2016 is coming!

Algebra Level 5

x 2016 + x 2015 + + x + 1 x^{2016} + x^{2015} + \cdots + x + 1

If x 1 , x 2 , x 3 , , x 2016 x_{1}, x_{2}, x_{3}, \ldots , x_{2016} are the roots of the expression above, find the value of

i = 1 2016 1 ( 1 x i ) 2 . \displaystyle\sum_{i=1}^{2016}\frac{1}{(1 - x_{i})^2}.


Inspired by many posts.


The answer is -338016.

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6 solutions

Otto Bretscher
Dec 24, 2015

Nice! We have not seen one of those in a while...

Let n = 2016 n=2016 and substitute y = 1 1 x y=\frac{1}{1-x} or x = y 1 y x=\frac{y-1}{y} . Since the x j x_j satisfy the equation x n + 1 = 1 x^{n+1}=1 , the corresponding y j y_j will satisfy ( y 1 ) n + 1 = y n + 1 (y-1)^{n+1}=y^{n+1} , or, simplified, y n n 2 y n 1 + n ( n 1 ) 6 y n 2 . . . = 0 y^n-\frac{n}{2}y^{n-1}+\frac{n(n-1)}{6}y^{n-2}-...=0 . It follows from Viete that j = 1 n y j 2 = ( n 2 ) 2 2 n ( n 1 ) 6 = n ( 4 n ) 12 \sum_{j=1}^{n}y_j^2=\left(\frac{n}{2}\right)^2-\frac{2n(n-1)}{6}=\frac{n(4-n)}{12} . For n = 2016 n=2016 this comes out to be 338016 \boxed{-338016}

Moderator note:

Great. The change of variables into roots of the intended form is often a better approach, rather than trying to figure out how to express the summation in terms of the symmetric polynomials.

Sir please explain the use of viete formula on y^2

Samarth Agarwal - 5 years, 5 months ago

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j = 1 n y j 2 = ( j = 1 n y j ) 2 2 ( i j y i y j ) \displaystyle \sum_{j=1}^n y_{j}^2 = (\displaystyle \sum_{j=1}^n y_{j})^2 - 2*(\displaystyle \sum_{i \neq j} y_{i}*y_{j})

Zeng Xunhao - 5 years, 5 months ago

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Ohh....thanx. :)

Samarth Agarwal - 5 years, 5 months ago

Its i <j for the second one

Mehdi Elkouhlani - 5 years, 5 months ago

P E R F E C T ! PERFECT!

Dev Sharma - 5 years, 5 months ago

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Please tell meHow to insert a picture in solution.

Akshay Sharma - 5 years, 5 months ago

Why do the x j x_j satisfy x n + 1 = 1 x^{n+1}=1 ? Is it a rule that roots of x n + x n 1 + . . . + 1 = 0 x^n + x^{n-1}+...+1 =0 satisfy x n + 1 = 1 x^{n+1}=1

First Last - 5 years, 5 months ago

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If you times the original equation by x, then subtract the original equation, you get x n + 1 = 1 x^{n+1}=1

Zeng Xunhao - 5 years, 5 months ago

The given polynomial is a polynomial with the 2016 complex 201 7 t h 2017^{th} roots of unity as it's roots.
x 2017 = 1 \therefore x^{2017} = 1

