Class 31 - Applications of power/Taylor series expansions (11/16/2017, Thursday)

Logistics

Final Exam

WebHW

Review of power series. We start by an exercise:

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

$\pi$ :

$\pi = 4\arctan(1) = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots.$

Log expansion. Consider the geometric series

$1 - t + t^{2} - t^{3} + \cdots = \frac{1}{1 + t}$
for

$-1 < t < 1.$ Apply

$\int_{t=0}^{t=x}( - )dt$ to the above to get

$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} = \ln(1 + x).$
Both sides make sense when

$x = 1$ and are continuous at

$x = 1,$ so taking the limit

$x \rightarrow 1,$ we get the above identity for

$-1 < x \leq 1.$

Exercise*. Do #9a Exam 3, Fall 2016.
Exercise*. Do #11 of Exam 3, Winter 2016.
Exercise*. Do #5 of Exam 3, Fall 2015.
Exercise*. Do #8 of Exam 3, Fall 2015.