Worksheet17: Dickson 086
Current Work (editable)
CHAPTER VII Solution of Numerical Equations 74. Horner's Method.[1] After we have isolated a real root of a real equation by one of the methods in Chapter VI, we can compute the root to any desired number of decimal places either by Horner's method, which is available only for polynomial equations, or by Newton's method (§75), which is applicable also to logarithmic, trigonometric, and other equations. To find the root between 2 and 3 of (1) x^{3} - 2x - 5 = 0, set x = 2 + p. Direct substitution gives the transformed equation for p: (2) p^{3} + 6p^{2} + 10p - 1 = 0. The method just used is laborious especially for equations of high degree. We next explain a simpler method. Since p = x - 2, x^{3} - 2x - 5 $$ (x - 2)^{3} + 6(x - 2)^{2} + 10(x - 2) - 1, identically in x. Hence -1 is the remainder obtained when the given polynomial x^{3} - 2x - 5 is divided by x - 2. By inspection, the quotient Q is equal to (x - 2)^{2} + 6(x - 2) + 10. Hence 10 is the remainder obtained when Q is divided by x - 2. The new quotient is equal to (x - 2) + 6, and another division gives the remainder 6. Hence to find the coefficients 6, 10, -1 of the terms following p^{3} in the transformed equation (2), we have only to divide the given polynomial x^{3} - 2x - 5 by x - 2, the quotient Q by x - 2, etc., and take the remainders in reverse order. However, when this work is performed by synthetic division (§15) as tabulated below, no reversal of order is 1 W. G. Horner, London Philosophical Transactions, 1819. Earlier (1804) by P. Ruffini. See Bulletin American Math. Society, May, 1911.
Answer Key (non-editable)
\Chapter{VII}{Solution of Numerical Equations} %[** Attn footnote] \Par{74. Horner's Method.}\footnote {W.~G. Horner, London Philosophical Transactions, 1819. Earlier (1804) by P.~Ruffini. See Bulletin American Math.\ Society, May, 1911.} After we have isolated a real root of a real equation by one of the methods in Chapter VI, we can compute the root to any desired number of decimal places either by Horner's method, which is available only for polynomial equations, or by Newton's method (§75), which is applicable also to logarithmic, trigonometric, and other equations. To find the root between $2$ and $3$ of \[ \Tag{(1)} x^{3} - 2x - 5 = 0, \] set $x = 2 + p$. Direct substitution gives the \emph{transformed equation} for $p$: \[ \Tag{(2)} p^{3} + 6p^{2} + 10p - 1 = 0. \] The method just used is laborious especially for equations of high degree. We next explain a simpler method. Since $p = x - 2$, \[ x^{3} - 2x - 5 \equiv (x - 2)^{3} + 6(x - 2)^{2} + 10(x - 2) - 1, \] identically in $x$. Hence $-1$ is the remainder obtained when the given polynomial $x^{3} - 2x - 5$ is divided by $x - 2$. By inspection, the quotient $Q$ is equal to \[ (x - 2)^{2} + 6(x - 2) + 10. \] Hence $10$ is the remainder obtained when $Q$ is divided by $x - 2$. The new quotient is equal to $(x - 2) + 6$, and another division gives the remainder $6$. Hence to find the coefficients $6$, $10$, $-1$ of the terms following $p^{3}$ in the transformed equation \Eq{(2)}, we have only to divide the given polynomial $x^{3} - 2x - 5$ by $x - 2$, the quotient $Q$ by $x - 2$, etc., and take the remainders in reverse order. However, when this work is performed by synthetic division (§15) as tabulated below, no reversal of order is
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