Class 32 - Binomial expansions + Probability distributions (11/17/2017, Friday)

Logistics

Final Exam

• 12/14 8:00 AM

Quiz 5.5

• Take-home quiz due 11/28 Tuesday. This will consist of some material that are not going to be tested in this class, but it will explain and use some facts that are crucial to this course. Anyone who shows up on 11/21 Tuesday will get much hint.

WebHW

• 10.2 due 11/20 Mon
• 10.3 due 11/27 Mon

Lecture

Binomial expansion. What is the coefficient of $x^{10}$ when you expand $(1 + x)^{100}$? If you think about the right-hand side of
\[(1 + x)^{100} = (1 + x)(1 + x) \cdots (1 + x)\]
whenever you go though one $(1 + x),$ you either choose $1$ or $x.$ You get $x^{10}$ if you end up choosing ten $x$'s when you multiply to get the terms. This means that the number of $x^{10}$'s appearing after the multiplication is equal to the number of $(1 + x)$'s you chose to select $x$ from. This is the same as the number of ways to select $10$ people out of $100$ people without order. Hence, it is equal to
\[\frac{100 \cdot 99 \cdot \cdots 91}{10!}.\]
Now, let's try calculus. Consider
\[f(x) = (1 + x)^{100}.\]
It is a polynomial, so in particular, it is already a power series with many zeros in the tail! Thus, the coefficient of $x^{10}$ must be equal to $f^{(10)}(0)/10!.$ Now, note that

• $f'(x) = 100(1 + x)^{99},$
• $f''(x) = 100 \cdot 99 (1 + x)^{98},$
• $\dots$
• $f^{(10)}(x) = (100 \cdot 99 \cdots 91)(1 + x)^{90}.$

Thus, we have $f^{(10)}(0) = 100 \cdot 99 \cdots 91,$ so we get the same answer
\[\frac{f^{(10)}(0)}{10!} = \frac{100 \cdot 99 \cdots 91}{10!}.\]
What's cooler about the power series method is that we can consider the expansion of
\[f(x) = (1 + x)^{r}\]
even when $r > 0$ is not an integer. However, we need to be careful about the legitimate $x$-values because for non-integer $r,$ the right-hand side of $f(x)$ is not a polynomial anymore, so even when we talk about the power series expansion, we may not have the infinite radius of convergence. What we are really looking at is:
\[f(x) = (1 + x)^{r} = e^{r\ln(1+x)}\]
for $-1 < x \leq 1,$ so we can differentiate for any $-1 < x < 1$ for the power series expansion of the right-hand side, and since we can compute the derivative in similarly to the particular case when $r > 0$ is an integer, we must have the same formula for the Taylor coefficients:
\[\frac{f^{(n)}(0)}{n!} = \frac{r(r-1) \cdots (r - (n-1))}{n!}.\]
Thus, for any $r > 0,$ we have
\[(1 + x)^{r} = \sum_{n=0}^{\infty}\frac{r(r-1) \cdots (r - (n-1))x^{n}}{n!}\]
where $-1 < x < 1,$ while any $x$ is valid if $r > 0$ is an integer.

Exercise*. Do #5 Exam 3, Fall 2016.
Exercise*. Do #4 of Exam 3, Fall 2015.

Introduction to probability distributions. The discussion here will be very brief, so I highly recommend you read through Section 8.7 and 8.8 of your book.

A continuous real-valued function $p(x)$ is called a probability density function (pdf) if the following two conditions are satisfied:

1. $p(x) \geq 0$ for all $0.$
2. $\int_{-\infty}^{\infty}p(x)dx.$

Thus, given such a function $p(x),$ you can model probability. For example, your model can design (if chosen at random) "the probability that a person's height $x$ is between $5$ feet and $6$ feet" as the quantity

\[\int_{x=5}^{x=6} p(x) dx.\]

The cumulative distribution function (cdf) of a pdf $p(x)$ is

\[P(t) := \int_{x= - \infty}^{x=t}p(x)dx.\]

Exercise. If $P(t)$ is the cdf of a pdf $p(x),$ then show that $P'(t) = p(t).$

Exercise. If $P(t)$ is the cdf of a pdf $p(x),$ then show that

1. $P(t)$ is non-decreasing,
2. $\lim_{t \rightarrow -\infty}P(t) = 0$,
3. $\lim_{t \rightarrow \infty}P(t) = 1$, and
4. $\int_{a}^{b}p(x)dx = P(b) - P(a).$

Exercise* Book 8.7: #1, #5, #14, #24

More exercises on Taylor series

Exercise*. Do #2 of Exam 3, Fall 2015.

Exercise*. Do #3 of Exam 3, Winter 2016.