Week 5 - Power Series

1. Definition

  • A power series is a series of the form n=0cnxn=c0+c1x+c2x2+c3x3+\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 cnc_n’s are constants called the coefficients of the series.
  • The sum of the series is a function f(x)=c0+c1x+c2x2++cnxn+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 ff resembles a polynomial.

2. Convergence of Power Series

2.1. Interval of convergence

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

2.2. Theorem

  • Suppose n=0anxn\displaystyle \sum_{n = 0}^{\infty} a_n x^n converge when x=x0x = x_0, then the series converge absolutely x between x0-x_0 and x0x_0.

  • Prove:

    • Suppose n=0anxn\displaystyle \sum_{n = 0}^{\infty} a_n x^n converge when x=x0x = x_0
    • So, limnanxon=0\lim_{n \to \infty} a_n x_o^n = 0,
    • There is an M, so that for all n, anx0nM|a_n x_0^n| \le M,
    • Pick x(x0,x0)x \in (-|x_0|, |x_0|),
    • anxn=anx0nxnx0nMxnx0n|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=xnx0n<1r = |\frac{x^n}{x_0^n}| < 1, we can get n=0Mrn\sum_{n = 0}^{\infty} M \cdot r^n converge,
    • So n=0anxn\sum_{n=0}^{\infty} |a_n x^n| converge.
  • Corollary:

    • Consider n=0anxn\sum_{n=0}^{\infty} a_n x^n.
    • There is an R, so that the series,
      • converges absolutely for x(R,R)x \in (-R, R),
      • diverges for x>R or x<Rx > R \text{ or } x < -R
    • The number R is called the radius of convergence of the power series.

2.3. Example

  • What's the interval of convergence of the series n=1xnn\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:
      • limnxn+1/n+1xn/n=x\displaystyle\lim_{n \to \infty}|\frac{x^{n+1}/{n+1} }{x^n/n}| = |x|
      • x<1|x|<1, the series converges,
      • x>1|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 = n=11n\sum_{n=1}^{\infty} \frac{1}{n}, which is harmonic series, also diverges.
    • if x = -1, then the series = n=1(1)nn\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 n=1xnn\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n} is [1,1)[-1, 1)

3. Differentiation and integration of power Series

3.1. Power Series Centered Around a

  • n=0cn(xa)n\displaystyle\sum_{n=0}^{\infty}c_n (x - a)^n
  • Use ratio test:
    • limnCn+1(xa)n+1Cn(xa)n=limnCn+1Cnxa\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 Cn+1Cn=1R|\frac{C_{n+1} }{C_n}| = \frac{1}{R}
      • 1Rxa<1\frac{1}{R} \cdot |x - a| < 1
      • aR<x<a+Ra - R < x < a + R

3.2. Differentiate a Power Series

  • Theorem:
    • f(x)=n=0anxn\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n, R = radius of convergence.
    • Then f(x)=n=1nanxn1 for x(R,R)\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)=0n = 0, f'(x) = 0

3.3. Integrate a Power Series

  • Theorem:
    • f(x)=n=0anxn\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n, R = radius of convergence.
    • Then 0tf(x)dx=n=1antn+1n+1 for x(R,R)\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: x=0tn=0xndx\displaystyle \int_{x=0}^{t} \sum_{n=0}^{\infty} x^n dx
    • First to prove: n=0xn=11x\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, x<1|x| < 1 => R=1R = 1
      • Use Geometric Series theorem, we know n=1xn=n=1xxn1=x1x\displaystyle \sum_{n=1}^{\infty} x^n = \sum_{n=1}^{\infty} x \cdot x^{n-1} = \frac{x}{1-x}
      • So n=0xn=x1x+x0=11x\displaystyle \sum_{n=0}^{\infty} x^n = \frac{x}{1-x} + x^0 = \frac{1}{1-x}
    • x=0t11xdx\displaystyle \int_{x = 0}^t \frac{1}{1-x} dx, use substitution rule, set u=1x,du=dxu = 1 - x, du = -dx
    • x=0t11xdx=x=0tduu\displaystyle \int_{x = 0}^t \frac{1}{1-x} dx = - \int_{x = 0}^t \frac{du}{u}
    • =logu]x=0t=log1x]x=0t=log1t= -\log|u| \rbrack_{x=0}^t = - \log|1-x|\rbrack_{x=0}^t = - \log |1 - t|
    • In another way, we can backwards the derivation:
      • log1t=x=0t11xdx\displaystyle - \log |1 - t| = \int_{x=0}^t \frac{1}{1-x} dx
      • =x=0tn=0xndx=n=0x=0txndx\displaystyle = \int_{x=0}^{t} \sum_{n=0}^{\infty} x^n dx = \sum_{n=0}^{\infty} \int_{x=0}^{t} x^n dx
      • =n=0tn+1n+1,t<1\displaystyle = \sum_{n=0}^{\infty} \frac{t^{n+1} }{n+1}, |t| < 1

3.4. e^x

  • To prove n=0xnn!=ex\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
  • f(x)=n=0xnn!\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}
  • prove f(0)=e0=0f(0) = e^0 = 0:
    • f(0)=n=00nn!=1\displaystyle f(0) = \sum_{n=0}^{\infty} \frac{0^n}{n!} = 1
      • 00=1,01=0,,0n=00^0 = 1, 0^1 = 0, \ldots, 0^n = 0
      • 0!=10! = 1. because: (n+1)!=(n+1)n!(n+1)! = (n+1) \cdot n!
    • e0=1e^0 = 1
  • after differentiation, both functions still themselves:
    • ddxn=0xnn!=n=0xn1(n1)!=n=0xnn!\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!}
    • ddxex=ex\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.

