Power Series
I. Return Midterm I
II. Definition of a Power Series
Let f(x) be the function represented by the series
f(x) = sum anxn
Then f(x) is called a power series function
More generally, if f(x) is represented by the series
f(x) = sum an(x - c)n
Then we call f(x) a power series centered at x = c.
The domain of f(x) is called the Interval of Convergence and half the length of the domain is called the Radius of Convergence.
III. The Radius of Convergence
To compute the radius of convergence, we use the ratio test.
Example: Find the radius of convergence of
f(x) = sum ((x - 3)n)/(2n)
Solution: We use the Ratio Test:
We solve
|(x - 3)/2| < 1 or |x - 3| < 2 so that
1 < x < 5
Since 1/2(5 - 1) = 2, the radius of convergence is 2
Exercise: Find the radius of convergence of
f(x) = sum (x - 2)n
IV. Interval of Convergence
To find the interval of convergence we follow the three steps:
1) Use the ratio test to find the interval where the series is absolutely convergent
2) Plug in the left endpoint to see if it converges at the left endpoint. (AST may be useful)
3) Plug in the right endpoint to see if it converges at the right endpoint. (AST may be useful)
Example:
Find the interval of convergence for the previous example:
f(x) = sum ((x - 3)n)/(2n)
Solution:
1) We have already done this step and found that the series converges absolutely for
1 < x < 5.
2) We plug in x = 1 to get
f(x) = sum ((1 - 3)n)/(2n) = sum (-1)n
This series diverges by the limit test.
3) We plug in x = 5 to get
f(x) = sum ((5 - 3)n)/(2n) = sum 1
This series also diverges by the limit test.
Hence the endpoints are not included in the interval of convergence. We can conclude that the interval of convergence is
1 < x < 5.
Exercise: Find the interval of convergence of the previous exercise:
f(x) = sum (x - 2)n
V. Differentiation and Integration of Power Series
Theorem:
Suppose that a function is given by the power series
f(x) = sum an(x - c)n
and that the interval of convergence is (c - R,c + R) (plus possible endpoints)
then f(x) is continuous, differentiable, and integrable on that interval (without the endpoints). To obtain the derivative or the integral of f(x) we can pass the derivative or integral through the sigma. In other words>
d/dx f(x) = d/dx (sum an(x - c)n) = sum d/dx(an(x - c)n) = sum nan(x - c)n-1 (here the sum starts at n = 1.
and
int f(x) dx = int(sum an(x - c)n)dx = sum int an(x - c)n dx = sum an/(n + 1)(x - c)n+1
Furthermore, the radius of convergence for the derivative and integral is R.
Example:
Consider the series
f(x) = sum xn
by the GST this series converges for |x| < 1
hence the center of convergence is 0 and the radius is 1.
By the above theorem
f'(x) = sum nxn-1
has center of convergence 0 and radius of convergence 1 also.
We can also say that
int f(x) dx = sum xn + 1/(n + 1)
also has center of convergence 0 and radius of convergence 1.
Exercise:
Show that
sum x2n/(2nn!)
satisfies the differential equation
y'' + xy' - y = 0