테일러 다항식과 오차

If $Q(x)$ is the quadratic approximation of $f$ about $3$, then $Q(3)=f(3)$, $Q'(3)=f'(3)$, and $Q''(3)=f''(3)$. There is no guarantee that $f(x)$ and $Q(x)$ share the same third derivative, though, so we do not have enough information to know $f'''(3)$.

$$\begin{aligned} f(3)&=-3^2+6(3)=9\\ f'(3)&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{-x^2+6x\right\}\right|_{x=3}=\left.-2x+6\right|_{x=3}=0\\ f''(3)&=\left.\frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\left\{-x^2+6x\right\}\right|_{x=3}= \left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{-2x+6\right\}\right|_{x=3}=-2 \end{aligned}$$

The quadratic approximation of $f(x)$ about $x=a$ is

$$\begin{aligned} f(x) &\approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2 \\ \text{We subsitute $f(a)=2a+5$, $f'(a)=2$, and $f''(a)=0$:} \\ f(x) &\approx (2a+5)+2(x-a)=2x+5 \end{aligned}$$

So, our approximation is $f(x) \approx 2x+5$.

Remark: Our approximation is exact for every value of $x$. This will always happen with a quadratic approximation of a function that is quadratic, linear, or constant.

Since $T_3(x)$ is the third order Taylor polynomial for $f(x)$ about $x=1$:

• $T_3(1)=f(1)$
• $T_3'(1)=f'(1)$
• $T_3''(1)=f''(1)$
• $T_3'''(1)=f'''(1)$

In particular, $f''(1)=T_3''(1)$.

$$\begin{aligned} T_3'(x)&=3x^2-10x+9\\ T_3''(x)&=6x-10\\ T_3''(1)&=6-10=-4 \end{aligned}$$

So, $f''(1)=-4$.

Since $f'(x)=6x$, when $x=10$, $f'(10)=60$. If $\Delta y=f(11)-f(10)$, and $\Delta x = 11-10$, then the linear approximation tells us $$\Delta y \approx 60\Delta x = 60$$ So, the linear approximates estimates the error in $f(x)$ to be about 60.

Since $f''(x)=6$, the quadratic approximation (using $f'(10)=60$, $f''(10)=6$, and $\Delta x = 1$) tells us $$\Delta y \approx f'(10)\Delta x+\frac{1}{2}f''(1)\left(\Delta x\right)^2= 60\cdot 1 +\frac{1}{2}(6)(1)^2= 63$$ So, the quadratic approximates estimates the error in $f(x)$ to be about 63. (Indeed, the exact value of $f(11)-f(10)$ is 63. It is not a fluke that our estimated error, using a quadratic approximation, is exactly the same as our actual error. It is a consequence of the fact that $f(x)$ is a quadratic function.)

Whenever you approximate a polynomial with a Taylor polynomial of greater or equal degree, your Taylor polynomial is exactly the same as the function you are approximating. So, the error is zero.

The constant approximation gives

$$\begin{aligned} \sin(33)&\approx \sin (0)=0 \\ \text{while the linear approximation gives} \\ f(x)&\approx f(0)+f'(0)x\\ \sin(x)& \approx \sin (0)+\cos(0) x\\&= x\\ \sin(33)&\approx 33 \end{aligned}$$

Since $-1 \leq \sin(33)\leq 1$, the constant approximation is better. (But both are a little silly.)

The third Maclaurin polynomial for $f(x)$ is

$$\begin{aligned} f(0) + f'(0)x +\frac{f''(0)}{2}\cdot x^2 + \frac{f'''(0)}{6}\cdot x^3 &=4+3x^2+\frac{1}{2}x^3. \end{aligned}$$

The coefficient of $x$ is $f'(0)$ on one side, and $0$ on the other, so $f'(0)=0$.

The coefficient of $x^2$ is $\dfrac{1}{2}f''(0)$ on one side, and $3$ on the other, so $f''(0)=6$.

The third Maclaurin polynomial for $h(x)$ is

$$\begin{aligned} h(0) + h'(0)x +\frac{h''(0)}{2}\cdot x^2 + \frac{h^{(3)}(0)}{3!}\cdot x^3 &= 1+4x-\frac{1}{3}x^2 + \frac{2}{3}x^3 \end{aligned}$$

The coefficient of $x^3$ is $\dfrac{1}{3!}h^{(3)}(0)$ on one side, and $\dfrac{2}{3}$ on the other. Thus $\dfrac{h^{(3)}(0)}{6}=\dfrac{2}{3}$, so $h^{(3)}(0)=6\cdot\dfrac{2}{3} = 4$.

The third order Taylor polynomial for $h(x)$ about $x=2$ is $$h(2) + h'(2)(x-2) +\frac{h''(2)}{2}\cdot (x-2)^2 + \frac{h'''(2)}{6}\cdot (x-2)^3$$ The coefficient of $(x-2)$ is $h'(2)$ in the definition, and $\dfrac{1}{2}$ in the given function, so $h'(2)=\dfrac{1}{2}$.

The coefficient of $(x-2)^2$ is $~\dfrac{1}{2}h''(2)~$ in the definition, and $0$ in the given function, so $h'(2)=0$.