Polynomial twister

$f(x)= x^{2015} + x^{2014} + x^{2013} + \ldots + x^2 + x + 1$

$f(1)=1+1+1+\ldots+1+1=1=2016$

$f'(x)=2015 x^{2014} + 2014x^{2013} +2012x^{2013}+ \ldots +2x+1$

$f'(1)=2015+2014+2013+\ldots+2+1=\frac{2015}{2}\times{(1+2015)}=\frac{2015\times{2016}}{2}$

We have other form of $f(x)$

$f(x)=(x-x_1)(x-x_2)\ldots(x-x_{2015})$

$\log{f(x)}=\log{(x-x_1)(x-x_2)\ldots(x-x_{2015})}$

$\large{\frac{f'(x)}{f(x)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}+\ldots+\frac{1}{x-x_{2015}}}$

$\large{\frac{f'(1)}{f(1)}=\frac{1}{1-x_1}+\frac{1}{1-x_2}+\ldots+\frac{1}{1-x_{2015}}}$

$\large{\frac{\frac{2015\times{2016}}{2}}{2016}=\frac{1}{1-x_1}+\frac{1}{1-x_2}+\ldots+\frac{1}{1-x_{2015}}}$

$\frac{1}{1-x_1}+\frac{1}{1-x_2}+\ldots+\frac{1}{1-x_{2015}}=\boxed{1007.5}$

We will find a polynomial with roots

Since is a root of

I think this I solved this in an easier way.

Since this is a monic polynomial, we can say that

$f(x) = x^{2015} + x^{2014} + \ldots + x^2 + x +1 = (x - x_1)(x-x_2)\ldots(x-x_{2015})$

$f(1) = \displaystyle \prod_{i=1}^{2015} (x-x_i) = 2016$

$\Rightarrow f'(x) = 1.(x-x_2).(x-x_3)\ldots(x-x_{2015}) + (x-x_1).1.(x-x_3)\ldots(x-x_{2015}) + \ldots + (x-x_1).(x-x_2)\ldots(x-x_{2014}).1$

The above step was achieved by product rule .

$f'(x) = 2015x^{2014} + 2014x^{2013} + \ldots + 2x + 1$

$\Rightarrow f'(1) = \displaystyle \sum_{j=1}^{2015} j= (1-x_2).(1-x_3)\ldots(1-x_{2015}) + (1-x_1).(1-x_3)\ldots(1-x_{2015}) + \ldots + (1-x_1).(1-x_2)\ldots(1-x_{2014})$

The expression can be written as:

$E = \displaystyle \frac{(1-x_2).(1-x_3)\ldots(1-x_{2015}) + (1-x_1).(1-x_3)\ldots(1-x_{2015}) + \ldots + (1-x_1).(1-x_2)\ldots(1-x_{2014})}{\prod_{i=1}^{2015} (1-x_i) }$

$\Rightarrow E = \dfrac{f'(1)}{f(1)}$

$\displaystyle \Rightarrow E = \dfrac{\sum_{j=1}^{2015} j}{2016} = \dfrac{\frac{2015.2016}{2}}{2016} = \boxed{1007.5}$

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Note:

$a.b$ is the same as $a \times b$

$x^{2015} + x^{2014} + x^{2013} + \ldots + x^2 + x + 1 = 0$

From the equation above we have

$a_1 = x_1 + x_2 + x_3 + \ldots + x_{2015} = -1$

$a_2 = x_1 x_2 + x_1 x_3 + x_1 x _4 + \ldots + x_{2014} x_{2015} = 1$

$a_3 = x_1 x_2 x_3 + x_1 x_2 x_4 + \ldots + x_{2013} x_{2014} x_{2015} = -1$

$\cdot$

$\cdot$

$\cdot$

$a_{2015} = x_1 x_2 x_3 x_4 \ldots x_{2014} x_{2015} = -1$

$a_{n} = \begin{cases} 1 & \quad \text{if } n \text{ is even} \\ -1 & \quad \text{if } n \text{ is odd} \\ \end{cases}$

Let

$A = \frac{1}{1- x_1} + \frac{1}{1- x_2} + \ldots + \frac{1}{1-x_{2015}}$

$A = \frac{(1- x_2)(1- x_3)\ldots (1- x_{2015}) + (1- x_1)(1- x_3) \ldots (1- x_{2015}) + \ldots + (1- x_1)(1- x_2) \ldots (1- x_{2014}) }{ (1- x_1)(1- x_2)(1-x_3) \ldots (1- x_{2015})}$

$A = \frac{2015 - 2014 a_1 + 2013 a_2 + 2012 a_3 + \ldots + (-1)^{n}(2015- n) a_n + \ldots + a_{2014} }{1 - a_1 + a_2 - a_3 + \ldots + (-1)^n a_n + \ldots - a_{2015} }$

$A = \frac{2015 + 2014 + 2013 + \ldots + 2 + 1 }{1 + 1 + 1 + \ldots + 1 + 1}$

$A = \frac{2031120}{2016}$

$A = \boxed{1007.5}$

My solution to this problem is just a matter of keen observation.

