# Sum of Roots of a polynomial

• Nov 30th 2010, 05:19 AM
EinStone
Sum of Roots of a polynomial
Let $\displaystyle p(z) = a_0 + a_1 z + \ldots + a_d z^d$ be a complex polynomial with coefficients $\displaystyle a_i \in \mathbb{C}$ of degree d.

1) Let $\displaystyle x_1,\ldots,x_d$ be the roots of p. Show that $\displaystyle \sum_{i=1}^d x_i = -\frac{a_{d-1}}{a_d}$.

2) Let $\displaystyle v \in \mathbb{C}$ and $\displaystyle y_1,\ldots,y_d$ satisfy $\displaystyle p(y_i) = v$ $\displaystyle \forall 1 \leq i \leq d$
What is $\displaystyle \sum_{i=1}^d y_i$ ?
• Nov 30th 2010, 06:26 AM
chisigma
1) is...

$\displaystyle \displaystyle p(z)= a_{d}\ z^{d} + a_{d-1}\ z^{d-1} + ... + a_{1}\ z + a_{0} = a_{d}\ \prod_{i=1}^{d} (x-x_{i})$ (1)

... so that...

$\displaystyle \displaystyle \frac{a_{d-1}}{a_{d}} = -\sum_{i=1}^{d} x_{i}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 30th 2010, 06:26 AM
tonio
Quote:

Originally Posted by EinStone
Let $\displaystyle p(z) = a_0 + a_1 z + \ldots + a_d z^d$ be a complex polynomial with coefficients $\displaystyle a_i \in \mathbb{C}$ of degree d.

1) Let $\displaystyle x_1,\ldots,x_d$ be the roots of p. Show that $\displaystyle \sum_{i=1}^d x_i = -\frac{a_{d-1}}{a_d}$.

This is almost trivial if you write $\displaystyle p(z) = a_0 + a_1 z + \ldots + a_d z^d=a_d(z-x_1)\cdot\ldots\cdot(z-x_d)$ and compare

coefficients of respective powers of z.

2) Let $\displaystyle v \in \mathbb{C}$ and $\displaystyle y_1,\ldots,y_d$ satisfy $\displaystyle p(y_i) = v$ $\displaystyle \forall 1 \leq i \leq d$
What is $\displaystyle \sum_{i=1}^d y_i$ ?

Use (1) with the polynomial $\displaystyle g(z) = p(z)-v$

Tonio

.