此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。
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Summary
- Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series associated to $(a_n)$ is $$sum_{n=0}^infty a_n \, x^n.$$ Note that $a_n$ does not depend on $x$.
- The set of values of $x$ for which the series $$sum_{n=0}^infty a_n \, x^n$$ converges is the interval of convergence. That is, by ratio test we have $$lim_{n oinfty}{|a_{n+1}cdot x^{n+1}|over|a_ncdot x^{n}|}=|x|cdotlim_{n oinfty}{|a_{n+1}|over|a_n|} < 1$$ it will converge. Technically, $${1over R}=lim_{n oinfty}{|a_{n+1}|over |a_n|}$$
- For a power series, the interval of convergence is, in fact, an interval. It has the form $(-R,R)$ or $[-R,R)$ or $(-R,R]$ or $[-R,R]$. In short, it is centered around $0$.
- In the interval of convergence of a power series, the value $R$ is called the radius of convergence of the series.
- Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series centered at $c$ and associated to $(a_n)$ is the series $$sum_{n=0}^infty a_n \, (x-c)^n.$$ That is, the interval of convergence is $I=(c-R, c+R)$ (or include the endpoints).
- Suppose the power series $$f(x)=sum_{n=0}^infty a_n(x-a)^n=a_0+a_1cdot(x-a)+a_2cdot(x-a)^2+cdots$$ has radius of convergence $R$. Then $$f'(x)=a_1+2a_2cdot(x-a)+cdots=sum_{n=1}^infty na_n(x-a)^{n-1}$$ $$int f(x)\,dx = C+sum_{n=0}^infty {a_nover n+1}(x-a)^{n+1}$$ for interval of $x$ in the interval $(a-R, a+R)$. These two new series have radius of convergence $R$, just like the original series.
Exercises 6.3
Find the radius and interval of convergence for each series. In exercises 3 and 4, do not attempt to determine whether the endpoints are in the interval of convergence.
1. $$sum_{n=0}^infty n x^n$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{n+1over n}=1$$ Thus $R=1$. When $x=pm1$, the series are $sum_{n=0}^{infty}n$ and $sum_{n=0}^{infty}(-1)^ncdot n$, which are diverge. Therefore the interval of convergence is $I=(-1, 1)$.
2. $$sum_{n=0}^infty {x^nover n!}$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{n!over (n+1)!}=lim_{n oinfty}{1over n+1}=0$$ Thus $R=infty$ and the interval of convergence is $I=(-infty, infty)$.
3. $$sum_{n=1}^infty {n!over n^n}x^n$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{(n+1)!over(n+1)^{n+1}}cdot{n^nover n!}=lim_{n oinfty}({nover n+1})^n={1over e}$$ Thus $R=e$ and the interval of convergence is $I=(-e, e)$.
4. $$sum_{n=1}^infty {n!over n^n}(x-2)^n$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{(n+1)!over(n+1)^{n+1}}cdot{n^nover n!}=lim_{n oinfty}({nover n+1})^n={1over e}$$ Thus $R=e$ and the interval of convergence is $I=(2-e, 2+e)$.
5. $$sum_{n=1}^infty {(n!)^2over n^n}(x-2)^n$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{[(n+1)!]^2over(n+1)^{n+1}}cdot{n^nover(n!)^2}=lim_{n oinfty}{(n+1)^2over(n+1)^{n+1}}cdot n^n$$ $$=lim_{n oinfty}(n+1)cdot({nover n+1})^n={1over e}cdotlim_{n oinfty}(n+1)=infty$$ Thus $R=0$ and it converges only on $x=2$ and diverges otherwise.
6. $$sum_{n=1}^infty {(x+5)^nover n(n+1)}$$
Solution: $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{n(n+1)over(n+1)(n+2)}=1$$ Thus $R=1$ and the endpoints are $x_1=-5-1=-6$ and $x_2=-5+1=-4$. Both of them are convergent. So the interval of convergence is $I=[-6, -4]$.
