
Power series are one of the most important topics in Calculus BC.
They allow us to represent functions as infinite polynomials and analyze them like series.
A power series centered at a is an infinite series of the form:
$$\sum^{\infty}_{n=0}c_{n}(x-a)^{n}$$
where:
$$c_{n}$$ = coefficients
a = center
x = variable
$$\sum^{\infty}_{n=0}\frac{(x-2)^{n}}{3^{n}}$$
This is a power series centered at a=2.
A power series either:
converges for all x, or
converges only within some interval around a, or
converges only at the center.
We find convergence using Ratio Test.
Given:
$$\sum{c_{n}}(x-a)^{n}$$
Compute:
$$L=\lim_{x\to\infty)\left|\frac{c_{n+1}}{c_{n}}\right|$$
Then the series converges when:
$$|x-a| < \frac{1}{L}$$
This number is the radius of convergence (R):
$$R=\frac{1}{L}$$
After finding R, check:
$$x=a – R and x=a + R$$
These endpoints may converge or diverge, and must be tested manually.
$$\sum\frac{x^{n}}{n}$$
Ratio test:
$$\lim_{n\to\infty}\left|\frac{x^{n+1}/(n+1)}{x^{n}/n}\right|=|x|\cdot\lim_{n\to\infty}\frac{n}{n+1}=|x|$$
Converges if:
$$|x|<1$$
Now test endpoints:
$$x=1 : \sum\frac{1}{n} \longrightarrow$$ diverges
$$x=-1 : \sun\frac{-1^{n}}{n} \longrightarrow$$ converges
(-1 , 1]
If a power series converges for some x, then:
It represents a function
It behaves like a polynomial
We can differentiate and integrate term by term
Given:
$$f(x)=\sum^{\infty}_{n=1}nc_{n}(x-a)^{n-1}$$
Differentiate term-by-term:
$$f^\prime(x)=\sum^{\infty}_{n=1}nc^{n}(x-a)^{n-1}$$
The radius of convergence stays the same.
Integrate term-by-term:
$$\int f(x)dx = \sum^{\infty}_{n=0}\frac{c_{n}}{n+1}(x-a)^{n+1}+C$$
Radius of convergence stays the same.
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