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此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。
PDF格式教材下载 Sequences and Series
本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution
Review
Determine whether the series converges.
1. $\displaystyle\sum_{n=0}^{\infty}{n\over n^2+4}$
Solution:
When $n > 4$ we have $${n\over n^2+4} > {n\over n^2+n} = {1\over n+1}\to\text{diverge}$$ By $p$-series test and comparison test, it diverges.
2. $\displaystyle{1\over1\cdot2}+{1\over3\cdot4}+{1\over5\cdot6} +{1\over7\cdot8}+\cdots\cdots$
Solution: $$s=\sum_{n=0}^{\infty}{1\over (2n+1)(2n+2)}$$ and $${1\over (2n+1)(2n+2)} = {1\over 4n^2+6n+2} < {1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
3. $\displaystyle\sum_{n=0}^{\infty}{n\over(n^2+4)^2}$
Solution: $${n\over(n^2+4)^2} < {n\over (n^2)^2} = {1\over n^3}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
4. $\displaystyle\sum_{n=0}^{\infty}{n!\over8^n}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n={(n+1)!\over8^{n+1}}\cdot{8^n\over n!}={n+1\over8}\to\infty$$ By ratio test, it diverges.
5. $\displaystyle1-{3\over4}+{5\over8}-{7\over12}+{9\over16}+ \cdots\cdots$
Solution: $$s=1+\sum_{n=1}^{\infty}(-1)^n{2n+1\over4n}$$ and $$\lim_{n\to\infty}{2n+1\over4n}={1\over2}\neq0$$ Thus it diverges.
6. $\displaystyle\sum_{n=0}^{\infty}{1\over\sqrt{n^2+4}}$
Solution: $$\lim_{n\to\infty}{1\over\sqrt{n^2+4}}\big/{1\over n} = \lim_{n\to\infty} {n\over\sqrt{n^2+4}}=1 > 0$$ Since ${1\over n}$ is harmonic series which is divergent, by limit comparison test, the original series is divergent.
7. $\displaystyle\sum_{n=0}^{\infty}{\sin^3n\over n^2}$
Solution: $${\sin^3n\over n^2}\leq{1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
8. $\displaystyle\sum_{n=0}^{\infty}{n\over e^n}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{n+1\over e^{n+1}}\cdot{e^n\over n}={1\over e} < 1$$ By ratio test, it converges.
9. $\displaystyle\sum_{n=0}^{\infty}{n!\over1\cdot3\cdot5\cdots(2n-1)}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{(n+1)!\over1\cdot3\cdot5\cdots(2n+1)}\cdot{1\cdot3\cdot5\cdot(2n-1)\over n!}$$ $$=\lim_{n\to\infty}{n+1\over (2n+1)\cdot n}=0 < 1$$ By ratio test, it converges.
10. $\displaystyle\sum_{n=1}^{\infty}{1\over n\sqrt{n}}$
Solution: $$\int_{1}^{\infty}{dx\over x\sqrt{x}}=\int_{1}^{\infty}x^{-{3\over2}}dx=-2x^{-{1\over2}}\Big|_{1}^{\infty}=2\to\text{converge}$$ By integral test, it converges.
11. $\displaystyle{1\over2\cdot3\cdot4}+{2\over3\cdot4\cdot5} +{3\over4\cdot5\cdot6}+{4\over5\cdot6\cdot7}+\cdots\cdots$
Solution: $$s=\sum_{n=0}^{\infty}{n+1\over(n+2)(n+3)(n+4)}$$ and $${n+1\over(n+2)(n+3)(n+4)} < {n+1\over(n+1)(n+1)(n+1)} = {1\over(n+1)^2} < {1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
12. $\displaystyle\sum_{n=1}^{\infty}{1\cdot3\cdot5\cdots(2n-1)\over(2n)!}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{1\cdot3\cdot5\cdots(2n+1)\over(2n+2)!} \cdot {(2n)!\over1\cdot3\cdot5\cdots(2n-1)}$$ $$=\lim_{n\to\infty}{2n+1\over(2n+2)(2n+1)}=0 < 1$$ By ratio test, it converges.
13. $\displaystyle\sum_{n=0}^{\infty}{6^n\over n!}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty} {6^{n+1}\over(n+1)!} \cdot {n!\over6^n}\lim_{n\to\infty}{6\over n+1}=0 < 1$$ By ratio test, it converges.