A Former Brilliant Member - 5 years, 5 months ago
Mark Hennings
Dec 25, 2015

This method is less elegant, but involves one of my favourite formulae! If we write ζ = e 2 π i 2 n + 1 \zeta = e^{\frac{2\pi i}{2n+1}} , then S = j = 1 2 n 1 ( 1 ζ j ) 2 = j = 1 2 n 1 ζ 2 j + 1 2 ζ j = j = 1 2 n ζ j 2 cos 2 π j 2 n + 1 2 = 1 4 j = 1 2 n ζ j sin 2 π j 2 n + 1 \begin{array}{rcl} S \; =\; \displaystyle\sum_{j=1}^{2n} \dfrac{1}{(1 - \zeta^j)^2} & = & \displaystyle\sum_{j=1}^{2n} \dfrac{1}{\zeta^{2j} + 1 - 2\zeta^j} \; = \; \sum_{j=1}^{2n} \frac{\zeta^{-j}}{2\cos \frac{2\pi j}{2n+1} - 2} \; = \; \displaystyle-\frac14\sum_{j=1}^{2n} \frac{\zeta^{-j}}{\sin^2\frac{\pi j}{2n+1}} \end{array} Replacing j j by 2 n + 1 j 2n+1-j , this last sum is also equal to 1 4 j = 1 2 n ζ j sin 2 π j 2 n + 1 -\dfrac{1}{4}\sum_{j=1}^{2n} \frac{\zeta^{j}}{\sin^2\frac{\pi j}{2n+1}} and hence S = 1 4 j = 1 2 n cos 2 π j 2 n + 1 sin 2 π j 2 n + 1 = 1 4 j = 1 2 n 1 2 sin 2 π j 2 n + 1 sin 2 π j 2 n + 1 = n 1 4 j = 1 2 n c o s e c 2 π j 2 n + 1 = 1 2 n 1 2 j = 1 n cot 2 π j 2 n + 1 \begin{array}{rcl} S & = & \displaystyle -\dfrac{1}{4}\sum_{j=1}^{2n}\frac{\cos\frac{2\pi j}{2n+1}}{\sin^2\frac{\pi j}{2n+1}} \; = \; -\dfrac{1}{4}\sum_{j=1}^{2n}\frac{1 - 2\sin^2\frac{\pi j}{2n+1}}{\sin^2\frac{\pi j}{2n+1}} \; = \; n - \displaystyle\dfrac14\sum_{j=1}^{2n} \mathrm{cosec}^2\,\frac{\pi j}{2n+1} \\ & = & \displaystyle \tfrac12n - \dfrac12\sum_{j=1}^n \cot^2\frac{\pi j}{2n+1} \end{array} Now sin ( 2 n + 1 ) θ = I m [ ( cos θ + i sin θ ) 2 n + 1 ] = j = 0 n ( 2 n + 1 2 j + 1 ) ( 1 ) j cos 2 n 2 j θ sin 2 j + 1 θ = sin 2 n + 1 θ j = 0 n ( 2 n + 1 2 j + 1 ) ( 1 ) j cot 2 n 2 j θ = sin 2 n + 1 θ f n ( cot 2 θ ) \begin{array}{rcl} \sin(2n+1)\theta & = & \displaystyle\mathrm{Im}\big[(\cos\theta + i\sin\theta)^{2n+1}\big] \; = \; \sum_{j=0}^n {2n+1 \choose 2j+1}(-1)^j \cos^{2n-2j}\theta \sin^{2j+1}\theta \\ & = & \displaystyle\sin^{2n+1}\theta \sum_{j=0}^n {2n+1 \choose 2j+1}(-1)^j \cot^{2n-2j}\theta \; = \; \sin^{2n+1}\theta f_n(\cot^2\theta) \end{array} where f n ( X ) = j = 0 n ( 2 n + 1 2 j + 1 ) ( 1 ) j X n j . f_n(X) \; = \; \sum_{j=0}^n {2n+1 \choose 2j+1} (-1)^j X^{n-j} \;. The roots of f n ( X ) f_n(X) are thus cot 2 π j 2 n + 1 \cot^2\dfrac{\pi j}{2n+1} for 1 j n 1 \le j \le n and hence j = 1 n cot 2 π j 2 n + 1 = ( 2 n + 1 3 ) ( 2 n + 1 1 ) 1 = 1 3 n ( 2 n 1 ) \sum_{j=1}^n \cot^2\dfrac{\pi j}{2n+1} \; = \; {2n+1 \choose 3} {2n+1 \choose 1}^{-1} \; = \; \dfrac13n(2n-1) This is the formula I always think remarkable! Thus S = 1 2 n 1 6 n ( 2 n 1 ) = 1 3 n ( n 2 ) . S \; = \; \tfrac12n - \tfrac16n(2n-1) \; = \; -\tfrac13n(n-2) \;. For n = 1008 n = 1008 , the answer is S = 338106 S = -338106 .

Good read!

