f(e))) de
As the eccentricity is sqrt(1+(b/a)^2), and the eccentricity of the conjugate interchanges the a and the b, we can convert the one to the other by squaring, subtracting 1, taking the reciprocal, adding 1 and taking the square root:
f(e) = sqrt(1+1/(e^2 - 1))
And f(f(e)) is just e, due to the mutual nature of the function.
So we're integrating e + sqrt(1+1/(e^2 - 1))
The following program integrates this numerically from 1 to sqrt(2):
h = .01
h = h / 10
t = 0
FOR x = 1 + h TO SQR(2) STEP h
t = t + h * (x + SQR(1 + 1 / (x * x - 1)))
with successively finer integration steps.
The successively better approximations are:
making it appear the answer is 3/2.
Posted by Charlie
on 2010-02-19 11:48:51