A real polynomial L(y) satisfies this recurrence relation:

2*L(y) + 1 = L(y-1) + L(y+1) - 1

It is also known that L'(5) = 15, and:

  6
 ∫ L(y-2) dy = 4*L(2)
 0

Determine the value of L(-5)

The recurrence relationship states that there is a constant difference between unit differences of the function.  This implies that the polynomial in question is quadratic, and the constant in front of the quadratic must be 1.

And there are two other statements that deal with the derivative and the integral of this polynomial.  The integral will therefore by cubic, and since it's a definite integral, we don't need to address the constant, so, for simplicity sake, let F'(X) = L(Y) and the cubic will be of the form:
F(X) = Ax³ + Bx² + Cx

its derivative (the polynomial in question) will have the form:
F'(X) = 3Ax² + 2Bx + C

and its derivative will, of course, have the form
F''(X) = 6Ax + 2B

The second statement in the problem tells us that if we substitute 5 for x, then [30A + 2B = 15] *** FORMULA 1

The third statement in the problem gives us a relationship between F and F' :
[64A + 16B + 4C] - [-8A + 4B - 2C] = 4[12A + 4B + C] (just evaluating the definite integral)
which simplifies to:
72A + 12B + 6C = 48A + 16B + 4C
24A - 4B = -2C *** FORMULA 2

As mentioned before the constant in front of the quadratic must be 1, and, looking at F'(X), this is given by 3A, therefore, A = 1/3.

We know by Formula 1, that 30A + 2B = 15... substituting for A, we see that B is 5/2.

And then substituting for A and B in Formula 2, we see that C is 1.

Therefore the polynomial in question is:
L(x) = F'(x) = 3(1/3)x² + 2(5/2)x + 1 = x² + 5x + 1

and solving for L(-5), we get the answer: 1

Posted by SilverKnight on 2008-02-18 10:12:37