테일러 다항식과 오차

(a) The area removed represents a proportion of $\dfrac{\theta}{2\pi}$ of the entire circle, whose area is $\pi(3^2)=9\pi$. So, the area of the sector removed is $$\dfrac{\theta}{2\pi} \cdot9\pi = \dfrac{9}{2}\theta$$

(b) To find $\theta$ from $d$, we cut our triangle (with angle $\theta$ opposite side of length $d$) into two equivalent right triangles, as shown below.

Using the information that the radius of the circle (also the hypotenuse of the right triangle) is 3,

$$\begin{aligned} \sin\left(\frac{\theta}{2}\right)&=\frac{\frac{d}{2}}{3}=\frac{d}{6} \\ \text{Since the question tells us the sector is no more than a quarter of the circle, we know $0 \leq \theta \leq \dfrac{\pi}{2}$, so $0 \leq \dfrac{\theta}{2} \leq \dfrac{\pi}{4}$. This puts $\dfrac{\theta}{2}$ well within the range of arcsine.} \\ \theta&=2\arcsin\left(\frac{d}{6}\right) \end{aligned}$$

(c) First, let's get an expression for the area of the sector in terms of $d$.

$$\begin{aligned} A&=\frac{9}{2}\theta = \frac{9}{2}\left(2\arcsin\left(\frac{d}{6}\right)\right) \\&=9\arcsin\left(\frac{d}{6}\right) \\ \text{Differentiating,} \\ A'(d)&=\frac{9}{\sqrt{1-\left(\frac{d}{6}\right)^2}}\cdot\frac{1}{6}\\ &=\frac{9}{\sqrt{36-d^2}} \\ \text{Let $\Delta A$ is the error in $A$ corresponding to an error of $\Delta d$ in $d$. Since we measured $d$ to be 0.7 instead of $0.68$, in the linear approximation we take $\Delta d = 0.02$ and $d_0=0.68$.} \\ \Delta A &\approx A'(d_0)\cdot\Delta d\\ &=A'\left(0.68\right)\cdot 0.02\\ &=\frac{9}{\sqrt{36-0.68^2}}\cdot0.02\\ &\approx 0.03 \end{aligned}$$

So, the error in $A$ is about 0.03.

Suppose we have a function $V(h)$ that gives the volume of water in the tank as a function of its height.

Let $h_0=0.5$ metres, $\Delta h = -0.05$, and $\Delta V = V(h_0+\Delta h)-V(h_0)=V(0.45)-V(0.5)$. Then, by the linear approximation, $$\Delta V \approx V'(0.5)\cdot \Delta h = -0.05V'(0.5)$$ In order to solve the problem, we will find a function $V(h)$ giving the volume of water in terms of the height, then find $V'(0.5)$, and finally approximate that the change in the volume of the water is $\Delta V \approx -0.05V'(0.5)$.

The water in the tank forms a cone. The volume of a cone of height $h$ and radius $r$ is $$V=\frac{1}{3}\pi r^2 h$$ We need to get rid of the variable $r$. We can do this using similar triangles. The diagram below shows the side view of the tank and the water.

The side view of the tank forms a triangle that is similar to the triangle formed by the side view of the water, so

$$\begin{aligned} \frac{1}{2}&=\frac{2r}{h}\\ r&=\frac{h}{4} \\ \text{Using this, we find our equation for the volume of the water in terms of $h$.} \\ V(h)&=\frac{1}{3}\pi r^2 h = \frac{\pi}{3}\left(\frac{h}{4}\right)^2h=\frac{\pi h^3}{48} \\ \text{Differentiating,} \\ V'(h)&=\frac{\pi h^2}{16}\\ V'(0.5)&=\frac{0.25\pi}{16}=\frac{\pi}{64} \\ \text{Finally, using the approximation $\Delta V \approx -0.05V'(0.5)$,} \\ \Delta V &\approx \frac{-0.05\pi}{64}=-\frac{\pi}{1280}\approx -0.00245 ~\mathrm{m}^3 \end{aligned}$$

We estimate that the volume decreased by about $0.00245$ cubic metres, or about 2450 cubic centimetres.

Let $q$ be the measured amount of the isotope remaining after 3 years. Let $h(q)$ be the half-life of the isotope that we calculate using $q$. We measured $q=0.9$, but we want to know what the change in $h$ is if $q$ moves by 0.05. So, let $\Delta q= \pm 0.05$, and let $\Delta h$ be the corresponding change in $h$. The linear approximation tells us

$$\begin{aligned}\Delta h& \approx h'(0.9)\Delta q \\ \text{So,} \\ \left|\Delta h\right|&\approx\left| h'(0.9)\right|\left|\Delta q\right|= h'(0.9)\cdot 0.05 \end{aligned}$$

This suggests a plan for solving the problem. We will find the equation $h(q)$ giving the calculated half-life of the isotope based on the measurement $q$. Then, we will find $h'(0.9)$. Finally, the equation $\left|\Delta h\right|=h'(0.9)\cdot0.05$ will tell us the change in $h$ that corresponds with a change of 0.05 in our measurement.

