03. 뉴턴의 방법Newton's Method

Solution. (1) If $f(x) = x^2 - 2$, then $f'(x) = 2x$, and Newton's iteration is $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^2-2}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right).$$ Starting from $x_0 = 1$ and iterating gives

Hence $x_n \to \sqrt{2} \approx 1.414214$, which is the approximate value of the positive root of $f(x)=x^2-2$.

(2) The $x$-coordinate of the intersection point of $y = x^3-x$ and $y=1$ is a root of $x^3-x = 1$, i.e., $x^3-x-1=0$. If $f(x) = x^3-x-1$, then $f'(x)=3x^2-1$, and the iteration is $$x_{n+1} = x_n - \frac{x_n^3-x_n-1}{3x_n^2-1}.$$ Starting from $x_0=1$ and iterating gives

Hence $x_n \to 1.324718$, which is the approximate $x$-coordinate of the intersection point.

Solution. (1) If $f(x) = -x^2+2x+1$, then $f'(x) = -2x+2$, and Newton's iteration is $$x_{n+1} = x_n - \frac{-x_n^2+2x_n+1}{-2x_n+2}.$$

Starting from $x_0=0$ gives

Hence $x_n \to 1-\sqrt{2} \approx -0.414214$.

Starting from $x_0=2$ gives

Hence $x_n \to 1+\sqrt{2} \approx 2.414214$. Indeed, the two roots of $-x^2+2x+1=0$, i.e., $x^2-2x-1=0$, are $x = 1\pm\sqrt{2}$ by the quadratic formula, so the two approximations above correspond exactly to these roots.

(2) The $x$-coordinate of the intersection point of $y=\tan x$ and $y=2x$ is a root of $f(x) = \tan x - 2x = 0$. Since $f'(x) = \sec^2 x - 2$, the iteration is $$x_{n+1} = x_n - \frac{\tan x_n - 2x_n}{\sec^2 x_n - 2}.$$ Starting from $x_0=1$ (which satisfies $0 \lt x_0 \lt \pi/2$) and iterating gives

Hence $x_n \to 1.165561$, which is the approximate $x$-coordinate of the intersection point lying between $0$ and $\pi/2$.

Solution. Let $f(x) = \tan x$, so $f'(x) = \sec^2 x$, and Newton's iteration is $$x_{n+1} = x_n - \frac{\tan x_n}{\sec^2 x_n} = x_n - \sin x_n \cos x_n = x_n - \frac{1}{2}\sin(2x_n).$$ Starting from $x_0 = 3$ and iterating gives

Hence $x_n \to \pi \approx 3.141593$. The roots of $\tan x = 0$ are of the form $x = k\pi$ for an integer $k$, and since $x_0=3$ is closest to $\pi \approx 3.14159$ among these, the iteration converges to that root.

Solution. If $f(x) = \dfrac{1}{x} - 3$, then $f'(x) = -\dfrac{1}{x^2}$, and Newton's iteration is $$x_{n+1} = x_n - \frac{\frac1{x_n}-3}{-\frac1{x_n^2}} = x_n + x_n^2\left(\frac1{x_n}-3\right) = 2x_n - 3x_n^2 = x_n(2-3x_n).$$ Since $f$ is defined only for $x \neq 0$, starting from $x_0 = 0.5$, which is close to the root $1/3$, and iterating gives

Hence $x_n \to \dfrac13 \approx 0.333333$, which agrees with the root of $f(x)=1/x-3=0$.

Proof. If $f(x) = x^{1/3}$, then $f'(x) = \dfrac13 x^{-2/3}$ for $x \neq 0$. For $x_n \neq 0$, Newton's iteration is $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^{1/3}}{\frac13 x_n^{-2/3}} = x_n - 3x_n^{1/3+2/3} = x_n - 3x_n = -2x_n.$$

Starting from $x_0=1$, one shows by induction that $x_n = (-2)^n x_0 = (-2)^n$: indeed $x_0 = 1 = (-2)^0$, and if $x_n=(-2)^n$ then $x_{n+1} = -2x_n = -2\cdot(-2)^n = (-2)^{n+1}$.

Hence $|x_n| = 2^n \to \infty$, so the sequence $\{x_n\}$ diverges and does not converge to the root $x=0$ of $f$. This happens because $f'(0)$ does not exist — the graph of $f$ has a nearly vertical tangent line near the root — so each Newton step moves the iterate to twice the distance from the origin instead of closer to it. Therefore Newton's method does not apply to $f(x)=x^{1/3}$ when $x_0=1$.

Proof. If $f(x) = 4x^4-4x^2 = 4x^2(x^2-1)$, then $f'(x) = 16x^3-8x = 8x(2x^2-1)$. For $x_n \neq 0$ and $x_n^2 \neq \frac12$, Newton's iteration simplifies to $$x_{n+1} = x_n - \frac{4x_n^2(x_n^2-1)}{8x_n(2x_n^2-1)} = x_n - \frac{x_n(x_n^2-1)}{2(2x_n^2-1)} = \frac{x_n\left[2(2x_n^2-1)-(x_n^2-1)\right]}{2(2x_n^2-1)} = \frac{x_n(3x_n^2-1)}{2(2x_n^2-1)}.$$

(1) If $x_0 = \sqrt{3/7}$, then $x_0^2 = 3/7$, so $$3x_0^2-1 = \frac97-1=\frac27, \qquad 2x_0^2-1=\frac67-1=-\frac17,$$ and hence $$x_1 = \frac{x_0 \cdot \frac27}{2\cdot\left(-\frac17\right)} = \frac{x_0\cdot\frac27}{-\frac27} = -x_0.$$ Since $x_1^2 = x_0^2 = 3/7$, applying the same computation to $x_1$ gives $x_2 = -x_1 = x_0$. That is, $x_n^2 = x_0^2 = 3/7$ holds for every $n$, so the coefficient $\dfrac{3x_n^2-1}{2(2x_n^2-1)}=-1$ persists throughout, giving $$x_{2k} = x_0, \qquad x_{2k+1} = -x_0 \qquad (k=0,1,2,\dots).$$ Note that $f(x_0) = 4x_0^2(x_0^2-1) = 4\cdot\frac37\cdot\left(\frac37-1\right) = -\frac{48}{49} \neq 0$, so $x_0$ is not a root of $f$. Hence the Newton iterates never converge to a root — they oscillate forever between $x_0$ and $-x_0$.

(2) If $x_0 = \sqrt{13/28} \approx 0.681385$, this exact cancellation no longer occurs. Computing directly from the iteration formula gives

Hence $x_n \to -1$, and $x=-1$ is indeed a root of $f(x)=4x^2(x^2-1)$ (since $f(-1)=4\cdot1\cdot0=0$). Thus a small change in the initial value — from $\sqrt{3/7}$ in part (1) to $\sqrt{13/28}$ in part (2) — can make Newton's method behave completely differently: perpetual oscillation versus convergence to an actual root.