Modified Newton’s Method for a Root and a Minimum: 04-BS-5, May 2016, Q6

Question 6: Modified Newton’s Method for a Root and a Minimum (A: 10, B: 10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x)=x^{4}-3x^{3}-5x^{2}-4x+6$; (A) a root near $x_0=4.0$; (B) a local minimum near $x_0=3$.

Find. (A) Two iterates $x_1,x_2$ of the given second-order (Chebyshev-type) formula. (B) The coordinates $(x^{*},f(x^{*}))$ of the minimum.

Approach. (A) Apply the supplied formula directly, using $f,f',f''$ evaluated at each iterate. (B) A smooth minimum is a root of $f'(x)=0$; apply ordinary Newton’s method to $f'$, i.e. $x_{i+1}=x_i-f'(x_i)/f''(x_i)$, iterating to convergence.

Derivatives. $f'(x)=4x^{3}-9x^{2}-10x-4$, $f''(x)=12x^{2}-18x-10$.
(A) First iterate from $x_0=4.0$. $f(4)=-26$, $f'(4)=124$, $f''(4)=110$: $$x_1=4-\frac{-26}{124}-\frac{(-26)^{2}(110)}{2(124)^{3}}=\boxed{4.2641080}$$
(A) Second iterate. Repeating from $x_1=4.2641080$ ($f(x_1)=2.749917$, $f'(x_1)=161.4919$, $f''(x_1)=137.7654$): $$x_2=\boxed{4.3027326}$$ The true root is $4.3027756$, so two iterates of this cubic-order method already agree to 4 decimal places — markedly faster than plain Newton would.
(B) Newton’s method on $f'(x)=0$ from $x_0=3$. Iterating $x_{i+1}=x_i-f'(x_i)/f''(x_i)$: $$3.0000000\to3.1590909\to3.1458670\to3.1457702\to3.1457702\ (\text{converged})$$ $$\boxed{x^{*}=3.1457702}$$
(B) Coordinates and classification. $f(x^{*})=-51.5245858$ and $f''(x^{*})=52.1265778\gt0$, confirming a genuine local minimum: $$\boxed{(x^{*},f(x^{*}))=(3.1457702,\ -51.5245858)}$$

Fig. Q6: $f(x)=x^4-3x^3-5x^2-4x+6$ near the root ($x\approx4.3028$, red) and the local minimum ($x\approx3.1458$, green).

Final results — Question 6
Quantity	Result
(A) $x_1$	$4.2641080$
(A) $x_2$	$4.3027326$ (true root $4.3027756$)
(B) minimum $x^{*}$	$3.1457702$
(B) $f(x^{*})$	$-51.5245858$

Topic: Higher-order (Chebyshev-type) root-finding, and locating a stationary point via Newton’s method on the derivative.

Key relations: plain Newton $x_{i+1}=x_i-f/f'$; the given second-order correction $-\frac{[f]^{2}f''}{2[f']^{3}}$; a smooth local min/max is a root of $f'(x)=0$, located by $x_{i+1}=x_i-f'(x_i)/f''(x_i)$.

Why this works: the extra curvature term cancels the next Taylor-series error term beyond plain Newton, giving cubic (not merely quadratic) convergence — visible here since two iterates already match the true root to four decimals; finding a minimum by Newton-on-the-derivative is the same algorithm one derivative order up, using $f',f''$ in place of $f,f'$.

Common pitfall: using the wrong sign on the correction term, or, in part (B), applying Newton’s method to $f(x)$ itself (which finds a root of $f$, not a minimum) instead of to $f'(x)$.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch.19.2 (Newton’s Method, iteration methods for equations).

Question 6 of 7: Modified Newton’s Method for a Root and a Minimum

Question 6: Modified Newton’s Method for a Root and a Minimum (A: 10, B: 10 marks)