Week 5 - Power Series #

Definition #

A power series is a series of the form $$\sum_{n=0}^{\infty}c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$$ where x is a variable and the $c_n$’s are constants called the coefficients of the series.
The sum of the series is a function $$f(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n + \cdots$$ whose domain is the set of all x for which the series converges. Notice that $f$ resembles a polynomial.

Convergence of Power Series #

Interval of convergence #

$$C = { x \in \mathbb{R} | \sum_{n = 0}^{\infty} a_n x^n\\ \text{converge}}$$
C is called the interval of convergence.

Theorem #

Suppose $\displaystyle \sum_{n = 0}^{\infty} a_n x^n$ converge when $x = x_0$, then the series converge absolutely x between $-x_0$ and $x_0$.
Prove:
- Suppose $\displaystyle \sum_{n = 0}^{\infty} a_n x^n$ converge when $x = x_0$
- So, $\displaystyle \lim_{n \to \infty} a_n x_0^n = 0$,
- There is an M, so that for all n, $|a_n x_0^n| \le M$,
- Pick $x \in (-|x_0|, |x_0|)$,
- $|a_n x^n| = |a_n x_0^n| \cdot |\frac{x^n}{x_0^n}| \le M \cdot |\frac{x^n}{x_0^n}|$
- With ratio test, $r = |\frac{x^n}{x_0^n}| < 1$, we can get $\sum_{n = 0}^{\infty} M \cdot r^n$ converge,
- So $\sum_{n=0}^{\infty} |a_n x^n|$ converge.
Corollary:
- Consider $\sum_{n=0}^{\infty} a_n x^n$.
- There is an R, so that the series,
  - converges absolutely for $x \in (-R, R)$,
  - diverges for $x > R \text{ or } x < -R$
- The number R is called the radius of convergence of the power series.

Example #

What’s the interval of convergence of the series $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n}$
First, we ask the easier question. What’s its radius of convergence?
- Think this series converge absolutely, and use ratio test:
  - $\displaystyle\lim_{n \to \infty}|\frac{x^{n+1}/{n+1} }{x^n/n}| = |x|$
  - $|x|<1$, the series converges,
  - $|x|>1$, the series diverges.
- So the radius of convergence is 1.
What about the endpoints 1 and -1?
- if x = 1, then the series = $\sum_{n=1}^{\infty} \frac{1}{n}$, which is harmonic series, also diverges.
- if x = -1, then the series = $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, that’s the alternating harmonic series, and converges.
To summarize this:
- The interval of convergence of series $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n}$ is $[-1, 1)$

Differentiation and integration of power Series #

Power Series Centered Around a #

$\displaystyle\sum_{n=0}^{\infty}c_n (x - a)^n$
Use ratio test:
- $\displaystyle \lim_{n \to \infty} |\frac{c_{n+1} \cdot (x - a)^{n+1} }{c_n \cdot (x - a)^n}| = \lim_{n \to \infty} |\frac{c_{n+1} }{c_n}| \cdot |x - a|$
- To get the interval of convergence, let’s set $|\frac{c_{n+1} }{c_n}| = \frac{1}{R}$
  - $\frac{1}{R} \cdot |x - a| < 1$
  - $a - R < x < a + R$

Differentiate a Power Series #

Theorem:
- $\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n$, R = radius of convergence.
- Then $\displaystyle f’(x) = \sum_{n=1}^{\infty} n \cdot a_n \cdot x^{n-1} \text{ for } x \in (-R, R)$
  - Notice the index of n start from 1, because when $n = 0, f’(x) = 0$

Integrate a Power Series #

Theorem:
- $\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n$, R = radius of convergence.
- Then $\displaystyle \int_0^t f(x) dx = \sum_{n=1}^{\infty} \frac{a_n \cdot t^{n+1} }{n+1} \text{ for } x \in (-R, R)$
Example: $\displaystyle \int_{x=0}^{t} \sum_{n=0}^{\infty} x^n dx$
- First to prove: $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$, $|x| < 1$ => $R = 1$
  - Use Geometric Series theorem, we know $\displaystyle \sum_{n=1}^{\infty} x^n = \sum_{n=1}^{\infty} x \cdot x^{n-1} = \frac{x}{1-x}$
  - So $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{x}{1-x} + x^0 = \frac{1}{1-x}$
- $\displaystyle \int_{x = 0}^t \frac{1}{1-x} dx$, use substitution rule, set $u = 1 - x, du = -dx$
- $\displaystyle \int_{x = 0}^t \frac{1}{1-x} dx = - \int_{x = 0}^t \frac{du}{u}$
- $= -\log|u| \rbrack_{x=0}^t = - \log|1-x|\rbrack_{x=0}^t = - \log |1 - t|$
- In another way, we can backwards the derivation:
  - $\displaystyle - \log |1 - t| = \int_{x=0}^t \frac{1}{1-x} dx$
  - $\displaystyle = \int_{x=0}^{t} \sum_{n=0}^{\infty} x^n dx = \sum_{n=0}^{\infty} \int_{x=0}^{t} x^n dx$
  - $\displaystyle = \sum_{n=0}^{\infty} \frac{t^{n+1} }{n+1}, |t| < 1$

e^x #

To prove $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$
- It’s the sum, n goes from 0 to infinity of x to the n over n factorial
$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
prove $f(0) = e^0 = 0$:
- $\displaystyle f(0) = \sum_{n=0}^{\infty} \frac{0^n}{n!} = 1$
  - $0^0 = 1, 0^1 = 0, \ldots, 0^n = 0$
  - $0! = 1$. because: $(n+1)! = (n+1) \cdot n!$
- $e^0 = 1$
after differentiation, both functions still themselves:
- $\displaystyle \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{n-1} }{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
- $\frac{d}{dx}e^x = e^x$
These two star-crossed functions agree at a single point, and they’re changing in the same way, and consequently, they must be the same function.

