อนุกรมเทย์เลอร์มาจากไหน? เฉลยกลไกการประมาณฟังก์ชันสุดมหัศจรรย์

$$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{1!}(x-a)^3+\cdots \\~\\ =\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

$$f(x)=\sum_{n=0}^{3}\frac{f^{(n)}(0)}{n!}x^n \\~\\ \downarrow \\~\\ f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3$$

$$f(0)= \sin0=0 \\ f'(0)=\cos0=1 \\ f''(0)=-\sin0=0\\ f^{(3)}(0)=-\cos0=-1 \\$$

$$f(x)=0+\frac{1}{1!}x+\frac{0}{2!}x^2+\frac{-1}{3!}x^3$$

$$f(x)=x-\frac{1}{3!}x^3$$

$$f(0.5)=f\bigg(\frac{1}{2}\bigg)=\frac{1}{2}-\frac{1}{3!}\cdot\bigg(\frac{1}{2}\bigg)^3 \\~\\ =\frac{1}{2}-\frac{1}{6}\cdot\frac{1}{8} \\~\\ =\frac{1}{2}-\frac{1}{48} \\~\\ =\frac{23}{48}\approx0.479167$$

$$p(x)=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4$$

$$p(0)=f(0) \\~\\ c_0+c_1(0)+c_2(0)^2+c_3(0)^3+c_4(0)^4 = e^0 \\~\\ c_0 = 1 \\~\\ \downarrow \\~\\ p(x)=1$$

$$p'(x)=c_1+2\cdot c_2x+3\cdot c_3x^2+4\cdot c_4x^3 \\~\\ f'(x)=e^x \\~\\ p'(0)=f'(0) \\~\\ c_1+2\cdot c_2(0)+3\cdot c_3(0)^2+4\cdot c_4(0)^3 = e^0 \\~\\ c_1 = 1 \\~\\ \downarrow \\~\\ p(x)=1+x$$

$$p''(x)=2\cdot c_2+3\cdot2\cdot c_3x+4\cdot3\cdot c_4x^2 \\~\\ f''(x)=e^x \\~\\ p''(0)=f''(0) \\~\\ 2\cdot c_2+3\cdot2\cdot c_3(0)+4\cdot3\cdot c_4(0)^2 = e^0 \\~\\ 2\cdot c_2 = 1 \\~\\ c_2 = \frac{1}{2} \\~\\ \downarrow \\~\\ p(x)=1+x+\frac{1}{2}x^2$$

$$p^{(3)}(x)=3\cdot2\cdot c_3+4\cdot3\cdot2\cdot c_4x \\~\\ f^{(3)}(x)=e^x \\~\\ p^{(3)}(0)=f^{(3)}(0) \\~\\ 3\cdot2\cdot c_3+4\cdot3\cdot2\cdot c_4(0) = e^0 \\~\\ 6\cdot c_3 = 1 \\~\\ c_3 = \frac{1}{6} \\~\\ \downarrow \\~\\ p(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3$$

$$p^{(4)}(x)=4\cdot3\cdot2\cdot c_4 \\~\\ f^{(4)}(x)=e^x \\~\\ p^{(4)}(0)=f^{(4)}(0) \\~\\ 4\cdot3\cdot2\cdot c_4= e^0 \\~\\ 24\cdot c_4 = 1 \\~\\ c_4 = \frac{1}{24} \\~\\ \downarrow \\~\\ p(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4$$

$$p(x)=c_0+c_1+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots+c_nx^n \\~\\ p^{(3)}(x)=3\cdot2\cdot c_3+4\cdot3\cdot2\cdot c_4x+5\cdot4\cdot3\cdot c_4x^2+\cdots \\~\\ p^{(3)}(0)=3\cdot2\cdot c_3+\cancel{4\cdot3\cdot2\cdot c_4(0)}+\cancel{5\cdot4\cdot3\cdot c_4(0)^2}+\cancel{\cdots} \\~\\ p^{(3)}(0)=3\cdot2\cdot c_3$$

$$p^{(3)}(0)=3\cdot2\cdot c_3 \\~\\ p^{(3)}(0)=3!\cdot c_3 \\~\\ p^{(n)}(0)=n!\cdot c_n$$

$$p^{(n)}(0)=n!\cdot c_n=f^{(n)}(0) \\~\\ n!\cdot c_n=f^{(n)}(0) \\~\\ c_n=\frac{f^{(n)}(0)}{n!}$$

$$p(x)=c_0+c_1+c_2x^2+c_3x^3+c_4x^4+c_5x^5+\cdots+c_nx^n \\~\\ p(x)=\frac{f(0)}{0!}+\frac{f'(0)}{1!}+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\cdots+\frac{f^{(n)}(0)}{n!}x^n \\~\\ p(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$

$$p(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$