All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
Derivative, Integral And Value Puzzle (Posted on 2008-02-17) Difficulty: 3 of 5
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)

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Ho Hum | Comment 2 of 3 |

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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information