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)

Let ay^p + by^(p-1) + cy^(p-2) + q be the first three terms and the constant term of L(y). Matching the terms on the left and right hand sides of the first equality gives:

( 2c + ap(p-1) )y^(p-2) - 1 = 2cy^(p-2) + 1

Requiring p=2 and a=1, so L(y) = y² + by + c. Then

L'(5) = 2*5 + b = 15, so b=5. Next, the integral gives

(64+8)/3 + (16-4)5/2 + 6c = 4(14 + c) so c = 1 and

L(y) = y² + 5y + 1, whence L(-5) = 1

 

Posted by FrankM on 2008-02-18 10:49:47