A 2015 degrees! what is it, mercury???

Algebra Level 4

If the roots of the polynomial:

P ( x ) = x 1007 + x 1006 + x 1005 + x 2 + x + 1 P(x) =x^{1007}+x^{1006}+x^{1005}\dots\dots+x^2+ x+1

are a 1 , a 2 , a 3 , a 4 a 1007 , a_1,a_2,a_3,a_4\dots\dots a_{1007}, what is the value of i = 1 1007 F ( a i ) \displaystyle{\sum_{i=1}^{1007} F(a_i)} when

F ( x ) = x 2015 + x 2014 + x 2013 + x 2012 + + x 2 + x + 1 ? F(x)=x^{2015}+x^{2014}+x^{2013}+x^{2012}+\dots\dots+x^2+x+1?


The answer is 0.

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2 solutions

Jon Haussmann
Dec 26, 2014

Note that P ( x ) = x 1008 1 x 1 , P(x) = \frac{x^{1008} - 1}{x - 1}, so a i 1008 = 1 a_i^{1008} = 1 for all i i .

Then F ( a i ) = a i 2016 1 a i 1 = ( a i 1008 1 ) ( a i 1008 + 1 ) a i 1 = 0 F(a_i) = \frac{a_i^{2016} - 1}{a_i - 1} = \frac{(a_i^{1008} - 1)(a_i^{1008} + 1)}{a_i - 1} = 0 for all i i , so i = 1 1007 F ( a i ) = 0. \sum_{i = 1}^{1007} F(a_i) = 0.

Aareyan Manzoor
Dec 24, 2014

first note that P ( a i ) = 0 P(a_i)=0 for i [ 1 , 2 , 3 , 1007 ] i \in [1,2,3\dots\dots ,1007]\quad and that F ( x ) = x 1008 ( x 1007 + x 1006 + x + 1 ) + x 1007 + x 1006 + x + 1 F(x)=x^{1008} ( x^{1007}+x^{1006}+\dots\dots x+1)+x^{1007}+x^{1006}+\dots\dots x+1 F ( x ) = x 1008 ( P ( x ) ) + P ( x ) F(x)=x^{1008}(P(x))+P(x) F ( a i ) = x 1008 ( P ( a i ) ) + P ( a i ) = x 2013 ( 0 ) + 0 = 0 F(a_i)=x^{1008}(P(a_i))+P(a_i) =x^{2013}(0)+0=0 now add it 1012 times i = 1 1006 F ( a i ) = 0 × 1007 = 0 \sum_{i=1}^{1006} F(a_i)=0\times 1007 =\boxed{\LARGE{0}}

Given that you changed the problem significantly, can you update your solution too? IE there is no a 1012 a_ {1012} .

I fail to understand what your F ( x ) F(x) in the solution is, and how it corresponds to the polynomial given in the question.

Calvin Lin Staff - 6 years, 5 months ago

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sir, F ( x ) F(x) is defined at the end of the question . sorry for the change, there was a silly mistake in the degrees.

Aareyan Manzoor - 6 years, 5 months ago

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1) Can you explain how your solution

F ( x ) = x 1008 ( x 2007 + x 2006 + x + 1 ) + x 2007 + x 2006 + x + 1 F(x)=x^{1008} ( x^{2007}+x^{2006}+\dots\dots x+1)+x^{2007}+x^{2006}+\dots\dots x+1

corresponds with the problem

F ( x ) = x 2015 + x 2014 + x 2013 + x 2012 + + x 2 + x + 1 F(x)=x^{2015}+x^{2014}+x^{2013}+x^{2012}+\dots\dots+x^2+x+1

2) Can you explain how your solution

F ( x ) = x 1008 ( x 2007 + x 2006 + x + 1 ) + x 2007 + x 2006 + x + 1 F(x)=x^{1008} ( x^{2007}+x^{2006}+\dots\dots x+1)+x^{2007}+x^{2006}+\dots\dots x+1

is equal to F ( x ) = x 1008 P ( x ) + P ( x ) F(x) = x^{1008} P(x) + P(x) ?

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin sorry again, it shoud be 1007 in the solution, not 2007

Aareyan Manzoor - 6 years, 5 months ago

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@Aareyan Manzoor Thank you, this now makes sense to me.

Calvin Lin Staff - 6 years, 5 months ago

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