14. $\displaystyle\sum_{n=1}^{\infty}{(-1)^{n-1}\over\sqrt{n}}$
Solution: $$\lim_{n\to\infty}{1\over\sqrt{n}}=0$$ and it is decreasing. By alternating series test, it converges (conditionally).
15. $\displaystyle\sum_{n=1}^{\infty}{2^n3^{n-1}\over n!}$
Solution: $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{2^{n+1} 3^{n}\over(n+1)!}\cdot {n!\over2^n3^{n-1}}=\lim_{n\to\infty}{6\over n+1}=0 < 1$$ By ratio test, it converges.
16. $\displaystyle1+{5^2\over2^2}+{5^4\over(2\cdot4)^2}+{5^6\over(2\cdot4 \cdot6)^2}+{5^8\over(2\cdot4\cdot6\cdot8)^2}+\cdots\cdots$
Solution: $$s=\sum_{n=0}^{\infty}{5^{2n}\over4^{n}\cdot(n!)^2}$$ and $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{25^{n+1} \over4^{n+1}\cdot((n+1)!)^2}\cdot{4^n\cdot(n!)^2\over25^n}$$ $$=\lim_{n\to\infty}{25\over4(n+1)^2}=0 < 1$$ By ratio test, it converges.
17. $\displaystyle\sum_{n=1}^{\infty}\sin{1\over n}$
Solution: $$\lim_{n\to\infty}{\sin{1\over n}\over{1\over n}}=1 > 0$$ Since ${1\over n}$ is harmonic series which is divergent, by limit comparison test, the original series is divergent. Find the interval and radius of convergence; you need not check the endpoints of the intervals.
18. $\displaystyle\sum_{n=0}^{\infty}{2^n\over n!}x^n$
Solution:
By ratio test, we have $$\lim_{n\to\infty}{2^{n+1}\over(n+1)!}\cdot{n!\over2^n}\cdot\big|{x^{n+1}\over x^{n}}\big|=\lim_{n\to\infty}{2|x|\over n+1}=0\Rightarrow x\in(-\infty, \infty)$$
19. $\displaystyle\sum_{n=0}^{\infty}{x^n\over1+3^n}$
Solution:
By ratio test, we have $$\lim_{n\to\infty} {1+3^n\over1+3^{n+1}}\cdot\big|{x^{n+1}\over x^{n}}\big|={1\over3}|x| < 1\Rightarrow x\in(-3, 3)$$
20. $\displaystyle\sum_{n=1}^{\infty}{x^n\over n\cdot3^n}$
Solution:
By ratio test, we have $$\lim_{n\to\infty}{n\cdot3^n\over(n+1)\cdot3^{n+1}}\cdot|x|={1\over3}|x| < 1\Rightarrow x\in(-3, 3)$$
21. $\displaystyle x+{1\over2}\cdot{x^3\over3}+{1\cdot3\over2\cdot4}\cdot {x^5\over5}+{1\cdot3\cdot5\over2\cdot4\cdot6}\cdot{x^7\over7} +\cdots$
Solution: $$s=\sum_{n=0}^{\infty}{(2n)!\big/2^n\cdot n!\over2^n\cdot n!}\cdot {x^{2n+1}\over2n+1}=\sum_{n=0}^{\infty}{(2n)!\over 4^n\cdot(n!)^2}\cdot {x^{2n+1}\over2n+1}$$ and by ratio test, we have $$\lim_{n\to\infty}{(2n+2)!\over4^{n+1}\cdot((n+1)!)^2}\cdot{1\over2n+3} \cdot {4^n\cdot(n!)^2\over(2n)!}\cdot(2n+1)\cdot\big|{x^{2n+3}\over x^{2n+1}}\big|$$ $$=\lim_{n\to\infty}{(2n+2)(2n+1)\cdot (2n+1)\over 4(n+1)^2\cdot(2n+3)}\cdot x^2=x^2 < 1\Rightarrow x\in(-1, 1)$$
22. $\displaystyle\sum_{n=1}^{\infty}{n!\over n^2}x^n$
Solution:
By ratio test, we have $$\lim_{n\to\infty}{(n+1)!\over(n+1)^2}\cdot{n^2\over n!}\cdot|x|=\lim_{n\to\infty}(n+1)\cdot|x|\to\infty$$ Thus $R=0$ and it converges only when $x=0$.
