Mooculus calculus-2: Sequence and progression study note 6. Power Series

This course (MOOCULUS-2 "sequences and Series") was taught by Ohio State University on the Coursera platform in 2014.

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Summary

• Let $ (a_n) $ is a sequence of real numbers starting with $a _0$. Then the power series associated to $ (a_n) $ are $$\sum_{n=0}^\infty a_n \, x^n.$$ 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 Convergenc E. is, by ratio test we have $$\lim_{n\to\infty}{|a_{n+1}\cdot X^{n+1}|\over|a_n\cdot x^{n}|} =|x|\cdot\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} < 1$$ it'll converge. Technically, $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over |a_n|} $$
• For a power series, the interval of convergence are, 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) $ is a sequence of real numbers starting with $a _0$. Then the power series centered in $c $ and associated to $ (A_n) $ are the series $$\sum_{n=0}^\infty a_n \, (X-C) ^n. $$ that's, 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_1\cdot (x-a) +a_2\cdot (x-a) ^2+\cdots$$ has radius of Convergence $R $. Then $ $f ' (x) =a_1+2a_2\cdot (x-a) +\cdots=\sum_{n=1}^\infty na_n (x-a) ^{n-1}$$ $$\int f (x) \,dx = c+\sum_{n=0}^\infty {A_n \over n+1} (x-a) ^{n+1}$$ for interval of $x $ in the interval $ (a-r, A+r) $. These the new series has 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, does not attempt to determine whether the endpoints is in the interval of convergence.

1. $$\sum_{n=0}^\infty n x^n$$

Solution: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{n+1\over n}=1$$ Thus $R =1$. When $x =\pm1$, the series is $\sum_{n=0}^{\infty}n$ and $\sum_{n=0}^{\infty} ( -1) ^n\cdot n$, which is diverge. Therefore the interval of convergence is $I = (-1, 1) $.

2. $$\sum_{n=0}^\infty {x^n\over n!} $$  

Solution: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{n!\over (n+1)!} =\lim_{n\to\infty}{1\over n+1}=0$$ Thus $R =\infty$ and the interval of convergence is $I = (-\infty, \infty) $.

3. $$\sum_{n=1}^\infty {n!\over n^n}x^n$$

Solution: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{(n+1)!\over (n+1) ^{n+1}}\cdot{n^n\over n!} =\lim_{n\to\infty} ({N\over n+1}) ^n={1\over e}$$ 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: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{(n+1)!\over (n+1) ^{n+1}}\cdot{n^n\over n!} =\lim_{n\to\infty} ({N\over n+1}) ^n={1\over e}$$ Thus $R =e$ and the interval of convergence is $I = (2-E, 2+e) $.

5. $$\sum_{n=1}^\infty {(n!) ^2\over n^n} (x-2) ^n$$

Solution: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{[(n+1)!] ^2\over (n+1) ^{n+1}}\cdot{n^n\over (n!) ^2}=\lim_{n\to\infty}{(n+1) ^2\over (n+1) ^{n+1}}\cdot n^n$$ $$=\lim_{n\to\infty} (n+1) \cdot ({n\over n+1}) ^n={1\over E }\cdot\lim_{n\to\infty} (n+1) =\infty$$ Thus $R =0$ and it converges only on $x =2$ and diverges otherwise.

6. $$\sum_{n=1}^\infty {(x+5) ^n\over N (n+1)}$$

Solution: $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{n (n+1) \over (n+1) (n+2)}=1$$ Thus $R =1$ and the endpoints are $x _1=-5-1=-6$ and $x _2=-5+1=-4$. Both of them are convergent. The interval of convergence is $I =[-6, -4]$.

7. Find a power series with radius of convergence $0$.  

Solution:

There is 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 $${1\over1-x}=\sum_{n=0}^{\infty}x^n\rightarrow \int {1\over1-x}dx=-\log|1-x|= \sum_{n=0}^{\infty}{1\over n+1} x^{n+1}$$ 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: $${1\over1-x}=\sum_{n=0}^{\infty}x^n\rightarrow ({1\over1-x}) ' ={1\over (1-x) ^2}=\sum_{n=1}^{\ infty}nx^{n-1}$$

3. Find a power series representation for $2/(1-x) ^3$.

Solution: $${1\over1-x}=\sum_{n=0}^{\infty}x^n\rightarrow ({1\over1-x}) ' ={1\over (1-x) ^2}=\sum_{n=1}^{\ infty}nx^{n-1}$$ $$\rightarrow ({1\over1-x}) ' ={2\over (1-x) ^3}=\sum_{n=2}^{\infty}n (n-1) x^{n-2}$$

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 $${1\over (1-x) ^3}=\sum_{n=2}^{\infty}{n (n-1) \over2}x^{n-2}=\sum_{n=0}^{\ infty}{(n+1) (n+2) \over2}x^{n}$$ and $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{(n+2) (n+3) \over (n+1) (n+2)}=1$$ Thus The radius is $R =1$.

