9.1 Zeroth approximation

An approximation should be something you can actually figure out–otherwise it's no use.

Since $f(0)$ is closer to $g(0)$ than it is to $h(0)$, you would probably want to estimate $f(0) \approx g(0)=1+2\sin (1)$ if you had the means to efficiently figure out what $\sin(1)$ is, and if you were concerned with accuracy. If you had a calculator, you could use this estimation. Also, later in this chapter we will learn methods of approximating $\sin (1)$ that do not require a calculator, but they do require time.

Without a calculator, or without a lot of time, using $f(0)\approx h(0)=0.7$ probably makes the most sense. It isn't as accurate as $f(0) \approx g(0)$, but you get an estimate very quickly, without worrying about figuring out what $\sin(1)$ is.

Since $f(0)$ is closer to $g(0)$ than it is to $h(0)$, you would probably want to estimate $f(0) \approx g(0)=1+2\sin (1)$ if you had the means to efficiently figure out what $\sin(1)$ is, and if you were concerned with accuracy. If you had a calculator, you could use this estimation. Also, later in this chapter we will learn methods of approximating $\sin (1)$ that do not require a calculator, but they do require time.

Without a calculator, or without a lot of time, using $f(0)\approx h(0)=0.7$ probably makes the most sense. It isn't as accurate as $f(0) \approx g(0)$, but you get an estimate very quickly, without worrying about figuring out what $\sin(1)$ is.

Remark: when you're approximating something in real life, there probably won't be an obvious “correct" way to do it. There's usually a trade-off between accuracy and ease.

You'll need some constant $a$ to approximation $\log(0.93) \approx \log(a)$. This $a$ should have two properties: it should be close to 0.93, and you should be able to easily evaluate $\log(a)$.

$\log(0.93)\approx \log(1)=0$

0.93 is pretty close to 1, and we know $\log(1)=0$, so we estimate $\log(0.93) \approx \log(1)=0$.

You'll need some constant $a$ to approximate $\arcsin(0.1) \approx \arcsin(a)$. This $a$ should have two properties: it should be close to 0.1, and you should be able to easily evaluate $\arcsin(a)$.

$\arcsin(0.1) \approx 0$

We don't know $\arcsin(0.1)$, but 0.1 is reasonably close to 0, and $\arcsin(0)=0$. So, we estimate $\arcsin(0.1) \approx0$.

You'll need some constant $a$ to approximate $\sqrt{3}\tan(1) \approx \sqrt{3}\tan(a)$. This $a$ should have two properties: it should be close to 1, and you should be able to easily evaluate $\sqrt{3}\tan(a)$.

$\sqrt{3}\tan(1) \approx 3$

We don't know $\tan(1)$, but we do know $\tan\left(\dfrac{\pi}{3}\right)=\sqrt{3}$. Since $\dfrac{\pi}{3}\approx 1.047$ is pretty close to 1, we estimate $\sqrt{3}\tan(1) \approx \sqrt{3}\tan\left(\dfrac{\pi}{3}\right)=\left(\sqrt{3}\right)^2=3.$

We could figure out $10.1^3$ exactly, if we wanted, with pen and paper. Since we're asking for an approximation, we aren't after perfect accuracy. Rather, we're after ease of calculation.

$10.1^3 \approx 10^3=1000$

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.