Class 31 - Applications of power/Taylor series expansions (11/16/2017, Thursday)
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. 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 #11 of Exam 3, Winter 2016.
Exercise*. Do #5 of Exam 3, Fall 2015.
Exercise*. Do #8 of Exam 3, Fall 2015.
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 $-1 < x \leq 1.$
More exercises
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.
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