The respective
eccentricities of the hyperbola x
^{2}/ a
^{2}  y
^{2}/ b
^{2} = 1 and its conjugate y
^{2}/ b
^{2}  x
^{2}/ a
^{2} = 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 AZ
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 20100219 11:48:51 