# Difference between revisions of "Vieta's Formulas"

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Let <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>, | Let <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>, | ||

− | where the coefficient of <math>x^{i}</math> is <math>{a}_i</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that | + | where the coefficient of <math>\displaystyle x^{i}</math> is <math>\displaystyle {a}_i</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>\displaystyle {r}_i</math> are the roots of <math>\displaystyle P(x)</math>. We thus have that |

− | + | <math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math> | |

Expanding out the right hand side gives us | Expanding out the right hand side gives us | ||

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<center><math> a_nx^n - a_n(r_1+r_2+\cdots+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n)x^{n-2} + \cdots + (-1)^na_n r_1r_2\cdots r_n.</math></center> | <center><math> a_nx^n - a_n(r_1+r_2+\cdots+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n)x^{n-2} + \cdots + (-1)^na_n r_1r_2\cdots r_n.</math></center> | ||

− | We can see that the coefficient of <math> x^k </math> will be the <math> k </math>th [[symmetric sum]]. The <math>k</math>th symmetric sum is just the sum of the roots taken <math>k</math> at a time. For example, the 4th symmetric sum is <math>\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.</math> Notice that every possible [[combination]] of four roots shows up in this sum. | + | We can see that the coefficient of <math>\displaystyle x^k </math> will be the <math> \displaystyle k </math>th [[symmetric sum]]. The <math>k</math>th symmetric sum is just the sum of the roots taken <math>k</math> at a time. For example, the 4th symmetric sum is <math>\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.</math> Notice that every possible [[combination]] of four roots shows up in this sum. |

− | We now have two different expressions for <math>P(x)</math>. These ''must'' be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that | + | We now have two different expressions for <math>\displaystyle P(x)</math>. These ''must'' be equal. However, the only way for two polynomials to be equal for all values of <math>\displaystyle x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math>\displaystyle x^n </math>, we see that |

<center><math>\displaystyle a_n = a_n</math></center> | <center><math>\displaystyle a_n = a_n</math></center> | ||

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<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center> | <center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center> | ||

− | More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>). | + | More commonly, these are written with the roots on one side and the <math>\displaystyle a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>\displaystyle a_n</math>). |

− | If we denote <math>\sigma_k</math> as the <math>k</math>th symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math> | + | If we denote <math>\displaystyle \sigma_k</math> as the <math>\displaystyle k</math>th symmetric sum, then we can write those formulas more compactly as <math>\displaystyle \sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>\displaystyle 1\le k\le {n}</math>. |

== See also == | == See also == |

## Revision as of 13:56, 24 July 2006

## Introduction

Let , where the coefficient of is . As a consequence of the Fundamental Theorem of Algebra, we can also write , where are the roots of . We thus have that

Expanding out the right hand side gives us

We can see that the coefficient of will be the th symmetric sum. The th symmetric sum is just the sum of the roots taken at a time. For example, the 4th symmetric sum is Notice that every possible combination of four roots shows up in this sum.

We now have two different expressions for . These *must* be equal. However, the only way for two polynomials to be equal for all values of is for each of their corresponding coefficients to be equal. So, starting with the coefficient of , we see that

More commonly, these are written with the roots on one side and the on the other (this can be arrived at by dividing both sides of all the equations by ).

If we denote as the th symmetric sum, then we can write those formulas more compactly as , for .