The respective eccentricities of the hyperbola x²/ a² - y²/ b² = 1 and its conjugate y²/ b² - x²/ a² = 1 are denoted by e and f(e).

Evaluate:

√2
∫ (f(e) + f(f(e))) de
1

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):

DEFDBL A-Z
h = .01
DO
  h = h / 10
  t = 0
  FOR x = 1 + h TO SQR(2) STEP h
    t = t + h * (x + SQR(1 + 1 / (x * x - 1)))
  NEXT
  PRINT t
LOOP

with successively finer integration steps.

The successively better approximations are:

1.496733581220134
1.467652052959196
1.489726658257067
1.496733584625637
1.498966670699274
1.49967334338193
1.499896743818623
1.499967188468245

making it appear the answer is 3/2.

 

Posted by Charlie on 2010-02-19 11:48:51