3.5. Multiply Two Power Series

  • (n=0anxn)(n=0bnxn)\displaystyle (\sum_{n=0}^{\infty} a_n x^n) \cdot (\sum_{n=0}^{\infty} b_n x^n)
  • =(a0+a1x+a2x2+)(b0+b1x+b2x2+)=(a_0 + a_1 x + a_2 x^2 + \cdots) \cdot (b_0 + b_1 x + b_2 x^2 + \cdots)
  • =a0b0+(a1+b1)x+(a0b2+a1b1+a2b0)x2+= a_0 b_0 + (a_1 + b_1) x + (a_0 b_2 + a_1 b_1 + a_2 b_0) x^2 + \cdots
  • (n=0anxn)(n=0bnxn)=n=0(i=0naibni)xn\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)=(n=0anxn)f(x) = \displaystyle (\sum_{n=0}^{\infty} a_n x^n), g(x)=(n=0bnxn)g(x) = \displaystyle (\sum_{n=0}^{\infty} b_n x^n), and their radius of convergence R\ge \mathbb{R}
  • Then we can get: f(x)g(x)=n=0(i=0naibni)xn\displaystyle f(x) g(x) = \sum_{n=0}^{\infty} (\sum_{i=0}^{n} a_i b_{n-i}) x^n, for x(R,R)x \in (-R, R)

  • Example:

    • ex=n=0xnn!=1+x+x22+x36+\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots
    • (ex)2=e(2x)(e^x)^2 = e^{(2x)}
    • e(2x)=1+2x+2x2+8x36+e^{(2x)} = 1 + 2x + 2x^2 + \frac{8x^3}{6} + \cdots
    • (ex)2=(1+x+x22+x36+)2(e^x)^2 = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots)^2
      • =1+2x+(12+1+12)x2+(16+12+12+16)x3+= 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

4. Fibonacci Numbers

4.1. Transform 1/(1-x)

  • n=0xn=11x\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}
  • Two ways to transfer 1(1x)2\frac{1}{(1-x)^2}
    • First:
      • 1(1x)2=ddx(11x)=ddxn=0xn=n=0ddx(xn)=n=1nxn1\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}
      • =1x0+2x1+3x2+= 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + \cdots
      • =n=0(n+1)xn\displaystyle = \sum_{n=0}^{\infty} (n+1) \cdot x^n
    • Second:
      • (n=0xn)2=(1+x+x2+x3+)(1+x+x2+x3+)\displaystyle (\sum_{n=0}^{\infty} x^n)^2 = (1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots)
      • =1+2x+3x2+4x3+= 1 + 2x + 3x^2 + 4x^3 + \cdots
      • =n=0(n+1)xn\displaystyle = \sum_{n=0}^{\infty} (n+1) \cdot x^n
      • Or we can use the theorem of Multiply Two Power Series
        • n=0(i=0naibni)xn\displaystyle \sum_{n=0}^{\infty} (\sum_{i=0}^{n} a_i b_{n-i}) x^n
        • In this case, ai=bni=1a_i = b_{n-i} = 1,
        • So =n=0(i=0n1)xn=n=0(n+1)xn\displaystyle = \sum_{n=0}^{\infty} (\sum_{i=0}^{n} 1) x^n = \sum_{n=0}^{\infty} (n+1) \cdot x^n

4.2. A Formula for the Fibonacci Numbers

  • Let's say, an{a_n} is Fibonacci Sequence, and f(x)=n=0anxnf(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n
    • =x+x2+2x3+3x4+5x5+ = x + x^2 + 2x^3 + 3x^4 + 5x^5 + \cdots
  • Then:
    • xf(x)=x2+x3+2x4+3x5+5x6+ x \cdot f(x) = x^2 + x^3 + 2x^4 + 3x^5 + 5x^6 + \cdots
    • x2f(x)=x3+x4+2x5+3x6+5x7+ x^2 \cdot f(x) = x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + \cdots
  • Conclude above equations, we get:
    • f(x)xf(x)x2f(x)=xf(x) - x \cdot f(x) - x^2 \cdot f(x) = x
  • So f(x)=x1xx2\displaystyle f(x) = \frac{x}{1-x-x^2}
  • set ϕ=1+52\displaystyle \phi = \frac{1+\sqrt{5} }{2}, after some calculation we get:
  • f(x)=1/51(xϕ)+1/51(x(1ϕ))=1511(xϕ)+1511(x(1ϕ))=15n=0(xϕ)n+15n=0(x(1ϕ))n=n=0(15ϕn15(1ϕ)n)xn=n=0ϕn(1ϕ)n5xn=n=0anxn\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, an=ϕn(1ϕ)n5\displaystyle a_n = \frac{\phi^n - (1 - \phi)^n}{\sqrt{5} }

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