23. $\displaystyle\sum_{n=1}^{\infty}{(-1)^n\over n^2\cdot3^n}x^{2n}$
Solution:
By ratio test, we have $$\lim_{n\to\infty} {n^2\cdot3^n\over(n+1)^2\cdot3^{n+1}}\cdot {x^{2n+2}\over x^{2n}}={x^2\over3} < 1\Rightarrow x\in(-\sqrt3, \sqrt3)$$
24. $\displaystyle\sum_{n=0}^{\infty}{(x-1)^n\over n!}$
Solution: $$\lim_{n\to\infty}{n!\over(n+1)!}\cdot|x-1|=0\Rightarrow x\in(-\infty, \infty)$$
Find a series for each function, using the formula for Maclaurin series and algebraic manipulation as appropriate.
25. $2^x$
Solution: $$f(0)=2^x\big|_{x=0}=1$$ $$f‘(0)=2^x\cdot\log2\big|_{x=0}=\log2$$ $$f‘‘(0)=2^x\cdot(\log2)^2\big|_{x=0}=(\log2)^2$$ $$\cdots\cdots\cdots$$ Thus $$2^x=1+{\log2\over1!}x+{(\log2)^2\over2!}x^2+\cdots =\sum_{n=0}^{\infty}{(\log2)^n\over n!}x^n$$
26. $\log(1+x)$
Solution: $$f(0)=\log(1+x)\big|_{x=0}=0$$ $$f‘(0)={1\over1+x}\big|_{x=0}=1$$ $$f‘‘(0)={-1\over(1+x)^2}\big|_{x=0}=-1=-1!$$ $$f‘‘‘(0)={2\over(1+x)^3}\big|_{x=0}=2=2!$$ $$f^{(4)}(0)={-6\over(1+x)^4}\big|_{x=0}=-6=-3!$$ $$\cdots\cdots\cdots$$ Thus $$\log(1+x)=x-{1!\over2!}x^2+{2!\over3!}x^3-{3!\over4!}x^4+\cdots =\sum_{n=0}^{\infty}{(-1)^n\over n+1}x^{n+1}$$
27. $\displaystyle\log\left({1+x\over1-x}\right)$
Solution: $$\log(1+x)=\sum_{n=0}^{\infty}{(-1)^n\over n+1}x^{n+1} = x - {1\over2}x^2+{1\over3}x^3-{1\over4}x^4+\cdots$$ $$\log(1-x)=\sum_{n=0}^{\infty}{(-1)^n\over n+1}(-x)^{n+1} = \sum_{n=0}^{\infty}{-1\over n+1}x^{n+1}=-x - {1\over2}x^2-{1\over3}x^3-{1\over4}x^4+\cdots$$ Thus $$\log\big({1+x\over1-x}\big)=\log(1+x)-\log(1-x)$$ $$=2x+{2\over3}x^3+{2\over5}x^5+\cdots =\sum_{n=0}^{\infty}{2\over 2n+1}x^{2n+1}$$
28. $\sqrt{1+x}$
Solution: $$f(0)=\sqrt{1+x}\big|_{x=0}=1$$ $$f‘(0)={1\over2}(1+x)^{-{1\over2}}\big|_{x=0}={1\over2}$$ $$f‘‘(0)=-{1\over4}(1+x)^{-{3\over2}}\big|_{x=0}=-{1\over4}$$ $$f‘‘‘(0)={3\over8}(1+x)^{-{5\over2}}\big|_{x=0}={3\over8}$$ $$f^{(4)}(0)=-{15\over16}(1+x)^{-{7\over2}}\big|_{x=0}=-{15\over16}$$ $$\cdots\cdots\cdots$$ Thus $$\sqrt{1+x}=1+{1\over2\cdot1!}x-{1\over4\cdot2!}x^2+{3\over8\cdot3!}x^3 -{15\over16\cdot4!}x^4+\cdots$$ $$=1+{x\over2}+\sum_{n=2}^{\infty}{(-1)^{n+1}\cdot1\cdot3\cdots(2n-3)\over 2^n\cdot n!}x^n$$
29. $\displaystyle{1\over1+x^2}$
Solution: $${1\over 1-x}=\sum_{n=0}^{\infty}x^n\Rightarrow {1\over1+x} = \sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^n\cdot x^n$$ $$\Rightarrow {1\over1+x^2}=\sum_{n=0}^{\infty}(-1)^n\cdot x^{2n}$$
30. $\arctan(x)$
Solution: $$\int{1\over1+x^2}dx=\arctan(x)$$ $$\Rightarrow \arctan(x)=\int\sum_{n=0}^{\infty}(-1)^n\cdot x^{2n}dx = \sum_{n=0}^{\infty}{(-1)^n\over2n+1}\cdot x^{2n+1}$$
