V. A. Maleev and Yu. V. Chistyakov.

On the computation of derivatives of sums of like powers of roots in terms of the coefficients of algebraic equations.

Let us have \(n\) variables \(x_1, x_2, x_3,\ldots,x_n\). We shall regard these variables as the roots of the algebraic equation of degree \(n\)

\[ f(x)=x^n-a_1x^{n-1}+a_2x^{n-2}-a_3x^{n-3}+\ldots+(-1)^{n-1}a_{n-1}x+(-1)^n a_n=0 \tag{1} \]

Denote

\[ S_k=\sum_{i=1}^{n}x_i^k=x_1^k+x_2^k+x_3^k+\ldots+x_n^k \ldots \tag{2} \]

Let us form a new function

\[ u_k=\frac{1}{k+1}\frac{\partial S_{k+1}}{\partial z} =\sum_{i=1}^{n} x_i^k \frac{\partial x_i}{\partial z} \ldots \tag{3} \]

where \(\zeta\) is for the time being an arbitrary variable. It is easy to show that the functions \(u_k\) determined by formula (3) satisfy the difference equation

\[ u_{n+h}-a_1u_{n+h-1}+a_2u_{n+h-2}-a_3u_{n+h-3}+\ldots+(-1)^n a_nu_h=0 \ldots \tag{4} \]

Indeed,

\[ \begin{aligned} &u_{n+h}-a_1u_{n+h-1}+a_2u_{n+h-2}-a_3u_{n+h-3}+\ldots+(-1)^n a_nu_h=\\ &=\sum_{i=1}^{n}x_i^h\left(x_i^n-a_1x_i^{n-1}+a_2x_i^{n-2}-a_3x_i^{n-3}+\ldots+(-1)^n a_n\right)\frac{\partial x_i}{\partial z}=\\ &=\sum_{i=1}^{n}x_i^h f(x_i)\frac{\partial x_i}{\partial z}=0 \end{aligned} \]

since we have agreed to regard the \(x_i\) as roots of equation (1).

We shall regard formula (4) as a linear difference equation, where \(n\) is a given quantity and \(h\) is variable.

Putting \(u_k=\omega^k\), we obtain

\[ \omega^{n+h}-a_1\omega^{n+h-1}+a_2\omega^{n+h-2}-a_3\omega^{n+h-3}+\ldots+(-1)^n a_n\omega^h=0 \]

whence

\[ \omega^n-a_1\omega^{n-1}+a_2\omega^{n-2}-a_3\omega^{n-3}+\ldots+(-1)^n a_n=f(\omega)=0. \]

Consequently, the values \(\omega\) sought by us coincide with the values of the roots of equation (1), i.e. \(\omega_1=x_1,\ \omega_2=x_2,\ \omega_3=x_3,\ldots,\omega_n=x_n\). Then the general solution of the difference equation will be

\[ u_k=\sum_{i=1}^{n}c_i x_i^k=c_1x_1^k+c_2x_2^k+c_3x_3^k+\ldots+c_nx_n^k \ldots \tag{5} \]

We shall determine the values of the constants \(c_i\) from the initial conditions. First of all, choose a definite value for the variable \(\xi\) with respect to which the differentiation is performed. Let, in our case, \(\xi\) be the first coefficient of equation (1), i.e. \(\xi=a_1\). As initial conditions for determining \(c_1,c_2,c_3,\ldots,c_n\), we compute the values \(u_k\), from formula (3), which correspond to the values \(k=0,-1,-2,\ldots,-n+1\).

Putting \(k=0\), we have
\[ u_0=\left[\frac{1}{k+1}\frac{\partial S_{k+1}}{\partial a_1}\right]_{k=0} =\frac{\partial S_1}{\partial a_1}=1,\quad \text{since } S_1=a_1 . \]

Putting \(k=-1\), we have
\[ u_{-1}=\left[\frac{1}{k+1}\frac{\partial S_{k+1}}{\partial a_1}\right]_{k=-1}=0,\quad \text{since } \lim_{k=-1} \frac{\dfrac{d}{dk}\left(\dfrac{\partial S_{k+1}}{\partial a_1}\right)} {\dfrac{d(k+1)}{dk}} = \lim_{k=-1}\frac{\partial}{\partial a_1}\left(\frac{dS_{k+1}}{dk}\right)= \]
\[ =\lim_{k=-1}\frac{\partial}{\partial a_1} \left(x_1^{k+1}\ln x_1+x_2^{k+1}\ln x_2+\cdots+x_n^{k+1}\ln x_n\right) = \frac{\partial}{\partial a_1}\ln(x_1x_2x_3\cdots x_n)= \]
\[ =\frac{\partial \ln a_n}{\partial a_1}=0. \]

