Taylor Polynomials & Newton's Method

$f = \log x$, $f' = \dfrac{1}{x}$, $f'' = -\dfrac{1}{x^2}$, $f''' = \dfrac{2}{x^3}$.

$f(1) = 0$, $f'(1) = 1$, $f''(1) = -1$, $f'''(1) = 2$.

Coefficients: $\dfrac{0}{1} = 0$, $\dfrac{1}{1} = 1$, $\dfrac{-1}{2}$, $\dfrac{2}{6} = \dfrac{1}{3}$.

$T_3(x) = (x - 1) - \dfrac{1}{2}(x - 1)^2 + \dfrac{1}{3}(x - 1)^3$.

$f = x^{-1}$, $f' = -x^{-2}$, $f'' = 2x^{-3}$, $f''' = -6x^{-4}$.

$f(2) = \tfrac{1}{2}$, $f'(2) = -\tfrac{1}{4}$, $f''(2) = \tfrac{2}{8} = \tfrac{1}{4}$, $f'''(2) = -\tfrac{6}{16} = -\tfrac{3}{8}$.

Coefficients: $\tfrac{1}{2}$, $-\tfrac{1}{4}$, $\tfrac{1}{4} \div 2 = \tfrac{1}{8}$, $-\tfrac{3}{8} \div 6 = -\tfrac{1}{16}$.

$T_3(x) = \tfrac{1}{2} - \tfrac{1}{4}(x - 2) + \tfrac{1}{8}(x - 2)^2 - \tfrac{1}{16}(x - 2)^3$.

$f = \cos x$, $f' = -\sin x$, $f'' = -\cos x$.

$f\!\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2}$, $f'\!\left(\tfrac{\pi}{3}\right) = -\tfrac{\sqrt{3}}{2}$, $f''\!\left(\tfrac{\pi}{3}\right) = -\tfrac{1}{2}$.

Coefficients: $\tfrac{1}{2}$, $-\tfrac{\sqrt{3}}{2}$, $-\tfrac{1}{2} \div 2 = -\tfrac{1}{4}$.

$T_2(x) = \tfrac{1}{2} - \tfrac{\sqrt{3}}{2}\left(x - \tfrac{\pi}{3}\right) - \tfrac{1}{4}\left(x - \tfrac{\pi}{3}\right)^2$.

$f(x) = \sqrt{x}$, $a = 25$.

$f(25) = 5$; $f'(x) = \dfrac{1}{2\sqrt{x}}$, $f'(25) = \dfrac{1}{10}$; $f''(x) = -\dfrac{1}{4x^{3/2}}$, $f''(25) = -\dfrac{1}{500}$.

$T_2(x) = 5 + \dfrac{1}{10}(x - 25) - \dfrac{1}{1000}(x - 25)^2$.

$T_2(26) = 5 + 0.1 - 0.001 = 5.099$.

$f(x) = e^x$, $a = 0$.

Every derivative is $e^x$, so $f(0) = f'(0) = f''(0) = f'''(0) = 1$.

$T_3(x) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$.

$T_3(0.2) = 1 + 0.2 + 0.02 + \dfrac{0.008}{6} = \dfrac{458}{375} \approx 1.22133$.

$f(x) = \log x$, $a = 1$.

$f(1) = 0$, $f'(1) = 1$, $f''(1) = -1$.

$T_2(x) = (x - 1) - \dfrac{1}{2}(x - 1)^2$.

$T_2(1.2) = 0.2 - \dfrac{1}{2}(0.04) = 0.18$.

$n = 2$, $a = 25$, $x = 26$.

$f'''(x) = \dfrac{3}{8}x^{-5/2}$.

$f'''$ is decreasing, so on $[25, 26]$ it is largest at $25$: $M = \dfrac{3}{8 \cdot 25^{5/2}} = \dfrac{3}{8 \cdot 3125} = \dfrac{3}{25000}$.

$|R_2(26)| \le \dfrac{3}{25000} \cdot \dfrac{1^3}{3!} = \dfrac{1}{50000} = 0.00002$.

$n = 3$, $a = 0$, $x = 0.2$.

$f^{(4)}(x) = e^x$.

For $0 \le c \le 0.2$: $e^c \le e^{0.2} < e < 3$, so $M = 3$.