7. Find a power series with radius of convergence $0$.
Solution:
There are many choices---for instance, see Exercise 5---alternatively $sum_{n=0}^infty n! cdot x^n$ also works.
Exercises 6.4
1. Find a series representation for $log 2$.
Solution:
Begin with the geometric series, namely $${1over1-x}=sum_{n=0}^{infty}x^nRightarrow int {1over1-x}dx=-log|1-x|= sum_{n=0}^{infty}{1over n+1} x^{n+1}$$ So $x=-1$ and the result is $$log2=-sum_{n=0}^{infty}{(-1)^{n+1}over n+1} = sum_{n=0}^{infty}{(-1)^{n} over n+1}$$
2. Find a power series representation for $1/(1-x)^2$.
Solution: $${1over1-x}=sum_{n=0}^{infty}x^nRightarrow ({1over1-x})'={1over(1-x)^2}=sum_{n=1}^{infty}nx^{n-1}$$
3. Find a power series representation for $2/(1-x)^3$.
Solution: $${1over1-x}=sum_{n=0}^{infty}x^nRightarrow ({1over1-x})'={1over(1-x)^2}=sum_{n=1}^{infty}nx^{n-1}$$ $$Rightarrow ({1over1-x})''={2over(1-x)^3}=sum_{n=2}^{infty}n(n-1)x^{n-2}$$
4. Find a power series representation for $1/(1-x)^3$. What is the radius of convergence?
Solution:
According to the above exercise, we have $${1over(1-x)^3}=sum_{n=2}^{infty}{n(n-1)over2}x^{n-2}=sum_{n=0}^{infty}{(n+1)(n+2)over2}x^{n}$$ And $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{(n+2)(n+3)over(n+1)(n+2)}=1$$ Thus the radius is $R=1$.
5. Find a power series representation for $intlog(1-x)\,dx$.
Solution: $$log(1-x)=-int {1over1-x}dx=-intsum_{n=0}^{infty}x^n dx=sum_{n=0}^{infty}{-1over n+1}x^{n+1}$$ $$Rightarrow intlog(1-x)dx=intsum_{n=0}^{infty}{-1over n+1}x^{n+1}dx=C+sum_{n=0}^{infty}{-1over(n+1)(n+2)}x^{n+2}$$
Additional Exercises
1. For which real number $x$ does the series $$sum_{m=4}^{infty}{({1over6})^mcdot x^mover7m}$$ converge.
Solution:
Let $a_m={({1over6})^mover7m}$, we have $${1over R}=lim_{m oinfty}{|a_{m+1}|over |a_m|}=lim_{m oinfty}{({1over6})^{m+1}over7(m+1)}cdot{7mover({1over6})^m}={1over6}$$ Thus $R=6$. The endpoints are $x_1=-6$ and $x_2=6$. When $x=-6$, we have $$sum_{m=4}^{infty}{({1over6})^mcdot x^mover7m}=sum_{m=4}^{infty}{(-1)^mover7m}$$ which is an alternating harmonic series, and it converges. When $x=6$, we have $$sum_{m=4}^{infty}{({1over6})^mcdot x^mover7m}=sum_{m=4}^{infty}{1over7m}$$ which is harmonic series, and it diverges. Thus the interval of converges is $I=[-6, 6)$.
2. Which is the radius of convergence of the series $$sum_{n=4}^{infty}{(8^n+n)cdot x^nover3n}$$
Solution:
Let $a_n={8^n+nover3n}$, we have $${1over R}=lim_{n oinfty}{|a_{n+1}|over|a_n|}=lim_{n oinfty}{8^{n+1}+n+1over3(n+1)}cdot{3nover 8^n+n}=8$$ Thus the radius is $R={1over8}$.
3. $$f(x)=sum_{n=0}^{infty}-{(5n+3)x^nover2n-3}$$ Consider $f'(x)$.
