Difference between revisions of "Vieta's Formulas"
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| − | == | + | == Introduction == |
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>. | + | 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 |
| − | + | <center><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></center> | |
| − | + | Expanding out the right hand side gives us | |
| − | <math>\ | + | <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. | |
| − | + | So 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 | |
| − | === See also | + | <center><math> a_n = a_n</math></center> |
| + | <center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center> | ||
| + | <center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center> | ||
| + | <center><math>\vdots</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 just divide both sides of all the equations by <math>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>{}1\le k\le {n}</math>. | ||
| + | |||
| + | == See also == | ||
* [[Algebra]] | * [[Algebra]] | ||
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* [[Newton's identities]] | * [[Newton's identities]] | ||
| − | + | == Related Links == | |
[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article] | [http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article] | ||
Revision as of 10:06, 22 June 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.
So 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 just divide 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 
.