To determine \(u_{-2},u_{-3},\ldots,u_{-n+1}\), consider equation (1) in the following form, dividing it by \((-1)^n a_n x^n\):
\[ \frac{f(x)}{(-1)^n a_n x^n} = x^{-n} -\frac{a_{n-1}}{a_n}x^{-n+1} +\frac{a_{n-2}}{a_n}x^{-n+2} -\cdots+ \]
\[ +(-1)^{-n+1}\frac{a_1}{a_n}x^{-1} +(-1)^{-n}\frac{1}{a_n}=0. \tag{6} \]

and find the sums of like powers of the roots of this equation (6), \(s_1,s_2,s_3,\ldots,s_{n-1}\), corresponding to \(S_{-1},S_{-2},S_{-3},\ldots,S_{-n+1}\) of the original equation (1).

It is not difficult to see that in the series of these sums, \(s_1=S_{-1}, s_2=S_{-2},\ldots,s_{n-1}=S_{-n+1}\) will be expressed in terms of the coefficients of equation (1) independently of \(a_1\), and consequently
\[ u_{-2}=-\frac{\partial S_{-1}}{\partial a_1},\quad u_{-3}=-\frac{1}{2}\frac{\partial S_{-2}}{\partial a_1},\ldots,\quad u_{-n+1}=-\frac{1}{n-2}\frac{\partial S_{-n+2}}{\partial a_1} \]
will all be equal to zero. Thus, as initial conditions, we have
\[ u_0=1,\quad u_{-1}=0,\quad u_{-2}=0,\ldots,u_{-n+1}=0. \]
Then, to determine the \(n\) constants \(c_1,c_2,c_3,\ldots,c_n\), we obtain the following system of linear equations:
\[ \left. \begin{aligned} u_0&=c_1+c_2+c_3+\cdots+c_n=1,\\ u_{-1}&=c_1x_1^{-1}+c_2x_2^{-1}+c_3x_3^{-1}+\cdots+c_nx_n^{-1}=0,\\ u_{-2}&=c_1x_1^{-2}+c_2x_2^{-2}+c_3x_3^{-2}+\cdots+c_nx_n^{-2}=0,\\ &\hspace{2.5em}\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ u_{-n+1}&=c_1x_1^{-n+1}+c_2x_2^{-n+1}+c_3x_3^{-n+1}+\cdots+c_nx_n^{-n+1}=0 \end{aligned} \right\}. \tag{7} \]

Solving this system with respect to \(c_i\), we obtain

\[ c_i=\frac{R_i}{R},\quad \text{where } R= \begin{vmatrix} 1&1&\ldots&1\\ x_1^{-1}&x_2^{-1}&\ldots&x_n^{-1}\\ x_1^{-2}&x_2^{-2}&\ldots&x_n^{-2}\\ \ldots&\ldots&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\\ x_1^{-n+1}&x_2^{-n+1}&\ldots&x_n^{-n+1} \end{vmatrix} \quad \text{and} \]

\[ R_i=(-1)^{i+1} \begin{vmatrix} x_1^{-1}&x_2^{-1}&\ldots&x_{i-1}^{-1}&x_{i+1}^{-1}&\ldots&x_n^{-1}\\ x_1^{-2}&x_2^{-2}&\ldots&x_{i-1}^{-2}&x_{i+1}^{-2}&\ldots&x_n^{-2}\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ x_1^{-n+1}&x_2^{-n+1}&\ldots&x_{i-1}^{-n+1}&x_{i+1}^{-n+1}&\ldots&x_n^{-n+1} \end{vmatrix}, \quad \text{whence} \]

\[ u_k=\sum_{i=1}^{n} c_i x_i^k =\frac{1}{R}\sum_{i=1}^{n} x_i^k R_i =\frac{1}{R} \begin{vmatrix} x_1^k&x_2^k&\ldots&x_n^k\\ x_1^{-1}&x_2^{-1}&\ldots&x_n^{-1}\\ x_1^{-2}&x_2^{-2}&\ldots&x_n^{-2}\\ \ldots&\ldots&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\\ x_1^{-n+1}&x_2^{-n+1}&\ldots&x_n^{-n+1} \end{vmatrix} \]

