Worksheet13: Dickson 068
Current Work (editable)
EXERCISES 1. Prove that 8x^{3} - 4x^{2} - 18x + 9 = 0 has a root between 0 and 1, one between 1 and 2, and one between -2 and -1. 2. Prove that 16x^{4} - 24x^{2} + 16x - 3 = 0 has a triple root between 0 and 1, and a simple root between -2 and -1. 3. Prove that if a < b < c ... < l, and \alpha, \beta, ..., \lambda are positive, these quantities being all real, \alpha/x - a + \beta/x - b + \gamma/x - c + ... + \lambda/x - l + t = 0 has a real root between a and b, one between b and c, ... one between k and l, and if t is negative one greater than l, but if t is positive one less than a. 4. Verify that the equation in Ex. 3 has no imaginary root by substituting r + si and r - si in turn for x, and subtracting the results. 5. Admitting that an equation f(x) $$ x^{n} + ... = 0 with real coefficients has n roots, show algebraically that there is a real root between a and b if f(a) and f(b) have opposite signs. Note that a pair of conjugate imaginary roots c ± di are the roots of (x - c)^{2} + d^{2} = 0 and that this quadratic function is positive if x is real. Hence if x_{1}, ..., x_{r} are the real roots and \phi(x) $$ (x - x_{1}) ... (x - x_{r}), then \phi(a) and \phi(b) have opposite signs. Thus a - x_{i} and b - x_{i} have opposite signs for at least one real root x_{i}. (Lagrange.) 64. Sign of a Polynomial. Given a polynomial f(x) = a_{0} x^{n} + a_{1} x^{n-1} + ... + a_{n} (a_{0} $$ 0) with real coefficients, we can find a positive number P such that f(x) has the same sign as a_{0} x^{n} when x > P. In fact, f(x) = x^{n} (a_{0} + \phi), \phi = a_{1}/x + a_{2}/x^{2} + ... + a_{n}/x^{n}. By the result in §62, the numerical value of \phi is less than that of a_{0} when 1/x is positive and less than a sufficiently small positive number, say 1/P, and hence when x > P. Then a_{0} + \phi has the same sign as a_{0}, and hence f(x) the same sign as a_{0} x^{n}. The last result holds also when x is a negative number sufficiently large numerically. For, if we set x = -X, the former case shows that f(-X) has the same sign as (-1)^{n} a_{0} X^{n} when X is a sufficiently large positive number.
Answer Key (non-editable)
\begin{Exercises} \begin{itemize} \item[1.] Prove that $8x^{3} - 4x^{2} - 18x + 9 = 0$ has a root between $0$ and $1$, one between $1$ and $2$, and one between $-2$ and $-1$. \item[2.] Prove that $16x^{4} - 24x^{2} + 16x - 3 = 0$ has a triple root between $0$ and $1$, and a simple root between $-2$ and $-1$. \item[3.] Prove that if $a < b < c \dots < l$, and $\alpha$, $\beta$, \dots, $\lambda$ are positive, these quantities being all real, \[ \frac{\alpha}{x - a} + \frac{\beta}{x - b} + \frac{\gamma}{x - c} + \dots + \frac{\lambda}{x - l} + t = 0 \] has a real root between $a$ and $b$, one between $b$ and $c$, \dots\ one between $k$ and $l$, and if $t$ is negative one greater than $l$, but if $t$ is positive one less than $a$. \item[4.] Verify that the equation in Ex.~3 has no imaginary root by substituting $r + si$ and $r - si$ in turn for $x$, and subtracting the results. \item[5.] Admitting that an equation $f(x) \equiv x^{n} + \dots = 0$ with real coefficients has $n$ roots, show algebraically that there is a real root between $a$ and $b$ if $f(a)$ and $f(b)$ have opposite signs. Note that a pair of conjugate imaginary roots $c ± di$ are the roots of \[ (x - c)^{2} + d^{2} = 0 \] and that this quadratic function is positive if $x$ is real. Hence if $x_{1}$, \dots, $x_{r}$ are the real roots and \[ \phi(x) \equiv (x - x_{1}) \dots (x - x_{r}), \] then $\phi(a)$ and $\phi(b)$ have opposite signs. Thus $a - x_{i}$ and $b - x_{i}$ have opposite signs for at least one real root $x_{i}$. (Lagrange.) \end{itemize} \end{Exercises} \Par{64. Sign of a Polynomial.} Given a polynomial \[ %[** Attn alignment] f(x) = a_{0} x^{n} + a_{1} x^{n-1} + \dots + a_{n} \qquad (a_{0} \neq 0) \] with real coefficients, we can find a positive number $P$ such that $f(x)$ has the same sign as $a_{0} x^{n}$ when $x > P$. In fact, \[ f(x) = x^{n} (a_{0} + \phi),\qquad \phi = \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} + \dots + \frac{a_{n}}{x^{n}}. \] By the result in §62, the numerical value of $\phi$ is less than that of $a_{0}$ when $1/x$ is positive and less than a sufficiently small positive number, say $1/P$, and hence when $x > P$. Then $a_{0} + \phi$ has the same sign as $a_{0}$, and hence $f(x)$ the same sign as $a_{0} x^{n}$. The last result holds also when $x$ is a negative number sufficiently large numerically. For, if we set $x = -X$, the former case shows that $f(-X)$ has the same sign as $(-1)^{n} a_{0} X^{n}$ when $X$ is a sufficiently large positive number.
Comparison of Current Work and Answer Key
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