01
Modifying geometric power series
Consider the geometric power series
for . For this problem, you should modify the series for
. (a) Write
as a power series and determine its interval of convergence. (b) Write
as a power series and determine its interval of convergence.
Solution
02
(a)
The geometric series for
converges when and diverges for . So ours will converge when , which is when , and diverge otherwise. The interval is therefore . One can check this in more detail by doing the ratio test:
But we must be careful: the ratio test will not tell us what happens at the endpoints of the interval. If we apply the ratio test here, we would have to check the endpoint separately. But if we use the known result for geometric series, we know it diverges at both endpoints.
(b)
The geometric series for
Link to originalconverges when . So our series will converge when , which is when , and diverges for . So the interval is .
02
Power series of a derivative
Suppose that a function
has power series given by: The radius of convergence of this series is
. What is the power series of
and what is its interval of convergence?
Solution
08
If
Link to originalfor , then we know for .
03
Finding a power series
Find a power series representation for these functions:
(a)
(b)
Solution
03
(a)
Another approach:
We know that:
Plug in
: Complete:
(b)
Notice:
Integrate:
Plug in
to solve and find . Now then:
Link to original
04
Modifying and integrating a power series
(a) Modify the power series
for to obtain the power series for . (b) Now integrate this series to find the power series for
.
Solution
09
(a)
(b)
Link to original