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MaiaMitchell

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UsernameMaiaMitchell
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Questions 44
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MaiaMitchell  Oct 3, 2015
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MaiaMitchell  Oct 3, 2015
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Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as $x^2+ax+b=(x-p)(x-q)$ (Note that the first term is $x^2$, not $ax^2$.) Using the distributive property to expand the right side we get $x^2+ax+b=x^2-(p+q)x+pq$ We know that two polynomials are equal if and only if their coefficients are equal, so $x^2+ax+b=x^2-(p+q)x+pq$ means that $a=-(p+q)$ and $b=pq$. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the $x$ term. A similar set of relations for cubics can be found by expanding $x^3+ax^2+bx+c=(x-p)(x-q)(x-r)$. We can state Vieta's formula's more rigorously and generally. Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where ${r}_i$ are the roots of $P(x)$. We thus have that $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$ Expanding out the right hand side gives us $a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$ The coefficient of $x^k$ in this expression will be the $k$th symmetric sum of the $r_i$. We now have two different expressions for $P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $x^n$, we see that $a_n = a_n$ $a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$ $a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$ $\vdots$ $a_0 = (-1)^n a_n r_1r_2\cdots r_n$ More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$). If we denote $\sigma_k$ as the $k$th symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Problems Beginner Let $r_1,r_2,$ and $r_3$ be the three roots of the cubic $x^3+3x^2+4x-4$. Find the value of $r_1r_2+r_1r_3+r_2r_3$. Suppose the polynomial $5x^3+4x^2-8x+6$ has three real roots $a,b$, and $c$. Find the value of $a(1+b+c)+b(1+a+c)+c(1+a+b)$. Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$. Find $(m + 7)(n + 7)$. Intermediate Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$ And The $k$-th elementary symmetric sum of a set of $n$ numbers is the sum of all products of $k$ of those numbers ($1 \leq k \leq n$). For example, if $n = 4$, and our set of numbers is $\{a, b, c, d\}$, then: 1st Symmetric Sum = $S_1 = a+b+c+d$ 2nd Symmetric Sum = $S_2 = ab+ac+ad+bc+bd+cd$ 3rd Symmetric Sum = $S_3 = abc+abd+acd+bcd$ 4th Symmetric Sum = $S_4 = abcd$ Notation The first elementary symmetric sum of $f(x)$ is often written $\sum_{sym}f(x)$. The $n$th can be written $\sum_{sym}^{n}f(x)$ Uses Any symmetric sum can be written as a polynomial of the elementary symmetric sum functions. For example, $x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - xz) + 3xyz = S_1^3 - 3S_1S_2 + 3S_3$. This is often used to solve systems of equations involving power sums, combined with Vieta's formulas. Elementary symmetric sums show up in Vieta's formulas. In a monic polynomial, the coefficient of the $x^1$ term is $e_1$, and the coefficient of the $x^k$ term is $e_k$, where the symmetric sums are taken over the roots of the polynomial.(done finally)

Aug 9, 2015