Let us find the half-life of the isotope, if after three years $q$ $\mu$g is remaining. The amount of the isotope present after $t$ years is given by

$$\begin{aligned} Q(t)&=Q(0)e^{-kt} \\ \text{for some constant $k$. Let's take $t=0$ to be the time when precisely one $\mu$g was present. Then} \\ Q(t)&=e^{-kt} \\ \text{After three years, $q$ is the amount of the isotope remaining, so} \\ q&=e^{-k\cdot 3}\\ q^{\frac{1}{3}}&=e^{-k}\\ Q(t)&=\left(e^{-k}\right)^t=q^{\frac{t}{3}} \\ \text{The half-life is the value of $t$ for which $Q(t)=\frac{1}{2}Q(0)=\frac{1}{2}$.} \\ \frac{1}{2}&=Q(t)=q^{\frac{t}{3}}\\ \log\left(\frac{1}{2}\right)&=\log\left(q^{\frac{t}{3}}\right)\\ -\log 2 &=\frac{t}{3}\log q\\ t&=\frac{-3\log 2}{\log q} \end{aligned}$$

So, we calculate the half-life to be $\dfrac{-3\log 2}{\log q}$. This gives us our first goal: a function $h(q)$ that tells us the calculated half-life of the element.

$$\begin{aligned}h(q)&=\dfrac{-3\log2}{\log q} \\ \text{Following our plan, we find $h'(0.9)$.} \\ h'(q)&=\frac{\mathrm{d}}{\mathrm{d}q}\left\{\dfrac{-3\log2}{\log q}\right\}\\ &=-3\log 2\cdot\frac{\mathrm{d}}{\mathrm{d}q}\left\{\left(\log q\right)^{-1}\right\}\\ &=-3\log 2\cdot(-1)\left(\log q\right)^{-2}\cdot\frac{1}{q}\\ &=\frac{3\log 2}{q\log^2q}\\ h'(0.9)&=\frac{3\log2}{0.9\log^2(0.9)}\approx 208 \\ \text{Finally, as outlined in our plan, } \\ \left|\Delta h\right|&=h'(0.9)\cdot0.05\\ &=\frac{3\log2}{18\log^2(0.9)}\approx 10.4 \end{aligned}$$

If our measurement changes by $\pm 0.05$ $\mu$g, then we estimate our calculated half-life changes by about $\pm$10.4 years. Since our measurement is accurate to within $0.05$ $\mu$g, that means we estimate our calculated half-life to be accurate to within about 10.4 years.

Remark: since $h(0.9)=\dfrac{-3\log 2}{\log 0.9} \approx 19.7$, an absolute error of 10.4 years corresponds to a percentage error of $100\dfrac{10.4}{19.7}\approx 53\%$. The question did not specify absolute or percentage error. Since both make sense, you can use either one.

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.)

Our error will have the form $\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-1)^{n+1}$ for some constant $c$, so let's find an equation for $f^{(n)}(x)$. This has been done before in the text, but we'll do it again here: we'll take several derivatives, then notice the pattern.

$$\begin{aligned} f(x)&=\log x\\ f'(x)&=x^{-1}\\ f''(x)&=-x^{-2}\\ f'''(x)&=2!\,x^{-3}\\ f^{(4)}(x)&=-3!\,x^{-4}\\ f^{(5)}(x)&=4!\,x^{-5} \\ \text{So, when $n \geq 1$,} \\ f^{(n)}(x)&=(-1)^{n-1}(n-1)!\cdot x^{-n} \end{aligned}$$

Now that we know the derivative of $f(x)$, we have a better idea what the error in our approximation looks like.

