# Theory of Equation – Symmetric Function of Roots

Theory of Equation / Wednesday, October 23rd, 2019

# Symmetric Function of Roots

Hello friends if you were already have gone through my previous post on Relation between Roots and Coefficients of an Equation, now is the time to learn about another most important topic Symmetric function of roots. Usually, more of the students find it tough but if you follow my post properly you will find that it is not so tough. So let’s start learning….

## Symmetric Function

Expression involving all the roots of a polynomial equation is said to be symmetric function if the expression remains unaltered when any two of the roots are interchanged.

 Example

If α, β, γ are the roots of a cubic equation, expression like

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma },~~{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}},~~\frac{\alpha }{\beta +\gamma }+\frac{\beta }{\gamma +\alpha }+\frac{\gamma }{\alpha +\beta },$

${{\alpha }^{2}}\beta +\alpha {{\beta }^{2}}+\alpha {{\gamma }^{2}}+{{\alpha }^{2}}\gamma +{{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}$

are all symmetric function of the roots α, β, γ of the cubic equation.

Symmetric function of roots are often written in abbreviated form, in sigma notation like

$\sum{\frac{1}{\alpha }=}\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$

$\sum{{{\alpha }^{2}}=}{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}$

$\sum{\frac{\alpha }{\beta +\gamma }=}\frac{\alpha }{\beta +\gamma }+\frac{\beta }{\gamma +\alpha }+\frac{\gamma }{\alpha +\beta }$

$\sum{{{\alpha }^{2}}\beta =}{{\alpha }^{2}}\beta +\alpha {{\beta }^{2}}+\alpha {{\gamma }^{2}}+{{\alpha }^{2}}\gamma +{{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}$

## Order and Weight of a Symmetric Function

The sum of the exponents of the roots of each term is a symmetric function is called the weight of the symmetric function, while the highest exponent in any term is called its order. For example, in the symmetric function Σα3β2, the weight is 3 + 2 = 5 and its order is 3.

## A few important symmetric functions

Let α1, α2, α3, …, αn be the roots of the polynomial equation

${{x}^{n}}+{{p}_{1}}{{x}^{n-1}}+{{p}_{2}}{{x}^{n-2}}+{{p}_{3}}{{x}^{n-3}}+….+{{p}_{n}}=0$

Then,

$\sum{{{\alpha }_{1}}=-{{p}_{1}}}$

$\sum{{{\alpha }_{1}}{{\alpha }_{2}}={{p}_{2}}}$

$\sum{{{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=-{{p}_{3}}}$

$\sum{\alpha _{1}^{2}=}{{\left( \sum{{{\alpha }_{1}}} \right)}^{2}}-2\sum{{{\alpha }_{1}}{{\alpha }_{2}}=p_{1}^{2}-2{{p}_{2}}}$

$\sum{\alpha _{1}^{3}=}\left( \sum{\alpha _{1}^{2}} \right)\left( \sum{{{\alpha }_{1}}} \right)-\sum{\left( \alpha _{1}^{2}{{\alpha }_{2}} \right)=}3{{p}_{1}}{{p}_{2}}-p_{1}^{3}-3{{p}_{2}}$

$\sum{\alpha _{1}^{2}{{\alpha }_{2}}=}\left( \sum{{{\alpha }_{1}}{{\alpha }_{2}}} \right)\left( \sum{{{\alpha }_{1}}} \right)-3\sum{{{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=3{{p}_{3}}}-{{p}_{1}}{{p}_{2}}$

 Example 01

If α, β, γ are the roots of the equation x3 + px + q = 0, find the value of

$\left( \beta +\gamma -\alpha \right)\left( \gamma +\alpha -\beta \right)\left( \alpha +\beta -\gamma \right)$

Solution:

Since α, β, γ are the roots of the equation x3 + px + q = 0

$\sum{\alpha =0,}~~\sum{\alpha \beta =p,~~\alpha \beta \gamma =-q}$

Now,

$\left( \beta +\gamma -\alpha \right)\left( \gamma +\alpha -\beta \right)\left( \alpha +\beta -\gamma \right)$

$=\left( \alpha +\beta +\gamma -2\alpha \right)\left( \alpha +\beta +\gamma -2\beta \right)\left( \alpha +\beta +\gamma -2\gamma \right)$