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Consider $\dfrac{1}{1-x_1}+\dfrac{1}{1-x_2}$ , which can be written as $\dfrac{2-(x_1+x_2)}{1-(x_1+x_2)+x_1x_2}$

Observe the coefficients of the $\text{Numerator}:$ $2$ and $-1$

Observe the coefficients of the $\text{Denominator}:$ $1$ , $-1$ and $1$

Please have a note of the terms in the expression too... $(e_1\text{and}e_2)$ are the terms

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Now consider $\dfrac{1}{1-x_1}+\dfrac{1}{1-x_2}+\dfrac{1}{1-x_3}$ which can be written as

$\dfrac{3-2(x_1+x_2+x_3)+(x_1x_2+x_2x_3+x_1x_3)}{1-(x_1+x_2+x_3)+(x_1x_2+x_2x_3+x_1x_3)-x_1x_2x_3}$

$\rightarrow$ Observe the coefficients of the $\text{Numerator}:$ $3$ , $-2$ and $1$

$\rightarrow$ Observe the coefficients of the $\text{Denominator}:$ $1$ , $-1$ , $1$ and $-1$

$\dfrac{3-2(x_1+x_2+x_3)+(x_1x_2+x_2x_3+x_1x_3)}{1-(x_1+x_2+x_3)+(x_1x_2+x_2x_3+x_1x_3)-x_1x_2x_3}$ can be written as $\dfrac{3-2e_1+1e_2}{1-e_1+e_2-e_3}$

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So for $2015$ such terms, the coefficients in $\text{Numerator}$ will be $2015,-2014,2013,-2012...-2,1$

And coefficients of $\text{denominator}$ will be $1,-1,1,-1....1,-1$

So obviously the the expression of the question could be written as

$\dfrac{2015-2014e_1+2013e_2-2012e_3+2011e_4-...-2e_{2013}+1e_{2014}-0}{1-e_1+e_2-e_3+...-e_{2015}}$

Where,

$\rightarrow$ $e_1=\sum{x_1}=-1$

$\rightarrow$ $e_2=\sum{x_1x_2}=1$

$\rightarrow$ $e_3=\sum{x_1x_2x_3}=-1$ and so on until

$\rightarrow$ $e_{2015}=x_1x_2...x_{2015}=-1$

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Therefore, we get $\dfrac{2015+2014+2013+...+2+1}{1+1+1....+1(2016\text{times)}}$

$=\dfrac{(2015)(2016)}{2(2016)}$ $=\dfrac{2015}{2}=\boxed{1007.5}$

For simplicity, let $N:=2015$ and define the monic polynomial $\begin{aligned} P(x) &:= \sum_{k=0}^N x^k=\prod_{k=1}^N(x-x_k), &&& x_k&:\text{ roots of }P(x), &&& P(1) &= N+1\neq 0 \end{aligned}$ If we take the derivative $P^{(1)}(x)$ , the product formula always cancels exactly one factor of $P(x)$ . With $P(1)\neq 0$ we may write $\sum_{k=1}^N\frac{1}{1-x_k}=\frac{P^{(1)}(1)}{P(1)} = \frac{ \sum_{k=0}^N k \cdot 1^{k-1} }{ N+1 }=\frac{N(N+1)}{2(N+1)}=\frac{N}{2}=\boxed{1007.5}$

note that x is 2016th root of unity which means that | x| = 1; now let x be a non-real root, then x conjugate must also be a root, now note pair all of the terms with their conjugate and upon summing each pair you will get 1007. Then note that for the last lone term its value is just 1/2 = 0.5 thus answer is 1007.5

Notice that x sub 1008 is -1 while x sub (1008+k) and x sub (1008-k) are conjugates in the complex plane with magnitude 1. We can then find out that 1/(1-z1) + 1/(1-z2) where z1 and z2 are conjugates is equal to 1. Since 1/(1-x_1008) is equal to 1/2 and there are 1007 pairs of conjugates we have the sum is equal to 1007 + 1/2 or 1007.5