$|R_3(0.2)| \le \dfrac{3}{4!}(0.2)^4 = \dfrac{3 \cdot 0.0016}{24} = 0.0002$.

$n = 3$, $a = 0$, $x = 0.3$. The estimate: $0.3 - \dfrac{0.027}{6} = 0.2955$.

$f^{(4)}(x) = \sin x$.

$|\sin c| \le 1$ for every $c$, so $M = 1$.

$|R_3(0.3)| \le \dfrac{1}{24}(0.3)^4 = \dfrac{0.0081}{24} = 0.0003375$.

$e^u$ with $u = -x^2$.

$e^u \approx 1 + u + \dfrac{u^2}{2} + \dfrac{u^3}{6}$ (degree $3$ in $u$ is degree $6$ in $x$).

$1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6}$.

Nothing more to do: that is the polynomial.

$\cos u$ with $u = 3x$, then multiply by $x$.

$\cos u \approx 1 - \dfrac{u^2}{2} + \dfrac{u^4}{24}$.

$\cos 3x \approx 1 - \dfrac{9x^2}{2} + \dfrac{81x^4}{24} = 1 - \dfrac{9x^2}{2} + \dfrac{27x^4}{8}$.

Multiply by $x$: $x - \dfrac{9x^3}{2} + \dfrac{27x^5}{8}$.

$\sin u$ with $u = 3x$.

$\sin u = u - \dfrac{u^3}{6} + \dots$

$\sin 3x = 3x - \dfrac{27x^3}{6} + \dots = 3x - \dfrac{9}{2}x^3 + (\text{terms in } x^5, x^7, \dots)$.

$\dfrac{\sin 3x - 3x}{x^3} = \dfrac{-\tfrac{9}{2}x^3 + \dots}{x^3} = -\dfrac{9}{2} + (\text{terms in } x^2, \dots) \to -\dfrac{9}{2}$.

$(1 + u)^{k}$ with $k = -\tfrac{1}{2}$, $u = -2x$.

$(1 + u)^k \approx 1 + ku + \dfrac{k(k - 1)}{2}u^2 = 1 - \dfrac{1}{2}u + \dfrac{3}{8}u^2$.

$1 - \dfrac{1}{2}(-2x) + \dfrac{3}{8}(4x^2) = 1 + x + \dfrac{3}{2}x^2$.

Nothing more to do: that is the polynomial.

$\sqrt{5}$ is the positive root of $f(x) = x^2 - 5$.

$f'(x) = 2x$, so $x_{n+1} = x_n - \dfrac{x_n^2 - 5}{2x_n}$.

$x_0 = 2$ (given; $2^2 = 4$ is close to $5$).

$x_1 = 2 - \dfrac{-1}{4} = \dfrac{9}{4}$. $x_2 = \dfrac{9}{4} - \dfrac{81/16 - 5}{9/2} = \dfrac{9}{4} - \dfrac{1/16}{9/2} = \dfrac{9}{4} - \dfrac{1}{72} = \dfrac{161}{72} \approx 2.23611$.

$f(x) = x^3 + x - 3$.

$f'(x) = 3x^2 + 1$, so $x_{n+1} = x_n - \dfrac{x_n^3 + x_n - 3}{3x_n^2 + 1}$.

$x_0 = 1$.

$f(1) = -1$, $f'(1) = 4$: $x_1 = 1 + \tfrac{1}{4} = \tfrac{5}{4}$. $f\!\left(\tfrac{5}{4}\right) = \tfrac{13}{64}$, $f'\!\left(\tfrac{5}{4}\right) = \tfrac{91}{16}$: $x_2 = \tfrac{5}{4} - \tfrac{13}{364} = \tfrac{5}{4} - \tfrac{1}{28} = \tfrac{17}{14} \approx 1.21429$.

$f(x) = \cos x - x$.

$f'(x) = -\sin x - 1$, so $x_{n+1} = x_n - \dfrac{\cos x_n - x_n}{-\sin x_n - 1}$.

$x_0 = 1$ (radians).

$f(1) \approx -0.459698$, $f'(1) \approx -1.841471$: $x_1 \approx 1 - 0.249636 = 0.750364$. $f(x_1) \approx -0.018923$, $f'(x_1) \approx -1.681905$: $x_2 \approx 0.750364 - 0.011251 = 0.739113$.