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

\[K = \lim_{n \rightarrow \infty}\frac{|a_{n+1}|}{|a_{n}|},\]

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:

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

• $f'(x) = a_{1} + 2a_{2}(x - c) + 3a_{3}(x - c)^{2} + 4a_{4}(x - c)^{3} + \cdots$,
• $f''(x) = 2a_{2} + (3 \cdot 2)a_{3}(x - c) + (4 \cdot 3)a_{4}(x - c)^{2} + (5 \cdot 4)(x - c)^{3} + \cdots$,

      $\vdots$

• $f^{(n)}(x) = n!a_{n} + ((n+1)!/1!)(x - c) + ((n+2)!/2!)(x - c)^{2} + ((n+3)!/3!)(x - c)^{3} + \cdots$.

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.