5. Find a power series representation for $\int\log (1-x) \,dx$.

Solution: $$\log (1-x) =-\int {1\over1-x}dx=-\int\sum_{n=0}^{\infty}x^n Dx=\sum_{n=0}^{\infty}{-1\over n+1} x^{n+1}$$ $$\rightarrow \int\log (1-x) dx=\int\sum_{n=0}^{\infty}{-1\over n+1}x^{n+1}dx=c+\sum_{n=0}^{\infty}{-1\ Over (n+1) (n+2)}x^{n+2}$$

Additional Exercises

1. For which real number $x $ does the series $$\sum_{m=4}^{\infty}{({1\OVER6}) ^m\cdot x^m\over7m}$$ converge.

Solution:

Let $a _m={({1\OVER6}) ^m\over7m}$, we have $${1\over r}=\lim_{m\to\infty}{|a_{m+1}|\over |a_m|} =\lim_{m\to\infty}{({1\OVER6}) ^{m+1}\over7 (m+1)}\cdot{7m\over ({1\over6}) ^m}={1\over6}$$ Thus $R =6$. The endpoints is $x _1=-6$ and $x _2=6$. When $x =-6$, we have $$\sum_{m=4}^{\infty}{({1\OVER6}) ^m\cdot x^m\over7m}=\sum_{m=4}^{\infty}{( -1) ^m\over7m}$$ which is an alternating harmonic series, and it converges. When $x =6$, we had $$\sum_{m=4}^{\infty}{({1\OVER6}) ^m\cdot x^m\over7m}=\sum_{m=4}^{\infty}{1\over7m}$$ 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 $$\sum_{n=4}^{\infty}{(8^n+n) \cdot x^n\over3n}$$

Solution:

Let $a _n={8^n+n\over3n}$, we have $${1\over r}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} =\lim_{n\to\infty}{8^{n+1}+n+1\over3 (n+1)}\cdot{3n\over 8^n+n}=8$$ Thus The radius is $R ={1\over8}$.

3. $ $f (x) =\sum_{n=0}^{\infty}-{(5n+3) x^n\over2n-3}$$ consider $f ' (x) $.

Solution: $ $f ' (x) =\sum_{n=1}^{\infty}-{5n+3\over2n-3}\cdot N\cdot x^{n-1}=\sum_{n=0}^{\infty}-{5 (n+1) +3\ Over2 (n+1) -3}\cdot (n+1) \cdot x^{n+1-1}$$ $$=\sum_{n=0}^{\infty}-{5n+8\over2n-1}\cdot (n+1) \cdot x^{n}$$

4. Suppose $$\sum_{n=1}^{\infty}b_n={2\over (2x-1) ^2}$$ Find An expression of $b _n$ (involve $x $).

Solution:

Let $F (x) =\sum_{n=1}^{\infty}b_n={2\over (2x-1) ^2}$, we have $$\int F (x) =\int {2\over (2x-1) ^2} dx=\int{d (2x-1) \over ( 2X-1) ^2}={1\over1-2x}=f (x) $$ on the other hand, $ $f (x) ={1\over1-2x}=\sum_{n=0}^{\infty} (2x) ^n$$ Thus $ $F (x) =f ' (x) =\ Sum_{n=1}^{\infty}2^n\cdot N\cdot x^{n-1}$$ $$\rightarrow b_n=2^n\cdot N\cdot x^{n-1}$$

5. Suppose $$\sum_{n=1}^{\infty}b_n={9x\over (9x^2-1) ^2}$$ Find An expression of $b _n$ (involve $x $).

Solution:

Let $F (x) =\sum_{n=1}^{\infty}b_n={9x\over (9x^2-1) ^2}$, we have $$\int F (x) =\int {9x\over (9x^2-1) ^2}dx={1\over2}\cdot \int {d (9x^2-1) \over (9x^2-1) ^2}={1\over2}\cdot{1\over1-9x^2}=f (x) $$ on the other hand, $ $f (x) ={1\over2}\cdot\sum_{n= 0}^{\infty} (9x^2) ^n$$ Thus $ $F (x) =f ' (x) ={1\over2}\cdot\sum_{n=1}^{\infty}9^n\cdot 2n\cdot x^{2n-1}$$ $$\Rightarrow b _n=9^n\cdot N\cdot x^{2n-1}$$

6. Consider the function $ $f (t) =\int_{0}^{t}e^{-x^2}dx$$ Compute $f ({3\over2}) $ to within ${1\over2}$.

Solution:

Note that the Power series (Taylor series) of $e ^x$ is $ $e ^x=\sum_{n=0}^{\infty}{x^n\over n!} $$ Thus We have $ $e ^{-x^2}=\sum_{n=0}^{\infty}{{(-x^2)}^n\over n!} =\sum_{n=0}^{\infty}{( -1) ^n\cdot X^{2n}\over n!} $$ $$\rightarrow F (t) =\int_{0}^{t}e^{-x^2}dx=\int_{0}^{t}\sum_{n=0}^{\infty}{( -1) ^n\cdot X^{2n}\over n!} dx$$ $$=\sum_{n=0}^{\infty}{( -1) ^n\over n!} \cdot\int_{0}^{t}x^{2n}dx=\sum_{n=0}^{\infty}{( -1) ^n\over n!} \cdot{1\over 2n+1}x^{2n+1}\big|_{0}^{t}$$ $$=\sum_{n=0}^{\infty}{( -1) ^n\over n!} \cdot{1\over 2n+1}t^{2n+1}$$ $$\rightarrow f ({3\over2}) =\sum_{n=0}^{\infty}{( -1) ^n\over n!} \cdot{1\over 2n+1} ({3\over2}) ^{2n+1}=a_n$$ Our aim was to find a $|a_n| < 0.5$ 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's, $ $a _0=1.5,\ a_1=-1.125,\ a_2=0.759375,\ a_3=-0.406808 \in ( -0.5, 0.5) $$ Thus The value within 0.5 is $${1\over2}\ CDOT (S_2+s_3) =0.930971$$

Mooculus calculus-2: Sequence and progression study note 6. Power Series