and finally

\[ u_k=\frac{1}{k+1}\frac{\partial S_{k+1}}{\partial a_1} = \frac{ \begin{vmatrix} 1&1&\ldots&1\\ x_1&x_2&\ldots&x_n\\ x_1^2&x_2^2&\ldots&x_n^2\\ \ldots&\ldots&\ldots&\ldots\\ x_1^{n-2}&x_2^{n-2}&\ldots&x_n^{n-2}\\ x_1^{k+n-1}&x_2^{k+n-1}&\ldots&x_n^{k+n-1} \end{vmatrix} }{ \begin{vmatrix} 1&1&\ldots&1\\ x_1&x_2&\ldots&x_n\\ \ldots&\ldots&\ldots&\ldots\\ x_1^{n-2}&x_2^{n-2}&\ldots&x_n^{n-2}\\ x_1^{n-1}&x_2^{n-1}&\ldots&x_n^{n-1} \end{vmatrix} }. \tag{8} \]

Example 1. For \(n=2\) we have

\[ u_k= \frac{ \begin{vmatrix} 1&1\\ x_1^{k+1}&x_2^{k+1} \end{vmatrix} }{ \begin{vmatrix} 1&1\\ x_1&x_2 \end{vmatrix} } = \frac{x_2^{k+1}-x_1^{k+1}}{x_2-x_1} = x_2^k+x_2^{k-1}x_1+x_2^{k-2}x_1^2+\ldots+x_1^k . \]

Example 2. For the case of three variables, the expressions \(u_k\) for \(k=1,2,3,4,5\) will be as follows:

\[ u_1=a_1 \]

\[ u_2=a_1^2-a_2 \]

\[ u_3=a_1^3-2a_1a_2+a_3 \]

\[ u_4=a_1^4-3a_1^2a_2+2a_1a_3+a_2^2 \]

\[ u_5=a_1^5-4a_1^3a_2+3a_1^2a_3+3a_1a_2^2-2a_2a_3 \]

Or, expressed through the roots of the equation,

\[ u_1=x_1+x_2+x_3 \]

\[ u_2=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3 \]

\[ u_3=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3 \]

\[ \begin{aligned} u_4={}&x_1^4+x_2^4+x_3^4+x_1^3x_2+x_1^3x_3+x_2^3x_1+x_2^3x_3+x_3^3x_1+x_3^3x_2+x_1^2x_2^2\\ &+x_1^2x_3^2+x_2^2x_3^2+x_1^2x_2x_3+x_2^2x_1x_3+x_3^2x_1x_2 . \end{aligned} \]

\[ \begin{aligned} u_5={}&x_1^5+x_2^5+x_3^5+x_1^4x_2+x_1^4x_3+x_2^4x_1+x_2^4x_3+x_3^4x_1+x_3^4x_2\\ &+x_1^3x_2^2+x_1^3x_3^2+x_2^3x_1^2+x_2^3x_3^2+x_3^3x_1^2+x_3^3x_2^2+x_1^3x_2x_3+x_2^3x_1x_3\\ &+x_3^3x_1x_2+x_1^2x_2^2x_3+x_1^2x_3^2x_2+x_2^2x_3^2x_1 . \end{aligned} \]

Let us prove the following properties of the expressions \(u_k\) represented by formula (8).

First property. If one considers the result of the algebraic division

\[ \frac{x^p}{f(x)}, \]

where \(p\) is any positive number, then the coefficients of the consecutive decreasing powers of \(x\) are the quantities \(u_0,u_1,u_2,\ldots\).

For the proof, form the expressions

\[ \frac{x^{p+n}}{f(x)}, \]

where \(n=0,1,2,\ldots,n\), and where \(f(x)\) is the function defined by equation (1).

