Worksheet04: Dickson 016
Current Work (editable)
to have n distinct roots \alpha_{1}, ..., \alpha_{n}. In f(x) $$ (x - \alpha_{1}) Q(x) take x = \alpha_{2}; then 0 = (\alpha_{2} - \alpha_{1}) Q(\alpha_{2}), whence Q(\alpha_{2}) = 0 and Q(x) = 0 has the root \alpha_{2}. Similarly, Q_{1}(x) = 0 has the root \alpha_{3}, etc. Thus all of the assumptions (each introduced by an "if") made in the above discussion have been justified and we have the conclusion (5). Hence if an equation f(x) = 0 of degree n has n distinct roots \alpha_{1}, ..., \alpha_{n}, f(x) can be expressed in the factored form (5). It follows readily that the equation can not have a root \alpha different from \alpha_{1}, ..., \alpha_{n}. For, if it did, the left member of (5) is zero when x = \alpha and hence one of the factors of the right member must then be zero, say \alpha - \alpha_{j} = 0, whence the root \alpha is equal to \alpha_{j}. We have now proved the following important result. Theorem. An equation of degree n cannot have more than n distinct roots. 17. Multiple Roots.[1] Equalities may occur among the \alpha's in (5). Suppose that exactly m_{1} of the \alpha's (including \alpha_{1}) are equal to \alpha_{1}; that \alpha_{2} $$ \alpha_{1}, while exactly m_{2} of the \alpha's are equal to \alpha_{2}; etc. Then (5) becomes (6) f(x) $$ c_{0}(x - \alpha_{1})^{m_{1}} (x - \alpha_{2})^{m_{2}} ... (x - \alpha_{k})^{m_{k}}, m_{1} + m_{2} + ... + m_{k} = n, where \alpha_{1}, ..., \alpha_{k} are distinct. We then call \alpha_{1} a root of multiplicity m_{1} of f(x) = 0, \alpha_{2} a root of multiplicity m_{2}, etc. In other words, \alpha_{1} is a root of multiplicity m_{1} of f(x) = 0 if f(x) is exactly divisible by (x - \alpha_{1})^{m_{1}}, but is not divisible by (x - \alpha_{1})^{m_{1} + 1}. We call \alpha_{1} also an m_{1}-fold root. In the particular cases m_{1} = 1, 2, and 3, we also speak of \alpha_{1} as a simple root, double root, and triple root, respectively. For example, 4 is a simple root, 3 a double root, -2 a triple root, and 6 a root of multiplicity 4 (or a 4-fold root) of the equation 7(x - 4) (x - 3)^{2} (x + 2)^{3} (x - 6)^{4} = 0 of degree 10 which has no further root. This example illustrates the next theorem, which follows from (6) exactly as the theorem in §16 followed from (5). Theorem. An equation of degree n cannot have more than n roots, a root of multiplicity m being counted as m roots. 18. Identical Polynomials. If two polynomials in x, a_{0} x^{n} + a_{1} x^{n-1} + ... + a_{n}, b_{0} x^{n} + b_{1} x^{n-1} + ... + b_{n}, 1 Multiple roots are treated by calculus in §58.
Answer Key (non-editable)
to have $n$ distinct roots $\alpha_{1}$, \dots, $\alpha_{n}$. In $f(x) \equiv (x - \alpha_{1}) Q(x)$ take $x = \alpha_{2}$; then $0 = (\alpha_{2} - \alpha_{1}) Q(\alpha_{2})$, whence $Q(\alpha_{2}) = 0$ and $Q(x) = 0$ has the root $\alpha_{2}$. Similarly, $Q_{1}(x) = 0$ has the root $\alpha_{3}$, etc. Thus all of the assumptions (each introduced by an ``if'') made in the above discussion have been justified and we have the conclusion \Eq{(5)}. Hence \begin{Thm}if an equation $f(x) = 0$ of degree $n$ has $n$ distinct roots $\alpha_{1}$, \dots, $\alpha_{n}$, $f(x)$ can be expressed in the factored form \Eq{(5)}\end{Thm}. It follows readily that the equation can not have a root $\alpha$ different from $\alpha_{1}$, \dots, $\alpha_{n}$. For, if it did, the left member of \Eq{(5)} is zero when $x = \alpha$ and hence one of the factors of the right member must then be zero, say $\alpha - \alpha_{j} = 0$, whence the root $\alpha$ is equal to $\alpha_{j}$. We have now proved the following important result. \begin{Theorem} An equation of degree $n$ cannot have more than $n$ distinct roots. \end{Theorem} % [** Attn: Footnote] \Par{17. Multiple Roots.}\footnote {Multiple roots are treated by calculus in §58.} Equalities may occur among the $\alpha$'s in \Eq{(5)}. Suppose that exactly $m_{1}$ of the $\alpha$'s (including $\alpha_{1}$) are equal to $\alpha_{1}$; that $\alpha_{2} \neq \alpha_{1}$, while exactly $m_{2}$ of the $\alpha$'s are equal to $\alpha_{2}$; etc. Then \Eq{(5)} becomes \[ \Tag{(6)} f(x) \equiv c_{0}(x - \alpha_{1})^{m_{1}} (x - \alpha_{2})^{m_{2}} \dots (x - \alpha_{k})^{m_{k}}, \quad m_{1} + m_{2} + \dots + m_{k} = n, \] where $\alpha_{1}$, \dots, $\alpha_{k}$ are distinct. We then call $\alpha_{1}$ a \emph{root of multiplicity} $m_{1}$ of $f(x) = 0$, $\alpha_{2}$ a root of multiplicity $m_{2}$, etc. In other words, $\alpha_{1}$ is a root of multiplicity $m_{1}$ of $f(x) = 0$ if $f(x)$ is exactly divisible by $(x - \alpha_{1})^{m_{1}}$, but is not divisible by $(x - \alpha_{1})^{m_{1} + 1}$. We call $\alpha_{1}$ also an \emph{$m_{1}$-fold root}. In the particular cases $m_{1} = 1$, $2$, and $3$, we also speak of $\alpha_{1}$ as a \emph{simple root}, \emph{double root}, and \emph{triple root}, respectively. For example, $4$ is a simple root, $3$ a double root, $-2$ a triple root, and $6$ a root of multiplicity $4$ (or a $4$-fold root) of the equation \[ 7(x - 4) (x - 3)^{2} (x + 2)^{3} (x - 6)^{4} = 0 \] of degree $10$ which has no further root. This example illustrates the next theorem, which follows from \Eq{(6)} exactly as the theorem in §16 followed from \Eq{(5)}. \begin{Theorem} An equation of degree $n$ cannot have more than $n$ roots, a root of multiplicity $m$ being counted as $m$ roots. \end{Theorem} \Par{18. Identical Polynomials.} \begin{Thm} If two polynomials in $x$, \[ a_{0} x^{n} + a_{1} x^{n-1} + \dots + a_{n},\qquad b_{0} x^{n} + b_{1} x^{n-1} + \dots + b_{n}, \] \end{Thm}%[** Continues]
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