Class 30 - Power/Taylor series expansion (11/14/2017, Tuesday)
Logistics
Final Exam- 12/14 8:00 AM
WebHW
- 10.1 due 11/16 Thurs
- 10.2 due 11/20 Mon
Lecture
Review of power series. Recall that a power series centered at x = c is f(x) = \sum_{n=0}^{\infty}a_{n}(x - c)^{n}. Suppose that we can compute
Derivative: term-by-term within radius of convergence. This is a VERY IMPORTANT FACT. You can take derivative term by term within the radius of convergence (i.e., in the in open interval (c - R, c + R)) as long as you have R > 0 (some room). Why should we care? Well, first, consider
f(x) = a_{0} + a_{1}(x - c) + a_{2}(x - c)^{2} + a_{3}(x - c)^{3} + \cdots.
Then we have
The above computation should let you guess
\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. We are missing some rigorous reasoning. If you are interested in filling the gap, read Section 10.4 of your book.
Exercise. Compute the power series representation of
f(x) = \dfrac{1}{1 + x^{2}}
near 0. (Hint: look at it as a geometric series with the initial term 1 and the ratio -x.)
Applying \int_{x=0}^{x=t}( - )dx to the answer to this exercise lets you see that
\arctan(t) = t - \frac{t^{3}}{3} + \frac{t^{5}}{5} - \frac{t^{7}}{7} + \cdots
where -1 < t < 1. Both sides make sense when t = 1 and you can actually check that we can take both sides t \rightarrow 1- to conclude that
\frac{\pi}{4} = \arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots.
Now, you no longer need to worry about approximating \pi:
\pi = 4\arctan(1) = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots.
Exercise*. Do #12 of Exam 3, Winter 2016.
Exercise*. Do #9 c, d of Exam 3, Fall 2016.
K = \lim_{n \rightarrow \infty}\frac{|a_{n+1}|}{|a_{n}|},
allowing the possibility that K = \infty. Then the radius of convergence of f(x) is R = 1/K, where we set R = \infty when K = 0, and R = 0 when K = \infty. Recall that this means that for x \in (c - R, c + R), the sum on the right-hand side below is absolutely convergent:
f(x) = \sum_{n=0}^{\infty}a_{n}(x - c)^{n}.
Exercise*. Do #8b of Exam 3, Winter 2016.
It's probably hard. Do it after understanding the following:
f(x) = a_{0} + a_{1}(x - c) + a_{2}(x - c)^{2} + a_{3}(x - c)^{3} + \cdots.
Then we have
- f'(x) = a_{1} + 2a_{2}(x - c) + 3a_{3}(x - c)^{2} + 4a_{4}(x - c)^{3} + \cdots,
- f''(x) = 2a_{2} + (3 \cdot 2)a_{3}(x - c) + (4 \cdot 3)a_{4}(x - c)^{2} + (5 \cdot 4)(x - c)^{3} + \cdots,
\vdots
- f^{(n)}(x) = n!a_{n} + ((n+1)!/1!)(x - c) + ((n+2)!/2!)(x - c)^{2} + ((n+3)!/3!)(x - c)^{3} + \cdots.
In particular, we have f^{(n)}(c) = n!a_{n}, so we must have
a_{n} = \frac{f^{(n)}(c)}{n!}.
Approximation. A function f(x) with a power series representation at x = 0 (or "near" 0) is precisely the one that can be written as
f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots
where the radius of convergence on the right-hand side is positive. We have observed that in this case, we necessarily have a_{n} = f^{(n)}(0)/n!. In particular, the partial sumes
P_{n}(x) = f(0) + f'(0)x + \frac{f^{(2)}(0)}{2!}x^{2} + \cdots + \frac{f^{(n)}(0)}{n!}x^{n}
must be a reasonable approximation of f(x) for x \in (-\epsilon, \epsilon) where \epsilon is the radius of convergence. We call P_{n}(x) the Taylor polynomial of degree n near 0 associated to f(x). Notice that the number of terms of P_{n}(x) is usually n + 1, not n.
When the center of power series shifts to say x = c, the Talyor polynomial of f(x) can change center to c, and the degree n polynomial is
P_{n}(x) = f(c) + f'(c)x + \frac{f^{(2)}(c)}{2!}(x - c)^{2} + \cdots + \frac{f^{(n)}(c)}{n!}(x - c)^{n}.
In particulare, we see that P_{1}(x) = f(c) + f'(c)x is the linear approximation of f(x), a.k.a. the tangent line of y = f(x) at x = c.
Power series representation of usual functions. Exercise. In class, we have already seen that
e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots.
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 guess
\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. We are missing some rigorous reasoning. If you are interested in filling the gap, read Section 10.4 of your book.
Exercise. Compute the power series representation of
f(x) = \dfrac{1}{1 + x^{2}}
near 0. (Hint: look at it as a geometric series with the initial term 1 and the ratio -x.)
Applying \int_{x=0}^{x=t}( - )dx to the answer to this exercise lets you see that
\arctan(t) = t - \frac{t^{3}}{3} + \frac{t^{5}}{5} - \frac{t^{7}}{7} + \cdots
where -1 < t < 1. Both sides make sense when t = 1 and you can actually check that we can take both sides t \rightarrow 1- to conclude that
\frac{\pi}{4} = \arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots.
Now, you no longer need to worry about approximating \pi:
\pi = 4\arctan(1) = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots.
More exercises
Exercise*. Do #6b of Exam 3, Winter 2016.Exercise*. Do #12 of Exam 3, Winter 2016.
Exercise*. Do #9 c, d of Exam 3, Fall 2016.
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