급수 심화 전체 문제

We use the function $f(x)=x^{1/3}$ and point $a=27$ as the centre of our approximation since we can easily compute

$$\begin{aligned} f(27)&=3 \end{aligned}$$

We compute $f'(x)=\dfrac{1}{3} x^{-2/3}$, so

$$\begin{aligned} f'(27)&= \frac{1}{3} \cdot (27)^{-2/3} = \frac{1}{27} \end{aligned}$$

So, the linear approximation of $26^{1/3} = f(26)$ is

$$\begin{aligned} 26^{1/3}&\approx T_1\left(26\right) = f(27) + f'(27)\cdot (26-27) \\ &= 3 - \frac{1}{27} = \frac{80}{27} \end{aligned}$$

We use the function $f(x)=x^5$ and point $a=10$ as the centre of our approximation since we know that $f(a)=f(10)=10^5$.

Since $f'(x) = 5x^4$ we have $f'(10) = 50,000$.

So, a linear approximation of $10.1^5$ is

$$\begin{aligned} T_1(10.1)&= f(10) + f'(10)\cdot (10.1-10) \\ &= 100,000 + 50,000 \cdot 0.1 = 105,000\,. \end{aligned}$$

We use the function $f(x)=\sin(x)$ and point $a=\pi$ as the centre of our approximation since we know that

$$\begin{aligned} \sin(a)&=f(\pi)=\sin{\pi}=0. \end{aligned}$$

and $\pi$ is reasonably close to $\dfrac{101\pi}{100}$.

We compute $f'(x)=\cos(x)$, so

$$\begin{aligned} f'(\pi)&=\cos(\pi) = -1. \\ \text{So, the linear approximation of $\sin\left(\frac{101\pi}{100}\right) $ is} \\ f\left(\frac{101\pi}{100}\right)&\approx T_1\left(\frac{101\pi}{100}\right)\\ &= f(\pi) + f'(\pi)\cdot \left(\frac{101\pi}{100}-\pi\right) \\ &=0+(-1)\cdot\frac{\pi}{100}=-\frac{\pi}{100}. \end{aligned}$$

Set $f(x)=\arctan(x)$. Then $f'(x)=\dfrac{1}{1+x^2}$, so $f'(1)=\dfrac{1}{2}$ and $$f(1.1)\approx f(1)+f'(1)(1.1-1)=\frac{\pi}{4}+\frac{1}{20}$$

Set $f(x)=(2+x)^3$, so we are approximating $f(0.001)$.

The obvious choice of $a$ is $a=0$.

Then $f'(x)=3(2+x)^2$, so $$(2.001)^3=f(0.001)\approx f(0)+f'(0)(0.001-0) =8+\frac{12}{1000}=\frac{8012}{1000}$$

Remark: if we had chosen $f(x)=x^3$ and $a=2$, the result would have been exactly the same.

We set $f(x)=(8+x)^{2/3}$, and choose $a=0$ as our centre.

Then $f'(x)=\dfrac{2}{3}(8+x)^{-1/3}$, so that

$$\begin{aligned} (8.06)^{2/3}=f(0.06)&\approx f(0)+f'(0)\cdot 0.06\\ &=8^{2/3}+\frac{2}{3}8^{-1/3}\cdot 0.06\\ &=\sqrt[3]{8}^2+\frac{2}{3\sqrt[3]{8}}\cdot 0.06\\ &=2^2+\frac{2}{3\cdot 2}\cdot 0.06\\ &=4+\frac{1}{3}\cdot 0.06\\&=4.02=\frac{402}{100}\end{aligned}$$

We begin by finding the derivatives of $f$ at $x=0$.

$$\begin{aligned} f(x)&=(1 - 3x)^{-1/3} & f(0) & = 1 \\ f'(x)&=(-3)\frac{-1}{3}(1 - 3x)^{-4/3}=(1-3x)^{-4/3} & f'(0) & = 1 \\ f''(x)&=(-3)\frac{-4}{3}(1-3x)^{-7/3}={4}(1 - 3x)^{-7/3} & f''(0) & = 4 \\ f^{(3)}(x)&=(-3)(4)\frac{-7}{3}(1-3x)^{-10/3}=28(1 - 3x)^{-10/3} & f^{(3)}(0) & = 28 \\ \end{aligned}$$

Plugging these into the definition of a Taylor Polynomial, we find that the third–order Taylor polynomial for $f$ around $x = 0$ is

$$\begin{aligned} T_3(x)&=1+x+\frac{4}{2!}x^2+\frac{28}{3!}x^3\\ &=1 + x +2x^2+\frac{14}{3}x^3 \end{aligned}$$

The fourth order Maclaurin polynomial for $f(x)$ is

$$\begin{aligned} T_4(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f^{(4)}(0)x^4 \\ \text{while the third order Maclaurin polynomial for $f(x)$ is} \\ T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3 \\ \text{So, we simply ``chop off" the part of $T_4(x)$ that includes $x^4$. Since that's already 0, in this case $T_3(x)=T_4(x)$.} \\ T_3(x)&=5x^2-9 \end{aligned}$$

Let $f(x)=\log x$ and $x_0=10$. Then

$$\begin{aligned} f(x)&=\log x & f'(x)&=\frac{1}{x} & f''(x)&=-\frac{1}{x^2} \\ f(10)&=\log 10\approx 2.30259 & f'(10)&=\frac{1}{10} & f''(10)&=-\frac{1}{100} \end{aligned}$$

so that, with $x=10.3$, $$\log 10.3=f(10.3)\approx f(10)+f'(10)(10.3-10) =2.30259+\frac{0.3}{10} =2.33259$$ The error in this approximation (excluding the error in the given data $\log 10\approx 2.30259$) is $\dfrac{1}{2} f''(z)(10.3-10)^2$ for some $z$ between $10$ and $10.3$. Because $f''(z)=-\dfrac{1}{z^2}$ increases as $z$ increases, it must be between $-\dfrac{1}{10^2}$ and $-\dfrac{1}{10.3^2}$. This forces $\dfrac{1}{2} f''(z)(10.3-10)^2$ to be between $-\dfrac{1}{2}\cdot \dfrac{1}{10^2}(0.3)^2=-0.00045$ and $-\dfrac{1}{2}\cdot \dfrac{1}{10.3^2}(0.3)^2<-0.00042$.

