April's Lagrange Interpolation

Algebra Level 5

Let P ( x ) P(x) be a monic polynomial of degree 2015 2015 , satisfying, for x Z + , x 2015 x\in \mathbb{Z}^+, x\le 2015 , P ( x ) = k = 1 x Q ( k ) 3 + 1 k 3 + 1 P(x)=\sum_{k=1}^x \dfrac{Q(k)^3+1}{k^3+1} where Q ( x ) = { i = 1 1 i x if the sum converges 1 x if the sum diverges Q(x)=\left\{ \begin{array}{l}-\sum\limits_{i=1}^{\infty}\dfrac{1}{i^{|x|}}\qquad \text{if the sum converges}\\ -\dfrac{1}{|x|}\qquad \text{if the sum diverges}\end{array}\right. Now, let a n a_n be the coefficient of the x n x^n term in P ( x ) P(x) . Finally, let S = i = 1 2015 a i 1 S=\sum_{i=1}^{2015}a_{i-1} Find the positive remainder when S 2015 S \lfloor S\rfloor^{2015-\lceil S\rceil} is divided by 1000 1000 .


The answer is 1.

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1 solution

Daniel Liu
Apr 1, 2015

Note that i = 0 2015 a i = P ( 1 ) \displaystyle\sum_{i=0}^{2015}a_i=P(1) and a 2015 = 1 a_{2015}=1 so S = P ( 1 ) 1 S=P(1)-1

But P ( 1 ) = Q ( 1 ) 3 + 1 2 = 1 + 1 2 = 0 P(1)=\dfrac{Q(1)^3+1}{2}=\dfrac{-1+1}{2}=0 so S = 1 S=-1 .

Finally, S 2015 S = ( 1 ) 2016 = 1 \lfloor S \rfloor ^{2015 - \lceil S \rceil}=(-1)^{2016}=1 and taking mod 1000 \text{mod }1000 we get the answer as 1 \boxed{1} .

This is a better trolling attempt than my problem , although definitely much less obvious. Well done.

Ivan Koswara - 6 years, 2 months ago

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I think you're problem is better; I still don't see how to do it at a glance. I will look into it deeper when I have the time.

Daniel Liu - 6 years, 2 months ago

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