$f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$

Question

(From OMITS Indonesian $2015$)

Known that $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$, if $x_1, x_2, x_3,..., x_{2015}$ is the roots of $f(x)$, Find :

$\sum_{n=0}^{2014}$ $\sum_{k=1}^{2015}$ $(x_k)^n$

$a. -2016$

$b. -2015$

$c. -2014$

$d. -2013$

$e. -2012$

What should i do? and where i can learn more about it?

have a look at https://en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence — Albus Dumbledore, Dec 20 '20 at 13:52
Further to @AlbusDumbledore's link, $f$ is an arithmetico-geometric sequence$$\sum_{k=1}^n[a+(k-1)d]br^{k-1}=\frac{ab-(a+nd)br^n}{1-r}+\frac{dbr(1-r^n)}{(1-r)^2}$$with $a=2016,,d=-1,,b=1,,r=x,,n=2016$, so$$f=\frac{2016-2017x+x^{2017}}{(1-x)^2}.$$ — J.G., Dec 20 '20 at 14:01
@J.G. The two equations are not equivalent. Notice that 1 is a root of your eqn, whereas it is not a root of the original eqn. — Ishraaq Parvez, Dec 20 '20 at 14:07
@IshraaqParvez when $=1$ f is not defined! actually $f=stuff$ when $x\neq 1$ — Albus Dumbledore, Dec 20 '20 at 14:08
@AlbusDumbledore Sorry what do you mean by f=stuff ? And why would f not be defined at x=1? f(1) is clearly 2033136. — Ishraaq Parvez, Dec 20 '20 at 14:10
@IshraaqParvez $f=$ (that function J.G wrote) i was lazy to write the same again — Albus Dumbledore, Dec 20 '20 at 14:11
@IshraaqParvez agreed i meant that the function J.G wrote will not be defined at x=1 — Albus Dumbledore, Dec 20 '20 at 14:16
Further to comments, yes, what I wrote only works if $x\ne1$. — J.G., Dec 20 '20 at 14:23

score 3 · Answer 1 · answered Dec 20 '20 at 16:25

First, we note that $f(x)-f(x)/x=\frac{-2016}{x}+\sum_{k=0}^{2015} x^k$ Dividing by $x$ and rearranging, we get $$\sum_{k=0}^{2014} x^k=f(x)/x-f(x)/x^2+2016/x^2 -1/x$$ Evaluating at the roots of $f(x)$ and summing yields

$$\sum_i\sum_{k=0}^{2014} x_i^k=\sum_i 2016/x_i^2 -1/x_i.$$

This reduces the problem to finding the sum of the reciprocals of the roots, and finding the sum of the squares of the reciprocals of the roots. This is an exercise in rewriting symmetric functions in terms of elementary symmetric polynomials to rewrite the expression in terms of the coefficients of $f(x)$.

+1, this is how to shorten my answer! – Peanut Dec 20 '20 at 16:40 — Peanut, Dec 20 '20 at 16:40

score 2 · Accepted Answer · answered Dec 20 '20 at 16:12

First notice that $$\sum_{n=0}^{2014}\sum_{k=1}^{2015}(x_k)^n = \sum_{k=1}^{2015}\frac{(x_k)^{2015}-1}{x_k-1}$$ since $x_k$ is always $\ne1$.

Now we will compute the sums $S_1 = \sum\frac1{x_k}$ and $S_2 = \sum\frac1{x_k^2}$ over all roots. Consider the function $f(1/x)$ for $x \ne 0$. $f(1/x) = 0 \iff 2016x^{2015}+2015x^{2014}+\ldots+2x+1 = 0$. Let $x_1', \ldots, x_k'$ be the roots of this last equation. By Vieta's formulas $x_1'+\ldots+x_k' = -2015/2016$. Moreover, the sum of the squares of the roots is $x_1'^2+\ldots x_k'^2 = (x_1'+\ldots x_k')^2-2\sum_{0\le i < j \le k}x_i'x_j' = (-\frac{2015}{2016})^2-2\frac{2014}{2016}$. It's quite obvious now that since each $x_k \ne 0$ the roots $x_1', \ldots, x_k'$ must be exactly $1/x_1, 1/x_2, \ldots 1/x_k$. Overall we find that $$S_1 = -\frac{2015}{2016}\ \ \text{and}\ \ S_2=\left(-\frac{2015}{2016}\right)^2-2\frac{2014}{2016}$$ Finally the comment by J.G. tells us that $$2016-2017x_k+x_k^{2017} = 0 \implies x_k^{2015} = -\frac{2016}{x_k^2}+\frac{2017}{x_k}\tag{1}$$and$$\frac{2016-2017x_k+x_k^{2017}}{1-x_k} = 0\implies 2016+\frac{x_k^{2017}-x_k}{1-x_k} = 0\implies \frac{x_k^{2016}-1}{x_k-1}=\frac{2016}{x_k}\tag{2}$$ Thus using $(1)$ and $(2)$ the solution is $$\sum_{k=1}^{2015}\frac{(x_k)^{2015}-1}{x_k-1} = \sum_{k=1}^{2015}\frac{(x_k)^{2016}-1}{x_k-1}-\sum_{k=1}^{2015}x_k^{2015} = -\sum_{k=1}^{2015}\frac1{x_k}+2016\sum_{k=1}^{2015}\frac1{x_k^2}=$$ $$= \frac{2015}{2016}+\frac{2015^2}{2016}-2\cdot 2014 = -2013$$

I think that this can be shortened, but these were my calculations... — Peanut, Dec 20 '20 at 16:18

Piquito · Answer 3 · 2020-12-20T22:08:13.457

$f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$.

Take this as a particular example of the class of polynomials $$f_d(x)=x^d+2x^{d-1}+3x^{d-2}+\cdots+dx+(d+1)$$ where $d$ is the degree. Here we have the values of the $d$ elementary polynomials in the roots $x_1,x_2,\cdots, x_d$ (Vieta's coefficients) which are $e_k=e_k(x_1,x_2,\cdots, x_d)=(-1)^k(k+1)$ and we have five answers to choose the correct: $$a)\space -(d+1)\hspace1cm b)\space -d\hspace1cm c)\space -(d-1)\hspace1cm d)\space -(d-2)\hspace1cm e)\space -(d-3)$$ The problem being to calculate $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)$$

We use Newton's identities to solve the problem for general $d$ degree.

Noting $P_i=x_1^i+x_2^i+\cdots+x_d^i$, Newton's identities give for the five first values $$\begin{cases}P_1=e_1(x_1,x_2,\cdots, x_d)=e_1\\P_2=e_1^2-2e_2\\P_3=e_1^3-3e_1e_2+3e_3\\P_4=e_1^4-4e_1^2e_2+4e_1e_3+2e_2^2-4e_4\\P_5=e_1^5-5e_2e_1^3+5e_3e_1^2+5e_2^2e_1-5e_4e_1-5e_3e_2+5e_5\end{cases}$$ Calculation gives now a constant value for each $P_i$ $$P_i=-2\text{ for } 1\le i\le 5$$ We assume as a consequence and without proof that $P_i=-2$ for all $i$ which we have also verified for $P_6$ but it is too long to write. Thus $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)= (d)+\underbrace{(-2)+(-2)+\cdots+(-2)}_{(d-1)\space\text{ times}}=d-2(d-1)=-(d-2)$$ Thus in this particular problem the answer is $\boxed{-2013}$ (the option d).

$f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$

3 Answers3