31. Use the answer to the previous problem to discover a series for a well-known mathematical constant $\pi$.
Solution: $${\pi\over4}=\arctan1=\sum_{n=0}^{\infty}{(-1)^n\over2n+1}\Rightarrow \pi=\sum_{n=0}^{\infty}(-1)^n\cdot{4\over2n+1}$$
Final
1. To say that the sequence $a_n$ converges to $L$ means what? In other words, what is the definition of the statement $\displaystyle\lim_{n\to\infty}a_n=L$?
Solution:
For every positive real number $\varepsilon > 0$ there exists an $N\in\mathbf{N}$ so that whenever $n \geq N$, we have $|a_n-L| < \varepsilon$.
2. To say that $\displaystyle\sum_{k=4}^\infty a_k = L$ means what? In order words, what does it mean to say that the ``value‘‘ of a series is $L$?
Solution:
The sequence of partial sums $s_n = \displaystyle\sum_{k=4}^n a_k$ converges to $L$.
3. Which of the following could be the initial terms of a monotonic sequence?
A. $3,\quad 1,\quad -5,\quad -7,\quad -10,\quad -6,\quad\ldots$
B. $3,\quad -2,\quad -4,\quad -7,\quad -13,\quad -18,\quad\ldots$
C. $3,\quad 0,\quad 3,\quad 2,\quad -4,\quad -5,\quad\ldots$
D. $3,\quad 8,\quad 4,\quad 2,\quad -4,\quad -10,\quad\ldots$
E. $3,\quad -2,\quad -5,\quad -4,\quad -6,\quad -8,\quad\ldots$
Solution:
A monotonic sequence is a sequence which is either increasing, decreasing, non-increasing, or non-decreasing. So B is correct.
4. Consider the sequence $b_m=-4m^2-m-2$. Is the sequence bounded above? Bounded below?
Solution: $$-4m^2-m-2=-4(m+{1\over8})^2-{31\over16} \leq -{31\over16}$$ Bounded above, but not bounded below.
5. Evaluate $\displaystyle\sum_{j=3}^\infty \left(-\displaystyle\frac{3}{10}\right)^{j}$.
Solution:
This is geometric series, $\displaystyle\sum_{n=k}^{\infty}r^n={r^k\over1-r}$, where $|r| < 1$. Thus $$\sum_{j=3}^\infty \left(-\frac{3}{10}\right)^{j}={(-{3\over10})^3\over1-(-{3\over10})}=-\frac{27}{1300}$$
6. Does the series $\displaystyle\sum_{n=7}^\infty \left( \frac{n^{3.0} - 3 \, n^{2}}{n^{5.0} + 3 \, n^{3}} \right)$ converge or diverge?
Solution: $$\frac{n^{3.0} - 3 \, n^{2}}{n^{5.0} + 3 \, n^{3}} < {n^3\over n^5+3n^3} < {n^3\over n^5} = {1\over n^2}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
7. Evaluate $\displaystyle\sum_{m=4}^\infty\frac{20}{16 \, m^{2} + 40 \, m + 21}$.
Solution: $$\frac{20}{16 \, m^{2} + 40 \, m + 21}={20\over(4m+3)(4m+7)} =5\cdot{(4m+7)-(4m+3)\over(4m+3)(4m+7)}=5({1\over4m+3}-{1\over4m+7})$$ Thus $$\sum_{m=4}^\infty\frac{20}{16 \, m^{2} + 40 \, m + 21}=5\cdot({1\over19}-{1\over23}+{1\over23}-{1\over27}+\cdots) ={5\over19}-\lim_{m\to\infty}{1\over4m+7}={5\over19}$$
8. Does the series $\displaystyle\sum_{m=0}^\infty \left(\frac{7}{5 \, m + 20} \right)$ converge or diverge?
Solution: $$\frac{7}{5m + 20} > {5\over5m+20} ={1\over m+4}\to\text{diverge}$$ By $p$-series test and comparison test, it diverges.
9. Suppose $(a_n)$ is a sequence involving both positive and negative numbers, and suppose that the series $\displaystyle\sum_{n=1}^{\infty}|a_n|$ converges. What can be known for certain about the series $\displaystyle\sum_{n=1}^{\infty}a_n$?
