10/07/2009, 02:04 AM
(10/06/2009, 12:05 AM)mike3 Wrote: I wouldn't know but I'm still trying to figure out how precisely he got those coefficients. It would seem a little good to be true wouldn't it, given how exotic this function is?
Essentially, by the method of undetermined coefficients.
Assume that the solution is a polynomial, say of the form \( EF(x) = a_0 + a_1x + a_2 x^2 \), and substitute that in the functional equation gives
\( \log_{x+1}(a_0 + a_1(x+1) + a_2(x+1)^2) = a_0 + a_1x + a_2 x^2 \)
and expanding the left-hand-side of this equation about x=1, and truncating the result at 3 or so terms gives an approximately equal equation \( O(x^3) \). Since this equation should hold for all \( 1 \le x \le 2 \) (by assumption*), you can match each coefficient of x on each side to form a system of 3 equations in 3 unknowns \( (a_0, a_1, a_2) \). Even though this system of algebraic equations is simpler (in a way) than the single functional equation, they are not linear in the unknowns, so use your favorite nonlinear system method, and solve for \( (a_0, a_1, a_2) \).
Does this help?
Andrew Robbins
* Assuming the functional equation holds for all \( 0 \le x \le 1 \) seems to lead to a contradiction, because \( EF(0) = EF'(1) \) and yet \( 0^x = 0 \) for all x, so you can't solve for x such that the functional equation holds.