Pi Han Goh - 5 years, 5 months ago
Ravi Dwivedi
Dec 31, 2015

P ( x ) = ( x x 1 ) ( x x 2 ) . . . . ( x x 2016 ) P(x)=\left(x-x_{1}\right)\left (x-x_{2} \right)....\left (x-x_{2016} \right)

log P ( x ) = log ( x x 1 ) + log ( x x 2 ) + . . . . + log ( x x 2016 ) \log P(x)=\log \left(x-x_{1}\right) +\log \left (x-x_{2} \right) +.... +\log \left (x-x_{2016} \right)

Taking derivative of both sides

1 P ( x ) d y d x = 1 1 x 1 + 1 1 x 2 + . . . + 1 1 x 2016 \frac{1}{P(x)} \frac{dy}{dx} = \frac{1}{1-x_{1}} + \frac{1}{1-x_{2}}+...+\frac{1}{1-x_{2016}}

Differentiate once again

P ( x ) P ( x ) ( P ( x ) P ( x ) ) 2 = [ 1 ( 1 x 1 ) 2 + 1 ( 1 x 2 ) 2 + . . . + 1 ( 1 x 2016 ) 2 ] \frac{P^{''} (x)}{P(x)}-\left(\frac{P^{'} (x)}{P(x)}\right)^2 = -\left[\frac{1}{(1-x_{1})^2} + \frac{1}{(1-x_{2})^2}+...+\frac{1}{(1-x_{2016})^2}\right]

Multiply by 1 -1

( P ( x ) P ( x ) ) 2 P ( x ) P ( x ) = [ 1 ( 1 x 1 ) 2 + 1 ( 1 x 2 ) 2 + . . . + 1 ( 1 x 2016 ) 2 ] \left(\frac{P^{'} (x)}{P(x)}\right)^2-\frac{P^{''} (x)}{P(x)} = \left[\frac{1}{(1-x_{1})^2} + \frac{1}{(1-x_{2})^2}+...+\frac{1}{(1-x_{2016})^2}\right]

Put x = 1 x=1 in above equation and thats the desired answer

P ( 1 ) = 2017 , P ( x ) = 2016 x 2015 + 2015 x 2014 + . . . + 2 x + 1 P(1)=2017, P'(x)=2016x^{2015}+2015x^{2014}+...+2x+1 so P ( 1 ) = 2016 + 2015 + . . . 1 = ( 2016 ) ( 2015 ) 2 P'(1)=2016+2015+...1=\frac{(2016)(2015)}{2}

P ( x ) = ( 2016 ) ( 2015 ) x 2014 + ( 2015 ) ( 2014 ) x 2013 + . . . + ( 3 ) ( 2 ) x + ( 2 ) ( 1 ) P''(x)=(2016)(2015)x^{2014}+(2015)(2014)x^{2013}+...+(3)(2)x+(2)(1)

P ( 1 ) = n = 1 2015 n ( n + 1 ) = n = 1 2015 n 2 + n = 1 2015 n = ( 2015 ) ( 2016 ) ( 4031 ) 6 + ( 2015 ) ( 2016 ) 2 = ( 2015 ) ( 2016 ) 6 ( 4031 + 3 ) = ( 2015 ) ( 2016 ) 6 ( 4034 ) P''(1)=\displaystyle \sum_{n=1}^{2015} n(n+1) =\displaystyle \sum_{n=1}^{2015} n^2+ \displaystyle \sum_{n=1}^{2015} n = \frac{(2015)(2016)(4031)}{6}+ \frac{(2015)(2016)}{2} = \frac{(2015)(2016)}{6}\left(4031+3\right) = \frac{(2015)(2016)}{6}\left(4034\right)

P ( 1 ) P ( 1 ) = ( 2016 ) ( 2017 ) 2 × 2017 = 2016 2 = 1008 \frac{P'(1)}{P(1)}=\frac{(2016)(2017)}{2 \times 2017} = \frac{2016}{2}= 1008

P ( 1 ) P ( 1 ) = ( 2015 ) ( 2016 ) 6 ( 4034 2017 ) = ( 2015 ) ( 2016 ) 6 × 2 = 2015 × 672 \frac{P''(1)}{P(1)}=\frac{(2015)(2016)}{6}\left(\frac{4034}{2017}\right)=\frac{(2015)(2016)}{6} \times 2 = 2015 \times 672

Now Required answer = ( P ( 1 ) P ( 1 ) ) 2 P ( 1 ) P ( 1 ) = ( 1008 ) 2 2015 × 6072 = 338016 \left(\frac{P'(1)}{P(1)}\right)^2 - \frac{P''(1)}{P(1)} = (1008)^2 - 2015\times 6072 = \boxed{-338016}

same methode

Mas Mus - 5 years, 5 months ago

@Ravi Dwivedi how on taking the derivative of log(x-x1) you got 1/(x-x1)??