Multiply Two Power Series #

$\displaystyle (\sum_{n=0}^{\infty} a_n x^n) \cdot (\sum_{n=0}^{\infty} b_n x^n)$
$=(a_0 + a_1 x + a_2 x^2 + \cdots) \cdot (b_0 + b_1 x + b_2 x^2 + \cdots)$
$= a_0 b_0 + (a_1 + b_1) x + (a_0 b_2 + a_1 b_1 + a_2 b_0) x^2 + \cdots$
$\displaystyle (\sum_{n=0}^{\infty} a_n x^n) \cdot (\sum_{n=0}^{\infty} b_n x^n) = \sum_{n=0}^{\infty} (\sum_{i=0}^{n} a_i b_{n-i}) x^n$

Theorem #

$f(x) = \displaystyle (\sum_{n=0}^{\infty} a_n x^n)$, $g(x) = \displaystyle (\sum_{n=0}^{\infty} b_n x^n)$, and their radius of convergence $\ge \mathbb{R}$
Then we can get: $\displaystyle f(x) g(x) = \sum_{n=0}^{\infty} (\sum_{i=0}^{n} a_i b_{n-i}) x^n$, for $x \in (-R, R)$
Example:
- $\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$
- $(e^x)^2 = e^{(2x)}$
- $e^{(2x)} = 1 + 2x + 2x^2 + \frac{8x^3}{6} + \cdots$
- $(e^x)^2 = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots)^2$
  - $= 1 + 2x + (\frac{1}{2} + 1 + \frac{1}{2}) x^2 + (\frac{1}{6} + \frac{1}{2} + \frac{1}{2} + \frac{1}{6}) x^3 + \cdots$

Fibonacci Numbers #

Transform 1/(1-x) #

$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$
Two ways to transfer $\frac{1}{(1-x)^2}$
- First:
  - $\displaystyle \frac{1}{(1-x)^2} = \frac{d}{dx} (\frac{1}{1-x}) = \frac{d}{dx} \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} \frac{d}{dx}(x^n) = \sum_{n=1}^{\infty} n \cdot x^{n-1}$
  - $= 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + \cdots$
  - $\displaystyle = \sum_{n=0}^{\infty} (n+1) \cdot x^n$
- Second:
  - $\displaystyle (\sum_{n=0}^{\infty} x^n)^2 = (1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots)$
  - $= 1 + 2x + 3x^2 + 4x^3 + \cdots$
  - $\displaystyle = \sum_{n=0}^{\infty} (n+1) \cdot x^n$
  - Or we can use the theorem of Multiply Two Power Series
    - $\displaystyle \sum_{n=0}^{\infty} (\sum_{i=0}^{n} a_i b_{n-i}) x^n$
    - In this case, $a_i = b_{n-i} = 1$,
    - So $\displaystyle = \sum_{n=0}^{\infty} (\sum_{i=0}^{n} 1) x^n = \sum_{n=0}^{\infty} (n+1) \cdot x^n$

A Formula for the Fibonacci Numbers #

Let’s say, ${a_n}$ is Fibonacci Sequence, and $f(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$
- $= x + x^2 + 2x^3 + 3x^4 + 5x^5 + \cdots$
Then:
- $x \cdot f(x) = x^2 + x^3 + 2x^4 + 3x^5 + 5x^6 + \cdots$
- $x^2 \cdot f(x) = x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + \cdots$
Conclude above equations, we get:
- $f(x) - x \cdot f(x) - x^2 \cdot f(x) = x$
So $\displaystyle f(x) = \frac{x}{1-x-x^2}$
set $\displaystyle \phi = \frac{1+\sqrt{5} }{2}$, after some calculation we get:
$$\begin{aligned} f(x) &= \frac{1/{\sqrt{5} } }{1-(x \cdot \phi)} + \frac{-1/{\sqrt{5} } }{1-(x \cdot (1-\phi))} \\ &= \frac{1}{\sqrt{5} } \cdot \frac{1}{1-(x \cdot \phi)} + \frac{-1}{\sqrt{5} } \cdot \frac{1}{1-(x \cdot (1-\phi))} \\ &= \frac{1}{\sqrt{5} } \sum_{n=0}^{\infty} (x \cdot \phi)^n + \frac{-1}{\sqrt{5} } \sum_{n=0}^{\infty} (x \cdot (1 - \phi))^n \\ &= \sum_{n=0}^{\infty} (\frac{1}{\sqrt{ 5 } } \cdot \phi^{n} - \frac{1}{\sqrt{ 5 } } \cdot (1 - \phi)^{n}) x^n \\ &= \sum_{n=0}^{\infty} \frac{\phi^n - (1 - \phi)^n}{\sqrt{5} } x^n = \sum_{n=0}^{\infty} a_n x^n \end{aligned}$$
So, $\displaystyle a_n = \frac{\phi^n - (1 - \phi)^n}{\sqrt{5} }$