Class 26 - Power/Taylor series + Problems (11/3/2017, Friday)
Logistics
Exam 2- 11/13 Mon
- Check Syllabus on the course website. Sections 9.1 - 9.5 were dealt in Final Exams in the past, so you want to take note of that when you do problems from past exams.
- 11/6 Tues
- 9.4 due 11/6 Mon
Lecture
A power series centered at x = c is f(x) = \sum_{n=0}^{\infty}a_{n}(x - c)^{n}. Using the ratio test, we will have to compute\lim_{n \rightarrow \infty}\frac{|a_{n+1}|}{|a_{n}|}|x - c|.
Suppose that we can compute
K = \lim_{n \rightarrow \infty}\frac{|a_{n+1}|}{|a_{n}|},
allowing the possibility that K = \infty. This means that
- if K|x - c| < 1, then \sum_{n=0}^{\infty}a_{n}(x - c)^{n} absolutely converges;
- if K|x - c| > 1, then \sum_{n=0}^{\infty}a_{n}(x - c)^{n} diverges;
- if K|x - c| = 1, then we have no clue!
Writing K = 1/R with the convention that R = 0 when K = \infty and R = \infty when K = 0, the above translates to:
- if |x - c| < R, (i.e., distance of x from the center is less than R) then \sum_{n=0}^{\infty}a_{n}(x - c)^{n} absolutely converges;
- if |x - c| > R, (i.e., distance of x from the center is greater than R) then \sum_{n=0}^{\infty}a_{n}(x - c)^{n} diverges;
- if |x - c| = R, (i.e., distance of x from the center is R) then we have no clue!
- (c-R, c+R);
- [c-R, c+R);
- (c-R, c+R];
- [c-R, c+R].
Exercise*. Do #5 of Exam 3, Winter 2015.
Exercise*. Do #1 of Exam 3, Fall 2016.
Exercise*. Do #2 of Exam 3, Fall 2016.
Exercise*. Do #8a of Exam 3, Winter 2016.
Taking the derivative of a power series at the center. Unless the radius of convergence of the power series is 0 (i.e., if R > 0), we can take the derivative of the power series f(x) = \sum_{n=0}^{\infty}a_{n}(x - c)^{n} term by term for x \in (c-R, c+R). Note that with this observation you can realize
- f(c) = a_{0},
- f'(c) = a_{1},
- f''(c)/2! = a_{2},
- \cdots
- f^{(n)}(c)/n! = a_{n}.
Exercise*. Do #8b of Exam 3, Winter 2016.
Exercise. Let f(x) = \sin(x). Compute f(0), f'(0), f''(0), f^{(3)}(0), f^{(4)}(0), f^{(5)}(0), \dots.
The above computation should let you realize
\sin(x) = f(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots.
Working similarly with \cos(x), you should get
\cos(x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots.
Here, it is important that the both of the power series have the radius of convergence = \infty.
Exercise*. Do #6b of Exam 3, Winter 2016.
Exercise*. Do #9 c, d of Exam 3, Fall 2016.
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