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.
Comments
Post a Comment