Worksheet11: Dickson 058
Current Work (editable)
In particular, if P is a bend point, the slope of the (horizontal) tangent at P is zero, whence 3x^{2} + 8x = 0, x = 0 or x = -8/3. Equation (3) gives the corresponding values of y. The resulting points M = (0, -11), M' = (-8/3, -41/27) are easily shown to be bend points. Indeed, for x > 0 and for x between -4 and 0, x^{2} (x + 4) is positive, and hence f(x) > -11 for such values of x, so that the function (3) has a relative minimum at x = 0. Similarly, there is a relative maximum at x = -8/3. We may also employ the general method of §59 to show that M and M' are bend points. Since these bend points are both below the x-axis we are now certain that the graph crosses the x-axis only once. The use of the bend points insures greater accuracy to the graph than the use of dozens of points whose abscissas are taken at random. 56. Derivatives. We shall now find the slope of the tangent to the graph of y = f(x), where f(x) is any polynomial (5) f(x) = a_{0} x^{n} + a_{1} x^{n-1} + ... + a_{n-1} x + a_{n}. We need the expansion of f(x + h) in powers of x. By the binomial theorem, a_{0} (x + h)^n = a_{0} x^{n} + na_{0} x^{n-1} h + n(n - 1)/2 a_{0} x^{n-2} h^{2} + ..., a_{1} (x + h)^{n-1} = a_{1} x^{n-1} + (n - 1) a_{1} x^{n-2} h + (n - 1)(n - 2)/2 a_{1} x^{n-3} h^{2} + ..., ......... a_{n-2} (x + h)^{2} = a_{n-2} x^{2} + 2a_{n-2} xh + a_{n-2} h^{2}, a_{n-1} (x + h) = a_{n-1} x + a_{n-1} h, a_{n} = a_{n}. The sum of the left members is evidently f(x + h). On the right, the sum of the first terms (i.e., those free of h) is f(x). The sum of the coefficients of h is denoted by f'(x), the sum of the coefficients of 1/2 h^{2} is denoted by f''(x), ..., the sum of the coefficients of h^{k}/1·2 ... k
Answer Key (non-editable)
In particular, if $P$ is a bend point, the slope of the (horizontal) tangent at $P$ is zero, whence $3x^{2} + 8x = 0$, $x = 0$ or $x = -\frac{8}{3}$. Equation \Eq{(3)} gives the corresponding values of $y$. The resulting points \[ M = (0, -11),\qquad M' = (-\tfrac{8}{3}, -\tfrac{41}{27}) \] are easily shown to be bend points. Indeed, for $x > 0$ and for $x$ between $-4$ and $0$, $x^{2}(x + 4)$ is positive, and hence $f(x) > -11$ for such values of $x$, so that the function \Eq{(3)} has a relative minimum at $x = 0$. Similarly, there is a relative maximum at $x = -\frac{8}{3}$. We may also employ the general method of §59 to show that $M$ and $M'$ are bend points. Since these bend points are both below the $x$-axis we are now certain that the graph crosses the $x$-axis only once. The use of the bend points insures greater accuracy to the graph than the use of dozens of points whose abscissas are taken at random. \Par{56. Derivatives.} We shall now find the slope of the tangent to the graph of $y = f(x)$, where $f(x)$ is any polynomial \[ \Tag{(5)} f(x) = a_{0} x^{n} + a_{1} x^{n-1} + \dots + a_{n-1} x + a_{n}. \] We need the expansion of $f(x + h)$ in powers of $x$. By the binomial theorem, %[** Attn alignment] \begin{align*} a_{0} (x + h)^n &= a_{0} x^{n} + na_{0} x^{n-1} h + \frac{n(n - 1)}{2} a_{0} x^{n-2} h^{2} + \dots, \\ a_{1} (x + h)^{n-1} &= a_{1} x^{n-1} + (n - 1)a_{1} x^{n-2} h + \frac{(n - 1)(n - 2)}{2} a_{1} x^{n-3} h^{2} + \dots, \\ \multispan{2}{\dotfill} \\ a_{n-2} (x + h)^{2} &= a_{n-2} x^{2} + 2a_{n-2} xh + a_{n-2} h^{2}, \\ a_{n-1} (x + h) &= a_{n-1} x + a_{n-1} h, \\ a_{n} &= a_{n}. \end{align*} The sum of the left members is evidently $f(x + h)$. On the right, the sum of the first terms (i.e., those free of $h$) is $f(x)$. The sum of the coefficients of $h$ is denoted by $f'(x)$, the sum of the coefficients of $\frac{1}{2} h^{2}$ is denoted by $f''(x)$, \dots, the sum of the coefficients of \[ \frac{h^{k}}{1·2 \dots k} \]
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