$=\left( -2\alpha \right)\left( -2\beta \right)\left( -2\gamma \right)~~~~~~as~~\sum{\alpha =0}$

$=-8\alpha \beta \gamma =8q$

 Example 02

α, β, γ are the roots of the equation x3 + x + 1 = 0, find the value of

$\left( {{\alpha }^{2}}+1 \right)\left( {{\beta }^{2}}+1 \right)\left( {{\gamma }^{2}}+1 \right)$

Solution:

Here,

$\sum{\alpha =0,}~~\sum{\alpha \beta =1,~~\alpha \beta \gamma =-1}$

Now,

$\sum{\alpha \beta =\alpha \beta +\beta \gamma +\beta \gamma =1}$

$\Rightarrow \alpha \beta +\gamma \left( \alpha +\beta \right)=1$

$\Rightarrow \alpha \beta +\gamma \left( -\gamma \right)=1~~~as~~\sum{\alpha =0}$

$\Rightarrow \alpha \beta ={{\gamma }^{2}}+1$

Similarly,

${{\alpha }^{2}}+1=\beta \gamma ~~~and~~~{{\beta }^{2}}+1=\gamma \alpha$

Hence,

$\left( {{\alpha }^{2}}+1 \right)\left( {{\beta }^{2}}+1 \right)\left( {{\gamma }^{2}}+1 \right)=\left( \beta \gamma \right)\left( \gamma \alpha \right)\left( \alpha \beta \right)$

$\Rightarrow \left( {{\alpha }^{2}}+1 \right)\left( {{\beta }^{2}}+1 \right)\left( {{\gamma }^{2}}+1 \right)={{\left( \alpha \beta \gamma \right)}^{2}}={{\left( -1 \right)}^{2}}=1$

 Example 03

α, β, γ are the roots of the equation x3 + px2 + qx + r = 0, find the value of

$\sum{\left( \frac{{{\beta }^{2}}+{{\gamma }^{2}}}{\beta +\gamma } \right)}$

Solution:

Here,

$\sum{\alpha =-p,}~~\sum{\alpha \beta =q,~~\alpha \beta \gamma =-r}$

We have,

$\frac{{{\beta }^{2}}+{{\gamma }^{2}}}{\beta +\gamma }=\frac{{{\left( \beta +\gamma \right)}^{2}}-2\beta \gamma }{\beta +\gamma }=\left( \beta +\gamma \right)-\frac{2\beta \gamma }{\beta +\gamma }$

$\therefore \sum{\left( \frac{{{\beta }^{2}}+{{\gamma }^{2}}}{\beta +\gamma } \right)}=\sum{\left( \beta +\gamma \right)-2\left\{ \frac{\beta \gamma }{\beta +\gamma }+\frac{\gamma \alpha }{\gamma +\alpha }+\frac{\alpha \beta }{\alpha +\beta } \right\}}$

$=2\sum{\alpha -\frac{2\sum{\beta \gamma \left( \gamma +\alpha \right)\left( \alpha +\beta \right)}}{\left( \beta +\gamma \right)\left( \gamma +\alpha \right)\left( \alpha +\beta \right)}}$

$=-2p-\frac{2\sum{\beta \gamma \left( \sum{\alpha \beta +{{\alpha }^{2}}} \right)}}{\left( -p-\alpha \right)\left( -p-\beta \right)\left( -p-\gamma \right)}$

$=-2p+\frac{2\sum{\beta \gamma \left( q+{{\alpha }^{2}} \right)}}{\left( p+\alpha \right)\left( p+\beta \right)\left( p+\gamma \right)}$

$=-2p+\frac{2\left( q.q+\alpha \beta \gamma \sum{\alpha } \right)}{{{p}^{3}}+{{p}^{2}}\sum{\alpha +p\sum{\alpha \beta +\alpha \beta \gamma }}}$

$=-2p+\frac{2\left( {{q}^{2}}+pr \right)}{{{p}^{3}}+{{p}^{2}}\left( -p \right)+pq-r}$

$=\frac{2{{q}^{2}}+4pr-2{{p}^{2}}q}{pq-r}$