From $x^4+4x^3+x^2-6x-1=0$ then find $S=\frac{m}{n}=\sum_{i=1}^{4}\frac{1}{\alpha_i^4-15\alpha_i^2-10\alpha_i+24}$

Question

Recently, I have found this problem:

Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4 \in \mathbb{R}$ the roots of the polynomial $p(x)=x^4+4x^3+x^2-6x-1=0$.

Calculate the value of the following sum: $$S=\sum_{i=1}^{4}\frac{1}{\alpha_i^4-15\alpha_i^2-10\alpha_i+24}=\frac{m}{n}$$ In particular, express the result in the form $\frac{m}{n}$ where $G.C.D(m,n)=1$ and $m,n \in \mathbb{N}$. Give $m+n$ as your answer.

I've tried to plot the graph of the function:

and then try to find some fraction to aproximate the real roots, but it seems no useful. Any idea?

Answer 1

Let

\begin{align*} f(z) &= z^4 + 4 z^3 + z^2 - 6 z - 1, \\ g(z) &= z^4-15z^2 - 10z + 24. \end{align*}

Then we have

$$ \lim_{R\to\infty} \frac{1}{2\pi i} \oint_{|z|=R} \frac{f'(z)}{f(z)g(z)} \, \mathrm{d}z = 0. $$

On the other hand, noting that all the zeros of $f(z)g(z)$ are simple, for any sufficiently large $R$ such that $|z| = R$ encloses all the zeros of $f(z)$ and $g(z)$, we get

\begin{align*} \frac{1}{2\pi i} \oint_{|z|=R} \frac{f'(z)}{f(z)g(z)} \, \mathrm{d}z &= \sum_{\zeta : f(\zeta)g(\zeta) = 0} \underset{z=\xi}{\mathrm{Res}}\, \frac{f'(z)}{f(z)g(z)} \\ &= \sum_{\alpha : f(\alpha) = 0} \frac{1}{g(\alpha)} + \sum_{\zeta: g(\zeta) = 0} \frac{f'(\zeta)}{f(\zeta)g'(\zeta)}. \end{align*}

So it follows that

$$ S = \sum_{\alpha : f(\alpha) = 0} \frac{1}{g(\alpha)} = -\sum_{\zeta: g(\zeta) = 0} \frac{f'(\zeta)}{f(\zeta)g'(\zeta)}. $$

By noting that $g(z) = (z - 4)(z - 1)(z + 2)(z + 3)$, we end up with

$$ S = \frac{f'(4)}{f(4)g'(4)} + \frac{f'(1)}{f(1)g'(1)} + \frac{f'(-2)}{f(-2)g'(2)} + + \frac{f'(-3)}{f(-3)g'(-3)} = \frac{212}{503}. $$

Answer 2

The hint:

Use $$x^4-15x^2-10x+24=(x-1)(x+2)(x+3)(x-4)$$ and the Viet's formulas

I got $$\frac{1}{x^4-15x^2-10x+24}=\frac{1}{252}\left(\frac{-7}{x-1}+\frac{14}{x+2}-\frac{9}{x+3}+\frac{2}{x-4}\right).$$

Answer 3

Here is a very general method for computing this kind of expressions using linear algebra.

Let $x_k$ (your $\alpha_k$s) be the roots of equation

$$p(x)=x^4+4x^3+x^2-6x-1=0\tag{1}$$

Therefore

$$x_k^4=-(4x_k^3+x_k^2-6x_k-1)\tag{2}$$

We have to compute $\sum y_k$, where, using (2), we can transform :

$$y_k=\frac{1}{x_k^4-15x_k^2-10x_k+24}=\frac{1}{-4x_k^3-16x_k^2-4x_k+25}\tag{3}$$

(this operation wasn't compulsory, but it is simpler to work with a lower degree)

Now, we will consider the generic constraint :

$$y=\frac{1}{-4x^3-16x^2-4x+25}$$

which can be written

$$q_y(x)=-4yx^3-16yx^2-4yx+(25y-1)=0\tag{4}$$

Consider the following determinant, called a resultant, such that $R=0$ expresses the existence of a common root to $p(x)=0$ and $q_y(x)=0$ (see remark below) :

$$R=Res(p,q_y)=\begin{vmatrix} 1& 4& 1& -6& -1& 0& 0\\ 0& 1& 4& 1& -6& -1& 0\\ 0& 0& 1& 4& 1& -6& -1\\ -4y& -16y& -4y& (25y - 1)& 0& 0& 0\\ 0& -4y& -16y& -4y& (25y - 1)& 0& 0\\ 0& 0& -4y& -16y& -4y& (25y - 1)& 0\\ 0& 0& 0& -4y& -16y& -4y& (25y - 1) \end{vmatrix}$$

where the coefficients of $p$ (resp $q_y$) are repeated three times (resp. 4 times) with shifts.

Expanded (using a Computer Algebra System !), we obtain :

$$R=R(y)=- 503y^4 + 212y^3 + 62y^2 - 28y + 1$$

As we are looking for the sum of the roots $\sum y_k$, we just have to use the first Vieta's formula (opposite of the ratio of the 2 first coefficients) to obtain it

$$\sum y_k=\dfrac{212}{503}$$

Remark : A reference to a similar problem. See also here for a different issue. You will find explanations there.