Denoting the coefficients of the consecutive decreasing powers of the variable \(x\) by \(v_0,v_1,v_2,\ldots\), we have:

\[ \frac{x^p}{f(x)} =v_0x^{p-n}+v_1x^{p-n-1}+v_2x^{p-n-2}+\cdots+v_hx^{p-n-h}+\cdots \]

\[ \frac{x^{p+1}}{f(x)} =v_0x^{p-n+1}+v_1x^{p-n}+v_2x^{p-n-1}+v_3x^{p-n-2}+\cdots+v_{h+1}x^{p-n-h}+\cdots \]

\[ \frac{x^{p+2}}{f(x)} =v_0x^{p-n+2}+v_1x^{p-n+1}+ \]

\[ \qquad\qquad +v_2x^{p-n}+v_3x^{p-n-1}+v_4x^{p-n-2}+\cdots+v_{h+2}x^{p-n-h}+\cdots \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \frac{x^{p+n}}{f(x)} =v_0x^p+v_1x^{p-1}+v_2x^{p-2}+\cdots+v_{n+h}x^{p-n-h}+\cdots \]

Multiplying the rows of the expression obtained respectively by \((-1)^n a_n\), \((-1)^{n-1}a_{n-1}\), \((1)^{n-2}a_{n-2},\ldots\), and the last by unity, and adding, we obtain:

\[ x^p=v_0x^p+(v_1-a_1v_0)x^{p-1}+(v_2-a_1v_1+a_2v_0)x^{p-2}+\ldots\ldots\ldots\ldots\ldots+ \]

\[ +\left[v_{n+h}-a_1v_{n+h-1}+a_2v_{n+h-2}-a_3v_{n+h-3}+\ldots+(-1)^n a_nv_h\right]x^{p-n-h}+\ldots \tag{9} \]

From comparison of the coefficients of the identity obtained (9) we find:

\[ v_0=1 \]

\[ v_1-a_1v_0=0 \]

\[ v_2-a_1v_1+a_2v_0=0 \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \]

\[ v_{n+h}-a_1v_{n+h-1}+a_2v_{n+h-2}-a_3v_{n+h-3}+\ldots+(-1)^n a_nv_h=0 \]

Consequently, the coefficients of the quotient in the division of \(x^p\) by \(f(x)\) satisfy the recurrence formula (4) and the corresponding initial conditions; hence \(v_k=u_k\), as was required to prove.

Second property. The function \(u_k\), represented by formula (8) as the quotient of two expressions in the form of determinants which are functions of \(n\) variables \(x_1,x_2,x_3,\ldots,x_n\), gives, as the result of division, a homogeneous integral symmetric function of the \(K\)-th degree of the roots of equation (1), composed in such a way that its terms represent all possible products of \(K\) factors of the roots of equation (1) (where equal factors may occur among these factors), and the coefficients of all terms are positive units.

For the proof let us consider the fraction

\[ \frac{x^p}{f(x)} \]

Since \(x_1,x_2,\ldots,x_n\) are the roots of the equation \(f(x)=0\), then

\[ \frac{x^p}{f(x)} = \frac{x^p}{(x-x_1)(x-x_2)(x-x_3)\ldots(x-x_n)} = \]

\[ =x^{p-n} \frac{1}{1-\frac{x_1}{x}}\cdot \frac{1}{1-\frac{x_2}{x}}\cdot \frac{1}{1-\frac{x_3}{x}}\cdots \frac{1}{1-\frac{x_n}{x}} \]

Actually carrying out the indicated operations in the right- and left-hand sides of the equality and using property 1, we obtain the identity:

\[ u_0x^{p-n}+u_1x^{p-n-1}+u_2x^{p-n-2}+u_3x^{p-n-3}+\ldots = \]

\[ =x^{p-n}\left[1+\frac{x_1}{x}+\left(\frac{x_1}{x}\right)^2+\ldots\right]\cdot \left[1+\frac{x_2}{x}+\left(\frac{x_2}{x}\right)^2+\ldots\right]\ldots \left[1+\frac{x_n}{x}+\left(\frac{x_n}{x}\right)^3+\ldots\right]\ldots \tag{10} \]

Comparing the coefficients of like powers of the variable \(x\) in the two sides of the identity (10), we obtain:

\[ u_0 = 1 \]

\[ u_1 = x_1 + x_2 + x_3 + \cdots + x_n \]

\[ u_2 = x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 + x_1x_2 + x_1x_3 + \cdots + x_{n-1}x_n \]

\[ u_3 = x_1^3 + x_2^3 + x_3^3 + \cdots + x_n^3 + x_1^2x_2 + x_1^2x_3 + \cdots + x_{n-2}x_{n-1}x_n \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

Thus, the coefficients of each successive power of the variable \(x\) on the right-hand side of the identity (10) are integral symmetric functions of the variables \(x_1, x_2, \ldots, x_n\), into which enter all possible combinations of these variables in the form of products with coefficients equal to positive unity, as was required to prove.