Solution: $$f'(x)=sum_{n=1}^{infty}-{5n+3over2n-3}cdot ncdot x^{n-1}=sum_{n=0}^{infty}-{5(n+1)+3over2(n+1)-3}cdot (n+1)cdot x^{n+1-1}$$ $$=sum_{n=0}^{infty}-{5n+8over2n-1}cdot (n+1)cdot x^{n}$$
4. Suppose $$sum_{n=1}^{infty}b_n={2over(2x-1)^2}$$ Find an expression of $b_n$ (involve $x$).
Solution:
Let $F(x)=sum_{n=1}^{infty}b_n={2over(2x-1)^2}$, we have $$int F(x)=int {2over(2x-1)^2} dx=int{d(2x-1)over(2x-1)^2}={1over1-2x}=f(x)$$ On the other hand, $$f(x)={1over1-2x}=sum_{n=0}^{infty}(2x)^n$$ Thus $$F(x)=f'(x)=sum_{n=1}^{infty}2^ncdot ncdot x^{n-1}$$ $$Rightarrow b_n=2^ncdot ncdot x^{n-1}$$
5. Suppose $$sum_{n=1}^{infty}b_n={9xover(9x^2-1)^2}$$ Find an expression of $b_n$ (involve $x$).
Solution:
Let $F(x)=sum_{n=1}^{infty}b_n={9xover(9x^2-1)^2}$, we have $$int F(x)=int {9xover(9x^2-1)^2}dx={1over2}cdotint {d(9x^2-1)over(9x^2-1)^2}={1over2}cdot{1over1-9x^2}=f(x)$$ On the other hand, $$f(x)={1over2}cdotsum_{n=0}^{infty}(9x^2)^n$$ Thus $$F(x)=f'(x)={1over2}cdotsum_{n=1}^{infty}9^ncdot 2ncdot x^{2n-1}$$ $$Rightarrow b_n=9^ncdot ncdot x^{2n-1}$$
6. Consider the function $$f(t)=int_{0}^{t}e^{-x^2}dx$$ Compute $f({3over2})$ to within ${1over2}$.
Solution:
Note that the power series (Taylor series) of $e^x$ is $$e^x=sum_{n=0}^{infty}{x^nover n!}$$ Thus we have $$e^{-x^2}=sum_{n=0}^{infty}{{(-x^2)}^nover n!}=sum_{n=0}^{infty}{(-1)^ncdot x^{2n}over n!}$$ $$Rightarrow f(t)=int_{0}^{t}e^{-x^2}dx=int_{0}^{t}sum_{n=0}^{infty}{(-1)^ncdot x^{2n}over n!}dx$$ $$=sum_{n=0}^{infty}{(-1)^nover n!}cdotint_{0}^{t}x^{2n}dx=sum_{n=0}^{infty}{(-1)^nover n!}cdot{1over 2n+1}x^{2n+1}Big|_{0}^{t}$$ $$=sum_{n=0}^{infty}{(-1)^nover n!}cdot{1over 2n+1}t^{2n+1}$$ $$Rightarrow f({3over2})=sum_{n=0}^{infty}{(-1)^nover n!}cdot{1over 2n+1}({3over2})^{2n+1}=a_n$$ Our aim is to find an $|a_n| < 0.5$ and by computing in R:
f = function(x) (-1)^x / factorial(x) * 1 / (2 * x + 1) * (3/2)^(2 * x + 1)
for (i in 0:100) {
if (f(i) < 0.5 & f(i) > -0.5) {
print(i)
print(f(x = 0:i))
print((sum(f(0:i)) + sum(f(0:(i-1)))) / 2)
break
}
}
# [1] 3
# [1] 1.500000 -1.125000 0.759375 -0.406808
# [1] 0.930971
That is, $$a_0=1.5, a_1=-1.125, a_2=0.759375, a_3=-0.406808 in (-0.5, 0.5)$$ Thus the value within 0.5 is $${1over2}cdot (s_2+s_3)=0.930971$$