$$\begin{aligned} \left|f(1.1)-T_n(1.1)\right|&=\left|\dfrac{f^{(n+1)}(c)}{(n+1)!}(1.1-1)^{n+1}\right|\\ &=\left|f^{(n+1)}(c)\right|\frac{0.1^{n+1}}{(n+1)!}\\ &=\left|\frac{n!}{c^{n+1}} \right|\frac{1}{10^{n+1}(n+1)!}\\ &=\frac{1}{ |c|^{n+1}\cdot10^{n+1}\cdot(n+1)} \\ \text{for some $c$ in $(1,1.1)$} \\ &<\frac{1}{(n+1)10^{n+1}\cdot 1^{n+1}}\\ &=\frac{1}{(n+1)10^{n+1}} \\ \text{What we've shown so far is } \\ |f(1.1)-T_n(1.1)|&<\dfrac{1}{(n+1)10^{n+1}} \\ \text{If we can show that $\dfrac{1}{(n+1)10^{n+1}}\le10^{-4}$, then we'll be able to conclude} \\ |f(1.1)-T_n(1.1)|&<\dfrac{1}{(n+1)10^{n+1}}\le10^{-4} \end{aligned}$$

That is, our error is less than $10^{-4}$.

So, our goal for the problem is to find a value of $n$ that makes $\dfrac{1}{(n+1)10^{n+1}}\le10^{-4}$. Certainly, $n=3$ is such a number. Therefore, any $n$ greater than or equal to 3 is an acceptable value.

We will approximate $f(x)=x^{\tfrac{1}{7}}$ using a Taylor polynomial. Since $3^7=2187$, we will use $x=2187$ as our centre.

We need to figure out which order Taylor polynomial will result in a small enough error.

If we use the $n$th Taylor polynomial, our error will be

$$\begin{aligned} \left|f(2200)-T_n(2200)\right|&=\left|\frac{f^{(n+1)}(c)}{(n+1)!}(2200-2187)^{n+1}\right|\\ &=\left|f^{(n+1)}(c)\right|\cdot\frac{13^{n+1}}{(n+1)!} \\ \text{for some $c$ in $(2187,2200)$. In order for this to be less than 0.001, we need} \\ \left|f^{(n+1)}(c)\right|\cdot\frac{13^{n+1}}{(n+1)!}&<0.001\\ \left|f^{(n+1)}(c)\right|&<\frac{ (n+1)!}{1000\cdot13^{n+1}} \end{aligned}$$

It's a tricky thing to figure out which $n$ makes this true. Let's make a table. We won't show all the work of filling it in, but the work is standard.

$$\begin{array}{c|c|c|c} n & \dfrac{(n+1)!}{1000\cdot 13^{n+1}} & \left|f^{(n+1)}(c)\right| & \textup{Is }\left|f^{(n+1)}(c)\right|<\dfrac{ (n+1)!}{1000\cdot13^{n+1}}\textup{?} \\~&~&~\\ \hline ~&~&~\\ 0&\dfrac{1}{1000\cdot 13}&|f'(c)|=\dfrac{1}{7c^{6/7}}<\dfrac{1}{7\cdot 3^6}& \\~&~&~\\ \hline ~&~&~\\ 1&\dfrac{2}{1000\cdot 13^2}&|f''(c)|=\dfrac{6}{7^2\cdot c^{\tfrac{13}{7}}}<\dfrac{6}{7^2\cdot 3^{13}}&\textup{Yes!} \end{array}$$

That is: if we use the first order Taylor polynomial, then for some $c$ between $2187$ and $2200$,

$$\begin{aligned} |f(2200)-T_1(2200)|&=\left|f''(c)\right|\cdot\frac{13^2}{2!}\\ &=\frac{6}{7^2\cdot c^{\tfrac{13}{7}}}\cdot\frac{13^2}{2}\\ &<\dfrac{6}{7^2\cdot 3^{13}}\cdot\dfrac{13^2}{2}\\ &=\dfrac{3\cdot 13^2}{7^2 \cdot 3^{13}}\approx 0.0000065 \end{aligned}$$

So, actually, the linear Taylor polynomial (or any higher order Taylor polynomial) will result in an approximation that is much more accurate than required. (We don't know, however, that the constant approximation will be accurate enough–so we'd better stick with $n \geq 1$.)

Now that we know we can take the first order Taylor polynomial, let's compute $T_1(x)$. Recall we are taking the Taylor polynomial for $f(x)=x^{\tfrac{1}{7}}$ about $x=2187$.

$$\begin{aligned} f(2187)&=2187^{\tfrac{1}{7}}=3\\ f'(x)&=\frac{1}{7}x^{-\tfrac{6}{7}}\\ f'(2187)&=\frac{1}{7\sqrt[7]{2187}^6}=\frac{1}{7\cdot3^6}\\ T_1(x)&=f(2187)+f'(2187)(x-2187)\\ &=3+\frac{x-2187}{7\cdot3^6}\\ T_1(2200)&=3+\frac{2200-2187}{7\cdot3^6}\\ &=3+\frac{13}{7\cdot 3^6}\\ &\approx 3.00255 \end{aligned}$$