Moulik Bhattacharya - 3 years, 2 months ago

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Apply chain rule.

Ravi Dwivedi - 3 years, 2 months ago

Given equation :
r = 0 2016 x r = 0 \displaystyle \sum_{r=0}^{2016}x^{r} = 0
Let y = 1 1 x x = y 1 y y = \dfrac{1}{1-x} \rightarrow x = \dfrac{y-1}{y}
Substituting in the given equation,
r = 0 2016 ( y 1 y ) r = 0 \displaystyle \sum_{r=0}^{2016}\left(\dfrac{y-1}{y}\right)^{r} = 0
Multiplying by y 2016 y^{2016}
r = 0 2016 ( y 1 ) r y 2016 r = 0 \displaystyle \sum_{r=0}^{2016}(y-1)^{r}y^{2016-r} = 0
y 2017 ( y 1 ) 2017 = ( y ( y 1 ) ) × ( r = 0 r = 2016 ( y 1 ) r y 2016 r ) y^{2017} - (y-1)^{2017} = \left(y-(y-1)\right) \times \left( \displaystyle \sum_{r=0}^{r=2016}(y-1)^{r}y^{2016-r}\right)
Using the above relation,




( y ) 2017 ( y 1 ) 2017 = 0 (y)^{2017} - (y-1)^{2017} = 0
Co-efficient of y 2016 y^{2016} (a) = ( 2017 1 ) \dbinom{2017}{1}
Co-efficient of y 2015 y^{2015} (b)= ( 2017 2 ) \displaystyle - \dbinom{2017}{2}
Co-efficient of y 2014 y^{2014} (c) = ( 2017 3 ) \dbinom{2017}{3}
Let S i S_{i} denote the symmetric sums of roots of the polynomial.
According to Vieta's,
S 1 = b a = ( 2017 2 ) ( 2017 1 ) = 1008 S_{1} = \dfrac{-b}{a} = \dfrac{\dbinom{2017}{2}}{\dbinom{2017}{1}} = 1008
S 2 = c a = ( 2017 3 ) ( 2017 1 ) = 2016 × 2015 6 S_{2} = \dfrac{c}{a} = \dfrac{\dbinom{2017}{3}}{\dbinom{2017}{1}} =\dfrac{2016\times2015}{6}
Applying Newton's identity,
P 2 = S 1 2 2 S 2 P_{2} = S_{1}^2 - 2S_{2}
P 2 = 100 8 2 2016 × 2015 3 P_{2} = 1008^2 - \dfrac{2016\times2015}{3}
P 2 = 1016064 1354080 = 338016 P_{2} = 1016064 - 1354080 = -338016

Kevin Catbagan
Dec 26, 2015

I used a calculus approach by labeling the given polynomial as p ( x ) . p(x). Taking derivatives twice gives

p ( x ) = p ( x ) t ( x ) p ( x ) s ( x ) p''(x) = p'(x)t(x) - p(x) s(x)

where t t is the sum of the reciprocals of the linear factors of p p and s s is the sum of squared reciprocals of linear factors of p p . Evaluate everything at x = 1 x=1 and s ( 1 ) s(1) gives the desired value. :)

Andreas Wendler
Dec 25, 2015

Using an analytical way with the help of Vieta's formula I got a closed formula for n (highest coefficient):

1 n + 1 [ n 2 ( n + 1 ) 4 2 i = 0 n 2 ( n i ) ! 2 ! ( n i 2 ) ! ] \frac{1}{n+1}[\frac{n^{2}(n+1)}{4}-2\sum_{i=0}^{n-2}\frac{(n-i)!}{2!(n-i-2)!}]

Simplified this expression results in:

n 12 ( 4 n ) \frac{n}{12}(4-n)

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