↑ Year 1 · Semester 1

MH1805

Calculus

Power Series

Purpose: Express complicated/abstract expressions into polynomials for easier calculations.

Power series: ∑n=0∞an(x−c)n=a0+a1(x−c)+a2(x−c)2+…\text{Power series: }\sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x - c) + a_2(x - c)^2 + \ldots

Common examples (Maclaurin expansions):

Geometric series: 11−x=∑n=0∞xn=1+x+x2+x3+…, for |x|<1\text{Geometric series: } \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \ldots, \text{ for } |x| < 1

ln⁡(1+x)=∑n=1∞(−1)n−1xnn=x−x22+x33−x44+⋯,x∈(−1,1].\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots,\quad x \in (-1,1].

−ln⁡(1−x)=∑n=1∞xnn=x+x22+x33+x44+…,x∈[−1,1).-\ln(1 - x) = \sum_{n=1}^{\infty} \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots, \quad x \in [-1, 1).

*can be derived from ln(1 + x).

ex=∑n=0∞xnn!=1+x+x22!+x33!+…,x∈(−∞,∞).e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots,\quad x \in (-\infty, \infty).

sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!=x−x33!+x55!−…,x∈(−∞,∞).\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots,\quad x \in (-\infty, \infty).

cos⁡x=ddx(sin⁡x)=∑n=0∞(−1)nx2n(2n)!=1−x22!+x44!−…,x∈(−∞,∞).\cos x = \frac{d}{dx}(\sin x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots,\quad x \in (-\infty, \infty).

*can be derived by differentiating sinx.

tan⁡−1x=∑n=0∞(−1)nx2n+12n+1=x−x33+x55−…,x∈[−1,1].\tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots, \quad x \in [-1, 1].

*can be derived from geometric series by integrating 1/(1 + x2).

Convergence

For the power series ∑n=0∞an(x−c)n:\text{For the power series } \sum_{n=0}^{\infty} a_n (x - c)^n:

3 scenarios

1.Only converges at x=c[Series =a0+0+0+…].1.\ \text{Only converges at } x = c \ [\text{Series }= a_0 + 0 + 0 + \ldots].

2.Converges for all x∈ℝ.2.\ \text{Converges for all } x \in \mathbb{R}.

3.Converges when |x−c|<R and diverges when |x−c|>R, for some R>0.3.\ \text{Converges when } |x - c| < R \text{ and diverges when } |x - c| > R, \text{ for some } R > 0.

Case3

If ∑n=0∞an(x−c)n converges for some x=x0≠c,\text{If } \sum_{n=0}^{\infty} a_n (x - c)^n \text{ converges for some } x = x_0 \neq c, then ∑n=0∞an(x−c)n converges absolutely for smaller width |x−c|<|x0−c|.\text{then } \sum_{n=0}^{\infty} a_n (x - c)^n \text{ converges absolutely for smaller width } |x - c| < |x_0 - c|.

Radius of convergence is R,\text{Radius of convergence is } R,

possible intervals of convergence are:

(c−R,c+R),(c−R,c+R],[c−R,c+R),[c−R,c+R].(c - R,\, c + R), (c - R,\, c + R], [c - R,\, c + R), [c - R,\, c + R].

Convergence at endpoints x=c±R must be determined separately.\text{Convergence at endpoints } x = c \pm R \text{ must be determined separately.}

Differentiation, integration

If f is convergent on (c−R,c+R),\text{If } f \text{ is convergent on } (c - R,\, c + R),

then ddxf(x)=∑n=0∞nan(x−c)n−1and ∫f(x)dx=∑n=0∞ann+1(x−c)n+1+C\text{then } \frac{d}{dx}f(x) = \sum_{n=0}^{\infty} n a_n (x - c)^{n - 1} \text{and } \int f(x)\,dx = \sum_{n=0}^{\infty} \frac{a_n}{n + 1}(x - c)^{n + 1} + C

both have the same interval of convergence (c−R,c+R).\text{both have the same interval of convergence } (c - R,\, c + R).

Differentiation and integration does not affect the radius of convergence.

AT ENDPOINTS, differentiation may lose convergence, integration may gain convergence.

Abel’s Theorem

A power series is continuous on its interval of convergence.

→ we can find the power series of some function by first differentiating/integrating it.

→ If we find that the series converges at the endpoints x=R of its interval of convergence, we can conclude that the function is continuous at x=R too.

Example (series to function)

We found the series

x22−x33⋅2+x44⋅3−x55⋅4+…=∑n=2∞(−1)nxnn(n−1)=xln⁡(1+x)−x+ln⁡(1+x),\frac{x^2}{2} - \frac{x^3}{3 \cdot 2} + \frac{x^4}{4 \cdot 3} - \frac{x^5}{5 \cdot 4} + \dots = \sum_{n=2}^{\infty} (-1)^n \frac{x^n}{n(n-1)} = x \ln(1+x) - x + \ln(1+x),

by integrating ln⁡(1+x)=∑n=1∞(−1)n−1xnn,for x∈(−1,1].\text{by integrating } \ln(1 + x) = \sum_{n=1}{\infty} (-1)^{n-1} \frac{x^n}{n}, \text{for } x \in (-1, 1].

We should check endpoint x=−1 for convergence.\text{We should check endpoint }x = -1 \text{ for convergence.}

The interval of convergence would be x∈[−1,1]x \in [-1,1].

However, the expression is undefined for x=−1x = -1.

So using Abel’s Theorem, by continuity,

f(−1)=limx→−1+f(x)=limx→−1(xln⁡(1+x)−x+ln⁡(1+x))=limx→−1((1+x)ln⁡(1+x)−x).f(-1) = \lim_{x \to -1^+} f(x) = \lim_{x \to -1} (x \ln(1+x) - x + \ln(1+x)) = \lim_{x \to -1} ((1+x) \ln(1+x) - x).

We get f(−1)=1f(-1) = 1.

Example (function to series)

We show that

ln⁡(1+x)=∑k=1∞(−1)k−1xkk\ln(1 + x) = \sum_{k = 1}^{\infty} (-1)^{k-1} \frac{x^k}{k}

by integrating 11−x=∑k=0∞xk,for x∈(−1,1)\text{by integrating } \frac{1}{1 - x} = \sum_{k=0}^{\infty} x^k, \text{for } x \in (-1, 1)

Although we may gain convergence at endpoints, ln⁡(1+x) is undefined for x=−1.\text{Although we may gain convergence at endpoints, } \ln(1 + x) \text{ is undefined for } x = -1.

So we only need to check for x=1.\text{So we only need to check for }x = 1.

Determining interval of convergence and sum of series given some series

Tests

Sum

E.g. ∑n=1∞n5n.\sum_{n=1}^{\infty} \frac{n}{5^n}.