Solution:
Since $$0\leq a_n+|a_n| < 2|a_n|\to\text{converge}$$ by comparison test we have $\displaystyle\sum_{n=1}^{\infty}(a_n+|a_n|)$ converges, and hence $$\displaystyle\sum_{n=1}^{\infty}(a_n+|a_n|)-\sum_{n=1}^{\infty}|a_n|=\displaystyle\sum_{n=1}^{\infty}a_n$$ converges. This shows that "absolute convergence implies convergence".
10. Does the series $\displaystyle\sum_{i=5}^\infty \left( \frac{\left(-1\right)^{i}}{i^{1.16}} \right)$ converge or diverge?
Solution:
This is $p$-series where $p=1.16 > 1$, thus it converges absolutely.
11. Does the series $\displaystyle\sum_{n=7}^\infty \left( \frac{10^{n} n!}{\left(2 \, n\right)^{n}} \right)$ converge or diverge?
Solution: $$\lim_{n\to\infty}{a_{n+1}\over a_n}=\lim_{n\to\infty} {10^{n+1}(n+1)!\over(2n+2)^{n+1}}\cdot{(2n)^n\over10^n\cdot n!}$$ $$=\lim_{n\to\infty}{10(n+1)\over2n+2}\cdot\left({2n\over2n+2}\right)^n = {5\over e} > 1$$ By ratio test, it diverges.
12. Does the series $\displaystyle\sum_{n=7}^\infty \left( \left(\frac{7 \, n + 3}{5 \, n + 9}\right)^{n} \right)$ converge or diverge?
Solution: $$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}{7n+3\over5n+9}={7\over5} > 1$$ By root test, it diverges.
13. For which real numbers $x$ does the series $$\sum_{n=1}^\infty \left(\frac{2 \, \left(-\frac{4}{7}\right)^{n}\cdot x^{n}}{5 \, n} \right)$$ converge?
Solution:
By ratio test, we have $$\lim_{n\to\infty}{|a_{n+1}\over |a_n|}\cdot|x| = \lim_{n\to\infty}{({4\over7})^{n+1}\over5(n+1)}\cdot{5n\over({4\over7})^n}\cdot|x| = {4\over7}\cdot|x| < 1\Rightarrow -{7\over4} < x < {7\over4}$$ When $x=\displaystyle-{7\over4}$, $\displaystyle\sum_{n=1}^{\infty}{2\over5n}$ diverges; When $x=\displaystyle{7\over4}$, $\displaystyle\sum_{n=1}^{\infty}{(-1)^n\cdot2\over5n}$ is an alternating series and it is convergent. Thus the interval of convergence is $\displaystyle(-{7\over4}, {7\over4}]$.
14. What is the radius of convergence of the series $\displaystyle\sum_{k=4}^\infty \left( \frac{2 \, {\left(k + 1\right)} x^{k}}{6^{k} + k + 4} \right)$?
Solution:
By ratio test, we have $$\lim_{k\to\infty}{2(k+2)\over6^{k+1}+k+1+4}\cdot{6^k+k+4\over2(k+1)}\cdot|x| ={1\over6}|x| < 1 \Rightarrow -6 < x < 6$$ Thus the radius is $R=6$.
15. Suppose $\displaystyle\sum_{n=1}^\infty b_n = \frac{x}{{\left(x^{2} - 1\right)}^{2}}$. What is the expression of $b_n$?
Solution: $$f(x)=\int{x\over(x^2-1)^2}dx={1\over2}\int{d(x^2-1)\over(x^2-1)^2} $$ $$=-{1\over2}\cdot{1\over x^2-1}={1\over2}\cdot{1\over1-x^2}={1\over2}\cdot\sum_{n=0}^{\infty}x^{2n}$$ Thus $$F(x)=f‘(x)=\sum_{n=1}^{\infty}n\cdot x^{2n-1}=\sum_{n=1}^{\infty}b_n\Rightarrow b_n=n x^{2n-1}$$
16. What are the first few terms of the Taylor series for $f(x) = e^{\left(e^{x} - 1\right)} - 1$ centered around the point $a=0$?
