# Find $x_1^n+x_2^n$ on any quadratic equation, general case.

I have a simple quadratic (with $x^2$) equation, x can Be complex too:

$$x^2+x+1=0$$

But it could be any equation, the equation above is just an example. I need to compute $x_1^{10}+x_2^{10}$, but it could have another exponents (ex: $x_1^{50}+x_2^{50}$).

I need to know, on a general case, how to find $x_1^n+x_2^n,\ n\in\mathbb{N}\ ax^2+bx+c=0,\ a\ is\ not\ 0$?

I ask this because I have to create a software which computes this (user writes the equation and the number n = exponent) and I can't find the roots always, because sometimes are complex. I think I should make use of Viete, but I don't know how to compute $x_1^n+x_2^n$.

Thank you very much!!

• For the given first equation, $x$ is a complex cube root of unity – lab bhattacharjee Jul 15 '16 at 18:40
• @labbhattacharjee Thank You! Please read the entire question. I am asking for computing $x_1^n+x_2^n$ on any quadratic equation. A method, other than finding roots. – MM PP Jul 15 '16 at 18:43
• @MMPP why are you avoiding finding roots? – Mark Jul 15 '16 at 18:43
• – lab bhattacharjee Jul 15 '16 at 18:45
• You can convert the roots from the quadratic formula to polar form and then compute the powers that way. – arctic tern Jul 15 '16 at 19:08

## 3 Answers

One approach is to represent the roots in polar form, and take advantage of their symmetry. For the sake of ease of exposition, let the quadratic be monic; that is, $a = 1$. If it's not already in that form, it's trivial to convert it to a monic form.

If the roots are real—if $b^2-4c \geq 0$—I assume you know how to handle that. So we'll just consider the case where they're not real. In that case, $b^2-4c < 0$, and the roots are given by

$$x_{1, 2} = \frac{-b \pm \sqrt{b^2-4c}}{2} = \frac{-b}{2} \pm \frac{\sqrt{4c-b^2}}{2}i$$

Note that $c > 0$ necessarily if the roots are not real. We can see, using the Pythagorean theorem, that $|x_1| = |x_2| = \sqrt{c}$. Now, find $\theta, 0 \leq \theta \leq \pi$ such that

$$\cos\theta = -\frac{b}{2\sqrt{c}}$$ $$\sin\theta = \sqrt{1-\frac{b^2}{4c}}$$

This can be done using the atan2 function in many programming languages. Then $x_{1, 2}$ can be represented as $\sqrt{c} \text{ cis } (\pm\theta) \equiv \sqrt{c} \left[\cos(\pm\theta)+i\sin(\pm\theta)\right]$. Then

$$x_1^n = \sqrt{c^n} \text{ cis }(n \theta)$$ $$x_2^n = \sqrt{c^n} \text{ cis } (-n\theta)$$

Since $x_1^n$ and $x_2^n$ form a conjugate pair (just as $x_1$ and $x_2$ do), we therefore have

$$x_1^n+x_2^n = 2\sqrt{c^n}\cos (n\theta)$$

Newton-Girard's relations are very general, so we can redo the computations in the case of two roots.

Set $x+y= s; \enspace xy=p$. We want to compute recursively the sums of powers $$P_n=x^n+y^n$$ as polynomials in $s$ and $p$.

Initialisation:

$$P_0=2,\quad P_1=s=-1, \quad P_2=x^2+y^2=s^2-2p=-1.$$

Recursion relation: \begin{align*} P_{n+1}&=(x^n+y^n)(x+y)-x^ny-xy^n=(x^n+y^n)(x+y)-xy(x^{n-1}+y^{n-1})\\ &=sP_n-pP_{n-1}. \end{align*} Thus, \begin{align*} P_3&=sP_2-pP_1=s^3-3ps\\ P_4&=sP_3-pP_2=s^4-4ps^2+2p^2\\ P_5&=sP_4-sP_3=s^4-5ps^3+5p^2s\\ \&c.&\;\&c. \end{align*}

However, for this precise equation, we have $s=-1,\enspace p=1$ and the recursion relation becomes $$P_{n+1}=-(P_n+P_{n-1}),$$ so the computation is very simple: \begin{align*} P_0&=2,&P_1&=-1,&P_2&=-1,& P_3&=2,&P_4&=-1,&P_5&=-1.\end{align*} This is more than enough to see the sequence $(P_n)$ is periodic and that $$P_n=\begin{cases}2&\text{if}\enspace n\equiv 0\mod 3,\\ -1&\text{if}\enspace n\not\equiv 0\mod 3.\end{cases}$$ So the sums are $$x_1^{10}+x_2^{10}=x_1^{50}+x_2^{50}=-1.$$

• It's simply $$x_{1}^{n}+x_{2}^{n}=\exp \frac{2n\pi i}{3}+\exp \frac{4n\pi i}{3}$$ – Ng Chung Tak Jul 17 '16 at 6:44
• @Ng Chung Tak: You're absolutely right, but my point was to illustrate the use of Newton-Girard's relations in a very simple case – Bernard Jul 17 '16 at 9:26

\begin{align*} \alpha+\beta &= -\frac{b}{a} \\ \alpha \beta &= \frac{c}{a} \\ \alpha^n+\beta^n &= (\alpha+\beta)^{n}+\sum_{k=1}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n-k}{k} \frac{n(-\alpha \beta)^{k} (\alpha+\beta)^{n-2k}}{n-k} \\ &= (\alpha+\beta)^{n}-n\alpha \beta (\alpha+\beta)^{n-2}+ \frac{n(n-3)}{2!} \alpha^2 \beta^2 (\alpha+\beta)^{n-4}-\ldots \\ \end{align*}