급수 심화 전체 문제

Recall that

$$T_{2n}(x)=\sum_{k=0}^{2n} \frac{f^{(k)}\left(\frac{\pi}{4}\right)}{k!}\left(x-\frac{\pi}{4}\right)^k$$

Let's start by taking some derivatives. Of course, since we're differentiating sine, the derivatives will repeat every four iterations.

$$\begin{aligned} f(x)&=\sin x & f\left(\frac{\pi}{4}\right)&=\frac{1}{\sqrt 2}\\ f'(x)&=\cos x & f'\left(\frac{\pi}{4}\right)&=\frac{1}{\sqrt{2}}\\ f''(x)&=-\sin x & f''\left(\frac{\pi}{4}\right)&=-\frac{1}{\sqrt 2}\\ f'''(x)&=-\cos x & f'''\left(\frac{\pi}{4}\right)&=-\frac{1}{\sqrt{2}}\\ \end{aligned}$$

So, the pattern of derivatives is $\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, etc. This is a little tricky to write in sigma notation. We can deal with the “doubles" by separating the even and odd powers. The first few terms of $T_{2n}$ that contain even powers of $\left(x-\frac{\pi}{4}\right)$ are

$$\begin{aligned} \underbrace{\frac{1}{\sqrt{2}}}_{k=0} -\underbrace{\frac{1}{2!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2}}_{k=2} +\underbrace{\frac{1}{4!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{4}}_{k=4} \end{aligned}$$

Observe that the signs alternate between successive terms. So if we rename $k$ to $2\ell$ these terms are

$$\begin{aligned} \underbrace{\frac{1}{\sqrt{2}}}_{\ell=0} -\underbrace{\frac{1}{2!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2}}_{\ell=1} +\underbrace{\frac{1}{4!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{4}}_{\ell=2} \end{aligned}$$

and the $\ell^{\rm th}$ term here is $\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell}$. To verify that this really is the $\ell^{\rm th}$ term, evaluate this for $\ell=0,1,2$ explicitly. When $k=2n$, $\ell=n$ so that

$$\begin{aligned} \sum_{\genfrac{}{}{0in}{}{0\le k\le 2n}{k{\rm\ even}}} \frac{f^{(k)}\left(\frac{\pi}{4}\right)}{k!}\left(x-\frac{\pi}{4}\right)^k =\sum_{\ell=0}^{n}\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell} \end{aligned}$$

Now for the odd powers. The first few terms of $T_{2n}$ that contain odd powers of $\left(x-\frac{\pi}{4}\right)$ are

$$\begin{aligned} \underbrace{\frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right)}_{k=1} -\underbrace{\frac{1}{3!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{3}}_{k=3} +\underbrace{\frac{1}{5!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{5}}_{k=5} \end{aligned}$$

Observe that the signs again alternate between successive terms. So if we rename $k$ to $2\ell+1$ these terms are

$$\begin{aligned} \underbrace{\frac{1}{\sqrt{2}}\left(x-\frac{\pi}{4}\right)}_{\ell=0} -\underbrace{\frac{1}{3!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{3}}_{\ell=1} +\underbrace{\frac{1}{5!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{5}}_{\ell=2} \end{aligned}$$

and the $\ell^{\rm th}$ term here is $\frac{(-1)^\ell}{(2\ell+1)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell+1}$. To verify that this really is the $\ell^{\rm th}$ term, evaluate this for $\ell=0,1,2$ explicitly. The largest odd integer that is smaller than $2n$ is $2n-1$ and when $k=2n-1=2\ell+1$, $\ell=n-1$ so that

$$\begin{aligned} \sum_{\genfrac{}{}{0in}{}{0\le k\le 2n}{k{\rm\ odd}}} \frac{f^{(k)}\left(\frac{\pi}{4}\right)}{k!}\left(x-\frac{\pi}{4}\right)^k =\sum_{\ell=0}^{n-1}\frac{(-1)^\ell}{(2\ell+1)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell+1} \end{aligned}$$

Putting the even and odd powers together

$$\begin{aligned} T_{2n}(x)&=\sum_{\ell=0}^{n}\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell} +\sum_{\ell=0}^{n-1}\frac{(-1)^\ell}{(2\ell+1)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell+1} \end{aligned}$$

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.)

Let $A$ be the area of a pen of radius $r$. Then

$$\begin{aligned} A(r)&=\pi r^2 \\ \text{Differentiating with respect to $r$,} \\ A'(r)&=2\pi r \\ \text{The exact area desired is $A_0$. Let the corresponding exact radius desired be $r_0$.} \\ \\ \text{Using the linear approximation formula, where $\Delta A$ is the change in $A$ corresponding to a change in $r$ of $\Delta r$,} \\ \Delta A &\approx A'(r_0)\Delta r = 2\pi r_0 \Delta r\\ \Delta r &\approx \dfrac{\Delta A}{2\pi r_0} \\ \text{What we're interested in is the percent error $r$ can have. The percent error is:} \\ 100\frac{\Delta r}{r_0}&\approx100\frac{\Delta A}{2\pi r_0\cdot r_0}\\ &=100\frac{\Delta A}{2(\pi r_0^2)}\\ &=100\frac{\Delta A}{2\cdot A_0}\\ &=\left( 100\frac{\Delta A}{A_0}\right)\frac{1}{2}\\ &\leq\left( 2\right)\frac{1}{2}=1 \end{aligned}$$

(To get the last line, we used the given information that the percent error in the area, $100\dfrac{\Delta A}{A_0}$, must be less than 2%.)

We conclude the error in $r$ cannot be more than 1%.

(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.