Solution: $$f(0)=e^{\left(e^x-1\right)}-1\big|_{x=0}=0$$ $$f‘(0)=e^{\left(e^x-1\right)}\cdot e^x\big|_{x=0}=1$$ $$f‘‘(0)=e^{\left(e^x-1\right)}\cdot e^{2x}+e^{\left(e^x-1\right)}\cdot e^x\big|_{x=0}=2$$ $$f‘‘‘(0)=e^{\left(e^x-1\right)}\cdot e^{3x}+2e^{\left(e^x-1\right)}\cdot e^{2x}+e^{\left(e^x-1\right)}\cdot e^{2x}+e^{\left(e^x-1\right)}\cdot e^x\big|_{x=0}=5$$ $$\cdots\cdots\cdots$$ Thus $$e^{\left(e^{x} - 1\right)} - 1=0+x+x^2+{5\over6}x^3+\cdots\cdots$$
17. By finding the Taylor series around $x=1$, rewrite the polynomial $p(x)=-3x^3-x^2+x-1$ as a polynomial in $(x-1)$.
Solution: $$p(1)=-3x^3-x^2+x-1\big|_{x=1}=-4$$ $$p‘(1)=-9x^2-2x+1\big|_{x=1}=-10$$ $$p‘‘(1)=-18x-2\big|_{x=1}=-20$$ $$p‘‘‘(1)=-18\big|_{x=1}=-18$$ Thus $$p(x)=-4-10(x-1)-10{(x-1)}^2-3{(x-1)}^3$$
18. By considering Taylor series, evaluate $$\lim_{x \to 0} \frac{{\left(\displaystyle\frac{\cos\left(x\right) - 1}{x} + \sin\left(x\right)\right)}^{2}}{\log\left(x + 1\right) \tan\left(x\right)}.$$
Solution: $$\cos(x)=1-{x^2\over2}+{x^4\over24}+O(x^6)\Rightarrow {\cos(x)-1\over x}=-{1\over2}x+{1\over24}x^3+O(x^5)$$ $$\sin(x)=x-{1\over6}x^3+O(x^5)$$ $$\log(x+1)=x-{1\over2}x^2+{1\over3}x^3-{1\over4}x^4+O(x^5)$$ $$\tan(x)=x+{1\over3}x^3+O(x^5)$$ Thus $$\lim_{x \to 0} \frac{{\left(\displaystyle\frac{\cos\left(x\right) - 1}{x} + \sin\left(x\right)\right)}^{2}}{\log\left(x + 1\right) \tan\left(x\right)} = \lim_{x\to0}{\displaystyle\left({1\over2}x-{1\over8}x^3+O(x^5)\right)^2\over \displaystyle x^2-{1\over2}x^3+O(x^4)}={1\over4}$$
19. Let $f(x)=\cos(x^3)$. By considering the Taylor series for $f$ around 0, compute $f^{(48)}(0)$.
Solution:
Since $$\cos(x)=\sum_{n=0}^{\infty}{(-1)^n\over (2n)!}x^{2n}\Rightarrow \cos(x^3)=\sum_{n=0}^{\infty}{(-1)^n\over (2n)!}x^{6n}$$ And hence we have $${f^{(48)}(0)\over48!}={1\over16!}\Rightarrow f^{(48)}(0)={48!\over16!}$$
20. You may remember that $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2}.$$ By computing some terms of the (alternating!) Taylor series for arctangent, approximate $\arctan \left(\displaystyle\frac{1}{3}\right)$ to within ${1\over100}$.
Solution: $$\arctan(x)=\int{1\over1+x^2}dx=\int{1\over1-(-x^2)}dx$$ $$=\int\sum_{n=0}^{\infty}(-x^2)^ndx=\int\sum_{n=0}^{\infty}(-1)^n\cdot x^{2n}dx=\sum_{n=0}^{\infty}{(-1)^n\over2n+1}x^{2n+1}$$ So $$\arctan\left({1\over3}\right)=\sum_{n=0}^{\infty}{(-1)^n\over2n+1}\cdot\left({1\over3}\right)^{2n+1}$$ We hope to find a term that $|a_n| \leq \displaystyle{1\over100}$. We can write the first few terms: $$\arctan\left({1\over3}\right)={1\over3}-{1\over81} +{1\over1215}+\cdots$$ Thus the first two terms are enough and the result is $\displaystyle{1\over3}-{1\over81}={26\over81}$ .
MOOCULUS微积分-2: 数列与级数学习笔记 Review and Final
标签：style http io ar color os sp for strong
原文地址：http://www.cnblogs.com/zhaoyin/p/4159939.html
