선형근사와 미분소

Since $10.1$ is pretty close to $10$, we estimate $10.1^3 \approx 10^3=1000$.

Remark: these kinds of approximations are very useful when you are doing computations. It's easy to make a mistake in your work, and having in mind that $10.1^3$ should be about a thousand is a good way to check that whatever answer you have makes sense.

If $f(x)=x^3$, then $f(10.1)=10.1^3$, which is the value we want to estimate. Let's take the linear approximation of $f(x)$ about $x=10$:

$$\begin{aligned} f(10)&=10^3=1000\\ f'(x)&=3x^2\\ f'(10)&=3(10^2)=300\\ f(a) &\approx f(10)+f'(10)(x-10)\\ &=1000+300(x-10)\\ f(10.1)& \approx 1000+300(10.1-10)=1030 \end{aligned}$$

We estimate $10.1^3 \approx 1030$. If we calculate $10.1^3$ exactly (which is certainly possible to do by hand), we get 1030.301.

Remark: in the previous subsection, we used a constant approximation to estimate $10.1^3 \approx 1000$. That approximation was easy to do in your head, in a matter of seconds. The linear approximation is more accurate, but not much faster than simply calculating $10.1^3$.

There are many possible answers. One is: $$f(x)=\sin x \qquad a=0 \qquad b=\pi$$ We know that $f(\pi)=0$ and $f(0)=0$. Using a constant approximation of $f(x)$ about $x=0$, our estimation is $f(\pi)\approx f(0)=0$, which is exactly the correct value. However, is we make a linear approximation of $f(x)$ about $x=0$, we get $$f(\pi) \approx f(0)+f'(0)(\pi-0)=\sin(0)+\cos(0)\pi=\pi$$ which is not exactly the correct value.

Remark: in reality, we wouldn't estimate $\sin(\pi)$, because we know its value exactly. The purpose of this problem is to demonstrate that fancier approximations are not always more accurate. At the end of this section, we'll talk about how to bound the error of your estimations, to make sure that you are finding something sufficiently accurate.

The linear approximation $L(x)$ of $f(x)$ about $x=a$ is chosen so that $L(a)=f(a)$ and $L'(a)=f'(a)$. So,

$$\begin{aligned} L'(a)&= f'(a)=\dfrac{1}{1+a^2}\\ \frac{1}{4}&=\frac{1}{1+a^2}\\ a&=\pm\sqrt{3} \\ \text{We've narrowed down $a$ to $\sqrt{3}$ or $-\sqrt{3}$. Recall the linear approximation of $f(x)$ about $x=a$ is $f(a)+f'(a)(x-a),$ so its constant term is $f(a)-af'(a)=\arctan(a)-\dfrac{a}{1+a^2}$. We compute this for $a=\sqrt{3}$ and $a=-\sqrt{3}$.} \\ a=\sqrt{3}:~&~\arctan\left(a\right)-\frac{a}{1+a^2} =\arctan\left(\sqrt{3}\right)-\frac{\sqrt{3}}{1+\left(\sqrt{3}\right)^2}=\frac{\pi}{3}-\frac{\sqrt{3}}{4}=\frac{4\pi-\sqrt{27}}{12}\\ a=-\sqrt{3}:~&~\arctan\left(a\right)-\frac{a}{1+a^2} =\arctan\left(-\sqrt{3}\right)-\frac{-\sqrt{3}}{1+\left(-\sqrt{3}\right)^2}=-\frac{\pi}{3}+\frac{\sqrt{3}}{4}=\frac{-4\pi+\sqrt{27}}{12} \\ \text{So, when $a=\sqrt{3}$,} \\ L(x)&=\frac{1}{4}x+\frac{4\pi-\sqrt{27}}{12} \end{aligned}$$

and this does not hold when $a=-\sqrt{3}$. We conclude $a=\sqrt{3}$.