(a) If Olaf is old, and I am even older, then I am old as well.

(b) If Olaf is old, and I am not as old, then perhaps I am old as well (just slightly less so), or perhaps I am young. There is not enough information to tell.

(c) If Yuan is young, and I am older, then perhaps I am much older and I am old, or perhaps I am only a little older, and I am young. There is not enough information to tell.

(d) If Yuan is young, and I am even younger, then I must also be young.

Another way to think about this is with a timeline of birthdates. People born before the threshold are old, and people born after it are young.

If I'm born before (older than) Olaf, I'm born before the threshold, so I'm old.

If I'm born after (younger than) Yuan I'm born after the threshold, so I'm young.

If I'm born after Olaf or before Yuan, I don't know which side of the threshold I'm on. I could be old or I could be young.

The integral test does not apply because $f(x)$ is not decreasing.

One possible answer: $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$. This series converges (it's a $p$–series with $p=2>1$), but if we take the ratio of consecutive terms: $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{n^2}{(n+1)^2}=1$$ The limit of the ratio is 1, so the ratio test is inconclusive.

By the divergence test, for a series $\sum a_n$ to converge, we need $\lim\limits_{n \to \infty} a_n=0$. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If $\displaystyle\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|<1$ or (equivalently) $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|>1$, then $|a_{n+1}|>|a_n|$ for sufficiently large $n$, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.

The terms of the series only see a small portion of the domain of the integral. We can try to think of a function $f(x)$ that behaves “nicely" when $x$ is a whole number (that is, it produces a sequence whose sum converges), but is more unruly when $x$ is not a whole number.

For example, suppose $f(x)=\sin(\pi x)$. Then $f(x)=0$ for every integer $x$, but this is not representative of the function as a whole. Indeed, our corresponding series has terms $\{a_n\}=\{0,0,0,\ldots\}$.

• $\displaystyle\int_{1}^\infty \sin(\pi x)\,\mathrm{d}x = \lim_{R \to \infty} \left[-\frac{1}{\pi}\cos (\pi x)\right]_1^R = \lim_{R \to \infty}\Big[-\cos (\pi R)\Big]-\frac{1}{\pi}$

Since the limit does not exist, the integral diverges.

• $\displaystyle\sum_{n=1}^{\infty}\sin(\pi n) = \sum_{n=1}^\infty 0 = 0$. The series converges.

First, we'll check the divergence test. It doesn't always work, but if it does, it's likely the easiest path.

$$\begin{aligned} \lim_{n\to\infty} \frac{n^2}{3n^2+\sqrt n} \left(\frac{\frac{1}{n^2}}{\frac{1}{n^2}}\right) = \lim_{n\to\infty} \frac{1}{3+\frac{1}{n\sqrt n}} = \frac{1}{3} \neq 0 \end{aligned}$$

Since the limit of the terms being added is not zero, the series diverges by the divergence test.

This precise question was asked on a 2014 final exam. Note that the $n^{\rm th}$ term in the series is $a_n=\frac{5^k}{4^k+3^k}$ and does not depend on $n$! There are two possibilities. Either this was intentional (and the instructor was being particularly nasty) or it was a typo and the intention was to have $a_n=\frac{5^n}{4^n+3^n}$. In both cases, the limit

$$\begin{aligned} \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^k}{4^k+3^k} = \frac{5^k}{4^k+3^k} \neq 0 \\ \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^n}{4^n+3^n} =\lim_{n\to\infty} \frac{(5/4)^n}{1+(3/4)^n} = +\infty \neq 0 \\ \end{aligned}$$

is nonzero, so the series diverges by the divergence test.

The terms of the series tend to 0, so we can't use the divergence test.

To generate a guess about its convergence, we do the following: $$\sum \frac{1}{\sqrt{k}\sqrt{k+1}}=\sum\frac{1}{\sqrt{k^2+k}} \approx \sum\frac{1}{\sqrt{k^2}}=\sum\frac{1}{k}$$We guess that our series behaves like the harmonic series, and the harmonic series diverges (which can be demonstrated by $p$-test or integral test). So, we guess that our series diverges. However, in order to directly compare our series to the harmonic series and show our series diverges, our terms would have to be bigger than the terms in the harmonic series, and this is not the case. So, we use limit comparison.

$$\begin{aligned} \frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\frac{\sqrt{k^2+k}}{k}=\frac{\sqrt{k^2+k}}{\sqrt{k^2}}= \sqrt{\frac{k^2+k}{k^2}}=\sqrt{1+\frac{1}{k}},\text{ so}\\ \lim_{k \rightarrow \infty}\frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\lim_{k \rightarrow \infty}\sqrt{1+\frac{1}{k}}=1 \end{aligned}$$