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

False. The linear approximation is an approximation. It tells us $$\Delta y \approx f'(x_0)\Delta x=f'(2)(1)=25$$ However, from our definition of $\Delta y$, $$\Delta y = f(x_0+\Delta x)-f(x_0)=f(2+1)-f(2)=58-26=32$$

Remark: this is to emphasize that the calculations in this subsection are estimations of error bounds, rather than actual error bounds. All we can say is that we estimate the error will be no more than some number–we don't guarantee it.

In the next subsection, we will introduce an error bound that is guaranteed to be accurate. It is usually harder to calculate than the estimations in this section.

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.)

From the given information, $$|R(10)|=|f(10)-F(10)|=|-3-5|=|-8|=8$$ So, (a) is false (since 8 is not less than or equal to 7), while (b), (c), and (d) are true.

Remark: $R(x)$ is the error in our approximation. As mentioned in the text, we almost never know $R$ exactly, but we can give a bound. We don't need the tightest bound–just a reasonable one that is easy to calculate. If we were dealing with real functions and approximations, we might not know that $|R(10)|=8$, but if we knew it was at most 9, that would be a pretty decent approximation.

Often in this section, we will make simplifying assumptions to get a bound that is easy to calculate. But, don't go overboard! It is a true statement to say that our absolute error is at most 100, but this statement would probably not be very helpful as a bound.

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 $h'(2)=0$.