The function G is such that each of G, G’ and G” exists and is continuous on the interval [0, e].

It is further known that G’(e) = G(e) = G’(1) = G(1) = 1, and:

e
∫ G’(y)* y^-2 dy = 0.5
1

Evaluate:

e
∫ G”(y)*ln y dy
1

Note: ln y denotes the natural logarithm of y.

Just a thought..

Here’s a fabricated, possible form for G(y). It contains a parameter k, but satisfies all four boundary conditions and the integral constraint for all values of k. (The exact form in Maple contains lots of e’s!)

G(y) =
(1.185193911 - 0.08062379520*k)y⁵ + (-8.485808618 + 0.7348992616*k)y⁴
+ (21.06791156 - 2.559419754*k)y³ + (-21.41917276 + 4.236636925*k)y²
+ (8.651876020 - 3.331492634*k)y + k

The required integral ‘evaluates’ to   0.0007799*k + 0.586745 ; in other words, to any value we like by choice of k.

Should we be suspicious?

Posted by Harry on 2010-02-21 13:37:57